Vector spaces are assumed to be finite-dimensional and over \(\bb{R}\). The grade of a multivector \(\alpha\) will be written \(\| \alpha \|\), while its magnitude will be written \(\Vert \alpha \Vert\). Bold letters like \(\b{u}\) will refer to (grade-1) vectors, while Greek letters like \(\alpha\) refer to arbitrary multivectors with grade \(\| \alpha \|\).

You may have noticed that the behavior of the wedge product on pure multivectors is to append them as lists: \(\b{wx} \^ \b{yz} = \b{wxyz}\), where some signs come in if you’d rather have the terms in a different order. This could also be interpreted as taking their union as sets. Either way, \(\^\) seems to act like a union or concatenation operation combined with a bonus antisymmetrization step which has the effect of making duplicate terms like \(\b{x \^ x}\) vanish.

In fact, if we forget about the scalars on some pure multivectors \(\alpha\), \(\beta\) and just consider them as sets of basis vectors, for example \(\b{x \^ y} \ra (\b{x, y})\), then:

\[\alpha \^ \beta= \begin{cases} \pm \alpha \cup \beta & \if \alpha \cap \beta = \emptyset \\ 0 & \text{ otherwise} \end{cases}\]

And the interior product acts like set subtraction:

\[\alpha \cdot \beta= \begin{cases} \pm \beta \setminus \alpha & \if \alpha \subseteq \beta \\ 0 & \text{ otherwise} \end{cases}\]

It turns out that a lot of identities are consistent with this model; for example, \((\alpha \^ \beta) \cdot \gamma = \beta \cdot (\alpha \cdot \gamma)\) expresses the same thing, more or less, as the set algebra identity \(C \setminus (A \cup B) = (C \setminus A) \setminus B\). These seem not to be coincidences: there turns out to be a sense in which exterior algebra is very nearly a “linearized set algebra”. I have seen mentions of this in lots of places. From “On the Exterior Calculus of Invariant Theory” (Barnabei, Brini, Rota 1985):

In the context of Peano spaces the theory of this operator acquires a crystalline simplicity, and the algebraic system obtained by meet, join and star operator is at last seen to be the much-sought-after linear analog of the Boolean algebra of sets with intersection, union and complement, an analogy which was first intuited by Grassmann and later pursued, with partial success, by Clifford, Peano, Burali-Forti and A. N. Whitehead.

A more direct quote of Rota: “Philosophically, the exterior algebra is a “linearization” of the Boolean algebra of all subsets of \(\{1,2,...,n\}\).” (from Combinatorics: The Rota Way). So the analogy is mentioned, but I haven’t come across much more of an explanation. I thought I would investigate.

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2. Set Algebra

A boolean algebra is a set \(A\) equipped with two binary options \(\^\) (“meet”) and \(\vee\) (“join”), a unary option \(\neg\) (“complement”) and two distinguished elements \(0\) (“bottom”) and \(1\) (“top”), with these axioms for all \(a,b,c \in A\):

\[\begin{aligned} a \vee (b \vee c) &= (a \vee b) \vee c & a \^ (b \^ c) &= (a \^ b) \^ c) \\ a \vee b &= b \vee a & a \^ b &= b \^ a \\ a \vee (a \^ b) &= a & a \^ (a \vee b) &= a \\ a \^ 0 &= a & a \vee 1 &= a \\ a \vee (b \^ c) &= (a \vee b) \^ (a \vee c) & a \^ (b \vee c) &= (a \^ b) \vee (a \^ c) \\ a \vee \neg a &= 1 & a \^ \neg a &= 0 \end{aligned}\]

The conventional algebra of \((true, false)\) is an example of this, where \(\^\) is “and”, \(\vee\) is “or”, \(\neg\) is “not”, and \(0 = false\) and \(1 = true\). The terminology of “meet” and “join” comes from lattices, of which this structure is an example.

Set algebra is an example of a boolean algebra on the subsets \(2^U\) of a “universe” set \(U\), with \(\vee\) is \(\cup\) (union), \(\^\) is \(\cap\) (intersection), \(\neg\) is \(a \ra a^c\) (complement with respect to \(U\), ie \(a^c = U \setminus a\)), \(0\) is \(\emptyset\), and \(1\) is \(U\).

