Abstract
We build on the description of left congruences on an inverse semigroup in terms of the kernel and trace due to Petrich and Rankin. The notion of an inverse kernel for a left congruence is developed. Various properties of the trace and inverse kernel are discussed, in particular that the inverse kernel is a full inverse subsemigroup and that both the trace and inverse kernel maps are onto \(\cap \)-homomorphisms. It is shown that a left congruence is determined by its trace and inverse kernel, and the lattice of left congruences is identified as a subset of the direct product of the lattice of congruences on the idempotents and the lattice of full inverse subsemigroups. We demonstrate that every finitely generated left congruence is the join of a finitely generated trace minimal left congruence and a finitely generated idempotent separating left congruence. Characterisations are given of inverse semigroups that are left Noetherian, or are such that Rees left congruences are finitely generated.
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1 Introduction
The study of congruence lattices for inverse semigroups has a rich history. In broad generality there are two ways to describe a congruence. The first uses the fact that a congruence is completely determined by the equivalence classes of idempotents, and describes such collections of sets. The second makes use of the fact that a congruence is determined by its trace (the restriction to the idempotents), and its kernel (the union of the congruence classes containing idempotents).
Following the former philosophy, in 1974 Meakin [8] developed left-kernel systems for inverse semigroups, these are exactly the collections of subsets that are the equivalence classes of a left congruence which contain idempotents. He showed that every congruence on the semilattice of idempotents arises as the restriction of a left congruence on the whole semigroup, and that the set of left congruences that agree on the idempotents forms an interval in the lattice of left congruences. In addition, he gave a description of the maximum and minimum elements in this interval.
Following the second philosophy, in 1992 Petrich and Rankin [11] developed the kernel trace approach for one-sided congruences on inverse semigroups. From a lattice perspective this approach is advantageous, as the ordering of left congruences is induced by the natural orderings on the sets of kernels and traces. The function taking a left congruence to its trace is a \(\cap \)-homomorphism but not a \(\vee \)-homomorphism, and the corresponding function for the kernel is neither. Further, the set of left congruences with a given kernel has a maximum element, but in general no minimum element.
In this paper we build on these foundations and introduce and characterise the inverse kernel for a left congruence on an inverse semigroup and give necessary and sufficient conditions for an inverse subsemigroup and a congruence on the idempotents to form an inverse congruence pair (that is, the trace and inverse kernel of a left congruence). We prove that the set of left congruences is in bijection with the set of inverse congruence pairs which identifies the set of left congruences as a subset of the direct product of the lattice of congruences on the idempotents and the lattice of full inverse subsemigroups. We show that the set of left congruences that share an inverse kernel has a minimum element. Also, we extend results of Meakin to describe the sublattice of left congruences having a fixed trace. The function taking a left congruence to its inverse kernel is shown to be a \(\cap \)-homomorphism, and a description of the join of two left congruences in terms of the trace and inverse kernel is given.
Finally we consider left congruences in terms of generating sets. We show that we may choose a generating set for a left congruence that corresponds to generating sets for the trace and inverse kernel. This naturally leads to a discussion of finitely generated left congruences, and we show that the description of left Noetherian inverse semigroups [7] is a natural consequence of the inverse kernel description for left congruences. We conclude by analysing Rees left congruences and classify those inverse semigroups for which every Rees left congruence is finitely generated.
2 Preliminaries
We assume a familiarity with standard results in inverse semigroup theory (see [5, Chapter 5]). Unless otherwise stated S will denote an inverse semigroup, and \(E=E(S)\) will be the semilattice of idempotents of S. Greek letters shall be used for equivalence relations, and [a] is the equivalence class of a, or \([a]_\rho \) if \(\rho \) is not clear.
The set of left congruences on S can be ordered by inclusion; we denote this lattice by \(\mathfrak {LC}(S)\). Also, we write \(\mathfrak {RC}(S)\) for the lattice of right congruences on S, and \(\mathfrak {C}(S)\) for the lattice of congruences on S. It is straightforward that the identity congruence \(\iota ,\) and the universal congruence \(\omega ,\) are minimum and maximum elements, respectively, in each of these three lattices. We recall that for an inverse semigroup the lattices \(\mathfrak {LC}(S)\) are \(\mathfrak {RC}(S)\) are isomorphic via the map
Bearing this is mind we may restrict our attention to left congruences for the majority of this paper.
Given a binary relation \(R\subseteq S\times S\) we denote by \(\langle R\rangle \) the smallest left congruence containing R. We recall that \(a\ \langle R\rangle \ b\) if and only if \(a=b\) or there is a sequence
where \(c_i\in S^1\) and \((x_i,y_i)\in R\cup R^{-1},\) for \(1\le i\le n\). We say that this is an R sequence from a to b. For a binary relation \(R\subseteq S\times S\) and a subset \(T\subseteq S,\) write \(R|_T\) for the restriction \(R\cap (T\times T)\) of R to T.
A subsemigroup of S is said to be full if it contains all idempotents of S. The set of all full subsemigroups of S forms a lattice under set inclusion which is denoted \(\mathfrak {K}(S)\). The set \(\mathfrak {V}(S)\) of full inverse subsemigroups of S is a sublattice of \(\mathfrak {K}(S)\).
We denote by \(\vee \) the join in a lattice, so if \(T_1,T_2\in \mathfrak {V}(S)\) then \(T_1\vee T_2\) is the full inverse subsemigroup generated by \(T_1\cup T_2,\) if \(\tau _1,\tau _2\in \mathfrak {C}(E)\) are congruences on E then \(\tau _1\vee \tau _2\) is the congruence on E generated by \(\tau _1\cup \tau _2\) and if \(\rho _1,\rho _2\in \mathfrak {LC}(S)\) then \(\rho _1\vee \rho _2\) is the left congruence on S generated by \(\rho _1\cup \rho _2\).
Definition 2.1
Let \(\kappa \) be an equivalence relation on S. The kernel and trace of \(\kappa \) are defined
The notions of kernel and trace will be applied to a left congruence \(\rho \). It is easily verified that the trace of \(\rho \) is a congruence on E. Furthermore, the kernel of \(\rho \) is a full subsemigroup of S; however in general it is not an inverse subsemigroup. The main result of [11] is that a left congruence is determined by its kernel and trace.
Theorem 2.2
[11, Theorem 3.5] Let \(\tau \) be a congruence on E, and let K be a full subsemigroup of S such that the following are true:
-
(i)
for all \(a\in S\) and \(b\in K\), if \(ab^{-1}b=b\) and \(a^{-1}a\ \tau \ b^{-1}b\) then \(a\in K\);
-
(ii)
for all \(a\in K\) and \(e,f\in E,\) if \(e\ \tau \ f \) then \(a^{-1}ea\ \tau \ a^{-1}fa\);
-
(iii)
for every \(a\in K\) there exists \(b\in S\) with \(ab^{-1}b=b,\ a^{-1}a\ \tau \ b^{-1}b\) and \(b^{-1}\in K\).
Define a relation
Then the relation \(\rho _{(K,\tau )}\) is a left congruence on S with kernel K and trace \(\tau \).
Conversely if \(\rho \) is a left congruence on S then the pair \(({{\,\textrm{ker}\,}}\rho ,{{\,\textrm{trace}\,}}\rho )\) obeys (i),(ii),(iii) and \(\rho =\rho _{({{\,\textrm{ker}\,}}\rho ,{{\,\textrm{trace}\,}}\rho )}\).
Hence we may realise \(\mathfrak {LC}(S)\) as a subset of \(\mathfrak {K}(S)\times \mathfrak {C}(E),\) and as previously mentioned the natural ordering of pairs \(({{\,\textrm{ker}\,}}\rho ,{{\,\textrm{trace}\,}}\rho )\) in \(\mathfrak {K}(S)\times \mathfrak {C}(E)\) (inclusion in both coordinates) coincides with the ordering on left congruences. Explicitly if \(\rho _1,\rho _2\) are left congruences on S, then
Consequently,
Definition 2.3
The trace map and the kernel map are the functions
If \(\rho \) is a left congruence on S then the trace class of \(\rho \) is
and the kernel class of \(\rho \) is
It is a straightforward observation that the trace and kernel maps are order-preserving functions. It is known (see [8, 11]) that the trace map is a \(\cap \)-homomorphism and that the trace classes are intervals in the lattice of left congruences. However the kernel map and the kernel classes are less well behaved; in [11] it is observed that in general the kernel map is not a \(\cap \)- or \(\vee \)-homomorphism and that the kernel classes may not be intervals in \(\mathfrak {LC}(S)\).
Definition 2.4
Let \(\tau \) be a congruence on E. Define \(N_L(\tau ),\) the left-normaliser of \(\tau \), by
and \(N_R(\tau ),\) the right normaliser, by
The normaliser \(N(\tau )\) is then defined as
If \(T\subseteq N(\tau )\) then we say that \(\tau \) is normal in T.
It is important to note that if \(\tau \) is a congruence on E then \(N(\tau )\) is a full inverse subsemigroup, indeed it is the largest inverse subsemigroup contained in \(N_L(\tau )\) (or indeed in \(N_R(\tau )\)). This follows from the observation that
It is worth noting that for a left congruence \(\rho \) on S with trace \(\tau ,\) we have that \({{\,\textrm{ker}\,}}\rho \subseteq N_L(\tau )\). Indeed, suppose that \(a\in {{\,\textrm{ker}\,}}\rho ,\) so there is \(g\in E\) with \(a\ \rho \ g\). For \(e,f\in E\) with \(e\ \tau \ f\) we observe
hence \(a\in N_L(\tau )\).
Theorem 2.5
([8, Theorem 3.1], [11, Propositions 4.1 and 4.2]) Let \(\tau \) be a congruence on E. Then there exists a left congruence \(\rho \) on S with \({{\,\textrm{trace}\,}}\rho =\tau \). Moreover the relations
and
are left congruences on S with trace \(\tau \). In addition, \(\nu _\tau \) is the minimum left congruence with trace \(\tau \) and \(\mu _\tau \) is the maximum left congruence with trace \(\tau \).
If we regard \(\tau \in \mathfrak {C}(E)\) as a binary relation on S, then \(\langle \tau \rangle \) is the minimum left congruence on S containing \(\tau \). Thus we have a closed form for \(\langle \tau \rangle =\nu _\tau ,\) and moreover we have that \(\langle \tau \rangle \cap (E\times E)=\tau \). In particular if \(\tau \) is finitely generated then so is \(\nu _\tau \).
In [8] Meakin discusses idempotent separating left congruences on S (a relation \(\rho \) on S is idempotent separating if \(e\ \rho \ f\) implies \(e=f\) for any \(e,f\in E,\) that is, \({{\,\textrm{trace}\,}}\rho =\iota \)), and demonstrates that the maximum idempotent separating left congruence is Green’s relation \(\mathcal {R}\). As is well known for inverse semigroups this relation is
Notice that if \(\rho \subseteq \mathcal {R}\) and \(a\ \rho \ e\) then \(e=aa^{-1},\) whence \(a\ \rho \ aa^{-1}\) and so \(a^{-1}a\ \rho \ a^{-1}\). Then \(a^{-1}\in {{\,\textrm{ker}\,}}\rho \). Thus if \({{\,\textrm{trace}\,}}\rho =\iota \) then \({{\,\textrm{ker}\,}}\rho \) is an inverse subsemigroup.
Theorem 2.6
[8, Theorem 4.2] The restriction of the kernel map to the set of idempotent separating left congruences is a lattice isomorphism onto \(\mathfrak {V}(S)\).
Hence for each full inverse subsemigroup T of S there is a unique idempotent separating left congruence with kernel T. We will use \(\chi _T\) to denote this left congruence.
3 The inverse kernel
As explained in the previous section it is possible to reconstruct a left congruence from the kernel and trace. We now show that we may define an inverse subsemigroup of the kernel, that we name the inverse kernel, from which we can also recover the original left congruence. In [11] it is remarked that the primary issue with the kernel is that given an element in the kernel it is not simple to determine to which idempotent it is related. To this end we make the following definition.
