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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeMar 31st 2015

wrote an entry Deligne’s theorem on tensor categories on the statement that every regular tensor category is equivalent to representations of a supergroup. Added brief paragraphs pointing to this to superalgebra and supersymmetry, added cross-links to Tannaka duality, Doplicher-Roberts reconstruction etc. Also created a disambiguation page Deligne’s theorem

• CommentRowNumber2.
• CommentAuthorDavid_Corfield
• CommentTimeMar 31st 2015

I see at superalgebra you write

This in turn may itself be understood from a more general perspective as follows.

and then pass to Kapronov’s ideas on superalgebra. Is there a way to see how the more general perspective makes sense of Deligne’s theorem?

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeMar 31st 2015

I should expand on that, I just meant to refer to the fact that at the heart of Deligne’s theorem is in turn the statement that there are, up to equivalence, only two non-trivial symmetric braidings on $\mathbb{Z}_2$-graded vector space, the trivial one and the super-one.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeMar 31st 2015
• (edited Mar 31st 2015)

I have expanded a good bit the preliminaries to the Statement making some of the details come out more, such as the definition of the inner parity.

There is often voiced some discomfort with the terminology of the proof of Doplicher-Roberts reconstruction in (Müger 06), which considers ordinary groups but equips them with central involutive elements and let’s them act on super-vector spaces, where these pairs of an ordinary group and a central element are called “supergroups”.

While that is maybe indeed not such a good idea, this structure is precisely what appears in Deligne’s proof for the case that the underlying group is ordinary (purely even): that central element is then Deligne’s inner parity operation.

In any case, I have added a remark on that, too.

• CommentRowNumber5.
• CommentAuthorDavid_Corfield
• CommentTimeMar 31st 2015

In Def 2., I corrected the middle expression so that the exponent is $S_n$

$S_{\lambda}(X) \coloneqq (V_\lambda \otimes X^{\otimes_n})^{S_n} \coloneqq \left( \underset{g\in S_n}{\sum} \rho(g) \right) \left( V_\lambda \otimes X^{\otimes_n} \right)$

That right hand term looks odd to me, and I couldn’t see it quite like that in the Deligne article.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeMar 31st 2015

That’s just expressing the invariants via group averaging. (At least when a factor $\frac{1}{# S_n}$ is included, which was missing.) It’s not much of a point though, I agree. I was first following the review arXiv:0401347 which likes to put it that way.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeMar 31st 2015
• CommentRowNumber8.
• CommentAuthorDavid_Corfield
• CommentTimeApr 1st 2015

Re #3, so you weren’t pointing to Kapranov’s ’superalgebra as arising from 2-truncation of the sphere spectrum’ idea?

As you probably picked up from comments I often make, I’m hoping that that idea will feature more. Here and elsewhere, such as ’supergeometry via Aufhebung’, you’re looking for ways to make supergeometry appear naturally. Is there something missing with an approach which explains it by truncation from something universal?

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeApr 1st 2015
• (edited Apr 1st 2015)

I am pointing to just that. The story is that ordinary superalgebra is governed by the structure of symmetric monoidal 1-categories. Kapranov effectively says that this is just the first stage of symmetric monoidal $\infty$-categories, and hence that one should ask for a higher kind of superalgebra which picks up its gradings from this more general situation. Deligne’s theorem fits in as a further amplification of the 1-categorical situation.

And yes, that’s precisely what I am after, to characterize a supergeometric topos internally, and that’s why I am interested in Deligne’s theorem and Kapranov’s perspective on the generalization. But while all this is great in itself, I still don’t see how to turn it into a neat internal characterization of supergeometric toposes.

So in parallel I am approaching it from the opposite direction. Given that the internal characterization should proceed by a system of modal operators, I am exploring to which extent the more-or-less-canonical system that one may consider knows about supergeometry. Remarkably, when driving it to level 3 then it does give something of which supergeometry is a faithful model.

If you see how that connects to Deligne and Kapranov, I’d be very interested. I am thinking about it, but I don’t see it yet.

• CommentRowNumber10.
• CommentAuthorDavid_Corfield
• CommentTimeApr 1st 2015

Sadly little time for most of today, but…

Might it possible to build up to the full untruncated version from the bottom up? Do the levels leading to supergeometry make any sense seen from the absolute?

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeOct 26th 2016
• (edited Oct 26th 2016)

I am adding more details to Deligne’s theorem on tensor categories.

