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Bijection between reduced words and balanced tableaux

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Bijection between reduced words and balanced tableaux

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Content BIJECTION BETWEEN REDUCED WORDS AND BALANCED TABLEAUX
by
John C. Borger
A Thesis Presented to the
FACULTY OF THE USC DORNSIFE COLLEGE OF LETTERS, ARTS, AND SCIENCES
UNIVERSITY OF SOUTHERN CALIFORNIA
In Partial Fulfillment of the
Requirements for the Degree
MASTER OF ARTS
(MATHEMATICS)
May 2024
Copyright 2024 John C. Borger

I dedicate this thesis to my wife Olivia,
who has always supported my wildest dreams,
and to my parents,
who made my life sweeter than fiction.
ii

Acknowledgments
I would like to acknowledge a few individuals among many who have made this thesis possible.
Professor Sami Assaf, who deserves all the credit but will not accept any of it...
Assistant Professor Joj Helfer, who introduced me to Professor Assaf...
Assistant Professor Greg Muller, who introduced me to combinatorics...
Mrs. Susan Sath Vaswani, who is the glue that keeps the whole department running...
Benjamin Gillen & Jonathan Michala, who taught me how to identify a Rothe Diagram...
Mr. Tom Stevens, who taught me hard work...
Dr. Thomas Zurbuchen, who taught me how to combine hard work with leadership...
LTC Shaw Yoshitani (Ret.), who guided me through graduate school...
Professor Eric M. Friedlander, who somehow made me like abstract algebra...
And to Joj again, without whom I could have never made it this far...
iii

Table of Contents
Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii
Chapter 1: Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 2: Rothe Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Chapter 3: Insertion Algorithm 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Chapter 4: Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Chapter 5: Insertion Algorithm 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Chapter 6: Insertion Algorithm 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Chapter 7: Bijection Between Reduced Words and Balanced Tableaux . . . . . . . . 20
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
iv

List of Figures
2.6.1 D(143625). The fourth column, labeled 6, has bubbles in rows 2 and 5 because 2
and 5 are inverted with 6 in w = 143625. . . . . . . . . . . . . . . . . . . . . . . . 5
2.6.2 Method 2, Step 1 : Place bubbles in every cell of the first quadrant. . . . . . . . . . 5
2.6.3 Method 2, Step 2 : Pop bubbles in cells (i,wi) so that these cells are now empty. . . 5
2.6.4 Method 2, Step 3 : Pop bubbles above and to the right of the (i,wi) cells. The fourth
column of D(w), also labeled 6, the cell of interest is (i,wi) = (4,w4) = (4,6). . . 5
3.2.1 Insertion Algorithm 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
4.3.1 The Rothe Diagram D(143625) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.3.2 A tableau of shape D(143625) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.6.1 A balanced tableau. Every bubble has zero column anomalies and zero row anomalies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.6.2 A balanced tableau. ⃝1 is a column anomaly for ⃝3 . The bubble ⃝4 is a row
anomaly for ⃝3 . All other bubbles have zero column anomalies and zero row
anomalies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.6.3 An unbalanced tableau. ⃝1 is a column anomaly of ⃝2 , but ⃝2 has no row anomalies. 9
5.2.1 Insertion Algorithm 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
5.4.1 The bubbles ⃝f and ⃝g where f < g by assumption. . . . . . . . . . . . . . . . . . 12
5.4.2 The bubble ⃝f cannot be balanced if f < g. . . . . . . . . . . . . . . . . . . . . . 13
5.5.1 Unbalanced tableau of shape D(17823456). The tableau cannot be balanced because ⃝7 and ⃝9 are an adjacent ascending pair of bubbles in columns 2 and 3 with
w2 = 7 < 8 = w3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
5.6.1 Partition of cells in Insertion Algorithm 2. . . . . . . . . . . . . . . . . . . . . . 14
5.6.2 Category 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
5.7.1 Classification of bubbles in U. Category 1 :⃝2 ,⃝4 . Category 2 : ⃝1 . Category 3 :
⃝6 . Category 4 : ⃝3 ,⃝5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
v

6.2.1 Insertion Algorithm 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
vi

Abstract
A bijection between the set of reduced words for a finite permutation and the set of balanced
tableaux of Rothe Diagram shape is presented. The bijection follows the insertion style of the
Schensted Insertion used in the Robinson-Schensted-Knuth correspondence.
vii

