Alex Kritchevsky

Exact Divergent Series

2026-08-17T00:00:00+00:00

I tend to believe that mathematical results which seem surprising or paradoxical are better interpreted as just being bad results—and they are really indicators of places that our foundations have gone wrong. For instance I think Banach-Tarski is an indictment of measure theory, not a surprising fact about the universe; the long list of pathological counterexamples in topology is an argument for switching to a finistic foundation which routes around all of them; and the surprising implications of the axiom of choice are evidence that we should discard set theory as a mathematical foundation. That sort of thing. (Unsurprisingly, the list of disagreeable results is almost entirely statements involving infinity.) I don’t think I can convincingly argue for these claims yet, but I do believe them.

A particularly salient example of the problem is found in the summation of divergent series. I have long held a vendetta against all those online math explainers who gave credence to nonsensical manipulations like the infamous \(1 + 2 + 3 + 4 + \ldots \? -\frac{1}{12}\) sum, which I wrote about in one of my first articles. That post was inspired by a Numberphile video which had a very disappointing discourse around it at the time: people were far too willing claim that the sum equaling \(-1/12\) was true in some sense, or even literally true, instead of what seems like to me like the obviously correct conclusion: that math has an unresolved hole in it, which any layperson can clearly see but which people trapped inside the ‘framework’ of modern mathematics somehow miss. Considerations of questions like this have led me personally to a version of finitism (although not a version I have seen anyone else exactly ascribed to). But that is for another time.

Anyway this post is something of a followup to that earlier one. In the intervening decade I have come to know a much more simple and pleasing exposition for those divergent sums than I knew at the time, one which I think dispels all possible objections. I thought I would write it out as a standalone article in case anyone else wants to see this laid to rest (or, I guess, thinks there’s something I’m still missing). Nothing in here is particularly deep, and that’s kind of the point. It is just a simple explanation for a simple thing. Unfortunately it is somewhat dry to explain all in detail… I imagine that it will be interesting if you specifically care about this particular puzzle, and not if you don’t, and that’s fine.

(Parts of this article were previously part of an earlier post about Taylor series. This section was much more likely to be interesting to anyone else than the rest so I decided to split it out, and have added some more material in the process.)

1. The Example of \(1/(1-x)\)

First we will discuss the objectively-coolest Taylor series, that of

\[f(x) = \frac{1}{1-x}\]

which is

\[S(x) = 1 + x + x^2 + x^3 + \ldots\]

This converges to the true value of \(f(x)\) for \(\| x \| < 1\). For example,

\[\begin{aligned} f(\frac{1}{2}) &= \frac{1}{1-\frac{1}{2}} = 2 \\ S(\frac{1}{2}) &= 1 + \frac{1}{2} + \frac{1}{4} + \ldots = 2 \end{aligned}\]

The equality holds specifically if the \(1+1/2+1/4+\ldots\) is regarded as a limit, which is standard: it’s equivalent to how \(1.99\overline{9} = 2\). In fact it’s the same phenomenon exactly, since the statement \(1+\frac{1}{2}+\frac{1}{4} + \ldots = 2\) becomes \(1.11\overline{1} = 10\) when written in binary. However, keep in mind for later that “regarded as a limit” is not the only way to interpret equality; it’s just the one we’re steeped in and therefore used to.

We can also write down a Taylor series for \(f(x)\) around points other than \(x=0\). A useful one is the series around \(x=\infty\), which is the same thing as a series in the variable \(1/x\) around \(1/x=0\). I’ll write this as \(S_{\infty}(x)\) to distinguish it from the other one:

\[\begin{aligned} f(x) &= \frac{1}{1-x} \\ &= (-\frac{1}{x}) \frac{1}{1 - \frac{1}{x}} \\ &= -\frac{1}{x} [1 + \frac{1}{x} + \frac{1}{x^2} + \ldots] \\[0.5em] S_{\infty}(x)&= -\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} - \ldots \end{aligned}\]

This converges for \(\| x \| > 1\), for example

\[\begin{aligned} f(2) &= \frac{1}{1-2} = -\frac{1}{2} \frac{1}{1 - \frac{1}{2}} = -1 \\ S_{\infty}(2) &= -\frac{1}{2}[1 + \frac{1}{2} + \frac{1}{4} + \ldots] = -1 \end{aligned}\]

There are also series around any \(x = k \neq 1\), but we won’t need tohs.e

Outside the radii of convergence each of these series gives an infinite result which is not equal to \(f(x)\). For instance \(S(x)\) converges for \(\| S \| \leq 1\), and outside of that it diverges to infinity:

\[S(2) = 1 + 2 + 4 + 8 + \ldots\]

The paradox arises because, although this value is infinite, there appears to be a sense in which it does seem to encode the value of \(f(2) = 1/(1-2) = -1\) somehow: when this divergent sum is evaluated by any of the various divergent series summation techniques, it gives the right answer! For the example the naivest way to evaluate a divergent sum is to algebraically manipulate it to be finite again:

\[\begin{aligned} S(2) &= 1 + 2 + 4 + 8 + \ldots \\ S(2) - 2 S(2) &= [1 + 2 + 4 + \ldots] - (2 + 4 + 8 + \ldots) \\ (1-2)S(2) &= 1 \\ S(2) &= -1 \end{aligned}\]

We are left with the puzzle of explaining how this is even possible.

One might interpret this to mean that, although the result of \(S(2)\) is not a number, it still contains the data \(-1\) somehow—maybe in a form like \(-1 + O(\infty)\)? I have seen this type of explanation a lot. People freqeuntly link Tao’s article on the subject, which shows how you can “smooth out” these sums to remove the jumps at each discrete integer. Doing this typically reveals a constant term which has the finite value that the summation techniques get. The implication is that the sum \(S(2) = 1+2+4+8+\ldots\) contains more data in than just its ‘value’ \(\infty\); in particular it encodes its growth rate, which has a finite part equal to \(-1\).

I’m sure that is true, but I’ve come to think that it’s making things too complicated. Also, it doesn’t really explain what is going on. Do divergent sums have values, or not? If they’re not numbers, then what are they, and what algebra do they obey? If this value \((-1)\) is in there, what else is in there?

So now I will describe a very simple way of thinking about this that is completely satisfactory for intuition. Depending on your background it might seem a bit elementary, so I feel almost weird making a big deal about it, but I think it’s important for staying sane. We are simply going to refuse to consider “infinite” sums at all and only allow ourselves to regard them finitistically, up to some unspecified position \(N\), and it turns out this explains everything.

2. Exact Divergent Sums

Recall that for any finite \(N\), the partial sum

\[S_N(x)= 1 + x + x^2 + \ldots + x^N\]

is not a multiplicative inverse of \(1/(1-x)\). If it were we would have \((1-x) S_N(x) = 1\), but we don’t. Instead we have

\[(1-x) S_N(x) = (1-x)(1 + x + x^2 + \ldots + x^N) = 1 - x^{N+1}\]

Evidently the real value of \(S_N(x)\) is

\[S_N(x) = \frac{1-x^{N+1}}{1-x} = \frac{1}{1-x} - \frac{x^{N+1}}{1-x}\]

Normally we think of the value of \(N\) as implicitly going to infinity, and then it’s clear why this converges only for \(\| x \|<1\), because that’s the case where the error \(x^{N}\) goes to \(0\). This is why \(S(1/2) = 2\) is correct, because \((1/2)^N\) becomes indistinguishable from \(0\) for large enough \(N\). And the reason that \(S_N(2) = -1\) diverges is that

\[S_N(2) = \frac{1}{1-2} - \frac{2^{N+1}}{1-2} = 2^{N+1} - 1\]

is not the value of \(\frac{1}{1-2}\), even though it is indeed the value of \((1+2+4+\ldots + 2^N)\).

When we find that \(S_N(2) \? -1\) under divergent summation techniques it’s because the algebraic manipulations used are somehow erasing the \(x^{N+1}\) term even though it should contribute to the result. Specifically it happens at the step where we subtracted the series from itself. It should have read:

\[\begin{aligned} (1-x)S_N(x) &= (1 + x + \ldots + x^N) \\ &\; \;\;\; - (x + x^2 + \ldots + x^{N+1}) \\ 1 &\stackrel{??}{\neq} 1 - x^{N+1} \\ \end{aligned}\]

Writing this as \(1\) requires arbitrarily dropping the \(O(x^N)\) term even though it is not necessarily small—but since it was hiding behind the ellipses we “got away with it”. In fact the computation that gave \(S(2)=-1\) should have read

\[\begin{aligned} (1-2) S_N(2) &= S_N (2) - 2S_N(2) \\ -S_N(2) &= -1 + 2^{N+1} - 2(-1 + 2^{N+1}) \\ &= 1 - 2^{N+1} \\ S_N(2) &= 2^{N+1} - 1 \end{aligned}\]

which is correct for all \(N\). When we wrote that \(S(2) = -1\), it’s not that the \(2^{N+1}\) term actually cancelled—it’s just that we dropped it. Which was aided by the fact that writing it as \(1+x+x^2+x^3+\ldots\) notation made it easy to do so. So regardless of what confusions you might have about \(S(x)\), there is no mystery with \(S_N(x)\), and one way of looking at the problem is just that we found some algebraic manipulations which allow one to pretend that the \(2^{N+1}\) terms cancel out when they really don’t.

My takeaway is that we ought to insist that our infinite sums are really partial sums: \(S(x)\) doesn’t need to have a value at \(N \ra \infty\), just at some finite \(N\) that suits our purposes. It’s fine if we want to omit it from our notation but we don’t get to also complain when that leads to confusing algebra, because the \(N\)-dependent term was really supposed to be there the whole time.

Just in case it still seems like you might generally be able to figure out what the constant terms of a series are without seeing their \(N\)-dependencies, here is a more complicated example which can only be resolved by including the \(x^{N+1}\) terms.

Consider the function \(g(x) = \frac{1+x}{1+x+x^2}\) which has \(g(1) = 2/3\). The summation for \(g(1)\) appears to give an unsummable divergent series that looks exactly like \(S(-1)\):

\[\begin{aligned} g(x) &= \frac{1+x}{1+x+x^2} \\ &= \frac{1-x^2}{1-x^3} \\ &= (1-x^2)(1 + x^3 + x^6 + \ldots) \\ &= 1 - x^2 + x^3 - x^5 + x^6 - x^8 + \ldots \\ g(1) &= 1 - 1 + 1 - 1 - 1 + 1 + \ldots \end{aligned}\]

The manipulation \(\frac{1+x}{1+x+x^2} = \frac{1-x^2}{1-x^3}\) is true if you regard the value at \(x=1\) as a limit, whereuon it has the value \(g(1) = 2/3\) by l’Hôpital’s rule. But there’s no way to get \(\frac{2}{3}\) from the series \(1-1+1-1\ldots\) alone—clearly, because it is the same series as \(S(-1)\), which we have to suppose equals \(f(-1) = 1/2\) if we are buying the divergent sum techniques. So how are we to argue that \(g(1) = 2/3\)? The ‘data’ for that, if it is found in the series at all, must be encoded in the powers of \(x\)—the \((+x^0 -x^2 +x^3 -x^5 +x^6 \ldots)\) pattern. But there’s no way to say that it is found in \(1-1+1-1\ldots\) as well. What to do?

The mystery gets resolved if we also look at the \(O(x^N)\) term:

\[\begin{aligned} \frac{1-x^2}{1-x^3} &= (1-x^2)(1 + x^3 + x^6 + \cdots + x^{3N} + \frac{(x^3)^{N+1}}{1-x^3}) \\ &= 1 - x^2 + x^3 - x^5 + \cdots + x^{3N} - x^{3N+2} + x^{3N + 3} \frac{1-x^2}{1-x^3} \\ &\stackrel{x=1}{\Ra}(1-1) + (1-1) + \cdots + (1-1) + \frac{2}{3} \end{aligned}\]

This time the remainder term which is otherwise dropped is doing all the work: every other pair of terms cancel, and then the remainder gives the full value \(2/3\). So in this case erasing the \(N\)-dependence would leave something unusable. (You can also get a version of \(1-1+1-1+\ldots\) which equals any rational \(m/n\) you want by computing \((1-x^m)/(1-x^n)\) instead.)

So that is how I think one can stay sane thinking about divergent sums. They were never divergent at all. Their algebraic properties follow entirely from treating them as partial sums to an arbitrary \(N\) and just making sure to be careful about tracking it.

This is an trick that turns out to work in a lot of settings, by the way: any time an infinite limit does something odd, just stop regarding it as infinite and do everything with a finite range instead, but keep the range as a free variable in the expression. Usually you will figure out what is going on pretty quickly.

One might object that this approach is really just a way of skirting around the issue. Wasn’t the actual mystery about the limit as \(N \ra \infty\)? After all that is the thing that has \(S(1/2) = 2\) and seems to also allo for \(S(1) = -1\). The mystery of divergent sums is in how they seem to work after taking limits, despite the limits being invalid—I hear you say.

I don’t buy it. I really think the limits don’t actually matter. What makes \(1+1/2 +1/4+\ldots = 2\) is that it is indistinguishable from \(2\) in a practical sense, not that they are literally equal. This is what is meant by the epsilon-delta definition of a limit anyway: for any accuracy (epsilon) you want, I can show a value that is close enough to \(1/2\) (by delta) to get that close to \(2\) (by epsilon). It was never needed that they be literally equal. It’s that they are equal enough that you can’t tell.

This is the ‘finitist’ perspective which I like to adopt anytime math does something weird, because it cuts through through all the mystery: you can’t complain about limits at infinity doing mysterious things if you refuse to take any limits at all. Although math traditionally assumes limits and infinitism everywhere, physical applications necessarily do not care about the difference. In a physical theory, any epsilon-delta limit you’re using doesn’t actually have to hold for all choices of \(\e\), but rather only for the choices of \(\e\) you actually have—which are finite values, characterizing the ‘resolution’ with which you can measure the number. Therefore one does not need to really consider limits, except insofar as they summarize the approximation process succinctly. If they start giving funny answers, the right thing to do is to make your \(N\)-dependencies explicit again—hopefully you did not forget where they are—and keep going. Which is not the usual way that limits are conceptualized, at least in pure mathematics, but I think it’s how they should be viewed.

3. Some other sums

\(S(x) = 1+x+x^2 +\ldots\) is, admittedly, the easiest divergent series to understand. Therefore, to demonstrate that this way of thinking works more generally, and also mostly just to have a record of these calculations, I will go through each of the other interesting divergent sums I know and show how this ‘finitistic’ perspective gives the right answer. (Well, most of them. Unfortunately the sums involving primes are beyond me because I have never managed to absorb the content of number theory.)

The first series to consider is (wiki):

\[P = 1 + 1 + 1 + 1 + \ldots\]

Which happens to be \(S(1)\). An easy manipulation shows that \(P\) equals… any integer you want?—

\[\begin{aligned} P - P &= 1 + 1 + 1 + 1 + \ldots \\ &\;\;\;\;\;\,-(1 + 1 + 1 + 1 + \ldots)\\ &= 1 \\ &= 1 + 1 + 1 + 1 + \ldots \\ &\;\;\;\;\;\;\;\;\;\;\;\;-( 1 + 1 + 1 + 1 + \ldots) \\ &= 2\ \end{aligned}\]

—which can’t be right.

If we fix \(P\) to have a finite number of terms \(N\) then its sum is clearly \(P = N\) To handle this better, we will label the sum with a subscript to indicate by how many terms were included in it, like this:¹

\[P_N = S_{N-1}(1) = \underbrace{1+1+1+1+\ldots}_{N \text{ times}} = \sum_1^N 1 = N\]

Then we can say concretely:

\[\begin{aligned} P_N - P_N &= 0 \\ P_N - P_M &= N - M \end{aligned}\]

Which dispels the mystery. You can get any number if you use two un-“correlated” copies of the value, but if you subtract the same value from itself you of course get \(0\), regardless of how you try to arrange the terms. So this just works and is basically trivial.

Next we have Grandi’s Series, which is \(S(-1)\):

\[Q = 1 - 1 + 1 - 1 + \ldots\]

It is ‘summable’ with the classic manipulation

\[\begin{aligned} Q + Q &= 1 - 1 + 1 - 1 + \ldots \\ &\;\;\;\;\;\,+(1 - 1 + 1 - 1 + \ldots) \\ &= 1 \\ \Ra Q &\? \frac{1}{2} \\ \end{aligned}\]

The actual value is clearly

\[\begin{aligned} Q_N &= S_{N-1}(-1) = \frac{1}{2}(1 - (-1)^N) \end{aligned}\]

If we repeat the calculation with the actual value,

\[Q_N + Q_N = 2Q_N = 1 - (-1)^N\]

it is clear that the \(N\)-dependency was inadvertently being dropped.

However, we saw before that \((1-x^2)/(1-x^3)\) also gives the same series, yet equals \(\frac{2}{3}\). How do we reconcile this? For that matter, how does a sum of \(\pm 1\)s ever equal a non-integer? Why, it is simple. Remember the actual equivalency was

So it is not the case that \((1-x^2)/(1-x^3)\) equals the sum \(1-1+1-1+\ldots\). Instead it has an additional term at the end, whose value is exactly \(2/3\). Whew. No “sum of integers equals a fraction” today.

Next up,

\[R = 1 - 2 + 3 - 4 + \ldots\]

(wiki) with the classic manipulation

\[\begin{aligned} R + R &= 1 - 2 + 3 - 4 + \ldots \\ &\;\;\;\;\;\,+(1 - 2 + 3 - 4 + \ldots) \\ &= 1 -1 + 1 - 1 + \ldots \\ &= Q \? \frac{1}{2} \\ \Ra R &\? \frac{1}{4} \end{aligned}\]

We already know the value of \(Q\) is wrong there, and probably so is the value of \(R\), since the \(N\) dependenceis are missing. How do we do it right? Well, the partial sums of \(R\) are

\[(1, -1, 2, -2, 3, -3, \ldots)\]

It is a bit harder to see how to write this as a closed form. It is almost \((1, \frac{3}{2}, 2, \frac{5}{2}, \ldots)\), except that every other term is wrong by a certain amount. Once you rewrite it like this

\[(\frac{1}{4} + \frac{3}{4}, \frac{1}{4} - \frac{5}{4}, \frac{1}{4} + \frac{7}{4}, \frac{1}{4} - \frac{9}{4}, \frac{1}{4} + \frac{11}{4}, \frac{1}{4} - \frac{13}{4}, \ldots)\]

the closed form becomes obvious:

\[R_N = \frac{1}{4} - (-1)^N \frac{2N+1}{4}\]

This is why the invalid computation that canceled out the \(N\)-dependency gave \(R = 1/4\): it was just dropping the \(N\)-dependence (again).

Another hacky way of getting to \(R=1/4\) that I’ve seen around is to square \(Q\) and then rearrange the terms in a certain way:

\[\begin{aligned} Q^2 &= (1-1+1-1+\ldots)(1-1+1-1+\ldots) \\ (\frac{1}{2})^2 &= 1 - 1 + 1 - 1 + \ldots \\ & \;\;\;\;\;\,-(1-1+1-1+\ldots) \\ & \;\;\;\;\;\;\;\;\;\;\;\;+(1-1+1-1+\ldots) \\ & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;- \text{(...etc...)} \\ \frac{1}{4} &\? 1 - 2 + 3 - 4 + \ldots\\ \end{aligned}\]

For getting \(R_N\) this obviously won’t do at all: once we switch to actually tracking the length of the partial sums there are going to be a bunch of floating \(\pm 1\)s at the end of each term which don’t accumulate to give a term in \(R_N\). However, we can fix this method. If it is going to compute a particular value of \(R_N\), the sum should actually be

\[R_N = Q_N - Q_{N-1} + Q_{N-2}-Q_{N-3} \ldots\]

such that every sum terminates on exactly the \(N\)th term of \(1-2+3-4 +\ldots\). Since \(Q_N\) is \(0\) for \(N\) even and \(1\) for \(N\) odd, it simplifies to

\[\begin{aligned} R_N = \underbrace{Q_N - Q_{N-1} + Q_{N-2}-Q_{N-3} + \ldots \mp Q_1}_{N \text{ terms}} &= \begin{Bmatrix} -\frac{N}{2} & N \text{ even} \\[0.5em] \frac{N+1}{2} & N \text{ odd} \end{Bmatrix} \end{aligned}\]

Which happens to be another way of writing \(R_N = \frac{1}{4} - (-1)^N \frac{2N+1}{4}\). So that’s good. The reason \(Q^2 \? R \? \frac{1}{4}\) seemed to work is because it is equivalent to the constant terms of this calculation: naturally all the \(Q_{N-K}\)s have constant term \(1/2\), and so did the number of terms in the sum since it oscillated between even and odd numbers.

4. -1/12

Finally we have the Ramunajan sum

\[T = 1 + 2 + 3 + 4 + \ldots\]

Which can be ‘summed’ by reducing to the \(R = \frac{1}{3}\) from above:

\[\begin{aligned} T - 4T &= 1 + 2 + 3 + 4 + 5 + 6 + \ldots \\ &\;\;\;\,- 4(1 \;\;\;+ \;\;\; 2 \;\;\;+\;\;\; 3 +\ldots) \\ &= 1 - 2 + 3 -4 + 5 -6 \\ -3 T &= R \\ &\? \frac{1}{4} \\ T &= - \frac{1}{12} \end{aligned}\]

This one uses a new trick, of ‘spacing out’ the series in order to act on every other term in the sum, which we will have to handle. There is also another problem, which is that we know the closed form for the partial sums for this series—it’s

\[\sum_{k=1}^N k = \frac{N(N+1)}{2}\]

which decidedly does not contain a \(-\frac{1}{12}\) term. So where does that value come from?

To get a closed form for \(T_N\) we will attempt to duplicate the summation logic while carefully tracking the length of the partial sums. The ‘spacing out’ step requires including only the terms up to exactly the \(N\)th term of \(T\). For this we need to use \(T_{N/2}\) if \(N\) is even or \(T_{(N-1)/2}\) if \(N\) is odd, like this:

\[\begin{aligned} R_N &= \begin{cases} T_N - 4 T_{N/2} & N \text{ even} \\ T_N - 4 T_{(N-1)/2} & N \text{ odd} \\ \end{cases} \end{aligned}\]

More concisely:

\[R_n = T_N - 4 T_{\lfloor N/2 \rfloor}\]

Using \(R_N = \frac{1}{4} - (-1)^N \frac{2N+1}{4}\) we can rework this into a recurrence for \(T_N\):

\[T_N = \frac{1}{4} - (-1)^N \frac{2N+1}{4} + 4 T_{\lfloor N/2 \rfloor}\]

(with \(T_0 = 0\)). It is not too hard to check that \(T_N = \frac{N(N+1)}{2}\) obeys this relation: if \(N\) is even then

\[\begin{aligned} \frac{N(N+1)}{2} &= \frac{N^2}{2} + \frac{N}{2} \\ &= -\frac{N}{2} + (N + \frac{N^2}{2}) \\ &= \underbrace{(\frac{1}{4} -\frac{2N+1}{4})}_{R_N} + 4(\underbrace{\frac{(N/2)(N/2+1)}{2}}_{T_{\lfloor N/2 \rfloor}}) \end{aligned}\]

and there’s something similar for \(N\) odd.

Okay, fine. But then where does the \(-1/12\) come from? Let’s go back to the fake derivation. The \(-3\) factor showed up because of this manipulation:

\[T - 4T = R\]

This is simply not valid if you’re tracking partial sums: \(T_N - 4T_N \? R_N\) is false; the actual relationship is \(T_N - 4T_{\lfloor N/2 \rfloor} = R_N\) But maybe that’s sufficient? After all the constant term of \(R_N\) is going to depend on the constant terms of \(T_N\) and \(T_{\lfloor N/2 \rfloor}\), which depends on \(T_{\lfloor \lfloor N/2 \rfloor/2 \rfloor}\), etc, all the way down to \(T_0 = 0\). So we need to sum over the whole recurrence to see if this is true. It looks like this:

\[T_N = \underbrace{R_N + 4 R_{\lfloor N/2 \rfloor} + 4^2 R_{\lfloor (\lfloor N/2 \rfloor)/2 \rfloor} + \ldots}_{\log_2 N + 1 \text{ terms}}\]

There are \(\log_2 (N) + 1\) terms, since that is the number of times you can repeat this \(N \mapsto \lfloor \frac{N}{2} \rfloor\) calculation (each one being equivalent to lopping off the lowest digit in binary; we would write \(N \gg 1\) if we were in a programming context).

So since \(T_N = R_N + 4T_{\lfloor N/2 \rfloor}\) is defined by a recurrence, there is a series of \(R_N\)s in the resulting sum, and therefore multiple copies of the \(\frac{1}{4}\) constant term from each \(R_N\). Here is an example calculation of \(T_5\) to make it easy to see:

\[\begin{aligned} T_5 &= \boxed{\frac{1}{4}} - (-1)^5 \frac{2(5)+1}{4} + \boxed{4 [\frac{1}{4}} - (-1)^2 \frac{2(2)+1}{4} + \boxed{4 [\frac{1}{4}} - (-1)^1 \frac{2(1)+1}{4}]] \\ &= \frac{1}{4} + (\frac{2(5)+1}{4}) + 1 - 4 \frac{2(2)+1}{4} + 4 + 16 \frac{2(1)+1}{4} \\ &= \frac{1}{4} + \frac{11}{4} + 1 - 5 + 4 + 12 \\ &= 15 \\ &= 1 + 2 + 3 + 4 + 5 \end{aligned}\]

The multiple copies of \(\frac{1}{4}\) form a series that looks like this:

\[\begin{aligned} T_N &\sim \frac{1}{4} + 4 \frac{1}{4} + 4^2 \frac{1}{4} + \ldots \\ &= (\underbrace{1 + 4 + 4^2 + \ldots}_{\approx \, \log_2 N + 1 \text{ terms}}) (\frac{1}{4}) \\ \end{aligned}\]

But we can (divergently) sum that—it’s

\[\begin{aligned} T &\? (1 + 4 + 4^2 + \ldots) (\frac{1}{4}) \\ &\? \frac{1}{1 - 4} (\frac{1}{4}) \\[0.8em] &= -(\frac{1}{3}) (\frac{1}{4}) \\[0.8em] &= -\frac{1}{12} \end{aligned}\]

Evidently this is where the number is coming from. Inside the overall series for \(T_N\) there are a series of \((\log_2 N + 1)\) of these \(4^k \frac{1}{4}\) terms which on their own can be have a constant part of \(-\frac{1}{12}\). They come from the fact that \(T_N\) can be written as a sum of a bunch of \(R_N\), each of which has a constant term. Evidently \(T_N\), with \(N\) terms, ‘contains’ another series inside of it with \((\log_2 N + 1)\) terms, which has constant part \(-\frac{1}{12}\).

The exact value of the whole \((1 + 4 + 4^2 + \ldots) (\frac{1}{4})\) series when we use the exact sum instead of the divergent one is

\[\begin{aligned} T_N &\? (1 + 4 + 4^2 + \ldots) (\frac{1}{4}) \\ &= \frac{1}{4} S_{\log_2 N + 1}(4) \\ &= (\frac{1}{4}) \frac{1 - 4^{\log_2 N + 1}}{1 - 4} \\ &= (\frac{1}{4}) \frac{4N^2-1}{3} \\ &= -\frac{1}{12} + \frac{N^2}{3} \\ \end{aligned}\]

Therefore a complete expression for \(T_N\) is that, plus all the \(\pm\) terms that we dropped:

\[T_N = -\frac{1}{12} + \frac{N^2}{3} + \text{(a bunch of oscillating terms)}\]

And the reason that the \(T - 4T = R\) manipulation appeared to work is that it was actually computing

\[\begin{aligned} T - 4T &= (R_N + 4 R_{\lfloor N/2 \rfloor} + 4^2 R_{\lfloor (\lfloor N/2 \rfloor)/2 \rfloor} + \ldots) - 4(R_N + 4 R_{\lfloor N/2 \rfloor} + \ldots) \\ &= R_N + 4 (\underbrace{\cancel{R_{\lfloor N/2 \rfloor} - R_N}}_{\text{??}}) + 4^2 (\underbrace{\cancel{R_{\lfloor (\lfloor N/2 \rfloor)/2 \rfloor} - R_{\lfloor N/2 \rfloor}}}_{\text{???}} ) + \ldots \\ &\? R_N \end{aligned}\]

Which, again, is definitely invalid in general, but does work if you consider only the constant terms of each of the \(R\) since they are all \(\frac{1}{4}\) with no \(N\)-dependence.

So I guess that’s why the \(-\frac{1}{12}\) shows up: it is really there, but only when you divergently sum up the series created by all \(R_N\)’s constant terms, ignoring their actual value. Including those gives the partial sums the overall shape of a parabola \(-\frac{1}{12} + \frac{N^2}{3}\), and adding back in all the oscillating terms as well gives the full \(T_N = N(N+1)/2\).

I am still not quite sure how to reconcile this expansion with the fact that the closed form for \(T_N\) is \(\frac{N(N+1)}{2}\), though. I guess there’s a sense in which these closed forms want to be written as a sum of oscillating- and non-oscillating terms? Wikipedia suggests that the sense in which it is correct is that, as \(N \ra \infty\), the best parabolic approximation to \(T_N\) is one which ‘smoothes out’ the jumps at each integer discontinuity. For whatever reason constructing \(T_N\) in this way correctly handles that, whereas just writing its value down directly does not.

Okay, that’s enough for now. Perhaps later I will come back and add some more worked-out sums on here. I am fairly confident this sort of technique continues to work on most other divergent sum results, but I’ll have to try them to be sure.

5. A philosophical note

What I have done with each series here demonstrate something that I have found to be a helpful technique for resolving a class of mathematical paradoxes, in particular anything that involves infinities and limits. The general approach: whenever there is a limit \(\lim_{N \ra a}\) taken in the course of demonstrating some paradoxical outcome, instead of thinking of the value of \(N\) canceling out, think of it as just going unstated in the rest of the problem. Basically think of it as still being there but you’re just electing not to write it anymore. If later on you produce a result which seems paradoxical, go back and restore the \(N\)s throughout the computation and see if they resolve the surprise. Usually they do.

I consider this to be a finitist approach, and the arguments in this paper to me demonstrate the merit of this type of finitism. I keep finding that very many of the things which are otherwise vexing in higher mathematics simply disappear if you insist on finitism: you can’t get surprising results from limits if you refuse to take limits at all! Finitism (or at least, my flavor of it, which I have not seen written down anywhere yet) is so often helpful for ensuring nothing bizarre happens that I have come to think that it will continue to work for anything else I study.

The downside to finitism and the rejection of limits is that much of the content of mainstream mathematics becomes basically unusable—not because it’s wrong or anything, but because it’s just about the wrong stuff. In particular topology and analysis are built entirely on the assumption that numbers can be compared with basically “infinite resolution”, and therefore many of their theorems do not really survive when you reject this assumption. As a result learning these subjects is a slog. On the other hand finitism can maybe save you time when studying physics, because physics is inherently finitistic: you can never have truly infinite quantities (except for self-induced coordinate singularities, which can be handled). Therefore, sticking with finitistic math means that you will never have to worry about mathematical paradoxes sneaking into your physical models which you would then have to deal with. Probably this is beneficial.

