I am reading a paper by James H. White which states that the Willmore functional is invariant under conformal mappings.

Let $M$ be a smooth compact surface and $f:M\to\mathbb{R}^3$ an immersion. The Willmore functional is given by $$\mathcal{W}(f)=\int_M\;H^2\;dA,$$ where $H$ is the mean curvature and $dA$ is the induced area form.

By Liouville's Theorem, all the conformal mappings can be written as combination of euclidean motions, homotheties and inversions in spheres. The fact that $\mathcal{W}$ is invariant under euclidean motions and homotheties is clear. So we have to see how $\mathcal{W}$ behaves under inversions.

Since, by Gauss-Bonnet, $$\int_M (H^2-K)dA=\int_M H^2 dA - 2\pi\chi(M)=\mathcal{W}(f) - 2\pi\chi(M)$$ and $\chi(M)$ is a geometric invariant, we can focus on the quantity $(H^2-K)dA$ and prove what is left for it.

So this is what I want to prove:

$(H^2-K)dA$ is invariant under inversions.

We can assume that the center of the inversion is $0$. If we denote the radius of the inversion by $r$, the position vector of a point on the inverted surface in terms of a point $x\in M$ is given by


First we observe that the surface form $dA$ changes under the inversion by $d\tilde{A}=\left({r\over\|x\|}\right)^4 dA$. This is easily done.

Now, the next step is to compute the principal curvatures of the inverted surface $\tilde{\kappa}_1, \tilde{\kappa}_2$ in terms of $\kappa_1, \kappa_2$. In the mentioned paper, they get the following:

$$\tilde{\kappa}_1=-{\|x\|^2\kappa_1-2h\over r^2},\;\tilde{\kappa}_2=-{\|x\|^2\kappa_2-2h\over r^2},$$

where we have set $h=x\cdot N$ with $N$ the surface normal vector.

With this calculation the proof is almost done, but I don't know how to get these values for the inverted curvatures.

My guess is that I should get first the inverted normal, say $\tilde{N}$, but my calculations get me to $\tilde{N}=\left({r\over\|x\|}\right)^4 N$, which I'm not sure is right. But even if this was the right value of $\tilde{N}$, how could I derive the principal curvatures?

Any help is really appreciated.



You must log in to answer this question.