Timeline for Witten's QFT and Jones Poly paper
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 21, 2011 at 20:55 | comment | added | Kevin Wray | Further, we care about elements of $H^n(G;\mathbb{R})$ where $n=3$ since we are integrating over the $3$-manifold $M$, and no other reason; i.e., why not care about $\pi_1(G)$ or $\pi_2(G)$ (I know they are trivial in these two cases, but still would like a nice description for why we don't care about them)? | |
| Feb 21, 2011 at 20:54 | comment | added | Kevin Wray | @ Konrad: Yes, but there is no "quantization" condition on $k$ for the $U(1)$ theory, right? And this has to do with $\pi_3\big(U(1)\big)=0$, right? Also, are you saying: If $G$ is compact and simple, then we know that $\pi_3(G)\cong \mathbb{Z}$. Hence, using the Hurewicz isom., we get that $H_3(G)\cong \mathbb{Z}$. Therefore, $H^3(G;\mathbb{R})$ must be nontrivial. | |
| Feb 21, 2011 at 20:16 | comment | added | Konrad Waldorf | By the way, Chern-Simons theory for $G=U(1)$ is non-trivial. That's because $H^4(BU(1),\mathbb{Z})$ is non-trivial. | |
| Feb 21, 2011 at 20:09 | history | edited | Konrad Waldorf | CC BY-SA 2.5 |
added 311 characters in body
|
| Feb 21, 2011 at 20:07 | comment | added | Konrad Waldorf | @ klw1026: $U(1)$ is not simple. It's a subtle terminology, but simple contains the assumption of being non-abelian. Also, I think we need the assumption of simply-connectedness, see my edit. | |
| Feb 21, 2011 at 19:19 | comment | added | Kevin Wray | Sorry, I didn't mean to say $U(1)$. | |
| Feb 21, 2011 at 19:14 | comment | added | Somnath Basu | @ klw1026 - $U(1)$ is the circle which is $1$-dimensional, whence $H^3(U(1);\mathbb{R})=0$. | |
| Feb 21, 2011 at 18:36 | comment | added | Kevin Wray | But, isn't true that $H^3\big(U(1),\mathbb{R}\big)\neq 0$? If so, then shouldn't we also have nontrivial gauge transformations in this case? However, we know that $k$ is not quantized for $U(1)$ theories - quantization comes directly from nontrivial gauge transformations. I'm sure that I am mixing something up here...just not sure what. | |
| Feb 21, 2011 at 14:21 | history | answered | Konrad Waldorf | CC BY-SA 2.5 |