One question on the definition of “finite $\otimes$-generation” of $k$-tensor categories $\mathcal{C}$:

Deligne 02, 0.1 requires of the finite $\otimes$-generator that every object of $\mathcal{C}$ is obtained by some iteration of any of the operations of forming direct sums, tensor products and subquotients.

But in his review, Ostrik 04, 2.1 gives the more restrictive condition that every object is a subquotient of a direct sum of tensor products of the generator (i.e. in this order and without iterating).

Why is this equivalent (if it is)?

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeNov 4th 2016

Maybe somebody could give me a double check on this, regarding the French:

on the bottom of his p. 3 (pdf) Pierre Deligne writes:

Une réduction assez pénible au cas de ⊗-génération finie permet de vérifier que 0.6 reste vrai sans hypothèse de ⊗-génération finie. Nous ne l’avons pas rédigée.

He genuinely claims that the statement remains true without assuming finite $\otimes$-generation, right?

• CommentRowNumber13.
• CommentAuthorDavid_Corfield
• CommentTimeNov 4th 2016

Yes, that’s my reading too. Reduction from the case without the condition to the case with the condition is ’penible’ (difficult, troublesome, painful), and he hasn’t written it out.

• CommentRowNumber14.
• CommentAuthorDavid_Corfield
• CommentTimeNov 4th 2016

Since the concept of tensor categories is “god given”, much like that of natural mubrs (being a natural categorification),

I can’t guess what misprint ’mubrs’ might be. ’Numbers’ doesn’t make sense there. Or is there some version of ’mutually unbiased bases’ around?

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeNov 4th 2016

Thanks for the double check!

In fact in the review in section 9.11 of EGNO 15 they don’t seem to mention the assumption of finite $\otimes$-generation.

Regarding the typo. I did mean to write “natural numbers”. Now how embarrassing is that? And now I don’t even get your comment in MUBs.

• CommentRowNumber16.
• CommentAuthorDavid_Corfield
• CommentTimeNov 4th 2016

I was just wondering whether just as mutually unbiased bases are abbreviated as MUBs, there might be a concept like ’mutually unbiased basis restriction’ with abbreviation MUBR.

But then back to why I thought it couldn’t be ’numbers’,

of natural numbers (being a natural categorification),

finite sets are a natural categorification of $\mathbb{N}$, but natural numbers themselves?

• CommentRowNumber17.
• CommentAuthorDavid_Corfield
• CommentTimeNov 4th 2016

Is there not a tension between

the concept of tensor categories is “god given”

and at tensor category

The precise definition associated with the term “tensor category” varies somewhat in the literature?

But then perhaps we mortals are rather good at not understanding clearly what the immortals send to us.

• CommentRowNumber18.
• CommentAuthorUrs
• CommentTimeNov 4th 2016
• (edited Nov 4th 2016)

Sorry that this remark of mine is causing headaches. Let me try to tell you in more detail what I want to express there, and maybe you could help me to find better wording for it:

So on the face of it the raison d’être of supersymmetry might seem very weak. You might wonder why to introduce any graded commutativity at all, and even if you accept that, you might wonder why $\mathbb{Z}/2$-grading as opposed to any other grading. In short, supersymmetry groups might feel like a random concept that plays no fundamental role and is only popular due to fashion.

On the other hand, there is no doubt about the fundamental role of monoidal categories. That’s just a fundamental concept whose nature is not to be argued about.

So we classify these. In the process of course we add lots of qualifiers, in order to get a chance to find classification results. There are lot of choices of qualifiers (braided, symmetric, rigid, linear, finite, etc. pp.) and so the literature disagrees about which word to use to subsume which of these qualifiers, but apart from choice of words, there is again no room to have an agument about any of these qualifiers. Nobody will disagree that the concept of a linear monoidal category is to be considered.

So that’s the force of the theorem, which I tried to highlight. It relates the “evidently god given” concept of tensor categories with the superficially much more random looking concept of supersymmetry. It says that the latter is really what you get from the former. Hence it teaches us that supersymmetry, even if we may have been surprised to encounter it, is there as soon as linear tensor categories enter the picture, which there is zero reason to be surprised about.

• CommentRowNumber19.
• CommentAuthorTodd_Trimble
• CommentTimeNov 4th 2016

I don’t think there’s a tension; it just means the same name was given to rather different concepts. The word “rather” signifies that it wasn’t a matter of fiddling with the (“god-given”) Deligne concept by altering a hypothesis ever so slightly: the Deligne notion (combining $k$-linearity and symmetric monoidal duals) seems more specific and sui generis.

