Timeline for Are metric isometries smooth at the boundary?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 8, 2016 at 9:34 | vote | accept | Asaf Shachar | ||
| Nov 7, 2016 at 16:37 | history | bounty ended | Asaf Shachar | ||
| Nov 6, 2016 at 1:57 | comment | added | Deane Yang | Asaf, I didn't say that carefully enough. Everything I said is for a manifold without boundary. It all applies, because you can extend the manifold and metric to an open Riemannian manifold containing $M$. | |
| Nov 5, 2016 at 19:43 | history | edited | Mizar | CC BY-SA 3.0 |
added 60 characters in body
|
| Nov 5, 2016 at 19:11 | comment | added | Asaf Shachar | @DeaneYang Thanks. I gues the only delicacy in the general case where $S$ is a submanifold of a manifold with boundary $M$, which intersects $\partial M$ is that the tubular neighbourhood of our submanifold "looks strange" around a point in $S \cap \partial M$. In our case $S=\partial M$ so the tubular neighbourhood looks locally like a closed half-ball. | |
| Nov 5, 2016 at 18:29 | comment | added | Deane Yang | The existence of a unique and smooth "exponential map" emanating from a submanifold and the fact that it is a diffeomorphism on a tubular neighborhood of the submanifold can be proved with essentially the same proof of the standard one. The one for a hypersurface has been used to prove isometric inequalities. It might be described in the books by Jost and Gallot-Hulin-Lafontaine. There is also a classic paper by Heintze and Karcher. | |
| Nov 5, 2016 at 18:22 | comment | added | Deane Yang | To prove uniqueness of the geodesic normal to the boundary, it suffices to extend the metric smoothly to a neighborhood of $\partial M$ and observe that there exists a unique geodesic through any point in $\partial M$ and normal to $\partial M$ in that open manifold. Since the geodesic is transversal to $\partial M$, there exists a sufficiently small interval starting from but not including that point, which lies in the interior of $M$. | |
| Nov 5, 2016 at 18:09 | comment | added | Mizar | (2) I have no reference, but the idea is that it suffices to prove $\alpha$ injective (by the inverse function theorem). You can cover $\overline{U}$ with finitely many open sets $V_i$ such that $\alpha$ is injective on $V_i\times [0,\epsilon]$ for a unique, very small $\epsilon>0$. Now, if $\alpha(y,s)=\alpha(y',t)$ and $\epsilon$ is very small, then $y$ is very close to $y'$, so they both lie in some $V_i$, i.e. $y=y'$ and $s=t$. | |
| Nov 5, 2016 at 18:07 | comment | added | Mizar | (1) By "geodesic" I mean a smooth curve $\gamma$ such that $\nabla_{\dot\gamma}\dot\gamma\equiv 0$. So a geodesic is uniquely determined by starting point and initial velocity (if you write the equation explicitly in a chart, you are solving a Cauchy problem). | |
| Nov 5, 2016 at 17:55 | comment | added | Asaf Shachar | 2) Do you have a refernce for the fact there is a small neighbourhood $U$, and $\epsilon$ such that $\alpha$ (that is the normal exponential map) is a diffeomorphism? I know this is true when there is no boundary, and what you have described sounds to me like a "half" of a tubular neighbourhood. Is there an analogue of theat theorem for here? | |
| Nov 5, 2016 at 17:54 | comment | added | Asaf Shachar | Thanks. I am trying to fill in some details which I am not sure about: 1) You implicitly assume there are unique geodesics emanating from the boundary and orthogonal to it. I guess the only situation of non-uniqeuness that can happen (i.e two different geodesics with identical initial velocity) is if the velocity is tangent to the bounday? Do you know a reference for this? | |
| Nov 5, 2016 at 16:46 | history | answered | Mizar | CC BY-SA 3.0 |