Tropical linear representations of the Chinese monoid

Abstract

We introduce a faithful tropical linear representation of the Chinese monoid, and thus prove that this monoid admits all the semigroup identities satisfied by upper triangular tropical matrices.

Appendices

Appendix A. Linear representation of $\textsf{Ch}_2$

The Chinese monoid $\textsf{Ch}_2$ of rank 2 is the presented monoid $\textsf{Ch}_2:= \langle a, b \rangle ,$ subject to the relations

$$\begin{aligned} aba = baa \qquad bba = bab, \end{aligned}$$

(CH2)

obtained from (CH) by first letting $a_i = a$, $a_j = a_k = b$, and then taking $a_i = a_j = a$, $a_k = b$. Each element $x \in \textsf{Ch}_2$ can be uniquely written in the canonical form

$$\begin{aligned} \mathfrak {C}(x) =a^i (ba)^j b ^k, \qquad i,j,k \in \mathbb N_0, \end{aligned}$$

with $i = k_{11}$, $j = k_{21}$, $k = k_{22}$ in the notation of (CFb).

Recall that $\mathcal U_2(\mathbb T)$ is the submonoid of $2 \times 2$ upper triangular tropical matrices. In the matrices below we write “$-$" for “$-\infty$", for short. We define the map $\rho : \textsf{Ch}_2 \rightarrow \mathcal U_2(\mathbb T)$ by the generators’ mapping

$$\begin{aligned} a \mapsto A = \begin{pmatrix} 1 &{} 0 \\ - &{} 0 \end{pmatrix} , \qquad b \mapsto B = \begin{pmatrix} 1 &{} 0 \\ - &{} 0 \end{pmatrix} \qquad e \mapsto E = \begin{pmatrix} 1 &{} 0 \\ - &{} 0 \end{pmatrix} . \end{aligned}$$

(M1)

A straightforward computation shows that

$$\begin{aligned} ABA = BAA = \begin{pmatrix} 2 &{} 1 \\ - &{} 1 \end{pmatrix}, \qquad BBA = BAB = \begin{pmatrix} 1 &{} 1 \\ - &{} 2 \end{pmatrix} \end{aligned}$$

(M2)

and thus $\rho$ is a representation.

For an arbitrary $x \in \textsf{Ch}_2$, a straightforward computation gives

$$\begin{aligned} a^i (ba)^j b ^k \longmapsto X= A^i (BA)^j B ^k&= \begin{pmatrix} i &{} i-1 \\ - &{} 0 \end{pmatrix} \begin{pmatrix} j &{} j-1 \\ - &{} j \end{pmatrix} \begin{pmatrix} 0 &{} k-1 \\ - &{} k \end{pmatrix}\\&= \begin{pmatrix} i+j &{} i+j+k-1 \\ - &{} j+k \end{pmatrix} \end{aligned}$$

so that i, j, and k can be uniquely determined by X. That is $i = X_{12}-X_{22} +1$, $j = X_{11} + X_{22} - X_{12} -1$, and $k = X_{12}- X_{11} +1$. Therefore, $\rho : \textsf{Ch}_2 \rightarrow \mathcal U_2(\mathbb T)$ is a faithful representation.

Appendix B. Linear representations of $\textsf{Ch}_3$

Concerning the Chinese monoid $\textsf{Ch}_3:= \langle a,b, c \rangle$ of rank 3, we begin by constructing three linear representations

$$\begin{aligned} \rho _\ell : \textsf{Ch}_3 \longrightarrow \mathcal M_2(\mathbb T), \qquad \ell = 1,2,3, \end{aligned}$$

in terms of generator maps.¹ These representations are later combined together to form a faithful representation of $\textsf{Ch}_3$. Let  x be an element of $\textsf{Ch}_3$, written in the canonical form

$$\begin{aligned} x = a^{k_{11}} (ba)^{ k_{21}} b^{ k_{22}} (ca)^{ k_{31}} (cb)^{k_{32}} c^{k_{33}}. \end{aligned}$$

(CF3)

