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Oct 26, 2010 at 1:49 answer added Dylan Thurston timeline score: 5
Oct 26, 2010 at 1:34 history edited Dylan Thurston
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Oct 26, 2010 at 1:13 comment added Tom Church 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"?
Oct 25, 2010 at 23:40 comment added Ryan Budney 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.
Oct 25, 2010 at 23:28 comment added Mark Bell 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.
Oct 25, 2010 at 23:19 comment added HJRW Isn't that the definition of a reducible map?
Oct 25, 2010 at 23:07 history asked Mark Bell CC BY-SA 2.5