MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathcal{C}$ be a cocomplete $R$-linear tensor category. Many notions of commutative algebra can be internalized to $\mathcal{C}$. For example a commutative 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 identifying $x_1 \otimes ... \otimes x_n = x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$, 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 a commutative algebra and in fact $\text{Sym}$ is left adjoint to the forgetful functor $\mathsf{CAlg}(\mathcal{C}) \to \mathcal{C}$.

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 $$... \otimes a \otimes ... \otimes a \otimes ... = 0,$$ for example because we want to have that $\Lambda^p(1^{\oplus n}) \cong 1^{\oplus \binom{n}{p}}$. 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 $M$, we have that $T \otimes_{R[\Sigma_n]} M^{\otimes n} \cong \Lambda^n M := M^{\otimes n}/(... \otimes x ... \otimes x ...)$? Because then we could define $\Lambda^n X := T \otimes_{R[\Sigma_n]} X^{\otimes n}$ for $X \in \mathcal{C}$.

Concerning the "hidden extra structure" in the case of modules: Let the base ring be $\mathbb{Z}$, or more generally a commutative 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 $M$ is an $R$-module, then there is a well-defined(!) homomorphism $M^{\otimes~ n-1} \to \text{ASym}^n(M), x_1 \otimes ... \otimes x_n \mapsto x_1 \wedge x_1 \wedge ... \wedge x_n$, and its cokernel is $\Lambda^n(M)$.

Concerning non-linear tensor categories, I've asked here a similar question.

share|cite|improve this question
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 '11 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 '11 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 '11 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 '11 at 15:43
@Theo: Surely there are some super settings where the "right" notion of $Sym^n$ (for the purpose at hand) involves the square of an odd-degree element being zero. The $\Lambda^n$ that is related to this by your equation is what Martin is calling the exterior power (in which the square of an even-degree element is zero). Likewise there are some settings where what you are calling $Sym^n$ is the "right" thing; and for these the $\Lambda^n$ that is given by your equation is what Martin is calling the skew-symmetric power. – Tom Goodwillie Jul 19 '11 at 17:05

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

share|cite|improve this answer
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 '11 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 '11 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 '11 at 20:39
My mistake. +1. – S. Carnahan Jul 20 '11 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 '11 at 7:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.