The Mystery of a Presentation of BS(1,2)

On Friday, Professor Alden Walker gave a lovely talk on the Baumslag-Solitar Groups. I was fascinated by the Cayley graph of $BS(1,2)$ . Also, he discussed geodesics between group elements (a shortest but not necessarily unique! path in its Cayley graph that travels one from another). The nice presentation (the focus of this post) of $BS(1,2)$ gives a very easy way of identifying geodesics so that we can even say a bit more about the spheres (the number of group elements a certain geodesic length away from a fixed element), which indeed involves his own research!

With the above said, $BS(1,n)$ is a very special group bearing many interesting properties. In this blog post, I will highlight some special features of the action of $BS(1,n)$ on $\mathbb{R}$ . This action is very different from the actions we used to see in the past three weeks, specifically reflection actions. Without further due, let me recall the definition of $BS(1,n)$ .

In the following definition of $BS(1,n)$ , we directly view them as a subgroup of Homeo( $\mathbb{R}$ ) so that the aspect of them as actions is straightforward. Like what Professor Walker discussed, given the vastness of Homeo( $\mathbb{R}$ ), we would like to find a subgroup that is not trivial, but also not too completely so that any concrete understanding is beyond our ability.

Definition 1. Note that $a(x)=nx,$ where $n\geq 2 \in \mathbb{Z}$ , and $b(x)=x+1$ are elements of Homeo( $\mathbb{R}$ ). Then, we say that the Baumslag-Solitar Group $BS(1,n)$ is the subgroup of Homeo( $\mathbb{R}$ ) generated by $a(x)=nx$ and $b(x)=x+1$ .

Then, we directly state a fact regarding $BS(1,2)$ .

Fact: $BS(1,2)=\langle a,b\mid ab=b^2a \rangle.$

Utilizing the above presentation of $BS(1,2)$ , we draw the Cayley graph of $BS(1,2)$ .

Diagram 1. A plane of the Cayley graph of $BS(1,2)$ . *From Wikipedia, Baumslag–Solitar group.

In the above diagram, we see the relator $ab=b^2a$ of the generating set gives each rectangle, in which we can either go red twice and blue once or go blue once and red once. In other words, this diagram manifests the “weird commutativeness” of $BS(1,2)$ . However, Diagram 1 is only part of the story since we are missing a red edge. Indeed, adding another red edge would result in generating another plane. Therefore, we form a “3-dimensional” picture of the Cayley graph.

Diagram 2. The Cayley graph of $BS(1,2)$ given by the default generating set. *From Wikipedia, Baumslag–Solitar group.

I know you will be amazed by the nice animation above! Thank you, Wikipedia! Clearly illustrated by the animation, we extend another copy of the plane whenever we introduce another red edge “skew” from the current plane.

After introducing $BS(1,2)$ rigorously, I discuss some “weridness” of $BS(1,2) \curvearrowright \mathbb{R}$ . For the linear function $a(x)=2x$ , it achieves the effect of “stretching” the real line. For the linear function $b(x)=x+1$ , we achieve the effect of translating the real line by a unit length. Combining two together, we are able to translate the real line to an arbitrarily small distance. Wait a minute, isn’t that strange? This means that the action is no longer “discrete” but extremely “continuous”.

We can make the idea of an action being discrete a bit more rigorous.

Definition 2. An action $G\curvearrowright X$ is called discrete if every orbit form a discrete subset of $X$ . In other words, for every $x\in X$ , there exists a neighborhood $N_x$ of $X$ such that the intersection of the orbit of $x$ and $N_x$ is exactly $x$ .

Notice that the reflection group $D_\infty \curvearrowright \mathbb{R}$ with reflections an unit distance apart is discrete, which agrees with our intuition that symmetries generated by reflections are rigid and discrete. First, we note that a fundamental domain of $D_\infty \curvearrowright \mathbb{R}$ is $[0,1]$ . From there, we conclude that the orbit of a point is $\{2n\pm r \mid n\in \mathbb{N}\}$ , where $r$ is the decimal value of $x.$ Therefore, we can definitely find a small enough neighborhood such that the intersection is exactly $x$ .

However, when we go back to $BS(1,2) \curvearrowright \mathbb{R}$ , it is a completely different story. Let’s consider the point $0$ , its orbit includes all points $1/2^k$ since

$a^{-k}b(0)=\frac{1}{2^k}.$

Thus, no matter how close we choose a neighborhood around $0$ , it would necessarily intersect with some point in the orbit other than $0$ . Therefore, this action is not discrete.

Another difference we can make between $BS(1,2) \curvearrowright \mathbb{R}$ and $D_\infty \curvearrowright \mathbb{R}$ is the cardinality of the stabilizer. Often times, we would think that the stabilizer of a point is finite, just like in the case of $D_\infty \curvearrowright \mathbb{R}$ . However, it is far from being true for $BS(1,2) \curvearrowright \mathbb{R}$ .

Definition 3. An action $G\curvearrowright X$ is called proper if the stabilizer of any point $x\in X$ is finite.

