Grammic monoids with three generators

Abstract

Young tableaux are combinatorial objects whose construction can be achieved from words over a finite alphabet by row or column insertion as shown by Schensted sixty years ago. Recently Abram and Reutenauer studied the action the free monoid on the set of columns by slightly adapting the insertion algorithm. Since the number of columns is finite, this action yields a finite transformation monoid. Here we consider the action on the set of rows. We investigate this infinite monoid in the case of a 3 letter alphabet. In particular we show that it is the quotient of the free monoid relative to a congruence generated by the classical Knuth rules plus a unique extra rule.

Full text

arXiv:2207.13043v1 [math.CO] 26 Jul 2022
Grammic monoids with three generators
Christian Choffrut,
IRIF, Universit´
e Paris Cit´
e July 27, 2022
Contents
1 Introduction 1
2 Preliminaries 2
2.1 Young tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2.2 Grammic congruence . . . . . . . . . . . . . . . . . . . . . . . 4
3 Congruence in the case of three generators 9
3.1 Characterization of pairs of congruent words . . . . . . . . . . 9
3.2 Projections on subalphabets . . . . . . . . . . . . . . . . . . . 10
4 The Grammic monoid over {1,2,3}11
5 Further works 13
AbstractYoung tableaux are combinatorial objects whose construction
can be achieved from words over a finite alphabet by row or column insertion
as shown by Schensted sixty years ago. Recently Abram and Reutenauer
studied the action the free monoid on the set of columns by slightly adapting
the insertion algorithm. Since the number of columns is finite, this action
yields a finite transformation monoid. Here we consider the action on the
set of rows. We investigate this infinite monoid in the case of a 3 letter
alphabet. In particular we show that it is the quotient of the free monoid
relative to a congruence generated by the classical Knuth rules plus a unique
extra rule.
Keywords Young tableaux, monoid congruences.
1 Introduction
Let Abe a finite ordered alphabet. A Young tableau is a finite labeling of the
right top quarter discrete plane justified to the left and to the bottom in such
a way that the rows are nondecreasing from left to right and the columns are
increasing from bottom to top, see Example 1. Using Schensted’s method
of inserting a letter in a row (or in a column), a Young tableau can be
associated to a word in such a way that all letter occurrences appear once
1

and only once in the tableau. Two word are associated to the same tableau
if and only if they are equivalent in a congruence whose rules were given by
the Knuth [3, Thm. 6.6]. The quotient thus defined is the plactic monoid
denoted by Plactic(A).
Schensted’s left insertion of a letter in a column results in a column
plus a possible letter. Ignoring this extra letter yields an action of the free
monoid on the (finite) set of columns. The quotient of the free monoid by
the nuclear congruence is a finite monoid Styl(A) - the stylic monoid, many
interesting properties of which are studied in [1], such as the Jclasses of
these finite monoids of transformation, the notion of N-tableaux as repre-
sentations of congruence classes in the same way Young tableaux represent
congruence classes of the plactic monoid and a minimum set of rules defin-
ing the stylic congruence. Here we consider the “dual” problem of acting
on the set of rows which are finite nondecreasing sequences. The main de-
parture is that this monoid, called Grammic monoid as suggested to us by
Christophe Reutenauer is clearly no longer finite. In this note we consider
a specific aspect, namely the presentation of the monoid when Ahas three
generators. Based on the characterization of the pairs of words defining the
same mapping, our main result states that these pairs of words are precisely
those that are equivalent in the congruence generated by the Knuth rules
and the single new rule cbab =bcab where a < b < c. This result is optimal
since on 3 generators the plactic congruence is strictly finer than the gram-
mic congruence. An open issue is to generalize it to an arbitrary number of
generators. We conjecture that for 4 letters the congruence is generated by
the Knuth rules and the single new rule dbac =bdac where a < b ≤c < d.
I would like to relate the present result with the main Theorem 2.4. of
[4] investigating the algebra of the Plactic monoid over a field Kand which is
a continuation of [2, Theorem 9]. It states that the principal ideal generated
by cbab −bcab is one of the two principal ideals that are minimal minimal
prime ideals.
2 Preliminaries
2.1 Young tableaux
The reader is referred to the chapter [6] for an introduction of the plactic
monoid but the basics is recalled now.
Let Abe a totally ordered alphabet with kelements a1<··· < ak.
Given a element uof the free monoid A∗and a letter a∈Awe let |u|a
denote the number of occurrences of ain uand |u|denote its length. We
2

