Tableau Stabilization and Rectangular Tableaux Fixed by Promotion Powers

TL;DR

This paper introduces tableau stabilization, a new concept based on jeu de taquin, and uses it to construct large rectangular tableaux fixed by promotion powers, connecting to cyclic sieving phenomena.

Contribution

It develops the theory of tableau stabilization and applies it to explicitly construct fixed points of promotion powers in rectangular tableaux.

Findings

Defined tableau stabilization and analyzed its properties.

Established bounds and shape characteristics of stabilized tableaux.

Constructed large rectangular tableaux fixed by promotion powers.

Abstract

We introduce tableau stabilization, a new phenomenon and statistic on Young tableaux based on jeu de taquin. We investigate bounds for tableau stabilization, the shape of stabilized tableaux, and tableau stabilization as a permutation statistic. We apply tableau stabilization to construct the sufficiently large rectangular tableaux fixed by powers of promotion, which were counted by Brendon Rhoades via the cyclic sieving phenomenon

Equations

$$S^{(\ast k)}\sim S^{(k)},\mathrm{Rect}(S^{(\ast k)})=\mathrm{Rect}(S^{(k)}),\text{ and }{\mathrm{Rect}}^{\ast }(S^{(\ast k)})={\mathrm{Rect}}^{\ast }(S^{(k)}).$$

$${\mathrm{Rect}}^{\ast }(S)=\mathrm{Rect}(S^{†}{)}^{†}=(e(\mathrm{Rect}(S)){)}^{†},$$

$${\mathrm{Rect}}^{\ast }(S)=\mathrm{Rect}(S^{†}{)}^{†}.$$

$$\mathrm{Rect}(S^{†}{)}^{†}=P(w_{0}w{w}_0{)}^{†}.$$

$$P(w_{0}w{w}_0{)}^{†}=e(P(w){)}^{†}=(e(\mathrm{Rect}(S)){)}^{†}$$

$$\mathrm{rv}({\mathrm{Rect}}^{\ast }(S^{(k)}){)}_j=\mathrm{rv}(\mathrm{Rect}(S^{(k)}){)}_{b+1-j}.$$

$${\mathrm{stab}}^{\ast }(S)=\mathrm{stab}(S).$$

$$\mathrm{rv}({\mathrm{Rect}}^{\ast }(S^{(k)}){)}_j=\mathrm{rv}(e(\mathrm{Rect}(S^{(k)}){)}^{†}{)}_j=\mathrm{rv}(\mathrm{Rect}(S^{(k)}{)}^{†}{)}_j=\mathrm{rv}(\mathrm{Rect}(S^{(k)}){)}_{b+1-j}$$

$$\mathrm{rv}(\mathrm{Rect}(S^{(k)}){)}_j-\mathrm{rv}(\mathrm{Rect}(S^{(k-1)}){)}_j=r\text{ for all }j=1,\dots ,b.$$

$$\mathrm{rv}({\mathrm{Rect}}^{\ast }(S^{(k)}){)}_j-\mathrm{rv}({\mathrm{Rect}}^{\ast }(S^{(k-1)}){)}_j=r\text{ for all }j=1,\dots ,b.$$

$${p^k(T)|}_{[1,n-k]}$$

$${p^k(T)|}_{[n-k+1,n]}$$

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$$R_a(S)\in \mathrm{SYT}((ar{)}^b{)}^{p^{br}}.$$

$$\mathrm{rv}(\mathrm{Rect}(S^{(k-1)}){)}_j$$

$$\mathrm{rv}({\mathrm{Rect}}^{\ast }(S^{(k-1)}){)}_j$$

$$\mathrm{rv}(R{)}_j$$

$$=(k-1)r+\sum _{i=j}^{b-j}{c}_i+(a-2k+2)r+(k-1)r+\sum _{i=b+1-j}^{j-1}{c}_i$$

$$=ar+\sum _{i=j}^{j-1}{c}_i$$

$$=ar,$$

$${R|}_{[1,jbr]}=\mathrm{Rect}(S^{(j)})\text{ for all }j\in [0,a-k+1].$$

$${R|}_{[jbr+1,abr]}={\mathrm{Rect}}^{\ast }(S^{(a-j)})+jbr\text{ for all }j\in [k-1,a].$$

$${p^{jbr}(R)|}_{[1,(a-j)br]}$$

$$=\mathrm{Rect}({\mathrm{Rect}}^{\ast }(S^{(a-j)})+jbr)-jbr\text{ by }\text{eqref}eq:stabstart1$$

$$=\mathrm{Rect}(S^{(a-j)})\text{ as rectification is well-defined }$$

$$={R|}_{[1,(a-j)br]}\text{ by }\text{eqref}eq:stabstart2.$$

$${p^{jbr}(R)|}_{[(a-j)br+1,abr]}$$

$$={\mathrm{Rect}}^{\ast }(\mathrm{Rect}(S^{(j)}))+(a-j)br\text{ by }\text{eqref}eq:stabstart2$$

$$={\mathrm{Rect}}^{\ast }(S^{(j)})+(a-j)br\text{ as anti-rectification is well-defined }$$

$$={R|}_{[(a-j)br+1,abr]}\text{ by }\text{eqref}eq:stabstart1.$$

$$p^{jbr}(R)=R\text{ for all }j\in [k-1,a-k+1].$$

$$p^{br}(R)=p^{kbr-(k-1)br}(R)=p^{kbr}(p^{-(k-1)br}(R))=p^{kbr}(R)=R.$$

$$\beta :=\underset{b\text{ times}}{\underbrace{(r)\cup (r)\cup \cdots \cup (r)}}.$$

$$\left\{R_a(S):S\in \mathrm{SYT}(\beta ),\mathrm{stab}(S)\le \frac{a+1}{2}\right\}\subseteq \mathrm{SYT}((ar{)}^b{)}^{p^{br}}.$$

$$S^{\prime }\sim {R_a(S^{\prime })|}_{[(k-1)br+1,kbr]}={R_a(S)|}_{[(k-1)br+1,kbr]}\sim S,$$

Full text

Department of Mathematics, Texas State University, San Marcos, TX 78666, USA; c_a518@txstate.edu

1. Introduction

In this paper, we introduce tableau stabilization, a new phenomenon we found in order to construct sufficiently large rectangular tableaux that are fixed by promotion powers. Central to defining and investigating tableau stabilization are Schütenberger’s jeu de taquin and the rectification operator, which are already well-established algorithms in the theory of Young tableaux. Tableau stabilization is the phenomenon that if we attach sufficiently many shifted copies of a skew tableau to its right and then rectify, some copy and all of those to its right only experience horizontal slides. We will investigate when the vertical slides stop, i.e. when the skew tableau stabilizes. We will also determine the shape of the stabilized tableau if the initial skew tableau has same size rows. The case where all rows have the same size is used to construct sufficiently large rectangular tableaux that are fixed by promotion powers. We leave the reader with open problems on tableau stabilization, most notably extending our bound from equal row sizes and finding its distribution as a permutation statistic. See LABEL:sec:bkgd for definitions and further background.

