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Let $M$ be a compact riemannian manifold with boundary $\partial M$. We have that $\partial M$ is also compact and I was able to show that there is some $a>0$ such that the map $F:[0,a]\times \partial M \to M$ given by $F(r,p)= \exp_p(r\nu(p))$ is a diffeomorphism onto its image, say $U$ (here $\nu$ is the inward unit normal vector to $\partial M$. So I want to pull back the metric from $U$ to $[0,a]\times \partial M$: $$ \tilde{g} (u,v)\dot{=}g(dF_{(r,p)}u,dF_{(r,p)}v)).$$

I want to prove that the new metric has the form $\tilde{g}=dr^2+g_r$ (this appears in the paper I'm reading), but I'm having some trouble with the computation.

What I've tried:

Each tangent vector in $T_{(r,p)}([0,a]\times \partial M)$ has the form $(s,u)$, with $s\in \Bbb{R}$ and $u\in T_p\partial M$. So, I THINK the expression $\tilde{g}=dr^2+g_r$ means (at the point $(r,p)$): $$\tilde{g}((s,u),(t,v))=st+g_r(u,v),$$ where each $g_r$ is a metric on $\partial M$.

Using the Gauß lemma, I came to, at $(r,p)$,

$$\tilde{g}((s,u),(t,v))=st+g(d(\exp_p)_{r\nu(p)}s\nu(p),dF_{(r,p)}(0,v))+g(dF_{(r,p)}(0,u)(d(\exp_p)_{r\nu(p)}t\nu(p))+g(dF_{(r,p)}(0,u),dF_{(r,p)}(0,v)).$$

The middle term $g(d(\exp_p)_{r\nu(p)}s\nu(p),dF_{(r,p)}(0,v))+g(dF_{(r,p)}(0,u)(d(\exp_p)_{r\nu(p)}t\nu(p))$ is smooth symmetric and bilinear on $((s,u),(t,v))$, but not necessarily positive definite. So, in order to this remaining term to be a metric, we need that $$g(d(\exp_p)_{r\nu(p)}s\nu(p),dF_{(r,p)}(0,v))+g(dF_{(r,p)}(0,u)(d(\exp_p)_{r\nu(p)}t\nu(p))=0,$$ or, by symmetry, $$g(d(\exp_p)_{r\nu(p)}s\nu(p),dF_{(r,p)}(0,v))=0,$$ for all $s\in\Bbb{R}$ and $v\in T_p\partial M$. So, we would have $$g_r(u,v)=g(dF_{(r,p)}(0,u),dF_{(r,p)}(0,v))$$ (that indeed depends only on $u$ and $v$).

The main difficulty is to manipulate the expression $dF_{(r,p)}(0,v)$, since $$dF_{(r,p)}(0,v)=\left.\frac{d}{dt}\right|_0 \exp_{\alpha(t)}(r\nu(\alpha(t))),$$ in which the base point is varying (here $\alpha(0)=p$ and $\alpha'(0)=v$).

Originally, I had posted this question without the context, but since I still hadn't any answer, I'm posting this new one, with the context (which makes possible other approaches that maybe avoids this difficulty).

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Is this answer correct?

As said in the question we only have to show that $$f_v(r)=g(d(\exp_p)_{r\nu(p)}\nu(p),dF_{(r,p)}(0,v))=0,$$ for every vector $v\in T_p\partial M$ and $0\leq r\leq a$.

We verify easily that $f_v(0)=0$. Now let's derivate $f_v$:

$$\frac{d}{dr}f_v(r)=g(\frac{D}{dr}\frac{d}{dr}(\exp_p(r\nu(p))), dF_{(r,p)}(0,v))+g(d(\exp_p)_{r\nu(p)}\nu(p),\frac{D}{dr}dF_{(r,p)}(0,v))=g(d(\exp_p)_{r\nu(p)}\nu(p),\frac{D}{dr}dF_{(r,p)}(0,v))=(\ast),$$

since $r\mapsto \exp_p(r\nu(p))$ is a geodesic (i.e. $\frac{D}{dr}\frac{d}{dr}(\exp_p(r\nu(p)))=0$).

On the other hand, if $t\mapsto(r,\alpha(t))$ is a curve with $(r,\alpha(0))=(r,p)$ and $\alpha'(0)=v$, we have that $$ \frac{D}{dr} dF_{(r,p)}(0,v)=\frac{D}{dr} \frac{d}{dt}|_0(F(r,\alpha(t)))=\frac{D}{dt}|_0\frac{d}{dr} F(r,\alpha(t))=\frac{D}{dt}|_0\frac{d}{dr}\exp_{\alpha(t)}(r\nu(\alpha(t)))=\frac{D}{dt}|_0 d(exp_{\alpha(t)})_{r\nu(\alpha(t))}\nu(\alpha(t))=\frac{D}{dt}|_0 X(t),$$

with $X(t):=d(exp_{\alpha(t)})_{r\nu(\alpha(t))}\nu(\alpha(t))$. Therefore,

$$(\ast)=g(X(0),\frac{D}{dt}|_0 X(t))=\frac{1}{2}\frac{d}{dt}|_0 g(X(t),X(t))=0,$$

since by the Gauß lemma we have $g(X(t),X(t))=g(\nu(\alpha(t)),\nu(\alpha(t)))=1$.

Therefore, $f_v(r)$ must be constant equal to $0$, and this holds for any $v$.

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  • $\begingroup$ Post this in MO and they will answer you thoroughly. $\endgroup$ – DeepSea Nov 29 '18 at 18:47

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