let the plactic congruence on A∗be denoted by ≡and we recall that it is
generated by the Knuth relations [3, Expression 6.7]
bac ≡bca where a < b ≤c,
acb ≡cab where a≤b < c (1) {eq:knuth}{eq:knuth}
A Young tableau is a labelling by occurrences of letters in Aof a lower-
order ideal of N2where the ordering is the natural product ordering on N.
Every row is non decreasing from left to right and every column is strictly
decreasing from top to bottom.
{ex:young-tableau}
Example 1. We assume A={a, b, c}with a < b < c. The following is a
Young tableau
c
bbc
aabb
The height of a Young tableau is the number of its rows. We will speak of
bottom, top rows and the like in the natural way. In example 1the tableau
has height 3, the bottom row is aabb and the top row is c.
Schensted gave in [7] an algorithm associating to a word wa Young
tableau P(w) in which every letter occurrence appears once and only once
in such a way that two words uand vare ≡-congruent if they are associated
to the same tableau, see e.g., [6, Proposition 6.2.3]. We recall it and re-
serve henceforth the term “row” for a nondecreasing sequence of letters and
“column” for a decreasing sequence of letters. This algorithm proceeds by
insertion of a letter bin a row b1···bp. If b > bpthen the insertion results in
the row b1···bpb. Otherwise, if biis the least letter greater than bthen the
insertion results in the row obtained by substituting bfor bi. More generally,
the insertion of bin a Young tableau consists of inserting bin the bottom
row. If bis greater than the greatest element in the row the procedure stops.
Otherwise bbumps the letter biwhich is inserted in the next upper row and
the process is iterated as long as necessary. The Young tableau associated
with a word wis obtained incrementally by starting from the empty tableau
and successively inserting the letters of w.
Dually the tableau can be constructed by defining how to insert a letter
cin a column c1···cp. If c > c1···cpthe resulting column is cc1···cp.
Otherwise if ciis the least letter greater than or equal to c(note the difference
between row and column insertion) then cis substituted for ciin the column.
The construction of the tableau follows the same pattern as in the case of row
3

insertion except that instead of proceeding form bottom to top, it proceeds
form let to right.
As a consequence of Schensted construction, there are two specific rep-
resentatives of an ≡-equivalence class, namely the row normal form which
is the sequence of the rows of the Young tableau from top to bottom and
the column normal form which is the sequence of the columns of the tableau
from left to right.
In example 1its row and column normal forms are c/bbc/aabb and
cba/ba/cb/b where we used the backslash symbol for visual convenience.
In particular cbbcaabb ≡cbabacbb.
2.2 Grammic congruence
We consider the action of a letter bon a row (we recall that it is a nonde-
creasing sequence of letters) which is the insertion of bas explained above
in Schensted procedure except that the possible letter which is expelled is
definitely lost. More precisely, we set
a1···an·b=


a1···an·bif an≤b
a1···ai−1bai+1 ···anif aiis the leftmost letter
greater than b,
The mapping extends to A∗and factors through the plactic monoid
because of the following equalities which can be readily verified.
a1···an·bac =a1···an·bca where a < b ≤c,
a1···an·acb =a1···an·cab where a≤b < c
We let u≡gram vdenote the grammic congruence between two words u
and vdefining the same mappings on the rows and call grammic monoid
the quotient of A∗by ≡gram. The congruence ≡gram is clearly coarser than
the congruence ≡. Alternatively, by identifying an1
1. . . ank
kwith the k-tuple
(n1,...,nk) we may consider that the free monoid acts on
ktimes
z}| {
N× · · · × Nin an
obvious way.
{le:action}
Lemma 2. Consider the alphabet a1, a2,...,akwith kletters. If the map-
pings are equal then the words have the same commutative image.
Proof. Consider a word u=α1···αn∈A∗of length nand the vector
λ= (n,n,...,n)∈Nk. By the definition of the action of letter on vectors,
for all proper prefixes wof u, no component of λ·uvanishes. Thus
4