Definition 1.1**.**

For any standard skew tableau $S$, let $S^{(k)}$ denote the result of attaching $(k-1)$ shifted copies of $S$ to the right of $S$ so that the result is a standard skew tableau. Let $m$ denote the size of $S$ and $k$ be a positive integer. We say $S$ stabilizes at $k$ if the entries in $[(k-1)m+1,km]$ lie in the same rows in $\operatorname{Rect}(S^{(k)})$ and $S^{(k)}$. Let $\operatorname{stab}(S)$ denote the minimum value at which $S$ stabilizes.

Example 1.2**.**

Let $\operatorname{Rect}$ denote the rectification operator. Consider

[TABLE]

Note that $2$ does not lie in the same row in $S^{(3)}$ and $\operatorname{Rect}(S^{(3)})$, but $7,8,\dots,12$ do. Hence, $\operatorname{stab}(S)=2$. As $13,14,\dots,18$ also stay in the same row in $S^{(3)}$ and $\operatorname{Rect}(S^{(3)})$, $S$ stabilizes at 3 as well. Note $10$ does not lie in the same row in $T^{(3)}$ and $\operatorname{Rect}(T^{(3)})$, but $13,14,\dots,18$ do. Hence, $\operatorname{stab}(T)=3$.

Also consider

Then,

[TABLE]

Since $\left.\operatorname{Rect}^{\ast}(S^{(\ast 3)})\right|_{[11,15]}\not\sim S$, but $\left.\operatorname{Rect}^{\ast}(S^{(\ast 3)})\right|_{[6,10]}\sim S$, we have $\operatorname{stab}^{\ast}(S)=2$.

Remark 5.3**.**

Similarly to LABEL:rem:rowsizes, if $S$ does not have increasing row vector, $S^{(\ast k)}$ need not be a standard skew tableau. Hence, the notions of stabilization and anti-stabilization only make sense simultaneously when the row vector is constant.

Anti-stabilization has many of the same properties as stabilization. If we apply the same reasoning using anti-rectification instead of rectification and $S^{(\ast k)}$ instead of $S^{(k)}$, we get the following analogues of LABEL:lem:stabrse, LABEL:thm:dualequivstab, LABEL:lem:stabcontinues, LABEL:thm:stabweaklydecrows, and LABEL:thm:stab for anti-stabilization.

Corollary 5.4**.**

Suppose $S$ is a standard skew tableau with $b$ rows and increasing row vector. Then,

1. (a)

Anti-stabilization is well-defined up to row shift equivalence. 2. (b)

$\operatorname{stab}^{\ast}$ is constant on dual equivalence classes of standard skew tableaux. 3. (c)

$S$ anti-stabilizes eventually, and if $S$ anti-stabilizes at $k$, it anti-stabilizes at any $k^{\prime}\geq k$. 4. (d)

If $b\geq 2$, then $S$ anti-stabilizes at $2b-2$. 5. (e)

If $S$ has constant row vector, then $S$ anti-stabilizes at $b$.

If $S$ is a standard skew tableau with constant row vector, then $S^{(k)}$ and $S^{(\ast k)}$ are both standard tableaux with the same reading word. Thus, by LABEL:lem:rectP,

[TABLE]

Lemma 5.5**.**

For any standard skew tableau $S$,

[TABLE]

where $e$ is Schützenberger’s evacuation operator.

Proof.

Let $w=w_{1}\dots w_{m}$ be the reading word of $S$, so $w$ is a permutation of size $m$. Anti-rectifying $S$ and rectifying $S^{\dagger}$ differ by a rotation of $180^{\circ}$, implying

[TABLE]

As $\dagger$ reverses the order of the cells in the reading word and reverses the entry values, the reading word of $S^{\dagger}$ is $w_{0}ww_{0}$, where $w_{0}=[m,m-1,\dots,2,1]$ is the reversal permutation of size $m$. Thus, by LABEL:lem:rectP,

[TABLE]

Moreover, $P(w_{0}ww_{0})=e(P(w))$ by [Sag01, Theorem 3.9.4], so

[TABLE]

by LABEL:lem:rectP. Combining (141), (142), and (143) gives 5.5.

∎

Lemma 5.6**.**

Suppose that $S$ is a standard skew tableau with row vector $(r^{b})$. Then, for all $k\geq 1$, and $j=1,\dots,b$,

[TABLE]

In addition,

[TABLE]

Proof.

For any $k\geq 1$, we have $S^{(k)}\sim S^{(\ast k)}$ by (140). Thus, by 5.5,

[TABLE]

for all $k\geq 1$ and $j=1,\dots,b$, proving (144). By LABEL:def:stab2(c), $S$ stabilizes at $k$ if and only if

[TABLE]

Similarly, $S$ anti-stabilizes at $k$ if and only if

[TABLE]

By (144), (146) holds if and only if (147) holds. Thus, $S$ stabilizes at $k$ if and only if $S$ anti-stabilizes at $k$, and (145) follows.

∎

6. Sufficiently Large Tableaux Fixed by Powers of Promotion

In this section, we first give an alternative method for doing multiple promotions at once. Recall $\operatorname{p}:\operatorname{SYT}(a^{b})\to\operatorname{SYT}(a^{b})$ denotes Schützenberger’s promotion operator, see LABEL:def:pr. We then construct the sufficiently large rectangular standard tableaux fixed by promotion powers. In particular, we construct the tableaux in $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$ for $a\geq 2b-1$ and prove LABEL:thm:prfixedtabs. LABEL:thm:stabshape plays a key role in showing these tableaux are rectangular, tableau stabilization is central in showing these tableaux are fixed by $\operatorname{p}^{br}$, and the bound in LABEL:thm:stab lets us control the size of these rectangular tableaux. We also prove LABEL:cor:prfixedpr, which describes the action of promotion on $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$.

Lemma 6.1**.**

Suppose $a,b\in\mathbb{Z}_{\geq 1}$ and $n\coloneqq ab$. For any tableau $T\in\operatorname{SYT}(a^{b})$ and $k=1,\dots,n$, we have

[TABLE]

Proof.