The main reason I have written this article is in order to have these calculations set down somewhere so I can not worry about them again. But I also wanted to demonstrate some of the merits of my finitist philosophy. Sticking with finite results whenever possible lets us say only things which are unobjectionably true. We see that for these (comparatively gentle) divergent sums, all of the special ‘techniques’—like analytic continuation, rules about when you can interchange or respace terms, smoothing with mollifiers, whatever else—are not necessary; careful math gives the same answers with less theory and with no question as to the interpretation of the result. I can’t confidently say that this will work in general since I’ve tackled only the easiest examples here, but I do think it is a worthy approach, and far more philosophically sound than the other things I see people trying.

Note that the indexing is off by one because \(S_N(x) = 1+x+x^2+\ldots+x^N\) has \(N+1\) terms whereas \(P_N = 1+1+1+1\ldots = N\) has \(N\). ↩

Linear Algebra with Frames

2026-07-10T00:00:00+00:00

I have a bunch of notations I like to use for linear algebra and vector calculus which are at this point scattered across various notes and posts. I wanted to consolidate them in one place so that I can refer to them later.

I am pretty sure that the ideas in here are useful and that various parts of math and physics would benefit from incorporating them. However, I’m not sure I have figured out the best ways of stating them all yet. Probably I will make some edits to this over time as I settle on better and less-confusing ways of writing things. Note that the concept of a frame, and its utility in interpreting linear algebra, is not coupled to the division notation which I like to use to write it, so if you don’t like the notation feel free to ignore it.

(Most of this crystallized for me a few years ago, but I got so “blocked” on getting it exactly right that I lost motivation and put the whole thing away every time I tried to write it up. Recently I decided it would be better to just publish the broken version so I can stop having it hanging over me, and I can fix it later if I figure it out, or if somebody clues me into what I’m missing.)

All vector spaces considered are finite-dimensional and over \(\bb{R}\) and come with the standard Euclidean dot product. Much of this works without those restrictions but it’s good to start with something easy to visualize. The linear span of vectors or vector spaces is denoted \(\< \x, \y, \ldots \> \equiv \span(\x, \y, \ldots)\).

1. Projections

First, some useful notations for talking about projections and subspaces. Even if the rest of this article seems silly you should still help me standardize these.

The projection operator (or ‘projector’) for a vector space \(X\) is written

\[I_X\]

I prefer the symbol \(I\) over the usual \(P\) because these operators are exactly the identity transformation for the spaces they project onto. This is why the idempotency condition \(P^2 = P\) which is often taken as a definition for projections holds—because identities are inherently idempotent:

\[I_X^2 = I_X\]

Given any basis for a space \(X\), the matrix representation of \(I_X\) will look like an identity operator. For example if \(X = \< \x, \y\> \sub \bb{R}^3 = \< \x, \y, \z \>\) and \(I_X\) acts on a vector \(\b{a} = a_x \x + a_y \y + a_z \z\), then¹

\[\begin{aligned} I_{X} \b{a} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} = \begin{pmatrix} a_x \\ a_y \\ 0 \end{pmatrix} \end{aligned}\]

In the case where the vector space labeled by a projection is one dimensional, like \(\<\b{x} \>\), we write just the non-bolded symbol for subspace instead for brevity:

\[I_x \equiv I_{\< \x\>}\]

If the vector space is spanned by multiple basis vectors which have letter names, like \(\<\x, \y\>\), then we also write the list of all of these as the subscript to label the space:²

\[I_{xy} \equiv I_{\< \x, \y \>}\]

If a vector space decomposes into subspaces \(V = X \oplus Y\), then its projections also decompose:

\[I_V = I_X + I_Y\]

For example, if \(V = \bb{R}^3\) with its \((\x, \y, \z)\) basis then we can write

\[\begin{aligned} I_{xy} &= I_x + I_y \\ I_{xyz} &= I_{xy} + I_z = I_x + I_y + I_z \\ \end{aligned}\]

Vector projections onto a subspace are shorthanded by writing the subscript on the vector itself:

\[\b{a}_X \equiv I_X \b{a}\]

Decompositions of a vector space and its projectors carry forward onto vector projections:

\[\b{a}_{V} = \b{a}_{X \oplus Y} = \b{a}_X + \b{a}_Y = (I_X + I_Y) \b{a}\]

Therefore in \(\bb{R}^3\), all of these ways of writing a vector are equivalent:

\[\b{a} = \b{a}_{xyz} = \b{a}_{xy} + \b{a}_z = \b{a}_x + \b{a}_y + \b{a}_z\]

Given a vector subspace \(X \sub V\), there is an orthogonal subspace \(X_{\perp}\) (which I pronounce “\(X\)-perp”) such that \(V = X \oplus X_{\perp}\). (The word ‘orthogonal’ here technically requires an inner product, but if one is not available you can call it the complementary subspace instead; the idea is the same.) The projection onto this subspace is

\[I_{\perp X} = I - I_X\]

In subscripts I always write \({\perp} X\) instead of \(X_{\perp}\), just because it is easier to read as the symbols become smaller.³

The projector for this orthogonal subspace is the rest of the vector after removing the projection onto \(X\):

\[\b{a}_{\perp X} = I_{\perp X} = (I - I_X) \b{a} = \b{a} - \b{a}_X\]

A vector space \(V\) which contains \(X\) generically factors into \(V = X \oplus X_{\perp}\), and each of its vectors can be split according to this factorization⁴

\[\b{a} = I_V \b{a} = (I_X + I_{\perp X}) \b{a} = \b{a}_X + \b{a}_{\perp X}\]

When we write a vector \(\b{a}\) without any subscript, we usually ambiently know that it is supposed to be an element of some explicit vector space, \(\b{a} \in V\). Sometimes it is elegant to regard this as a shorthand for explicitly projecting it into that space:

\[\b{a} \stackrel{\text{means?}}{=} \b{a}_V = I_V \b{a}\]

This doesn’t affect any results, but it can be a nice way to think about things. In particular it implies the interpretation that \(\b{a}\) really refers to what we know about \(\b{a}\), based on the space in which we are viewing it, but that perhaps there is more data in there which we are missing by restricting it to this space.

Similarly, the symbol \(I\) with no subscript at all means “the identity on the ambient space \(V\)”, but we can think of it as a shorthand for an explicit \(I_V\). A scalar \(1\) is often interpreted as a sort of “identity on everything” operation as well; I tend to think that this is a mistake, and we should think of all scalars as being scoped to some particular space that they act on. But this is a philosophical distinction and mostly should not affect anything.⁵

2. Vector Division

Projectors like \(I_{xy}\) and \(I_V\) are linear transformations, which look like the identity on their respective subspaces. Their matrix representations are sparse: they have \(1\)s along the diagonal corresponding to the subspace and \(0\)s elsewhere. A very compact way of writing linear transformations of this form is by listing only the terms that are nonzero. My preferred way of writing this is as vector division:

\[\begin{aligned} I_x &= \frac{\x}{\x} \\ I_y &= \frac{\y}{\y} \\ I_{xy} &= I_x + I_y = \frac{\x}{\x} + \frac{\y}{\y} \end{aligned}\]

Here \(\frac{\x}{\x}\) is interpreted as the linear transformation \(\x \o \x^{-1} = \x \o \frac{\x^*}{\| \x \|^2}\). More generally we write

\[\frac{\b{b}}{\b{a}} \equiv \b{b} \o \b{a}^{-1} = \b{b} \o \frac{\b{a}^*}{\| \b{a} \|^2}\]

Theses are the rules for interpreting a vector division expression:

Fractions do not cancel with themselves. Instead \(\x / \x\) is interpreted as a shorthand for \(\x \o \x^{-1}\).
\(\x^{-1}\) or \(\frac{1}{\x}\) means \(\frac{\x^*}{\| \x \|^2}\), the unique vector in \(\< \x \>\) which has \(\x^{-1} \cdot \x = 1\).
For convenience, juxtaposition of an inverse vector on the left with a numerator or bare vector on the right implies a dot product: \(\frac{1}{\x} \b{a} = \x^{-1} \b{a} \equiv \frac{\x}{\| \x \|^2} \cdot \b{a} = a_x\).
No contraction occurs if a denominator is on the left of a numerator. Instead, a tensor product is implied: \(\b{a} \b{a}^{-1} \equiv \b{a} \o \b{a}^{-1}\)
The dual symbol \(\b{a}^*\) is used whenever we write a denominator as a numerator instead, to remind us that it is ‘acting like’ a denominator. We mostly don’t use it in its normal sense of dualization, but I am including it to connect to the way things are usually written.

In the case of unit \(\x, \y\), we have

\[\frac{\x}{\y} = \x \o \y^* = \x \y^*\]

In the case of non-unit \(\b{a}\) we have

\[\frac{\b{b}}{\b{a}} = \b{b} \o \frac{\b{a}^*}{\| \b{a} \|^2} = \frac{1}{\| \b{a} \|^2} \b{b} \b{a}^*\]

The implication of these rules is that a denominator \(\x\) selects only the numerator \(\x\)s from the thing it acts on:

\[\begin{aligned} I_x \b{a} &= \frac{\x}{\x} (a_x \x + a_y \y + a_z \z) \\ &= a_x \frac{\x}{\x} \x + a_y \frac{\x}{\bcancel{\x}} \bcancel{\y} + a_z \frac{\x}{\bcancel{\x}} \bcancel{\z} \\ &= a_x \x \\ &= \b{a}_x \\ I_{xy} \b{a} &= (\frac{\x}{\x} + \frac{\y}{\y}) (a_x \x + a_y \y + a_z \z) \\ &=a_x \x + a_y \y \\ &= \b{a}_{xy} \end{aligned}\]

Hopefully it is clear that this might be a useful way of writing things. Let me emphasize, though, that these notations to be understood as shorthands for explicit operations on the underlying tensors. The reason \(\x/\x\) does not cancel out to give \(1\) is because it is such a common object that it’s convenient to regard it as a tensor, and \(\x^{-1} \b{a}\) implies a dot product because it’s useful, not because we are defining a canonical way of doing algebra that cancels inverses but not other terms. We are always free to insert dot and tensor product symbols to make more complicated expressions like \((\x \o \y^{-1} \o \z) \cdot \b{a}\) unambiguous. Also, although there is nothing stopping one from writing higher-degree tensors with more fractions, like \(\Gamma = \x \y \frac{1}{\x}\), the division notation stops be as helpful at that point because there is not a good way to do anything with the term in the middle.

This interpretation of \(1/\x\) is as a pseudoinverse of the linear transformation given by dotting with \(\x\), \(\x \cdot: \b{a} \mapsto \x \cdot \b{a}\), evaluated at \(1\). Being a pseudoinverse it is nonzero on the image of \(\x \cdot\) and \(0\) elsewhere.

\[\begin{aligned} \x^{-1} \cdot \x &= 1 \\ \x^{-1} \cdot \y &= 0 \\ \x^{-1} \cdot \z &= 0 \\ \end{aligned}\]

The job of division by a vector \(\x\) is to extract the \(\x\)-component of the vector it acts on,

\[\frac{1}{\x} \b{a} = \frac{1}{\x}(a_x \x + \b{a}_{\perp x}) = a_x\]

Multiplying by \(\x\) again therefore gives the vector projection:

\[\b{a}_x = \frac{\x}{\x} \b{a} = \frac{\b{a} \cdot \x}{\| \x \|^2} \x\]

In my opinion these definition of the scalar and vector projections are the most useful ones. Note that the scalar projection is usually defined as the component of \(\b{a}\) along the unit vector \(\hat{\x} = \x / \| \x \|\) instead,

\[\proj_{x} (\b{a}) \? \b{a} \cdot \frac{\x}{\| \x \|}\]

such that \(\b{a}_x = a_{\hat{x}} \hat{\b{x}}\), and it gives the “shadow” of \(\b{a}\) along the direction of \(\b{x}\). I think this definition strictly less useful, since it can’t express the general concept of the “\(\x\)-component of \(\b{a}\)”. You can always recover it by manually projecting onto a unit vector if you want.

Vector division notations are helpful for doing symbolic algebra on linear transformations, especially sparse ones. For example the operator of rotation in the \((x,y)\) plane of \(\bb{R}^3\) can be written

\[R_{xy} = (\frac{\y}{\x} - \frac{\x}{\y}) + \frac{\z}{\z} = r_{xy} + I_z\]

Where \(R_{xy}\) is the entire operator and \(r_{xy} = \y/\x - \x/\y\) is the generator (also related by \(R_{xy} = e^{r_{xy} \frac{\pi}{2}}\).) Mirroring across the \(x=y\) axis is given by:

\[P_{xy} = \frac{\y}{\x} + \frac{\x}{\y} + I_z\]

Et cetera. Calculations with linear transformations written in this form consist of cancelling terms, which is fairly easy to do by hand (certainly compared to multiplying matrices). Here the surviving three terms are those where the denominator on the left term matches the numerator on the right:

\[\begin{aligned} R_{xy}^2 &= (\frac{\y}{\x} - \frac{\x}{\y} + \frac{\z}{\z}) (\frac{\y}{\x} - \frac{\x}{\y} + \frac{\z}{\z}) \\ &= (\frac{\y}{\x}) (- \frac{\x}{\y} ) + (-\frac{\x}{\y})(\frac{\y}{\x}) + (\frac{\z}{\z})(\frac{\z}{\z}) \\ &= -\frac{\y}{\y} - \frac{\x}{\x} + \frac{\z}{\z} \\ &= - I_{xy} + I_z \end{aligned}\]

More general calculations proceed by factoring the term on the right:

\[\frac{1}{\x} \b{a} = \frac{1}{\x} (\b{a}_x + \b{a}_{\perp x}) = \frac{1}{\x} (a_x \x + \b{a}_{\perp x}) = a_x\]

A projector like \(I_{xy}\) is the identity on the space it describes regardless of the choice of basis for that space. If \(\< \x, \y \>\) is also spanned by some other orthogonal basis \(( \b{e}^1, \b{e}^2 )\), then in that basis \(I_{xy}\) takes the form

\[I_{xy} = \frac{\x}{\x} + \frac{\y}{\y} = \frac{\b{e}^1}{\b{e}^1} + \frac{\b{e}^2}{\b{e}^2}\]

which still has matrix form \(\diag(1,1)\) in this basis. (The orthogonality of \((\b{e}^1, \b{e}^2)\) is important for this; more on that later.)

Another useful thing you can do is talk about the projections and rejections for an arbitrary vector in a compact way. Suppose \(\b{n}\) is a normal vector to a surface \(S\) in \(V = \bb{R}^3 = \< \x, \y, \z \>\). Then we can write the rejection and projection operators for tangent vectors for this surface as

\[\begin{aligned} I_n =& I_{\perp S} = \frac{\b{n}}{\b{n}} \\ I_S &= I_{\perp n} = I_{xyz} - I_n = \frac{\x}{\x} + \frac{\y}{\y} + \frac{\z}{\z} - \frac{\b{n}}{\b{n}} \end{aligned}\]

If we happen to know that orthogonal vectors \((\b{u}, \b{v})\) span the surface \(S\), we can also write these as

\[\begin{aligned} I_S &= I_{uv} = \frac{\b{u}}{\b{u}} + \frac{\b{v}}{\b{v}} \\ I_{\perp S} &= I_n = I_{xyz} - I_{uv} \\ &= (\frac{\x}{\x} + \frac{\y}{\y} + \frac{\z}{\z}) - (\frac{\b{u}}{\b{u}} + \frac{\b{v}}{\b{v}}) \end{aligned}\]

These are quite a bit easier to work with than their matrix representations, especially on paper. If you want you can factor them out into individual terms to get something equivalent to the matrix form in a particular basis. For example, if \(\b{n} = \x + \y + \z\), then this becomes

\[\begin{aligned} I_n &= \frac{\x + \y + \z}{\x + \y + \z} \\ &= (\x + \y + \z) \o \frac{\x^* + \y^* + \z^*}{\| \x + \y + \z \|^2} \\ &= \frac{1}{3} (\x + \y + \z) \o (\x^* + \y^* + \z^*) \\ &= \frac{1}{3} [\frac{\x}{\x} + \frac{\x}{\y} + \frac{\x}{\z} + \frac{\y}{\x} + \frac{\y}{\y} + \frac{\y}{\z} + \frac{\z}{\x} + \frac{\z}{\y} + \frac{\z}{\z}] \\ I_S &= \frac{\x}{\x} + \frac{\y}{\y} + \frac{\z}{\z} - \frac{1}{3} (\text{...all that stuff...}) \end{aligned}\]

where the complicated formulas’ components are exactly the components of the equivalent matrix in the \((\x, \y, \z)\) basis. Usually it is preferable not to do this, though.

In some fields you will see dyadics like \(\x\x\) used to express the same idea as \(\x/\x\), since for unit \(\x\) they are equal: \(\x\x \cdot \b{a} = \x (\x \cdot \b{a}) = a_x \x\). However, I think this notation is strictly inferior to my vector division. You really do want to represent the ‘input’ as a denominator, since for most purposes it acts like one. In the case of \(\x\x\), the \(\x/\x\) notation correctly handles the case where \(\x\) is of non-unit length. It also generalizes in useful ways that dyadics cannot, to be described shortly. Juxtaposition notation is useful for constructing tensors, especially of degree \(>2\) like \(\x \y \z\). It just doesn’t work as well for linear transformations.

A helpful thing about the \(I_x = \frac{\x}{\x}\) notation is that it explicitly labels the “input” and “output” vectors of the linear transformation: denominators are inputs; numerators are outputs. This also reflects the way the two components transform under coordinate changes: numerators are covariant and denominators are contravariant, i.e. transform with the inverse transformation. The more standard way to express this is to rely on the vector space dual operation, \(I_x = \x \o \x^*\) where the asterix indicates a contravariant dual vector \(\x^* \in V^*\). I don’t like this as much. Dual vectors are awkward, and it doesn’t automatically create a projector if \(\x\) has non-unit length, since under the transformation \(\x \ra 2\x\) this needs to transform as \(I_x = (2 \x ) \o (\frac{1}{2} \x^*)\). I think that since the second term transforms like an inverse we may as well write it as one. The more complicated case where we transform coordinates by mixing them together can also be handled elegantly, as we will see later.

Generally speaking, dualization and conjugation operations across linear algebra have the behavior of inverting the direction of vectors while preserving the magnitude, e.g. \(\overline{re^{i \theta}} = re^{-i\theta}\) (which is well-defined for \(\bb{R}^{> 2}\) by the way, \(\overline{re^{\vec{r} \cdot \vec{\theta}}} = re^{-\vec{r} \cdot \vec{\theta}}\)). This is fine… unless what you actually want to be doing is measuring one vector in terms of another, like \(\x^{-1} (a_x \x + \b{a}_{\perp x}) = a_x\), in which case, shouldn’t you just actually be able to invert “all” of \(\x\) to make the division work? This is basically the argument for vector division: it is the operation of measuring one vector in terms of another. And very rarely in geometry do you have to deal with the problem of thinking about vectors with norm zero (which are not really a problem anyway, if you allow for the generalized-inverse perspective, to be discussed later).⁶

3. Frames

We saw that \(\frac{1}{\b{x}} \b{a} = a_x\) gives the “\(x\)-coordinate of \(\b{a}\)”, that is, the object which when multiplied by \(\x\) gives \(\b{a}_x\) again:

\[\b{a}_x = a_x \x = \frac{\x}{\x} \b{a} = I_x \b{a}\]

The same calculation can be extended to subspaces which are more than one-dimensional. For example, we can project \(\b{a}\) onto \(\< \x, \y \>\) with

\[I_{xy} \b{a} = (\frac{\x}{\x} + \frac{\y}{\y}) \b{a} = a_x \x + a_y \y\]

It is helpful to think of this as the product of a tuple with another tuple,

\[I_{xy} \b{a} = (a_x, a_y) \cdot (\x, \y) = a_{xy} (\x, \y)\]

such that \(a_{xy} = (a_x, a_y)\) can be thought of as the “\((xy)\)-coordinate of \(\b{a}\)”. Of course, the word ‘coordinate’ is not quite valid anymore, since it is not a scalar but a tuple of scalars. Nevertheless it is a good label. Describing vectors in this way turns out to be nice way of thinking about things.

The object \((\x, \y)\) here is what I like to call a frame, which is a generalization of a vector. A frame is a tuple of linearly-independent vectors, or a ‘vector of vectors’—isomorphic to a matrix, but with a particularly useful interpretation. We will also say that a \(k\)-frame is a tuple of exactly \(k\) such vectors, and for most purposes a single vector can be regarded as a \(1\)-frame. A basis for an \(N\)-dimensional vector space is therefore an \(N\)-frame. An orthogonal frame is a frame whose vectors are mutually orthogonal, and an orthonormal frame is one whose vectors are orthonormal. A surprising amount of linear algebra generalizes to work on frames instead of vectors without much modification.⁷

I usually label frames with capital letters, and when possible choose the same letter as the vector space they span, for example \(X = (\x, \y)\) is a frame which spans the vector space \(X = \< \x, \y \>\). This just turns out to be convenient in practice, but there are cases where you want to distinguish the two. For instance you might have two distinct frames for the same subspace, like \((\x, \y)\) and \((\x + \y, \x - \y)\) which are both bases for \(\< \x, \y \>\), in which case you will want to give them different names to distinguish them. But for the most part conflating frames and the vector spaces they span is a useful convention. (Sometimes I think it might be better to use bold capital letters for frames, so that “bold” always means “vector”, but for now I won’t.)

Frames allow us to use the same notations for components of vectors along subspaces of any dimension. If a vector space \(V = X \oplus Y\), then a vector \(\b{a} \in V\) decomposes as \(\b{a} = \b{a}_X + \b{a}_Y\), and we can write each component as a product of a coordinate and a frame for that subspace:

\[\begin{aligned} \b{a} &= I_X \b{a} + I_Y \b{a} \\ &= a_X X + a_Y Y \\ &= (a_{x_1}, a_{x_2}, \ldots) (\x^1, \x^2, \ldots) + (a_{y_1}, a_{y_2}, \ldots) (\y^1, \y^2, \ldots) \end{aligned}\]

A particularly common usage for this is that it lets us generically factor \(\b{a}\) into its components on and off of a subspace:

\[\b{a} = \b{a}_X + \b{a}_{\perp X} = a_X X + a_{\perp X} X_{\perp}\]

which implies bases for \(X\) and \(X_{\perp}\) without having to actually write down what they are.

As with vectors, we use division notation for extracting the projections onto frames:

\[\b{a}_X = \frac{X}{X} \b{a} = X (\frac{1}{X} \cdot \b{a}) = a_x X\]

To make sense of this, consider a frame \(X = (\x, \y)\), which for now we assume to be orthogonal (we’ll discuss the non-orthogonal case in the next section). The inverse frame consists of the componentwise inverses of each vector in the frame:

\[\frac{1}{X} = \frac{1}{(\x, \y)} = (\frac{1}{\x}, \frac{1}{\y})\]

When this is juxtaposed with a vector, we take it to imply a dot product which “distributes over the comma” in the frame, producing a tuple of coordinates:

\[\frac{1}{X} \b{a} \equiv \frac{1}{X} \cdot \b{a} = (\frac{1}{\x}, \frac{1}{\y}) \cdot \b{a} = (\frac{1}{\x} \cdot \b{a}, \frac{1}{\y} \cdot \b{a}) = (a_x, a_y)\]

The projection \(I_X = \frac{X}{X}\) then ‘dots’ the frame with this vector again, producing the vector projection again:

\[\frac{X}{X} \b{a} = X (\frac{1}{X} \cdot \b{a}) = (\x , \y) (a_x, a_y) = a_x \x + a_y \y = \b{a}_X\]

To clarify the general meaning of this operation I’ll translate it into index notation. A generic frame is given by an upper-indexed list of vectors

\[X = (\x^1, \x^2, \ldots) = \x^i\]

Such that an arbitrary vector written out in terms of the frame’s vectors pairs a lower-indexed coordinate with an upper-indexed frame:

\[\b{a}_X = a_X X = a_i \x^i = (a_1, a_2, \ldots) (\x^1, \x^2, \ldots) = a_1 \x^1 + a_2 \x^2 + \ldots\]

Given a choice of basis \(\b{e}^j\), the frame is equivalent to a matrix action on that basis,

\[X = \x^i = X^i_{\, j} \b{e}^j\]

That is, \(X\) can be thought of as a linear transformation from any choice of basis into the vectors \((\x^i)\), with \(X^i_{\, j} = \x^i \cdot \b{e}_j\). But this just kicks the interpretive can down the road, because now you have to make sense of the basis frame \(E = \b{e}^j\). So we write \(X = \x^i\) and leave it at that. The individual vectors of \(X\) are to be understood as ‘geometric’ vectors, quantities you can measure which act like vectors, rather than numerical data in some particular coordinate system.

The inverse frame is written with lowered indices:

\[\frac{1}{X} = (\x_1, \x_2, \ldots) = \x_i\]

Such that

\[X \cdot \frac{1}{X} = \x^i \cdot \x_j = 1_{i=j}\]

Which is the sense in which it is the ‘inverse’ of \(X\). Equivalently, their matrix representations in any basis are pseudoinverses:

\[X \cdot \frac{1}{X} = (X^i_{\, j} \b{e}^j) \cdot (X_i^{\, k} \b{e}_k) = \b{e}^j \o \b{e}_k \mid_X = I_X\]

The inverse frame acting on a vector produces the lower-indexed vector components in that frame:

\[\begin{aligned} \frac{1}{X} \b{a} &= \x_i \cdot \b{a} \\ &= (\x_1 \cdot \b{a}, \x_2 \cdot \b{a}) \\ &= (a_1, a_2, \ldots) \\ &= a_X \end{aligned}\]

Which follows from noting that it holds if you write \(\b{a}\) in its \(X\) components:

\[\x_i \cdot \b{a} = \x_i \cdot (a_j \x^j) = a_j 1_{i=j} = a_i\]

The rest of the terms cancel because of \(\x_i \cdot \x^j = 1_{i=j}\). Note how \(a_X\) is a shorthand for \(\frac{1}{X} \cdot \b{a}\): lower indexes imply division, act like denominators, and are contravariant; upper indexes and numerators are covariant.

In index notation, the projector \(X/X\) contains a contraction over the frame index:

\[\frac{X}{X} = \x^i \x_i\]

And its action on a vector is given by first extracting the \(a_X\) components, then multiplying by the frame again:

\[\frac{X}{X} \b{a} = \x^i \x_i \cdot \b{a} = a_i \x^i = a_X X = \b{a}_X\]

Fully written out in matrices over some other \((\b{e}^j)\) basis this becomes

\[\frac{X}{X} \b{a} = (X^i_{\, j} \b{e}^j X_i^{\, k} \b{e}_k) \cdot (a_l \b{e}^l) = (X^i_{\, j} \b{e}^j) X_i^{\, k} a_k = \x^i (X_i^{\, k} a_k)\]

Frames can be thought of as a choice of coordinate system for the vector space they span. For example, \((\x, \y)\) is a basis for the vector space \(X = \< \x, \y \> \subset V\). Division by \((\x, \y)\) is therefore a linear transformation from the vector space \(X\) itself to some other vector space—a vector space of coordinates for \(X\) in this particular frame, which I like to write as \(\cal{I}_X\) (which I call the “index space”, for lack of a better name). Then the frame \(X = (\x, \y)\) itself is a linear transformation \(\cal{I}_X \ra X\), so it takes a vector of coordinates for \(X\) and maps them back to \(X\) itself. This is what is happening when we write out a vector in terms of coordinates:

\[\b{a}_X = a_X X \in X \subset V\]

In this sense we see that \(I_X = X/X\) amounts to contracting \(X\) and \(X^{-1}\) on their \(\cal{I}_X\) index, resulting in the identity on the space \(X\). Contracting on their \(X\) index on the other hand gives the identity on the index space,

\[\frac{1}{X} \cdot X = \x_i \cdot \x^j = I_{\cal{I}_X}\]

That is, it’s the identity operator \(\diag(1, 1, \ldots)\) but in the basis for the coordinates \(\cal{I}_X\), rather than \(X\) itself. This works like you’d expect: if \(a_X = (1/X) \b{a}\) then \((I_{\cal{I}_X}) a_X = (1/X \o X) \cdot (1/X) \b{a} = (1/X) \b{a} = a_X\) again.

Frame division, therefore, involves contracting two frames on their frame index. We can write this out explicitly by labeling a basis frame for \(\cal{I}_X\) as \((e^1, e^2, \ldots)\), where these vectors are not in bold because they are vectors in the index space \(\cal{I}_X\), not the geometric space \(X\). Then

\[\begin{aligned} \frac{Y}{X} &= \frac{(\y^1, \y^2, \ldots)}{(\x^1, \x^2, \ldots)} \\ &= (\y^1 e_1 + \y^2 e_2 + \ldots) \cdot (\x_1 e^1 + \x_2 e^2 + \ldots) \\ &= \y^1 \x_1 + \y^2 \x_2 + \ldots \\ &= \y^i \x_i \end{aligned}\]

Since \(Y/X\) implicitly contracts on an index, it only makes sense when the numerator and denominator have the same number of components. The vector division notation \(\b{b} / \b{a} = \b{b} \o \b{a}^{-1}\) can be thought of as the \(n=1\) case of this, since vectors are \(1\)-frames.

Note that this is really what is meant by the \(X = (\x^1, \x^2, \ldots)\) tuple notation: it is a ‘vector’ over a choice of basis for \(\cal{I}_X\)

\[X = \x^1 e_1 + \x^2 e_2 + \ldots\]

and the dual frame is a vector over \(\cal{I}_X^*\)

\[\frac{1}{X} = e^1 \x_1 + e^2 \x_2 + \ldots\]

A vector’s coordinates in a particular frame \(X\) form a vector in the index space \(\cal{I}_X\):

\[\begin{aligned} \frac{1}{X}\b{a} &= (e^1 \x_1 + e^2 \x_2 + \ldots) \cdot \b{a} \\ &= a_1 e^1 + a_2 e^2 + \ldots \\ &= (a_1, a_2, \ldots) \in \cal{I}_X \end{aligned}\]

So a frame’s coordinates \(X = \x^1 e_1 + \x^2 e_2 + \ldots\) are a linear transformation \(\cal{I}_X \ra X\). The choice to write a linear transformation \(X \ra Y\) out in terms of list of vectors \(X = (\x^1, \x^2, \ldots)\) and \(Y = (\y^1, \y^2, \ldots)\) is actually the choice to insert an intermediate vector space inside the transformation, \(X \ra \cal{I}_X = \cal{I}_Y \ra Y\), which we then use to match up the basis projections of the inputs and outputs.

Each of these objects make sense without reference to the \(\cal{I}_X\) coordinate system at all, however, and we could clearly have chosen any other list of vectors \(\in X\) as a basis instead, producing a different choice of basis \((e^i)\) for \(\cal{I}_X\). The expansion of frames into bases is sort of philosophically superfluous: it is a choice of a useful way to measure things in order to write them out in a way that is computationally tractable. Even the choice of the number of vectors in the basis is a bit superfluous (although you might need to fiddle with the precise definitions of linear algebra if you allow this to vary).

