# What does the Tannakian formalism reconstruct when fed the category of chain complexes?

I've recently realized that there is a gap in my understanding of the Tannakian formalism for reconstructing an (algebraic) group from its category of (finite-dimensional) representations. To warm up, here is a version I do understand. Let $(\mathcal C,\otimes,\mathrm{flip})$ be a symmetric monoidal category $\mathbb C$-linear category satisfying some technical conditions (the details of which depend on exactly what you're trying to do, but e.g. "abelian and every object is dualizable" should suffice), and suppose that $F: \mathcal C \to \mathrm{Vect}_{\mathbb C}$ is a faithful symmetric monoidal $\mathbb C$-linear (exact, ...) functor. Then $G = \mathrm{End}_\otimes(F)$, the monoid of monoidal natural transformations of $F$, is a (affine algebraic over $\mathbb C$) group, and $\mathcal C$ is equivalent to the category of $G$-representations.

Versions of the above statement are due to Deligne, and some are older and some are newer. A very important theorem of Deligne's is that (depending on the precise set-up of the problem) the fiber functor $F$ isn't needed. Indeed, a result that I continue to find amazing is that there exists a unique-up-to-(necessarily nonunique!)-isomorphism fiber functor $\mathcal C \to \mathrm{Vect}_{\mathbb C}$ iff every object in $\mathcal C$ has nonnegative-integer dimension. (In case it's not clear, everywhere I write "$\mathrm{Vect}$" and so on, I mean to imply the category of finite dimensional vector spaces.)

I'm not so worried about the above version of Tannakian theory (although I'm sure I missed some details), as about the "super" version. Let me start with the generalization of the second paragraph. Recall that the category $\mathrm{SVect}_{\mathbb C}$ of super vector spaces is the free (technical conditions) monoidal category generated by $\mathrm{Vect}_{\mathbb C}$ and an object $X$ that $\otimes$-squares to the unit object, with the symmetric structure uniquely determined by the request that $\mathrm{flip} : X \otimes X \to X\otimes X$ is minus the identity. (As a monoidal category, but not as a symmetric monoidal category, $\mathrm{SVect}_{\mathbb C}$ is equivalent to the category of representations of $\mathbb Z/2$.) The generalization of the second paragraph, also due to Deligne, is that a rigid symmetric monoidal (technical conditions) category has a fiber functor to $\mathrm{SVect}_{\mathbb C}$ iff every object has (possibly negative) integer dimension is annihilated by some Schur functor.

So let's take some category $\mathcal C$, like $\mathcal C = \mathrm{SVect}_{\mathbb C}$ itself. My problem is to understand the generalization of the first statement. See, the identity functor $\mathrm{SVect}_{\mathbb C} \to \mathrm{SVect}_{\mathbb C}$ has a nontrivial symmetric monoidal automorphism, which acts by $-1$ on the "fermionic" generating object $X$. So the naive theorem fails: it is not true that $\mathrm{SVect}_{\mathbb C}$ is equivalent to the category of $\mathrm{End}_\otimes(\mathrm{id})$-representations in $\mathrm{SVect}_{\mathbb C}$.

One potential fix is to strengthen the condition that $\mathcal C$ be $\mathbb C$-linear to the condition that it be tensored over $\mathrm{Vect}_{\mathbb C}$ or over $\mathrm{SVect}_{\mathbb C}$ (depending on where the fiber functor lands). Then the identity functor $\mathrm{id} : \mathrm{SVect}_{\mathbb C} \to \mathrm{SVect}_{\mathbb C}$ has only the identity automorphism when thought of as a functor of $\mathrm{SVect}_{\mathbb C}$-tensored categories.

But my actual interest is in the category $\mathrm{DGVect}$ of chain complexes of (non-super) vector spaces. This category is not tensored over $\mathrm{SVect}$, but it does have a fiber functor to it. When you calculate the endomorphism supergroup group of this fiber functor, you get the group $G = \mathbb G_m \ltimes \mathbb G_a^{0|1}$, which is the group of affine transformations of the odd line. But the category of $G$-representations in $\mathrm{SVect}$ is the category of chain complexes of supervector spaces, not the category of chain complexes of regular vector spaces. Not that this latter thing is a bad category — it's just that I'd like to describe the other one too. And on the other hand, $G$ isn't defined over $\mathrm{Vect}$, so I can't say that $\mathrm{DGVect}$ is "the category of $G$-modules in $\mathrm{Vect}$".

So my question is: how does Tannakianism treat the category of chain complexes? More generally, how does it treat a category with a fiber functor to $\mathrm{SVect}$? I guess I could have also asked an even easier example (although it's not my main motivation): how can I describe $\mathrm{Vect}$ as a Tannakian category over $\mathrm{SVect}$ (with the unique inclusion)? In all cases, I have a fiber functor $F : \mathcal C \to \mathrm{SVect}$ and I can construct the supergroup $G = \mathrm{End}_\otimes(F)$, but to get $\mathcal C$ back I seem to need some way to mandate how $G$ interacts with the canonical $\mathbb Z/2$ that acts on $\mathrm{SVect}$ — how, and what is this generalization of supergroup?

