Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My question is an exercise in Peter Petersen "Riemannian Geometry" Chapter 5 #10

Let $N \subset M$ be a submanifold of Riemannian manifold $(M,g)$.

(a) The distance from N to $x \in M$ is defined as $d(x,N) = \inf\{ d(x,y)\ |\ p \in N\}$. A unit speed curve $\sigma : [a,b] \to M$ with $\sigma(a) \in N,\sigma(a)$ and $l(\sigma) = d(x,N)$ is called a segment from $x$ to $N$. Show that $\sigma$ is also a segment from $N$ to any $\sigma(t),t<b$. Show that $\sigma'(a)$ is perpendicular to $N$.

(b) Show that if $N$ is a closed subspace of $M$ and $(M,g)$ is complete, then any point in $M$ can be joined to $N$ by segments.

(d) Show that $d(\dot \ ,N)$ is smooth on a neighborhood of $N$ and that the integral curves for its gradient are the geodesics that perpendicular to $N$.

Please give me a answer as complete as possible,. Thank you very much!

share|cite|improve this question
A hint for the part b can be: you can use the first variation formula for the energy function (see for instance DoCarmo or either Petersen). The variation to be considered is that wich have "final vector" tangent to the submanifold. The condition on minimal distance will be useful because a minimal of the energy makes the derivative of energy zero, and hence you can get "zero inner-products" in order to prove orthogonality. – matgaio Apr 21 '12 at 23:35

a) Assume $\sigma_{[a,t]}$ is not a segment from $\sigma(t)$ to $N$ for some $a < t < b$. Then $d(\sigma(t),N) < l(\sigma_{[a,t]})$. Then there exists some point $q \in N$ such that $d(q,\sigma(t) < l(\sigma_{[a,t]})$. So $$l(\sigma) = d(\sigma(b),N) \leq d(\sigma(b),q) \leq d(q,\sigma(t)) + d(\sigma(t),\sigma(b)) < l(\sigma_{[a,t]}) + l(\sigma_{[t,b]}) = l(\sigma),$$ a contradiction. Assume $\sigma'$ is not perpendicular to $N$. Then there is a smooth curve $\alpha$ in $N$ having an angle strictly less than $\pi/2$ to $\sigma$. It follows from the first variation formula that $\sigma$ is not the shortest connection from $\sigma(t)$ to $\alpha$ for $t$ close to $a$.

b) Let $p \in M$ and $x_n$ be a sequence in $N$ such that $r := d(N,p) = \lim_{n \to \infty}d(x_n,p).$ Then $x_n \in \overline{B}_{r + 1}(p)$ for all $n \geq n_o$ and some $n_0$. Since closed balls are compact there exists some convergent subsequence, which will be called $x_n$ to. So let $x_n \to x$. Choose shortest arc length geodesics $\gamma_n$ from $p$ to $x_n$. Since $M$ is complete there exists a uniformly convergent subsequence $\gamma_{\tilde n} \to \gamma$ (one needs Hopf Rinow or Arzela Ascoli here). Since $l(\gamma) = \lim l(\gamma_n)$ this $\gamma$ will be a segment.

c) This needs one well known fact in order to be easy: For a smooth submanifold $N$ there exists a neighborhood $U$ of the zero section of the normal bundle such that the normal exponential map, i call it just $\exp$, is a diffeomorphism onto it's image. One may also assume that $M$ is complete since the questions are local. Now let $p \in U$ and $\gamma$ be a shortest connection from $p$ to $N$. Then this segment is perpendicular to $N$, so it is of the form $t \mapsto \exp(tv)$ for some vector normal to $N$. Since $\exp$ is injective this segment is unique. To figure out the gradient without calculation one can argue like this: For the gradient $X$ of a distance function $f$ one always has $||X|| = 1$. Since the level sets are the sets of constant distance one has $df(v) = 0$ for all $v$ tangent to a level set. And by the above one has $df(\dot \gamma (t)) = 1$ for an arc length segment $\gamma$. Since the level sets are of codimension $1$ we see that $\dot \gamma$ must equal the gradient.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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