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I would like to know if it is possible to prove the Willmore Conjecture over the particular case of a revolution torus having the knowledge of a first course in differential geometry.

Suppose $0 < r < a$ and let $T(a,r)$ be a revolution torus made by rotating over the Z axis the circumference $\mathbb{S}^1(a,r)$ in the plane $\{y = 0\}$. There is an almost everywhere parameterization $X\colon ]0,2\pi[\times ]0,2\pi[ \to \mathbb{R}^3$, given by

$$ X(u,v) = ((a+r\cos u)\cos v, (a + r \cos u) \sin v, r \sin u). $$

So, the Willmore energy over $T(a,r)$ will be given by the expression

$$ \int_{T(a,r)}H^2 = \int_{]0,2\pi[\times ]0,2\pi[}(H^2 \circ X) |X_u \wedge X_v|, $$

where $H$ is the mean curvature function. We can calculate the coefficientes of the first fundamental form. $$ E(u,v) = |X_u|^2 = r^2 $$ $$ F(u,v) = \langle X_u, X_v \rangle = 0 $$ $$ G(u,v) = |X_v|^2 = (a + r\cos u)^2 $$

We call $\alpha(u,v) = \sqrt{EG-F^2}(u,v) = r(a+r\cos u)$. We can then calculate the coefficients of second fundamental form.

$$e(u,v) = \frac{1}{\alpha}\det(X_u,X_v,X_{uu})(u,v) = r$$ $$f(u,v) = \frac{1}{\alpha}\det(X_u,X_v,X_{uv})(u,v) = 0$$ $$g(u,v) = \frac{1}{\alpha}\det(X_u,X_v,X_{vv})(u,v) = \cos u(a + r\cos u).$$

We have then, $$ (H \circ X)(u,v) = \frac{1}{2}\frac{eG+gE-2fF}{EG-F^2}(u,v) = \frac{1}{2}\frac{r(a+r\cos u)^2+r^2\cos u (a + r \cos u)}{r^2 (a + r \cos u)^2} = \frac{a+2r\cos u}{r(a+r\cos u)}. $$

Since $(H^2 \circ X) = (H \circ X)^2$ and $|X_u \wedge X_v|^2 = |X_u|^2 |X_v|^2 - \langle X_u,X_v \rangle^2 = EG - F^2$, and then $|X_u \wedge X_v| = \alpha$, we finally obtain

$$ \int_{T(a,r)} H^2 = \int_{]0,2\pi[\times]0,2\pi[} \frac{(a + 2r\cos u)^2}{r(a+r\cos u)} d(u,v) = \frac{2\pi}{r}\int_{0}^{2\pi}\frac{(a + 2r\cos u)^2}{a+r\cos u} du .$$

I'm stuck at this point. I have tried the change $u = 2 \arctan t$, but I haven't been able to compute that integral. Is this way right to calculate Willmore energy for this kind of torus? If so, how can I continue? Thanks in advance.

utbutnut
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1 Answers1

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Use $a+2r\cos u = 2(a+ r\cos u) - a$, so

$$\frac{(a+ 2r\cos u)^2}{a+r\cos u}=4(a+r\cos u) -4a + \frac{a^2}{a+r\cos u}.$$

So

$$\int_0^{2\pi} \frac{(a+ 2r\cos u)^2}{a+r\cos u} du = a^2 \int_0^{2\pi} \frac{1}{a+r\cos u} du.$$

Using the tangent half angle substitution, we have

$$ \int_0^{2\pi} \frac{1}{a+r\cos u} du = \frac{\pi}{\sqrt{a^2- r^2}}.$$

So

$$\int_{T(a,r)} H^2 d\mu = \frac{8\pi a^2}{r\sqrt{a^2 - r^2}} = \frac{2\pi^2}{c\sqrt{1-c^2}},$$

where $c = r/a$. Using basic calulus, the minimum is attained at $c = 1/\sqrt 2$, and in that case the Willmore energy is $2\pi^2$.

It sounds naive, but the above calculation is exactly what Willmore did in 1965 and gave birth to the Willmore conjecture.