$\frac{f(\gamma/\mu)}{F(\gamma/\mu)}\,\leq\,{C_{d,\varepsilon}}\,\Phi(\gamma/\mu)\,,\quad\frac{f(\lambda/\gamma)}{F(\lambda/\gamma)}\,\leq\,{C_{d,\varepsilon}}\,\Phi(\lambda/\gamma)\quad\text{
and
}\quad\frac{f(\lambda/\mu)}{F(\lambda/\mu)}\,\geq\,\frac{1}{C_{d,\varepsilon}}\,\Phi(\lambda/\mu)\hskip
1.70709pt,$ (10.5)

for some absolute constant $C_{d,\varepsilon}>0$. Note that the last two
inequalities follow immediately from Lemma 4.4. We now prove that the first
inequality holds.

By the progressive assumption on
$(\lambda,\gamma,\mu)\in\Lambda(n,d,\varepsilon)$, we have:

$\gamma_{i}-\gamma_{i+1}\,\geq_{\eqref{eq:progressive-def}}\ p\hskip
1.70709pt(\lambda_{i}-\lambda_{i+1})\hskip 1.70709pt+\hskip
1.70709pt(1-p)(\mu_{i}-\mu_{i+1})\hskip 1.70709pt-\hskip 1.70709pt2\hskip
0.85355ptn^{\frac{3}{4}}\,\geq\,p\hskip
1.70709pt(\lambda_{i}-\lambda_{i+1})\hskip 1.70709pt-\hskip 1.70709pt2\hskip
0.85355ptn^{\frac{3}{4}}\,\geq\,p\hskip 1.70709pt\varepsilon\hskip
1.70709pt|\lambda|\hskip 1.70709pt-\hskip 1.70709pt2\hskip
0.85355ptn^{\frac{3}{4}}\hskip 0.85355pt,$

for every $1\leq i\leq d-1$. Similarly, by the $\varepsilon$-separation
assumption on $(\lambda,\gamma,\mu)$, we have:

$p\,=\,\frac{|\gamma|-|\mu|}{n}\,=\,\sum_{i=1}^{d}\frac{\gamma_{i}-\mu_{i}}{n}\,\geq_{\eqref{eq:separated-
def}}\
\left(\frac{d\varepsilon^{3}}{2}\right)\frac{|\lambda|}{n}\,\geq\,\frac{d\hskip
0.85355pt\varepsilon^{3}}{2}\,.$

Thus, for sufficiently large $n$, we have:

$\displaystyle\gamma_{i}-\gamma_{i+1}\,\geq\,\frac{d\hskip
0.85355pt\varepsilon^{4}}{2}|\lambda|\hskip 1.70709pt-\hskip 1.70709pt2\hskip
0.85355ptn^{\frac{3}{4}}\,\geq\,\frac{d\hskip
0.85355pt\varepsilon^{4}}{4}\hskip 1.70709pt|\lambda|\,\geq\,\frac{d\hskip
0.85355pt\varepsilon^{4}}{4}\hskip 1.70709pt|\gamma|\hskip 0.85355pt.$ (10.6)

By the same argument as above, for sufficiently large $n$, we have:

$\displaystyle\gamma_{d}\,\geq\,\frac{d\hskip
0.85355pt\varepsilon^{4}}{4}\hskip 1.70709pt|\gamma|\hskip 0.85355pt.$ (10.7)

Conditions (10.6) and (10.7) imply that $\gamma/\mu$ is $(d\hskip
0.85355pt\varepsilon^{4}/4)$-smooth, for $n$ large enough. Applying Lemma 4.4,
we obtain the first inequality in (10.5). This completes the proof. $\square$

##  11 The case of TVK skew shapes

In this section we give upper and lower bounds for the number of standard
Young tableaux corresponding to TVK pairs. We then prove Lemma 4.5 and Theorem
1.4.

