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Prove that there exists a universal constant $K$ such that given any three geodesics in the $\gamma_1, \gamma_2, \gamma_3$ in the hyperbolic plane forming a triangle, there exists three points $A \in \gamma_1$, $B \in \gamma_2$, $C \in \gamma_3$ such that the geodesic triangle on vertexes $A,B,C$ has its diameter bounded by $K$

My attempt was the following: Given any geodesic triangle, there exists a inscribed circle $\Gamma$, which provides me a natural candidate for the triple $A,B,C$ as the three points that intersect such circle. Also, the diameter of the geodesic triangle is bounded by the radius of $\Gamma$. I thought then of using Gauss-Bonnet Theorem to get a bound of the area of $\Gamma$

$$ \int_\Gamma K dA + \int_{\partial \Gamma} k_g dt + 3\pi - \sum_{i=1,2,3} \phi_i = 2\pi $$ where $k_g$ is the geodesic curvature of the the curve $\gamma$ the boundary of $\Gamma$ and $\phi_i$’s are the inner angles.

So, I get $$ Area(\Gamma) = \pi - \sum_{i=1,2,3} \phi_i + \int_{\partial \Gamma} k_g dt $$

$$ Area(\Gamma) \le \pi + \int_{\partial \Gamma} k_g dt $$

So now, I would just need to be able to get a universal bound on the term $ \int_{\partial \Gamma} k_g dt $ and use the formula of the area of a circle to get an upper bound on the radius. How could I get this upper bound on $ \int_{\partial \Gamma} k_g dt $?

On the other hand, I could try to use Gauss-Bonnet in the region given by the geodesic triangle given by $A,B,C$ to avoid needing to bound the geodesic curvature, but it does not seem to be easy to get back an upper bound on the length of the longest side of the triangle.

I would appreciate any hints. Thank you.

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  • $\begingroup$ I’m sorry, could you elaborate a bit more? I don’t really understand. $\endgroup$ – Kernel Oct 31 '17 at 15:22
  • $\begingroup$ Sorry I'm not totally familiar with hyperbolic spaces. But it seems to me from physics that it is locally euclidean but then there can't be a $K$ such that all groups of three lines forming a triangle have smaller diameter. Hmm maybe im not familiar enough with the hyperbolic plane. $\endgroup$ – marshal craft Oct 31 '17 at 16:33
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A geodesic triangle with angles $\alpha,\beta,\gamma $ has area $\pi-\alpha-\beta-\gamma \le \pi$, proof here (but you already knew this since you know Gauss-Bonnet).

Hence, the inscribed circle has area $\le \pi $.

Hence, the radius of inscribed circle is bounded by some constant $K$ (the area of a hyperbolic circle can be computed explicitly, but it suffices to know that by increasing the radius $\to\infty$ we cover the hyperbolic plane which has infinite area).

Hence, the diameter of the triangle formed by the points where the inscribed circle touches the original triangle is bounded by $2K$.

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