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.
Can you please help me in understanding this equation?
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?
 A: 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.
