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 % <![CDATA[ \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 % <![CDATA[ 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} )
% <![CDATA[ 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
% <![CDATA[ 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.

  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 % <![CDATA[ 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} )
% <![CDATA[ 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}, %]]> % <![CDATA[ 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.

  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