7
$\begingroup$

For an immersed closed surface $f: \Sigma \rightarrow \mathbb R^3$ the Willmore functional is defined as $$ \cal W(f) = \int _{\Sigma} \frac{1}{4} |\vec H|^2 d \mu_g, $$ where $\vec H$ is the mean curvature vector in $\mathbb R^3$and $g$ is the induced metric.

If $\Sigma$ is closed we have the estimate $$ \cal W(f) \geq 4 \pi $$ with equality only for $f$ parametrizing a round sphere.

Recently, the Willmore conjecture was proved (the paper can be found on arxiv), which states that for closed surfaces $\Sigma$ of genus $g \geq 1$ this estimate can be improved: $$ \cal W(f) \geq 2 \pi^2 $$ with equality only for the Cilfford torus.

Are there any conjectures about the minimizers in the case of genus $g \geq 2$? And what happens if we consider surfaces immersed in some $\mathbb R^n$ instead of $\mathbb R ^3$?

$\endgroup$
2
  • $\begingroup$ Can you give an explici citing of the archiv paper? $\endgroup$ Mar 23, 2012 at 18:39
  • 1
    $\begingroup$ It is the following paper by Marques and Neves: arXiv:1202.6036 $\endgroup$
    – Sebastian
    Mar 23, 2012 at 18:52

2 Answers 2

10
$\begingroup$

First of all, by a result of Bauer and Kuwert, there exists a smooth minimizer of the Willmore functional in the class of compact surfaces with fixed genus g, for any g. They have Willmore functional below $8\pi$ and by a result of Kuwert, Li and Schaetzle, the Willmore functional of the minimzers for genus $g$ tends to $8\pi$ when $g$ goes to infinity. Not much more is known about higher genus surfaces, but there is a vague conjecture, that the minimzers are the so called Lawson surface $\xi_{g,1}.$

$\endgroup$
1
3
$\begingroup$

I remember there is a paper by Kusner named: comparison surfaces for the Willmore problem in which the author conjectured that the Lawson surface(see Sebastian's answer) minimizes the Willmore energy of genus g surface. For surfaces immersed in R^n, it is also conjectured the Clifford torus should be the minimizer, but it seems to me that this is still an open question.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.