Peter Krautzberger on the web

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

  • 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 $\pm 1$.
  • For $A$ of rank $1$ we may assume (by base change) that $A$ is one of the two matrices

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 the solutions for $B$ being of rank one consist of four one – dimensional families, namely (for $x\in \mathbb{Q}$)

Additionally, we have four special solutions

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

We can also describe size and the algebraic structure.

  1. $A$ with $F_1$ ($F_2$) generates a right (left) zero semigroup (hence of size $2$, except for $x=0$).
  2. $A$ 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 $3$).
  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 the solutions for $B$ being of rank one consist of five one – dimensional families namely (for $x\in \mathbb{Q}$)

As before we can describe size and structure.

  1. $A$ with $H_1$ ($H_2$) generates a right (left) zero semigroup (as before).
  2. $A$ with $H_3$ or $H_4$ generates a semigroup with $AB$ nilpotent (as before).
  3. $A$ 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.

It is striking to see that the orders of all finite semigroups generated by rational idempotent two by two matrices are either $2^k,2^k + 1$ or $2^k + 2$.

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. pdf