ON KISELMAN QUOTIENTS OF 0HECKE MONOIDS
International Electronic Journal of Algebra
Volume
ON KISELMAN QUOTIENTS OF 0HECKE MONOIDS Olexandr Ganyushkin and Volodymyr Mazorchuk
0 Volodymyr Mazorchuk Department of Mathematics Uppsala University , Box 480 SE75106, Uppsala , SWEDEN
Combining the de nition of 0Hecke monoids with that of Kiselman semigroups, we de ne what we call Kiselman quotients of 0Hecke monoids associated with simply laced Dynkin diagrams. We classify these monoids up to isomorphism, determine their idempotents and show that they are J trivial. For type A we show that Catalan numbers appear as the maximal cardinality of our monoids, in which case the corresponding monoid is isomorphic to the monoid of all orderpreserving and orderdecreasing total transformations on a nite chain. We construct various representations of these monoids by matrices, total transformations and binary relations. Motivated by these results, with a mixed graph we associate a monoid, which we call a HeckeKiselman monoid, and classify such monoids up to isomorphism. Both Kiselman semigroups and Kiselman quotients of 0Hecke monoids are natural examples of HeckeKiselman monoids. Mathematics Subject Classi cation (2010): 20M10, 20M20, 20M30

articles where the emphasis is made on its semigroup algebra and not its structure
as a monoid. Therefore semigroup properties of H are not really spelled out
in the above papers. However, with some efforts one can derive from the above
literature that the monoid H is J trivial (we will show this in Subsection 2.1) and
has 2n idempotents, where n is the number of vertexes in ? (we will show this in
Subsection 2.2).
Another example of an idempotent generated J trivial monoid with 2n
idempotents (where n is the number of generators) is Kiselman?s semigroup Kn, defined as
follows: it is generated by idempotents ei, i = 1, 2, . . . , n, subject to the relations
eiej ei = ej eiej = eiej for all i > j (see [
12
]). This semigroup was studied in [
18,1
].
In particular, in [
18
] it was shown that Kn has a faithful representation by n ? n
matrices with nonnegative integer coefficients.
The primary aim of this paper is to study natural mixtures of these two
semigroups, which we call Kiselman quotients of H . These are defined as follows:
choose any orientation ?? of ? and define the semigroup KH? as the quotient of H
obtained by imposing the additional relations ?i?j ?i = ?j ?i?j = ?i?j in all cases
when ?? contains the arrow i / j . These relations are natural combinations
of the relations defining H and Kn. Our first result is the following theorem:
Theorem 1. (i) The semigroup KH? is J trivial. (ii) The set E := {?i : i ? ?0} is the unique irreducible generating system for
KH? .
(iii) The semigroup KH? contains 2n idempotents, where n is the number of
vertexes in ?.
(iv) The semigroups KH? and KH? are isomorphic if and only if the directed
graphs ? and ? are isomorphic.
(v) The semigroups KH? and KH? are antiisomorphic if and only if the directed
graphs ? and ? are antiisomorphic.
(vi) If ? is a Dynkin diagram of type An, then KH?  ? Cn+1, where Cn :=
1 (2n)is the nth Catalan number.
n+1 n
(vii) If ? is a Dynkin diagram of type An, then KH?  = Cn+1 if and only if ?? is
isomorphic to the graph
?
/ ?
/ ?
(viii) If ?? is as in (vii), then the semigroup KH? is isomorphic to the semigroup
Cn+1 of all orderpreserving and orderdecreasing total transformations of
{1, 2, . . . , n, n + 1} (see [10, Chapter 14]).
The semigroup Cn+1 appears in various disguises in [
26,25,17,10
]. Its presentation
can be derived from [26], however, in the present paper this semigroup appears in a
different context and our proof is much less technical. In [
26
] it is also observed that
the cardinality of the semigroup with this presentation is given by Catalan numbers.
Classically, Catalan numbers appear in semigroup theory as the cardinality of the
socalled TemperleyLieb semigroup TLn, see [27, 6.25(g)]. That Catalan numbers
appear as the cardinality of Cn+1 was first observed in [
13
] (with an unnecessarily
difficult proof, see [27, 6.19(u)] for a straightforward argument). In [
8
] it was
shown that Catalan numbers also appear as the maximal cardinality of a nilpotent
subsemigroup in the semigroup IOn of all partial orderpreserving injections on
{1, 2, . . . , n} (see also [
9
] for an alternative argument).
