3-Dimensional Space

Fiber bundle

Hyperbolic surface suspension

Consider a closed compact surface $\Sigma$ of genus $g \geq 2$. Given a homeomorphism $f$ of $\Sigma$, one can produce a 3-dimensional manifold $M$ by suspending $\Sigma$ according to $A$. More precisely $M$ is the quotient of $\Sigma \times [0,1]$ by the equivalence relation identifying $(x,0)$ with $(f(x),1)$. In some cases, the universal cover $X$ of $M$ can be identified with one of the eight Thurston geometries.

if $f$ has finite order, then $X = \mathbb H^2 \times \mathbb E$.
if $f$ is pseudo-Anosov, then $X = \mathbb H^3$.

For a more general homeomorphism $f$, the geometrization conjecture applies. That is one can cut $M$ along tori so that each resulting piece admits a geodesic structure. Here the tori correspond to the suspension of simple closed curved fixed by $f$ (???)

Isotropic geometry

All the Thurston geometries are homogenous (i.e. all the points play the same role). In an isotropic geometry $X$, all the directions also play the same role. Said differently, the stabilizer of a point $x \in X$ in the geometry is isomorphic to $O(3)$ acting by isometries on the tangent space $T_xX$.

Lie group

Those geometries are themselves a Lie group $G$, so that the left action of $G$ on itself is an action by isometries.

Torus suspension

Given a matrix $A \in {\rm SL}(2, \mathbb R)$, one can produce a 3-dimensional manifold $M$ by suspending the 2-torus $T$ according to $A$. More precisely $M$ is the quotient of $T \times [0,1]$ by the equivalence relation identifying $(x,0)$ with $(Ax,1)$. The universal cover $X$ of $M$ can be identified with one of the eight Thurston geometries

if $A$ has finite order, then $X = \mathbb E^3$.
if $A$ is a Dehn twist, then $X = {\rm Nil}$.
if $A$ is Anosov, then $X = {\rm Sol}$.

Unit tangent bundle

Given a 2-dimensional compact manifold $M$, its unit tangent bundle $T^1M$ is a 3-dimensional compact manifold. Depending on the curvature of $M$, this new space carries one of the eight Thurston geometries.

If $M$ has positive curvature, then $T^1M$ is modelled on $S^3$.
If $M$ has zero curvature, then $T^1M$ is modelled on $\mathbb E^3$.
If $M$ has negative curvature, then $T^1M$ is modelled on $\widetilde{\rm SL}(2, \mathbb R)$.

Tags