Equation (148) follows from LABEL:def:pr and rectification being well-defined. By LABEL:thm:prnid, $\operatorname{p}^{n}=\operatorname{id}$ on $\operatorname{SYT}(a^{b})$, so $\operatorname{p}^{k}=(\operatorname{p}^{-1})^{n-k}$. Then, (149) follows from the fact that $T$ is rectangular, meaning demotion and promotion are inverses, LABEL:def:pr, and anti-rectification being well-defined.

∎

Definition 6.2**.**

Suppose $S$ is a standard skew tableau with row vector $(r^{b})$, and let $k\coloneqq\operatorname{stab}(S)$. For any $a\geq 2k-1$, let $R_{a}(S)$ denote the filling of rectangular shape formed by row-concatenating $\operatorname{Rect}(S^{(k-1)}),S^{(a-2k+2)}+(k-1)br,$ and $\operatorname{Rect}^{\ast}(S^{(k-1)})+(a-k+1)br$ together from left to right.

Example 6.3**.**

Let

[TABLE]

which has $k=\operatorname{stab}(S)=3$, and let $a=6$. Observe that

[TABLE]

so

[TABLE]

Theorem 6.4**.**

Suppose that $S$ is a standard skew tableau with row vector $(r^{b})$. Then, for any integer $a\geq 2\operatorname{stab}(S)-1$,

[TABLE]

Proof.

Let $k\coloneqq\operatorname{stab}(S)$ and $R\coloneqq R_{a}(S)$. First, we check that $R$ has rectangular shape $((ar)^{b})$. By LABEL:cor:stabshapeearly and (144),

[TABLE]

for all $j=1,\dots,b$. Hence, for $j=1,\dots,b$,

[TABLE]

which shows $\operatorname{rv}(R)=((ar)^{b})$. Since $R$ is left-justified, it is a filling of shape $((ar)^{b})$.

Secondly, we check that $R$ is a standard tableau. We break $R$ into three pieces:

• •

$R_{1}\coloneqq\left.R\right|_{[1,kbr]}=\operatorname{Rect}(S^{(k)})$, because $S$ stabilizes at $k$

• •

$R_{2}\coloneqq\left.R\right|_{[kbr+1,(a-k+1)br]}\sim S^{(a-2k+1)}$,

• •

$R_{3}\coloneqq\left.R\right|_{[(a-k+1)br+1,abr]}=\operatorname{Rect}^{\ast}(S^{(k-1)})+(a-k+1)br$.

The fillings $R_{1}$ and $R_{3}$ are skew tableaux because they are the rectification and anti-rectification of skew tableaux, respectively. The filling $R_{2}$ is a skew tableau since it is formed by row-concatenating shifted copies of $\left.R_{1}\right|_{[(k-1)br+1,kbr]}$. Since $R_{1},R_{2},R_{3}$ use the entries $[1,kbr]$, $[kbr+1,(a-k+1)br]$, and $[(a-k+1)br+1,abr]$ respectively, $R$ uses each of $1,2,\dots,abr$ exactly once. Because $R_{2}$ adjoins to the right of $R_{1}$ and $R_{3}$ adjoins to the right of $R_{2}$, the rows and columns of $R$ are increasing. Hence, $R$ is a standard tableau.

Thirdly, we check that $R$ is fixed by $br$ promotions. Since $S$ stabilizes at $k$,

[TABLE]

By 5.6, $S$ also anti-stabilizes at $k$, so similarly,

[TABLE]

Note that both sides of (156) are empty when $j=a$. Then, for all $j\in[k-1,a]$,

[TABLE]

Similarly, for all $j\in[1,a-k+1]$,

[TABLE]

Putting (157) and (158) together,

[TABLE]

In particular, because $a\geq 2k-1$, we have $k-1,k\in[k-1,a-k+1]$, so

[TABLE]

∎

Notation 6.5**.**

Fix $b,r\in\mathbb{Z}_{\geq 1}$, and let $\beta$ be the block anti-diagonal skew shape

[TABLE]

Although 6.4 holds for any standard skew tableau $S$ with row vector $(r^{b})$, $S$ is row shift equivalent to a standard skew tableau $S^{\prime}$ of shape $\beta$ by performing horizontal slides. Thus, by LABEL:lem:stabrse, $\operatorname{Rect}(S^{(k)})=\operatorname{Rect}((S^{\prime})^{(k)})$ and $\operatorname{stab}(S)=\operatorname{stab}(S^{\prime})$, which means $R_{a}(S)=R_{a}(S^{\prime})$ for all $a\geq 2\operatorname{stab}(S)-1$. Hence, any rectangular tableau of the form $R_{a}(S)$ for some standard skew tableau $S$ with row vector $(r^{b})$ is also of the form $R_{a}(S^{\prime})$ for some $S^{\prime}\in\operatorname{SYT}(\beta)$.

Corollary 6.6**.**

For all $a\in\mathbb{Z}_{\geq 1}$,

[TABLE]

Proof.

For fixed $S\in\operatorname{SYT}(\beta)$, $R_{a}(S)$ is defined for $a\geq 2\operatorname{stab}(S)-1$. Thus, for fixed $a\in\mathbb{Z}_{\geq 1}$ and all $S\in\operatorname{SYT}(\beta)$ with $\operatorname{stab}(S)\leq\frac{a+1}{2}$, $R_{a}(S)$ is defined and lies in $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$ by 6.4.

It remains to show the elements of $\left\{R_{a}(S):S\in\operatorname{SYT}(\beta)\right\}$ are distinct. Suppose that $R_{a}(S)=R_{a}(S^{\prime})$ for $S,S^{\prime}\in\operatorname{SYT}(\beta)$. Letting $k=\operatorname{stab}(S)$, we have $\operatorname{rv}(\left.R_{a}(S)\right|_{[(k-1)br+1,kbr]})=(r^{b})$ and hence $\left.\operatorname{rv}(R_{a}(S^{\prime})\right|_{[(k-1)br+1,kbr]})=(r^{b})$ as well. This and 6.2 mean

[TABLE]

forcing $S=S^{\prime}$ because $S,S^{\prime}\in\operatorname{SYT}(\beta)$.

∎

Example 6.7**.**

Consider

[TABLE]

which have

[TABLE]

Thus, the smallest $R_{a}(S),R_{a^{\prime}}(T),R_{a^{\prime\prime}}(U)$ we can construct with 6.2 are

[TABLE]

On the other hand, neither

[TABLE]

is of the form $R_{a}(S)$ for $S\in\operatorname{SYT}((1)\cup(1)\cup(1)\cup(1))$ because $\left.V\right|_{[4(j-1)+1,4j]}$ and $\left.W\right|_{[4(j-1)+1,4j]}$ never use all 4 rows for any value of $j$.

