Timeline for What does the Tannakian formalism reconstruct when fed the category of chain complexes?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 28, 2017 at 6:40 | vote | accept | Theo Johnson-Freyd | ||
| Jan 27, 2017 at 5:32 | answer | added | Artur Jackson | timeline score: 1 | |
| Apr 3, 2012 at 5:32 | comment | added | Noah Snyder | Basically the problem is that if you have objects of dimension 0, then the length of $X^{\otimes n}$ could grow superexponentially even though its dimension grows only exponentially. On the other hand, you can often apply Deligne using only dimensional consideration. For example, if you have any dimension function which is bounded away from zero (it doesn't need to be categorical dimension) then length grows only exponentially and you can apply Deligne's theorem. This applies to symmetric fusion categories using FP dimension. | |
| Apr 3, 2012 at 2:55 | answer | added | Ben Wieland | timeline score: 8 | |
| Mar 31, 2012 at 3:23 | comment | added | Theo Johnson-Freyd | @Pavel: Thanks! Corrected. I was going from memory from the appendix to arxiv.org/abs/math-ph/0602036 by Michael Muger. | |
| Mar 31, 2012 at 3:21 | history | edited | Theo Johnson-Freyd | CC BY-SA 3.0 |
corrected an error
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| Mar 29, 2012 at 19:51 | comment | added | Pavel Etingof | The condition that Deligne needs in the supercase is that for every object, some Schur functor of this object is zero. This can be replaced by the condition that the category has exponential growth (the length of $X^{\otimes n}$ is at most $C(X)^n$ for any $X$). | |
| Mar 29, 2012 at 19:50 | comment | added | Pavel Etingof | @Theo: You say: "The generalization of the second paragraph, also due to Deligne, is that a rigid symmetric monoidal (technical conditions) category has a fiber functor to SVectC iff every object has (possibly negative) integer dimension." I don't think it is true. The Deligne category of representations of the symmetric group with -1 element is a counterexample (see Deligne's paper on representations of $S_t$ when t is not a positive integer). | |
| Nov 14, 2011 at 8:43 | comment | added | Torsten Ekedahl | @Theo: You are absolutely right, I am glad for the qualifications I put in... I leave my comment in case someone else makes the same mistake or more precisely "enom till straff, androm till varnagel" ("punishment for one, warning to others" a phrase used in old Swedish laws). | |
| Nov 13, 2011 at 17:41 | comment | added | Theo Johnson-Freyd | @Torsten: One invariant of a (braided) monoidal category is its (abelian) group of invertible objects-up-to-isomorphism. For the usual category of chain complexes of vector spaces, this group is $\mathbb Z$. For the category of $G = \mathbb G_m\ltimes \mathbb G_a^{0|1}$-representations in SVect, this group is $\mathbb Z \times \mathbb Z/2$. In particular, the trivial $G$-rep on $\mathbb C^{0|1}$ is its own $\otimes$-inverse, and DGVect does not contain such an object. | |
| Nov 13, 2011 at 9:37 | comment | added | Torsten Ekedahl | To me it seems that the $G$-representations in $\mathrm{SVect}$ consist of a graded supervector space together with a differential of superdegree $1$ and internal degree $1$. Such a graded vector spaces can be thought of as a graded vector space where the even vectors of degree $n$ correspond to new degree $2n$ and the odd vectors of degree correspond to new degree $2n+1$. In the new degree the differential is of degree $1$. This looks like the category of complexes (with the Koszul rule built into the monoidal structure as it should). | |
| Nov 13, 2011 at 5:57 | history | asked | Theo Johnson-Freyd | CC BY-SA 3.0 |