Why do we use cosh to define the angle between two vectors in hyperbolic geometry?

Question

I can kind of see why this works when we use the regular dot product, but I don't understand why this is still true when we use the dot product adapted for hyperbolic geometry?

Answer 1

Let's consider the hyperbolic (Minkowski) and complex (Euclidean) planes.

In the Euclidean plane, you can describe any point by $k(\cos(\theta),\sin(\theta))$. Analogously, in the hyperbolic plane, you can describe points by $k(\cosh(u),\sinh(u))$. A unit hyperbolic vector (in the hyperbolic plane) is then $(\cosh(u),\sinh(u))$.

The actual inner product for the hyperbolic plane (and for higher dimensional hyperbolic spaces it generalizes in the obvious way) is defined as $\langle (c,d),(e,f)\rangle = ce-df$. We can thus verify that $(\cosh(u),\sinh(u))$ is a unit vector because $\langle (\cosh(u),\sinh(u)),(\cosh(u),\sinh(u))\rangle = \cosh^2(u)-\sinh^2(u)=1$.

Even though above is how we really define the inner product, let's see if we can't figure out an analog to the cosine formula for the inner product on the complex plane.

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Just to recall, the cosine formula $\langle \vec a, \vec b\rangle=\|\vec a\|\|\vec b\|\cos(\theta)$ is derivable as:

$$\left\langle a\left(\cos(\theta_1),\sin(\theta_1)\right),b\left(\cos(\theta_2),\sin(\theta_2)\right)\right\rangle \\ = a\cos(\theta_1)b\cos(\theta_2)+a\sin(\theta_1)b\sin(\theta_2) \\ = \frac {ab}2\left[\cos(\theta_1-\theta_2)+\cos(\theta_1+\theta_2)\right] + \frac {ab}2\left[\cos(\theta_1-\theta_2)-\cos(\theta_1+\theta_2)\right] \\ = ab\cos(\theta_1-\theta_2)$$

where $\theta_1-\theta_2$ is just the angle between $\vec a$ and $\vec b$.

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Now let's try to do the same thing for two vectors in the hyperbolic plane:

$$\left\langle a\left(\cosh(u_1),\sinh(u_1)\right),b\left(\cosh(u_2),\sinh(u_2)\right)\right\rangle \\ = a\cosh(u_1)b\cosh(u_2)-a\sinh(u_1)b\sinh(u_2) \\ = \frac {ab}2\left[\cosh(u_1+u_2)+ \cosh(u_1-u_2)\right] - \frac {ab}2\left[\cosh(u_1+u_2)-\cosh(u_1-u_2)\right] \\ = ab\cosh(u_1-u_2)$$

where $u_1 - u_2$ is the hyperbolic angle between hyperbolic vectors $\vec a$ and $ \vec b$.

Answer 2

Consider $\mathbb{H}^1$ as one branch of the hyperbola $x_{1}^2 - x_{2}^2 = -1$ in $\mathbb{R}^{1,1}$. This is a curve which can be parameterized as $\gamma(s) = (\sinh s, \cosh s)^{T}$, because $\cosh^2 s - \sinh^2 s = 1$. Now $\gamma^\prime (s) = (\cosh s, \sinh s)^{T}$, so $\langle \gamma^\prime (s), \gamma^\prime (s)\rangle = 1$ and $$ \mathrm{length}(\gamma_{|[s_1, s_2]}) = \int\limits_{s_1}^{s_2} \sqrt{\langle \gamma^\prime (s), \gamma^\prime (s)\rangle} = s_2 - s_1. $$ On the other hand, $$ \langle \gamma (s), \gamma (s)\rangle = \sinh s_1 \sinh s_2 - \cosh s_1 \cosh s_2 = - \cosh (s_1 - s_2). $$ Hence the hyperbolic distance $\mathrm{d}(p,q)$ of two points $p,q \in \mathbb{H}^1$ is given by $$ - \cosh\mathrm{d}(p,q) = \langle p, q \rangle. $$ Now define $a := \mathrm{d}(p,q)$ and one obtain: $$ -\cosh a = \langle p, q \rangle. $$