(λ·u)i=n+|ua1|if i= 1
n+|uai| − |uai−1|if 1 < i ≤k
Therefore (λ·u)i= (λ·v)i= implies |u|ai=|v|ai.
{le:bottom-row}
Lemma 3. The bottom row of the Young tableau uis (0,0. . . , 0) ·u.
Proof. Indeed, we make two preliminary remarks. From the very defini-
tion of the action of the plactic monoids on the rows, it is clear that the
image of (0,0. . . , 0) by the product of pcolumns with the same right-
most letter equal to ai, is the vector with all components equal to 0 ex-
cept component iwhich is equal to p. Furthermore, the action of aion a
vector in Nkaffects no components smaller than to i. In other words for
u∈(A\ {a1,...,ai−1})∗and (n1,...,nk)∈Nkthere exists (mi,...,mk)
such that (n1,...,nk)·u= (n1,...,ni−1, mi,...,mk). The claim of the
lemma results from the previous observations and the fact that an element
of the plactic monoid has a representative as a product of non decreasing
columns in the following partial ordering over decreasing sequences of letters
a1···ap≥b1···bqif p≥qand ap−q+1 ≥b1,··· , ap≥bq(2)
(e.g., with Example 1we have cba ≥ba ≥cb ≥b).
Lemmas 2and 3have simple consequences for the case of two- and three-
letter alphabets.
{cor:two-letters}
Corollary 4. Over a two letter alphabet we have
u≡v⇔u≡gram v
Corollary 5. Iu≡gram vand u6≡plactic vthen either uor vor both have
height at least 3.
Proof. The bottom rows are equal. If they have only two rows, then the top
rows are also equal
{cor:a+c}
Corollary 6. Over a three-letter alphabet, if u≡gram vand u6≡plactic vthen
the row representatives of uand vin the plactic monoid are of the form
3a2b3c1d2e3fand 3a′2b′3c′1d′2e′3f′(3) {eq:general-form}{eq:general-form}
with a6=a′,0< b =b′,e=e′,0< d =d′,f=f′,a, a′≤b≤d,
b+c, b +c′≤d+eand a+c=a′+c′.
5

Proof. By the previous corollary the two words contain an occurrence of
each of the three letters and the bottom rows of their Young tableaux are
equal, hence e=e′,d=d′>0 and f=f′. The rows above the bottom
one contain the same number of occurrences of 2. If this number is null
then the row normal forms are u= 3c′1d2e3fand v= 3c′1d2e3fwhich
implies c=c′because the words are commutatively equivalent and hence
u≡plactic v. Equality a+c=a′+c′is yet another consequence of the
commutative equivalence. The remaining inequalities result from the fact
that the expressions of 3are row normal forms.
{le:non-vanishing}
Lemma 7. Let 0< i ≤kand u∈A∗in row normal form, u1/u2/···/up.
Set uj=bjwjwhere b1> b2>··· > bpare the initial letters of the rows.
Consider
mi=X
i<j X
a<bi
|uj|a
Then for all vectors x∈Nkwith xi> miand for all proper prefixes wof u
it holds (x·w)i>0.
Proof. Let rbe the greatest integer less than ior 0 if such an integer does
not exist. Since it is clear that (x·u1u2···up)i= (x·ur+1 u2···up)iwe may
assume r= 0 and compute (x·u1···up)i.
Rewrite mi=Pi<j µjand assume xi>Pi<j µj. Then by induction on
j= 1,...,p for all prefixes vof ujwe have
(x·u1···uj−1v)i−X
ℓ>j
µℓ>0
With example 1the previous computation yields x1>0, x2>2 (actually
it can be checked that x3>6 and thus (x1, x2+2, x3+6).w = (x1+|w|a, x2+
2 + |w|b− |w|a, x3+ 6) + |w|c− |w|b).
In the next lemma given a formal vector X= (ω1,...,ωi−1, xi, ωi+1,...,ωk)
where ωj,j6=iis an integer constant and xia variable and a word u∈A∗,
we view the expression X·uas a function of Ninto Nk.
{le:one-variable}
Lemma 8. Let u, v ∈A∗be two words of length n. If for all |xi| ≤ n+ 1
the functions X·uand X·vare equal, then these functions are equal for all
values of xi.
6

