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Jul 18, 2016 at 21:24 history edited Asaf Shachar CC BY-SA 3.0
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Jun 15, 2016 at 13:48 comment added Willie Wong ... the tangential components are determined by $J(0)$ (which is tangent) by the shape operator of $S$).
Jun 15, 2016 at 13:46 comment added Willie Wong from $S$, based at $\tilde{s}(x)$ with appropriate velocity, has surjective differential, and use the local inverse function theorem. Since now the "constrained" surface $S$ is on the side of the initial data,, and not the final data, you avoid the free boundary problem. And now at the initial data level you have the right degree of freedom: $v \in T_{\tilde{s}} S$ has $\dim(S)$, and $J'(0)$ has $\dim(M) - \dim(S)$ after using the constraint that your Jacobi field corresponds to a family of geodesics orthogonal to $S$. (You can pick the components of $J'(0)$ orthogonal to $S$ freely, but...
Jun 15, 2016 at 13:40 comment added Willie Wong The problem with $D_t J_v$ is real, however, because you have to choose it so that the perturbed geodesic is the minimizing geodesic from the nearby point to $S$. Deane mentioned Sturm-Liouville in his comments: a slight problem is that the endpoint "1" is not always "1"; you have a bit of a free boundary issue there due to what I wrote in my previous comment. These can all be sorted out using some sort of inverse function theorem argument. But if you are going that way you might as well just directly prove that for every $x\in M$ and $\tilde{s}(x)\in S$, that the normal exponential map..
Jun 15, 2016 at 13:35 comment added Willie Wong A few comments. First en.wikipedia.org/wiki/Jacobi_field#Solving_the_Jacobi_equation may help. For the argument to work you probably need $v$ to be orthogonal to the minimizing geodesic to ensure $J(1)$ is parallel to $S$. (Think about the case where $\alpha$ is equal to the minimizing geodesic.)
Jun 15, 2016 at 2:39 comment added Deane Yang I'll take a look at your answer and might try to rewrite my answer with more details.
Jun 14, 2016 at 21:30 history answered Asaf Shachar CC BY-SA 3.0