Recall that the category $\mathrm{SuperVect}$, as a category, consists of pairs of vector spaces, thought of as formal direct sums $V \oplus W\,\Pi$, where $\Pi$ is the "odd line". (Called "$\Pi$" because tensoring with it is "parity reversal".) Indeed, there is a good notion of "direct sum of categories" in which case, as categories, we have $$ \mathrm{SuperVect} = \mathrm{Vect} \oplus \mathrm{Vect}\, \Pi$$ As a monoidal category, we declare that the one-dimensional vector space $\mathbf 1 \in \mathrm{Vect}$ is the monoidal unit and that $\Pi \otimes \Pi = \mathbf 1$, and give it the trivial associator (so that as a monoidal category $\mathrm{SuperVect}$ is the category of sheaves of vector spaces on $\mathbf Z/2$ with the convolution product; when $2$ is invertible, which I assume it is, this is also the category of representations of $\mathbf Z/2$).

The interesting part of $\mathrm{SuperVect}$ is its braiding/symmetry. A braiding $\sigma$ on this monoidal category is uniquely determined by its value on $$ \mathbf 1 = \Pi \otimes \Pi \overset \sigma \longrightarrow \Pi \otimes \Pi = \mathbf 1 $$ which is just a number $\sigma_{\Pi,\Pi}$. The axioms of symmetric monoidal category force $\sigma_{\Pi,\Pi}$ to square to $1$, but do not force it to be $1$ itself. The symmetric monoidal category $\mathrm{SuperVect}$ is determined by declaring that $\sigma_{\Pi,\Pi}$ is the other number that squares to $1$, namely $-1$.

At least over $\mathbb C$, $\mathrm{SuperVect}$ has the following important property, due to Deligne. Not every symmetric monoidal category is tannakian over $\mathrm{Vect}$ --- in particular, there does not exist a symmetric monoidal functor $\mathrm{SuperVect} \to \mathrm{Vect}$ --- but every symmetric monoidal category (satisfying some technical bounds on growth rates of objects) is tannakian over $\mathrm{SuperVect}$.

My question is whether this trick can be repeated one dimension higher. Let $\mathrm{Vect}_{\mathrm{SuperVect}}$ denote the 2-category of all "supercategories", i.e. categories with an action by $\mathrm{SuperVect}$. (I'd rather not make this precise. Probably I should fill in some words like "abelian". I'd be perfectly happy just working with the 2-category whose objects are the natural numbers and whose morphisms are matrices filled in with supervector spaces, just like one can model $\mathrm{Vect}$ as the category whose objects are natural numbers and whose morphisms are matrices of numbers.) Then $\mathrm{Vect}_{\mathrm{SuperVect}}$ is symmetric monoidal (because $\mathrm{SuperVect}$ is) with unit object $\mathbf 1$, and with $\mathrm{End}(\mathbf 1) = \mathrm{SuperVect}$.

Then there's a perfectly good 2-category $$ \mathrm{SuperDuperVect} = \mathrm{Vect}_{\mathrm{SuperVect}} \oplus \mathrm{Vect}_{\mathrm{SuperVect}} \, \Xi$$ where the letter $\Xi$ is just a formal symbol playing the role of $\Pi$ above. I would like to give this category a symmetric monoidal structure in which $\Xi \otimes \Xi = \mathbf 1$ but the braiding on $\Xi$ is the endomorphism $$ \sigma_{\Xi,\Xi} = \Pi \in \mathrm{End}(\mathbf 1) = \mathrm{End}(\Xi \otimes \Xi). $$ The idea is that this is the other object that squares to $\mathbf 1 \in \mathrm{SuperVect}$.

Now, I should be a bit careful. Symmetric monoidal 2-categories, when written out in full detail, consist of quite a lot of data and coherence conditions. Perhaps there's some rule that says that $\sigma_{\Xi,\Xi}$ not only has to square to $\mathbf 1$ but also has to braid trivially with itself. I don't know, and I'm not sure where to look up the axioms.

Does such a symmetric monoidal 2-category "$\mathrm{SuperDuperVect}$" exist?

Is it symmetric-monoidally equivalent to some more basic thing, say the 2-category of sheaves of supercategories on something with convolution product, or the 2-category of supercategorical representations of something?

Is there a symmetric monoidal 2-functor $\mathrm{SuperDuperVect} \to \mathrm{Vect}_{\mathrm{SuperVect}}$?

Of course, if the answer to the first question is "no", then the other two are moot. If the answer to the first question is "yes", then I can't imagine positive answers to the other two, but maybe my imagination is faulty.

