Realizing braid group by homeomorphisms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:49:21Z http://mathoverflow.net/feeds/question/55555 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55555/realizing-braid-group-by-homeomorphisms Realizing braid group by homeomorphisms Scott P 2011-02-15T21:34:13Z 2011-02-15T23:07:33Z <p>Markovich and Saric proved the following remarkable theorem. Let $S$ be a compact surface of genus at least $2$ and let $MCG(S)=\pi_0(Homeo^{+}(S))$ be the mapping class group of $S$. There is then no right inverse to the natural projection $Homeo^{+}(S) \rightarrow MCG(S)$. This should be contrasted with the solution to the Nielsen realization problem by Kerckhoff, who proved that any finite subgroup of $MCG(S)$ can be realized by homeomorphisms of the surface.</p> <p>My question is whether this is known for braid groups. Let me make this a little more precise. Let $X_n$ be a $2$-dimensional disc with $n$ punctures. By $Homeo^{+}(X_n)$, we mean homeomorphisms of $X_n$ that are the identity on the boundary and extend over the punctures (but are allowed to permute the punctures). The group $\pi_0(Homeo^{+}(X_n))$ is then the $n$-strand braid group $B_n$. My question is if it is known whether or not there is a right inverse to the natural projection $Homeo^{+}(X_n) \rightarrow B_n$.</p> <p>Since $B_1=1$ and $B_2=\mathbb{Z}$, the first nontrivial case is $n=3$.</p> <p>I would find it shocking if this is not known for $n=3$. The braid group then has two generators $a$ and $b$ and only one relation $aba=bab$. The problem is then asking whether or not we can find homeomorphisms $f,g \in Homeo^{+}(X_n)$ in the homotopy classes of $a$ and $b$ such that $fgf=gfg$. Surely this cannot be open!</p> http://mathoverflow.net/questions/55555/realizing-braid-group-by-homeomorphisms/55563#55563 Answer by Johannes Ebert for Realizing braid group by homeomorphisms Johannes Ebert 2011-02-15T23:07:33Z 2011-02-15T23:07:33Z <p>The problem is much easier, but still interesting and nontrivial, if you replace homeomorphisms by diffeomorphisms, so it might be worth to study this case. Logically, because any action by diffeomorphisms is an action by homeomorphisms and if no action by homeomorphisms exists, then certainly there cannot be any action by diffeomorphisms. But that is not the point. </p> <p>The point is that one can use cohomology theory (which is impossible in the homeomorphism case, see below). The case of closed surfaces and diffeomorphisms was solved by Morita around 1985, and here is a punchline.</p> <p>Consider a closed oriented $n$-manifold $M$ and the orientation-preserving diffeomorphism group $Diff(M)$. Let $Diff(M)^{\delta}$ be the diffeomorphism group with the discrete topology. There is an induced map $B Diff(M)^{\delta} \to B Diff(M)$, inducing a map in rational cohomology. We now describe cohomology classes that automatically lie in the kernel. Let $f:A Diff(M) \to B Diff(M)$ be the universal $M$-bundle. It has a vertical tangent bundle $T \to A Diff(M)$ of the universal $M$-bundle. It has Pontrjagin classes $p_i \in H^{4i}$, and with the help of the Gysin map, you produce classes $f_! (q(p_1,p_2,..))$ for each polynomial in the Pontrjagin classes (let us call them Mumford-Morita-Miller or MMM-classes).</p> <p>Now let $X$ be a smooth manifold and $X \to B Diff (M)^{\delta}$ a map, classifying a manifold bundle $E \to X$. Since the structural group is discrete, there would be a foliation of $E$ by leaves that are transverse to the fibres.The normal bundle of the foliation is the vertical tangent bundle.</p> <p>But there is a theorem of Bott. He showed that the Pontrjagin classes of the normal bundle of a foliation are zero in a range of degrees (I forgot the range, but it is sufficient to guarantee that in the case of surfaces $p_{1}^{k}$ is zero for $k \geq 2$). This relies on Chern-Weil theory. Upshot of the discussion: in a range of degrees, the MMM-classes map to zero in $H^* (B Diff(M)^{\delta})$.</p> <p>Now assume that there is a section $s: MCG(M) \to Diff(M)$. It would, clearly, factor through $Diff(M)^{\delta}$. In particular, a portion of the MMM-classes map to zero under the section.</p> <p>In the surface case, there is a theorem by Earle and Eells, saying that the components of $Diff$ are contractible. Consequence: the induced map $Bs: B MCG (M) \to B Diff (M)$ is a homotopy equivalence, but it maps many MMM-classes to zero. This is a contradiction once you know that enough MMM classes are nonzero. This is precisely what Morita did.</p> <p>The question for homeomorphisms cannot be attacked using these methods, because of the following, really awesome, theorem by Dusa Mc Duff (I think, first stated by Bill Thurston, and big portions of the proof were done by Mather and Segal). It says that for each (maybe PL is necessary) manifold $M$, the natural map $B Homeo(M)^{\delta} \to B Homeo (M)$ is a homology equivalence!! So you will never get a contradiction in that way.</p> <p>Now if you wish to study the Braid group case (and diffeomorphisms), then some of these things will work (and may be easier): The components off the corresponding diffeomorphism groups are contractible, the (rational) cohomology of the braid groups is explicitly known; and it should be relatively easy to figure out the Pontrjagin classes of the vertical tangent bundle. Once some of them are nontrivial, you are done.</p>