For $BS(1,2) \curvearrowright \mathbb{R}$ , we claim that the stabilizer is infinite since $\langle a \rangle$ (elements are scalar functions) would keep $0$ fixed. Therefore, the action of $BS(1,2)$ on the real line is neither discrete nor proper. Indeed, similar arguments can be applied on any $BS(1,n)$ so that the action of $BS(1,n)$ on the real line is neither discrete nor proper!

Following the “weridness” of $BS(1,2) \curvearrowright \mathbb{R}$ , can we categorize the orbits of any given points? The answer is yes, and is implied from the following proposition.

Proposition 4. The elements of $BS(1,2)$ are the linear functions $g:\mathbb{R} \to \mathbb{R}$ are in the form

$g(x)=2^n\cdot x + \frac{m}{2^k},$

where $n,m,k$ are all integers.

I sketch a proof of the above proposition. First, we note that the elements of $BS(1,2)$ are all linear functions since composition of linear functions are linear. Then, we use induction on the length of words. For the length 1 case, we note that the formula holds for all $a,a^{-1},b,b^{-1}$ . Assume that the formula holds for all length $\leq N-1$ . Then, we look at the first word. Indeed, there are only four possibilities $a,a^{-1},b,b^{-1}$ again. No matter which one we consider, composing it with the rest (length $N-1$ now so that we can use the hypothesis) would again yield the form mention in Proposition 4. Therefore, the statement follows.

As a final remark in this post, the above proposition can be proved for any $BS(1,n)$ with similar lines of arguments. You can give yourself a small challenge to supply a proof to yourself!

7 Responses to The Mystery of a Presentation of BS(1,2)

Akash Ganguly says:

October 10, 2022 at 3:59 pm

Okay Shuhang this Post was extremely Lit. I loved the animations but especially appreciated the differences you highlighted between BS(1,2) and the infinite dihedral group. Throughout the lecture, I was thinking that there were a lot of similarities between the groups but your examples quickly showed me the difference. Did you come up with these ideas on your own? In that case, I have a question — what is your process? How do you experiment with groups to find out information like that?

I also really loved the animation.

- Shuhang Xue says:
  
  October 17, 2022 at 2:42 pm
  
  Thank you, Akash! I looked at the definition of discrete and proper action in the textbook. In topology, the notion of discrete means exactly the orbit of any point is not a limit point. Then, I tried to come up with an argument of why BS(1,2) does not satisfy either of those things. The fundamental reason is that we can cook up a map g(x)=1/2x, which would then mean that we can make the factor arbitrarily small. In analysis, as soon as we have all powers of 1/2, we have a sequence that converges to zero, suggesting the existence of a limit point in the topological sense.
  
  Indeed, this action is very different from all the other actions we have seen, because it is “dense” at zero and any coordinates in the form m/2^k by a translation related to b. I hope that this mind process helps!
  
Michaela Polley says:

October 13, 2022 at 5:34 am

Nice job, Shuhang! I liked how you really took the topic in a different direction from the lecture. I feel like I have a much more well-rounded understanding of BS(1,2) now. I especially liked how you discussed the idea of discrete vs. continuous actions. It’s fascinating that we can combine two discrete actions to get a continuous one!

Osip Surdutovich says:

October 13, 2022 at 11:19 pm

Great post Shuhang!! I really like your pictures and explanations. One thing I have been wondering is if we can represent the Baumslag Solitar groups in higher dimensionality. It seems that when we take it in three dimensions we are able to to see more of the structure of the group as it builds on itself, what if we go to a fourth dimension? Is there a way to represent that?

- Shuhang Xue says:
  
  October 17, 2022 at 2:54 pm
  
  This is a great question. I suppose that you are asking what if we have one more a=nx or b=x+m? The first thing to notice is that if we have two a_1=n_1x and a_2=n_2x so what we really have in the group would be a=gcd(n_1,n_2)x. If we have two b_1=x+m_1 and b_2=x+m_2, then we can replace them by b=x+min{m_1,m_2,m_1+m_2,m_1-m_2}. Therefore, we should still get a generating set of two elements. However, I might be wrong…
  
- MurphyKate says:
  
  October 19, 2022 at 3:38 pm
  
  I think they will all embed in $\mathbb{R}^3$ . For example, in BS(1, 2) we saw that the Cayley graph is made by glueing together a bunch of “tiles,” each of which is a rectangle with a top of length 1 and a bottom of length 2. In general, BS(n, m) will have a Cayley graph made of tiles that are rectangles with a top of length n and a bottom of length m. So we should get more overlapping tiles, but still have this structure of a tree when viewed from one “side”, and lots of planes when viewed from another. But in this case our “tree” will have a higher valence. (If you Google “Cayley graph of BS(2, 3)” you will find some pictures that show how, at least locally, this embeds in $\mathbb{R}^3$ .)
  
John Eichelberger says:

October 14, 2022 at 2:13 am

I am amazed by the animation! Another great post. As you pointed out at the end, it’s quite a nice fact that BS $(1, n)$ gives you the $n$ -adic numbers, which are dense in $\mathbb{R}$ . In fact this is related to one of the homework problems!

Homeo $(\mathbb{R})$ is vast and terrifying. I wonder if there are more manageable fields whose homeomorphism groups are a little less terrifying.

7 Responses to The Mystery of a Presentation of BS(1,2)

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