Bull. Aust. Math. Soc. 83 (2011), 273–288 doi:10.1017/S0004972710001851 ON THE PARTITION MONOID AND SOME RELATED SEMIGROUPS D. G. FITZGERALD ˛ and KWOK WAI LAU (Received 16 June 2010) Abstract The partition monoid is a salient natural example of a *-regular semigroup. We find a Galois connection between elements of the partition monoid and binary relations, and use it to show that the partition monoid contains copies of the semigroup of transformations and the symmetric and dual-symmetric inverse semigroups on the underlying set. We characterize the divisibility preorders and the natural order on the (straight) partition monoid, using certain graphical structures associated with each element. This gives a simpler characterization of Green’s relations. We also derive a new interpretation of the natural order on the transformation semigroup. The results are also used to describe the ideal lattices of the straight and twisted partition monoids and the Brauer monoid. 2010 Mathematics subject classification: primary 20M20; secondary 20M17, 05A18, 06A15. Keywords and phrases: partition monoid, Brauer monoid, ideal structure, natural order. 1. Diagrams and products Partition algebras, which are twisted semigroup algebras of the partition monoids, are important in the theory of group representations, combinatorics, and statistical mechanics, and have an extensive literature including significant studies in T7, 13, 14U. Generators and relations for partition monoids have been studied in T3U, and their endomorphisms in T15U. Wilcox T20, Section 7U studied the structure of the partition monoid in an application of his quite general theorem about the cellularity of twisted semigroup algebras of regular semigroups. It is our intention to investigate the structure of partition monoids further. We use this first section to describe the elements of the partition monoid PX and their multiplication, and to draw attention to some of the subsemigroups of PX which are interesting in their own right. Let X be a set. A diagram over X is an equivalence class of graphs on a vertex set X [ X 0 (consisting of two copies of X). Two such graphs are regarded as equivalent if they have the same connected components. We define PX as the set of all diagrams over X. Also, if X is finite, say X D f1; 2;:::; ng, we conventionally write Pn in place of PX , and similarly for other families of semigroups. c 2010 Australian Mathematical Publishing Association Inc. 0004-9727/2010 $16.00 273 Downloaded from https://www.cambridge.org/core. IP address: 170.106.35.234, on 27 Sep 2021 at 05:15:21, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S0004972710001851 274 D. G. FitzGerald and K. W. Lau [2] FIGURE 1. Multiplication of a and b in P6 (left, expanded form; right, contracted form). We need to have a standard representative of each equivalence class; for this purpose, we choose a graph with maximal edge set, so that each of its components is a complete graph. Proofs below will generally use standard representatives. When it comes to drawing the graphs, however, it is convenient to use a minimal number of edges, as in Figure1. We draw the graph of a 2 PX so that the undashed vertices (elements of X) are arranged in a horizontal row, and the corresponding dashed vertices (those in X 0) are directly below. Thus we will refer to the undashed elements as the upper vertices, and the dashed elements as the lower vertices. To multiply two elements a; b of PX , their diagrams are first drawn stacked vertically, with a above b. Then the lower vertices of a are identified with the upper vertices of b, in an ‘interior’ row called an interface; we call this diagram the expanded form of the product. Next, the connected components of the expanded form are constructed. Finally, we ignore the vertices in the interface, and any components using only these vertices. This results in another member of PX , which defines the product ab; we call it the contracted form in contrast to the expanded form. Note too that a path in the expanded form becomes an edge in the standard representative of the contracted form. An example with a; b 2 P6 is