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In "Automorphisms of Surfaces after Nielsen & Thurston" by Casson & Bleiler (on pages 75 - 80) they discuss classifying automorphisms of a surface. They show that, if $S$ is a closed orientable surface, $f \colon S \to S$ an automorphism and $c$ is a geodesic 1-submanifold of $S$ such that $f(c) \simeq c$ then $f$ is reducible map.

Suppose $S = T^2 \sharp D^2 \sharp D^2$ (the twice punctured torus) and $\delta$ is a loop around one of the boundary components. Then $\delta$ is non-trivial in $H_1(S, \mathbb{Z})$ but $\forall [\phi] \in \mathcal{MCG}(S)$, $\phi(\delta) \simeq \delta$ or $\phi(\phi(\delta)) \simeq \delta$. Hence this statement doesn't hold for $S$.

Is there a similar result for $S$ (or indeed general surfaces with 2 or more boundary components)?

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Isn't that the definition of a reducible map? –  HJRW Oct 25 '10 at 23:19
    
Not quite, f is reducible iff f is homotopic to an automorphism g which leaves invariant an essential 1-submanifold. Their remark says that it is equivalent to say that there exists a geodesic 1-submanifold which is isotopic to its image under f. –  Mark Bell Oct 25 '10 at 23:28
    
So $S$ isn't just a surface (i.e. 2-manifold) but it's a surface endowed with a complete Riemann metric, as you're talking about geodesics. You haven't specified what your set-up is exactly. And you want your geodesics to be closed? Certainly there's corresponding results for surfaces with boundary and/or punctures but what kind of setting do you want the result to live in? We don't have Casson and Bleiler here (isn't it out of print?) so it would be helpful if you were more clear about what you're looking for. –  Ryan Budney Oct 25 '10 at 23:40
    
One natural thing to do is to take $S$ to have two punctures rather than boundary components, in which case you can't realize $\delta$ by a geodesic on a finite-volume complete hyperbolic surface. Does this resolve your question? Any theorem like this can be corrected by incanting the appropriate spell ("essential", or "not boundary-homotopic", etc.) once you know what the result should be. Can you clarify what you want the theorem to say? Are you looking for a version of "reducible <=> fixes a nontrivial curve" or "curve nontrivial <=> geodesic representative"? –  Tom Church Oct 26 '10 at 1:13
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1 Answer 1

First off, the mapping class group of a surface with boundary is generally taken to mean the group of diffeomorphisms that fix the boundary, up to isotopies fixing the boundary. In this context, a Dehn twist around one boundary component is not isotopic to the identity. What you're describing is usually called the mapping class group of a surface with punctures.

Under many definitions (including Casson and Bleiler's) the identity automorphism is also reducible (as well as periodic). So I don't see a counterexample in what you write.

If you're asking, for which simple closed curves is the Dehn twist around that curve not the identity (and also not periodic), the answer is simple: it's those curves that do not surround a single puncture. These are the same as the curves that are homotopic to a geodesic in any hyperbolic representative with cusps at the punctures. (For most curves, you look at a shortest representative in its homotopy class and you automatically get a geodesic. For a curve that surrounds a puncture, there is no shortest representative: you can make the curve arbitrarily short by pushing it out the cusp.)

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