Ternary rings of operators arising from inverse semigroups

Abstract

We are interested in properties, especially injectivity (in the sense of category theory), of the ternary rings of operators generated by certain subsets of an inverse semigroup via the regular representation. We determine all subsets of the extended bicyclic semigroup which are closed under the triple product $xy^*z$ (called semiheaps) and show that the weakly closed ternary rings of operators generated by them are injective operator spaces.

1 Introduction

Ternary rings of operators (TROs) originated in the work of M. R. Hestenes in 1962 [4]. These are linear spaces of operators from one Hilbert space to another which are stable under the triple product $XY^*Z$, which he called ternary algebras. By their nature, these spaces satisfied an associativity condition involving five elements, namely,

$$\begin{aligned} (XY^*Z)U^*W=XY^*(ZU^*W)=X(UZ^*Y)^*W. \end{aligned}$$

(1)

These were subsequently axiomatized and named associative triple systems [12]. A milestone in their development in the realm of functional analysis was a Gelfand-Naimark type representation theorem for associative triple systems equipped with an operator type norm [16].

At around the same time as Hestenes’ work, unbeknownst to the researchers in the West due partially to the Cold War [5], the concept of semiheap was introduced in the Soviet Union [8]. A semiheap is a set together with a single three-variable operation satisfying an abstract version of (1), and akin to the known concepts of ternary group and inverse semigroup.

Since the concept of semiheap is central to this paper, we provide the formal definition, as stated in [8, p. 56]. By a semiheap, we mean a set K together with a singled-valued, everywhere defined ternary operation $[\cdot \cdot \cdot ]$, satisfying the condition

$$\begin{aligned}{}[[k_1k_2k_3]k_4k_5]=[k_1[k_4k_3k_2]k_5]=[k_1k_2[k_3k_4k_5]]. \end{aligned}$$

An inverse semigroup is a semigroup S in which for every element x there exists a unique element $x^*$, called the inverse or generalized inverse of x, such that $x = x x^* x$ and $x^* = x^* x x^*$. For the basic facts on inverse semigroups, see [11, Chapter 1] or [7, Chapter 5].

Semiheaps and their associated structures are closely related to inverse semigroups. In turn, inverse semigroups, together with groupoids, give rise to operator algebras [13]. A ubiquitous example of an inverse semigroup is the bicyclic semigroup, given abstractly ([11, Section 3.4]) by the presentation $\langle p,q:pq=1\rangle ,$ and concretely ([7, p. 144]) as $\mathbb {N}\times \mathbb {N}$ with the multiplication

$$\begin{aligned} (m,n)(p,q)=(m-n+\max (n,p),q-p+\max (n,p)) \end{aligned}$$

(2)

We shall use the following notation: $\mathbb {N}=\{1,2,\ldots \}$; $\mathbb {N}_0=\mathbb {N}\cup \{0\}$; $\mathbb {Z}=\mathbb {N}_0\cup -\mathbb {N}$.

In this paper, we analyze the extended bicyclic semigroup, which we call E throughout this paper, in such a way that exhibits its semiheap structure. This inverse semigroup E, which is the set $\mathbb {Z}\times \mathbb {Z}$ together with the multiplication (2), was defined originally in [15, p. 367]; however, in that and most other papers, only binary structures are considered.

Unlike the bicyclic semigroup, the extended bicyclic semigroup is not finitely generated, nor does it have an identity element. Nevertheless, they share the same semigroup identities ([1, Corollary 4.3]).

We shall use the representation of the extended bicyclic semigroup which is based on the realization of the bicyclic semigroup by the unilateral shift ([13, p. 188]), as follows.

Let $E_{22}$ be the bicyclic semigroup, as realized by the unilateral shift; that is,

$$\begin{aligned} E_{22}=\{a_{ij}=\sum _{k\ge 0}e_{i+k,j+k}:i,j\in \mathbb {N}_0\}, \end{aligned}$$

where for any $i,j\in \mathbb {Z}$, $e_{ij}$ is the matrix over $\mathbb {Z}$ with 1 in the i, j position and zeros elsewhere, and the $\aleph _0$ by $\aleph _0$ matrix $a_{ij}$ acts as a linear operator on column vectors of complex numbers. ($a_{ij}$ is a bounded operator on the Hilbert space $\ell ^2(\mathbb {Z})$.)

Set

$$\begin{aligned} E=E_{11}\cup E_{12}\cup E_{21}\cup E_{22} \end{aligned}$$

where $E_{21}=\{a_{ij}:i\in \mathbb {N}_0, j\in -\mathbb {N}\}$, $E_{11}=\{a_{ij}:i,j\in -\mathbb {N}\}$ and $E_{12}=\{a_{ij}:i\in -\mathbb {N}, j\in \mathbb {N}_0\}$.

We note that for $i,j,p,q\in \mathbb {Z}$, $a_{ij}^*=a_{ji}$, $a_{ij}$ is a partial isometry on $\ell ^2(\mathbb {Z})$ and

$$\begin{aligned} a_{ij}a_{pq}=\left\{ \begin{array}{cc} a_{i,q+j-p}, &{} p\le j\\ a_{i+p-j,q},&{} p\ge j \end{array}, \right. \end{aligned}$$

(3)

equivalently

$$\begin{aligned} a_{ij}a_{pq}=a_{i+p-\min (j,p),j+q-\min (j,p)}. \end{aligned}$$

In particular, $a_{ij}a_{pq}\ne 0$, $a_{ij}a_{jq}=a_{iq}$, and $a_{ii}a_{pp}=a_{mm}$ with $m=\max (i,p)$.

Remark 1.1

Thus E is an inverse semigroup consisting of partial isometries with inverse $a_{ij}^*$ equal to the adjoint of $a_{ij}$. E is isomorphic to the extended bicyclic semigroup, and when convenient notationally, we represent $a_{ij}$ in formulas and diagrams simply by $(i,j)\in \mathbb {Z}\times \mathbb {Z}$.

We shall analyze the extended bicyclic semigroup E toward the aims of finding all of the subsemiheaps of E, and showing that the associated W*-TROs, that is, weakly closed TROs, are injective operator spaces.

In our main and only theorem, Theorem 1.2, we classify all of the subsemiheaps of this extended bicyclic semigroup. We then show in Corollary 4.3, via a general result applying to all inverse semigroups [13, Theorem 4.5.2], that each of the examples resulting from this classification has the property that the weakly closed ternary ring of operators it generates is an injective operator space. It is worth pointing out that, although the injectivity of the W*-TROs generated by the classification of subsemiheaps uses deep results in functional analysis ([3, Theorem 2.5], [13, Theorem 4.5.2]), the classification itself is self-contained using only elementary arguments.

Our results are summarized in the following theorem, listing all of the subsemiheaps of the extended bicyclic semigroup. The proof is contained in the references in each statement to later results of this paper.

Theorem 1.2

The subsemiheaps of the extended bicyclic semigroup are the inductive limits (see Remark 3.1) of sequences of the following semiheaps K:

• K is a single point $\{a_{pq}\}$ (Lemma 3.2 and Example 2.4)

• $K= \{a_{\alpha _0,\beta _0}\}\cup \{a_{\alpha _0+k_j,\beta _0+k_j}:1\le j\le n_0\}$, where $1\le k_1<k_2<\cdots <k_{n_0}$ and $n_0\in \mathbb {N}$ (Proposition 3.3 and Examples 2.4)

• $K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_j,\beta _0+k_j}:j\in \mathbb {N}\}$ where $1 \le k_1<k_2<\cdots<k_j<\cdots <\infty$ (Proposition 3.4 and Example 2.4)

• There exist $\sigma ,\ell _0\in \mathbb {N}$ such that

  $$\begin{aligned} K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_j,\beta _0+k_j}:j=1,\ldots \ell _0-1\}\cup K_{\alpha _0+k_{\ell _0},\beta _0+k_{\ell _0}}^\sigma \end{aligned}$$

  or

  $$\begin{aligned} K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_i,\beta _0+k_i}:1\le i<\ell _0\}\cup \left( \bigcup _{i=\ell _0}^{j} K^{\sigma }_{\alpha _0+k_i,\beta _0+k_i}\right) \end{aligned}$$

  where $1\le k_1<k_2<\cdots <k_{\ell _0}$. (Proposition 3.4 and Examples 2.5 and 2.6)

• $K=K_{\alpha _0,\beta _0}^p$ for some $p>0$ (Proposition 3.18 and Example 2.7)

• There exist $p>0$ and $q > 0$ such that

  $$\begin{aligned} K= \bigcup _{i=0}^n K_{\alpha _0+q_i, \beta _0+q_i}^p. \end{aligned}$$

  where $a_{\alpha _0+q_i,\beta _0+q_i}$, $0\le i<\infty$, are the points of K lying on the diagonal, such that

  $$\begin{aligned} q=q_0<q_1<q_2<\cdots<q_n<p\quad \hbox { and }\quad p<q_{n+1}<q_{n+2}<\cdots . \end{aligned}$$

  (Proposition 3.18 and Example 2.7)

All of the subsemigroups of the bicyclic semigroup have been determined in [2]. Those subsemigroups which are inverse subsemigroups, which were determined earlier in [14] and later in [6], were also identified in [2, Theorem 7.1]. Since inverse subsemigroups are semiheaps, our results give a new approach to the description of the inverse subsemigroups of the (extended) bicyclic semigroup.

2 Diagrams 1–10 and Examples

In order to analyze the subsemiheaps of the extended bicyclic semigroup E, we prepare some material.

The idempotents of E are the elements $a_{ii}$ with $i\in \mathbb {Z}$ and $a_{ii}\le a_{jj}$, that is, $a_{ii}a_{jj}=a_{ii}$, if and only if $j\le i$. From (3), we calculate and find that for $p,q\in \mathbb {Z}$,

$$\begin{aligned}&a_{ii}a_{pq}=\left\{ \begin{array}{ll} a_{i,q+i-p}\quad \quad &{}p\le i\\ a_{pq}\quad \quad &{}p\ge i \end{array}\right. \\&a_{pq}a_{jj}=\left\{ \begin{array}{ll} a_{pq}\quad \quad &{}j\le q\\ a_{p+j-q,j}\quad \quad &{}j\ge q \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} a_{ii}a_{pq}a_{jj}= \left\{ \begin{array}{ll} a_{i,q+i-p}\quad \quad &{}p\le i\hbox { and }j\le q+i-p\\ a_{j-q+p,q}\quad \quad &{}p\le i\hbox { and }j\ge q+i-p\\ a_{pq}\quad \quad &{}p\ge i\hbox { and }j\le q\\ a_{p+j-q,j}\quad \quad &{}p\ge i\hbox { and }j\ge q \end{array}\right. \end{aligned}$$

In particular, $a_{00}E=E_{21}\cup E_{22}$ and $Ea_{00} =E_{12}\cup E_{22}$. Also, $a_{ii}E$, $Ea_{jj}$ and $a_{ii}Ea_{jj}$ are subsemigroups and semiheaps, and $a_{ii}Ea_{jj}$ is an inverse semigroup if $i=j$. Also,

$$\begin{aligned} Ea_{jj}=\{a_{pq}:j\le q\},\quad a_{ii}E=\{a_{pq}:p\ge i\}, \end{aligned}$$

and

$$\begin{aligned} a_{ii}E\cap Ea_{jj}=a_{ii}Ea_{jj}=\{a_{pq}:p\ge i, q\ge j\}. \end{aligned}$$

From (3), we have the following lemma.

Lemma 2.1

For any $a_{ij}, a_{pq}, a_{rs}$ in E, we have

$$\begin{aligned} a_{ij}a_{pq}^*a_{rs}=\left\{ \begin{array}{lrl} \hbox {(i) }a_{i,s+p+j-q-r}&{}\quad r\le p+j-q,&{}q\le j\\ \hbox {(ii) }a_{i+r-p-j+q,s}&{}\quad r\ge p+j-q,&{}q\le j\\ \hbox {(iii) } a_{i+q-j+r-p,s}&{}\quad r\ge p,&{}q\ge j\\ \hbox {(iv) }a_{i+q-j,s+p-r}&{}\quad r\le p,&{}q\ge j\\ \end{array}\right. \end{aligned}$$

It is worth noting, as will be evident in the ten diagrams that follow, all triple products in E which involve only two elements, produce new elements which do not propagate to the left of, or up from the diagram.

Lemma 2.2 and Diagrams 1–5 describe the case in which the slope of the line connecting the two points is negative (or zero or infinite). Lemma 2.3 and Diagrams 6-10 describe the case in which the slope of the line connecting the two points is positive (or zero or infinite).

Lemma 2.2

If K is a subsemiheap of E, and if $a_{\alpha \beta },a_{\gamma \delta }\in K$ with $\gamma \ge \alpha$ and $\delta \ge \beta$, then the following elements belong to K:

• $x_1=a_{\alpha +\delta -\beta ,\beta +\gamma -\alpha }$

• $x_2=a_{\alpha +\delta -\beta ,\delta }$

• $x_3=a_{\gamma ,\beta +\gamma -\alpha }$

• $x_4=a_{\gamma ,\delta +(\delta -\beta )-(\gamma -\alpha )}$   if $\gamma -\alpha \le \delta -\beta$

• $x_5=a_{\gamma +(\gamma -\alpha )-(\delta -\beta ),\delta }$    if $\gamma -\alpha \ge \delta -\beta$

Proof

The following are the eight possible triple products containing two distinct elements, and thus belong to K. They are calculated using Lemma 2.1.

