Matrices vs. idempotent ultrafilters part 2

07 Jan 2010

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

\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\,.

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 = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}

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

F_1(x) = \begin{pmatrix} 1 & x \\ 0 & 0 \end{pmatrix}, F_2(x) = \begin{pmatrix} 1 & 0 \\ x & 0 \end{pmatrix}, F_3(x) = \begin{pmatrix} 0 & x \\ 0 & 1 \end{pmatrix}, F_4(x) = \begin{pmatrix} 0 & 0 \\ x & 1 \end{pmatrix}.

Additionally, we have four special solutions

G_1 = \begin{pmatrix} – 1 & 1 \\ – 2 & 2 \end{pmatrix}, G_2 = \begin{pmatrix} – 1 & – 1 \\ 2 & 2 \end{pmatrix}, G_3 = \begin{pmatrix} – 1 & 2 \\ – 1 & 2 \end{pmatrix}, G_4 = \begin{pmatrix} – 1 & – 2 \\ 1 & 2 \end{pmatrix}.

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 = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}

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

H_1(x) = \begin{pmatrix} 1 & x \\ 0 & 0 \end{pmatrix},

H_2(x) = \begin{pmatrix} x+1 & x \\ ( – x – 1) & – x \end{pmatrix},

H_3(x) = \begin{pmatrix} 0 & x \\ 0 & 1 \end{pmatrix},

H_4(x) = \begin{pmatrix} ( – x+1) & ( – x+1) \\ x & x \end{pmatrix},

H_5(x) = \begin{pmatrix} ( – x+1) & ( – x – 1 – \frac{2}{x – 2}) \\ x – 2 & x \end{pmatrix} , x \neq 2.

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.

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