# Why does it appear that Willmore energy is always zero?

The answer is "because I'm being sloppy," but the problem is I don't know exactly where I'm being sloppy. Here's my sloppy argument:

Let $M$ be a smooth compact surface without boundary in $\mathbb{R}^3$ and let $H$ be its mean curvature. If $\langle \cdot, \cdot \rangle$ denotes the $L^2$ inner product, then the Willmore energy can be expressed as $$W = \langle H, H \rangle.$$ Equivalently, since mean curvature can be expressed as $H = \nabla \cdot N$ where $N$ is the unit normal field, we have $$W = \langle \nabla \cdot N, H \rangle.$$ But by Stokes' theorem $$\langle \nabla \cdot N, H \rangle = -\langle N, \nabla H \rangle.$$ And since $\nabla H$ is always tangent to $M$, this inner product vanishes, i.e., the Willmore energy is always zero!

Where did I go wrong? There are several potential flaws -- I suspect that the basic problem is I'm not thinking correctly about how quantities get extended to the ambient space. But I'm having trouble putting my finger on the precise problem.

Thanks!

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Could you tell me exactly what you are quoting as Stokes' Theorem? Edit in a few lines from the book. – Will Jagy Nov 9 '12 at 20:36
Hey Will - no book, but here's how I derived it. Stokes' says $\int_M d\alpha = \int_{\partial M} \alpha$ for $\alpha$ an $(n-1)$-form on an $n$-manifold. Then $\int_M df \wedge \star \alpha = \int_M d(f\star\alpha)-fd\star\alpha=\int_{\partial M} f\star\alpha - \int_M fd\star\alpha$ and the boundary integral in the final expression vanishes because there is no boundary. Letting $f=H$ and $\alpha=N^\flat$ we get the statement made in the original post. (Recall that $\nabla \cdot N = \star d\star N^\flat$ and $\nabla H = (dH)^\sharp$.) – PolyKnowMeAll Nov 9 '12 at 20:43
So, what is $N^\flat?$ If it means the metric dual pairing one-forms and vector fields, that is for vector fields tangent to the (sub)manifold. – Will Jagy Nov 9 '12 at 20:49
Well, just do your exact calculation for the standard unit sphere, see what happens. – Will Jagy Nov 9 '12 at 20:55
This was a good exercise -- thanks! – PolyKnowMeAll Nov 15 '12 at 0:59

The mistake becomes clear once you ask yourself what "$\nabla \cdot N$" means. In other words, what is the divergence of the unit normal field? Since the surface is embedded in $\mathbb{R}^3$, one possibility is to consider the function $\phi$ representing the distance to $M$, and let $N = \nabla \phi$. (Since $M$ is smooth, $\nabla \phi$ will be well-defined in a sufficiently small neighborhood around $M$.) In other words, $N$ is no longer just the normal to the surface, but also gives the normals to every level set of the distance function.

At this point you're just working with vector fields on $\mathbb{R}^3$ and everything becomes easy. Most importantly, it becomes clear that $\nabla H$ is not tangent to $M$ -- consider for instance the unit sphere centered around the origin (as suggested by Will Jagy). For any point $p \in \mathbb{R}^3$ (excluding the origin) we have

$$N(p) = p / |p|,$$

i.e., the normal is just the normalized position. One can easily show that

$$\nabla \cdot N = 1/|p|,$$

or in other words, the mean curvature of a sphere is equal to the reciprocal of its radius, $r$. Sounds pretty good. From there we have

$$\nabla H = -p/|p|^3 = -N/r^2,$$

which means that the gradient of mean curvature is in fact parallel to the normal in this case. The Willmore energy of a sphere of radius $r$ would then be

$$W = \langle H, H \rangle = \langle \nabla \cdot N, H \rangle = -\langle N, \nabla H \rangle = \frac{1}{r^2} \langle N, N \rangle = \frac{4\pi r^2}{r^2} = 4\pi.$$

Good thing, because Willmore energy is supposed to be scale-invariant and equal to $4\pi$ for (round) spheres.

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+1 Great answer to your own question, with all the details. This is what this site is all about. – Jesse Madnick Nov 15 '12 at 2:09