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Euclidean, hyperbolic, discrete, convex, coarse geometry, metric spaces, comparisons in Riemannian geometry, symmetric spaces.
4
votes
2
answers
396
views
Is there a contractible bounded homogeneous space?
Let us phrase the question in the title in more detail: I wonder if there exists a metric space $X$ which has at least two points, has finite diameter (in the sense that there is an upper bound for th …
8
votes
2
answers
488
views
How close can closed geodesics be?
A consequence of the famous Jørgensen inequality is that there is a lower bound for the distance between closed geodesics in hyperbolic three-manifolds: for any $R>0$ there is a c>0 such that for any …
7
votes
1
answer
435
views
A criterion for loxodromicity in Gromov-hyperbolic spaces
Recall that an isometry of a Gromov-hyperbolic space $X$ is called loxodromic if it has exactly two fixed points on the Gromov boundary $\partial X$, one being "attracting" and the other "repelling". …
2
votes
Accepted
How does Siegel's Hilbert-Blumenthal fundamental domain differ from Götsky's?
I'm not familiar with the Götsky--Cohn construction but Siegel's (as explained in van der Geer's book) seems clear:
there is a "height function" $y$ on $X = \mathbb H^2 \times \mathbb H^2$ (the "di …
3
votes
A criterion for loxodromicity in Gromov-hyperbolic spaces
An answer was given in Ashot Minasyan's comment, i'm writing it here so the question does not remain unanswered.
In Lemma 2.5 of "Group actions on metric spaces: fixed points and free subgroups" by Ma …
8
votes
Just how close can two manifolds be in the Gromov-Hausdorff distance?
This is only an answer to one point of your question: for surfaces of large genus $g$ the distance should be
$$
d(S, \mathrm{point}) \asymp \log(g).
$$
The lower bound should follow from volume esti …
8
votes
1
answer
273
views
Length and curvature for closed curves in negatively curved spaces
In the Euclidean plane, for a closed smooth curve of length $\ell$ whose curvature is bounded above by $\epsilon$ we have the inequality
$$ \ell \ge 2\pi \epsilon^{-1} $$
which follows from the fact …