Introduction - reiner/REU/GaetzMastrianniPatriasPeck… · K-KNUTH EQUIVALENCE FOR INCREASING...

K-KNUTH EQUIVALENCE FOR INCREASING TABLEAUX

CHRISTIAN GAETZ, MICHELLE MASTRIANNI, REBECCA PATRIAS, HAILEE PECK,COLLEEN ROBICHAUX, DAVID SCHWEIN, AND KA YU TAM

Abstract. A K-theoretic analogue of RSK insertion and Knuth equivalence

relations was first introduced in 2006 [1]. The resulting K-Knuth equivalencerelations on words and increasing tableaux on [n] has prompted investigation

into the equivalence classes of tableaux arising from these relations. Of par-

ticular interest are the tableaux that are unique in their class, which we referto as unique rectification targets (URTs). In this paper we give several new

families of URTs and a bound on the length of intermediate words connect-

ing two K-Knuth equivalent words. In addition, we describe an algorithm todetermine if two words are K-Knuth equivalent and to compute all K-Knuth

equivalence classes of tableaux on [n].

1. Introduction

In 2006, Buch et al. introduced a new combinatorial algorithm called Heckeinsertion, used to insert a word into an increasing tableau [1]. The algorithm is aK-theoretic analogue of the well-known Schensted algorithm for the insertion of aword into a semistandard Young tableau.

If two words insert into the same tableau via Schensted’s insertion algorithm,they are said to be Knuth equivalent and can be connected via the Knuth equiv-alence relations. Knuth equivalence has a K-theoretic analogue referred to as K-Knuth equivalence, also introduced in [1]. An important difference between Knuthequivalence and K-Knuth equivalence is that, while insertion equivalence via theSchensted algorithm (resp. the Hecke algorithm) implies Knuth equivalence (resp.K-Knuth equivalence), the converse holds for the standard version but not for theK-theoretic version. Hence two words can be K-Knuth equivalent but insert intodifferent tableaux via the Hecke insertion algorithm.

A K-Knuth equivalence class typically contains words from different insertionclasses. There are some K-Knuth classes, however, for which all words in theclass insert into the same tableau. A class with this property is called a uniquerectification class, and its corresponding insertion tableau is a unique rectificationtarget (URT). In both [1] and [6], Hecke insertion and K-Knuth equivalence wereused to rederive a K-theoretic version of the Littlewood-Richardson rule for thecohomology rings of Grassmanians. In order to get a working version of this rule,non-URTs needed to be avoided. Hence Patrias and Pylyavskyy [6] posed thefollowing natural question, an open problem.

Problem 1. Characterize all URTs or at least provide an efficient algorithm todetermine if a given tableau is a URT.

Date: September 19, 2014.

1

2 GAETZ, MASTRIANNI, PATRIAS, PECK, ROBICHAUX, SCHWEIN, AND TAM

This paper makes partial progress towards answering Problem 1. In more detail,we will extend previous results, many from a paper of Buch and Samuel [2], aboutURTs and K-Knuth equivalence. In Section 2 of the paper, we provide more back-ground on Hecke insertion and K-Knuth equivalence and discuss the K-theoreticextension of the jeu de taquin algorithm of Thomas and Yong, which provides an-other way of determining whether two tableaux are in the same equivalence class.We also summarize Buch and Samuel’s results about URTs and give invariants forclasses of K-Knuth equivalent tableaux. In Section 3, we give an algorithm tocompute all K-Knuth classes of tableaux on a given alphabet [n] and a run timefor that algorithm. From this we derive a finite-time algorithm to determine if twogiven words are K-Knuth equivalent. Section 4 shows that every two K-Knuthequivalent tableaux on [n] can be connected by intermediate words of length atmost 13n(n + 1)(n + 2) + 3. The proof of this bound also includes a useful lemmastating that any words of length ` in an insertion class can be connected to therow word of the corresponding insertion tableau by moving through intermediatewords of length at most `. In Sections 5 and 6, we give new families of URTs. Weintroduce a new technical invariant for K-Knuth classes called T-compatibility anduse it to prove that all right-alignable tableaux are URTs. We also give a methodfor easily determining whether any given hook-shaped tableau is a URT. Finally,in Section 7, we discuss various findings on the number of K-Knuth equivalenceclasses of tableaux on an alphabet [n] and the number of unique rectification classesamong them. We then give additional conjectures and related results.

2. Background

The goal of this section is to acquaint the reader with the language of K-Knuthequivalence relations on increasing tableaux, which for the most part parallels thebetter-known Knuth equivalence relations [5].

2.1. Increasing Tableaux. In this section we will define in more detail increasingtableaux, the main subject of this paper, as well as related terminology, followingthe formalization of [2, Section 3.1]. Throughout this paper, N will denote the setof positive integers.

Elements of the set N × N are called boxes, and will form the building blocksof increasing tableaux. We will visualize N × N as an infinite matrix comprised ofboxes: the box (i, j) appears in row i and column j.

Suppose α = (i1, j1) and β = (i2, j2) are boxes. We say that α is strictly northof β if i1 < i2 and weakly north of β if i1 ≤ i2; we say that α is strictly northeastof β if i1 < i2 and j1 > j2, and we say that α is weakly northeast of β if i1 ≤ i2and j1 ≥ j2. The reader can formulate the analogous definitions for the remainingcardinal directions, which we omit. In addition, we say α is above β to mean α isnorth of β, we say α is directly above β to mean i1 = i2− 1 and j1 = j2, and so on.

A shape λ is any finite subset of N×N. We say λ is a straight shape if wheneverλ contains the box α it contains all boxes weakly northwest of α. A skew shape ν/µis the set difference of two straight shapes ν ⊇ µ.

K-KNUTH EQUIVALENCE FOR INCREASING TABLEAUX 3

Example 2.1. Of the shapes below, the first is neither straight nor skew, thesecond is skew but not straight, and the third is straight.

We can identify a straight shape with a partition as follows. Given a straightshape λ, let λi denote the number of boxes in row i. If λ has ` nonempty rows thenλ is uniquely determined by the tuple (λ1, λ2, · · · , λ`). By definition, λ1 ≥ λ2 ≥· · · ≥ λ`. The straight shape given in Example 2.1, for instance, corresponds to thepartition (4, 4, 2).

A filling of a shape λ is any map T : λ → N, which assigns an integer to eachbox of λ. The image of a box α under T is called the label of α. We say that thefilling T is an increasing tableau (of shape λ) if the entries of T strictly increasedown columns and from left to right along rows, that is, if T (α) < T (β) wheneverα is weakly northwest of β. In this paper all tableaux are increasing tableaux, andin particular we will not consider semistandard tableaux. A tableau T of shape λis straight if λ is straight and skew if λ is skew. Unless otherwise mentioned, wewill write “tableau” to mean “straight tableau.”

Example 2.2. Of the fillings below, only the third is an increasing tableau.

1 2 2 5

3 4 5 5

7 7

1 2 3 6

3 4 5 6

3 7

1 2 3 5

3 4 5 6

6 7

As with matrices, let λt denote the transpose of λ, defined by

λt = {(j, i) : (i, j) ∈ ν}.Let T t : λt → N denote the transpose of T , defined by T t(j, i) = T (i, j). Thetranspose of a tableau or shape is sometimes referred to as its conjugate.

Example 2.3. The tableau1 2 3 5

3 4 5 6

6 7

has transpose1 3 6

2 4 7

3 5

5 6

Definition 2.4. A tableau T is initial if the set of labels of T is [n] for some n ∈ N.A word w is initial if the set of letters appearing in w is [n] for some n ∈ N.Example 2.5. The word 124335 is initial but the word 14355 is not. Of the twotableaux below, the lefthand tableau is initial and the righthand tableau is not.

1 2 5

2 3

4

5

2 4 8

4 6

7

8


Initial tableaux are often easier to work with, and for this reason we will usu-ally restrict our attention to initial tableaux. This restriction comes at no loss ofgenerality, as the following definition will make clear.

Definition 2.6. Let w = w1w2 . . . wk be a word and let a1 < a2 < · · · < a` be theordered list of letters appearing in w. The standardization of w is the word formedby replacing ai with i in w.

Similarly, let T be a tableau and let a1 < a2 < · · · < a` be the ordered list ofletters appearing in T . The standardization of T is the tableau formed from T byreplacing every entry ai with i.

Example 2.7. The standardization of the word 35822 is 23411. The standardiza-tion of the tableau 2 5 6 8

5 9is 1 2 3 4

2 5.

2.2. Hecke Insertion. Hecke insertion is an algorithm for inserting a positiveinteger into an increasing tableau, resulting in another increasing tableau, whichmay or may not be the same as the original. The elementary step of the Heckeinsertion algorithm is the insertion of a positive integer into a row of the tableau.After the row is modified, either a new positive integer is inserted into the next rowor the algorithm terminates.

The rules for Hecke inserting a positive integer x into row R of a tableau T areas follows. Suppose first that x ≥ y for all y ∈ R.

(1) If adjoining a box containing x to the end of R results in a valid increas-ing tableau T ′, then T ′ is the result of the insertion and the algorithmterminates.

(2) If adjoining a box containing x to the end of R does not result in a validincreasing tableau, then R is unchanged and the algorithm terminates.

Otherwise, let y be the smallest integer in R that is strictly larger than x.

(3) If replacing y with x results in an increasing tableau, then replace y with xand insert y into the next row.

(4) If replacing y with x does not result in an increasing tableau, then insert yinto the next row and do not change R.

We write T ← x to denote the final tableau resulting from the Hecke insertion of xinto T .

It will occasionally be convenient to consider the column insertion of x into T ,which is computed by performing Hecke insertion with columns playing the role ofrows. Formally, the column insertion of x into T is given by (T t ← x)t. From nowon, “insertion” will always refer to Hecke insertion.

Example 2.8. [6, Example 2.3]

1 2 3 52 3 4 667

← 3 =1 2 3 52 3 4 667

In this example, inserting 3 into the first row invokes rule (4), so we insert 5 intothe second row. This invokes (4) again, so we insert 6 into the third row. By (2),we get the tableau shown.


Example 2.9. [6, Example 2.4]

2 4 63 6 87

← 5 =2 4 53 6 87 8

We first insert 5 into the first row, which by (3) replaces the 6 in the rightmost box,bumping the 6 into the second row. By rule (4), the second row is unchanged, andwe insert an 8 into the third row. Rule (1) gives the resulting tableau.

Let w = w1 · · ·wn be a word. The insertion tableau of w, written P (w), isformed by recursively Hecke inserting the letters of w from left to right:

P (w) = (· · · ((∅← w1)← w2) · · · ← wn).We define the insertion class of w to be {w′ : P (w′) = P (w)}.

