## Exterior powers in tensor categories

Let $\mathcal{C}$ be a cocomplete $R$-linear tensor category. Many notions of commutative algebra may be internalized to $\mathcal{C}$. For example an algebra is an object $A$ in $\mathcal{C}$ together with morphisms $e : 1 \to A$ (unit) and $m: A \otimes A \to A$ (multiplication) satisfying the usual laws. The $n$-th symmetric power $\text{Sym}^n(X)$ of an object $X$ is the quotient of $X^{\otimes n}$ by $x_1 \otimes ... \otimes x_n = x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$, so formally it is defined as a coequalizer of the $n!$ symmetries $X^{\otimes n} \to X^{\otimes n}$. Then $\text{Sym}(X) = \bigoplus_{n\geq 0} \text{Sym}^n(X)$ is an algebra object and $\text{Sym}$ is in fact left adjoint to the forgetful functor $\mathsf{CAlg}(\mathcal{C}) \to \mathsf{CAlg}(\mathcal{C})$.

But now what about the exterior power $\Lambda^n(X)$? It is clear how to define $X^{\otimes n}$ modulo $x_1 \otimes ... \otimes x_n = \text{sgn}(\sigma) \cdot x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$ in this context, which one might call the anti-symmetric power $\mathrm{ASym}^n(X)$. The correct definition of the exterior power also has to mod out $... \otimes a \otimes ... \otimes a \otimes ... = 0$. But I have no idea how to internalize this to $\mathcal{C}$, even for $n=2$. The reason is that there is no morphism $X \to X \otimes X$ which acts like $a \mapsto a \otimes a$. Another idea would be to define $\Lambda(X)$ as a graded-commutative algebra object with the usual universal property, classifying morphisms $f$ on $X$ which satisfy something like $f(x)^2=0$, but again it is unclear how to formulate this in $\mathcal{C}$.

If this is not possible at all, which additional structure on $\mathcal{C}$ do we need in order to define exterior powers within them? Is this some categorified $\lambda$-ring structure? This structure should be there in the case of usual module categories (over rings or even ringed spaces). Of course there is no problem when $2 \in R^*$, because then the exterior power equals the anti-symmetric power. The question was also discussed in a blog post.

Here is a more specific (and a bit stronger) formulation: Is there some $R[\Sigma_n]$-module $T$, such that for every $R$-module $X$, we have that $T \otimes_{R[\Sigma_n]} X^{\otimes n} \cong \Lambda^n X := X^{\otimes n}/(... \otimes x ... \otimes x ...)$?

Concerning the "hidden extra structure" in the case of modules: Let the base ring be $\mathbb{Z}$, or more generally a ring $R$ in which $r^2 - r \in 2R$ for all $r \in R$; this includes boolean rings such as $\mathbb{F}_2$ and also $\mathbb{Z}/n$. If $X$ is an $R$-module, then there is a well-defined(!) homomorphism $X^{\otimes~ n-1} \to \text{ASym}^n(X), x_1 \otimes ... \otimes x_n \mapsto x_1 \wedge x_1 \wedge ... \wedge x_n$, and its cokernel is $\Lambda^n(X)$.

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Surely you meant to include the condition that your tensor category is symmetric? – Bruce Westbury Jul 18 2011 at 11:01
Why do you believe "the correct one has to mod out be $\dots \otimes a\otimes \dots \otimes a \otimes \dots = 0$"? As you say, such a map does not exist in the categorical context, and it's not at all clear to me that it's the natural condition. For example, in the super/graded world, it can be very useful to play off the fact that $\wedge^n(X) = \pi^{-n}\operatorname{Sym}^n(\pi X)$, where $\pi$ is the "shift" functor. This fails when $X$ has odd part if you try to use a "mod out by squares" definition. – Theo Johnson-Freyd Jul 18 2011 at 11:46
Agree. I mean with the wrong definition we have $\wedge^n A = A/2A$ for $n \geq 2$, but it should be $0$. – Martin Brandenburg Jul 18 2011 at 13:14
There is no "right" and "wrong" definition. There simply exist two different mathematical objects, that happen to agree when 2 is invertible. One of them happens to produce 2-torsion in some cases. – André Henriques Jul 18 2011 at 15:34
@Martin: What is the simplest example of a category in which you'd like to interpret ᐽ(X), and where you don't know how to do it? – André Henriques Jul 18 2011 at 15:43

Deligne (Categories Tannakiennes, 1990, p165) defines it to be the image of the antisymmetrisation $a=\sum(-1)^{\epsilon(\sigma)}\sigma\colon X^{\otimes n}\rightarrow X^{\otimes n}$.

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This is what Martin calls the skew-symmetric power in the question. It appears that he is looking for a different functor (but often given the same name), for which high-order exterior powers of the unit object vanish even when 2 is not invertible. – S. Carnahan Jul 19 2011 at 16:22
No. This map from $X^{\otimes n}$ to itself kills tensors of the form $\dots \otimes a\otimes\dots\otimes a\otimes\dots$, and at least for free modules its image coincides with $\Lambda^nX$. The expression "antisymmetrization map", like many other terms in this area, can be misleading, because one could also use it for the map from $X^{\otimes n}$ to the largest quotient on which the action (with signs) of the symmetric becomes trivial. – Tom Goodwillie Jul 19 2011 at 16:51
a) Deligne talks about abelian tensor categories. In my case, images don't exist in general. b) Does this image has a universal property at all (which resembles the usual one)? – Martin Brandenburg Jul 19 2011 at 20:39
My mistake. +1. – S. Carnahan Jul 20 2011 at 8:10
@André: The image is usually defined to be the kernel of the cokernel; whereas the cokernel of the kernel is called the coimage. There is always a morphism from the coimage to the image, and abelian categories are those additive categories where this morphism is always an isomorphism. Anyway, in may case kernels don't have to exist. – Martin Brandenburg Jul 21 2011 at 7:44
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