# The integral of the mean curvature vector over a closed immersed surface

Suppose we have a closed, orientable, smooth surface $\Sigma$ immersed smoothly in $\mathbb R^n$ via $f:\Sigma \rightarrow \mathbb R^n$. Impose a Riemannian structure on $\Sigma$ by taking $g_{ij} = \partial_if\cdot\partial_jf$, the metric induced on $\Sigma$ by the immersion $f$. The inner product here is just the usual inner product from $\mathbb R^n$.

The mean curvature vector is $$\vec H = \Delta f,$$ where $\Delta$ is the Laplace-Beltrami operator on $(\Sigma,g)$.

Consider the integral of the mean curvature vector over the surface $\Sigma$: $$\int_\Sigma \vec H\ d\mu.$$ It seems rather plausible that this ought to be zero in the case where $\Sigma$ is closed, embedded, and has only one codimension. Is this known? Is it easy to prove?

If it is not zero in the generality above, as a surface immersed in $\mathbb R^n$, is it equal to some expression involving topological information of $\Sigma$?

• I found this (paywalled) just from googling mean curvature vector. – anon Aug 25 '11 at 9:01
• @anon My university doesn't have a subscription to that journal, and it would be nice to see what kind of conditions they impose on $f$ and $\Sigma$ to obtain their result. – Glen Wheeler Aug 25 '11 at 9:04
• The preview shows that the authors impose the conditions closed, orientable, and $C^2$, but the paper only considers surfaces in $\mathrm{R}^3$. – anon Aug 25 '11 at 9:07
• @anon Since I can't see the proof, it may also be that they are assuming the surface is embedded and has genus 0. Sometimes this went without saying. – Glen Wheeler Aug 25 '11 at 9:09
• What do you mean by the phrase "Laplace-Beltrami operator on $f$"? $f$ is a function, not a Riemannian manifold. And the domain of $f$ is $\Sigma$, which you did not assume to have a Riemannian structure. – Willie Wong Aug 25 '11 at 11:22

The result is in fact true for arbitrary codimensions. See Lemma 2.1 in this paper. A very quick proof taken from that paper:

Let $\Sigma \subset \mathbb{R}^n$ be some immersed submanifold. Let $X$ be a vector field on $\mathbb{R}^n$. We can decompose locally $X = X_t + X_n$ the tangential and normal componens to $\Sigma$. By definition we have that for $Y$ tangent to $\Sigma$

$$\partial_Y X_t = \nabla_Y X_t + h(X_t,Y)$$

and

$$\partial_Y X_n = - A_{X_n}(Y) + \nabla^\perp_Y X_n$$

where $\nabla$ is the induced Levi-Civita connection, and $\nabla^\perp$ is the induced normal connection. $h$ is the second fundamental form and $A$ is the Weingarten map associated to $X_n$: $\langle Z, A_{X_n}(Y)\rangle = \langle - h(Y,Z), X_n\rangle$

Suppose $X$ is a parellel vector field on $\mathbb{R}^n$. $\partial_Y X = 0$. This implies that $\nabla_Y X_t = - A_{X_n}(Y)$. Using the definition of the Weingarten map, we have that the $g$-trace of $\nabla X_t = \operatorname{div} X_t$ is equal to $\langle H, X\rangle$. So we have that if $\Sigma$ is a closed manifold, by the divergence theorem, $\int_\Sigma \langle H,X\rangle d\mu = 0$ if $X$ is a parallel, hence constant, vector field on $\mathbb{R}^n$.

One could also note the following: while the notion of $\int_\Sigma H d\mu$ is not well-defined for $\Sigma$ isometrically immersed in an arbitrary Riemannian manifold $M$, because there is no canonical vector space in which the $H$, evaluated at different points in $\Sigma$, all live. But if instead we consider the version where instead we treat $\int_\Sigma \langle H,X\rangle d\mu$, we see that for any $X$ a vector field in $M$ defined along $\Sigma$ such that $X$ is parallel along $\Sigma$, we have the same conclusion.

• Hi Willie, thanks for the answer. Could you elaborate a little more on why there is no canonical vector space in which the mean curvature of a submanifold lives? I would have naively thought the Euclidean analogue works under mild conditions on the ambient space. – Glen Wheeler Aug 26 '11 at 8:49
• @Glen: the same reason that there is no canonical vector space in which the tangent vectors to an abstract Riemannian manifold lives. A choice of decomposition $TM = M \times V$ for some vector space $V$ would imply that (a) $M$ is parallelisable and (b) it admits a flat connection. To make sense of differential and integration, you need that this connection agree with the Levi-Civita one. Now, of course you don't need the whole $TM$ for there to be a canonical vector space associated to the normal bundle of $\Sigma$, but you'd need at very least sufficiently large parallel subbundle of $TM$. – Willie Wong Aug 26 '11 at 11:26
• In other words, you need a distribution $E\subset TM$ such that $E$ contains the normal bundle of $\Sigma$, and that $E$ is invariant under parallel transport in $M$ along $\Sigma$ with trivial holonomy. This puts a restriction on $M$ and $\Sigma$ that I wouldn't necessarily call "mild". As an illustration: for the case of an orientable hypersurface, the normal bundle is of course trivialisable. However, it is parallel only when the second fundamental form vanishes. – Willie Wong Aug 26 '11 at 11:40
• Thanks for the reply. I'll have to think a bit more before I can digest this fully. – Glen Wheeler Aug 26 '11 at 12:39

For immersions into $R^3$ (rather than arbitrary codimension) there is basically a one-line proof:

$$2\int_\Sigma HN dA = \int_\Sigma df \wedge dN = \int_\Sigma d(fdN) = \int_{\partial\Sigma = \emptyset} f dN = 0.$$

Here $N$ is the Gauss map associated with the immersion $f: M \to \mathbb{R}^3$. The relationship $df \wedge dN = 2HNdA$ is also not hard: let $\kappa_i$ be the curvatures associated with principal directions $X_i$, so that $dN(X_i)=\kappa_i df(X_i)$. Then

$$df \wedge dN(X_1,X_2) = df(X_1) \times dN(X_2) - df(X_2) \times dN(X_1) = (\kappa_1 + \kappa_2) df(X_1) \times df(X_2) = 2HNdA(X_1,X_2).$$

(Note that here the wedge product uses the cross product $\times$ to define an algebra on $\mathbb{R}^3$, i.e., $\alpha \wedge \beta(X,Y) := \alpha(X) \times \beta(Y) - \alpha(Y) \times \beta(X)$.)

• An even simpler way: constants are not in the kernel of the Laplacian; just apply the divergence theorem: $\int_\Sigma \vec{H}\ dA = \int_\Sigma \Delta f\ dA = \int_\Sigma \nabla \cdot \nabla f\ dA = \int_{\partial\Sigma = \emptyset} N \cdot \nabla f\ dA = 0.$ – Omega Tree Aug 10 '17 at 2:25
• Yes, I didn't come back to this question but you are right, it does follow very quickly from Stokes. – Glen Wheeler May 1 '18 at 14:23