###  11.1 Conditions for interval decomposition

We define three types of conditions for interval decomposition
$\mathcal{B}=(B_{1},\ldots,B_{r})$ of $[d]$. These conditions will be used in
combinations, to cover all possible TVK pairs. Formally, consider:

$\lambda_{i}-\lambda_{j}\,\geq\,\varepsilon\hskip
0.85355pt|\lambda|\quad\text{for all}\quad i\operatorname{\
\overset{\mathcal{B}}{\nsim}\ }j\,,\ \ 1\leqslant i<j\leqslant d\hskip
0.85355pt,$ (11.1) $\lambda_{i}-\lambda_{j}\ \leq\ 1\quad\text{for all}\quad
i\operatorname{\ \overset{\mathcal{B}}{\sim}\ }j\,,\ \ 1\leqslant i<j\leqslant
d\hskip 0.85355pt,$ (11.2) $\mu_{i}-\mu_{j}\ \leq\ 1\quad\text{for all}\quad
i\operatorname{\ \overset{\mathcal{B}}{\sim}\ }j\,,\ \ 1\leqslant i<j\leqslant
d\hskip 0.85355pt.$ (11.3)

The motivation behind these conditions for TVK $(\alpha,\beta)$-shapes will
become apparent later in this section. For now, we treat them as abstract
constraints on the interval decompositions,

###  11.2 Upper bounds

We start with estimating each term in the definition of $\Phi(\lambda/\mu)$,
see (4.1), and we collect these estimates in the next three lemmas.

######  Lemma 11.1.

Fix $d\geq 2$ and $\varepsilon>0$. Let $\lambda,\mu\in\mathbb{P}_{d}$, such
that $\mu\subseteq\lambda$. Suppose (11.1) holds for the interval
decomposition $\mathcal{B}$ of $[d]$. Then:

$1\,\leq\,\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\,\leq\,\frac{d}{\varepsilon}\,,\quad\text{for
all \ \ $i\operatorname{\ \overset{\mathcal{B}}{\nsim}\ }j$, \ \hskip
1.70709pt$1\leqslant i<j\leqslant d$. }$ (11.4)

In particular, we have:

$\frac{\varepsilon}{d}\,\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\,\leq\,\min\left\\{\mu_{i}-\mu_{j}+j-i,\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\right\\}\,\leq\,\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\,.$
(11.5)

###### Proof.

The lower bound in (11.4) follows from (4.2). The upper bound in (11.4),
follows verbatim (10.1). For the lower bound in (11.5), we have

$\displaystyle\min\left\\{\mu_{i}-\mu_{j}+j-i,\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\right\\}\
\geq_{\eqref{eq:lambdai-lambdaj is greater than 1},\hskip
1.70709pt\eqref{eq:mui-muj is greater than 1}}\ 1\,\geq_{\eqref{eq:condition 1
min}}\,\left(\frac{\varepsilon}{d}\right)\hskip
0.85355pt\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\,,$

as desired. ∎

######  Lemma 11.2.

Fix $d\geq 2$ and $\varepsilon>0$. Let $\lambda,\mu\in\mathbb{P}_{d}$, such
that $\mu\subseteq\lambda$. Suppose (11.2) holds for the interval
decomposition $\mathcal{B}$ of $[d]$. Then:

$\frac{1}{d}\hskip
1.70709pt(\mu_{i}-\mu_{j}+j-i)\,\leq\,\min\left\\{\mu_{i}-\mu_{j}+j-i,\hskip
1.70709pt\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\right\\}\,\leq\,\mu_{i}-\mu_{j}+j-i\hskip
0.85355pt,$ (11.6)

for all $i\operatorname{\ {\sim}\ }j$, $1\leqslant i<j\leqslant d$.

###### Proof.

The upper bound is straightforward. For the lower bound, it follows from
(11.2), that

$\begin{split}\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\,\geq\,\frac{\lambda_{i}+d-i}{1+j-i}\,\geq\,\frac{\lambda_{i}+d-i}{d}\,\geq\,\frac{\mu_{i}-\mu_{j}+j-i}{d}\hskip
1.70709pt.\end{split}$

It then follows from the equation above that

$\displaystyle\min\left\\{\mu_{i}-\mu_{j}+j-i,\hskip
1.70709pt\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\right\\}\,\geq\,\min\left\\{\mu_{i}-\mu_{j}+j-i,\frac{\mu_{i}-\mu_{j}+j-i}{d}\right\\}\,=\,\frac{\mu_{i}-\mu_{j}+j-i}{d}\,,$

as desired. ∎

######  Lemma 11.3.

Fix $d\geq 2$ and $\varepsilon>0$. Let $\lambda,\mu\in\mathbb{P}_{d}$, such
that $\mu\subseteq\lambda$. Suppose (11.3) holds for the interval
decomposition $\mathcal{B}$ of $[d]$. Then (11.6) holds for all
$i\operatorname{\ {\sim}\ }j$, $1\leqslant i<j\leqslant d$.