The containment in 6.6 can be strict in general, as illustrated by $V,W$ in 6.7. However, equality in 6.6 holds for $a\geq 2b-1$, when $\operatorname{stab}(S)\leq\frac{a+1}{2}$ holds for all $S\in\operatorname{SYT}(\beta)$ by LABEL:thm:stab. Thus, 6.6 specializes to

[TABLE]

The fact that equality holds in (172) is the content of LABEL:thm:prfixedtabs.

Remark 6.8**.**

In LABEL:thm:prfixedtabs, $R_{a}(S)$ is defined by row concatenating shifted copies of $\operatorname{Rect}(S^{(b-1)})$, $S^{(a-2b+2)}$, and $\operatorname{Rect}^{\ast}(S^{(b-1)})$ instead of shifted copies of $\operatorname{Rect}(S^{(k-1)}),$ $S^{(a-2k+2)},$ and $\operatorname{Rect}^{\ast}(S^{(k-1)})$ where $k=\operatorname{stab}(S)$ as in 6.2. These 2 constructions agree by the definition of tableau stabilization, see 1.1, and because $\operatorname{stab}(S)\leq b$ for $S\in\operatorname{SYT}(\beta)$, see LABEL:thm:stab.

Proof of LABEL:thm:prfixedtabs.

Fix a positive integer $a\geq 2b-1$. Equality in (172) will follow from the fact that both sides have the same size. By LABEL:lem:rectquotients, we have the quotient

[TABLE]

Thus, $Q_{a}((ar)^{b})=\beta$ because $a\geq 2b-1\geq b$. By LABEL:cor:prfixedribbontab and (173),

[TABLE]

which shows both sides of (172) have the same size. Therefore,

[TABLE]

Finally, we can choose $S\in\operatorname{SYT}(\beta)$ by choosing the sets of $r$ elements that go in each of $b$ rows, which determines $S$ uniquely since each row of $S$ is increasing. Thus,

[TABLE]

∎

Not only can we describe the elements of $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$ for $a\geq 2b-1$, but we can also describe the action of promotion on $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$, which is closed under promotion. In fact, using the definition of promotion for skew shapes, see LABEL:def:pr, promotion commutes with the $R_{a}$ operator:

[TABLE]

as in LABEL:cor:prfixedpr.

Proof of LABEL:cor:prfixedpr.

For any $T\in\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$, we have

[TABLE]

Thus, $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$ is closed under promotion. So, fixing $S\in\operatorname{SYT}(\beta)$,

[TABLE]

by LABEL:thm:prfixedtabs. It suffices to show $S^{\prime}=\operatorname{p}(S)$. As rectification is well-defined, we get the same result whether we perform an inner slide at 1’s cell before or after rectification, so

[TABLE]

Since $S\in\operatorname{SYT}(\beta)$, the inner slide started by removing 1’s cell in $S^{(b)}$ consists of only horizontal slides, so

[TABLE]

Then,

[TABLE]

By LABEL:thm:stab, $\left.R_{a}(S^{\prime})\right|_{[(b-1)br+1,b^{2}r]}$ and $\left.R_{a}(\operatorname{p}(S))\right|_{[(b-1)br+1,b^{2}r]}$ each have row vector $(r^{b})$, which together with (176), forces $b^{2}r$ to be in the same cell in $R_{a}(S^{\prime})$ and $R_{a}(\operatorname{p}(S))$. Combining this with (176) again,

[TABLE]

In particular,

[TABLE]

forcing $S^{\prime}=\operatorname{p}(S)$ since $S^{\prime},\operatorname{p}(S)\in\operatorname{SYT}(\beta)$. ∎

Corollary 6.9**.**

Let $\lambda=((b+1)r,br,\dots,2r,r)$ and $\mu=(br,(b-1)r,\dots,r,0)$. Then, for all integers $a\geq 2b-1$, $\nu\vdash br$, and $T_{0}\in\operatorname{SYT}(\nu)$,

[TABLE]

the Littlewood-Richardson coefficient from LABEL:def:LRCs.

Proof.

Note that $\beta=\lambda/\mu$. By LABEL:thm:prfixedtabs and $\left.R_{a}(S)\right|_{[1,br]}=\operatorname{Rect}(S)$, we have

[TABLE]

by LABEL:def:LRCs. ∎

7. Other Tableaux Fixed by Promotion Powers

In this section, we present the construction of $\operatorname{SYT}((2r)^{b})^{\operatorname{p}^{br}}$ by White and Rhee [Whi06] [Rhe12], see LABEL:thm:fixedbyab/2prom, using a similar construction to LABEL:thm:prfixedtabs. Then, we describe the action of promotion on $\operatorname{SYT}((2r)^{b})^{\operatorname{p}^{br}}$ in 7.4. Next, we characterize the block diagonal skew tableaux fixed by a power of promotion in terms of tableaux of straight shape fixed by certain powers of promotion, see 7.7, inspired by White [Whi06]. This has consequences for describing the tableaux in $\operatorname{SYT}((2r)^{b})^{\operatorname{p}^{br/k}}$ in terms of rectangular tableaux fixed by smaller promotion powers, see 7.9.

We next describe the tableaux in $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$ when $a=2$. Using LABEL:lem:rectquotients and that $\left\lceil\frac{b}{2}\right\rceil=\left\lfloor\frac{b}{2}\right\rfloor$ when $b$ is even, we have the quotient

[TABLE]

Therefore, by LABEL:cor:prfixedribbontab,

[TABLE]

Definition 7.1**.**

Fix $b,r\in\mathbb{Z}_{\geq 1}$ and let $\gamma\coloneqq(r^{\left\lceil\frac{b}{2}\right\rceil})\cup(r^{\left\lfloor\frac{b}{2}\right\rfloor})$. For $S\in\operatorname{SYT}(\gamma)$, let $R_{2}(S)$ be the filling formed by row-concatenating $\operatorname{Rect}(S)$ and $\operatorname{Rect}^{\ast}(S)+br$ together from left to right.

Example 7.2**.**

Suppose $b=4,r=2$, and choose

[TABLE]

Then,

[TABLE]

We use the name $R_{2}(S)$ because the construction is similar to the definition of $R_{a}(S)$ if we set $a=2$. Note that we cannot just set $a=2$ in 6.2 because 6.2 requires $a\geq 2\operatorname{stab}(S)-1$. Similar to 6.4, we will show $R_{2}(S)$ has shape $((2r)^{b})$, is a standard tableau, and is fixed by $\operatorname{p}^{br}$.

Lemma 7.3**.**

The map

[TABLE]

is bijective.

Proof.