• $a_{\alpha \beta }a_{\alpha \beta }^*a_{\alpha \beta }=a_{\alpha \beta }$

• $a_{\alpha \beta }a_{\gamma \delta }^*a_{\alpha \beta }=a_{\alpha +\delta -\beta ,\beta +\gamma -\alpha }=x_1$

• $a_{\alpha \beta }a_{\alpha \beta }^*a_{\gamma \delta }=a_{\gamma \delta }$

• $a_{\alpha \beta }a_{\gamma \delta }^*a_{\gamma \delta }=a_{\alpha +\delta -\beta ,\delta }=x_2$

• $a_{\gamma \delta }a_{\alpha \beta }^*a_{\alpha \beta }=a_{\gamma \delta }$

• $a_{\gamma \delta }a_{\gamma \delta }^*a_{\alpha \beta }=a_{\gamma ,\beta +\gamma -\alpha }=x_3$

• $a_{\gamma \delta }a_{\alpha \beta }^*a_{\gamma \delta }= \left\{ \begin{array}{ll} a_{\gamma ,\delta +\alpha +\delta -\beta -\gamma }=x_4,&{}\quad \gamma -\alpha \le \delta -\beta \\ a_{\gamma +\gamma -\alpha -\delta +\beta ,\delta }=x_5, &{}\quad \gamma -\alpha \ge \delta -\beta \end{array}\right.$

• $a_{\gamma \delta }a_{\gamma \delta }^*a_{\gamma \delta }=a_{\gamma \delta }$

$\square$

Diagram 1

Diagram 2

Diagram 3

Diagram 4

Diagram 5

Lemma 2.3

If K is a subsemiheap of E, and if $a_{\alpha \beta },a_{\gamma \delta }\in K$ with $\gamma \ge \alpha , \delta \le \beta$, then the following elements belong to K:

• $x_1=a_{\alpha ,\beta +(\gamma -\alpha )+(\beta -\delta )}$

• $x_2=a_{\gamma +\beta -\delta ,\beta }$

• $x_3=a_{\gamma ,\beta +\gamma -\alpha }$

• $x_4=a_{\gamma +(\beta -\delta )+(\gamma -\alpha ),\delta }$

Proof

The following eight products belong to K and can be calculated using Lemma 2.1.

• $a_{\alpha \beta }a_{\alpha \beta }^*a_{\alpha \beta }=a_{\alpha \beta }$

• $a_{\alpha \beta }a_{\gamma \delta }^*a_{\alpha \beta }=a_{\alpha ,\beta +(\gamma -\alpha )+(\beta -\delta )}=x_1$

• $a_{\alpha \beta }a_{\alpha \beta }^*a_{\gamma \delta }=a_{\gamma \delta }$

• $a_{\alpha \beta }a_{\gamma \delta }^*a_{\gamma \delta }=a_{\alpha \beta }$

• $a_{\gamma \delta }a_{\alpha \beta }^*a_{\alpha \beta }=a_{\gamma +\beta -\delta ,\beta }=x_2$

• $a_{\gamma \delta }a_{\gamma \delta }^*a_{\alpha \beta }=a_{\gamma ,\beta +\gamma -\alpha }=x_3$

• $a_{\gamma \delta }a_{\alpha \beta }^*a_{\gamma \delta }=a_{\gamma +\beta -\delta +\gamma -\alpha ,\delta }=x_4$

• $a_{\gamma \delta }a_{\gamma \delta }^*a_{\gamma \delta }=a_{\gamma \delta }$

$\square$

Example 2.4

For $\alpha ,\beta \in \mathbb {Z}$, and $J\subset \mathbb {N}_0$, $D_{\alpha ,\beta }(J):= \{a_{\alpha +j,\beta +j}: j\in J\}$ is a subsemiheap of E.

Example 2.5

For $\alpha ,\beta \in \mathbb {Z}$, and $\sigma \in \mathbb {N}$, $K_{\alpha ,\beta }= \{a_{\alpha +\ell ,\beta +m}: \ell ,m\in \mathbb {N}_0\}$, and more generally, $K_{\alpha ,\beta }^\sigma := \{a_{\alpha +\ell \sigma ,\beta +m\sigma }: \ell ,m\in \mathbb {N}_0\}$ are subsemiheaps of E.

Special cases of Lemma 2.3 and their diagrams are as follows:

Diagram 6

Diagram 7

Diagram 8

Diagram 9

Diagram 10

Example 2.6

$K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_j,\beta _0+k_j}:j=1,\ldots \ell _0-1\}\cup K_{\alpha _0+k_{\ell _0},\beta _0+k_{\ell _0}}^\sigma$ is a subsemiheap of E, where $\alpha _0,\beta _0\in \mathbb {Z}$, $\ell _0,\sigma ,k_i\in \mathbb {N}$, $k_1<k_2<\cdots <k_{\ell _0-1}$, and

$$\begin{aligned} K_{\alpha _0+k_{\ell _0},\beta _0+k_{\ell _0}}^\sigma =\{a_{\alpha _0+k_{\ell _0}+m\sigma ,\beta _0+k_{\ell _0}+n\sigma }:m,n\in \mathbb {N}_0\}. \end{aligned}$$

Proof

Let $x_j=a_{\alpha _0+k_j,\beta _0+k_j}$ for $1\le k_j<\ell _0$ and $y_{mn}=a_{\alpha _0+\ell _0+m\sigma ,\beta _0+\ell _0+n\sigma }$ for $m,n\in \mathbb {N}_0$.

The following eight products belong to K, as calculated by Lemma 2.1.

• 1. $x_jy_{mn}^*y_{pq}=\left\{ \begin{array}{ll} y_{n+p-m,q}&{}\hbox { if } p\ge m \quad \quad { (\hbox {Lemma}~2.1\hbox {(iii))}}\\ y_{n,q+m-p}&{}\hbox { if } p\le m \quad \quad { (\hbox {Lemma}~2.1\hbox {(iv))}}\\ \end{array}\right.$

• 2. $y_{mn}x_j^*y_{pq}=\left\{ \begin{array}{ll} y_{m,n+q-p}&{}\hbox { if } p\le n \quad \quad { (\hbox {Lemma}~2.1\hbox {(i))}}\\ y_{m+p-n,q}&{}\hbox { if } p\ge n \quad \quad { (\hbox {Lemma}~2.1\hbox {(ii))}}\\ \end{array}\right.$

• 3. $y_{mn}y_{pq}^*x_j=\left\{ \begin{array}{ll} y_{m,p+n-q}&{}\hbox { if } q\le n \quad \quad { \hbox {(Lemma}~2.1\hbox {(i))}}\\ y_{m+q-n,p}&{}\hbox { if } q\ge n \quad \quad { \hbox {(Lemma}~2.1\hbox {(iv)}}\\ \end{array}\right.$

• 4. $y_{mn}x_i^*x_j=y_{mn} \quad \quad { \hbox {(Lemma}~2.1\hbox {(i))}}$

• 5. $x_iy_{mn}^*x_j=y_{nm} \quad \quad { \hbox {(Lemma}~2.1\hbox {(iv))}}$

• 6. $x_ix_j^*y_{mn}=y_{mn} \quad \quad { \hbox {(Lemma}~2.1\hbox {(ii))}}$

• 7. $x_ix_j^*x_\ell =x_{\max (i,j,\ell )}\quad { \hbox {(Lemma}~2.1\hbox {(i)-(iv))}}$

• 8. $y_{mn}y_{pq}^*y_{rs}=\left\{ \begin{array}{ll} y_{m,s+p+n-r-q}&{}\hbox { if } q\le n\hbox { and }r+q\le p+n\quad { \hbox {(Lemma}~2.1\hbox {(i))}}\\ y_{ m+r+q-p-n,s}&{}\hbox { if } q\le n\hbox { and }r+q\ge p+n\quad { \hbox {(Lemma}~2.1\hbox {(ii))}}\\ y_{m+q-n+r-p,s}&{}\hbox { if } q\ge n\hbox { and }r\ge p\quad \quad \quad \quad { \hbox {(Lemma}~2.1\hbox {(iii))}}\\ y_{ m+q-n+p-r,s}&{}\hbox { if } q\ge n\hbox { and }r\le p\quad \quad \quad \quad { \hbox {(Lemma}~2.1\hbox {(iv))}}\\ \end{array}\right.$

We provide some details for cases 1 and 8.

For case 1, by Lemma 2.1(iii),

$$\begin{aligned}&x_jy_{mn}^*y_{pq}= a_{\alpha _0+k_j,\beta _0+k_j} a_{\alpha _0+\ell _0+m\sigma ,\beta _0+\ell _0+n\sigma }^* a_{\alpha _0+\ell _0+p\sigma ,\beta _0+\ell _0+q\sigma }\\&\quad =a_{\alpha _0+\ell _0+(n+p-m)\sigma ,\beta _0+\ell _0+q\sigma }, \end{aligned}$$

if $p\ge m$ and $\ell _0+n\sigma \ge k_j$; and by Lemma 2.1(iv),

$$\begin{aligned}&a_{\alpha _0+k_j,\beta _0+k_j} a_{\alpha _0+\ell _0+m\sigma ,\beta _0+\ell _0+n\sigma }^* a_{\alpha _0+\ell _0+p\sigma ,\beta _0+\ell _0+q\sigma }\\&\quad =a_{\alpha _0+\ell _0+n\sigma ,\beta _0+\ell _0+(q+m-p)\sigma }, \end{aligned}$$

if $p\le m$ and $\ell _0+n\sigma \ge k_j$.

For case 8, with $q\le n$, by Lemma 2.1(i),

$$\begin{aligned}&a_{\alpha _0+\ell _0+m\sigma ,\beta _0+\ell _0+n\sigma } a_{\alpha _0+\ell _0 +p\sigma ,\beta _0+\ell _0+q\sigma }^* a_{\alpha _0+\ell _0+r\sigma ,\beta _0+\ell _0+s\sigma }\\&\quad =a_{\alpha _0+\ell _0+m\sigma ,\beta _0+\ell _0+(s+p+n-r-q)\sigma }, \end{aligned}$$

if $q\le n$ and $r+q\le p+n$, so that $s+p+n-r-q\in \mathbb {N}_0$; and by Lemma 2.1(ii),

$$\begin{aligned}&a_{\alpha _0+\ell _0+m\sigma ,\beta _0+\ell _0+n\sigma } a_{\alpha _0+\ell _0 +p\sigma ,\beta _0+\ell _0+q\sigma }^* a_{\alpha _0+\ell _0+r\sigma ,\beta _0+\ell _0+s\sigma }\\&\quad =a_{\alpha _0+\ell _0+(m+r+q-p-n)\sigma ,\beta _0+\ell _0+s\sigma }, \end{aligned}$$

if $q\le n$ and $r+q\ge p+n$, so that $m+r+q-p-n\in \mathbb {N}_0$.

The subcases of case 8 for which $q\ge n$ are as follows. By Lemma 2.1(iii),

$$\begin{aligned}&a_{\alpha _0+\ell _0+m\sigma ,\beta _0+\ell _0+n\sigma } a_{\alpha _0+\ell _0 +p\sigma ,\beta _0+\ell _0+q\sigma }^* a_{\alpha _0+\ell _0+r\sigma ,\beta _0+\ell _0+s\sigma }\\&\quad =a_{\alpha _0+\ell _0+(m+q-n+r-p)\sigma ,\beta _0+\ell _0+s\sigma }, \end{aligned}$$

if $q\ge n$ and $r\le p$, so that $m+q-n\in \mathbb {N}_0$; and by Lemma 2.1(iv),

$$\begin{aligned}&a_{\alpha _0+\ell _0+m\sigma ,\beta _0+\ell _0+n\sigma } a_{\alpha _0+\ell _0 +p\sigma ,\beta _0+\ell _0+q\sigma }^* a_{\alpha _0+\ell _0+r\sigma ,\beta _0+\ell _0+s\sigma }\\&\quad =a_{\alpha _0+\ell _0+(m+q-n)\sigma ,\beta _0+\ell _0+(s+p-r)\sigma }, \end{aligned}$$

if $q\ge n$ and $r\le p$, so that $m+q-n\in \mathbb {N}_0$ and $s+p-r\in \mathbb {N}_0$. $\square$

Example 2.7

Let $K= \bigcup _{k\in A} K_{\alpha _0+k,\beta _0+k}^p$, where $\alpha _0,\beta _0\in \mathbb {Z}$, $p>0$, $A\subset \{0,1,\ldots , p-1\}$ and $a_{\alpha _0+k,\beta _0+k}$, $k\in A$, denote the elements of K lying on the diagonal with $k<p$. (See the following Diagram and Proposition 3.15.) In fact, K is an inverse subsemigroup of E.

Proof

We note first that setting $\alpha _0=\beta _0=0$ for convenience (see Remark 2.8), and changing notation (see Remark 1.1),

$$\begin{aligned} K=\{(k+\ell p,k+mp):k\in A,\ell ,m\in \mathbb {N}_0\} \end{aligned}$$

and it suffices to show that

$$\begin{aligned} (k_1+\ell _1p,k_1+m_1p)(k_2+\ell _2 p,k_2+m_2p)^*(k_3+\ell _3p,k_3+m_3p) \end{aligned}$$

belongs to K. We calculate this triple product using the four cases in Lemma 2.1.

By Lemma 2.1(i), if $k_2+m_2p\le k _1+m_1p$, and $k_3+\ell _3p\le k_1+(\ell _2+m_1-m_2)p$, then

$$\begin{aligned}&(k_1+\ell _1p,k_1+m_1p)(k_2+\ell _2 p,k_2+m_2p)^*(k_3+\ell _3p,k_3+m_3p)\\&\quad =(k_1+\ell _1p,k_1+(m_3+\ell _2+m_1-m_2-\ell _3)p), \end{aligned}$$

and it is required to show that $m_3+\ell _2+m_1-m_2-\ell _3\ge 0$.

Following the argument in [2, Lemma 4.5], we have

$$\begin{aligned} k_1+(\ell _2+m_1-m_2-\ell _3)p\ge k_3\ge 0 \end{aligned}$$

so that $(\ell _2+m_1-m_2-\ell _3)p\ge -k_1>-p$ and therefore $\ell _2+m_1-m_2-\ell _3\ge 0$ and $m_3+ \ell _2+m_1-m_2-\ell _3\ge 0$, as required.

By Lemma 2.1(ii), if $k_2+m_2p\le k _1+m_1p$, and $k_3+\ell _3p\ge k_1+(\ell _2+m_1-m_2)p$, then

$$\begin{aligned}&(k_1+\ell _1p,k_1+m_1p)(k_2+\ell _2 p,k_2+m_2p)^*(k_3+\ell _3p,k_3+m_3p)\\&\quad =(k_3+(\ell _1+\ell _3-\ell _2-m_1+m_2)p,k_3+m_3p) \end{aligned}$$

and it is required to show that $\ell _3-\ell _2-m_1+m_2\ge 0$.