Chapter 1
Introduction
In 1970, Knuth published what is now known as the Robinson-Schensted-Knuth correspondence.
This correspondence is a bijection between permutations and pairs of standard Young tableaux
of the same shape [6]. The bijection relies on a beautifully simple insertion algorithm called
Schensted Insertion where one ‘inserts’ each letter of a word into a tableau. The simplicity of the
algorithm, detailed in the first pages of [4], allows even the most casual reader an insight into a
powerful tool of combinatorics.
In 1987, Edelman and Greene introduced the balanced tableau in [2] which allows greater
flexibility in the entries of a tableau compared to the standard and semi-standard tableaux Knuth
was working with. Then in 1997, a group of four mathematicians led by Fomin published [3] which
included a bijection between the set of reduced words for a permutation and the set of balanced
tableaux of Rothe Diagram shape.
Here we present a new bijection between these two sets. This new bijection, called Insertion
Algorithm 3, provides a more intuitive approach than the bijection presented by Fomin in 1997.
The algorithm echos the simplicity and style of the Schensted Insertion.
First, we introduce Rothe Diagrams (Chapter 2) and manipulate Rothe Diagrams using Insertion Algorithm 1 (Chapter 3). Balanced tableaux are then introduced (Chapter 4) and letters are
inserted into balanced tableaux using Insertion Algorithm 2 (Chapter 5). Words are inserted into
balanced tableaux using Insertion Algorithm 3 (Chapter 6) and it is shown that this insertion is a
1

bijection between the set of reduced words for a permutation and the set of balanced tableaux of
Rothe Diagram shape (Chapter 7).
2

Chapter 2
Rothe Diagrams
Let Sn be the set of all permutations π : {1,2,3,...,n} to {1,2,3,...,n}. Then for any w ∈ Sn, we
can express w as w = w1w2 ...wn, and Sn is generated by si which transposes wi with wi+1.
Definition 2.1. Given a permutation w ∈ Sn, we define l(w), the length of w, to be the number of
inversions in w given by the cardinality of {(i, j) | i < j,wi > wj}.
Definition 2.2. A reduced word ρ = ρ1ρ2 ...ρl(w)
for w is a sequence of transpositions such that
w = sρ1
sρ2
...sρl(w)
and R(w) is the set of all reduced words for w.
Example 2.3. ρ = 23432 is a reduced word for w = 15342 because
12345
13245 - s2
13425 - s2s3
13452 - s2s3s4
13542 - s2s3s4s3
15342 - s2s3s4s3s2
Letting ρk = ρ1ρ2 ···ρl(w)k and wsk = sρ1
sρ2
···sρl(w)
sk
, we know from [1] :
1. l(wsk) = l(w)±1
2. ρk is a reduced word for wsk ⇐⇒ l(wsk) = l(w) +1 ⇐⇒ wk < wk+1
3

Definition 2.4. For a permutation w, the Rothe Diagram D(w) is the subset of Z
+ ×Z
+ defined
as
D(w) = {(wj
,i) | i < j,wi > wj}
Graphically we display a Rothe Diagram with bubbles in the first quadrant. We label the
vertical axis with i coordinates and the horizontal axis with wj coordinates. A bubble is placed at
every (wj
,i) ∈ D(w). We say that every (x, y) ∈ Z
+ ×Z
+ is a cell, and a particular cell (x0, y0) is
empty if (x0, y0) ∈/ D(w). Cells are either empty or contain a bubble. Observe that the number of
bubbles in D(w) is equal to the length of w.
Definition 2.5. Using the same coordinate system, we define another diagram D(w) in the following way.
1. Place bubbles in every cell of the first quadrant.
2. Pop bubbles in cells (i,wi) so that these cells are now empty.
3. Pop bubbles above and to the right of the (i,wi) cells.
• (i,l) for l > wi
• (k,wi) for k > i
We claim that for any permutation w, the two methods of producing D(w) yield equivalent
Rothe Diagrams.
4

