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Let $M$ be an Riemannian manifold with boundary $\partial M$ and $e_n$ be a unit normal vector on $\partial M$. With respect to $e_n$, around a point $p$ on boundary, we have the usual normal coordinate $(x,t)$, where $x$ is the coordinate on $\partial M$ while $t$ is the coordinate for the geodesic $exp_p(te_n)$. Now, the question is whether we can assume in such coordinate, the metric has following form $g=dt^2+g^t_{\partial M}$, where $g^t_{\partial M}$ is just a familiy of metric on $\partial M$ depending on the normal coordinate $t$. It seems to be a folklore for people working on APS index theorem (non-product case). But I cannot find a proof for this result on any Riemannian geometry textbook. In fact, I suspect certain condition should be added for the existence of such coordinate. I think $(\nabla e_n)|_{\partial M}=0$ will work.

Edit: I make a rather simple mistake, the condition we need is $\langle \nabla_{\partial x_i} e_n, e_n \rangle =0$ for any local coordinate $x_i$ on $\partial M$, which is right clearly. I'm quite depressed that several months were wasted for such foolish mistake. For the people who are concerned, thank you all the same.

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  • $\begingroup$ You want geodesic normal or exponential coordinates relative to a hypersurface or submanifold. It should be in many texts. I would try Jost or Gallot-Hulin-Lafontaine. $\endgroup$
    – Deane Yang
    Apr 15, 2014 at 2:51
  • $\begingroup$ Usually, we only need existence of such geodesic normal coordinates which many books do mention. However, it is untrivial that the metric have the indicated form, especially we note that the claim is about the metric AROUND the boundary not just ON the boundary. @DeaneYang $\endgroup$
    – xiangsheng
    Apr 15, 2014 at 3:58
  • $\begingroup$ A normal coordinate chart (if it means that $\nabla g(p)=0$ for some $p\in\partial M$) exists only when the second fundamental form of $\partial M$ vanishes, that is what you mentioned in the last sentence. But "normal" sometimes does not indicate the meaning above, it only means "perpendicular". $\endgroup$ Apr 15, 2014 at 6:22
  • $\begingroup$ A correction: What I want to say is $\partial g(p)$ instead of $\nabla g(p)$ in the previous comment. $\endgroup$ Apr 15, 2014 at 15:56
  • $\begingroup$ As a special case, there are Fermi coordinates for submanifolds as discussed in chapter 2 of Gray's book Tubes. $\endgroup$
    – user7868
    Aug 30, 2021 at 8:27

1 Answer 1

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Let $d(x)$ be the distance from $x$ to $\partial M$ and for each $t \ge 0$ let $H_t = d^{-1}(t)$. If $M$ is compact with boundary, then for any $t$ sufficiently close to $0$, $H_t$ is a smooth hypersurface with metric $g_t$ and outer unit normal $\nabla d$. It follows from this that $g = dt^2 + g_t$. You can write $g_t$ in terms of inner products of Jacobi fields obtained by varying geodesics normal to $\partial M$. This is a great exercise in understanding the relationship between Jacobi fields and the metric tensor along a geodesic.

This is used and perhaps explained in this classic and important paper: Heintze, Ernst; Karcher, Hermann A general comparison theorem with applications to volume estimates for submanifolds. Ann. Sci. École Norm. Sup. (4) 11 (1978), no. 4, 451–470

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