Following the argument in [2, Lemma 4.5], we have

$$\begin{aligned} k_3+(\ell _3-\ell _2-m_1+m_2)p\ge k_1\ge 0 \end{aligned}$$

so that $(\ell _3-\ell _2-m_1+m_2)p\ge -k_3>-p$ and therefore $\ell _3-\ell _2-m_1+m_2\ge 0$ and $\ell _1+\ell _3-\ell _2-m_1+m_2\ge 0$, as required.

By Lemma 2.1(iii), if $k_3+\ell _3p\ge k _2+\ell _2p$, and $k_2+m_2p\ge k_1+m_1p$, then

$$\begin{aligned}&(k_1+\ell _1p,k_1+m_1p)(k_2+\ell _2 p,k_2+m_2p)^*(k_3+\ell _3p,k_3+m_3p)\\&\quad =(k_3+(\ell _1+m_2-m_1+\ell _3-\ell _2)p,k_3+m_3p) \end{aligned}$$

and it is required to show that $\ell _1+m_2-m_1+\ell _3-\ell _2\ge 0$.

Following the argument in [2, Lemma 4.5], we have

$$\begin{aligned} k_3+(\ell _3-m_1+m_2-\ell _2)p\ge k_1\ge 0 \end{aligned}$$

so that $(\ell _3-m_1+m_2-\ell _2)p\ge -k_3>-p$ and therefore $\ell _3-m_1+m_2-\ell _2\ge 0$ and $\ell _1+\ell _3-m_1+m_2-\ell _2\ge 0$, as required.

By Lemma 2.1(iv), if $k_3+\ell _3p\le k _2+\ell _2p$, and $k_2+m_2p\ge k_1+m_1p$, then

$$\begin{aligned}&(k_1+\ell _1p,k_1+m_1p)(k_2+\ell _2 p,k_2+m_2p)^*(k_3+\ell _3p,k_3+m_3p)\\&\quad =(k_2+(\ell _1+m_2-m_1)p,k_2+(m_3+\ell _2-\ell _3)p) \end{aligned}$$

and it is required to show that $\ell _1+m_2-m_1\ge 0$. and $m_3+\ell _2-\ell _3\ge 0$.

Following the argument in [2, Lemma 4.5], we have

$$\begin{aligned} k_2+(\ell _2-\ell _3)p\ge k_3\ge 0 \end{aligned}$$

so that $(\ell _2-\ell _3)p\ge -k_3>-p$ and therefore $\ell _2-\ell _3\ge 0$ and $m_3 +\ell _2-\ell _3\ge 0$, as required.

Following the argument in [2, Lemma 4.5], we have

$$\begin{aligned} k_2+(m_2-m_3)p\ge k_1\ge 0 \end{aligned}$$

so that $(m_2-m_1)p\ge -k_1>-p$ and therefore $m_2-m_1\ge 0$ and $\ell _1+m_2-m_1\ge 0$, as required. $\square$

Remark 2.8

The adjoint operation $a_{ij}\mapsto a_{ij}^*=a_{ji}$ on the extended bicyclic semigroup E is an anti-isomorphism of a subsemiheap K of E onto the subsemiheap $K^*$, that is, $(ab^*c)^*=c^*ba^*$. As another application of Lemma 2.1, the translation map on the extended bicyclic semigroup is a triple isomorphism, that is, if $\varphi _{\alpha ,\beta }(a_{ij})=a_{i+\alpha ,j+\beta }$, then

$$\begin{aligned} \varphi (a_{ij}a_{pq}^*a_{rs})=\varphi ( a_{ij})\varphi (a_{pq})^*\varphi (a_{rs}). \end{aligned}$$

Hence, if K is a subsemiheap of $K_{\alpha ,\beta }$, then $\varphi _{-\alpha ,-\beta }(K)$ is a subsemiheap of the bicyclic semigroup $K_{0,0}$. At the very least, this fact can simplify notation in parts of this paper.

3 Subsemiheaps of the extended bicyclic semigroup

In this section, we shall determine all of the subsemiheaps of the extended bicyclic semigroup. We shall proceed as follows. First, for an arbitrary subsemiheap K of E, we define

$$\begin{aligned} \alpha _0=\inf \{\alpha \in \mathbb {Z}:\exists \beta \in \mathbb {Z}, a_{\alpha \beta }\in K\}, \end{aligned}$$

and

$$\begin{aligned} \beta _0=\inf \{\beta \in \mathbb {Z}:\exists \alpha \in \mathbb {Z}, a_{\alpha \beta }\in K\}. \end{aligned}$$

We have four mutually exclusive and exhaustive cases, namely,

• 1. Quadrant $\alpha _0\ne -\infty , \beta _0\ne -\infty$

• 2. Right Half Plane $\alpha _0=-\infty , \beta _0\ne -\infty$

• 3. Lower Half Plane $\alpha _0\ne -\infty , \beta _0= -\infty$

• 4. Full Plane $\alpha _0=-\infty , \beta _0= -\infty$

Remark 3.1

We only need to find all of the subsemiheaps of E which are in case (1), since the other cases can be reduced to this case in steps, as follows, which shows that every subsemiheap of the extended bicyclic semigroup is the inductive limit of subsemiheaps in case (1) in the category of semiheaps and semiheap homomorphisms.¹

• If a subsemiheap K of E is in case (2), then $K\subset \{a_{ij}:i\in \mathbb {Z}, j\ge \beta _0\}$ and $K=\cup _{\alpha \in \mathbb {Z}} K^\alpha$, where $K^{\alpha }=K\cap \{a_{ij}:i\ge \alpha , j\ge \beta _0\}$ (which we have denoted by $K_{\alpha ,\beta _0}$) is in case (1).

• If a subsemiheap K of E is in case (3), then $K\subset \{a_{ij}:i\ge \alpha _0, j\in \mathbb {Z}\}$ and $K=\cup _{\beta \in \mathbb {Z}} K_\beta$, where $K_\beta =K\cap \{a_{ij}:i\ge \alpha _0, j\ge \beta \}$ (=$K_{\alpha _0,\beta }$) is in case (1).

• If a subsemiheap K of E is in case (4), then $K\subset \{a_{ij}:i, j\in \mathbb {Z}\}$ and $K=\cup _{\alpha \in \mathbb {Z}} K_{(\alpha )}$, where $K_{(\alpha )}=K\cap \{a_{ij}:i\ge \alpha , j\in \mathbb {Z}\}$ is in case (3).

  Alternatively, if a subsemiheap K of E is in case (4), then $K=\cup _{\beta \in \mathbb {Z}} K^{(\beta )}$, where $K^{(\beta )}=K\cap \{a_{ij}:i\in \mathbb {Z}, j\ge \beta \}$ is in case (2).

Therefore we shall concentrate only on case (1). Suppose then that $\alpha _0\ne -\infty$ and $\beta _0\ne -\infty$. Then $K\subset K_{\alpha _0,\beta _0}=\{a_{pq}:p\ge \alpha _0,q\ge \beta _0\}.$ We define three parameters as follows:

$$\begin{aligned}&\overline{\beta }=\sup \{\beta \in \mathbb {Z}:a_{\alpha _0\beta }\in K\} \\&\overline{\alpha }=\sup \{\alpha \in \mathbb {Z}:a_{\alpha \beta _0}\in K\} \\&\overline{\gamma }=\sup \{k\in \mathbb {N}_0:a_{\alpha _0+k,\beta _0+k}\in K\}. \end{aligned}$$

.

We shall consider three primary cases:

$$\begin{aligned} \mathbf{1.}\ \overline{\beta }=\beta _0\quad \quad \mathbf{2.}\ \beta _0<\overline{\beta }<\infty \quad \quad \mathbf{3.}\ \overline{\beta }=\infty \end{aligned}$$

Each of the cases 1, 2, 3, consists of three further subcases.

$$\begin{aligned}&\mathbf{1.1}\ \overline{\beta }=\beta _0, \overline{\alpha }=\alpha _0\quad \quad \quad ~\mathbf{1.2}\ \overline{\beta }=\beta _0, \alpha _0<\overline{\alpha }<\infty \quad \quad \quad ~\mathbf{1.3}\ \overline{\beta }=\beta _0, \overline{\alpha }=\infty . \\&\mathbf{2.1}\ \overline{\beta }=\beta _0<\infty , \overline{\alpha }=\alpha _0~\mathbf{2.2}\ \beta _0<\overline{\beta }<\infty ,\alpha _0<\overline{\alpha }<\infty ~\mathbf{2.3}\ \beta _0<\overline{\beta }<\infty , \overline{\alpha }=\infty . \\&\mathbf{3.1}\ \overline{\beta }=\infty , \overline{\alpha }=\alpha _0\quad \quad \quad ~\mathbf{3.2}\ \overline{\beta }=\infty , \alpha _0<\overline{\alpha }<\infty ~\quad \quad \quad \mathbf{3.3}\ \overline{\beta }=\infty , \overline{\alpha }=\infty . \end{aligned}$$

Each of these nine cases consists of three further subcases. Thus, in order to account for the quadrant case (1), and hence the other three cases, it will be necessary to consider 27 cases. We summarize the results in the table Classification Scheme below. It is worthy to note that by Diagram 2, if $\overline{\beta }$ is finite, by which we mean, $\beta _0<\overline{\beta }<\infty$, then $a_{\alpha _0,\overline{\beta }}$ is the only point of K of the form $a_{\alpha _0,\beta }$. A similar statement holds for $\overline{\alpha }$. Also, if $\overline{\alpha }=\alpha _0$, or if $\overline{\beta }=\beta _0$, then $a_{\alpha _0,\beta _0}\in K$. Thus in cases 2.2, 2.3, 3.2, and 3.3, it is necessary to consider the two possibilities: $a_{\alpha _0,\beta _0}\in K$, and $a_{\alpha _0,\beta _0}\not \in K$.

Classification Scheme

	Case	subcase	$\overline{\beta }$	$\overline{\alpha }$	$\overline{\gamma }$	Exists?	result
		1.1.1.	$\beta _0$	$\alpha _0$	0	Yes	Lemma 3.2
	1.1	1.1.2.	$\beta _0$	$\alpha _0$	Finite	Yes	Proposition 3.3
		1.1.3.	$\beta _0$	$\alpha _0$	$\infty$	Yes	Proposition 3.4
		1.2.1.	$\beta _0$	Finite	0	No	Lemma 3.7
1	1.2	1.2.2.	$\beta _0$	Finite	Finite	No	Lemma 3.7
		1.2.3.	$\beta _0$	Finite	$\infty$	No	Lemma 3.7
		1.3.1.	$\beta _0$	$\infty$	0	No	Lemma 3.7
	1.3	1.3.2.	$\beta _0$	$\infty$	Finite	No	Lemma 3.7
		1.3.3.	$\beta _0$	$\infty$	$\infty$	No	Lemma 3.7
		2.1.1.	Finite	$\alpha _0$	0	No	Lemma 3.8
	2.1	2.1.2.	Finite	$\alpha _0$	Finite	No	Lemma 3.8
		2.1.3.	Finite	$\alpha _0$	$\infty$	No	Lemma 3.8
		2.2.1.	Finite	Finite	0	No	Lemma 3.9
2	2.2	2.2.2.	Finite	Finite	Finite	No	Lemma 3.9
		2.2.3.	Finite	Finite	$\infty$	No	Lemma 3.9
		2.3.1.	Finite	$\infty$	0	No	Lemma 3.10
	2.3	2.3.2.	Finite	$\infty$	Finite	No	Lemma 3.10
		2.3.3.	Finite	$\infty$	$\infty$	No	Lemma 3.10
		3.1.1.	$\infty$	$\alpha _0$	0	No	Lemma 3.11
	3.1	3.1.2.	$\infty$	$\alpha _0$	Finite	No	Lemma 3.11
		3.1.3.	$\infty$	$\alpha _0$	$\infty$	No	Lemma 3.11
		3.2.1.	$\infty$	Finite	0	No	Lemma 3.11
3	3.2	3.2.2.	$\infty$	Finite	Finite	No	Lemma 3.11
		3.2.3.	$\infty$	Finite	$\infty$	No	Lemma 3.11
		3.3.1.	$\infty$	$\infty$	0	No	Proposition 3.12
	3.3	3.3.2.	$\infty$	$\infty$	Finite	No	Proposition 3.12
		3.3.3.	$\infty$	$\infty$	$\infty$	Yes	Proposition 3.18

We now proceed to analyze all 27 cases.

Lemma 3.2

In case 1.1.1 ($\overline{\beta }=\beta _0,\ \overline{\alpha }=\alpha _0,\ \overline{\gamma }=0$), we have $K=\{a_{\alpha _0\beta _0}\}$.

Proof

In this case, the diagram is the following, where the bullet represents the element $a_{\alpha _0\beta _0}$, and the circles indicate that no element of K occupies that position. (Ignore, for the moment, the symbols $\blacksquare ,\blacktriangle , \triangle$)

Suppose that $a_{\alpha _0+2,\beta _0+1}$, denoted by $\blacktriangle$, belonged to K. Then by Diagram 3 applied to the points $a_{\alpha _0\beta _0}$ and $\blacktriangle$, the point $a_{\alpha _0+1,\beta _0+1}$, denoted by $\triangle$, would belong to K, a contradiction. So $\blacktriangle$ does not belong to K. By the same argument, no element of K resides in the second column of the diagram.

Suppose that $a_{\alpha _0+1,\beta _0+2}$, denoted by $\blacksquare$, belonged to K. Then by Diagram 1 applied to the points $a_{\alpha _0\beta _0}$ and $\blacksquare$, the point $a_{\alpha _0+1,\beta _0+1}$, denoted by $\triangle$, would belong to K, a contradiction. So $\blacksquare$ does not belong to K. By the same argument, no element of K resides in the second row of the diagram.