Say $\gamma=\lambda/\mu$ as a skew shape. By [Sta86, Lemma 3.3], we have the Littlewood-Richardson coefficient

[TABLE]

for all $\nu=(\nu_{1},\dots,\nu_{b})\vdash br$. 7.3 follows immediately from (183) and LABEL:def:LRCs.

∎

Proof of LABEL:thm:fixedbyab/2prom.

[Whi06] [Rhe12] Fix $S\in\operatorname{SYT}(\gamma)$. For all $j=1,\dots,b$ and

[TABLE]

Indeed, $R_{2}(S)$ is a filling of shape $((2r)^{b})$. The fillings $\left.R_{2}(S)\right|_{[1,br]}=\operatorname{Rect}(S)$ and $\left.R_{2}(S)\right|_{[br+1,2br]}=\operatorname{Rect}^{\ast}(S)+br$ are skew tableaux using the entries $[1,br]$ and $[br+1,2br]$, repsectively. Thus, their row-concatenation from left to right, $R_{2}(S)$, is a standard tableau.

Similar to the proof of 6.4,

[TABLE]

and

[TABLE]

Putting (184) and (185) together gives $\operatorname{p}^{br}(R_{2}(S))=R_{2}(S)$. Thus,

[TABLE]

The fact that $\{R_{2}(S):S\in\operatorname{SYT}(\gamma)\}$ is a set and not a multiset is a consequence of 7.3. By (178), both sides of (186) have the same size, so

[TABLE]

Finally, we count these tableaux by first partitioning $\{1,2,\dots,br\}$ into two blocks of size $r\cdot\left\lceil\frac{b}{2}\right\rceil$ and $r\cdot\left\lfloor\frac{b}{2}\right\rfloor$. Then, we choose a filling of $(r^{\left\lceil\frac{b}{2}\right\rceil})$ with the first block and a filling of $(r^{\left\lfloor\frac{b}{2}\right\rfloor})$ with the second block so that both fillings have increasing rows and columns. This yields

[TABLE]

∎

We also describe the action of promotion on $\operatorname{SYT}((2r)^{b})^{\operatorname{p}^{br}}$, which is closed under promotion. Like $R_{a}$ in LABEL:cor:prfixedpr, the $R_{2}$ operator also commutes with promotion.

Corollary 7.4**.**

[Whi06] [Rhe12] For any $S\in\operatorname{SYT}(\gamma)$,

[TABLE]

Proof.

Similar to LABEL:cor:prfixedpr, $\operatorname{SYT}((2r)^{b})^{\operatorname{p}^{br}}$ is closed under promotion. Thus, fixing $S\in\operatorname{SYT}(\gamma)$,

[TABLE]

Our reasoning from the proof of LABEL:cor:prfixedpr still holds through (176), so

[TABLE]

By 7.3, $\left.R_{2}(S^{\prime})\right|_{[1,br]}$ and $\left.R_{2}(\operatorname{p}(S))\right|_{[1,br]}$ each have shapes $\nu=(\nu_{1},\dots,\nu_{b})$ satisfying $\nu_{j}+\nu_{b+1-j}=2r$, which together with (187), forces $br$ to be in the same cell in $R_{2}(S^{\prime})$ and $R_{2}(\operatorname{p}(S))$. Thus,

[TABLE]

Finally, 7.3 forces $S^{\prime}=\operatorname{p}(S)$. ∎

For any $k\mid br$,

[TABLE]

By LABEL:cor:prfixedpr and 7.4, we have, for $a=2$ or $a\geq 2b-1$,

[TABLE]

For $a=2$, we will describe $\{S\in\operatorname{SYT}(\gamma):\operatorname{p}^{br/k}(S)=S\}$ in terms of rectangular tableaux fixed by smaller promotion powers.

Definition 7.5**.**

For a standard tableau $T$ of size $n$, $m\in\mathbb{Z}_{\geq 1}$, and $A=\{a_{1},\dots,a_{k}\}\subseteq[m]$ with $a_{1}<\dots<a_{k}$ and $k\mid n$, let $I_{m,A}(T)$ denote the skew tableau obtained from $T$ by replacing $ik+j$ by $im+a_{j}$ for all $i=0,1,\dots,n/k-1$ and $j=1,\dots,k$. For example, if

[TABLE]

then

[TABLE]

Definition 7.6**.**

On a tableau with entries $i_{1}<\dots<i_{n}$, let promotion act on the indices as it would on a standard tableau. For example, since

[TABLE]

We realize there are other variations on how promotion acts on tableaux with distinct non-standard entries, such as in [Rho10], but this will prove convenient for our purposes.

Theorem 7.7**.**

(inspired by [Whi06]) For any partitions $\lambda^{(1)},\dots,\lambda^{(b)}$ with total size $n$, and $k\mid n$,

[TABLE]

if $k\mid\gcd(|\lambda^{(1)}|,\dots,|\lambda^{(b)}|)$. Else,

[TABLE]

Example 7.8**.**

Suppose $b=2,\lambda^{(1)}=(3,3),\lambda^{(2)}=(2,2),k=2$. Let

[TABLE]

Then,

[TABLE]

Proof.

Suppose $\lambda^{(1)},\dots,\lambda^{(b)}$ are partitions with total size $n$ and $k\mid\gcd(|\lambda^{(1)}|,\dots,|\lambda^{(b)}|)$. Let

[TABLE]

Consider $T_{j}\in\operatorname{SYT}(\lambda^{(j)})^{\operatorname{p}^{m_{j}}}$ for $j=1,\dots,b$ and $(M_{1},\dots,M_{b})\in\binom{[m]}{m_{1},\dots,m_{b}}$. Let

[TABLE]

and

[TABLE]

Also, write

[TABLE]

For each $j$, the entries in $S_{j}$ are $\cup_{i=1}^{k}(M_{j}+(i-1)m)$. Thus, since $\operatorname{p}^{m}$ performs a total of $m$ decrements modulo $n$, the entries in $\operatorname{p}^{m}(S)_{j}$ are

[TABLE]

Thus, the same set of entries appears in $S_{j}$ and $\operatorname{p}^{m}(S)_{j}$. By this and the fact that $m_{j}$ of the $m$ promotions on $S$ slide through $S_{j}$,

[TABLE]

Now,

[TABLE]

for all $j=1,\dots,b$. Combining (198), (199), and (200) yields $\operatorname{p}^{m}(S)=S$, so $\supseteq$ holds in (196).