Proposition 2.10. [2, Theorem 6.2] If w and w′ are in the same insertion class,then w ≡ w′.

Hence Hecke insertion equivalence implies K-Knuth equivalence. The converse,however, is false – K-Knuth equivalent words may not be insertion equivalent. Sounlike Knuth equivalence classes, K-Knuth equivalence classes may contain morethan one tableau.

Example 2.11. Let w = 1342 and w′ = 13422. Clearly w and w′ are K-Knuthequivalent. Notice, however, that

P (w) = 1 2 43

and

P (w′) = 1 2 43 4

Patrias and Pylyavskyy [6] define a recording tableau Q(w) for the Hecke inser-tion of a word w, in analogy to the recording tableau for Schensted insertion, whichallows one to uniquely recover an inserted word w from the pair (P (w), Q(w)) viareverse Hecke insertion. We will not use this notion, but note it for completeness.

2.3. K-Knuth equivalence. Just as Hecke insertion is a K-theoretic analogue ofthe standard RSK insertion, K-Knuth equivalence is the corresponding analoguefor Knuth equivalence. Recall that the Knuth equivalence relations are as follows:

xzy ∼ zxy, (x < y < z)yxz ∼ yzx, (x < y < z)xyx ∼ yxx, (x < y)yxy ∼ yyx, (x < y).

Two words are said to be Knuth equivalent if one can be obtained from the othervia a finite series of applications of the above Knuth relations.

In the K-theoretic case, the first two rules are precisely the same. However, thiscase has two additional rules with important consequences. The K-Knuth relationsare as follows:


xzy ≡ zxy, (x < y < z)yxz ≡ yzx, (x < y < z)p ≡ pp,

pqp ≡ qpq.

Again, two words are said to be K-Knuth equivalent if one can be obtained fromthe other via a series of applications of the above K-Knuth relations.

The third and fourth relations have some important implications. The third ruleimplies that each K-Knuth equivalence class of words has infinitely many words ofarbitrarily large length. Because Hecke insertion results in an increasing tableau,there are only finitely many tableaux into which words on an alphabet [n] (wordscontaining at least one of each letter from {1, 2, ..., n}) can be inserted. Hence thereare finitely many equivalence classes on any alphabet [n] with infinitely many wordsin each class. This is in contrast to the standard version, in which there are onlyfinitely many words in each class, but infinitely many classes on any given alphabet.

The fourth rule implies that two words can be equivalent, but each letter couldappear a different number of times in the two words. For example, 121 ≡ 212, but1 appears twice in the first word and once in the second.

2.4. Reading Words. The Hecke insertion algorithm associates to each word anincreasing tableau. In this section, we describe a way to associate to each increasingtableau a certain set of words, called reading words for the tableau. A tableau canhave many reading words, but they will all be K-Knuth equivalent.

Let T be an increasing tableau. The most commonly used reading word for T isthe row word for T , written row(T ), which is obtained by reading the entries of Tfrom left to right along each row, starting from the bottom row and moving upward.Similarly, the column word for T , written col(T ), is obtained by reading the entriesof T from bottom to top along each column, starting from the first column andmoving rightward.

More generally, let σ be a sequence of boxes of T . If for any two boxes α and β,we have that α appears before β in σ whenever α is southwest of β in T , then σ issaid to be a reading order for T . If σ is a reading order for T , we write rowσ(T ) todenote the word produced by reading the entries of T in the order given by σ.

Example 2.12. If

T =1 3 4 52 4 54

then row(T ) = 42451345 and col(T ) = 42143545. A third valid reading word for Tis 42145345.

Proposition 2.13. [2, Lemma 5.4] If σ and π are two reading orders of an in-creasing tableau T , then rowσ(T ) ≡ rowπ(T ).

Let T and T ′ be two increasing tableaux. If row(T ) ≡ row(T ′), we say that T isK-Knuth equivalent to T ′ and write T ≡ T ′, giving a K-Knuth equivalence relationon the set of increasing tableaux.


2.5. K-Jeu de Taquin. The classical jeu de taquin (jdt) algorithm defines anequivalence relation on standard skew tableaux. Recall a tableau T is standard ifit has entries exactly 1, 2, . . . , n for some n.

Example 2.14. Of the fillings below, only the second is standard.

1 2 3 5

3 4 5 6

7 8

1 2 3 5

4 6 7 9

8 10

In this section we give a K-theoretic extension of jdt to increasing tableaux,K-jdt, closely following [2]. The K-jdt algorithm gives an alternative method fortesting tableaux equivalence.

Definition 2.15. We say two boxes α, β ∈ N × N are neighbors if α is directlyabove, below, left, or right of β. Given a tableau T and two entries s and s′ of T ,define a new tableau swaps,s′(T ) of the same shape by

swaps,s′(T ) : α 7→

s′ if T (α) = s and T (β) = s′ for some neighbor β of α;

s if T (α) = s′ and T (β) = s for some neighbor β of α;

T (α) otherwise.

Example 2.16. Two examples of swaps may be found below.

1 2 • 4• 33

swap•,3−−−−−→

1 2 • 43 ••

1 2 43 4 ◦◦

swap◦,4−−−−−→

1 2 ◦3 ◦ 4◦

To define K-jdt, we will extend the labels of a tableau to the unordered setN∪ {•, ◦}. The symbols • and ◦ are undefined and will move across the tableau asswaps are performed during K-jdt. A complete K-jdt move is composed of a seriesof these swaps.

Definition 2.17 ([2]). For a tableau T of shape λ on letters in the interval [a, b]and a subset C ⊂ λ of the maximally southeast boxes in λ, we define the forwardslide of T starting from C as

jdtC(T ) = (swapa,• swapa+1,• · · · swapb−1,• swapb,•([C → •] ∪ T )),where [C → •] denotes the constant tableau of shape C that puts • in each box.Similarly, if Ĉ ⊂ N×N\ν is a subset of the maximally northwest boxes of N×N\ν,then the reverse slide of T starting from Ĉ is

ĵdtĈ(T ) = (swapb,◦ swapb−1,◦ · · · swapa+1,◦ swapa,◦(T ∪ [Ĉ → ◦])),

where [Ĉ → ◦] denotes the constant tableau of shape Ĉ that puts ◦ in each box.

It is apparent from the definitions of swap that tableaux resulting from forwardand reverse K-jdt slides remain increasing along rows and columns. It is alsoapparent that, by design, forward and reverse moves are inverses. Furthermore,one can use forward moves to transform a skew shape into a straight shape, andreverse moves to do the opposite.


Example 2.18. The examples below give one complete forward jdt slide and onecomplete reverse jdt slide, showing the sequence of swaps performed during theslide.

• 1 3• 2 42 3

→1 • 3

• 2 42 3

→1 • 3

2 • 4• 3

→1 3 •

2 3 43 •

→1 3 •

2 3 43 •

=1 3

2 3 43

1 2 53 5 ◦4

→1 2 ◦3 ◦ 54

→1 2 ◦3 ◦ 54

→1 2 ◦◦ 3 54

→1 ◦ 2◦ 3 54

→◦ 1 21 3 54

=1 2

1 3 54

In the same way that the K-Knuth “moves” give an equivalence relation onwords, K-jdt slides give an equivalence relation on tableaux.

Definition 2.19 ([2]). We say that two increasing tableaux S and T are K-jeude taquin equivalent if S can be obtained by applying a sequence of forward andreverse K-jeu de taquin slides to T .

The importance of K-jdt equivalence lies in the following theorem, proved in [2].

Theorem 2.20. [2, Theorem 6.2] For tableaux T and T ′, row(T ) ≡ row(T ′) if andonly if T and T ′ are K-jdt equivalent.

Therefore, K-jdt equivalence is the same as K-Knuth equivalence.

2.6. Unique Rectification Targets. After applying enough forward K-jdt slidesto a skew tableau T , a straight tableau called a K-rectification of T will eventuallyresult. The rectification order is the choice of the placements of •’s (resp. ◦’s) foreach forward (resp. reverse) K-jdt slide. In contrast to the classical theory of jdt,different rectification orders may result in different K-rectifications; in other words,varying the placements of •’s during K-jdt slides may result in different straighttableaux.

Example 2.21. Here is an example how different rectification orders may producedifferent K-rectifications. The tableau

• 22

1 3 4

has the rectifications

• 22

1 3 4→

2• 4

1 3→

2• 3 41

→ • 21 3 4

→ • 2 41 3

→ 1 2 43

and

2• 2

1 3 4→

2• 2 41 3

→• 2

1 2 43

→• 2

1 2 43

→• 2 41 43

→ 1 2 43 4

,

resulting in different tableaux.

In some instances the K-rectification may be unique, motivating the followingdefinition.


Definition 2.22. [2, Definition 3.5] An increasing tableau U is a unique rectificationtarget (URT) if, for every increasing tableau T that has U as a rectification, U isthe only rectification of T .

Equivalently, an increasing tableau is a URT if it is the only tableau in its K-Knuth equivalence class. The literature gives several classes of URTs, which wesummarize below.

Definition 2.23. A minimal tableau is a tableau in which each box is filled withthe smallest positive integer that will make the filling a valid increasing tableau.

Example 2.24. The following tableau is minimal of shape (4, 3, 3, 1):

1 2 3 42 3 43 4 54

Proposition 2.25. [2, Corollary 4.7] Every minimal tableau is a URT.

Definition 2.26. A superstandard tableau is a standard tableau that fills the firstrow with 1, 2, . . . , λ1, the second row with λ1 + 1, λ1 + 2, . . . , λ1 + λ2, etc., whereλi is the length of the i

th row of the tableau.

Example 2.27. The following tableau is superstandard of shape (4, 3, 3, 1):

1 2 3 45 6 78 9 1011

Proposition 2.28. Every superstandard tableau is a URT.

Proposition 2.28 is a corollary of [9, Theorem 3.7]; it will also follow from The-orem 5.3.

Buch and Samuel also proved in [2] that certain URTs can be added to minimalhooks to generate new URTs. We introduce this result with a few preliminarydefinitions.

Definition 2.29. A fat hook is a partition of the form (ab, cd), where a, b, c, d arenonnegative integers with a ≥ c.Example 2.30. The tableau below is a fat hook of shape (42, 23).

Definition 2.31. Let Mλ be the minimal increasing tableau corresponding to a fathook λ = (ab, cd), and let U be an increasing tableau. We say U fits in the cornerof Mλ if U has at most d rows, at most a− c columns, and all integers containedin U are strictly larger than all integers contained in Mλ.

In other words, U fits in the corner of Mλ if the entries of U are strictly greaterthan the entries of Mλ and if positioning U in the corner of the hook results in anincreasing tableau.