Proof. We set ∆u(xi) = X·u−Xand we show that, viewed as a function
of xi, it is constant for all xi>|u|. We set u=b1···bnand observe that by
Lemma 7for all prefixes w=b1···br,r= 0,...,n, we have (X·w)i≥n−r.
We show by induction that the function ∆b1···br(X) is a constant for all
xi≥n. If r= 0 then ∆1(xi) is the zero vector thus we assume r > 0. We
set w=vbrand br=ajand we compute the action of von X.
Case 1 j < i and (X·v)h= 0 for all j < h < i then
(X·w)j= (X·v)j+ 1,
(X·w)i= (X·v)i−1,(Lemma 7)
(X·w)ℓ= (X·v)ℓ, ℓ 6=i, j
Case 2 j < i and (X·v)h6= 0 for some least j < h < i then
(X·w)j= (X·v)j+ 1,
(X·w)h= (X·v)h−1,
(X·w)ℓ= (X·v)ℓ, ℓ 6=j, h
Case 3 j=iand (X·v)h6= 0 for some least i < h ≤nthen
(X·w)j= (X·v)j+ 1,
(X·w)h= (X·v)h−1,
(X·w)ℓ= (X·v)ℓ, ℓ 6=j, h
Case 4 j=iand (X·v)h= 0 for all h≤nthen
(X·w)j= (X·v)j+ 1,
(X·w)ℓ= (X·v)ℓ, ℓ 6=i
As a result for xi> n we have
X(xi)·u=X(n)·u+ (xi−n)[X(n+ 1) ·u−X(n)·u
In particular if uand vsatisfy the hypothesis of the Lemma, the two func-
tions X(xi)·uand X(xi)·vare equal for all xi∈N.
The next claim is a refinement of Lemma 2and shows that the equiva-
lence ≡gram can be computed. {pr:equivalence}
Proposition 9. Two words vand vdefine the same mappings on the rows
if and only if they have the same length and their restrictions coincide over
the subset {x1,...,xk)∈Nk|0≤xi≤max{|u|,|v|} + 1,1≤i≤k}.
7

Proof. Observe that if they have the same restriction over the set of vectors
of maximum coordinate max{|u|,|v|} + 1 then by Lemma 2they have the
same length n. Let Fbe the set of partial functions f:{1,...,n} 7→ Nand
let supp(f) be their domain of definition. With f∈Flet Xf= (α1,...,αk)
be a formal vector where αjis a constant f(j) when j∈supp(f) and αjis
a variable otherwise. The set of variables is {xi1,...,xip}where i1,...,ipis
the ordered subset {1,...,n} \ supp(f). Given a word u∈A∗we interpret
Xf·uas a function of Npinto Nkin the natural way. For (ξi1,...,ξip)∈Np
the value of the function is denoted Xf(ξi1,...,ξip)·u. We claim that for
all f∈F, for all formal vectors Xfand all words u, v, it holds
∀ξi1,...,ξip:Xf(ξi1,...,ξip)·u=Xf(ξi1,...,ξip)·v
⇔
∀ξi1,...,ξip≤n+ 1 : Xf(ξi1,...,ξip)·u=Xf(ξi1,...,ξip)·v(4) {eq:restriction}
We show that the bottom statement 4implies the top statement. We
proceed by induction on pby skipping the trivial case p= 0 and the case
p= 1 which is covered by Lemma 8. Define f′∈Fby the condition
f′(i) = 