Example 2.6. D(w) for w = 143625
1
1
2
3
4
5
6
7
8
1 4 3 6 2 5 7 8
Method 1
2
1
2
3
4
5
6
7
8
1 4 3 6 2 5 7 8
3
1
2
3
4
5
6
7
8
1 4 3 6 2 5 7 8
Method 2
4
1
2
3
4
5
6
7
8
1 4 3 6 2 5 7 8
Figure 2.6.1: D(143625). The fourth column, labeled 6, has bubbles in rows 2 and 5 because 2
and 5 are inverted with 6 in w = 143625.
Figure 2.6.2: Method 2, Step 1 : Place bubbles in every cell of the first quadrant.
Figure 2.6.3: Method 2, Step 2 : Pop bubbles in cells (i,wi) so that these cells are now empty.
Figure 2.6.4: Method 2, Step 3 : Pop bubbles above and to the right of the (i,wi) cells. The fourth
column of D(w), also labeled 6, the cell of interest is (i,wi) = (4,w4) = (4,6).
In Example 2.6 we used red arrows to demonstrate the popping of bubbles that occurs in the
building of D(w). We call these arrows death rays and refer to the cells of the form (i,wi) as death
ray origins (DROs).
For every permutation w, each row and column of the Rothe Diagram D(w) has exactly one
DRO. The placement of DROs in cells of the form (i,wi) allows us to recover w from D(w) by
reading the height (second coordinate) of the DROs from left to right.
5

Chapter 3
Insertion Algorithm 1
In this section, we define Insertion Algorithm 1 which takes the Rothe Diagram D(w) and a
number k such that wk < wk+1 as inputs and then outputs the Rothe Diagram D(wsk).
Definition 3.1. Insertion Algorithm 1
Given the Rothe Diagram D(w) for a permutation w and a letter k such that wk < wk+1, let
p = wk be the height of the DRO in column k and let q = wk+1 be the height of the DRO in column
k +1.
• Swap : Swap the heights of the DROs in columns k and k + 1 so that the DRO in (k, p)
moves up to (k,q) and the DRO in (k +1,q) moves down to (k +1, p).
• Preserve : If there are bubbles in column k+1 in between rows p and q, move those bubbles
left to column k.
• Add : Add a new bubble to (k, p), the cell that the DRO in column k vacated in Swap.
6

Example 3.2. w = 146325, k = 2, ws2 = 164325
D(w)
q
p
k
k+1
1
2
3
4
5
6
7
8
1 4 6 3 2 5 7 8
Swap
1
2
3
4
5
6
7
8
1 6 4 3 2 5 7 8
Preserve
1
2
3
4
5
6
7
8
1 6 4 3 2 5 7 8
Add, D(ws2)
1
2
3
4
5
6
7
8
1 6 4 3 2 5 7 8
Figure 3.2.1: Insertion Algorithm 1
Proposition 3.3. The Rothe Diagram produced by Insertion Algorithm 1 is D(wsk).
Proof. If suffices to show that the Rothe Diagram output by Insertion Algorithm 1 has a DRO in
every cell of the form (i,(wsk)i).
Case 1 i ̸= k,i ̸= k +1 The input diagram D(w) has DROs in (i,wi). Insertion Algorithm 1 does
not alter the DROs in these columns and wi = (wsk)i
.
Case 2 i = k Swap places a DRO in (k,q). Applying wk+1 = (wsk)k and i = k we see that (k,q) =
(k,wk+1) = (k,(wsk)k) = (i,(wsk)i).
Case 3 i = k +1 Swap places a DRO in (k+1, p). Applying wk = (wsk)k+1 and i = k+1 we see
that (k +1, p) = (k +1,wk) = (k +1,(wsk)k+1) = (i,(wsk)i).
Steps Preserve and Add do not play a role in the proof of Proposition 3.3. Their utilization
will become apparent in our construction of Insertion Algorithm 2 on tableaux.
7

Chapter 4
Tableaux
In this section, we define balanced tableaux of Rothe Diagram shape.
Definition 4.1. A tableau T is a subset {(x, y)} ∈ Z
+ × Z
+ combined with a numbering. The
numbering assigns each member of the set with a positive integer. A standard tableau is a tableau
with the additional requirement that the numbering assigns each member of the set a unique value
{1,...,n} where n is the cardinality of the set.
Definition 4.2. The shape of a tableau T, shape(T), is the set of its bubbles {(x, y)} ∈ Z
+ × Z
+
without their assigned numbering. We say that T is of the shape D(w) if :
shape(T) = {(x, y)} = {(wj
,i)} = D(w)
All tableaux in this paper will be standard tableaux of shape D(w) for some finite permutation
w. We use ⃝b to refer to the unique bubble with value b. As with Rothe Diagrams, we display
tableaux graphically as bubbles in the first quadrant and use the same coordinate system as we did
with Rothe Diagrams. However the bubbles of a tableau are labeled with their individual values
whereas the bubbles of a Rothe Diagram are unlabelled. All properties of DROs in Rothe Diagrams
apply to tableaux of shape D(w).
8

