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$\DeclareMathOperator\sVect{sVect}\DeclareMathOperator\Vect{Vect}$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. 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?, and the "algebraic closure" idea even appears there in a comment 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.

LSpice
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Tim Campion
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  • I think I'm confused by your definition-- is this the standard category of super vector spaces? If so, then it is $\mathbb Z/2$ graded, and there are many more even objects than you claim. – Phil Tosteson Apr 22 '20 at 15:57
  • @PhilTosteson Sorry -- by "even object $V$", I mean that the braiding $V \otimes V \to V \otimes V$ is the identity, in analogy to "odd object $W$" meaning that the braiding $W \otimes W \to W \otimes W$ is the scalar $-1$. I now see that this clashes horribly with using "even" / "odd" to simply mean that the grading is purely even or odd. I'm not sure where I picked up this terminology -- I think I'll try to think of something better to replace it with. – Tim Campion Apr 22 '20 at 16:07
  • So are you intentionally excluding all objects whose underlying vector space is not one dimensional? I.e. under this terminology there is a unique even and a unique odd object? – Phil Tosteson Apr 22 '20 at 16:18
  • @PhilTosteson Exactly. It's definitely suboptimal terminology. – Tim Campion Apr 22 '20 at 16:24
  • @PhilTosteson Okay, I've replaced "even" with "even line" and "odd" with "odd line". Still not great, but hopefully less misleading. – Tim Campion Apr 22 '20 at 16:29
  • @TimCampion The braiding $V \otimes V \to V\otimes V$ is never the identity except when $V \cong \mathbb C^{1|0}$. Rather, you should look at the canonical automorphism of the identity functor called "$(-1)^f$", which is related (via Crane-Yetter TQFT) to Dehn twists. It can be defined in any symmetric monoidal category with enough dualizable objects as the partial trace of the braiding $V \otimes V \to V \otimes V$. An "even object" is one which on which this natural automorphism is the identity. – Theo Johnson-Freyd Apr 22 '20 at 17:20
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    One thing that has always confused me is the following: adding this odd vector space works well when k = R (the reals) because R/(R)^2 = Z/2, where R* = units in R. (In other words, there are two possible choices for what an odd automorphism can square to, and you pick the nontrivial one.) So maybe the analogue of super vector spaces over other base fields is to add in "odd" automorphisms parametrized by elements of k/(k)^2? – skd Apr 22 '20 at 17:25
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    When you look at the characteristic-$p$ case, is there any reason to expect issues with separability (whatever that might mean) outside of the $p = 2$ case? (EDIT: Yes, in the sense made precise by @TheoJohnson-Freyd.) – LSpice Apr 22 '20 at 17:31
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    @LSpice Actually, I left something out of my answer. Ostrik's paper shows that sVec is always "separably closed" for characteristic $\neq 2$, but not "algebraically closed" for characteristic $\geq 5$. He conjectures an "algebraic closure", which is sVec in the $p=3$ case. – Theo Johnson-Freyd Apr 23 '20 at 21:49
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    I am a bit confused about the $p=2$ case. The conjecture from Ostrik's paper is that Vec is already "algebraically closed" when $p=2$, and he proves that it is "separably closed". I was told that Etingof, based on a hint from Deligne, found a counterexample to Ostrik's conjecture when $p=2$. – Theo Johnson-Freyd Apr 23 '20 at 21:51
  • My memory of the counterexample was the following. Let $\mathbb{k}$ be an algebraically closed field of characteristic $p=2$. Consider the category of $\mathbb Z/2\mathbb Z$-modules over $\mathbb{k}$. Note that, in characteristic $2$, the group algebra of $\mathbb Z/2\mathbb Z$ can be written as $H = \mathbb{k}[x] / (x^2)$. Instead of the standard comultiplication, give it the comultiplication determined by the Hopf structure $\Delta(x) = x \otimes 1 + 1 \otimes x$. This is cocommutative, but instead I'll twist the braiding by writing down a nontrivial R-matrix $R = x \otimes x$. – Theo Johnson-Freyd Apr 23 '20 at 21:54
  • This is the symmetric monoidal category that is supposed not to map to $\mathrm{Vec}$. What confuses me is the following. If I am doing it right, I think that this category does admit a symmetric monoidal functor to $\mathrm{Vec}$. The issue is that that functor is neither left- nor right-exact. – Theo Johnson-Freyd Apr 23 '20 at 21:59
  • So you see that questions about whether something counts as "algebraically closed" do depend a lot on your "categorified linear algebra": how exact do you require your functors? If you only care about separable closures, then you should only test against very finite extensions, and semisimplicity is a finiteness condition, so you don't see this issue. But for algebraic rather than separable closures, you should test against a larger class of categories, and then it does matter. – Theo Johnson-Freyd Apr 23 '20 at 22:00
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    @skd The fact that Galois cohomology $\mathrm{H}^1(\mathbb{R}; \mu_2) = \mathbb{R}^\times/(\mathbb{R}^\times)^2 \cong \mathbb{Z}/2\mathbb{Z}$ matters, but not in the way you say. Indeed, the extension from Vec to sVec is highly nontrivial also over $\mathbb{C}$. – Theo Johnson-Freyd Apr 23 '20 at 22:04
  • @TheoJohnson-Freyd, what does "very finite" mean? – LSpice Apr 23 '20 at 23:21
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    @LSpice Let's fix an ambient world of categorified linear algebra, and write $\mathcal{K} = Vec$ for the "ground field". An "extension" of $\mathcal{K}$ is then a commutative $\mathcal{K}$-algebra $\mathcal{L}$, i.e. a symmetric monoidal linear category. A good definition of "very finite" is that underlying monoidal category of $\mathcal{L}$ should be fully dualizable in the 3-category of (noncommutative) associative $\mathcal{K}$-algebras (and their bimodules). Equivalently, the underlying category of $\mathcal{L}$ should be 2-dualizable, and $\mathcal{L}$ should be separable. – Theo Johnson-Freyd Apr 24 '20 at 14:58
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    @LSpice The expectation is that the collection of very finite extensions is independent of your choice of world of "categorified linear algebra". This expectation is a theorem for some cases, and open in others. Here's a hint for why this is a good expectation. For many, but not all, versions of categorified linear algebra, it is known (c.f. "bestiary" linked in a comment to my answer below) that a linear category of 2-dualizable if and only if it is the category of modules of a finite-dimensional separable associative algebra. – Theo Johnson-Freyd Apr 24 '20 at 15:00
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    "separability" then forces $\mathcal{L}$ to be fusion and of nonzero global dimension. – Theo Johnson-Freyd Apr 24 '20 at 15:01
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    Ostrik was interested in a more general class of extensions, where the underlying category of $\mathcal{L}$ is fusion but not separable, or even when the underlying category of $\mathcal{L}$ is not 2-dualizable. But once you start dropping finiteness conditions, your choice of categorified linear algebra really matters (e.g. my comment earlier where it matters whether you demand that functors be exact). – Theo Johnson-Freyd Apr 24 '20 at 15:03