Theorem 2.32. [2, Theorem 6.9] Let λ be a fat hook, and let U be any unique rec-tification target that fits in the corner of Mλ. Then Mλ∪U is a unique rectificationtarget.

Example 2.33. We may conclude that the tableau

T =

1 2 3 42 3 4 53 6 74 75

is a URT because T = M(42,13) ∪ U , where

M(42,13) =

1 2 3 42 3 4 5345

and U = 6 77

2.7. K-Knuth invariants. Now that we have the notion of an equivalence classof tableaux, we will provide several invariants under the K-Knuth equivalence rela-tion. These will aid in proving results concerning the relations between tableaux inequivalence classes. A comprehensive list will be provided at the end of the sectionfor ease of reading.

Definition 2.34. For a word w, let lis(w) (resp. lds(w)) denote the length of thelongest strictly increasing (resp. decreasing) subsequence of w.

If w is the row word of a tableau T then lis(w) is the length of the first row of Tand lds(w) is the length of the first column of T .

Example 2.35. We will use the reading word of the tableau T from Example 2.33to illustrate this concept. We see that if w = row(T ) = 54736723451234, thenlis(w) = 4, and lds(w) = 2.

Proposition 2.36. [6, Lemma 2.17] If w1 ≡ w2 then lis(w1) = lis(w2) andlds(w1) = lds(w2).

The above equalities follow easily from the K-Knuth equivalence relations.

Theorem 2.37. [8, Theorem 1.3] For any word w, the size of the first row andfirst column of P (w) are given by lis(w) and lds(w), respectively.

See Definition 5.9 for a detailed description of the northeast-hook-closed shapesin part (b) of the following definition and proposition.

Definition 2.38. For a word w, let w|[a,b] denote the word obtained from w bydeleting all integers not contained in the interval [a, b]. Likewise, let T be an in-creasing tableau, not necessarily straight, and let T |[a,b] denote the tableau obtainedfrom T by removing all boxes not in [a, b].

Proposition 2.39. [2, Lemma 5.5]Let [a, b] be an integer interval.

(a) Let w1 and w2 be K-Knuth equivalent words. Then w1|[a,b] and w2|[a,b] areK-Knuth equivalent words.


(b) Let T1 and T2 be increasing tableaux of northeast-hook-closed shapes withT1 ≡ T2. Then T1|[a,b] and T2|[a,b] are K-Knuth equivalent tableaux.

Example 2.40. We have that T1 ≡ T2 for

T1 =

1 2 3 4 52 335

T2 =

1 2 3 4 52 3 53 55

Therefore T1|[3,5] ≡ T2|[3,5], so that

3 4 53

35

≡

3 4 53 5

3 55

Part (b) of Proposition 2.39 gives the second invariant for equivalent tableaux,their restriction to an interval subalphabet. We can now form equivalent sub-tableaux by deleting the boxes of equivalent tableaux that are not included in theinterval subalphabet.

Definition 2.41. Let T be a straight tableau. The outer hook of T is the sub-tableaux of T consisting of the first row and the first column.

Example 2.42. The outer hook of the tableau below is shaded gray.

1 2 4 53 4 86 7

Our third invariant for K-Knuth classes is the outer hook.

Proposition 2.43. Let T and T ′ be initial tableau such that T ≡ T ′. Then Tand T ′ have the same outer hook.

Proof. Suppose T and T ′ have letters in the alphabet [n]. Consider the two tableausequences

T |[1], T |[2], . . . , T |[n]T ′|[1], T ′|[2], . . . , T ′|[n].

We have already seen that T |[i] ≡ T ′|[i] because T ≡ T ′. Now proceed by inductionon n. The tableaux T |[1] and T ′|[1] are equivalent and contain one box, meaningT |[1] = T ′|[1]. At the ith step of the sequence, if T |[i] and T ′|[i+1] have the same outerhook then T |[i+1] and T ′|[i+1] must have the same outer hook as well because theouter hook of T |[i] (resp. T ′|[i]) differs from the outer hook of T |[i+1] (resp. T ′|[i+1])by the addition of at most two boxes, which must be labeled with i+ 1. �

Our fourth invariant is simple and involves the transpose of a tableau: if T1 ≡ T2,then T t1 ≡ T t2 . Invariance under the transpose follows from the fact that if T1 andT2 are K-Knuth equivalent then they are K-jdt equivalent, and any sequence ofK-jdt moves connecting the two may be applied to their transposes.

Example 2.44. We have that

1 2 43

≡ 1 2 43 4


implying that

1 324

≡1 32 44

Another invariant is the Hecke permutation, defined in [1] to provide a coarserequivalence than K-Knuth equivalence. Define Σ to be the group of bijective maps

{w : N→ N | w(x) = x for all but finitely many x ∈ N}.The adjacent transpositions si = (i, i + 1) ∈ Σ generate Σ and give the group anatural presentation as a Coxeter group. We will use this Coxeter group structureto define a new product on Σ that makes Σ into a monoid.

Given a permutation u, let `(u) denote the shortest length of a factorization u =si1 · · · sik of u as a product of the si transpositions. Equivalently, `(u) equals thenumber of inversions of u, that is, the number of tuples i < j such that u(i) > u(j)(see [4, Sec. 1.6 Exercise 2]).

Definition 2.45. Let u ∈ Σ be a permutation. The Hecke product of u and atransposition si is defined by

u · si =

{usi if `(usi) > `(u);

u otherwise.

Given a second permutation v = si1 · si2 · · · sil ∈ Σ, the Hecke product of u and vis defined by

u · v = u · si1 · si2 · · · sil ,multiplying in left to right order.

The Hecke product is associative and gives a monoidal structure on Σ, allowingus to introduce the following concept.

Definition 2.46. The Hecke permutation of an increasing tableau T is w(T ) =w(a) = sa1 · sa2 · · · sak , where a = a1 · · · ak is a reading word of T .

Proposition 2.47. The Hecke permutation w(T ) of an increasing tableau is in-variant under K-Knuth moves.

Proposition 2.47 is equivalent to Corollary 6.5 in [2], using the fact that K-jdtequivalence implies K-Knuth equivalence.

Having the same Hecke permutation is a necessary but not sufficient conditionfor two tableaux to appear in the same K-Knuth class, meaning that the numberof K-Knuth equivalence classes on [n] is at least as large as Sn+1, the symmetricgroup on n+ 1 elements. Hence there are a minimum of (n+ 1)! K-Knuth classesof tableaux on [n] letters. Proposition 7.2 will make use of this fact.

Example 2.48. The Hecke permutation for the word 21231 is given by

1 7→ 3, 2 7→ 2, 3 7→ 4, 4 7→ 1.

In summary, the following are invariant under theK-Knuth equivalence relations:

(1) the longest strictly increasing (or decreasing) subword of a word,(2) the restriction of a word (or a tableau) to an interval subalphabet, up to

K-Knuth equivalence,(3) the outer hook of a tableau,


(4) the transpose of a tableau, up to K-Knuth equivalence, and(5) the Hecke permutation.

3. Algorithms

This section deals with computational aspects of the K-Knuth equivalence rela-tion. In particular, we will describe algorithms to answer the following two prob-lems:

(1) Determine if two words are K-Knuth equivalent.(2) Compute all K-Knuth classes of tableaux on [n].

Let Tn be the set of (not necessarily initial) increasing tableaux on [n], and let Nnbe its cardinality. We will prove the following theorem.

Theorem 3.1. If w1 and w2 are words of length at most ` on [n], then we candetermine whether they are K-Knuth equivalent in O(` + n3Nnα(n)) time, whereα(n) is the inverse Ackermann function.

Recall that the inverse Ackermann function is very slowly-growing and satisfiesα(n) = o(log∗ n), where the iterated logarithm log∗ n is, roughly speaking, thenumber of times the logarithm function is applied iteratively to n until we arriveat a number less than or equal to 1. More precisely, log∗ is defined recursively by

log∗ n =

{0 if n ≤ 11 + log∗(log n) if n > 1

Surprisingly enough, the way to achieve the efficiency is to determine all K-Knuth equivalence classes. More precisely, we will prove the following.

Theorem 3.2. We can divide all increasing tableaux on [n] into K-Knuth equiva-lence classes in O(n3Nnα(n)) time.

We first remark that it is fairly easy to construct the set Tn of all (not necessarilyinitial) tableaux on [n]. Indeed, one can construct this set recursively. Given Tn−1,one can check for each T ∈ Tn−1 where one can add boxes with entry n to T toget T ′ ∈ Tn. This takes only O(Nn) time. Therefore, we will assume that we havethe set Tn at our disposal at our disposal in the algorithms for Theorem 3.1 andTheorem 3.2.

We also note the following simple lemma. Consider the following directed graphG with labelled edges: the set of vertices of G is Tn, and for 1 ≤ y ≤ n, there is anedge from T to T ′ with label y if and only if T ′ = T ← y.

Lemma 3.3. We can construct the directed graph G in O(n2Nn log n) time.

Proof. The graph G contains O(nNn) edges. Given a row of length L and a lettery, we can use binary search to find the smallest entry in the row greater than yand then check whether replacing the entry with y violates the increasing tableaucondition in constant time. Therefore, insertion of a letter into a row of lengthL takes O(logL) time, and each Hecke insertion takes O(n log n) time. Thus theconstruction of G takes O(n2Nn log n) time. �

Using Lemma 3.3, we will assume that we have the graph G at our disposal inthe algorithms for Theorem 3.1 and Theorem 3.2. We now show that Theorem 3.1follows from Theorem 3.2.


Proof of Theorem 3.1. Note that w1 ≡ w2 if and only if P (w1) ≡ P (w2). Hencewe can compute P (w1) and P (w2) and then check whether P (w1) ≡ P (w2) bycomputing all K-Knuth eqivalence classes on [n]. Hecke insertion can be doneby going through a path with corresponding labels in the directed graph G, socomputing P (w1) and P (w2) takes O(`) time. �

Before describing an algorithm to prove Theorem 3.2, We first recall two resultsabout algorithms.

Proposition 3.4. [3, Section 21.3] Let X be a set of cardinality k. The data struc-ture disjoint-set forest can store a set partition P of X and perform the followingoperations:

(1) Given x, y ∈ X, we can check whether x and y are in the same set of P;(2) If x, y ∈ X are in different sets in P, we can merge the sets containing x

and y;(3) Output all sets in P.

Moreover, if we perform m operations of type (1) and (2) and one operation oftype (3), the total runtime is O((m+ k)α(k)).

Proposition 3.5. [3, Section 10.1] The data structure queue can perform the fol-lowing operations:

(1) Insert an element into the queue;(2) Remove some element from the queue and output the removed element.

