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Let $A$ be an irreducible nonnegative $N\times N$ integer matrix with constant row sum $D$. Let $A_1, \dots, A_D$ be nonnegative integer matrices, each with constant row sum $1$, such that $\sum_k A_k = A$. Each $A_k$ defines a mapping $f_k: [N] \to [N]$ where $f_k(i) = j$ iff $(A_k)_{ij} = 1$. Let $S \subset [N]^{[N]}$ denote the transformation semigroup generated by the $f_k$'s; equivalently, the finite semigroup of matrices with row sum $1$ generated by the $A_k$'s. Let $S^*$ denote the minimal ideal of $S$. As matrices, the elements $S^*$ are the matrices with minimal rank $r$; as mappings, these are the $f$ with $|f([N])|=r$.

I am interested in bounds on the possible values of $r$ and $|S^*|$ in terms of $D$ and $N$. There are some trivial bounds, but I am wondering if someone who knows e.g. Krohn-Rhodes theory could point me in a suitable direction.

I am particularly interested in using such bounds to address the following question. For $f \in S^*$, let $\alpha_f$ be the partition of $[N]$ where $\alpha_f(i) = \alpha_f(j) \Leftrightarrow f(i) = f(j)$. Let $\alpha^* = \bigvee_{f \in S^*} \alpha_f$, the coarsest common refinement of the $\alpha_f$'s (I'm aware that in some settings, the opposite convention for $\vee$ vs. $\wedge$ is used). So $\alpha^*(i) = \alpha^*(j)$ iff $\alpha_f(i) = \alpha_f(j)$ for all $f \in S^*$.

I would really like to show that there is a choice of the $A_k$'s such that $\alpha^*$ is not the discrete partition, i.e. such that $|\alpha^*| \leq N-1$. (This is equivalent to Trahtman's Road Colouring Theorem, but I would like to avoid using that result, as I'm trying to generalize the result to a setting where his method does not work well.)

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  • $\begingroup$ Just to translate to my language you have a strongly connected graph with constant out-degree D and you want to turn it into an automaton on a D letter alphabet and ask about bounds on the minimum rank and the size of the minimal ideal of the transition monoid. I'm confused on alpha*. So you want to identify states is any two elements of the minimal ideal agree on them or if some element agrees on them? Krohn-Rhodes theory probably won't help you with this problem because it is about increasing state size to get a nicer action. Also only your last question involves A. $\endgroup$ Commented Mar 26, 2021 at 18:21
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    $\begingroup$ It seems that you are asking a question about a known result that isn't really the result in which you're interested, in order to get idas for a result in which you are interested. If that's so, then you'll probably get better responses if you specify what the actual result is in which you're interested. $\endgroup$
    – LSpice
    Commented Mar 26, 2021 at 20:28
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    $\begingroup$ @LSpice well, perhaps, but I wasn't deliberately obfuscating things. I am looking to re-prove an existing result (the Road Colouring Theorem) by new methods, and I have an idea of how to do it but don't know how to make a key step work. So I'm describing what I'm trying to do in that step, and what I don't about it, in the hope that someone more familiar with the relevant objects can offer ideas. If this makes sense and you have suggestions about a way to go about things, I would be glad to edit the question further to make it more appropriate. $\endgroup$
    – Sophie M
    Commented Mar 26, 2021 at 20:36
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    $\begingroup$ At least to me, moving your explanation that this is a known result earlier, and editing in your explanation from your comment that you are trying to re-prove the result by different means, would make things clearer, especially about what you are actually asking; but that may just be personal taste. $\endgroup$
    – LSpice
    Commented Mar 26, 2021 at 20:40
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    $\begingroup$ Krohn-Rhodes is about writing a finite semigroup as a quotient of a subsemigroup of a nice wreath product so it will typically blow up the minimal ideal to do this. That's why I don't see it helping to bound S*. $\endgroup$ Commented Mar 26, 2021 at 20:46

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