Motivated by both Kiselman semigroups and Kiselman quotients of 0Hecke
monoids, we propose the notion of HeckeKiselman semigroups associated with an
arbitrary mixed (finite) graph. A mixed graph is a simple graph in which edges can
be both oriented and unoriented. Such graph is naturally given by an antireflexive
binary relation ? on a finite set (see Subsection 5.1). The corresponding
HeckeKiselman semigroup HK is generated by idempotents ei indexed by vertexes of
the graph, subject to the following relations:
? if i and j are not connected by any edge, then eiej = ej ei;
? if i and j are connected by an unoriented edge, then eiej ei = ej eiej ;
? if i and j are connected by an oriented edge i ? j, then eiej ei = ej eiej =
eiej .
Theorem 2. Let ? and ? be two antireflexive binary relations on finite sets. Then
HK ?= HK if and only if the corresponding mixed graphs are isomorphic.
The paper is organized as follows: Theorem 1 is proved in Section 2. In
Section 3 we construct representations of KH? by total transformations, matrices
with nonnegative integral coefficients and binary relations. We also describe
simple and indecomposable projective linear representations of KH? over any field. In
Section 4 we give an application of our results to combinatorial interpretations of
Catalan numbers. Finally, in Section 5 we present a general definition of
HeckeKiselman semigroups and prove Theorem 2. As a corollary, we obtain a formula
for the number of isomorphism classes of HeckeKiselman semigroups on a given
set. We complete the paper with a short list of open problems on HeckeKiselman
semigroups.
2. Proof of Theorem 1
As usual, we denote by ?0 the set of vertexes of the graph ? and set n =
?0. Consider the free monoid Wn generated by a1, . . . , an and the canonical
epimorphism ? : Wn ? KH? , defined by ?(ai) = ?i, i ? ?0. We identify KH?
with the quotient of Wn by Ker(?).
For w ? Wn the content c(w) is defined as the set of indexes for which the
corresponding generators appear in w. For any relation v = w used in the definition
of KH? we have c(v) = c(w). This implies that for any ? ? KH? (which we
interpret as an equivalence class in Ker(?)) and any v, w ? ? we have c(v) = c(w).
Hence we may define c(?) as c(v) for any v ? ?.
2.1. Proof of statement (i). We start with the following statement, which we
could not find any explicit reference to.
Lemma 3. The monoid H is J trivial.
Proof. For w ? W denote by Hw ? H the corresponding element (if w =
si1 si2 ? ? ? sik is a reduced decomposition of w into a product of simple reflections,
then Hw = ?i1 ?i2 ? ? ? ?ik ). Let l : W ? {0, 1, . . . } denote the classical length
function. Then the usual multiplication properties of the Hecke algebra ([20,
Lemma 1.12]) read as follows:
?
?iHw = ?Hsiw
?Hw
l(siw) > l(w);
otherwise;
?
Hw?i = ?Hwsi l(wsi) > l(w);
?Hw otherwise.
Hence for any w ? W the twosided ideal H HwH consists of Hw and, possibly,
some elements of strictly bigger length. In particular, for any x ? H HwH such
that x ?= Hw we have H HwH ?= H xH . The claim follows.
As any quotient of a finite J trivial semigroup is J trivial (see e.g. [19,
Chapter VI, Section 5]), statement (i) follows from Lemma 3.
2.2. Proof of statement (ii). The set E generates KH? by definition. We claim
that this generating system is irreducible. Indeed, if we can write ei as a product w
of generators, then c(w) = {i}, implying w = ei. Hence E is irreducible. Further,
we know that KH? is J trivial from statement (i). Uniqueness of the irreducible
generating system in a J trivial monoid was established in [5, Theorem 2]. This
implies statement (ii).
2.3. Proof of statement (iii). Identify ?0 with {1, 2, . . . , n} such that i / j
implies i > j for all i and j. Then the mapping ei 7? ?i, i ? {1, 2, . . . , n}, extends to
an epimorphism ? : Kn KH? (as all relations for generators of Kn are satisfied
by the corresponding generators of KH? ).