Definition 3.1
For a left congruence \(\rho \) on S the inverse kernel of \(\rho \) is the set
We immediately note that the inverse kernel of a left congruence \(\rho \) is contained in the kernel, and that while given \(a\in {{\,\textrm{Iker}\,}}\rho \) and \(e\in E\) it is not possible to determine from \({{\,\textrm{Iker}\,}}\rho \) alone whether \(a\ \rho \ e\), we are able to specify one idempotent to which a is \(\rho \)-related, namely \(aa^{-1}\).
Proposition 3.2
Let \(\rho \) be a left congruence on S, let \(\tau ={{\,\textrm{trace}\,}}\rho ,\) and let \(K={{\,\textrm{ker}\,}}\rho \). Then the following hold:
-
(i)
\({{\,\textrm{Iker}\,}}\rho \) is a full inverse subsemigroup of S;
-
(ii)
\({{\,\textrm{Iker}\,}}\rho ={{\,\textrm{ker}\,}}(\rho \cap \mathcal {R});\)
-
(iii)
\({{\,\textrm{Iker}\,}}\rho =\{ a\in K\mid a^{-1}\in K\};\)
-
(iv)
\({{\,\textrm{Iker}\,}}\rho =K \cap N(\tau )\).
Proof
-
(i)
First we observe that \({{\,\textrm{Iker}\,}}\rho \) is a subsemigroup. Indeed suppose that \(a,b\in {{\,\textrm{Iker}\,}}\rho ,\) so \(a\ \rho \ aa^{-1},\) and \(b\ \rho \ bb^{-1}\). Then
$$\begin{aligned} ab\ \rho \ abb^{-1} = abb^{-1}a^{-1}a\ \rho \ abb^{-1}a^{-1}aa^{-1}=abb^{-1}a^{-1} = (ab)(ab)^{-1}. \end{aligned}$$Further, as \(e\ \rho \ e=ee^{-1}\) we have that \(E\subseteq {{\,\textrm{Iker}\,}}\rho ,\) so \({{\,\textrm{Iker}\,}}\rho \) is full. Also if \(a\ \rho \ aa^{-1}\) then by multiplying on the left by \(a^{-1}\) we observe that \( a^{-1}a\ \rho \ a^{-1},\) hence \({{\,\textrm{Iker}\,}}\rho \) is closed under taking inverses. Thus \({{\,\textrm{Iker}\,}}\rho \) is a full inverse subsemigroup of S.
-
(ii)
This is immediate from the definition of the inverse kernel.
-
(iii)
We have already noted that \({{\,\textrm{Iker}\,}}\rho \) is an inverse subsemigroup contained in the kernel, thus
$$\begin{aligned} {{\,\textrm{Iker}\,}}\rho \subseteq \{ a\in K\mid a^{-1}\in K\}. \end{aligned}$$For the reverse inclusion we suppose that \(a,a^{-1}\in K\) so there are \(e,f\in E\) with \(a\ \rho \ e\) and \(a^{-1}\ \rho \ f\). We note that since \(a^{-1}\ \rho \ f\) we have \(fa^{-1}\ \rho \ f\ \rho \ a^{-1}\). We then observe that
$$\begin{aligned} a=aa^{-1}a\ \rho \ aa^{-1}e = eaa^{-1}\ \rho \ eafa^{-1} = afa^{-1}e\ \rho \ afa^{-1}a = af\ \rho \ aa^{-1}. \end{aligned}$$ -
(iv)
We recall that if \({{\,\textrm{trace}\,}}\rho =\tau \) then \(K\subseteq N_L(\tau )\). We have observed that \(N(\tau )=\{a\in N_L(\tau )\mid a^{-1}\in N_L(\tau )\}\). Then (iii) gives \({{\,\textrm{Iker}\,}}\rho \subseteq N(\tau )\). Thus we have that \({{\,\textrm{Iker}\,}}\rho \subseteq {N(\tau )}\cap {K}\). For the reverse inclusion suppose \(a\in N(\tau )\cap K\). As \(a\in K\) there is \(e\in E\) with \(a\ \rho \ e\) from which we obtain that \(a^{-1}ea\ \rho \ a^{-1}e\ \rho \ a^{-1}a\). As \(a^{-1}a,a^{-1}ea\) are idempotents this implies \(a^{-1}a\ \tau \ a^{-1}ea,\) which we may conjugate by \(a\in N(\tau )\) to obtain that
$$\begin{aligned} aa^{-1}= a(a^{-1}a)a^{-1}\ \tau \ a(a^{-1}ea)a^{-1} = aa^{-1}e. \end{aligned}$$As \(a\ \rho \ e\) it follows that \((aa^{-1})a\ \rho \ (aa^{-1})e\) and so
$$\begin{aligned} a=aa^{-1}a\ \rho \ aa^{-1}e\ \rho \ aa^{-1}. \end{aligned}$$Thus \(a\in {{\,\textrm{Iker}\,}}\rho ,\) and so we have that \({{\,\textrm{Iker}\,}}\rho =K\cap N(\tau )\). \(\square \)
From Proposition 3.2(iii) we observe that if the kernel of a left congruence is inverse then the inverse kernel is equal to the kernel. For instance, for two-sided congruences and idempotent separating left congruences the notion of kernel and inverse kernel coincide. For certain classes of inverse semigroups the kernel of a left congruence is always an inverse subsemigroup, including Clifford semigroups (see [11]). The description of one-sided congruences on inverse semigroups given in this paper coincides with the kernel trace description from [11] for such classes of semigroups.
We mentioned previously that the normaliser of a congruence on E is the maximum inverse subsemigroup contained in the left normaliser. Proposition 3.2(iii) informs us that that the inverse kernel of a left congruence is the largest inverse subsemigroup contained in the kernel. Applying this to the maximum left congruence with a fixed trace we have the following corollary.
Corollary 3.3
For every congruence \(\tau \) on E, we have
Proof
From the description of \(\mu _\tau \) from Theorem 2.5 we get that if \(a\in N(\tau )\) then \(a\ \mu _\tau \ aa^{-1}\). Hence \(N(\tau )\subseteq {{\,\textrm{Iker}\,}}\mu _\tau \). However Proposition 3.2(iv) gives that \({{\,\textrm{Iker}\,}}\mu _\tau \subseteq N(\tau )\). \(\square \)
Following on from Definition 3.1 we make a further natural definition.
Definition 3.4
The inverse kernel map is the function
For \(\rho \) a left congruence the inverse kernel class of \(\rho \) is
We recall from Theorem 2.6 that for each full inverse subsemigroup T there is a unique idempotent separating left congruence for which T is the kernel, and we write \(\chi _T\) for this left congruence. As remarked previously, the kernel map is not a \(\cap \)-homomorphism (see [11]). It follows immediately from the definition that the inverse kernel map is a \(\cap \)-homomorphism.
Proposition 3.5
The inverse kernel map is a complete surjective \(\cap \)-homomorphism of \(\mathfrak {LC}(S)\) onto \(\mathfrak {V}(S)\).
Moreover, the inverse kernel class \(\{\kappa \in \mathfrak {LC}(S)\mid {{\,\textrm{Iker}\,}}(\kappa )=T\}\) is closed under \(\cap \) and has a minimum element, which is \(\chi _T\). In particular, if \(\rho \) is a left congruence then the minimum left congruence in \([\rho ]_{{{\,\textrm{Iker}\,}}}\) is \(\rho \cap \mathcal {R}\).
Proof
That the inverse kernel map is a complete \(\cap \)-homomorphism follows immediately from the definition, and that it is surjective follows from Theorem 2.6.
That each inverse kernel class is closed under \(\cap \) is now immediate. Furthermore, if \(\rho \) is a left congruence then \(\rho \cap \mathcal {R}\) is an idempotent separating left congruence. By Proposition 3.2, \({{\,\textrm{Iker}\,}}(\rho )={{\,\textrm{ker}\,}}(\rho \cap \mathcal {R})={{\,\textrm{Iker}\,}}(\rho \cap \mathcal {R}),\) so \(\rho \cap \mathcal {R}\in [\rho ]_{{{\,\textrm{Iker}\,}}}\). It follows from Theorem 2.6 that \(\rho \cap \mathcal {R}\) is the unique idempotent separating left congruence with inverse kernel equal to \({{\,\textrm{Iker}\,}}(\rho )\). The same argument implies that for any \(\kappa \in [\rho ]_{{{\,\textrm{Iker}\,}}},\) \(\kappa \cap \mathcal {R}\) is also an idempotent separating congruence with inverse kernel \({{\,\textrm{Iker}\,}}(\kappa ),\) which is equal to \({{\,\textrm{Iker}\,}}(\rho )\). Then \(\kappa \cap \mathcal {R}=\rho \cap \mathcal {R},\) so in particular \(\rho \cap \mathcal {R}\subseteq \kappa \). Thus we have that \(\rho \cap \mathcal {R}\) is minimum in the inverse kernel class of \(\rho \). \(\square \)
Our next objective is to show how a left congruence is characterised by its trace and inverse kernel.
Definition 3.6
Let \(\tau \) be a congruence on E, and let \(T\subseteq S\) be a full inverse subsemigroup. We say that \((\tau ,T)\) is an inverse congruence pair for S if \((\tau ,T)\) satisfies the following conditions:
-
(D1)
\(T\subseteq N(\tau );\)
-
(D2)
for \(x\in S\), if there exist \(e,f\in E\) such that \(x^{-1}x\ \tau \ e,\ xx^{-1}\ \tau \ f\) and \(xe,fx\in T\), then we have \(x\in T\).
For an inverse congruence pair \((\tau ,T),\) define the relation
Proposition 3.7
If \((\tau ,T)\) is an inverse congruence pair for S, then \(\rho _{(\tau ,T)}\) is a left congruence on S such that \({{\,\textrm{Iker}\,}}(\rho _{(\tau ,T)})=T\) and \({{\,\textrm{trace}\,}}(\rho _{(\tau ,T)})=\tau \).
Proof
Let \(\rho =\rho _{(\tau ,T)}\). First we show that \(\rho \) is a left congruence. It is immediate that \(\rho \) is reflexive and symmetric. We next show left compatibility. Suppose that \(a\ \rho \ b\), so \(a^{-1}b\in T\), \(a^{-1}bb^{-1}a\ \tau \ a^{-1}a\) and \(b^{-1}aa^{-1}b\ \tau \ b^{-1}b\). Since T is a full inverse subsemigroup we then have
We also note that
Thus, as \(a^{-1}bb^{-1}a\ \tau \ a^{-1}a,\)
and similarly we obtain \((cb)^{-1}(ca)(ca)^{-1}(cb)\ \tau \ (cb)^{-1}(cb)\). Thus \(ca\ \rho \ cb\) and we have shown that \(\rho \) is left compatible.
We now show that \(\rho \) is transitive, to which end suppose that \(a\ \rho \ b\) and \(b\ \rho \ c\). Thus \(a^{-1}b,\ b^{-1}c\in T,\) and \(a^{-1}bb^{-1}a\ \tau \ a^{-1}a,\) \(b^{-1}aa^{-1}b\ \tau \ b^{-1}b,\) \(b^{-1}cc^{-1}b\ \tau \ b^{-1}b\) and \(c^{-1}bb^{-1}c\ \tau \ c^{-1}c\). We need to show that \(c^{-1}a\in T,\) and that \(c^{-1}aa^{-1}c\ \tau \ c^{-1}c\) and \( a^{-1}cc^{-1}a\ \tau \ a^{-1}a\). For the latter claim note that as \(\tau \) is a congruence,
Also, as \(T\subseteq N\) and \(a^{-1}b\in T,\) we conjugate \(b^{-1}cc^{-1}b\ \tau \ b^{-1}b\) by \(a^{-1}b\) to obtain
We now have that
and the dual argument gives that \(c^{-1}aa^{-1}c\ \tau \ c^{-1}c\).