Moreover, each insertion and removal takes O(1) time.

Throughout the algorithm we will maintain a set partition P of Tn and a queue Qthat stores pairs (T1, T2) of tableaux in Tn.

We say that a pair (a, b) of words is a primitive pair if either

(1) a = p, b = pp for some letter p;(2) a = pqp, b = qpq for some letters p 6= q;(3) a = xzy, b = zxy for some letters x < y < z; or(4) a = yxz, b = yzx for some letters x < y < z.

Algorithm 1 Algorithm for Computing all K-Knuth Classes

1: Initialization: P := {{T} : T ∈ Tn}; Q empty.2: for all T ∈ Tn do3: for all primitive pair (a, b) do4: Compute T1 := T ← a and T2 := T ← b using G;5: if T1 and T2 are in different sets of P then6: Merge the sets in P containing T1 and T2 respectively;7: Insert the pair (T1, T2) into Q.8: end if9: end for

10: end for11: while Q is non-empty do12: Remove a pair (U1, U2) from Q.13: for all 1 ≤ y ≤ n do14: Compute T1 := U1 ← y and T2 := U2 ← y using G;


15: if T1 and T2 are in different sets of P then16: Merge the sets in P containing T1 and T2 respectively;17: Insert the pair (T1, T2) into Q.18: end if19: end for20: end while21: Output all sets S ∈ P.

We claim that the set partition P gives the K-Knuth equivalence classes in Tnat the termination of the algorithm.

Theorem 3.6. If T1 and T2 are K-Knuth equivalent, then they lie in the same setS ∈ P at the end of the algorithm.

To prove Theorem 3.6 we require two preliminary lemmas.

Lemma 3.7. Fix 1 ≤ y ≤ n. Let T1 and T2 be two tableaux and let T ′i = Ti ← y.If T1 and T2 lie in the same set of P at the end, then T ′1 and T ′2 lie in the same setof P at the end.

Proof. We first prove the assertion for the special case where, at some point of theexecution, (T1, T2) ∈ Q. In this case, (T1, T2) is eventually removed from Q inline 12. If T ′1 and T

′2 are in the same S ∈ P at this point, then the assertion holds.

Otherwise, we will merge the sets containing T ′1 and T′2.

Now we consider the general case. We need a simple claim to aid us in the proof.

Claim 3.8. There exists a sequence T1 = U0, U1, · · · , Ur = T2 of tableaux suchthat, for each i, either (Ui, Ui+1) or (Ui+1, Ui) is in Q at some point.

We shall see how this claim proves the general case. For each i, let U ′i = Ui ← y.Claim 3.8 shows that, at the end of the program, U ′i and U

′i+1 are in the same set

of P for each i, i.e., all U ′i are in the same S ∈ P. In particular, T ′1 and T ′2 are inthe same S ∈ P.

It remains to prove the claim. Assume that T1 and T2 are in the same set of Pafter the kth merge but not before. We will proceed by induction on k. The casek = 0 is trivial. Suppose k > 0. Let S1 and S2 be the sets containing T1 and T2,respectively, before the kth merge. By assumption, the sets S1 and S2 merge at thekth merge. After that merge, we insert into Q the pair (U1, U2), for some U1 ∈ S1and U2 ∈ S2. We know from the induction hypothesis that there is a chain of pairsin Q connecting T1 and U1 as well as T2 and U2. The result then follows. �

Lemma 3.9. Let T ∈ Tn and let (a, b) be a primitive pair. Then T1 := T ← a andT2 := T ← b are in the same S ∈ P at the end of the algorithm.

Proof. At some point we check whether T1 and T2 are in the same set S ∈ P.If they are in the same set, then we are fine. Otherwise, we will merge the setscontaining them on the next line. �

Proof of Theorem 3.6. We start with a special case. Suppose there exist wordsw1 and w2 that differ by one K-Knuth move and such that P (w1) = T1 andP (w2) = T2. Write w1 = uav and w2 = ubv where a and b form a primitive pair,so that by Lemma 3.9 the tableaux P (ua) and P (ub) are in the same set of P.


Applying Lemma 3.7 multiple times, we conclude that T1 and T2 are in the sameset of P.

For the general case, there is a sequence T1 = U0, U1, · · · , Ur = T2 of tableauxsuch that for each i there exist two words differing by one K-Knuth move andinserting into Ui and Ui+1 respectively. By the previous case Ui and Ui+1 are inthe same set of P, so at the end all Ui’s are in the same S ∈ P. In particular, T1and T2 are in the same set of P. �

Having shown the correctness of the algorithm, we will now analyze the runtime.

Theorem 3.10. Algorithm 1 runs in O(n3Nnα(n)) time, where Nn is the numberof increasing tableaux on [n].

Proof. There are O(n3) primitive pairs, so lines 4 – 7 will be executed O(n3Nn)times. There are at most Nn − 1 merges, so lines 14 – 17 will be executed O(nNn)times.

By Proposition 3.4 and Proposition 3.5 the runtime for the whole algorithm isO(n3Nnα(Nn)). Since every tableau on [n] has at most n

2 boxes and each box is

either empty or contains an entry in [n], we have Nn ≤ (n+1)n2

. This implies thatα(Nn) = Θ(α(n)), giving the desired asymptotic bound. �

4. Length of Intermediate Words

If w ≡ w′, then there exists a sequence w = w0, w1, · · · , wr = w′ of words suchthat wi and wi+1 differ by one K-Knuth move. It is natural to ask whether it isalways possible to find such a sequence so that the intermediate words wi alwayshave length at most that of the longer one. Surprisingly, the answer is no, as onecan check by computer that 4235124 ≡ 4523124 cannot be connected by words oflength at most 7. However, it is possible to give an upper bound in terms of thesize of the alphabet.

To put the result on a more formal footing, we will make the following definition.

Definition 4.1. Let w and w′ be words and let k be a positive integer. We say

that wk≡ w′ if there exists a sequence w = w0, w1, · · · , wr = w′ of words such that

wi and wi+1 differ by one K-Knuth move, and each word wi has length at most k.

We will prove the following result.

Theorem 4.2. Suppose T1 ≡ T2 are tableaux on [n]. Let N = 13n(n+1)(n+2)+3.Then row(T1)

N≡ row(T2).

Computer evidence suggests that the bound in Theorem 4.2 can be tightened tothe largest size of a tableau in the K-Knuth equivalence class, where the size of atableau T of shape λ is the number of boxes contained in λ.

Conjecture 4.3. Let T and T ′ be two tableaux with T ≡ T ′, and let k be the largestsize of a tableau K-Knuth equivalent to T or T ′. Then row(T )

k≡ row(T ′).

Conjecture 4.3 has been verified for tableaux on [n] with n ≤ 5.


4.1. Proof of Theorem 4.2. We will use the following lemma, which concerns K-Knuth equivalence within an insertion class, to prove Theorem 4.2. Let |w| denotethe number of letters in a word w.

Lemma 4.4. If w is a word and P (w) = T , then w|w|≡ row(T ).

We defer the proof of Lemma 4.4 to Section 4.2 because it is fairly technical.Assuming Lemma 4.4, our first step towards the theorem is the reduction to thecase where there exist words w1 and w2 such that w2 differs from w1 by one singleK-Knuth move, P (w1) = T1, and P (w2) = T2. Suppose the result had been shownfor this special case. Let T1 and T2 be any two K-Knuth equivalent tableaux on [n].Let T1 = U0, U1, U2, · · · , Ur = T2 be the tableaux in the insertion classes visitedby the sequence of words connecting row(T1) and row(T2). By the result for the

special case row(Ui)N≡ row(Ui+1), and the general case follows.

To prove the result for the special case, we can just construct words w′1 andw′2 with |w′1|, |w′2| ≤ N such that w′2 differs from w′1 by a single K-Knuth move,P (w′1) = T1 and P (w

′2) = T2. Indeed, by Lemma 4.4, we have w

′1

N≡ row(T1) andw′2

N≡ row(T2), and the result then follows.The construction of the words w′1 and w

′2 relies on the following observation: if t

is a letter of a word w = u1tu2 for which P (u1t) = P (u1), i.e., if t “does nothing” inthe insertion of w, then P (u1tu2) = P (u1u2). More precisely, write w1 = uav andw2 = ubv where a is of the form x (resp. xx, xyx, xzy, yxz) and b is of the formxx (resp. x, yxy, zxy, yzx). Let u′ (resp. v′) be the word obtained by deletingall letters in u (resp. v) which “do nothing” in the insertion of both w1 and w2.The words w′1 = u

′av′ and w′2 = u′bv′ then satisfy P (w′1) = T1 and P (w

′2) = T2.

The theorem thus follows from the following upper bound on the number of letterswhich “do something” in the insertion.

Lemma 4.5. If w is a word on [n], then there are at most 16n(n+ 1)(n+ 2) letterst such that, when we write w = utv, we have P (ut) 6= P (u).

We may now conclude the proof of Theorem 4.2.

Proof of Lemma 4.5. Fix i and j with i + j = k + 1. If P (ut) and P (u) havedifferent (i, j)th entries, then the insertion of t either creates a new entry at the(i, j)th position or decreases the (i, j)th entry. At the end the (i, j)th entry mustbe at least k, so there are at most n − k + 1 letters in w that change the (i, j)thentry. There are exactly k pairs of (i, j) with i + j = k + 1, so the result followsfrom

n∑k=1

k(n− k + 1) = 16n(n+ 1)(n+ 2). �

4.2. Proof of Lemma 4.4. The proof of Lemma 4.4 will consist of a carefulanalysis of the Hecke insertion algorithm. In essence, computing an insertiontableau P (w) is the same as making a sequence of K-Knuth moves to the word w,and none of these moves lengthens the word.

We first require a simple technical result.

Lemma 4.6. If a1 < a2 < · · · < ai < b, then a1a2 · · · ai−1baii+1≡ ba1a2 · · · ai. If

b < a1 < a2 < · · · < ai, then a1ba2 · · · aii+1≡ a1a2 · · · aib.


Proof. For the first (resp. second) statement, apply the K-Knuth moves zxy → xzy(resp. yzx→ yxz) for x < y < z as many times as needed to move the b. �

The first step of the proof of Lemma 4.4 is the reduction to the special casewhere w = row(T ′)y for some tableau T ′ and some letter y. Assuming this case hasbeen proved, write w = w1w2 · · ·wk. Let Ti = P (w1w2 · · ·wi), ti = row(Ti), andui = tiwi+1 · · ·wk. The assertion in the special case implies that ti+1

i+1≡ tiwi+1, sothat ui+1

k≡ ui. The general case then follows.We further reduce the special case w = row(T ′)y to proving the following asser-

tion.

