A Munn type representation of abundant semigroups with a multiplicative ample transversal

Abstract

The celebrated construction by Munn of a fundamental inverse semigroup $T_E$ from a semilattice E provides an important tool in the study of inverse semigroups and ample semigroups. Munn’s semigroup $T_E$ has the property that a semigroup is a fundamental inverse semigroup (resp. a fundamental ample semigroup) with a semilattice of idempotents isomorphic to E if and only if it is embeddable as a full inverse subsemigroup (resp. a full subsemigroup) into $T_E$. The aim of this paper is to extend Munn’s approach to a class of abundant semigroups, namely abundant semigroups with a multiplicative ample transversal. We present here a semigroup $T_{(I,\Lambda , E^{\circ }, P)}$ from a so-called admissible quadruple $(I,\Lambda , E^{\circ }, P)$ that plays for abundant semigroups with a multiplicative ample transversal the role that $T_E$ plays for inverse semigroups and ample semigroups. More precisely, we show that a semigroup is a fundamental abundant semigroup (resp. fundamental regular semigroup) having a multiplicative ample transversal (resp. multiplicative inverse transversal) whose admissible quadruple is isomorphic to $(I,\Lambda , E^{\circ }, P)$ if and only if it is embeddable as a full subsemigroup (resp. full regular subsemigroup) into $T_{(I,\Lambda , E^{\circ }, P)}$. Our results generalize and enrich some classical results of Munn on inverse semigroups and of Fountain on ample semigroups.

1 Introduction

Let S be a semigroup. We denote the set of all idempotents of S by E(S) and the set of all inverses of $x\in S$ by V(x). Recall that $V(x)=\{a\in S|xax=x, axa=a\}$ for all $x\in S$. A semigroup S is called regular if $V(x)\not =\emptyset$ for any $x\in S$, and a regular semigroup S is called inverse if E(S) is a commutative subsemigroup (i.e. a subsemilattice) of S, or equivalently, the cardinal of V(x) is equal to 1 for all $x\in S$.

Recall that a regular semigroup S is fundamental if the largest congruence contained in ${\mathcal {H}}$ on S is the identity congruence. Structure theorems for certain important subclasses of the class of fundamental regular semigroups are already known. The first initiating the work in this direction is due to Munn [19]. He proved that given a semilattice E, the Munn semigroup $T_E$ of all isomorphisms of principal ideals of E is “maximal” in the class of all fundamental inverse semigroups whose semilattices of idempotents are E, that is, every semigroup belonging to this class is isomorphic to a full inverse subsemigroup of $T_E$. Further from Munn [19] if S is an inverse semigroup such that E(S) is isomorphic to a given semilattice E, then there exists a homomorphism $f: S\rightarrow T_E$ and the kernel of f is the largest congruence contained in ${\mathcal {H}}$ on S.

The pioneering work of Munn was generalized first by Hall in 1971 to orthodox semigroups (i.e. regular semigroups whose idempotents form subsemigroups) in [17] in which the Hall semigroup $W_B$ of a band B was constructed. Recall that a band is a semigroup in which every element is idempotent. The Hall semigroup $W_B$ has properties analogous to those described above for $T_E$ (see Hall [17] for details). As another direction, Fountain [10] generalized Munn’s result to a class of non-regular semigroup, namely adequate semigroups, by considering Green’s $*$-relations ${\mathcal {L}}^{*}$ and ${\mathcal {R}}^{*}$ on semigroups. Let S be a semigroup and $a, b\in S$. Then a and b are ${\mathcal {L}}^{*}$-related if and only if they are ${\mathcal {L}}$-related in an oversemigroup of S; the relation ${\mathcal {R}}^{*}$ can be defined dually. It is obvious that ${\mathcal {L}}^{*}$ and ${\mathcal {R}}^{*}$ are a left congruence and a right congruence, respectively. We denote the ${\mathcal {L}}^{*}$-class (resp. ${\mathcal {R}}^{*}$-class) of S containing $a\in S$ by $L^{*}_a(S)$ (resp. $R^{*}_a(S)$). A semigroup S is abundant if each ${\mathcal {L}}^{*}$-class and each ${\mathcal {R}}^{*}$–class of S contains an idempotent and an abundant semigroup S is called adequate if E(S) is subsemilattice of S. If S is adequate, the ${\mathcal {L}}^{*}$-class (resp. ${\mathcal {R}}^{*}$-class) of $a\in S$ contains a unique idempotent, denoted by $a^{*}$ (resp. $a^+$). From Proposition 1.6 in [10], if S is adequate, then for all $a,b\in S$, we have $a{\mathcal {L}}^{*} b$ (resp. $a{\mathcal {R}}^{*} b$) if and only if $a^{*}=b^{*}$ (resp. $a^+=b^+$), moreover,

$$\begin{aligned} (ab)^{*}=(a^{*} b)^{*},\quad (ab)^+= (ab^+)^+,\quad a^+ (ab)^+=(ab)^+, \quad (ab)^{*} b^{*}=(ab)^{*}. \end{aligned}$$

(1.1)

If S is a regular semigroup, then ${\mathcal {L}}^{*}={\mathcal {L}}$ and ${\mathcal {R}}^{*}={\mathcal {R}}$. Obviously, regular semigroups are abundant and inverse semigroups are adequate. Moreover, for an inverse semigroup S and $a\in S$, we have $a^{*}=a^{-1}a$ and $a^+=aa^{-1}$. For an abundant semigroup, let ${{\mathcal {H}}}^{*}={\mathcal {L}}^{*}\cap {\mathcal {R}}^{*}$ and let ${\mathcal {D}}^{*}$ be the smallest equivalence containing ${\mathcal {L}}^{*}$ and ${\mathcal {R}}^{*}$. An abundant semigroup S is called fundamental if the largest congruence $\mu _S$ contained in ${{\mathcal {H}}}^{*}$ is the identity congruence. From Proposition 2.1 in [7], for an abundant semigroup S and $a,b\in S$, we have

$$\begin{aligned} a\mu _S b \quad \text{ if } \text{ and } \text{ only } \text{ if } ea{\mathcal {L}}^{*} eb \quad \text{ and } \quad ae{\mathcal {R}}^{*} be \quad \text{ for } \text{ all } e\in E(S). \end{aligned}$$

(1.2)

In [10], Fountain shows that if S is an adequate semigroup and satisfies the following condition

$$\begin{aligned} (\forall a\in S)(\forall e\in E(S))\ \ ea=a(ea)^{*}, \quad ae=(ae)^+ a, \end{aligned}$$

(1.3)

then there is a homomorphism $f: S\rightarrow T_{E(S)}$ whose kernel is the largest congruence contained in ${{\mathcal {H}}}^{*}$ on S. Such a semigroup is called type A by Fountain in [10] and called ample by Gould in [14]. Obviously, an inverse semigroup is ample.

Following the above direction, Fountain et al. [11] and Gomes and Gould [12] investigate other classes of non-regular semigroups having a semilattice of idempotents by using Munn’s approach. More recent developments in this area can be found in the survey articles of Gould [15] and Hollings [18]. Furthermore El-Qallali et al. [9], Gomes and Gould [13] and Wang [21] go a step further to extend Hall’s approach for orthodox semigroups to some classes of non-regular semigroups having a band of idempotents. From the above texts, we can see that the approaches of Munn and Hall can be extended to some generalizations of inverse semigroups and orthodox semigroups, respectively.

On the other hand, Blyth and McFadden [3] introduced the concept of inverse transversals for regular semigroups. A subsemigroup $S^{\circ }$ of a regular semigroup S is called an inverse transversal of S if $V(x)\cap S^{\circ }$ contains exactly one element for all $x\in S$. Clearly, in this case, $S^{\circ }$ is an inverse subsemigroup of S. Since an inverse semigroup can be regarded as an inverse transversal of itself, the class of regular semigroups with inverse transversals contains the class of inverse semigroups as a proper subclass. Regular semigroups with inverse transversals are investigated extensively by many authors (see Blyth [4] and Tang [20] for details), and some generalizations of inverse transversals are proposed (see [5, 8, 22, 23]). In particular, El-Qallali introduced adequate transversals and ample transversals for abundant semigroups in [8]. It is well known that both adequate semigroups (resp. ample semigroups) and regular semigroups with inverse transversals are abundant semigroups having adequate transversals (resp. ample transversals). Adequate transversals of abundant semigroups have been studied by several researchers and some meaningful results are obtained, see [1, 2, 6, 16] and their references.

Inspired by the above facts, the following problem is natural: Can we study abundant semigroups with an adequate transversal by Munn’s approach? In this paper, we shall initiate the investigation of the above question by extending Munn’s approach to abundant semigroups with a multiplicative ample transversal. We present here a semigroup $T_{(I,\Lambda , E^{\circ }, P)}$ from an admissible quadruple $(I,\Lambda , E^{\circ }, P)$ (see Definition 3.1) that plays for abundant semigroups with a multiplicative ample transversal the role that the Munn semigroup $T_E$ plays for inverse semigroups and ample semigroups. More precisely, for a given admissible quadruple $(I,\Lambda , E^{\circ }, P)$, we show that a semigroup is a fundamental abundant semigroup (resp. fundamental regular semigroup) having a multiplicative ample transversal (resp. multiplicative inverse transversal) whose admissible quadruple is isomorphic to $(I,\Lambda , E^{\circ }, P)$ if and only if it is embeddable as a full subsemigroup (resp. full regular subsemigroup) into $T_{(I,\Lambda , E^{\circ }, P)}$. Moreover, some further properties of admissible quadruples are also explored.

2 Preliminaries

This section gives some useful results related to ample transversals which will be used throughout the paper. We begin with the following alternative description of ${\mathcal {L}}^{*}$, which may be found in Fountain [10].

Lemma 2.1

Elements a, b of a semigroup S are ${\mathcal {L}}^{*}$-related if and only if, for all $x, y\in S^1$, $ax=ay$ if and only if $bx=by$.

Let S be an abundant semigroup and U an abundant subsemigroup of S. If there exist an idempotent $e\in L_a^{*}(S)\cap U$ and an idempotent $f\in R_a^{*}(S)\cap U$ for all $a\in U$, then U is called a $*$ -subsemigroup of S. It is well known that U is a $*$-subsemigroup if and only if

$$\begin{aligned} {\mathcal {L}}^{*}(U)={\mathcal {L}}^{*}(S)\cap (U\times U), \quad {\mathcal {R}}^{*}(U)={\mathcal {R}}^{*}(S)\cap (U\times U). \end{aligned}$$

It is obvious that a regular subsemigroup of a regular semigroup S is always a $*$-subsemigroup of S. From El-Qallali [8], an adequate $*$-subsemigroup $S^{\circ }$ of an abundant semigroup S is called an adequate transversal of S if for each element $a\in S$, there are a unique element $\overline{a}$ in $S^{\circ }$ and $u,v\in E(S)$ such that

$$\begin{aligned} a=u\overline{a}v, \quad \text{ where } u{\mathcal {L}}\overline{a}^+, v{\mathcal {R}}\overline{a}^{*} \quad \text{ and } \quad \overline{a}^+, \overline{a}^{*}\in E(S^{\circ }). \end{aligned}$$

(2.1)

In this case, u, v are uniquely determined by a and so we denote them by $u_a$ and $v_a$, respectively. An adequate transversal $S^{\circ }$ of S is called an ample transversal if $S^{\circ }$ is also an ample semigroup. Let S be an abundant semigroup with an ample transversal $S^{\circ }$. Denote

$$\begin{aligned} I^S=\{u_a|a\in S\}, \quad \Lambda ^S=\{v_a|a\in S\}. \end{aligned}$$

The following lemma characterizes the relations ${\mathcal {L}}^{*}$ and ${\mathcal {R}}^{*}$ on an abundant semigroups having an ample transversal.

