Timeline for Is Yetter's invariant multiplicative under connected sum?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 16, 2018 at 4:57 | comment | added | Theo Johnson-Freyd | In the comments to the original post, Kevin Walker corrected my mistake. This is not the reduction of DW theory along S^1. It is simply |G| times the DW invariant. | |
| Mar 15, 2018 at 16:13 | history | edited | Arun Debray | CC BY-SA 3.0 |
oops! I forgot to use group orders instead of groups
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| Mar 15, 2018 at 14:54 | vote | accept | Manuel Bärenz | ||
| Mar 15, 2018 at 14:49 | comment | added | Manuel Bärenz | Oh, I see, but if we multiply Yetter by $Z_\mathcal{T}(S^n)$, we arrive at a multiplicative invariant once again! (Or, as per Kevin's suggestion, with an Euler characteristic theory if $n$ is even) | |
| Mar 15, 2018 at 11:20 | comment | added | Manuel Bärenz | I think this is a misunderstanding. The weights only need to show up in a different formulation of Dijkgraaf-Witten. See e.g. this article. | |
| Mar 14, 2018 at 21:21 | comment | added | Theo Johnson-Freyd | In particular, note that OP's invariant is not multiplicative for connect sums when $\dim M = 1$. Then $S^1 \# S^1 = S^1$, and $Z(S^1) =$ number of conjugacy classes in $G$, which is typically nonzero. | |
| Mar 14, 2018 at 21:20 | comment | added | Theo Johnson-Freyd | What you've explained is that the answer to OP's question is "yes" if certain homotopy groups of $\mathcal T$ vanish. Which homotopy groups? It depends on the dimension of $n$. | |
| Mar 14, 2018 at 21:19 | comment | added | Theo Johnson-Freyd | As I said in the comments, I think OP's weightings are your waitings for $M \times S^1$. Reduction along $S^1$ takes TFTs to TFTs. | |
| Mar 14, 2018 at 19:15 | history | answered | Arun Debray | CC BY-SA 3.0 |