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Igor Pak's argument takes care of the general convex case. So to complete the proof (without assuming any smoothness) it remains to consider the nonconvex case. To this end it is enough to note that i …

Here is a general way to construct a large family of bendings of the cylinder $M$, with nonplanar boundaries, via Alexandrov's isometric embedding theorem. All these examples will be convex. First no …

A natural definition for convex space curves is given by requiring that the curve lies on the boundary of its convex hull. In this sense, the curves that you describe will be convex. Convex space cur …

First of all, we should assume that the original knot $K(t)$ is twice differentiable and has non vanishing curvature, so that its principal normal $N(t)$ exists and the curve $c(t)$ you want is well …

In a recent paper, Centers of disks in Riemannian manifolds, Igor Belegradek and I study whether it is possible to extend to nonconvex objects the notion of center of mass or other classical centers a …

How can one construct examples of Riemannian manifolds which have nonpositive radial curvature about some point, but are not nonpositively curved everywhere? (I presume that they exist, but do not kn …

Let $pxy$ be a triangle in a CAT(0) space $X$, and $p' x' y'$ be a triangle in $\mathbf{R}^2$ such that the lengths $|px|=|p'x'|$, $|py|=|p'y'|$ and the angle $\angle(xpy)=\angle(x'p'y')$. Let $z\in x …

I think that 8 might be possible, by interlocking two Star Trek symbols as shown below. Adendum: This candidate may not work, as quarague points out, but I leave it as a potential "how not to" examp …

Here is another example with 8 vertices: a short fat Star Trek symbol and a square in orthogonal planes. Since the distance between the base points of the red figure is greater than its height, one c …

This is just a bit more elaboration on Anton's nice example, and subsequent comment. Here is the picture of the two triangles: Note that $p'x'y'$ is obtained from $pxy$ by rotating the side $\overlin …

Let $M$ be a compact $3$-dimensional manifold diffeomorphic to a ball. Suppose that $M$ has nonpositive (sectional) curvature and its boundary $\partial M$ is convex, or even that $M$ is a Riemannian …

Answer to question 1: If a convex surface is not closed, then generally it is far from rigid as it might admit infinitely many isometric deformations; however, if the surface has $\mathcal{C}^{2,1}$ r …

If the high school students are taking, or have already taken, Calculus I, and know how to differentiate, then I would use the notion of acceleration. I would start by asking: what is a straight line …

The following paper develops the outline described by Anton Petrunin to solve Gromov's problem on total absolute curvature for simply connected surfaces: Convexity and rigidity of hypersurfaces in Car …

A Cartan-Hadamard 3-space $M$ is a complete simply connected 3-dimensional Riemannian manifold with nonpositive sectional curvature. A (smooth) convex surface $\Gamma\subset M$ is an embedded topologi …