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Given an isometric (in the Riemannian way) immersion $f:N\rightarrow M$ between complete, smooth riemannian manifolds, are there conditions on $M$, $N$, $f$, such that the normal exponential map $\mathrm{exp}^{\nu}:\nu(N)\rightarrow M$ is surjective?

I'm interested in the case of $f$ being not closed. An example of non surjectivity is given by $f:\mathbb{R}\rightarrow\mathbb{R}^2$, where f is the logarithmic spiral. In this case, the normal exponential map misses the origin.

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(1) in your example with logarithmic spiral $N$ is not complete, but you can build a spiral to make it complete. (2) It seems not sufficient to ask that you have a bound on second fundamental form; a similar spiral-like example can be build which approach a circle. (3) embedding is clearly enough, but it is not what you want. --- BUT what do you want...? – Anton Petrunin Jan 25 '10 at 5:48
(1-2) Thank you, i was actually thinking about a spiral approaching a circle, as you said. (3) First, i would like to know if there are papers about this topic. Secondly, i look at cases where $N$ doesn't "get closer and closer to itself", as in the spiral case. In other words, $N$ is a leaf of a Riemannian foliation, defined on an open set $V\supset N$. In particular, for any $x\in \overline{N}\cap V$, there is a neighbourhood $U$ such that the connected components of $N\cap U$ are costant distance apart. But since this riemannian foliation is not defined everywhere, it's hard to be precise.. – Marco Radeschi Jan 25 '10 at 14:26
actually, embedding is not enough, since the spiral approaching a circle is, in fact, embedded in the plane (the induced topology is the topology of the real line). I think that the "clear" condition, is $f$ closed. – Marco Radeschi Jan 25 '10 at 14:33
I think a stricter condition you may want to try for noncompact preimage N is that the embedding or immersion is a proper map, for Riemannian manifolds sometimes abbreviated "proper embedding" or "proper immersion." See and surprising… – Will Jagy Jan 25 '10 at 21:13

Here is a proof of Will Jagy's suggestion: $\def\dist{\operatorname{dist}}$

  • Let $M,N$ be connected complete smooth Riemannian manifolds, and let $f:N\to M$ be a proper smooth mapping. Then the normal exponential map is surjective.

Proof: Note that $f(N)$ is closed in $M$. For $y\in M$ choose $x_n\in N$ such that $\dist^N(y,f(x_n))< \dist^N(y,f(N))+\frac1n$. Then $x_n$ is in the inverse image under $f$ of the closed geodesic ball with radius 2 centered at $f(x_1)$, which is a compact set in $N$. Thus $(x_n)$ contains a subsequence which converges to $x\in N$. Then $\dist^N(y,f(x))=\dist(y,f(N))$. Choose a minimal geodesic from $f(x)$ to $y$. It hits $f(N)$ orthogonally at $f(x)$, because otherwise one can shorten it locally by a broken geodesic using Gauss' lemma. For that note that $f$ is an embedding on a neighborhood of $x$. Thus the normal exponential map has $y$ in its image.

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Do you mean that $x_n$ is in the inverse image under $f$ of the closed geodesic ball about $y$ of radius $\mathrm{dist}(y, f(N)) + 2$? Because otherwise, one can construct a counterexample with a circle of radius $>1$ embedded in the plane. – Matthias Ludewig May 22 '13 at 20:34

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