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 $(r,p)$,


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).


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 $v$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.