Example 4.3. w = 143625
1
1
2
3
4
5
6
7
8
1 4 3 6 2 5 7 8
2
1
2
3
4
5
6
7
8
1 4 3 6 2 5 7 8
3
2
4 1
5
Figure 4.3.1: The Rothe Diagram D(143625)
Figure 4.3.2: A tableau of shape D(143625)
Definition 4.4. For any bubble ⃝b in a tableau T, a column anomaly of ⃝b , notated CA[⃝b ], is a
bubble in the same column as ⃝b , in a row above ⃝b , and assigned a value smaller than b. A row
anomaly of ⃝b , notated RA[⃝b ], is a bubble in the same row as ⃝b , located to the right of ⃝b , and
assigned a value larger than b.
Definition 4.5. A tableau T is balanced [2] if for every bubble ⃝ ∈ b T :
#{RA[⃝b ]} = #{CA[⃝b ]}
Example 4.6. Tableaux of shape D(w) for w = 143625
1
1
2
3
4
5
6
7
8
1 4 3 6 2 5 7 8
3
4
2 1
5
2
1
2
3
4
5
6
7
8
1 4 3 6 2 5 7 8
3
1
4 2
5
3
1
2
3
4
5
6
7
8
1 4 3 6 2 5 7 8
4
5
3 2
1
Figure 4.6.1: A balanced tableau. Every bubble has zero column anomalies and zero row anomalies.
Figure 4.6.2: A balanced tableau. ⃝1 is a column anomaly for ⃝3 . The bubble ⃝4 is a row anomaly
for ⃝3 . All other bubbles have zero column anomalies and zero row anomalies.
Figure 4.6.3: An unbalanced tableau. ⃝1 is a column anomaly of ⃝2 , but ⃝2 has no row anomalies.
9

Chapter 5
Insertion Algorithm 2
In this section, we define Insertion Algorithm 2 which takes a balanced tableau T of shape D(w)
and a letter k such that wk < wk+1 as inputs and outputs a balanced tableau U of shape D(wsk).
Definition 5.1. Insertion Algorithm 2
Given a balanced tableau T of shape D(w) and letter k such that wk < wk+1, let p = wk
, the
height of the DRO in column k, and let q = wk+1, the height of the DRO in column k +1.
• Boost : Add 1 to the value of each bubble.
• Swap : Swap the heights of the DROs in columns k and k+1 so that the DRO in (k, p) moves
up to (k,q) and the DRO in (k +1,q) moves down to (k +1, p).
• Preserve : If there are bubbles in column k+1 in between rows p and q, move those bubbles
left to column k.
• Add : Add a new bubble ⃝1 to (k, p), the cell that the DRO in column k vacated in Swap.
• Flip : For any row l below p, if there are bubbles both in column k and column k +1, swap
these two bubbles.
The resulting tableau is U.
10

Example 5.2. w = 13542, k = 2, ws2 = 15342
T
q
p
k
k+1
1
2
3
4
5
6
7
8
1 3 5 4 2 6 7 8
4 2 3
1
Boost
1
2
3
4
5
6
7
8
1 3 5 4 2 6 7 8
5 3 4
2
Swap
1
2
3
4
5
6
7
8
1 5 3 4 2 6 7 8
5 3 4
2
Preserve
1
2
3
4
5
6
7
8
1 5 3 4 2 6 7 8
5 3 4
2
Add
1
2
3
4
5
6
7
8
1 5 3 4 2 6 7 8
5 3 4
2
1
Flip, U
1
2
3
4
5
6
7
8
1 5 3 4 2 6 7 8
3 5 4
2
1
Figure 5.2.1: Insertion Algorithm 2
Observe that the Insertion Algorithm 2 does not move bubbles vertically from one row to
another. Additionally, all horizontal movement of bubbles is limited to columns k and k +1.
Since Insertion Algorithm 2 and Insertion Algorithm 1 induce the same movement of DROs,
we know by Proposition 3.3 that the tableau U produced by Insertion Algorithm 2 is of shape
D(wsk).
To prove that S is balanced, we require two preliminary lemmas.
Lemma 5.3. The cells in Flip (cells in columns k and k +1 for any row l below row p) are either
both empty or both contain bubbles.
Proof. There cannot be a DRO in (k,l) or (k + 1,l) because Swap places DROs in (k,q) and
(k + 1, p). Therefore the DRO in row l is either left of column k or right of column k + 1. In
the former case, the cells of interest are both empty, and in the latter case, the cells both contain
bubbles.
11