On the other hand, suppose $S\in\operatorname{SYT}(\lambda^{(1)}\cup\dots\cup\lambda^{(b)})^{\operatorname{p}^{m}}$. Let $S=S_{1}\cup\dots\cup S_{b}$, $M_{j}$ denote the entries in $S_{j}$ that lie in $[m]$ and $m_{j}\coloneqq\#M_{j}$. Since $\operatorname{p}^{m}(S)=S$, the set of entries in $S_{j}$ must be fixed by $m$ decrements modulo $n$. When applying $\operatorname{p}^{m}$ to $S$, we apply $m_{j}$ inner slides and $m$ decrements modulo $n$ to $S_{j}$, so letting $\{i_{1},\dots,i_{|\lambda^{(j)}|}\}$ denote the set of entries of $S_{j}$ in increasing order, we must have

[TABLE]

Equation (201) means that shifting the indices by $m_{j}$ modulo $|\lambda^{(j)}|$ corresponds to shifting the entries by $m$ modulo $n$. This means $|\lambda^{(j)}|=k_{j}m_{j}$ for some $k_{j}\in\mathbb{Z}_{\geq 1}$, $M_{j}=\{i_{1},\dots,i_{m_{j}}\}$, and that the set of entries in $S_{j}$ is

[TABLE]

As $S\in\operatorname{SYT}(\lambda^{(1)}\cup\dots\cup\lambda^{(b)}),$ $S$ must use exactly the entries $1,2,\dots,n$, forcing

[TABLE]

Now we have

[TABLE]

Hence, $\operatorname{SYT}(\lambda^{(1)}\cup\dots\cup\lambda^{(b)})^{\operatorname{p}^{n/k}}=\varnothing$ unless $k\mid\gcd(|\lambda^{(1)}|,\dots,|\lambda^{(b)}|)$.

Now, assume that $k\mid\gcd(|\lambda^{(1)}|,\dots,|\lambda^{(b)}|)$. Since the set of entries in $S_{j}$ is $\bigcup_{\ell=1}^{k_{j}}(M_{j}+(\ell-1)m)$, we can write $S_{j}=I_{m,M_{j}}(T_{j})$ for some $T_{j}\in\operatorname{SYT}(\lambda^{(j)})$. Then,

[TABLE]

for all $j=1,\dots,b$. This proves that $\subseteq$ holds in (196) and hence proves 7.7.

∎

Corollary 7.9**.**

For any $r,b\in\mathbb{Z}_{\geq 1}$, and $k\mid br$,

[TABLE]

if $k\mid\gcd(r^{\left\lfloor\frac{b}{2}\right\rfloor},r^{\left\lceil\frac{b}{2}\right\rceil})$. Else,

[TABLE]

Proof.

By LABEL:thm:fixedbyab/2prom and (189),

[TABLE]

7.9 follows from (202) and 7.7. ∎

Example 7.10**.**

Continuing 7.8,

[TABLE]

Remark 7.11**.**

One can generalize 7.9 to give a similar description of $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br/k}}$ for $a\geq 2b-1,k\geq 2$. However, such a generalization would not give us anything new. In order for $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br/k}}$ to be nonempty, we must have $b\mid\frac{br}{k}$ or $ar\mid\frac{br}{k}$ by (LABEL:eq:prfixednonempty). Since $ar\geq(2b-1)r>\frac{br}{k}$, $ar\not|\frac{br}{k}$, so

[TABLE]

Note also that $ak>a\geq 2b-1$. Then, by LABEL:thm:prfixedtabs,

[TABLE]

These tableaux were already constructed in LABEL:thm:prfixedtabs.

8. Stabilization as a Permutation Statistic

We can identify each permutation $w\in S_{n}$ with the unique skew tableau with shape $(1)\cup\dots\cup(1)$ whose reading word is $w$. Under this identification, we translate the stabilization statistic to the symmetric group. We realize stabilization is invariant on dual equivalence classes and is bounded below strictly by the number of ascents of $w$. We characterize the permutations with stabilization statistic 1 and 2 in terms of their recording tableaux.

Definition 8.1**.**

For $w=w_{1}\dots w_{n}\in S_{n}$, let $T(w)$ be the unique standard skew tableau with shape $(1)\cup\dots\cup(1)$ with reading word $w$, and define $\operatorname{stab}(w)\coloneqq\operatorname{stab}(T(w))$. Let

[TABLE]

denote the number of ascents of $w$.

Example 8.2**.**

[TABLE]

Recall from (LABEL:eq:DEQ) that $v,w\in S_{n}$ are dual equivalent if and only if $Q(v)=Q(w)$. As in LABEL:thm:dualequivstab, dual equivalence plays well with stabilization.

Lemma 8.3**.**

If $v,w\in S_{n}$ are dual equivalent, then $\operatorname{stab}(v)=\operatorname{stab}(w)$.

Proof.

Since $v,w$ are dual equivalent, $T(v)$ and $T(w)$ are dual equivalent, see [Hai92, Lemma 2.11]. It follows by LABEL:thm:dualequivstab that

[TABLE]

∎

Lemma 8.4**.**

For all $w\in S_{n}$, $\operatorname{stab}(w)>\operatorname{asc}(w)$.

Proof.

Suppose $w=w_{1}\dots w_{n}\in S_{n}$, and let $r=\operatorname{stab}(w)$, so $T(w)$ stabilizes at $r$. In particular, this means $\operatorname{Rect}(T(w)^{(r)})$ has $n$ rows. The reading word of $T(w)^{(r)}$ is

[TABLE]

so

[TABLE]

by LABEL:lem:rectP. By LABEL:thm:Greene, $w_{1}^{(r)}\dots w_{n}^{(r)}$ must have a decreasing subsequence of size $n$. Yet, at most one of $w_{j},(w_{j}+n),\dots,(w_{j}+(r-1)n)$ can be in such a decreasing subsequence. Thus, $w_{1}^{(r)}\dots w_{n}^{(r)}$ must have a decreasing subsequence of the form

[TABLE]

for some $1\leq i_{1}\leq i_{2}\dots\leq i_{r-1}\leq n$. Since this subsequence is decreasing,

[TABLE]

Therefore, $w$ can only have ascents at possibly $i_{1},\dots,i_{r-1}$, so $w$ has at most $r-1$ ascents. Hence, $\operatorname{stab}(w)=r>r-1\geq\operatorname{asc}(w)$.

∎

8.3 and 8.4 give us some information about the distribution of $\operatorname{stab}$ on $S_{n}$, but what else can we say about this distribution? By LABEL:thm:stab, we know $1\leq\operatorname{stab}(w)\leq n$ for $w\in S_{n}$. In Figure 3, we give the distribution of $\operatorname{stab}$ on $S_{1},\dots,S_{8}$. We have a formula for the number of permutations in $S_{n}$ with $\operatorname{stab}$ 1 or $\operatorname{stab}$ 2. In fact, we characterize exactly which permutation has $\operatorname{stab}$ 1 and which permutations have $\operatorname{stab}2$ in terms of their recording tableaux.