To show \(\rho \) is a left congruence it remains to show that \(c^{-1}a\in T\). As \(\rho \) is left compatible we have that \(a^{-1}b\ \rho \ a^{-1}c\) and \(c^{-1}a\ \rho \ c^{-1}b\) and hence \(c^{-1}aa^{-1}b \in T\) and \(b^{-1}cc^{-1}a\in T\). We also have \(b^{-1}a,\ c^{-1}b\in T,\) and since T is a subsemigroup, \((c^{-1}aa^{-1}b)(b^{-1}a),\ (c^{-1}b)(b^{-1}cc^{-1}a)\in T\). Also,
and
Thus by (D2) with \(x=c^{-1}a,\) \(e=a^{-1}bb^{-1}a,\) and \(f=c^{-1}bb^{-1}c\) we have that \(c^{-1}a\in T\). Hence \(\rho \) is transitive and thus a left congruence on S.
Finally we show that \({{\,\textrm{trace}\,}}\rho =\tau \) and \({{\,\textrm{Iker}\,}}\rho =T\). Suppose that we have \(e,f\in E\) with \(e\ \rho \ f\). Then, since \(e^{-1}=e,\) and \(f^{-1}=f,\) we have \(e\ \tau \ ef\ \tau \ f\) and so \(e\ \tau \ f\). Conversely, if \(e\ \tau \ f\) then it is immediate that \(e\ \rho \ f,\) so \({{\,\textrm{trace}\,}}\rho =\tau \). To see that the inverse kernel is T we note that if \(a\in T\) then \(a\ \rho \ aa^{-1},\) so \(T\subseteq {{\,\textrm{Iker}\,}}\rho \). Conversely if \(a\ \rho \ aa^{-1},\) then \(a^{-1}(aa^{-1})=a^{-1}\in T\) and since T is inverse we get \(a\in T,\) thus \({{\,\textrm{Iker}\,}}\rho =T\). \(\square \)
Shortly we shall see that every left congruence is of the form \(\rho _{(\tau ,T)}\) for an inverse congruence pair \((\tau ,T)\). However, it is beneficial to first consider how it is possible to recover a left congruence from the minimum left congruence with the same trace and the minimum left congruence with the same inverse kernel.
Theorem 3.8
For every left congruence \(\rho \) we have
Moreover for any \(a\in {{\,\textrm{ker}\,}}\rho \) there is some \(f\in E\) such that \(f\ \chi _{{{\,\textrm{Iker}\,}}\rho }\ fa\ \nu _{{{\,\textrm{trace}\,}}\rho }\ a\).
Proof
We write \(\nu \) for \(\nu _{{{\,\textrm{trace}\,}}\rho }\) and \(\chi \) for \(\chi _{{{\,\textrm{Iker}\,}}\rho }\) and recall from Proposition 3.5 that \(\chi =\rho \cap \mathcal {R}\). Certainly \(\nu ,\chi \subseteq \rho \), so \(\nu \vee \chi \subseteq \rho \). As \(\nu \subseteq \nu \vee \chi \subseteq \rho \) and using that the trace map is order preserving we have that \({{\,\textrm{trace}\,}}(\nu \vee \chi )={{\,\textrm{trace}\,}}\rho \). Also, as the kernel map is order preserving we have that \({{\,\textrm{ker}\,}}(\nu \vee \chi )\subseteq {{\,\textrm{ker}\,}}\rho \). To complete the proof it therefore suffices to prove the final claim of the theorem, whence it is immediate that \({{\,\textrm{ker}\,}}(\nu \vee \chi )\supseteq {{\,\textrm{ker}\,}}\rho \) and thus that \({{\,\textrm{ker}\,}}\rho ={{\,\textrm{ker}\,}}(\nu \vee \chi )\) so, by Theorem 2.2, \(\rho =\nu \vee \chi \).
To this end suppose \(a\in {{\,\textrm{ker}\,}}\rho \), let e be an idempotent in the \(\rho \)-class of a and let \(f=eaa^{-1}\). As \(e\ \rho \ a\) we have
and as \(\rho \) is a left congruence this implies \(fa\ \rho \ f\). We also note that \((fa)(fa)^{-1}=faa^{-1}f=faa^{-1}=f\), so \(fa\ \mathcal {R}\ f\) and therefore \(f\ \chi \ fa\). As \(fa\ \rho \ a\) we have that \(a^{-1}fa\ \rho \ a^{-1}a,\) so \(a^{-1}fa\ {{\,\textrm{trace}\,}}\rho \ a^{-1}a\). Thus \(a^{-1}fa\ \nu \ a^{-1}a,\) and as \(\nu \) is a left congruence we have \(fa\ \nu \ a,\) completing the proof. \(\square \)
Theorem 3.8 is reminiscent of [4, Theorem 3.8] which states that a two-sided congruence is the join of the minimum congruences with the same trace and the same kernel, as well as the meet of the maximum congruences with the same trace and the same kernel. While Theorem 3.8 states that a left congruence is the join of the minimum left congruences with the same trace and inverse kernel we shall see that in general there is no maximum left congruence with a given inverse kernel.
We now complete the fundamental result in the inverse kernel approach to left congruences, that a left congruence is uniquely determined by its trace and inverse kernel.
Theorem 3.9
If \((\tau ,T)\) is an inverse congruence pair for S, then \(\rho _{(\tau ,T)}\) is a left congruence on S with trace \(\tau \) and inverse kernel T. Conversely, if \(\rho \) is a left congruence on S then \(({{\,\textrm{trace}\,}}\rho ,{{\,\textrm{Iker}\,}}\rho )\) is an inverse congruence pair for S and \(\rho =\rho _{({{\,\textrm{trace}\,}}\rho ,{{\,\textrm{Iker}\,}}\rho )}\).
Proof
The first statement is precisely that of Proposition 3.7, so we want to prove the second statement. To this end suppose that \(\rho \) is a left congruence and let \(\tau ={{\,\textrm{trace}\,}}\rho \) and \(T={{\,\textrm{Iker}\,}}\rho \). By Proposition 3.2(iv) we have \(T=N(\tau )\cap {{\,\textrm{ker}\,}}\rho \), so \(T\subseteq N(\tau )\). Thus to show that \((\tau ,T)\) is an inverse congruence pair we need to verify that (D2) holds.
Suppose that \(a\in S\) and that there are \(e,f\in E\) with \(ae,fa\in T\) and \(a^{-1}a\ \tau \ e,\ aa^{-1}\ \tau \ f\). Since \(fa\in T\), we may conjugate \(a^{-1}a\ \tau \ e\) by fa to obtain that
As \(aa^{-1}\ \tau \ f\) we have \(aa^{-1}\ \tau \ aa^{-1}f\) and as \(a^{-1}a\ \tau \ e\) we have \(ae\ \rho \ aa^{-1}a = a\). As \(ae\in T\) we know that \(ae\ \rho \ (ae)(ae)^{-1} = aea^{-1}\). We then observe that
Hence \(a\in {{\,\textrm{Iker}\,}}\rho =T,\) so (D2) is satisfied and \((\tau ,T)\) is an inverse congruence pair.
It remains to show that \(\rho =\rho _{(\tau ,T)}\). We know that \({{\,\textrm{trace}\,}}(\rho _{(\tau ,T)})=\tau ={{\,\textrm{trace}\,}}\rho \). By Theorem 3.8 we have that \(\rho =\chi _{{{\,\textrm{Iker}\,}}\rho }\vee \nu _\tau ,\) and also \(\rho _{(\tau ,T)}=\chi _{{{\,\textrm{Iker}\,}}(\rho _{(\tau ,T)})}\vee \nu _\tau \). However \({{\,\textrm{Iker}\,}}\rho =T={{\,\textrm{Iker}\,}}(\rho _{(\tau ,T)})\) and since idempotent separating congruences are uniquely determined by their inverse kernel \(\chi _{{{\,\textrm{Iker}\,}}\rho }=\chi _{{{\,\textrm{Iker}\,}}(\rho _{(\tau ,T)})}\). Hence \(\rho =\rho _{(\tau ,T)}\). \(\square \)
We have shown that left congruences on inverse semigroups are determined by their trace and inverse kernel, and thus we may realise the lattice of left congruences as a subset of \(\mathfrak {C}(E)\times \mathfrak {V}(S)\). We denote the set of inverse congruence pairs by \(\mathfrak {ICP}(S)\). As in the case of the kernel trace description the ordering of left congruences coincides with the natural ordering in the lattice \(\mathfrak {C}(E)\times \mathfrak {V}(S)\).
Corollary 3.10
If \(\rho _1\) and \(\rho _2\) are left congruences on S, then
Consequently,
The following is an important corollary, and is the primary method with which the idea of the inverse kernel characterisation of left congruences will be applied in the rest of the paper.
Corollary 3.11
Let \(\rho \) be a left congruence on S. If \(T={{\,\textrm{Iker}\,}}\rho \) and \(\tau ={{\,\textrm{trace}\,}}\rho \) then \((\tau ,T)\) is the unique element in \(\mathfrak {C}(E)\times \mathfrak {V}(S)\) such that \((\tau ,T)\) is an inverse congruence pair, and
Now that it has been established that left congruences may be described in terms of their trace and inverse kernel it is a good opportunity to reflect on how the inverse kernel approach slots into the mathematical framework which has been built to describe one and two-sided congruences on inverse semigroups.
First we comment on the relationship between the kernel and the inverse kernel. From Proposition 3.2 it follows that if two left congruences have the same kernel then they have the same inverse kernel. Thus the inverse kernel classes are unions of the kernel classes. In the other direction, starting from the inverse kernel of a left congruence it is not possible to construct the kernel. However, by Theorem 3.8, we see that for a left congruence \(\rho \) on S with \({{\,\textrm{trace}\,}}\rho =\tau \) the kernel is
Further, with regard to one-sided congruences, it is worth mentioning the connection between the inverse kernel approach and the natural isomorphism between the lattices of left and right congruences. As previously mentioned this isomorphism is
All results and discussion thus far have analogues for right congruences, in particular the inverse kernel of a right congruence \(\rho \) is defined as \({{\,\textrm{Iker}\,}}\rho =\{a\in S\mid a\ \rho \ a^{-1}a\}\). We remark that by Proposition 3.2(iii) the inverse kernel of a left congruence \(\rho \) is equal to \(\{a\in S\mid \exists e,f\in E,\ a\ \rho \ e,\ a^{-1}\ \rho \ f\}\). The right-sided analogue of Proposition 3.2(iii) states that this is also an expression for the inverse kernel of a right congruence. In this fashion it is possible to “unify” the definitions of inverse kernel for a left congruence and a right congruence.
It is straightforward to see that \({{\,\textrm{trace}\,}}\rho ={{\,\textrm{trace}\,}}\rho _{-1}\) and noting that \({{\,\textrm{ker}\,}}\rho _{-1}=\{a\in S\mid a^{-1}\in {{\,\textrm{ker}\,}}\rho \}\) it follows from Proposition 3.2(iii) that
The next result then follows and makes clear the link between the inverse kernel approach and the isomorphism between \(\mathfrak {LC}(S)\) and \(\mathfrak {RC}(S)\).
Corollary 3.12
The pair \((\tau ,T)\) is the trace and inverse kernel of a left congruence if and only if it is the trace and inverse kernel of a right congruence. Moreover if \(\rho \) is a left congruence then the right congruence with the same trace and inverse kernel is \(\rho _{-1}\).
The kernel trace description for two-sided congruences on inverse semigroups is well known (see, for instance, [5, Chapter 5]), and states that \(\mathfrak {C}(S)\) is isomorphic to the lattice of congruence pairs. We say that a subsemigroup \(T\subseteq S\) is self conjugate if \(aTa^{-1}\subseteq T\) for all \(a\in S\).