Claim 4.7. Suppose R = r1r2 · · · rk is the row word of a row of T ′. If R is thefirst row, let S be the empty word; otherwise, let S = s1s2 · · · s` (with ` ≥ k) be therow word of the previous row. Suppose the insertion of y into R changes R to R′.Let m = k + `+ 1.

(1) If y ≥ rk, then we have RySm≡ R′S.

(2) If ri ≤ y < ri+1 for some i, then we have RySm≡ ri+1R′S.

We first examine how this claim implies the special case w = row(T ′)y ofLemma 4.4. Let R′i (resp. Ri) (1 ≤ i ≤ n) be the row word of the ith row ofT ′ (resp. T ) (if the tableau has no such row, let the corresponding row word bethe empty word). Suppose we only insert letters into the first through the ath rowwhen we insert the letter y into the tableau T ′, and let yi be the letter inserted intothe ith row (so that y1 = y). Note that R

′i = Ri for i > a. It then follows from

Claim 4.7 that

row(T ′)y = R′nR′n−1 · · ·R′3R′2R′1y1

m≡ R′nR′n−1 · · ·R′3R′2y2R1m≡ R′nR′n−1 · · ·R′3y3R2R1

m≡ · · · m≡ R′nR′n−1 · · ·R′a+1R′ayaRa−1 · · ·R1m≡ R′nR′n−1 · · ·R′a+1RaRa−1 · · ·R1 = RnRn−1 · · ·R1 = row(T ).

Proving Lemma 4.4 is therefore reduced to proving Claim 4.7. We will first provethe first part of the claim, the case where y does not replace any entry. The firstpart separates into the following cases.

(a) k = 0.(i) s1 6= y.

(ii) s1 = y.(b) k > 0 and rk = y.(c) k > 0, rk < y and sk+1 6= y.(d) k > 0, rk < y and sk+1 = y.

(i) k = 1.(ii) k > 1.

Define

α = r1r2r3 · · · rk−1.Now we tackle individual cases.

(a) (i) We have RyS = yS = R′S and the result is trivial.(ii) Let γ = s2s3s4 · · · s`. In this case,

RyS = yyγk+1≡ yγ = S = R′S.


(b) We have R = R′ = αy. Thus Ry = αyyk+1≡ αy = R′. Hence RyS m≡ R′S.

(c) We have RyS = αrkS = R′S and the result is trivial.

(d) Let

β = s1s2s3 · · · sk−2,γ = sk+2sk+3 · · · s`.

(i) We have RyS = r1ys1yγ and R′S = r1s1yγ. Hence it suffices to

prove that r1ys1y4≡ r1s1y. By standardization this is equivalent to

23134≡ 213, which follows from the following computation:

2313 ≡ 2133 ≡ 213.

(ii) Applying Lemma 4.6 twice, we have

RyS = αrkyβsk−1skyγm≡ αβrksk−1yskyγ.

Similarly, applying Lemma 4.6 gives us

R′S = αrkβsk−1skyγm≡ αβsk−1rkskyγ.

Hence it suffices to prove that rksk−1ysky5≡ sk−1rksky. By stan-

dardization this is equivalent to 314245≡ 3124, which follows from the

following computation:

31424 ≡ 31242 ≡ 13242 ≡ 13422 ≡ 1342 ≡ 1324 ≡ 3124.

Next, we prove the second part of Claim 4.7, where y replaces a letter. Again,there are several cases to consider.

(e) i = 0.(i) s1 6= y.(ii) s1 = y.

(f) i > 0 and ri = y.(g) i > 0, ri < y and si+1 6= y.(h) i > 0, ri < y and si+1 = y.

(i) i = 1.(ii) i > 1.

Defineα = r1r2r3 · · · ri−1,β = ri+2ri+3ri+4 · · · rk.

By Lemma 4.6, we have

(1) ri+1βyk−i+1≡ ri+1yβ.

Now we tackle individual cases.

(e) By (1), RySm≡ r1yβS.

(i) We have r1R′S = r1yS and the result follows.

(ii) Let χ = s2s3s4 · · · s`. In this case,

RyS = r1βyyχm+1≡ r1βyχ = r1r1βyχ = r1R′S.


(f) By equation (1),

RySm≡ αyri+1yβS

m≡ αri+1yri+1βS.By Lemma 4.6, we have

ri+1R′S = ri+1αyri+1βS

m≡ αri+1yri+1βS.

Hence RySm≡ ri+1R′S.

(g) By Lemma 4.6 and (1), we have

ri+1R′S = ri+1αriyβS

m≡ αriri+1yβSm≡ αriri+1βyS = RyS.

(h) Letγ = s1s2s3 · · · si−2,δ = si+2si+3 · · · sk,ε = ri+2si+2ri+3si+3 · · · rksk,χ = sk+1sk+2sk+3 · · · s`.

Applying Lemma 4.6 multiple times, we have

(2)

βγsi−1siyδ = ri+2ri+3 · · · rks1s2 · · · sk2k−i−1≡ ri+2ri+3 · · · rk−1s1s2 · · · sk−1rksk2k−i−1≡ · · ·2k−i−1≡ s1s2 · · · siyri+2si+2 · · · rksk= γsi−1siyε.

(i) By (1) and (2),

RyS = r1r2βys1yδχm≡ r1r2yβs1yδχ

m≡ r1r2ys1yεχ.Similarly, by (2),

r2R′S = r2r1r2βs1yδχ

m≡ r2r1r2s1yεχ.

Hence it suffices to prove that r1r2ys1y5≡ r2r1r2s1y. By standard-

ization this is equivalent to 243135≡ 42413, which follows from the

following computation:

24313 ≡ 42313 ≡ 42133 ≡ 4213 ≡ 4231 ≡ 2431≡ 22431 ≡ 24231 ≡ 24213 ≡ 42413.

(ii) By (1) and (2),

RyS = αriri+1βyγsi−1siyδχm≡ αriri+1yβγsi−1siyδχ

m≡ αriri+1yγsi−1siyεχ.Applying Lemma 4.6 then gives us

RySm≡ αγrisi−1ri+1ysiyεχ.

Similarly, by Lemma 4.6 and (2),

ri+1R′S = ri+1αriri+1βγsi−1siyδχ

m≡ αri+1riri+1βγsi−1siyδχm≡ αri+1riri+1γsi−1siyεχ

m≡ αγsi−1ri+1risiri+1yεχ.


Hence it suffices to prove that risi−1ri+1ysiy6≡ si−1ri+1risiri+1y. By

standardization this is equivalent to 3154246≡ 153254, which follows

from the following computation:

315424 ≡ 315242 ≡ 312542 ≡ 132542 ≡ 135242 ≡ 135424≡ 153424 ≡ 153244 ≡ 15324 ≡ 15342 ≡ 13542≡ 133542 ≡ 135342 ≡ 135324 ≡ 153524 ≡ 153254.

This completes the proof of Lemma 4.4.

5. Right-Alignable Tableaux

In this section we give a new family of URTs, the right-alignable tableaux. Sincesuperstandard and rectangular tableaux are right-alignable, we will in particularshow that all tableaux in those two classes are URTs. Although we have assumed sofar that tableaux are straight unless otherwise mentioned, in this section we relaxthat requirement. It will be useful to consider tableaux of more general shape.

Definition 5.1. Let λ = (λ1, . . . , λ`) be a straight shape and let T : λ → N be astraight tableau of shape λ. The right-justification of λ is the shape {(i, j + (λ1 −λi)) : (i, j) ∈ λ}. The right-justification of T is the filling TR : λR → N defined byTR(i, j + (λ1 − λi)) = T (i, j).

A tableau T is right-alignable if the right-alignment TR strictly increases downcolumns and left-to-right across rows.

Example 5.2. The tableau

T =1 2 33 54

has right justification TR =1 2 3

3 54.

Hence T is not right-alignable.

We will prove the following theorem in the next two sections.

Theorem 5.3. Every right-alignable tableau is a URT.

The following two corollaries follow easily from Theorem 5.3.

Corollary 5.4. Every superstandard tableau is a URT.

Corollary 5.5. Every rectangular tableau is a URT.

Theorem 5.3 is by no means sharp: there are URTs that are not right-alignable.For example, a minimal tableaux is right-alignable if and only if it is rectangular.

5.1. T -Compatibility. In this section we define a technical invariant for K-Knuthclasses, T -compatibility, which will be used to show that right-alignable tableauxare URTs.

Definition 5.6. Given a word w, let w∗ denote the set of tuples (x, i), where xis the letter in w appearing at position i. We say that (x, i) appears before (resp.after) (y, j) if i < j (resp. i > j).


For example, in 7556∗ the tuple (5, 2) appears before (5, 3) and after (7, 1). Ingeneral the set w∗ is identical to the word w, except that the same letters of w∗ aredifferent if they occupy different positions. For clarity, we will abuse notation andidentify a tuple (x, i) ∈ w∗ with its first component x. There is no risk of confusionunless the letter x appears twice in w∗, in which case we will distinguish the twoby subscripts: x1 and x2, say.

Definition 5.7. Let λ be any shape and let T be an increasing tableau of shape λ.We say that a word w is T -compatible if there exists a function f : w∗ → λ suchthat

(1) f is surjective,(2) the box f(x) has label x in T ,(3) if f(x) is above f(y) then x appears after y in w∗, and(4) if f(x) is to the right of f(y) then x appears after y in w∗.

A straight tableau U is T -compatible if the row word row(U) is T -compatible.

The tableau T in Definition 5.7 can have any shape, and in particular couldequal either of the following tableaux:

1

3 4 5

4 6

5

7

1 4 7

3 5

At the same time, we require the tableau U in Definition 5.7 to have straight shape.The requirements for the function f in Definition 5.7 could be reformulated as

follows. If the letters of w∗ are ordered under the relation “y ≤ x if y appearsbefore x in w∗”, and if the boxes of T are ordered under the relation “α ≤ β if α isweakly southwest of β,” then f satisfies the conditions of Definition 5.7 if and onlyif f is a surjective, monotonic map of posets.

Example 5.8. Let T = 1 2 43 5

. The word 35124 is T -compatible since the map

f : 35124∗ → given by

3 7→ (2, 1), 5 7→ (2, 2), 1 7→ (1, 1), 2 7→ (1, 2), 4 7→ (1, 3)is surjective, sends each letter to a box labeled with that letter, and respects theorder of the letters in 35124∗ In this case 35124 = row(T ), and it is true in generalthat any reading word for a tableau T is T -compatible: indeed, T -compatibilitycan be construed to generalize the reading word. Hence the word 31524 = col(T )is also T -compatible since the map f : 31524∗ → λ given by

3 7→ (2, 1), 1 7→ (1, 1), 5 7→ (2, 2), 2 7→ (1, 2), 4 7→ (1, 3)satisfies the necessary properties. However, the word 31254 is not T -compatible.There is only one map 31254∗ → λ sending each letter to a box labeled with thatletter: under such a map, 5 7→ (2, 2) and 2 7→ (1, 2). But (1, 2) is above (2, 2) while5 comes before 2 in 31254∗.

