2 corrected an error

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?

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# 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.

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?