###### Proof.

The upper bound is straightforward. For the lower bound, it follows from
(11.3) that

$\begin{split}\mu_{i}-\mu_{j}+j-i\,\leq\,1+j-i\,\leq\,d.\end{split}$ (11.7)

Therefore,

$\displaystyle\min\left\\{\mu_{i}-\mu_{j}+j-i,\hskip
1.70709pt\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\right\\}\,\geq_{\text{\eqref{eq:lambdai-
lambdaj is greater than 1},\hskip 1.70709pt\eqref{eq:mui-muj is greater than
1}}}\ 1\ \geq_{\eqref{eq:condition 3 min}}\ \frac{1}{d}\hskip
1.70709pt\bigl{(}\mu_{i}-\mu_{j}+j-i\bigr{)},$

as desired. ∎

We now combine these three lemmas to give an estimate for the quantity
$\Phi(\lambda/\mu)$ if (11.1) holds and either (11.2) or (11.3) holds. Denote

$K_{\mathcal{B}}(\lambda/\mu)\,:=\,\prod_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ {\sim}\ }j\end{subarray}}\hskip
1.70709pt(\mu_{i}-\mu_{j}+j-i)\,\prod_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ {\nsim}\ }j\end{subarray}}\hskip
1.70709pt\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\,.$ (11.8)

######  Lemma 11.4.

Fix $d\geq 2$ and $\varepsilon>0$. Let $\lambda,\mu\in\mathbb{P}_{d}$, such
that $\mu\subseteq\lambda$. Suppose (11.1) and (11.2) hold for the interval
decomposition $\mathcal{B}$ of $[d]$. Then:

$\left(\frac{\varepsilon}{d^{2}}\right)^{\frac{d(d-1)}{2}}K_{\mathcal{B}}(\lambda/\mu)\,\leq\,\Phi(\lambda/\mu)\,\leq\,K_{\mathcal{B}}(\lambda/\mu)\hskip
0.85355pt.$

The same conclusion holds if condition (11.2) is replaced with (11.3).

###### Proof.

By definition of $K_{\mathcal{B}}(\lambda/\mu)$, we have:

$\displaystyle\frac{\Phi(\lambda/\mu)}{K_{\mathcal{B}}(\lambda/\mu)}\,=\,\prod_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ {\nsim}\
}j\end{subarray}}\frac{\min\left\\{\mu_{i}-\mu_{j}+j-i,\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\right\\}}{\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}}\prod_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ {\sim}\
}j\end{subarray}}\frac{\min\left\\{\mu_{i}-\mu_{j}+j-i,\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\right\\}}{\mu_{i}-\mu_{j}+j-i}\,\leq\,1.$

For the lower bound, note that each term in the first product is bounded from
below by $\varepsilon/d$, by Lemma 11.1 and condition (11.1). Also note that
each term in the second product is bounded from below by $1/d$, by Lemma 11.2
when (11.2) holds, or by Lemma 11.3 when (11.3) holds. This implies the
result. ∎

The main result of this subsection is the following upper bound.

######  Lemma 11.5.

Fix $d\geq 2$ and $\varepsilon>0$. Let $\lambda,\mu\in\mathbb{P}_{d}$, such
that $\mu\subseteq\lambda$. Suppose (11.1) and (11.2) hold for the interval
decomposition $\mathcal{B}$ of $[d]$. Then:

$\frac{f(\lambda/\mu)}{F(\lambda/\mu)}\,\leq\,C_{d,\varepsilon}\Phi(\lambda,\mu)\hskip
0.85355pt.$

where $C_{d,\varepsilon}>0$ is an absolute constant. The same conclusion holds
if condition (11.2) is replaced with (11.3).