f(i) if i∈supp(f)
ξipif i=ip
undefined otherwise
If ξip≤n+ 1 then we have
Xf(ξi1,...,ξip)·u
=Xf′(ξi1,...,ξip−1)·uinduction
=Xf′(ξi1,...,ξip−1)·v
=Xf(ξi1,...,ξip)·v
Now define g∈Fby the condition
g(i) = 


f(i) if i∈supp(f)
ξisif 1 ≤s < r
undefined otherwise
Observe that supp(g) = {1,...,n} \ {ir}. Then we have
Xf(ξi1,...,ξip)·u
=Xg(ξip)·uLemma 8
=Xf′(ξip)·v
=Xf(ξi1,...,ξip)·v
The bound n+ 1 is not sharp and could probably be improved by using
Lemma 8at the price of a more confuse statement for Proposition 9.
8

3 Congruence in the case of three generators
We specialize Lemma 2to three generators. The set of rows is identified with
the set of triples (x1, x2, x3)∈N3and the alphabet is renamed as {1,2,3}
(x1, x2, x3)·1 = 


(x1+ 1, x2−1, x3) if x2>0
(x1+ 1, x2, x3−1) if x2= 0 and x3>0
(x1+ 1, x2, x3) otherwise
(x1, x2, x3)·2 = (x1, x2+ 1,max{0, x3−1})
(x1, x2, x3)·3 = (x1, x2, x3+ 1)
(5) {eq:rules}{eq:rules}
3.1 Characterization of pairs of congruent words
Because of Corollary 4we may concentrate on the pairs of words given by
Corollary 6.{pr:cns}
Proposition 10. With the notations of Corollary 6, two words uand vare
≡gram-congruent if and only if c, c′≤e.
Proof. We compute the action of 3a2b3c1d2e3fon (x1, x2, x3)∈N3with
a≤b≤dand b+c≤d+e+f. Observe that
(x1, x2, x3)·3a2b3c1d2e3f= (x1, x2, x3)·3a2b3c1d2e+ (0,0, f )
Thus we may suppose f= 0 and it thus suffices to compute (x1, x2, x3)·
3a2b3c1d2e. Because of Lemma 6it all amounts to determine under which
conditions, for fixed b, d, e, there exist different pairs (a, c) which define the
same mappings on the rows. We compute via the rules (5)
(x1, x2, x3)
3a
−→ (x1, x2, x3+a)
2b
−→ (x1, x2+b, max(0, x3+a−b)
3c
−→ (x1, x2+b, max(c, x3+a−b+c))
Case 1 If x2+b≥dthen
(x1, x2+b, max(c, x3+a−b+c))
1d
−→ (x1+d, x2+b−d, max(c, x3+a−b+c))
2e
−→ (x1+d, x2+b−d+e, max(0,max(c, x3+a−b+c)) −e)
= (x1+d, x2+b−d+e, max(0, c −e, x3+a−b+c−e))
9

