Matrices vs idempotent ultrafilters, part 2

In an earlier post I gave a short introduction to an interesting finite semigroup. This semigroup could be found in the $2\times 2$ matrices over $\mathbb{Q}$.

When I met with said friend, one natural question came up: what other semigroups can we find this way?

The first few simple observations we made were

Remark

• If either $A$ or $B$ is the identity matrix $I_2$ or the zero matrix $0_2$ the resulting semigroup will contain two elements with an identity or a zero element respectively.
• In general, we can always add $I_2$ or $0_2$ to the semigroup generated by $A$ and $B$ and obtain a possibly larger one.
• $A,B$ generate a finite semigroup iff $AB$ is of finite order (in the sense that the set of its powers is finite).
• $AB$ has finite order iff its (nonvanishing) eigenvalue is $+-1$.
• For $A$ of rank $1$ we may assume (by base change) that $A$ is one of the two matrices
  $$\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right).$$

So, as a first approach we thought about the following question.

Question If we take $A$ to be one of the above, what kind of options do we have for $B$, i.e., if $B$ is idempotent and $A,B$ to generate a finite semigroup.

Thinking about the problem a little and experimenting with Macaulay 2 we ended up with the following classification

Proposition For $A=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ the solutions for $B$ being of rank one consist of four one-dimensional families, namely (for $x\in \mathbb{Q}$)

$$\begin{array}{lll}\hfill F_1\left(x\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill x\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),F_2\left(x\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill x\hfill & \hfill 0\hfill \end{array}\right),& & \hfill \end{array}$$

$$\begin{array}{lll}\hfill F_3\left(x\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill x\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),F_4\left(x\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill x\hfill & \hfill 1\hfill \end{array}\right).& & \hfill \end{array}$$ Additionally, we have four special solutions $$\begin{array}{lll}\hfill G_1=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 2\hfill \end{array}\right),G_2=\left(\begin{array}{cc}\hfill -1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 2\hfill \end{array}\right),& & \hfill \end{array}$$

$$\begin{array}{lll}\hfill G_3=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right),G_4=\left(\begin{array}{cc}\hfill -1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right).& & \hfill \end{array}$$

Note: due to technical problems, this post continues here.

We can also describe size and the algebraic structure.

• $A$ with $F_1$ ($F_2$) generates a right (left) zero semigroup (hence of size $2$, except for $x=0$).
• $A$ with $F_3$ or $F_4$ generates a semigroup with $AB$ nilpotent (of size $4$, except for $x=0$, where we have the null semigroup of size $3$).
• $A$ with $G_i$ generate (isomorphic) semigroups of size $8$. These contain two disjoint right ideals, two disjoint left ideals generated by $A$ and $B$ respectively.

Luckily enough, we get something very similar from our alternative for $A$.

Proposition In case $A=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ the solutions for $B$ being of rank one consist of five one-dimensional families namely (for $x\in \mathbb{Q}$)

$$\begin{array}{llll}\hfill & H_1\left(x\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill x\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),H_2\left(x\right)=\left(\begin{array}{cc}\hfill x+1\hfill & \hfill x\hfill \\ \hfill -x-1\hfill & \hfill -x\hfill \end{array}\right),& \hfill & \\ \hfill & H_3\left(x\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill x\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),H_4\left(x\right)=\left(\begin{array}{cc}\hfill -x+1\hfill & \hfill -x+1\hfill \\ \hfill x\hfill & \hfill x\hfill \end{array}\right),& \hfill & \\ \hfill & H_5\left(x\right)=\left(\begin{array}{cc}\hfill -x+1\hfill & \hfill -x-1-\frac{2}{x-2}\hfill \\ \hfill x-2\hfill & \hfill x\hfill \end{array}\right),x\ne 2.& \hfill & \end{array}$$

As before we can describe size and structure.

• $A$ with $H_1$ ($H_2$) generates a right (left) zero semigroup (as before).
• $A$ with $H_3$ or $H_4$ generates a semigroup with $AB$ nilpotent (as before).
• $A$ with $H_5$ generates the same $8$ element semigroup (as before).

Finally, it might be worthwhile to mention that the seemingly missing copies of the $8$ element semigroup are also dealt with; e.g. $-G_i$ generates the same semigroup as $G_i$ etc.

At first sight it seems strange that we cannot find other semigroups with two generators like this. As another friend commented, there’s just not enough space in the plane. I would love to get some geometric idea of what’s happening since my intuition is very poor. But that's all for today.

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