Exterior algebra is related to set algebra. If we set \(\emptyset \ra 1\), \(U \ra \omega\), \(\cup \ra \^\), and \(a^c \ra \star a\), then these axioms are clearly similar:

\[\begin{aligned} \alpha \^ (\beta \^ \gamma) &= (\alpha \^ \beta) \^ \gamma \\ \alpha \^ \beta &= - \beta \^ \gamma \\ \alpha \^ 1 &= \alpha \\ \alpha \^ \star \alpha &= \omega \end{aligned}\]

Each of these looks a lot like an axiom up above. Here are a few other observations:

1. The wedge product of two pure multivectors \(\alpha \^ \beta\) gives either the correct union of their basis components except that it’s \(0\) in the case where there are duplicate vectors in each term, such as \((\b{xy}) \cup (\b{yz}) = \b{xyz}\).
2. The interior product of two pure multivectors \(\alpha \cdot \beta\) gives either the correct set subtraction of their basis components \(\beta \setminus \alpha\), except that it gives \(0\) in the case where every term in \(\alpha\) is not present in \(\beta\), such as in \((\b{xy}) \setminus (\b{yz}) = (\b{x})\).
3. The naive way of linearizing the union operation, by just enforcing bilinearity via \(a(\b{xy}) + b(\b{yz})) \cup (\b{x}) = a (\b{xy}) + b (\b{xyz})\), is not well-behaved:
   • It is not a graded algebra (the cardinalities don’t add)
   • As such, it cannot be made basis-invariant, because \((c \b{x}) \cup (c \b{x}) = c^2 \b{x} \neq c \b{x}\).
4. There are a few obvious ways to replace the union with a related operation that might make it into a graded algebra:
   • Concatenate lists, without removing duplicates (the tensor algebra)
     • Problem: it requires picking some ordering on the underlying sets, but it does keep all the information, which could be mapped to a union afterwards. Plus, we can have duplicates in a set, which is weird.
   • Concatenate sets, without removing duplicates (the symmetric algebra)
     • Problem: we can have duplicates in a set. Still weird.
   • Concatenate sets, removing duplicates (the exterior algebra)
5. The wedge product \(\^\) has the wrong symbol, because it corresponds to \(\vee\) (“or”, “join”) in boolean algebra and \(\cup\) (“union”) in the set algebra. Whoops! Fortunately we can probably keep calling it the ‘wedge’ product either way.

Maybe in order to “linearize” the set algebra in a useful way you have to adjust the operations so that they always respect grading on their arguments. The union must always add cardinalities; set subtraction must always subtract cardinalities; intersection must do… something. And any time this rule is violated, the result is \(0\) instead of what the set operation would give. This is certainly not a homomorphism of the algebras formed by \(\cup\) or \(\cap\), and shortly, when we discuss exterior algebra’s version of ‘intersection’, it will turn out that the distributivity axioms of set algebra do not carry over and must be weakened. Nevertheless it all seems a bit helpful? We can kinda think of multivectors as vectors of sets and intuit what identities should hold from that fact, up to a sign or so.

Here’s the glossary:

• Boolean algebra’s Or, And, Not are
• Lattice theory’s Join, Meet, Complement, which are
• Set algebra’s Union, Intersection, and Complement, which are related to
• Exterior Algebra’s Wedge, Antiwedge (?), and Hodge Star

(We should probably just call the Hodge Star the “complement”, anyway. Meanwhile, the exterior algebra version of intersection/meet doesn’t have a good name at all: it feels weird to call it the “meet” without calling wedge products the “join”, but “regressive product” (Grassmann’s name) isn’t great either, and “antiwedge” sucks too. Overall “meet” and “join” seem like the best set of names. Maybe I’ll use those in the future.)

Let’s round out our theory with the exterior-algebra version of “intersection”, which, for lack of a better name, we’ll call the “meet”.

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3. The Meet

The meet, denoted \(\vee\) and pronounced ‘vee’ if you want, is the linearization of set intersection \(\cap\). It’s also occasionally called the “antiwedge” or the “regressive product” (because it reduces grades. these people also call \(\^\) the ‘progressive’ product). It feels slightly strange to call it the ‘meet’ if we don’t also call \(\^\) the ‘join’, but, whatever.

The meet is the dual of the wedge product with respect to the complement \(\star\):

\[\begin{aligned} \alpha \vee \beta &= \star((\star \alpha) \^ (\star \beta)) \\ \alpha \^ \beta &= \star((\star \alpha) \vee (\star \beta)) \tag{1} \end{aligned}\]

(Compare to De Morgan’s laws in set algebra: \((a \^ b)^c = a^c \vee b^c\). We would write these as \(\star^{-1} (\alpha \^ \beta) = \star \alpha \vee \star \beta\).)

This definition is quite useable. It means that any wedge product on vectors has a parallel equation on pseudovectors:

\[\begin{aligned} (\star \b{x}) \vee (\star \b{y}) &= \star(\b{x \^ y}) \\ (\b{x}) \^ (\b{y}) &= \star((\star \b{x}) \vee (\star \b{y})) \end{aligned}\]

In general, this is what the meet does to grades:

\[\un{n-p}{\alpha} \vee \un{n-q}{\beta} = \underbrace{(\alpha \vee \beta)}_{n - p - q}\]

It can be useful to think of these things as being ‘graded modulo \(n\)’: \(\un{-p}{\alpha} \vee \un{-q}{\beta} \equiv \underbrace{(\alpha \vee \beta)}_{-(p+q)}\).

