Still more investigation into why \((-\frac{1}{2})! = \sqrt{\pi}\). 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 existing mathematics at all and instead just meandering around my own thoughts. 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.

Previously:¹

• Investigations on n-Spheres
• Factorials as Multiplicative Integrals
• More on √π

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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: if we can come up with a reason why \(\sqrt{\pi}\) is the value, it should also give 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 (we are not going to actually talk about the lemniscate itself; just this constant that comes from its area. But “locating the lemniscate constant” was a lamer title). Perhaps it would also help just some of the other nightmarish expressions that 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 of the other daunting expressions for values of of \(\Gamma(x)\) on Wikipedia under Particular values of the Gamma function; notice that closed expressions are only known for a few of them. And really most of those are cheating a bit, because if you’re counting an elliptic integral (equivalently, an arithmetic-geometric mean) as a closed expression, you may as well count \(\Gamma(z) = \int_0^{\infty} z^{t-1} e^{-t} \d t\), which is silly. So we’re not crossing our fingers for finding any closed expressions here: probably \(\Gamma(x)\) can be regarded as the definition of many constants; who’s to say we shouldn’t be defining \(\pi\) as \(\Gamma(1/2)^2\) anyway? But still, we would like some simpler expressions, if possible.

So far the most compelling clue has been that the volume of an \(N\)-ball is given by

\[V_{2k} = \frac{(\pi r^2)^k}{k!}\]

Some 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 \(k\)-dimensional triangle (i.e. a \(k\)-simplex) with side-lengths \(a\): point, line segment, right triangle, right tetrahedron, etc.

\[T_k(a) = \frac{a^k}{k!}\]

Except, when evaluated at \(a = \pi r^2\), it also gives the correct volume of the \(k\)-ball even at half-dimensions:

\[T_k(\pi r^2) = V_{2k}(r)\]

This seems to strongly 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 \(3/2\)-ball”

That is at least a bit concrete.

Here’s a way of thinking about why it works. 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 the coordinates such that our 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 \([0,a]^3\). This the same procedure you’re doing when you split an integral based on the even/odd-ness of an integrand (e.g. \(\int_{-a}^a \cos(x) \d x = 2 \int_0^a \cos(x) \d x\)), but the factorization is over a (slightly) more complicated symmetry group.

Now suppose we are thinking of the same tetrahedron as “actually” being a \(6\)-ball, whatever that means. We regard \((x,y,z)\) as being the areas of three circles with radii \((r_x, r_y, r_z)\), whose overall radius is constrained to be less than \(R = \sqrt{a/\pi}\)

\[\begin{aligned} 0 \leq \pi r_1^2 + \pi r_2^2 + \pi r_3^2 \leq \pi R^2 \end{aligned}\]

And we can expand this into the six underlying coordinates of the \((x_1, x_2)\), \((y_1, y_2)\), and \((z_1, z_2)\) planes.

\[\begin{aligned} 0 \leq x_1^2 + x_2^2 + y_1^2 + y_2^2 + z_1^2 + z_2^2 &\leq R^2 \\ \end{aligned}\]

Which is the equation for a \(6\)-ball with volume

\[V_6(R) = \frac{(\pi R^2)^3}{3!} = V_6(\sqrt{a/\pi}) = \frac{a^3}{3!} = T_3(a)\]

Here are a few ways to expand it as an integral:

\[\begin{aligned} V_6(R) &= \int_0^{\pi R^2} \int_0^{\pi R^2-x} \int_0^{\pi R^2-y-x} \d x \d y \d z \\[1em] &= \int_0^{\pi R^2} \int_0^{\pi R^2 - \pi r_3^2} \int_0^{\pi R^2 - \pi r_3^2 - \pi r_2^2} \d V_2(r_1) \d V_2(r_2) \d V_2(r_3) \\[1em] &= \int_0^R \int_0^{R - r_3} \int_0^{R - r_3 - r_2} (2 \pi dr_1) (2 \pi dr_2) (2 \pi dr_3) \\ &= \int_0^R \int_0^{\sqrt{R^2 - x_1^2}} \int_0^{\sqrt{R^2 - x_1^2 - x_2^2}} \int_0^{\sqrt{R^2 - x_1^2 - x_2^2 - y_1^2}} \int_0^{\sqrt{R^2 - x_1^2 - x_2^2 - y_1^2 - y_2^2}} \\[1em] &\mskip{7em}\int_0^{\sqrt{R^2 - x_1^2 - x_2^2 - y_1^2 - y_2^2 - z_1^2}} \d x_1 \d x_2 \d y_1 \d y_2 \d z_1 \d z_2 \end{aligned}\]

It is somewhat enlightening to sit and contemplate how all of these can mean the same thing.

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 \(0 \leq x_1^2 + x_2^2 + y_1^2 + y_2^2 + z_1^2 < R^2\), with no \(z_2^2\) term. Once everything is written as spheres, I suppose it’s not that mysterious that \(V_5\) comes between \(V_4\) and \(V_6\). The weird part is the doubling of the dimension to get \(2k\)-ball in the first place.

We can invert the formula for \(V_{2k}\) into something which gives a “definition” of the factorial in terms of it:

\[\begin{aligned} k! &= \frac{T_k(a)}{a^k} = \frac{V_{2k}(r)}{(\pi r^2)^k} = \frac{V_{2k}(r)}{V_2(r)^k} \end{aligned}\]

Maybe we simplify 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 \(V_{2k} = V_{2k(r)}\),

\[k! = T_k = \frac{V_{2k}}{\pi^k} = \frac{V_{2k}}{V_2^k}\]

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, by dividing by \(\pi^k\), they only have \(k!\) permutations instead of \(2k\)? It is odd to think about.