Definition 3.13
(Congruence Pair [4]) Let T be a self conjugate full inverse subsemigroup of S and let \(\tau \) be a congruence on E with \(N(\tau )=S\). Then \((\tau ,T)\) is a congruence pair if it satisfies
-
(P1)
for \(x\in S\) and \(e\in E\) if \(xe\in T\) and \(e\ \tau \ x^{-1}x\) then \(x\in T;\)
-
(P2)
for each \(x\in T\) we have \(xx^{-1}\ \tau \ x^{-1}x\).
The two-sided kernel trace approach may be deduced from the inverse kernel approach to one-sided congruences via the following elementary proposition, the proof of which is a straightforward verification exercise.
Proposition 3.14
Let \((\tau ,T)\) be an inverse congruence pair and let \(\rho ,\rho _{-1}\) be the corresponding left and right congruences respectively. Then \((\tau ,T)\) is a congruence pair if and only if \(\rho =\rho _{-1}\).
The kernel trace description for two-sided congruences on inverse semigroups then follows noting that a left congruence \(\rho \) is a two-sided congruence precisely when \(\rho =\rho _{-1}\).
Theorem 3.15
[4] Let \((\tau ,T)\) be a congruence pair for S, and define
Then \(P_{(\tau ,T)}\) is a congruence on S with trace \(\tau \) and kernel T. Moreover if \(\rho \) is a congruence on S then \(({{\,\textrm{trace}\,}}\rho ,{{\,\textrm{ker}\,}}\rho )\) is a congruence pair for S and \(\rho =P_{({{\,\textrm{trace}\,}}\rho ,{{\,\textrm{ker}\,}}\rho )}\).
4 Trace classes
Motivated by results describing the lattice of idempotent separating left congruences (for example Theorem 2.6) we describe the trace class for an arbitrary trace, including a description of the maximum and minimum elements in each trace class in terms of the inverse kernel and trace.
We define the centre of a congruence \(\tau \) on E as
The centre of \(\tau \) is a full inverse subsemigroup of S. It is straightforward to show this directly, or alternatively, after applying Theorem 2.5 to obtain that \({{\,\textrm{ker}\,}}\nu _\tau = \{a\in S\mid \exists e\in E,\ e\ \tau \ a^{-1}a,\ ae=e\}\), it is clear that \(C(\tau )=N(\tau )\cap {{\,\textrm{ker}\,}}\nu _\tau ={{\,\textrm{Iker}\,}}\nu _\tau \), from which it follows that \(C(\tau )\) is a full inverse subsemigroup.
We can also observe that \(C(\tau )\) is self conjugate in \(N(\tau )\). Indeed, suppose \(a\in C(\tau )\) and \(b\in N(\tau ),\) so there is \(e\in E\) with \(e\ \tau \ a^{-1}a\) and \(ae=e\). Then \(beb^{-1}\in E\) and
Further, it then follows that \(beb^{-1}=(bab^{-1})^{-1}(bab^{-1})(beb^{-1})\) (in an inverse semigroup S, if \(a\in S,\) \(e\in E\) and \(ae=e\) then \(a^{-1}ae=e\)), and as \(b\in N(\tau )\) we may conjugate \(a^{-1}a\ \tau \ e\) by b to obtain \(ba^{-1}ab^{-1}\ \tau \ beb^{-1}\). Thus
Hence \(bab^{-1}\in C(\tau )\) and thus \(C(\tau )\) is a self conjugate full inverse subsemigroup of \(N(\tau )\).
Proposition 4.1
([11, Proposition 6.4]) If \(\tau \) is a congruence on E then \(\nu _\tau |_{N(\tau )}\) is a two-sided congruence on \(N(\tau )\), and
Moreover, \(\nu _\tau |_{N(\tau )}\) is the minimum congruence on \(N(\tau )\) with trace \(\tau \).
Proof
This can be deduced from [11, Proposition 6.4] and the usual kernel trace description of a two-sided congruence on an inverse semigroup. However, as we shall rely on this result it is beneficial to include a direct proof. We write N for \(N(\tau )\) and \(\psi \) for the expression given on the right-hand side of the expression in the statement. Our objective is to prove that \(\langle \tau \rangle \cap (N\times N)=\psi ,\) with \(\langle \tau \rangle \) the left congruence on S generated by \(\tau \).
Suppose \((a,b)\in \psi ,\) so \(a^{-1}a\ \tau \ b^{-1}b\) and, as \(ab^{-1}\in C(\tau )\), there is \(e\in E\) such that \(ba^{-1}ab^{-1}\ \tau \ e\) and \(ab^{-1}e=e\). As \(b\in N\) we may conjugate \(ba^{-1}ab^{-1}\ \tau \ e\) by b, so we have
We then observe that
Thus \((a,b)\in \langle \tau \rangle ,\) so \(\psi \subseteq \langle \tau \rangle \).
Next we show that \(\langle \tau \rangle \cap (N\times N)\subseteq \psi \). We shall show that if \(a,b\in N\) and there is a \(\tau \)-sequence from a to b then \((a,b)\in \psi \). To this end suppose that \(a\ \langle \tau \rangle \ b\) so there is a \(\tau \)-sequence from a to b so we have \(c_1,\ldots ,c_n\in S\) and \(\{(e_1,f_1),\ldots ,(e_n,f_n)\}\subseteq \tau \) such that
We note that then \(a^{-1}a=e_1c_1^{-1}c_1,\) \(f_ic_i^{-1}c_i=e_{i+1}c_{i+1}^{-1}c_{i+1}\) for \(1\le i\le n-1\) and \(f_nc_n^{-1}c_n=b^{-1}b\). For each i, as \(e_i\ \tau \ f_i\) we have that \(e_ic_i^{-1}c_i\ \tau \ f_ic_i^{-1}c_i,\) thus
Therefore \(a^{-1}a\ \tau \ b^{-1}b\).
To show \(ab^{-1}\in C(\tau )\) we will proceed by induction on the length of the \(\tau \)-sequence. Suppose that \((c_ie_i)a^{-1}\in C(\tau ),\) so there is \(g\in E\) with \(ae_ic_i^{-1}c_ia^{-1}\ \tau \ g\) and \(c_ie_ia^{-1}g=g\). We shall show that \(a(c_if_i)^{-1}\in C(\tau ),\) whence \(a(c_{i+1}e_{i+1})^{-1}\in C(\tau ),\) from which it follows by induction that \((c_nf_n)a^{-1}=ba^{-1}\in C(\tau ),\) and as \(C(\tau )\) is inverse we then have that \(ab^{-1}\in C(\tau )\). Using that \(e_i\ \tau \ f_i\) we have that \(f_ic_i^{-1}c_i\ \tau \ e_if_ic_i^{-1}c_i\). Since \(\tau \) is normal in N we may conjugate this relation by a to obtain
We then observe that
Also,
Thus we have that \((c_if_i)a^{-1}\in C(\tau )\) completing the induction step, so we have shown that \(ab^{-1}\in C(\tau )\). Thus \(\langle \tau \rangle \cap (N\times N)=\psi ,\) so \(\psi \) is a left congruence on N.
To complete the proof it suffices to show that \(\psi \) is also a right congruence on N. Suppose that \(a\ \psi \ b,\) so \(a^{-1}a\ \tau \ b^{-1}b\) and \(ab^{-1}\in C(\tau )\) and let \(c\in N\). Then we conjugate the relation \(a^{-1}a\ \tau \ b^{-1}b\) by c to obtain
Since \(C(\tau )\) is a full subsemigroup and \(ab^{-1}\in C(\tau )\) we have that
Thus \(ac\ \psi \ bc\) and so \(\psi \) is a right congruence. \(\square \)
It is worth noting that in the statement and the proof of Proposition 4.1 no prior knowledge about \(\nu _\tau \) is assumed. In fact, it is straightforward to deduce directly from Proposition 4.1 that \(C(\tau )={{\,\textrm{Iker}\,}}\nu _\tau \). Descriptions of the maximum and minimum elements in a trace class are available in both [8] and [11]. Using the previous observation and the fact that \((\tau ,N(\tau ))\) is an inverse congruence pair (this follows immediately from the definition) we present the analogous results in terms of the inverse kernel approach.
Corollary 4.2
Let \(\tau \) be a congruence on E. The minimum and maximum left congruences with trace \(\tau \) are respectively
and
We now use the description of \(\nu _\tau \) on \(N(\tau )\) to proceed with an alternate characterisation of what it means for a full inverse subsemigroup to form an inverse congruence pair with a given congruence on the idempotents.
Proposition 4.3
Let \(\tau \) be a congruence on E and let \(T\subseteq S\) be a full inverse subsemigroup. Then \((\tau ,T)\) is an inverse congruence pair if and only if T is a union of \(\nu _\tau |_{N(\tau )}\)-classes.
Proof
Let \(\psi =\nu _\tau |_{N(\tau )}\) and first suppose that \((\tau ,T)\) is an inverse congruence pair. We want to show that T is a union of \(\psi \)-classes, to which end we suppose that \(a\in N(\tau )\) and \(a\ \psi \ b\) for some \(b\in T\). From the description of \(\psi \) in Proposition 4.1 we have \(a^{-1}a\ \tau \ b^{-1}b\) and \(ab^{-1}\in C(\tau )\). Since \(a\in N(\tau )\) we can conjugate \(a^{-1}a\ \tau \ b^{-1}b\) by a to obtain \(aa^{-1}\ \tau \ ab^{-1}ba^{-1}\). Then letting \(e=b^{-1}b,\ f=ab^{-1}ba^{-1}\) we have \(ae=ab^{-1}b=fa\). From Corollary 4.2, we know that \(\rho _{(\tau ,C(\tau ))}\) is the minimum left congruence with trace \(\tau ,\) so \(\rho _{(\tau ,C(\tau ))}\subseteq \rho _{(\tau ,T)}\). As the inverse kernel map is order preserving this implies that \(C(\tau )\subseteq T\). We then have that \(b,ab^{-1}\in T,\) so \(ab^{-1}b\in T,\) and by (D2) we have \(a\in T\). Thus T is a union of \(\psi \)-classes.
Conversely suppose that T is a union of \(\psi \)-classes. Then certainly \(T\subseteq N(\tau )\), so we need to verify that (D2) holds. Suppose that \(a\in S\) and there exist \(e,f\in E\) such that \(a^{-1}a\ \tau \ e,\ aa^{-1}\ \tau \ f,\) and \(ae,fa\in T\). We first show that \( a\in N(\tau )\). Suppose that \(g,h\in E\) with \(g\ \tau \ h\). Then as \(fa\in N(\tau )\) we conjugate \(g\ \tau \ h\) by fa to get \(faga^{-1}\ \tau \ faha^{-1}\). Since \(aa^{-1}\ \tau \ f\) we then have
Conjugating \(g\ \tau \ h\) by \((ae)^{-1}\), and using that \(a^{-1}a\ \tau \ e\), a similar argument gives that \(a^{-1}ga\ \tau \ a^{-1}ha,\) hence \(a\in N(\tau )\). To see that in fact \(a\in T\) we note that \(a^{-1}a\ \tau \ e\) implies that \(a\ \nu _\tau \ ae\) since \({{\,\textrm{trace}\,}}\nu _\tau =\tau \). As \(ae\in T\) and T is saturated by \(\psi \) it follows that \(a\in T\). \(\square \)
We can now formulate the main result of this section, namely an extension of Theorem 2.5 to an arbitrary trace class.
Theorem 4.4
For every congruence \(\tau \) on E the lattice of left congruences on S with trace \(\tau \) is isomorphic to the lattice of full inverse subsemigroups of \(N(\tau )/\nu _\tau |_{N(\tau )}\).
