# level set of function f coincides with that of |▽f|,M splits?

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?

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## 1 Answer

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 $N$.

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