Lemma 2.2

(Proposition 2.3 in [6]) Let S be an abundant semigroup with an ample transversal $S^{\circ }$ and $a,b\in S$. Then $a{\mathcal {R}}^{*} b$ (resp. $a{\mathcal {L}}^{*} b$) if and only if $u_a=u_b$ (resp. $v_a=v_b$).

Recall that a subsemigroup U of a semigroup S is full if $E(S)\subseteq U$.

Lemma 2.3

Let S be an abundant semigroup with an ample transversal $S^{\circ }$ and U a full subsemigroup of S. Then U is an abundant semigroup with an ample transversal $S^{\circ } \cap U$. Moreover, we have $I^S=I^U$ and $\Lambda ^S=\Lambda ^U$.

Proof

Since S is abundant, $S^{\circ }$ is an ample $*$-subsemigroup of S and U is full, by Lemma 2.1 and its dual, we can easily show that U is abundant and $S^{\circ }\cap U$ is an ample $*$-subsemigroup of U. Now, let $a\in U$. Since U is full, $a=u_a\overline{a}v_a$ and $u_a,v_a, \overline{a}^{*},\overline{a}^+\in E(S)$, we have

$$\begin{aligned} \overline{a}=\overline{a}^+ \overline{a}\ \overline{a}^{*}=\overline{a}^+ u_a\overline{a}v_a \overline{a}^{*}=\overline{a}^+ a \overline{a}^{*}\in U \end{aligned}$$

(2.2)

by (2.1). This yields that the equality $a=u_a \overline{a} v_a$ holds in U, whence $S^{\circ } \cap U$ is an ample transversal of U and $I^S=I^U, \Lambda ^S=\Lambda ^U$. $\square$

An ample transversal $S^{\circ }$ of an abundant semigroup S is multiplicative if $fg\in E(S^{\circ })$ for all $f\in \Lambda ^S$ and $g\in I^S$, or equivalently, if $v_a u_b\in E(S^{\circ })$ for all $a,b\in S$.

Corollary 2.4

Let S be an abundant semigroup with an ample transversal $S^{\circ }$ and U a full subsemigroup of S.

• (1) If $S^{\circ }$ is multiplicative, then $S^{\circ } \cap U$ is also multiplicative.

• (2) If $S^{\circ }$ is fundamental, then U is fundamental.

Proof

Item (1) follows from the fact $I^S=I^U$ and $\Lambda ^S=\Lambda ^U$ obtained by Lemma 2.3. Now we prove the item (2). Let $a,b\in U$ and $a\mu _U b$. Then $a{{\mathcal {H}}}^{*} b$ in U. By Lemma 2.2 and (2.1), we have $\overline{a}^+ {\mathcal {L}}u_a=u_b{\mathcal {L}} \overline{b}^+$ and $\overline{a}^{*} {\mathcal {R}} v_a=v_b {\mathcal {R}} \overline{b}^{*}$ whence $\overline{a}^+=\overline{b}^+$ and $\overline{a}^{*}=\overline{b}^{*}$. Since $a\mu _U b$, we have

$$\begin{aligned} \overline{a}=(\overline{a}^+ a \overline{a}^{*}) \mu _U (\overline{b}^+ b \overline{b}^{*})=\overline{b} \end{aligned}$$

by (2.2). In view of the fact (1.2), we have $e\overline{a}{\mathcal {L}}^{*} e\overline{b}$ and $\overline{a}e{\mathcal {R}}^{*} \overline{b}e$ in U for all $e\in E(U)$. Since U is full, we have $E(S^{\circ })\subseteq E(U)$ and so $e\overline{a}{\mathcal {L}}^{*} e\overline{b}$ and $\overline{a}e{\mathcal {R}}^{*} \overline{b}e$ in U for all $e\in E(S^{\circ })$. This implies that $e\overline{a}{\mathcal {L}}^{*} e\overline{b}$ and $\overline{a}e{\mathcal {R}}^{*} \overline{b}e$ in $S^{\circ } \cap U$ for all $e\in E(S^{\circ })$ since $S^{\circ }\cap U$ is a $*$-subsemigroup of U. Observe that $S^{\circ }\cap U$ is an ample subsemigroup of the ample semigroup $S^{\circ }$, it follows that $e\overline{a}{\mathcal {L}}^{*} e\overline{b}$ and $\overline{a}e{\mathcal {R}}^{*} \overline{b}e$ in $S^{\circ }$ for all $e\in E(S^{\circ })$. In view of the fact (1.2) again, we have $\overline{a}\mu _{S^{\circ }}\overline{b}$. This gives $\overline{a}=\overline{b}$ since $S^{\circ }$ is fundamental. This implies that $a=u_a\overline{a}v_a =u_b\overline{b}v_b=b$. Thus, U is fundamental. $\square$

To give further properties of abundant semigroups with an ample transversal, we need the following notion. Let S be an abundant semigroup and B the subsemigroup generated by E(S). For $e\in E(S)$, we denote the subsemigroup of eBe generated by the idempotents of eBe by $\langle e\rangle$. From El-Qallali and Fountain [7], we say S is idempotent-connected (IC for short) if for all $u \in R_a^{*}(S)\cap E(S)$ and $v \in L_a^{*}(S)\cap E(S)$, there is an isomorphism $\alpha : \langle u\rangle \rightarrow \langle v \rangle$ satisfying $xa =a(x\alpha )$ for all $x\in \langle u\rangle$. On IC abundant semigroups, we have the following.

Lemma 2.5

Let S be an IC abundant semigroup and $a\in S$. If $u,v\in E(S)$ and $u{\mathcal {R}}^{*} a {\mathcal {L}}^{*} v$, then there exists a unique isomorphism $\alpha$ from $\langle u \rangle$ onto $\langle v \rangle$ such that $a(x\alpha )=xa$ for all $x\in \langle u \rangle$. This isomorphism is called the idempotent-connected isomorphism from $\langle u \rangle$ onto $\langle v \rangle$.

Proof

Let $\alpha$ and $\beta$ be two isomorphisms satisfying the conditions given in the lemma. Then for all $x\in \langle u \rangle$, we have $a(x\alpha )=xa=a(x\beta )$. Since $a{\mathcal {L}}^{*} v$ and $x\alpha , x\beta \in \langle v\rangle$, we have $x\alpha =v (x\alpha )=v(x\beta )=x\beta$ by Lemma 2.1. $\square$

Combining Lemmas 2.1 and 4.1 in El-Qallali [8] and Proposition 3.1, Lemma 6.12 in Guo [16], we have the following lemma.

Lemma 2.6

([8, 16]) Let S be an abundant semigroup with a multiplicative ample transversal $S^{\circ }$. Then S is an IC abundant semigroup. Moreover, for $a,b\in S$ and $x\in E(S)$, we have

• (1) $u_a{\mathcal {R}}^{*} a {\mathcal {L}}^{*} v_a$.

• (2) $u_{ab}=u_a({\overline{a}}v_a u_b )^+, {\overline{ab}}={\bar{a}}v_a u_b {\bar{b}}, v_{ab}=(v_a u_b {\overline{b}})^{*} v_b$, $\overline{x}=v_xu_x\in E(S^{\circ })$.

• (3) $u_{u_a}=u_a, \overline{u_a}=\overline{a}^+=v_{u_a}, u_{v_a}=\overline{a}^{*}=\overline{v_a}, v_{v_a}=v_a.$

Corollary 2.7

Let S be an abundant semigroup with a multiplicative ample transversal $S^{\circ }$ and $x\in E(S)$. Then $x=u_xv_x$.

Proof

If $x\in E(S)$, then by Lemma 2.6(2), $\overline{x}\in E(S^{\circ })$. This implies that $\overline{x}^+=\overline{x}$. Therefore $x=u_x\overline{x} v_x=u_x \overline{x}^+v_x=u_xv_x$ by (2.1). $\square$

Recall that a band B is called left normal (resp. right normal, normal) if $efg=egf$ (resp. $efg=feg$, $efge=egfe$) for all $e,f,g\in B$.

Lemma 2.8

(Lemma 2.1 and Proposition 2.6 in [6]) Let S be an abundant semigroup with a multiplicative ample transversal $S^{\circ }$. Then

• (1) $I^S\cap \Lambda ^S=E(S^{\circ })$.

• (2) $I^S=\{e\in E(S)| \text{ there } \text{ exists } \text{ a } \text{ unique } e^{\circ }\in E(S^{\circ }) \text{ such } \text{ that } e{\mathcal {L}}e^{\circ }\}$, $\Lambda ^S=\{f\in E(S)| \text{ there } \text{ exists } \text{ a } \text{ unique } f^{\circ }\in E(S^{\circ }) \text{ such } \text{ that } f{\mathcal {R}}f^{\circ }\}$.

• (3) $I^S$ (resp. $\Lambda ^S$) is a left normal band (resp. a right normal band).

To end this section, we explore the relationship between transversals. As usual, if S is a regular semigroup with an inverse transversal $S^{\circ }$, then we denote the unique element in $V_{S^{\circ }}(a)=V(a)\cap S^{\circ }$ by $a^{\circ }$ for all $a\in S$, moreover, let $(a^{\circ })^{\circ }=a^{\circ \circ }$. From the remarks before Example 2.2 in [8] and Corollary 2.7 in [22], we have the fact below.

Lemma 2.9

([8, 22]) Let S be a regular semigroup and $S^{\circ }$ a subsemigroup of S. Then $S^{\circ }$ is an inverse transversal of S if and only if $S^{\circ }$ is an ample transversal of S. In this case, $u_a=aa^{\circ }, \overline{a}=a^{\circ \circ }$ and $v_a=a^{\circ } a$ for all $a\in S$. As consequence, we have $I^S=\{aa^{\circ }|a\in S\}$ and $\Lambda ^S=\{a^{\circ } a|a\in S\}$.

For multiplicative inverse transversals of bands, we have the following result.

Lemma 2.10

([3]) Let $B^{\circ }$ be an inverse transversal of a band B. Then $B^{\circ }$ is multiplicative if and only if B is normal.

3 The Munn semigroups of admissible quadruples

In this section, a generalization of the Munn semigroup of a semilattice, namely the Munn semigroup of an admissible quadruple, is constructed. Moreover, we show that this semigroup is a fundamental regular semigroup with a multiplicative inverse transversal. We first introduce admissible quadruples, which is inspired by Lemma 2.8.

Definition 3.1

Let I (resp. $\Lambda$) be a left normal band (resp. a right normal band), $E^{\circ }=I\cap \Lambda$ a subsemilattice of I and $\Lambda$, and $P=(P_{f,g})_{\Lambda \times I}$ be a $\Lambda \times I$-matrix over $E^{\circ }$. The quadruple $(I,\Lambda , E^{\circ }, P)$ is called admissible if for all $g\in I$ and $f\in \Lambda$, there exist $g^{\circ }, f^{\circ }\in E^{\circ }$ such that $g{\mathcal {L}}g^{\circ }$, $f{\mathcal {R}}f^{\circ }$ and for all $i,j\in E^{\circ }$,

$$\begin{aligned} iP_{f,g}=P_{if, g}, \quad P_{f,g}j=P_{f, gj}, \quad P_{f,j}=fj, \quad P_{i,g}=ig. \end{aligned}$$

(3.1)

Remark 3.2

On admissible quadruples, we have the following remarks.

• (1) Since $E^{\circ }$ is a subsemilattice, the elements $g^{\circ }$ and $f^{\circ }$ in Definition 3.1 are uniquely determined by g and f, respectively. In particular, $i\in E^{\circ }$ if and only if $i^{\circ }=i$. Thus, $P^{\circ }_{f,g}=P_{f,g}$ for all $f\in \Lambda$ and $g\in I$.