The reader may have noticed that the three words we gave in example 5.8 wereK-Knuth equivalent, but only the first and second were T -compatible: applyingthe K-Knuth move 31524 → 31254 fails to preserve T -compatibility. This failurereflects the fact that the shape (2, 2, 1) is missing a box in the bottom right corner.


More specifically, in order for T -compatibility to be an invariant forK-Knuth classesthe tableau T must contain enough boxes to be closed under forming certain hooks.

Recall that the (i, j)th entry of a tableau lies in the ith row and jth column.

Definition 5.9. A shape λ is northwest-hook-closed if it is closed under formingnorthwest hooks: whenever x = (i1, j1) and y = (i2, j2) are boxes of λ such thati1 ≥ i2 and j1 ≤ j2 then λ contains the boxes (r, j1) for i1 ≥ r ≥ i2 and (i2, c) forj1 ≤ c ≤ j2.

y

x

A shape λ is northeast-hook-closed if it is closed under forming northeast hooks:whenever x = (i1, j1) and y = (i2, j2) are boxes of λ such that i1 ≤ i2 and j1 ≤ j2then λ contains the boxes (r, j2) for i1 ≤ r ≤ i2 and (i1, c) for j1 ≤ c ≤ j2.

x

y

A shape λ is southeast-hook-closed if it is closed under forming southeast hooks:whenever x = (i1, j1) and y = (i2, j2) are boxes of λ such that i1 ≥ i2 and j1 ≤ j2then λ contains the boxes (r, j2) for i1 ≥ r ≥ i2 and (i1, c) for j1 ≤ c ≤ j2.

x

y

Definition 5.9 generalizes hook closure as defined in [2, section 5]: Buch andSamuel’s hook closure is the same as our northeast-hook closure.

Example 5.10. Of the shapes below, only the first is northwest-hook-closed, onlythe second is northeast-hook-closed, and only the third is southeast-hook-closed.The fourth satisfies none of the three hook-closure properties we defined.

Lemma 5.11 will concern shapes that are northwest- and southeast-hook closed;such shapes are the reflections of skew shapes across a vertical axis, like the examplesshown below.

Lemma 5.12 will concern shapes that are northeast-, northwest-, and southeast-hook closed; such shapes are the reflections of straight shapes across a verticalaxis.


Lemma 5.11. Let λ be a northwest- and southeast-hook-closed shape and let Tbe an increasing tableau of shape λ. If w ≡ w′ and w is T -compatible then w′ isT -compatible.

Proof. It suffices to assume that w′ differs from w by one K-Knuth move. Letf : w∗ → λ be the surjection showing that w is T -compatible. For each K-Knuthmove we will construct a function f ′ : (w′)∗ → λ showing that w′ is T -compatibleand differing from f only on the modified letters. Throughout the proof, we willwrite indices on letters that are the same but that appear in different positionsin w∗.

If w′ is obtained from w by replacing p1 with p2p3 then define f′(p2) = f

′(p3) =f(p1).

If w′ is obtained from w by replacing p1p2 by p3 then it suffices to show thatf(p1) = f(p2), in which case we may define f

′(p3) = f(p1). For the sake ofcontradiction, assume that f(p1) = (i1, j1) 6= f(p2) = (i2, j2). Since A increasesalong rows and columns, either f(p1) is strictly northeast of f(p2) or f(p2) is strictlynortheast of f(p1). In either case, using the fact that T is northwest- and southeast-hook closed, let u be a letter of w∗ with f(u) = (i1, j2) and let v be a letter of w

∗

with f(v) = (i2, j1).

u p1p2 v

The letters u and v must lie between p1 and p2 in w∗, contradicting the assumption

that the letters p1 and p2 are consecutive.If w′ is obtained from w by replacing p1q1p2 by q2p3q3 then it suffices to show

that f(p1) = f(p2), in which case we way define f′(p3) = f(p1) and f

′(q2) =f ′(q3) = f(q1). Arguing as in the previous case, we assume that f(p1) 6= f(p2) anddeduce that there are two letters between p1 and p2, a contradiction.

If w′ is obtained from w by replacing xzy with zxy, provided x < y < z, then itsuffices to show that f(z) is strictly southeast of f(x), in which case we may definef ′(x) = f(x) and f ′(z) = f(z). It is impossible for f(z) to appear weakly southwestof f(x) because z comes after x in w∗, and it is impossible for f(z) to appear weaklynorthwest of f(x) because x < z. Assume, then, for the sake of contradiction, thatf(z) = (i2, j2) is weakly northeast of f(x) = (i1, j1). Since there are no lettersbetween x and z in w∗, f(z) must be directly right of f(x), so that i2 = i1 andj2 = j1 + 1.

x z

The box f(y) = (i3, j3) cannot be weakly northwest of f(x) or weakly southeastof f(z) because x < y < z. And since y appears after x and z in w∗, the box f(y)must be directly above f(z), so that i3 = i2 − 1 and j3 = j2.

y

x z

Let u be a letter of w∗ such that f(u) = (i3, j1), which exists because T is northwest-hook closed.

u y

x z

Then u < z and u lies between x and y in w∗, a contradiction.Analogous arguments apply to the other three K-Knuth moves. �


5.2. Proof of Theorem 5.3. We have not yet specified why T -compatibility isuseful: if, in addition to the hypotheses of Lemma 5.11, T is northeast-hook closed,then there is at most one tableau which is T -compatible. By Lemma 5.11 thattableau must be a URT.

Lemma 5.12. Let λ be a northeast-, northwest-, and southeast-hook-closed shape.If T is a tableau of shape λ, then there is at most one straight tableau U that isT -compatible.

Proof. Let f : row(U) → T denote the surjective map certifying that U is T -compatible. Given a shape λ let rowi(λ) denote the ith row of λ; given a tableauT let rowi(T ) denote the subword of row(T ) consisting of the letters in the ith rowof T .

The proof consists of showing that f(rowi(U)∗) = rowi(T ), meaning that the

letters in the ith row of U are exactly the letters in the ith row of T . Hence U isunique among straight tableau that are T -compatible. We first prove an easy claim.

Claim 5.13. If w is T -compatible and x is the first letter of a strictly decreasingsubword of w∗ of length k, then f(x) appears in or below the kth row of λ.

Proof. Write the decreasing subword as x = x1 > x2 > · · · > xk. The box f(xi+1)cannot be weakly southeast of the box f(xi) because xi+1 < xi, and f(xi+1) cannotbe weakly southwest of f(xi) because xi+1 appears after xi in w

∗. Hence f(xi+1)is strictly north of f(xi). �

To finish the proof of lemma 5.12, we’ll show by backward induction on i thatf(rowi(U)

∗) = rowi(T ). Claim 5.13 implies that T has at least as many rowsas U . And the rightmost column of T gives a strictly decreasing subword of U withmaximal length, meaning that T and U have the same number of rows. Call thatnumber m. For any box α ∈ rowm(U) with f(x) = α, the letter x appears as thefirst letter of a decreasing subword of row(U)∗ of length m, and hence lies in themth row of U . Then f(rowm(U)

∗) = rowm(T ).Now assume that f(rowi(U)

∗) = rowi(T ). Suppose α ∈ rowi−1(U), and let x beany letter of w∗ for which f(x) = α. If α does not lie above a box of rowi(T ), thenx is less than every entry of rowj(U)

∗ for i ≤ j ≤ m. If α lies directly above abox of rowi(T ), then for every j with i ≤ j ≤ m the letter x comes after a lettery ∈ rowj(T )∗ that is greater than x. In either case we conclude that x cannot liebelow the (i− 1)th row of U , so that x ∈ rowi−1(U)∗. �

Under what circumstances does the tableau U mentioned in Lemma 5.12 actuallyexist? The proof of Lemma 5.12 shows that if a tableau T is closed under formingnorthwest, northeast, and southeast hooks and if U is T -compatible then T and Uhave the same entries in each row. Hence T and U have the same number of entriesin each row and differ only by the relative positions of each row. The rows of Umust be aligned along their left edges because U is straight, and the rows of Tmust be aligned along their right edges because, as Example 2.1 mentioned, T isthe reflection of a straight shape about a vertical axis. In other words, we mayobtain T from U by right-aligning the rows of U . In this way we complete theproof of Theorem 5.3.

Proof of Theorem 5.3. If TR is an increasing tableau then we may speak of TR-compatible tableaux. In particular row(T ) is TR-compatible, as we may show by


defining the map f : row(T )∗ → λR by f(T (i, j)) = (j + (λ1 − λi)), where T hasshape λ = (λ1, . . . , λ`). If T is K-Knuth equivalent to some other tableau T

′ thenT ′ is TR-compatible by Lemma 5.11 and so T

′ = T by Lemma 5.12. Hence T is aURT. �

5.3. Shapes of Non-URTs. Theorem 5.3 is not sharp: there are many URTs,most notably non-rectangular minimal tableaux, that do not satisfy the theorem’shypotheses. However, the corollary that every rectangular tableau is a URT isalmost sharp, as we will show in this section: every straight shape λ that is not arectangle, except for , has an increasing filling T that is not a URT.

Proposition 5.14. If λ is a straight shape that is not a rectangle or then thereis an increasing tableau of shape λ that is not a URT.

Proof. Write λ = (λ1, . . . , λ`). The proof will consist of expanding the prototypical

non-URTs 1 2 43

≡ 1 2 43 4

from Example 2.11.Suppose λ has two parts, λ2 = 1, and λ1 ≥ 3. Row inserting 2 into the tableau

1 3 4 · · · λ1+1

one or two times yields the following respective tableaux, which are equivalentbecause 2 ≡ 22:

1 2 4 · · · λ1+1

3

≡1 2 4 · · · λ1+1

3 4

.

For the remaining cases we require a family of tableaux parameterized by twonatural numbers, j and k. Consider the tableau

T =1 · · · k−1 k k+2 k+3 k+4 · · · j+1

2 · · · k(j ≥ 4, k ≥ 2).

Row inserting k + 1 into T yields the tableau

Tj,k =1 · · · k−1 k k+1 k+3 k+4 · · · j+1

2 · · · k k+2,

of shape (j, k), and row inserting k + 1 into Tj,k yields the tableau

T ′j,k =1 · · · k−1 k k+1 k+3 k+4 · · · j+1

2 · · · k k+2 k+3,

of shape (j, k + 1).Continuing with our case analysis, suppose λ has two parts, λ1 > λ2, and λ2 ≥ 2.