Proof
We know that the set of full inverse subsemigroups of \(N(\tau )\) that form an inverse congruence pair with \(\tau \) consists precisely of those saturated by \(\nu _\tau |_{N(\tau )}\). Therefore to complete the proof it suffices to note that by standard universal arguments this set is exactly the preimage of the set of full inverse subsemigroups of \(N(\tau )/\nu _\tau |_{N(\tau )}\) under the natural homomorphism \(N(\tau )\mapsto N(\tau )/\nu _\tau |_{N(\tau )}\). \(\square \)
5 Inverse kernel classes
As left congruences are determined by their trace and inverse kernel it is reasonable to seek analogues for inverse kernel classes of results concerning trace classes. We have previously seen (Proposition 3.5) that given a full inverse subsemigroup T there is an idempotent separating left congruence such that T is the inverse kernel and that this is the minimum element in the inverse kernel class. In this section we shall see that in general there is no maximum element in an inverse kernel class. We shall present a description of those elements of \(\mathfrak {C}(E)\) that form inverse congruence pairs with a given inverse subsemigroup in the case when E has the descending chain condition.
For illustrative purposes we describe a simple example of the use of the inverse kernel approach for the lattice of left congruences. We consider \(\mathcal {I}_2\), the symmetric inverse monoid on a 2-element set. We label the elements of \(\mathcal {I}_2\) by \(I_{1,2},I_1,I_2,I_\emptyset ,\alpha ,\beta ,\beta ^{-1}\) where \(I_X\) is the idempotent with domain \(X\subseteq \{1,2\},\) \(\alpha \) is the non-identity invertible element, and \(\beta \) has domain \(\{1\}\) and image \(\{2\}\). The semilattice of idempotents is isomorphic to the powerset of a 2-element set under intersection. The lattice of congruences on the idempotents is illustrated in Fig. 1, in which the partitions of the idempotents are shown. The semigroup \(\mathcal {I}_2\) has 3 distinct full inverse subsemigroups: \(E=E(\mathcal {I}_2),\) \(\mathcal {I}_2\) and \(T=E\cup \{\beta ,\beta ^{-1}\}\). The lattice of full inverse subsemigroups is displayed in Fig. 2.
The lattice of left congruences is then realised as a subset of the direct product of these two lattices. After elementary calculations to determine which elements of the direct product are inverse congruence pairs we obtain the lattice of left congruences as shown in Figs. 3 and 4, which both show the direct product \(\mathfrak {C}(E)\times \mathfrak {V}(\mathcal {I}_2)\) with the inverse congruence pairs indicated by circled vertices. Figure 3 shows the left congruences grouped into trace classes, and Fig. 4 shows them grouped by inverse kernel. It is then easy to observe that the inverse kernel class of E contains no maximum elements, indeed the labelled inverse congruence pairs \(\rho _{(\tau _3,E)}\) and \(\rho _{(\tau _5,E)}\) both have inverse kernel E, however the join of these left congruences is \(\omega ,\) which has inverse kernel \(\mathcal {I}_2\).
\(\mathfrak {C}(E(\mathcal {I}_2))\)
\(\mathfrak {V}(\mathcal {I}_2)\)
\(\mathfrak {LC}(\mathcal {I}_2)\subseteq \mathfrak {C}(E)\times \mathfrak {V}(S)\)
\(\mathfrak {LC}(\mathcal {I}_2)\subseteq \mathfrak {V}(S)\times \mathfrak {C}(E)\)
We now turn to describing the set of congruences on E that are the traces of an inverse kernel class. For a congruence \(\tau \) on E and a subset \(T\subseteq S\) the \(\tau \)-closure of T is defined as
This is similar to the well-known definition of the closure of a subset T of an inverse semigroup, which is \(\{s\in S\mid \exists t\in T, t\le s\}\). The closure is usually written \(T\omega \) and coincides with the \(\omega \)-closure for our definition of closure. It is possible to describe traces of left congruences \(\tau \) with a given inverse kernel T in terms of the \(\tau \)-closure. The following is a straightforward consequence of Theorem 3.9, as (D2) is equivalent to T being equal to \(T\tau \).
Corollary 5.1
Let T be a full inverse subsemigroup of S. A congruence \(\tau \) on E is the trace of a left congruence with inverse kernel T if and only if \(\tau \) is normal in T and \(T=T\tau \).
To obtain a more precise description of those \(\tau \in \mathfrak {C}(E)\) which are the trace of a left congruence with a given inverse kernel we restrict attention to inverse semigroups which have some additional structure.
A partial order is said to have the descending chain condition if it contains no infinite descending chains. Any partially ordered set with the descending chain condition has minimal elements; if a meet-semilattice has the descending chain condition then we note that it contains a minimum element. Let \(\tau \) be a congruence on a semilattice E. Since a congruence class is a subsemilattice we note that if a semilattice has the descending chain condition then each \(\tau \)-class has a minimum element. We observe that when E has the descending chain condition then the usual partial order on S (we recall that this is defined by \(a \le b\) when there is \(e\in E\) such that \(a=be\)) also has the descending chain condition, in this case we say that S has the descending chain condition.
Proposition 5.2
Let S be an inverse semigroup with the descending chain condition, and let \(T\subseteq S\) be a full inverse subsemigroup. If \(\tau \) is a congruence on E which is normal in T, then \((\tau ,T)\) is an inverse congruence pair if and only if for each minimal element \(a\in S\backslash T\) at least one of \(aa^{-1}\) and \(a^{-1}a\) is the minimum in its \(\tau \)-class.
Proof
Suppose that \(a\in S\backslash T\) is minimal, then it is clear that \(a^{-1}\) is also minimal. We note that if \(e< a^{-1}a\) then \(ae< a\). We assume initially that \((\tau ,T)\) is an inverse congruence pair, and that both \(a^{-1}a,\ aa^{-1}\) are not minimum in their \(\tau \)-class. So there are \(e< a^{-1}a,\ f< aa^{-1},\) with \(e\ \tau \ a^{-1}a,\ f\ \tau \ aa^{-1}\). Then as \(a,a^{-1}\) are minimal we get \(ae,a^{-1}f\in T\). But since \((\tau ,T)\) is an inverse congruence pair (D2) gives that \(a\in T\), which is a contradiction. Thus at least one of \(aa^{-1}\) and \(a^{-1}a\) is minimum in its congruence class.
For the converse suppose that for any minimal \(x\in S\backslash T\) at least one of \(xx^{-1}, x^{-1}x\) is minimum in its congruence class. We need to verify (D2). Let \(a\in S\) and suppose there are \(e,f\in E\) such that \(e\ \tau \ a^{-1}a,\ f\ \tau \ aa^{-1}\) and \(ae,fa\in T\). Suppose that \(a\in S\backslash T\). Then since S satisfies the descending chain condition we have that there is some \(h\in E\) such that \(b=ah\) and b is minimal in \(S\backslash T\). By assumption at least one of \(bb^{-1}\) and \(b^{-1}b\) is minimum in its \(\tau \)-class. If \(b^{-1}b\) is a minimum, then \(b^{-1}b=b^{-1}ba^{-1}a\ \tau \ b^{-1}be\) and the minimality of \(b^{-1}b\) implies that \(b^{-1}b=b^{-1}be\) so we have that \(b=be\). Then \(b=be=aeh\in T,\) a contradiction. Similarly, \(bb^{-1}\) cannot be minimum it its \(\tau \)-class. It follows that \(a\in T\) and (D2) holds. \(\square \)
Proposition 5.2 suggests the potential for computational application of the inverse kernel approach to describing left congruences. If the lattices \(\mathfrak {V}(S)\) and \(\mathfrak {C}(E)\) are known — both of which are much easier computational problems than computing \(\mathfrak {LC}(S)\) — and we also know \(N(\tau )\) for each \(\tau \in \mathfrak {C}(E),\) then Proposition 5.2 provides a mechanism to calculate \(\mathfrak {LC}(S)\) using the partial order structure as opposed to the multiplicative structure of S.
6 The lattice of left congruences
We now consider the lattice of left congruences on S from the perspective of the inverse kernel approach, regarding \(\mathfrak {LC}(S)\) as a subset of \(\mathfrak {C}(E)\times \mathfrak {V}(S)\). We shall describe the meets and joins of left congruences in terms of the trace and inverse kernel.
Recall that given \(\tau \in \mathfrak {C}(E)\) the left congruence \(\nu _\tau \) is generated by \(\tau \) regarded as a relation on S. It follows that given traces \(\tau _1,\tau _2\) we have that \(\nu _{\tau _1}\vee \nu _{\tau _2}=\langle \tau _1\vee \tau _2\rangle \).
Corollary 6.1
The map \(\tau \mapsto \nu _\tau \) is a \(\vee \)-semilattice embedding \(\mathfrak {C}(E)\hookrightarrow \mathfrak {LC}(S)\).
The image of this map is \(\{\nu _\tau \mid \tau \in \mathfrak {C}(E)\},\) and we refer to this set as the set of trace minimal left congruences.
Our first objective in this section is to describe the meet of left congruences via the inverse kernel approach. It is shown in [11] that the trace map is a complete surjective \(\cap \)-homomorphism. Combining this fact with Proposition 3.5 we have the following.
Corollary 6.2
The trace and kernel maps are complete surjective \(\cap \)-homomorphisms. Furthermore, regarded as a \(\cap \)-semilattice, \(\mathfrak {ICP}(S)\) is a complete \(\cap \)-subsemilattice of the direct product \(\mathfrak {C}(E)\times \mathfrak {V}(S)\).
In particular, if \(\{\rho _i\mid i\in I\}\) is a family of left congruences then
On the other hand it is a nontrivial question to determine the join of two left congruences on S. We now show that the inverse kernel approach provides a mechanism to handle this problem smoothly. We begin by making a straightforward observation about how the normalisers of congruences on E are related to normalisers of the joins and meets of these congruences.
Lemma 6.3
Let \(\{\tau _i\mid i\in I\}\) be a set of congruences on E with normalisers \(N_i=N(\tau _i)\) respectively. Then
Proof
The first part is straightforward: suppose \(a\in \bigcap _{i\in I} N_i\) and \( e\ (\bigcap _{i\in I} \tau _i)\ f\). Then we have \(aea^{-1}\ \tau _i\ afa^{-1}\) for each \(i\in I,\) so \(aea^{-1}\ (\bigcap _{i\in I} \tau _i)\ afa^{-1}\). Similarly we obtain \(a^{-1}ea\ (\bigcap _{i\in I} \tau _i)\ a^{-1}fa,\) so \(a\in N(\bigcap _{i\in I} \tau _i)\).
For the second claim we suppose that \(e\ (\bigvee _{i\in I} \tau _i)\ f\) and \(a\in \bigcap _{i\in I} N_i\). As \((\bigvee _{i\in I} \tau _i)\) is the transitive closure of the union of \(\{\tau _i\mid i\in I\}\) there are elements \(g_1,\ldots ,g_m\in E\) and \(i_1,\ldots ,i_{m+1}\in I\) such that
At each stage in this sequence we can conjugate by a and thus obtain that
By symmetry in a and \(a^{-1}\) we obtain that \(a\in N(\bigvee _{i\in I} \tau _i)\). \(\square \)
We note that these may both be strict inclusions, if we consider again the symmetric inverse monoid \(\mathcal {I}_2\) and consider \(\tau _3\) and \(\tau _5\) in Fig. 1 then we see that \(N(\tau _3)=E=N(\tau _5),\) but \(N(\tau _3\cap \tau _5)=N(\iota )=\mathcal {I}_2\) and also \(N(\tau _3\vee \tau _5)=N(\omega )=\mathcal {I}_2\).
Lemma 6.4
Let \(\tau \) be a congruence on E and let \(T\subseteq N(\tau )\) be a full inverse subsemigroup. Then \(V=\bigcup _{t\in T} [t]_{\nu _\tau |_{N(\tau )}}\) is a full inverse subsemigroup and \((\tau ,V)\) is an inverse congruence pair. Furthermore, \(\rho _{(\tau ,V)}=\nu _\tau \vee \chi _T\).