Repetition of these two arguments shows that no element of K resides in any column or row of the diagram, other than the first row and column, and therefore $K=\{a_{\alpha _0\beta _0}\}$ contains exactly one element. $\Box$

Proposition 3.3

In case 1.1.2 ($\overline{\beta }=\beta _0,\ \alpha _0=\overline{\alpha },\ 0<\overline{\gamma }<\infty$), we have

$$\begin{aligned}K= \{a_{\alpha _0,\beta _0}\}\cup \{a_{\alpha _0+k_j,\beta _0+k_j}:1\le j\le n_0\}, \end{aligned}$$

where $1\le k_1<k_2<\cdots <k_{n_0}=\overline{\gamma }$ and $n_0\in \mathbb {N}$.

Proof

In this case, in Diagram 11, the bullets represent some of the elements of K residing on the diagonal, the circles indicate that no element of K occupies that position, and the dots represent both the finite number of points of K on the diagonal together with some positions on the diagonal not containing points of K (Ignore for the moment, the symbols $\circledcirc , \boxdot$ which represent two elements of K lying on the diagonal, and the symbols $\blacksquare ,\square ,\blacktriangle , \triangle$). The symbol $\clubsuit$ represents the element $a_{\alpha _0+k_{n_0},\beta _0+k_{n_0}}$. We shall show that all off-diagonal positions are not occupied by elements of K, which means that $K=\{a_{\alpha _0,\beta _0}\}\cup \{a_{\alpha _0+k_j,\beta _0+k_j}:1\le j\le n_0\}$. $\square$

Suppose that for $k_j\le \ell <k_{j+1}$, the point $a_{\alpha _0+k_{j+1}, \beta _0+\ell }$, denoted by $\blacksquare$ in Diagram 11, belonged to K. Then by Diagram 2, starting with $\blacksquare$ and $a_{\alpha _0+k_{j+1},\beta _0+k_{j+1}}$, denoted by $\boxdot$ in Diagram 11, shows that

$$\begin{aligned} K\supset K_{\alpha _0+k_{j+1},\beta _0+\ell }^{k_{j+1}-\ell }=\{a_{\alpha _0+k_{j+1}+m(k_{j+1}-\ell ),\beta _0+\ell +n(k_{j+1}-\ell )}:m,n\in \mathbb {N}_0\}. \end{aligned}$$

Then choosing $m=n-1$, so that

$$\begin{aligned}k_{j+1}+m(k_{j+1}-\ell )=\ell +n(k_{j+1}-\ell ) \end{aligned}$$

and letting $n\rightarrow \infty$ shows that there are infinitely many points of K on the diagonal, a contradiction, so $\blacksquare \not \in K$. The same argument applies to every point on each row determined by $\alpha _0+k_{j+1}$ to the left of $a_{\alpha _0+k_{j+1},\beta _0+k_{j+1}}$ for $0\le j\le n_0-1$.

Suppose now that for $k_j<\ell \le k_{j+1}$, the point $a_{\alpha _0+k_{j}, \beta _0+\ell }$, denoted by $\blacktriangle$ in Diagram 11, belonged to K. Then by Diagram 2, starting with $a_{\alpha _0+k_{j},\beta _0+k_{j}}$, denoted by $\circledcirc$ in Diagram 11, and $\blacktriangle$, shows that

$$\begin{aligned} K\supset K_{\alpha _0+k_{j},\beta _0+\ell }^{\ell -k_{j}}= \{a_{\alpha _0+k_{j}+m(\ell -k_j),\beta _0+k_j+n(\ell -k_j)}:m,n\in \mathbb {N}_0\}. \end{aligned}$$

Then choosing $m=n$, and letting $n\rightarrow \infty$ show that there are infinitely many points of K on the diagonal, a contradiction, so $\blacktriangle \not \in K$. The same argument applies to every point on each row $\alpha _0+k_j$ to the right of $a_{\alpha _0+k_{j},\beta _0+k_{j}}$ for $0\le j\le n_0$.

Thus all rows containing an element of K on the diagonal do not contain any other elements of K, as in Diagram 12.

Diagram 11

Diagram 12

A parallel argument, using Diagram 4 shows that all columns containing an element of K on the diagonal do not contain any other elements of K. For completeness, we include the details.

Suppose that for $k_j\le \ell <k_{j+1}$, the point $a_{\alpha _0+\ell , \beta _0+k_{j+1}}$, denoted by $\square$ in Diagram 11, belonged to K. Then by Diagram 4, starting with $\square$ and $a_{\alpha _0+k_{j+1},\beta _0+k_{j+1}}$, denoted by $\boxdot$, shows that

$$\begin{aligned} K\supset K_{\alpha _0+\ell ,\beta _0+k_{j+1}}^{k_{j+1}-\ell }= \{a_{\alpha _0+\ell +m(k_{j+1}-\ell ),\beta _0+k_{j+1}+n(k_{j+1}-\ell )}:m,n\in \mathbb {N}_0\}. \end{aligned}$$

Then choosing $m=n+1$, so that

$$\begin{aligned}k_{j+1}+n(k_{j+1}-\ell )=\ell +m(k_{j+1}-\ell )\end{aligned}$$

and letting $n\rightarrow \infty$ shows that there are infinitely many points of K on the diagonal, a contradiction, so $\square \not \in K$. The same argument applies to every point on each column determined by $\beta _0+k_{j+1}$ above $a_{\alpha _0+k_{j+1},\beta _0+k_{j+1}}$ for $0\le j\le n_0-1$.

Suppose now that for $k_j\le \ell <k_{j+1}$, the point $a_{\alpha _0+\ell , \beta _0+k_j}$, denoted by $\triangle$ in Diagram 11, belonged to K. Then by Diagram 4, starting with $\triangle$ and $a_{\alpha _0+k_j,\beta _0+k_j}$, denoted by $\circledcirc$, shows that

$$\begin{aligned} K\supset K_{\alpha _0+k_{j},\beta _0+k_j}^{\ell -k_j}= \{a_{\alpha _0+k_{j}+m(\ell -k_j),\beta _0+k_j+n(\ell -k_j)}:m,n\in \mathbb {N}_0\}, \end{aligned}$$

Then choosing $m=n$, so that

$$\begin{aligned}k_j+n(\ell -k_j)=k_j+m(\ell -k_j)\end{aligned}$$

and letting $n\rightarrow \infty$ shows that there are infinitely many points of K on the diagonal, a contradiction, so $\triangle \not \in K$. The same argument applies to every point on each column determined by $\beta _0+k_j$ above $a_{\alpha _0+k_j,\beta _0+k_j}$ for $0\le j\le n_0$.

To complete the proof, we now show that no point $a_{\alpha _0+m,\beta _0+n}$, with $m\ne n$ can belong to K. By what was just proved, it suffices to consider points which are not on a row or column containing a point of K, that is, $m\ne k_j$ for all j and $n\ne k_\ell$ for all $\ell$.

We shall refer to the following diagram, which depicts the eight possible locations for the element $a_{\alpha _0+m,\beta _0+n}$, reflecting the cases $m>n$ and $m<n$, namely,

• 1. $m> k_{n_0}\ge k_{\ell +1}>n> k_\ell$, denoted by $\blacksquare$

• 2. $k_{\ell +1}>m> k_\ell \ge k_{j+1}>n> k_j$, denoted by $\blacktriangle$

• 3. $k_{j+1}>m>n> k_j$, denoted by $\blacktriangledown$

• 4. $m>n> k_{n_0}$, denoted by $\blacktriangleleft$

• 5. $n> k_{n_0}\ge k_{\ell +1}>m> k_\ell$, denoted by $\square$

• 6. $k_{\ell +1}>n> k_\ell \ge k_{j+1}>m> k_j$, denoted by $\triangle$

• 7. $k_{j+1}>n>m> k_j$, denoted by $\triangledown$

• 8. $n>m> k_{n_0}$, denoted by $\triangleleft$

Suppose first that $m>n$, for example case (2), $k_{\ell +1}>m_2>k_\ell \ge k_{j+1}>n_2>k_j$. We consider the two points $a_{\alpha _0+k_j,\beta _0+k_j}$ and $\blacktriangle =a_{\alpha _0+m_2,\beta _0+n_2}$. These two points are vertices of a triangle with height $h=m_2-k_j$ greater than the base $b=n_2-k_j$, so $h-b=m_2-n_2$. Then by Lemma 2.2 (see Diagram 3), the point $x_2=a_{\alpha _0+k_j+\beta _0+m_2-\alpha _0-k_j,\beta _o+m_2}=a_{\alpha _0+m_2,\beta _0+m_2}$ would belong to K, which is a contradiction since $k_\ell<m_2<k_{\ell +1}$.

Suppose next that $m<n$, for example, case (6), $k_j<m_6<k_{j+1}\le k_\ell<n_6<k_{\ell +1}$ We again consider the two points $a_{\alpha _0+k_j,\beta _0+k_j}$ and $\triangle =a_{\alpha _0+m_6,\beta _0+n_6}$. These two points are vertices of a triangle with height $h=m_6-k_j$ less than the base $b=n_6-k_j$ and $b-h=n_6-m_6$. Then by Lemma 2.2 (see Diagram 1), the point $x_1=a_{\alpha _0+k_j+\beta _0+n_6-\beta _0-k_j,\beta _0+n_6 }=a_{\alpha _0+n_6,\beta _0+n_6}$ would belong to K, which is a contradiction since $k_\ell<n_6<k_{\ell +1}$.

The same two-part argument works in all the other cases, more precisely, as follows:

• For case (3), $k_j<n_3<m_3<k_{j+1}$, Diagram 3 applied to $a_{\alpha _0+k_j,\beta _0+k_j}$ and $\blacktriangledown$ yields $x_2=a_{\alpha _0+n_3,\beta _0+n_3}$

• For case (7), $k_j<m_7<n_7<k_{j+1}$, Diagram 1 applied to $a_{\alpha _0+k_j,\beta _0+k_j}$ and $\triangledown$ yields $x_3=a_{\alpha _0+m_7,\beta _0+m_7}$

• For case (4), $m_4> n_4> k_{n_0}$, Diagram 3 applied to $a_{\alpha _0+k_{n_0},\beta _0+k_{n_0}}$ and $\blacktriangleleft$ yields $x_2=a_{\alpha _0+n_4,\beta _0+n_4}$

• For case (8), $n_8>m_8> k_{n_0}$, Diagram 1 applied to $a_{\alpha _0+k_{n_0},\beta _0+k_{n_0}}$ and $\triangleleft$ yields $x_3=a_{\alpha _0+m_8,\beta _0+m_8}$

• For case (1), $m_1> k_{n_0}\ge k_{\ell +1}>n_1> k_\ell$, Diagram 3 applied to $a_{\alpha _0+k_\ell ,\beta _0+k_\ell }$ and $\blacksquare$ yields $x_2=a_{\alpha _0+n_1,\beta _0+n_1}$

• For case (5), $n_5> k_{n_0}\ge k_{\ell +1}>m_5> k_\ell$, Diagram 1 applied to $a_{\alpha _0+k_\ell ,\beta _0+k_\ell }$ and $\square$ yields $x_3=a_{\alpha _0+m_5,\beta _0+m_5}$. This completes the proof of Proposition 3.3. $\Box$

Proposition 3.4

In case 1.1.3: ($\overline{\beta }=\beta _0,\ \overline{\alpha }=\alpha _0, \ \overline{\gamma }=\infty$), either $K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_j,\beta _0+k_j}:j\in \mathbb {N}\}$, or there exist $\sigma ,\ell _0\in \mathbb {N}$ such that (see Examples 2.6 and 2.7)

$$\begin{aligned}&K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_i,\beta _0+k_i}:i=1,\ldots \ell _0-1\}\cup K_{\alpha _0+k_{\ell _0},\beta _0+k_{\ell _0}}^\sigma ,\hbox { or} \end{aligned}$$

(4)

$$\begin{aligned}&K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_i,\beta _0+k_i}:1\le i<\ell _0\}\cup \left( \bigcup _{i=\ell _0}^{j} K^{\sigma }_{\alpha _0+k_i,\beta _0+k_i}\right) \end{aligned}$$

(5)

where $1\le k_1<k_2<\cdots<k_i<\cdots <\infty$, and in (5), $\sigma =k_j-k_{\ell _0}$ where $k_j$ is such that no point $a_{\alpha _0+k_{\ell _0},\beta _0+k_p}$ belongs to K for $1\le p<j$. If $\ell _0=1$, then the term $\{a_{\alpha _0+k_i,\beta _0+k_i}:1\le i<\ell _0\}$ is missing in (4) and (5).

Proof

In this case, the diagram is the following, where the bullet represents the element $a_{\alpha _0\beta _0}$, the circles indicate that no element of K occupies that position, and the dots indicate that infinitely many elements of K reside in the diagonal.

We consider a point $a_{\alpha _0+m,\beta _0+n}$, denoted by $\blacksquare$ in Diagram 13, with $m\ne n$ and $m\ne k_j, n\ne k_j$ for every $j\ge 1$.

Suppose first that $m>n$, more precisely, $k_{j+1}>m>k_j\ge k_{\ell +1}>n>k_{\ell }$. We consider the two points $a_{\alpha _0+k_\ell ,\beta _0+k_\ell }$ (an element of K on the diagonal), denoted by $\square$ in Diagram 13, and $a_{\alpha _0+m,\beta _0+n}=\blacksquare$. These two points are vertices of a right triangle with height $h=m-k_\ell$ greater than the base $b=n-k_\ell$, so $h-b=m-n$. Then by Lemma 2.2 (see Diagram 3), the point $x_2=a_{\alpha _0+k_\ell +\beta _0+n-\beta _0-k_\ell ,\beta _o+m}=a_{\alpha _0+n,\beta _0+n}$, denoted by $\circledcirc$, would belong to K, which is a contradiction since $k_\ell<n<k_{\ell +1}$. Therefore all elements below the diagonal which are neither located on a row nor on a column containing a diagonal point of K, do not belong to K.

