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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
10
votes
Geodesics in $\mathbb{R}^2 \times \mathbb{S}^1$ under "segment" metric
In addition to Anton Petrunin's answer, here is a trick to simplify (and in some sense solve) the geodesic equation.
Since the metric has three-dimensional group of isometries (generated by rigid moti …
7
votes
Accepted
Non-affine, projective vector field on $\mathbb{R}^n$
I don't know what was meant in that exercise, but your revised conjecture is certainly true and well-known. Here is an elementary proof.
The assumptions (local injectivity, continuity and segment-to-s …
29
votes
Accepted
Counterexample to Sard's theorem for a non-C1 map
No, such functions do not exist. More precisely, let $f:\mathbb R\to\mathbb R$ be an arbitrary function, $\Sigma$ is the set of $x\in\mathbb R$ such that $f'(x)$ exists and equals 0. Then $f(\Sigma)$ …
6
votes
Accepted
Questions about local triviality
Here is a counter-example to Q1. Consider a Reeb-like foliation on the annulus $E=S^1\times\mathbb R$. Namely fix a point $p\in S^1$, let $I=S^1\setminus \{p\}$ and fix a smooth function $h:I\to\mathb …
6
votes
Vanishing Gaussian curvature
Yes.
Consider segments of $u$-coordinate curves between two $v$-coordinate curves $\{u=u_1\}$ and $\{u=u_2\}$. These segments are a family of geodesics of the same length $|u_1-u_2|$. By the first var …
17
votes
Accepted
Are there unique geodesics in the NIL and SOL geometry?
The geodesics between points are not unique in both cases. Moreover the following is true: if $M$ is a universal cover of a compact Riemannian manifold whose fundamental group is virtually solvable bu …
9
votes
1
answer
315
views
Convex body with affine-equivalent cross-sections
I recently discovered the following fact: Let $K\subset\mathbb R^3$ be an origin-symmetric convex body with smooth and strictly convex boundary. Suppose that all central cross-sections of $K$ (that is …
8
votes
Accepted
Is displacement controled by stable norm?
If you allow an additive term $C(n)diam(g)$ rather than $2diam(g)$, then yes, the statement is true. In the paper D.Burago, "Periodic metrics", Adv. Soviet Math. 9, (1992), 205-210, he proves that for …
11
votes
A Converse to the Gauss Bonnet Theorem
First of all, the identity holds for any degree 1 map $F:\mathbb S^2\to\mathbb S^2$. Moreover, for any $F=(f,g,h):\mathbb S^2\to\mathbb S^2$,
$$
\int_{\mathbb S^2} f\,dgdh = \frac43\pi \deg F.
$$
Thi …
5
votes
Accepted
Preservation of injectivity radius
This is an expansion of Anton Petrunin's comment.
Let me describe how to perturb the standard metric of the plane so that the resulting metric is bi-Lipschitz to the original with Lipschitz constant …
3
votes
Accepted
Does such an operator commutes with the whole torus action?
No. The assumption is coordinate-independent (i.e., preserved by self-diffeomorphisms) but the desired conclusion is not.
Begin with $R$ being the standard irrational flow and $\mathcal O$ a coordina …
24
votes
3
answers
1k
views
Isoperimetric inequality on a Riemannian sphere
Consider a two-dimensional sphere with a Riemannian metric of total area $4\pi$. Does there exist a subset whose area equals $2\pi$ and whose boundary has length no greater than $2\pi$?
(To avoid tec …
3
votes
Ball-Box Theorem and Sequence of Distributions
Dealing with such a low regularity is a tricky business. However in dimension 3 you can get away with your set of assumptions.
First, if the distributions are uniformly Lipschitz and converge in $C^0 …
5
votes
Accepted
Stability of Pu's isosystolic inequality
There is no Lipschitz or even Gromov-Hausdorff stability - just consider a round metric with long hairy tails of small area.
One can hope for stability with respect to intrinsic flat distance in the …
4
votes
Accepted
Zoll Flat Finsler tori and convex bodies on a starry night
No such a set is not always a polytope. Consider the convex hull of the set of points of the form $(1/n.1/n^2)$ and $(0,0)$ in $\mathbb R^2$. Its boundary is a union of infinitely many segments with r …