Proof
Let \(\psi =\nu _\tau |_{N(\tau )}\). We first show that V is a full inverse subsemigroup. Recall from Proposition 4.1 that \(\psi \) is a two-sided congruence on \(N(\tau )\). If \(a,b\in V\), there exist \(x,y\in T\) with \(a\ \psi \ x,\ b\ \psi \ y\). Then, as \(\psi \) is a two-sided congruence we obtain \(ab\ \psi \ xy,\) and since \(xy\in T\) we have that \(ab\in V\). Again as \(\psi \) is two-sided, if \(a\ \psi \ b\) then \(a^{-1}\ \psi \ b^{-1},\) so as T is inverse it follows that V is inverse. We also note that as \(E\subseteq T\subseteq V\) it is immediate that V is full. Thus V is a full inverse subsemigroup. By definition V is saturated by \(\psi ,\) so applying Proposition 4.3 we have that \((\tau ,V)\) is an inverse congruence pair.
For the final claim we first recall from Corollary 3.11 that \(\rho _{(\tau ,V)}=\nu _\tau \vee \chi _V\). As \(T\subseteq V\), by applying Theorem 2.6 we obtain that \(\chi _T\subseteq \chi _V\). Then we have
As the trace map is order preserving this implies that \({{\,\textrm{trace}\,}}(\nu _\tau \vee \chi _T)=\tau \). For the inverse kernel we note that it is immediate that \(T\subseteq {{\,\textrm{Iker}\,}}(\nu _\tau \vee \chi _T)\subseteq V\). By Proposition 4.3 we know that \({{\,\textrm{Iker}\,}}(\nu _\tau \vee \chi _T)\) is a full inverse subsemigroup of \(N(\tau )\) and is saturated by \(\psi \). By definition V is the minimum inverse subsemigroup of \(N(\tau )\) containing T which is saturated by \(\psi ,\) so we have that \(V\subseteq {{\,\textrm{Iker}\,}}(\nu _\tau \vee \chi _T)\). We have shown that \({{\,\textrm{trace}\,}}(\nu _\tau \vee \chi _T)=\tau \) and \({{\,\textrm{Iker}\,}}(\nu _\tau \vee \chi _T)=V,\) so by Corollary 3.10 we have that \(\nu _\tau \vee \chi _T=\rho _{(\tau ,V)}\). \(\square \)
Proposition 6.5
Let \((\tau _1,T_1),\ (\tau _2,T_2)\) be inverse congruence pairs and denote by \(\xi \) the least congruence on E such that \(\xi \supseteq \tau _1\vee \tau _2\) and \(N(\xi )\supseteq T_1\vee T_2\). Then
where \(V=\bigcup _{t\in T_1\vee T_2} [t]_{\nu _\xi |_{N(\xi )}}. \)
Proof
We first note that \(\xi \) is well defined. Indeed, let \(\{\xi _i\mid i\in I\}\subseteq \mathfrak {C}(E)\) be the set of all congruences on E such that \(\tau _1\vee \tau _2\subseteq \xi _i\) and \(T_1\vee T_2\subseteq N(\xi _i)\), and let \(\xi =\bigcap _{i\in I} \xi _i\). Then it is immediate that \(\tau _1\vee \tau _2\subseteq \xi ,\) and, by Lemma 6.3, \(T_1\vee T_2\subseteq \bigcap _{i\in I} N(\xi _i)\subseteq N(\xi )\). Moreover, \(\xi \) is the least congruence such that \(\tau _1\vee \tau _2\subseteq \xi \) and \(T_1\vee T_2\subseteq N(\xi )\). From Lemma 6.4 we observe that \((\xi ,V)\) is an inverse congruence pair, and by appeal to Corollary 3.10 we have that \(\rho _{(\tau _1,T_1)}\vee \rho _{(\tau _2,T_2)}\subseteq \rho _{(\xi ,V)}\).
We now show that \(\rho _{(\xi ,V)}\subseteq \rho _{(\tau _1,T_1)}\vee \rho _{(\tau _2,T_2)}\). Let \((\zeta ,W)\) be the inverse congruence pair such that \(\rho _{(\tau _1,T_1)}\vee \rho _{(\tau _2,T_2)}=\rho _{(\zeta ,W)}\). By Corollary 3.10 we have that \(T_1\vee T_2\subseteq W\subseteq V\) and \(\tau _1\vee \tau _2 \subseteq \zeta \subseteq \xi \). As \((\zeta ,W)\) is an inverse congruence pair, \(W\subseteq N(\zeta )\) so it follows that \(T_1\vee T_2\subseteq N(\zeta )\). However, by definition \(\xi \) is the least congruence on E that has these properties. Therefore \(\xi \subseteq \zeta ,\) and thus \(\xi =\zeta \). Finally we can note that \(T_1\vee T_2\subseteq W,\) and, by Proposition 4.3, W is saturated by \(\nu _\xi |_{N(\xi )}\). It is then clear that \(V\subseteq W\) so \(V=W\). We have shown that \((\xi ,V)= (\zeta ,W),\) so \(\rho _{(\xi ,V)}=\rho _{(\tau _1,T_1)}\vee \rho _{(\tau _,T_2)}\). \(\square \)
We know that we may regard \(\mathfrak {LC}(S)\) as a subset of \(\mathfrak {C}(E)\times \mathfrak {V}(S)\). We have considered the trace and inverse kernel maps, which map \(\mathfrak {LC}(S)\) onto the components of the direct product. We can combine these maps naturally to obtain the function
We recall that there is a lattice embedding \(\mathfrak {V}(S)\rightarrow \mathfrak {LC}(S),\ T\mapsto \chi _T,\) and a \(\vee \)-semilattice embedding \(\mathfrak {C}(E)\rightarrow \mathfrak {LC}(S),\ \tau \mapsto \nu _\tau \). We consider the function
Theorem 6.6
The function \(\Phi \) is a \(\cap \)-homomorphism, and \(\Theta \) is an onto \(\vee \)-homomorphism. Moreover, \(\Phi \Theta : \mathfrak {LC}(S)\rightarrow \mathfrak {LC}(S)\) is the identity map.
Proof
That \(\Phi \) is a \(\cap \)-homomorphism is immediate as the trace and inverse kernel maps are \(\cap \)-homomorphisms. Suppose \((\tau _1,T_1),(\tau _2,T_2)\in \mathfrak {C}(E)\times \mathfrak {V}(S)\). Then, utilising that the trace minimal left congruences and the idempotent separating left congruences form \(\vee \)-subsemilattices (by Corollary 6.1 and Theorem 2.6, respectively), we obtain
Thus \(\Theta \) is a \(\vee \)-homomorphism. From Corollary 3.11 we know that a left congruence \(\rho _{(\tau ,T)}\) is equal to \(\nu _\tau \vee \chi _T\). Thus it is clear both that \(\Phi \) is onto, and that the function \(\Phi \Theta \) is the identity map. \(\square \)
We remark that in general neither is \(\Phi \) a \(\vee \)-homomorphism, nor is \(\Theta \) a \(\cap \)-homomorphism. To see that \(\Phi \) does not preserve join recall that for \(\mathcal {I}_2\) there are distinct left congruences (\(\rho _{(\tau _3,E)}\) and \(\rho _{(\tau _5,E)}\) in Fig. 3) with inverse kernel equal to E and join equal to the universal congruence, which has inverse kernel \(\mathcal {I}_2\).
The example \(\mathcal {I}_2\) also suffices to show that \(\Theta \) is not a \(\cap \)-homomorphism. Indeed consider \(\tau _3,\tau _5\) from Fig. 3, then \(N(\tau _3)=E=N(\tau _5)\). We observe that \(\tau _3\cap \tau _5=\iota ,\) the trivial congruence. We also note that \((\tau _3,\mathcal {I}_2)\Theta = \omega = (\tau _5,\mathcal {I}_2)\Theta \). Thus
On the other hand
Since \(\omega \ne \mathcal {R}\) we have that \(\Theta \) is not a \(\cap \)-homomorphism.
7 Finitely generated left congruences
In this section we consider left congruences defined by a generating set. In a slight change of notation from previously we now write \(\langle R\rangle _{{{\,\textrm{LC}\,}}(S)}\) and \(\langle R\rangle _{{{\,\textrm{C}\,}}(S)}\) for, respectively, the left congruence and congruence generated by the binary relation R on S (usually we shall refer to congruences on the semilattice of idempotents and left congruences on the whole semigroup). We make this change to enable us to write \(\langle Z\rangle _{{{\,\textrm{IS}\,}}}\) for the inverse subsemigroup generated by a set \(Z\subseteq S\). We shall use this notation for the rest of the paper. We also observe that without changing whether Z or R is finite or infinite we may assume that Z is an inverse subset (i.e. Z contains all inverses of elements in Z) and that R is symmetric (i.e., if \((a,b)\in R\) then \((b,a)\in R\)).
Several properties of semigroups are related to which one-sided congruences are finitely generated, for example whether a semigroup is left Noetherian, or is left coherent. We will see that for inverse semigroups finite generation of left congruences is closely tied to finite generation of the trace and the inverse kernel. We write \(\mathfrak {LC}_{{{\,\textrm{FG}\,}}}(S)\) and \(\mathfrak {C}_{{{\,\textrm{FG}\,}}}(S)\) for the sets of finitely generated left congruences, and finitely generated congruences respectively, on S.
Initially we have a technical lemma regarding generating sets for one-sided congruences.
Lemma 7.1
Let \(H\subseteq S\times S\) be a subset and let \(\rho =\langle H\rangle _{{{\,\textrm{LC}\,}}(S)}\). Then there exists a (symmetric) set \(H^\prime \subseteq S\times S\) such that \(\rho =\langle H^\prime \rangle _{{{\,\textrm{LC}\,}}(S)}\) and
Moreover, if H is finite then \(H^\prime \) is also finite.
Proof
Suppose that \((a,b)\in H,\) so certainly \(a\ \rho \ b\). Then \(a^{-1}a\ \rho \ a^{-1}b\) and \(b^{-1}a\ \rho \ b^{-1}b\). Also, \(a^{-1}b\in {{\,\textrm{Iker}\,}}\rho \) so \(a^{-1}b\ \rho \ a^{-1}bb^{-1}a\). Consequently we have that \(a^{-1}a\ \rho \ a^{-1}bb^{-1}a\) and dually that \(b^{-1}aa^{-1}b\ \rho \ b^{-1}b\).
Conversely we note that if the three relations: \(a^{-1}b\ \rho \ a^{-1}bb^{-1}a,\ a^{-1}a\ \rho \ a^{-1}bb^{-1}a\) and \(b^{-1}aa^{-1}b\ \rho \ b^{-1}b\) hold then we also have \(a\ \rho \ b\). Indeed, from \(a^{-1}b\ \rho \ a^{-1}bb^{-1}a\) we have \(b^{-1}a(a^{-1}b)\ \rho \ b^{-1}a(a^{-1}bb^{-1}a)=b^{-1}a\). We then observe
Hence we may replace each pair \((a,b)\in H\) with the 3 pairs \((a^{-1}a,a^{-1}bb^{-1}a),\) \((b^{-1}b,b^{-1}aa^{-1}b)\) and \((a^{-1}b,a^{-1}bb^{-1}a)\) which have the required form, and the left congruence generated by this new set is the same as the original.
To complete the proof it suffices to note that when H is finite \(|H^\prime |\le 3|H|,\) so \(H^\prime \) is certainly finite. Furthermore, if necessary we make \(H^\prime \) symmetric by adding (b, a) to \(H^\prime \) for each \((a,b)\in H^\prime \) and the resulting set is still finite if H is finite. \(\square \)
Corollary 7.2
Every finitely generated left congruence \(\rho \) on S can be written as the join \(\nu _\tau \vee \chi \), where \(\tau \) is a finitely generated congruence on E and \(\chi \) is a finitely generated idempotent separating left congruence on S.
