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Is there some sense in which the category $sVect$ of super-vector spaces is the "maximal non-trivial extension" of $Vect$ as a symmetric monoidal category?

Is the $\mathbb Z/2$ that shows up in the definition of $sVect$ some kind of homology group of $Vect$?

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I am sure I don't understand the question, but $Vect$ has lots of symmetric monoidal extensions (for example, representations of some affine group scheme), that are not contained in $sVect$. –  Angelo Dec 8 '11 at 17:46
All the symmetric monoidal extensions that you name are split: they admit a symmetric monoidal functor back to $Vect$ (the functor takes a representation to its underlying vector space). In that sense they're not "non-trivial extensions". –  André Henriques Dec 8 '11 at 18:22
$sVect$ is an extension of $Vect$ by what category, exactly? –  Konrad Waldorf Dec 8 '11 at 20:05
Konrad: it's an analogy. A central extension of a group $G$ is a map $\tilde G\to G$. Here, the situation is different: $sVect$ admits an inclusion functor $Vect \to sVect$. So, you see, the arrow goes the other way. I wouldn't know how to answer your question, but I think that it's safe to say that $sVect$ is twice as big as $Vect$. That's why I mentioned $\mathbb Z/2$ in my question. –  André Henriques Dec 8 '11 at 20:28
Maybe "universal central extension" is a bit misleading. It was suggested to me by Alexandru Chirvasitu that $sVect$ might be more like an "algebraic closure" of $Vect$. –  André Henriques Dec 9 '11 at 17:51

2 Answers 2

up vote 15 down vote accepted

This will be a little imprecise, but hopefully if you need a precise result like this you can fill in the details.

First of all Vect has not only the symmetric monoidal structure but also the direct sum. If you don't look for extensions which are linear for the direct sum, then I think you can form crazy extensions which behave differently depending on the dimension of the vector space.

But let's assume that your extension is linear and that everything in the extension E is a direct sum of elements in the picard category pic(E) inside your symmetric monoidal category.

Under these assumptions the extension of symmetric monoidal categories

$$ Vect \to E $$

is completely determined by an extension of symmetric monoidal 2-groups:

$$ pic(Vect) \to pic(E)$$

Now a symmetric monoidal 2-group is classified by two abelian groups: the objects $\pi_0$, the automorphisms of the identity $\pi_1$, and a (stable) k-invariant: $$k : \mathbb{Z}/2 \otimes \pi_0 \to \pi_1 $$

This later is the same as a map from $\pi_0$ into the 2-torsion of $\pi_1$.

For $pic(vect) = lines$ we have $\pi_0 = 0$, and $\pi_1 = K^{\times}$, the group uf units of your ground field. The universal extension then has the same $\pi_1$, but has $\pi_0$ equal to the (multiplicative) 2-torsion of $\pi_1$ with the canonical $k$-invariant.

If you are not in characteristic 2, this means $\pi_0 \; pic(E)= \mathbb{Z}/2$, and E is super vector spaces.

If you are in characteristic 2, then sVect is the same as Z/2-graded vector spaces and is no longer universal (the extension splits in the sense you described in your comment).

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The assumption in the third paragraph that all objects are sums of invertible objects seems to be very strong. Is this a notion of universal central extension that doesn't exist for, e.g., representation categories of nonabelian groups? –  S. Carnahan Dec 9 '11 at 9:04
It is not at all clear to me that such universal central "extensions" exist. Even in the classical setting of finite groups, a universal central extension only exists for perfect groups, not in general. Now there is homotopy theory governing this more general extension theory, but it is very complicated and similar to the homotopy theory of topological commutative monoids. The homotopy theory of topological commutative monoids itself is extremely difficult, more difficult then ordinary unstable homotopy theory of spaces. (... cont...) –  Chris Schommer-Pries Dec 9 '11 at 14:10
(... cont .. ) On the other hand, under the strong assumption of the third paragraph, the above extension problem essentially reduces to a problem in stable homotopy theory, and not a very hard problem at that. –  Chris Schommer-Pries Dec 9 '11 at 14:12

I am not sure if that's what you want, but let me try though. When I hear of extensions, one of the pictures I have in my mind is something like crossed products, and I want to push that analogy. There is a natural action of $Vect$ on itself given by the monoidal structure (moreover, there is a left and a right action, but since the category is symmetric, natural isomorphisms allow us to identify those). Let me, to emphasize where I am going with that, denote the category that is acting by $Vect_0$, and the category that is being acted on by $Vect_1$. When trying to use our action of $Vect_0$ on $Vect_1$ to form an extension, we need to define the monoidal structure $Ob(Vect_1)\otimes Ob(Vect_1)\to Ob(Vect_0)$, and the symmetry isomorphisms. I think the compatibility with what we already have will almost probably force the monoidal structure to be the tensor product, and then there is a question of what sort of freedom you have to define the symmetry isomorphisms. Altogether this should naturally lead to super vector space. Unfortunately, you do a very particular case of an extension, something like a crossed product with the regular module.

Maybe it's too silly though, in which case you have my sincere apologies.

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