jcreed blog > How to Invent Hyperbolic Geometry

How to Invent Hyperbolic Geometry

Rotation

Rotations are nice maps from a space to itself. Well, but "nice" is subjective, and there's lots of arguably nice maps that aren't rotations: translations and scaling are pretty well-behaved, too. What makes rotations special? Let's focus on the fact that they What do we mean by length? In euclidean space $\R^n$, the length of a vector $\bx$ is $\sqrt{x_1^2 + \cdots + x_n^2}$. For simplicity, let's think about $\R^2$ and only focus on squared length \[|\bx|^2 = x_1^2 + \cdots + x_n^2\] What linear maps $\R^2 \to \R^2$ preserve squared length? We're looking for a matrix \[ M = \begin{array}{cc} a&b\\c&d \end{array}\] such that whenever we have a vector \[ \bx = \begin{array}{c} x\\ y \end{array}\] we find that $|M\bx|^2 = |\bx|^2$. This means \[ x^2 + y^2 = \left| \left(\begin{array}{cc} a&b\\c&d \end{array}\right)\left(\begin{array}{c} x\\ y \end{array}\right) \right|^2 \] \[ = \left|\begin{array}{c} ax + by \\ cx + dy \end{array}\right|^2 \] \[ = (ax+by)^2 + (cx + dy)^2 \] \[ = a^2x^2 + 2abxy + b^2y^2 + c^2x^2 + 2cdxy + d^2y^2 \] \[ = (a^2 + c^2)x^2 + 2(ab+cd)xy + (b^2+d^2)y^2 \] Since this needs to be true for every $x,y$, we have \[ a^2 + c^2 = b^2 + d^2 = 1 \] \[ ab+ cd = 0\] This means that $(a,c)$ and $(b,d)$ are some points on the unit circle, so we can find a $\theta, \phi$ such that $a = \cos \theta$ and $c = \sin \theta$ and $b = \sin \theta$ and $d = \cos \theta$. The condition that $ab+cd = 0$ means \[ \sin \phi \cos \theta + \cos \phi \sin \theta = 0\] which by trig identities means \[ \sin (\phi + \theta) = 0\] which means $\phi = 2\pi n - \theta$ for some $n \in \Z$, so without loss we might as well set $\phi = - \theta$. This means our rotation matrix must look like \[ R_\theta = \left(\begin{array}{cc} \cos \theta & - \sin\theta \\ \sin \theta &\cos \theta \end{array}\right) \] for some $\theta$.

Generalizing Length

What would happen if we instead said that the squared length of a vector was \[ \left|\begin{array}{c} x \\ y \end{array}\right|^2 = k^2 x^2 + y^2 \] for some constant $k$? Then the requirement that rotations preserve length would look like \[ k^2x^2 + y^2 = \left| \left(\begin{array}{cc} a&b\\c&d \end{array}\right)\left(\begin{array}{c} x\\ y \end{array}\right) \right|^2 \] \[ = \left|\begin{array}{c} ax + by \\ cx + dy \end{array}\right|^2 \] \[ = k^2(ax+by)^2 + (cx + dy)^2 \] \[ = k^2a^2x^2 + 2k^2abxy + k^2b^2y^2 + c^2x^2 + 2cdxy + d^2y^2 \] \[ = (k^2a^2 + c^2)x^2 + 2(k^2ab+cd)xy + (k^2b^2+d^2)y^2 \] Let's define $C = c/k$ and $B = kb$. This is equivalent to \[ k^2x^2 + y^2 = (k^2a^2 + k^2C^2)x^2 + 2k(aB+Cd)xy + (B^2+d^2)y^2 \] which means \[ a^2 + C^2 = B^2 + d^2 = 1\] \[ aB + Cd = 0\] We can repeat the reasoning from the last section to conclude that there exists a $\theta$ such that \[ \left(\begin{array}{cc} a&B\\C&d \end{array}\right) = \left(\begin{array}{cc} \cos \theta & -\sin \theta\\ \sin\theta&\cos \theta \end{array}\right) \] but that's not what our actual transform is: it's \[ R_{\theta;k} = \left(\begin{array}{cc} a&b\\c&d \end{array}\right) = \left(\begin{array}{cc} a&B/k\\kC&d \end{array}\right) = \left(\begin{array}{cc} \cos \theta & -(1/k)\sin \theta\\ k\sin\theta&\cos \theta \end{array}\right) \]

Adjustment for $\theta$

Let's consider the derivative \[{d\over d\theta} R_{\theta;k} \bx \adjust_{\theta=0} = {d\over d\theta}\left(\begin{array}{cc} \cos \theta & -(1/k)\sin \theta\\ k\sin\theta&\cos \theta \end{array}\right) \left(\begin{array}{c} x \\ y \end{array}\right) \adjust_{\theta=0}\] \[ = {d\over d\theta}\left(\begin{array}{c}x \cos \theta -(y/k)\sin \theta \\ kx \sin\theta + y \cos \theta \end{array}\right) \adjust_{\theta=0}\] \[ = \left(\begin{array}{c} -y/k \\ kx \end{array}\right) \] Let's suppose we'd like to adjust the choice of $\theta$ so that the derivative when $(x,y) = (0,1)$ is constant, instead of varying with $k$. This means considering instead \[R_{k\theta;k} = \left(\begin{array}{cc} \cos (k\theta) & -(1/k)\sin (k\theta)\\ k\sin(k\theta)&\cos (k\theta) \end{array}\right)\] and then we find \[{d\over d\theta} R_{k\theta;k} \bx = \left(\begin{array}{c} -y \\ k^2x \end{array}\right) \]