That said, Grassmann lumped \(\^\) and \(\vee\) together into one operation that basically acted on grades modulo \(n\) and in my opinion it did not work very well.

Because \(( \^ V, \vee)\) is isomorphic to the exterior algebra \((\^V, \^)\) under \(\star\), it is itself an exterior algebra, except that the ‘vectors’ are the pseudovectors \(\^^{n-1} V\) and the product is \(\vee\). This algebra, which we call \(\vee V\), has exactly the same relations as \(\^V\), with \(\omega\) as the identity and \(1 = \star \omega\) as the ‘pseudoscalar’. Taken together, the Join and Meet algebras are occasionally called the “Double Algebra” on \(V\), which we might write as \(\^ {\vee} V\).

When \(\| \alpha \| + \| \beta \| = n\), the meet is just a scalar product, and some authors use this to define the inner and interior products:¹

\[\begin{aligned} \un{k}{\alpha} \vee \un{n-k}{\beta} &= \star(\star \alpha \^ (\star \beta)) \\ &= \star \<\star \alpha, \beta\> \omega \\ &= \< \star \alpha, \beta \> \end{aligned}\]

We can also define the interior product and meet in terms of each other:²

\[\begin{aligned} \alpha \cdot \beta &= \star^2 \beta \vee (\star^{-1} \alpha) \\ \alpha \vee \beta &= (\star \beta) \cdot (\star^2 \alpha) \end{aligned}\]

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Unfortunately, meets are confusing. The operation almost acts like intersection, but is very dependent on the grade of the space you’re working in. It’s only nonzero when the grades of its arguments sum to \(\geq n\), which is easily seen from the duality \((1)\):

\[\un{p}{\alpha} \vee \un{q}{\beta} = \begin{cases} \underbrace{(\alpha \vee \beta)}_{p + q - n} & \if p + q \geq n \\ 0 & \if p + q < n \end{cases}\]

As a result, meets are mostly not useful as an intersection operation in practice, and an equation which gives a nonzero result in \(\^ \bb{R}^n\) may give zero in \(\^ \bb{R}^{m > n}\).

Here are some examples of Meets in \(\bb{R}^3\):

\[\begin{aligned} \b{x \vee y} &= \star ((\b{x \^ y}) \^ (\b{z\^x})) = 0 \\ \b{(x \^ y) \vee (y \^ z)} &= \star ( \b{z \^ x} ) = \b{y} \\ \b{x \vee (x \^ y)} &= \star(\b{y \^ z} \^ \b{z}) = 0 \\ \b{x \vee (y \^ z)} &= \star(\b{y \^ z \^ x}) = 1 \end{aligned}\]

In \(\bb{R}^4\) with \(\omega = \b{w \^ x \^ y \^ z}\):

\[\begin{aligned} \b{(x \^ y) \vee (y \^ z)} &= \star(\b{z \^ w \^ w \^ x}) = 0 \\ \b{(w \^ x) \vee (y \^ z)} &= 1 \\ \b{(w \^ x \^ y) \vee (z)} &= 1 \end{aligned}\]

It’s sort of computing intersections, but not very well. \(\b{(x \^ y) \vee (y \^ z)} = \b{y}\) but \(\b{x \vee (x \^ y)} = 0\)? What are we supposed to do with that? The reason is that, for the meet to be nonzero, the result must contain an entire copy of \(\omega\). For instance, \((\alpha) \vee (\star \alpha \^ \beta) \propto \beta\): the meet removes the \(\alpha \^ (\star \alpha) \propto \omega\) part, leaving a ‘remainder’ of \(\pm \beta\). It is way too much to keep track of.

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3. Problems

The “exterior algebra as a linearization of set algebra” analogy is just not very good.

A few reasons: we know that \(\alpha \^ \alpha = 0\) unless \(\alpha \in \^^0 V\), yet \(a \cup a = a\). This means lots of identities don’t work. We might try work around this by saying: well, identities from set algebra hold in exterior algebra so long as both sides are nonzero, which we could write as \(\stackrel{\neq 0}{=}\). But it’s still not true that \(\alpha \^ \alpha \stackrel{\neq 0}{=} \alpha\) in any sense – if it’s a scalar it would be equal to \(\| \alpha \|^2 1\); otherwise it would be equal to \(0\).

I tried for a long time to make it work and didn’t get anywhere. It’s just not that useful to have set operations that kinda work but constantly give \(0\) in sneaky ways.

The analogy to set algebra was supposed to make everything simpler, but instead we got a weird new operation and no simplicity at all. Guess we have to keep looking.

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My other articles about Exterior Algebra:

0. Oriented Areas and the Shoelace Formula
1. Matrices and Determinants
2. The Inner product
3. The Hodge Star
4. The Interior Product
5. EA as Linearized Set Theory?
6. Oriented Projective Geometry
7. Simplex Volumes and Boundaries
8. All the Exterior Algebra Operations