# Analogy between the exterior power and the power set

The symmetric algebra of an object exists in every cocomplete $$\otimes$$-category. For the category of sets $$\mathrm{Sym}(X)$$ is the set of multi-subsets of $$X$$.

The usual definition of the exterior power works in every cocomplete linear $$\otimes$$-category in which $$2$$ is invertible. But what about the non-linear case? Are there also "exterior powers" in $$\otimes$$-categories which are not linear? Of course the usual definition using alternating maps does not work. But isn't it striking that for the cartesian category of sets there is a quite natural candidate, namely the power set? Here are some analogies (here $$P(X)$$ denotes the power set of $$X$$ if $$X$$ is finite; in general it is the set of all finite subsets of $$X$$; $$P_n(X)$$ is the set of all subsets of $$X$$ with $$n$$ elements):

• $$P(X) = \coprod_n P_n(X)$$ and $$\Lambda(M) = \oplus_n \Lambda^n(M)$$

• $$P(X \sqcup Y) = P(X) \times P(Y)$$ and $$\Lambda(M \oplus N) = \Lambda(M) \otimes \Lambda(N)$$

It follows the "categorified Vandermonde identity":

• $$P_n(X \sqcup Y) = \coprod_{p+q=n} P_q(X) \times P_q(Y)$$ and $$\Lambda^n(M \oplus N) = \oplus_{p+q=n} \Lambda^p(M) \otimes \Lambda^q(N)$$

• $$(P(X),\cup)$$ is a commutative monoid and $$(\Lambda(M),\wedge)$$ is a graded-commutative algebra, i.e. commutative monoid object in the tensor category of graded modules equipped with twisted symmetry.

• If $$M$$ is free with (ordered) basis $$X$$, then $$\Lambda(M)$$ is free with basis $$P(X)$$, and $$\Lambda^n(M)$$ is free with basis $$P_n(X)$$. In particular, $$\dim \Lambda^n(M)=\dim P_n(X)$$.

• If $$T$$ is a commutative monoid, then homomorphisms $$P(X) \to T$$ correspond to maps $$f : X \to T$$ with $$f(x)^2=f(x)$$, and if $$A$$ is a graded-commutative algebra, then homomorphisms $$\Lambda(M) \to A$$ correspond to homomorphisms of modules $$f : M \to A_1$$ with $$f(x)^2=0$$ or rather $$f(x)f(y)+f(y)f(x)=0$$ in the context of $$\otimes$$-categories (so these conditions are not the same, but both use $$f(x)^2$$).

Therefore I would like to ask: Is there a notion of exterior algebra for certain cocomplete $$\otimes$$-categories, including categories of modules and the category of sets? In the latter case, do we get the power set?

• Maybe the power set is more like a Clifford algebra. Commented Apr 13, 2013 at 19:27
• To make the algebraic structure on $P(X)$ closer to that on $\Lambda(X)$, you can use the disjoint union rather than the union. Then the last bullet looks a little better. But note that the $\Lambda$ side of the last bullet doesn't make sense in arbitrary categories --- rather, it has something special to do with usual modules over a ring in which $2$ is invertible. Commented Apr 13, 2013 at 22:24
• I wasn't really thinking of a quadratic form. I was just thinking that, like a Clifford algebra, the power set has a filtration such that the associated graded object is (like) an exterior algebra. Commented Apr 14, 2013 at 0:26
• Meanwhile I've been able to formalize the analogy, using "$\pm$-enriched" categories. Commented Mar 31, 2014 at 10:45
• Sketch: Consider $C_2\mathsf{-Set}$-enriched categories, i.e. each hom-set has a specified involution which we denote by $-$. If $R$ is a ring, $\mathsf{Mod}_R$ is an example. The category of sets is not an example but $C_2\mathsf{-Set}$ is an example, and we may use the free functor $\mathsf{Set}\to C_2\mathsf{-Set}$ etc. to reduce to the $C_2\mathsf{-Set}$-enriched case. Now, given a cocomplete tensor $C_2\mathsf{-Set}$-category, we call a map $\omega : X^{\otimes n} \to Y$ skew-symmetric if $\omega \circ \tau = - \omega$ for any swap $\tau$. There is a universal example $\Lambda^n(X)$ etc. Commented Dec 9, 2020 at 22:06

To a set $$X$$ associate the free vector space $$M(X)$$ over $$X$$; conversely, for a vector space $$M$$ let $$X$$ be the index set of a basis of $$M$$. Then the analogy is just how one does exterior algebra in terms of a basis.
This fits into the way how representation theory for $$GL(n)$$ and for the symmetric group $$S(n)$$ are related to each other, both using Young projectors in interated tensor products of $$\mathbb C^n$$. This becomes more striking even if we take the direct limit for $$n\to \infty$$. See books and papers by Yuri Neretin (in arXiv).
Edit: For modules $$M$$ over an algebra $$A$$, one could consider the corresponding algebra of dual numbers $$A\circledS M$$ (i.e., $$A\oplus M$$ with multiplication $$(a,m).(a',m') = (a.a', a.m' + m.a')$$ and the Kaehler differentials over this algebra. See 2.3 of here.
• I agree with Qiaochu. See also the fourth $\bullet$. Commented Apr 13, 2013 at 19:18
• The algebra of Kaehler differentials of $A\circledS M$ generalizes the exterior algebra from vector spaces to modules over a commutative algebra, or even to bimodules over a non-commutative algebra. Commented Apr 14, 2013 at 12:41