# Uniqueness of distance realizing geodesic in hyperbolic surface. [duplicate]

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Hyperbolic surfaces

Given a hyperbolic surface S with geodesic boundary. Let a and b be two distinct simple closed geodesic boundaries. Does there exist a unique distance realizing geodesic in S? (1) for S is a pair of pants. (2) S is any hyperbolic surface with boundary.

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## marked as duplicate by S. Carnahan♦May 2 '12 at 7:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If you want to refine your question, you should edit the first one. There is an "edit" link below the tags. – S. Carnahan May 2 '12 at 7:16

## 1 Answer

For the pants, yes. In general, no. To prove this for the pants, classify all geodesic arcs and just observe the result. There are many ways to find a "no" example in the general case; the first one that came to my mind was taking a double cover.

EDIT - I see that this is a near-duplicate of a closed question. You could improve your question by giving some motivation. Reading the FAQ will be very useful in writing questions that get good answers. In particular please see http://mathoverflow.net/faq#whatnot

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well, I have got an example of hyperbolic surface with boundary where more than one (at least two) distance realizing geodesics between two distinct geodesic boundaries will exist. I have a further question: Suppose p and q are two distance realizing geodesics between the boundary geodesics. Is it true that p and q are always disjoint? – Bidyut Sanki May 2 '12 at 8:26
They are always disjoint. This is proved by the usual "exchange and round-off" technique. See, if $p$ and $q$ intersect, then at the intersection point you can do a little cut and paste to get two new arcs $p'$ and $q'$, still connecting the same boundary components, and slightly shorter, which is a contradiction. – Sam Nead Jan 3 '14 at 11:00