Do the following set of Dehn twists generate the mapping class group? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T10:30:32Zhttp://mathoverflow.net/feeds/question/104308http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/104308/do-the-following-set-of-dehn-twists-generate-the-mapping-class-groupDo the following set of Dehn twists generate the mapping class group?unknown (google)2012-08-08T21:28:21Z2012-08-09T10:22:23Z
<blockquote>
<p>If $S$ is the surface illustrated below, do the Dehn twists about the red curves generate the mapping class group $\operatorname{MCG}(S,\partial S)$?</p>
</blockquote>
<p><img src="http://math.stanford.edu/~pardon/img/dehntwists.png"/></p>
http://mathoverflow.net/questions/104308/do-the-following-set-of-dehn-twists-generate-the-mapping-class-group/104311#104311Answer by Andy Putman for Do the following set of Dehn twists generate the mapping class group?Andy Putman2012-08-08T21:52:57Z2012-08-08T22:03:37Z<p>No. If they did, then they would still generate the mapping class group of the closed surface that results from gluing a disc to the boundary component. However, in that surface they all commute with the hyperelliptic involution, which is not central for $g$ at least $3$.</p>
<p>In fact, Humphries proved that you need at least $2g+1$ Dehn twists to generate the mapping class group. This is contained somewhere in Farb-Margalit's primer on mapping class groups.</p>
<p>In the closed surface, these Dehn twists actually generate the centralizer of the hyperelliptic involution, which is known as the hyperelliptic mapping class group. This is a theorem of Birman and Hilden, which can also be found somewhere in Farb-Margalit.</p>
http://mathoverflow.net/questions/104308/do-the-following-set-of-dehn-twists-generate-the-mapping-class-group/104337#104337Answer by Ronnie Brown for Do the following set of Dehn twists generate the mapping class group?Ronnie Brown2012-08-09T10:22:23Z2012-08-09T10:22:23Z<p>Humphries arguments are in </p>
<p>Humphries, Stephen P.
Generators for the mapping class group. Topology of low-dimensional manifolds (Proc. Second Sussex Conf., Chelwood Gate, 1977), pp. 44–47,
Lecture Notes in Math., 722, Springer, Berlin, 1979. </p>
<p>The key argument for the $\ne 2g$ case is that the symplectic group $Sp(2g, \mathbf F_2)$ is not generated by $2g$ transvections (which maybe was well known). Steve and I joined up to present this argument and more in our paper </p>
<p>``Orbits under symplectic
transvections II: the case $K=\bf F_2$'', <em>Proc. London
Math. Soc.</em> (3) 52 (1986) 532-556.</p>
<p>and Steve followed this up with much more work in this area. </p>