1.1 Representations I and II

Applying Corollary  1.2 with $\ell = 1,2$ to the faithful linear representation $\rho : \textsf{Ch}_2 \rightarrow \mathcal U_2(\mathbb T)$ defined by  (M1) yields the two (non-faithful) representations $\rho _1, \rho _2: \textsf{Ch}_3 \rightarrow \mathcal U_2(\mathbb T)$, given by:

$$\begin{aligned} \rho _1(a) = \rho _1(b) = A, \ \rho _1(c) = B, \qquad \rho _2(a) = A, \ \rho _2(b) = \rho _2(c) = B. \end{aligned}$$

Then, using Lemma 1.4 (or just a direct calculation), one can show that the matrix image of (CF3) with respect to $\rho _1$ and $\rho _2$ is

$$\begin{aligned} \rho _1(x)&= \begin{pmatrix} k_{11}+2 k_{21}+k_{22}+k_{31}+k_{32} &{} k_{11}+2 k_{21}+k_{22} + k_{31}+k_{32} + k_{33}- 1 \\ - &{} k_{31}+k_{32} +k_{33}, \end{pmatrix} \end{aligned}$$

(R1)

$$\begin{aligned} \rho _2(x)&= \begin{pmatrix} k_{11}+k_{21}+k_{31} &{} k_{11}+k_{21}+ k_{22}+k_{31} +2 k_{32}+k_{33}-1\\ - &{} k_{21}+ k_{22}+k_{31}+2 k_{32}+k_{33} \end{pmatrix} \end{aligned}$$

(R2)

1.2 Representation III

Define $\rho _3: \textsf{Ch}_3 \rightarrow \mathcal U_2(\mathbb T)$ by the generators’ mapping

$$\begin{aligned} a \mapsto A = \begin{pmatrix} 1 &{} 1 \\ - &{} 0 \end{pmatrix} \quad b \mapsto B = \begin{pmatrix} 0 &{} 0 \\ - &{} 0 \end{pmatrix} \quad c \mapsto C = \begin{pmatrix} 0 &{} 1\\ - &{} 1 \end{pmatrix}, \quad e \mapsto E = B. \end{aligned}$$

Note that the matrix B is idempotent, i.e., $B^2 = B$, which acts identically on the left and right of both A and C. So, to verify the relations (CH) it suffices to check that $ACA = CAA$, $BCA = CBA = CAB$ and $CCA = CAC$ (the other relations $BBA = BAB$, $ABA = BAA$, $B C B = CBB$, $CCB = CBC$ are immediate). Indeed, for these matrices we have the products

$$\begin{aligned} \begin{array}{c} BCA = CBA = CAB \begin{pmatrix} 1 &{} 1 \\ - &{} 1 \end{pmatrix}, \\ A C A = CAA = \begin{pmatrix} 2 &{} 2 \\ - &{} 1 \end{pmatrix}\quad CCA = CAC = \begin{pmatrix} 1 &{} 2 \\ - &{} 2 \end{pmatrix} \end{array} \end{aligned}$$

(M3)

so that the relations (CH) are satisfied. Then, an element $x \in \textsf{Ch}_3$ of the form (CF3) is mapped to

$$\begin{aligned} \rho _3(x)&= A^{k_{11}}(BA)^{k_{21}}B^{k_{22}}(CA)^{k_{31}}(CB)^{k_{32}}C^{k_{33}} \\ {}&= A^{k_{11}}(A)^{k_{21}}(CA)^{k_{31}}(C)^{k_{32}}C^{k_{33}}\\&=A^{k_{11}+k_{21}}(CA)^{k_{31}}(C)^{k_{32}+k_{33}}\\&= \begin{pmatrix} k_{11}+k_{21} &{} k_{11}+k_{21} \\ - &{} 0 \end{pmatrix} \begin{pmatrix} k_{31} &{}k_{31}\\ - &{} k_{31} \end{pmatrix} \begin{pmatrix} 0 &{}k_{32}+k_{33} \\ - &{} k_{32}+k_{33} \end{pmatrix}\\&= \begin{pmatrix} k_{11}+k_{21}+k_{31} &{} k_{11}+k_{21}+ k_{31}+k_{32} + k_{33} \\ - &{} k_{31}+k_{32} + k_{33} \end{pmatrix} \end{aligned}$$

