User scott p - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T10:01:00Z http://mathoverflow.net/feeds/user/13001 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 Comment by Scott P Scott P 2011-02-16T17:40:43Z 2011-02-16T17:40:43Z This is probably as good an answer as I'm going to get, so I'll accept it. Thanks! http://mathoverflow.net/questions/55555/realizing-braid-group-by-homeomorphisms Comment by Scott P Scott P 2011-02-16T17:40:24Z 2011-02-16T17:40:24Z @Bill Thurston : Thanks! http://mathoverflow.net/questions/55555/realizing-braid-group-by-homeomorphisms Comment by Scott P Scott P 2011-02-15T23:17:36Z 2011-02-15T23:17:36Z (continued) If you blowup this &quot;special&quot; puncture to a boundary component, then the resulting action of the braid group will rotate the new boundary component. I want to know if you can make this action fix this new boundary component on the nose. http://mathoverflow.net/questions/55555/realizing-braid-group-by-homeomorphisms Comment by Scott P Scott P 2011-02-15T23:16:10Z 2011-02-15T23:16:10Z @Bill Thurston : That lifting argument doesn't quite give you the three-strand braid group; rather, it gives you a lift of the three-strand braid group modulo its center (namely, $PSL_2(\mathbb{Z})$) to homeomorphisms of a sphere with 4 punctures. There is a &quot;special&quot; puncture that stays fixed under this action of $PSL_2(\mathbb{Z})$ (regarding the 4-punctured sphere as the quotient of $T^2$ by the hyperelliptic involution minus the fixed points, this &quot;special&quot; puncture is the image of the basepoint on $T^2$). http://mathoverflow.net/questions/55555/realizing-braid-group-by-homeomorphisms Comment by Scott P Scott P 2011-02-15T21:48:11Z 2011-02-15T21:48:11Z Ryan - Notice the phrase &quot;in the homotopy classes of $a$ and $b$&quot; in the last paragraph =).