Laplacian of a restricted ambient function (${\Delta }_{\mathrm{geom}}=\mathrm{div}\mathrm{grad}$)

Use the divergence-of-gradient convention ${\Delta }_{\mathrm{geom}}=\mathrm{div}\mathrm{grad}$. For an embedded submanifold $M\subset {\mathbb{R}}^{n+m}$ and a smooth ambient function $f$,where $\mathbf{H}=\mathrm{tr}II$ is the vector-valued mean curvature.

At a point, choose a local orthonormal frame whose intrinsic covariant derivatives vanish there. Differentiate $f{|}_M$ twice and insert the Gauss formula. The tangent derivative terms give the ambient Hessian matrix trace, and the normal terms sum to $⟨\overline{\nabla }f,\mathbf{H}⟩$.

For a coordinate function $x^j$, the ambient Hessian matrix is zero, so ${\Delta }_{\mathrm{geom},M}{x}^j=H^j$. coordinates of a minimal submanifold are therefore harmonic. On the round unit sphere, $\mathbf{H}=-nX$, giving ${\Delta }_{\mathrm{geom}}{x}^j=-n{x}^j$. The nonnegative Laplace-Beltrami operator uses the opposite sign; declaring the convention fixes the signs in these formulas.

For the squared distance $r^2(x)=|x{|}^2$, a minimal $n$-submanifold satisfies ${\Delta }_M{r}^2=2n$. On a compact boundaryless manifold, $r^2$ attains a maximum where the Laplacian is nonpositive, a contradiction for $n>0$.