# Interpretation of the equation of a horocycle in the hyperboloid model.

Let $$v$$ be non-zero with $$B(v,v)=0$$. I would like to show that an equation $$B(u,v)=1$$ defines a horocycle in the hyperboloid model.

A horocycle in the Poincare disk model has a touching point $$v$$ on the boundary of the unit circle. And if we see this $$v$$ in the hyperboloid model, $$v$$ will be a point at infinity on the asymptotic cone. In the hyperboloid model, a horocycle with $$v$$ is the intersection of the hyperboloid ($$-x^2-y^2+z^2=1$$) and some plane that is parallel to the vector $$v$$. Plus, as the asymptotic cone is defined by the equation $$x^2+y^2=z^2$$, we have $$B(v,v)=0$$ and $$v \neq 0$$.

If I have understood the equation correctly, for any $$u$$ in a horocycle that is (Euclidean Sense) parallel to the vector $$v$$, we have $$B(u,v)=1$$.

However, I'm not sure why we should have $$B(u,v)=1$$ for any $$u$$ on a horocycle.

EDIT:

I found that $$B(u,v)=1$$ gives the equation of the plane. And this plane must have the normal vector $$(v_1,v_2,-v_3)$$ if $$v=(v_1,v_2,v_3)$$. Thus, by wikipedia definition of horocycle in the hyperboloid model, the equation $$B(u,v)=1$$ defines a horocycle if we just let $$u \in H$$ where $$H$$ stands for the hyperboloid.

Did I get it right?

• I'm pretty sure the Wikipedia definition of horocycle is not the definition in your course or textbook. Even though it might be equivalent, this exercise is about proving this equivalence. If you want to solve the problem properly, you will need to show that your course/textbook's definition of horocycle is satisfied. Apr 29 at 0:07

This answer assumes that the definition of a horocycle is in terms of constant curvature, which would be implied by the existence of isometric motion translating along it.

To show that the curve $$C_v := \{u: \langle u,v\rangle = 1\}$$ you define has constant curvature, you want to find a continuous family of hyperbolic isometries that preserves this curve. Once you find this family, all that remains is to show that it is the correct kind of constant-curvature curve, so not a circle or a hypercycle or a geodesic.

To find such a family of isometries preserving the given curve, you have basically two approaches:

Option 1: Pick your favourite $$v$$, e.g. $$v = (1,0,1)$$, and just find a family of isometries in $$O(2,1)$$ preserving this particular $$v$$, probably as a parametrized matrix. Isometries preserve $$B$$ by definition, so if they also preserve $$v$$, they automatically preserve $$C_v$$ too since it was defined in terms of $$B$$ and $$v$$.

Once you've showed that $$C_v$$ is a horocycle, by symmetry of the hyperboloid model you immediately obtain that all $$C_w$$ are horocycles too for any other $$w \neq 0$$ with $$B(w,w)=0$$: if an isometry maps $$w$$ to $$v$$, it will map $$C_w$$ to the horocycle $$C_v$$ (again since $$C_w$$ was defined in terms of $$B$$ and $$w$$), so $$C_w$$ is also a horocycle.

Option 2: Let $$v$$ be arbitrary instead of fixing it, and extend $$v$$ to a basis whose pairwise $$B$$-products have nice known values like $$0$$ or $$1$$ or $$-1$$ by applying some Gram-Schmidt analogous stuff. Now again, find a family of isometries preserving $$v$$, but instead of writing them in matrix form, try to write them in terms of your basis vectors and $$B$$. If you manage to do this, the part where you prove they are in fact $$v$$-preserving isometries will be a very simple term rewriting exercise which does not require any coordinate math.

• Except constant curvature is not enough. A better definition is that a horocycle is an orbit of a 1-parameter parabolic (unipotent) subgroup. Apr 29 at 0:57
• @MoisheKohan Yes, that is what I meant by "correct kind of curve". There are several ways to define which kind of constant-curvature curve is a horocycle, and I didn't want to make more assumptions about definitions than necessary in my answer. Apr 29 at 9:30