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For a certain commutative semigroup of integer size $n$, $G=(\{1,2,\dots,n\},\circ: x\circ y\mapsto \min(n,x+y))$, consider all complex functions on it, denoted by $\mathbb C[G]$ or $\mathbb CG$. Where $\mathbb C[G]=\{f:G\rightarrow \mathbb C\}$. For functions in $\mathbb C[G]$, $G$ is called indices domain and $\mathbb C$ is called value range.

$\mathbb C[G]$ is also regarded as a vector/linear space with scalar space $\mathbb C$, with natural addition/subtraction and scalar multiplication.

Besides, a multiplication $\times$ is defined between two functions in $\mathbb C[G]$. $(i\mapsto f(i))\times (i\mapsto g(i))=(i\mapsto \sum_{\forall j,k,\mathrm{s.t.} j\circ k=i} f(j)g(k))$.

Then we consider matrix representations of $\mathbb C[G]$. A matrix representation is a injective homomorphism between $\mathbb C[G]$ and $\mathbb C^{m\times m}$ ($m\ge n$, remind that $n=\lvert G\rvert$). Where multiplication in $\mathbb C[G]$ corresponds to matrix multiplication.

Notice that a matrix representation of $\mathbb C[G]$ corresponds to a unique matrix representation of $G$ with same order of matrices. Since $\mathbb C[G]$ has a basis of $\{(j\mapsto [j=i]):i=1,2,\dots,n\}$. Where $[~]$ is Iverson bracket, $[P]=1$ if $P$ is true and $[P]=0$ if $P$ is false.

A naive representation is that let $m=n+1$, extend $G$ to $G'=G\cup \{e\}$ by adding an extra unitary element $e$ (also denoted by integer $n+1$) where $e\circ a=a\circ e=e, \forall a\in G'$. Then define the representation of $G'$ by $u\mapsto A^{(u)}$ for $u\in G'$. Where $A^{(u)}_{ij}=[u\circ i=j]$.

Now the problem is: is there any non-trivial representation of $\mathbb C[G]$? That is, is there any representation such that all the matrices are in a block-diagonal space with the most number of blocks? That means, the representation is not only in $C^{m\times m}$, but also in $C^{d_1\times d_1}\times C^{d_2\times d_2}\times \dots \times C^{d_k\times d_k}$ where $\sum_{i=1}^k d_i=m$. And the target is to maximize $k$.

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    $\begingroup$ Your question is a bit confusing. First of all a semigroup should be denoted $S$ and not $G$. In any event, there is no invertible homomorphism $\mathbb C[G]\to \mathbb C^{m\times m}$ because $\mathbb C[G]$ is not a unital algebra and also it is commutative and for many other reasons. Do you mean just an injective homomorphism? If I define representation to mean homomorphism then up to equivalence every representation of $G$ sends $k$ to $diag(A^k,I)$ where $A$ is a square matrix with $A^n=0$ and $I$ is an identity matrix. You can of course assume $A$ as a strictly upper triangular. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 26, 2022 at 15:22
  • $\begingroup$ The above representation will be injective on $\mathbb C[G]$ iff $A^{n-1}\neq 0$. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 26, 2022 at 15:25
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    $\begingroup$ Note that $\mathbb C[G]\cong \mathbb C\times I$ where $I$ is the ideal of $\mathbb C[t]/(t^n)$ generated by $t+(t^n)$. The isomorphism takes $k$ to $(1,t^k)$ where I identify $G$ with the basis for $\mathbb C[G]$. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 26, 2022 at 15:27

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