If λ1 > λ2 + 1 then the tableau Tλ1,λ2 has shape λ and is not a URT. If λ1 = λ2 + 1then T ′λ1,λ2 has shape λ and is not a URT.

We will handle the remaining cases, which concern tableaux having three or moreparts, using the following observation. Given a tableau T of shape λ = (λ1, . . . , λ`),let T (r,s) denote the restriction of T to the rows r, r + 1, . . . , s of λ.


Claim 5.15. If T and T ′ are two tableaux that differ only in the ith and (i+ 1)throws, and if T (i,i+1) ≡ (T ′)(i,i+1), then T ≡ T ′.

Proof. The row words row(T ) and row(T ′) may be connected by K-Knuth movesthat only use letters from the ith and (i+ 1)th rows. �

To continue with the proof of Proposition 5.14, suppose λ has more than twoparts and there is an index i ≥ 1 such that λi ≥ λi+1+2. Given a tableau T , let T [n]denote the tableau formed by increasing the entries of T by n: formally, T [n](i, j) =T (i, j) + n. There exists a tableau T of shape λ such that T (i,i+1) = Tλi,λi+1 [n]for some n. Let T ′ denote the tableau of shape (. . . , λi, λi+1 + 1, λi+2, . . . ) thathas the same labels as T in every row except the ith and (i + 1)th, and such that(T ′)(i,i+1) = T ′λi,λi+1 [n]. Then T ≡ T

′ but T 6= T ′, as desired.By constructing an analogous argument using the transposes T tj,k and (T

′j,k)

t,

we can show that the proposition holds if either λ or λt has more than one entryin the second row and has two consecutive rows of different lengths. This exhaustsall cases. �

6. Hook-Shaped Tableaux

In this section, we examine a class of tableaux known as the hook-shaped tableauxand characterize which hook-shaped tableaux are URTs.

Definition 6.1. A straight shape λ is hook-shaped if λ = (m, 1n) for some m ≥ 1and n ≥ 0. An increasing tableau T of shape λ is hook-shaped if λ is hook-shaped.

Of the tableaux below, the tableau on the left is hook-shaped and the twotableaux on the right are not.

Theorem 6.2. Suppose T is a hook-shaped, increasing tableau such that

(1) the first row of T is labeled 1, 2, . . . , n, and(2) the first column is labeled 1, `2, `3, . . . , `m, where `i+1 = `i + 1 for 2 ≤ i ≤

m− 1 and `2 ≥ 2.Then T is a URT.

Proof. We proceed by induction on n. Given a tableau T satisfying the hypothesesof the claim, call n the rightmost entry of T . If n = 1, the tableau T is a URT byvirtue of being rectangular. For the induction step, assume that every tableau withrightmost entry n is a URT. We’ll deduce a contradiction from the existence of atableau T with rightmost entry n + 1 that is not a URT. Since minimal tableauxare URTs, we may assume that `2 ≥ 3. Let T ′ be a tableau distinct from T andlying in the same K-Knuth equivalence class. Then the skew tableaux T |[2,n] andT ′ |[2,n] are K-Knuth equivalent by Proposition 2.39. Perform K-jdt on T |[2,n] andT ′ |[2,n] at position (1, 1), that is, at the position vacated by 1 upon restriction tothe subalphabet [2, n]. Because the first rows of T |[2,n] and T ′ |[2,n] are labeledconsecutively and `2 ≥ 3, performing K-jdt has the effect of translating the firstrow of each tableaux one box to the left.

Let S and S′ denote the tableaux resulting from T |[2,n] and T ′ |[2,n], respectively.Since S and S′ are distinct and K-Knuth equivalent we conclude that S is not


a URT. But the tableau formed from S by decreasing each of its entries by 1has rightmost entry n and satisfies the hypotheses of the claim, contradicting theinduction hypothesis. �

Example 6.3. These hook-shaped tableaux satisfy the hypotheses of Theorem 6.2and are therefore URTs:

1 2 3 4345

1 2 3 4 53

1 2 3 4 556

.

These hook-shaped tableaux do not satisfy the hypotheses of Theorem 6.2:

1 2 3 54

1 2 3 424

1 2 3 4235

.

Next, we will show that Theorem 6.2 is sharp.

Claim 6.4. Let T be a hook-shaped, increasing initial tableau with the followingproperties:

(1) the first row of T is labeled 1, `2, `3, ..., `s,(2) the first column of T is labeled 1,m2,m3, ...,mt, and(3) at least one of `2, ...`s and m2, ...mt is a non-consecutive sequence.

Then T is not a URT.

Proof. We consider two cases. For simplicity, we assume without loss of generalitythat every tableau T in consideration is an initial tableau. Also for simplicity, wemake a distinction between the first row and the first column of the tableaux inconsideration, but this distinction is arbitrary due to the fact that transposes ofnon-URTs are non-URTs.

For simplicity of notation, [r(1, `2, `3, ..., `s), c(1,m2,m3, ...,mt)] will denote thehook-shaped tableau whose first row has entries 1, `2, `3, ..., `s and whose first col-umn has entries 1,m2,m3, ...,mt:

1 `2 `3 ... `sm2m3...mt

We will often show that such a hook-shaped tableau is equivalent to a tableauwith the same outer hook and one extra entry in position (2, 2). Tableaux of thatshape will be denoted [r(1, `2, `3, ..., `s), c(1,m2,m3, ...,mt),n], where n is the newentry in (2, 2):

1 `2 `3 ... `sm1 km2...mt


Case 1. T has the following properties:

(1) The first row of T is of the form 1, `2, . . . , `s, where `2, ..., `s is any strictlyincreasing sequence, and

(2) The first column of T is of the form 1, 2,m3, . . . ,mt, where m3, ...,mt is anon-consecutive sequence (such that T is initial).

It will suffice to consider the tableau [r(1, k), c(1, 2, ..., k − 1, k + p)], where k =ì ≥ 3 and p ≥ 1. This tableau is K-equivalent to [r(1, k), c(1, 2, ..., k − 1, k +p),k+p]. The equivalence can be seen easily and directly by performing K-Knuthmoves on the row words of the two tableaux.

The above simplified example of a hook-shaped non-URT generalizes to all initialtableaux with the properties specified in Case 1 via row and column insertion,explained below.

If we row-insert some sequence of letters greater than k to the first row of eachtableau, the equivalence of the two tableaux does not change. Hence we have

[r(1, k, k + 1, ..., k + (p− 1), ì+p, ..., `s), c(1, 2, ..., k − 1, k + p)]

≡ [r(1, k, k + 1, ..., k + (p− 1), ì+p, ..., `s), c(1, 2, ..., k − 1, k + p),k+p].Similarly, we can column-insert integers less than k + 1 into the first column

of each tableau and maintain equivalence. Thus we can obtain any sequence ofintegers between 1 and k in the first row. For example, let the sequence we want toobtain between 1 and k be `1, ..., ì−1. We can first column-insert ì−1 − 1 into thefirst column. This has the effect of shifting everything to the right of 1 in the firstrow one box to the right, and inserting ì−1 into the position (1, 2). We can repeatthis process, inserting ì−2 − 1 into the first column, and so on, until we have thefollowing equivalence:

[r(1, `1, ..., ì−1, k, k + 1, ..., k + (p− 1), ì+p, ..., `s), c(1, 2, ..., k − 1, k + p)]

≡ [r(1, `1, ..., ì−1, k, k + 1, ..., k + (p− 1), ì+p, ..., `s), c(1, 2, ..., k − 1, k + p),k+p].Finally, we can column-insert any sequence of integers larger than k + 1 in the

first column. Hence we have the equivalence

[r(1, `1, ..., ì−1, k, k+ 1, ..., k+ (p− 1), ì+p, ..., `s), c(1, 2, ..., k− 1, k+ p,mj , ...,mt)]

≡ [r(1, `1, ..., ì−1, k, k+1, ..., k+(p−1), ì+p, ..., `s), c(1, 2, ..., k−1, k+p,mj , ...,mt),k+p].This completes the proof because the hook-shaped tableau above (assumed to

be initial based on our construction) represents the general version of the tableauwith properties specified in Case 1. Hence any tableau of that form is not a URT.

Case 2. T has the following properties:

(1) The first row of T is of the form 1, 2, ...n.(2) The first column of T is of the form 1,m1, ...,mt, where the sequence

m1, ...,mt is non-consecutive.

Note that the case where m1 = 2 was already covered by Case 1.

It will suffice to consider the tableau [r(1, 2, ..., k), c(1, k − 1, k + p)]. We showby induction that this is not a URT for all k ≥ 3. When k = 3 the tableau is not


a URT due to the below equivalence, which can be easily checked by performingK-Knuth equivalence moves on the row words of two tableaux:

[r(1, 2, 3), c(1, 2, 4)] ≡ [r(1, 2, 3), c(1, 2, 4),4].Assume that for some k, the following tableaux are equivalent:

[r(1, 2, ..., k), c(1, k − 1, k + p)] ≡ [r(1, 2, ..., k), c(1, k − 1, k + p),k+p].We show that

[r(1, 2, ..., k+1), c(1, k, k+(p+1))] ≡ [r(, 1, 2, ..., k+1), c(1, k, k+(p+1)),k+(p+1)]by examining K-Knuth equivalence moves on their row words.

First, consider the row word of the tableau on the lefthand side of the aboveequivalence. We have, with commas separating distinct letters for ease of reading:

k + (p+ 1), k, 1, 2, ..., k + 1 ≡ 1, k + (p+ 1), k, 2, ..., k + 1.

In the word above, consider everything to the right of 1. Standardize it to obtainthe word

k + 1, k − 1, 1, · · · , k.By assumption, this word is K-Knuth equivalent to

k + 1, k − 1, k + 1, 1, · · · , k.Thus

1, k + (p+ 1), k, 2, · · · , k + 1 ≡ 1, k + (p+ 1), k, k + (p+ 1), 2, · · · , k + 1.Since these two words respectively insert into the two tableaux depicted above, thetableaux are equivalent.

