Question on ideal triangle and hyperbolic distance

Let's consider a hyperbolic triangle (I'm considering the Poincarè disc model of the hyperbolic plane) and from each one of the three vertices let's foliate the triangle with horocycles untill we reach the central zone bounded by three horocycles. I made a (bad) picture: the geodesics are in red and the horocycles in blu. Now let's choose one spike on the triangle and the horocycle at hyperbolic distance $t$ from the central, unfoliated zone. My question is:

How do I compute the hyperbolic length of this horocycle? Does it vary linearly with $t$? I'm new to hyperbolic geometry so I apoligise if my question is trivial. And I'm sorry if I'm not explaining my attempt to solve the problem, but this seems rather complicated to me, so I really need some help understanding where to start to solve it.

Thank you!

While I like the Poincaré disk for aesthetic reasons, sometimes the half-plane model is easier, and I'd say this is such a case. Consider the ideal triangle with vertices at $(\pm1,0)$ and at infinity. It's boundary is the upper half of the unit circle, together with two vertical tangents to that circle. The horocycles incident with the point at infinity are horizontal lines. So you are asking about the hyperbolic distance from $(-1,y)$ to $(1,y)$ for some $y$ which should be computable from your $t$. To find the relation between $y$ and $t$, you need the “midpoints” of the triangle edges. More precisely, you could describe them as the points where the incircle of the ideal triangle (drawn in cyan) touches its edges. The incenter in our case is the circle of radius $1$ with center at $(0,2)$. So the distance $t$ would be the distance between $(1,2)$ and $(1,y)$.

Now Wikipedia tells you that the distance from the incircle contact point is

$$t=\ln\frac{y}{2}\quad\implies\quad y=2e^t$$

and the path length along the horocycle would be

$$s=\int_S\sqrt{\mathrm (ds)^2}= \int_{-1}^{1}\sqrt{\frac{(\mathrm dx)^2}{y^2}}=\frac2y=e^{-t}$$

So no, the relation between $t$ and $s$ is not linear but exponential.

It's nice to see that this answer agrees with what D. Thomine found independently. Their maximal horocycle arc length $s_0$ would be what you get for $y=2$, i.e. $s_0=1$.

• Thank you! that was really clear! – Rick21 Apr 19 '16 at 11:44

The horocycles (seen in $T^1 \mathbb{H}$, by orienting then normally) are the stable leaves (or unstable leaves, depending on the orientation) of the geodesic flows. If $(h_s)$ is the horocyclic flow, then:

• $(h_s)$ has unit speed (it is easy to check that it has unit speed when acting on $i$, and since its action elsewhere is by conjugation by isometries...);

• $g_t \circ h_s = h_{e^{-t}s} \circ g_t$ (it can be seen with the matrices corresponding to the geodesic and horocyclic flow).

Let $s_0$ be the length of the arc of horocycle at the center of the triangle. Let $u$, $v$ be the end-points of this arc, with unit tangent vector pointing towards the cusp. Then, up to exchanging $u$ and $v$, we have $h_{s_0} (u) = v$ (that's because the horocycle flow has unit speed). Hence:

$$g_t (v) = g_t \circ h_{s_0} (u) = h_{e^{-t}s_0} \circ g_t (u),$$

so the length of the arc of horocycle between $g_t (u)$ and $g_t (v)$ is $e^{-t}s_0$. All is left is to compute $s_0$, for which I don't see any other way than computing with an explicit parametrization of the arc.