Examensarbete i matematik, 30 hp
Handledare och examinator: Volodymyr Mazorchuk Maj 2012
Effective representations of Hecke-Kiselman monoids of type A
Love Forsberg
Department of Mathematics
Uppsala University
MONOIDS OF TYPE A
LOVE FORSBERG
Abstract. We prove eectiveness of certain representations of Hecke-Kiselman monoids of type A constructed by Ganyushkin and Mazorchuk and also con- struct further classes of eective representations for these monoids. As a con- sequence the eective dimension of monoids of type A is determined. We also show that odd Fibonacci numbers appear as the cardinality of certain bipar- tite HK-monoids and count the number of multiplicity free elements in any HK-monoid of type A.
1. Introduction
In [3], Kiselman noted that certain operators c, l, m in convexity theory form a monoid that can be presented as
G =⟨c, l, m : c2= c, l2= l, m2= m,
clc = lcl = lc, cmc = mcm = mc, lml = mlm = ml⟩.
It can be noted that G is generated by three idempotents and that it has 8 = 23 idempotents.
To avoid confusion with for example representations of semigroup algebras (which do not appear in this thesis) we say that a semigroup representation is eective if it is injective. Note that a faithful semigroup algebra representation induces an eective semigroup representation, but that the converse is not true. The term is borrowed from [6], which deals with the (in general) very hard problem of nding a minimal eective representation for any given semigroup.
Kiselman proved that G has an eective representation by 3 × 3 matrices with positive integer coe cients. This was generalised by Ganyuskin and Mazorchuk to a series Kn of monoids with n generators called Kiselmans semigroups (unpublished)
Kn=⟨c1,· · · , cn: c2i = ci∀i, cicjci= cjcicj = cicj∀i ≤ j⟩.
In this setting we have G = K3.
Kudryavtseva and Mazorchuk later proved that Kn is generated by n idempo- tents and contains 2n idempotents. Moreover each Kn has an eective representa- tion by n × n matrices with positive integer coe cients [5]. The cardinality is in general only known to be nite [5], but we have |K1|=2, |K2|=5, |K3|=18 [3] and
|K4|=115, |K5|=1710 [1]. Some more terms (|K6| − |K14|) are claimed on oeis.org [8], but are not contained in any of the references.
1
On the other hand we can associate to every (disjoint union of) simply laced Dynkin diagram(s) Γ the 0−Hecke monoid HΓ which can be presented as
HΓ=⟨v ∈ V (Γ) : v2= v∀v ∈ V (Γ),
vwv = wvw∀{v, w} ∈ E(Γ), vw = wv∀{v, w} ̸∈ E(Γ)⟩.
It can be shown that the semigroup algebra of HΓ is isomorphic to the special- ization of the Hecke algebra Hq(WΓ)at q = 0, with WΓ the Weyl group of Γ. This explains the name.
The notion of a Hecke-Kiselman monoid was introduced in [4] and is de ned as follows
De nition 1. Let Γ = (V (Γ), E(Γ)) be a simple directed graph. The Hecke- Kiselman monoid HKΓof Γ is the free semigroup (V (Γ))∗generated by the vertices quotioned by the relations
(1) a² = a for all a ∈ V (Γ).
(2) If there is no edge between a and b, then ab = ba.
(3) If a → b, then aba = bab = ab.
(4) If a b, then aba = bab
We will call these relations (including a2= a) edge relations.
The monoids HKΓ play a central role in a paper by Grensing [2, 2012] and also appear in her Ph.D. thesis [7, 2011]. Her interest is in projection functors that turn out to obey the edge relations.
In the broader setting of Hecke-Kiselman moniods, Kn comes from the graph with V (Γ) = {1, 2, · · · .n} and E(Γ)={(i,j)|i<j}. As HKΓ can be viewed as a generalisation of Kn, it is reasonable to ask the following questions
(1) Does HKΓ have 2n idempotents, where n is the number of vertices? If no, then how many?
(2) Does HKΓhave an eective representation by n×n matrices? With positive integer coe cients?
