 # Sol geometry and the tangent bundle of a hyperbola

Posted by in math

If you try to make rigid motions in some dimension act as transformations of a lower-dimensional space, you might hope that it respects some structure in the lower-dimensional space. Unfortunately this is guaranteed for neither distances nor angles. Here are some pictures—one of which is interactive (!)—for a particularly pretty example using Thurston’s Sol geometry.

$$\newcommand{\R}{\mathbb{R}} \newcommand{\H}{\mathbb{H}} \newcommand{\SO}{\operatorname{SO}} \newcommand{\Sol}{\operatorname{Sol}}$$

## Setup

Definition
$$\Sol^3$$ is the Lie group $$\R^2 \rtimes \SO(1,1)^0$$.

Visualize $$\Sol^3$$ as a line’s worth of copies of $$\R^2$$, where moving up or down in the stack stretches them in one direction and shrinks them by the same factor in the other. $$\Sol^3$$ is a stack of planes. (Interactive)

We are interested in the action of $$\Sol^3$$ on the tangent bundle $$T\H^1$$ of one branch of a hyperbola.

## Description of the action

Let $$\mathcal{F}$$ be the 1-D foliation spanned by $$(\cosh z, \sinh z, 0)$$ at $$(x,y,z)$$. Under the metric $$dx^2 - dy^2$$ on a plane $$\{(*,*,z)\}$$, the line tangent to $$x^2 - y^2 = 1$$ at $$(\cosh z, \sinh z, 0)$$ is orthogonal to $$\mathcal{F}$$. This identifies $$\Sol^3/\mathcal{F}$$ with the tangent bundle $$T\H^1$$ of one branch of $$x^2 - y^2 = 1$$.

Since $$\mathcal{F}$$ is left-invariant (being spanned by a left-invariant vector field), the left action of $$\Sol^3$$ on itself descends to $$T\H^1$$. Letting $$(p,q)$$ denote a tangent velocity of $$p$$ at $$q \in \H^1$$, you can check that this action is given by \begin{align*} (0,0,z)(p,q) &= (p,q+z) \\ (x,y,0)(p,q) &= (p+y \cosh q - x \sinh q, q) . \end{align*}

## No preservation of distances or angles

Were we hoping to preserve either in $$T\H^1$$, we’d be out of luck—the stabilizer of $$(0,0)$$ consists of maps $(p,q) \overset{(x,0,0)}{\mapsto} (p - x \sinh q, q) ,$ whose action on the tangent space $$T_{(0,0)}(T\H^1)$$ is by matrices $\begin{pmatrix} 1 & * \\ & 1 \end{pmatrix}.$

## Diffy queues

You can check using the formula that $$\Sol^3$$ acts faithfully on $$T\H^1$$. In fact, here’s a nice description of $$\Sol^3$$ as its image in $$\operatorname{Diff} T\H^1$$.

Proposition
The image of $$\Sol^3$$ in $$\operatorname{Diff} T\H^1$$ consists of the diffeomorphisms $$f = (f_p, f_q)$$ that satisfy \begin{align*} \frac{\partial f_p}{\partial p} &= 1 & \frac{\partial^2 f_p}{\partial q^2} &= f_p \\ \frac{\partial f_q}{\partial p} &= 0 & \frac{\partial f_q}{\partial q} &= 1 . \end{align*}

## Recovering the group

To put it another way, the image consists of those diffeomorphisms expressible as $$P \circ Q_z$$ where

• $$Q_z$$ is translation by $$(0,z)$$, and
• $$P$$ adds to $$p$$ some linear combination of $$\cosh q$$ and $$\sinh q$$ (or of $$e^q$$ and $$e^{-q}$$).

To see that this is $$\Sol^3$$, parametrize the space $$V$$ of images of the $$q$$-axis as $V = \{ \text{graph of } (q \mapsto xe^q + ye^{-q}) \mid x,y \in \R \} .$ Then $$Q_z$$, sending $$q$$ to $$q+z$$, acts on $$V$$ by $$Q_z (x,y) = (xe^z, ye^{-z})$$, so $\Sol^3 = V \rtimes \{Q_z \mid z \in \R\} .$

Remark
This is an alternative parametrization of $$\Sol^3$$ under which the action on $$T\H^1$$ becomes $(x,y,z)(p,q) = (p + ye^{q+z} - xe^{-q-z}, q+z) .$