Deck transformations, revisted

A long time ago, I had animated the deck transformations of the covering space of a pair of circles. I’m talking about this blog post.

A lot of time has passed and I couldn’t help but rethink it in the light of some insights that I later had after the post. The deck transformations that I had created in my first attempt are quite unnatural. The reason is simply that a Cayley graph of a free group of two generators cannot be embedded inside {\mathbb{R}^2} with the geometry intact. Therefore, any attempt to visualize the Cayley graph in that setting means forgetting about the geometry.

However, there is one setting in which the Cayley graph can naturally exist, without lying about its geometry. And that place is the hyperbolic 2-space (or hyperbolic n-space)! I was inspired by this idea when I found about this project called Walrus. This program is a way to visualize exponentially growing graphs using tools of hyperbolic geometry. The Cayley graph discussed in the previous blog post is also an exponentially growing graph.

Earlier, we had considered the universal covering space of the following topological space (think of it as a subspace of {\mathbb{R}^2}).

circles

The covering space looked like the following. It is a graph, better expressed with a bunch of line segments quotiented away at the nodes that connect them.

 

cayley_graph_of_f2

 

Notice that the above graph is the same as the following graph. All the vertices are tetravalent like how they should be. As some of you may be able to guess, this graph is actually embedded inside the Poincare disc, with the line segments now being the Hyperbolic geodesics of the negative curvature space.

sametree

For clearer understanding, let us glance over what I mean. As a set, the Poincare disc {D} is given by

\displaystyle D= \{ z \in \mathbb{C} \ | \ |z| < 1\}. \ \ \ \ \ (1)

However, I would be kidding you if I told you that this is it. The Poincare disc is a metric space with the following metric. For two points {z_1,z_2 \in D}, the distance {d(z_1,z_2)} between them is given by

\displaystyle \cosh(d(z_1,z_2)) = 1 + 2\frac{|z_1-z_2|^2}{(1-|z_1|^2)(1-|z_2|^2)} . \ \ \ \ \ (2)

Of course, this bizarre formula is not very telling. It’s not even clear why this is a metric (for instance, why does it satisfy the triangle inequality?). Questions like these, and why this metric space is interesting will require a lot of discussion and we’ll not get into it. Interested readers can find this in plenty of books. To get a short introduction, you can watch this numberphile video.

The group of isometries (set of distance preserving homeomorphisms {D \rightarrow D}) of the space {D} also happen to have a very neat and natural description. It is the group {PSU(1,1)} which is the group defined as

\displaystyle PSU(1,1):= \{ \left( \substack{a\ b \\ \bar{b}\ \bar{a}} \right)\ ,|a|^2-|b|^2=1 , (a,b,c,d) \in \mathbb{C}^4 \} / \{ \pm 1\} . \ \ \ \ \ (3)

The action of this group on {D} is via

\displaystyle z \mapsto \frac{az+b}{cz+d}. \ \ \ \ \ (4)

Enthusiastic readers of my blog (if there are any) will be able to relate the above formula to the modular group action in this blog post. They’re both examples of fractional linear transformations. In fact, the group {PSU(1,1)} is isomorphic to the group {PSL(2,\mathbb{C})} via a conjugation by some matrix (there are many such). Talking about this will lead us to the upper-half plane model of Hyperbolic geometry, the discussion of which we will save for some other time.

I am interested in two particular matrices:

\displaystyle A:= \pm \Big{(} \substack{\sqrt{2}\ \ 1 \\ 1 \ \ \sqrt{2}}\Big{)} ,\ B:= \pm \Big{(} \substack{\sqrt{2}\ \ i \\ -i\ \ \sqrt{2}} \Big{)}. \ \ \ \ \ (5)

These matrices are carefully chosen, so that the following happens. Here is the action of {A} on the picture produced previously.

treeright

Here is the action of {B}.

treenotdown.gif

How were those specific matrices calculated? It’s enough for me to show this picture to you perhaps. This is the fundamental region of the group {\langle A,B \rangle \subset PSU(1,1)}, which is the subgroup of {PSU(1,1)} generated by the matrices {A} and {B}.stillfundregionGenerating the entire tiling using the above fundamental region gives us the following pictures (the pointy edges in the four corners are something that is unnatural. It is because I could only calculate finitely many tiles. Those pointy corners would look denser, if it were calculated by a computer that had more time).

stilltiling

Here are the actions of {A} and {B} on the above picture.

tilingrightup.gif

For clarity, I will also demonstrate the Cayley tree with the tiling appearing faintly in the background.

skillskeleton

This group {\langle A, B\rangle} is, in fact, isomorphic to the free group of two generators, as you would expect. It must be isomorphic to the group of deck transformations of the covering space of {B}. Another direct way to prove it is to use the Ping-Pong lemma.

Here are some more trippy visuals of the action of {A} and {B}.

 

christmastree.gif

somethingcool.gif

 

Although in the previous post I had made a visualization of the covering map itself, I don’t think there is a natural way that map can be visualized using hyperbolic isometries, without which the purity of the Hyperbolic space will be ruined. If any of my readers gets an insight of having a way to map the Cayley tree to the wedge of two circles in a natural way, I would be excited to listen to your opinions in the comments.

One important discussion here may be the fact that {\frac{D}{\langle A, B\rangle}}, which is the topological space of orbits of {D} under the action of {\langle A, B\rangle}, is homeomorphic to a torus with a point removed. A torus with a point removed has the same homotopy equivalence (and hence the same fundamental group) as that of a wedge of two circles (How? I wouldn’t ruin it for you). Do you smell a connection? It’s an interesting bedtime thought, trying to put this picture together. Someday maybe in a future project, I will try to make a visualization about it.

Here’s an exercise for you. Can you guess which element of {\langle A, B\rangle} does the following transformation correspond to?

 

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