1 Answers1

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$\newcommand\sVec{\mathrm{sVec}}\newcommand\Vec{\mathrm{Vec}}$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. (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. 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.

Theo Johnson-Freyd
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  • Of course, you should add appropriate words to my answer like linear, cocontinuous, or what have you, to specify your ambient "categorified linear algebra". The details of how you add those words don't matter too much, because you expect (co)completion operations that move you between worlds. When defining "separable extension", you should demand very strong finiteness (i.e. dualizability) conditions. With those conditions in place, the different worlds for categorified linear algebra all match; see the appendix A bestiary of 2-vector spaces. – Theo Johnson-Freyd Apr 22 '20 at 17:15
  • Do you mean 'admit' or 'emit' in "which does not emit a symmetric monoidal functor"? Both makes sense to me, but I'd expect the former. – LSpice Apr 22 '20 at 17:34
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    Side comment: even in ordinary algebra, I think a lot of people would consider "the" algebraic closure to be something of an abuse of language, because there is no preferred way of comparing two such. Too many automorphisms. – Todd Trimble Apr 22 '20 at 19:38
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    @LSpice I meant “emit”: I want to quantify over all (nonzero) symmetric monoidal functors with specified domain. – Theo Johnson-Freyd Apr 22 '20 at 20:11
  • @Todd I can’t argue with that – Theo Johnson-Freyd Apr 22 '20 at 20:12
  • @TheoJohnson-Freyd This is great, thanks! I don't suppose you know anything about the relative version of being weakly terminal? That is, the property of every map $A \to sVect$ extending along various $A \to B$'s to yield $B \to sVect$'s? – Tim Campion Apr 23 '20 at 02:50
  • @TimCampion I wrote my answer without really reading your question. I apologize deeply for that. I'll add some comments in the question above. – Theo Johnson-Freyd Apr 23 '20 at 21:36