$\DeclareMathOperator\sVect{sVect}\DeclareMathOperator\Vect{Vect}$The category $\sVect\_k$ of (let's say finite-dimensional) super vector spaces can be obtained from the category $\Vect\_k$ of (finite-dimensional) vector spaces by formally adjoining an "odd line square root" $\Pi k$ to the unit object $k \in \Vect\_k$ -- see Prop 2.6 in [Rezk - The congruence criterion for power operations in Morava E-theory](https://arxiv.org/abs/0902.2499). Here "square root" means that $\Pi k \otimes \Pi k \cong k$, and "odd line" means that the braiding $\Pi k \otimes \Pi k \to \Pi k \otimes \Pi k$ is given by the scalar $(-1)$.


It's not hard to see that $\sVect\_k$ has *odd line* square roots for all *even line* objects (where "even line" means that the braiding is the identity)—the only even line object being $k$ itself again. So $\sVect\_k$ can be characterized as the closure of $\Vect\_k$ under the operation of adding odd line square roots for even line objects. This is analogous to $\mathbb C$ being the closure of $\mathbb R$ under the operation of adding square roots for all elements.


But in the case of $\mathbb C$ and $\mathbb R$, much more can be said—$\mathbb C$ is in fact *algebraically closed*, i.e., closed under the operation of adding roots for all polynomials. Can something analogous be said for the case of $\sVect\_k$?


**Question 1:** Is there a reasonable sense in which the symmetric monoidal $k$-linear category $\sVect\_k$ is "algebraically closed"?


I'm primarily interested in the case $k = \mathbb C$.


Here is an attempt to make the question more precise. One way of saying that $\mathbb C$ is algebraically closed is that for every injective map of finitely-generated commutative $\mathbb R$-algebras $A \to B$ and every map $A \to \mathbb C$, there is an extension $B \to \mathbb C$. This motivates the following somewhat more precise question:


**Question 2:** Is there a reasonably large class of symmetric monoidal $k$-linear functors $A \to B$ between $k$-linear symmetric monoidal categories with the property that any symmetric monoidal $k$-linear functor $A \to \sVect\_k$ extends to $B \to \sVect\_k$?


Finally, here's a guess at a class of maps $A \to B$ which might possibly do the trick:


**Question 3:** In particular, let $A \to B$ be a conservative strong symmetric monoidal $k$-linear functor where $A$, $B$ are symmetric monoidal $k$-linear categories with duals for all objects. Then does any strong symmetric monoidal $k$-linear functor $A \to \sVect\_k$ admit a lift $B \to \sVect\_k$?


This question bears some similarities to [Is super-vector spaces a "universal central extension" of vector spaces?](https://mathoverflow.net/questions/82985/is-super-vector-spaces-a-universal-central-extension-of-vector-spaces), and the "algebraic closure" idea even appears there in a [comment](https://mathoverflow.net/questions/82985/is-super-vector-spaces-a-universal-central-extension-of-vector-spaces#comment213477_82985) of André Henriques, attributed to Alexandru Chirvasitu.


**Remark:** It might be better to assume that the $k$-linear categories under consideration are also abelian (with bicocontinuous $\otimes$) and that the functors under consideration are exact. Or perhaps some other variation of this flavor.


**Edit:** I'm mostly interested in characteristic zero, but my intuition is that in characteristic $p$, it would be reasonable to replace "algebraically closed" above with "separably closed", though I don't really know what that would mean in this categorified context.



Is the category $\operatorname{sVect}$ an "algebraic closure" of $\operatorname{Vect}$?
$\newcommand\sVec{\mathrm{sVec}}\newcommand\Vec{\mathrm{Vec}}$Yes. Over an algebraically closed field of characteristic $0$, $\sVec$ is the algebraic closure of $\Vec$. By "algebraic closure" of $K$ I mean a weakly-terminal object of the category of not-too-large non-zero commutative $K$-algebras. (An object is *weakly terminal* if it receives maps from all other objects, and *terminal* if that map is unique.) With this definition, the statement "$\sVec$ is the algebraic closure of $\Vec$" is a summary of Deligne's theorem on the existence of super fibre functors. This interpretation of Deligne's theorem is due to my paper [Spin, statistics, orientations, unitarity](https://arxiv.org/abs/1507.06297). (I had the opportunity to ask Deligne last fall if he had been aware of this interpretation of his theorem. He said no, he had been focused on the question "what distinguishes categories of representations of groups?", but that he liked my interpretation.)