So vectors and frames both acquire their components when they are ‘measured’ in a particular basis, with frames being general lists of (linearly-independent) vectors and vectors being the \(k=1\) case of frames. The value of ‘thinking in frames’ is that often the geometric objects you want to think about really are lists of vectors, rather than individual vectors, which transform as a unit. Linear transformations really act like ratios of frames, contravariant bases really act like denominators, etc.

4. Non-orthogonal frames

So far we have discussed projections onto frames under the assumption that the vectors in the frames are orthogonal. This is the condition under which fractions expand termwise:

\[\begin{aligned} \frac{Y}{X} &= \frac{(\y^1, \y^2, \ldots)}{(\x^1, \x^2, \ldots)} \\ \stackrel{\text{orthogonal}}{\implies} &= \frac{\y^1}{\x^1} + \frac{\y^2}{\x^2} + \ldots \\ &= \y^1 \o \x_1 + \y^2 \o \x_2 + \ldots \\ &= \y^1 \o \frac{\x^{1*}}{\| \x^1 \|^2} + \y^2 \o \frac{\x^{2*}}{\| \x^2 \|^2} + \ldots \end{aligned}\]

In the case where the denominator \(X\) is not orthogonal, we can still perform a similar expansion, but it is somewhat more complicated, for the same reason that inverting a general matrix is a lot more complicated than inverting a diagonal one.

The complexity is very practical. When we start with a frame \(F = (\b{e}^1, \b{e}^2, \ldots) = (\b{e}^i)\) and write a vector in terms of it

\[\b{a} = a_1 \b{e}^1 + a_2 \b{e}^2 + \ldots = a_i \b{e}^i\]

We want to be able to use the inverse frame \(1/F = (\b{e}_1, \b{e}_2, \ldots) = (\b{e}_i)\) to extract these components:

\[\b{e}_i \cdot \b{a} = \b{e}_i \cdot (a_j \b{e}^j) = a_i\]

which requires that \(\b{e}_i \cdot \b{e}^j = 1_{i=j}\). This is more restrictive than just the requirement that \(\b{e}_i \cdot \b{e}^i = 1\), which is the stipulation filled by \(\b{e}^i / \| \b{e}^i \|^2\); we also require that \(\b{e}_i \cdot \b{e}^j = 0\) for \(j \neq i\). We will basically accomplish this by adding a term to \(\b{e}^i / \| \b{e}^i \|^2\) which cancels out its dot product with all of the other \(\b{e}^i\).

Consider a two dimensional frame as an example. Suppose we pick the non-orthogonal frame \(F = \b{f}^i = (\x, \x+\y)\) to describe the vector space \((\x, \y)\). It is still possible to decompose vectors in terms of this basis, \(\b{a} = a_1 \x + a_2 (\x + \y)\). However, the actual projections are no longer independent of each other: the value of \(a_1\) and \(a_2\) are not given by \(a_x = (\x/\x) \b{a}\) and \(a_y = (\y/\y) \b{a}\). We can still write

\[\b{a}_F = I_{xy} \b{a} = \frac{F}{F} \b{a} = \frac{(\x, \x+\y)}{(\x, \x+\y)} \b{a}\]

But we cannot factor it along each component, as

\[\frac{(\x, \x+\y)}{(\x, \x+\y)} \neq \frac{\x}{\x} + \frac{\x + \y}{\x + \y}\]

Instead, the correct projector has to use what is called the “dual” basis for the frame \(X\). It turns out to be⁸

\[\frac{1}{F} = (\x - \y, \y)\]

Because

\[\begin{aligned} (\x - \y) \cdot (\x) &= 1 \\ (\x - \y) \cdot (\x + \y) &= 0 \\ (\y) \cdot (\x + \y) &= 1 \\ (\y) \cdot (\x) &= 0 \\ \end{aligned}\]

That is,

\[F^{-1} \cdot F = 1_F\]

So the factorization of \(F/F\) happens componentwise by pairing the vectors of the numerator with the vectors of the denominator:

\[\begin{aligned} \b{a} &= \frac{F}{F} \b{a} \\ &= [\x \o (\x^* - \y^*) + \y \o \y^*] (a_1 \x + a_2 (\x + \y)) \\ &= a_1 \x + a_2 (\x + \y) \end{aligned}\]

Note that the dual frame is not an explicit denominator. Rather they are the dual vector components \(\b{e}_i\) of the denominator \(1/F\). True, we could write them as explicit denominators as well,

\[\begin{aligned} \frac{(\x, \x+\y)}{(\x, \x+\y)} &\? \frac{\x}{(\x - \y)/2} + \frac{\x + \y}{\y} = \x \o (\x^* - \y^*) + (\x + \y) \o \y^* \end{aligned}\]

But there’s little point to doing so. The inverses of the individual vectors like \(\frac{1}{2}(\x - \y)\) are not really useful in this geometry at all. The only thing that it makes any sense to dualize is the entire frame at once, \(1/F = (\x, \x + \y)^{-1} = (\x^* - \y^*, \y^*)\).

In general the inverse frame \(F^{-1}\) must have matrix elements corresponding to the inverse of the matrix elements of \(F\). The matrix representations are

\[\begin{aligned} F &= \begin{pmatrix} \x & \y \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} e_1 \\ e_2 \end{pmatrix} \\ &= \x(e_1 + e_2) + \y (e_2) \\ &= (\x) e_1 + (\x + \y) e_2 \\ &= (\x, \x + \y)\\[1em] F^{-1} &= \begin{pmatrix} e^1 & e^2 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \x^* \\ \y^* \end{pmatrix} \\ &= e^1 (\x^* - \y^*) + e^2 ( \y^*) \\ &= (\x^* - \y^*, \y^*) \end{aligned}\]

(where I have explicitly written out the bases for each index space as a demonstration). The coordinate matrices are of course inverses:

\[\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\]

The actual construction of the dual frame \(F^{-1}\) is therefore a matrix inverse, and has to be constructed using the same exterior algebra construction as all matrix inverses. The procedure should look something like this:

\[\frac{F}{F} = (F \frac{1}{F}) \cdot \frac{F^{\^n}}{F^{\^n}} = F \frac{F^{\^n-1}}{F^{\^n}}\]

This is supposed to be equivalent to the usual \(\adj(F)/\det(F)\) construction in, um, untyped linear algebra. However I have so far been unable to a simple set of rules for doing doing this calculation purely symbolically. The problem is that it’s not clear how to keep track of which index is left ‘unbound’ in the denominator (that produces the alternating minus signs when translated into the matrix inverse). It refers to a correct calculation on tensors of course, but I’m not satisfied with the shorthand. I’ll update this if I come up with something.⁹

In any case: if a denominator frame is not orthogonal, then we cannot factor the frames apart term-by-term:

\[\begin{aligned} \frac{F}{E} = \frac{(\b{f}^1, \b{f}^2)}{(\b{e}^1, \b{e}^2)} \stackrel{!}{\neq} \frac{\b{f}^1}{\b{e}^1} + \frac{\b{f}^2}{\b{e}^2} \end{aligned}\]

Instead we have to factor them according to the dual frame \(E^{-1} = 1/(\b{e}^i) = (\b{e}_i)\), always written with lower indexes when it is not in a denominator:

\[\begin{aligned} \frac{F}{E} = \frac{(\b{f}^1, \b{f}^2)}{(\b{e}^1, \b{e}^2)} = \b{f}^1 \b{e}_1 + \b{f}^2 \b{e}_2 \end{aligned}\]

The values of the \(\b{e}_i\) are functionally dependent on the entire frame \(F\), rather than just on \(\b{e}^i\) alone.

Once we have defined \(\b{e}_i\) in this way, everything we did with division continues to work. For example,

\[\frac{F}{E} \b{a} = \b{f}^i \b{e}_i \cdot (a_j \b{e}^j) = \b{f}^i (\b{e}_i \cdot a_j \b{e}^j) = \b{f}^i a_j 1_{i=j} = a_i \b{f}^i\]

We just can’t use \(\b{e}_i = \b{e}^{i*}/\| \b{e}^i \|^2\) anymore.

5. Linear Transformations

Linear transformations can generally be written out with a different frame in the numerator and denominator:

\[A = \frac{F}{E} = \frac{(\b{f}^1, \b{f}^2, \ldots)}{(\b{e}^1, \b{e}^2, \ldots)}\]

For example a rotation operator in \(V = \< \x , \y, \z \>\) might be written:

\[R_{xy} = \frac{(\y, -\x, \z)}{(\x, \y, \z)} = \frac{R_{xy}V}{V}\]

Its action on a vector in the \(V\) frame is to rewrite it in the \(R_{xy}V\) frame:

\[R_{xy} (\b{a}) = \frac{R_{xy} V}{V} (a_V V) = a_V (R_{xy} V) = (a_x, a_y, a_z)(\y, -\x, \z)\]

It is interesting that when you write a linear transformation like \(A \equiv \frac{AV}{V}\) as a standalone symbol, it sorta looks like you are canceling out the \(V\)s. This is probably a good way of thinking about things: your abstract transformation \(A\) is being written out in terms of its action on the chosen frame \(V\).

So long as the vectors of the denominator are orthogonal, frame division expands by pairing elements in the numerator and denominator together and then inverting them individually:

\[\frac{F}{E} = \b{f}^1 \o (\b{e}^1)^{-1} + \b{f}^2 \o (\b{e}^2)^{-1} + \ldots = \sum_i \b{f}^i \o \frac{\b{e}^{i*}}{\| \b{e}^{i} \|^2} \tag{orthogonal}\]

which is still plenty useful in many common situations:

\[R_{xy} = \frac{(\y, -\x, \z)}{(\x, \y, \z)} = \frac{\y}{\x} + \frac{-\x}{\y} + \frac{\z}{\z}\]

But even when the denominator is not orthogonal, we can still write things out termwise using the dual frame, which in index notation corresponds to using lower-indices instead of upper ones (with an implicit summation):

\[\begin{aligned} \frac{F}{E} &= \frac{(\b{f}^1, \b{f}^2, \ldots)}{(\b{e}^1, \b{e}^2, \ldots)}\\ &= \b{f}^1 \o \b{e}_1 + \b{f}^2 \o \b{e}_2 + \ldots \\ &= \b{f}^i \b{e}_i \end{aligned} \tag{general}\]

So the behavior of a linear transformation is to select each basis vector \(\b{e}^i\) and replace it with \(\b{f}^i\).

\[\frac{F}{E} \b{a} = (\b{f}^i \b{e}_i) \cdot (a_j \b{e}^j) = a_j \b{f}^j\]

A linear transformation is invertible if both the numerator and denominator are nonsingular, which are the only frames we’ve been considering so far. The inverse is constructed by just swapping numerators and denominator frames, aka swapping upper and lower indexes:

\[A^{-1} = (\frac{F}{E})^{-1} = \frac{E}{F} = \b{e}^i \b{f}_i\]

Such that

\[\frac{E}{F} (\frac{F}{E} \b{a}) = (\b{e}^i \b{f}_i) \cdot (\b{f}^j \b{e}_j) \cdot (a_k \b{e}^k) = (a_j \b{f}^j) = a_i \b{e}^i = \b{a}\]

When the numerator and denominator frames are both orthogonal, we can construct the inverse by swapping each term in the expansion individually:

\[\begin{aligned} R_{xy} &= \frac{(\y, -\x, \z)}{(\x, \y, \z)} = \frac{\y}{\x} + \frac{-\x}{\y} + \frac{\z}{\z} \\ R_{xy}^{-1} &= \frac{(\x, \y, \z)}{(\y, -\x, \z)} = \frac{\x}{\y} - \frac{\y}{\x} + \frac{\z}{\z} \end{aligned}\]

By expanding both the numerator and denominator in coordinates we get a matrix representation of the transformation. For example \(R_{xy} = \frac{\y}{\x} - \frac{\x}{\y} + \frac{\z}{\z}\) as a matrix is

\[R_{xy} = (1) \frac{\y}{\x} + (-1) \frac{\x}{\y} + (1) \frac{\z}{\z} = \begin{pmatrix} 0 & -1 & 0\\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\]

All linear transformations on a space can be written in a form where the denominator is any particular choice of canonical basis frame like \(V = (\x, \y, \z)\). So an arbitrary linear transformation \(A\) may be written

\[\frac{AV}{V} = \frac{A(\x, \y, \z)}{(\x, \y, \z)} = \frac{(A\x, A\y, A\z)}{(\x, \y, \z)}\]

And its inverse can be written

\[A^{-1} = \frac{V}{AV} = \frac{A^{-1} V}{A^{-1} A V} = \frac{A^{-1} V}{V}\]

(it seems like it should be valid to multiply by \(A^{-1} / A^{-1}\) here, but I’m not sure how to think about it)

It is nice to think of \(AV/V\) as giving the procedure by which \(A\) acts on a vector: first it projects the vector into its \(V\) components, then it maps each of these to a new basis vector.

\[A \b{b} = \frac{AV}{V} \b{b} = \frac{AV}{V} (b_V V) = b_V (AV)\]

Or in coordinates,

\[[ (A \b{v}^i) \b{v}_i ] \cdot (b_j \b{v}^j) = b_i (A \b{v}^i)\]

If \(AV \sub \<W\>\), then we can expand it in matrix elements \(A_j^{\, i}\) in terms of the frames \(V\) and \(W\):

\[\frac{W}{W} A \frac{V}{V} = W (\frac{1}{W} A V) \frac{1}{V} = A_W^{\, V} \frac{W}{V}\]

Which in coordinates is

\[\b{w}^j (\b{w}_j \cdot A \cdot \b{v}^i) \b{v}_i = A_j^{\, i} \b{v}_i \b{w}^j\]

with \(\frac{1}{W} A V = A^V_W = A_j^{\, i}\) being the matrix ‘coordinates’ of \(A\) in the \(W\) and \(V\) frames.¹⁰

This is essentially the same as index notation, like \(A_{i}^{\, k}\), just with capital letters instead of lowercase ones. Basically we thinking of the indexes as explicitly labeling what frame they come from, rather than just being placeholders for summations over a sequence of numbers. This would not work if you thought of the indexes as labeling a vector space rather than a frame, because you really do need to keep track of a basis: for example, \(R_{xy} V\) and \(V\) describe the same vector space, but \(A_{\, R_{xy} V}^V\) is different than \(A_V^V\), because \((R_{xy})_{\, R_{xy} V}^V = \diag(1,1,1)\) while \((R_{xy})_{\,V}^V\) is the antisymmetric matrix shown above. So in a sense index notations have been frame notations all along. I prefer to conceptualize them as such.

If two frames span the same space, \(\< E \> = \< F \>\) then we can write \(F\) in terms of \(E\)

\[F = \frac{E}{E} F = (\frac{1}{E} F) E = EP\]

With \(P = \frac{1}{E} F\) as a change-of-basis matrix. (I would rather just leave it as \(E^{-1} F\) but we should connect this to how it’s usually written.) Note, we can write \(EP\) or \(PE\) for the result, because the index that connects them is implicit; I am writing \(EP\) because it’s more useful later.

In coordinates this is

\[\b{f}^i = (\b{e}^j \b{e}_j) \cdot \b{f}^i = (\b{e}_j \cdot \b{f}^i) \b{e}^j = P_j^{\, i} \b{e}^j\]

So \(P\) is

\[P = \frac{1}{E} F = \b{e}_j \cdot \b{f}^i = P_j^{\, i}\]

Since \(F: \cal{I}_F \ra F = E\) and \(E^{-1}: E \ra \cal{I}_E\), it must be that \(P: \cal{I}_F \ra \cal{I}_E\). Given a vector’s components \(a_F \in \cal{I}_F\), we can use this \(P\) to write

\[\b{a} = a_F (EP)\]

(Strictly speaking that should be \(EP a_F\) to match up all the spaces right-to-left, but I’m not too concerned with that; I like to keep vectors as \(a_F F\) and not worry about ordering on the \(\cal{I}_F\) component.)

The inverse frame is

\[\frac{1}{F} = \frac{1}{F} \frac{E}{E} = (\frac{1}{F} E) \frac{1}{E} = P^{-1} E^{-1} = \frac{1}{EP}\]

Which in coordinates is

\[\b{f}_i = (\b{f}_i \cdot \b{e}^j) \b{e}_j = P_i^{\, j} \b{e}_j\]

(Note that \(P_i^{\, j}\) is \(P^{-1}: \cal{I}_E \ra \cal{I}_F\), the matrix inverse of \(P^i_{\, j}\).)

So we have

\[\b{a} = \frac{1}{F} \b{a} F = \frac{1}{PE} \b{a} (EP) = (EP) P^{-1} (\frac{1}{E} \b{a}) = a_E E\]

Similarly a matrix transforms as

\[\begin{aligned} A &= \frac{F}{F} A \frac{F}{F} \\ &= F (\frac{1}{F} A F) \frac{1}{F} \\ &= (EP) (\frac{1}{EP} A EP) \frac{1}{EP} \\ &= (EP) (P^{-1} \frac{1}{E} A E P) (P^{-1} E^{-1}) \end{aligned}\]

Extracting the matrix elements \(\frac{1}{F} AF = A_F^{\, F}\) gives

\[A_F^{\, F} = P^{-1} A_E^{\, E} P\]

This is all a bit messy, admittedly, but I think it is good to see some of the things that are often written in indexes written with frames instead. There is a certain logic to it, although it takes some getting used to.

I’ll mention a few more linear algebra things to show them in frame notation.

The transpose of an operator is literally the same operator but acting on a denominator on the left instead of a numerator on the right:

\[\begin{aligned} \frac{E}{F} \b{v} = A \b{v} \\ \b{u}^* \frac{E}{F} = A^T \b{u}^* \end{aligned}\]

Which is why transposes are sort of silly. Note that this is a lot like

\[\frac{1}{\b{v}} \frac{E}{F} = (\frac{F}{E} \b{v})^{-1}\]

Except without using the actual inverse. As I touched on earlier, transposition/dualization/conjugation operations all consist of inverting the direction of an object but not its magnitude (e.g. \(re^{i \theta} \mapsto re^{-i \theta}\)). I believe the reason for this is that people are usually used to doing linear algebra, so they want linear maps, but they still want inverses at the same time, so inverting the direction is the best you can do. Mostly I suspect that this is a mistake and they should just fully invert objects including their scalars instead, but probably there are cases for both. In any case: the transpose is just what you get when want to treat \(A = E/F\) as acting on a dual vector to the left instead of a vector to the right. I suppose it might be written

\[A^T = \frac{F^{-1}}{E^{-1}}\]

to keep the notation consistent. In any case, it really refers to the same object \(A\), just used in a different way.

If an operator maps a space into itself, \(A: X \ra X\), we also have the operation of taking its trace by dotting the input and output vectors together.

\[\tr A = \x_i \cdot A \cdot \x^i = A^i_{\, i}\]

Sometimes this is written with a ‘double dot product’ to indicate that it takes two operators and contracts them on both components:

\[\tr A = A : I_X = A : \frac{X}{X} = A : \x_i \x^i\]

The fact that it is basis-independent follows from the fact that \(I_X\) is itself basis independent; writing \(I_X = \x_i \x^i\) just reflects a choice of coordinates on the frame \(X\).

The trace of the projection operator \(I_X = X/X\) itself gives the dimension of the space it projects onto

\[\begin{aligned} I_X &= \frac{(\x^1, \x^2, \ldots)}{(\x^1, \x^2, \ldots)} \\ &= \x^i \o \x_i \\ \tr I_X &= \x^i \cdot \x_i \\ &= \x^1 \cdot \x_1 + \x^2 \cdot \x_2 + \ldots \\ &= \dim X \end{aligned}\]

More generally the trace of a transformation times its inverse is its rank, the number of terms in each frame:

\[\tr (A A^{-1}) = \frac{F}{E} : \frac{E}{F} = \b{f}^i \b{e}_j : \b{e}^k \b{f}_l = 1_{i=l} 1_{j=k} = \text{rank}(A)\]

After all \(A A^{-1} = \frac{F}{E} \frac{E}{F} = \frac{F}{F} = I_F\) and \(A^{-1} A = \frac{E}{E} = I_E\). The rank is the dimension of both of these spaces.

6. Singular Decompositions

If we are generally given a linear transformation \(A = F/E\), we can decompose whatever it acts on into its \(E\) and \(E_{\perp}\) components, with the \(E_{\perp}\) component being in the kernel:

\[\begin{aligned} A \b{a} &= \frac{F}{E} \b{a} \\ &= \frac{F}{E} (\b{a}_E + \b{a}_{\perp E}) \\ &= \frac{F}{E} (a_E E + a_{\perp E} E_{\perp}) \\ &= \frac{F}{E} (a_E E) + \cancel{\frac{F}{E} (a_{\perp E}) E_{\perp}} \\ &= a_E F \\ &= (\b{a} \cdot \b{e}_i) \b{f}^i \end{aligned}\]

Here \(E_{\perp}\) is a frame referring to the orthogonal subspace to \(E\). Even though we don’t have a basis for it, we can use it symbolically as if we did, and \(a_{\perp E}\) refers to the coordinates in that basis.

Similarly a frame can be split into a list of other frames, like

\[E = (E_X, E_{\perp X})\]

where we assume that creating frames as tuples of other frames works the same way it does with vectors \(E = (\x, \y)\). However the notation starts to break a bit here. We would like to implement \(E_X\), the “restriction of the frame \(E\) to the frame \(X\)”, with something like a projection, \(E_X \? \frac{X}{X} E\), and such that \(E = E_X + E_{\perp X} = \frac{X}{X} E + \frac{X_{\perp}}{X_\perp} E\). But this doesn’t really work, because, for instance, \(\frac{\x}{\x}(\x, \y) \? (\x, 0)\) gives a ‘singular’ frame (a frame which is not linearly independent), and \(\frac{\x}{\x}(\x, \y) + \frac{\y}{\y}(\x, \y) \? (\x, 0) + (0, \y) \? (\x, \y)\) requires the “\(+\)” sign to act in a way that is hard to make consistent with everything else.¹¹

Maybe there is a way to get explicit algebra on frames to work in the same way that it works on vectors; I’m not sure. But it is still useful to think in frames, and their decompositions along other frames, because it gives a slightly more concrete way of writing things in linear algebra that are normally expressed in subspaces.

For example if \(E = E_X + E_{\perp X}\) then a linear transformation \(A = E/F\) factors as

\[A = \frac{F}{E} = \frac{AE_X}{E_X} + \frac{AE_{\perp X}}{E_{\perp X}}\]

if it is additionally the case that \(AE = AE_X + AE_{\perp X}\), that is, the two output frames are orthogonal, then the inverse inverts them separately:

\[A^{-1} \underbrace{=}_{\text{orthogonal}} \frac{E_X}{AE_X} + \frac{E_{\perp X}}{AE_{\perp X}}\]

We can use this to make the structure of a linear transformation \(A: V \ra W\) algebraically explicit. Suppose \(E \sub V\) and \(A\) is bijective on \(E\), so \(\ker A = E_{\perp}\). Writing \(n = \dim V\) and \(k = \dim E\), then we can factor \(A\) as

\[A = \frac{F}{E} + \frac{0^{\oplus n-k}}{E_{\perp}}\]

where \(0^{\oplus n-k} = \underbrace{(0, 0, \ldots)}_{n-k\text{ times}}\) is a ‘singular’ frame that describes the image of the kernel.

We can also add in a third term for the cokernel. Set \(m = \dim W\), and write

\[A = \frac{F}{E} + \frac{0^{\oplus n-k}}{E_{\perp}} + \frac{F_{\perp}}{\bot^{\oplus m-k}}\]

Here \(\bot\) is a symbol that I’m using to represent a ‘null’ frame that matches nothing in a denominator—the ‘empty frame’, if you will. We have to have a different symbol than \(0\) here, because it does not act the same as a zero frame in a denominator, as we will see in a second. You can think of \(\bot\) as being an empty frame, \(\bot = ()\), whereas \(0^{\oplus n-k} = (0, 0, \ldots)\) is a frame of zeroes. It is raised to the \((m-k)\) power because that is how many dimensions it is annihilating (I’m also not sure if this is necessary, but it’s nice for bookkeeping).¹²

Here is perhaps a better way to write this. Write \(A_{\parallel}\) for \(F/E\), the ‘parallel part’ of \(A\) (as opposed to the orthogonal part). Then

\[A = A_{\parallel} + \frac{0^{\oplus n-k}}{\ker A} + \frac{\text{coker} A}{\bot^{\oplus m-k}}\]

I have written \(A_{\parallel} = F/E\) for the ‘non-trivial’ part of \(A\), the part that does something (corresponding to its nonzero singular values in its SVD, incidentally). Then I’m using \(\ker A\) and \(\text{coker} A\) as frames for their respective subspaces. We can also write \(A_{\parallel}\) out as a ratio of frames as well:

\[A = \frac{\text{col}(A)}{\text{row}(A)} + \frac{0^{\oplus n-k}}{\ker A} + \frac{\text{coker} A}{\bot^{\oplus m-k}}\]

Note that \(\text{col}(A)\) and \(\text{row}(A)\) refer to frame which span the column and row spaces of \(A\), but they don’t correspond to the colums and rows exactly (which wouldn’t work, anyway, because it would be double-counting all the coordinates), and there are many ways to pick these frames that give the same transformation.

This decomposition roughly corresponds to the SVD of \(A\) (although we have not singled out the singular values). If \(A = U \Sigma V^T\) is the SVD, we might write it in frames as \(A = \frac{U \Sigma}{V}\). The nonzero singular values correspond to \(A_{\parallel}\), and the remaining projections onto the kernel and cokernel correspond to the other two terms. To fully recover the SVD we would need to also write out \(A\) as a list of vectors \(A = \frac{\b{f}^1}{\b{e}^1} + \frac{\b{f}^2}{\b{e}^2} + \ldots\), and then factor out the singular values as \(A = \sigma_1 \frac{\widehat{\b{f}^1}}{\widehat{\b{e}^1}} + \sigma_2 \frac{\widehat{\b{f}^2}}{\widehat{\b{e}^2}} + \ldots\). You can also join terms together if they share a singular value, e.g. \(A = \sigma_1 \frac{F^1}{E^1} + \sigma_2 \frac{F^2}{E^2} + \ldots\), which suppresses the degrees of freedom in picking bases for those frames.

Incidentally, the SVD tells us that it is always possible to pick the numerators and denominators in \(A = F/E = (\b{f}^1, \b{f}^2, \ldots)/(\b{e}^1, \b{e}^2, \ldots)\) to be simultaneously orthogonal. I haven’t yet thought of a way to see intuitively that this must be true.

Now consider trying to invert \(A\). Only the non-trivial part \(A_{\parallel}\) is invertible. Inverting only this part, and setting the rest to zero, gives the Moore-Penrose Pseuodinverse of \(A\), which I have also written about here (I don’t care for the \(A^+\) notation so I’m only going to use it this once):

\[A^+ = (A_{\parallel})^{-1} = ( \frac{\text{col} (A)}{\text{row} (A)})^{-1} = \frac{\text{row}(A)}{\text{col}(A)}\]

Incidentally this is why it corresponds to inverting only the non-zero singular values in the SVD: because it inverts onto \(A_{\parallel}\), the nonzero part of the operator.

Since the pseudoinverse invertibly maps \(\text{row}(A) \ra \text{col}(A)\), we can use it to construct projectors for each subspace:

\[\begin{aligned} I_{\text{row}(A)} &= (A_{\parallel}^{-1}) (A_{\parallel}) = \frac{\text{row}(A)}{\text{row}(A)} \\ I_{\text{col}(A)} &= A_{\parallel} A_{\parallel}^{-1} = \frac{\text{col}(A)}{\text{col}(A)} \\ I_{\text{ker}(A)} &= I_V - I_{\text{row}(A)} = I_V - A_{\parallel}^{-1} A_{\parallel} = \frac{\text{ker}(A)}{\text{ker}(A)} \\ I_{\text{coker}(A)} &= I_W - I_{\text{col}(A)} = I_W - A_{\parallel} A_{\parallel}^{-1} = \frac{\text{coker}(A)}{\text{coker}(A)} \end{aligned}\]

We can also talk about the generalized inverse of \(A\), which is the broadest way of thinking about what is meant by inverting \(A\). This consists conceptulaly of inverting all three terms separately:

\[\begin{aligned} A^{-1} &= (\frac{\text{col} (A)}{\text{row} (A)})^{-1} + (\frac{0^{\oplus n-k}}{\text{ker} (A)})^{-1} + (\frac{\text{coker} (A)}{\bot^{\oplus m-k}})^{-1} \\ &= \frac{\text{row} (A)}{\text{col} (A)} + \frac{\text{ker} (A)}{0^{\oplus n-k}} + \frac{\bot^{\oplus m-k}}{\text{coker} (A)} \\ &= A_{\parallel}^{-1} + \frac{\text{ker} (A)}{0^{\oplus n-k}} + \frac{\bot^{\oplus m-k}}{\text{coker} (A)} \end{aligned}\]

The kernel and cokernel have swapped roles. Now \(\text{coker}(A) \subset W\) is the set of vectors which are mapped to \(0_V\), since they have no preimage in \(V\), while \(\text{ker}(A) \subset V\) is the cokernel of \(A^{-1}\), consisting of the vectors that were all mapped to \(0 \in W\) by \(A\). However, the cokernel and kernel still act differently. The kernel vectors are all in the preimage, so they are all accessible. We represent this by dividing by zero.

But now we have a different interpretation: instead of saying that these vectors are inaccessible from \(A^{-1}\), we can think of them as all being accessible. The correct preimage \(A^{-1}(0_W)\), in a generalized inverse sense, is that it is all of the vectors in the kernel of \(A\). The \(1/0^{\oplus n-k}\) tells us that we would need \((n-k)\) additional pieces of data to describe the precise vector in the preimage, similar to inverting \(0 \times a = 0\) to get \(0^{-1}(0) \? a\) requires recovering the missing information of the value \(a\).

The implication is that the preimage of any point \(\in A^{-1}(W)\) includes the entire subspace \(\text{ker}(A)\) at once, which would take \((n-k)\) additional parameters to describe an individual point in, if we had them. We can write this as a linear transformation from an additional \((n-k)\) pieces of data \(\Lambda^{n-k} = (\lambda_1, \lambda_2, \ldots \lambda_{n-k})\) into the kernel:

\[\frac{\text{ker} (A)}{0^{\oplus n-k}} \? \frac{\text{ker} (A)}{\Lambda^{n-k}}\]

So we think of this “unbound” linear transformation as a description of the entire kernel at once. (Note, we could also have written \(0/\ker A\) and forgotten the \((n-k)\) information, and that would probably characterize \(A\) more correctly. But I like to keep it around because it elucidates the structure better.)

The inverse cokernel behaves slightly differently than the kernel did. To correctly write the inverse \(A^{-1}\) we still want the image of \(\text{coker} A\) to be literally deleted, not mapped to zero: it is not even in the domain of the operator \(A^{-1}\). This is not how linear algebra typically thinks about it, but I think it is the cleanest approach. What we have constructed is \(A^{-1}(\text{im A})\), so it has no action on \(\text{coker} A\) whatsoever. If you want \(A^{-1}: W \ra V\) to exist, you kinda have to explicitly extend it to be defined on \(W\) as well. Of course, if you don’t care about this level of pedantry, you can replace \(\bot\) with \(0\) here without losing anything.