seen in Figure1. Here a has components f1; 30g; f2; 3; 10g; f4; 5; 40g; f6g; f20g; f50; 60g and b has components f1; 2; 20g; f3; 4g; f5; 6g; f10g; f30; 60g; f40; 50g: We call PX , with this multiplication, the partition monoid on the set X because the components of a diagram are the blocks of a partition of X [ X 0. An edge of 0 a member a of PX is called transversal if it is of the form fi; j g with i 2 X and j0 2 X 0. Likewise a component of a is transversal if its vertices include both upper and lower elements (and so it includes a transversal edge), and otherwise the component is nontransversal. For example, in Figure1, the product ab has one transversal component, f2; 3; 20g. A Galois connection with binary relations. DEFINITIONS. Let RelX denote the set of binary relations on X: We define a mapping F V PX −! RelX as follows. With a 2 PX in standard form (union of complete Downloaded from https://www.cambridge.org/core. IP address: 170.106.35.234, on 27 Sep 2021 at 05:15:21, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S0004972710001851 [3] On the partition monoid and some related semigroups 275 aF : aFG : bF : bFG : aF ° bF : (aFG)(bFG) : (a) (b) FIGURE 2. The action of the maps F and G V (a) F applied to diagrams a and b in Figure1, and the composite relation aF ◦ bF; (b) images under G of the relations in (a), and their product as diagrams. subgraphs), put aF D f.i; j/ 2 X × X V fi; j0g 2 a for i 2 X and j0 2 X 0g; the relation on X induced by the transversal edges of a. In the reverse direction, we define G V RelX −! PX thus: 0 for ρ 2 RelX , ρG is the graph on X [ X with edge set being all (finite- length) paths produced from the edges ffi; j0g V .i; j/ 2 ρg. In consequence, ρG is an element of PX , in fact in standard representative form. These definitions are illustrated in Figure2. We say a diagram is earthed if all its nontransversal components are singletons. −1 We remind the reader that ρ 2 RelX is bifunctional if ρ ◦ ρ ◦ ρ ⊆ ρ, where ◦ means composition of binary relations and ρ−1 is the inverse relation of ρ. We shall show that F; G constitute a Galois connection between RelX and PX , in which the closed elements are the earthed diagrams and the bifunctional relations. THEOREM 1.1. (i) F and G are monotone with respect to the usual inclusion orders; and (ii) GFG D G and F D FGF. For all ρ 2 RelX and a 2 PX : (iii) ρG ⊆ a if and only if ρ ⊆ aFI Downloaded from https://www.cambridge.org/core. IP address: 170.106.35.234, on 27 Sep 2021 at 05:15:21, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S0004972710001851 276 D. G. FitzGerald and K. W. Lau [4] (iv) aFG ⊆ a, with equality if and only if a is earthed; and (v) ρ ⊆ ρG F, with equality if and only if ρ is bifunctional. 0 PROOF. (i) Suppose that a; b 2 PX and a ⊆ b. If .i; j/ 2 aF, then fi; j g is an edge of a and hence of b. So .i; j/ 2 bF, and thus aF ⊆ bF. Suppose that ρ; σ 2 RelX and ρ ⊆ σ . If fi; jg is an edge in ρG, there is a path .i; i1; i2;:::; im; j/ with successive −1 −1 pairs .i; i1/, .i1; i2/; : : : ; .im; j/ in ρ or ρ and so in σ [ σ , whence i; j are in the same component of σ G. So ρG ⊆ σ G. (ii) This will be proved last. (iii) Suppose that ρG ⊆ a and .i; j/ 2 ρ. Then fi; j0g is a transversal edge of ρG, hence of a. So .i; j/ 2 aF. Conversely, suppose that ρ ⊆ aF and fi; jg is an edge of ρG. Then there is a path .i; i1; i2;:::; im; j/ with successive pairs .i; i1/, −1 −1 .i1; i2/; : : : ; .im; j/ in ρ or ρ , hence in aF or .aF/ . Thus i; j are in the same component of a, so ρG ⊆ a. (iv) Taking ρ D aF in (iii) implies that aFG ⊆ a. For any ρ 2 RelX , ρG is earthed by definition, so a D aFG implies that a is earthed. Conversely, suppose 0 that a 2 PX is earthed and fi; jg is an edge of a, with i; j 2 X [ X . There is a path in a from i to j using only transversal edges. Therefore either fi; jg is transversal, or there are transversal edges fi; kg and f j; kg in a. Then .i; j/ 2 aF [ .aF/−1 or .i; k/; . j; k/ 2 aF [ .aF/−1. In either case, fi; jg is an edge of aFG. Thus a ⊆ aFG. Together with aFG ⊆ a, this gives a D aFG. (v) Taking a D ρG in (iii) implies that ρ ⊆ ρGF. If .i; k/; . j; k/; .