Mean curvature computed as $H=-\frac{1}{n}\mathsf{div}_\widetilde{g}\left(\frac{\mathsf{grad\ } F}{\lVert\mathsf{grad\ } F\rVert}\right)$

Question

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:

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.

Answer 1

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.

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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 $(**)$.

Answer 2

$\newcommand\tr{\operatorname{trace}}$My original explanation below did not require using the hint. However, if you want to derive the equation for mean curvature from the formula in (a), you do need the hint. The point is that $\widetilde\nabla^2 F$ is a symmetric $2$-tensor on $T_p\widetilde{M}$ but the second fundamental form is the symmetric $2$-tensor $h$ obtained by restricting $\widetilde\nabla^2 F$ to $T_pM \subset T_p\widetilde{M}$. If $(e_1, \dots, e_{n+1})$ is an orthonormal basis of $T_p\widetilde M$ such that $e_{n+1}$ is normal to $T_pM$, then $$ \tr_{\tilde g} \widetilde\nabla^2F = \sum_{i=1}^{n+1}\widetilde\nabla^2F(e_i,e_i)$$ but $$ H = \tr_g h = \sum_{k=1}^{n} h(e_k,e_k) = \sum_{k=1}^{n} \widetilde\nabla^2F(e_k,e_k)$$ The two need not be equal. You should apply the hint to $A = \widetilde\nabla^2F$.

Original explanation

The formula is a direct consequence of the following:

• The vector field $$N = \frac{\widetilde\nabla F}{|\widetilde\nabla F|},$$ restricted to $M$ is the Gauss map
• The second fundamental form is given by the covariant derivative of the Gauss map. Using your definition, if $X,Y$ are tangent to $M$, then \begin{align*} h(X,Y) &= \langle N,\nabla_XY\rangle\\ &= -\langle\widetilde\nabla_XN,Y\rangle\\ &= -\left\langle\widetilde\nabla_X\left(\frac{\widetilde\nabla F}{|\widetilde\nabla F|}\right),Y\right\rangle \end{align*}
• The mean curvature is the trace of $h$: If $(e_1, \dots, e_n)$ is an orthonormal basis of $T_pM$, then \begin{align*} H &= \sum_{i=1}^n h(e_i,e_i)\\ &= -\sum_{i=1}^n \left\langle\widetilde\nabla_{e_i}\left(\frac{\widetilde\nabla F}{|\widetilde\nabla F|}\right),e_i\right\rangle\\ &= -\operatorname{div}_g\left(\frac{\widetilde\nabla F}{|\widetilde\nabla F|}\right) \end{align*} Note that although I've written $\widetilde\nabla F$, the gradient of $F$ depends only on the metric, so it does not depend on on the connection. Since $g = \tilde g$ on $M$, the gradient is the same whether you use $g$ or $\tilde g$