TITLE: Gradient vector field and level sets
QUESTION [5 upvotes]: So assume we have a complete Riemannian manifold $M$, and $f\in C^\infty(M)$. Suppose that $|\nabla f|=1$. Then if we let $p\in f^{-1}(0)$ does that imply that $f(\exp_p(t\nabla f))=t$. I asked an earlier question regarding the Cheeger-Gromell splitting theorem and the Busemann function, and it more or less reduces to this question.

REPLY [7 votes]: Suppose $|\nabla f| = 1$. This implies
$$ g( \nabla f, \nabla f) = 1 $$
which implies
$$ \nabla g(\nabla f, \nabla f) = 0 $$
Using compatibility of the Levi-Civita connection we have (now using index notation to compress notation a bit)
$$ \nabla_a f \nabla_b \nabla^a f = 0 $$
Using that $f$ is scalar, and Levi-Civita is torsion free, we have that $\nabla_b\nabla_a f = \nabla_a \nabla_b f$. So we conclude finally
$$ \nabla_a f \nabla^a\nabla_b f = 0 $$
or in index-free notation
$$ \nabla_{\nabla f} \nabla f = 0 $$
In particular the gradient vector field is geodesic. 
This means that the flow w.r.t. $\nabla f$ and the geodesic flow starting from $\nabla f(p)$ coincide. And hence it is true that $f(\exp_p (t\nabla f)) = t$.