• To me it seems that the $G$-representations in $\mathrm{SVect}$ consist of a graded supervector space together with a differential of superdegree $1$ and internal degree $1$. Such a graded vector spaces can be thought of as a graded vector space where the even vectors of degree $n$ correspond to new degree $2n$ and the odd vectors of degree correspond to new degree $2n+1$. In the new degree the differential is of degree $1$. This looks like the category of complexes (with the Koszul rule built into the monoidal structure as it should). – Torsten Ekedahl Nov 13 '11 at 9:37
• @Torsten: One invariant of a (braided) monoidal category is its (abelian) group of invertible objects-up-to-isomorphism. For the usual category of chain complexes of vector spaces, this group is $\mathbb Z$. For the category of $G = \mathbb G_m\ltimes \mathbb G_a^{0|1}$-representations in SVect, this group is $\mathbb Z \times \mathbb Z/2$. In particular, the trivial $G$-rep on $\mathbb C^{0|1}$ is its own $\otimes$-inverse, and DGVect does not contain such an object. – Theo Johnson-Freyd Nov 13 '11 at 17:41
• @Theo: You are absolutely right, I am glad for the qualifications I put in... I leave my comment in case someone else makes the same mistake or more precisely "enom till straff, androm till varnagel" ("punishment for one, warning to others" a phrase used in old Swedish laws). – Torsten Ekedahl Nov 14 '11 at 8:43
• @Theo: You say: "The generalization of the second paragraph, also due to Deligne, is that a rigid symmetric monoidal (technical conditions) category has a fiber functor to SVectC iff every object has (possibly negative) integer dimension." I don't think it is true. The Deligne category of representations of the symmetric group with -1 element is a counterexample (see Deligne's paper on representations of $S_t$ when t is not a positive integer). – Pavel Etingof Mar 29 '12 at 19:50
• Basically the problem is that if you have objects of dimension 0, then the length of $X^{\otimes n}$ could grow superexponentially even though its dimension grows only exponentially. On the other hand, you can often apply Deligne using only dimensional consideration. For example, if you have any dimension function which is bounded away from zero (it doesn't need to be categorical dimension) then length grows only exponentially and you can apply Deligne's theorem. This applies to symmetric fusion categories using FP dimension. – Noah Snyder Apr 3 '12 at 5:32

Deligne does not do what you seem to want, which is give a theory internal to super vector spaces. To do so is probably an open question. The obvious version of Hopf algebras in SVect does not work: the category of comodules admits an action by the lines in SVect (commuting with the fiber functor), and thus cannot recover categories like ordinary vector spaces.

Deligne's theory is external, mixing Vect and SVect. A supergroup is an algebraic group internal to the algebraic geometry of super vector spaces (that is, a cocommutative Hopf algebra in that symmetric monoidal category, subject to some conditions). An "external group" is a supergroup with a distinguished element (of $G(k)$) of order at most 2. We do not look at all representations of the group, but only ones on which the distinguished element acts as parity (and we only allow distinguished elements such that the adjoint representation is allowed). Deligne's theorem is that every nice category is this subcategory of representations of an external group. We recover the group as automorphisms of the fiber functor to SVect. As you note, the automorphisms of the identity functor of SVect is $Z/2$; that is its external group, with the nontrivial element distinguished. Vect is representations of the trivial group. These are the initial and final external groups, and they are not isomorphic, so external groups are not group objects in some category. In general, internal representations of a supergroup are representations of an external group which is the semidirect product of the two element group acting on the original supergroup. Representations of ordinary algebraic groups reuse the group as the external group, with the trivial element distinguished.

A fairly easy example is that graded vector spaces with the signed symmetry are representations the external group $G_m$, with $-1$ distinguished. Your calculations of the automorphisms of the fiber functor on chain complexes and of the full category of representations of $G=G_m \ltimes G^{0|1}_a$ each imply that the relevant subcategory of its representations are chain complexes of ordinary vector spaces.

This is all based on just the first few pages of Deligne's paper. I don't think he has a name for what I call "external groups"; the Mueger appendix you cite calls them supergroups, but I think that's a bad choice.

Somehow I've stumbled on this very old post. There is a very general form of this question: How much can we relax the definition of Tannakian and still have a meaningful theory?–––And how would it work?'

I'm currently writing a paper on this and I'll give an idea of the answer. We can relax quite a bit. We can even drop additivity. The catch is, we must simultaneously let our theory of schemes relax: There is a notion of Tannakian category $T$ relative' to a cosmos $\mathcal{C}$. The algebraic geometry that is then exploited is that of "schemes relative to the cosmos $\mathcal{C}$," as introduced in the very nice paper "Au dessous de Spec $\mathbb{Z}$," by Toën and Vaquié. I'll point out, that the definition is a bit wordy, but the salient point is that $T$ is enriched over $\mathcal{C}$.

The analogous theorem is then that $T$ is the category of compact representations for a `cosmic' algebraic group $G$. Something nontrivial to notice here is that this realizes the objects of $T$ as coming from module categories for commutative algebra objects in $\mathcal{C}$.

One also extends the theory in a very straightforward way to involving gerbes, etc., just as Deligne did in the Feitschrift chapter.

Coming back to your specific question concerning chain complexes, if you work this out, then in this picture, you will output an algebraic group built from a scheme theory constructed from simplicial rings (if you were working with positive complexes).

• Could you explain what is the structure involved in the case of complexes? OK it is something like an algebraic group in the sense of cosimplicial schemes or whatever, but what group? At least "ideologically", could one identify it with something like $\Bbb G_m$, or, say, say what functor from simplicial rings to groups does it represent? – მამუკა ჯიბლაძე Jan 27 '17 at 5:41
• Hi. Yes! The theory of cosimplicial schemes certainly arises as Sch($\mathcal{C}$) for the obvious cosmos. I think you are asking how to think of the group valued functors that the theory will spit out? I'll speak generally: one could talk about fpqc group valued sheaves on the one hand, but on the other hand, also saying that an affine algebraic group rel $\mathcal{C}$ is given by a Hopf algebra object in $\mathcal{C}$ is a completely correct description (of affine algebraic groups). There isn't much more happening besides replacing the word algebra many places with 'algebra object,' etc. – Artur Jackson Jan 27 '17 at 16:07
• So for chain complexes you get certain simplicial Hopf algebra? – მამუკა ჯიბლაძე Jan 27 '17 at 19:17