## Donnerstag, 7. Januar 2010

### 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 $2×2$ matrices over $ℚ$.

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
$\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\phantom{\rule{0.3em}{0ex}}.$

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 $x\in ℚ$)

$\begin{array}{lll}\hfill {F}_{1}\left(x\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill x\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),{F}_{2}\left(x\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill x\hfill & \hfill 0\hfill \end{array}\right),& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

$\begin{array}{lll}\hfill {F}_{3}\left(x\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill x\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),{F}_{4}\left(x\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill x\hfill & \hfill 1\hfill \end{array}\right).& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$ Additionally, we have four special solutions $\begin{array}{lll}\hfill {G}_{1}=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 2\hfill \end{array}\right),{G}_{2}=\left(\begin{array}{cc}\hfill -1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 2\hfill \end{array}\right),& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

$\begin{array}{lll}\hfill {G}_{3}=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right),{G}_{4}=\left(\begin{array}{cc}\hfill -1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right).& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

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 $x\in ℚ$)

$\begin{array}{llll}\hfill & {H}_{1}\left(x\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill x\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),{H}_{2}\left(x\right)=\left(\begin{array}{cc}\hfill x+1\hfill & \hfill x\hfill \\ \hfill -x-1\hfill & \hfill -x\hfill \end{array}\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & {H}_{3}\left(x\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill x\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right),{H}_{4}\left(x\right)=\left(\begin{array}{cc}\hfill -x+1\hfill & \hfill -x+1\hfill \\ \hfill x\hfill & \hfill x\hfill \end{array}\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & {H}_{5}\left(x\right)=\left(\begin{array}{cc}\hfill -x+1\hfill & \hfill -x-1-\frac{2}{x-2}\hfill \\ \hfill x-2\hfill & \hfill x\hfill \end{array}\right)\phantom{\rule{0.3em}{0ex}},\phantom{\rule{3.33151pt}{0ex}}x\ne 2.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

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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