Lemma 8.5**.**

The permutation $w=n\,(n-1)\,\dots\,2\,1$ is the only permutation with $\operatorname{stab}(w)=1$.

Proof.

If $w=n\,(n-1)\,\dots\,2\,1$, then $\operatorname{Rect}(T(w))=P(w)$ has a single column, so $\operatorname{stab}(w)=1$. If $w\in S_{n}$ with $\operatorname{stab}(w)=1$, then $P(w)$ consists of $n$ rows, forcing $w=n\,(n-1)\,\dots\,2\,1$.

∎

Notation 8.6**.**

Fix $n\in\mathbb{Z}_{\geq 1}$. For $k=1,\dots,n$, let

[TABLE]

where $x=2k+1,y=n-k+1$. This also means

[TABLE]

Theorem 8.7**.**

For all $n\in\mathbb{Z}_{\geq 1}$,

[TABLE]

Consequently,

[TABLE]

See OEIS entry A201686.

We break the proof of 8.7 into 3 steps. First, we use 8.4 to prove (213). Secondly, we calculate the number of standard tableaux of a given size with 1 or 2 columns. Thirdly, we use this result to prove (214).

Proof of (213).

Suppose $w\in S_{n}$ has $\operatorname{stab}(w)=2$. By 8.4, $\operatorname{asc}(w)\leq 1$, so we can write $w=w_{1}\dots w_{k}w_{k+1}\dots w_{n}$ where

[TABLE]

for some $k=1,\dots,n$. We must also have $w_{k}<w_{k+1}$ or else $w=n\,(n-1)\,\dots\,2\,1$, which has $\operatorname{stab}(w)=1$. Thus, $w_{k}<w_{k+1}$. Since $\operatorname{stab}(w)=2$, $\operatorname{Rect}(T(w)^{(2)})$ must have $n+1,\dots,2n$ in distinct rows. Note that we can perform inner slides to $T(w)^{(2)}$ to get

[TABLE]

where $x^{\prime}=x+n$ for all $x$. In order for $T(w)$ to stabilize at 2, $w_{1}^{\prime},\dots,w_{n}^{\prime}$ can only slide horizontally. Since $w_{k}<w_{k+1}$ and thus $w_{k}^{\prime}<w_{k+1}^{\prime}$, $w_{k}^{\prime}$ will slide vertically if the cell above it is vacated while it is in the second column. Since any entry above $w_{k}^{\prime}$ sliding into column 1 forces $w_{k}^{\prime}$ to slide up, if $w_{k}^{\prime}$ slides into column 1, it must do so before any of the entries above it. Hence, $w_{k},\dots,w_{1}$ must slide up either above $w_{k}^{\prime}$ or to the top before $w_{n},\dots,w_{k+1}$ experience any horizontal slides.

If $k\geq\left\lfloor\frac{n}{2}\right\rfloor$, then $w_{k},\dots,w_{1}$ slide to the top and $w_{n},\dots,w_{k+1}$ stay still, so

[TABLE]

where $x=w_{2k-n+1},y^{\prime}=w_{n-k+1}^{\prime}$. Since $w=w_{1}\dots w_{k}w_{k+1}\dots w_{n}$, this means during RSK on $w$, the whole first column of $P(w)$ was created first. Thus,

[TABLE]

If $k<\left\lfloor\frac{n}{2}\right\rfloor$, then $w_{k},\dots,w_{1}$ slide up above $w_{k}^{\prime}$ before $w_{n},\dots,w_{k+1}$ experience any horizontal slides, so

[TABLE]

In particular, based on the position of $w_{1}\dots w_{2k}$, $\operatorname{RSK}(w_{1}\dots w_{2k})$ began with a column of size k followed by a second column of size $k$. Thus,

[TABLE]

As $w$ has a decreasing subsequence of size $n-k$, the first column of $Q(w)$ must have size at least $n-k$ by LABEL:thm:Greene. As $Q(w)$ contains $Q(w_{1}\dots w_{2k})$, we must have

[TABLE]

This shows that if $\operatorname{stab}(w)=2$, then $Q(w)=T_{k}$ for some $k$, and hence

[TABLE]

In order to show the reverse containment of (262) and thus complete the proof of (213), suppose $Q(w)=T_{k}$ for some $k$. To show $\operatorname{stab}(w)=2$, it suffices to verify $\operatorname{stab}(w)=2$ for only 1 such word $w$ in each dual equivalence class by 8.3. As the recording tableau characterizes dual equivalence classes, it suffices to check that

[TABLE]

by (212). Let $w=k\dots 1\,n\dots(k+1)$ so that by (211),

[TABLE]

where $x=n-k+1,y=n+k+1,z=n+2k+1$. Either way, $\operatorname{stab}(w)=2$.

∎

Lemma 8.8**.**

Let $\operatorname{SYT}^{(2)}(n)$ denote the set of size $n$ standard tableaux with at most 2 columns. Then,

[TABLE]

Proof.

Because RSK: $S_{n}\to\cup_{\lambda\vdash n}\operatorname{SYT}(\lambda)\operatorname{\times}\operatorname{SYT}(\lambda)$ is a bijection,

[TABLE]

For all $k=0,1,\dots,\left\lfloor\frac{n}{2}\right\rfloor$, let

[TABLE]

which has shape $(2^{k},1^{n-2k})$. Now, $\{Q_{k}:k=0,1,\dots,\left\lfloor\frac{n}{2}\right\rfloor\}$ is the set of size $n$ standard tableaux with descents at $1,\dots,\left\lfloor\frac{n}{2}\right\rfloor-1,\left\lfloor\frac{n}{2}\right\rfloor+1,\dots,n-1$. Thus,

[TABLE]

by choosing the set $\{w_{1},\dots,w_{\left\lfloor\frac{n}{2}\right\rfloor}\}$ from $[n]$, which determines $w$ uniquely.

∎

Proof of (214).

We can add $(n+1)$ to 2 positions to tableaux in $\operatorname{SYT}(2^{r},1^{n-2r})$ for $r=0,\dots,\left\lfloor\frac{n-1}{2}\right\rfloor$, but only 1 position to tableaux in $\operatorname{SYT}\left(2^{n/2}\right)$, meaning

[TABLE]

Note $\operatorname{SYT}\left(2^{n/2}\right)=\varnothing$ if $n$ is odd. Therefore,

[TABLE]

∎

9. Open Problems

We finish by discussing related open questions about tableau stabilization and promotion. While we have proven some of the important properties of tableau stabilization, much remains unknown. The most glaring open problem is LABEL:conj:stab.