###### Proof.

For the first part, it follows from Lemma 11.4 that $\Phi(\lambda/\mu)$ is
equal to $K_{B}(\lambda/\mu)$ up to a multiplicative constant. Therefore, it
suffices to show that

$\frac{f(\lambda/\mu)}{F(\lambda/\mu)}\,\leq\,C_{d,\varepsilon}\hskip
1.70709ptK_{\mathcal{B}}(\lambda/\mu)\hskip 0.85355pt,$

for some absolute constant $C_{d,\varepsilon}>0$. Let $N(\ell,B)$ be as in
(8.2). Then

$\displaystyle N(\ell,\mathcal{B})\,=\,\max_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ {\nsim}\ }j\end{subarray}}\hskip
1.70709pt\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}\,\leq_{\text{\eqref{eq:lambda
big gap},\hskip 1.70709pt\eqref{eq:condition 1
min}}}\,\frac{d}{\varepsilon}\,.$

Substituting this into Corollary 8.5, we get

$\displaystyle\frac{f(\lambda/\mu)}{F(\lambda/\mu)}\ $
$\displaystyle\leq\,C_{d}\prod_{\begin{subarray}{c}1\leqslant i<j\leqslant
d\\\ i\operatorname{\ {\sim}\
}j\end{subarray}}\left(\mu_{i}-\mu_{j}+j-i+\hskip
1.70709pt\frac{d}{\varepsilon}\right)\,\prod_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ {\nsim}\
}j\end{subarray}}\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}$
$\displaystyle\leq\,C_{d}\prod_{\begin{subarray}{c}1\leqslant i<j\leqslant
d\\\ i\operatorname{\ {\sim}\
}j\end{subarray}}\left(1+\frac{d}{\varepsilon}\right)\bigl{(}\mu_{i}-\mu_{j}+j-i\bigr{)}\,\prod_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ {\nsim}\
}j\end{subarray}}\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}$
$\displaystyle\leq\,C_{d}\left(1+\frac{d}{\varepsilon}\right)^{\frac{d(d-1)}{2}}\prod_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ {\sim}\
}j\end{subarray}}\bigl{(}\mu_{i}-\mu_{j}+j-i\bigr{)}\,\prod_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ {\nsim}\
}j\end{subarray}}\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}$
$\displaystyle\leq_{\eqref{eq:HB(lambda/mu)}}\ C_{d,\varepsilon}\hskip
1.70709ptK_{\mathcal{B}}(\lambda/\mu)\hskip 0.85355pt,$

for some absolute constants $C_{d}$, $C_{d,\varepsilon}>0$. This finishes the
proof of the first part. The second part follows verbatim; we omit the
details. ∎

###  11.3 Lower bounds

Our first ingredient is the following estimate on the hook-lengths.

######  Lemma 11.6.

Fix $d\geq 2$ and $\varepsilon>0$. Let
$(\lambda,\mu)\in\Lambda(n,d,\varepsilon)$. Let $\mathcal{B}$ be an interval
decomposition of $[n]$ such that (11.2) holds. Then, for all $(i,j)\in\mu$,
and all $k\geq 0$ such that $i\operatorname{\ \overset{\mathcal{B}}{\sim}\
}(i+k)$, we have:

$\frac{h_{\lambda}(i+k,j+k)}{h_{\lambda}(i,j)}\,\geq\,1\hskip 1.70709pt-\hskip
1.70709pt\frac{2\hskip 0.85355ptd}{\varepsilon\hskip 0.85355pt|\lambda|}\hskip
1.70709pt.$

###### Proof.

By the definition (2.6) of the hook lengths, we have:

$\begin{split}\frac{h_{\lambda}(i+k,j+k)}{h_{\lambda}(i,j)}\,&\geq\,\frac{\lambda_{i+k}-j-k}{\lambda_{i}-i+d-j+1}\,\geq_{[\text{since}\
i\hskip 0.85355pt\sim\hskip
0.85355pt(i+k)]}\,\frac{\lambda_{i}-1-j-k}{\lambda_{i}-i+d-j+1}\,=\,1\hskip
1.70709pt-\hskip 1.70709pt\frac{d+k+2-i}{\lambda_{i}-i+d-j+1}\\\
&\geq\,1\hskip 1.70709pt-\hskip
1.70709pt\frac{2d}{\lambda_{i}-j+(d-i)+1}\,\geq_{[\text{since}\ (i,j)\hskip
0.85355pt\in\hskip 0.85355pt\mu]}\ 1\hskip 1.70709pt-\hskip
1.70709pt\frac{2d}{\lambda_{i}-\mu_{i}}\,\geq_{\eqref{eq:separated-
def}}\,1\hskip 0.85355pt-\hskip
1.70709pt\frac{2d}{\varepsilon|\lambda|}\,,\end{split}$

as desired. ∎

We apply Lemma 11.6 to get a lower bound for the product of hooks of a flagged
tableau, see (2.10). Let $\mathcal{B}$ be an interval decomposition of $[d]$.
Denote by