(3) Is HKΓ nite? If so, can we nd the number of elements?
Every graph in this thesis is assumed to have no double edges a b, unless ex- plicitely stated otherwise. This is mainly becuse the type of representation we will study is useless for graphs with double edges a b (see Remark 6). We will use the notation n = {1, 2, · · · , n}.
Remark 2. It is easy to see that if Γ = Γ1⊔ Γ2 we have that every vertex from Γ1
commutes with every vertex from Γ2and hence HKΓ= HKΓ1⊕HKΓ2as monoids.
Hence we only need to study connected graphs.
Since elements in HKΓ are equivalence classes of words in the alphabet V (Γ) of vertices, we will denote by [w] ∈ HKΓ an element and w′ ∈ [w] a word, i.e.
a representative of that equivalence class. We will use bold fonts to distuinguish general words from vertices, i.e. a ∈ V (Γ) is a vertex, but w ∈ (V (Γ))∗ is a word.
The empty word will be denoted ε. We will make use of two binary relations on (V (Γ))∗. Let w ≈ w′ be de ned as w and w′dier by at most one edge relation, and w ∼ w′ :⇐⇒ [w] = [w′]. Note that the relation ∼ is the transitive closure of the relation ≈ and thus in particular w ≈ w′ is a stronger statement than w ∼ w′.
It is easy to see that all edge relations preserve existence of a given vertex in a word (while multiplicity will vary). Thus existence of a particular vertex is also a property of elements in HKΓ.
De nition 3. The set of vertices that are present in an element [w] ∈ HKΓ is denoted by c[w] and is called the content of [w]. The set of elements [w] which have c[w] = V (Γ) will be denoted by MaxΓ and it's cardinality by m(Γ).
Let R be a unital commutative ring without zero divisors and W =⊕
v∈V (Γ)Rv.
Let vi, i∈ n, be an enumeration of the vertecis in Γ and let
θfi(vj) =
{ v∑j, i̸= j
vk→vif (vk→ vi)vk, i = j
for a xed function f : E(Γ) → R. For readability, we assume the notation fki:=f(vk → vi) and θi= θif, when f is clear from context. We de ne the function Rf : V (Γ) → EndR(W )by vi 7→ θif and extend it to words in the obvious way, namely that juxtaposition becomes composition Rf(vw) = Rf(v)◦ Rf(w). The endomorphisms θi will be referred to as atomic.
The following theorem is a straightforward generalisation of linerar integral representations in [4, 3.2] that nevertheless needs to be proved.
Theorem 4. Rf is well-de ned as a map from HKΓ→ EndR(W )and is a repre- sentation.
Proof. Since Rf is de ned as a composition of atomic endomorphisms, it is a rep- resentation if it is wellde ned as a map HKΓ→ EndR(W ). We need to show that Rf respects the edge relations.
Relation a² = a. θi◦ θi(vj) =
{∑ vj, j̸= i
vk→vifkiθi(vk), j = i
But since Γ is loop free, vk → vi⇒ k ̸= i and thus θi(vk) = vk. Thus θi² = θi
for all i ∈ n.
Relation ab = ba for a and b non-adjacent.
θi◦ θj(vk) =
{∑ θi(vk), k̸= j
vl→vjfljθi(vl), k = j =
vk, k̸= j, i
∑
vl→viflivl, k = i
∑
vl→vjfljvl, k = j
The reason that θi(vl) = vlis that vl→ vjand vi ̸→ vj. Since the right hand side is symmetrical we get θi◦ θj = θj◦ θi for all i, j such that vi and vj non-adjacent.
Relation aba = bab = ab for a → b.
