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In R.C.Penner "Decorated Teichmuller theory of boarded surface", on Page 7 and 8, it says that (without proof) the Teichmuller space of surface with $s$ labelled punctures and $r$ labelled boundary components and one marked point on each boundary is homeomorphic to an open ball of dimension $6g-6+2s+4r$, where is the proof of that?I never see the version of T space that has marked points on the boundary, is also that they all homeomorphic to a ball like usual? Is there any reference about this?Thank you!

edited Sep 23 2010 at 22:47

asked Sep 23 2010 at 6:10

Hao
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Kevin Lin · Answer 1 · 2010-09-23 07:19:22Z UTC

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Just look up Teichmuller theory on the Internet; there are plenty of references. For example see these notes by Curtis McMullen.

answered Sep 23 2010 at 7:19

Kevin Lin
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@Kevin, I know Teichmuller space, but I don't see the one could have marked points on the boundary, what's the proof that it is a ball? – Hao Sep 23 2010 at 10:01

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Ordinary Teichmüller space is (6g-6)-dimensional. Each puncture contributes 2 dimensions, so that accounts for the 2s. A boundary circle comes from removing a disc: a disc has a center point (2 dimensions) and a radius (1 dimension). A marked point on a boundary circle contributes 1 more dimension. 4r = (2+1+1)r. – Kevin Lin Sep 23 2010 at 13:42

@Kevin, that is the dimension, but why topologically it is a ball?Any proof? – Hao Sep 23 2010 at 21:32

Take the proof you know for when there is no boundary, and then add boundary and marked points. It's the same proof. – Ryan Budney Sep 24 2010 at 2:48

A question in R.C.Penner’s paper about Teichmuller space

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