By [
18
], the semigroup Kn has exactly 2n idempotents, all having different
contents. As ? preserves the content, we obtain 2n different idempotents in KH? . As
any epimorphism of finite semigroups induces an epimorphism on the corresponding
sets of idempotents, the statement (iii) follows.
For completeness, we include the following statement which describes
idempotents in H in terms of longest elements for parabolic subgroups of W (this claim
can also be deduced from [22, Lemma 2.2]).
Lemma 4. For any X ? ?0 left wX denote the longest element in the parabolic
subgroup of W associated with X (w? = e). Then HwX ? H is an idempotent,
and every idempotent of H has the form HwX for some X as above. In particular,
H has 2n idempotents.
Proof. Let w ? W . Assume that Hw is an idempotent. From (1) it follows that
HwHw = Hw implies that ?iHw = Hw?i = Hw for any i ? c(Hw). In particular,
for any i ? c(Hw) we have l(siw) < l(w) and l(wsi) < l(w), in other words, both
the left and the right descent sets of w contain all simple reflections appearing in
any reduced decomposition of w. From [3, 2.3] it now follows that w is the longest
element of the parabolic subgroup of W , generated by all si, i ? c(Hw).
On the other hand, if w is the longest element from some parabolic subgroup of
W , then the same arguments imply ?iHw = Hw?i = Hw for any i ? c(Hw) and
hence HwHw = Hw. The claim follows.
2.4. Proof of statement (iv). This statement follows from a more general
statement of Theorem 16, which will be proved in Subsection 5.3.
2.5. Proof of statement (v). By Proposition 13, which will be proved in a
more general situation in Subsection 5.1, the semigroups KH? and KH? are
antiisomorphic if and only if KH? and KH? op are isomorphic. By statement (iv), the
latter is the case if and only if ?? and ??op are isomorphic, which implies statement
(v).
2.6. Proof of statement (vi). Since ? is now of type An, the group W is
isomorphic to the symmetric group Sn+1. Consider the canonical projection H KH? .
Then any equivalence class of the kernel of this projection contains some element
of minimal possible length (maybe not unique). Let Hw be such an element and
w = si1 si2 . . . sik be a reduced decomposition in W . Then this reduced
decomposition cannot contain any subword of the form sisj si (where i and j are connected in
?), in other words, w is a shortbraid avoiding permutation. Indeed, otherwise Hw
would be equivalent to Hw? , where w? is a shorter word obtained from w by
changing sisj si to either sisj or sj si depending on the direction of the arrow between i
and j in ??, which would contradict our choice of w.
Therefore the cardinality of KH? does not exceed the number of shortbraid
avoiding elements in Sn+1. These are known to correspond to 321avoiding
permutations (see e.g. [2, Theorem 2.1]). The number of 321avoiding permutations in
Sn+1 is known to be Cn+1 (see e.g. [27, 6.19(ee)]). Statement (vi) follows.
2.7. Proof of statement (vii). Assume first that ?? coincides with
From (vi) we already know that KH?  ? Cn+1. For i = 1, 2, . . . , n denote by Ti
the following transformation of {1, 2, . . . , n + 1}:
( 1
1
2 . . . i ? 1 i i + 1 i + 2 . . .
2 . . . i ? 1 i i i + 2 . . .
n
n
n + 1 )
n + 1
.
The semigroup Cn+1, generated by the Ti?s is the semigroup of all orderdecreasing
and orderpreserving total transformations on the set {1, 2, . . . , n + 1}, see [10,
Chapter 14]. One easily checks that the Ti?s are idempotent, that TiTj = Tj Ti if
i ? j > 1 and that TiTi+1Ti = Ti+1TiTi+1 = TiTi+1 for all i = 1, 2, . . . , n ? 1.
Therefore, sending ?i to Tn+1?i for all i defines an epimorphism from KH? to
Cn+1. As Cn+1 = Cn+1 by [27, 6.25(g)], we obtain that KH?  ? Cn+1 and hence
KH?  = Cn+1.