θi◦ θj◦ θi(vk) =
{∑ θi◦ θj(vk), k̸= i
vl→vifliθi◦ θj(vl), k = i =
vk, k̸= i, j
∑
vl→vjfljθi(vl) k = j
∑
vl→viflivl, k = i Since vi→ vj we have
∑
vl→vj
fljθi(vl) = ∑
vl→vj,l̸=i
fljvl+ fij ∑
vl→vi
flivl
On the other hand, we have θj◦θi◦θj(vk) =
{∑ θj◦ θi(vk), k̸= j
vl→vjfljθj◦ θi(vl), k = j =
vk, k̸= i, j
∑
vl→viflivl k = i
∑
vl→vjfljθj◦ θi(vl), k = j
But ∑
vl→vj
fljθj◦ θi(vl) = ∑
vl→vj,l̸=i
fljθj(vl) + fij ∑
vl→vi
fliθj(vl).
Since (vl→ vj or vl→ vi)⇒ vl̸= vj we have
∑
vl→vj
fljθj◦ θi(vl) = ∑
vl→vj,l̸=i
fljvl+ fij
∑
vl→vi
flivl.
θi◦ θj(vk) =
{∑ θi(vk), k̸= j
vl→vjfljθi(vl), k = j =
vk, k̸= i, j
∑
vl→viflivl k = i
∑
vl→vjfljθi(vl), k = j
But ∑
vl→vj
fljθi(vl) = ∑
vl→vj,l̸=i
fljvl+ fij
∑
vl→vi
flivl.
Thus θi◦ θj◦ θi = θj◦ θi◦ θj= θi◦ θj for vi→ vj. In the case Γ = 1 → 2 → · · · → n, fi(i+1) ≡ 1 and R = Z we have that Rf is eective, as proved in [4]. We will by Rk denote the unique (for each graph and each ring) representation Rf with f(vi→ vj)≡ k.
Lemma 5. Let Γ = 1 → 2 → · · · → n. Then any choice of f : E(Γ) → R such that fi(i+1)̸= 0 for all i ∈ n − 1 gives an eective representation Rf of HKΓ. Proof. First observe that Rf[w](x) = cx′ is a sum of at most one non-zero terms.
Let x′ = xi → xi+1 → · · · → xj = x be the unique oriented path from x′ to x.
Then Rf[w](x) = cx′⇒ Rf[w](x) = θi◦ θi+1◦ · · · ◦ θj(x). Thus the only thing that dier between Rf[w]and θi◦ θi+1◦ · · · ◦ θj is endomorphisms that act like unity on x.
Since this is done using only fij ̸= 0 the map cx 7→ x, c ̸= 0 is injective. Indeed, for R1[w](x) = x′we have Rf[w](x) = (Πjk=i−1fk(k+1))x′which given cx,x′=Πjk=i−1fk(k+1)
constructed from the unique path gives the inverse x 7→ cx,x′x′ which proves the claim. Recall that R is assumed to have no zero divisors. Remark 6. If Γ contains a double edge we have that Rf in general is not a represen- tation, and if it is, it is not eective. Indeed, if the graph v1 v2 is a subgraph of Γwe should have that θ1◦ θ2◦ θ1= θ2◦ θ1◦ θ2. Let us simplify notation by setting Si=∑
vk→vi,k̸=1,2fkivk for i ∈ {1, 2}. Note that θi(Sj) = Sj for i, j ∈ {1, 2}.
θ1◦θ2◦θ1(v1) = θ1◦θ2(f21v2+S1) = θ1(f12f21v1+S1+S2) = f212f12v2+2S1+S2
and
θ2◦ θ1◦ θ2(v1) = θ2◦ θ1(v1) = θ2(c21v2+ S1) = f12f21v1+ S1+ S2
Since the left hand sides coincide, so do the right hand sides.
f212f12v2+ 2S1+ S2= f12f21v1+ S1+ S2 ⇐⇒ f212f12v2+ S1= f12f21v1
This means that f12f21 = 0 = S1, which since R has no zero divisors implies that f12= 0or f21= 0. By symmetry we get S2= 0. W.l.o.g. assume f12= 0. Then
θ2◦ θ1(v1) = θ2(f21v2+ S1) = f12f21v1= 0 = θ1◦ θ2◦ θ1(v1) and
θ2◦ θ1(v2) = θ2(v2) = f12v1+ S2= 0 = θ1◦ θ2◦ θ1(v2)
Since θi(vk) = vk for i ∈ {1, 2} and k ̸= 1, 2 we have that Rf[v1v2v1] = θ1◦ θ2◦ θ1= θ2◦ θ1= Rf[v2v1].
