# Hessian of the square of the distance function

Let $(M,g)$ be a Riemannian manifold and $\Sigma$ be a compact smooth hypersurface in $M$. I would like to know what is the geometric meaning of the eigenvalues of the Hessian of the square of the distance function $\nabla^2 dist^2(x,\Sigma)$ for $x$ in a small tubular neighborhood of $\Sigma$. Can we relate these eigenvalues or bounds on them to some extrinsic and intrinsic curvatures of $\Sigma$ e.g. principal curvature, mean curvature, Ricci curvature etc.

## 1 Answer

If $\Sigma$ is two-sided (has trivial normal bundle), it is convenient to instead study the signed distance function $\delta$ which is plus or minus the distance to $\Sigma$ depending on which side of $\Sigma$ you're on. If $\Sigma$ is one-sided then you can construct such a $\delta$ locally.

The advantage of using this function is that the normal vector field to $\Sigma$ is simply $\nabla \delta$, and thus the second fundamental form ${\rm II}_\Sigma$ is the restriction of the Hessian $\nabla^2 \delta$ to $T \Sigma$; so the eigenvalues of $\nabla^2 \delta |_{T\Sigma}$ are exactly the principal curvatures of $\Sigma$.

If you really do want to study $f=\delta^2$ instead, then we have $\nabla^2f = 2(\nabla \delta \otimes \nabla \delta + \delta\ \nabla^2 \delta)$; so for points close to (but not on) $\Sigma$, restricting to the complement of $\nabla \delta$ will give you $\nabla^2 f|_{\nabla\delta^\perp} = {\rm \sqrt f\ II}$.