• (2) If S is an abundant semigroup with a multiplicative ample transversal $S^{\circ }$, then it is obvious that $(I^S,\Lambda ^S, E(S^{\circ }), P^S)$ is an admissible quadruple by Lemma 2.8, where $P^S_{f,g}$ is equal to the product of f and g in S for all $f\in \Lambda$ and $g\in I$. In this case, $(I^S,\Lambda ^S, E(S^{\circ }), P^S)$ is called the admissible quadruple of S. If U is a full subsemigroup of S, then by Lemma 2.3 and Corollary 2.4(1), $S^{\circ }\cap U$ is a multiplicative ample transversal of U and the admissible quadruples of S and U are equal.

To construct the Munn semigroup of an admissible quadruple, we need some preliminaries. First, we have the following basic facts on admissible quadruples.

Lemma 3.3

Let $(I,\Lambda , E^{\circ }, P)$ be an admissible quadruple and $e,g\in I, f,h\in \Lambda$. Then

$$\begin{aligned} eg=eg^{\circ }, \quad (eg)^{\circ }=e^{\circ } g^{\circ }, \quad fh=f^{\circ } h, \quad (fh)^{\circ }=f^{\circ } h^{\circ }. \end{aligned}$$

Moreover, we have $eE^{\circ } e=eE^{\circ }$ and $fE^{\circ } f=E^{\circ } f$, which are subsemilattices of I and $\Lambda$, respectively.

Proof

Since $g{\mathcal {L}}g^{\circ }$ and I is a left normal band, we have $eg=egg^{\circ }=eg^{\circ } g=eg^{\circ }$. This implies that $e^{\circ } g=e^{\circ } g^{\circ }$, and so $eg{\mathcal {L}}e^{\circ } g=e^{\circ } g^{\circ }\in E^{\circ }$ by the fact that $e{\mathcal {L}}e^{\circ }$. This yields that $(eg)^{\circ }=e^{\circ } g^{\circ }$. Finally, it follows that $eE^{\circ } e=eE^{\circ }$ by the fact that I is a left normal band. Moreover, for $i,j\in E^{\circ }$, we have

$$\begin{aligned} (ei)(ej)=(eie)j=eij=eji=(eje)i=(ej)(ei), \end{aligned}$$

whence $eE^{\circ }$ is a subsemilattice of I. The remaining facts of this lemma can be proved by symmetry. $\square$

Remark 3.4

Let I (resp. $\Lambda$) be a left normal band (resp. a right normal band), and $E^{\circ }=I\cap \Lambda$ be a subsemilattice of I and $\Lambda$. Suppose that for each $e\in I$ and $f\in \Lambda$ there exist $e^{\circ }, f^{\circ }\in E^{\circ }$ such that $e{\mathcal {L}}e^{\circ }$ and $f{\mathcal {R}}f^{\circ }$, respectively. Then by Lemma 3.3 it is easy to see that $(I,\Lambda , E^{\circ }, Q)$ forms an admissible quadruple, where $Q_{f,g}=f^{\circ } g^{\circ }$ for all $f\in \Lambda$ and $g\in I$. This admissible quadruple is called the normal admissible quadruple determined by $I, \Lambda$ and $E^{\circ }$.

Let $(I,\Lambda , E^{\circ }, P)$ be an admissible quadruple and $e\in I, f\in \Lambda$. If $eE^{\circ }$ is isomorphic to $E^{\circ } f$, we write $eE^{\circ }\cong E^{\circ } f$, and denote the set of isomorphisms from $eE^{\circ }$ to $E^{\circ } f$ by $T_{e,f}$. Moreover, we denote

$$\begin{aligned} {\mathcal {U}}=\{(e,f)\in I\times \Lambda |eE^{\circ }\cong E^{\circ } f\},\quad T_{(I,\Lambda , E^{\circ }, P)}=\bigcup _{(e,f)\in {\mathcal {U}}}T_{e,f}. \end{aligned}$$

Obviously, the elements in the Munn semigroup $T_{E^{\circ }}$ of the semilattice $E^{\circ }$ are contained in $T_{(I,\Lambda , E^{\circ }, P)}$. The following proposition provides some other elements in $T_{(I,\Lambda , E^{\circ }, P)}$. As usual, we use $\iota _M$ to denote the identity transformation on the non-empty set M.

Proposition 3.5

Let $(I,\Lambda , E^{\circ }, P)$ be an admissible quadruple and $g\in I, f\in \Lambda$. Define

$$\begin{aligned} \pi _{f, g}: gP_{f,g}E^{\circ } \rightarrow E^{\circ } P_{f,g}f,\quad x\mapsto x^{\circ } P_{f,g}f. \end{aligned}$$

Then $\pi _{f, g}\in T_{gP_{f,g},P_{f,g}f}$ and the inverse mapping of $\pi _{f, g}$ is

$$\begin{aligned} \pi ^{-1}_{f, g}: E^{\circ } P_{f,g}f \rightarrow gP_{f,g}E^{\circ },\quad y\mapsto gP_{f,g}y^{\circ }. \end{aligned}$$

In particular, we have

$$\begin{aligned} \pi _{g^{\circ }, g}{:}\, gE^{\circ }\rightarrow E^{\circ } g^{\circ },\quad x\mapsto x^{\circ } g^{\circ },\ \ \pi _{f, f^{\circ }}: f^{\circ } E^{\circ }\rightarrow E^{\circ } f,\quad x\mapsto x^{\circ } f \end{aligned}$$

and $\pi _{i,j}=\iota _{ijE^{\circ }}, \pi _{i,i}=\iota _{iE^{\circ }}$ for all $i,j\in E^{\circ }$.

Proof

Clearly, $\pi _{f,g}$ is well defined. Let $x\in gP_{f,g}E^{\circ }$ and $y\in E^{\circ } P_{f,g}f$. Then by condition (3.1) and Lemma 3.3, we have

$$\begin{aligned} gP_{f,g}(x^{\circ } P_{f,g}f)^{\circ }=gP_{f,g}x^{\circ } P_{f,g}f^{\circ } =gf^{\circ } P_{f,g}x^{\circ }= g P_{f^{\circ } f,g}x^{\circ }=g P_{f,g}x^{\circ }=gP_{f,g}x=x. \end{aligned}$$

Dually, we can obtain that $(gP_{f,g}y^{\circ })^{\circ } P_{f,g}f=y$. Moreover, for $x_1,x_2\in gP_{f,g}E^{\circ }$, by Lemma 3.3 and $f{\mathcal {R}}f^{\circ }$, we have $(x_1x_2)\pi _{f,g}=(x_1x_2)^{\circ } P_{f,g}f=x^{\circ }_1 x^{\circ }_2P_{f,g}f$ and

$$\begin{aligned} (x_1\pi _{f,g})(x_2\pi _{f,g})= & {} (x^{\circ }_1 P_{f,g}f)(x^{\circ }_2 P_{f,g}f) =x^{\circ }_1 P_{f,g}(fx^{\circ }_2) P_{f,g}f\\= & {} x^{\circ }_1 P_{f,g}(f^{\circ } x^{\circ }_2) P_{f,g}f =x^{\circ }_1 x^{\circ }_2 P_{f,g} (f^{\circ } f)=x^{\circ }_1 x^{\circ }_2 P_{f,g} f=(x_1x_2)\pi _{f,g}. \end{aligned}$$

This implies that the first part of this lemma holds. The remaining part follows from the fact that

$$\begin{aligned} P_{g^{\circ },g}=g^{\circ } g=g^{\circ }, \quad P_{f,f^{\circ }}=ff^{\circ }=f^{\circ }, \quad P_{i,j}=ij, x^{\circ }=x \end{aligned}$$

for all $g\in I, f\in \Lambda$ and $i,j,x\in E^{\circ }$. $\square$

The following results give some simple but useful properties of the elements in the set $T_{(I,\Lambda , E^{\circ }, P)}$.

Lemma 3.6

Let $(I,\Lambda , E^{\circ }, P)$ be an admissible quadruple, $\alpha \in T_{e,f}$ and $x\in eE^{\circ }, y\in E^{\circ } f$.

• (1) $e \alpha =f$ and $\alpha ^{-1}$ is an isomorphism from $E^{\circ } f$ onto $eE^{\circ }$.

• (2) $(xE^{\circ })\alpha =E^{\circ }(x\alpha )$ and $(E^{\circ } y)\alpha ^{-1}=(y\alpha ^{-1})E^{\circ }$.

• (3) $(x\alpha )^{\circ }=(x\alpha )f^{\circ }$ and $x\alpha = (x\alpha )^{\circ } f$.

• (4) $(y\alpha ^{-1})^{\circ }=e^{\circ }(y\alpha ^{-1})$ and $y\alpha ^{-1}=e(y\alpha ^{-1})^{\circ }$.

Proof

• (1) This is obvious.

• (2) Let $xi\in xE^{\circ }, i\in E^{\circ }$. Since I is a left normal band, we have $xix=xi$ and so

  $$\begin{aligned} (xi)\alpha =(xix)\alpha =(xi)\alpha \cdot (x\alpha )=((xi)\alpha )^{\circ } (x\alpha )\in E^{\circ } (x\alpha ) \end{aligned}$$

  by Lemma 3.3. Conversely, let $u'\in E^{\circ } (x\alpha )$. Since $x\alpha \in E^{\circ } f$, we obtain that $u'\in E^{\circ } f$ whence $u'=u\alpha$ for some $u\in eE^{\circ }$. Observe that I is a left normal band and $\Lambda$ is a right normal band, it follows that

  $$\begin{aligned} u\alpha =(u\alpha )(x\alpha )=(x\alpha )(u\alpha )(x\alpha )=(xux)\alpha =(xu)\alpha . \end{aligned}$$

  Noticing that $\alpha$ is injective, we get $u=xu=xu^{\circ }\in xE^{\circ }$ by Lemma 3.3. This implies that $u'=u\alpha \in (xE^{\circ })\alpha$. The other identity can be proved by similar methods.

• (3) Since $x\alpha \in E^{\circ } f \subseteq \Lambda$, we have $x\alpha =(x\alpha )f=(x\alpha )^{\circ } f$ whence

  $$\begin{aligned} (x\alpha )^{\circ }=((x\alpha )f)^{\circ }=(x\alpha )^{\circ } f^{\circ }=(x\alpha ) f^{\circ } \end{aligned}$$

  by Lemma 3.3.

• (4) This is the dual of item (3). $\square$

Now, let $(I,\Lambda , E^{\circ }, P)$ be an admissible quadruple, $\alpha \in T_{e,f}, \beta \in T_{g,h}$. Consider the composition $\alpha \pi ^{-1}_{f,g}\beta$ in the symmetric inverse semigroup on the set $I\cup \Lambda$. Since

$$\begin{aligned} \mathrm{dom}(\pi ^{-1}_{f,g}\beta )=(gP_{f,g}E^{\circ }\cap gE^{\circ }) \pi _{f,g}=(gP_{f,g}E^{\circ })\pi _{f,g}=E^{\circ } P_{f,g}f, \end{aligned}$$

it follows that

$$\begin{aligned} \mathrm{dom}(\alpha \pi ^{-1}_{f,g}\beta )=(E^{\circ } f\cap E^{\circ } P_{f,g}f)\alpha ^{-1} =(E^{\circ } P_{f,g}f)\alpha ^{-1}=(P_{f,g}f)\alpha ^{-1}E^{\circ } \end{aligned}$$

and

$$\begin{aligned} \mathrm{ran}(\alpha \pi ^{-1}_{f,g}\beta )=(E^{\circ } P_{f,g}f)\pi ^{-1}_{f,g}\beta =(gP_{f,g}E^{\circ })\beta =E^{\circ }(gP_{f,g})\beta \end{aligned}$$

by Lemma 3.6(2). Thus, we have

$$\begin{aligned} \alpha \pi ^{-1}_{f,g}\beta \in T_{j,k}, \ \ \ j=(P_{f,g}f)\alpha ^{-1},\quad k=(gP_{f,g})\beta . \end{aligned}$$

(3.2)

In view of the above discussions, we can define a multiplication $``\circ ''$ on $T_{(I,\Lambda , E^{\circ }, P)}$ as follows: For $\alpha \in T_{e,f}, \beta \in T_{g,h}$,

$$\begin{aligned} \alpha \circ \beta =\alpha \pi ^{-1}_{f,g}\beta , \end{aligned}$$

where $\pi ^{-1}_{f,g}$ is defined as in Proposition 3.5.