But since v1v2v1̸= v2v1 ([v2v1] ={v2k+1vl+11 |k, l ∈ N} does not include v1v2v1) we have that Rf is not eective.
Theorem 7. Let C ⊂ Γ be an oriented cycle of length l ≥ 3 and let [w] ∈ HKΓ be an element such that V (C) ⊂ c[w]. Then [wk]are dierent elements for all k ∈ N.
Proof. Let p : HKΓ→ HKC be the projection that removes all vertices and egdes not in C. If we prove that p[wk] ∈ HKC are dierent for all k ∈ N, then so are [wk]∈ HKΓ, so it is su cient to prove the claim for Γ=C=v1→ v2→ · · · → vl→ v1. For simple notation, we adopt cyclic indexing, i.e the indicies i ∈ Zl add modulo l.
We will need a representation to separate dierent elements, and we choose R2
with the ground ring R = Z. Let [w] ∈ MaxC. Since each atomic representation θi takes vi to 2vi−1 and [w] contains every vertex we get that R2[w](vi)=2mivt(i), where mi ≥ 1 and t is a transformation on the set l. By forgetting momentarily the transformation in the indicies and concentrating on the coe cient we see that R2[wk](vi) = 2mi12mi2· · · 2mikvT (i), i.e. that each application of R2[w]raises the coe cient. In particular we have that R2[wk1](vi)̸= R2[wk2](vi) for k1̸= k2, which implies R2[wk1]̸= R2[wk2] for k1̸= k2. Corollary 8. The number of idempotents in HKΓ equals the number of (directed, length l ≥ 3) cycle free full subgraphs of Γ.
Proof. For a subgraph Γ′ not containing any directed cycle of length l ≥ 3 we have that HKΓ′ is a quotient of the Kiselman monoid by certain edge relations. Indeed, if we view the graph as a relation on the edges and take the transitive closure, we get a partial order ≺. Since any partial order can be extended to a total order, this induces an inclusion Γ′ ⊂ Gm, where Gmis the graph of the Kiselman semigroup Km(the graph of a Hecke-Kiselman monoid is unique up to isomorphism by [4]).
Since edge relations respect content and Km has precisely one idempotent for each content [5] it follows that HKΓ′ has precisely one idempotent of maximal content. Thus HKΓ has precicely one idempotent with content V (Γ′).
For a subgraph Γ′ containing a directed cycle of length l ≥ 3 we are in the situation described by the previous theorem and thus HKΓ′ has no idempotent of maximal content. Equivalently, HKΓ has no idempotent with content V (Γ′). De nition 9. Let Γ be obtained from the Dynkin diagram An by choosing a direction on every edge. Then Γ is said to be of type An. Because of the bijectivity (up to isomorphism) between graphs and their Hecke-Kiselman monoids we can de ne a Hecke-Kiselman monoid to be of type An if its corresponding graph is of type An.
The same de nition is used in [4], but with the notation Γ is of type A instead of Γ is of type An.
a1 // a2oo a3 oo · · · // an
Figure 1.1. An example of a graph of type An
a //
>
>>
>>
>>
> b
c d
Figure 1.2. In this graph deg+(a) = 2,deg+(b) = 1,deg+(c) = deg+(d) = 0
De nition 10. For any directed graph Γ we have the indegree function deg−(v) : V (Γ)→ N, which counts how many edges that point towards v, and the outdegree function deg+(v) : V (Γ)→ N, which counts how many edges that point from v.
De nition 11. A vertex v satisfying deg+(v)deg−(v) = 0is called a kink. Equiv- alently v is a kink i v is a source or v is a sink.
Lemma 12. Any element of [w′]∈ HKΓcontains a word w where all kinks appear with multiplicity at most 1. This will later be phrased as w is multiplicity free with respect to a or a is multiplicity free in w.
