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Suppose you have a map $g:\Sigma \rightarrow G$ from a Riemann surface $\Sigma$ to a compact Lie group $G$. What is the obstruction to finding a $3$-manifold $W$, such that $\partial W = \Sigma$, and an extension of $g$ to a map $\tilde{g}:W\rightarrow G$? In the paper I'm reading they say it lies in $H_2(G,\mathbb{Z})$. Why is this true? I mean, the obstruction class to extending $g$ to $\tilde{g}$ is an element in $H^3(W,\pi_2(G))$, which vanishes since $\pi_2(G)=0$ for compact $G$, right? So does this mean that the obstruction to finding a $3$-manifold $W$ with boundary $\Sigma$ lies in $H_2(G,\mathbb{Z})$? By the way, the paper I'm referring is here , see section 4.1 (page 405). Thanks.

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I added the mp tag b/c the question arises from a paper written by two mathematical physicists. Hence, maybe the solution relies on some ideas from physics. – Kevin Wray Apr 27 '11 at 0:55
up vote 10 down vote accepted

This has nothing to do with Lie groups. Let $X$ be any space, let $S$ be a closed, orientable surface and let $f : S \rightarrow X$ be a map. We then get a canonical homology class $f_{\ast}([S]) \in H_2(X;\mathbb{Z})$. If there exists a closed orientable $3$-manifold $M$ with boundary $S$ such that $f$ extends over $S$, then we are done : the manifold $M$ maps into $X$ to provide a homology between $f_{\ast}([S])$ and $0$. Conversely, assume that $f_{\ast}([S]) = 0$. This implies that $f_{\ast}([S])$ is the boundary of a singular chain mapping into $X$. This singular chain can be thought of as a collection of tetrahedra glued together. Aside from the faces of the tetrahedra which lie in $S$, the faces of these tetrahedra are glued together in pairs. Let $M'$ be the result of gluing these tetrahedra together. We thus have $\partial M' = S$ and a map $F' : M' \rightarrow X$ extending $f$. It is a fun exercise to show that $M'$ is a 3-manifold except possibly at finitely many points $x_1,\ldots,x_n$. A small neighborhood of $x_i$ is homeomorphic to the cone on an oriented surface. Cut out these neighborhoods and glue in handlebodies. We get a $3$-manifold $M$ with $\partial M = S$, and from the construction it is clear that we can modify $F'$ to give a function $F : M \rightarrow X$ which still extends $f$.

The point here is that in low degree, bordism agrees with homology.

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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? – Dan Ramras Apr 27 '11 at 2:20
@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. – Andy Putman Apr 27 '11 at 2:28
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. – Dan Ramras Apr 27 '11 at 2:41
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). – Andy Putman Apr 27 '11 at 2:43
@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 $H_2(G)=\pi_2(G)=0$. – Andy Putman Apr 27 '11 at 2:47

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