Well, I suppose that there might be such a tension, depending on what we assume of $k$. But that particular type of variation isn’t listed in the nLab.

• CommentRowNumber20.
• CommentAuthorDavid_Corfield
• CommentTimeNov 4th 2016

Since the concept of linear tensor categories arises very naturally in mathematics, the theorem gives a purely mathematical “reason” for the relevance of superalgebra and supergeometry. One might have wondered why of all possible generalizations of commutative algebra, it is supercommutative superalgebras that are to be singled out (from alternatives such as plain $\mathbb{Z}_2$-graded algebras, or in fact $\mathbb{Z}_n$-graded algebras or general noncommutative algebras or, the like), as they are notably in theoretical physics (“supersymmetry”), but also in mathematical fields such as spin geometry and K-theory.

But with $k$-linear tensor categories appearing on general abstract grounds as the canonical structure to consider in representation theory, Deligne’s theorem serves to base supercommutative superalgebra on this same general abstract foundation, showing that this is precisely the context in which full $k$-linear tensor categories exhibit full Tannaka duality.

We’ll reserve the real god-given explanation (truncation of the sphere spectrum) for later.

• CommentRowNumber21.
• CommentAuthorUrs
• CommentTimeNov 4th 2016

Okay, thanks.

• CommentRowNumber22.
• CommentAuthorTodd_Trimble
• CommentTimeNov 4th 2016

IMHO it reads really well, David.

1. I am not convinced that Deligne’s theorem is actually indicating anything particularly special about $\mathbb{Z}/2\mathbb{Z}$. That we have a theorem for $\mathbb{Z}/2\mathbb{Z}$ does not mean that there is no theorem for some other choice of grading, for instance: $\mathbb{Z}/2\mathbb{Z}$ is just the simplest possible choice. I have taken a look at the proof, and I suspect that it can be generalised at least to $\mathbb{Z}/n\mathbb{Z}$ for any $n$. It is just harder to figure out the correct choices to obtain an analogue of Proposition 2.9. The rest is very formal.

• CommentRowNumber24.
• CommentAuthorCharles Rezk
• CommentTimeNov 5th 2016

I don’t see how that can be. Proposition 2.9 seems to depend crucially on properties of Schur functors (ultimately, on the representation theory of symmetric groups), and symmetric and exterior powers play a distinguished role here (ultimately, because symmetric groups don’t have very many abelian quotients).

Here’s a related fact: suppose I take a symmetric monoidal category $A$, and look at it’s picard category $\mathrm{Pic}(A)$ of $\otimes$-invertible objects. For any $L$ in $\mathrm{Pic}(A)$, there is an invariant $\chi(L) \in \mathrm{Hom}(\mathbb{Z}/2, \mathrm{Aut}(1))$, given by the symmetry isomorphism $L\otimes L\rightarrow L\otimes L$ and the fact that canonically $\mathrm{End}(L\otimes L)\approx \mathrm{End}(1)$. In Deligne’s set up he assumes $\mathrm{End}(1)=k$ with $k$ a field of characteristic 0, so that $\chi(L)\in \{\pm1\}$. This means that the symmetric group acts on $L^{\otimes n}$ either trivially or via the sign representation. I.e., invertible objects only come in even or odd flavors. Deligne’s theorem feels to me like a fancier version of that observation. (From the homotopy point of view, $\chi$ is the function $\pi_0\underline\mathrm{Pic}(A)\to \pi_1\underline\mathrm{Pic}(A)$ (where $\underline\mathrm{Pic}(A)$ is the Picard spectrum) given by composition with the stable hopf map $\eta\in \pi_1\mathbb{S}\approx \mathbb{Z}/2$.)

2. Thanks for your thoughts. I could have been a little clearer. What is needed is that there is some object $\overline{1}$ as in the hypotheses of Proposition 2.9 such that (i) in this proposition holds. What I am principally getting at is that I suspect that one can produce an analogue of $\overline{1}$ in the $\mathbb{Z}/n\mathbb{Z}$ setting, at least in some cases; and that one can formulate an analogue of the statement of Proposition 2.9 (i); and that if one assumes that one can prove this, then one prove the corresponding version of Deligne’s theorem. The challenge would then be to demonstrate that the analogue of Proposition 2.9 (i) holds in some good degree of generality. Deligne’s approach is to relate it to Schur functors. Maybe this could still work, maybe not. But my guess is that it is possible somehow.