Proof. Let ←−
Γ be the graph obtained from Γ by reversing all arrows. By [4] we have that HKΓ is anti-isomorphic with HK←−Γ. Thus by reversing every word w we may assume any given kink to be a source. Let w be a word with more than one occurance of a kink a. For any x ∈ V (Γ) we have one of the following
(1) x = a ⇒ ax = aa = aaa = axa (2) a → x ⇒ ax = axa
(3) a and x are nonadjacent ⇒ ax = aax = axa Thus ax = axa for all x ∈ V (Γ)
w = w1ax1x2· · · xnaw2≈ w1ax1ax2· · · xnaw2≈ w1ax1ax2a· · · xnaw2≈ · · ·
≈ w1ax1ax2a· · · axnaaw2≈ w1ax1ax2a· · · axnaw2≈ · · · ≈ w1ax1x2· · · xnw2
This gives us an equivalent word which up to deletion of one a is precisely w.
As long as there are multiple copies we repeat the procedure. What remains in the end is w with all but the leftmost occurance of a removed (for sources) or all but the rightmost occurance of a (for sinks). Since no other letter was changed, we can
delete kinks independantly of each other.
De nition 13. Let A ⊂ V (Γ) be a set of vertices. We de ne M FA⊂ (V (Γ))∗as the set of words which are multiplicity free with respect to all vertices in A.
Lemma 14. Let A be a set of kinks (not necessarily all) in Γ. Given two words w∼ w' such that w, w′∈ M FAwe can nd a series of words w=w1≈ · · · ≈ wk=w' where wi∈ M FA for all i.
Proof. By the de nition of the relation ∼ there exists a series of words w = w1≈ w2≈ · · · ≈ wk = w′,
so what remains to be shown is that wi can be assumed to be in M FA.
Let us rst concentrate on one kink a ∈ A. As can be seen in the previous lemma we may remove all but the leftmost or all but the rightmost appearances
a
AAAAAAAAA b
///
////
////
////
/ c
~~}}}}}}}}}
a, 2
""EEEEEEEE c, 2
||yyyyyyyy d
= =
==
==
==
= e
f
d, 1
B B BB BB BB
B e, 1
g ONMLHIJKh i ONMLHIJKh, 0
~~|||||||||
x // y // z
^^>>>>>>>>>
x, 3 // y, 2 // z, 1
``BBBB BBBBB
Γ Sh
Figure 1.3. The source graph Shof a vertex h ∈ Γ. The numbers at each vertex in Shindicate the rst Viwhich the vertex appeared in. In this case, both a directed cycle h ← z ← y ← x ← h and an oriented cycle a → e → h ← d ← a appear.
of a depending on if a is a sink or a source. Let ψa be a function on words which removes all super uous appearances of a accordingly. W.l.o.g. assume we keep the leftmost a. Note that ψa(w) = w and ψa(w′) = w′. We need to split into cases depending on the relation used between wi and wi+1
(1) x2 = x, xy = yxor xyx = yxy = xy for x, y ̸= a will give wi ≈ wi+1 ⇒ ψa(wi)≈ ψa(wi+1)with the same relation as the dierance.
(2) a2 = a, ax = xa or axa = xax = ax when the leftmost a is not involved gives ε = ε, x = x or x = x2= xunder ψaand thus wi≈ wi+1⇒ ψa(wi)≈ ψa(wi+1).
(3) a2= a, ax = xa or axa = xax = ax when the leftmost a is involved gives a = a, ax = xa or ax = xax = ax under ψa and thus wi ≈ wi+1 ⇒ ψa(wi)≈ ψa(wi+1).
To get a series which is multiplicity free for all chosen kinks ai, i∈ I we form ψai
in the same manner and set ψ = ψa1◦ ψa2◦ · · · ◦ ψaj. Then
w = ψ(w) = ψ(w0)≈ ψ(w1)≈ · · · ≈ ψ(wk) = ψ(w′) = w′
is a series which satis es our claim.
