# Short map between hyperbolic triangles

Given two (Euclidean or hyperbolic) triangles $T = ABC$ and $T' = A'B'C'$, the natural map is the one that sends $A' \mapsto A$, $B' \mapsto B$, $C' \mapsto C$ and maps affinely each side of $T'$ onto the corresponding side of $T$. We say that the triangle $T'$ dominates the triangle $T$ if the natural map is a short map (Lipschitz with constant $1$) with respect to the distance in the (Euclidean or hyperbolic) plane.

My question is, given a triangle $T$ with side lengths $(a, b,c)$, is it true that the triangle $T'$ with side lengths $(a+\epsilon, b+\epsilon, c+\epsilon)$ dominates $T$ for all $\epsilon>0$ small enough?

I can prove this statement for Euclidean triangles by some calculations involving the law of cosines, but I couldn't manage to do the same in the hyperbolic plane.

• Your map is between the boundaries of the triangles, and not defined in the interior, right? – Ivan Izmestiev Sep 28 '16 at 11:12
• Yes, the map is only defined on the boundary. If we know it is a short map we can extend it to the interior of the triangle with the Kirszbraun theorem. – Florentin MB Sep 28 '16 at 11:16
• Can you compute the derivative with respect to $\epsilon$ of the distance between pairs of corresponding points (could be easier than comparing distances)? – Ivan Izmestiev Sep 28 '16 at 11:19
• This is what I did for the Euclidian case, but the calculation of the derivatives for the hyperbolic gives messy formulas involving hyperbolic sines and cosines that I was unable to manipulate. – Florentin MB Sep 28 '16 at 11:24
• Perhaps try working in barycentric coordinates using metric pulled back from the Klein model. – Neal Sep 28 '16 at 12:54

No, this is not true. Here is an indirect argument (if I made no mistake).

If the statement would be true for $\epsilon > 0$ small enough, then it would be true for all $\epsilon > 0$ (since the maps commute, the set of "good" epsilons for a given $(a,b,c)$ is open; this set is also closed because the Lipschitz condition is closed). We will construct two triangles $(a,b,c)$ and $(a+M, b+M, c+M)$ for which the latter does not dominate the former.

Take $a=b=1$ and $c=2$ (or slightly smaller if you need). Take points $A_1$ and $B_1$ on the sides $BC$ and $AC$ at distance $\frac13$ from $C$. The distance between $A_1$ and $B_1$ is $\frac23$. Now consider a triangle $(x, x, x+1)$ for a very large $x$. I claim that the distance between the corresponding points $A'_1$ and $B'_1$ tends to $0$ as $x$ tends to $\infty$.

By the sine law in a right triangle we have $$\frac{\sinh \frac{A'_1B'_1}{2}}{\sinh\frac{x}{3}} = \sin\frac{\gamma'}{2} = \frac{\sinh\frac{x+1}2}{\sinh x}$$ which implies that $\sinh\frac{A'_1B'_1}2$ goes down as $e^{-\frac{x}6}$.

This means that somewhere inbetween there is an isosceles triangle for which the map is not $1$-Lipschitz even for small $\epsilon$...

• Thank you, that is a nice argument. Does your argument still work if we allow arbitrary perturbation of the vertices of the triangle ? – Florentin MB Sep 28 '16 at 16:21
• I don't know. Maybe there is a "side elongation" formula different from $(a,b,c) \mapsto (a+\epsilon, b+\epsilon, c+\epsilon)$ that works. I wouldn't be surprised. – Ivan Izmestiev Sep 28 '16 at 17:06

We can answer the question positively if we adapt the definition of domination to the hyperbolic setting. Given two hyperbolic triangles $T = ABC$ and $T' = A'B'C'$, the natural map is the one that sends $A' \mapsto A$, $B' \mapsto B$, $C' \mapsto C$ and maps each side of $T'$ onto the corresponding side of $T$ with the following parametrization:

A geodesic $[x,y]$ of length $l$ is in bijection with the interval $[0,1]$ via the map that sends $t$ to the point in $[x,y]$ at distance $\sinh^{-1} ( t \sinh (l))$ from $x$; now maps each side of $T'$ to the corresponding side of $T$ with the mapping which sends a point to the point with the same parameter in $[0,1]$.

With this definition of domination it can be check directly using the hyperbolic law of cosines that for small epsilon, the triangle with sides $(a+\epsilon, b+\epsilon, c+\epsilon)$ dominates the triangle with sides $(a,b,c)$.

• Nice! Do you have an explicit description of an "expanding" extension of this map to the interiors of the triangles? – Ivan Izmestiev Nov 21 '16 at 15:59