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.² 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”.

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: some sense in which \(S_k\), the symmetric group on \(k\) elements, can be identified perhaps with a quotient \(D^{2k}/(D^1)^k\), where \(D^n\) is an \(n\) disk (the interior of an \(n\)-ball), and then our equation is expressing its volume.

Well, 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 \([0,1]\) takes \(n\) points from the unit interval without ordering, that is, each equivalence class can be identified by sorting the points–which is exactly the \(k\)-simplexes we were discussing 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 of the \(2\)-disc (with volume \(V_2\)) by the symmetric group on \(k\) elements.

I have no idea in what sense it is valid to take a “volume” of this, but since \(V^{2k} = V_2^k/k!\) also holds, I’m assuming it works in some sense. (Also, 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:

1. 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 \(D^2\)?
2. Can we “invert” this somehow to describe the “symmetric group on \(k\) elements” via \(S_k = (D^2)^k/D^{2k}\)?. (\(D^n\) is of course not a label, not an exponent, so this is non-trivial.)

For (2), the LLM suggests that this is true if you regard \(S_k\) 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. Let me go do a few years of graduate school and I’ll get back to you.

We’ll stop this line of investigation there since I don’t know enough about topology to continue. Anyway, even if it works, it only gets you values of \(S_{n/2}\), not any other fractions. We will need to keep searching for produce larger denominators.

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2. 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 \(2k\)-balls look like \(k\)-simplexes is because their constraint

\[\begin{aligned} x_1^2 + x_2^2 + y_1^2 + y_2^2 + \ldots = R^2 \end{aligned}\]

Can be regarded as a linear equation in pairs of variables

\[(\pi r_1^2) + (\pi r_2^2) + \ldots + (\pi r_k^2) = \pi R^2\]

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 \(x_1^1\) is still a bit miraculous. Regardless, though, it seems clear that to get a \(1/4\) interpolation we want an expression which groups factors four at a time instead of two.

We could try using the \(4\)-disc instead of the \(2\)-disc? Starting with

\[w^2 + x^2 + y^2 + z^2 = R^2\]

The volume is \(V_4(R) = \frac{1}{2} \pi^2 R^4\). But this doesn’t work. 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 \(\frac{1}{2}\pi^2 R^4\).

What we need to do instead is consider a figure defined by fourth powers:

\[w^4 + x^4 + y^4 + z^4 < S^4\]

These are the “\(n\)-balls in other norms” (ref). In this case the \(L_4\) norm \(\| x \| = \sqrt[4]{\sum x_i^4}\).

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 postulate that a similar formula will hold as for spheres:

\[W_{4k}(S) \? \frac{W_4(S)^k}{k!}\]

And this is trivially the case for integer \(k\), since the constraint is not equivalent to

\[c \sum_{i=1}^k x_{i1}^4 + x_{i2}^4 + x_{i3}^4 + x_{i4}^4 = \sum_{i=1}^k cs_k^4 = \sum_{i=1}^k W_4(s_i) = cS^4\]

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 puts it in the same linear form as for the spheres and gives the volume \(W_{4k}(S) = W_4(S)^{k}/k!\).

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.Apparently \(\varpi = 2 \sqrt{2} \int_0^1 \sqrt[4]{1-x^4} \d x\), 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 defined by \((x^2 + y^2)^2 = x^2 - y^2\) with volume \(2 \varpi\). The volume constant is related to ours because both geometries involve fourth powers and are apparently connected by a 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.

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This seems to lend some 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. There’s no new algebraic material here, 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):

1. To construct half-integer permutations, we interpolate sets of \(k\) elements with balls of \(2k\)-dimensions
2. To construct quarter-integer permutations, we interpolate those \((2k)\)-balls with \((4k)\)-\(L_4\)-balls (whatever you call those).
3. 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.
4. Indeed, the regular permutations \(n!\) at integers may be seen as (proportional to) the volumes of \(V^{1}\) balls, which are defined by \(\| x \| = \sum \| x_i \| < R\) and are therefore given by copies of stamdard simplexes rotated around the origin.

Writing \(V_n^p\) for the volume of a unit \(n\)-ball in the \(p\)-norm, 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

\[V_m^k = \frac{ (2 (\dfrac{1}{k})!)^{m/k}}{(\dfrac{m}{k})!}\]

(Again, none of this is novel; this formula is on Wikipedia, and the same formulas can be reached by manipulating \(\Gamma(x)\)’s integral form \(\int_0^{\infty} t^{x-1} e^{-t} \d t\) directly; here.)

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

\[\begin{aligned} [\int_0^1 \sqrt[k]{1 - x^k} dx] \times (\frac{2}{k})! &= (\dfrac{1}{k})!^{2} \end{aligned}\]

Showing that the algebraic relationship between \((1/k)!\) and \((2/k)!\) is non-trivial, involving the value of this 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 want to try to write 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.

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. But I thought it was an interesting way of reaching the existing ones, 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 interpolation alluded to by the Gamma function, and even though a concrete interpretation of it is hard. My curiosity about this was in the first place piqued by looking for answers to these sorts of questions and finding lots of “no-go” replies on e.g. StackExchange: arguments of the form, “based on what a dimension is, this won’t work”. I still don’t have an answer, but I’m pretty sure there’s more going on here: there is some valid notion of fractional permutations, fractional sets, fractional spheres, etc in which these concepts are going to make some kind of 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…