De nition 15. Given a graph Γ and a vertex a we de ne the source graph Saas the full subgraph of Γ with vertices V (Sa) ={v ∈ Γ|there is a directed path from v to a}.
Alternatively we can de ne the vertex set V (Sa)recursively as (1) V0={a}
(2) Vi+1= Vi∪ {v ∈ Γ|∃u ∈ Vi: v→ u}
For some N ≤ n = |Γ| we have that VN +k = . . . = VN +1= VN for all k ∈ N and we set V (Sa) = VN. Note that the source graph can contain cycles.
Lemma 16. For a ∈ Γ let pa : HKΓ → HKSa be the projection which removes vertices not in the source graph of a. Then for [w] ∈ HKΓ and a function f : E(Γ)→ R we have that Rf[w](a) =Rfpa[w](a).
Proof. By the construction of Rf we have Rf[w](a) = Σv∈Sacvvfor some constants cv ∈ R for any element [w] ∈ HKΓ. Let w = w1xw2be a word with x ̸∈ Sa. Then
Rf[w1xw2](a) = Rf[w1]Rf[x]Rf[w2](a) =
Rf[w1]Rf[x]Σv∈Sacvv = Rf[w1]Σv∈SacvRf[x](v).
But since x ̸∈ Sa and v ∈ Sa we have x ̸= v and thus Rf[x](v) = v by de nition of Rf. Thus
Rf[w1]Σv∈SacvRf[x](v) = Rf[w1]Σv∈Sacvv = Rf[w1]Rf[w2](a) = Rf[w1w2](a).
As long as we have x ̸∈ Sa present in w we can cancel it with the same resulting action on a. Eventually we will get to pa[w]and have Rf[w](a) = Rfpa[w](a). De nition 17. For Γ′⊂ Γ a full subgraph and a representation Rf:HKΓ →EndR(W) there is one canonical representation R′f : HKΓ′ → EndR(⊕
v∈V (Γ′)Rv) de ned on the atomic representations as
θ′i(vj) =
{ v∑j, i̸= j
vk→vi,vk∈V (Γ′)fkivk, i = j We say that R′f is restricted from Rf.
If Γ =∪
i∈IΓiis an edge disjoint union and Rfi: HKΓi → EndR(⊕
v∈V (Γi)Rv) are representation there is one unique representation Rf : HKΓ → EndR(W ) de ned by f(vj → vk) = fi(vj → vk)for the unique i such that the edge vj → vk
lies in Γi. We say that Rf is up induced from {Rfi}i∈I.
Theorem 18. For Γ′ ⊂ Γ a full subgraph, let pV (Γ′) : W → ⊕
v∈Γ′Rv be the canonical projection in modules and p : HKΓ → HKΓ′ the canonical projection in monoids. Assume that every directed path that starts and ends in Γ′ is completely contained in Γ′. Then pV (Γ′)◦ Rf[w]|V (Γ′)= R′fp[w], where R′f is restricted from Rf (as in the de nition above).
Proof. We prove the claim by showing that it holds for atomic representations and compositions with one of the representations beeing atomic.
pV (Γ′)θi(vj) =
{ pV (Γ′)(vj), i̸= j pV (Γ′)(∑
vk→vifkivk), i = j
vj, i̸= j, vj∈ Γ′
0, i̸= j, vj̸∈ Γ′
∑
vk→vi,vk∈V (Γ′)fkivk, i = j But we also have
= R′fp[vi](vj) =
{ R′f[vi](vj), vi∈ Γ′ 0, vi̸∈ Γ′
=
vj, i̸= j, vi ∈ Γ′
∑
vk→vi,vk∈V (Γ′)fkivk, i = j, vi ∈ Γ′
0, i = j, vi ̸∈ Γ′
Fix some a ∈ Γ′ and let Sain:= Sa∩ Γ′and Saout := Sa\ Γ′. Every edge between vout ∈ Saout and vin∈ Sainhas to be in the direction vout→ vin. Indeed, if there is an edge vin→ vout we can add it to the path vout → v1 → · · · → vk → a (part of the de nition of Sa) to obtain vin → vout → v1 → · · · → vk → a, a directed path which starts and ends in Γ′, but has at least one vertex vout not in Γ′, which would contradict our assumption. Thus θi(vout) =∑
vk→vout,vk̸∈Γcvkvkfor all vout∈ Saout. Now assume pV (Γ′)◦ Rf[w]|V (Γ′)= R′fp[w]for some [w] ∈ HKΓ. Then
pV (Γ′)◦ Rf[vjw]|V (Γ′)(vi) = pV (Γ′)( ∑
vk→vi
fkiθj(vk)) =
∑
vk→vi
fkipV (Γ′)(θj(vk)) = ∑
vk→vi,vk∈Γ
fkipV (Γ′)(θj(vk)).