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    $\begingroup$ My guess would be that the 'symmetry structure' in SuperDuperVect comes from an appropriate truncation of the sphere spectrum, since that is where it comes from for SuperVect. $\endgroup$ Jun 20, 2015 at 1:03
  • $\begingroup$ @DavidRoberts I don't think I know that story. Will you elaborate? $\endgroup$ Jun 21, 2015 at 0:21
  • $\begingroup$ As to where to look up the axioms, see ncatlab.org/nlab/show/monoidal+bicategory#references $\endgroup$ Jun 21, 2015 at 15:30
  • $\begingroup$ @MikeShulman Thanks! I should have remembered that Chris's thesis has a definition. Also, I think you have some work along these lines for the Morita bicategory? $\endgroup$ Jun 21, 2015 at 19:00
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    $\begingroup$ @Theo The free Picard category on one object is the category of Z-graded Z-lines. This is the truncation of the sphere spectrum to dimensions 0 and 1. If one instead takes the looping of the sphere spectrum and then truncates to dimensions 0 and 1 (so dimensions 1 and 2 in the unlooped case) then one has a similar structure, but now with Z/2-graded (=super) lines. The categorified sign representation takes values 'in this category' (see Ganter-Kapranov arXiv/1110.4753). The symmetry structure I think extends to that on supervector spaces more generally. Work of Osorno-Gurski is also relevant. $\endgroup$ Jun 21, 2015 at 23:19

1 Answer 1


Yes, the symmetric monoidal 2-category you are looking for does exist.

I think that there is a slightly different 2-category which is better, but yours embedds inside the one I will describe, which differs in that there are interesting "cross-terms" from $\Xi$ to 1, i.e. morphisms between these objects. This 2-category has a more familiar description. It is the Morita category of finite dimensional semisimple superalgebras (over $\mathbb{C}$).

Here a superalgebra is just an algebra object in SuperVect. The theory of semisimple modules and algebras mirrors that for ordinary algebras, but with a few subtleties (it is super subtle). A very nice treatment which covers the case that the ground field is algebraically closed is this paper:

Semisimple Superalgebras Tadeusz Jozefiak Volume 1352 of the series Lecture Notes in Mathematics pp 96-113.

There you see that there is a classification of semisimple superalgebras over $\mathbb{C}$. They are finite sums of simple superalgebras. The simple superalgebras are classified as either

  1. $End(\mathbb{C}^{p|q})$ (which is super Morita equivalent to $\mathbb{C}$)
  2. $Q(n) = M_n(\mathbb{C}) \otimes Cl_1$ where $Cl_1$ is the Clifford algebra on a one-dimensional complex vector space.

They have multiplications (using the super-tensor product, of course):

  1. $End(\mathbb{C}^{p|q}) \otimes End(\mathbb{C}^{m|n}) = End(\mathbb{C}^{pm + qn|pn + qm})$
  2. $End(\mathbb{C}^{p|q}) \otimes Q(n) = Q(pn + qn)$
  3. $Q(m) \otimes Q(n) = End(\mathbb{C}^{mn|mn})$

Now the 2-category I want to consider is the Morita 2category whose objects are fin. dim. semisimple superalgebras and whose 1-morphisms are superbimodules between them.

The isomorphism classes of objects in this 2-category are given by pairs of natural numbers which count the number of $End(\mathbb{C}^{p|q})$ and $Q(n)$ factors.

The categories of morphisms are exactly what you describe, as long as you throw out the cross-term morphisms. For example the category of morphisms from $\mathbb{C}$ to $Q(n)$ is equivalent to the category of $Q(n)$-modules, where in your 2-category (if I understand your notation) you would have the zero category here. The 2-category you describe sits inside as the sub-2-category with only the zero cross term morphisms.

Now for your final question. There is no symmetric monoidal functor $$ SuperDuperVect \to Vect_{SuperVect} $$ where SuperDuperVect is either the Morita category I describe or the subcategory you mention. You can see this by passing to maximal Picard sub-2-categories (i.e. the max 2-category in which all morphisms and objects are invertible).

These Picard 2-categories are equivalent to spectra with 3 consecutive homotopy groups. The target gives a spectrum with $$\pi_0 = 0, \; \pi_1 = \mathbb{Z}/2, \; \pi_2 = \mathbb{C}^\times$$ while the source (in either case) has $$\pi_0 = \mathbb{Z}/2, \; \pi_1 = \mathbb{Z}/2, \; \pi_2 = \mathbb{C}^\times$$. Moreover the k-invariants of these are well-known. The latter looks like a truncated variant of the Brown-Comenetz dual of the sphere. It is closely related to real K-theory KO. The former (the target) looks like the connective cover. The k-invariant connecting the bottom $\mathbb{Z}/2$ to the other homotopy groups (eg. the next $\mathbb{Z}/2$) obstructs the existence of your map, and it is known to be non-zero.

  • $\begingroup$ See also the references at ncatlab.org/nlab/show/super+line+2-bundle $\endgroup$ Nov 17, 2015 at 16:35
  • $\begingroup$ This answer reminded me of a lot of things that I knew about but had forgotten. But now I'm not sure I believe it. Actually, the problem is probably my fault: either I am simply confused, or I chose bad notation. My notation "Vect_SuperVect" is not intended to mean "categories that are direct sums of copies of SuperVect", but rather "module categories for SuperVect" (so perhaps I should have used "Mod_SuperVect"), with the balanced tensor product. Now, surely the Morita category of superalgebras embeds in here just as in the usual case? E.g. Cliff(1) corresponds to Vect as a SuperVect-module. $\endgroup$ Nov 18, 2015 at 15:27
  • $\begingroup$ But I will say: it had never occurred to me to ask what the symmetry of Cliff(1) with itself is. That it is the odd Cliff(2)-bimodule is interesting! $\endgroup$ Nov 18, 2015 at 15:30
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    $\begingroup$ This is a super answer. $\endgroup$ Nov 28, 2015 at 20:49

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