A quick example. Consider \(A: \< \x, \y \> \ra \< \b{a}, \b{b} \>\) given by \(A = \frac{\b{a}}{\x}\). The full expansion is

\[A = \frac{\b{a}}{\x} + \frac{(0)}{\y} + \frac{\b{b}}{\bot}\]

and the generalized inverse is

\[A^{-1} = \frac{\x}{\b{a}} + \frac{\y}{(0)} + \frac{\bot}{\b{b}}\]

Which is “correct” in the sense that it’s the best you can do to invert \(A\). \(A\) maps \(\x \lra \b{a}\) so those get swapped, and it maps \(\y \ra 0\) so the preimage contains a the whole one-dimensional subspace \(\< \y \>\), and it maps nothing to \(\b{b}\) so the preimage does not produce a factor of \(\b{b}\) at all. Note that the separate use of \(0\) and \(\bot\) allows us to express some slightly different objects pretty succinctly as well. For example, consider \(A \mid_{\< \x \>}\), the restriction of \(A\) to \(\< \x \>\). Now we have no action on \(\y\) at all:

\[A \mid_{\< \x \>} = \frac{\b{a}}{\x} + \frac{\b{b}}{\bot}\]

We could also wonder about the case where we really do map \(\y \ra \bot\) instead of \(0\), so we “erase” the component rather than zeroing it. This strays away from the usual formulation of linear algebra, but why not?

\[A' = \frac{\b{a}}{\x} + \frac{\bot}{\y} + \frac{\b{b}}{\bot}\]

Etc. It is interesting to think about.

Another case to explore is to let the denominator frame vary with a parameter in order to see how the formulas ‘switch over’ when it becomes singular. Write

\[A = \frac{(\x , \x + c \y)}{(\b{a}, \b{b})}\]

whose action is

\[A \b{v} = A(v_a \b{a} + v_b \b{b}) = v_a \x + v_b (\x + c \y)\]

The inverse frame for \(c \neq 0\) is \((\x^* - \frac{1}{c} \y^*, \frac{1}{c} \y^*)\), so the inverse is

\[A^{-1} =\frac{(\b{a}, \b{b})}{ (\x, \x + c \y)} = \b{a} \o (\x^* - \frac{1}{c} \y^*) + \b{b} \o (\frac{1}{c} \y^*)\]

whose action is

\[A^{-1} (v_a \x + v_b (\x + c \y)) = v_a \b{a} + v_b \b{b}\]

When \(c = 0\), the action is to take both components to \(\x\):

\[A(v_a \b{a} + v_b \b{b}) = v_a \x + v_b \x\]

How does the inverse work? \(\lambda = 1/c \? 1/0\) can be thought of as allowing any possible value. The inverse looks like

\[A^{-1} = \b{a} \o (\x^* - \lambda \y^*) + \b{b} \o (\lambda \y^*)\]

Its action includes a free parameter that selects the actual preimage:

\[A^{-1} (v_a \x + v_b \x) = (v_a + v_b) \b{a} + \lambda (\b{b} - \b{a})\]

Which for the value \(\lambda = v_b\) becomes

\[A^{-1} (v_a \x + v_b \x) = v_a \b{a} + v_b \b{b}\]

Leaving the \(\lambda\) unspecified, however, produces the entire preimage of that point, which is the best \(A^{-1}\) can be asked to do.

I don’t know what use this is, but I think it’s a really nice way of looking at things, and everything seems to work out algebraically. It is probably the closest you can get to a true division by zero without actually changing the mathematics of your problem by inventing some structure that allows it (which is my complaint about a lot of the division-by-zero things out there, e.g. the Riemann sphere, meadows, etc).

Let me also mention that I did not invent this way of constructing inverses. It’s something that comes up in literature about the Moore-Penrose pseudoinverse, among other places. They usually write it in the form \(A^{-1} = A^+ + (I - A^+ A) \vec{\lambda}\), with \(\vec{\lambda}\) a vector of free parameters; \(I - A^+ A = I_{\ker A}\) is the projector for the kernel of \(A\). But I do prefer my frame notations for understanding it.

Anyway, there is much more that could be investigated about singular frames, but it gets rather confusing. Often it seems like there are some ‘missing’ algebraic operations that are needed to make sense of it For example, can you add two frames together? Does \((\x, \y) \oplus (\z) = (\x, \y, \z)\)? If so, then what does \((\x, \x)\) do? Thinking about questions like these made my head hurt, so I’m not going to look into it anymore today. But I’m convinced there’s some kind of ‘fluent’ calculus for all this, where every concept has a corresponding algebraic expression, and every expression corresponds to a meaningful concept. I just haven’t found it yet. So let’s move on.

7. Differentials

Frames work the same way on differentials, if you regard \((dx, dy)\) and \((\p_x, \p_y)\) as dual frames to each other. I think the best way to do this is to pick one symbol and use it for both. In this section I will use \(dx\) for differentials and \(1/dx\) for their pseudoinverses, the partial derivatives (not total derivatives). So differentials correspond to \(dx^i\) and partial derivatives to \(dx_i\).¹³

\[\begin{aligned} \frac{df}{(dx, dy)} &\equiv (\frac{\p f}{\p x}, \frac{\p f}{\p y}) = (f_x, f_y) = f_{x, y} \\ \frac{df}{(dx, dy)} \cdot (dx, dy) &\equiv f_x \d x + f_y \d y = f_{x, y} (dx, dy) \end{aligned}\]

Admittedly \(f_{x,y} = (f_x, f_y)\) is pretty awkward. Ideally we would write \(f_{xy}\) in analogy with the vector version \(\b{a}_{xy}\) earlier, but unfortunately that notation is already taken by the second partial derivative, so \(f_{x,y}\) will have to do. All of this is a bit of a mess because the shorthand expression for derivatives should really be \(df_{dx}\) (after all \(\p_x f\) is the \(dx\) component of \(df\)!), but even I can’t bring myself to do that yet.

If we write \(d \b{x} = (dx, dy, dz)\) as a symbol for the frame then this gives a literal algebraic meaning to the somewhat-common expressions \(f_{\b{x}}\) and \(df/d\b{x}\) for the gradient:

\[f_{\b{x}} = \frac{df}{d \b{x}} = (f_x, f_y, f_z) = \del f\]

So the gradient is the partial derivative with respect to the frame \(d \b{x}\), a single component of the geometric vector \(d f\). Note, writing \(d \b{x}\) here might be bit awkward because of the overlap with the unit vector \(\b{x}\). Usually it’s okay because you don’t use both objects at the same time. But an alternate and more unambiguous notation would be to use \(X=(x,y,z)\) and \(dX = (dx, dy, dz)\) for the differential version, and then \(df(X)/dX = df(x,y,z)/(dx,dy,dz)\) is pretty clean also. Regardless, when thinking with frames is we try to make all the notations “decomposition-invariant”: the same notations should be used for dealing with single vectors as for dealing with subspaces of more than one dimension.

We can put a frame in the numerator as well. Suppose \(\b{u} = (u, v) = (u(x,y,z), v(x,y,z))\) is a change of coordinates, for example one that parameterizes a 2d surface in 3-space. Then

\[\frac{d \b{u}}{d \b{x}} = \frac{(du, dv)}{(dx, dy, dv)} = \begin{pmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \end{pmatrix}\]

is the first derivative / Jacobian matrix (hate that name) of the coordinate transformation. These show up, for example, in line integrals:

\[\begin{aligned} \int df &= \int \frac{df}{(dx, dy, dz)} (dx, dy, dz) \\ &= \int (f_x, f_y, f_z) (dx, dy, dz) \\ &= \int \frac{df}{(du, dv)} \frac{(du, dv)}{(dx, dy, dz)} (dx, dy, dz) \\ &= \int (f_u, f_v) \frac{(du, dv)}{(dx, dy, dz)} (dx, dy, dz) \\ &= \int \frac{df}{d \b{u}} \frac{d \b{u}}{d \b{x}} d \b{x} \\ &= \int f_{\b{u}} \frac{d \b{u}}{d \b{x}} d \b{x} \end{aligned}\]

Today this is often written as \(\frac{\p(u, v)}{\p(x,y,z)}\) which is essentially equivalent, but as far as I know that is thought of as a shorthand, not an actual algebraic operation.

I have mostly not discussed exterior algebra to keep this from getting any bigger, but in this section I have to mention it. Frames and ratios of frames can be acted on with the wedge product, like \((\x, \y, \z) \^ (\x, \y, \z) = (\x \^ \y, \y \^ \z, \z \^ \x)\). The result is another frame of vectors, now drawn from \(\^^2 \bb{R}^3\) instead of \(\bb{R}^3\). As with a basis on \(\bb{R}^3\), the order in which we write the powers doesn’t matter, as long as we expand anything that multiplies the frame in the same order. In the case of differentials this looks like \(d \b{x} \^ d \b{x} = d \b{x}^{\^ 2} = (dx \^ dy, dy \^ dz, dz \^ dx)\). These exterior powers give a simple algebraic interpretation for many operations of multivariable calculus, such as reparameterizing a surface integral:

\[\begin{aligned} \iint dA &= \iint du \^ d v \\ &= \iint d \b{u}^{\^2} \\ &= \iint \frac{d \b{u}^{\^2}}{d \b{x}^{\^2}} d \b{x}^{\^2} \\ &= \iint \big[ \frac{du \^ dv}{dx \^ dy} dx \^ dy + \frac{du \^ dv}{dy \^ dz} dy \^ dz + \frac{du \^ dv}{dz \^ dx} dz \^ dx\big] \\ &= \iint \big[ (u_x v_y - u_y v_x) (dx \^ dy) + (u_y v_z - u_z v_y) (dy \^ dz) \\ &+ (u_z v_x - u_x v_z) (dx \^ dz) \big] \\ \end{aligned}\]

In the special case where \(d \b{u}\) and \(d \b{x}\) are both \(k\)-frames then \((d \b{u}/d\b{x})^{\^k}\) is the (pseudoscalar) Jacobian determinant. Otherwise the ratio gives some vector of wedge products that cover all the ways of projecting the numerator onto the denominator. In any case, the operation is just rewriting one frame of two-forms in terms of another, really just a rote application of linear algebra when expressed in terms of the pseudoinverse.

Notice that in the above we replaced \(dA\) with \(d \b{u}^{\^2}\). This does not require actually having a coordinate system for the surface of integration: since we know the integral is two-dimensional we can simply assume that there exist two coordinates for the surface and write the integral out in terms of them. We are also free to invent a normal coordinate \(n\), such that \((u, v, n)\) forms an orthogonal coordinate system in the neighborhood of the surface, and \((du, dv, dn)\) forms an orthogonal basis for the vectors on the surface. Then

\[d_{uv} = du \frac{1}{du} + dv \frac{1}{dv} = d \b{u} \frac{1}{d \b{u}}\]

is a projection onto the surface (normally written \(d \b{u} \frac{1}{\p_\b{u}}\)) and

\[d_n = dn \frac{1}{dn} = d - d_{uv}\]

is a projection off the surface. These are equivalent to \(I_{xy}\) and \(I_z\) on vectors, but when we write differentials as projections then it’s better to use the \(d\) notation to imply that they still are, after all, derivative operators. For example

\[d_{uv} f = f_u du + f_v dv = I_{uv} (df)\]

If \((u, v)\) is the coordinate system for a surface \(\sigma\), then this is the (classical) covariant derivative for the surface. It gives the gradient of \(f\) but projected onto the surface (which is not very interesting because it’s just acting on a scalar, but still). Line integrals of \(f\) that remain on the surface can then use this object without knowing about any of the other variation of \(f\) along the orthogonal \(n\) direction:

\[\int_{\gamma \in \sigma} d_{\sigma} f = \int_{\gamma \in \sigma} d_{uv} f = \int_{\gamma \in \sigma} f_u du + f_v dv\]

The \((u,v)\) coordinate system does not know that the orthogonal \(n\) coordinate exists, so calculus has to work the same way without it. The only way you can “detect” the surface is if you locate some non-trivial topology in which the \((u,v)\) coordinates double back on themselves. Indeed, the symbol \(d\) in \(df\) refers to the overall differential of \(f\), but really it is just the differential with respect to the coordinates we know about: \(d \equiv d_{xyz}\), say. ¹⁴

Now I’m going to briefly mix the vector notations and differential notations in order to demonstrate the covariant derivative of tensor fields in terms of frames. Let \(\b{a} = \b{a}_X = a_X X\) be a vector field in space and \(U\) be the tangent frame of some surface, as above. Then \(\b{a}_{U} = a_U U = a_X (X/U) U\) is as above. We can take the covariant derivative of this with respect to the surface frame \(U\):

\[\begin{aligned} d_U \b{a}_U &= d_U (a_U U) \\ &= d_U (a_U) U + a_U d_U (U) \\ &= (\p_i a_j) du^i \b{u}^j + a_j (\p_i \b{u}^j) du^i \\ \end{aligned}\]

The two terms here correspond to the fact that a vector field defined on a surface varies in space as you move around the surface. The first term corresponds to the “naive” partial derivative, which differentiates only the coordinates of \(\b{a}\) in terms of the surface coordinates \(U\). The second term is the differential of the vector field itself, which also changes as you move around the surface. In differential geometry this notation quickly becomes annoying, because the only basis for defining the unit vectors \(\b{u}^j\) that makes any sense is either the differentials \(d\b{u}\) or the derivatives \(1/d\b{u}\). So a more normal way to write this is

\[\begin{aligned} d_U \b{a}_U &= (\p_i a_j) du^i du^j + a_j (\p_i du^j) du^i \\ &= (\p_i a_j) du^i du^j + a_j (\Gamma_{ik}^j du^k) du^i \end{aligned}\]

Where we’ve relabeled the indexes and used the Christoffel symbol \(\Gamma_{ik}^j du^k = (\p_i du^j) du^k\) for the derivative of the frame with respect to itself. My version is much easier to read!

One can similarly take the derivative of a tangent vector field \(d_U (\b{a}_U) = d_U (a^U \frac{1}{U})\), or any higher-degree tensor. Although my presentation here is a bit muddled because of jumping between \(d\b{u}\) and \(U\), the reason I mention it is that I find the whole expression so much easier to expand without indices, in terms of frames and the pseudoinverses which project everything onto the surface again. In particular, it allows you to do calculus with vector fields without having to immediately write them out in a basis of derivatives or differentials. If you always tihnk of your vector fields \(a_U = (a_1, a_2, \ldots)\) as “really” being written in frames \(a_U U\), then it is perfectly natural that \(d (a_U U)\) has a product rule term \((d a_U) U + a_U (d U)\).

Another interesting application of calculus with frames is the case of Lagrange multipliers (see this other article for a more detailed exposition):

Suppose you have a set of constraints \(G_{\perp}(\b{x}) = (g_1(\b{x}), g_2(\b{x}), \ldots, g_k(\b{x}))\) in \(n\)-dimensional space. The surface \(G_{\perp}(\b{x}) = \vec{C} = (c_1, c_2, \ldots, c_k)\) gives a partial coordinate system on the space: given \(k\) values for the constraints you have eliminated \(k\) degrees of freedom, defining an \((N{-}k)\)-dimensional surface of points which satisfy the constraints, which we call \(G\). (The symbol \(G_{\perp}\) for the constraints is of course because they define the frame perpendicular to \(G\), rather than \(G\) itself.) Then we can talk about \(d_{\perp G} = dg_1 \frac{1}{dg_1} + dg_2 \frac{1}{dg_2} + \ldots\), the derivative with respect the constraint equations, as well as \(d_G = d - d_{\perp G}\), the derivative with respect to the surface \(G\). The important part is that we don’t have to actually have coordinates for the surface in order to compute this: knowing the ambient space and an expression for \(d_{\perp G}\) is enough.

Now suppose you want to maximize a function \(f(\b{x})\) subject to these constraints, that is, given a particular value of \(\vec{C}\) that defines the surface \(G\), you want to find the maximum value of \(f\) on \(G\). Since we already know how to compute derivatives on this surface, we can simply use the usual calculus rule that maxima are points at which the derivative is zero, then expand into \(d \x\) coordinates.

\[d_G f = (d - d_{\perp G}) f = \frac{df}{d \x} - \frac{d f}{d G_{\perp}} \frac{dG_{\perp}}{d \b{x}} = 0\]

But this means that at the solutions,

\[\frac{df}{d \x} = \frac{d f}{d G_{\perp}} \frac{dG_{\perp}}{d \x} = \vec{\lambda} \frac{dG_{\perp}}{d \x}\]

The coefficient is \(\vec{\lambda} = \frac{d f}{d G_{\perp}}\), what is normally called the vector of Lagrange multipliers corresponding to the constraints. Conceptually they represent the partial derivatives of \(f\) with respect to the frame spanned by the constraints; overall the equation says that at the solutions, the entire derivative of \(f\) is orthogonal to the surface \(G\), which is why its projection onto the surface is zero. Actually computing the value of \(\vec{\lambda}\) is messy because the \((dg_i)\) differentials in the \(dG_{\perp}\) frame do not have to be orthogonal to each other, meaning that \(df/dG_{\perp} \neq (df/dg_1, df/dg_2, \ldots)\), so a full pseudoinverse of the \((dg^i)\) differentials is needed. On the other hand the \(G\) and \(G_{\perp}\) coordinate systems are orthogonal, which is why it’s possible to separate the derivative \(d = d_G + d_{\perp G}\) cleanly.

I find this idea of arbitrarily factoring a space into orthogonal coordinate systems, and then doing calculus on those coordinates without writing them down explicitly, to be very nice to work with. Similar to how frames allow us to write vector equations where each basis element is free to be multi-dimensional, our approach to calculus is to treat individual coordinates \((x)\) as special cases of tuples of coordinates \((x,y,z)\), so that the same equations can be used on both. In both cases this means that certain properties no longer always hold: for example one-forms have \(d x \^ d x = 0\) but it’s not necessarily the case that \(d \b{x} \^ d \b{x} \neq 0\) for frames of one-forms. So all of our techniques have to be generalizes to handle the case of \((k > 1)\)-frames. But it’s worth it: a great deal of mathematical knowledge is just duplicating the same concepts for \(k=1\) and \(k > 1\) because the notations aren’t expressive enough to handle both cases at once. Frames follow philosophically from the idea that we should do something about that.

(Of course a great deal of mathematics, e.g. category theory, is concerned with factoring out commonalities across subjects at a higher level of abstraction, and all of the tools I’ve described here can be found in various parts of higher math, particularly differential geometry. My claim is that actually there are some very basic ways of refactoring things like this which all of mathematics would benefit from: actually many people are inventing the same things over and over in different notations because things are not sufficiently unified in their foundations. I am not aware of anyone else trying to do this at the elementary level of linear algebra and calculus; I’d be very interested to learn of any similar approaches.)

The transformation laws for differential forms are the same as those for frames and inverse frames that I discussed earlier, but just to connect things that might be familiar I’ll rehash them in differential notation here.

If space is parameterized by coordinates \(\b{x} = (x^1, x^2, \ldots)\), then the frame is the differentials

\[d\b{x} = (dx^1, dx^2, \ldots)\]

and the inverse frame is

\[\frac{1}{d \b{x}} = (\frac{1}{d x^1}, \frac{1}{dx^2}, \ldots)\]

Such that the partial derivative operators are given by \(d/d\b{x} = (\p_{x^1}, \p_{x^2}, \ldots) = \p_{\b{x}}\). Under a change of coordinates \(\b{u}(\b{x})\), the two change with opposite variance.

\[\begin{aligned} d \b{u} &= \frac{d \b{u}}{d \b{x}} d \b{x} \\[1em] \frac{1}{d \b{u}} &= \frac{d \b{x}}{d \b{u}} \frac{1}{d \b{x}} \\ &= (\frac{d \b{u}}{d \b{x}})^{-1} \frac{1}{d \b{x}} \end{aligned}\]

For example, if \((u, v) = (x, x+y)\), then their differentials are \((du, dv) = (dx, dx + dy)\), and the dual basis is

\[\begin{aligned} \frac{1}{du} &= \frac{1}{dx} - \frac{1}{dy} \\ \frac{1}{dv} &= \frac{1}{dy} \end{aligned}\]

Such that \(du (1/du) = 1\) and \(du(1/dv) = 0\), etc.

This is the reason that derivatives and differentials have opposite variance: one is a numerator and the other is a denominator. Which one is which is somewhat arbitrary. Normally in differential geometry it’s standard to think of the partial derivatives as the basis for “vectors” (I think since the tangent space is easier to define intrinsically?) and then the differential forms are the covectors, their dual space. In that case you would say that something that changes like a vector is covariant while something that changes like a differential is contravariant. But it doesn’t really matter which is which. As far as I can tell everything is the same if you swap the numerators and denominators. The only time it makes a difference is when you want to replace dividing by a vector \(\b{a}/\b{b}\) with dotting with a vector \(\b{a} \cdot \b{b}^*\), and we are not doing that with differentials at all here; everything is intrinsic to the linear algebra itself.

In any case I think the interpretation as numerators and denominators is much simpler than the notions of covariance and contravariance. Of course they’re of opposite variance; one is a numerator and the other a denominator! And it makes it clear why the combination of the two forms an intrinsically coordinate-invariant object: all of the variance simply cancels out.

\[df = \frac{df}{d \b{x}} d \b{x} = f_{\b{u}} \frac{d \b{u}}{d \b{x}} \frac{d \b{x}}{d \b{u}} d \b{u} = \frac{df}{d \b{u}} d \b{u} = df\]

It’s all just linear algebra. But like, actual algebra where you can just cancel things out? Mostly.

8. Conclusion

This concludes my long-winded narrative of how to rewrite things in terms of frames. Let me add that I am certainly not convinced I’m doing things in the best way possible here. A lot of it feels clunky or confusing, and it seems like there are some operations hinted at that I’m missing. The dream is to get to a point where a lot of multivariable calculus manipulation follow from rote cancellations like in high-school algebra, or things like that—but it’s definitely not quite there yet.

Still, I’m convinced there’s something here. There are too many things that work out for. The act of projecting a vector onto a frame is as fundamental as projecting it onto a single coordinate. And in many ways, the algebra behaves the same. The reason for using a division notation at all, for that matter, is because in many ways it really does act like division: if you rescale the frame by a coefficient \(c\), the coordinates divide by \(c\).

I am very intrigued also by how much frames sorta act like numbers. Combining vectors into a frame acts in some places like multiplication. Linear transformations act like ratios of frames. Free parameters in inverses act like division by zero. Etc. It is all cute to not be meaningful, yet too confusing to crisply articulate a way of doing arithmetic on it without keeping track of the precise index expression that you’re referring to. Maybe there’s a way to fix this, but I haven’t found it yet.

Although generalizing from vectors to frames adds complexity in some places, it also seems to reveal a big symmetry that underlies a lot of linear algebra: that most of the things we do with it are somewhat agnostic as to whether the individual directions in our vector spaces are actually one-dimensional. Decompositions like \(\b{a} = \b{a}_X + \b{a}_{\perp X}\) work fine regardless of the dimensionality of the components. Often I get the impression that the choice to view a particular system as \(n\)-dimensional is a significant choice of coordinate system, just like how defining the \((\x, \y, \z)\) axes is, and a ‘covariance’ with respect to this choice is going to be necessary to correctly reckon with the its applications in (among other places) physics. (You have no choice about how to do this in the normal way that linear algebra is formalized over a particular field; however, I think this is up for refactoring if necessary.) It is also necessary to solve my ongoing quest to make sense of ‘fractional’-dimensional objects; for instance a vector space of dimension \(1/2\) is probably going to be equivalent to a vector space of dimension \(1\), but “measured” in a basis of dimension \(2\). Something like that. I don’t know yet! But I like to think about it.

One can instead think of \(I_X\) as actually erasing the third component, such that it is a homomorphism \(\bb{R}^3 \ra \bb{R}^2\) like \(I_X (a_x, a_y, a_z)^T = (a_x, a_y)\). Still another option is to interpret \(I_X\) as a projection onto the quotient, \(\bb{R}^3 \ra \bb{R}^3 / \< \z \>\), such that \(I_X (a_x \x + a_y \y + a_z \z) = a_x \x + a_y \y + \< \z \>\). These are all isomorphic in linear algebra, but they are distinctly different at the ‘type’ level, and generalize differently (I presume). We will stick with the version I have described, though, because I want to keep everything concrete and familiar. ↩
The direct sum operation \(\oplus\) acts more like a product than a sum (and, I understand, is equivalent for finite-dimensional spaces to the direct product \(X \times Y\)). It would be most accurate to explicitly write sense to write \(I_{x \oplus x}\), but that’s unwieldy. Another option is \(I_{x,y}\), since in many cases commas are used to represent elements of direct sums, e.g. \((v_x, v_y) \in \< x \> \oplus \< y \>\), but \(I_{xy}\) is unambiguous enough that I haven’t seen a reason to switch. ↩
Sometimes this operator is called the ‘rejection’, but I don’t realy like that name, because it’s hard to use in a sentence: \(I_X\) is the projection onto \(X\), so \(I_{\perp X}\) is the… rejection… off of… \(X\)? There’s not really a good way to say it. Calling \(I_{\perp X}\) the ‘rejector’ for \(X\) works, however. ↩
An alternative notation for \(I_{\perp X}\) might be \(I_{V/X}\), since the quotient space \(V/X\) is isomorphic to \(X_{\perp}\). One advantage of this is that it makes clear what \(X_{\perp}\) is perpendicular with respect to. Also, when projectors are written out using basis vectors like \(I_{xyz}\), the division symbol seems to helpfully invert the implied ‘multiplication’ operation inside \(xyz\), like \(I_{yz} = I_{xyz/x}\). On the other hand, one normally thinks of elements of \(V/X\) as being cosets of elements modulo \(X\), rather than just regular elements of \(V\), and although they are isomorphic, they are distinctly different. I haven’t been able to decide if this is worth it. Still another notation might be \(I_{\perp X} = I_{V \ominus X}\), where \(\ominus\) is interpreted as a formal inverse to the direct sum \(V = X \oplus X_{\perp}\). ↩
Aside: it is possible to write down expressions with the projectors for subspaces which resemble set algebra. For example, \(I_{xyz} = I_{xy} + I_{yz} - I_{xy} I_{yz}\) resembles the inclusion-exclusion of sets \(A = \{ x, y \}\) and \(B = \{ y, z \}\): \(I_{A \cup B} = I_A + I_B - I_{A \cap B}\). However, this gets messy as soon as you try it on more complicated bases. For instance \(I_{x}\) and \(I_{x+y}\) do not even commute as operators. Also, \(I_{xy} I_{yz} = I_y\) must hold when \(I_{xy}\) is replaced with \(I_{x+y, x-y}\), but then it no longer looks like a simple intersection. I think there is something interesting here, but it it is too complicated to explore in this article. ↩
Although I have defined \(\b{b} / \b{a} = \b{b} \o \b{a}^{-1}\) here, it is often useful to actually cancel them out, letting \(\b{b} / \b{a}\) mean \(\b{a}^{-1} \b{b}\). There are arguments for both. I won’t do that in this article just to avoid confusing things, but in other places I tend to use them interchangeably. The implied contraction is just a shorthand anyway. By the way: the basic reason this comes up for vectors but not scalars is that, in scalar algebra, \(b/a = a^{-1}(b)\) and \(b/a = b \o a^{-1}\) mean the same thing and act the same way in all calculations. Every term has one implicit index anyway, so there’s no difference whether you trace over it or not. In vector algebra you have to actually distinguish the two cases. (In fact \(\b{a}^{-1} \cdot \b{b}\) can be viewed as “really” being \(\b{b} \o \b{a}^{-1}\) but implicitly viewed through its projection onto a scalar/spin-\(0\) subrepresentation, so “measured” along the “basis” operator \(I_x = (\x \x^* + \y \y^* + \ldots)\); this is a pretty good perspective actually but it’s too far afield for this article.) ↩
Note, there are several things out there called “frame” and they’re slightly different. This concept is most similar to \(k\)-frames and moving frames, and less similar (but more important than) what Wikipedia currently describes on the Frame (linear algebra) article, which is instead a generalization of a basis to be overdetermined (to contain more basis vectors than the dimension). Although this is a useful concept, it is more useful most of the time to talk about linearly independent frames, so if I want to talk about one with that restriction relaxed I would add a qualifier like ‘singular frame’, since the associated matrix is singular, i.e. non-invertible. (Strictly speaking the word ‘singular’ only applies to square matrices but I like to use as a term for all non-linearly-independent frames, because ‘not full rank’ is a terrible term.) ↩
Aside: if we do not assume the Euclidean dot product, then it is also possible to define an inner product which makes \((\x, \x + \y)\) appear to be orthogonal, in which case \(a_1 (\x) + a_2 (\x + \y)\) works the same way as \(a_x \x + a_y \y\) did before. We are not considering this case. I think it is always easier to think about there being two sets of vectors and one unchanging product, rather than one set of vectors but with the product changing (contrary to the usual approach in differential geometry…). Basically it is nice for \(\x^{-1} \cdot \x\) to always equal \(1\), since we are interpreting it as “measuring \(\x\) in terms of \(\x\)” and what else would that be? ↩
This is the part that I got really stuck on a few years ago… it is morally correct, ish, but I don’t have a good notation for it. ↩
The expression \(\frac{1}{W} AV\) seems to break the rule of denominators contracting with numerators, since \(AV\) is two numerators in a row. This is unfortunate. The problem is that \(A\) is “really” a pairing of a numerator and a denominator in the same term. It would be better if we always wrote linear transformations as ratios like \(A=F/E\) to symbolize the fact that they have an input and output component. ↩
It feels like there must be an elegant way to do this, but I haven’t found it yet. ↩
I can’t really justify this notation; it just feels right. By the way, \(\bot\) here is inspired not so much by the perpendicularity symbol, actually, but rather by ‘bottom types’ in type theory, which are types with no members (although it’s convenient that it makes sense without this). I vaguely recall getting the notion from the graphical linear algebra website, but I can no longer quite remember why, because they use the symbol differently than I am. Anyway, I can’t see a way around using it. There really are two “kinds” of zero here. Which is curious. ↩
It does not seem to matter which one you pick as the numerator and denominator, as far as I can tell—similar to how index notation / Ricci calculus doesn’t care about your convention for numerators vs. denominators. The only place it would matter is if you were using the ‘naive’ (one-dimensional) pseudoinverse division \(1/\x = \x / \| \x \|^2\), but with differentials you don’t want to do that, because you have to always invert them using the metric. ↩
A trivial example is that \(f(x) = ax\) can also be thought of as having an \(a\) dependence, if we like, which means \(d_x f = f_x dx\) is just restriction of \(d_{x,a} f = f_x dx + f_a da\). It could also have a \(b\) dependence, with \(f(x,b) = abx\) evaluated at \(b= 1\). We want to use notations that always leave such interpretations open to us. This is not trivial, either: very many mathematical proofs proceed by viewing an object as a projection out of a larger object. An example that comes to mind is adding variables to integrals to perform Feynman’s integration trick (e.g. integrating \(\int \frac{\sin x}{x} \d x\) by solving \(I'(a) = \p_a \int e^{ax} [\frac{\sin x}{x}] dx\), then solving for \(I(a) = \int I'(a) da\), then setting \(a = 0\)). This seems to be a common theme when thinking in frames: we try to think of all mathematical objects as being viewed through projections onto some known coordinate system, so any equation like \(df\) that we write is really interpreted as being relative to a projection like \(d_{xyz} f\), leaving open the possibility that there is other information about \(f\) which we are ignoring. ↩

Everything Is Logarithms

2026-05-25T00:00:00+00:00

Some connections between things, which I have not seen elsewhere. Maybe they mean something?