On the other hand,
R′fp[vjw](vi) = ∑
vk→vi,vk∈V (Γ′)
fkiθj′(vk),
and pV (Γ′)◦ θj=θj', so we have the desired equality.
2. Serial decomposition
We may try to understand a large graph and it's corresponding semigroup in terms of subgraphs.
De nition 19. Let Γ = ∪
i∈nΓi be a graph which can be desribed as an edge disjoint union of graphs that each share one vertex with the next (a cut vertex) and such that each such vertex is a kink. We call such a graph a chain and the subgraphs Γi links. We enumerate the cut vertecis as ai ∈ Γi∩ Γi+1. By Rfi we mean a representation corresponding to the link Γi. By Γ′i = Γi\ {ai−1, ai} we mean the interior of a link.
· · ·
· · · · · ·
↑↓ Γ1 a1
00..
nnpp
Γ2 a2
nnpp
// · · · // an−1 Γn ↑↓
· · ·
OO
· · · · · ·
OO\\
Figure 2.1. An example of a chain. The signs s(ai)may come in any order.
For each link we have the projection pi : HKΓ → HKΓi which is obtained by introducing the relations x = ε for all x ∈ V (Γ \ Γi). We combine them to the pro- jection p : HKΓ→ HKΓ1× HKΓ2× · · · × HKΓn, p[w] = (p1[w], p2[w], . . . , pn[w]). Theorem 20. Let Γ be a chain and p as above. Then p is injective.
· · ·
· · ·
· · ·
↑↓ Γ1 a1 Γ2 a2 Γ3 a3
ppnn
// · · ·
· · ·
OO BB
· · ·
\\ BB
· · ·
\\
Figure 2.2. 2-chains as links in larger 2-chains
Proof. By grouping together links we see that it is enough to show the claim for a chain with two links (for short a 2-chain). Indeed, if we can join two links Γ1, Γ2
and have p injective we can say that this new chain Γ1∪ Γ2 is a link in a larger chain (Γ1∪ Γ2)∪ Γ3and continue as long as we have links left. Eventually we reach Γ = (· · · (Γ1∪ Γ2)∪ Γ3)∪ · · · Γn−1)∪ Γn. Now
˜
p[w] = (p(1)[w], pn[w]) = ((p(2)[w], pn−1), pn[w]) = . . .
= ((· · · ((p1[w], p2[w])· · · ), pn−1[w]), pn[w]) be obtained from composition of the projections of each 2-chain. It is clear that ˜p is injective i p is injective.
Thus let Γ = Γ1∪ Γ2with a the joining kink. We either have a∈ p1[w]∨a∈ p2[w]
or a̸∈ p1[w]∨a̸∈ p2[w]. By letting α = a if the rst holds and α = ε otherwise we get that there are words w1, w2 such that w1 = x1αx2, w2 = y1αy2 and [w1] = p1[w], [w2] = p2[w], since a is a kink.
Now we de ne w = x1y1αx2y2. Note that x1, x2 commute with y1, y2. In fact this is wellde ned as a function between elements since if x1αx2= ˜x1α˜x2 and y1αy2= ˜y1α˜y2we have
x1y1αx2y2= x1y1αy2x2= x1y˜1α˜y2x2= ˜y1x1αx2y˜2= ˜y1x˜1α˜x2y˜2= ˜x1y˜1α˜x2y˜2, so equivalent words give an equivalent word. Let us denote this function by φ : Im(p)→ HKΓ.