Diagram 13

Suppose next that $m<n$, more precisely, $k_\ell<m<k_{\ell +1}\le k_j<n<k_{j+1}$ We again consider the two points $a_{\alpha _0+k_\ell ,\beta _0+k_\ell }$, denoted by $\square$ in Diagram 13, and $a_{\alpha _0+m,\beta _0+n}$, denoted by $\blacktriangle$ in Diagram 13. These two points are vertices of a right triangle with height $h=m-k_\ell$ smaller than the base $b=n-k_\ell$ and $b-h=n-m$. Then by Lemma 2.2 (see Diagram 1), the point $x_1=a_{\alpha _0+k_l+\beta _0+n-\beta _0-k_\ell ,\beta _0+n }=a_{\alpha _0+n,\beta _0+n}$, denoted by $\circledcirc$, would belong to K, which is a contradiction since $k_j<n<k_{j+1}$.

The same two-part argument works in the cases $k_j<m<n<k_{j+1}$ and $k_j<n<m<k_{j+1}$. Therefore all elements above or below the diagonal which are not located on a row or on a column containing a diagonal point of K, do not belong to K.

We now have Diagram 14. Thus the only off-diagonal points that can possibly belong to K are those that are located either on a row containing a diagonal point of K, indicated by $\cdots$, or on a column containing a diagonal point of K, indicated by vertical dots.

We next show that points which lie on a row or column, but not both, cannot belong to K.

Consider first, for any row determined by $k_j$ and any n with $n \ne k_m$ for all m, the element $a_{\alpha _0+k_j,\beta _0+n}$, denoted by $\blacksquare$ in Diagram 15, and suppose it belonged to K. Then by Lemma 2.1(iii),

$$\begin{aligned} a_{\alpha _0\beta _0}a^*_{\alpha _0+k_j,\beta _0+n}a_{\alpha _0+k_j,\beta _0+n}=a_{\alpha _0+n,\beta _0+n}\hbox { (denoted by} \square {\hbox {)}} \end{aligned}$$

would belong to K, a contradiction since $n \ne k_m$ for every m.

Next, for any column determined by $k_j$ and any n with $n \ne k_m$ for all m, consider the element $a_{\alpha _0+n,\beta _0+k_j}$, denoted by $\blacktriangle$ in Diagram 15, and suppose it belonged to K. Then by Lemma 2.1(i),

$$\begin{aligned} a_{\alpha _0+n,\beta _0+k_j}a^*_{\alpha _0+n,\beta _0+k_j}a_{\alpha _0\beta _0}=a_{\alpha _0+n,\beta _0+n} \hbox { (denoted by} \square \hbox {)} \end{aligned}$$

would belong to K, a contradiction since $n \ne k_m$ for every m.

We now have Diagram 16. Thus the only off-diagonal points that can possibly belong to K are those, denoted by $\blacksquare$, that are located simultaneously on a row containing a diagonal point of K and a column containing a diagonal point of K.

If none of the off-diagonal elements $\blacksquare$ belong to K, then obviously $K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_j,\beta _0+k_j}:j\in \mathbb {N}\}$. We next consider the case that some of the elements $\blacksquare$ in Diagram 16 belong to K.

Diagram 14

A row determined by $k_j$ for which no element other than $a_{\alpha _0+k_j,\beta _0+k_j}$ belongs to K, that is, $a_{\alpha _0+k_j,\beta _0+k_j}\in K$, and $a_{\alpha _0+k_j,\beta _0+k_p}\not \in K$ for all $p\in \mathbb {N}-\{k_j\}$, will be called a null row. More precisely, a right-null (respectively left-null) row determined by $k_j$ is one that satisfies $a_{\alpha _0+k_j,\beta _0+k_j}\in K$, and $a_{\alpha _0+k_j,\beta _0+k_p}\not \in K$ for all $p> k_j$ (respectively $p<k_j$). The row determined by $\alpha _0$ is a null row.

Similarly, a row determined by $k_j$ which contains an element of K other than $a_{\alpha _0+k_j,\beta _0+k_j}$, that is, there exists $\ell >j$ (respectively $\ell <j$), such that $a_{\alpha _0+k_j,\beta _0+\ell }\in K$, will be called a right-ample (respectively left-ample) row. A row that is either left-ample or right-ample (or both), will be called simply ample. By Diagram 2, a left-ample row is also right-ample, but not conversely (see the sentence following Lemma 2.1). For the same reason, a right-null row is also left-null. As noted above, if all rows of K are null, then $K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_j,\beta _0+k_j}:j\in \mathbb {N}\}$. Thus we have the following lemma.

Lemma 3.5

If all the rows of K are right-null, then $K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_j,\beta _0+k_j}:j\in \mathbb {N}\}.$

Diagram 15

Diagram 16

Lemma 3.6

All right-null rows lie above all ample rows. Hence, if there is at least one ample row, then there are only finitely many right-null rows.

Proof

Suppose $k_j$ determines a right-null row, and $k_\ell$ determines an ample row, which we may assume to be right-ample, say containing the element $a_{\alpha _0+k_\ell ,\beta _0+k_m}$, and suppose by way of contradiction that $\ell <j$. Since $\ell<j<m$, the diagram is the following, where $\blacksquare$ denotes the element $a_{\alpha _0+k_j,\beta _0+k_m}$.

The two points $\blacksquare$ and $a_{\alpha _0+k_j,\beta _0+k_j}$ are vertices of a triangle with base $b=k_m-k_j$ and height $h=k_j-k_\ell$. Then by Diagrams 6, 8, or 10, depending on the relative sizes of b and h, the point $x_3=a_{\alpha _0+k_j,\beta _0+k_m+k_j-k_\ell }$ would belong to K, which is a contradiction since $k_m+k_j-k_\ell > k_j$. $\square$

Let $\ell _0\in \mathbb {N}$ be such that the first ample row is determined by $k_{\ell _0}$, and assume without loss of generality, that this row is right-ample. Assume also, temporarily, that $\ell _0>1$. The rows lying above the row determined by $\alpha _0+k_{\ell _0}$ do not contain any elements of K above the diagonal. It follows from Diagram 4 that the columns lying to the left of the column determined by $\beta _0+k_{\ell _0}$ do not contain any elements of K below the diagonal. By Diagram 2, K contains infinitely many elements to the right of $a_{\alpha _0+k_{\ell _0},\beta _0+k_{\ell _0}}$, and then by Diagrams 2 and  4, K contains infinitely many elements below $a_{\alpha _0+k_{\ell _0},\beta _0+k_{\ell _0}}$. Since $\overline{\gamma }=\infty$, the subsemiheap $K\cap K_{\alpha _0+k_{\ell _0},\beta _0+k_{\ell _0}}$ falls into subcase 3.3.3 below (see Propositions 3.15 and  3.16, and Diagram 17), and therefore

$$\begin{aligned} K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_i,\beta _0+k_i}:1\le i<\ell _0\}\cup K_{\alpha _0+k_{\ell _0},\beta _0+k_{\ell _0}}^{k_j-k_{\ell _0}}, \end{aligned}$$

(6)

or

$$\begin{aligned} K=\{a_{\alpha _0\beta _0}\}\cup \{a_{\alpha _0+k_i,\beta _0+k_i}:1\le i<\ell _0\}\cup \left( \bigcup _{i=\ell _0}^{j} K^{k_j-k_{\ell _0}}_{\alpha _0+k_i,\beta _0+k_i}\right) \end{aligned}$$

(7)

where $k_j$ is such that no point $a_{\alpha _0+k_{\ell _0},\beta _0+k_p}$ belongs to K for $1\le p<j$. If $\ell _0=1$, then the term $\{a_{\alpha _0+k_i,\beta _0+k_i}:1\le i<\ell _0\}$ is missing in (6) and (7). (In Diagram 17, $\sigma =k_{\ell _0+1}+(k_j-k_{\ell _0})$). This completes the proof of Proposition 3.4 and hence of case 1.1. $\square$

In each of the six subcases of cases 1.2 and 1.3, the diagram is the following:

Then applying Diagram 4 to the points $a_{\alpha _0\beta _0}$ and $a_{\alpha _0+r,\beta _0}$ shows that $x_1=a_{\alpha _0,\beta _0+r}\in K$, a contradiction. Hence we have the following lemma.

Lemma 3.7

Cases 1.2 and 1.3 do not occur.

Diagram 17

Lemma 3.8

Case 2.1 does not occur.

Proof

Diagram 18 includes each of the three subcases of case 2.1. Then applying Diagram 2 to the points $a_{\alpha _0\beta _0}$ and $a_{\alpha _0,\beta _0+p}$ shows that $x_1=a_{\alpha _0+p,\beta _0}\in K$, a contradiction.

$\square$

Lemma 3.9

Case 2.2 does not occur.

Proof

Note first that by Diagrams 2 and  4, only one element of K can reside on the row determined by $\alpha _0$ or on the column determined by $\beta _0$. Thus, Diagram 19 depicts this case (if $r<p$).

Then by Diagrams  6 and  8 (depending on whether $r<p$ or $p\ge r$, K would contain elements $a_{\alpha _0,\beta _0+\ell }$ with $\ell >p$, a contradiction. $\square$

Diagram 18

Diagram 19

Lemma 3.10

Case 2.3 does not occur, hence case 2 does not occur.

Proof

In case 2.3.1 ($\beta _0<\overline{\beta }<\infty , \overline{\alpha }=\infty ,\overline{\gamma }=0$), note that $a_{\alpha _0,\beta _0+p}$ is the only point of K on the row determined by $\alpha _0$. From the following diagram we see that if $p<r$, we get a contradiction using Diagram 8, and if $r<p$ we get a contradiction using Diagram 6, whereas if $p=r$, we get a contradiction using Diagram 10.

The same proof applies to cases 2.3.2 ($\beta _0<\overline{\beta }<\infty , \overline{\alpha }=\infty ,0<\overline{\gamma }<\infty$) and 2.3.3 ($\beta _0<\overline{\beta }<\infty , \overline{\alpha }=\infty , \overline{\gamma }=\infty$).$\square$

Lemma 3.11

Cases 3.1 and 3.2 do not occur.

Proof

Cases 3.1.1, 3.1.2, and 3.1.3 do not occur by Diagram 2. Cases 3.2.1, 3.2.2, and 3.2.3 do not occur by Diagram 6. $\square$

Proposition 3.12

Cases 3.3.1 and 3.3.2 do not occur.

Proof

By Diagram 2 or  4, we may assume that $a_{\alpha _0,\beta _0}\not \in K$. The subsets $K\cap K_{\alpha _0,\beta _0+k_1}$ and $K\cap K_{\alpha _0+\ell _1,\beta _0}$ are subsemiheaps of K which fall into case 3.3.3, which is described below in Proposition 3.18. However, as shown in the proof of Lemma 3.17 below, the four possible situations each lead to $a_{\alpha _0,\beta _0}\in K$. $\square$

Case 3.3.3 ($\overline{\beta }=\infty , \overline{\alpha }=\infty ,\overline{\gamma }=\infty$)

Let $a_{\alpha _0,\beta _0+p}\in K$ with $p\ge 1$ and $a_{\alpha _0,\beta _0+p'}\not \in K$ for $1\le p'< p$. Similarly, let $a_{\alpha _0+r,\beta _0}\in K$ with $r\ge 1$ and $a_{\alpha _0+r',\beta _0}\not \in K$ for $1\le r'< r$ and let $a_{\alpha _0+q,\beta _0+q}\in K$ with $q\ge 1$ and $a_{\alpha _0+q',\beta _0+q'}\not \in K$ for $1\le q'< q$

In the diagram below, the bullets represent the three points of K which were just defined, the symbol $\circledcirc$ means that the element $a_{\alpha _0\beta _0}$ may or may not belong to K, and the circles indicate that no element of K occupies that position. The diagram represents just one of 13 possible cases (namely case (5) below), and is for illustration purposes only.

Of course, we must consider the various relations between the three elements $p,q,r\in \mathbb {N}$, some of which can be equal, of which there are six, namely

• $r\le p\le q$

• $p\le r\le q$

• $p\le q\le r$

• $r\le q\le p$

• $q\le r\le p$

• $q\le p\le r$

But for our purposes, it is necessary to distinguish 13 more refined cases, namely

• 1. $r<p<q$

• 2. $p<r<q$

• 3. $p<q<r$

• 4. $r<q<p$

• 5. $q<r<p$

• 6. $q<p<r$

• 7. $r=p<q$

• 8. $p=q<r$

• 9. $r=q<p$

• 10. $r<p=q$

• 11. $p<r=q$

• 12. $q<r=p$

• 13. $r=p=q$

Lemma 3.13

Cases (3) to (11) do not occur. If $a_{\alpha _0,\beta _0}\in K$, then cases (1) and (2) do not occur. If $a_{\alpha _0\beta _0}\not \in K$, then case (12) does not occur. In case (13), $a_{\alpha _0\beta _0}\in K$.

Proof

By Lemma 2.1, we have

$$\begin{aligned} a_{\alpha _0,\beta _0+p}a^*_{\alpha _0+q,\beta _0+q}a_{\alpha _0+r,\beta _0}= \left\{ \begin{array}{ll} \hbox {(i) } a_{\alpha _0,\beta _0+p-r} &{} \hbox { if }r\le p\hbox { and }q\le p\\ \hbox {(ii) } a_{\alpha _0+r-p,\beta _0} &{} \hbox { if }r\ge p\hbox { and }q\le p\\ \hbox {(iii) } a_{\alpha _0+r-p,\beta _0} &{} \hbox { if }r\ge q\hbox { and }q\ge p\\ \hbox {(iv) } a_{\alpha _0+q-p,\beta _0+q-r} &{} \hbox { if }r\le q\hbox { and }q\ge p. \end{array} \right. \end{aligned}$$

(8)

In case (1) with $a_{\alpha _0\beta _0}\in K$, we obtain a contradiction by Diagram 4.

In case (2) with $a_{\alpha _0\beta _0}\in K$, we obtain a contradiction by Diagram 2.

In case (3), we obtain a contradiction by (8(iii)).

In case (4), we obtain a contradiction by (8(i)).

In case (5), we obtain a contradiction by (8(i)).

In case (6), we obtain a contradiction by (8(ii)).

In case (7), we obtain a contradiction by (8(iv)).

In case (8), we obtain a contradiction by (8(ii)).

In case (9), we obtain a contradiction by (8(i)).

