Timeline for Extending maps on Riemann surfaces
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 27, 2011 at 2:50 | vote | accept | Kevin Wray | ||
| Apr 27, 2011 at 2:50 | comment | added | Kevin Wray | AH OK, that is nice to know, plus I understand the statement better now. So that explains why when $G$ is simply connected they have no problem extending this map to a map over $W$. Thanks. | |
| Apr 27, 2011 at 2:47 | comment | added | Andy Putman | @klw1026 : 1. I am claiming that $H_2(X)$ is the same as $\Omega_2(X)$. In fact, I basically gave a proof of it!, and 2. If $G$ was simply-connected, then by Hurewitz we would have $H_2(G)=\pi_2(G)=0$. | |
| Apr 27, 2011 at 2:45 | comment | added | Kevin Wray | @Andy So, if $G$ was simply connected then we could also extend over the $2$-cells we attached first? | |
| Apr 27, 2011 at 2:43 | comment | added | Kevin Wray | Thanks for the responses. Andy, could you further elaborate on your last statement about bordism and homology? Are you saying that $H_2(X)\cong \Omega_2(X)$? I have heard that $\Omega_3$ and $H_3$ are equivalent up to torsion, but nothing relating $\Omega_2$ and $H_2$. | |
| Apr 27, 2011 at 2:43 | comment | added | Andy Putman | To construct a handlebody from a surface, you can't just attach $3$-cells. You have to first attach some $2$-cells (to kill off half of the first homology group). | |
| Apr 27, 2011 at 2:41 | comment | added | Dan Ramras | Okay, I agree, but I'm still confused... Let $G=SO(3)\times SO(3)$ so that $H_2 (G; Z) = Z/2$ and take a map $g:\Sigma\to G$ representing the non-trivial class. Why doesn't $\pi_2 (G) = 0$ imply that we can extend $g$ over a handlebody? The handlebody is a 3-complex and the attaching maps of the 3-cell, when composed with $g$, is a (nullhomotopic) map $S^2\to G$, which says $g$ extends over the 3-cell. Surely I'm just being dumb. But this really gets to the heart of the original question, I think. | |
| Apr 27, 2011 at 2:28 | comment | added | Andy Putman | @Dan : If $v \in H_2(X;\mathbb{Z})$ is any homology class, then there exists a closed orientable surface $S$ and a map $f : S \rightarrow X$ such that $f_{\ast}([S]) = v$ (think about piecing together the simplices in a singular 2-cycle). Thus there will be something like I describe if and only $H_2(X;\mathbb{Z})$ is nonzero. | |
| Apr 27, 2011 at 2:20 | comment | added | Dan Ramras | Andy, does this actually give rise to an example of a map $\Sigma \to G$ that can't be extended over, say, a handlebody? I think klw1026 correctly explained that the (homotopical) obstruction to that extension problem is trivial, because $\pi_2 (G) = 0$ and we're only adding 3-cells to form the handlebody. So I'm really confused by Dijkgraaf and Witten's statement that sometimes there's a non-trivial obstruction (of the sort you're describing). It's not very consequential to their paper, though, so maybe they just didn't notice that the obstruction always vanishes? | |
| Apr 27, 2011 at 2:10 | history | answered | Andy Putman | CC BY-SA 3.0 |