degenerating surface - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:44:50Z http://mathoverflow.net/feeds/question/81325 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81325/degenerating-surface degenerating surface Paul 2011-11-19T08:42:03Z 2011-12-04T01:00:32Z <p>Hi, i have a sequence of immersed disc $u_n: \mathbb{D} \rightarrow \mathbb{R}^3$ which converge to a singular cover of the disc: $z^k$ for $k\geq 2$, moreprecisely $u_n \rightarrow z^k$ in $C^2(\mathbb{D})$. Of course the Gauss curvature of the image $\Sigma_n=u_n(\mathbb{B})$ blows up thanks to Gauss-Bonnet formula : $\int_{\Sigma_n}K= (1-k)2\pi + o(1)$. My questions are the following:</p> <p>1) can the Gauss curvature be bounded form above? i.e the blow-up come only from necks and there is no pinching region... my feeeling is no, since you have to close the surface.</p> <p>2) Extra bonus: same question with only a convergence in $C^2_{loc}(\mathbb{D}\setminus { 0})$, here we allow the closing of the surface be made by a big sphere for example.</p> http://mathoverflow.net/questions/81325/degenerating-surface/81504#81504 Answer by Vitali Kapovitch for degenerating surface Vitali Kapovitch 2011-11-21T14:37:43Z 2011-12-04T01:00:32Z <p>for question 1) it's not clear to me why $C^2$ approximating $z \mapsto z^k$ by immersions is possible at all. do you have an example of such a sequence?</p> <p>edit: Ok, I thought some more on the topological issue of whether it's always possible to deform $z \mapsto z^k$ to an immersion in $C^2$ and I can say that it is definitely <strong>NOT possible</strong> if $k$ is even. In particular, it's impossible when $k=2$. There is an easy necessary condition for an existence of such deformation. An immersion $f$ of a disk gives a parallelization of the tangent bundle along $f$, i.e we get a map $f'\colon D^2\to V_2(\mathbb R^3)$ where $V_2(\mathbb R^3)$ is the Stiefel manifold of orthonormal 2-frames in $\mathbb R^3$ which is of course just $SO(3)$. This map ought to extend the map on the boundary of the disk $S^1$ which being an immersion already won't change much under a small deformation. That map is essentially given by $z\to kz^{k-1}$ which disregarding the conformal factor $k$ can be thought of as a map $S^1\to SO(2)=V_2^{or}(\mathbb R^2)\subset V_2(\mathbb R^2)=O(2)$. The natural map $V_2(\mathbb R^2)\to V_2(\mathbb R^3)$ corresponds to the standard inclusion $SO(2)\to SO(3)$ on identity component. Now, $\pi_1(SO(2))\cong \mathbb Z$ and $\pi_1(SO(3))\cong \mathbb Z/2\mathbb Z$ and the map $\pi_1(SO(2))\to \pi_1(SO(3))$ is well-known to be onto (as is immediate from the long exact sequence for the fibration $SO(2)\to SO(3)\to S^2$).</p> <p>That means that the map $z\mapsto z^{k-1}$ gives a generator of $\pi_1(SO(3))$ when $k$ is even and thus can not be extended to $D^2$.</p> <p>(Note that this takes care of <strong>both</strong> 1) and 2) when $k$ is even.)</p> <p>I think the above necessary condition should also be sufficient and therefore when $k$ is odd such a deformation should always be possible (at least in $C^0$) as evidenced by $k=1$ of course. I'm not at all an expert on immersions but the subject is very well understood and I hope someone who knows more about it will chime in.</p> <p>OK, Robert Bryant answered this (see below). Moreover, a direct calculation shows that his example produces a family of immersions $f_t$ with with the induced metrics $ds_t^2$ having $sec\le 0$ for all $t>0$. This settles the original question in the positive for $k$ odd.</p> <p><strike> For question 2) you can not expect any upper curvature bounds. Given an immersion you can locally perturb it on an arbitrary small neighborhood of 0 by adding a small thin "finger" to your surface. this will introduce some arbitrary positive (and negative) curvature. Doing this along a given sequence converging in <code>$C^2_{loc}(\mathbb D\backslash \{0\})\cap C^0(\mathbb D)$</code> will keep such convergence.</strike></p> http://mathoverflow.net/questions/81325/degenerating-surface/81671#81671 Answer by Robert Bryant for degenerating surface Robert Bryant 2011-11-23T00:32:04Z 2011-11-23T01:00:35Z <p>When $k$ is odd, there does exist such a family of immersions satisfying Paul's requirements. (When $k$ is even, Vitali Kapovich has shown, using a clever topological argument, that it's not possible to have such a family of degenerating immersions. Please see his answer for the details.)</p> <p>Set $k=2m+1$, and consider the (complex) $1$-parameter family of maps $u_t:\mathbb{C}\to\mathbb{R}^3$ given by $$u_t(z) = \bigl(Re(z^{2m+1}-(2m{+}1)t^2z),\ Im(z^{2m+1}+(2m{+}1)t^2z),\ \tfrac{4m+2}{m+1} Re(t z^{m+1})\ \bigr).$$ These smooth maps converge smoothly to $u_0$ as $t\to0$, and $u_t$ induces the metric $$ds_t^2 = (2m{+}1)^2\bigl(|z|^{2m}+|t|^2\bigr)^2 |dz|^2.$$ Thus, $u_t$ is an immersion for $t\not=0$, while $u_0(z) = \bigl(Re(z^{2m+1}),\ Im(z^{2m+1}),\ 0\ \bigr)$.</p> <p>The family $u_t$ was constructed using the Weierstrass formula for minimal immersions, so the image $u_t({\mathbb{C}})\subset \mathbb{R}^3$ is an immersed minimal surface, and, as a result, the Gauss curvature is everywhere non-positive. In fact, for $t\not=0$, the Gauss curvature only vanishes at $z=0$, and then only when $m>1$. In particular, all of these degenerating immersions have curvature bounded above.</p>