User - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T14:35:17Z http://mathoverflow.net/feeds/user/18589 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92013/willmore-minimizers-for-genus-geq-2 Willmore minimizers for genus $\geq 2$ jo1o 2012-03-23T14:43:39Z 2012-11-29T16:19:25Z <p>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.</p> <p>If $\Sigma$ is closed we have the estimate $$\cal W(f) \geq 4 \pi$$ with equality only for $f$ parametrizing a round sphere. </p> <p>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. </p> <p>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 $\mathbb R ^3$?</p> http://mathoverflow.net/questions/101463/properties-of-natural-numbers-such-that-there-is-a-very-large-largest-number-wi Properties of natural numbers such that there is a "very large largest" number with that property jo1o 2012-07-06T07:51:46Z 2012-07-09T08:45:24Z <p>I'm looking for properties (P) such that you <strong>would assume</strong> that there are infinitely many natural numbers with property (P) (for example because there are very large numbers with that property) but where it turns out that there are only finitely many.</p> <p>EDIT: Note that the <a href="http://mathoverflow.net/questions/15444/the-phenomena-of-eventual-counterexamples" rel="nofollow">eventual counterexamples</a> question asks for (P) such that the <strong>smallest</strong> $n$ with property (P) is large; the current question asks for (P) such that the <strong>largest</strong> $n$ with property (P) is large.</p> http://mathoverflow.net/questions/85802/hopf-tori-in-s3 Hopf Tori in $S^3$ jo1o 2012-01-16T12:31:07Z 2012-01-18T08:09:30Z <p>By means of the Hopf fibration $\pi: S^3 \rightarrow S^2$, U. Pinkall showed that every compact Riemannian surface of genus one can be conformally embedded in $S^3$.</p> <p>More precisely: Let $p$ be a closed curve on $S^2$ of length $L$. Lifting it to $S^3$ yields a torus isometric to $R^2 / \Gamma$, with $\Gamma$ generated by $(2\pi, 0)$ and $(A/2, L/2)$, where $A$ is the area enclosed by $p$. (This is Proposition 1 in Pinkall's paper.) </p> <p>Now it's claimed that if you lift a great-circle you should get the Clifford-torus. The above proposition then yields that the Clifford-torus is isometric to $R^2 / \Gamma _1$, with $\Gamma _1$ generated by $(2\pi, 0)$ and $(\pi, \pi)$. The usual definition of the Clifford torus is $R^2 / \Gamma _c$ with $\Gamma _c$ generated by $(2\pi, 0 )$ and $(0, 2 \pi)$. </p> <p>Who know's how this fits together?</p> http://mathoverflow.net/questions/90579/does-the-space-stt-2-g-of-symmetric-tracefree-divergence-free-2-0/91363#91363 Comment by 2012-03-20T09:23:01Z 2012-03-20T09:23:01Z Let me know if this satisfies you. http://mathoverflow.net/questions/90579/does-the-space-stt-2-g-of-symmetric-tracefree-divergence-free-2-0/91363#91363 Comment by 2012-03-16T13:52:25Z 2012-03-16T13:52:25Z I wrote so shortly since I thought no one was interested in my question anyways. By fairly simple means the space of symmetric trace-free divergence-free 2-tensors can be identified with the holomorphic quadratic differentials on $\Sigma$, and their dimension is given by the Riemann-Roch-Theorem. For details you could have a look at Tromba's book &quot;Teichmueller Theory&quot;. Hence the dimension does not depend on $g$. (Actually the dimension is $2$ if $\Sigma$ is a genus-$1$-surface and $6p - 6$ for genus- $p &gt;1$-surfaces). Still, the dimension alone doesn't anwser the question. http://mathoverflow.net/questions/85802/hopf-tori-in-s3 Comment by 2012-01-18T08:14:16Z 2012-01-18T08:14:16Z Thanks for the advice, I corrected it. http://mathoverflow.net/questions/85802/hopf-tori-in-s3/85811#85811 Comment by 2012-01-16T15:30:28Z 2012-01-16T15:30:28Z Thanks for your answer. You are right about the factor $1/\sqrt{2}$, of course. I forgot about it since I'm actually just interested in the conformal type. Still, I don't see how a rotation by $\pi/4$ yields a isometry between $\R^2/ \Gamma _c$ and $\R^2 / \Gamma _1$, since $\Gamma _c$ is a square and $\Gamma _1$ is a parallelogram, hence the corresponding tori can't have the same conformal type. Where am I wrong?