Why is the half-torus rigid? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T23:25:48Z http://mathoverflow.net/feeds/question/77760 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77760/why-is-the-half-torus-rigid Why is the half-torus rigid? Joseph O'Rourke 2011-10-10T23:01:04Z 2011-10-14T01:44:49Z <p>The half-torus surface that results from slicing a torus like a bagel, depicted below (left), is isometrically rigid. <br /> &nbsp; &nbsp; &nbsp; <img src="http://cs.smith.edu/~orourke/MathOverflow/HalfTorus.jpg" alt="HalfTori" /> <br /> I know this from a remark of Alexandrov in <a href="http://store.doverpublications.com/0486409163.html" rel="nofollow">Mathematics: Its Content, Methods and Meaning</a> (Chapter 7. Curves and surfaces, p.101):</p> <blockquote> <p>For example, it has been shown that a surface in the form of a circular trough (...), does not admit continuous deformations (this explains, among other things, the familiar fact that a pail with a curved rim is considerably stronger than one with a plain rim) ...</p> </blockquote> <p>He gives no hint of a proof, nor a reference. Could someone supply either? I would like to understand this enough to generalize to, e.g., the bottom quarter of a torus (above, right), or to cross-sections or sweep curves other than circles (ellipses, smooth convex curves, ...). Are there general conditions known for a surface in $\mathbb{R}^3$ with boundary to be isometrically rigid, i.e., to admit no continuous deformation that "preserves the length of all curves on the surface" (to quote Alexandrov's definition)? </p> http://mathoverflow.net/questions/77760/why-is-the-half-torus-rigid/77773#77773 Answer by Anton Petrunin for Why is the half-torus rigid? Anton Petrunin 2011-10-11T01:10:00Z 2011-10-14T01:44:49Z <p>I've got a letter from Idjad Sabitov which answer the question completely. Here is a short extract from it:</p> <ol> <li><p>half-torus has rigidity of second order (Rembs' theorem, see Е. Rembs. Verbiegungen hoeherer Ordnung und ebene Flaechenrinnen. Math. Zeitschrift 36 (1932) or <a href="http://www.mathnet.ru/links/24e21b3682b6a64bfabeced47dff4a16/rm8695.pdf" rel="nofollow">Ефимов, УМН, 1948, т.3, вып.2, стр. 135</a>)</p></li> <li><p>Any second order rigid surface does not admit <em>analytic deformation</em> (i.e., the deformation $h_t(u,v)$ which is analytic on $t$).</p></li> <li><p>For the surfaces of revolution, the assumption of analyticity can be removed. </p></li> </ol> <hr> <p><em>Below is the best part of my original post. It contains an idea which was not used by Rembs.</em></p> <hr> <p>Let $h(u,v)$ be a small perturbation of the standard embedding; $u\in (-\varepsilon,\varepsilon)$ and $v\in\mathbb S^1$.</p> <p>Consider convex hull $K$ of $\mathop{\rm Im}h$ and look at the closed curve $\gamma_0$ which is formed by boundary of $\partial K\cap \mathop{\rm Im}h$. I claim that $\gamma_0=h(0,{*})$ i.e. the Gauss curvature at points of $\gamma_0$ has to be $0$. Indeed since $\gamma_0$ lies on convex part, the Gauss curvature at the points of $\gamma_0$ has to be nonnegative. On the other hand $\gamma_0$ bounds a flat disc in $\partial K$; therefore its integral intrinsic curvature (in $\partial K$ and in the torus) has to be $2{\cdot}\pi$. If the Gauss curvature is positive at some point of $\gamma_0$ then total intrinsic curvature of $\gamma_0$ has to be $&lt;2{\cdot}\pi$, a contradiction.</p> <p>Note that if the asymptotic direction goes transversally to $\gamma_0$ at $\gamma_0(v)$ then $\gamma_v$ can not lie on the $\partial K$. I.e., $\gamma_0=h(0,{*})$ is an asymptotic curve.</p> <p>WLOG we can assume that the length of $\gamma_0=h(0,{*})$ is $2{\cdot}\pi$ and its intrinsic curvature is $\equiv 1$. In the space $\gamma_0$ has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.</p> http://mathoverflow.net/questions/77760/why-is-the-half-torus-rigid/77776#77776 Answer by Q.Q.J. for Why is the half-torus rigid? Q.Q.J. 2011-10-11T01:35:31Z 2011-10-11T22:39:51Z <p>For any surface patch with the first fundamental form $$g(u,v) = \left[\begin{array}{cc} (c+a\cos{v})^2 &amp; 0\\ 0 &amp; a^2\end{array}\right]$$ The Gauss and Codazzi Equations are</p> <p>\begin{align} ac\cos(v)+a^2\cos^2(v)-h_{11}h_{22}+h_{12}^2&amp;=0\\ h_{11,v} - h_{12,u} + \frac{a\sin(v)}{c+a\cos{v}}h_{11}+\frac{\sin(v)(c+a\cos(v))}{a}h_{22}&amp;=0\\ h_{22,u} - h_{12,v} + \frac{a\sin(v)}{c+a\cos{v}}h_{12}&amp;=0 \end{align} If we can show that the solution for the functions $h_{ij}$ is the same as that for the torus patch $(c+a\cos(v))\cos(u), (c+a\cos(v))\sin(u), a\sin(v))$, then we are done by uniqueness part of the Fundamental Theorem of Surfaces (patches with the same $g$ and $h$ differ only by a rigid motion).</p> <p>Remark: If we make an overly strong assumption that $h$ is diagonal then this gives the result, but otherwise, as Deane comments, it is not immediately clear how/if we can prove the uniqueness of the $h_{ij}$ in the general case. </p> <p><strong>Update:</strong> Consider a particular local isometry of a patch on the torus that is small enough to not create any umbillic points. We can reparametrise in the neighbourhood of any non-umbillic point to a principal patch where <strong>both</strong> $g$ and $h$ are diagonal. The first fundamental form for the reparametrised isometric patch will have the form</p> <p>$$g(u,v) = \left[\begin{array}{cc} \lambda(u,v)^2(c+a\cos{v})^2 &amp; 0\\ 0 &amp; \mu(u,v)^2a^2\end{array}\right]$$</p> <p>for known $\lambda,\mu$ and then the Codazzi equations are now a linear system for $h_{11},h_{22}$: $$h_{11,v} = \frac12\partial_v(\lambda^2(c+a\cos{v})^2)(\frac{h_{11}}{\lambda^2(c+a\cos{v})^2} + \frac{h_{22}}{\mu^2a^2})$$</p> <p>$$h_{22,u} = \frac12\partial_u(\mu^2a^2)(\frac{h_{11}}{\lambda^2(c+a\cos{v})^2} + \frac{h_{22}}{\mu^2a^2})$$</p> <p>and the Gauss equation is $$h_{11}h_{22} = \lambda^2\mu^2a\cos(v)(c+a\cos(v)).$$</p>