Lemma 3.7

The set $T_{(I,\Lambda , E^{\circ }, P)}$ forms a semigroup with respect to the multiplication “$\circ$” defined above.

Proof

Now, let $\alpha \in T_{e,f}$, $\beta \in T_{g,h}$, $\gamma \in T_{s,t}$ and

$$\begin{aligned} \alpha \circ \beta \in T_{j,k}, \quad (\alpha \circ \beta )\circ \gamma \in T_{m,n}, \quad \beta \circ \gamma \in T_{p,q}, \quad \alpha \circ (\beta \circ \gamma )\in T_{a,b}, \end{aligned}$$

where

$$\begin{aligned} j= & {} (P_{f,g}f)\alpha ^{-1},\quad k=(gP_{f,g})\beta ,\quad p=(P_{h,s} h)\beta ^{-1},\quad q=(sP_{h,s})\gamma ,\\ m= & {} (P_{k,s} k)(\alpha \circ \beta )^{-1},\quad n=(sP_{k,s})\gamma , \quad a=(P_{f,p} f)\alpha ^{-1},\quad b=(pP_{f,p})(\beta \circ \gamma ). \end{aligned}$$

On one hand, in view of the fact $k\in E^{\circ } h$ and Lemma 3.3, we have $k=kh=k^{\circ } h$. By condition (3.1) and $h^{\circ } {\mathcal {R}}h$,

$$\begin{aligned} P_{k,s}=P_{k^{\circ } h, s}=k^{\circ } P_{h,s}=k^{\circ } P_{h^{\circ } h,s}=k^{\circ } h^{\circ } P_{h,s}=k^{\circ } P_{h,s}h^{\circ }. \end{aligned}$$

(3.3)

Since $h^{\circ } k=hk$ (by Lemma 3.3), $k^{\circ } {\mathcal {R}} k$ and $\Lambda$ is a right normal band, this implies that

$$\begin{aligned} P_{k,\,s}k=k^{\circ } P_{h,s} h^{\circ } k=k^{\circ } P_{h,\,s} h k=k^{\circ }(p\beta )k=(p\beta )k^{\circ } k =(p\beta )k=(pgP_{f,g})\beta . \end{aligned}$$

By Lemma 3.3, condition (3.1) and the fact that $g{\mathcal {L}}g^{\circ }$, we have

$$\begin{aligned} m= & {} (P_{k,s}k)(\alpha \circ \beta )^{-1}=(P_{k,s}k)\beta ^{-1}\pi _{f,g}\alpha ^{-1}\\= & {} ((pgP_{f,g})\beta )\beta ^{-1}\pi _{f,g}\alpha ^{-1}=(pgP_{f,g})\pi _{f,g}\alpha ^{-1}=((pgP_{f,g})^{\circ } P_{f,g}f)\alpha ^{-1}\\= & {} (p^{\circ } g^{\circ } P_{f,g} P_{f,g}f)\alpha ^{-1}=(p^{\circ } P_{f,g}g^{\circ } f) \alpha ^{-1}=(p^{\circ } P_{f,gg^{\circ } } f) \alpha ^{-1}=(p^{\circ } P_{f,g} f) \alpha ^{-1}. \end{aligned}$$

On the other hand, by Lemma 3.3 and the fact $p\in gE^{\circ }$, we have

$$\begin{aligned} gp^{\circ }=gp=p, \quad P_{f,p}=P_{f, gp^{\circ }}=P_{f, g}p^{\circ }=p^{\circ } P_{f, g} \end{aligned}$$

(3.4)

by condition (3.1), which implies that $P_{f, p}f=p^{\circ } P_{f, g} f$ and so

$$\begin{aligned} a=(P_{f, p} f)\alpha ^{-1}=(p^{\circ } P_{f, g}f)\alpha ^{-1}=m. \end{aligned}$$

Dually, we can obtain that $n=b$.

Take $x\in aE^{\circ }=mE^{\circ }$ and denote $y=x\alpha$. On one hand,

$$\begin{aligned} x(\alpha \circ \beta )=(x\alpha )\pi ^{-1}_{f,g}\beta = y\pi ^{-1}_{f,g}\beta =(gP_{f, g}y^{\circ })\beta . \end{aligned}$$

(3.5)

Since both k and $x(\alpha \circ \beta )$ are in $\Lambda$, and $\mathrm{ran}(\alpha \circ \beta )=E^{\circ } k$, by Lemma 3.3 we have

$$\begin{aligned} k^{\circ }(x(\alpha \circ \beta ))^{\circ }=(x(\alpha \circ \beta )\cdot k)^{\circ }= (x(\alpha \circ \beta ))^{\circ } \end{aligned}$$

(3.6)

Combining the identities (3.3), (3.6) and (3.5), we obtain

$$\begin{aligned} x[(\alpha \circ \beta )\circ \gamma ]&=(x(\alpha \circ \beta ))\pi ^{-1}_{k,s}\gamma = (sP_{k, s}\cdot (x(\alpha \circ \beta ))^{\circ })\gamma \\&=(sk^{\circ } P_{h, s}\cdot (x(\alpha \circ \beta ))^{\circ })\gamma =(sP_{h, s}\cdot k^{\circ }(x(\alpha \circ \beta ))^{\circ })\gamma \\&=(sP_{h, s}\cdot (x(\alpha \circ \beta ))^{\circ })\gamma =(sP_{h, s}\cdot ((gP_{f, g}y^{\circ })\beta )^{\circ } )\gamma . \end{aligned}$$

On the other hand,

$$\begin{aligned} x[\alpha \circ (\beta \circ \gamma )]= (x\alpha )\pi ^{-1}_{f,p}\beta \pi ^{-1}_{h,s}\gamma =y\pi ^{-1}_{f,p}\beta \pi ^{-1}_{h,s}\gamma = (pP_{f, p}y^{\circ })\beta \pi ^{-1}_{h,s}\gamma . \end{aligned}$$

Observe that $gp^{\circ }=p$ (by (3.4)) and $y^{\circ } p=y^{\circ } p^{\circ }$ (by Lemma 3.3), it follows by the identity (3.4) that

$$\begin{aligned} (pP_{f, p}y^{\circ })\beta&=(gp^{\circ } P_{f, g}y^{\circ })\beta =(gP_{f, g} y^{\circ } p^{\circ })\beta \\&=(gP_{f, g} y^{\circ } p)\beta =(gP_{f, g}y^{\circ })\beta (p\beta )=(gP_{f, g}y^{\circ })\beta \cdot P_{h, s}h. \end{aligned}$$

By Lemma 3.3 and condition (3.1),

$$\begin{aligned} x[\alpha \circ (\beta \circ \gamma )]= & {} ((pP_{f, p}y^{\circ })\beta )\pi ^{-1}_{h,s}\gamma =((gP_{f, g}y^{\circ })\beta \cdot P_{h, s}h)\pi ^{-1}_{h,s}\gamma \\= & {} (sP_{h, s}\cdot ((gP_{f, g}y^{\circ })\beta \cdot P_{h, s}h)^{\circ } )\gamma =(sP_{h, s}\cdot h^{\circ } P_{h, s} \cdot ((gP_{f, g}y^{\circ })\beta )^{\circ } )\gamma \\= & {} (sP_{h, s}\cdot P_{ h^{\circ } h, s} \cdot ((gP_{f, g}y^{\circ })\beta )^{\circ } )\gamma =(sP_{h, s}\cdot ((gP_{f, g}y^{\circ })\beta )^{\circ } )\gamma . \end{aligned}$$

Hence, $(\alpha \circ \beta )\circ \gamma =\alpha \circ (\beta \circ \gamma )$ and so $T_{(I,\Lambda , E^{\circ }, P)}$ forms a semigroup with respect to the multiplication $``\circ$”. $\square$

Let $(I,\Lambda , E^{\circ }, P)$ be an admissible quadruple. If $I=\Lambda =E^{\circ }$, then it is easy to check that the semigroup $T_{(I,\Lambda , E^{\circ }, P)}$ coincides with the Munn semigroup $T_{E^{\circ }}$ since $\pi ^{-1}_{f,g}=\iota _{fgE^{\circ }}$ for all $f\in \Lambda$ and $g\in I$ in the case. Thus, the semigroup $T_{(I,\Lambda , E^{\circ }, P)}$ can be regarded as a generalization of the Munn semigroup of a semilattice and will be called the Munn semigroup of the admissible quadruple $(I,\Lambda , E^{\circ }, P)$.

Theorem 3.8

The semigroup $T_{(I,\Lambda , E^{\circ }, P)}$ is a regular semigroup with a multiplicative inverse transversal

$$\begin{aligned} T_{E^{\circ }}=\{\alpha \in T_{(I,\Lambda , E^{\circ }, P)} |\alpha \in T_{p,q}, p,q\in E^{\circ }\}. \end{aligned}$$

Proof

For $\alpha \in T_{e,f}$, let $\alpha ^{\circ }=\pi _{f,f^{\circ }}\alpha ^{-1} \pi _{e^{\circ }, e},$ where $\pi _{f,f^{\circ }}$ and $\pi _{e^{\circ }, e}$ are defined as in Proposition 3.5. It is routine to check that $\alpha ^{\circ }\in T_{f^{\circ }, e^{\circ }}$ and so $\alpha ^{\circ }\in T_{E^{\circ }}$. Furthermore, we have

$$\begin{aligned} \alpha \circ \alpha ^{\circ }=\alpha \pi ^{-1}_{f,f^{\circ }}\alpha ^{\circ } =\alpha \pi ^{-1}_{f,f^{\circ }}\pi _{f,f^{\circ }}\alpha ^{-1} \pi _{e^{\circ }, e}=\pi _{e^{\circ },e}. \end{aligned}$$

(3.7)

Similarly, we can prove that $\pi _{e^{\circ },e}\circ \alpha =\alpha$ and $\alpha ^{\circ } \circ \pi _{e^{\circ },e}=\alpha ^{\circ }$, which implies that $\alpha ^{\circ }$ is an inverse of $\alpha$ in $T_{E^{\circ }}$.