Case 2 If x2+b < d then
(x1, x2+b, max(c, x3+a−b+c))
1d
−→ (x1+d, 0,max(0,max(c, x3+a−b+c)) −(d−(x2+b)))
= (x1+d, 0,max(0, c −(d−(x2+b)), x3+a−b+c−(d−(x2+b)))
= (x1+d, 0,max(0, x2+c−d+b, x2+c−d+x3+a))
2e
−→ (x1+d, e, max(0,max(0, x2+c−d+b, x2+c−d+x3+a)−e)
= (x1+d, e, max(0, x2+c−d+b−e, x2+c−d+x3+a−e))
The same computation holds with a′and c′substituted for aand c.
Observe that in both cases 1 and 2 the first two components do not depend
on aand c.
Consider the third component. In expression max(0, c −e, x3+a−b+
c−e)) of case 1, for x3= 0, because of a−b≤0 the condition c−e > 0
yields max(0, c −e, x3+a−b+c−e)) = c−e. Thus if c > e the two words
are equivalent if and only if c=c′and a=a′. If c−e < 0 then for whatever
values for x3we have max(0, c−e, x3+a−b+c−e)) = max(0, x3+a−b+c−e)).
Then the two words are ≡gram-equivalent if and only if e′< c because
max(0, x3+a−b+c−e)) = max(0, x3+a′−b+c′−e)).
Example 11.
ccc/bbb/aaabbbbb ≡gram cc/bbbc/aaabbbbb ≡gram
c/bbbcc/aaabbbbb ≡gram bbbccc/aaabbbbb
Observe that for fixed values of b, d, e, f the maximum number of occur-
rences of 3 in w= 3a2b3c1d2e3fis so that wis not unique in its class is
b+e+f−1.
3.2 Projections on subalphabets
Given B(Awe let πbdenote the projection of A∗over B∗, i.e., the mor-
phism defined by πB(a) = aif a∈Band 1 if a6∈ B. Routine computations
show that over A={1,2,3}two ≡gram-equivalent words uand vso are their
projections over all subalphabets B⊆A. However, the converse does not
hold. Indeed, we have 23311223 6≡gram 23331122 but the row normal forms
of their projections over the subalphabets {1,2},{1,3}and {2,3}are equal,
respectively 21122, 33113 and 332223.
10

With |A|= 4 the projections of ≡gram-equivalent words are no longer nec-
essarily ≡gram-equivalent. Indeed, we have 4213 ≡gram 2413 1but 421 6≡gram
241 because the bottom rows of the tableaux are respectively 1 and 14.
4 The Grammic monoid over {1,2,3}
Theorem 12. Let ∼be the relation consisting of the pairs (xuy, xvy )∈
{0,1,2}∗where (u, v)is one of the rules in (1) or the new rule
(3212,2132) (6) {eq:relation2}{eq:relation2}
Then ≡gram is the transitive closure ∗
∼of the relation ∼. Furthermore,
if the words are in the form of Corollary 6, the number of applications of
the rule (6) to prove that they are ≡gram-equivalent is equal to |c−c′|.
Proof. We have 3212 ≡gram 2132 because for both words the image of
(x1, x2, x3) is (x1+ 1, x2+ 1,max{x3−1,0}thus u∗
∼vimplies u≡gram v.
We prove the converse. By Corollary 4we may assume that uand vhave an
occurrence of each of the three letters. We start with two different ≡gram-
congruent row normal forms
w(c) = 3a2b3c1d2e3fand w(c′) = 3a′2b3c′1d2e3f
with
0≤c, c′≤e, a +c=a′+c′.(7) {ex:abcdef}{ex:abcdef}
Observe that it suffices to consider the case c′=c+1 since if c′=c+t≤e′
then we have w(c)∼w(c+ 1) ∼ · · · ∼ w(c+t). We want to show that by
successive substitutions of one handside of equation 6for the other side we
can rewrite w(c) into w(c′).
We proceed by case study where Case 1 assumes d≤b+c+ 1 and Case
2 assumes d > b +c+ 1. We implicitly and repeatedly use the fact that two
columns commute if and only if one is a subset of the other.
Case 1. The two words are of the following column normal form.
(321)α(21)β(31)γ(32)δ2ǫ3fand (321)α−1(21)β+1(31)γ(32)δ+12ǫ−13fǫ > γ
We may factorize the words respectively as
1for both words the image of (x1, x2, x3, x4) is (x1+ 1, x 2, x3+1,max{x4−1,0}if x3= 0
and (x1+ 1, x2, x3, x4) otherwise
11

