Integral of the square of the mean curvature of a torus in R^3 is at least 2 pi^2

Geometry and Topology 1 Minute

You need to know: Curvature of a curve, closed surface S immersed in {\mathbb R}^3, integration over S, torus.

Background: Let S be a smooth immersed surface in {\mathbb R}^3. For any point x\in S, we can build a vector u perpendicular to S, choose any plain P containing u (called a normal plain), and measure the curvature \kappa(x,P) at x of a curve which is the intersection of S and P. Let k_1(x) and k_2(x) be the minimum and maximum values of \kappa(x,P) over all choices of normal plain P. The mean curvature of S at x is H(x)=\frac{1}{2}(k_1(x)+k_2(x)). The Willmore energy of S is W(S)=\int_S H^2(x) dx.

The Theorem: On 27th February 2012, Fernando Marques and André Neves submitted to arxiv a paper in which they proved that W(T)\geq 2\pi^2 for every torus T immersed in {\mathbb R}^3.

Short context: The Willmore energy is a way to measure the “total curvature” of a surface. It has nice mathematical properties and appears naturally in some physical contexts. It is known that W(S)\geq 4\pi for every closed surface S, with equality if S is a sphere. In 1965, Willmore conjectured that stronger lower bound W(T)\geq 2\pi^2 holds for every torus T. The Theorem confirms this conjecture. The bound is the best possible, because there is a torus for which the equality holds.

Links: Free arxiv version of the original paper is here, journal version is here.

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Published by Bogdan Grechuk

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