Now, let $\beta \in T_{p,q}\ (p,q\in E^{\circ })$ be an inverse of $\alpha \in T_{e,f}$ in $T_{E^{\circ }}$. We have to show that $\beta =\alpha ^{\circ }$. Let $\alpha \circ \beta \in T_{j,k}$ and $\beta \circ \alpha \in T_{u,v}$. Then $\alpha \circ \beta \circ \alpha =\alpha \in T_{e,f}$. Moreover,

$$\begin{aligned} j=(P_{f, p}f)\alpha ^{-1}\in \mathrm{dom}\alpha =eE^{\circ },\ \ \ k=(pP_{f, p})\beta \in \mathrm{ran}\beta =E^{\circ } q\subseteq E^{\circ }. \end{aligned}$$

Similarly, we can obtain that $v\in \mathrm{ran}\alpha =E^{\circ } f$ and $u\in pE^{\circ }\subseteq E^{\circ }$. It follows that

$$\begin{aligned} e=(P_{f, u}f)\alpha ^{-1}=(fuf)\alpha ^{-1}=(uf)\alpha ^{-1},\ \ f=(eP_{k, e})\alpha =(eke)\alpha =(ek)\alpha \end{aligned}$$

by condition (3.1) and the fact that I is a left normal band and $\Lambda$ is a right normal band, respectively. This implies that $uf=e\alpha =f$ and $ek=f\alpha ^{-1}=e$. Since $u\in pE^{\circ }$ and $k\in E^{\circ } q$, we have $pu=u$ and $kq=k$ whence $puf=uf=f$ and $ekq=ek=e$. This yields that $pf=f$ and $eq=e$. Consider $\beta \circ \alpha \circ \beta =\beta \in T_{p,q}$. Then we have

$$\begin{aligned} p=(P_{q,j}q)\beta ^{-1}=(qjq)\beta ^{-1}=(qj)\beta ^{-1}, \ \ \ q=(pP_{v,p})\beta =(pvp)\beta =(vp)\beta . \end{aligned}$$

This implies that $qj=p\beta =q$ and $vp=q\beta ^{-1}=p$. Observe that $j\in eE^{\circ }$ and I is a left normal band, it follows that $j=ej=eje$ whence $je=j$ and $qe=qje=qj=q$. Similar discussion gives $fp=p$. Therefore $f{\mathcal {R}} p$ and $e{\mathcal {L}}q$. Observe that $p,q\in E^{\circ }$, it follows that $p=f^{\circ }$ and $q=e^{\circ }$ by the definition of admissible quadruple, whence $\beta \in T_{f^{\circ },e^{\circ }}$. Moreover, by simple calculations, we can see that

$$\begin{aligned} \beta =\beta \pi ^{-1}_{e^{\circ }, e}\pi _{e^{\circ }, e}= \beta \circ \pi _{e^{\circ }, e}=\beta \circ \alpha \circ \alpha ^{\circ } \end{aligned}$$

by the identity (3.7) and so

$$\begin{aligned} \alpha ^{\circ }=\alpha ^{\circ }\circ \alpha \circ \beta \circ \alpha \circ \alpha ^{\circ }=\alpha ^{\circ } \circ \alpha \circ \beta . \end{aligned}$$

This implies that $\alpha ^{\circ }{\mathcal {L}} \beta$ in $T_{(I,\Lambda , E^{\circ }, P)}$. Dually, we can obtain $\alpha ^{\circ }{\mathcal {R}} \beta$ in $T_{(I,\Lambda , E^{\circ }, P)}$ and so $\alpha ^{\circ }{{\mathcal {H}}} \beta$ in $T_{(I,\Lambda , E^{\circ }, P)}$. However, both $\alpha ^{\circ }$ and $\beta$ are the inverses of $\alpha$, and hence $\beta =\alpha ^{\circ }$.

By the above discussions, we have shown that $T_{E^{\circ }}$ is an inverse transversal of $T_{(I,\Lambda , E^{\circ }, P)}$. To see $T_{E^{\circ }}$ is multiplicative, we take $\alpha \in T_{e,f}$ and $\beta \in T_{g,h}$. Then by the identity (3.7) and its dual, and $v_\alpha =\alpha ^{\circ }\circ \alpha , u_\beta =\beta \circ \beta ^{\circ }$ (by Lemma 2.9), we have

$$\begin{aligned} v_\alpha \circ u_\beta =\alpha ^{\circ } \circ \alpha \circ \beta \circ \beta ^{\circ }=\pi _{f, f^{\circ }}\circ \pi _{g^{\circ },g}\in T_{j,k} \end{aligned}$$

where

$$\begin{aligned} j=(P_{f, g}f)\pi ^{-1}_{f, f^{\circ }}\in \mathrm{ran}\pi ^{-1}_{f, f^{\circ }}=f^{\circ } E^{\circ }\subseteq E^{\circ } \end{aligned}$$

and

$$\begin{aligned} k=(gP_{f, g})\pi _{g^{\circ },g}\in \mathrm{ran}\pi _{g^{\circ },g}=E^{\circ } g^{\circ }\subseteq E^{\circ }, \end{aligned}$$

whence $v_\alpha \circ u_\beta \in T_{E^{\circ }}$. This implies that $T_{E^{\circ }}$ is a multiplicative inverse transversal of $T_{(I,\Lambda , E^{\circ }, P)}$. $\square$

The following corollary characterizes the idempotents in $T_{(I,\Lambda , E^{\circ }, P)}$.

Corollary 3.9

Let $\alpha \in T_{e,f}$. Then $\alpha \in E(T_{(I,\Lambda , E^{\circ }, P)})$ if and only if

$$\begin{aligned} f^{\circ }=P_{f,e}=e^{\circ } \quad \text{ and } \quad e(x\alpha )^{\circ }=x \quad \text{ for } \text{ all } x\in eE^{\circ }. \end{aligned}$$

Proof

Let $\alpha \in E(T)$. Then $\alpha \circ \alpha =\alpha$, whence $\mathrm{dom}(\alpha \circ \alpha )=\mathrm{dom}\alpha$ and $\mathrm{ran}(\alpha \circ \alpha )=\mathrm{ran}\alpha$. This implies that $(P_{f,e}f)\alpha ^{-1}=e$ and $(eP_{f,e})\alpha =f$, which gives $P_{f,e}f=e\alpha =f$ and $eP_{f,e}=f\alpha ^{-1}=e$. Moreover, by Lemma 3.3 and condition (3.1) we have

$$\begin{aligned} fP_{f,e}=f^{\circ } P_{f,e}=P_{f^{\circ } f,e}=P_{f,e} \end{aligned}$$

and

$$\begin{aligned} P_{f,e}e= P_{f,e}e^{\circ } =P_{f,ee^{\circ }}=P_{f,e}. \end{aligned}$$

Therefore $e{\mathcal {L}}P_{f,e}\in E^{\circ }$ and $f{\mathcal {R}}P_{f,e}\in E^{\circ }$ and so $f^{\circ }=P_{f,e}=e^{\circ }$. On the other hand, for $x\in eE^{\circ }$, we have

$$\begin{aligned} x\alpha =x(\alpha \circ \alpha )=((x\alpha )\pi ^{-1}_{f,e})\alpha . \end{aligned}$$

Since $\alpha$ is bijective, it follows that

$$\begin{aligned} x=(x\alpha )\pi ^{-1}_{f,e}=(eP_{f,e})(x\alpha )^{\circ }=ee^{\circ } (x\alpha )^{\circ }=e(x\alpha )^{\circ }. \end{aligned}$$

Conversely, if the given condition in the corollary holds, then we can deduce that $\alpha \in E(T_{(I,\Lambda , E^{\circ }, P)})$ by the above discussions. $\square$

Corollary 3.10

If $(I,\Lambda , E^{\circ }, Q)$ is the normal admissible quadruple determined by $I, \Lambda$ and $E^{\circ }$, then the Munn semigroup $NT=T_{(I,\Lambda , E^{\circ }, Q)}$ is an orthodox semigroup with a multiplicative inverse transversal $T_{E^{\circ }}$ such that E(NT) forms a normal band.

Proof

Let $\alpha , \beta \in E(NT)$ where $\alpha \in T_{e,f}$ and $\beta \in T_{g,h}$. Then by Corollary 3.9, we have

$$\begin{aligned} f^{\circ }=e^{\circ } \quad \text{ and } \quad e(x\alpha )^{\circ }=x \quad \text{ for } \text{ all } x\in eE^{\circ } \end{aligned}$$

(3.8)

and

$$\begin{aligned} h^{\circ }=g^{\circ } \quad \text{ and } \quad g(y\beta )^{\circ }=y \quad \text{ for } \text{ all } y\in gE^{\circ } \end{aligned}$$

(3.9)

Denote $\alpha \circ \beta \in T_{j,k}$, where

$$\begin{aligned} j=(Q_{f,g}f)\alpha ^{-1}=(f^{\circ } g^{\circ } f)\alpha ^{-1} =(g^{\circ } f)\alpha ^{-1} \end{aligned}$$

and $k=(gQ_{f,g})\beta =(gf^{\circ } g^{\circ })\beta =(gf^{\circ })\beta .$ Take $x=(g^{\circ } f)\alpha ^{-1}$ in (3.8). Then by Lemma 3.6(4), we have

$$\begin{aligned} j=(g^{\circ } f)\alpha ^{-1}=e(((g^{\circ } f)\alpha ^{-1})\alpha )^{\circ }=e(g^{\circ } f)^{\circ }=ef^{\circ } g^{\circ }=ee^{\circ } g^{\circ }=eg^{\circ }, \end{aligned}$$

(3.10)

which implies that $j^{\circ }=(eg^{\circ })^{\circ }=e^{\circ } g^{\circ }$. Similarly, we can show that $k^{\circ }=e^{\circ } g^{\circ }$ by (3.8) and (3.9). This gives that $j^{\circ }=Q_{k,j}=j^{\circ } k^{\circ }=k^{\circ }$. On the other hand, for $x\in jE^{\circ }=eg^{\circ } E^{\circ }\subseteq eE^{\circ }$, we have

$$\begin{aligned} x(\alpha \circ \beta )= & {} (x\alpha )\pi ^{-1}_{f,g}\beta =(gQ_{f,g}(x\alpha )^{\circ })\beta =(gf^{\circ } g^{\circ } (x\alpha )^{\circ })\beta \\= & {} (g g^{\circ } f^{\circ }(x\alpha )^{\circ })\beta =(g\cdot f^{\circ }(x\alpha )^{\circ })\beta =(g((x\alpha )f)^{\circ })\beta =(g(x\alpha )^{\circ })\beta . \end{aligned}$$

Since $eg^{\circ }=eg$ (by Lemma 3.3), $ege=eg$ (as I is a left normal band) and $jx=x$, this implies that

$$\begin{aligned} j(x(\alpha \circ \beta ))^{\circ }&=eg^{\circ } ((g(x\alpha )^{\circ })\beta )^{\circ } =e\cdot g((g(x\alpha )^{\circ })\beta )^{\circ }\\&=eg(x\alpha )^{\circ }=ege(x\alpha )^{\circ } =egx=eg^{\circ } x=jx=x \end{aligned}$$

by (3.10), (3.9) and (3.8). Again by Corollary 3.9, we have $\alpha \circ \beta \in E(NT)$. This implies that NT is orthodox, and so $E(T_{E^{\circ }})$ is a multiplicative inverse transversal of E(NT) by Theorem 3.8. In view of Lemma 2.10, E(NT) is a normal band. $\square$

The following example illustrates Theorem 3.8 and Corollary 3.10.

Example 3.11

Let $I=\{0,e,g\}$ and $\Lambda =\{0,e,f\}$ be a left normal band and a right normal band, respectively, and their multiplication tables are:

Denote $E^{\circ }=I\cap \Lambda =\{0,e\}$ and define a $\Lambda \times I$-matrix P over $E^{\circ }$ by

Then it is routine to check that $(I,\Lambda , E^{\circ }, P)$ is an admissible quadruple and

$$\begin{aligned} 0E^{\circ }=\{0\}, \quad eE^{\circ }=\{0,e\}, \quad gE^{\circ }=\{0,g\}; \quad E^{\circ } 0=\{0\}, \quad E^{\circ } e=\{0,e\}, \quad E^{\circ } f=\{0,f\}. \end{aligned}$$

This implies that

$$\begin{aligned} {\mathcal {U}}=\{(0,0),(e,e),(e,f),(g,e), (g,f)\} \end{aligned}$$

and $|T_{i,\lambda }|=1$ for all $i\in I$ and $\lambda \in \Lambda$. Moreover, if we denote the unique element in $T_{i,\lambda }$ by $\alpha _{i,\lambda }$, then we have the multiplication table of $T_{(I,\Lambda , E^{\circ }, P)}$

In this case, we have

$$\begin{aligned} V_{T_{E^{\circ }}}(\alpha _{0,0})=\{\alpha _{0,0}\}, \quad V_{T_{E^{\circ }}}(\alpha _{e,e})=V_{T_{E^{\circ }}}(\alpha _{e,f})=V_{T_{E^{\circ }}}(\alpha _{g,e}) =V_{T_{E^{\circ }}}(\alpha _{g,f})=\{\alpha _{e,e}\} \end{aligned}$$

and

$$\begin{aligned} I^T=\{\alpha _{0,0},\alpha _{e,e}, \alpha _{g,e}\}, \quad \Lambda ^{T}=\{\alpha _{0,0}, \alpha _{e,e},\alpha _{e,f}\}, \quad \Lambda ^T I^T=T_{E^{\circ }}, \end{aligned}$$

this shows that $T_{E^{\circ }}=\{\alpha _{0,0}, \alpha _{e,e}\}$ is a multiplicative inverse transversal of $T_{(I,\Lambda , E^{\circ }, P)}$. Observe that $E(T_{(I,\Lambda , E^{\circ }, P)})$ is not a band (since $\alpha _{g,e} \circ \alpha _{e,f}=\alpha _{g,f}\not \in E(T_{(I,\Lambda , E^{\circ }, P)})$).

