[Math] A smooth function f satisfies $\left|\operatorname{ grad}\ f \right|=1$ ,then the integral curves of $\operatorname{grad}\ f$ are geodesics

Question

$M$ is riemannian manifold, if a smooth function $f$ satisfies $\left| \operatorname{grad}\ f \right|=1,$ then prove the integral curves of $\operatorname{grad}\ f$ are geodesics.

Answer 1

I'll use $\nabla$ for the gradient.

If $|\nabla f| = 1$, we have that $g(\nabla f,\nabla f) = 1$ where $g$ is the metric. Taking the covariant derivatve of the expression you have

$$ 0 = \nabla (1) = \nabla\left( g(\nabla f,\nabla f)\right) = 2 g(\nabla f, \nabla^2 f) = 2 \nabla_{\nabla f} (\nabla f) $$

The third equality used that $\nabla g = 0$ for the Levi-Civita connection of a Riemannian metric, and the fourth inequality uses that the Hessian of a scalar function is symmetric.

Since $\nabla_{\nabla f} \nabla f = 0$, we have that the vector field $\nabla f$ is geodesic, and hence the integral curves are geodesic curves.