$\mathcal{D}$-classes

Question

Let $$\alpha = \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 1&3&3 \end{array}\right) \in \mathcal{T}_3\text{.}$$ (a) Show that the $\mathcal{D}$-class of $\alpha$ contains all those elements of $\mathcal{T}_3$ which have the same rank (cardinality of their image) as $\alpha$.

(b) Show that the $\mathcal{D}$-class of $\alpha$ contains 3 $\mathcal{R}$-classes and 3 $\mathcal{L}$-classes and that it has 2 element $\mathcal{H}$-classes.

How to do (b)?

From (a) we have that $D_{\alpha} = \{ \beta \in \mathcal{T}_3 : | \text{Im}( \beta )| = 2 \}$.

Explicitly then:

$$D_{\alpha} = \left\{ \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 1&1&2 \end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 1&2&1 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 2&1&1 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 2&2&1 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 2&1&2 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 1&2&2 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 3&3&1 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 3&1&3 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 1&3&3 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 1&1&3 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 1&3&1 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 3&1&1 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 2&2&3 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 2&3&2 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 3&2&2 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 3&3&2 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 3&2&3 \end{array}\right) , \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 2&3&3 \end{array}\right) \right\}$$

Answer 1

So you're fine with a)? By the way, this is a special case of a more general fact: the ${\cal D}$-classes in a full transformation semigroup are always the sets of all transformations with a given rank.

For b), try playing around with the elements of the ${\cal D}$-class to find out when they are ${\cal R}$- or ${\cal L}$-related. Hint: it has something to do with images and kernels.

In response to your question below: Okay, so you already knew the thing I was hinting at. An ${\cal R}$-class is just an equivalence class under the relation ${\cal R}$. So you need to show that there are are three different possible kernels and three different possible images for elements of ${\cal D}_\alpha$.

Answer 2

I've had to change accounts as I can't access my email account so I can't reply to any posts because I'm not using the account I originally asked the question from!

So:

$\beta \in R_{\alpha} \iff \text{Ker}(\beta)$ has classes $\{1\},\{2,3\}$ and $\beta \in L_{\alpha} \iff \text{Im}(\beta) = \{1,3\}$

$$R_{\alpha} = \left\{ \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 1&2&2 \end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 2&1&1 \end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 1&3&3 \end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 3&1&1 \end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 2&3&3 \end{array}\right), \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}} 1&2&3\\ 3&2&2 \end{array}\right) \right\}$$