Differential Geometry – Mean Curvature Computed as $H=-\frac{1}{n}\mathsf{div}_\widetilde{g}\left(\frac{\mathsf{grad\ } F}{\lVert\mathsf{grad\ } F\rVert}\right)$

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I'd like to derive a formula for the mean curvature $H$ in John Lee's IRM book:
$$H=-\frac{1}{n}\mathsf{div}_\widetilde{g}\left(\frac{\mathsf{grad\ } F}{\lVert\mathsf{grad\ } F\rVert}\right),$$
where $(M,g)$ is a Riemannian hypersurface in a Riemannian manifold $(\widetilde{M},\widetilde{g})$ and $F$ is a local defining function for $M$.

The classical version of a derivation can be found in Prove the curvature of a level set equals divergence of the normalized gradient, and I'm having a hard time generalizing it to the Riemannian version as above.

My definition of $H$ would be that if $h$ is the scalar-valued second fundamental form given by $h(X,Y)=\langle N,\widetilde{\nabla}_X Y\rangle$ for vector fields $X,Y$ and a unit normal vector field $N$, then
$$H:=\frac{1}{n}\mathsf{tr}_g h.$$
And I know from a previous exercise that given a vector field $X$ on $M$,
$$\mathsf{div\ }X=\mathsf{tr}(\nabla X),$$
which seems to suggest I should express $h$ as the total covariant derivative of the normalized gradient. Does it make sense?

On the other hand, I was hinted that I should first prove a linear-algebra lemma:

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Obviously, it is suggesting the unit vector be replaced by the normalized gradient. But how could I possibly define the operator $A$ needed to apply the lemma?

Thank you.

Best Answer

Your formula for mean curvature is wrong; there should be an extra $-\frac{1}{n}$ on the right where $n:=\dim M$. For the hint:

  • you fix a point $p\in M$.
  • let $V=T_p\widetilde{M}$, and consider $A:T_p\widetilde{M}\to T_p\widetilde{M}$ given by $x\mapsto -\widetilde{\nabla}_xN$, where $N:=\frac{\text{grad } F}{\|\text{grad }F\|}$ is the unit normal to level sets of $F$. The reason this makes sense is that $N$ is defined on a neighborhood of $M$ in $\widetilde{M}$ (which by shrinking $\widetilde{M}$, you may assume $N$ is defined on all of $\widetilde{M}$). Note that if $N$ was only defined on $M$, then you would only be able to take covariant derivatives along tangent vectors $x\in T_pM$, not all $x\in T_p\widetilde{M}$. Notice that the restriction of $A$ to $T_pM\to T_pM$ is exactly the shape operator/Weingarten map.

Anyway, even if you didn’t know the hint, you could still easily ‘follow your nose’. For any orthonormal basis $\{e_1,\dots, e_n\}$ of $T_pM$, we have \begin{align} nH_p:=\text{trace}(A|_{T_pM})=\sum_{i=1}^n\langle A(e_i),e_i\rangle.\tag{$*$} \end{align} The first equal sign is definition, the second is very trivial linear algebra. Next, note that $\{e_1,\dots, e_n,N_p\}$ is an orthonormal basis for $T_p\widetilde{M}$, and that \begin{align} \langle A(N_p),N_p\rangle=-\langle D_NN,N\rangle(p)=-\frac{1}{2}D_N\bigg(\langle N,N\rangle\bigg)(p)=0,\tag{$**$} \end{align} since we’re differentiating the constant function $1$. So, we can add $0$ to $(*)$ in the form of $(**)$ to see that \begin{align} nH_p=\sum_{x\in\{e_1,\dots, e_n,N_p\}}\langle A(x),x\rangle=\text{trace}(A)=\text{trace}\left(-\widetilde{\nabla}_{(\cdot)}N\right)=:-\text{div}_{\widetilde{g}}N, \end{align} and thus $H=-\frac{1}{n}\text{div}_{\widetilde{g}}(N)$, as expected.

Barring complicated notation, observe that $99\%$ of this is trivial; the only mathematical content in this calculation is (a simple linear algebra fact about traces and) $(**)$ where we used metric compatibility of the connection, and that the normal $N$ is defined everywhere and has unit length, so its derivative along the normal direction also vanishes. Why is this a ‘follow-your nose’ proof? Simply because you write out the definition of $nH$ as the trace of the shape operator, and compare with the definition of divergence of $-N$, and you’ll see that there is potentially only one term which could mess things up; but a moments’ consideration shows this term vanishes $(**)$.

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