$\operatorname{\mathcal{D}}_{\mathcal{B}}\hskip 1.70709pt:=\hskip
1.70709pt\bigl{\\{}T\in\operatorname{\mathrm{FT}}(\lambda/\mu)\ \mid\
i\operatorname{\ \overset{\mathcal{B}}{\sim}\ }T(i,j)\,\text{ for all
}\,(i,j)\in\mu\bigr{\\}},$

the set of flagged tableaux of $\lambda/\mu$, for which the entries for each
row $i$ are drawn from the block of $\mathcal{B}$ that contains $i$.

######  Lemma 11.7.

Fix $d\geq 2$ and $\varepsilon>0$. Let
$(\lambda,\mu)\in\Lambda(n,d,\varepsilon)$, and let $\mathcal{B}$ be an
interval decomposition of $[d]$, such that (11.2) holds. Then, for all
$T\in\operatorname{\mathcal{D}}_{B}$, we have:

$\prod_{(i,j)\in\mu}\hskip 1.70709pt\frac{h_{\lambda}\bigl{(}T(i,j),\hskip
0.85355ptj+T(i,j)-i\bigr{)}}{h_{\lambda}(i,j)}\,\geq\,C_{d,\varepsilon}\,,$

for some absolute constant $C_{d,\varepsilon}>0$.

###### Proof.

We have:

$\displaystyle\prod_{(i,j)\in\mu}\frac{h_{\lambda}\bigl{(}T(i,j),\hskip
0.85355ptj+T(i,j)-i\bigr{)}}{h_{\lambda}(i,j)}\ \geq_{\text{Lem~{}\ref{l:hook
cells lower bound}}}\ \left(1-\frac{2\hskip 0.85355ptd}{\varepsilon\hskip
0.85355pt|\lambda|}\right)^{|\mu|}\ \geq\ \left(1-\frac{2\hskip
0.85355ptd}{\varepsilon\hskip
0.85355pt|\lambda|}\right)^{|\lambda|}\,\geq\,\left(\frac{1}{3}\right)^{2d/\varepsilon}\,,$

for sufficiently large $|\lambda|$. This implies the result. ∎

Our second ingredient is the following lower bound on the cardinality of
$\operatorname{\mathcal{D}}_{B}$.

######  Lemma 11.8.

Fix $d\geq 2$ and $\varepsilon>0$. Let
$(\lambda,\mu)\in\Lambda(n,d,\varepsilon)$. Let $\mathcal{B}$ be an interval
decomposition of $[n]$ such that (11.2) holds. Then there exists an absolute
constant $C_{d,\varepsilon}>0$ such that

$\bigl{|}\operatorname{\mathcal{D}}_{\mathcal{B}}\bigr{|}\,\geq\,C_{d,\varepsilon}\hskip
1.70709pt\prod_{\begin{subarray}{c}1\leqslant i<j\leqslant d\\\
i\operatorname{\ \overset{\mathcal{B}}{\sim}\
}j\end{subarray}}({\mu_{i}-\mu_{j}+j-i}).$

###### Proof.

Let
$\operatorname{\mathcal{D}}_{\mathcal{B}}^{\prime}:=\operatorname{\mathcal{D}}^{\prime}_{\mathcal{B}}(\mu)$
be the set of semistandard Young tableau of shape $\mu$ given by

$\operatorname{\mathcal{D}}_{\mathcal{B}}^{\prime}\hskip 1.70709pt:=\hskip
1.70709pt\bigl{\\{}T\in\text{SSYT}(\mu)\ \mid\ i\operatorname{\
\overset{\mathcal{B}}{\sim}\ }T(i,j)\,\text{ for all
}\,(i,j)\in\mu\bigr{\\}}.$

Note that
$\operatorname{\mathcal{D}}_{\mathcal{B}}=\operatorname{\mathcal{D}}^{\prime}_{\mathcal{B}}\cap\operatorname{\mathrm{FT}}(\lambda/\mu)$.
We will estimate $|\operatorname{\mathcal{D}}_{\mathcal{B}}|$ via
$|\operatorname{\mathcal{D}}^{\prime}_{\mathcal{B}}|$.