Proof
Let \(\rho \) be a left congruence on S generated by a finite set H. By Lemma 7.1 we may assume that \(H\subseteq (E\times E)\cup \{ (aa^{-1},a)\mid a\in S\} \cup \{ (a,aa^{-1})\mid a\in S\}\). Let \(\tau =\langle H\cap (E\times E)\rangle _{{{\,\textrm{C}\,}}(E)}\) and let \(\chi =\langle H\cap (\{ (aa^{-1},a)\mid a\in S\} \cup \{ (a,aa^{-1})\mid a\in S\})\rangle _{{{\,\textrm{LC}\,}}(S)}\). Then, as H is finite, \(\tau \) and \(\chi \) are finitely generated. Further, \(\chi \) is idempotent separating as the generating set is certainly contained in \(\mathcal {R}\). Finally, it is clear that \(\rho =\nu _\tau \vee \chi \). \(\square \)
Definition 7.3
A full inverse subsemigroup \(T\subseteq S\) is said to be almost finitely generated if there exists a finite set X such that \(T=\langle X\cup E \rangle _{{{\,\textrm{IS}\,}}}\). We write \(\mathfrak {V}_{{{\,\textrm{AFG}\,}}}(S)\) for the lattice of almost finitely generated full inverse subsemigroups.
The notion of almost finitely generated exactly captures which full inverse subsemigroups are inverse kernels of finitely generated idempotent separating left congruences.
Proposition 7.4
Let \(\chi \) be an idempotent separating left congruence on S and let \(T={{\,\textrm{Iker}\,}}\chi \). If \(T=\langle X\cup E\rangle _{{{\,\textrm{IS}\,}}}\) for \(X\subseteq S\) then \(\chi =\langle \{(a,aa^{-1})\mid a\in X\}\rangle _{{{\,\textrm{LC}\,}}(S)}\). Furthermore, T is almost finitely generated if and only if \(\chi \) is finitely generated.
Proof
We initially note that \({{\,\textrm{Iker}\,}}\chi =T\) precisely says that \(\chi =\chi _T,\) and we suppose that \(T=\langle X\cup E\rangle _{{{\,\textrm{IS}\,}}}\). We write R for the set \(\{(a,aa^{-1})\mid a\in X\}\).
As \(X\subseteq T\) it is immediate that \(R\subseteq \chi ,\) so we certainly have that \(\langle R\rangle _{{{\,\textrm{LC}\,}}(S)}\subseteq \chi \). For the reverse inclusion we observe that \(X\subseteq {{\,\textrm{Iker}\,}}(\langle R\rangle _{{{\,\textrm{LC}\,}}(S)})\). Since the inverse kernel is a full inverse subsemigroup it follows that \(T=\langle X\cup E\rangle _{{{\,\textrm{IS}\,}}}\subseteq {{\,\textrm{Iker}\,}}(\langle R\rangle _{{{\,\textrm{LC}\,}}(S)})\). As the lattice of idempotent separating left congruences is isomorphic to the lattice of full inverse subsemigroups (Theorem 2.6), we have that \(\chi =\chi _T\subseteq \langle R\rangle _{{{\,\textrm{LC}\,}}(S)}\). Thus we have that the two are equal.
We now show that \(\chi \) is finitely generated if and only if T is almost finitely generated. First, suppose that T is almost finitely generated, so with X as above we now suppose that X is a finite set. Then we have shown that \(\chi \) is generated by \(\{(a,aa^{-1})\mid a\in X\},\) so \(\chi \) is certainly finitely generated.
For the converse we suppose that \(\chi \) is finitely generated. By Lemma 7.1, we can choose a finite generating set Q for \(\chi \) such that \(Q=\{(p,pp^{-1})\mid p\in P\}\) for some finite set \(P\subseteq S\). We claim that \({{\,\textrm{Iker}\,}}\chi =\langle P\cup E\rangle _{{{\,\textrm{IS}\,}}}\). It is immediate that \(P\cup E\subseteq {{\,\textrm{Iker}\,}}\chi \). Let \(\zeta =\chi _{\langle P\cup E\rangle _{{{\,\textrm{IS}\,}}}},\) then it suffices to show that \(\chi \subseteq \zeta \). We note that \((p,pp^{-1})\in \zeta \) for each \(p\in P\). Thus \(Q\subseteq \zeta ,\) and hence \(\langle Q\rangle _{{{\,\textrm{LC}\,}}(S)} =\chi \subseteq \zeta \). Thus we have that \({{\,\textrm{Iker}\,}}\chi \) is almost finitely generated. \(\square \)
For ease of notation if Y is an inverse subsemigroup then we write \(\chi _Y\) for the idempotent separating left congruence with inverse kernel \(\langle Y\cup E\rangle _{{{\,\textrm{IS}\,}}}\). If Y is an inverse subsemigroup of S then \(\langle Y\cup E\rangle _{{{\,\textrm{IS}\,}}} =YE\cup E\). Indeed, this follows from the observation that if \(a\in S\) and \(e\in E\) then \(ae=(aea^{-1})a\) and \(aea^{-1}\in E,\) so \(YE=EY\).
We note that \(\mathfrak {C}_{{{\,\textrm{FG}\,}}}(E)\) and \(\mathfrak {V}_{{{\,\textrm{AFG}\,}}}(S)\) are \(\vee \)-subsemilattices of \(\mathfrak {C}(E)\) and \(\mathfrak {V}(S)\), respectively. In general neither is a \(\cap \)-subsemilattice; there are principal congruences on countable semilattices that have non-finitely generated intersection, and it is possible for finitely generated subgroups of a group to have non-finitely generated intersection.
We recall from Theorem 6.6 the function
We have shown the following.
Theorem 7.5
Let \(\rho \) be a finitely generated left congruence on S. Then there are \(T\in \mathfrak {V}_{{{\,\textrm{AFG}\,}}}(S)\) and \(\tau \in \mathfrak {C}_{{{\,\textrm{FG}\,}}}(E)\) such that
In particular, \(\mathfrak {LC}_{{{\,\textrm{FG}\,}}}(S)\) is the image \((\mathfrak {C}_{{{\,\textrm{FG}\,}}}(E)\times \mathfrak {V}_{{{\,\textrm{AFG}\,}}}(S))\Theta \).
One avenue of interest is to consider when every left congruence is finitely generated.
Definition 7.6
A semigroup S is left Noetherian if every left congruence on S is finitely generated.
Left Noetherian inverse semigroups have been classified [7]. A partial order P is said to have the ascending chain condition if every increasing sequence is eventually constant. As in many cases it is possible to describe the left Noetherian condition in terms of the ascending chain condition; a semigroup is left Noetherian if the lattice of left congruences has the ascending chain condition. Theorem 7.5 states that S is left Noetherian if and only if every left congruence is join of a finitely generated trace minimal left congruence and a finitely generated idempotent separating left congruence.
The following is a straightforward observation about the ascending chain condition on partial orders.
Lemma 7.7
Let P, Q be partial orders that have the ascending chain condition, and let \(R\subseteq P\) be a suborder. Then R and \(P\times Q\) both have the ascending chain condition.
Theorem 7.8
Let S be an inverse semigroup. The lattice \(\mathfrak {LC}(S)\) has the ascending chain condition if and only if \(\mathfrak {V}(S)\) and \(\mathfrak {C}(E)\) have the ascending chain condition.
Proof
This follows directly from Lemma 7.7. We have injective order-preserving maps \(\mathfrak {V}(S)\hookrightarrow \mathfrak {LC}(S)\) and \(\mathfrak {C}(E)\hookrightarrow \mathfrak {LC}(S)\) onto the sets on idempotent separating and trace minimal left congruences, respectively. This implies that if \(\mathfrak {LC}(S)\) has the ascending chain condition then so do both \(\mathfrak {V}(S)\) and \(\mathfrak {C}(E)\). For the converse, the inverse kernel approach describes \(\mathfrak {LC}(S)\) as a suborder of \(\mathfrak {V}(S)\times \mathfrak {C}(E),\) thus if \(\mathfrak {V}(S)\) and \(\mathfrak {C}(E)\) have the ascending chain condition then so does \(\mathfrak {LC}(S)\). \(\square \)
It is easily seen that the lattice \(\mathfrak {V}(S)\) has the ascending chain condition if and only if every full inverse subsemigroup is almost finitely generated. For the lattice \(\mathfrak {C}(E)\) we have the following from [7], which is straightforward to show directly.
Lemma 7.9
[7, Proposition 3.4] If E is a semilattice then \(\mathfrak {C}(E)\) has the ascending chain condition if and only if E is finite.
In the case that E is finite the notions of finitely generated and almost finitely generated full inverse subsemigroups coincide. The usual formulation for the classification of left Noetherian inverse semigroups is immediate.
Theorem 7.10
[7, Proposition 4.3] An inverse semigroup is left Noetherian if and only if every full inverse subsemigroup is finitely generated.
8 Rees left congruences
An important family of congruences on any semigroup are the Rees congruences. In this section we discuss the left analogue of Rees congruences, with particular focus on when these are finitely generated. Given a left ideal \(A\subseteq S\) we define a relation \(\rho _A\) on S by
Since A is a left ideal it is immediate that \(\rho _A\) is a left congruence which we call the Rees left congruence; if A is an ideal then \(\rho _A\) is a two-sided congruence. For the remainder of this section we shall assume that A is a left ideal of S. For inverse semigroups the principal left ideals are \(\{Se\mid e\in E\}\) and we recall that a left ideal is finitely generated if it is the union of finitely many principal left ideals. The following is an elementary alternate characterisation of finitely generated left ideals for inverse semigroups.
Lemma 8.1
Let A be a left ideal of S. Then A is finitely generated if and only if E(A) has finitely many maximal idempotents and every idempotent in A is below a maximal idempotent in E(A).
We are interested in computing which left ideals A correspond to finitely generated Rees left congruences. We note that
We first note that if \(\rho _A\) is finitely generated then it follows in a straightforward fashion that A is finitely generated. However the converse is not true. Indeed, consider an infinite descending chain of idempotents indexed by \(\mathbb {N}\). This has all ideals principal, but no Rees congruence is finitely generated. We shall classify those left ideals \(A\subseteq S\) such that \(\rho _A\) is finitely generated. Our first step is a pair of technical lemmas the first of which may be considered a partial refinement of Proposition 6.5 in terms of generating sets for a left congruence.
Lemma 8.2
Let \(Z\subseteq E\times E\) be a binary relation, let \(Y\subseteq S\) be an inverse subsemigroup, and let \(\rho =\chi _Y\vee \nu _{\langle Z\rangle _{{{\,\textrm{C}\,}}(E)}}\). Then
Proof
Let \(\tau =\langle Z\rangle _{{{\,\textrm{C}\,}}(E)}\) and let \(X=Z\cup \{(aea^{-1},afa^{-1}) \mid (e,f)\in Z,\ a\in Y\}\). It is immediate that \(\tau \subseteq \langle X\rangle _{{{\,\textrm{C}\,}}(E)}\). Further, if \(e\ \tau \ f\) and \(a\in Y\) then it is straightforward that \((aea^{-1},afa^{-1})\in \langle X\rangle _{{{\,\textrm{C}\,}}(E)}\). Therefore
It follows that without loss of generality we may assume that \(Z=\tau ,\) which we do for the remainder of the proof.
Let \(\xi ={{\,\textrm{trace}\,}}\rho \). As \((\xi ,{{\,\textrm{Iker}\,}}\rho )\) is an inverse congruence pair we know \({{\,\textrm{Iker}\,}}\rho \subseteq N(\xi )\) and as \(Y\subseteq {{\,\textrm{Iker}\,}}\rho \) it follows that \(Y\subseteq N(\xi )\). By definition \(\tau \subseteq \xi \), and since \(\xi \) is normal in Y we have that \(X\subseteq \xi ,\) whence \(\langle X\rangle _{{{\,\textrm{C}\,}}(E)}\subseteq \xi \).