To show that the pairs of bubbles in Flip always descend in value from left to right, we prove
a broader statement about balanced tableaux.
Lemma 5.4. Let T be a balanced tableau of shape D(w) for some permutation w. If wk < wk+1
then any pairs of adjacent bubbles in cells (k,l),(k + 1,l), for any l, must descend in value from
left to right.
Proof. Assume there exists at least one pair of such bubbles that ascend in value from left to right.
Let l0 be the highest row with such a pair of bubbles. As shown, let ⃝f be the bubble in (k,l0) and
let ⃝g be the bubble in (k +1,l0).
l0
k
k+1
f g
Figure 5.4.1: The bubbles ⃝f and ⃝g where f < g by assumption.
Observe that any row anomaly of ⃝g is also a row anomaly of ⃝f simply because a bubble right
of ⃝g with value greater than g will also be right of ⃝f with value greater than f . The bubble ⃝g is
an additional row anomaly for ⃝f so we have
#{RA[⃝f ]} ≥ 1+#{RA[⃝g ]} (5.1)
Now consider the set of column anomalies of ⃝f , {CA[⃝f ]}. We know {CA[⃝f ]} is not empty
because #{CA[⃝f ]} = #{RA[⃝f ]} ≥ 1. Let ⃝ei be the members of {CA[⃝f ]}. By Lemma 5.3,
any ⃝ei in column k must be accompanied by an adjacent bubble in column k +1. Let ⃝di be these
bubbles in column k +1.
12

l0
k
k+1
f g
−→
l0
k
k+1
f
ei
g
−→
l0
k
k+1
f
ei di
g
Figure 5.4.2: The bubble ⃝f cannot be balanced if f < g.
By our assumption that ⃝f and ⃝g are the highest pair of ascending bubbles in columns k
and k + 1, we know di < ei
. This forces all ⃝di bubbles to be column anomalies of ⃝g because
di < ei < f < g. Therefore every column anomaly of ⃝f forces the existence of a column anomaly
for ⃝g . This means #{CA[⃝f ]} ≤ #{CA[⃝g ]} and since ⃝f and ⃝g are assumed to be balanced, this
same equation holds for their row anomalies as well.
#{RA[⃝f ]} ≤#{RA[⃝g ]}
This contradicts Equation 5.1 and we conclude that the tableau T cannot be balanced.
In particular, this means that the bubbles in Flip descend in value from left to right before
Insertion Algorithm 2 is applied.
Example 5.5. w = 17823456
1
2
3
4
5
6
7
8
1 7 8 2 3 4 5 6
1
7
8
3
5
10
9
6
2
4
Figure 5.5.1: Unbalanced tableau of shape D(17823456). The tableau cannot be balanced because
⃝7 and ⃝9 are an adjacent ascending pair of bubbles in columns 2 and 3 with w2 = 7 < 8 = w3.
Theorem 5.6. The tableau U produced by Insertion Algorithm 2 is balanced.
13

Proof. We will check that all bubbles in U are balanced. Our arguments will implicitly rely on the
following facts.
• All bubbles in T are balanced by assumption.
• Boost does not change the relative values of any bubbles and therefore does not affect the
balanced condition.
• Insertion Algorithm 2 does not move any bubbles vertically from row to row.
• Horizontal movement in Insertion Algorithm 2 is limited to columns k and k +1.
• All bubbles ⃝b in U other than ⃝1 originated as a bubble with value b−1 in T.
We partition all cells of U into four categories and check the balanced condition for any bubbles
that appear in each category.
Category 1
p
q
k
k+1
1
Category 2
p
q
k
k+1
1
Category 3
p
q
k
k+1
1
Category 4
p
q
k
k+1
1
Figure 5.6.1: Partition of cells in Insertion Algorithm 2.
Category 1 Any bubbles ⃝b in columns other than k and k+1. Consider any such bubble and call
it ⃝b .
Column Anomalies +0 ⃝b and any bubbles above ⃝b did not move, preserving any column
anomalies.
Row Anomalies +0 ⃝b does not move. Bubblesright of ⃝b may have moved within columns
k and k+1, but this movement does not affect their status as row anomalies of ⃝b . The
addition of ⃝1 to U did not create a row anomaly for ⃝b because Boost ensures b ≥ 2.
Category 2 ⃝1 , the bubble added to (k, p).
Column Anomalies = 0 Every bubble other than ⃝1 in U has value greater than 1 because
of Boost.
Row Anomalies = 0 There are no bubbles right of ⃝1 because Swap placed a DRO in (k+
1, p).
14

