Timeline for How to calculate the Witten-Reshetikhin-Turaev invariants from a triangulation?
Current License: CC BY-SA 3.0
9 events
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| Jul 29, 2013 at 11:41 | comment | added | Manuel Bärenz | This talk tries to clarify (amongst other things) which compositions (horizontal or vertical) have to be used for which parts of the 0-, 1- and 2-handles. Is there really a straightforward generalisation of doing this in the n-dimensional case, where you have n different compositions? | |
| Jun 24, 2011 at 18:33 | comment | added | Kevin Walker | I think state sums can be derived more simply and directly than is outlined in the talk you link to. (In the 2d case, I'm assuming Z(pt) is a semisimple 1-cat with finitely many simple objects. I'm not sure how much G-equivariance complicates things, but I suspect not much.) You just apply gluing formulas in a straightforward way to the handle decomposition associated to the cell decomposition, choosing orthogonal bases and idempotents at every step. | |
| Jun 24, 2011 at 15:13 | comment | added | Noah Snyder | On the one hand, given an extended TFT you can certainly compute its values using a triangulation. However, this process is a bit more complicated than you might expect (see pps. 4-5 of math.utexas.edu/users/odavidovich/Talks/BerkleyNov2010 for some discussion of these issues). On the other hand, I don't think you get the state sum model directly from an extended TFT. Instead to get the state sum model directly you really want an open-closed theory. In 2d you remove a little window around every vertex of your triangulation, compute that and then normalize. | |
| Jun 23, 2011 at 20:14 | comment | added | Kevin Walker | Thanks Noah. Some part of my brain knew that already, but apparently that part was not in communication with the parts which were doing the typing yesterday. So an extended 2d TQFT, plus some representations of the 1-category associated to a point (to label non-gluing boundaries), gives rise to an open-closed 2d TQFT. I suppose all open-closed TQFTs might arise this way, depending on the details of one's definition of an open-closed TQFT. | |
| Jun 23, 2011 at 18:19 | comment | added | Noah Snyder | "Open-closed" means you allow a different kind of boundary which you're not allowed to glue along. This is definitely a different concept than extending. It's more closely related to allowing boundaries between two different TFTs, but where one of the TFTs is trivial. I'm still a little confused though about the relationship of each of these notions (extended and open-closed) to triangulations, so I'm going to think some more about that before saying anything. | |
| Jun 22, 2011 at 23:31 | comment | added | Kevin Walker | Like André, I thought that an open-closed 2d TQFT was the same thing as a fully extended 2d TQFT. And I'm not sure what "open-closed" would mean in higher dimensions. I'm curious to hear what Noah had in mind. | |
| Jun 22, 2011 at 21:40 | comment | added | André Henriques | Isn't "open-closed" the same thing as "extended"?... (please tell me what you have in mind, because I'm not quite following your comment) The real problem is that you need to have framings around (or p_1-structures). If we were just talking about oriented TQFT's then triangulations should be just fine as input data. | |
| Jun 22, 2011 at 20:51 | comment | added | Noah Snyder | Does an extended TQFT really give a way of computing invariants from triangulations? Don't you need an open-closed extended TQFT? Just think about 2 dimensions. | |
| Jun 22, 2011 at 20:37 | history | answered | André Henriques | CC BY-SA 3.0 |