Assume now that ?? is not isomorphic to (2). Then either ?? or ??op must contain
the following full subgraph:
Using (v) and the fact that KH?  = KH?op, without loss of generality we may
assume that ?? contains (4). It is easy to see that the element sj sisksj ? W is
shortbraid avoiding. On the other hand, because of the arrows i / j and
k / j we have
?j ?i?k?j = ?j ?i?j ?k?j = ?j ?i?j ?k = ?j ?i?k.
Note that sj sisk is again shortbraid avoiding. It follows that in this case some
different shortbraid avoiding permutations correspond to equal elements of KH? .
Hence KH?  is strictly smaller than the total number of shortbraid avoiding
permutations, implying statement (vii).
2.8. Proof of statement (viii). Statement (viii) follows from the observation
that the epimorphism from KH? to Cn+1, constructed in the first part of our proof
of statement (vii), is in fact an isomorphism as KH?  = Cn+1 = Cn+1.
3. Representations of KH?
In this section ? is a disjoint union of Dynkin diagrams and ?? is obtained from
? by orienting all edges in some way.
3.1. Representations by total transformations. In this subsection we gener
alize the action described in Subsection 2.8. In order to minimize the cardinality of
the set our transformations operate on, we assume that ?? is such that the indegree
of the triple point of ?? (if such a point exists) is at most one. This is always
satisfied either by ?? or by ??op. In type A we have no restrictions. Using the results of
Subsection 2.5, we thus construct either a left or a right action of KH? for every
??.
Consider the set M defined as the disjoint union of the following sets: the set
?
?1 of all edges in ??, the set ??00 of all sinks in ?? (i.e. vertexes of outdegree zero),
the set ??10 of all sinks in ?? of indegree two, and the set ??20 of all sources in ?? (i.e.
vertexes of indegree zero). Fix some injection g : ??00 ? ??10 ? ??1 which maps a vertex
to some edge terminating in this vertex (this is uniquely defined if the indegree of
our vertex is one, but there is a choice involved if this indegree is two). Note that
under our assumptions any vertex which is not a sink has indegree at most one.
For i ? ?0 define the total transformation ?i of M as follows:
???y, y / i x / ;
?
?i(x) = ????i, i x / and i is a source;
?g(i), x = i is a sink;
?
?
?
???x, otherwise.
Proposition 5. Formulae (5) define a representation of KH? by total
transformations on M .
Proof. To prove the claim we have to check that the ?i?s satisfy the defining
relations for KH? . Relations ?i2 = ?i and ?i?j = ?j ?i if i and j are not connected follow
directly from the definitions. So, we are left to check that ?i?j ?i = ?j ?i?j = ?i?j if
we have
?
i
/ j
?? .
Every point in M coming from ? or ?? is invariant under both ?i or ?j , so on such
elements the relations are obviously satisfied.
The above reduces checking of our relation to the elements coming from the
following local situations:
,
/ .
In all these cases all relations are easy to check (and the nontrivial ones reduce
to the corresponding relations for the representation considered in Subsection 2.8).
This completes the proof.
Question 6. Is the representation constructed above faithful?
If ?? is given by (2), then M contains n + 1 elements and it is easy to see that
it is equivalent to the representation considered in Subsection 2.8. In particular,
as was shown there, this representation is faithful. So in this case the answer to
Question 6 is positive.
3.2. Linear integral representations. Let V denote the free abelian group gen
erated by vi, i ? ?0. For i ? ?0 define the homomorphism ?i of V as follows:
?i(vj ) = ???vj , i ?= j;
?? vk, i = j.
?k?i
Proposition 7. Mapping ?i to ?i extends uniquely to a homomorphism from KH?
to the semigroup EndZ(V ).
Proof. To prove the claim we have to check that the ?i?s satisfy the defining
relations for KH? . We do this below.
Relation ?i2 = ?i. If j ?= i, then ?i2(vj ) = ?i(vj ) = vj by definition. As ? contains
no loops, we also have
?i2(vi) = ?i(? vk) =
k?i
? ?i(vk) k=?=i ? vk = ?i(vi).
k?i k?i
Relation ?i?j = ?j ?i if i and j are not connected. If k ?= i, j, then ?i?j (vk) =
?j ?i(vk) = vk by definition. By symmetry, it is left to show that ?i?j (vi) = ?j ?i(vi).
