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.

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    $\begingroup$ Is a $Y$-piece what's also known as a pair of pants? $\endgroup$ Jan 8, 2011 at 4:14
  • $\begingroup$ I believe that $Y$-piece is the classical name for this surface -- it has been completely superceded by Bill Thurston's "pair of pants" terminology. $\endgroup$
    – Igor Rivin
    Jan 8, 2011 at 4:42
  • $\begingroup$ I noticed that you tag it "hyperbolic-geometry"; if curvature $\le 0$ then it is true, and your argument works. If some components coincide then you get a triangle with sum of angles $>\pi$. $\endgroup$ Jan 8, 2011 at 4:51
  • $\begingroup$ Yes, my question assumed here that we are working with a hyperbolic metric ( curvature -1 ) on Y, which I probably should have mentioned . $\endgroup$ Jan 8, 2011 at 5:17
  • $\begingroup$ @Salp. Did you understand my the proof in my comment? $\endgroup$ Jan 8, 2011 at 21:12

2 Answers 2


Let $h$ be the "common perpendicular". Assume $c\sim h$ with ends on $A$ and $B$. Choose a homotopy and let $\alpha(t)$, $\beta(t)$ be trajectories of its ends (in $A$ and $B$ resp.). Pass to universal cover, let $\tilde\beta*\tilde c*\tilde\alpha^{-1}$ a joint $\beta*c*\alpha^{-1}$. Consider of geodesics $\tilde c_t$ from $\tilde\alpha(t)$ to $\tilde\beta(t)$. Let $c_t$ be its projection to Y-piece, it is a homotopy of geodesics from $h$ to a geodesic $c_1$ and $c_1(0)=c(0)$, $c_1(1)=c(1)$. Thus, $c\sim c_1$ rel ends. Since curvature is negative, $c\sim c_1$ rel ends implies $c=c_1$. (If doubt pass to universal cover.)

Note that, the number of self intersections of $c_t$ stays constant; i.e. $c_1=c$ is simple.


The thing about homotopies of curves is that they preserve the endpoints (on $S^1_\infty$) of the lifts. Whether or not two such lifts intersect depends only on the sign of the cross-ratio of the four points involved, whence the result.

  • $\begingroup$ Sorry, but I could not follow how that proves it is not possible that A˜=B˜ and/or A*= B* . Could you explain a bit more ? Was my argument for the answer correct at all ? $\endgroup$ Jan 8, 2011 at 5:34

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