(R3)

Remark 2.2

Note that, $k_{11}, k_{21}, k_{22}, k_{31}, k_{32}, k_{33} \in \mathbb N_0$, which represent $x \in \textsf{Ch}_3$ in the canonical form (CF3), are mapped (injectively) by $\rho _1$, $\rho _2$, $\rho _3$ to three matrices (R1), (R2), (R3), having together 9 entries with values in $\mathbb N_0$. Thus, by the standard operations of $\mathbb R$, the exponent sequence $\chi _3(x) = (k_{11}, \dots , k_{33})$ of $x \in \textsf{Ch}_3$, and therefore its canonical form (CF3), can be recovered from the matrices (R1), (R2), (R3).

1.3 A faithful representation

Together, the three representations $\rho _1, \rho _2, \rho _3$ in (R1), (R2), (R3), provide the combined representation

$$\begin{aligned} \widetilde{\rho }:= (\rho _1, \rho _2, \rho _3) : \textsf{Ch}_3 \longrightarrow \big (\mathcal M_2(\mathbb T)\big )^3, \end{aligned}$$

(R)

given in terms of generators’ mapping. Since $\rho _1, \rho _2, \rho _3$ are homomorphisms, it is easily seen from (M2) and (M3) that

$$\begin{aligned} \begin{array}{c} \widetilde{\rho }(bca) = \widetilde{\rho }(cba) = \widetilde{\rho }(cab), \qquad \widetilde{\rho }(aba) = \widetilde{\rho }(baa), \qquad \widetilde{\rho }(bba) = \widetilde{\rho }(bab), \\ \widetilde{\rho }(ccb) = \widetilde{\rho }(cbc), \qquad \widetilde{\rho }(bcb) = \widetilde{\rho }(cbb), \qquad \widetilde{\rho }(cca) = \widetilde{\rho }(cac), \qquad \widetilde{\rho }(aca) = \widetilde{\rho }(caa),\end{array} \end{aligned}$$

where each is different from one another. Moreover, each is also different from $\widetilde{\rho }(bac) \ne \widetilde{\rho }(abc) \ne \widetilde{\rho }(acb)$, which are also different one from another. The product $\big (\mathcal M_2(\mathbb T)\big )^3$ embeds as diagonal blocks in $\mathcal M_6(\mathbb T)$, so that $\widetilde{\rho }$ can be realized as a map $\textsf{Ch}_3 \rightarrow \mathcal M_6(\mathbb T)$.

Lemma 2.3

The map (R) is an injective semigroup homomorphism.

Proof

Let x be an element of $\textsf{Ch}_3$, written in the canonical from (CF3). The entries $X_{ij}$, $Y_{ij}$, $Z_{ij}$ of the matrix images $X = \rho _1(x)$, $Y = \rho _2(x)$, $Z = \rho _3(x)$ give the following system of equations

$$\begin{aligned} \begin{array}{lrll} &{} X_{11} &{} = k_{11} +2 k_{21}+k_{22} + k_{31}+k_{32}, \\ &{} X_{12} &{} = k_{11}+2 k_{21}+ k_{22} + k_{31}+k_{32} + x_{33} - 1, \\ &{} X_{22} = Z_{22} &{} = k_{31}+k_{32} + k_{33}, \\ &{} Y_{11} = Z_{11}&{} = k_{11}+k_{21}+k_{31}, \\ &{} Y_{12} &{} = k_{11}+ k_{21}+k_{22}+ k_{31} +2 k_{32} +k_{33}-1, \\ &{} Y_{22} &{} = k_{21}+k_{22} +k_{31}+2 k_{32} +k_{33}, \\ &{} Z_{12} &{} =k_{11} +k_{21} + k_{31} +k_{32} + k_{33}, \end{array} \end{aligned}$$

cf. (R1), (R2),(R3) respectively. This system has a unique solution, so that the exponents $k_{11}$, $k_{21}$, $k_{22}$, $k_{31}$, $k_{32}$, $k_{33}$ are uniquely determined (cf. Remark 2.2). Hence the homomorphism $\widetilde{\rho }$ in (R) is injective. $\square$