Category 3 Any bubbles in between rows p and q that Preserve moved from column k +1 to k.
Consider any such bubble and call it ⃝r .
Column Anomalies +0 The bubble ⃝r and the set {CA[⃝r ]} ∈ U moved together to column k from column k +1 during Preserve. Their relative values and positions did not
change.
Row Anomalies +0 The bubbles in the set {RA[⃝r ]} ∈ U started and finished in columns
right of k +1. They remain row anomalies.
Category 4 Any bubbles in columns k and k+1 below row p. By Lemma 5.3 we know each row
in this category either contains a pair of bubbles or none at all. Consider any such row with
a pair of bubbles. As shown below, let ⃝g be the bubble that flipped into column k and let
⃝f be the bubble that flipped into column k+1 as shown below. Note that f > g by Lemma
5.4.
Before Swap
p
q
k
k+1
f g
−→
After Flip
p
q
k
k+1
1
g f
Figure 5.6.2: Category 4
Column Anomalies of ⃝f +0 Before Swap, the only bubbles above ⃝f belonged to Category 4 because there was a DRO in (k, p). By Lemma 5.3, all such bubbles also
Flipped to column k +1, preserving the column anomalies of ⃝f .
Row Anomalies of ⃝f +0 The set of bubbles right of ⃝f lost ⃝g during Flip and otherwise
remained unchanged. Since f > g, the bubble ⃝g was never a row anomaly of ⃝f .
Column Anomalies of ⃝g +1 The set of bubbles above ⃝g before Swap are all moved to
column k along with ⃝g . (Preserve moved bubbles between rows p and q, Flip moved
bubbles below row q.) The addition of ⃝1 to (k, p) creates a column anomaly for ⃝g
because g > 1.
Row Anomalies of ⃝g +1 The set of bubbles right of ⃝g gained ⃝f and otherwise remained
unchanged. Since f > g, this creates an additional row anomaly for ⃝g .
15

Example 5.7. w = 14362578, k = 3, ws3 = 14632578
T
p
q
k
k+1
1
2
3
4
5
6
7
8
1 4 3 6 2 5 7 8
3
1
4 2
5
U
p
q
k
k+1
1
2
3
4
5
6
7
8
1 4 6 3 2 5 7 8
4
2
3 5
6
1
Figure 5.7.1: Classification of bubbles in U. Category 1 :⃝2 ,⃝4 . Category 2 : ⃝1 . Category 3 : ⃝6 .
Category 4 : ⃝3 ,⃝5
.
16

Chapter 6
Insertion Algorithm 3
In this section, we define Insertion Algorithm 3. This insertion algorithm inputs a reduced word
τ = τ1τ2 ... τl(w)
for permutation w and outputs a balanced tableau T(τ) of shape D(w). We will
prove that Insertion Algorithm 3 is a bijection between the set of reduced words R(w) and the
set of balanced tableaux of shape D(w) in the following section.
Definition 6.1. Insertion Algorithm 3
Let τ = τ1τ2 ... τl(w) ∈ R(w) and let T(/0) be an empty tableau, meaning all cells of T(/0) are
empty. Using Insertion Algorithm 2, insert the letters of τ from left to right, starting with τ1 and
ending with τl(w)
. The letter τ1 is inserted into T(/0) to produce T(τ1). In this way τi
is inserted
into T(τ1τ2 ... τi−1) until τl(w)
is inserted into T(τ1τ2 ... τl(w)−1
), yielding T(τ).
17

