What is the Dehn twist factorization of the hyperelliptic involution on an oriented surface of genus g (with one boundary component)?
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$\begingroup$ What exactly do you mean by the hyperelliptic involution on a surface with boundary? If you want it to be an involution, then it cannot fix the boundary pointwise (it has to rotate it by 180 degrees). What are your conventions on how mapping classes interact with the boundary? $\endgroup$– Andy PutmanAug 22, 2013 at 3:12
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$\begingroup$ Without boundary, I believe an expression can be found in Joan Birman's book. $\endgroup$– Igor RivinAug 22, 2013 at 3:19
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$\begingroup$ Also in Farb-Margalit's book and Ivanov's survey. $\endgroup$– Andy PutmanAug 22, 2013 at 3:32
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$\begingroup$ I just meant that it has a boundary component, so involution acts on the boundary as well. $\endgroup$– nikitaAug 22, 2013 at 14:13
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$\begingroup$ Does it look something like this: $c_{2g+1}...c_1c_1....c_{2g+1}$? $\endgroup$– nikitaAug 22, 2013 at 15:47
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1 Answer
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This is almost covered by Proposition 4.12 of Farb and Margalit's book "The primer on mapping class groups". Another useful reference is the paper "Presentations for the punctured mapping class groups in terms of Artin groups" by Labruere and Paris. See the second displayed formula in Proposition 2.12 of that paper.
Here is an outline of the solution. Let $S$ be a surface of genus $g$ with one boundary component. Let $a_i$ be a "chain" of $2g$ curves in $S$. That is, $a_i$ and $a_{i+1}$ meet in exactly one point and otherwise the $a_i$ are disjoint from each other. This is not done cyclically - that is, $a_1$ and $a_{2g}$ are disjoint. Note that the boundary of a regular neighborhood of the union of the $a_i$ is isotopic to $\partial S$. Let's also use the symbol $a_i$ to denote a Dehn twist about the curve $a_i$. Consider the following product of Dehn twists:
$$H = ( a_1 a_2 \ldots a_{2g - 1} a_{2g} )^{2g + 1}$$
Claim: $H$ is the desired mapping class. Proof: $H$ fixes the curve $a_i$ (up to isotopy) but reverses the orientation of $a_i$. Since the chain $\cup a_i$ fills the surface, we are done. QED
Proposition 4.12 of the primer and Proposition 2.12 of [LP] says that $H^2$ is isotopic to a full Dehn twist about $\partial S$. Since you allow isotopies that move boundary points, for you $H^2$ is isotopic to the identity.
All of these claims can be proved by drawing a few pictures, applying induction, and using the "Alexander method". Another approach (that is less general) is to consider $S$ as a double branched cover of the $2g + 1$ times marked disk, and then to think about the generator of the center of the braid group.
One final remark - there are many ways to choose the chain of curves $a_i$ and this might seem like a problem... however, any two ways of choosing the chain differ by a mapping class: thus the apparent choice is no choice at all. I believe that Farb and Margalit call this the "change of coordinates principle".