• CommentRowNumber26.
• CommentAuthorzskoda
• CommentTimeNov 6th 2016
• (edited Nov 6th 2016)

I do not understand the definition 3.1. Condition 4 is that it is symmetric, and condition 5 that it is braided. This looks foolish – first stating standard notion of symmetric, and then asking to simultaneously satisfy complicated relaxation of this.

By the way, some of the arguments in the Deligne’s paper Catégories Tannakiennes (about the conditions for an existence a fiber functor, i.e. inner characterization of Tannakian categories) have been simplified in the seminar and the accompanying short article (in English)

• A. L. Rosenberg, The existence of fiber functors, The Gelfand Mathematical Seminars, 1996–1999, 145–154, Gelfand Math. Sem., Birkhäuser Boston, Boston, MA, 2000 doi
• CommentRowNumber27.
• CommentAuthorUrs
• CommentTimeNov 6th 2016

@Richard: the theorem says that there is a fiber functor to super-vector spaces as soon as the tensor category has subexponential growth.

@Charles, thanks for this. How do you prove the very last statement?

• CommentRowNumber28.
• CommentAuthorDavidRoberts
• CommentTimeNov 6th 2016

@Charles I was thinking similar thoughts, though less coherently, after this form: there is a canonical $\mathbb{Z}/2$ floating around in any symmetric monoidal category. The truncation of the sphere spectrum of course plays a rôle, but thanks for pinning it down.

@Zoran thanks for the reference!

3. Urs: yes…what is the point that you wished to make? I am saying that I suspect that one could also prove (though I am not saying that it is easy) that there is a fibre functor to the analogue of super vector spaces for more general $\mathbb{Z}/n\mathbb{Z}$, under some reasonable conditions. The purpose of ’subexponential growth’ is only to give a condition that is reasonably easily checked; what is actually important for constructing the fibre functor is that Proposition 2.9 (i) holds. Any nice condition which implies Proposition 2.9 (i) would do equally well as Schur functors, subexponential growth, etc.

• CommentRowNumber30.
• CommentAuthorDavidRoberts
• CommentTimeNov 6th 2016
• (edited Nov 6th 2016)

@Richard - which Proposition 2.9 (i) and where? You never said, but I’m guessing in Deligne’s paper? Or Ostrik? Or the EGNO book?

EDIT: Ah, I see, it’s in Deligne, and takes as basic assumption that every object is of finite length, and there is a “$\otimes$-square root” of the tensor unit $I$ with the induced automorphism of $I$ given by the symmetry map being just $-1\times(-)$.

4. Deligne’s original paper.

• CommentRowNumber32.
• CommentAuthorUrs
• CommentTimeNov 6th 2016

@Richard, given super $\mathbb{Z}/2$-braiding then it may happen to lift through some group homomorphism $A \to \mathbb{Z}/2$, for instance to integer grading under mod 2 reduction. That’s trivial and doesn’t change the statement. But it won’t lift to an incompatible braiding.

• CommentRowNumber33.
• CommentAuthorCharles Rezk
• CommentTimeNov 6th 2016

@Urs If you believe that Picard groupoids are the same as (-1)-connected, 1-truncated spectra, it’s easy. The operation $\chi\colon \pi_0 X\to \pi_1 X$ is a group homomorphism natural in $X$, so is represented by some stable map $S^1\to S^0$. Since $\chi$ is non-trivial, it has to be the non-trivial one. Probably somebody proves this somewhere, but I don’t know where.

• CommentRowNumber34.
• CommentAuthorzskoda
• CommentTimeNov 6th 2016

There is a sentence in the related entry symmetric monoidal category stating “which means that changing the order of a product twice, from $a \otimes b$ to $b \otimes a$ back to $a \otimes b$, indeed does yield a result equal to the original”. I find this sentence very misleading as the symmetry may be in general very complicated and not at all of the sort “changing order”.

• CommentRowNumber35.
• CommentAuthorRichard Williamson
• CommentTimeNov 6th 2016
• (edited Nov 6th 2016)

I agree completely that the use of the Koszul sign rule is crucial for Deligne’s result.