Actually, I'm not sure that the weak terminality condition that I use deserves the name "algebraic closure". The issue is that $\sVec$ is *not* weakly terminal among finitely generated symmetric monoidal categories: you need to include some growth conditions on powers of a generating object. In my paper, I only look at "finite dimensional" extensions of $\Vec$, which is good enough for the usual theory of algebraic closures of fields, but doesn't use the full strength of Deligne's theorem.


In positive characteristic $p\geq 5$, $\sVec$ is not weakly terminal among finite-dimensional extensions of $\Vec$, as observed by Ostrik in [On symmetric fusion categories in positive characteristic](https://arxiv.org/abs/1503.01492). But Ostrik does show that $\sVec$ is weakly terminal among *separable* extensions of $\Vec$, and so is the "separable closure" but not the "algebraic closure". So the category of vector spaces over an algebraically closed field of positive characteristic is not "perfect".


In unpublished work joint with Mike Hopkins, I have also established the 2-categorical version of the statement. Namely, the symmetric monoidal 2-category "$2{\sVec}$" of supercategories and superfunctors is the "separable closure" of the 2-category "$2{\Vec}$" of (linear) categories and functors. The 3-categorical version of the statement is false: we know a separable symmetric monoidal 3-category which does not emit a symmetric monoidal functor to the 3-category of super-2-categories.


Actually, there is one important piece of the story that I haven't worked out. In my paper cited above, I gave a quick-and-dirty definition to the word "field": I said a symmetric monoidal category is a "field" if all the symmetric monoidal functors that it emits are faithful and essentially injective. Under this definition, $\Vec$ and $\sVec$ are fields, so I felt it was good enough. But if you are not working over an algebraically closed base, then $2{\Vec}$ is not a field for this definition, which I do not like. I am still in the process of working out a good higher-categorical version of the word "field".


In the meantime, I would say that yes, $\sVec$ is "algebraically closed", but I would not say that it is "the" algebraic closure, since without a definition of "field", the weak-terminality definition does not characterise a unique object.




---


Added in response to comments:


Deligne proves the following stronger result than mere existence. Suppose that $C$ is a reasonable (i.e. linear over your algebraically-closed characteristic-zero ground field, some size constraints, rigid, etc.) symmetric monoidal category category. Then the category of all symmetric monoidal functors $C \to \sVec$ is a groupoid (this requires that $C$ is rigid), and $\pi\_0$ of this groupoid is $\operatorname{Spec}(\operatorname{End}(1\_{C}))$, where $1\_{C}$ is the unit object in $C$. I will write $\operatorname{Spec}(C)$ for the whole groupoid. (A better name would be $\operatorname{Spec}(C)(\sVec)$.)


In particular, if $A \to B$ is a functor of reasonably small symmetric monoidal categories, then you get a map $\operatorname{Spec}(B) \to \operatorname{Spec}(A)$ of groupoids. The question Tim asks above is whether a point in $\operatorname{Spec}(A)$ can be lifted against this map to $\operatorname{Spec}(B)$. This is a question that can be asked just in terms of $\pi\_0$ of these groupoid.


Said another way, a functor $F : A \to \sVec$ extends a long a functor $A \to B$ if and only if the induced map $F(1) : \operatorname{End}(1\_A) \to \mathbb{C}$ extends along $\operatorname{End}(1\_A) \to \operatorname{End}(1\_B)$. The answer is "not always": the point $F(1) \in \operatorname{Spec}(\operatorname{End}(1\_A))$ might not be in the image of $\operatorname{Spec}(\operatorname{End}(1\_B))$. But this is the only obstruction.


