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