Example 6.2. T(τ) for reduced word τ = 23432 of permutation w = 153426
T(/0)
1
2
3
4
5
6
7
8
1 2 3 4 5 6 7 8
T(2)
1
2
3
4
5
6
7
8
1 3 2 4 5 6 7 8
1
T(23)
1
2
3
4
5
6
7
8
1 3 4 2 5 6 7 8
2 1
T(234)
1
2
3
4
5
6
7
8
1 3 4 5 2 6 7 8
3 2 1
T(2343)
1
2
3
4
5
6
7
8
1 3 5 4 2 6 7 8
4 2 3
1
T(23432)
1
2
3
4
5
6
7
8
1 5 3 4 2 6 7 8
3 5 4
2
1
Figure 6.2.1: Insertion Algorithm 3
Proposition 6.3. If τ ∈ R(w) then T(τ) is a balanced tableau of shape D(w).
Proof. By Proposition 3.3 T(τ) is balanced and by Theorem 5.6 T(τ) is of the correct shape.
Lemma 6.4. For any finite permutation w, Insertion Algorithm 3 is an injection from the set of
reduced words R(w) to the set of balanced tableaux of shape D(w).
Proof. Let τ,σ ∈ R(w) and τ ̸= σ. It suffices to show that T(τ) ̸= T(σ)
Let m be the first position such that τm ̸= σm. Thus the two tableaux are equivalent after the
first m−1 letters are inserted, T(τ1τ2 ... τm−1) = T(σ1σ2 ...σm−1).
Inserting τm into T(τ1τ2 ... τm−1) places ⃝1 into the cell previously occupied by the DRO in
column τm. But inserting σm into T(σ1σ2 ...σm−1) places ⃝1 into the cell currently occupied by
the DRO in column σm. These cells must have different heights because two DROs can never
occupy the same row.
Therefore T(τ1τ2 ... τm) has ⃝1 in a different row than the ⃝1 of T(σ1σ2 ...σm). Since Insertion Algorithm 3 does not change the height of any bubbles, ⃝ ∈ 1 T(τ1τ2 ... τm) and ⃝1
18

∈ T(σ1σ2 ...σm) will remain in different rows than each other. Since τ and σ have l(w) letters,
both bubbles will be Boosted l(w) − m times and will finish Insertion Algorithm 3 with value
1+ (l(w)−m).
Therefore T(τ) ̸= T(σ) because the bubble with value 1+ (l(w)−m) occupies a different cell
in each tableau.
19

Chapter 7
Bijection Between Reduced Words and Balanced Tableaux
Now we have all the tools to conclude that Insertion Algorithm 3 is indeed a bijection between
the set of reduced words of a permutation and the set of balance tableaux of Rothe Diagram shape.
Theorem 7.1. For any finite permutation w the set of reduced words R(w) is equinumerous with
the set of balanced tableaux of shape D(w).
Proof. See [3, Theorem 2.4]
Theorem 7.2. Insertion Algorithm 3 is a bijection between the set of reduced words R(w) for
permutation w and the set of balanced tableaux of shape D(w).
Proof. This follows immediately from Lemma 6.4 and Theorem 7.1.
20

References
1. Bjorner, A. & Brenti, F. ¨ Combinatorics of Coxeter groups xiv+363 (Springer, New York,
2005).
2. Edelman, P. & Greene, C. Balanced tableaux. Adv. in Math. 63, 42–99. doi:10.1016/0001-
8708(87)90063-6 (1987).
3. Fomin, S., Greene, C., Reiner, V. & Shimozono, M. Balanced labellings and Schubert polynomials. European J. Combin. 18, 373–389. doi:10.1006/eujc.1996.0109 (1997).
4. Fulton, W. Young tableaux With applications to representation theory and geometry, x+260
(Cambridge University Press, Cambridge, 1997).
5. Gillen, B. & Michala, J. Characterizing Rothe Diagrams arXiv : 2303.11392. 2023.
6. Knuth, D. E. Permutations, matrices, and generalized Young tableaux. Pacific J. Math. 34,
709–727 (1970).
7. Stanley, R. P. On the number of reduced decompositions of elements of Coxeter groups. European J. Combin. 5, 359–372. doi:10.1016/S0195-6698(84)80039-6 (1984).
21

Abstract (if available)

Abstract A bijection between the set of reduced words for a finite permutation and the set of balanced tableaux of Rothe Diagram shape is presented. The bijection follows the insertion style of the Schensted Insertion used in the Robinson-Schensted-Knuth correspondence.

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