# The riemannian metric of a neighborhood of the boundary of a compact manifold

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).

## 1 Answer

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 $$v$$.

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