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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.


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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
up vote 1 down vote accepted

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

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