We have
?i?j (vi) j=?=i ?i(vi) = ? vk k=?=j ? ?j (vk) = ?j (? vk) = ?j ?i(vi).
k?i k?i k?i
Relation ?i?j ?i = ?j ?i?j = ?i?j if we have i / j . If k ?= i, j, then ?i?j (vk) =
?j ?i(vk) = vk by definition and our relation is satisfied. Further we have
The representation given by Proposition 7 is a generalization of Kiselman?s
representation for Kn, see [18, Section 5]. Using the canonical antiinvolution
(transposition) for linear operators and Subsection 2.5, from the above we also obtain a
representation for KH?op.
Question 8. Is the representation constructed above faithful (as semigroup
representation)?
If ?? is given by (2), then the linear representation of KH? given by
Proposition 7 is just a linearization of the representation from Subsection 3.1. Hence from
Subsection 2.8 it follows that the answer to Question 8 is positive in this case.
If we identify linear operators on V with n ? n integral matrices with respect to
the basis {vi : i ? ?0}, we obtain a representation of KH? by n ? n matrices with
nonnegative integral coefficients. Call this representation ?.
Lemma 9. The representation ? is a representation of KH? by (0, 1)matrices
(i.e. matrices with coefficients 0 or 1).
Proof. For ? ? KH? we show that ?(?) is a (0, 1)matrix by induction on the
length of ? (that is the length of the shortest decomposition of ? into a product of
canonical generators). If ? = ?, the claim is obvious. If ? is a generator, the claim
follows from the definition of ? (as ? is a simple graph).
Let ?? denote the homomorphism of V corresponding to ?. To prove the
induction step we consider some shortest decomposition ? = ?i1 ?i2 ? ? ? ?ip and set
? = ?i1 ?i2 ? ? ? ?ip?1 . Then for any j ?= ip we have ??(vj ) = ?? ?ip (vj ) = ?? (vj ),
which is a (0, 1)linear combination of the vk?s by the inductive assumption.
For vip we use induction on p to show that ??(vip ) is a (0, 1)linear combination
of vk such that there is a path from k to ip in ??. In the case p = 1 this follows
from the definition of ?. For the induction step, the part that ??(vip ) is a linear
combination of vk such that there is a path from k to ip in ?? follows from the
definition of ?. The part that coefficients are only 0 or 1 follows from the fact that
? contains no loops. This completes the proof.
3.3. Representations by binary relations. Consider the semigroup B(?0) of
all binary relations on ?0. Fixing some bijection between ?0 and {1, 2, . . . , n}, we
may identify B(?0) with the semigroup of all n ? nmatrices with coefficients 0 or
1 under the natural multiplication (the usual matrix multiplication after which all
nonzero entries are treated as 1). This identifies B(?0) with the quotient of the
semigroup Matn?n(N0) (here N0 = {0, 1, 2, . . . }) modulo the congruence for which
two matrices are equivalent if and only if they have the same zero entries.
As the image of the linear representation ? (and also of its transpose) constructed
in Subsection 3.2 belongs to Matn?n(N0), composing it with the natural projection
Matn?n(N0) B(?0) we obtain a representation ?? of KH? by binary relations on
?0. As matrices appearing in the image of ? are (0, 1)matrices, the representation
?? is faithful if and only if ? is.
3.4. Regular actions of Cn+1. The semigroup Cn+1 (which is isomorphic to the
semigroup KH? in the case ?? is of the form (2)) admits natural regular actions on
some classical sets of cardinality Cn+1. For example, consider the set M1 consisting
of all sequences 1 ? x1 ? x2 ? ? ? ? ? xn+1 of integers such that xi ? i for all i (see
[27, 6.19(s)]). For j = 1, . . . , n define the action of Ti (see (3)) on such a sequence
as follows:
Ti(x1, x2, . . . , xn+1) = (x1, . . . , xi?1, xi, xi, xi+2, . . . , xn+1).
It is easy to check that this indeed defines an action of Cn+1 on M1 by total
transformations and that this action is equivalent to the regular action of Cn+1.
