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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
21
votes
Accepted
A very torsioned closed curve in space
If by "skew" we mean nonparallel and nonintersecting, then the answer is NO. Every smooth closed curve in $\mathbf{R^3}$ has uncountably many pairs of intersecting tangent lines. This follows from Poi …
6
votes
Must a bending of the cylinder leave the bases planar?
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 …
3
votes
$C^0$-approximation by smooth curves with prescribed curvature
This has been established in a recent paper by Micha Wasem:
h-principle for curves with prescribed curvature.
Geom. Dedicata 184 (2016), 135–142.
In particular Wasem showed that one can prescribe c …
13
votes
Accepted
Tubular Neighborhood Theorem for $C^1$ Submanifold
The answer to your question depends on whether you are looking for a tubular neighborhood in the general differential topological sense or the more restrictive geometric sense. The answer in the topol …
5
votes
1
answer
329
views
Manifolds with nonpositive radial curvature
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 …
1
vote
2
answers
219
views
A triangle comparison in CAT(0) spaces
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 …
4
votes
Accepted
Can every surface be realized as a mean convex hypersurface in $\mathbb{R}^3$?
Yes, every closed orientable surface can be embedded in $R^3$ with positive mean curvature.
One way to construct higher genus examples is by gluing thin tori, which are mean convex. For instance we ca …
8
votes
The geometry of Nadirashvili's complete, bounded, negative curvature surface
The conjecture attributed to Hadamard, if one regards that as being concerned with the existence of a complete embedded negatively curved surface in a ball, is still open. I have read the correspondin …
40
votes
What is the best way to draw curvature?
The best way I know to illustrate the notion of curvature is via Toponogov's theorem. We can compare any (geodesic) triangle in a Riemannian manifold $M$ with one with the same edge lengths in Euclid …
3
votes
A triangle comparison in CAT(0) spaces
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 …
9
votes
1
answer
380
views
Perturbing metrics with nonpositive curvature
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 …
3
votes
Uniformisation for non simple closed curves
It sounds like, to some extent, what you are asking is whether an immersed closed curve in the plane can be extended to an immersion of a disk. This would yield the multiplicities that you seek; howev …
2
votes
Approximating a compact $C^1$ hypersurface without boundary
It seems that what you are asking is essentially how one can approximate the area of a surface via triangulations. This is a classical topic stretching back more than a century. The famous example of …
6
votes
What is known about sufficient conditions for the rigidity of a convex surface?
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 …
15
votes
Geodesics on the sphere
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 …