Timeline for Reshetikhin-Turaev and links with a distinguished component
Current License: CC BY-SA 3.0
8 events
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| Oct 28, 2011 at 10:25 | history | edited | Adrien | CC BY-SA 3.0 |
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| Oct 27, 2011 at 21:43 | comment | added | Adrien | @Theo: I agree of course that all trace are 0, it was just not obvious to me that in such a situation the cut open invariant is non constant. I'm aware of the situations you mention, but don't know so much about them.. | |
| Oct 27, 2011 at 21:14 | comment | added | Theo Johnson-Freyd | @Adrien: Note that if a simple module has zero (quantum) dimension, then all traces are $0$, and so the closed invariant has no information at all. Standard examples of simple zero-dimensional modules arrise when working in characteristic $p$, when working with quantum groups at roots of unity, and when working with (maybe quantum) super Lie algebras. I don't know the details, but I do know that some of the theory is developed by Nathan Geer. Development is important, because you do want an invariant that doesn't care about which component is distinguished. | |
| Oct 27, 2011 at 17:35 | comment | added | Adrien | For your second comment, the quantum dimension being 0 or not, it's always true that an endormophism could potentially carry more information than a scalar, but it's not obvious to me that it's indeed the case in this context. Do you have an example ? | |
| Oct 27, 2011 at 17:33 | comment | added | Adrien | Thanks, I agree that if V is simple it works fine, and in that case you get a numerical invariant is itself compatible with the connected sum. Note however that in the context of finite type invariant you want to prove something like "every weight system can be integrated to a knot invariant", and it seems to me that you also need non simple object. | |
| Oct 27, 2011 at 16:35 | comment | added | Noah Snyder | Two quick comments. First, if V is simple (which it often is in applications), then End(V) is canonically identified with the scalars via Schur's lemma. Second, sometimes the quantum dimension of V is 0, in which case you get substantially more information from the cut open invariant. | |
| Oct 27, 2011 at 15:50 | history | edited | Adrien | CC BY-SA 3.0 |
Typo
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| Oct 27, 2011 at 14:39 | history | asked | Adrien | CC BY-SA 3.0 |