But I didn’t understand your remark, Urs. The same Koszul sign rule defines a symmetric monoidal structure (let’s, as Zoran suggested, remove the use of the word ’braiding’ here, which is completely redundant) on $\mathbb{Z}/3\mathbb{Z}$ graded vector spaces (interpreting -1 as 2, or in other words working with multiplication in $(\mathbb{Z}/3\mathbb{Z})^{\times}$, of course). What I am getting at is that I suspect it is to possible to prove a theorem analogous to Deligne’s using this algebraic geometry built upon this symmetric monoidal category instead of super algebraic geometry, and using a too-be-determined appropriate analogue of Proposition 2.9 (i). Surely you are not claiming that this is trivial? It seems to me that such a result would be completely independent of the super one.

• CommentRowNumber36.
• CommentAuthorRichard Williamson
• CommentTimeNov 6th 2016
• (edited Nov 6th 2016)

And indeed, having briefly looked into it slightly more, I think that there is a very natural candidate for Proposition 2.9 (i). We suppose that there is an object $\overline{1}$ of $\mathcal{A}$ such that $\overline{1} \otimes \overline{1} \otimes \overline{1}$ is isomorphic to $1$, and such that the symmetry isomorphism $\overline{1} \otimes \overline{1} \rightarrow \overline{1} \otimes \overline{1}$ is multiplication by $-1$. Then our analogue of Proposition 2.9 (i) is exactly the same, for this choice of $\overline{1}$.

The isomorphism $\overline{1} \otimes \overline{1} \otimes \overline{1} \rightarrow \overline{1}$ allows us to argue in exactly the same way as in the crucial paragraph between the statement of Proposition 2.9 and its proof, taking $F(V)$ to be $(V^{0} \otimes 1) \oplus (V^{1} \otimes \overline{1}) \oplus (V^{2} \otimes \overline{1} \otimes \overline{1})$.

And, since we have this fact, the proof in 2.11 goes through unchanged, except for replacing $\mathbb{Z}/2\mathbb{Z}$ by $\mathbb{Z}/3\mathbb{Z}$. At the end of 2.11, we take the object $\overline{1}$ to be the object $1$ in degree $2$ (it is slightly miraculous that this works, but it does). The only gap is the appeal to 0.5 (ii), and that should be filled if it is possible to find an implication similar to Proposition 2.9 (ii) $\Rightarrow$ (i).

5. And, whilst I would certainly need to look more into it, a quick glance suggests to me that it may well be possible to adapt the Schur functor magic from Deligne’s paper to the $\mathbb{Z}/3\mathbb{Z}$-graded case, which would complete the story.

• CommentRowNumber38.
• CommentAuthorCharles Rezk
• CommentTimeNov 17th 2016
• (edited Nov 17th 2016)

I’ve been trying to understand Deligne’s theorem, and it feels to me that it is a version of the “splitting principle” that we find in $K$-theory; or a categorification of the algebraic splitting principle in a $\lambda$-ring.

Suppose you have a $\lambda$-ring $A$ (e.g., the $K$-theory ring of something, equipped with operators $\lambda^k$ and $\sigma^k$ which behave like exterior and symmetric powers). Classically, an element $a\in A$ is finite dimensional if $\lambda^k(a)=0$ for all large $k$. The splitting principle says that there exists a faithfully flat $A\to B$ of $\lambda$-rings, such that $a=\ell_1+\cdots+\ell_r$ where $\ell_i\in B$ are one-dimensional elements.

Call the above elements even finite dimensional. There is a corresponding notion of odd finite dimensional: $\sigma^k(a)=0$ for all large $k$. Analogously, there exists faithfully flat $A\to B$ such that $a$ is a sum of odd one-dimensional elements of $B$. (This is less exciting when you realize that $a$ is odd finite dimensional if and only if $-a$ is even finite dimensional.) More generally, you have a splitting principle for elements of the form (even f.d.)-(odd f.d.). This is done in Mazza and Weibel, “Schur finiteness in $\lambda$-rings”, rXiv:1011.1444v2, which is a nice read; I’m not sure if they are aware of the connection to Deligne’s theorem.

So you can think of Deligne’s theorem as saying, given his hypotheses, including especially the Schur functor hypothesis, that every object in his tensor category is “locally” a sum of one dimensional objects (locally meaning after applying a suitable fiber functor), where being one dimensional comes in even and odd flavors.

• CommentRowNumber39.
• CommentAuthorDavidRoberts
• CommentTimeNov 17th 2016

One could even interpret Deligne’s theorem as saying that in the internal logic every objects is merely a sum of graded “lines”. It even reminds me somewhat of Jim Dolan’s dimensional theories.