As another example consider the set M2 of sequences of 1?s and ?1?s, each
appearing n + 1 times, such that every partial sum is nonnegative (see [27, 6.19(r)]).
For j = 1, . . . , n define the action of Ti on such a sequence as follows: Ti moves the
i + 1st occurrence of 1 to the left and places it right after the ith occurrence, for
example,
T3(11 ? 1 ? ?11 ? ?) = 11 ? 11 ? ?1 ? ?
(here ?1 is denoted simply by ? and the element which is moved is given in bold).
It is easy to check that this indeed defines an action of Cn+1 on M2 by total
transformations and that this action is equivalent to the regular action of Cn+1.
3.5. Projective and simple linear representations. As KH? is a finite J
trivial monoid, the classical representation theory of finite semigroups (see e.g. [
11
]
or [10, Chapter 11]) applies in a straightforward way. Thus, from statement (iii)
it follows that KH? has exactly 2n (isomorphism classes of) simple modules over
any field . These are constructed as follows: for X ? ?0 the corresponding simple
module LX =  and for i ? ?0 the element ?i acts on LX as the identity if i ? X
and as zero otherwise.
The indecomposable projective cover PX of LX is combinatorial in the sense
that it is the linear span of the set
PX := {? ? KH? : for all i ? ?0 the equality ??i = ? implies i ? X}
with the action of KH? given, for ? ? KH? and ? ? PX , by
?
? ? ? = ???, ?? ? PX ;
?0, otherwise.
Remark 10. Both Theorem 1(i)(v) and Subsections 3.2, 3.3 and 3.5 generalize
mutatis mutandis to the case of an arbitrary forest ? (the corresponding Coxeter
group W is infinite in general). To prove Theorem 1(i) in the general case one
should rather consider KH? as a quotient of Kn (via the epimorphism ? from
Subsection 2.4).
4. Catalan numbers via enumeration of special words
The above results suggest the following interpretation for shortbraid avoiding
permutations. For n ? N consider the alphabet {a1, a2, . . . , an} and the set Wn of
all finite words in this alphabet. Let ? denote the minimal equivalence relation on
Wn such that for any i, j ? {1, 2, . . . , n} satisfying i ? j > 1 and any v, w ? Wn
we have vaiaj w ? vaj aiw.
A word v ? Wn will be called strongly special if the following condition is
satisfied: whenever v = v1aiv2aiv3 for some i, the word v2 contains both ai+1 and ai?1.
In particular, both a1 and an occur at most once in any strongly special word. It
is easy to check that the equivalence class of a strongly special word consists of
strongly special words.
Proposition 11. The number of equivalence classes of strongly special words in
Wn equals Cn+1.
Proof. We show that equivalence classes of strongly special words correspond
exactly to shortbraid avoiding permutations in Sn+1. After that the proof is
completed by applying arguments from Subsection 2.6.
If v = ai1 ai2 . . . aik is a strongly special word, then the corresponding
permutation si1 si2 . . . sik ? Sn+1 is obviously shortbraid avoiding.
On the other hand, any reduced expression of a shortbraid avoiding permutation
corresponds to a strongly special word. Indeed, assume that this is not the case.
Let si1 si2 . . . sik ? Sn+1 be a reduced expression for a shortbraid avoiding element
and assume that the corresponding word v = ai1 ai2 . . . aik is not strongly special.
Then we may assume that k is minimal possible, which yields that we can write
v = aiwai such that w contains neither ai nor one of the elements ai?1. Without
loss of generality we may assume that w does not contain ai+1.
First we observe that w must contain ai?1, for otherwise si would commute with
all other appearing simple reflections and hence, using si2 = e we would obtain that
our expression above is not reduced, a contradiction. Further, we claim that ai?1
occurs in w exactly once, for w does not contain ai and hence any two occurrences of
ai?1 would bound a proper subword of v that is not strongly special, contradicting
the minimality of k.
Since si commutes with all simple reflections appearing in our product but si?1,
which, in turn, appears only once, we can compute that siasi?1bsi = asisi?1sib,
which contradicts our assumption of shortbraid avoidance. The claim of the
proposition follows.
