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Consider a compact orientable Riemannian manifold $M$ (without boundary) isometrically immersed into $\mathbb{R}^3$. The Willmore energy of $M$ is the functional

$$\mathcal{W} = \int_M H^2 dA$$

where $H$ is the induced mean curvature. I am looking for a short, sweet, and utterly convincing (though not necessarily utterly formal) way of demonstrating that the Willmore energy is Möbius invariant. Of course, $H$ itself is rigid motion invariant, so the only thing that really needs to be explained is scale invariance and invariance w.r.t. sphere inversions.

My oddball way of seeing scale invariance is that if you think of the surface as a conformal immersion $f:M \rightarrow \mathbb{R}^3$ then the Willmore energy is simply the (squared) $L^2$ norm of the so-called mean curvature half-density $H|df|$, where $|df|: TM \rightarrow \mathbb{R}; X \mapsto |df(X)|$ can be thought of as the (isotropic) length element. And since mean curvature times length is scale invariant, so is the Willmore energy. But that's an oddball way of seeing things, and certainly not the simplest explanation.

As for sphere inversions, my only thought is that inversions are reflections in hyperbolic geometry. And reflections are isometries... So I have a sneaking suspicion that hyperbolic space provides a cute explanation -- perhaps for Möbius invariance on the whole -- but such an explanation eludes me.

Other perspectives are, of course, very welcome!

Update: it is tempting to try to show that the Willmore energy is more generally conformally invariant, but this statement is not true -- in two dimensions conformal structure is much more flexible than in dimensions three or higher. In particular, given a smooth surface $M$ equipped with a conformal structure there are many immersions $f: M \rightarrow \mathbb{R}^3$ such that the induced metric is compatible with the conformal structure, and not all of these immersions will have the same Willmore energy. A concrete example is the Dirac spheres, which are conformal immersions of $S^2$ with progressively larger constant mean curvature-half density, hence progressively larger Willmore energy (some pictures here, unfortunately low-resolution). But since there is only one conformal structure on $S^2$, $\mathcal{W}$ cannot be conformally invariant.

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Scaling invariance is obvious: Mean curvature scales like 1/length, and the area of a surface scales like length squared... – Robert Haslhofer Jun 14 '12 at 18:06
Good point! Thanks. – TerronaBell Jun 14 '12 at 18:45

The Willmore energy $\mathcal{W} = \int_M H^2 dA$ differs from the functional $$\widetilde{\mathcal{W}} = \int_M (H^2-K) dA$$ just by a constant as one can see from the Gauss - Bonnet theorem ($K$ here is the Gaussian curvature of $M$).

The expression $H^2-K$ in $\widetilde{\mathcal{W}}$ is the half of the square of the length of the trace-free part of the second fundamental form which is a (pointwise) conformally invariant density of conformal weight $-2$, while "dA" can be seen as a density with conformal weight $2$, so the entire integrand $(H^2-K) dA$ is independent of a choice of a metric.

Thus $\widetilde{\mathcal{W}}$ is manifestly conformally, and in particular, Möbius invariant. So is $\mathcal{W}$.

(A Liouville's theorem ensures that conformal maps of $\mathbb{R}^n$, $n\ge 3$, are restrictions of Möbius transformations.)

Edit. The above is an attempt to address the original request for a "tweetable" argument.

Of course, the precise statement is that the Willmore energy is conformally invariant with respect to the conformal transformations of the ambient space. (Otherwise we would not be able to invoke the Liouville's theorem).

The correct definition of the Willmore energy involves an immersion $f\colon M \rightarrow \mathbb{R}^3$ and the induced conformal structure on the immersed manifold. The Dirac spheres show, in particular, that the Willmore energy does depend on the immersion.

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Something is not quite right here - you would claim that Willmore energy is not only Möbius invariant but actually conformally invariant? But that is not true - consider two different conformal immersions of the sphere. They will not in general have the same Willmore energy, but one is certainly a conformal transformation of the other since there is only one conformal structure on $S^2$. – TerronaBell Jun 14 '12 at 16:07
@fuzzytron, conformal transformations of $\mathbb R^3.$ These come from stereographic projection to $\mathbb S^3,$ some isometry there, then stereographic projection back. – Will Jagy Jun 15 '12 at 3:46
@Will, fuzzytron. It is even true for conformal changes of the metric, not only Moebius transformations, as was shown in "Weiner, J., (1978) On a problem of Chen, Willmore, et al., Indiana University Mathematical Journal." – Sebastian Jun 15 '12 at 20:20
The Willmore energy is not invariant under inversions centred at some point on the image -- this is well-known. Some energy is lost depending on density. Similarly, images which stretch off to infinity gain energy under inversion. – Glen Wheeler Jun 16 '12 at 22:08

There exist the notion of the mean curvature sphere for surfaces $f\colon M\to R^3$ in 3-space: for $p\in M$ the sphere $S(p)$ is defined to be the unique sphere which goes through $f(p),$ which has at $f(p)$ the same tangent space, and which has the same mean curvature. Then one can show that this notion is Moebius invariant. Moreover, the energy of the mean curvature sphere in the space of spheres is the exactly the Willmore functional. Of course, one has to do some computations for this, but the mean curvature sphere is clearly an important object in the field of Moebius invariant surface geometry.

Some Literature: Bryant: A duality theorem for Willmore surfaces. Journal of Dif- ferential Geometry 20, Burstal, Ferus, Leschke, Pedit, Pinkall: Conformal geometry of surfaces in S4. Lec- ture Notes in Mathematics 1772

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what do you mean by "the energy of the sphere in the space of spheres"? – Ian Agol Jun 14 '12 at 7:21
A conformal model of the 3-sphere is given by the space of null-lines in the Minkowski 5 space $L^5.$ Consider $Q\subset L^5$ the set of unit length vectors. This can be identifies with the space of (oriented, round) two spheres in $S^3$ as for $v\in Q,$ $Pv^\perp\subset S^3$ is sphere. Then equip $Q$ with its induced pseudo-Riemannian structure from $L^5.$ – Sebastian Jun 14 '12 at 7:46
Sorry, I was unclear: I should have written mean curvature sphere, considered as a map from $M$ to the space of 2-spheres. – Sebastian Jun 14 '12 at 7:49

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