LABEL:conj:stab: Any standard skew tableau with $b$ rows and decreasing row vector stabilizes at $b$.

We have proven this bound is tight for skew tableaux with constant row vectors, see LABEL:thm:stab and LABEL:ex:stabatb. Due to LABEL:ex:samecjsdiffhape, our approach to proving LABEL:thm:stab does not readily generalize to proving LABEL:conj:stab.

The distribution of the stabilization statistic remains to be explored as well.

Open Problem 9.1**.**

What is the distribution of $\operatorname{stab}$ on tableaux of a fixed skew shape?

We expect 9.1 to be especially difficult. The permutation case, i.e. shape $(1)\cup\dots\cup(1)$, could be more tractable. The triangular array in Figure 3 describes stabilization’s distribution on permutations in $S_{1},\dots,S_{8}$. The distribution is unimodal and log-concave for these small cases. While we know the leftmost entry is always 1, is the rightmost entry is always 1 as well? In addition, since $\operatorname{stab}$ is invariant on dual equivalence classes, $Q(w)$ determines $\operatorname{stab}(w)$.

Conjecture 9.2**.**

The permutation $w=1\,2\,\dots\,(n-1)\,n$ is the only permutation with $\operatorname{stab}(w)=n$.

Open Problem 9.3**.**

What is the distribution of $\operatorname{stab}$ on $S_{n}$?

Open Problem 9.4**.**

Is the distribution of $\operatorname{stab}$ on $S_{n}$ unimodal? Is it log-concave?

Open Problem 9.5**.**

Characterize $\operatorname{stab}(w)$ directly in terms of $Q(w)$ for all $w\in S_{n}$.

We have made substantial progress on the problem of specifying the fixed points of the powers of promotion on rectangular tableaux, but some cases remain open. We constructed the tableaux in $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$ for $a\geq 2b-1$ in Section 6 and for $a=2$ in Section 7. The case $a=1$ is trivial, and the complete case $r=1$ was solved by Purbhoo and Rhee, see [PR17]. To completely solve the problem of specifying the fixed points of the powers of promotion on rectangular tableaux, 9.6 remains.

Open Problem 9.6**.**

Fix $b,r\in\mathbb{Z}_{\geq 1}$. For each $a\in[3,2b-2]$ ( $a\in[3,b-1]$ for $r=1$), describe the tableaux in $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$.

Generalizing Purbhoo and Rhee’s construction for $r=1,a\geq b$ to $r\geq 2,a\geq b$ is a nontrivial potential future task. By LABEL:cor:prfixedribbontab and (173), we have

[TABLE]

Purbhoo and Rhee’s proof relies on the fact that for all $T\in\operatorname{SYT}(b^{b})^{\operatorname{p}^{b}}$, the entries in the anti-diagonal cells $(b,1),(b-1,2),\dots,(1,b)$ of $T$ form a permutation in $S_{b}$ when taken modulo $b$. For example,

[TABLE]

has the anti-diagonal entries $[16,14,15,12,8]\equiv_{5}[1,4,5,2,3]\in S_{5}$. One might hope that for any $T\in\operatorname{SYT}((b\cdot r)^{b})^{\operatorname{p}^{br}}$, the cells $(b,1),\dots,(b,r),(b-1,r+1),\dots,(b-1,2r),\dots,(1,(b-1)r+1),\dots(1,br)$ have entries that form a permutation in $S_{br}$ when taken modulo $br$. But, considering such examples as

[TABLE]

this is not the case, since $[3,7,9,5,6,8]\equiv_{6}[3,1,3,5,6,2]\notin S_{6}$ and $[4,10,9,11,8,14]\equiv_{6}[4,4,3,5,2,2]\notin S_{6}$. In the first case, there is not even a symmetric choice of 3 consecutive entries in row 1 and 3 consecutive entries in row 2 which reduces to a permutation modulo $6$. We want a symmetric choice with respect to reflection so we can attach 2 symmetric shapes together to make a rectangle like we did for tableau stabilization.

The case $a<b$ is even more challenging since $\#\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}=\#\operatorname{SYT}(Q_{a}((ar)^{b}))$ changes according to LABEL:lem:rectquotients, using LABEL:cor:prfixedribbontab. Since now the pieces of the quotient are rectangles, one might have to do some tableau stabilization-like procedure with rectangles to find $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$, and it is so far unclear how this would work without just reducing to tableau stabilization. When $a=2$, row-concatenating the rectification to the anti-rectification works, but this does not easily generalize to $a\in[3,2b-1]$. Row-concatenating $\operatorname{Rect}(S^{\left(\lceil\frac{a}{2}\rceil\right)})$ and $\operatorname{Rect}^{\ast}(S^{\left(\lfloor\frac{a}{2}\rfloor\right)})+\left(\lceil\frac{a}{2}\rceil\right)br$ will not produce a rectangular tableaux in general.

For example, consider $a=3,r=2,b=6$, whence

[TABLE]

by LABEL:cor:prfixedribbontab and LABEL:lem:rectquotients. Choose

[TABLE]

Then, we have

[TABLE]

Row-concatenating these along 6 rows gives

[TABLE]

which is not even partitioned-shaped, let alone rectangular.

In LABEL:thm:prfixedtabs and LABEL:thm:fixedbyab/2prom, our construction of $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$ for $a\geq 2b-1$ and White and Rhee’s construction for $a=2$ included a natural bijection $R_{a}:\operatorname{SYT}(Q_{a}((ar)^{b}))\to\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$. Moreover, promotion commuted with $R_{a}$ as in LABEL:cor:prfixedpr and 7.4. One would hope these properties extended to a construction of $\operatorname{SYT}((ar)^{b})^{\operatorname{p}^{br}}$ for $a\in[3,2b-2]$ as well.

Open Problem 9.7**.**

Fix $b,r\in\mathbb{Z}_{\geq 1}$ and $a\in[3,2b-2]$. Find a bijection

[TABLE]

satisfying $\operatorname{p}(R_{a}(S))=R_{a}(\operatorname{p}(S))$ for all $S\in\operatorname{SYT}(Q_{a}((ar)^{b}))$, or show no such bijection exists.

Acknowledgments

I sincerely thank my advisor, Sara Billey, for helpful discussions and extensive comments on the manuscript. I thank Dennis White very much for his work on the problem of rectangular tableaux fixed by promotion powers, notes that inspired 7.7, helpful notes on proper citation, and for conjecturing LABEL:thm:rhoadesCSP in the first place. We would also like to thank Brendon Rhoades for proving LABEL:thm:rhoadesCSP and posing [Rho10, Problem 9.4], which inspired this work. I would also like to thank Kevin Purbhoo for making and pointing out to me previous progress on this problem in [PR17].