1
$\begingroup$

Summary: How to proove that a reunion of triangles in a $CAT(-1)$ space is isometric to the reunion of coresponding comparisons triangles ?

Context and notations:

Le $X$ be a $CAT(-1)$ metric space. Let $\left( y_{i}\right)_{i=1, \dots, n}$ be points in $X$ and $x$ their center of mass. We denote by $T_{i}$ the geodesic triangle with vertices $(y_{i}, x, y_{i+1})$ and $\tilde{T_{i}}$ the corresponding comparison triangle in $\mathbb{H}^{2}$, with vertices $(\tilde{y_{i}}, \tilde{x}, \tilde{y_{i+1}})$ .

Let $\alpha_{i}$ be the angle of $T_{i}$ at the vertex $x$ and $\tilde{\alpha_{i}}$ the angle of $\tilde{T_{i}}$ at the vertex $\tilde{x}$.

We denote by $\Sigma_x X$ the space of directions at $x$ and by $C_x X$ the tangent cone at $x$. We denote by $<c_1, c_2>$ the "inner product" on $C_x X$.

Fact: Because $x$ is the center of mass of $(y_i)$ we have: $ \sum_i \alpha_i \ge 2 \pi$.

Problem: Suppose now that $ \sum_i \tilde{\alpha_i} = 2 \pi$. Is the union $\bigcup_i T_i$ isometric to the gluing of the hyperbolic triangles $\bigcup_i \tilde{T_i}$ ?

What I did:

The hypothesis implies that $\tilde{\alpha_i} = \alpha_i$ and so each triangle $T_i$ is individually isometric to $\tilde{T_i}$. I prooved that it is then enough to show that $\alpha_i + \alpha_{i+1} = \angle_x (y_i, y_{i+2})$ for $T_i \cup T_{i+1}$ to be isometric to $\tilde{T_i} \cup \tilde{T_{i+1}}$. So I interpreted the problem in the space of directions: Let $P$ be the geodesic polygon in $\Sigma_x X$ with vertices $p_i = [x, y_i]$ the class of the geodesic going from $x$ to $y_i$. My problem is now equivalent to: is $P$ a closed geodesic ?

What could be useful:

The fact that $x$ is the center of mass of the $(y_i)$ implies that for every non-trivial geodesic $v$ starting at $x$: $ \sum_i d(x, y_i) \cos \angle (v, p_i) \le 0$.

A "short loop" theorem: a closed (geodesic polygonal) curve in a $CAT(1)$, whose length is less than $2 \pi$ is contained in a ball of radius $\pi/2$

All of my definitions are from "Metric spaces of non-positive curvature" by Bridson and Haefliger but feel free to ask me to precise something.

$\endgroup$
1
$\begingroup$

Yes it is true.

Note that the directions to the triangles at $x$, say $\gamma$, form a closed geodesic in $\Sigma_x$. If not one could short it a bit, apply Reshetnyak's majorization for $\gamma$ in $\Sigma_x$ and pass to the limit --- this way you get a short map from a convex polygon (which has to be digon since its perimeter is $2\cdot\pi$) to $\Sigma_x$ which is path isometric on the boundary. It follows that for some point $z$ we have $\measuredangle[x^{y_i}_z]\le \tfrac\pi2$ and for some $i$ the inequality is strict. The latter contradicts that $x$ is the barycenter.

Once it is done apply exponential map in the direction of $\gamma$ and your argument shows that it is isometry until the end of triangles.

(I assumed above without saying that $\Sigma_x$ is locally compact, but it is generalizable using ultralimits.)

$\endgroup$
  • $\begingroup$ More generally, you may think of it as a rigidity case in Kirszbraun's theorem --- it gives existence of isometric copy of of convex hyperbolic polyhedron in your space. $\endgroup$ – Anton Petrunin May 15 '16 at 21:18
  • $\begingroup$ Thank you for your response. Some things are still unclear: why the inequality would be strict for some $i$ ? where do you use the local compacity ? Could you also explained how this is related to Kirzsbraun's theorem ? I've looked at your paper about this theorem but didn't find a rigidity case. $\endgroup$ – Florentin MB May 16 '16 at 7:36
  • $\begingroup$ 1. since the majorizer is digon, it contains a point on distance $<\tfrac\pi2$ from all but two points in the digon. $\endgroup$ – Anton Petrunin May 16 '16 at 12:46
  • $\begingroup$ 2. There is 1+n lemma and it has a rigidity case, this is not in the paper, but it is easy to prove --- simply take the map provided by Kirszbraun theorem and check that it is distance preserving on the convex hull of $\{y_i\}$. $\endgroup$ – Anton Petrunin May 16 '16 at 12:49
  • $\begingroup$ 1. Is it clear that the majorizer can't be a great circle ? $\endgroup$ – Florentin MB May 17 '16 at 10:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.