This interpretation is closely connected with Kn. A word v ? Wn is called
special provided that the following condition is satisfied: whenever v = v1aiv2aiv3
for some i, then v2 contains both some aj with j > i and some aj with j < i.
In particular, every strongly special word is special. The number of special words
equals the cardinality of Kn (see [
18
]). So far there is no formula for this number.
5. HeckeKiselman semigroups
5.1. De nitions. Kiselman quotients of 0Hecke monoids suggest the following
general construction. For simplicity, for every n ? N we fix the set Nn := {1, 2, . . . , n}
with n elements. Let Mn denote the set of all simple digraphs on Nn. For ? ? Mn
define the corresponding HeckeKiselman semigroup HK (or an HKsemigroup
for short) as follows: HK is the monoid generated by idempotents ei, i ? Nn,
subject to the following relations (for any i, j ? Nn, i ?= j):
eiej =
eiejei =
eiejei =
eiejei =
Relations
ejei
ejeiej
ejeiej
ejeiej
= eiej
= ejei
Edge between i and j
i j
i h ( j
i / j
i o j
(6)
The elements e1, e2, . . . , en will be called the canonical generators of HK .
Example 12. (a) If ? has no edges, the semigroup HK is a commutative band
isomorphic to the semigroup (2Nn , ?) via the map ei 7? {i}.
(b) Let ? ? Mn be such that for every i, j ? Nn, i > j, the graph ? contains
the edge i / j . Then the semigroup HK coincides with the Kiselman
semigroup Kn as defined in [
18
]. This semigroup appeared first in [
12
] and was
also studied in [1].
(c) Let ? be a simply laced Dynkin diagram. Interpret every edge of ? as a pair of
oriented edges in different directions and let ? denote the corresponding simple
digraph. Then HK is isomorphic to the 0Hecke monoid H as defined in
Section 1.
(d) Let ? be an oriented simply laced Dynkin diagram and ? the corresponding
mixed graph. Then HK is isomorphic to the Kiselman quotient KH of the
0Hecke monoid as defined in Section 1.
For ? ? Mn define the opposite graph ?op ? Mn as the graph obtained from
? ? Mn by reversing the directions of all oriented arrows.
Proposition 13. For any ? ? Mn, mapping ei to ei extends uniquely to an
isomorphism from HKop to HK op .
Proof. This follows from (6) and the easy observation that the two last lines of
(6) are swapped by changing the orientation of the arrows and reading all words in
the relations from the right to the left.
5.2. Canonical maps.
Proposition 14. Let ?, ? ? Mn and assume that ? is obtained from ? by deleting
some edges. Then mapping ei to ei extends uniquely to an epimorphism from HK
to HK .
Proof. Note that for two arbitrary idempotents x and y of any semigroup the
commutativity xy = yx implies the braid relation
xyx = x(yx) = x(xy) = (xx)y = xy = x(yy) = (xy)y = (yx)y = yxy.
Therefore, by (6), in the situation as described above all relations satisfied by
canonical generators of HK are also satisfied by the corresponding canonical generators
of HK . This implies that mapping ei to ei extends uniquely to an homomorphism
from HK to HK . This homomorphism is surjective as its image contains all
generators of HK .
We call the epimorphism constructed in Proposition 14 the canonical projection
and denote it by p , .
For ? and ? as above we will write ? ? ?. Then ? is a partial order on
Mn and it defines on Mn the structure of a distributive lattice. The maximum
element of Mn is the full unoriented graph on Nn, which we denote by max. The
minimum element of Mn is the empty graph (the graph with no edges), which we
denote by min. By Example 12(a), the semigroup HKmin is a commutative band
isomorphic to (2Nn , ?). Further, for any ? ? Mn we have the canonical projections
pmax, : HKmax HK and p ,min : HK HKmin.
For w ? HK we define the content of w as c(w) := p ,min(w). This should
be understood as the set of canonical generators of HK appearing in any
decomposition of w into a product of canonical generators. Under the identification of
HKmin and (2Nn , ?), by c(w) we understand the number of generators used to
obtain w. In particular, c(e) = 0 and c(ei) = 1 for all i.
