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I know that I can decompose an hyperbolic closed surface of genus $g>1$ into $2(g−1)$ pants bounded by $3$ geodesics. It seems reasonable to think the same can be done for a closed surface of genus $g>1$ carrying an arbitrary metric: i.e. to decompose it into $2(g−1)$ pieces which are all homeomorphic to a disc with two holes and bounded by geodesic. is any one have a good reference for that ? Thx

This question have been first posted on mathstack exchange, without any success...

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It's the same proof. Take a topological pants decomposition as before, and look for a minimal-length representative on your given Riemannian metric. Then you invoke the theorem that if you have a simple multi-curve without trivial or parallel components on a Riemannian surface, then it has a minimal length representative without intersections.

This is one of a number of theorems in surface topology that seem obvious but have a surprising number of subtleties. I think this one appears in a thesis by Shepard at UC Berkeley.

The main thing that fails with respect to the hyperbolic case is that the minimal length representative need not be unique.

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  • $\begingroup$ Thanks for the answer. Can you give me more references. Notably,about the theorem you invoke. it is not totally clear to me since an homology class is like a open cylinder, and without the negative curvature assumption, perhaps the minimum is not achieved? And I didn't also find a Shepard at UCB... Thanks again for your help!! $\endgroup$ – Paul Jun 24 '15 at 20:59
  • $\begingroup$ Homology doesn't really come in. I should have mentioned homotopy, not homology. Any non-trivial homotopy class has a minimal-length representative on any compact Riemannian manifold (not just surfaces). $\endgroup$ – Dylan Thurston Jun 25 '15 at 4:22
  • $\begingroup$ yes I mean homotopy. have you a reference for the pants decomposition? because I need minimizing representant without intersection... who is Shepard?Thx $\endgroup$ – Paul Jun 25 '15 at 4:56

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