Any element [w′]∈ HKΓhas a word on the form w = w1αw2, but since Γ\{a} is disconnected each wi= xiyi, i∈ 2 where [x1], [x2]∈ HKΓ′1 and [y1], [y2]∈ HKΓ′2.
φ◦ p([x1y1αx2y2]) = φ([x1αx2], [y1αy2]) = [x1y1αx2y2]and
p◦ φ([x1αx2], [y1αy2]) = p([x1y1αx2y2]) = ([x1αx2], [y1αy2]) so φ and p are
inverses of each other.
Corollary 21. If Γ is a chain and Rfi is an eective representation of the monoid corresponding to each link, then Rf up induced from {Rfi} is an eective represen- tation of HKΓ.
Proof. Given Rf[x] for some [x] ∈ HKΓ we can get Rfipi[x] by lemma 18. But since all Rfi are eective this uniqely gives pi[x] for all i. Now apply φ from the
previous theorem to obtain [x].
Remark 22. Since the Hecke-Kiselman monoid of the graph v1 → v2 → · · · → vn
is known (see remark 5) to have the eective representations Rf, fij ̸= 0 we get that any graph of type An has (by abuse of notation) the eective representations Rf, fij ̸= 0. This answers the question 8 posed in [4]: Is the representation (R1) constructed above faithful (as semigroup representation)(when Γ is of type An)?
a rmatively. Here faithful means precisely eective, i.e. that the semigroup representation is injective.
Corollary 23. Let Γ be a chain. Then m(Γ) = Πnk=1m(Γi).
Proof. The only property that separates Im(p) from general elements in HKΓ1× HKΓ2× · · · × HKΓn is that kinks are simoultaneously in or not in neighbouring elements. For elements with maximal content this retraint is void and we may choose elements independantly from the links. Thus we may use the multiplicative
principle.
Let Γ be of type Anwith k links and let libe the lenght of link i. By introducing a few help functions we can nd a somewhat explicit formula for |HKΓ|. Let δi(A) = 1 if i ∈ A and = 0 otherwise for i ∈ k − 1. Let
ci(A) =
Cli−1, if δi−1(A) = δi(A) = 0 Cli− Cli−1, if δi−1(A) + δi(A) = 1 Cli+1− 2Cli+ Cli−1, if δi−1(A) = δi(A) = 1 for i ∈ k − 1 \ {1}, let
c1(A) =
{ Cl1, if δ1(A) = 0 Cl1+1− Cl1, if δ1(A) = 1 and let
ck(A) =
{ Clk, if δk−1(A) = 0 Clk+1− Cl1, if δk−1(A) = 1
Corollary 24. Let Γ be of type An and let li be the length of the i:th link. Then m(Γ) = Πni=1Cli is a product of catalan numbers, and |HKΓ| =∑
A⊂n(∏k
i=1ci(A)). Proof. For the rst claim, note that it su ces to know m(Γi)for every link by the previuos corollary. By [4] there is an isomorphism f between HKΓi and the semi- group of order preserving and order decreasing transformations on the set li+ 1. In this setting f(vi)(j) = j if j ̸= i + 1 and i otherwise. It follows that the image of an element of maximal content is a transformation that decreases all numbers (except 1) with at least 1.
On the other hand, any order preserving and order decreasing transformation which decreases all numbers (except 1) with at least 1has to have maximal content since the only way to change any number i is to include vi−1 in the element. By a simple indexshift it is easy to see that
(1) the number of order decreasing and order preserving transformations on lk+ 1 that decrease every number (except 1) with at least 1
(2) the number of order decreasing and order preserving transformations on lk
are equal and equal Clk.
For the second claim we need know how many elements in HKΓi that have /do not have ai−1, airespectively. We may start with the number of elements that have