In case (10), we obtain a contradiction by (8(i)).

In case (11), we obtain a contradiction by (8(iii)).

In case (12) with $a_{\alpha _0\beta _0}\not \in K$, we obtain a contradiction by (8(i)) or (ii).

In case (13), $a_{\alpha _0\beta _0}\in K$ by (8(iii)). $\square$

It remains to consider cases (1) and (2), with $a_{\alpha _0\beta _0}\not \in K$, and the cases (12) and (13), with $a_{\alpha _0\beta _0}\in K$. The latter two will be resolved in Propositions 3.15 and 3.16 and the former two in Lemma 3.17.

We start with some properties in case (12). The basic diagram for case (12) is the following.

Lemma 3.14

In case (12), with (necessarily) $a_{\alpha _0\beta _0}\in K$,

• (a) The rows $1, 2,\ldots , q-1$ contain no elements of K above the diagonal

  The columns $1, 2,\ldots , q-1$ contain no elements of K below the diagonal

• (b) The points $a_{\alpha _0+q,\beta _0+q+i}$, for $1\le i\le p-q$ do not belong to K.

  The points $a_{\alpha _0+q+i,\beta _0+q}$, for $1\le i\le p-q$ do not belong to K.

• (c) The points $a_{\alpha _0+q,\beta _0+p}$, and $a_{\alpha _0+p,\beta _0+q}$ do not belong to K.

• (d) The points $a_{\alpha _0+q,\beta _0+j}$, for $mp<j<mp+q$, with $m\in \mathbb {N}$ do not belong to K.

  The points $a_{\alpha _0+i,\beta _0+q}$, for $mp<i<mp+q$, with $m\in \mathbb {N}$ do not belong to K.

• (e) $a_{\alpha _0+q,\beta _0+p+q}, a_{\alpha _0+p+q,\beta _0+q}\in K$.

• (f) The points $a_{\alpha _0+q,\beta _0+j}$, for $mp+q<j\le (m+1)p$, with $m\in \mathbb {N}$ do not belong to K.

  The points $a_{\alpha _0+i,\beta _0+q}$, for $mp+q<i\le (m+1)p$, with $m\in \mathbb {N}$ do not belong to K.

• (g) The points $a_{\alpha _0,\beta _0+j}$, for $mp<j< (m+1)p$, with $m\in \mathbb {N}$ do not belong to K.

  The points $a_{\alpha _0+i,\beta _0}$, for $mp<i< (m+1)p$, with $m\in \mathbb {N}$ do not belong to K.

Proof

In what follows, we shall elaborate on the above diagram. In the next diagram, the symbols $\blacksquare ,\blacktriangle ,\blacktriangledown , \blacktriangleleft , \blacktriangleright , \blacklozenge , \spadesuit , \clubsuit$, and their blank versions, represent points which, a priori, do not belong to K. They should be temporarily ignored. (The locations of (a)–(g) are indicated in the diagram preceding Proposition 3.16.)

(a) Consider the two points $a_{\alpha _0\beta _0}$ and $a_{\alpha _0+i,\beta _0+j}$, the latter indicated by $\blacksquare$ in the next diagram, with $1\le i<q$, $2\le j<\infty$ and $i<j$, and suppose that $a_{\alpha _0+i,\beta _0+j}$ belongs to K. Then by Diagram 1, $x_3=a_{\alpha +i,\beta _0+i}$, indicated by $\square$, belongs to K, a contradiction. Therefore the rows $1,2,3,\ldots q-1$ contain no elements of K above the diagonal.

Consider the two points $a_{\alpha _0\beta _0}$ and $a_{\alpha _0+i,\beta _0+j}$, the latter indicated by $\blacktriangle$ in the preceding diagram, with $1\le j<q$, $2\le i<\infty$ and $i>j$, and suppose that $a_{\alpha _0+i,\beta _0+j}$ belongs to K. Then by Diagram 3, $x_2=a_{\alpha +j,\beta _0+j}$, indicated by $\triangle$, belongs to K, a contradiction. Therefore columns $1,2,3,\ldots q-1$ contain no elements of K below the diagonal.

(b) Assuming that $a_{\alpha _0+q,\beta _0+q+j}$, indicated by $\blacktriangledown$ in the preceding diagram, with $1\le j<p-q$, belongs to K, we have that

$$\begin{aligned}K\supset K_{\alpha _0+q,\beta _0+q}^j =\{a_{\alpha _0+q+\ell j,\beta _0+q+mj}: \ell ,m\ge 0\}. \end{aligned}$$

Then by Lemma 2.1(i),

$$\begin{aligned} a_{\alpha _0,\beta _0+p}a^*_{\alpha _0+q+\ell j,\beta _0+q+mj}a_{\alpha _0\beta _0}=a_{\alpha _0,\beta _0+(\ell -m)j+p}\in K, \end{aligned}$$

provided that $0\le (\ell -m)j+p$ and $q+mj\le p$. Then with $\ell =0$ and $m=1$, we have that $a_{\alpha _0,\beta _0-j+p}$, indicated by $\triangledown$, belongs to K, which is a contradiction.

For the second statement of (b), the proof is the same, namely, assuming that $a_{\alpha _0+q+i,\beta _0+q}$, indicated by $\spadesuit$, with $1\le i<p-q$, belongs to K, we have that

$$\begin{aligned}K\supset K_{\alpha _0+q,\beta _0+q}^i =\{a_{\alpha _0+q+\ell i,\beta _0+q+mi}: \ell ,m\ge 0\}. \end{aligned}$$

Then by Lemma 2.1(i),

$$\begin{aligned} a_{\alpha _0,\beta _0+p}a^*_{\alpha _0+q+\ell i,\beta _0+q+mi}a_{\alpha _0\beta _0}=a_{\alpha _0,\beta _0+(\ell -m)i+p}\in K, \end{aligned}$$

provided that $0\le (\ell -m)i+p$ and $q+mi\le p$. Then with $\ell =0$ and $m=1$, we have that $a_{\alpha _0,\beta _0-i+p}$, indicated by $\triangledown$, belongs to K, which is a contradiction.

(c) Assuming that $a_{\alpha _0+q,\beta _0+p}$, indicated by $\blacktriangleleft$, belongs to K, then by Lemma 2.1(iii),

$$\begin{aligned} a_{\alpha _0,\beta _0+p}a^*_{\alpha _0+q,\beta _0+p}a_{\alpha _0+q,\beta _0+q}= a_{\alpha _0,\beta _0+q}, \end{aligned}$$

indicated by $\triangleleft$, belongs to K, a contradiction. Assuming that $a_{\alpha _0+p,\beta _0+q}$, indicated by $\blacktriangleright$, belongs to K, then applying Diagram 4 to the two points $a_{\alpha _0+q,\beta _0+q}$ and $\blacktriangleright$ we have $x_1=a_{\alpha _0+q,\beta _0+p}\in K$, a contradiction to the previous paragraph.

(d) Suppose $mp<j<mp+q$ and assume that $a_{\alpha _0+q,\beta _0+j}$, indicated by $\blacklozenge$ in the preceding diagram (with $m=1$), belongs to K. Then by Lemma 2.1(iii),

$$\begin{aligned} a_{\alpha _0,\beta _0+mp}a^*_{\alpha _0+q,\beta _0+j}a_{\alpha _0+q,\beta _0+q}=a_{\alpha _0+j-mp,\beta _0+q}, \end{aligned}$$

indicated by $\lozenge$, belongs to K, a contradiction to (i), since $j-mp<q$.

Suppose $mp<i<mp+q$ and assume that $a_{\alpha _0+i,\beta _0+q}$, indicated by $\boxplus$ in the preceding diagram (with $m=1$), belongs to K. Then by Lemma 2.1(iv),

$$\begin{aligned} a_{\alpha _0+q,\beta _0+q}a^*_{\alpha _0+i,\beta _0+q}a_{\alpha _0+mp,\beta _0}=a_{\alpha _0+q,\beta _0+i-mp}, \end{aligned}$$

indicated by $\boxminus$, belongs to K, a contradiction to (i$^\prime$), since $i-mp<q$.

(e) By Diagrams 6 or  8, applied to the vertices $a_{\alpha _0+q,\beta _0+q}$ and $a_{\alpha _0+p,\beta _0}$, $x_1=a_{\alpha _0+q,\beta _0+p+q}\in K$, and then by Diagram 7, $a_{\alpha _0+p+q,\beta _0+q}\in K$.

In the proofs of (f) and (g), and in the rest of this section, we can assume (by Remark 2.8), with no loss of generality, that $\alpha _0=\beta _0=0$.

(f) Suppose that $(q,j)\in K$ with $mp+q<j\le (m+1)p$. By Lemma 2.1(iii), $(0,mp)(q,j)^*(q,q)=(j-mp,q)$. This is a contradiction to (b) since $p\ge j-mp>q$.

Suppose that $(i,q)\in K$ with $mp+q<i\le (m+1)p$. By Lemma 2.1(iv), $(q,q)(i,q)^*(mp,0)=(q,i-mp)$. This is a contradiction to (b) since $p\ge i-mp>q$.

(g) Supposing that $a_{\alpha _0,\beta _0+j}$, with $p< j<\infty$ and $j\not \in \{2p,3p,\ldots \}$, denoted by $\clubsuit$ in the preceding diagram, belongs to K, we apply Diagram 2 to $a_{\alpha _0,\beta _0+mp}$ and $a_{\alpha _0,\beta _0+j}$, where $mp<j<(m+1)p$ to get $x_2=a_{\alpha _0+j-mp,\beta _0+j}\in K$, a contradiction since $j-mp<p$. Hence no element of K occupies any position in the row determined by $\alpha _0$ except for the points $a_{\alpha _0,\beta _0+mp}$ for $m\in \mathbb {N}_0$.

Supposing that $a_{\alpha _0+i,\beta _0}$, with $p< i<\infty$ and $i\not \in \{2p,3p,\ldots \}$, denoted by $\bigstar$, belongs to K, we apply Diagram 4 to $a_{\alpha _0+mp,\beta _0}$ and $a_{\alpha _0+i,\beta _0}$, where $mp<i<(m+1)p$ to get a contradiction. Hence no element of K occupies any position in the column determined by $\beta _0$ except for the points $a_{\alpha _0+\ell p,\beta _0}$ for $\ell \in \mathbb {N}_0$.$\square$

We now have the following diagram for case (12) with $a_{\alpha _0,\beta _0}\in K$, and it is clear that $K\cap K_{q,q}$ is also in subcase (12), so it follows that $K=\bigcup _{i=0}^\infty K_{q_i,q_i}^p$, where $a_{\alpha _0+q_i,\beta _0+q_i}$ are the points of K lying on the diagonal with

$$\begin{aligned} q=q_0<q_1<q_2<\cdots<q_n<q_{n+1}<\cdots . \end{aligned}$$

Proposition 3.15

In case (12), with (necessarily) $a_{\alpha _0,\beta _0}\in K$, let $a_{\alpha _0+q_i,\beta _0+q_i}$, $0\le i<\infty$,be the points of K lying on the diagonal, such that

$$\begin{aligned} q=q_0<q_1<q_2<\cdots<q_n<p\quad \hbox { and }\quad p<q_{n+1}<q_{n+2}<\cdots . \end{aligned}$$

Then

$$\begin{aligned} K= \bigcup _{i=0}^n K_{q_i,q_i}^p. \end{aligned}$$

Proof

We know that $K=\bigcup _{i=0}^\infty K_{q_i,q_i}^p$. We need to show that $\bigcup _{i=n+1}^\infty K_{q_i,q_i}^p\subset \bigcup _{i=0}^n K_{q_i,q_i}^p$. For this it suffices to show that each $q_{n+j}$ with $j\ge 1$ is congruent to some element of $\{q_0,q_1,\ldots q_n\}$, modulo p.

Let $(q_k+\ell p,q_k+mp)\in K_{q_k,q_k}^p$ for some $k\ge n+1$ with fixed $\ell ,m$, and let $(\ell ' p,m'p)\in K_{0,0}^p$ with variable $\ell ',m'$. By Lemma 2.1(i),

$$\begin{aligned} (q_k+\ell p,q_k+mp)(\alpha _0,\beta _0)^*(\ell ' p,m'p)=(q_k+\ell p,q_k+(m'+m-\ell ')p)\in K \end{aligned}$$

as long as $q_k+mp\ge 0$ and $\ell ' p\le q_k+mp$, We now choose $\ell '$ such that $q_k=(\ell '-m)p+d$, where $\ell '-m\ge 1$ and $0\le d<p$. To check that $\ell ' p\le q_k+mp$, we note that $(\ell '-m) p=q_k-d\le q_k$. We now have

$$\begin{aligned} (q_k+\ell p,q_k+(m'+m-\ell ')p)=(d+(\ell +\ell '-m)p,d). \end{aligned}$$

Thus $(d+tp,d)\in K=\bigcup _{i=0}^\infty K_{q_i,q_i}^p$ for some $t\ge 0$, so that $(d+tp,d)=(q_i+rp,q_i+sp)$ for some $i,r,s\ge 0$. Hence $d+tp=q_i+rp$ and $d=q_i+sp$, so by subtraction $tp=(r-s)p$ and $d+(r-s)p=q_i+rp$ so that $d=q_i+sp$. Since $d<p$, $s=0$ and $d=q_i$ with $i\le n$. $\square$

Proposition 3.16

In case (13), $K=K_{\alpha _0,\beta _0}^p$

Proof

The diagram for case (13) is Diagram 20. (Temporarily ignore the symbols $\blacksquare ,\blacktriangle ,\blacktriangledown ,\blacktriangleleft$).

In the first place, we notice that by Diagrams 2 and  4,

$$\begin{aligned} K\supset K_{\alpha _0,\beta _0}^p=\{a_{\alpha _0+\ell p,\beta _0+mp}:\ell ,m\in \mathbb {N}_0\}. \end{aligned}$$

Diagram 20

The next four paragraphs refer to Diagram 20.