• CommentRowNumber40.
• CommentAuthorCharles Rezk
• CommentTimeNov 19th 2016

Can you? Deligne’s theorem involves a condition on objects, either (i) vanishing of some Schur functor, or (ii) subexponential growth of length. Can these be stated internally?

To spell out what I was thinking above, consider a $k$-tensor category (as here). Say an object $X$ is a line if $X\otimes X^\vee\approx 1$. Then $\operatorname{length}(X\otimes Y)=\operatorname{length}(Y)$ for all $Y$. In particular, $X\otimes X= S_{(2)}(X)\oplus S_{(1,1)}(X)$ has length 1, whence either $S_{(1,1)}(X)=0$ (trivial exterior square) or $S_{(2)}(X)=0$ (trivial symmetric square). Thus, lines necessarily come in even or odd flavors, and this distinction is preserved by fiber functors.

Note that if $X$ is a sum (or composition series) of $N$ lines, then $\operatorname{length}(X^{\otimes r}) = N^r$.

I would like to say that Deligne’s theorem mainly amounts to the following: if $X$ is an object with $\operatorname{length}(X^{\otimes r}) \leq N^r$ for some $N$, then there exists a fiber functor $F$ such that $F(X)$ is a sum of lines. (ideally, you would even be able to say, at most $N$ lines.) If the category is tensor generated by such an $X$, then every object is a sum of lines after applying $F$. If so, then there should be a further fiber functor that “trivializes” all the lines, i.e., one with target super-vector spaces.

Naive question: If all representations of an affine supergroup $(G,\epsilon)$ are sums of lines, is that the same as saying that $G$ is abelian?

• CommentRowNumber41.
• CommentAuthorDavid_Corfield
• CommentTimeNov 21st 2016

Charles’s messages got me wondering whether and how constructions for groups have been taken over to supergroups. I came across, and added to orbit method,

• Alexander Alldridge, Joachim Hilgert, Tilmann Wurzbacher, Superorbits (arXiv:1502.04375)

They quote Kostant:

Lie supergroups are “likely to be [… ] useful [objects] only insofar as one can develop a corresponding theory of harmonic analysis”.

The thought then is that it is necessary to extend to orbits through generalized elements.

• CommentRowNumber42.
• CommentAuthorDavid_Corfield
• CommentTimeNov 21st 2016

Since we deal with actions on generalized elements at Stabilizers of shapes, perhaps this superorbit material is quite ’HoTT-able’.

• CommentRowNumber43.
• CommentAuthorzskoda
• CommentTimeNov 21st 2016

People really say stage of definition for the domain of a generalized elements ? Who ?

• CommentRowNumber44.
• CommentAuthorDavid_Corfield
• CommentTimeNov 21st 2016
• (edited Nov 21st 2016)

I first encountered that in people talking about the Beth-Kripke-Joyal semantics of a topos, e.g., p. 123 of Lambek and Scott’s Introduction to Higher-Order Categorical Logic. I wonder if it goes back to that idea of interpreting the semantics of intuitionistic logic as changing states of knowledge of an ’ideal knower’ as they pass in time through a poset. [That’s what p. 123 of Lambek and Scott may be alluding to.]

• CommentRowNumber45.
• CommentAuthorDavid_Corfield
• CommentTimeNov 23rd 2016

I was reading through the entry and tidying up some typos. Just to check, it had at 3.17

$(\widetilde{Line}(sVect), \otimes, k, \tau^{super}) \;\simeq\; \tau_1 \Omega \mathbb{S} \,$

and I thought you wanted $\pi_1$, but I see in that Kapranov 15 paper he is using $\tau$ for truncation of spectra to intervals (e.g., Conjecture 3.3.1). On the other hand, $\tau$ is being used in the entry for the symmetric grading, so that would be confusing.

• CommentRowNumber46.
• CommentAuthorUrs
• CommentTimeNov 23rd 2016

Thanks. I am still in the process of writing the full exposition that I planned on.

Yes, the $\tau_1$ here is for 1-truncation. Since this is the only point where that concept gets used, I’ll replace the notation with something like “$trunc$”.

• CommentRowNumber47.
• CommentAuthorDavid_Corfield
• CommentTimeNov 23rd 2016

Kapranov’s notation would have required $\tau_{[1,1]}$ anyway.

