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Sol geometry and the tangent bundle of a hyperbola

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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.

Sol3 is a stack of planes.
\(\Sol^3\) 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 \(x^2 - y^2 = 1\).

A tangent line to a hyperbola
A tangent line to a hyperbola

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 \[ \begin{align*} (0,0,z)(p,q) &= (p,q+z) \\ (x,y,0)(p,q) &= (p+y \cosh q - x \sinh q, q) . \end{align*} \]

A diffeomorphism and the image of the q-axis
A diffeomorphism and the image of the q-axis

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

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 \[ (x,y,z)(p,q) = (p + ye^{q+z} - xe^{-q-z}, q+z) . \]
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