1. The Baseless Logarithm

Normally one writes a logarithm with a base, \(\log_b (x)\), to mean

\[y = \log_b (x) \Lra b^y = x\]

And then you can change the base of the logarithm with

\[\log_b (x) = \frac{\log_a (x)}{\log_a(b)}\]

Which follows from rearranging \(\log_a (x) = \log_a (b^{\log_b x}) = \log_b (x) \times \log_a (b)\).

One way of thinking about what this formula does is that it is a change of units, akin to writing \(2 \text{ km} = 2000 \text{ m} / \frac{1000 \text{ m}}{1 \text{ km}}\) or \(5 \text{ bytes} = 40 \text{ bits}/\frac{8 \text{ bits}}{1\text{ byte}}\). It says: how many copies of \(b\) are in \(x\)? It’s the number of copies of \(a\) in \(x\), divided by the number of copies of \(a\) that are in \(b\).

This is perfectly simple, but for some reason it’s hard to think about logarithms that way. The notation kind of… obfuscates things? Specifically it is hard to read \(\log_b x\) as “how many copies of \(b\) are in \(x\)”, because that English expression should correspond to the notation \(x/b\), not \(\log_b x\). “How many factors of \(b\) are in \(x\)” is a bit better, but it still feels off.

I found a way of thinking about logarithms which I think makes this clearer, but you have to allow a sort of odd object that I will call the baseless logarithm. It is simply a logarithm without a base:

\[\log N\]

which we regard as an abstract object, not a number. Then we write our normal “based” logarithm as a ratio of two of these baseless logarithms:

\[\log_2 N = \frac{\log N}{\log 2}\]

Note, this is already a thing people do colloquially, e.g. leaving out the base of logarithms in asymptotic formulas. But I do not mean it as a shorthand; it is more useful to regard it as an actual algebraic object.

We interpret \(\log 2\) as being the unit “bits”. To write \(\log N\) in bits is to factor it as a multiple of \(\log 2\):

\[\log N = \frac{\log N}{\log 2} \log 2 = \log_2 (N) \log 2 = \log_2 (N) \text{ bits}\]

Then the change-of-base for logarithms follows from just writing the same geometric quantity in different units. For example \(\log e\) as a unit is sometimes called “nats”:

\[\begin{aligned} \log N = \frac{\log N}{\log 2} \log 2 = \log_2 (N) \text{ bits} = \frac{\log N}{\log e} \log e = \ln (N) \text{ nats} \end{aligned}\]

The baseless \(\log N\) is sort of the multiplicative version of an object that might be familiar from doing geometry with vectors. It is common in geometry to distinguish between points and displacements as two different ‘kinds’ of vectors. A point is represented by a abstract symbol \((a)\), which cannot be added to another point: \((a) + (b)\) is not defined. Subtraction between points is allowed, however: The result is a displacement vector: \(\v = (b) - (a)\). We are allowed to add displacement vectors together, \(\b{u} + \b{v}\), and we are add to displacement vectors to points to produce new points: \(((b) - (a)) + (a) = (b)\), or \(((b) - (a)) + (c) = (d)\), a new point given by following the vector \((b) - (a)\) but starting at \((c)\) instead of \((a)\).

When we actually go and add a coordinate system to our geometry we then choose an explicit origin \(\O\), whereupon we represent all points as displacement vectors relative to this origin: \(\b{a} \equiv (a) - \O\) and \(\b{b} \equiv (b) - \O\). Then a displacement vector is constructed by subtracting off the factors of \(\O\), \(\b{v} = \b{b} - \b{a} = [(b) - \O] - [(a) - \O] = (b) - (a)\).

The baseless logarithm implements the same thing but with multiplicative translations instead of additive ones. The value \(\log N\) may be thought of as \(\log N / \log \O\) for an unspecified choice of origin. Turning it into an actual numeric value involves dividing two such logarithms to cancel out the origin, \(\log_M N = \log N / \log M = (\log N / \log \O) / (\log M / \log O)\), exactly like \((b) - (a) = \b{b} - \b{a}\) cancels out the origin of vectors to produce a displacement. I think of \(\log N\) as the point corresponding to \(N\) and \(\log N / \log \O\) as its corresponding displacement vector once you pick a coordinate system. So these ‘points’ \(\log N\) are the more fundamental object, from which their ‘displacements’ \(\log N / \log \O\) are built.

So logarithms act kinda like multiplicative vectors, in the sense that they have to be defined relative to an ‘origin’, that is, a choice of base, and they follow mostly the same arithmetic as vectors and their origins do, if you replace multiplication with addition. In fact there are many other surprising similarities between logarithms and vectors which we will now talk about.

2. Logarithms are Vectors

When doing vector algebra and differential geometry in a properly covariant way, we distinguish between abstract vectors and vectors in a particular coordinate system. My personal convention for this is to refer to the abstract vectors as “geometric” vectors and always write them in bold, \(\v\), whereas “coordinate” vectors, tuples of their values in coordinates, are written with an arrow over them like \(\vec{v} = (v_x, v_y, v_z)\). Boldface geometric vectors are always coordinate-free, whereas coordinate vectors are just collections of numbers or other objects. The geometric vector \(\b{v}\) can be written as a dot product of its coordinates with a ‘frame’ \(X = (\x, \y, \z)\) of basis vectors

\[\b{v} = \vec{v} \cdot X = (v_x, v_y, v_z) \cdot (\x, \y, \z) = v_x \x + v_y \y + v_z \z\]

The projection of \(\v\) onto a basis vector \(\x\) is then given by ‘measuring’ the vector against the basis vector (which does not have to be of unit length). I like to write this as division because it acts a lot like division (although it’s technically pseudodivision instead):

\[\frac{\v}{\x} = v_x\]

That’s in my own very nonstandard notation for vector division here. The more common way to write this is to project a component of a differential \(df = f_x dx + f_y dy + f_z dz\) with a partial derivative, which is also the pseudodivision operation (which is incidentally the sense in which partial derivatives kinda work like division but not really):

\[\frac{\p f}{\p x} = f_x\]

I will write things in both forms to make it easy to translate between them; I do prefer my vector-division version because it avoids bringing in the irrelevant notations of differential calculus, but since the latter is actually standard I ought to include it for comparison.

Suppose \(\b{v}\) is one-dimensional, \(\b{v} = v_x \x\). Then the projection onto a ‘measuring stick’ \(\b{m} = m \x\) measures its length in terms of multiples of \(m\):

\[\frac{\v}{\b{m}} = \frac{v_x \x}{m \x} = \frac{v_x}{m}\]

Multiplying by \(\b{m}\) again is what we mean by “writing \(\b{v}\) in units of \(\b{m}\)”:

\[\frac{\b{v}}{\b{m}} \b{m} = (\frac{v_x}{m}) (m \x)\]

Here \(m\) is the unit “meters” and \(v_x/m\) is the value of \(v_x\) written in meters. Of course to actually compute \(v_x/m\) you have to have it in units in the first place—but clearly it’s the same kind of thing as in the logarithm case, where you can think of \(\b{v}\) and \(\b{m}\) as “unitless” concepts that are compared geometrically, and then \(v_x/m\) as their projections into an arbitrary coordinate system.¹

The baseless logarithm is performing the same operation on logarithms, where \(\log N\) is filling the role of the geometric vector \(\v\) and \(\log 2 = \text{bits}\) is the unit vector or measuring stick, which takes the role of \(\x\).

\[\begin{aligned} \frac{\log N}{\log 2} &= \log_2 N \\ \frac{\log N}{\log 2} \log 2 &= \log_2 N \text{ bits} \end{aligned}\]

In this sense baseless logarithms write numbers in coordinates in exactly the same way that measuring sticks write vectors in coordinates.

The equivalence of logarithms in different units

\[\begin{aligned} \log N &= \frac{\log N}{\log 2} \log 2 = \log_2 (N) \text{ bits} \\ &= \frac{\log N}{\log e} \log e = \ln (N) \text{ nats} \end{aligned}\]

is the same as the equivalence of geometric vectors in different units

\[\begin{aligned} \v &= \frac{\v}{\x} \x = v_x \x \\[1em] &= \frac{\v}{\x'} \x' = v_{x'} \x' \\ \end{aligned}\]

\[\begin{aligned} df &= \frac{\p f}{\p x} dx = f_x dx \\ &= \frac{\p f}{\p x'} dx' = f_{x'} dx' \end{aligned}\]

And the change of base formula that computes a ratio of logarithms in different bases

\[\begin{aligned} \log_2 N \text{ bits}&= \ln N \text{ nats} \\ \log_2 N &= \frac{\text{nats}}{\text{bits}} \ln N\\ &= \frac{\log e}{\log 2} \ln N \\ &= \log_2 (e) \ln N \end{aligned}\]

is exactly like the change of coordinates for a vector, where \(\x\) and \(\x\) are two units for the same quantity.

\[\begin{aligned} v_x \x &= v_{x'} \x' \\ v_x &= \frac{\x'}{\x} v_{x'} \\ \end{aligned}\]

or²

\[\begin{aligned} f_x dx &= f_{x'} dx' \\ f_x &= \frac{dx'}{dx} f_{x'} \end{aligned}\]

What logarithms don’t allow that vector division and differential notations easily do is to talk about a partial projection operation or a partial derivative in isolation. For example, if \(N = 2^a 3^b\), you can only talk about the “total” logarithm, the ratio with respect to a single unit \(\log 2\)

\[\frac{\log N}{\log 2} = a \frac{\log 2}{\log 2} + b \frac{\log 3}{\log 2} = a + b \log_2 3\]

which is equivalent to writing a vector as a multiple of a single basis vector (like in Clifford/geometric algebra)

\[\frac{\v}{\x} = v_x + v_y \frac{\y}{\x}\]

or to a total derivative

\[\frac{df}{dx} = f_x + f_y \frac{dy}{dx}\]

But there is no equivalent of the operation of partial differentiation, a “partial logarithm”, which would let you factor a number like

\[N \? (\log_{\p 2} N) \log 2 + (\log_{\p 3} N) \log 3\]

However I keep finding that people have gone and invented the projection / partial derivative operation on logarithms anyway. For example, the p-adic valuation in number theory

\[\nu_p (n) = \max \{ k \in \bb{N} \mid p^k \mid n \}\]

corresponds to extracting the coefficient of \(\log p\) of a natural number in a logarithmic basis:

\[\begin{aligned} \log n &= \log 2^{n_2} 3^{n_3} 5^{n_5} \cdots \\ &= n_2 \log 2 + n_3 \log 3 + n_5 \log 5 + \ldots \\ &= \sum n_p \log p \\ \nu_p (n) &= n_p \end{aligned}\]

Each coefficient is a positive integer times a \(\log p\) factor and \(\nu_p\) just extracts the component corresponding to \(\log p\). Clearly \(\log n\) is like a vector (well since the coefficients are in \(\bb{N}\) it is not technically a vector, more like a commutative monoid or something… but it has the familiar structure of a vector). Also, since \(\nu_p\) is some kind of ‘projection’ out of this logarithm, it still obeys logarithmic identities like \(\nu_p(m/n) = \nu_p(m) - \nu_p(n)\). But there is not really a good notation for actually expressing it as a projection, so sadly it gets a whole separate nomenclature that you have to learn.³

The same thing also works for rational \(n\) or radical \(n\), meaning it is the product of radicals of prime factors, with the coefficients becoming integers or rationals respectively. In the case where the coefficients are rational the logarithms are actual vectors in a vector space (over \(\bb{Q}\)).

Another example of these logarithmic projections: in complex analysis the “order of vanishing” \(\text{ord}_a f(z)\) of a meromorphic function \(f(z)\) at a point \(z=a\) is the order of the pole or zero at a point (where zeroes are like negative poles). That is, it is the degree \(n\) of the lowest-degree term in the Laurent series of the function around the point \(z=a\),

\[f(z) = f_{-n} (z-a)^{-n} + f_{-n+1} (z-a)^{-n+1} + \cdots + f_{-1} (z-a)^{-1} + f_0 + f_1 (z-a) + \cdots\]

(that is, the value of \(n\) such that \((z-a)^n f(z)\) is holomorphic around \(a\)). This is extracted with a logarithm:

\[\text{ord}_a f(z) = \lim_{z \ra a} \frac{\log f(z)}{\log (z-a)} = -n\]

since for \(z \approx a\), \(f(z) \sim f_{-n} (z-a)^{-n}\) which dominates the other terms that blow up less quickly. If we write \(g(z)\) for the rest of \(f(z)\) which has \(\text{ord}_a (g(z)) > -n\):

\[\begin{aligned} \lim_{z \ra a} \frac{\log f(z)}{\log (z-a)} &= \lim_{z \ra a} \frac{\log (f_{-n} (z-a)^{-n} + g(z))}{\log (z-a)}\\ &= \lim_{z \ra a} \frac{\log f_{-n} (z-a)^{-n} (1 + \frac{g(z)}{f_{-n}} (z-a)^n)}{\log (z-a)} \\ &= \lim_{z \ra a} \frac{\log f_{-n}}{\log (z-a)} -n \frac{\log (z-a)}{\log (z-a)} + \frac{\log (1 + c (z-a))}{\log (z-a)} \\ &= -n \end{aligned}\]

So this is a very similar operation: the limit \(\lim_{z \ra a} \log (z-b)/\log(z-a) = 1_{a=b}\) serves to cancel out the rest of the terms, like how \(\p_j dx^i \sim (\p x^i)/(\p x^j) = 1_{i=j}\) serves to cancel out the terms in a partial derivative, extracting the \(dx\) component of \(df = f_x dx + f_y dy + \ldots\).

(I’m not very good at complex analysis so that’s all I’m going to say about that. Still, it seems clear that this is basically the same operation.)

We see that the baseless logarithm \(\log n\) works a lot like a vector \(\v\) or differential \(df\), and then expressing a logarithm in a base like \(\log_2 n = \log n / \log 2\) is a lot like a total derivative \(df/dx\) or Clifford division \(\v \ast \b{x}^{-1}\). What is missing is some equivalent of the partial derivative / projection operator that projects only onto that component… but various fields have gone and found a way to invent that anyway, either in the form of a partial derivative \(\p f/\p x\), or just by making up the \(p\)-adic valuation \(\nu_p\), or by the limits \(\lim_{z\ra a} \log f(z) / \log (z-a)\) in complex analysis. The similiarities are all suspicious, though, and I can’t help but think there is some unifying theory here that ties all this together… but I can’t see what it is yet.

One thing that we might try in order to invent a \(\log_2 N\) that acts like \(\p_x f\) or \(\b{v}/\x\) is to somehow restrict the values of the logarithms to certain spaces, e.g. integers or rationals. Since the \(\{\log p_i\}\) are linearly indepedent (which is essentially equivalent to prime factorizations being unique), you would end up with objects like \(\log_2 3 = \log_3/\log_2\) which have no value in \(\bb{Q}\). “Zeroing” those out then gives something that acts like a partial derivative but for logarithms. But I don’t know if that’s useful. Certainly it doesn’t help in any numeric context.

Anyway, onto more things that are logarithms.

3. Vectors are also Logarithms?

In differential geometry one interprets vectors like \(\v = v_x \x + v_y \y\) being written in a basis of partial derivative operators, \(\v = v_x \p_x + v_y \p_y\). These can then be used to create discrete translations which move around in the various coordinates,

\[T^{\v} = e^{\v} = e^{v_x \p_x + v_y \p_y }\]

The partial derivatives are here in order to make it operate on functions

\[e^{v_x \p_x + v_y \p_y} f(x,y) = f(x + v_x, y + v_y)\]

which is true at the level Taylor expansions as well. I often find it easier to dispense with the partial derivatives and just think of these as translation operators on the space \((x,y)\) directly

\[e^{v_x \p_x + v_y \p_y} (x, y) = (x + v_x, y + v_y)\]

(You can think of this acting on the function \(f(x,y) = (x,y)\) also, but that feels like overkill.)

In any case, all this is really doing (in flat space, at least) is rewriting the additive vector \(\b{v}\) into a multiplicative form \(T^{\b{v}}\) which corresponds to the same operation. Things are just being written differently: its terms are multiplied instead of added, and scalar coefficients are applied via exponentiation instead of multiplication. A basis for the vector space now consists of translation operators in each coordinate:⁴

\[T^{\v} = e^{v_x \p_x} e^{v_y \p_y} = T_x^{v_x} T_y^{v_y}\]

(In non-flat space this is not so simple because the translations in different coordinates may not commute; you can still write it in this form but it’s a lot more complicated.)

What this means for us is: look, vectors are logarithms too!

\[\begin{aligned} \ln T^{\v} &= \ln T_x^{v_x} T_y^{v_y} \\ &= v_x \ln T_x + v_y \ln T_y \\ &= v_x \p_x + v_y \p_y \end{aligned}\]

I can’t exactly say why, but it seems preferable to have this written in terms of baseless logarithms also. We do this by realizing that \(T_x = e^{\p_x} = T^{\p_x}\) and thinking of this symbol \(T\) as a sort of ‘generic’ base for translations, absent the numeric meaning of the symbol \(e\), which has \(\log T_x = \log T^{\p_x} = \p_x \log T\). Then

\[\log T^{\v} = \v \log T = v_x \p_x \log T + v_y \p_y \log T\]

And then we can write \(\v = \log_T T^{\v} = \log T^{\v} / \log T\). This is equivalent to the natural log version but it avoids explicitly depending on the numeric value of \(e\): any choice of base for the logarithm \(T\) gives the same concept of a vector, written in terms of the exponentiation of \(T\), but now we make explicit that the ‘units’ on \(\v\) come in part from the units on \(\log T\) itself.

So vectors in differential geometry may also be thought of as logarithms, specifically, the logarithms of translation operators.

Regular multiplication can even be viewed as an example of this. A product like \(xa\) can be rewritten as “translation” in the \(\ln a\) coordinate:

\[xa = e^{\ln x} e^{\ln a} = e^{(\ln x) \p_{\, \ln a}} a = x^{\p_{\, \ln a}} a\]

I mention this because it’s cute, but I can’t imagine how it would ever be useful.

4. Logarithms are Derivatives?

This part doesn’t really connect to the rest; I just thought I would mention it so that this article contains every fun fact about logarithms that I know.

One way of defining the natural logarithm is

\[\ln x = \lim_{a \ra 0} \frac{x^a - 1}{a}\]

Which can be found by rewriting \(x^a = e^{a \ln x}\) and then Taylor expanding:

\[\frac{x^a - 1}{a} = \frac{e^{a \ln x} - 1}{a} = \frac{(1 + a \ln x + \ldots) - 1}{a} \stackrel{a \ra 0}{=} \ln x\]

Plugging in \((1+x)\) reproduces the Taylor series for \(\ln\):

\[\begin{aligned} \ln (1+x) &= \frac{(1+x)^a -1}{a} \\ &= \frac{\sum \binom{a}{k} 1^{n-k} x^k - 1}{a} \\ &= \frac{(1 + ax + \frac{a(a-1)}{2} x^2 + \ldots) - 1}{a} \\ &\stackrel{a \ra 0}{=} x - \frac{1}{2} x^2 + \frac{1}{3} x^3 - \ldots \end{aligned}\]

The \(\lim_{a \ra 0} (x^a - 1)/a\) formula for \(\ln x\) resembles a derivative. To make it explicit, we can write it as

\[\ln x = \lim_{dy \ra 0} \frac{x^{y + dy} - x^y}{dy} \mid_{y=0} = \p_{y} x^y \mid_{y =0}\]

What I like about this form is that it explains what \(\ln\) is doing in calculus comes from by connecting \(\ln x\) with \(x^0\). It always struck me as strange that \(\int x^{k} = \ln x\) for \(k=-1\), whereas it is a polynomial for all other values of \(k\). Why is a logarithm like a polynomial? Turns out it’s because in a lot of ways \(\ln x\) acts like \(x^0\). Specifically it’s the ‘interesting’ part of \(x^0\), its first order approximation around \(x=1\):

\[\ln x \sim \frac{x^0 - 1}{0}\]

Just for fun, try using \(\p_x x^k = k x^{k-1}\) on it:

\[\p_x \ln x = \p_x \frac{x^0 - 1}{0} = \frac{0 x^{-1}}{0} = \frac{1}{x}\]

That’s all I really have to say about this. But I wonder if some of the other ideas on this page would benefit from being interpreted via the \(\ln x = \p_y x^y \mid_{y=0}\) form.

5. Dimensions are Logarithms

Another thing which clearly acts like a logarithm is the dimension operator \(\dim\) in linear algebra.

Let \(\dim_K V\) be the dimension of \(V\) as a vector space over the base field \(K\), and assume we’re only talking about finite-dimensional spaces here. Now compare:

\[\begin{aligned} \dim_{K} K^n &= n \dim_K K = n \\ \dim_K U \oplus V &= \dim_K U + \dim_K V \\ \dim_K U/V &= \dim_K U - \dim_K V \\ \dim_K U \o V &= (\dim_K U) \times (\dim_K V) \\ \end{aligned}\]

and

\[\begin{aligned} \log_k k^n &= n \log_k k = n \\ \log_k u \times v &= \log_k u + \log_k v \\ \log_k u/v &= \log_k u - \log_k v \\ \log_k k^{\log_k u \times \log_k v} &= (\log_k u) \times (\log_k v) \end{aligned}\]

The direct sum \(\oplus\) corresponds to multiplication \(\times\), which is really just a notational accident, since it is the same as the direct product on finite-dimensional vector spaces; the \(\oplus\) symbol reflects the fact that it adds bases as sets.⁵ Meanwhile the tensor product \(\otimes\) multiplies bases on sets, but corresponds in arithmetic to a sort of “commutative exponentiation” \(e^{\log_k u \log_k v} = u^{\log_k v}\) that you don’t see very much, sometimes called a commutative hyperoperation. (The next ‘displacement’ operation after \(b-a\) and \(b/a\) is therefore \(e^{\ln b / \ln a} = b^{1/\ln a}\).)

I am a bit upset that I have never seen anyone point out that \(\dim\) is a logarithm, since it’s so obviously the case. Maybe I’m missing something? After all I am ignoring the infinite-dimensional cases entirely. But I suspect it’s just that math likes to stay on more solid rigorous ground than I do, and this is all too handwavey to be precise. I have no such qualms and I love to speculate about underappreciated connections between things, so I have no problem saying: dimension is a logarithm.

The simple reason why \(\dim_K\) acts like \(\log_k\) in the case of finite \(K\) is as follows. We need three observations:

One, the dimension of a vector space is defined as the cardinality of its basis. An individual vector \(\b{v} = v_1 \x_1 + v_2 \x_2 + \ldots + v_n \x_n \in K^n \simeq V\) can be thought of as a choice of function \(\{\x_1, \x_2, \ldots, \x_n \} \ra K\), since it assigns a coefficient \(v_i \in K\) to each basis vector \(\x_i\).

Two, the cardinality of the functions between sets \(B \ra A\) is given by \(\| A \|^{\| B \|}\), which is why we use the symbol \(A^B\) for the sets \(B \ra A\). For example the powerset of \(A\), that is, the set of all possible subsets of \(A\), is notated \(2^A\) because it is equivalent to the functions \(A \ra \{ 0, 1 \} \equiv \b{2}\), where a given subset is identified with the elements that map to \(1\).

Three: applying that to a vector space \(V \simeq K^n\), we can interpret \(K^n\) as describing the literal set of functions from a particular choice of basis \(\b{n} = \{ \x_1, \x_2, \ldots, \x_n \}\) into the underlying field \(K\), which naturally has cardinality \(\| V \| = \|K\|^{\| \dim_K V \|}\). Therefore the logarithm of this is the dimension of \(V\) over \(K\):

\[\dim_K V = \log_{\| K \|} \| V \| = \log_{\| K \|} \|K \|^{\dim_K V}\]

This is literally true in the case where \(V\) is finite dimensional and the field \(K\) is also finite. It’s less solid if either is infinite; however, I tend to think that expressions of this form are also literally true in the case of infinite dimensions, if you define things in a slightly better way. In particular you have to use a concept other than cardinality to measure the size of things if you want infinite expressions like \(\log_{\| \bb{R} \|} \| \bb{R}^2 \| = 2\) to make any sense. I have thoughts on this, but they’re for some other day.

Anyway, even if you only take this as meaningful on cardinalities of finite-dimensional sets over finite fields, I think it’s strange that it never really comes up. It is such a natural construction! Or maybe it does and I’ve missed it. But anyway, I like it, and I happen to think the correspondence here is much stronger and more significant than what I’ve just described.

If we repeat the above with ‘baseless’ logarithms, we get expressions like

\[\dim K^n = n \dim K\]

such that

\[\dim_K V = \frac{\dim V}{\dim K}\]

This seems mostly fine to me. The one place we have to be careful is in the definition of a tensor product. We want it to be the case that

\[\dim_K K^a \o K^b = \dim_K K^a \times \dim_K K^b = a \times b\]

But the naive approach has an extra factor of \(\dim K\):

\[\dim_K (K^a \o K^b) = \frac{\dim K^a \dim K^b}{\dim K} = \frac{(a \dim K) (b \dim K)}{\dim K} = ab (\dim K)\]

The problem is that the definition of the tensor product is a bit more complicated than just multiplying bases. A vector \(\b{u} \o \b{v} \in K^a \o K^b\) is not the Cartesian product of vectors \(\b{u}\) and \(\b{v}\), but rather the Cartesian product modulo a quotient on its scalar coefficient which combines two scalars \((k_1, k_2)\) into one \((k_1 \times k_2)\). Since this divides out a factor of \(K\), we have to do the same with our \(\o\) operation in order to make the cardinalities work out. This is done by specifying an \(\o_K\) operation, the “tensor product with respect to the field \(K\)”, as

\[U \o_K V = K^{\dim_K U \dim_K V} = K^{\dim U \dim V / \dim K}\]

which allows \(\dim_K K^a \o_K K^b = \dim_K K^{ab} = ab\) to work. (I suspect sometimes that the quotient in the definition of \(\o_K\) is not actually needed for most purposes, which would have the nice side effect of making this all work out more simply, but let’s not get into that.)

The definition

\[\dim_K K^a = \frac{\dim K^a}{\dim K} = \frac{a \dim K}{\dim K}\]

seems to imply that one could take the dimension/logarithm of a vector space with respect to a different underlying object, not the field \(K\), and get a meaningful result. For example it is my dream to be able to say that this is how you construct a vector space with a “fractional dimension” of \(\frac{1}{2}\):

\[\dim_{K^2} K \? \frac{\dim K}{2 \dim K} = \frac{1}{2}\]

This works fine at the level of cardinalities, more or less (if you allow that the rationals are invented precisely to let you make objects like \(1/2\) which interpolate between ratios of non-divisible integers). But it is hard to imagine how it should work if you want anything like a “field” or a “vector space” with its usual axioms to be meaningful. Maybe a vector \(\b{v} \in \bb{R}^4\) is viewed as a vector over \(\bb{R}^2\) via \(\b{v} = (v_w, v_x) \cdot (\w, \x) + (v_y, v_z) \cdot (\y, \z)\)? But then how does scalar multiplication work? If the scalars are \(\in K^2\), they have zero divisors, so you are not working in a field anymore. And what is meant by a vector with dimension \(\frac{1}{2}\)? Is it spanned by ‘half’ a basis vector over that pseudo-field? Maybe its elements look like \(\u = (u_x, \bullet) \cdot (\x, \bullet)\)? One must attempt to define versions of the theorems of linear algebra which are compatible with this sort of decomposition. No idea how to do that at the moment, but I suspect it can be done, with sufficient imagination, and I hope to attempt it in a future article.

6. Bases are Logarithms

The dimension of a vector space is the cardinality of its basis. But just like we use expressions like \(B^A\) for functions between sets because they are respected at the level of cardinalities \(\| B \|^{\| A \|}\), we may as well interpret the \(\dim\) operator in the same way: if \(\dim\) returns the cardinality of the basis, then let’s say that \(\log\) returns the basis itself, which happens to have that cardinality. For instance if a vector space \(V \simeq K^3\) has basis \((\x, \y, \z)\), we might write

\[\begin{aligned} \log_K V &= (\x, \y, \z) \\ \end{aligned}\]

And then define \(\dim_K\) as the cardinality of this:

\[\begin{aligned} \dim_K V &= \| \log_K V \| \\ &= \| (\x, \y, \z) \| \\ &= 3 \end{aligned}\]

Why not? After all \((\x, \y, \z)\) is an object for which \(K^{(\x, \y, \z)} \simeq V\), sorta, therefore \(\log K^{(\x, \y, \z)} = (\x, \y, \z)\). (One could also just let \(\dim_K\) refer to both operations, perhaps, or maybe write capital \(\text{Dim}_K V\) for the same thing.) Perpaps it’s a bit weird to treat \(K^{(\x, \y, \z)}\) as a set exponentiation when the exponent is an tuple / Cartesian product, but it should be easy to adjust things to make it work.

There is an obvious issue, though. Why would this particular choice of basis be the value of \(\log_K V\), since \(V\) has very many possibly valid bases and no reason to choose one a particular one?

Maybe it is more correct to \(\log_K V\) as really being an object which refers to all possible bases of \(V\) at once (I’m not sure what it’s called. Sort of a frame bundle but with only one base point?) We can give it coordinates: the space \(X = \log_K V\) is parameterizable by coordinates \((X_0, \Lambda)\), where \(X_0 = (\x, \y, \z)\) is an arbitrary ‘origin’ frame and \(\Lambda\) is an arbitrary linear transformation \(\in GL(V)\), the automorphisms of \(V\).⁶ I guess we should just write

\[X = \{ \Lambda X_0 \mid \Lambda \in GL(V) \}\]

and then the dimension itself is the cardinality of the quotient of this by \(\Lambda\), which will be a sort of generic object that represents the size of any choice of basis.

\[\dim_K V = \| \frac{\log_K V}{\Lambda} \| = \| \frac{X}{\Lambda} \|\]

If \(\log_K V = X\), then there ought to be an operation which goes the other way, that reconstructs a vector space from its basis. We may as well equate this with the linear span operation;

\[\span(X) = K^X = V\]

This is not quite how span is normally defined. Usually it’s something like: “\(\span(\x, \y, \z)\) is subspace of the (ambient) vector space \(V\) over the (ambient) field \(K\) which contains the vectors \((\x, \y, \z)\) and is of minimal dimension”. To interpret it algebraically, though we don’t really want to make reference to an “ambient” vector space or field, because it should just be an operation on the vectors itself. For this we need to at least explicitly indicate the underlying field, by writing \(\span_K\) with a subscript:

\[\span_K(X) = K^X = V\]

All of this is definitely rife with abuses of notation, and I’m not sure that it’s quite the best way to think about things. But I still wanted to mention it because it’s nice to think of the operators \(\dim\) and \(\span\) as being linear algebra analogues of \(\log\) and \(\exp\).

