# Why is the half-torus rigid?

The half-torus surface that results from slicing a torus like a bagel, depicted below (left), is isometrically rigid.

I know this from a remark of Alexandrov in Mathematics: Its Content, Methods and Meaning (Chapter 7. Curves and surfaces, p.101):

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) ...

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)?

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The rigidity of the punctured torus is listed as an open problem (number 13) by Ghomi here: people.math.gatech.edu/~ghomi/Papers/op.pdf . I guess either the list is wrong, or I am missing a distinction between things I don't understand. –  Ramsay Nov 2 '11 at 19:07

I've got a letter from Idjad Sabitov which answer the question completely. Here is a short extract from it:

1. half-torus has rigidity of second order (Rembs' theorem, see Е. Rembs. Verbiegungen hoeherer Ordnung und ebene Flaechenrinnen. Math. Zeitschrift 36 (1932) or Ефимов, УМН, 1948, т.3, вып.2, стр. 135)

2. Any second order rigid surface does not admit analytic deformation (i.e., the deformation $h_t(u,v)$ which is analytic on $t$).

3. For the surfaces of revolution, the assumption of analyticity can be removed.

Below is the best part of my original post. It contains an idea which was not used by Rembs.

Let $h(u,v)$ be a small perturbation of the standard embedding; $u\in (-\varepsilon,\varepsilon)$ and $v\in\mathbb S^1$.

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 $<2{\cdot}\pi$, a contradiction.

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.

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.

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That's great start, but I'm not so sure about the "easily from here" part. Maybe for you but not for me. Do you consider the positively curved part separately from the negatively curved part? –  Deane Yang Oct 11 '11 at 1:17
Ouch. If you can't finish it, it's definitely not easy. –  Deane Yang Oct 11 '11 at 2:53
Why does $h\circ\gamma_0$ have curvature 1 in space? –  Sergei Ivanov Oct 11 '11 at 4:19
Anton : could you elaborate a little on the "central circle in a cone" part ? Is it something like isoperimetry on the sphere that forces $n(\gamma_t)$ to be a circle there ? Thanks. –  BS. Oct 12 '11 at 15:34
I now have the Rembs paper. For the record, here is a link: springerlink.com/content/t51345521u56u84l –  Joseph O'Rourke Oct 17 '11 at 16:02

For any surface patch with the first fundamental form $$g(u,v) = \left[\begin{array}{cc} (c+a\cos{v})^2 & 0\\\\ 0 & a^2\end{array}\right]$$ The Gauss and Codazzi Equations are

\begin{align} ac\cos(v)+a^2\cos^2(v)-h_{11}h_{22}+h_{12}^2&=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}&=0\\\\ h_{22,u} - h_{12,v} + \frac{a\sin(v)}{c+a\cos{v}}h_{12}&=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).

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.

Update: 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 both $g$ and $h$ are diagonal. The first fundamental form for the reparametrised isometric patch will have the form

$$g(u,v) = \left[\begin{array}{cc} \lambda(u,v)^2(c+a\cos{v})^2 & 0\\\\ 0 & \mu(u,v)^2a^2\end{array}\right]$$

for known $\lambda,\mu$ and then the Codazzi equations are now a linear system for $h_{11},h_{22}$: $$h_{11,v} = \frac12\partial_v(\lambda^2(c+a\cos{v})^2)(\frac{h_{11}}{\lambda^2(c+a\cos{v})^2} + \frac{h_{22}}{\mu^2a^2})$$

$$h_{22,u} = \frac12\partial_u(\mu^2a^2)(\frac{h_{11}}{\lambda^2(c+a\cos{v})^2} + \frac{h_{22}}{\mu^2a^2})$$

and the Gauss equation is $$h_{11}h_{22} = \lambda^2\mu^2a\cos(v)(c+a\cos(v)).$$

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Any tips on how to get the matrix to go over two lines? –  Q.Q.J. Oct 11 '11 at 1:37
@Q.Q.J: I edited it. Hope that matches what you were intending. –  George Lowther Oct 11 '11 at 1:57
Thanks George! So quadruple backslash is the trick. –  Q.Q.J. Oct 11 '11 at 1:59
@Q.Q.J: Yes. For each latex backslash you need an extra backslash to stop it getting absorbed by the parser –  George Lowther Oct 11 '11 at 2:01
You've restated the question as a nonlinear system of PDE's for the second fundamental form. It's well known that for a surface, this system is elliptic where the Gauss curvature is positive and hyperbolic where it is negative. Proving uniqueness for such a system is, as far as I know, not easy. What's particularly troublesome is that the boundary between elliptic and hyperbolic (i.e., where the Gauss curvature vanishes) is a characteristic curve, so standard uniqueness theorems for boundary value problems do not apply. Do you have some ideas about how to do this? –  Deane Yang Oct 11 '11 at 2:52