# Laplacian on minimal surface

Suppose $\Sigma$ is minimal hypersurface in $R^n$, consider following function $f(x)=\frac{1}{2}r^2(x)$ where $r(x)$ is the distance to the origin, let $t$ be the distance to the hypersurface, so $$n=\Delta_{R^n}f=\frac{\partial^2f}{\partial^2 t}+\Delta_\Sigma f=1+\Delta_\Sigma f$$.

I understand first equality it is just usual laplacian calculation, I also know that for $k$ dimensional minimal surface, $\Delta_\Sigma (\frac{1}{2}r^2(x))=k$, so I guess he is trying to show that for hypersurface that $\Delta_\Sigma f=n-1$. But why do we have last two equalities?

## 1 Answer

It is a local calculation. Suppose $$p\in \Sigma$$, and $$\{e_1,\cdots,e_{n-1},N\}$$ is local O.N. basis of $$R^n$$ such that $$\{e_1,\cdots,e_{n-1}\}$$ is local O.N. basis of $$\Sigma$$, $$N$$ is unit normal vector of $$\Sigma$$. Note that if $$t$$ represents distance to $$\Sigma$$, then $$\frac{\partial}{\partial t}\big|_{p}=N_p$$. Thus by definition of Laplacian we have $$\begin{eqnarray*}\Delta_{R^n}f &=&\Delta_\Sigma f+H(f)+(\bar{\nabla}^2f)(N,N)\end{eqnarray*}=\Delta_\Sigma f+g_{R^n}(N,N)=\Delta_\Sigma f+1$$ where $$\bar{\nabla}$$ is connection in $$R^n$$, $$g_{R^n}$$ is the standard metric of $$R^n$$. Here $$H=0$$ since it is minimal.