It is also interesting consider what might be meant by the baseless logarithm in the sense of bases. In the expression

\[\log_K K^X = \frac{\log K^X}{\log K} = \frac{X \log K}{\log K}\]

what would be meant by \(X \log K\) as a ‘basis’? Presumably the division by \(\log K\) corresponds to some sort of quotient… but we will need a way of interpreting \(\log K\) itself. Perhaps as a “basis for \(K\)”? I’m not sure. I do think there’s something here, but it gets much more speculative so I will leave it for another time.

7. Functions are Logarithms?

Treating \(\log_K K^n = n\) as returning a basis for \(K^n\) as a set is an example of a general procedure which doesn’t quite have a name as far as I know. It is sort of like categorification, but not quite. Rather than locating categories for set operations, we’re locating sets for algebraic operations, and not making any reference to categories really. So I’m not sure. Maybe ‘setification’? Or ‘structurization’? I dunno.

The standard example of this ‘setification’ is to treat arithmetic operations on natural numbers like \(A+B\), \(AB\) and \(B^A\) as being projections out of set operations \(A \sqcup B\), \(A \times B\), and \(B^A\) (the functions \(A \ra B\)). This works nicely for finite sets because the operations respect cardinalities. (As mentioned earlier, I think you have to replace ‘cardinality’ with something like ‘numerosity’ to make this work elegantly on infinite sets, and I don’t know how that works yet.)

A compelling reason for thinking this way is that the setified arithmetic operations in fact explicitly enumerate the sets they describe. For example, given sets \(A = \{ a, b \}\) and \(X = \{ x, y \}\), you can expand \(A^X\) algebraically, using the fact that all the variables will later equal \(1\):

\[(a+b)^{x+y} = (a+b)^x (a+b)^y = (a^x + b^x)(a^y + b^y) = a^x b^x + a^x b^y + a^y b^x + a^y b^y\]

Then upon actually setting the variables to \(1\) this correctly describes the relationship in cardinalities: \(2^2 = 1 + 1 + 1 + 1\), so the number of functions \(X \ra A\) is \(4\).

What’s interesting is that it also describes the sets themselves. Each term in the expanded sum is one of the four possible functions \(X \ra A\) exactly when we interpret \(a^x b^y\) as the function which maps \(x \ra a\) and \(y \ra b\). Also, evaluation of these variables corresponds to evaluating the functions, e.g. setting e.g. \(x=1\) and \(y=0\) to get \(a^x b^y \mapsto a^1 b^0 = a\), and setting one variable but leaving the other gives restriction, e.g. \(y=0\) sets \(a^x b^y \mapsto a^x\).

All of this basically also works if the variables have values other than \(1\), in which case they represent unlabeled sets of whatever cardinality; however, the algebraic manipulation \((a+b)^x = a^x + b^x\) is not valid and you have to use a binomial expansion instead.

You can do similar constructions with a lot of combinatoric objects, although they don’t always so cleanly correspond to algebraic manipulations. Factorials look like

\[\begin{aligned} (a+b+c)! &= a^a b^b c^c + a^a b^c c^b + a^b b^a c^c + a^b b^c c^a + a^c b^b c^a + a^c b^a c^b \end{aligned}\]

which enumerates the \(3! = 6\) permutations of \(3\) elements. Combinations looke like

\[\begin{aligned} \binom{a+b+c}{x+y} &= \frac{1}{x^x y^y+x^y y^x}[ a^x b^y + a^y b^x + a^x c^y + a^y c^x + b^x c^y + b^y c^x] \\ &= a^{q} b^q + b^q c^q + c^q a^q \end{aligned}\]

which enumerates the \(\binom{3}{2} = 3\) possible \(2\)-element combinations of \(3\) elements; here the \(\frac{1}{x^x y^y+x^y y^x}\) corresponds to the quotient by \((x+y)!\). Dividing through by the number of permutations implements the quotient \(x \sim y\) that avoids double counting, and \(q\) is a new variable that represents carrying out this quotient (I’m not sure if this is the best way to write this). Note that although all these variables will end up equaling \(1\), by leaving them as independent variables they track meaningful information from step to step.

I suspect that every arithmetic identity has some equivalent set-ified expression like this. I also notice that a lot of information is lost when you map these set expressions back onto arithmetic. For example, you elide the distinctions between all possible quotients that lead to the same cardinality. Probably there is a lot of interesting structure there.

Anyway, for our purposes, I want to observe one thing about these. When thinking of functions as sets we usually picture them as ‘relations’: a function \(f: X \ra A\) is modeled as the set

\[\begin{aligned} f = \{ (x, f(x)) \mid x \in X \} \subset X \times A \\ \end{aligned}\]

E.g. \(\{ (x,a), (y, b) \} = xa + yb\) in our example. This set happens to have the cardinality \(\| f \| = \| X \|\), although it’s not clear what use that is.

Now consider \((a+b)^{x+y} = a^x b^x + a^x b^y + a^y b^x + a^y b^y\) from earlier. If \(a^x b^y\) is supposed to describe a single function from \(X = \{ x, y\}\) to \(A = \{ a,b \}\) , then why doesn’t it setify to something like \(\{ (x, a), (y, b) \}\), with cardinality \(2\)?

Maybe you see where I’m going with this. \(f = a^x b^y\) has cardinality \(1\), because it’s one function. Its logarithm, however, looks closer to its representation as a relation:

\[\log f \? x \log (a) + y \log (b)\]

It’s a lot like \(xa + yb\), but it’s also suspiciously different. Also, it doesn’t have a cardinality since we need to divide by a base, but when we do it seems like any choice we make has to give the cardinality \(\log_b f = \log f / \log b = x \log_b a + y \log_b b = x (0) + y(0) = 0\). What do we make of this?

After thinking this for a while I still don’t really feel like I have a good explanation for it, but I think we are supposed to think of it as equivalent to \(x a + yb\), just with the \(a\) and \(b\) written in a different basis, so it is more like a comparison between \(a \o x + b \o y\) and \(a \o \log x + b \o \log y\) than between numeric expressions. The cardinality being \(0\) doesn’t matter, because it’s not meaningful to talk about the cardinality of a function. And the role of \(\log x\) is just to change algebras for \(x\) from multiplicative to additive, but the two objects are supposed to be isomorphic and regarded as the same, at least in this case where the cardinality doesn’t mean anything.

I’m not sure about this part, and might come back and rewrite it later if I find a better interpretation. In any case I think it is interesting (or amusing, maybe) that \(\log f = \log a^x b^y\) gives something that at least resembles the function’s representation as a relation. Everything is logarithms?

8. Everything is Logarithms

What we have been discussing is the most simple and well-behaved version of a logarithm in mathematics, the isomorphism between the additive real algebra \((\bb{R}, +)\) and the multiplicative one \((\bb{R}^{\geq 0}, \times)\). Of course there are logarithms in mathematics which are more complicated than that, such as the complex logarithm \(\log z = \text{Log } z + 2\pi i k \mid k \in \bb{Z}\), or its messier cousins like the logarithm of a matrix. But I suspect these are a confusion of concepts. What’s really going on in the logarithm on \(\bb{C}\), for instance, is that angles really take their values \(\in S_1\), not \(\bb{R}\), which has a different topology, and the weird behavior follows from not respecting this. A different set of conventions would move the problem out of the logarithm and into the definitions of the values themselves. Unfortunately that’s not how things are are defined today so you have to deal with it—but, still, it doesn’t seem like the logarithm’s fault to me.

Anyway the discussion in this article ignores those cases and assumes that \(\log\) really is an isomorphism: it’s just a way of taking something expressed in a multiplicative form and re-expresses it in an additive form. This is turns out corresponds to many operations that one learns in math, such as the \(\dim\) operator in linear algebra and the \(\nu_p\) operation in number theory (sorta) and the total derivative in calculus (also sorta).

All of these things which appear to be very different seem to in some way be instances of the same basic primitives. And although these associations arise from my sort of… numerology… I can’t shake the feeling that it’s all too clean to not matter. Perhaps math needs to clean this all up: we are somehow missing the forest for the trees by keeping all this redundancy buried in the notations; actually there are only a few basic operations which are being written differently everywhere, and with all the patterns disguised everything is a lot harder than it needs to be. I suspect the patterns I’ve written about in this article should not feel like things I had to rediscover for myself. They follow naturally from the material that everybody learns.

I also keep finding that the math of physics seems to end up at a lot of the same structure. I first noticed these patterns in the operator formulation of quantum mechanics because it seems to insist on a certain ontology for its mathematics. I wonder if this is because physics is telling us how things “should be done”. Since in physics the mathematics is a human lens through which we view reality, the math must not impose its own views on how things are done, and any views you accidentally impose eventually clash with the requirements of the physics.

This is the idea behind the concept general covariance, that the properties of objects are independent of the coordinates we use to express them, and so the meaningful theorems about reality end up being expressed in coordinate-free ways. The same philosophy applied to linear algebra or differential geometry leads to their covariant formulations that are indisputably ‘better’ than the forms in coordinates.

The baseless logarithm, which seems somewhat nonsensical mathematically, is an example of this applied on purely mathematical terms. It basically says that the isomorphism from multiplicative to additive algebraic representations of the same thing is separate from the choice of units on those algebras, but most of its properties are unrelated to the units. Just like how the concept of a geometric vector is distinct from its projection onto a particular coordinate system. Meanwhile a bunch of other things with other notations are basically the same operation as the logarithm, or closely related to it.

When you take general covariance to its extreme you end up asking that all of your mathematics be formulated in a covariant way, as explicit relations between one thing you measure to another thing you measure. For example we think of say having a certain cardinality, but in fact cardinality is a property of the set that we measure, and we have to be clear about how we do that because it’s all relative to the “coordinate system” for those measurements. Such a formulation is necessary to find the answers to why questions about how mathematics works, about what is ‘actually’ going on independent of the human definitions and frameworks of set or category theory or whatever. The observations in this article are not very deep, but they seem to me to be among the many clues which point towards that formulation. I still can’t see it, though.

In differentials, this operation the differential of \(f\) but restricted to its \(dx\) component: \(\frac{\p f}{\p x} dx = f_x dx = df \mid_{x}\). This is a perfectly interesting object, a covariant derivative on the foliations of the \(x\) coordinate, I believe (if I have that right), but it’s not normally written this way. ↩
The \(f_x = \p_x f = \p f / \p x\) notation for partial derivatives is unfortunate; it should be \(df_{dx}\), to indicate that it is the “\(dx\) component” of the vector \(df\), or \(d_x f\), meaning the \(x\) component of \(d\) acting on \(f\). Better yet it would be \((\p f)_{\p x} = \p_x f\) and the \(d\) symbol would be retired, but that seems like a tall order. ↩
There is a somewhat-related thing called an arithmetic derivative \(D(n)\) which has \(D(p^k) = k p^{k-1}\) in analogy to a regular derivative and obeys the product rule \(D(mn) = D(m) n + m D(n)\). It has a corresponding partial derivative \(\p_p(n) = \nu_p(n)/p\) which does extract the \(\nu_p\) coefficient, albeit in a roundabout way. I find this unsettling, because I can’t see a way to interpret \(D(n)\) as an actual differential—the derivative \(D(2^k) = k 2^{k-1}\) would have to be something like \((2+\e)^k - 2^k\), which seems weird? There is also a version of the logarithmic derivative \(\text{ld}(2^a 3^b \cdots) = a/2 + b/3 + \ldots\) which acts a lot like \(\log n\) with \(\log 2 \mapsto 1/2\)… I dunno what to make of any of these, but they do seem to live in the same region of mysterious-connection-space as everything else. ↩
If you happen to have a vector in a polar form like \(\b{v} = v_r e^{R v_{\theta}}\), that refers to a second layer of exponential representation, via \(T^{\v} = T^{v_r e^{\p_{\theta} v_{\theta}}(\p_x)}\), where \(\p_x\) is a choice of origin for the rotational \(\theta\) coordinate (which may be multidimensional as well). ↩
Apparently the \(\oplus\) symbol is due to Bourbaki because everything was a mess prior to that. Also it happens to be a coproduct (which came later) and those do correspond to addition on sets, so there is at least a connection to addition… but at present I think it is largely a mistake. ↩
The technical term is that it is a \(GL(V)\)-torsor since the choice of origin \(X\) is arbitrary. The concept is easier to understand from Baez. This is one of those mathematical terms which I don’t like because it is so simple that it should not really have a special name (nor such a technical Wikipedia article). ↩

Further Meditation on Taylor Series

2026-02-05T00:00:00+00:00

I have spent a lot of time trying to think intuitively about calculus, Taylor series, divergent series, and things like that. Here are a couple things I realized at some point which I would like to have written down. (This is sort of a sequel to a much more elementary post some years ago. I know a lot more now, and have also apparently gotten a lot more verbose.) Maybe they are well-known to some people, or maybe not, but they were at least interesting to me.

As usual: don’t expect anything even resembling rigor in here; I’m just poking around.

1. Taylor Series Five Ways

The general form of a Taylor series is

\[\begin{aligned} f(x+a) &= f(x) + f'(x) a + f''(x) \frac{a^2}{2!} + f'''(x) \frac{a^3}{3!} + \ldots \\ &= \sum_{k=0}^{\infty} f^{(k)}(x) \frac{a^k}{k!} \\ \end{aligned}\]

For example consider the expansion around \(x=0\) of \(\frac{1}{1-x}\), my favorite function (it’s so mysterious!):

\[S_0(x) = 1 + x + x^2 + x^3 + \ldots\]

The approximation converges for \(\| x \|<1\), at which point the \(x^k\) terms never go to \(0\) and the sum diverges. If you want a version that works for \(\| x \|>1\) you can use the expansion around \(x=\infty\) instead. Simply write \(y=1/x\) and expand everything in terms of \(y\) instead:

\[\begin{aligned} \frac{1}{1-x} &= -\frac{1}{x} \frac{1}{1 - \frac{1}{x}} \\ &= -\frac{1}{x} [1 + \frac{1}{x} + \frac{1}{x^2} + \ldots] \\ S_{\infty}(x)&= -\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} - \ldots \end{aligned}\]

Some algebra gives a version around any \(x=k \neq 1\), although it’s a bit unwieldy:

\[S_k(x) = \frac{1}{1-k} \frac{1}{1-\frac{x-k}{1-k}} = \frac{1}{1-k}[1 +(\frac{x-k}{1-k}) + (\frac{x-k}{1-k})^2 + \ldots]\]

These are always good to keep around as an example.

Now here are six ways of thinking about where the Taylor series for \(1/(1-x)\) “comes from”. Probably there is some insight to be gleaned from the fact that they all lead to the same thing.

Way 1: Repeated integration

The most elementary way is to repeatedly apply the fundamental theorem of calculus. Assuming \(f(x)\) has all of its derivatives and things like that, we repeatedly expand and then factor out all the constant terms:

\[\begin{aligned} f(x+a) &= f(x) + \int_x^{x+a} f'(x_1) d x_1 \\ &= f(x) + \int_{x}^{x+a} [f'(x) + \int_x^{x_1} f''(x_2) \d x_2] \d x_1 \\ &= f(x) + \int_x^{x+a} f'(x) + [\int_x^{x_1} [f''(x) + \int_{x}^{x_2} f'''(x_3) \d x_3] \d x_2] \d x_1 \\ &= \text{(etc...)} \\ &= f(x) + \int f'(x) \d x_1 + \int \int f''(x) \d x_2 \d x_1 + \int \int \int f'''(x) \d x_3 \d x_2 \d x_1 + \ldots \\ &= f(x) + f'(x) a + f''(x) \frac{a^2}{2} + f'''(x) \frac{a^3}{3!} + \ldots \end{aligned}\]

Easy enough. In the case of \(1/(1-x)\) you get \(\p^k_x (1-x)^{-1}= (1-x)^{-(k+1)}\), so all the derivatives are \(1\) at \(x=0\).

Way 2: Exponentiation of a derivative

Since \(f(x + dx) \approx f(x) + f'(x) dx = (I + dx \, \p_x) f(x)\), we can interpret \((I + dx \, \p_x)\) as an operator \(T^{dx}\) which translates \(f(x) \mapsto f(x+dx)\):

\[T^{dx} = I + dx \, \p_x\]

To translate over some finitesimal¹ distance \(a\) we then apply this repeatedly, up to \(a/dx\) times, so that we will have translated by \(a\) in total. When we take the limit as \(dx \ra 0\) this becomes exactly the construction of the exponential function:

\[\begin{aligned} \lim_{dx \ra 0} [T^{dx}]^{a/dx} f(x) &= \lim_{dx \ra 0} [I + dx \, \p_x]^{a/dx} f(x) \\ &= [\lim_{dx \ra 0} [I + dx \, \p_x]^{1/dx}]^a f(x) \\ f(x+a) &= e^{a \p_x} f(x) \\ &= [1 + a \p_x + \frac{(a \p_x)^2}{2!} + \frac{(a \p_x)^3}{3!} + \ldots] f(x) \end{aligned}\]

So \(T^{dx}\) exponentiates to give \(T^a = (T^{dx})^{a/dx}\). (Also why we write \(T^{dx}\) instead of the \(T_{dx}\) you sometimes see.²) Now you might call that circular because we end up using the Taylor series for \(\exp\) to produce the general form. But, we can also come up with the series from a simple combinatoric argument (well, simple in the sense that it’s simple to follow. It still involves a sketchy limit.):

Suppose that the product \((1 + \e x)^{1/\e}\) has an integer number of terms (so we’re assuming \(n= 1/\e\) an integer). Then the binomial theorem gives

\[\begin{aligned} (1 + \e x )^n &= \sum_{k=0}^n (\e x )^k 1^{n-k} \\ &= \binom{n}{0} 1 + \binom{n}{1} (\e x ) + \binom{n}{2} (\e x )^2 + \binom{n}{3} (\e x )^3 + \ldots \\ &= 1 + \frac{n!}{(n-1)!} (\e x x) + \frac{n!}{(n-2)! 2!} (\e x )^2 + \frac{n!}{(n-3)! 3!} (\e x )^3\ldots \\ &= 1 + n \e x + \frac{n(n-1)}{2!} (\e x )^2 + \frac{n(n-1)(n-2)}{3!} (\e x)^3 \ldots \\ &\approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \end{aligned}\]

Another way of thinking about it is that in the giant product

\[(1 + \e x)^n = (1 + \e x)(1 + \e x)(1 + \e x)(1 + \e x)\cdots\]

the linear term \(n \e x = x\) consists of \(n\) ways of having one factor of \(\e x\), the quadratic term consists of all \(\binom{n}{2} = n(n-1)/2!\) ways of having two factors of \(\e x\), etc. As \(n\ra \infty\) these are dominated by their highest-order term \(n^k/k!\), which is where the terms in the Taylor series come from. (I guess this is obvious when you think about it but I remember not realizing it for a long time.)

Note that there is an approximation going on: we end up collapsing terms like \(n(n-1) \e^2 = (n \e)^2 [1 - \frac{1}{n}] = n^2 \e^2 - n \e^2 = 1 - \e\) to just \((n\e)^2 = 1\), dropping a factor of \(\e\). A completely-correct series (for integer \(n\)) would be \((1 + \e x)^n = 1 + x + (1-e) \frac{x^2}{2!} + (1 - 3 \e + 2 \e^2)\frac{x^3}{3!} + \ldots\), which is kinda interesting. But anyway.

Writing \(e^{a \p_x} f(x)\) perspective treats \(\p_x\) as a generator in the sense of groups, specifically in the Lie algebras of Lie groups. It ‘generates’ translation in the \(x\) coordinate,³ according to

\[e^{a \p_x} f(x) = f(x+a)\]

A Taylor series in any other variable can be produced by writing \(f(x) = f(g^{-1}(g(x)))\); then

\[e^{a \p_{g(x)}} f(x) = f(g^{-1}(g(x) + a))\]

For example,

\[e^{(\ln a) x \p_x} f(x) = e^{(\ln a)\p_{\, \ln x}} f(e^{\ln x}) = f(e^{\ln x + \ln a}) = f(a x)\]

since \(\frac{\p}{\p (\ln x)} = \frac{\p}{(\p x)/x} = x \p_x\). More generally, if we come across a derivative operator \(\p_{h(x)}\), we can interpret it as translation in the \(g(x) = \int \frac{1}{h(x)} dx\) “coordinate”. Interestingly this can also be written

\[a^{\p_{\, \ln x}} f(x) = f(ax)\]

See the Wiki on shift operators for more examples.

Logarithms work in the same framework by using the approximation \(\log (1+x) \approx x\):

\[\log T^{dx} = \log (1 + dx \p_x) \approx dx \p_x\]

and more generally

\[\log T^{a} = \log e^{a \p_x} = a \p_x\]

which is neat. It also works in more dimensions, in which case vectors are generally the logarithms of finite translation operators:

\[\log T^{a \x + b \y} = \log T^{a \x} T^{b \y} = a \p_x + b \p_y\]

although things get more complicated if the coordinates have curvature.

Something I like about this approach is that it is valid even outside the radius of convergence of the series, in a certain sense. In the case of \(f(x) = \frac{1}{1-x}\), for example, the derivative stops existing at \(a=1\), but it is still meaningful to talk about, say, \(T^{2} f(x) \vert_{x=0} = [e^{2 \p_x} f(x)] \vert_{x=0} = f(2) = -1\). Whatever goes wrong with the Taylor series is an artifact of how we write out the series, but the translation operator ought to still be valid everywhere. Perhaps the ‘true’ Taylor series for \(1/(1-x)\) ought to be something like

\[\frac{1}{1-x} \? 1 + x + x^2 + \ldots - 2 \delta(1-x)\]

where that term on the end is sort of like an ‘infinite-order’ delta-function derivative that only applies for the \(T^{dx}\) which crosses \(x=0\) and serves to precisely ‘jump’ the function up to its correct value on the other side of the asymptote. After all in the finite approximations, \(f'(x)\) involves \(f(x+dx)\), \(f''(x)\) involves \(f(x + 2 \d x)\), etc. So it must be that the terms involving \(f(x+1)\) or higher only show up after the \(1/dx\)‘th entry in the series—way off at infinity. Now it may seem hard to write such at thing down correctly. But in fact we do know what the series is at that point. The result must be

\[\begin{aligned} \frac{1}{1-x} &= S_0(x) + \theta_{1-x} [S_{\infty}(x) - S_0(x)] \\ &= S_0(x) - \theta_{1-x}[\ldots + x^{-2} + x^{-1} + 1 + x + x^2 + \ldots] \end{aligned}\]

with \(\theta_{1-x}\) as a step function for \(x>1\). Which is maybe an interesting way of thinking about it?

Way 3: direct combinatorics

This is related to the previous argument, but somewhat suggestive in a different direction.

For mysterious but significant reasons it is mathematically valid to apply the binomial theorem for fractional or negative exponents. Every term in \((a+b)^n\) still looks like \(\binom{n}{k} a^k b^{n-k}\), but the rule is that extend the sum to infinity instead of stopping at \(k=n\). In fact there is no need to think of it as a rule: since \(\binom{n}{k} = \frac{n!}{k! (n-k)!}\), we simply observe that if \(n\) is negative or fractional, then it is no longer the case that \(\binom{n}{k} = 0\) for \(k > n\). This requires doing some dubious things with “negative” factorials”, but they… seem to work?

\[\begin{aligned} \binom{-1}{k} &= \frac{(-1)!}{k!(-1 - k)!} \\ &= \frac{(-1) (-2) (-3) \times \ldots \times (-k) \cancel{(-1-k)!}}{k! \times \cancel{(-1-k)!}} \\ &= \frac{k! (-1)^k}{k!} \\ &= (-1)^k \\ \end{aligned}\]

So this gives, for \(1/(1-x) = (1-x)^{-1}\):

\[\begin{aligned} \frac{1}{1-x} &\? \sum_{k=0}^{\infty} \binom{-1}{k} 1^k (-x)^{-1 - k} \\ &= - \sum_k (-1)^k (-1)^k x^{-1-k} \\ &= -\frac{1}{x} [1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \ldots] \end{aligned}\]

which is the Taylor series around \(x=\infty\). Strangely, if you swap the roles of the two variables in the summation, you get the series around \(x=0\) instead:

\[\begin{aligned} \frac{1}{1-x}&= \sum_{k=0}^{\infty} \binom{-1}{k} (-x)^k 1^{-1-k} \\ &= \sum (-1)^k (-1)^k x^k \\ &= 1 + x + x^2 + x^3 + \ldots \end{aligned}\]

This I find very mysterious. More generally, there are two series for \((x+y)^{-1}\) depending on whether you divide through by \(x\) or \(y\) first:

\[\begin{aligned} \frac{1}{x+y} &=\frac{1}{x} (1 + \frac{y}{x})^{-1} = x^{-1} - y x^{-2} + y^2 x^{-3} - y^3 x^{-4} + \ldots \\ &= \frac{1}{y} (1 + \frac{x}{y})^{-1} = y^{-1} - x y^{-2} + x^2 y^{-3} - x^3 y^{-4} + \ldots \end{aligned}\]

Observe that when you multiply one of these series by \((x+y)\) it does indeed seem to give \(1\):

\[\begin{aligned} (x+y)(x+y)^{-1} &= (x+y) (x^{-1} - y x^{-2} + y^2 x^{-3} - y^3 x^{-4} + \ldots) \\ &= (x+y) x^{-1} - (x+y) yx^{-2} + (x+y) y^2 x^{-3} - (x+y) y^3 x^{-4} + \ldots \\ &= (1 + \cancel{y x^{-1}}) - (\cancel{x y^{-1}} + \cancel{y^2 x^{-2}}) + (\cancel{x^{-2} y^2} - \cancel{y^3 x^{-3}}) + \ldots \\ &= 1 \end{aligned}\]

And likewise for the other (as it must, by symmetry). Interestingly you can also use linear combinations of both solutions. If we write \(S_1 = x^{-1} - y x^{-2} + y^2 x^{-3} - \ldots\) and \(S_2 = y^{-1} - x y^{-2} + x^2 y^{-3} + \ldots\) then

\[(x+y) (\lambda S_1 + (1-\lambda) S_2) = 1\]

for any \(\lambda\) (technically this is not allowed because both are not going to converge at the same time, but I suspect it’s still meaningful somehow).

It’s intriguing that the series you can get from these expansions are the same choices as the Taylor series that you get as you vary the point you’re expanding around. Whatever mysterious meaning underlies these “unnatural combinatorics” (unnatural in the sense that… well… the exponents are fractions or negative, so not \(\in \bb{N}\)…), it is the same mystery as to why Taylor series break at poles (and then sorta keep working anyway if you squint).

Although I’m not gonna fill up space showing it, the same sort of thing happens for series like \((x+y)^{1/2}\) and any other non-natural exponent. And if you feel like splitting a product up into pieces

\[(x+y) = (x+y)^{3/2} (x+y)^{-1/2}\]

You will find that the two series perfectly cancel out however you do it and whichever variable you expand in, even if the series don’t converge. So I am inclined to believe, just for elegances’ sake, that all of these series are really valid at all inputs, and that it is really the notion of “convergence”, or maybe just the notion of “addition”, which is not useful so useful for understanding them.⁴

Way 4: Long Division

Another way of producing the combinatoric series is polynomial long division. The normal approach: to compute divide \(g(x)\) by \(f(x)\) we repeatedly compute \(g_i(x) = f(x) q_i(x) + r_i(x)\), setting \(g_{i+1} = r_i\) on each step:

\[\begin{aligned} \frac{g(x)}{f(x)} &= \frac{f(x) q_0(x) + r_0(x)}{f(x)} \\ &= q_0(x) + \frac{r_0(x)}{f(x)} \\ &= q_0(x) + \frac{f(x) q_1(x) + r_1(x)}{f(x)} \\ &= q_0(x) + q_1(x) + \frac{r_1(x)}{f(x)} \\ &= q_0(x) + q_1(x) + q_2(x) + \ldots \end{aligned}\]

Simply enough. Now consider calculating \(1/(1-x)\), that is, \(g(x) = 1\) and \(f(x) = 1-x\).

\[\begin{aligned} \frac{1}{1-x} &= \frac{1(1-x) + x}{1-x} \\ &= 1 + x \frac{1(1-x) + x}{1-x} \\ &= 1 + x + x^2 \frac{1(1-x) + x}{1-x} \\ &= 1 + x + x^2 + x^3 + \ldots \\ &= S_0(x) \end{aligned}\]

The algorithm doesn’t terminate, for the same reason it doesn’t terminate when you divide non-divisible numbers like \(22/7\), but for \(\| x \| < 1\) it does converge in the sense that the infinite-order terms become negligible.⁵

But we can also do the division on the other term:

\[\begin{aligned} \frac{1}{1-x} &= (-\frac{1}{x})(1-x) + \frac{1}{x} \\ &= -\frac{1}{x} + \frac{1}{x}[ (-\frac{1}{x})(1-x) + \frac{1}{x}] \\ &= -\frac{1}{x} - \frac{1}{x^2} - \frac{1}{x^3} - \ldots \\ &= S_{\infty}(x) \end{aligned}\]

By the way, the long division algorithm is equivalent to a series of nested substitutions into the definition of \(S(x)\):

\[\begin{aligned} S &= \frac{1}{1-x} \\ 1 &= (1-x) S \\ S &= 1 + x S \\ &= 1 + x (1+ xS) \\ &= 1 + x (1 + x(1+xS)) \ldots \\ &= 1 + x + x^2 + x^3 + \ldots \end{aligned}\]

You see these substitutions done sometimes when fiddling with divergent series and it’s good to realize that they are just long division.

Way 5: Orthonormal Bases

A standard operation in linear algebra is to project a vector onto an orthornomal basis, as in \(\b{v} = v_x \b{x} + v_y \b{y} + v_z \b{z}\). The same basically works for functions, although a lot more analytic details are involved in talking about when it’s possible (Something like: your function has to be an element of a Hilbert space, meaning an infinite-dimensional inner-product space which is closed under limits of sequences). When the construction is valid you end up with a an expansion like

\[f(x) = \sum_k \<f, \phi_k\> \phi_k(x)\]

which expresses \(f(x)\) in terms of its projections onto some family of functions \(\{ \phi_k \}\). which has \(\< \phi_i, \phi_j \> = 1_{i=j}\). The inner product \(\< f, g \>\) usually looks like infinite-dimensional dot product—that is, an integral like \(\int_X f(x) \overline{g(x)} \d x\). The overline indicates complex conjugation, which is invariably involved whenever the functions are allowed to be complex.

