## 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 ${\mathbb{R}^2}$). 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. 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. 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. Here is the action of ${B}$. 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 ${B}$. Generating 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). Here are the actions of ${A}$ and ${B}$ on the above picture. For clarity, I will also demonstrate the Cayley tree with the tiling appearing faintly in the background. 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 ${B}$.  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? Advertisements

## Modular group in action

I’m sorry for the long gap, though there’s nothing really to be sorry about. I know I wasn’t missed.

This posts will present some of my newly created gifs. Before showing the outcome, first let’s delve into some theory (the hasty reader is free to scroll to the juicier part, of course). As you may have perhaps deduced from the title, the theme of discussion today is the Modular group.

By the modular group ${PSL(2,\mathbb{Z})}$, we almost mean the following group (fondly known as ${SL(2,\mathbb{Z})}$). $\displaystyle SL(2,\mathbb{Z}) := \{ \left( \substack{a\ b \\ c\ d} \right)\ ,ad-bc=1 , (a,b,c,d) \in \mathbb{Z}^4 \}. \ \ \ \ \ (1)$

However, in reality, the modular group is just slightly away from being that group. It’s the group given by the quotient group $\displaystyle PSL(2,\mathbb{Z}):=SL(2,\mathbb{Z})/\{\pm 1_{SL(2,\mathbb{Z})}\}. \ \ \ \ \ (2)$

There is a natural action of the modular group on the complex upper half plane, which is $\displaystyle \mathbf{H}=\{z \in \mathbb{C}, \ \Im(z) > 0 \}. \ \ \ \ \ (3)$

The action of an element ${\pm (\substack{a\ b \\ c\ d})}$, is given by (check that the sign does not matter) $\displaystyle z \mapsto \frac{az+b}{cz+d}. \ \ \ \ \ (4)$

What is almost so magical about the above action is that it actually is a group action. That is, there is associativity with the group product and group identity fixes everything. The former of these two facts is not immediately trivial, but to witness it, look at the following calculation: $\displaystyle (\substack{a_1\ b_1 \\ c_1\ d_1})\Big{(}(\substack{a_2\ b_2 \\ c_2 \ d_2}) z\Big{)} = (\substack{a_1 \ b_1 \\ c_1 \ d_1})(\frac{a_2 z + b_2}{ c_2 z + d_2}) \ \ \ \ \ (5)$ $\displaystyle = \frac{(a_1 a_2 + b_1 c_2)z + (a_1 b_2 + b_1 d_2)}{ (c_1 a_2 + d_1 c_2) z + (c_1 b_2 + d_1 d_2)} \ \ \ \ \ (6)$

The group action of the modular group on the upper half plane has inspired a lot of mathematics. It has deep connections with number theory and hyperbolic geometry. There are a lot of reasons to talk about these groups. There are complete books written on these subjects.

We will also witness another concept called the fundamental domain. Suppose we have a group ${G}$ acting on a topological space ${X}$. Then a fundamental domain of the action of ${X/G}$ (which is the set of ${G}$-orbits endowed with the Quotient topology), is a certain set ${D}$ inside ${X}$ which is in a set-bijection with ${X/G}$. More precisely, it is a region ${D\subset X}$ satisfying

• For any ${\varphi \in G}$, ${\varphi D \cap D \neq \phi \Leftrightarrow \varphi = 1_G}$,
• ${X= \bigsqcup_{\varphi \in G} \varphi D}$ .

If you’ve ever googled these things, the most commonly given example is always of the modular action, and how there is a fundamental domain ${D \subset \mathbf{H}}$ such that the following holds true $\displaystyle U \subset D \subset \bar{U}, \ \ \ \ \ (7)$

where ${U}$ is the set $\displaystyle U : = \{ z \in \mathbb{C},\ |z| > 1, \ |\Re(z)|<\frac{1}{2} \}. \ \ \ \ \ (8)$

A picture of this region (accurate upto boundary, beyond which it will be impossible for a computer to be accurate), is given here: When acted upon by different elements of the modular group, the following “tiling” can be generated: It is widely known that the modular group ${PSL(2,\mathbb{Z})}$ can be generated by the generators ${S,T}$ as described by $\displaystyle S:= \pm \Big{(} \substack{0\ -1 \\ 1 \ \ 0}\Big{)} ,\ T:= \pm \Big{(} \substack{1 \ 1 \\ 0 \ 1} \Big{)}. \ \ \ \ \ (9)$

In the language of the above matrices, it is known that the modular group ${PSL(2,\mathbb{Z})}$ is given as the presentation $\displaystyle \langle S , T \ | \ S^2=1,\ (ST)^3=1 \rangle. \ \ \ \ \ (10)$

Here is the action of ${T}$ (a little uneventful, unfortunately): Here is the action of the element the element ${S}$ upon our favorite tiling: To keep track of the flipping around, here is the same with each tile randomly colored (it was difficult to color the inside of a non-convex tile on the computer without triangulating it, so I chose the following scheme instead): Here is the action of ${T}$ on the Poincare Disk model Here is the action of ${S}$ on the same (looking much less dramatic): While proving a landmark result (the Monstrous moonshine conjecture) related to modular forms, which are related to our discussion, Richard Borcherds had recounted his experience by saying, “I sometimes wonder if this is the feeling you get when you take certain drugs. I don’t actually know, as I have not tested this theory of mine.”

Inspired by this, I have created the following imagery (the colours of the tiles are a smooth function of time): 