I'm trying to figure out how to prove the following claim:
Suppose that $S$ is the graph of a function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ and every plane containing the $z$-axis intersects $S$ along a geodesic of $S$. Suppose in addition that $f(0)=0$ and $\nabla f(0)=0$. Then, $\nabla f$ is a radial vector field.
I've tried to use some kind of Gauss lemma, in order to verify that the level sets of $f$ are circles, but I could not make any progress. if someone could help me, I'd appreciate. This is not homework. Thanks in advance.
The most direct (and tedious) way to do this is probably by computing the Christoffel symbols, etc, but here's a geometric argument. I'll use cylindrical coordinates $(r,\theta,z)$ and the fact that the covariant derivative in a submanifold is just the orthogonal projection of the ambient covariant derivative.
If the radial curve $\gamma_\theta : t \mapsto (r(t), \theta, f(r(t),\theta))$ is a geodesic, then its covariant acceleration in $S$ is zero; so its acceleration in $\mathbb R^3$ must be orthogonal to $S$ - i.e. $T_p S = \ddot \gamma_\theta |_p^\perp$ whenever $p$ is on $\gamma_\theta$. Since this curve stays in the plane given by the constant $\theta$, its acceleration must have no $\theta$ component - that is $\langle \partial_\theta, \ddot \gamma_\theta\rangle = 0$, and so $\partial_\theta \in T_p S$. (I used here that the cylindrical coordinate system has orthogonal coordinate vectors.)
This implies $\nabla f$ has no component in the $\theta$ direction - one way to see this formally is noting that $S$ is a level set of $z - f(x,y)$ and thus $T_p S$ is the set of vectors $v$ such that $\langle v , \partial_z - \nabla f\rangle = 0$. Thus $\partial_\theta \in T_pS \implies \langle \nabla f , \partial_\theta \rangle = 0$.
Since every point on $S$ lies on some radial curve, this shows $\nabla f$ is everywhere radial.