Now, we consider the Munn semigroup $T_{(I,\Lambda , E^{\circ }, Q)}$ of the normal admissible quadruple determined by $I,\Lambda$ and $E^{\circ }$, where the $\Lambda \times I$-matrix Q over $E^{\circ }$ is defined by

Then we can obtain the multiplication table of $T_{(I,\Lambda , E^{\circ }, Q)}$:

In this case, $T_{(I,\Lambda , E^{\circ }, Q)}$ forms a normal band with a multiplicative inverse transversal $T_{E^{\circ }}=\{\alpha _{0,0}, \alpha _{e,e}\}$.

The following corollary gives some useful information about the semigroup $T_{(I,\Lambda , E^{\circ }, P)}$ which will be used in the next sections frequently.

Corollary 3.12

Let $\alpha \in T_{e,f}$, $\beta \in T_{g,h}$ and $T=T_{(I,\Lambda , E^{\circ }, P)}$.

• (1) $\alpha ^{\circ }=\pi _{f,f^{\circ }}\alpha ^{-1}\pi _{e^{\circ },e}\in T_{f^{\circ }, e^{\circ }}$, $\overline{\alpha }=\alpha ^{\circ \circ }=\pi ^{-1}_{e^{\circ },e}\alpha \pi ^{-1}_{f,f^{\circ }}\in T_{e^{\circ }, f^{\circ }}$.

• (2) $u_{\alpha }=\alpha \circ \alpha ^{\circ }=\pi _{e^{\circ }, e}, v_{\alpha }=\alpha ^{\circ } \circ \alpha =\pi _{f,f^{\circ }}$ and so $I^{T}=\{\pi _{e^{\circ }, e}|e\in I\}$ and $\Lambda ^{T}=\{\pi _{f,f^{\circ }}|f\in \Lambda \}$.

• (3) $\alpha {\mathcal {R}} \beta$ (resp. $\alpha {\mathcal {L}} \beta$) in T if and only if $e=g$ (resp. $f=h$).

Proof

Item (1) follows directly from the proof of Theorem 3.8 and Lemma 2.9, and item (2) follows from Lemma 2.9 and the identity (3.7) and its dual. Item (3) follows from Lemma 2.2, Lemma 2.9, item (2) above and the fact that ${\mathcal {L}}={\mathcal {L}}^{*}$ and ${\mathcal {R}}={\mathcal {R}}^{*}$ on a regular semigroup.

$\square$

We say that two admissible quadruples $(I,\Lambda , E^{\circ }, P)$ and $(J,\Pi , F^{\circ },R)$ are isomorphic if there exist an isomorphism $\varphi$ from I onto J and an isomorphism $\psi$ from $\Lambda$ onto $\Pi$ such that

$$\begin{aligned} \varphi |_{E^{\circ }}=\psi |_{E^{\circ }}, \quad E^{\circ }\varphi =F^{\circ }, \quad P_{f,g}\varphi =R_{f\varphi , g\psi } \end{aligned}$$

for all $f\in \Lambda$ and $g\in I$. If this is the case, then one can easily show that $T_{(I,\Lambda , E^{\circ }, P)}$ is isomorphic to $T_{(J,\Pi , F^{\circ },R)}$. Moreover, we have the following.

Corollary 3.13

Let $(I,\Lambda , E^{\circ }, P)$ be an admissible quadruple. Then $(I,\Lambda , E^{\circ }, P)$ is isomorphic to the admissible quadruple of $T=T_{(I,\Lambda , E^{\circ }, P)}$. In particular, if $(I,\Lambda , E^{\circ }, Q)$ is the normal admissible quadruple determined by $I, \Lambda$ and $E^{\circ }$, then $(I,\Lambda , E^{\circ }, Q)$ is isomorphic to the admissible quadruple of the normal band E(NT), where NT is the Munn semigroup $T_{(I,\Lambda , E^{\circ }, Q)}$.

Proof

By Corollary 3.12(2), we can define the mappings

$$\begin{aligned} \varphi : I\rightarrow I^T, \quad e\mapsto \pi _{e^{\circ }, e},\ \ \ \psi : \Lambda \rightarrow \Lambda ^T, \quad f\mapsto \pi _{f, f^{\circ }}. \end{aligned}$$

By Lemma 3.3, condition (3.1) and Proposition 3.5, it is routine to check that the above mappings are isomorphisms such that

$$\begin{aligned} \varphi |_{E^{\circ }}=\psi |_{E^{\circ }}, \quad E^{\circ } \varphi =E(T_{E^{\circ }}), \quad P_{f,g}\psi =P^T_{f\varphi ,g\psi }. \end{aligned}$$

For the normal admissible quadruple $(I,\Lambda , E^{\circ }, Q)$, E(NT) is a normal band by Corollary 3.10 and so E(NT) is a full subsemigroup of NT. By Remark 3.2(2), $E(T_{E^{\circ }})=T_{E^{\circ }}\cap E(NT)$ is an inverse transversal of E(NT) and the admissible quadruples of NT and E(NT) are equal. By the first part of this corollary, $(I,\Lambda , E^{\circ }, Q)$ is isomorphic to the admissible quadruple of E(NT). $\square$

Remark 3.14

The above Corollary 3.13 shows that admissible quadruples come from regular semigroups with multiplicative inverse transversals and normal admissible quadruples come from normal bands with inverse transversals, respectively.

Now, we are a position to give the main result of this section.

Theorem 3.15

Let $(I,\Lambda , E^{\circ }, P)$ be an admissible quadruple and U a full subsemigroup of $T_{(I,\Lambda , E^{\circ }, P)}$. Then U is a fundamental abundant semigroup with a multiplicative ample transversal $T_{E^{\circ }}\cap U$ whose admissible quadruple is isomorphic to $(I,\Lambda , E^{\circ }, P)$. In particular, if U is also regular, then U is a fundamental regular semigroup with a multiplicative inverse transversal $T_{E^{\circ }}\cap U$. As a direct consequence, $T_{(I,\Lambda , E^{\circ }, P)}$ itself is fundamental.

Proof

By Theorem 3.8 and Lemma 2.9, $T_{E^{\circ }}$ is a multiplicative ample transversal of $T_{(I,\Lambda , E^{\circ }, P)}$. Since the Munn semigroup $T_{E^{\circ }}$ of the semilattice $E^{\circ }$ is fundamental, it follows that U is a fundamental abundant semigroup with a multiplicative ample transversal $T_{E^{\circ }}\cap U$ whose admissible quadruple is isomorphic to $(I,\Lambda , E^{\circ }, P)$ by Lemma 2.3, Corollary 2.4, Remark 3.2(2) and Corollary 3.13. The remaining result now follows from Lemma 2.9. $\square$

4 A Munn type representation of abundant semigroups with a multiplicative ample transversal

In this section, we always assume that S is an abundant semigroup with a multiplicative ample transversal $S^{\circ }$. By the previous section, we have the Munn semigroup $T_{(I^S,\Lambda ^S, E(S^{\circ }),P^S)}$ of the admissible quadruple of S, where $P^S_{f,g}$ is equal to the product of f and g in S for all $f\in \Lambda ^S$ and $g\in I^S$ (see Lemma 2.8 and Remark 3.2). The aim of this section is to show that there is a homomorphism $\rho : S\rightarrow T_{(I^S,\Lambda ^S, E(S^{\circ }),P^S)}$ whose kernel is $\mu _S$. For simplicity, we write $E(S^{\circ })$ as $E^{\circ }$. To accommodate with the notations of Sect. 3, we use the notations in Sect. 3 for the admissible quadruple $(I^S,\Lambda ^S, E(S^{\circ }),P^S)$ throughout this section.

In view of Lemmas 2.5 and 2.6, S is IC and for every $a\in S$, there exists a unique idempotent-connected isomorphism from $\langle u_a \rangle$ onto $\langle v_a \rangle$. We denote this isomorphism by $\lambda _a$ in the sequel. For all $a\in S$, denote the restriction of $\lambda _a$ to $u_aE^{\circ }$ by $\rho _a$, that is, $\rho _a=\lambda _a|_{u_aE^{\circ }}$. Recall that $a(x\lambda _a)=xa$ for all $a\in S$ and $x\in \langle u_a\rangle$. By the definition of $\rho _a$, we have $a(x\rho _a)=xa$ for all $a\in S$ and $x\in u_aE^{\circ }$.

Lemma 4.1

$\rho _a\in T_{u_a,v_a}$ for all $a\in S$.

Proof

Clearly, we have $u_aE^{\circ }=u_aE^{\circ } u_a\subseteq \langle u_a \rangle =\mathrm{dom}\lambda _a.$ Take $x=u_a i\in u_aE^{\circ }, i\in E^{\circ }$. Then $a(x\lambda _a)=xa$. Since $a=u_a\overline{a}v_a$, we have $u_a\overline{a}v_a(x\lambda _a)=u_aiu_a\overline{a}v_a$. Observe that $u_a{\mathcal {L}}\overline{a}^+{\mathcal {R}}^{*} \overline{a}$ and $\overline{a}^+ i=i\overline{a}^+$, it follows that

$$\begin{aligned} \overline{a} v_a(x\lambda _a)=\overline{a}^+ u_a\overline{a}v_a(x\lambda _a)=\overline{a}^+ u_aiu_a\overline{a}v_a=i\overline{a}v_a. \end{aligned}$$

Because $S^{\circ }$ is ample, we have $i\overline{a}=\overline{a}(i\overline{a})^{*}, (i\overline{a})^{*}\in E^{\circ }$ by the identity (1.3). This implies that $\overline{a} v_a(x\lambda _a)=\overline{a}(i\overline{a})^{*} v_a$. By Lemma 2.1 and the fact that $\overline{a}{\mathcal {L}}^{*} \overline{a}^{*}{\mathcal {R}}v_a$ (see (2.1)) and $x\lambda _a\in \langle v_a \rangle$, we obtain that

$$\begin{aligned} x\lambda _a=v_a(x\lambda _a)=\overline{a}^{*} v_a(x\lambda _a)=\overline{a}^{*}(i\overline{a})^{*} v_a =(i\overline{a})^{*}\overline{a}^{*} v_a=(i\overline{a})^{*} v_a\in E^{\circ } v_a. \end{aligned}$$

(4.1)

Dually, we can see that $\lambda ^{-1}_a|_{E^{\circ } v_a}$ is a mapping from $E^{\circ } v_a$ to $u_aE^{\circ }$. Thus, $\rho _a\in T_{u_a,v_a}$.   $\square$

Lemma 4.2

$\rho _a\circ \rho _b=\rho _{ab}$ for all $a,b\in S$.

