# Rigidity in a $CAT(-1)$ space

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.

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.)

• 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. May 15, 2016 at 21:18
• 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.
– FMB
May 16, 2016 at 7:36
• 1. since the majorizer is digon, it contains a point on distance $<\tfrac\pi2$ from all but two points in the digon. May 16, 2016 at 12:46
• 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 $\{y_i\}$. May 16, 2016 at 12:49
• 1. Is it clear that the majorizer can't be a great circle ?
– FMB
May 17, 2016 at 10:01