Regarding that paper, there still something odd about that use of the “coincidence” that $\pi^{st}_1$ and $\pi^{st}_2$ are equal, leading to

physical applications exploiting the parallelism between the 1st and the 2nd levels.

He seems to be saying that physics exploits both $\tau_{[0,1]} \mathbb{S}$ and $\Omega \tau_{[1,2]} \mathbb{S}$, and that each gives rise to the Koszul sign rule. But is it the case that any single use is attributable to one of these spectra rather than the other?

• CommentRowNumber48.
• CommentAuthorUrs
• CommentTimeNov 23rd 2016

Okay, as David noticed, I have been adding a fair bit of material to Deligne’s theorem on tensor categories.

I created a section “Background” that means to introduce in expositional style (supposed to be usable as lecture notes to an audience with little to no background) all the relevant concepts and facts from algebra in tensor categories, as well as the relevant concepts of internal automorphisms of fiber functors from Deligne 90

Then I made a section “Proof” with an outline of the three main steps. (1. From subexponential growth to Schur annihilation, 2. from Schur annihilation to super fiber functors, 3. from super fiber functors to Tannaka reconstruction).

I wanted to do much more, but on my little web-book it gets to the point of being impossible to edit. If I don’t find some workaround, I don’t know how to continue editing the nLab. Maybe I need to abandon it after all.

• CommentRowNumber49.
• CommentAuthorDavid_Corfield
• CommentTimeDec 1st 2016

Added this reference for an attempt to generalize Deligne’s theorem to positive characteristic, see

• CommentRowNumber50.
• CommentAuthorUrs
• CommentTimeDec 1st 2016

Thanks. Interesting.

• CommentRowNumber51.
• CommentAuthorTodd_Trimble
• CommentTimeDec 1st 2016

Urs, can you write just short articles or article sections which you or someone else can merge into longer articles later? It would probably be simple enough if you just put your desired text on the nForum, then maybe someone can paste it in wherever you want. Just an idea.

• CommentRowNumber52.
• CommentAuthorUrs
• CommentTimeDec 1st 2016

Hi Todd, thanks, I suppose you are now replying to #48. Meanwhile I have found a way to proceed: I am editing all entries in a separate editor and then paste them back in. Actually now I realize that this is much more convenient for various other reasons, too.

So myself I am fine now. But I do wonder what’s going on with the nLab software that even the edit window, where no math formatting takes place, completely stalls on small devices as soon as there is a non-trivial amount of text in the edit window. I am wondering what on earth the computer is being asked to do by the software while characters typed into the window appear on screen at the rate of one per second.

• CommentRowNumber53.
• CommentAuthorTodd_Trimble
• CommentTimeDec 1st 2016

Yes, I was replying to #48. Glad you’ve found a way to proceed.

• CommentRowNumber54.
• CommentAuthorjib
• CommentTimeNov 28th 2018
In the proof of 3.15 (that there are only two symmetric monoidal structures on super vector spaces), the full subcategory of invertible objects is employed. There are two things about it that I believe need intervention, but I am not sure what to do, so I decided to comment here.

First, it is claimed that it is sufficient to show that there is an essentially unique non-trivial symmetric braiding on this full subcategory. It might be so but I don't see it. Theoretically it might happen that several different symmetric braidings restrict to the same braiding on this subcategory.

Second, it is said that this subcategory is necessarily a groupoid. Although from the subsequent text it becomes clear that actually a 2-group is meant, I still believe that as stated it is confusing. This category is not a groupoid since e.g., being a full subcategory of an abelian category, it contains the zero maps. Only when viewed as a 2-category with single object, it becomes a groupoid (up to somethings).
• CommentRowNumber55.
• CommentAuthorUrs
• CommentTimeNov 29th 2018

Thanks. Would have to go back to this. Do you think you can fix it? Then please go ahead. Otherwise I try to come back to it later.

• CommentRowNumber56.
• CommentAuthorUrs
• CommentTimeNov 29th 2018
• (edited Nov 29th 2018)

Fixed the issue with $Line(\cdots)$ not being a groupoid as stated, by writing $Line(\cdots)_{iso}$ instead and declaring it to be the groupoid core.

This now makes the proof run as intended. But it still leaves the gap to show that the restriction of symmetric monoidal structures on all of super vector spaces to just its Picard groupoid is injective.

Need to think about this. Back when I wrote the proof it seems I found this evident, but maybe it’s not.