Supposing that $a_{\alpha _0+i,\beta _0+j}$ for $1\le i<p$ and $2\le j<\infty$, denoted by $\blacksquare$, belongs to K, we apply Diagram 1 to $a_{\alpha _0\beta _0}$ and $a_{\alpha _0+i,\beta _0+j}$ to get $x_3=a_{\alpha _0+i,\beta _0+i}\in K$, a contradiction. Hence no element of K occupies any position above the diagonal in the rows determined by $\alpha _0+i$, for $1\le i<p$.

Supposing that $a_{\alpha _0+i,\beta _0+j}$ for $2\le i<\infty$ and $1\le j<p$, denoted by $\blacktriangle$, belongs to K, we apply Diagram 3 to $a_{\alpha _0\beta _0}$ and $a_{\alpha _0+i,\beta _0+j}$ to get $x_2=a_{\alpha _0+j,\beta _0+j}\in K$, a contradiction. Hence no element of K occupies any position below the diagonal in the columns determined by $\beta _0+j$, for $1\le j<p$

Supposing that $a_{\alpha _0,\beta _0+j}$, with $p< j<\infty$ and $j\not \in \{2p,3p,\ldots \}$, denoted by $\blacktriangledown$, belongs to K, we apply Diagram 2 to $a_{\alpha _0,\beta _0+kp}$ and $a_{\alpha _0,\beta _0+j}$, where $kp<j<(k+1)p$ to get $x_2=a_{\alpha _0+j-kp,\beta _0+j}\in K$, a contradiction since $j-kp<p$. Hence no element of K occupies any position in the row determined by $\alpha _0$ except for the points $a_{\alpha _0,\beta _0+mp}$ for $m\in \mathbb {N}_0$.

Supposing that $a_{\alpha _0+i,\beta _0}$, with $p< i<\infty$ and $i\not \in \{2p,3p,\ldots \}$, denoted by $\blacktriangleleft$, belongs to K, we apply Diagram 4 to $a_{\alpha _0+kp,\beta _0}$ and $a_{\alpha _0+i,\beta _0}$, where $kp<i<(k+1)p$ to get a contradiction. Hence no element of K occupies any position in the column determined by $\beta _0$ except for the points $a_{\alpha _0+\ell p,\beta _0}$ for $\ell \in \mathbb {N}_0$.

We now have

We next consider what happens in the row defined by $\alpha _0+p$.

Supposing that $a_{\alpha _0+p,\beta _0+p+i}$ belongs to K, with $1\le i<p$, then applying Diagram 3 to $a_{\alpha _0,\beta _0+p}$ and $a_{\alpha _0+p,\beta _0+p+i}$ we obtain $x_2=a_{\alpha _0+i,\beta _0+p+i}\in K$, which is a contradiction, and repeating this argument shows that no element of K occupies any position in the row determined by $\alpha _0+p$ except for the points $a_{\alpha _0+ p,\beta _0+mp}$ for $m \in \mathbb {N}_0$.

We next consider what happens in the column defined by $\beta _0+p$.

Supposing that $a_{\alpha _0+p+i,\beta _0+p}$ belongs to K, with $1\le i<p$, then applying Diagram 1 to $a_{\alpha _0+p,\beta _0}$ and $a_{\alpha _0+p+i,\beta _0+p}$ we obtain $x_3=a_{\alpha _0+p+i,\beta _0+i}\in K$, which is a contradiction, and repeating this argument shows that no element of K occupies any position in the column determined by $\beta _0+p$ except for the points $a_{\alpha _0+ \ell p,\beta _0+p}$ for $\ell \in \mathbb {N}_0$.

We now have (ignore temporarily the symbol $\blacksquare$)

Finally, we consider what happens along the diagonal.

Supposing that $a_{\alpha _0+p+i,\beta _0+p+i}$, denoted by $\blacksquare$, with $1\le i<p$, belongs to K, we apply Diagram 1 to $a_{\alpha _0+p,\beta _0}$ and $a_{\alpha _0+p+i,\beta _0+p+i}$, we obtain $x_3=a_{\alpha _0+p+i,\beta _0+i}\in K$, which is a contradiction, and repeating this argument shows that no element of K occupies any position in the diagonal except for the points $a_{\alpha _0+ \ell p,\beta _0+\ell p}$ for $\ell \in \mathbb {N}_0$.

We now have

We are now in the position at the beginning of the proof, namely, the semiheap $K\cap K_{p,p}$ is in subcase (13) of case 3.3.3, and the result follows by applying successively what has already been proved. $\square$

We shall now consider cases (1) and (2) with $a_{\alpha _0,\beta _0}\not \in K$ (See Lemma 3.13), and assume with no loss of generality, that $\alpha _0=\beta _0=0$. We consider the following diagram for case (1) and establish the following notation. The points of K on the row determined by $\alpha _0$, indicated by $\blacktriangle$, are $a_{\alpha _0,\beta _0+m_i}$, with $1\le m_1<m_2< \cdots$, and the points on the column determined by $\beta _0$, indicated by $\boxtimes$, are $a_{\alpha _0+\ell _i,\beta _0}$, with $1\le \ell _1<\ell _2<\cdots$. We denote $\sigma =m_2-m_1$ and $\rho =\ell _2-\ell _1$. For example, in that diagram, $\sigma =6$ and $\rho =r=4$.

We consider first $K_{r,0}$. By Diagram 2, the points (r, i), $1<i<\rho$, indicated by $\circledcirc$, do not belong to K. By Diagram 4, $K\supset K_{r,0}^\rho$. The semiheap $K\cap K_{r,0}$ falls into case 3.3.3, more precisely, either cases (7), (12) or (13), but case (7) does not occur. In case (13), the points $(r+j,j)$, $1\le j<\rho$, indicated by $\circleddash$, do not belong to K, so by Proposition 3.16, $K\cap K_{r,0}=K_{r,0}^\rho$, and therefore in this case,

$$\begin{aligned} K_1:=K_{r,0}^\rho \cap K_{r,p}=\{(\alpha _0+r+\ell \rho , \beta _0+m\rho : \ell \in \mathbb {N}_0,m\rho \ge p \}. \end{aligned}$$

By the same argument applied to $K_{0,p}$, assuming that $K\cap K_{0,p}$ is also in case (13), we have

$$\begin{aligned} K_2:=K_{0,p}^\sigma \cap K_{r,p}=\{(\alpha _0+\ell '\sigma , \beta _0+p+m'\sigma ): \ell '\sigma \ge r, m'\in \mathbb {N}_0 \}. \end{aligned}$$

$K_1$ is depicted by the symbols $\square$ in $K_{r,p}$ and $K_2$ is depicted by the symbols $\blacktriangle$ in $K_{r,p}$, and we must have $K_1=K\cap K_{r,p}=K_2$.

As suggested by the diagram, we now show that $\sigma =\rho$, and that p and r are divisible by $\sigma$.

• Taking $m'=0$ and $\ell '\sigma \ge r$, $(\ell '\sigma ,p)\in K_2$ so that $(\ell '\sigma ,p)=(r+\ell \rho ,m\rho )$ for some $\ell ,m\in \mathbb {N}_0$ with $m\rho \ge p$. Therefore $p=m\rho$ and $r=\ell '\sigma -\ell \rho$.

• Taking $\ell =0$ and $m\rho \ge p$, $(r,m\rho )\in K_1$ so that $(r,m\rho )=(\ell '\sigma ,p+m'\rho )$ for some $\ell ',m'\in \mathbb {N}_0$ with $\ell '\sigma \ge r$. Therefore $r=\ell '\sigma$ and $m\rho =p+m'\sigma$.

  Thus p is divisible by $\rho$, say $p=m_0\rho$, and r is divisible by $\sigma$, say $r=\ell _0\sigma$.

• Taking $\ell =0$ and $(m_0+1)\rho =p+\rho >p$, $(r,(m_0+1)\rho )\in K_1$ so that $(r,(m_0+1)\rho )=(\ell ''\sigma ,p+m''\rho )$ for some $\ell '',m''\in \mathbb {N}_0$ with $\ell ''\sigma \ge r$. Therefore $r=\ell ''\sigma$ and $(m_0+1)\rho =p+m''\sigma$. So $p+\rho =p+m''\sigma$, and $\rho =m''\sigma$.

• Taking $m'=0$ and $(\ell _0+1)\sigma =r+\sigma > r$, $((\ell _0+1)\sigma ,p)\in K_2$ so that $((\ell _0+1)\sigma ,p)=(r+\ell \rho ,m\rho )$ for some $\ell ,m\in \mathbb {N}_0$ with $m\rho \ge p$. Therefore $r+\sigma =r+\ell \rho$, so that $\sigma =\ell \rho$

  Thus $\rho$ is divisible by $\sigma$ and $\sigma$ is divisible by $\rho$, hence $\sigma =\rho$.

Since p and r are each a multiple of $\rho$, it follows that $(r,p)\in K$, so that $(0,p)(r,p)^*(r,0)=(0,0)\in K$, which is a contradiction. We conclude that if both semiheaps $K\cap K_{r,0}$ and $K\cap K_{0,p}$ are in case (13), then case (1) does not occur.

It remains to show that case (1) does not occur in the three other possible cases, namely,

• $K\cap K_{r,0}$ is in case (12) and $K\cap K_{0,p}$ is in case (13)

• $K\cap K_{r,0}$ is in case (13) and $K\cap K_{0,p}$ is in case (12)

• $K\cap K_{r,0}$ is in case (12) and $K\cap K_{0,p}$ is in case (12)

Let us now suppose that $K\cap K_{r,0}$ is in case (12), and $K\cap K_{0,p}$ is in case (13) and refer to the following diagram. Recall that the points of K on the row determined by $\alpha _0$, indicated by $\blacktriangle$, are $a_{\alpha _0,\beta _0+m_i}$, with $1\le m_1<m_2< \cdots$, and the points on the column determined by $\beta _0$, indicated by $\boxtimes$, are $a_{\alpha _0+\ell _i,\beta _0}$, with $1\le \ell _1<\ell _2<\cdots$. We denote $\sigma =m_2-m_1$ and $\rho =\ell _2-\ell _1$. For example, in that diagram, $\sigma =6$ and $\rho =r=4$.

Since $K\cap K_{r,0}$ is assumed in case (12), by Proposition 3.15, there exist $0=j_0<1\le j_1<j_2<\cdots<j_n<\rho$ such that

$$\begin{aligned} K\cap K_{r,0}=\bigcup _{i=0}^n K_{r+j_i,j_i}^\rho , \end{aligned}$$

and therefore in this case,

$$\begin{aligned}&K_1:=K_{r,0}^\rho \cap K_{r,p}=\bigcup _{i=0}^n\left( K_{r+j_i,j_i}^\rho \cap K_{r,p}\right) \\&\quad =\bigcup _{i=0}^n\{(r+j_i+\ell \rho ,j_i+m\rho ):\ell ,m\in \mathbb {N}_0, j_i+m\rho \ge p \}. \end{aligned}$$

In the diagram, we indicate the points of $K_{r+j_1,j_1}^\rho =K_{r+3,3}^4$ with the symbols $\heartsuit$, and the points of $K_{r+j_2,j_2}^\rho =K_{r+6,6}^4$ with the symbols $\oplus$.

As before, assuming that $K\cap K_{0,p}$ is in case (13), we have

$$\begin{aligned} K_2:=K_{0,p}^\sigma \cap K_{r,p}=\{(\alpha _0+\ell '\sigma , \beta _0+p+m'\sigma ): \ell '\sigma \ge r, m'\in \mathbb {N}_0 \} \end{aligned}$$

Also in the diagram, $K_1$ is depicted by the symbols $\square ,\heartsuit ,\oplus$ in $K_{r,p}$ and $K_2$ is depicted by the symbols $\blacktriangle$ in $K_{r,p}$, and we must have $K_1=K\cap K_{r,p}=K_2$.

We show first that $\rho =\sigma$. Since $K_1\subset K_2$, for $\ell ,m,i\in \mathbb {N}_0$, with $j_i+m\rho \ge p$, there exist $\ell ',m'\in \mathbb {N}_0$ with $\ell '\sigma \ge r$ and

$$\begin{aligned} (r+j_i+\ell \rho ,j_i+m\rho )=(\ell '\sigma ,p+m'\sigma ). \end{aligned}$$

(9)

Fix i such that $j_i\ge p$. Then for all $\ell ,m\in \mathbb {N}_0$, there exist $\ell ',m'\in \mathbb {N}_0$ with $\ell '\sigma \ge r$ such that

$$\begin{aligned} r+j_i+\ell \rho =\ell '\rho \hbox { and }j_i+m\rho =p+m'\sigma . \end{aligned}$$

Eliminating $j_i$ from these two equations results in

$$\begin{aligned} r+p=(m-\ell )\rho +(\ell '-m')\sigma , \end{aligned}$$

(10)

with $(\ell ',m')$ depending on $(\ell ,m)$ and satisfying $\ell '\sigma \ge r$.

Since $K_2\subset K_1$, for $\ell ,m\in \mathbb {N}_0$, with $\ell \sigma \ge r$, there exist $\ell ',m',i\in \mathbb {N}_0$ with $j_i+m'\rho \ge p$ such that

$$\begin{aligned} r+j_i+\ell '\rho =\ell \sigma \hbox { and }j_i+m'\rho =p+m\sigma . \end{aligned}$$

Eliminating $j_i$ from these two equations results in

$$\begin{aligned} r+p=(m'-\ell ')\rho +(\ell -m)\sigma , \end{aligned}$$

(11)

with $(\ell ',m')$ depending on $(\ell ,m)$, provided $\ell \sigma \ge r$.

With $\ell \ge 0$ and $m\ge 0$, from (10), there exist $\ell _1,m_1$ such that

$$\begin{aligned} r+p=(m-\ell )\rho +(\ell _1-m_1)\sigma \end{aligned}$$

and there exist $\ell _2,m_2$ such that

$$\begin{aligned} r+p=(m+1-\ell )\rho +(\ell _2-M_2)\sigma , \end{aligned}$$

so by subtraction, $0=\rho +[(\ell _2-m_2)+(\ell _1-m_1)]\sigma$ and $\sigma$ divides $\rho$.