For the reverse inclusion suppose that \(e\ \langle X\rangle _{{{\,\textrm{C}\,}}(E)}\ f\) so there are \((p_i,q_i)\in X,\ h_i\in E\) such that there is an X-sequence
from e to f. We observe for \(a\in Y\) and \(x,y\in E\) that \(axya^{-1}=(axa^{-1})(aya^{-1})\). Therefore at each stage of the X-sequence we may conjugate by \(a\in Y\) to obtain an X-sequence from \(aea^{-1}\) to \(afa^{-1}\). Thus \(Y\subseteq N(\langle X\rangle _{{{\,\textrm{C}\,}}(E)})\). Since \((\langle X\rangle _{{{\,\textrm{C}\,}}(E)},N(\langle X\rangle _{{{\,\textrm{C}\,}}(E)}))\) is an inverse congruence pair, and \(\tau \subseteq \langle X\rangle _{{{\,\textrm{C}\,}}(E)}\) and \(Y\subseteq N(\langle X\rangle _{{{\,\textrm{C}\,}}(E)})\) it follows that \(\rho \subseteq \rho _{(\langle X\rangle _{{{\,\textrm{C}\,}}(E)},N(\langle X\rangle _{{{\,\textrm{C}\,}}(E)})}\). Hence by Corollary 3.10 we have \(\xi \subseteq \langle X\rangle _{{{\,\textrm{C}\,}}(E)}\). Thus we have shown that \(\xi =\langle X\rangle _{{{\,\textrm{C}\,}}(E)}\). \(\square \)
Our second technical lemma states that every convex subsemilattice of E is a congruence class of some congruence on E. This may be thought of as a simplified version of the congruence extension property, which semilattices satisfy (see [14]).
Lemma 8.3
Let E be a semilattice and \(F\subseteq E\) be a subsemilattice. Let \(\overline{F}=\{ e\in E\mid \exists f_1,f_2\in F \text { with } f_1\le e\le f_2 \}\) be the convex closure of F. Then \(\overline{F}\) is a congruence class of \(\tau =\langle F\times F\rangle _{{{\,\textrm{C}\,}}(E)}\).
Proof
We first note that certainly for \(k_1,k_2\in \overline{F}\) we have \(k_1\ \tau \ k_2,\) hence \(\overline{F}\times \overline{F}\subseteq \tau \). For the reverse inclusion we suppose \(g\ \tau \ h\) with \(h\in \overline{F},\) so there is a \(\tau \)-sequence from h to g. Thus there are \(p_i,q_i\in F,\) \(y_i\in E\) such that
We prove \(g\in \overline{F}\) by induction. Suppose \(y_ip_i\in \overline{F},\) then there is \(f\in F\) such that \(f\le y_ip_i\). Then \(fq_i\le y_ip_iq_i\le y_iq_i\le q_i\). As \(f_i,q_i\in F\) and F is a subsemilattice we have \(f_iq_i\in F\). Therefore \(y_iq_i\in \overline{F}\). Since \(y_iq_i=y_{i+1}p_{i+1}\) this completes the inductive step, and as \(y_1p_1=h\in \overline{F}\) it follows that \(g=y_nq_n\in \overline{F}\). \(\square \)
We may now proceed with the main result of this section.
Theorem 8.4
Let S be an inverse semigroup, and \(A\subseteq S\) a left ideal. Then \(\rho _A\) is a finitely generated left congruence on S if and only if A is finitely generated, and there is a finitely generated subsemigroup \(W\subseteq A\) such that for each \(a \in A\) there is some \(f\in E(W)\) with \(af\in W\).
Proof
Suppose first that \(\rho _A\) is generated by a finite set. Applying Theorem 7.5 we obtain that there is a finite set \(Y\subseteq A\) and a finitely generated congruence \(\tau \) on E such that \(\rho _A=\nu _\tau \vee \chi _{\langle Y\rangle _{{{\,\textrm{IS}\,}}}}\). Further, as the only nontrivial \({{\,\textrm{trace}\,}}\rho _A\) class is E(A) we may assume that there is a finite set \(X\subseteq E(A)\) such that \(\tau =\langle X\times X\rangle _{{{\,\textrm{C}\,}}(E)}\). We have previously observed that if \(\rho _A\) is finitely generated then A is finitely generated. By Lemma 8.2 there are then finitely many maximal idempotents in A and each idempotent in A is below some maximal idempotent. Without loss of generality we assume that all maximal idempotents are in X.
We now show that there is a finitely generated subsemigroup W such that for \(a\in A\) there is some \(f\in E(W)\) with \(af\in W\). Let \(W=\langle X\cup Y\rangle _{{{\,\textrm{IS}\,}}}\) and note that, as Y, X are finite, W is finitely generated, and, as \(X,Y\subseteq A\), we have that \(W\subseteq A\). Let \(Z= X\cup \{ aea^{-1}\mid a\in \langle Y\rangle _{{{\,\textrm{IS}\,}}},\ e\in X\}\) and observe that \(Z\subseteq W\). By Lemma 8.2,
It follows that \({{\,\textrm{trace}\,}}\rho _A=\langle Z\times Z\rangle _{{{\,\textrm{C}\,}}(E)}\). We then note that as \(Z\subseteq E(W)\subseteq E(A)\) we have that \({{\,\textrm{trace}\,}}\rho _A=\langle E(W)\times E(W)\rangle _{{{\,\textrm{C}\,}}(E)}\). As W is a subsemigroup E(W) is certainly a subsemilattice of E(S), so we may apply Lemma 8.3 to obtain that \(\overline{E(W)}=\{ e\in E(S)\mid \exists f_1,f_2\in E(W) \text { with } f_1\le e\le f_2\}\) (the convex closure of E(W)) is a congruence class of \({{\,\textrm{trace}\,}}\rho _A\). The only nontrivial congruence class is E(A), so \(\overline{E(W)}=E(A)\). Thus given \(e\in E(A)\) there is some \(f\in E(W)\) such that \(f\le e\).
It is straightforward to see that \(\chi _{\langle Y\rangle _{{{\,\textrm{IS}\,}}}}\vee \nu _{{{\,\textrm{trace}\,}}\rho _A}=\rho _A\). Therefore, by Lemma 6.4
Using that \({{\,\textrm{ker}\,}}\rho _A=A\cup E,\) and using the remark after Corollary 3.11 we have that
Combining these observations we obtain that
Hence for each \(a\in A\) there is \(t\in \langle Y\cup E\rangle _{{{\,\textrm{IS}\,}}}\) such that \(a\ \nu _{{{\,\textrm{trace}\,}}\rho _A}\ t\). By Theorem 2.5 this says that there is \(e\in E\) such that \(ae=te\) and
Since \(a^{-1}a\in A\), we obtain that \(e\in A,\) and we also note that \(ae=te\in \langle Y\cup E\rangle _{{{\,\textrm{IS}\,}}}\).
If \(ae\in E(A)\) then as \(E(A)=\overline{E(W)}\) there is \(f\in E(W)\) such that \(f\le ae\). Then certainly \(f\le e\) so \(af=aef=f\) so \(af\in W\). Suppose now that \(ae\notin E\). As \(\langle Y\cup E\rangle _{{{\,\textrm{IS}\,}}}=E \cup \langle Y\rangle _{{{\,\textrm{IS}\,}}} E\) this implies that \(ae\in \langle Y\rangle _{{{\,\textrm{IS}\,}}}E,\) so there are \(b\in \langle Y\rangle _{{{\,\textrm{IS}\,}}}\) and \(e^\prime \in E\) such that \(ae=be^\prime \) (note that we may choose \(e^\prime \in E(A)\)). Then as \(E(A)=\overline{E(W)}\) we may take \(f\in E(W)\) with \(f\le ee^\prime \). Then \(af=aef=be^\prime f=bf\in W\). This completes the proof that A satisfies the properties claimed.
We now prove the converse, so we assume that A is as claimed, with \(W=\langle V\rangle _{{{\,\textrm{IS}\,}}}\) for some finite set \(V\subseteq A\) closed under taking inverses. Without loss of generality we may assume that all maximal idempotents are in V. Let \(T=\langle V\cup E\rangle _{{{\,\textrm{IS}\,}}},\) and \(X=\{ vv^{-1}\mid v\in V\}\). Let \(\tau =\langle X\times X\rangle _{{{\,\textrm{C}\,}}(E)}\) and \(\rho =\chi _T\vee \nu _\tau \). We shall show that \(\rho =\rho _A\). We first note that for any \(u\in V\) we have \(u\ \chi _T\ uu^{-1}\) and, as V is closed under taking inverses, for any \(u,v\in V\) we have \(u^{-1}u\ \tau \ vv^{-1}\). Therefore, for \(a=v_1\cdots v_n\in W,\) we have
It follows that for every pair \(a,b\in W\) we have that \(a\ \rho \ b\). In particular this implies that for \(e,f\in E(W)\) we have \(e\ \rho \ f\).
By assumption for \(a\in A\) there is \(f\in E(W)\) with \(af\in W\). Choose m a maximal idempotent in A with \(m\ge a^{-1}a\). As \(m,f\in E(W)\) we then have \(m\ \rho \ f,\) and thus \(a=am\ \rho \ af\). We have shown that \(\rho _A\subseteq \rho \). The reverse inclusion follows immediately from the observation that if \(Z\subseteq A\) for a left ideal A then \(\langle Z\times Z\rangle _{{{\,\textrm{LC}\,}}(S)} \subseteq \rho _A\). Thus we have \(\rho =\rho _A\) as claimed. Finally we note that \(\rho _A\) is finitely generated. Indeed, X is finite by definition and T is almost finitely generated, so by Proposition 7.4\(\chi _T\) is finitely generated. \(\square \)
It is of interest to consider semigroups for which the universal relation \(\omega \) is finitely generated as a left congruence. For monoids this is equivalent to the monoid being of type left \(\text {FP}_1\) (see, for example, [1]). This has been studied for several classes of semigroups, and a classification for inverse semigroups is available [1]. Since \(\omega =\rho _S\) for the ideal S, we may deduce the classification of which inverse semigroups are such that the universal relation is finitely generated.
Corollary 8.5
[1, Theorem 5.1] Let S be an inverse semigroup. Then \(\omega \) is finitely generated as a left congruence on S if and only if E contains finitely many maximal elements and every idempotent is below a maximal idempotent, and there is a finitely generated subsemigroup \(W\subseteq S\) such that for each \(a \in S\) there is some \(f\in E(W)\) with \(af\in W\).
Finally we will classify those inverse semigroups such that for every left ideal A the left congruence \(\rho _A\) is finitely generated. Before we state this result we provide a pair of elementary lemmas from which the result follows.
Lemma 8.6
Let S be an inverse semigroup. Then the following statements are equivalent.
-
(i)
For all finitely generated left ideals \(A\subseteq S\) the left congruence \(\rho _A\) is finitely generated.
-
(ii)
For all principal left ideal \(B\subseteq S\) the left congruence \(\rho _B\) is finitely generated.
The second lemma is a standard result about which inverse semigroups have every left ideal finitely generated.
Lemma 8.7
Let S be an inverse semigroup. Then the following statements are equivalent.
-
(i)
Every left ideal of S is finitely generated.
-
(ii)
Every ideal of E is finitely generated.
-
(iii)
The semilattice E contains no infinite antichains and no infinite ascending chains.
We may now state our classification.
Theorem 8.8
Let S be an inverse semigroup. Every Rees left congruence is finitely generated if and only if E contains no infinite antichains and no infinite ascending chains and for all principal left ideals B there is a finitely generated subsemigroup \(W\subseteq B\) such that for all \(b\in B\) there is \(f\in E(W)\) such that \(bf\in W\).
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Acknowledgements
This project forms part of the work toward my PhD at the University of York, supported by EPSRC grant EP/N509802/1. I would like to thank my supervisor, Professor Victoria Gould, for all her guidance. I would also like to thank the referee for their diligent review and constructive feedback.
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Brookes, M.D.G.K. The lattice of one-sided congruences on an inverse semigroup. Period Math Hung 87, 1–26 (2023). https://doi.org/10.1007/s10998-022-00497-z
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DOI: https://doi.org/10.1007/s10998-022-00497-z