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$(M^n,g)$ is a compact $n$ dimensional manifold of negative curvature with n>2 . let $\alpha$ be a simple closed geodesic loop in $M$ based at a point $p$

1) will the geodesic in the free homotopy class of alpha be simple ?

2) can $\alpha$ be homotopic ( with respect to $p$ ) to a power of another closed curve at $p$

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in 1), do you mean $\alpha$ is a simple closed loop? –  Ian Agol Apr 14 '12 at 2:59
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Assuming Ian's reading of your question, take hyperbolic n-manifold $M$ containing a closed totally-geodesic subsurface $S$. Then most closed geodesics in $S$ are not simple but are are homotopic to simple loops in $M$ provided $n>2$. For part 3 answer is also negative for $n>2$. You should proofread your posts. –  Misha Apr 14 '12 at 4:34
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No, the question is still misstated, currently you are asking if A implies A. –  Misha Apr 14 '12 at 12:04

1 Answer 1

2) Yes $\alpha$ can be homotopic to power of another loop.

Choose a manifold with a closed geodesic $\gamma$, say 1-periodic. Assume that the curvature tensor along $\gamma$ is generic (this can be arranged by small perturbation).

If $n\ge 3$ then there is a Jacobi field $J(t)$, $t\in[0,2]$ such that $J(0)=J(2)$, but $J(t)\ne J(t+1)$ for any $t\in[0,1]$.

Take $p=\exp_{\gamma(0)}(\epsilon\cdot J(0))$ for small $\epsilon>0$. Then the geodesic loop $\alpha$ based at $p$ which goes around $\gamma$ twice will have no self-intersections.

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