Hyperbolic Geometry

The conceptual leap at this point is the question: what happens if we consider imaginary numbers and set $k = i$? Then our notion of squared length becomes \[ \left|\begin{array}{c} x \\ y \end{array}\right|^2 = y^2 - x^2 \] And our rotation becomes \[R_{i\theta;i} = \left(\begin{array}{cc} \cos (i\theta) & i\sin (i\theta)\\ i\sin(i\theta)&\cos (i\theta) \end{array}\right) = \left(\begin{array}{cc} \cosh \theta & -\sinh \theta\\ -\sinh\theta&\cosh \theta \end{array}\right)\] So just to clear the minus sign we could define \[ H_{\theta} = R_{-i\theta;i} = \left(\begin{array}{cc} \cosh \theta & \sinh \theta\\ \sinh\theta&\cosh \theta \end{array}\right)\]

The Beltrami-Klein model

We could say that the hyperbolic plane is the set of all points in $\R^3$ such that $z > 0$ and $z^2 - x^2 - y^2 = 1$. We can see that \[ H_{\theta;x} = \left(\begin{array}{cc} \cosh \theta & 0 &\sinh \theta\\ 0 & 0 & 0 \\ \sinh\theta&0 &\cosh \theta \end{array}\right)\] \[ H_{\theta;y} = \left(\begin{array}{cc} 0 & 0 & 0 \\ 0 & \cosh \theta & \sinh \theta\\ 0 & \sinh\theta&\cosh \theta \end{array}\right)\] both map the hyperbolic plane into itself. Although we've talked about these as generalizations of rotation, we can think of them as translations in the $y$ and $x$ directions.

We say the Beltrami-Klein disk is the interior of the unit circle.

We can go back and forth between the hyperbolic plane and the disk like so:

If we have a point $(x, y, \sqrt{1 + x^2 + y^2})$ in the hyperbolic plane, then we output the point $(x/\sqrt{1 + x^2 + y^2}, y/\sqrt{1 + x^2 + y^2})$. If we have a point $(u,v)$ in the disk, then we are trying to find a point in the hyperbolic plane $(x, y, \sqrt{1 + x^2 + y^2})$ such that \[ u = x/\sqrt{1 + x^2 + y^2} \] \[ v = y/\sqrt{1 + x^2 + y^2} \] Trying to solve for $x,y$ leads us to \[ u^2 = x^2/(1 + x^2 + y^2) \qquad v^2 = y^2/(1 + x^2 + y^2) \] \[ u^2(1 + x^2 + y^2) = x^2 \qquad v^2(1 + x^2 + y^2) = y^2 \] \[ x^2 (u^2-1) + y^2u^2 = -u^2\] \[ x^2 v^2 + y^2(v^2-1) = -v^2\] \[\left(\begin{array}{c}x^2\\y^2 \end{array}\right) = - \left(\begin{array}{cc}u^2-1& u^2 \\ v^2 & v^2-1 \\ \end{array}\right)^{-1} \left(\begin{array}{c}u^2\\v^2 \end{array}\right)\] \[ = {1\over u^2 + v^2 - 1} \left(\begin{array}{cc}v^2-1& -u^2 \\ -v^2 & u^2-1 \\ \end{array}\right) \left(\begin{array}{c}u^2\\v^2 \end{array}\right)\] \[ = {1\over u^2 + v^2 - 1} \left(\begin{array}{c}v^2u^2 - u^2 - u^2v^2 \\ -u^2v^2 + u^2v^2 - v^2 \end{array}\right)\] \[ = \left(\begin{array}{c} {u^2 \over 1 - u^2 - v^2} \\ {v^2 \over 1 - u^2 - v^2} \end{array}\right)\] \[ \left(\begin{array}{c}x\\y \end{array}\right) = \left(\begin{array}{c} {u \over \sqrt{1 - u^2 - v^2}} \\ {v \over \sqrt{1 - u^2 - v^2}} \end{array}\right)\] What happens if we take a point in the Beltrami-Klein disk, map it into the hyperbolic plane, hit it with $H_{\theta;x}$, then map it back? We start with $(u,v)$, produce the point \[ {1\over\sqrt{1-u^2 - v^2}} \left(\begin{array}{c}u\\v\\1 \end{array}\right)\] which gets turned by $H_{\theta;x}$ into \[ {1\over\sqrt{1-u^2 - v^2}} \left(\begin{array}{c}cu+s\\v\\su+c \end{array}\right)\] for $c = \cosh\theta$ and $s = \sinh \theta$. Transforming this back into the Beltrami-Klein disk gives \[ {1\over su+c} \left(cu+s , v \right) \] \[= \left({cu+s \over su+c} , {v \over su+c}\right) \] \[= \left({u+s/c \over us/c+1} , {v/c \over us/c+ 1}\right) \] \[= \left({u+\tanh \theta \over 1 + u\tanh \theta} , {v \mathrel\mathrm{sech} \theta \over 1 + u\tanh \theta}\right) \]