(321)α·(321)(21)β(31)γ(32)δ2·2ǫ−13f
(321)α·(21)β+1(31)γ(32)δ+1 ·2ǫ−13fwith ǫ≥γ+ 1, δ ≥0
Therefore it suffices to show how to pass from w1= (321)(21)β(31)γ(32)δ2γ+1
to w2= (21)β+1(31)γ(32)δ+12γ. This is obtained by applying the elemen-
tary rules concerning the commutation of two columns and equalities of the
form 1γ(32)γ= (31)γ2γ
(321)(21)β(31)γ(32)δ2γ2
≡(321)(21)β1γ(32)δ+γ2
≡(21)β1γ(321)(32)δ+γ2
≡(21)β1γ(321)2(32)δ+γ
∼(21)β1γ(21)(32)(32)δ+γ
≡(21)β+11γ(32)δ+γ+1
≡(21)β+1(31)γ2γ(32)δ+1
≡(21)β+1(31)γ(32)δ+1 2γ
thus w1
∗
∼w2. Furthermore the computation used the substitution 3212 ∼
(21)(32) on a unique occurrence which means that there exist two word u, v
such that
w1≡u3212v∼u2312v≡w2
Case 2 The two words are of the following form.
(321)α(21)β(31)γ1δ2ǫ3f
(321)α−1(21)β(31)γ+11δ2ǫ3fǫ≥γ+ 1, δ > 0
and as above is suffices to show how to pass from w1= (321)(21)β(31)γ1δ2γ+1
to w2= (21)β+1(31)γ+1 1δ−12γ+1. This results from the following boring
computation
(321)(21)β(31)γ1δ2γ+1
≡(321)(21)β1δ(31)γ2γ2
≡(321)(21)β1γ+δ(32)γ2
≡(21)β1γ+δ(321)2(32)γ
∼(21)β1γ+δ(21)(32)(32)γ
≡(21)β+11γ+δ(32)γ+1
≡(21)β+11δ(31)γ+1 2γ2
≡(21)β+1(31)γ+1 1δ2γ+1
As in the previous case we have w1
∗
∼w2and the computation used the
substitution 3212 ∼2132 on a unique occurrence.
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5 Further works
Our characterization of ≡gram-congruent words over a three-letter alphabet
allows one to answer some questions concerning arbitrary alphabets. E.g.,
in [1] the authors show the following property for all alphabets B⊆A: if
two words in B∗have the same action of all columns labelled by Bthey have
the same action on all columns labeled by A. It is routine to check that the
result holds when |B| ≤ 3 and when “row” is substituted for “column”.
A modest amount of computation led us to conjecture that in the case
of four letters, the congruence is generated by the Knuth rules plus the
rule dbac =badc with a < b ≤c < d which could probably be proved or
disproved by case study performed by a theorem prover. However, it is not
clear how far a brute force approach could be drawn and a more conceptual
approach is desirable.
References
[1] A. Abram and C. Reutenauer. The stylic monoid. The stylic monoid,
Semigroup Forum 105 (2022), https://doi.org/10.1007/s00233-022-
10285-3, 2022
[2] F. C´edo and J. Okni´nski Plactic algebras. J. of Algebra, 274, 97–117,
2002
[3] D. Knuth. Permutations, matrices and generalized Young tableaux.
Pacific J. Math., 34, 709–727, 1970.
[4] L. Kubat and J. Okni´nski. Plactic algebra of rank 3. Semigroup Forum,
84, 241–266, 2012
[5] A. Lascoux and M.-P. Sch¨utzenbeger Le mono¨ıde plaxique. Quaderni
della ricerca scientifica, Roma, CNR, n. 109, 1981
[6] A. Lascoux, B. Leclerc and J.-Y. Thibon. The plactic monoid. in:
M. Lothaire, Algebraic Combinatorics on Words, Cambridge , 144-172,
2002
[7] C. Schensted. Longest increasing and decreasing subsequences. in:
Canad. J. Math., 13, 179–191, 1961.
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