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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 -- one must remember that in two dimensions a given 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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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 -- one must remember that in two dimensions a given 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 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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# Tweetable way to see that Willmore energy is Möbius invariant?

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!