Proof

Since $\rho _a\in T_{u_a,v_a}$ and $\rho _b\in T_{u_b,v_b}$, we can assume that $\rho _a\circ \rho _b\in T_{j,k}$ where

$$\begin{aligned} j=(P^S_{v_a, u_b}v_a)\rho ^{-1}_a=(v_au_bv_a)\rho ^{-1}_a,\quad k=(u_bP^S_{v_a, u_b})\rho _b=(u_bv_au_b)\rho _b. \end{aligned}$$

We first show that $j=u_{ab}$ and $k=v_{ab}$. In fact, by Lemma 2.6(2), we have $u_{ab}=u_a({\bar{a}}v_a u_b )^+\in u_aE^{\circ }$. Since $S^{\circ }$ is an ample semigroup and $v_au_b\in E^{\circ }$, we get $({\overline{a}}v_a u_b)^+ {\overline{a}}={\overline{a}}v_a u_b$ by (1.3). In view of the identity (4.1), it follows that

$$\begin{aligned} u_{ab}\rho _a=(({\overline{a}}v_a u_b )^+ \overline{a})^{*} v_a=({\overline{a}}v_a u_b)^{*} v_a =({\overline{a}}^{*} v_a u_b)^{*} v_a=(v_a u_b)^{*} v_a=v_au_bv_a \end{aligned}$$

by the identity (1.1) and the fact that $v_a {\mathcal {R}}^{*} \overline{a}^{*}$ and $v_au_b\in E^{\circ }$. This implies that $j=(v_au_bv_a)\rho ^{-1}_a=u_{ab}$. Dually, we can prove that $k=v_{ab}$.

Finally, let $x\in \mathrm{dom}\rho _{ab}$. On one hand,

$$\begin{aligned} x(\rho _a\circ \rho _b)=x\rho _a\pi ^{-1}_{v_a,u_b}\rho _b= (u_bP^S_{v_a, u_b}(x\rho _a)^{\circ })\rho _b=(u_b v_a u_b (x\rho _a)^{\circ })\rho _b. \end{aligned}$$

On the other hand, since $v_au_b\in E^{\circ }\subseteq \Lambda ^S, x\rho _a\in \Lambda ^S$, we have

$$\begin{aligned} (x\rho _a)^{\circ }\in E^{\circ }, \quad (x\rho _a)^{\circ }\cdot v_au_b=v_au_b\cdot (x\rho _a)^{\circ },\quad (x\rho _a)(v_au_b)=(x\rho _a)^{\circ }(v_au_b) \end{aligned}$$

by Lemma 3.3. Observe that $\Lambda ^S$ is a right normal band, it follows that $v_au_b(x\rho _a)v_au_b=(x\rho _a)v_au_b$, whence

$$\begin{aligned} ab\cdot x(\rho _a\circ \rho _b)= & {} ab\cdot (u_b v_a u_b (x\rho _a)^{\circ })\rho _b=a\cdot [b\cdot (u_b v_a u_b (x\rho _a)^{\circ })\rho _b]\\= & {} a\cdot u_b v_a u_b (x\rho _a)^{\circ } b=a\cdot u_b (x\rho _a)^{\circ } v_a u_b b =av_a\cdot u_b (x\rho _a)^{\circ } v_a u_b b\\= & {} a\cdot v_a u_b (x\rho _a) v_a u_b b \!=\!a (x\rho _a)\cdot v_a u_b b=xa\cdot v_a u_b b=xab=ab\cdot (x\rho _{ab}). \end{aligned}$$

Since $ab{\mathcal {L}}^{*} v_{ab}$ and

$$\begin{aligned} x(\rho _a\circ \rho _b),\quad x\rho _{ab}\in \mathrm{ran}\rho _{ab}=E^{\circ } v_{ab}=v_{ab}E^{\circ } v_{ab}, \end{aligned}$$

we have

$$\begin{aligned} x(\rho _a\circ \rho _b)=v_{ab}\cdot x(\rho _a\circ \rho _b)=v_{ab}\cdot x\rho _{ab}=x\rho _{ab} \end{aligned}$$

by Lemma 2.1. Thus $\rho _a\circ \rho _b=\rho _{ab}$. $\square$

Lemma 4.3

$\mu _S=\{(a,b)\in S\times S|\rho _a=\rho _b\}$.

Proof

Denote $\delta =\{(a,b)\in S\times S|\rho _a=\rho _b\}$. By Lemma 4.2, $\delta$ is a congruence on S. If $\rho _a=\rho _b$, then $\mathrm{dom}\rho _a=\mathrm{dom}\rho _b$ and $\mathrm{ran}\rho _a=\mathrm{ran}\rho _b$, which implies that $a{\mathcal {R}}^{*} u_a=u_b {\mathcal {R}}^{*} b$ and $a{\mathcal {L}}^{*} v_a=v_b {\mathcal {L}}^{*} b$ by Lemma 2.6, and so $a{{\mathcal {H}}}^{*} b$. Thus $\delta \subseteq {{\mathcal {H}}}^{*}$.

On the other hand, let $\sigma$ be a congruence on S such that $\sigma \subseteq {{\mathcal {H}}}^{*}$ and $(a,b)\in \sigma$. Then $(a,b)\in {{\mathcal {H}}}^{*}$ and so

$$\begin{aligned} v_a{\mathcal {L}}^{*} a{\mathcal {L}}^{*} b{\mathcal {L}}^{*} v_b, \quad u_a{\mathcal {R}}^{*} a{\mathcal {R}}^{*} b{\mathcal {R}}^{*} u_b \end{aligned}$$

by Lemma 2.6. Since $I^S$ is a left normal band and $\Lambda ^S$ is a right normal band, we have $\overline{a}^+ {\mathcal {L}}u_a=u_b {\mathcal {L}} \overline{b}^+$ and $\overline{a}^{*} {\mathcal {R}}v_a=v_b{\mathcal {R}} \overline{b}^{*}$ by (2.1), and so $\overline{a}^+=\overline{b}^+$ and $\overline{a}^{*}=\overline{b}^{*}$. Since $a\sigma b$, it follows that $\overline{a}=\overline{a}^+ a \overline{a}^{*} \sigma \overline{b}^+ b \overline{b}^{*}=\overline{b}$ by (2.2). Let

$$\begin{aligned} x\in \mathrm{dom}\rho _a=\mathrm{dom}\rho _b=u_aE^{\circ }=u_bE^{\circ }. \end{aligned}$$

Observe that $a(x\rho _a)=xa$, $b(x\rho _b)=xb$ and $a\sigma b$, it follows that $a(x\rho _a)= xa\ \sigma \ xb= b(x\rho _b)$ and so $\overline{a}^+ a(x\rho _a) \sigma \overline{b}^+ b(x\rho _b)$. Because

$$\begin{aligned} x\rho _a, x\rho _b\in E^{\circ } v_a=E^{\circ } v_b= \overline{a}^{*} E^{\circ } v_a=\overline{b}^{*} E^{\circ } v_b, \overline{a} \sigma \overline{b}, \end{aligned}$$

we have

$$\begin{aligned} \overline{a}(x\rho _a)=\overline{a}^+ a\overline{a}^{*} (x\rho _a)=\overline{a}^+ a(x\rho _a) \sigma \overline{b}^+ b(x\rho _b)=\overline{b}^+ b\overline{b}^{*}(x\rho _b)=\overline{b}(x\rho _b)\sigma \overline{a}(x\rho _b), \end{aligned}$$

which implies that $\overline{a}(x\rho _a)\sigma \overline{a}(x\rho _b)$ and so $\overline{a}(x\rho _a){\mathcal {L}}^{*} \overline{a}(x\rho _b)$ by $\sigma \subseteq {{\mathcal {H}}}^{*}$. Since ${\overline{a}}{\mathcal {L}}^{*}{\overline{a}}^{*}$, we have

$$\begin{aligned} x\rho _a=\overline{a}^{*}(x\rho _a){\mathcal {L}} \overline{a}^{*}(x\rho _b)=x\rho _b. \end{aligned}$$

Observe that $\Lambda ^S$ is a right normal band, it follows that $x\rho _a=x\rho _b$. This implies that $\rho _a=\rho _b$. Thus, $\delta$ is the largest congruence contained in ${{\mathcal {H}}}^{*}$ on S. That is to say, $\delta =\mu _S$.   $\square$

Theorem 4.4

Define $\rho : S\rightarrow T=T_{(I^S,\Lambda ^S, E(S^{\circ }), P^S)}, a\mapsto \rho _a$. Then $\rho$ is a homomorphism whose kernel is $\mu _S$. Moreover, $\rho$ satisfies the following conditions:

• (1) $\rho |_{I^S}$ (resp. $\rho |_{\Lambda ^S}$) is an isomorphism from $I^S$ onto $I^{T}$ (resp. $\Lambda ^{T}$).

• (2) $S^{\circ } \rho \subseteq T_{E^{\circ }}$ and $\rho |_{E^{\circ }}$ is an isomorphism from $E^{\circ }$ onto $E(T_{E^{\circ }})$.

• (3) $\rho |_{E(S)}$ is a bijection from E(S) onto E(T).

Proof

The first part follows from Lemmas 4.1, 4.2 and 4.3.

• (1) By Corollary 3.12, $I^T=\{\pi _{e^{\circ }, e}|e\in I^S\}$. We first show that $\rho _e=\pi _{e^{\circ }, e}$ for all $e\in I^S$. In fact, since $e=ee^{\circ } e^{\circ }$ and $e{\mathcal {L}}e^{\circ } {\mathcal {R}}e^{\circ }\in E^{\circ }$, we have $u_e=e,v_e=e^{\circ }$ by (2.1), and so $\rho _e\in T_{u_e,v_e}=T_{e,e^{\circ }}\ni \pi _{e^{\circ }, e}$. Let $x\in \mathrm{dom}\rho _e=\mathrm{dom}\pi _{e^{\circ },e}=eE^{\circ }$. Then we have $e(x\rho _e)=xe, x\rho _e\in \mathrm{ran}\rho _e=E^{\circ } e^{\circ }$ and so

  $$\begin{aligned} x\rho _e=e^{\circ } (x\rho _e)=e^{\circ } e(x\rho _e)=e^{\circ } xe. \end{aligned}$$

  Since $I^S$ is a left normal band, it follows that

  $$\begin{aligned} e^{\circ } xe=e^{\circ } e x=e^{\circ } x=e^{\circ } x^{\circ }=x^{\circ } e^{\circ }=x\pi _{e^{\circ },e} \end{aligned}$$

  by Lemma 3.3 and Proposition 3.5. This yields that $\rho _e=\pi _{e^{\circ },e}$ for all $e\in I^S$. By the above discussions and Lemma 4.2, $\rho |_{I^S}$ is a homomorphism from $I^S$ onto $I^{T}$. Since the kernel of $\rho$ is $\mu _S$, it follows that $\rho |_{I^S}$ is injective. The result for $\rho |_{\Lambda ^S}$ can be proved by symmetry.

• (2) If $a\in S^{\circ }$, then $u_a, v_a\in E^{\circ }$. This implies that $\rho _a\in T_{u_a,v_a}\subseteq T_{E^{\circ }}$ by Lemma 4.1 and Theorem 3.8. The remaining result follows from item (1) by considering the restriction of $\rho |_{I^S}$ to $E^{\circ }$.

• (3) Since $\rho$ is a homomorphism whose kernel is $\mu _S$, we have $E(S)\rho \subseteq E(T)$ and $\rho |_{E(S)}$ is injective. Let $\alpha \in T_{e,f}, e\in I^S, f\in \Lambda ^S$ and $\alpha \in E(T)$. Then $\alpha \circ \alpha =\alpha$, whence $\mathrm{dom}(\alpha \circ \alpha )=\mathrm{dom}\alpha$. This implies that $(P^S_{f,e}f)\alpha ^{-1}=e$, which gives $fef=P^S_{f,e}f=e\alpha =f$ by Lemma 3.6(1). Thus, $(ef)^2=e(fef)=ef\in E(S)$. Moreover, by Corollaries 2.7,  3.12, the fact that $\rho _e=\pi _{e^{\circ },e}, \rho _f=\pi _{f,f^{\circ }}$ and Lemma 4.2, we have

  $$\begin{aligned} \alpha =u_\alpha \circ v_\alpha =\pi _{e^{\circ },e}\circ \pi _{f,f^{\circ }}=\rho _e\circ \rho _f=\rho _{ef}, \end{aligned}$$

  whence $\rho |_{E(S)}$ is also surjective.$\square$

Combining Theorem  3.15 and Theorem 4.4, we obtain the main result of this paper.

