Extending maps on Riemann surfaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:36:16Z http://mathoverflow.net/feeds/question/63096 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63096/extending-maps-on-riemann-surfaces Extending maps on Riemann surfaces Kevin Wray 2011-04-27T00:47:42Z 2011-04-27T02:17:00Z <p>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 <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.cmp/1104180750" rel="nofollow">http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.cmp/1104180750</a> , see section 4.1 (page 405). Thanks.</p> http://mathoverflow.net/questions/63096/extending-maps-on-riemann-surfaces/63103#63103 Answer by Andy Putman for Extending maps on Riemann surfaces Andy Putman 2011-04-27T02:10:09Z 2011-04-27T02:10:09Z <p>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 $f$.</p> <p>The point here is that in low degree, bordism agrees with homology.</p>