Standard examples (which are more-or-less equivalent to each other if you gloss over all the analysis) include:

The Fourier transform, with \(\phi_k = e^{ikx}\), writes \(f(x) = \int_{\bb{R}} f_k e^{ikx} dx\)
The Laplace Transform, with \(\phi_k = e^{st}\), writes \(f(x) = \int_{c + i\bb{R}} f_s e^{sx} ds\) (ish)
The Cauchy Integral Formula, with \(\phi_k = z\), writes \(f(z) = \sum_{k \in \bb{Z}} f_k z^k\), giving the Laurent series for \(f(z)\) (which generalizes the Taylor series to allow negative coefficients.)
Function evaluation can also be included on this list: what is \(f(x)\) but the ‘coefficient’ of \(f\) at the point \((x)\), that is, the value of \(\int f(x) \delta(x) \d x\)?

We could certainly say that (for analytic \(f\)) that the Taylor series terms are all given by the Cauchy integral formula

\[f^{(n)}(0) = \frac{n!}{2\pi i} \oint \frac{f(z)}{z^{n+1}} dz\]

Which after a change of variables to \(z = r e^{i \theta}\) becomes

\[\begin{aligned} f^{(n)}(0) &= \frac{n!}{2\pi i} \oint \frac{f(z)}{(re^{i \theta})^{n+1}} (\cancel{dr e^{i\theta}} + i r e^{i \theta} d \theta) \\ &=\frac{n!}{2\pi} \oint \frac{f(z)}{(re^{i\theta})^n} d\theta \end{aligned}\]

When \(f(z) = z^m\) this cancels out unless \(m-n = 0\), in which case it gives \(2\pi\), which is why it extracts the \(f^{(m)}(0)\) term from \(f(z)\) when it has a Laurent series. (This constuction essentially takes the Fourier transform of \(f(r e^{i \theta})\) in the \(\theta\) coordinate.) So the Taylor/Laurent series is a sort of projection onto the orthonormal basis given by \(z^n\), computed by changing coordinates on \(\bb{C}\).

I happen to disprefer anything on principle anything that involves \(\bb{C}\), so here’s an attempt at a strictly real version of the same projection. We would want to treat \(\phi_k = \frac{x^k}{k!}\) as a basis and then come up with some sort of dual basis \(\phi^j\) such that \(\< \phi_k, \phi_j \> = \int \phi^j \phi_k dx = 1_{1=j}\). The obvious thing to try is just the inverse, \(\phi^j = \frac{j!}{x^j}\), but it doesn’t work

\[\begin{aligned} \< \phi_j, \phi_k \> &= \int_{\bb{R}} \phi^j \phi_k dx \\ &= \int \frac{j!}{x^j} \frac{x^k}{k!} \d x \\ &= \int \frac{j!}{k!} x^{k-j} \d x\\ &\? \begin{cases} 0 & k > j \\ 1 & k = j \\ \text{???} & k<j \end{cases} \end{aligned}\]

So, not quite. What does work, although it feels like cheating, is using derivatives of delta functions,

\[\phi^j(x) = (-\p_x)^j \delta(x)\]

because they pass over to anything they’re intergrated against via integration by parts: \(\int \delta' (x) f(x) dx = - \int \delta(x) f'(x) \d x = -f'(0)\).

\[\< \phi_j , \phi_k \> =\int (-\p_x)^j \delta(x) \frac{x^k}{k!} dx = \int \delta(x) (\p_x)^j \frac{x^k}{k!} \d x = 1_{j=k}\]

It’s cheating because \((-\p_x)^j \delta(x)\) is just a trick for writing derivatives of \(f(x)\) inside an integral. But, okay, it sorta lets us write a \(\phi_k\) which is ‘actually’ just the derivative operator \(\p_k\), such that the inner product \(\< \phi_j, \phi_k \>\) with this definition just ends up computing \(\p_x^j \phi_k \|_{x=0}\), which is exactly the definition of a Taylor series we already knew. Note that this also does not work if \(f(x)\) has any negative-order Laurent series terms, which makes it sorta strictly less powerful than the Cauchy integral form.

An interesting but highly-questionable modification comes from rewriting the delta derivative with a polynomial, like \(\delta(x)/x = -\delta'(x)\). This is based on how it behaves in an integral: \(\int (\delta(x)/x) f(x) \d x = \int \delta(x) f'(x) \d x\) if \(f(x) = x g(x)\), that is, if it does not have a constant term.

\[\phi^j(x) = (-\p_x)^j \delta(x) \? \frac{j!}{x^j} \delta(x)\]

You see this operation around sometimes, but it feels really sketchy; it really relies on the “not having a constant term”, though (I wrote about its sketchiness some here). Maybe there’s a principal value / finite part-type-justification for it even in cases where it doesn’t literally apply? But it’s hard to imagine how you could get \(\int (\delta(x)/x^2) dx = 0\) to ever work on the reals alone without a “finite part” operation, which is to say, freely dropping the divergent part of a divergent result without using an algebraic operation to do it. I mention it only because the presence of \((x^j/j!)^{-1}\) in this form seems suspiciously relevant.

In any case, treating \(\phi^k = (-\p_x)^k \delta(x)\) as the “dual basis” for \(\phi_k = x^k/k!\), lets us regard the Taylor series (or the Laurent series) as a projection onto a basis like any other vector:

\[f(x) = \sum \< \phi_k, f \> \phi_k = f(0) + f'(0) x + f''(0) \frac{x^2}{2!} + \ldots\]

It does seem like this is ‘morally’ what a Taylor series is doing, projecting a function onto a basis like this, even if the literal algebraic operations don’t quite seem to do that. Residue integrals are the standard way of working around the questionableness, but I wonder if there is simpler way to think about it.

Bonus way 6: More Basis Projections

Here is another way of producing an orthonormal-basis-projection-type formula for the Taylor series, which is kind of equivalent to Way 5, but weird. It’s also kinda circular; let’s ignore that.

In the exponential series \(e^{a \p_x} = 1 + a \p_x + \frac{a^2}{2!} \p_x^2 + \ldots\), each of the \(a\) terms is \(a^k/k!\), which we can write as a sort of inverse of a derivative operator

\[\frac{a^k}{k!} = \int_0^a \int_0^{a'} \int_0^{a''} \cdots (1) \cdots da'' da' da \? \p_a^{-k} (1)\]

Using these each term \(\frac{a^k}{k!} f^{(k)}(0)\) results from one of each type of operator, both evaluated at \(x=0\). I will write as a fraction shorthand, like this:

\[\begin{aligned} \frac{a^k}{k!} f^{(k)}(0) &= (\p_a^{-k} \p_x^k) f(x) \\ &= (\p_a^{-1} \p_x)^{k} f(x) \\ &= (\frac{ \p_x}{ \p_a})^k f(x) \end{aligned}\]

Then the whole Taylor series is

\[[1 + (\frac{ \p_x}{ \p_a}) + (\frac{ \p_x}{ \p_a})^2 + \ldots] f(x)\]

which…. we can then write as a Taylor series in this \(\p_x/\p_a\) thing, I guess?

\[\frac{1}{1-\frac{\p_x}{\p_a}} f(x) \? f(x+a) = e^{a \p_x} f(x)\]

This operator \((\frac{\p_x}{\p_a})^k\) acts like a projection onto polynomials of degree \(k\), so it’s a lot like a projection onto an orthonormal basis as well. In this case each power of the same operator produces the projection onto the next term. This approach doesn’t seem to clearly do anything about the negative powers that Laurent series are able to handle, but that’s ok. Weird, I know. But I think it’s another interesting way of looking at things.

Also, fun observation: \((1-\p_x/\p_a)^{-1} = e^{a \p_x}\) is basically a valid differential equation.

\[\begin{aligned} (1 - \frac{\p_x}{\p_a}) e^{a \p_x} &= e^{a \p_x} - \p_x [\int_0^a e^{a \p_x} \d a] \\ &\? e^{a \p_x} - \p_x \frac{1}{\p_x}[e^{a \p_x} - 1] \\ &\? e^{a \p_x} - e^{a \p_x} + 1 \\ &= 1 \end{aligned}\]

Maybe this could be thought of as a definition of \(e^{a \p_x}\)? But, uhhhh, don’t look too hard at that step in the middle.

So those are all the ways I know of to invent a Taylor series. I like to see them all laid out side-by-side, since I am looking for, among other things, some elegant logic behind the fractional calculus and fractional combinatorics in general. Basically everything seems connected somehow in a way that I am not aware of math describing, and there are clues everywhere.

I have one other thing I want to talk about which feels closely related to all of the above, which is the mysteriously blurriness of the distinction between integrals and infinite series in general.

2. Connection to Divergent Series

There used to be a section here about divergent series summation techniques, but I decided to remove it to its own article so that I could expand on it.

The short version is that

\[S_0(x) = 1+x+x^2+x^3 = \frac{1}{1-x}\]

should actually be thought of as

\[S_0(x) = 1 + x + x^2 + \ldots + x^n = \frac{1}{1-x} - \frac{x^{N+1}}{1-x}\]

which works always, and most of the divergent series trickery people do consists of finding ways to pretend like that second term doesn’t exist so they can magically compute \(S_0(2) = -1\).

Now, back to the original:

Evidently the way that divergent series techniques work is that we are implicitly performing a projection that removes the \(N\)-dependence, and which is valid only in cases where that removed part goes to zero. Let us write \(\pi_{\perp N}\) for this ‘projection’ operator that deletes the \(N\) dependence from a function, where \(N\) is some ambient variable that is implicitly being used in some of the internal limits. Then \(\pi_{\perp N} [S_0(x)] = \pi_{\perp N}[\frac{1}{1-x}]\) and \(\pi_{\perp N} [S_0(2)] = -1\), yet \(S_0(2) = -1 + 2^{N+1}\) is simply a different value than \(\frac{1}{1-2} = -1\). Probably a good way of thinking about it is to view the algebra of functions on \(x\) as having a basis in terms of their \(N\) dependencies, and then \(\pi_{\perp N}\) is the projector onto the \(N^0\) term. I think this is a good way of thinking about what’s going on with a lot of the standard divergent series examples.

This idea of factoring equations in terms of their ambient variables shows up in other places and is, I think, generally under-appreciated. It’s the idea behind the entropy of a continuum, in which you separate out the part of the entropy which depends on your fine-graining of space from the constant part \(H[p(x)] = - \int p(x) \log p(x) dx + H[\mathcal{U}(0, 1)]\). It’s also implied (I believe) in the concepts of mixed volumes and numerosity and perhaps renormalization, each of which involves making normal quantities depending on hidden “scale” variables, although I haven’t had to study the connection in detail.

The general theme is that when you use series expansions there are limits, and when you take limits there is often an ambient variable involved in how the limit is taken, and even if you suppress it in the notation, it can end up mattering later if two expressions couple the limit together. A standard example would be the sum of two divergent integrals in a Cauchy principle value:

\[\P[\frac{1}{x}] = \P \int_{-1}^1 \frac{dx}{x} = \lim_{a \ra 0} \int_{-1}^a \frac{dx}{x} + \lim_{b \ra 0} \int_{b}^1 \frac{dx}{x}\]

whose value is really a function of \((a,b)\), and in some cases it is possible to get this to come out to a finite value. Only in the case where you assume something about their relationship does it have a definite value. In particular the Cauchy principal value is the case where you assume \(a=b\); since this assumption is hidden you end up with an expression that is implicitly dependent on coordinates (see here for more). The fact that this is not written explicitly always seemed like a hack to me. I don’t like hacks, so I think we need a standard way of dealing with things this, and remembering the dependence on values that are normally hidden feels like a good way of doing it.

Something else to notice here is that, if we allow limits to go off to infinity, then it is the case that

\[(1-x)(\sum_{k = -\infty}^{\infty} x^k) = (1-x) (\ldots + x^{-2} + x^{-1} + 1 + x + x^2 + \ldots) = 0\]

This means that at least in infinite-divergent-series-land, \((1-x)\) is a zero divisor: there exists a nonzero divergent series \(h(x)\) with \((1-x)h(x) = 0\) even if \(x\neq 0\). Therefore an “inverse” of multiplication by \((1-x)\) has a free term: if \(S(x) = \frac{1}{1-x}\) then \(S(x) + \lambda h(x)\) is also a solution.

This is sort of the sense in which \(S_0(x)\) and \(S_{\infty}(x)\) are “both” solutions to \((1-x)^{-1}\), if you ignore their \(N\)-dependence: they differ by a factor of \(h(x)\), which has no \(N^0\) part:

\[\pi_{\perp N} [h(x)] = \pi_{\perp N} [S_0(x) - S_{\infty}(x)] = 0\]

So arithmetic like \((1-x)h(x) = 0\) is sorta illegitimate when you track the limits at infinity, but if you don’t, it’s a good way of interpreting what’s going on when multiple divergent series solve the same equation. They differ by a zero-divisor that you’re canceling out when you multiply through by \((1-x)\).

3. Integration as an Infinite Series

(This section rehashes something I talked about in More on \(\sqrt{\pi}\) because it’s relevant here also. I first learned of this from this pdf by Nicholas Wheeler.)

Earlier I talked about the translation-by-\(dx\) operator \(T^{dx} = 1 + dx \, \p_x\), which has \(T^{dx} f(x) = f(x+dx)\). Flipping this around we can write a “finitesimal” version of the derivative operator

\[\p_x = \frac{T^{dx} - 1}{dx}\]

which in the limit gives the usual derivative

\[\lim_{dx \ra 0} f(x) = \lim_{dx \ra 0} \frac{f(x+dx) - f(x)}{dx} = f'(x)\]

A fun thing to do is to expand this to the \(-1\) power and see if it works like an integral. After all an integral is the more-or-less the inverse of a derivative operator, right?

\[\begin{aligned} \p_x^{-1} &= [\frac{T^{dx} - 1}{dx}]^{-1} \\ &= -dx \frac{1}{1-T^{dx}}\\ &\? -dx (1 + T^{dx} + T^{2 dx} + T^{3 dx} + \ldots) \\ \end{aligned}\]

In fact that looks very much like a Riemann integral. Acting on a function it seems to give

\[\begin{aligned} \p_x^{-1} f(x) &= -dx [f(x) + f(x+dx) + f(x+2 \d x) + \ldots] \\ &\? - \int_{x}^{\infty} f(x') \d x'\\ &\? \int_{\infty}^x f(x') \d x' \end{aligned}\]

It makes sense that the integration bound has to end at \(x\): after all, we would want \(\p_x \p_x^{-1} f(x) = f(x)\) again. But the fact that we didn’t get to choose an integration bound seems like it might be a problem.

We can control the integration bound like this. In the expression \(\p_x^{-1} = -dx (1 + T^{dx} + T^{2 dx} + \ldots)\) there are really two limits being implied at once: the limit as \(dx \ra 0\) and the limit of the number of terms \(N\) as the series is extended to infinity. By writing them out explicitly we can couple how the limits are taken together. In particular if we want to limit the integral to a distance \(a\), then we need for the final term in the sum to correspond to \(T^{(N-1) dx} = T^{a - dx}\). Is this reasonable to do? I dunno: normally one does not get to pick a “stopping” point for their infinite series. But we’ll do it anyway.⁶

\[\begin{aligned} \p_x^{-1} f(x) &= \lim_{dx \ra 0} (- dx )[\sum_{k=0}^{a/dx - 1} T^{k \d x}] f(x) \\ &= - \int_{x}^{x+a} f(x') \d x' \\ &= \int_{x+a}^x f(x') \d x' \end{aligned}\]

We get a different result if we Taylor expand \(\p_x^{-1}\) around \(T^{-dx}\), which is the equivalent of expanding around \(x=\infty\) earlier except now the object is an operator so it’s really not that weird. (Well… okay, it’s still kinda weird, because I’m pretending like I can divide by it like a variable. sorry.)

\[\begin{aligned} (\p_x^{-1}) f(x) &= \frac{dx}{T^{dx} - 1} \\ &= \frac{1}{T^{dx}} \frac{dx}{1 - T^{-dx}} f(x) \\ &\? T^{-dx} [1 + T^{-dx} + T^{-2dx} + \ldots + T^{-(a-1)d x}] f(x) dx \\ &= [T^{-dx} + T^{-2 dx} + T^{-3 dx} + \ldots + T^{-a \, d x}] f(x) dx \\ &= \int_{x-a}^x f(x') \d x' \end{aligned}\]

So it seems like:

different choices of how to take your Taylor-expand the operator and how to take the limits corresponds to different integration bounds
but, the bounds always end at \(x\).
and the choice of whether to extend the integral to \(x+a\) vs \(x-a\) is probably equivalent to the choice of the specific bound \(a\): it corresponds to taking the limit in a positive vs. negative direction, relative to the limit of \(dx\).

So inverting \(\p_x\) using the Taylor series leads to this object which does initially seem to describe \(\p_x^{-1}\). Unfortunately it is not so simple. The resulting object \(\p_x^{-1}\) is not actually the inverse of \(\p_x\), in the sense of \(\p_x \circ \p_x^{-1} = 1\).

\[\begin{aligned} \p_x \circ \p_x^{-1} &= \frac{ T^{dx} - 1 }{dx} [-(dx)(1 + T^{dx} + T^{2 dx} + \ldots + T^{(N-1)dx})] \\ &= (T^{dx}-1)(1 + T^{dx} + T^{2 dx} + T^{3 dx} + \ldots + T^{(N-1)dx}) \\ &= 1 - T^{N \d x} = 1 - T^a\\ [\p_x \circ \p_x^{-1}] f(x) &= [1 - T^{N \d x}]f(x) = f(x) - f(x+a) \end{aligned}\]

This is the correct derivative of the integral \(\int_{x+a}^x f(x') dx'\) because both integration bounds have \(x\) dependences and according to the Leibniz integral rule:

\[\p_x \int_{a(x)}^{b(x)} f(x) \d x = f(b(x)) b'(x) - f(a(x)) a'(x)\]

which in this case is just \(f(x) - f(x+a)\). But it’s not \(f(x)\) again. What went wrong?

Well it is the same thing again. The problem is that \(1+x+x^2+x^3 + \ldots\) is not equal to \(1/(1-x)\) unless \(\| x \| < 1\), or abstractly, if the terms at infinity can be ignored. So our use of the series expansion to turn \(-dx/(1 - T^{dx})\) into an integral is not valid unless \(T^{a} f(x) = 0\). A correct way to write the equivalence for all \(f(x)\) is,

\[\begin{aligned} \p_x^{-1} f(x) &= \p_x^{-1} f(x+a) - \int_x^{x+a} f(x') \d x' \\ &= F(x + a) - \int_x^{x+a} f(x') \d x' \end{aligned}\]

where \(F(x)\) is an (imagined) explicit antiderivative of \(f(x)\), if we have one (or a local approximation \(f(x^*+a) x\) around \(f(x+a)\), where \(x^*\) means the value is treated as a constant). The same fix that made divergent sums correct in the previous section is required to have \(\p_x \circ \p_x^{-1} = 1\).

(I suppose another option is to allow the limits to explicitly depend on \(x\) as well: to pick \(N = N(x, dx) = (a-x)/dx\) such that \(a = x + N \d x\) is no longer a function of \(x\) at all, giving \(\p_x^{-1} f(x) = \int_x^a f(x) \d x\). But then you have to contend with \(\p_x [\p_x^{-1} f(x)]\) taking the derivative of the new definition of \(\p_x^{-1}\) as well… seems like a hack to me.)

To summarize, we can see (Riemann) integration as equivalent in a sense to the Taylor series of \(1/\p_x\), but only if we do not discard the term at infinity, like

\[\begin{aligned} \p_x^{-1} f(x) &= \frac{-dx}{1 - T^{dx}} f(x) \\ &= -dx (1 + T^{dx} + \ldots + T^{N}) f(x) dx + \frac{T^{N+1}}{1 - T^{dx}} f(x) dx \\ &= \int_{x+a}^x f(x) \d x + \underbrace{\p_x^{-1} f(x+a + dx) \d x}_{\text{equivalent to the } \frac{x^{N+1}}{1-x} \text{ from earlier}}\\ \end{aligned}\]

and a “free parameter” shows up in the relative rates of the limits of \(N\) and \(dx\) which corresponds to the choice of integration bound. There also ought to be a term corresponding to the kernel of \(\p_x\), that is, an integration constant.⁷ In particular if we treat \(N\) and \(dx\) as uncoupled, then there are two valid interpretations, which correspond to the Taylor expansions \(S_0(T^{dx})\) and \(S_{\infty}(T^{dx}) = S_0(T^{-dx})\), that is, the limits as \(dx \ra \pm 0\).

\[\p_x^{-1} f(x) \? \begin{aligned} \int_\infty^x f(x) \d x &= F(x) - F(\infty) \\ \int_{-\infty}^x f(x) \d x &= F(x) - F(-\infty) \end{aligned}\]

where \(F(\infty)\) is the imagined value of actually summing that integral forever (even though it may not converge). These are only valid inverses of \(\p_x\) in cases where the integral converges such that the result does not have an \(x\)-dependence from how the \(N \ra \infty\) limit is taken, which is the integral equivalent of cases where the Taylor series converges.

Despite this there is still a way to produce an antiderivative that has \(\p_x \p_x^{-1} f(x) = f(x)\) again. The ‘true’ inverse \(\p_x^{-1}\) comes with an integration constant corresponding to terms which have \(\p_x c = 0\). One such term is \(\p_x^{-1} f(y) = F(y) - F(\infty)\), which (assuming we use the same \(dx\) variable in the definition) corresponds to the same \(F(\infty)\) as the one in \(F(x) - F(\infty)\). So we can freely say that

\[\p_x^{-1} f(x) = [F(x) - F(\infty)] - [F(y) - F(\infty)] = F(x) - F(y) = \int_y^x f(x') dx'\]

recovering what we expect to be the true value of \(\p_x^{-1} f(x)\). This only works because we allow ourselves to cancel out the two \(F(\infty)\) terms since they ought to be exactly the same.

Flipping this around we might imagine interpreting the Taylor series \(1/(1-x) = 1 + x + x^2 + \ldots\) as performing a sort of (finite) integral. Suppose we think of as the monomials \(x^k\) as representing the points of \(\bb{Z}\). Then multiplication by \(x\) looks like translation in the \(k\) coordinate

\[x \times x^k = x^{k+1} = e^{(1) \p_k} x^k = T_k^1 x^k\]

Then the operation \((1-x)\) is a finite difference \(\Delta_k x^k = (1 - T^{1}_k) x^k\) and its inverse is (like) an integral

\[\frac{1}{1-x} x^k = \frac{1}{\Delta_k} x^k \? \begin{cases} \sum_{-\infty}^x x^k \\ \sum_{\infty}^x x^k \end{cases}\]

No idea if it’s useful, but, I often get a vague impression that a lot of the analysis on \(\bb{R}\) which works some of the time but in other cases fails to converge or is ill-defined in other cases would be better-behaved if interpreted as geometric operations on points like this. And regardless it’s always fun to think of one thing as a different kind of thing. So there you go.

Before concluding this section, I was also curious what this looks like if used a different Taylor expansion \(\p_x^{-1}\) other than the ones corresponding to \(S_0\) or \(S_\infty\). As before we have

\[S_k(x) = \frac{1}{1-k} \frac{1}{1-\frac{x-k}{1-k}} = \frac{1}{1-k}[1 +(\frac{x-k}{1-k}) + (\frac{x-k}{1-k})^2 + \ldots]\]

For instance around \(x=2\) it’s \(1/(1-x) = -[1 - (x-2) + (x-2)^2 - (x-2)^3 + \ldots]\).

If we plug \(T^{dx}\) into that we get something that does not seem especially usable as an integral: \(\p_x^{-1} \? -dx [1 - (T^{dx}-2) + (T^{dx}-2)^2 - \ldots]\). Maybe it corresponds to integrating \(f(x)\) in a diferent variable, or something like that? But I doubt it. More likely this manipulation is invalid. After all the \(1\) in the equation is not really a ‘number’ so much as the identity operator \(T^{0 dx} = I\). So maybe no actual ‘numbers’ should anywhere; perhaps we have to consistently treat these as manipulations of operators, not numbers.

In that case we should really be expanding around a different translation operator, like

\[\begin{aligned} \frac{-dx}{1 - T^{dx}} &\? \frac{-dx}{1 - T^{2 d x}}[1 + (\frac{T^{dx} - T^{2 dx}}{1 - T^{2 dx}}) + (\frac{T^{dx} - T^{2 dx}}{1 - T^{2 dx}})^2 + \ldots] \\ \end{aligned}\]

That seems more correct, but I’m not sure how to interpret it. It seems to be an infinite series of nested derivatives and integrals? First, we have

\[\frac{T^{2 dx} - 1}{dx} f(x) = 2 \frac{f(x + 2 dx) - f(x)}{2 dx} = 2 \p_{2 x}\]

which ought to equal \(\p_x\), but maybe it will be relevant that there it will translate twice as far: should it be that \(\p_{2x}^{-1} f(x) = \int_{x+2a}^x f(x) \d x\)? I’m not sure. But anyway, proceeding naively and treating everything like a variable, the terms in the sum are

\[\frac{T^{dx} - T^{2 dx}}{1 - T^{2 dx}} = T^{dx} \frac{(T^{dx} - 1)}{dx} \frac{1}{2} \frac{2 dx}{T^{2 dx} - 1} \? \frac{1}{2} T^{dx} \p_x \p_{2x}^{-1}\]

Making the overall series something like

\[\p_x^{-1} \? \frac{1}{2} \p_{2x}^{-1} [1 + (\frac{1}{2} T^{dx} \p_x \p_{2x}^{-1}) + (\frac{1}{2} T^{dx} \p_x \p_{2x}^{-1})^2 + \ldots]\]

if I did not make any algebra mistakes, and if my calculation is legitimate. I have no idea what to make of that, but I guess it is exactly what it seems like, which is a Taylor expansion of an integral operator.

Presumably there is a way to interpret it under a Fourier transform, whereupon it becomes the literal Taylor series of \(\p_x^{-1} = -dx/(1- T^{dx}) \mapsto -dx/ik\). But Fourier transforms involving \(1/x\) and (equivalently) integration are fraught with principal-value-type dilemmas that seem too messy to make clear sense of, so I’ll skip that. But anyway it (sorta) makes sense that infinite series of regular variables in Fourier space correspond to infinite series of integral and derivative operators in position space and the constructions more-or-less pass through the transform. Everything is the same.

I’ll mention in passing that all of the above also gives one of the versions (among many) of the fractional calculi when you allow yourself to do fractional combinatorics on the \(\p_x^k\) operator. For instance Binomial-expanding/Taylor-expanding \(\p_x^{1/2}\) gives a working version of a half-derivative that has \([(\p_x)^{1/2}]^{\circ 2} = \p_x\) (the Grünwald derivative, specifically). Good luck figuring out what it means, though.

4. Conclusion

What have we learned?

There are several ways to produce the series \(1/(1-x) = 1 + x + x^2 + \ldots\) or \(1/(1-x) = -x^{-1} - x^{-2} - \ldots\).
However, those series are not ‘exact’ except in their radii of convergence. On the other hand, the identity \(1/(1-x) = 1 + x + x^2 + \ldots + x^{N+1}/(1-x)\) is exact, and equivalent to any other exact series. This approach is a good way of ignoring any trickery involving divergent series.
The inverse of a (finitesimal) derivative \(\p_x = (T^{dx}-1)/dx\) can be sort of interpreted using the same series. It gives an integral \(\int_{x+a}^x f(x) \d x\), but the integration bound is an artifact of using the Taylor approximation (and taking limits in a certain way). With the exact series, the term corresponding to \(x^{N+1}/(1-x)\) serves to endure that the integral goes off to infinity (and diverges, outside of certain cases).
However, the difference of two \(\p_x^{-1} f(x) - \p_x^{-1} f(y)\) will cancel out those infinite parts, which is why it’s possible to end up with \(\p_x (\p_x^{-1} f(x)) = f(x)\) again. \(\p_x^{-1} f(x) = F(x) - F(\infty)\), perhaps, but we can freely add a constant, such as \(F(\infty) - F(y)\), to get \(\p_x^{-1} f(x) = F(x) - F(y)\) as well.

I can’t help but think that there is a sort of underlying logic to all of this which connects summations, integrals, derivatives, combinatorics, and everything else into a simple framework with no surprises—and is not the logic that you acquire in, say, complex analysis, which does not really make sense, as far as I know, of all these similarity. The closest thing I’m aware of is in Fourier/Laplace/etc analysis, since the transforms turn derivatives into variables and vice-versa, showing some of the similarities here. But I don’t think they go as far as I want. If what I’m describing does turn out to exist in some obscure parts of higher math, apologies, I was not aware of it, and I would like to know more!

Anyway, towards the goal of figuring out how it all works, I feel like my approach of making everything ‘finitesimal’ and trying to write down exact expressions wherever possible is going to be important. For everything to play well together, we can’t take the usual shortcuts of analysis, because hiding in all those constructions are simplifications and approximations that don’t always hold in other cases. Keeping everything literal and exact for as long as possible avoids this.

In particular a lot of things that are mysterious in calculus (divergences of series for instance) seem more natural when implemented on translation operators or derivative operators, which do not have the property that their norm blows up on higher powers. E.g. \(x^k\) blows up for \(\| x \| \geq 1\), whereas \(T^{k \d x}\) just corresponds to translation in another space entirely. So I keep getting the sense that the most correct way to think about everything is to keep it all in the operator notation for as long as possible. Once you project onto numbers a lot of the structure disappears and is hard to recover. For example, it is really a lot more interesting to me to view an infinite sum as a sort of integral (which is always defined, in a geometric sense) compared to a literal sum of numbers (which may sporadically diverge depending on the situation).

Okay, that’s all I’ve got for now. Hope it was interesting. And please join me in telling people to stop telling other people that \(\sum 2^k = -1\). Thanks.