With $\ell \sigma \ge r$ and $m\ge 0$, from (11), there exist $\ell _3,m_3$ such that

$$\begin{aligned} r+p=(m_3-\ell _3)\rho +(\ell -m)\sigma \end{aligned}$$

and there exist $\ell _4,m_4$ such that

$$\begin{aligned} r+p=(m_4-\ell _4)\rho +(\ell +1-m)\sigma , \end{aligned}$$

so by subtraction, $0=[(m_4-\ell _4)-(m_3-\ell _3)]\rho +\sigma$ and $\rho$ divides $\sigma$.

Hence $\rho =\sigma$ and from (10) or (11), $\rho$ divides $r+p$. Now, taking $i=0,\ell =0$ in (9), $r=\ell '\sigma$, so that also $\rho$ divides p. Hence, as in the previous case, $(r,p)\in K$, so that $(0,p)(r,p)^*(r,0)=(0,0)\in K$, which is a contradiction. We conclude that if the semiheap $K\cap K_{r,0}$ is in case (12) and the semiheap $K\cap K_{0,p}$ is in case (13), then case (1) does not occur.

Since the adjoint mapping is an anti-isomorphism of the extended bicyclic semigroup (See Remark 2.8), it follows that if the semiheap $K\cap K_{r,0}$ is in case (13) and the semiheap $K\cap K_{0,p}$ is in case (12), then case (1) does not occur.

It remains to consider the case when both semiheaps $K\cap K_{r,0}$ and $K\cap K_{0,p}$ are in case (12). After this, again since the adjoint mapping is an anti-isomorphism, and case (1) has been shown to not occur, it will follow that case (2) also does not occur, so we will have the following lemma.

Lemma 3.17

Cases (1) and (2) with (necessarily) $(0,0)\not \in K$, do not occur.

Proof

It suffices to show that if both semiheaps $K\cap K_{r,0}$ and $K\cap K_{0,p}$ are in case (12), then $(0,0)\in K$ and therefore case (1) does not occur. We have

$$\begin{aligned} K_1:=K\cap K_{r,0}\cap K_{r,p}=\bigcup _{i=0}^n\{(r+j_i+\ell \rho ,j_i+m\rho ): \ell ,m\in \mathbb {N}_0, j_i+m\rho \ge p\} \end{aligned}$$

and

$$\begin{aligned} K_2:=K\cap K_{0,p}\cap K_{r,p}=\bigcup _{i=0}^{n'}\{(k_i+\ell \sigma ,p+k_i+m\sigma ):\ell ,m\in \mathbb {N}_0, k_i+\ell \sigma \ge r\}, \end{aligned}$$

where $0=k_0<1\le k_1<k_2<\cdots<k_{n'}<\sigma$.

Since $K_1\subset K_2$, for $i,\ell ,m\in \mathbb {N}_0$ with $j_i+m\rho \ge p$, there exist $i',\ell ',m'\in \mathbb {N}_0$ satisfying $k_{i'}+\ell '\sigma \ge r$, such that

$$\begin{aligned} (r+j_i+\ell \rho ,j_i+m\rho )=(k_{i'}+\ell '\sigma ,p+k_{i'}+m'\sigma ) \end{aligned}$$

so that

$$\begin{aligned} r+j_i+\ell \rho =k_{i'}+\ell '\sigma \hbox { and } j_i+m\rho =p+k_{i'}+m'\sigma \end{aligned}$$

Fix i such that $j_i\ge p$. Then for every $\ell ,m\in \mathbb {N}_0$, by subtraction, we have

$$\begin{aligned} r+p=(m-\ell )\rho +(\ell '-m')\sigma \end{aligned}$$

(12)

with $\ell ',m'$ depending only on $\ell ,m\in \mathbb {N}_0$ (and $\ell '$ satisfying $k_{i'}+\ell '\sigma \ge r$ for some $i'$).

Since $K_2\subset K_1$, for $i,\ell ,m\in \mathbb {N}_0$ with $k_i+\ell \sigma \ge r$, there exist $i',\ell ',m'\in \mathbb {N}_0$ satisfying $j_{i'}+m'\rho \ge p$, such that

$$\begin{aligned} (k_i+\ell \sigma , p+k_i+m\sigma )=(r+j_{i'}+\ell '\rho ,j_{i'}+m'\rho ) \end{aligned}$$

so that

$$\begin{aligned} k_i+\ell \sigma =r+j_{i'}+\ell '\rho \hbox { and } p+k_i+m\sigma =j_{i'}+m'\rho \end{aligned}$$

Fix i such that $k_i\ge r$. Then for every $\ell ,m\in \mathbb {N}_0$, by subtraction, we have

$$\begin{aligned} r+p=(\ell -m)\sigma +(m'-\ell ')\rho \end{aligned}$$

(13)

with $\ell ',m'$ depending only on $\ell ,m\in \mathbb {N}_0$ (and $m'$ satisfying $j_{i'}+m'\rho \ge p$ for some $i'$).

With $\ell \ge 0$ and $m\ge 0$, from (12), there exist $\ell _1,m_1$ such that

$$\begin{aligned} r+p=(m-\ell )\rho +(\ell _1-m_1)\sigma \end{aligned}$$

and there exist $\ell _2,m_2$ such that

$$\begin{aligned} r+p=(m+1-\ell )\rho +(\ell _2-M_2)\sigma , \end{aligned}$$

so by subtraction, $0=\rho +[(\ell _2-m_2)+(\ell _1-m_1)]\sigma$ and $\sigma$ divides $\rho$.

With $\ell \ge 0$ and $m\ge 0$, from (13), there exist $\ell _1,m_1$ such that

$$\begin{aligned} r+p=(\ell -m)\sigma +(m_1-\ell _1)\rho \end{aligned}$$

and there exist $\ell _2,m_2$ such that

$$\begin{aligned} r+p=(\ell +1-m)\sigma +(m_2-\ell _2)\rho , \end{aligned}$$

so by subtraction, $0=\sigma +[(m_2-\ell _2)+(m_1-\ell _1)]\rho$ and $\rho$ divides $\sigma$.

Hence $\rho =\sigma$ and from (12) or (13), $\sigma$ divides $p+r$. In fact, $\sigma$ divides both p and r. Indeed, since $(q,q)\in K_1$ and $K_1=K_2$, there exist $\ell ,m,i$ and $\ell ',m',i'$, such that

$$\begin{aligned} (q,q)=(r+j_i+\ell \sigma ,j_i+\sigma ) = (k_{i'}+\ell '\sigma ,p+k_{i'}+m'\sigma ), \end{aligned}$$

so that $r+\ell \sigma =m\sigma$ and $p+m'\sigma =\ell '\sigma$. We now have $(r,p)\in K$, so that $(0,0)=(0,p)(r,p)^*(r,0)\in K$, a contradiction. $\square$

We summarize the results of Lemma 3.13 to Lemma 3.17 in the following proposition.

Proposition 3.18

If the semiheap K is in case 3.3.3, then either $K=K_{\alpha _0,\beta _0}^p$ for some $p>0$, or there exist $p>0$ and $q > 0$ such that

$$\begin{aligned} K= \bigcup _{i=0}^n K_{q_i,q_i}^p. \end{aligned}$$

where

$$\begin{aligned} q=q_0<q_1<q_2<\cdots<q_n<p\quad \hbox { and }\quad p<q_{n+1}<q_{n+2}<\cdots . \end{aligned}$$

Proof

By Lemma 3.17, cases (1) and (2) do not occur. By Lemma 3.13, cases (3)–(11) do not occur. Cases (12) and (13) are described in Propositions 3.15 and 3.16. $\square$

4 Injectivity of W*-TROs

The notation for this section is the following.

S is an inverse semigroup with generalized inverse $x^*$.

K is a subset of S closed under the triple product $xy^*z$ (semiheap).

$\pi$ is the left regular representation of S on $H:=\ell ^2(S)$ so that S is an orthonormal basis for H and $\pi (x)$ is the partial isometry defined by $\pi (x)y=xy$ if $yy^*\le x^*x$ and $\pi (x)y=0$ otherwise.

$C^*_{\hbox {red}}(S)$ is the C*-algebra generated by $\{\pi (x):x\in S\}$ and is the norm closure of $\hbox {span}\, \pi (S)$.

TRO(K) is the TRO generated by $\pi (K)$ and is the norm closure of $\hbox {span}\, \pi (K)$.

VN(S) is the von Neumann algebra generated by $\pi (S)$ and is the weak closure of $C^*_{\hbox {red}}(S)$

VNTRO(K) is the W*-TRO generated by $\pi (K)$ and is the weak closure of TRO(K).

Details of the left regular representation are as follows ([13, pp. 25–27]). We have

$$\begin{aligned} \pi (a_{ij})a_{pq}=\left\{ \begin{array}{cc} a_{ij}a_{pq}, &{} a_{pq}a_{pq}^*\le a_{ij}^*a_{ij}\\ 0,&{} \hbox {otherwise} \end{array} \right. , \end{aligned}$$

that is,

$$\begin{aligned} \pi (a_{ij})a_{pq}=\left\{ \begin{array}{cc} a_{ij}a_{pq}, &{} a_{pp}\le a_{jj}\\ 0,&{} \hbox {otherwise} \end{array} \right. , \end{aligned}$$

or

$$\begin{aligned} \pi (a_{ij})a_{pq}=\left\{ \begin{array}{cc} a_{ij}a_{pq}, &{} p\ge j\\ 0,&{} \hbox {otherwise} \end{array} \right. , \end{aligned}$$

or

$$\begin{aligned} \pi (a_{ij})a_{pq}=\left\{ \begin{array}{cc} a_{i+p-j,q}, &{} p\ge j\\ 0,&{} \hbox {otherwise} \end{array} \right. \end{aligned}$$

Define provisionally a linear map $\Phi _0:\hbox {span}\, \pi (S)\rightarrow \hbox {span}\, \pi (K)$ as follows: $\Phi _0(0)=0$, and for $x_1,\ldots x_n\in S$,

$$\begin{aligned} \Phi _0\left( \sum _{i=1}^n\lambda _i\pi (x_i)\right) = \sum _{x_i\in K}\lambda _i\pi (x_i). \end{aligned}$$

Proposition 4.1

The idempotent map $\Phi _0$ is well-defined and contractive, and therefore extends to a contractive projection $\Phi$ on $C^*_{\hbox {red}}(S)$ with range TRO(K). Moreover, $\Phi$ extends to a completely contractive projection on VN(S) with range VNTRO(K). Hence, if VN(S) in an injective von Neumann algebra, then VNTRO(K) is an injective operator space.

Proof

Let $a=\left\| \sum _{i=1}^n\lambda _i\pi (x_i) \right\|$ and $b=\left\| \sum _{x_i\in K}\lambda _i\pi (x_i) \right\|$. With $\xi =\sum _{z\in S}(\xi ,z)z\in \ell ^2(S)$,

$$\begin{aligned}\pi (x_i)\xi =\sum _{zz^*\le x_i^*x_i}(\xi ,z)x_iz \end{aligned}$$

so that

$$\begin{aligned} b^2=\sup _{\Vert \xi \Vert \le 1}\left\| \sum _{x_i\in K}\lambda _i\pi (x_i) \xi \right\| ^2=\sup _{\Vert \xi \Vert \le 1}\sum _{x_i\in K,z\in S,zz^*\le x_i^*x_i}|\lambda _i(\xi ,z)|^2 \end{aligned}$$

and by the same calculation

$$\begin{aligned} a^2=\sup _{\Vert \xi \Vert \le 1}\left\| \sum _{x_i\in S}\lambda _i\pi (x_i) \xi \right\| ^2=\sup _{\Vert \xi \Vert \le 1}\sum _{x_i\in S,z\in S,zz^*\le x_i^*x_i}|\lambda _i(\xi ,z)|^2. \end{aligned}$$

Therefore $\Phi _0$ is contractive and extends to a contractive projection on $C^*_{\hbox {red}}(S)$ with range TRO(K).

Let $A=C^*_{\hbox {red}}(S)$, $U=TRO(K)$, so that $\Phi ^{**}$ is a contractive projection on the von Neumann algebra $A^{**}$ with range $U^{**}$. By [10, Lemma], $U^{**}$ is isomorphic to VNTRO(K), and by [3, Theorem 2.5], $\Phi ^{**}$ is a completely contractive projection with range VNTRO(K). Therefore, the restriction $\overline{\Phi }$ of $\Phi ^{**}$ to VN(S) is a completely contractive projection of VN(S) onto VNTRO(K). If VN(S) is injective, then there is a completely contractive projection P of B(H) onto VN(S), so that $\overline{\Phi }\circ P$ is a completely contractive projection with range VNTRO(K). $\square$

Example 4.2

Suppose that e and f are idempotents in the inverse semigroup S and that $K=eSf$, which is a subsemiheap of S. The corresponding induced map takes the form $\pi (x)\mapsto \pi (exf)$ (with $0\rightarrow 0$) and is contractive since

$$\begin{aligned} \left\| \sum _i \lambda _i \pi (ex_if)\right\| \le \Vert \pi (e)\Vert \left\| \sum _i\lambda _i\pi (x_i)\right\| \Vert \pi (f)\Vert =\left\| \sum _i\lambda _i\pi (x_i)\right\| . \end{aligned}$$

Hence Proposition 4.1 applies. This also applies to maps of the form $x\mapsto ex$ and $x\mapsto xf$.

The maximal subgroups of any inverse semigroup S are of the form

$$\begin{aligned} S_e^e=\{s\in S:ss^*=s^*s=e\} \end{aligned}$$

for some idempotent e (See [13, p. 198]). Thus, the maximal subgroups of the extended bicyclic semigroup E reduce to one-element groups, so are trivially amenable and hence by [13, Theorem 4.5.2], VN(E) is injective.²

Since VN(E) is injective, where E is the extended bicyclic semigroup, it follows from Proposition 4.1 and Example 4.2, that $VNTRO(a_{ii}Ea_{jj})$ is an injective operator space, as are $VNTRO(Ea_{jj})$ and $VNTRO(a_{ii}E)$. More generally, we have

Corollary 4.3

All of the subsemiheaps of the extended bicyclic semigroup E (which were determined in Theorem 1.2) give rise to injective W*-TROs.