In order to see how this simplified example of a hook-shaped non-URT generalizesto all tableaux with the properties specified in Case 1, we again utilize row andcolumn insertion. We have the equivalence

[r(1, 2, .., k), c(1, k − 1, k + p)] ≡ [r(1, 2, ..., k), c(1, k − 1, k + p),k+p)]As in the first case, row- and column-inserting letters into each of the tableaux

will not change their equivalence. Thus the following equivalence holds:

[r(1, 2, ..., k, k + 1, ..., k + (p− 1), k + p, k + (p+ 1), ..., n),c(1,m1, ...,mi, k − 1, k + p,mi+3, ...,mt)]≡ [r(1, 2, ..., k, k + 1, ..., k + (p− 1), k + p, k + (p+ 1), ..., n),

c(1,m1, ...,mi, k − 1, k + p,mi+3, ...,mt),k+p]Clearly, the hook-shaped tableau above is a general tableau (again assumed to

be initial) such that the first row has entries 1, 2, ..., n, and the first column hasnon-consecutive entries. Hence, every tableau of this form is not a URT. �

An important takeaway from the above proof is that a hook-shaped initialtableau

1 a2 ... anb2...bn


is equivalent to

1 a2 ... anb2 ai...bn

whenever ai − ai−1 > 1.

7. Conjectures and Related Results

7.1. Sizes of Tableaux Classes. In the course of studying the K-Knuth equiva-lence relation on tableaux, we computed all equivalence classes of tableaux on [n]for 0 ≤ n ≤ 7. We were unable to obtain asymptotic bounds on the size of K-Knuthequivalence classes, but they seem to grow at least as quickly as n!.

Table 1. Sets of Initial Tableaux

Alphabet SizeInitial Increas-ing Tableaux

K-Knuth Classesof Initial Tableaux

URTs

0 1 1 11 1 1 12 3 3 33 13 13 134 87 79 715 849 620 4596 11915 6036 33137 238405 70963 25904

Table 1 shows that the ratio of unique rectification classes of tableaux on [n] toall K-Knuth classes of tableaux on [n] decreases monotonically, and we expect theratio to asymptotically tend to zero.

Conjecture 7.1. Let In denote the number of K-Knuth equivalence classes ofinitial tableaux on [n], and let Un denote the number of URTs on [n]. Thenlimn Un/In = 0.

7.2. Composition of K-Knuth Classes of Tableaux.

Proposition 7.2. For every n ≥ 2, there is an equivalence class of tableaux on[2n] containing at least n! distinct tableaux.

Proof. The comments following the proof of Claim 6.4 have as a corollary that forevery k = 2, 3, . . . , n there is an equivalence

1 2 4 · · · 2n

2

3

...

2n

≡

1 2 4 · · · 2n

2 2k

3

...

2n

.


Let T denote the tableau on the left, let T ′ denote the tableau on the right, and letw = row(T ). A simple computation shows that P (row(T )) = T , and row inserting2k − 2 into T shows that T ′ = P (w(2k − 2)). Hence w = row(T ) ≡ row(T ′) ≡w(2k − 2).

It follows that if we Hecke insert any of the positive integers 2, 4, 6, · · · , 2(n−1) inany order into the first row of T , the resulting tableau lies in the same equivalenceclass as T . Hence any tableau T ′ with the following two properties is K-Knuthequivalent to T .

(1) The first row and the first column of T ′ agree with the first row and thefirst column of T , respectively.

(2) Let U denote the restriction of T ′ to the (2n− 1)× n rectangle consistingof the boxes (i, j) satisfying 2 ≤ i ≤ 2n and 2 ≤ j ≤ n + 1. Then thetableau U uses the letters 4, 6, . . . , 2n

There are at least n! possibilities for the tableau U , as we saw in Section 2.7. Thusthe class of tableaux K-Knuth equivalent to T contains at least n! tableaux. �

The process for generating tableaux described in the proof of Proposition 7.2can produce many tableaux in the equivalence class of a hook-shaped tableaux. Itsuggests an important relationship betweenK-Knuth equivalence and row insertion,a relationship we have yet to fully understand. For completeness we include thefollowing proposition.

Proposition 7.3. Let T be an initial, hook-shaped tableaux whose first row is labeled1, a1, a2, . . . and whose second row is labeled 1, b1, b2, . . . . Let

A = {ai : i ≥ 1, ai+1 − ai ≥ 2}B = {bi : i ≥ 1, bi+1 − bi ≥ 2}.

Then the set of straight tableaux that are K-Knuth equivalent to T includes all thetableaux that may be obtained by making the following insertions into T , in anyorder: (1) row inserting elements of A into the first row of T and (2) columninserting elements of B into the first column of T .

Example 7.4. The following tableaux are K-Knuth equivalent. They may all beobtained by row inserting 2 and column inserting 3.

1 2 4 52356

1 2 4 52 4356

1 2 4 52 5356

1 2 4 52 43 556

Proposition 7.3 does not give all tableaux in a class, however: the tableaux aboveare also equivalent to the tableau

1 2 4 52 4 5356

,

which cannot be obtained by making the described row and column insertions.


Consequently, the size of a K-Knuth equivalence class of tableaux on [n] has noconstant upper bound as n increases. In fact, the number of standard tableaux ina K-Knuth equivalence class is also unbounded.

Proposition 7.5. For every n > 0, there exists a K-Knuth class containing atleast 2n standard tableaux.

Proof. We first consider the tableaux

T =1 2 53 4 76

and T ′ =1 2 53 46 7

.

Note that 36731452 ≡ 36734152 and the two words insert into T and T ′ respectively,so that T ≡ T ′.

We will build, using 3× 3 blocks, a standard tableau U whose equivalence classcontains at least 2n standard tableaux. For each i, j ≥ 1 with i + j ≤ n + 1, weconstruct an interval Ii,j of positive integers with the following properties:

(1) The intervals are pairwise disjoint.(2) The union of all the Ii,j is an interval of the form [1, N ] for some N > 0.(3) For every i and j, the interval Ii,j is after Ii+1,j and before Ii,j+1.(4) If i+ j ≤ n, then Ii,j has length 9; if i+ j = n+ 1, then Ii,j has length 7.

Now we construct the tableaux Ti,j with three rows and three columns as follows. Ifi+ j ≤ n, let Ti,j be an arbitrary increasing tableau on Ii,j with shape (3, 3, 3) andno repeated entries (so Ti,j is a 3×3 square). If i+ j = n+1, let Ti,j (resp. T ′i,j) bethe unique increasing tableau on Ii,j such that std(Ti,j) = T (resp. std(T

′i,j) = T

′).Finally, we build a tableau U with 3n rows and 3n columns such that the 3× 3

block Ti,j occupies the (3i − 2)th through 3ith rows and the (3j − 2)th through3jth columns. The above construction guarantees that U is a standard tableau.

For each i, define

Ri = row(Ti,1)row(Ti,2)row(Ti,3) · · · row(Ti,n−i)row(Ti,n−i+1),R′i = row(Ti,1)row(Ti,2)row(Ti,3) · · · row(Ti,n−i)row(T ′i,n−i+1).

The fact T ≡ T ′ implies row(Ti,n−i+1) ≡ row(T ′i,n−i+1). This means that Ri ≡ R′i.Observe that w = RnRn−1 · · ·R1 is a reading word of U . Let S be a subset of [n].If U ′ is the tableau obtained by replacing Ti,n−i+1 with T

′i,n−i+1 for all i ∈ S, then

the word w′ obtained by replacing Ri with R′i for all i ∈ S is a reading word for U ′.

Hence it follows that U ≡ U ′ for every choice of S ⊆ [n]. Since [n] has 2n subsets,the K-Knuth equivalence class contains at least 2n standard tableaux. �

For instance, if n = 3, one possible construction of U is

1 2 3 10 11 12 19 20 23

4 5 6 13 14 15 21 22 25

7 8 9 16 17 18 24

26 27 28 35 36 39

29 30 31 37 38 41

32 33 34 40

42 43 46

44 45 48

47


7.3. Shapes of Tableaux. Which shapes appear in a K-Knuth class of tableaux?We initially suspected that each tableau class contains a minimum and maximumshape, ordering the shapes under inclusion, but the following class disproves ourconjecture:

1 2 52 3 6345

1 2 52 33 645

1 2 52 3 63 645

1 2 52 33 54 65

However, it seems – and we have not been able to find a counterexample – that iftwo shapes λ1 ⊆ λ2 appear among the shapes in an equivalence class, then everyshape in the interval [λ1, λ2] of Young’s lattice appears among the shapes in thatclass.

Conjecture 7.6. Let T1, T2, . . . , Tk be a K-Knuth class of straight tableaux and letTi have shape λi. Then the set Σ = {λ1, λ2, . . . , λk} has the following property: ifλi, λj ∈ Σ then [λi, λj ] ⊆ Σ.

This conjecture has been verified for K-Knuth classes on [n] for n ≤ 7.

7.4. Changes in Tableau Shape. Initially, we had conjectured that if two wordsw and w′ differ by just one K-Knuth move, then the shapes of P (w) and P (w′) differby just one box. However, we found a counterexample: 5451342154 ≡ 54513422154because 2 ≡ 22, but the two words insert into the following tableaux, respectively:

1 2 4 5

2 5

3

4

5

≡

1 2 4 5

2 4 5

3 5

4

5

Acknowledgments

This research was conducted at the 2014 REU program at the University ofMinnesota, Twin Cities, and was supported by NSF grant DMS-1148634. For theirmentorship and support, we would like to thank Vic Reiner, Joel Lewis, and GreggMusiker. We are especially grateful to Pavlo Pylyavskyy for introducing us to thisproblem and guiding us throughout the project.

References

[1] A. Buch, A. Kresch, M. Shimozono, H. Tamvakis, and A. Yong, Stable Grothendieck Polyno-

mials and K-Theoretic Factor Sequences, preprint arXiv:math/0601514v1, 2006.[2] A. Buch and M. Samuel, K-Theory of Minuscule Varieties, preprint arXiv:1306.5419v1, 2013.[3] T. Cormen, C. Leiserson, R. Rivest, and C. Stein, Introduction to Algorithms, 3rd Edition,

MIT Press, 2009.[4] J. Humphreys, Reflection Groups and Coxeter Groups, Cambridge University Press, 1992.

[5] A. Lascoux, B. Leclerc, J. Thibon, The Plactic Monoid, in Algebraic Combinatorics on Words,

Cambridge University Press, 2001.[6] R. Patrias and P. Pylyavskyy, K-Theoretic Poirer-Reutenauer Bialgebra, preprint

arXiv:1404.4340v1, 2014.[7] H. Thomas and A. Yong, A Jeu de Taquin Theory for Increasing Tableaux, with Applications

to K-Theoretic Schubert Calculus, preprint arXiv:0705.2915v3, 2007.


[8] H. Thomas and A. Yong, Longest Increasing Subsequences, Plancherel-Type Measure and the

Hecke Insertion Algorithm, preprint arXiv:0801.1319v2, 2008.[9] H. Thomas and A. Yong, The Direct Sum Map on Grassmannians and Jeu de Taquin for

Increasing Tableaux, preprint arXiv:1002.1662v1, 2010.

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