Theorem 4.5

Let $(I,\Lambda , E^{\circ }, P)$ be a given admissible quadruple. Then a semigroup S is a fundamental abundant semigroup (resp. fundamental regular semigroup) having a multiplicative ample transversal (resp. multiplicative inverse transversal) whose admissible quadruple is isomorphic to $(I,\Lambda , E^{\circ }, P)$ if and only if it is isomorphic to a full subsemigroup (resp. full regular subsemigroup) of $T_{(I,\Lambda , E^{\circ }, P)}$.

5 Properties of some special admissible quadruples

In this section, we consider some special admissible quadruples. An admissible quadruple $(I,\Lambda , E^{\circ }, P)$ is called rigid if $|T_{e,f}|=1$ for all $(e,f)\in {\mathcal {U}}$.

Proposition 5.1

An admissible quadruple $(I,\Lambda , E^{\circ }, P)$ is rigid if and only if ${\mathcal {H}}^{*}$ is a congruence on every abundant semigroup with a multiplicative ample transversal whose admissible quadruple is isomorphic to $(I,\Lambda , E^{\circ }, P)$.

Proof

Let S be an abundant semigroup with a multiplicative ample transversal $S^{\circ }$ whose admissible quadruple $(I^S,\Lambda ^S, E(S^{\circ }), P^S)$ is rigid. By Theorem 4.4,

$$\begin{aligned} \rho : S\rightarrow T_{(I^S,\Lambda ^S, E(S^{\circ }), P^S)}, \quad a\mapsto \rho _a \end{aligned}$$

is a homomorphism whose kernel is $\mu _S$. If $(a,b)\in {{\mathcal {H}}}^{*}$, then $u_a=u_b$ and $v_a=v_b$ by Lemma 2.2. This implies that $\rho _a,\rho _b\in T_{u_a,v_a}=T_{u_b,v_b}$ and $\rho _a=\rho _b$ since $(I^S,\Lambda ^S, E(S^{\circ }), P^S)$ is rigid. This gives ${{\mathcal {H}}}^{*}\subseteq \mu _S$ and so ${{\mathcal {H}}}^{*}=\mu _S$ is a congruence on S.

On the other hand, by Theorem 3.8, Lemma 2.9 and Corollary 3.13, $T_{(I,\Lambda , E^{\circ }, P)}$ is an abundant semigroup with a multiplicative ample transversal whose admissible quadruple is isomorphic to $(I,\Lambda , E^{\circ }, P)$. If the relation ${{\mathcal {H}}}^{*}$ on $T_{(I,\Lambda , E^{\circ }, P)}$ is a congruence, then we have $\mu _S={{\mathcal {H}}}^{*}$. Observe that $T_{(I,\Lambda , E^{\circ }, P)}$ is fundamental by Theorem 3.15, it follows that ${\mathcal {H}}^{*}$ is the identity congruence on $T_{(I,\Lambda , E^{\circ }, P)}$. This gives $|T_{e,f}|=1$ for all $(e,f)\in {\mathcal {U}}$ by Corollary 3.12(3). That is, $(I,\Lambda , E^{\circ }, P)$ is rigid.$\square$

We call an admissible quadruple $(I,\Lambda , E^{\circ }, P)$ uniform if $(e,f)\in {\mathcal {U}}$ for all $e\in I$ and $f\in \Lambda$. On uniform admissible quadruples, we have the following.

Proposition 5.2

Let S be an abundant semigroup with a multiplicative ample transversal $S^{\circ }$. If S is ${\mathcal {D}}^{*}$-simple (i.e. any two elements in S are ${\mathcal {D}}^{*}$-related), then its admissible quadruple is uniform. On the other hand, if an admissible quadruple $(I,\Lambda , E^{\circ }, P)$ is uniform, then $T_{(I,\Lambda , E^{\circ },P)}$ is ${\mathcal {D}}$-simple (and also ${\mathcal {D}}^{*}$-simple since $T_{(I,\Lambda , E^{\circ },P)}$ is regular).

Proof

Let $u_a\in I^S$ and $v_b\in \Lambda ^S$ where $a,b\in S$. If S is ${\mathcal {D}}^{*}$-simple, then $u_a{\mathcal {D}}^{*} v_b$ and so there exist $c_1,c_2,\ldots , c_n\in S$ such that $u_a {\mathcal {R}}^{*} c_1 {\mathcal {L}}^{*} c_2 {\mathcal {R}}^{*} c_3\cdots c_n {\mathcal {R}}^{*} v_b$. By Lemma 2.2 and Lemma 2.6(3), we have

$$\begin{aligned} u_a=u_{u_a}=u_{c_1},\quad v_{c_1}=v_{c_2},\quad u_{c_2}=u_{c_3},\ldots , v_{c_{n-1}}=v_{c_n},\quad u_{c_n}=u_{v_b},\quad v_{v_b}=v_b. \end{aligned}$$

This implies that $\rho _{c_1}\rho ^{-1}_{c_2}\rho _{c_3}\cdots \rho ^{-1}_{c_n}\rho _{v_b}$ is an isomorphism from $u_aE^{\circ }$ onto $E^{\circ } v_b$ by Lemma 4.1 and so $(u_a,v_b)\in {\mathcal {U}}$. Thus, the admissible quadruple of S is uniform.

On the other hand, let $(I,\Lambda , E^{\circ },P)$ be uniform and $\alpha \in T_{e,f}$ and $\beta \in T_{g,h}$ be two elements in $T_{(I,\Lambda , E^{\circ },P)}$. Since $(I,\Lambda , E^{\circ },P)$ is uniform, we can take $\gamma \in T_{e,h}$. Then we have $\alpha {\mathcal {R}} \gamma {\mathcal {L}} \beta$ and so $\alpha {\mathcal {D}} \beta$ by Corollary 3.12(3). Thus $T_{(I,\Lambda , E^{\circ },P)}$ is ${\mathcal {D}}$-simple. $\square$

Finally, an admissible quadruple $(I,\Lambda , E^{\circ }, P)$ is called left anti-uniform (resp. right anti-uniform) if $(e,f)\in {\mathcal {U}}$ implies that $f=e^{\circ }$ (resp. $e=f^{\circ }$) for all $e\in I$ and $f\in \Lambda$. To give some properties of left anti-uniform and right anti-uniform admissible quadruples, we need some notions and facts. Recall the an abundant semigroup S is called superabundant if every ${{\mathcal {H}}}^{*}$-class of S contains an idempotent. We call a superabundant semigroup S is a left normal (resp. right normal) superabundant semigroup if E(S) forms a left normal band (resp. right normal band).

Lemma 5.3

Let S be an abundant semigroup with a multiplicative ample transversal $S^{\circ }$ and $a\in S$. If S is a left normal (resp. right normal) superabundant semigroup, then $a{{\mathcal {H}}}^{*} u_a$ (resp. $a{{\mathcal {H}}}^{*} v_a$).

Proof

Since E(S) is a left normal band and $v_a{\mathcal {R}}\overline{a}^{*}$ by (2.1), we have $v_a=\overline{a}^{*} v_a=\overline{a}^{*} v_a\overline{a}^{*}=\overline{a}^{*}$. On the other hand, since S is superabundant, it follows that $a {{\mathcal {H}}}^{*} e$ for some $e\in E(S)$. Since $\overline{a}^+{\mathcal {L}}u_a{\mathcal {R}}^{*} a {\mathcal {L}}^{*} v_a=\overline{a}^{*}$ by (2.1) and Lemma 2.6(2), we have $\overline{a}^+{\mathcal {L}}u_a{\mathcal {R}}e{\mathcal {L}}v_a=\overline{a}^{*}$. This implies that $\overline{a}^{*} {\mathcal {R}}\overline{a}^{*} u_a{\mathcal {L}}u_a{\mathcal {L}}\overline{a}^+$. Observe that $\overline{a}^{*} u_a\in E(S^{\circ }) I^S\subseteq I^S\subseteq E(S)$ by Lemma 2.8, it follows that $\overline{a}^{*} {\mathcal {L}} \overline{a}^+ \overline{a}^{*}{\mathcal {R}}\overline{a}^+$ whence $\overline{a}^+ =\overline{a}^{*}$. Thus, $u_a{\mathcal {R}}^{*} a {\mathcal {L}}^{*} v_a=\overline{a}^{*} =\overline{a}^+{\mathcal {L}}u_a$, which gives $a{{\mathcal {H}}}^{*} u_a$. Dually, we can prove that $a{{\mathcal {H}}}^{*} v_a$ if S is a right normal superabundant semigroup.$\square$

Proposition 5.4

An admissible quadruple $(I,\Lambda , E^{\circ }, P)$ is left anti-uniform (resp. right anti-uniform) if and only if every abundant semigroup with a multiplicative ample transversal whose admissible quadruple is isomorphic to $(I,\Lambda , E^{\circ }, P)$ is a left normal (resp. right normal) superabundant semigroup.

Proof

Let $(I,\Lambda , E^{\circ }, P)$ be left anti-uniform and S be an abundant semigroup with a multiplicative ample transversal whose admissible quadruple is isomorphic to $(I,\Lambda , E^{\circ }, P)$. Then $(I^S,\Lambda ^S, E(S^{\circ }), P^S)$ is left anti-uniform. By Lemma 4.1, for every $a\in S$, we have $(u_a,v_a)\in {\mathcal {U}}$ and $v_a=u^{\circ }_a=\overline{a}^+$ since $u_a{\mathcal {L}}\overline{a}^+$ by (2.1). This implies that $u_a{\mathcal {R}}^{*} a {\mathcal {L}}^{*} v_a=u^{\circ }_a=\overline{a}^+ {\mathcal {L}}u_a$ by (2.1) and Lemma 2.6(1), and hence $a{{\mathcal {H}}}^{*} u_a$. Therefore S is superabundant. Moreover, for $e\in E(S)$, we have $\overline{e}\in E(S^{\circ })$ by Lemma 2.6(2), and so $v_e=\overline{e}^+=\overline{e}$. This implies that $e=u_e\overline{e}v_e=u_e\overline{e}\in I^S$. Thus $E(S)=I^S$ is a left normal band by Lemma 2.8.

Conversely, let $(I,\Lambda , E^{\circ }, P)$ be not left anti-uniform. Then there exist $e\in I$ and $f\in \Lambda$ such that $(e,f)\in {\mathcal {U}}$ and $f\not =e^{\circ }$. Thus there exists $\alpha \in T_{e,f}$ such that $u_\alpha =\alpha \circ \alpha ^{\circ }=\pi _{e^{\circ }, e}\in T_{e,e^{\circ }}$ in $T_{(I,\Lambda , E^{\circ }, P)}$ by Corollary 3.12(2). Since $\alpha \in T_{e,f}, u_\alpha \in T_{e,e^{\circ }}$ and $f\not =e^{\circ }$, it follows that $\alpha$ and $u_\alpha$ are not ${\mathcal {L}}^{*}$-related in $T_{(I,\Lambda , E^{\circ }, P)}$ by Corollary 3.12(3). In view of Lemma 5.3, $T_{(I,\Lambda , E^{\circ }, P)}$ is not a left normal superabundant semigroup. The case for right anti-uniform admissible quadruples can be proved by symmetry. $\square$