Skip to main content
MathOverflow

Return to Question

added 7 characters in body
Source Link
Analysis Now
Analysis Now
  • 1.5k
  • 13
  • 25

This is a basic question, still I dare to ask :

Let Y be the Y-piece with geodesic boundaries A,B, C and ( if possible ) c the non simple geodesic from A to B intersecting itself at a point p. I want to prove that c can NOT be homotopic to the common perpendicular $\gamma $, which is the shortest geodesic joining A and B .Here is the answer I was thinking of, it would be great to have your opinion :

Since c intersects itself at p, lifting c in the universal cover $ \tilde{Y} $ of Y , we get two lifts $\tilde{c_1} $ and $\tilde{c_2}$ intersecting each other at say p~ and we also get two different lifts A~,A* and B~, B* such that end points of $\tilde{c_1}$ are on A~, B~ and endpoints of c2~ are on A*,B* respectively.

My doubt is : why is it not possible that $ \tilde{A} = \tilde{B} $ and/or A*= B* ? ]

Now lift the homotopy between c and the shortest ( simple ) geodesic ( common perpendicular to A and B ) g$\gamma$ to have two lifts of $\gamma$, say $\gamma_1$ with end points on $\tilde{A},\tilde{B} , \gamma_2 $ with end points on A*,B* respectively. But then $ \tilde{\gamma_1}, \tilde{\gamma_2} $ intersect each other transversally, and so $\gamma $ will not be simple, which is a contradiction.

This is a basic question, still I dare to ask :

Let Y be the Y-piece with geodesic boundaries A,B, C and ( if possible ) c the non simple geodesic from A to B intersecting itself at a point p. I want to prove that c can NOT be homotopic to the common perpendicular $\gamma $, which is the shortest geodesic joining A and B .Here is the answer I was thinking of, it would be great to have your opinion :

Since c intersects itself at p, lifting c in the universal cover $ \tilde{Y} $ of Y , we get two lifts $\tilde{c_1} $ and $\tilde{c_2}$ intersecting each other at say p~ and we also get two different lifts A~,A* and B~, B* such that end points of $\tilde{c_1}$ are on A~, B~ and endpoints of c2~ are on A*,B* respectively.

My doubt is : why is it not possible that $ \tilde{A} = \tilde{B} $ and/or A*= B* ? ]

Now lift the homotopy between c and the shortest ( simple ) geodesic ( common perpendicular to A and B ) g to have two lifts of $\gamma$, say $\gamma_1$ with end points on $\tilde{A},\tilde{B} , \gamma_2 $ with end points on A*,B* respectively. But then $ \tilde{\gamma_1}, \tilde{\gamma_2} $ intersect each other transversally, and so $\gamma $ will not be simple, which is a contradiction.

This is a basic question, still I dare to ask :

Let Y be the Y-piece with geodesic boundaries A,B, C and ( if possible ) c the non simple geodesic from A to B intersecting itself at a point p. I want to prove that c can NOT be homotopic to the common perpendicular $\gamma $, which is the shortest geodesic joining A and B .Here is the answer I was thinking of, it would be great to have your opinion :

Since c intersects itself at p, lifting c in the universal cover $ \tilde{Y} $ of Y , we get two lifts $\tilde{c_1} $ and $\tilde{c_2}$ intersecting each other at say p~ and we also get two different lifts A~,A* and B~, B* such that end points of $\tilde{c_1}$ are on A~, B~ and endpoints of c2~ are on A*,B* respectively.

My doubt is : why is it not possible that $ \tilde{A} = \tilde{B} $ and/or A*= B* ? ]

Now lift the homotopy between c and the shortest ( simple ) geodesic ( common perpendicular to A and B ) $\gamma$ to have two lifts of $\gamma$, say $\gamma_1$ with end points on $\tilde{A},\tilde{B} , \gamma_2 $ with end points on A*,B* respectively. But then $ \tilde{\gamma_1}, \tilde{\gamma_2} $ intersect each other transversally, and so $\gamma $ will not be simple, which is a contradiction.

Source Link
Analysis Now
Analysis Now
  • 1.5k
  • 13
  • 25

Why a non-simple geodesic in a Y-piece is NOT homotopic to a common perpendicular to the geodesic boundaries ?

This is a basic question, still I dare to ask :

Let Y be the Y-piece with geodesic boundaries A,B, C and ( if possible ) c the non simple geodesic from A to B intersecting itself at a point p. I want to prove that c can NOT be homotopic to the common perpendicular $\gamma $, which is the shortest geodesic joining A and B .Here is the answer I was thinking of, it would be great to have your opinion :

Since c intersects itself at p, lifting c in the universal cover $ \tilde{Y} $ of Y , we get two lifts $\tilde{c_1} $ and $\tilde{c_2}$ intersecting each other at say p~ and we also get two different lifts A~,A* and B~, B* such that end points of $\tilde{c_1}$ are on A~, B~ and endpoints of c2~ are on A*,B* respectively.

My doubt is : why is it not possible that $ \tilde{A} = \tilde{B} $ and/or A*= B* ? ]

Now lift the homotopy between c and the shortest ( simple ) geodesic ( common perpendicular to A and B ) g to have two lifts of $\gamma$, say $\gamma_1$ with end points on $\tilde{A},\tilde{B} , \gamma_2 $ with end points on A*,B* respectively. But then $ \tilde{\gamma_1}, \tilde{\gamma_2} $ intersect each other transversally, and so $\gamma $ will not be simple, which is a contradiction.