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 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?

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By $P_n(X)$ do you mean the set of $n$-element subsets? In that case your first identity holds only when $X$ is finite, while $\Lambda(M)$ makes sense for $M$ of any dimension and has the decomposition that you give. Interesting question! –  MTS Apr 13 '13 at 17:26
Ah sorry, I should write "finite power set" everywhere, i.e. the set of finite subsets. –  Martin Brandenburg Apr 13 '13 at 18:07
Maybe the power set is more like a Clifford algebra. –  Tom Goodwillie Apr 13 '13 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. –  Theo Johnson-Freyd Apr 13 '13 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. –  Tom Goodwillie Apr 14 '13 at 0:26

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 $\bullet$. –  Martin Brandenburg Apr 13 '13 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. –  Peter Michor Apr 14 '13 at 12:41