I like to use this word ‘finitesimal’, as opposed to ‘finite’, to refer to objects which are normally infinitesimal (so going to zero in a limit), but which are instead being treated as literally finite. Like I’ve done here by using \(dx\) as a variable without a limit sign in front of it. I’m not sure if anyone else uses the word this way, or where I got it, but I kinda like it—the suffix ‘-esimal’ refers to the division into parts, e.g. centisaml means (in Latin) division into a hundred parts and infinitesimal means division into infinite parts. So finitesimal means… division into finite parts. Pretty accurate, right? ↩
Also the reason why one should write \(R^{\theta} = (e^{r})^{\theta}\) for rotation operators (with \(r\) as the generator) instead of \(R_{\theta}\). ↩
I learned this sense of generators in quantum mechanics, which refers to them as the “generators of translation” in some coordinate, e.g. the energy operator \(\hat{E} = i \hbar \p_t\) generates time translation. These are almost the same idea except for extraneous factors of \(i\) and \(\hbar\); when plugged into \(e^{-i/\hbar \hat{E} t} = e^{t \p_t}\) they become the same thing. The idea is that a quantum systems’ evolution in time can be implemented by \(e^{-i/\hbar \hat{E} \Delta t} \ket{x, t} = \ket{x, t+ \Delta t}\), and the Schrödinger equation then says that time and space evolution are coupled by \(E^2 = m^2c^4 + p^2 c^2\). In my opinion the \(i\)s and \(\hbar\)s are very distracting and the whole thing would be easier to learn without them. ↩
Miscellaneous breadcrumb: in Donald’s Knuth’s Two Notes on Notation he discusses the fact that the two kinds of Stirling numbers (…‘of the first kind’ and ‘of the second kind’) are really the same kind, if you allow their domains to extend into negatives despite that not making a lot of sense for their combinatoric interpretation. In the same way it is clear that factorials and binomial coefficients also have some validity in the negatives. I expect that the Stirling numbers can also be happily interpolated into fractions or imaginary numbers as well, just like the factorials can. It seems likely to be case that all of these continuations are instances of the same general construction, a single way of thinking about combinatorics on negative and fractional sets that is shows up all over math. ↩
You might view \(1 + x + x^2 + \ldots\) is the “base-\(x\) representation” of \(1/(1-x)\). ↩
The \(a-dx\) is in order to have the last partition be \((a-dx, a)\), to match the partitions in a Riemann integral over the same range. Maybe that’s not that important though. Also, I’m not sure if we should track what happens to the partitions when we negate the orientation of the integral. I’m gonna pretend like it doesn’t matter since it should be immaterial after the limit. ↩
Thought experiment: consider how an integration constant could be equivalent to an infinite series \(c = \ldots + T^{-2 \d x} + T^{-dx} + 1 + T^{dx} + T^{2 \d x} + \ldots = \sum_{\bb{N}} T^{k \, dx}\), since both are (plausibly) in \(\ker \p_x\). ↩

Locating the Lemniscate

2025-11-03T00:00:00+00:00

Still more investigation into why \((-\frac{1}{2})! = \sqrt{\pi}\). Now we are circling in on the real question, which is: what could it possibly mean to take a permutation of a fractional number of elements?

At this point I am not recounting regular mathematics at all but instead just meandering around my own thoughts on the subject. Read on if that’s interesting to you, but be warned, this is more-or-less a stream of consciousness. If I find anything good I’ll summarize it later.

Part four in a series:

1. An Interpolation of Spheres

A fruitful explanation for \(\Gamma(\frac{1}{2}) = (-\frac{1}{2})! = \sqrt{\pi}\) is going to be one that generalizes well. Iif we can come up with a reason why \(\sqrt{\pi}\) is the value, it should also give, at minimum, answers to why the other values of \(\Gamma\) are what they are. For example,

\[\Gamma(\frac{1}{4}) = (-\frac{3}{4})! = \sqrt{2 \varpi \sqrt{2 \pi}}\]

where \(\varpi\) is the lemniscate constant. (The name is because \(2 \varpi\) is the perimeter of a shape called Bernoulli’s lemniscate, defined by the equation \((x^2 + y^2)^2 = x^2 - y^2\), similar to \(2 \pi\) being the perimeter of the circle defined by \(x^2 + y^2 = 1\). We are only interested in the constant, though, not the shape.) Perhaps it could also simplify some of the nightmarish expressions which are known for other values of \(\Gamma\), such as

\[\begin{aligned} \Gamma(\frac{1}{3}) &= 2^{7/9} \frac{\sqrt[3]{\pi K(\frac{\sqrt{3}-1}{2 \sqrt{2}})}}{\sqrt[12]{3}} \end{aligned}\]

where \(K(k) = \int_0^1 (1-t^2)^{-1/2} (1-k^2 t^2)^{-1/2} \d t\) is a certain type of elliptic integral. You can see some others on Wikipedia under Particular values of the Gamma function; notice that closed expressions are only known for a few of them, and most are in terms of these unwieldy integrals (either elliptic integrals or arithmetic-geometric means).

So far the most compelling clue has been that the volume of an \(N\)-ball is given by

\[V_{2k} = \frac{(\pi r^2)^k}{k!}\]

For both integer and half-integer \(k\). Examples:

\[\begin{aligned} V_0(r) &= \frac{1}{0!} &=& \, 1 \\ V_1(r) &= \frac{(\pi r^2)^{1/2}}{(\frac{1}{2})!} &=& \, 2 r \\ V_2(r) &= \frac{\pi r^2}{1!} &=& \, \pi r^2 \\ V_3(r) &= \frac{(\pi r^2)^{3/2}}{(\frac{3}{2})!} &=& \, \frac{4}{3} \pi r^3 \\ V_4(r) &= \frac{(\pi r^2)^2}{2!} &=& \, \frac{1}{2} \pi^2 r^4 \end{aligned}\]

This is notable because it is is pretty much the formula for the volume \(T_k(a)\) of a \(k\)-simplex with side-lengths \(a\). That is, the series of shapes defined by \((k+1)\)-points: a point, line segment, right triangle, right tetrahedron, etc.

\[T_k(a) = \frac{a^k}{k!}\]

But apparently we are also able to evaluate this formula at half-dimensions, for \(a = \pi r^2\):

\[T_k(\pi r^2) = V_{2k}(r)\]

Which seems to hint at what’s going on. Nature’s answer to

“What shape lies halfway between a triangle and a tetrahedron?”

appears to be:

“Regard the triangle as a \(4\)-ball and the tetrahedron as a \(6\)-ball. Then there is a \(5\)-ball halfway between, which can thought of as being a sort of squared \(3/2\)-simplex.”

This is at least a little bit concrete?

Here’s a way of thinking about what’s going on. One possible set of constraints to describe a right tetrahedron with side lengths \(a\) is

\[\begin{aligned} 0 < x, y, z &< a \\ x + y + z &< a \end{aligned}\]

The volume integral over this is

\[T_3(a) = \int_0^a \int_0^{a-z} \int_0^{a-y-z} \d x \d y \d z = \frac{a^3}{3!}\]

The factor \(1/3!\) here indicates that inside the volume \([0,a]^3\) there are six total tetrahedrons, of which we have described one. (The precise number is easier to see if you change coordinates such that the tetrahedron is described by \(0 < x' < y' < z' < a\), via \((x',y',z') = (x, x+y,x+y+z)\). Then there is clearly one tetrahedron for each possible ordering of \((x',y',z')\), hence \(3! = 6\) total inside the volume, and we can write \(\int_{[0,a]^3} dV = 3! \cdot T_3(a)\). This is the same procedure as when you split an integral based on the even/odd-ness of an integrand, such as \(\int_{-a}^a \cos(x) \d x = 2 \int_0^a \cos(x) \d x\), except that the factorization is over a slightly more complicated symmetry.)

Now suppose we are thinking of the same tetrahedron as “actually” being a \(6\)-ball, whatever that means. We regard \(x + y + z < a\) as describing the relationship between the areas of three circles with radii \((r_1, r_2, r_3)\), whose overall area is constrained to be less than \(\pi R^2\)

\[\begin{aligned} 0 \leq \pi r_1^2 + \pi r_2^2 + \pi r_3^2 < \pi R^2 \end{aligned}\]

And we can expand this into six individual coordinates, with each radius being the radius in the plane given by a pair of them. Call these coordinates \((x_1, x_2)\), \((y_1, y_2)\), and \((z_1, z_2)\):

\[\begin{aligned} 0 \leq x_1^2 + x_2^2 + y_1^2 + y_2^2 + z_1^2 + z_2^2 &< R^2 \\ \end{aligned}\]

This of course describes a \(6\)-ball. The volume is

\[V_6(R) = \frac{(\pi R^2)^3}{3!} = V_6(\sqrt{a/\pi}) = \frac{a^3}{3!} = T_3(a)\]

And here are a few ways to write that volume as an integral that show the various ways of thinking about as either having threefold or sixfold symmetry:

\[\begin{aligned} V_6(R) &= \int_0^{\pi R^2} \int_0^{\pi R^2-z} \int_0^{\pi R^2-z-y} \d x \d y \d z \\[1em] &= \int_0^{\pi R^2} \int_0^{\pi R^2 - \pi r_3^2} \int_0^{\pi R^2 - \pi r_3^2 - \pi r_2^2} \d V_2(r_1) \d V_2(r_2) \d V_2(r_3) \\[1em] &= \int_0^R \int_0^{R - r_3} \int_0^{R - r_3 - r_2} (2 \pi dr_1) (2 \pi dr_2) (2 \pi dr_3) \\ &= \int_0^R \int_0^{\sqrt{R^2 - z_2^2}} \int_0^{\sqrt{R^2 - z_2^2 - z_1^2}} \int_0^{\sqrt{R^2 - z_2^2 - z_1^2 - y_2^2}} \int_0^{\sqrt{R^2 - z_2^2 - z_1^2 - y_2^2 - y_1^2}} \ldots \\[1em] &\mskip{7em} \ldots \int_0^{\sqrt{R^2 - z_2^2 - z_1^2 - y_2^2 - y_1^2 - x_2^2}} \d x_1 \d x_2 \d y_1 \d y_2 \d z_1 \d z_2 \end{aligned}\]

I found it somewhat enlightening to sit and contemplate how all of these can mean the same thing despite the fact that we started out describing a tetrahedron.

Anyway, the Gamma function tells us that the interpolation between \(T_2(a)\) and \(T_3(a)\) looks like removing one of the internal coordinates of one of these circles: it’s a figure defined by, say, \(0 \leq x_1^2 + x_2^2 + y_1^2 + y_2^2 + z_1^2 < R^2\), with the \(z_2^2\) term removed.

I guess it might be helpful to picture a tetrahedron with three sides of length \(\pi R^2\)… and then imagine “square-rooting” one of them, getting two and a half sides? Which would sound absurd except that \(\Gamma\) is telling us that it’s not. To draw something like this, one idea I had is to draw a figures with doubled lines to represent when a single direction is referring to multiple “internal” dimensions. Then the relationship we’re positing might be illustrated as

Can’t tell you why it works yet, but nature seems to say it does. This is apparently what is meant by \((5/2)\)-fold symmetry.

2. Fractional Permutations

We can invert the formula for \(V_{2k}\) into something which gives a “definition” of the factorial in terms of the volume:

\[\begin{aligned} k! &= \frac{T_k(a)}{a^k} = \frac{V_{2k}(r)}{(\pi r^2)^k} = \frac{V_{2k}(r)}{V_2(r)^k} \end{aligned}\]

Maybe we simplify now by dropping the radial factors. After all the factorial is only really describing the dimensionality of the space, how many permutations of its axes there are, and the radii only serve to turn this into a volume. With \(T_k = T_k(1)\) and \(V_{2k} = V_{2k}(1)\),

\[k! = T_k = \frac{V_{2k}}{\pi^k} = \frac{V_{2k}}{V_2^k}\]

The first equality is perfectly agreeable: for integer \(k\), the factorial is the number of permutations of \(k\) directions in space, equivalent to the permutations of \(k\) elements. The others are still weird. A \((2k)\) ball also has \(2k\) “axes”, but somehow, by dividing by \(\pi^k\), they only have \(k!\) symmetry, instead of \(2k\)? Or maybe a better way of putting it is: if you want to locate \((\frac{5}{2})!\) between \(2!\) and \(3!\), you can double both dimensions and treat them as symmetries of squares… but you have to rescale \(a \mapsto \pi r^2\) to keep everything having the same volume.

One attempt at a combinatoric interpretation might be: suppose we had a list which appeared to us to contain three elements \((x,y,z)\), but each element was “actually” a label for two elements, so the list is really \((x_1, x_2, y_1, y_2, z_1, z_2)\), with \(6!\) permutations.¹ We can trivially recover the sense of there being \(3!\) permutations by demanding that the elements stay in pairs: \(((x_1, x_2), (y_1, y_2), (z_1, z_2))\), where the permutations keep each pair together. But the above formula suggests that this is not what’s happening: in \(k! = V_{2k}/\pi^k\), all six internal elements are really treated symmetrically in the “permutation”. Permuting pairs produces \((2k)!!\), not \(k!\), and doesn’t have any interpretation on an odd number of elements.

The only way I can imagine making actual sense of a formula like that is if we can think of a geometric space in which it is actually the case. That would be some situation in which \(S_k\), the symmetric group on \(k\) elements, can be identified perhaps with a quotient \(D^{2k}/(D^2)^k\), where \(D^n\) is an \(n\) disk (the interior of an \(n\)-ball), and then our equation is expressing its volume.

Actually, turns out that’s a thing: the symmetric product \(\text{Sym}^n(X)\) of a topological space. I regret that I know almost nothing about topology, but the basic idea is pretty straightforward:

Starting from a space \(X\), one constructs the space \(\text{Sym}^n(X)\) which takes \(n\) points from \(X\) and then forgets their ordering via the quotient \(X^n/S_n\) (with \(S_n\) being the symmetric group on \(n\) elements). For example, \(\text{Sym}^k(I)\) where \(I\) is the unit interval \([0,1]\) consists of sets of \(n\) points from the unit interval without ordering. Then each equivalence class can be identified by sorting the points. This describes the \(k\)-simplexes: it is exactly the construction we saw for the volume of a tetrahedron above.

An LLM informs me that in the case where \(X = D^2\), the \(2\)-disc (that is, the interior of the \(2\)-ball). Then \(D^2\) turns out to have symmetric product

\[D^{2k} = \text{Sym}^k(D^2) = \frac{(D^2)^k}{S_k}\]

That is: the \(2k\)-disc (which has volume \(V_{2k}\)) is exactly what you get if you quotient \(k\) copies of the \(2\)-disc (each with volume \(V_2\)) by the symmetric group on \(k\) elements.

I have no idea how valid to take a “volume” of this, but since \(V^{2k} = V_2^k/k!\) holds, I’m assuming it works in some sense. (Despite this the construction is not as simple as it looks. Apparently the boundary of this turns out to be \(\p \text{Sym}^k(\p D^2) = \text{Sym}^k(S^1) \cong S^1\) rather than \(\p D^{2k} = S^{2k-1}\), so the sense of “equivalence” here is a bit subtle.)

Here then are two natural questions that I’m also not equipped to answer:

Can we make sense of \(D^{2k-1} = (D^2)^{k-1/2} / S_{k-1/2}\) somehow? What would it mean to talk about the product of \((k-1/2)\) copies of \(D^2\) in the symmetric product?
Can we “invert” this somehow to describe the symmetry group as \(S_k = (D^2)^k/D^{2k}\)?, giving something that acts like a permutation group but is defined at half-integers?

For (2), the LLM suggests that you can extract \(S_k\) from the spaces by thinking of it as the deck transformation group (not a great name) of the covering given by \((D^2)^k\) over \(D^{2k}\). I haven’t found any reference which suggests that this respects volumes, but I don’t see why it wouldn’t. But still we would need to make it make sense on half-integers.

We’ll stop this line of investigation there since I don’t know enough about topology to continue. Let me do a few years of grad school and get back to you. Anyway, even if it works, it only gets you values of \(S_{n/2}\), not any other fractions, so we will need to keep looking for a way to produce larger denominators.

3. The \(k=1/4\) Case

Now we look for an equivalent formula to \(V_{2k}(r) = V_2(r)^k/k!\) that involves \((\frac{1}{4})! = \frac{1}{4} \Gamma(\frac{1}{4})\) instead. From the earlier discussion we can say that the reason the volume \(2k\)-balls looks like the volume of \(k\)-simplexes is because their constraint

\[\begin{aligned} x_1^2 + y_1^2 + x_2^2 + y_2^2 + \ldots + x_k^2 + y_k^2 < R^2 \end{aligned}\]

Can be regarded as a linear equation in the volume of each pair of variables

\[\begin{aligned} \vol(x_1^2 + y_1^2)/\pi + \vol(x_2^2 + y_2^2)/\pi + \ldots + \vol(x_k^2 + y_k^2)/\pi &< R^2 \\ \end{aligned}\]

And then these obey permutation symmetry. because they’re a list of identical linear factors that add up to a constant:

\[(\pi r_1^2) + (\pi r_2^2) + \ldots + (\pi r_k^2) < \pi R^2\]

That is,

\[a_1 + a_2 + \ldots + a_k = a\]

(There are presumably many other ways to describe the same property; this is just one that I find easy to think about.)

The fact that this interpolates to half-integers by removing one factor like \(z_1^2\) is still a bit miraculous. But anyway, the critical feature here is that the volume of a pair of terms \(x_1^2 + y_1^2\) is proportional to their sum: if \(x_1^2 + y_1^2 = r_1^2\), then \(\vol(x_1^2 + y_1^2 < r_1^2) = \pi r_1^2\).

It seems clear that to get a \(1/4\) interpolation we want an expression which groups factors four at a time instead of two, such that the volume of four terms is proportional to their sum, and then we can try removing three terms to produce a “\(1/4\)” permutation.

A first guess would be to try using the \(4\)-disc instead of the \(2\)-disc? But this doesn’t work. The volume of

\[w^2 + x^2 + y^2 + z^2 = R^2\]

is \(V_4(R) = \frac{1}{2} \pi^2 R^4\), which is not proportional to the sum of the four values. If it was going to work, we would be able to summarize a list of \(4k\) coordinates into \(k\) copies of \(4\)-discs, e.g. \(\sum_{i=1}^8 x_i^2 = V_4(r_1) + V_4(r_2) \? V_4(R)\). Then the volume would be \(V_8(R) \? V_4(R)^2/2!\) and would correspond to an integral

\[V_8(R) \? \int_0^{V_4(R)} \int_0^{V_4(R) - V_4(r_2)} \d V_4(r_1) \d V_4(r_2)\]

But that’s not the case: we know that \(V_8(R) = (\pi R^2)^4/4! = \frac{1}{3!} V_4(R)^2\), not \(\frac{1}{2} V_4(R)^2\). The problem is that \(V_4(r_1) + V_4(r_2) \neq V_4(R)\): Each term expands like \(V_4(r_1) = \frac{1}{2} \pi^2 r_1^4 = \frac{1}{2} \pi^2 (x_1^2 + x_2^2 + x_3^3 + x_4^2)^2\), and there are a bunch of cross terms so the pair does not add up to \(\frac{1}{2}\pi^2 R^4\).

Put differently: the reason that balls let us interpolate to half-integers is because, in the linear sum \(a_1 + a_2 + \ldots + a_k = a\), we can interpret each \(a_i\) as the volume of a pair of variables \(a_i = \pi r_i^2 = \pi (x_i^2 + y_i^2)\), and the resulting \(\pi x_1^2 + \pi y_1^2 = \pi x_2^2 + \ldots < \pi R^2\) is symmetric in every term, but now can be interpolated to an odd number of terms. Therefore, to interpolate to \(1/4\) terms, we need each \(a_i\) to expand as the volume of four terms, such that the volume of those four terms is proportional to their sum. And since the volume of a four-dimensional figure is going to be proportional to a fourth power, we need consider the figure defined by fourth powers instead:

\[w^4 + x^4 + y^4 + z^4 < S^4\]

These are the “\(n\)-balls in other norms” (ref), in this case the \(L_4\) norm \(\| x \| = \sqrt[4]{\sum x_i^4}\). The volume should now be proportional to \(S^4\), since it defines a four-dimensional figure, although I don’t know the proportionality constant without looking it up. However, we can try to figure it out.

Suppose we define \(W_4(S) = cS^4\) to be the volume of the region satisfying that constraint, where \(c\) is a constant akin to \(\pi\) that we don’t know yet. Then we deduce that a similar formula works for these \(L_4\)-spheres as worked for regular spheres:

\[W_{4k}(S) = \frac{W_4(S)^k}{k!}\]

This is definitely true for integer \(k\), since the constraint is equivalent to

\[c \sum_{i=1}^k x_{i1}^4 + x_{i2}^4 + x_{i3}^4 + x_{i4}^4 = \sum_{i=1}^k cs_k^4 = \sum_{i=1}^k W_4(s_i) = cS^4\]

and by symmetry this is \(1/k!\) of the volume if the individual \(W_4(s_i)\) volumes were unconstrained. For example,

\[x_1^4 + x_2^4 + x_3^4 + x_4^4 + y_1^4 + y_2^4 + y_3^4 + y_4^4 = cW_4(s_1) + cW_4(s_2) < cS^4\]

which is in the same linear form as for the spheres and gives the volume \(W_{8}(S) = W_4(S)^{2}/2!\).

Our postulate, then, is that this also makes sense for fractional values of \(k\).

To find the value of \(c\) we could try integrating over four variables, which is hard, or we can assume that our interpolation formula works and plug in \(k=1/2\). Then the constraint is

\[x^4 + y^4 < S^4\]

Which has volume given by the integral

\[\begin{aligned} \text{vol}(x^4 + y^4 < S^4) &= 4 \int_0^S \sqrt[4]{S^4 - x^4} \d x \\ &= 4 S^2 \int_0^1 \sqrt[4]{1-x^4} \d x \\ \end{aligned}\]

We can find this among the identities for the lemniscate constant (ref), which after all is just a name given to the volume of a particular region, which can be defined by an integral. Just like \(\pi\) could be defined by \(\pi \equiv 4 \int_0^1 \sqrt{1-x^2} dx\), it turns out that \(\varpi = 2 \sqrt{2} \int_0^1 \sqrt[4]{1-x^4} \d x\), so

\[\text{vol}(x^4 + y^4 < S^4) = 4S^2 \frac{\varpi}{2 \sqrt{2}} = \sqrt{2} S^2 \varpi\]

(Note: the actual lemniscate figure is a different shape, defined by \((x^2 + y^2)^2 = x^2 - y^2\), with volume \(2 \varpi\), and so everything involving \(\varpi\) has a weird constant in fron it. But its volume is related to ours. Both geometries involve fourth powers and I suppose the simple proportionality factor implies that they are connected by a simple change-of-variables.)

Assuming our interpolation \(W_{4k}(S) = W_4(S)^k/k!\) is true at multiples of \(1/4\), then

\[W_2(S) = \frac{W_4(S)^{1/2}}{(\frac{1}{2})!} = \frac{\sqrt{c} S^2}{\frac{1}{2} \sqrt{\pi}} = \sqrt{2} S^2 \varpi\]

Which gives the value of \(c\) as

\[c = \frac{1}{2} \pi \varpi^2\]

And the volume of \(W_{4k}(S)\) as

\[W_{4k}(S) = \frac{[\frac{1}{2} \pi \varpi^2 S^4]^{k}}{k!}\]

Which is the \(W\)-version of \(V_{2k}(R) = (\pi R^2)^k / k!\).

Now we check \(W_1 \equiv W_1(1)\), which must equal \(2\) since that is the volume of \(x^4 < 1\), to get the interpolated value of \(\Gamma(\frac{1}{4})!\)

\[\begin{aligned} W_1 &= \frac{W_4^{1/4}}{(\frac{1}{4})!} \\ 2 &= \frac{\sqrt[4]{c}}{\frac{1}{4} \Gamma(\frac{1}{4})} \\ \Gamma(\frac{1}{4}) &= 2 \sqrt[4]{\frac{1}{2} \pi \varpi^2} \\ &= \sqrt[4]{8\pi \varpi^2} \\ &= \sqrt{2 \varpi \sqrt{2 \pi}} \end{aligned}\]

Which is correct. We can also double-check our calculation of \(W_2\):

\[W_2 = \frac{W_4^{1/2}}{(\frac{1}{2})!} = \frac{\sqrt{\frac{1}{2} \pi \varpi^2}}{\frac{1}{2} \sqrt{\pi}} = \sqrt{2} \varpi\]

As expected.

Since \(V_{2k}(R) = T_k(\pi R^2)\), we have in general:

\[\begin{aligned} W_{4k}(S) &= V_{2k}(\frac{1}{\sqrt{2}} \varpi S^2) = T_k(\frac{1}{2} \pi \varpi^2 S^4) \end{aligned}\]

Therefore we can write \(V_{2k}(R) = W_{4k}(\sqrt{\sqrt{2} R/\varpi})\). For instance \(W_2(\sqrt{\sqrt{2} R/\varpi}) = W_2(1) (\sqrt{\sqrt{2} R/\varpi})^2 = 2R = V_1(R)\). So \(W_k(\sqrt{\sqrt{2} R/\varpi})\) is producing the volumes of spheres in a way that interpolates to half-integers, whatever that means.

4. The General Case

That everything worked up there seems to lend credence to this sense of interpolation—we picked \(W_4\) based on the pattern set by \(V_2\) and came up with the right answer with a (relatively) simple argument. Not a proof, but I think it’s pretty convincing. It seems clear that what we did here should work for any \(L_p\) norm: to interpolate to \((1/k)!\), define the figure \(x_1^k + x_2^k + \ldots + x_k^k < R^k\) which has volume \(V^{(k)}_k\), and then extrapolate to non-integer values. Not that there’s any new material here; it’s just a rephrasing of what we already know about \(\Gamma\). But I think this way of proceeding through things sheds some light on what’s going on, and I haven’t seen anything like this really described anywhere else (everyone seems very nervous about trying to literally talk about permutations of fractional dimensions since they seem so nonsensical). Apparently:

To construct half-integer permutations, we interpolate sets of \(k\) elements with balls of \(2k\)-dimensions
To construct quarter-integer permutations, we interpolate those \((2k)\)-balls with \((4k)\)-\(L_4\)-balls (whatever you call those).
Presumably this continues to hold for all the other rationals: \(V^{(ab)}\) balls are used to interpolate \(b\) values between each of the \(V^{(a)}\) balls.
Indeed, the regular permutations \(n!\) at integers may be seen as (proportional to) the volumes of \(L_1\)-balls, which are defined by \(\| x \| = \sum \| x_i \| < R\) and are therefore given by copies of standard simplexes rotated around the origin. (Specifically, \(n!! = 2^n n!\) is \(L_1\)-ball in \(n\) dimensions, since the ball is within \([-1,1]^n\) instead of \([0, 1]^n\).)

Writing \(V_n^{(p)}\) for the volume of a unit \(n\)-ball in the \(p\)-norm (so the spheres \(V_n\) become \(V_n^{(2)}\)), we extrapolate to a formula for all rationals \(m/k\):

\[V_{m}^{(k)} =\frac{(V^{(k)}_k)^{m/k}}{(\dfrac{m}{k})!}\]

Since in all norms the \(1\)-ball has volume \(2\), we can write

\[\begin{aligned} 2 = V^{(k)}_1 &= \frac{\sqrt[k]{V^{(k)}_k}}{(\dfrac{1}{k})!} \\ V^{(k)}_k &= (2 (\dfrac{1}{k})!)^k \\[1.5em] \end{aligned}\]

Giving the general \(V^k_m\) volume as

\[V_m^{(k)} = \frac{ (2 (\dfrac{1}{k})!)^{m/k}}{(\dfrac{m}{k})!}\]

And conversely the factorial as

\[(\frac{m}{k})! = \frac{ (2 (\dfrac{1}{k})!)^{m/k}}{V_m^{(k)}} = \frac{(V^{(k)}_k)^{m/k}}{V_m^{(k)}}\]

These relations are all well-known (see Wikipedia or this SE question); I just like my way of finding them.

For completeness let me mention an entirely different way of reaching this formula, which is to start from \(\Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} \d t\). First substitute \(t=y^{1/x}\), then evaluate at \(1+1/k\) and use the same \(I^2 = \pi\) trick as for solving the Gaussian, but for an \(L_k\) volume instead:

\[\begin{aligned} \Gamma(x) &= \int_0^{\infty} t^{x-1} e^{-t} \d t \\ &= \int_0^{\infty} \cancel{y^{(x-1)/x}} e^{-y^{1/x}} \frac{1}{x} \cancel{y^{(1-x)/x}} \d y \\ \Gamma(1+x) &= \int_0^{\infty} e^{-y^{1/x}} \d y \\ \Gamma(1+\frac{1}{k}) &= \int_0^{\infty} e^{-y^k} \d y \\ \Gamma(1+\frac{1}{k})^k &= \frac{1}{2^k} \int_{\bb{R}^k} e^{-y_1^k - y_2^k - \ldots - y_k^k} \d y\\ &= \frac{1}{2^k} V^{(k)}_k \int_{0}^{\infty} r^{k-1} e^{-r^k} \d r \\ (\frac{1}{k})!^k &= \frac{1}{2^k} V^{(k)}_k \end{aligned}\]

One other interesting form follows from the fact that we know how to write \(V_2^{(k)}\) as an integral:

\[V_2^{(k)} = 4 \int_0^1 \int_0^{\sqrt[k]{1-x^k}} \d y \d x = \int_0^1 \sqrt[k]{1 - x^k} dx\]

Which after plugging into \(V_2^{(k)} = (V_k^{(k)})^{2/k} / (\frac{2}{k})!\) gives

\[\begin{aligned} [\int_0^1 \sqrt[k]{1 - x^k} dx] \times (\frac{2}{k})! &= (\dfrac{1}{k})!^{2} \end{aligned}\]

This shows that the algebraic relationship between \((1/k)!\) and \((2/k)!\) is non-trivial, since it always involves the value of this elliptic integral. Similar arguments would work for \(V_3^{(k)} = 2^3 \int_0^1 \int_0^{\sqrt[k]{1-z^k}} \int_0^{\sqrt[k]{1-z^k-y^k}} \d x \d y \d z\), etc, but I don’t think there’s much point in writing them all out.

Suffice to say, the general relationship between \((\frac{a}{k})!\) and \((\frac{a+1}{k})!\) is nested integration. Maybe that works as a concise explanation of why the values of the factorial function are so complicated. It also gives way of making sense why \(\Gamma\) shows up in the fractional calculus: \(\p_x^{-2} f(x) = \int \int f(x) \d x \d x\) was already a nested integral; turns out when you fractionalize \(\p_x\) you have to split each single integral into more integrals to do it; in fact the factorial is interpolating between integers in the same way.

5. Conclusion

So far I can’t tell if there’s any way to derive any new relationships from this way of looking at things. My guess is no: I think what we’re doing here is using a geometric interpretation in order to manipulate the integral form of \(\Gamma\) indirectly. But I thought it was an interesting way of doing things, which at least I have not come across anywhere else. It really is saying something about what comes “in between” integer permutations, at least in the specific interpolation of \(n!\) which is alluded to by the Gamma functiom.

My curiosity about this was in the first place piqued when I was searching for answers to questions like “is it possible to interpret \(\Gamma\) at non-integers as some kind of fractional permutation?” And almost everything you find in response is some sort of “no-go” reply (e.g. on StackExchange): arguments of the form, “based on what a dimension is, no”. Perhaps with some vague nod to fractals that can exhibit qualities akin to non-integer dimension. I’ve always found those answers to be somewhat close-minded and unimaginative. Not that I have the answer myself, but isn’t it pretty clear there’s something concrete going on here? \(\Gamma\) is not “just” an interpolation; there’s far too much structure for that. There is some valid notion of fractional permutations, fractional sets, fractional spheres, etc in which these concepts are going to make sense.

A good thing to do next would be to look at some examples of how fractional factorials actually arise and see if we can locate the interpolating spherical geometry within them. But I don’t have any good ideas, yet. I’d also like to try to make sense of the interpolations at negative or imaginary numbers, since \(\Gamma\) is well-defined at most of those. I expect that even the negative integers have a concrete meaning waiting to be found.

This is also what happened in the previous article when we compared \(\p^{1/2}_x x^n\) with \(\p_{\sqrt{x}} x^n\): the square-root just treats \(x^n\) as \(y^{2n}\) and produces factors of \((2n)!\) instead, while \(\p^{1/2}_x\) products factors of \(n!\). ↩