Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

M is a complete Riemannian manifold,f is a function on M with no critical points.If the level set of f coincides with the level set of |▽f|,then M must be topologically the product RXN?

share|improve this question

1 Answer 1

The hypothesis implies that $|\nabla f|^2=F(f)$ for some function $F:R\to R$ which is never zero on the image of $f:M\to R$. Then, if $G:R\to R$ is a primitive of $1/\sqrt{F}$ we have $\nabla [G(f)]=\frac{\nabla f}{\sqrt{F(f)}}$, hence $\vert\nabla G(f)\vert=1$ and $G(f)=z:M\to R$ is a smooth solution of the "eikonal" equation on the whole $M$, hence $z$ (up to a constant) must be the "signed" distance function to a fixed sublevel $N$ of $f$. Using "adapted" coordinates $M$ given by $(z,x_1,\dots,x_n)$, where $\{x_i\}$ are coordinates on $N$, it is well defined a correspondence between the points of any two different sublevels which is a diffeomorphism and the metric on $M$ can be written as $g(s,x)=dz^2+g_s(x)$ where $g_s$ is the metric on the sublevel $\{z=s\}$. It is then clear that topologically $M$ is the product of $R$ with $N$.

share|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.