Let m, n ? N, ? ? Mm and ? ? Mn. Assume that f : ? ? ? is a full
embedding of graphs, meaning that it is an injection on vertexes and edges and its
image in ? is a full subgraph of ?.
Proposition 15. In the situation above mapping ei to ef(i) induces a
monomorphism from HK to HK .
Proof. From (6) and our assumptions on f it follows that ef(i)?s satisfy all the
corresponding defining relations satisfied by ei?s. This implies that mapping ei to
ef(i) induces a homomorphism ? from HK to HK .
To prove that this homomorphism is injective it is enough to construct a left
inverse. Similarly to the previous paragraph, from (6) and our assumptions on f
it follows that mapping ef(i) to ei and all other canonical generators of HK to
e induces a homomorphism ? from HK to HK . It is straightforward to verify
that ? ? ? acts as the identity on all generators of HK . Therefore ? ? ? coincides
with the identity. The injectivity of ? follows.
We call the monomorphism constructed in Proposition 15 the canonical injection
and denote it by if .
5.3. Classi cation up to isomorphism. The main result of this subsection is
the following classification of HeckeKiselman semigroups up to isomorphism in
terms of the underlying mixed graphs.
Theorem 16. Let m, n ? N, ? ? Mm and ? ? Mn. Then the semigroups HK
and HK are isomorphic if and only if the graphs ? and ? are isomorphic. In
particular, if HK and HK are isomorphic, then m = n.
Proof. Let f : ? ? ? be an isomorphism of graphs with inverse g. By
Proposition 15 we have the corresponding natural injections if : HK ? HK and
ig : HK ? HK . By definition, both ig ? if and if ? ig act as identities on the
generators of HK and HK , respectively. Hence if and ig are mutually inverse
isomorphisms. This proves the ?if? part of the first claim of the theorem.
Lemma 17. We have Irr(HK ) = {e1, e2, . . . , en} = {w ? HK : c(w) = 1}.
Proof. From the definitions we see that Irr(HK ) is contained in any generating
system for HK , in particular, in {w ? HK : c(w) = 1}.
Since all canonical generators of HK are idempotents, it follows that {w ?
HK : c(w) ? 1} ? {e, e1, e2, . . . , en}. It is straightforward to verify that
{e1, e2, . . . , en} ? Irr(HK ), which completes the proof.
Assume that ? : HK ? HK is an isomorphism. Then ? induces a bijection
from Irr(HK ) to Irr(HK ), which implies m = n by comparing the cardinalities
of these sets (see Lemma 17). This proves the second claim of the theorem.
Let ei and ej be two different canonical generators of HK . By (6), in the case
when the graph ? contains no edge between i and j the elements ei and ej commute
in HK . As ? is an isomorphism, we get that ?(ei) = es and ?(ej ) = et commute
in HK . Using (6) again we obtain that the graph ? contains no edge between s
and t.
Similarly, comparing the subsemigroup of HK generated by ei and ej with the
subsemigroup of HK generated by ?(ei) and ?(ej ) for all other possibilities for
edges between i and j, we obtain that ? induces a graph isomorphism from ? to ?.
This proves the ?only if? part of the first claim of the theorem and thus completes
the proof.
Corollary 18. For ? ? Mn the set {e1, e2, . . . , en} is the unique irreducible
generating system of HK .
Proof. That {e1, e2, . . . , en} is an irreducible generating system of HK follows
from the definitions. On the other hand, that any generating system of HK
contains {e1, e2, . . . , en} follows from the proof of Lemma 17. This implies the
claim.
From the above it follows that the number of isomorphism classes of semigroups
HK , ? ? Mn, equals the number of simple digraphs. The latter is known as the
sequence A000273 of the OnLine Encyclopedia of Integer Sequences.
5.4. Some open problems. Here is a short list of some natural questions on
HeckeKiselman semigroups:
? For which ? is HK finite?
? For which ? is HK J trivial?
? For a fixed ?, what is the smallest n for which there is a faithful
representation of HK by n ? n matrices (over Z or C)?
? For a fixed ?, how to construct a faithful representation of HK by
(partial) transformations?
? What is a canonical form for an element of HK ?
Acknowledgment. We thank Ganna Kudryavtseva for her comments.
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