Recall the definition (8.1) of interval decompositions. For each
$k\in\\{1,\ldots,r\\}$, denote by $\mu^{(k)}$ the partition obtained from
$\mu$ by restricting to rows indexed by $B_{k}$:

$\mu^{(k)}\,=\,\bigl{(}\mu^{(k)}_{1},\mu^{(k)}_{2},\hskip
0.85355pt\ldots\hskip
0.85355pt,\mu^{(k)}_{b_{k}-b_{k-1}}\bigr{)}\,:=\,\bigl{(}\mu_{b_{k-1}+1},\mu_{b_{k-1}+2},\hskip
0.85355pt\ldots\hskip 0.85355pt,\mu_{b_{k}}\bigr{)}.$ (11.9)

In this notation,

$\operatorname{\mathcal{D}}_{\mathcal{B}}^{\prime}\hskip 1.70709pt=\hskip
1.70709pt\bigl{\\{}T\in\text{SSYT}(\mu)\ \mid\ b_{k-1}\hskip 1.70709pt<\hskip
1.70709ptT(i,j)\hskip 1.70709pt\leq\hskip 1.70709ptb_{k}\,\text{ for all
}\,(i,j)\in\mu,\,i\in B_{k}\bigr{\\}}.$ (11.10)

Therefore, that the following map is a bijection:

$\displaystyle\psi:\hskip
1.70709pt\operatorname{\mathcal{D}}_{\mathcal{B}}^{\prime}\,\to\,\operatorname{SSYT}\bigl{(}\mu^{(1)}\bigr{)}\times\ldots\times\operatorname{SSYT}\bigl{(}\mu^{(r)}\bigr{)},\quad\psi(T)\hskip
1.70709pt:=\hskip 1.70709pt\bigl{(}T^{(1)},\ldots,T^{(r)}\bigr{)},$ (11.11)
$\displaystyle\qquad\text{where}\ \ \,T^{(k)}(i,j)\hskip 1.70709pt=\hskip
1.70709ptT(i+b_{k-1},j)-b_{k-1}\quad\text{for all}\ \
\,(i,j)\in\mu^{(k)}\hskip 1.70709pt.$

In other words, the semistandard Young tableaux $T^{(k)}$ is obtained by
restricting $T$ to rows indexed by $B_{k}$ and normalizing the smallest
entries to start from $1$. It now follows from (11.11) and (2.5), that

$\begin{split}|\operatorname{\mathcal{D}}^{\prime}_{\mathcal{B}}|\ &=\
\prod_{k=1}^{r}\,\prod_{\begin{subarray}{c}1\leqslant i<j\leqslant d\\\ i,j\in
B_{k}\end{subarray}}\,\frac{\mu_{i}-\mu_{j}+j-i}{j-i}\ \geq\
\prod_{k=1}^{r}\,\prod_{\begin{subarray}{c}1\leqslant i<j\leqslant d\\\ i,j\in
B_{k}\end{subarray}}\,\frac{\mu_{i}-\mu_{j}+j-i}{d-1}\\\ \ &\geq\
(d-1)^{-\frac{d(d-1)}{2}}\hskip 1.70709pt\prod_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ \overset{\mathcal{B}}{\sim}\
}j\end{subarray}}\hskip
1.70709pt\bigl{(}{\mu_{i}-\mu_{j}+j-i}\bigr{)}.\end{split}$ (11.12)

We claim that
$\operatorname{\mathcal{D}}_{\mathcal{B}}=\operatorname{\mathcal{D}}_{\mathcal{B}}^{\prime}$
for sufficiently large $|\lambda|$. It suffices to show that each
$T\in\operatorname{\mathcal{D}}_{\mathcal{B}}^{\prime}$ is a flagged tableau
of $\lambda/\mu$, for sufficiently large $|\lambda|$. Let $(i,j)\in\mu$, and
let $k$ be the index such that $B_{k}$ is the block of $\mathcal{B}$ that
contains $i$. We have:

$\displaystyle j+T(u)-1\ $ $\displaystyle\leq\ \mu_{i}+d-1\ =\
\lambda_{i}-(\lambda_{i}-\mu_{i})+d-1\ \leq_{\eqref{eq:separated-def}}\
\lambda_{i}-\varepsilon\hskip 0.85355pt|\lambda|+(d-i)$
$\displaystyle\leq_{\text{\eqref{eq:lambda small gap},\hskip
1.70709pt$i\operatorname{\ \overset{\mathcal{B}}{\sim}\ }T(u)$}}\
(\lambda_{T(u)}+1)-\varepsilon\hskip 0.85355pt|\lambda|+(d-i)\ \leq\
\lambda_{T(u)}\hskip 1.70709pt,$

for sufficiently large $\lambda_{T(u)}$. This proves the claim.

By (4.7), we have:

$\lambda_{1}\ \geq\ \ldots\ \geq\ \lambda_{d}\ \geq\ \lambda_{d}-\mu_{d}\
\geq\ \varepsilon\hskip 0.85355pt|\lambda|,$ (11.13)

so the claim above assumes only that $|\lambda|$ is large enough. We conclude:

$|\operatorname{\mathcal{D}}_{\mathcal{B}}|\ =_{\eqref{eq:Dc-lower-sep}}\
|\operatorname{\mathcal{D}}_{\mathcal{B}}^{\prime}|\ \geq_{\eqref{eq:Dc lower
bound 3}}\ (d-1)^{-\frac{d(d-1)}{2}}\hskip
1.70709pt\prod_{\begin{subarray}{c}1\leqslant i<j\leqslant d\\\
i\operatorname{\ \overset{\mathcal{B}}{\sim}\ }j\end{subarray}}\hskip
1.70709pt\bigl{(}{\mu_{i}-\mu_{j}+j-i}\bigr{)},$ (11.14)

for all $|\lambda|$ sufficiently large. This completes the proof. ∎

The main result of this subsection is the following lower bound for
$f(\lambda/\mu)$.

######  Lemma 11.9.

Fix $d\geq 2$ and $\varepsilon>0$. Let
$(\lambda,\mu)\in\Lambda(n,d,\varepsilon)$. Let $\mathcal{B}$ be an interval
decomposition of $[n]$ such that (11.1) and (11.2) hold. Then there exists an
absolute constant $C_{d,\varepsilon}>0$ such that

$\frac{f(\lambda/\mu)}{F(\lambda/\mu)}\,\hskip 1.70709pt\geq\,\hskip
1.70709ptC_{d,\varepsilon}\,\Phi(\lambda/\mu)\hskip 0.85355pt.$

###### Proof.

We have:

$\displaystyle\frac{f(\lambda/\mu)}{F(\lambda/\mu)}\,$
$\displaystyle=_{\text{Thm~{}\ref{t:NHLF}}}\
\sum_{T\in\operatorname{\mathrm{FT}}(\lambda/\mu)}\,\prod_{(i,j)\in\mu}\frac{h_{\lambda}(T(i,j),j+T(i,j)-i)}{h_{\lambda}(i,j)}\,\geq\,\sum_{T\in\operatorname{\mathcal{D}}_{B}}\,\prod_{(i,j)\in\mu}\frac{h_{\lambda}(T(i,j),j+T(i,j)-i)}{h_{\lambda}(i,j)}$
$\displaystyle\geq_{\text{Lem~{}\ref{l:hook products lower bound}}}\
\sum_{T\in\operatorname{\mathcal{D}}_{B}}\hskip
1.70709ptC_{d,\varepsilon}\,=\,C_{d,\varepsilon}\hskip
0.85355pt|\operatorname{\mathcal{D}}_{B}|\,\geq_{\text{Lem~{}\ref{l:Dc lower
bound}}}\ C_{d,\varepsilon}\prod_{\begin{subarray}{c}1\leqslant i<j\leqslant
d\\\ i\operatorname{\ {\sim}\ }j\end{subarray}}\hskip
1.70709pt({\mu_{i}-\mu_{j}+j-i})\hskip 1.70709pt.$

This implies that

$\displaystyle\frac{f(\lambda/\mu)}{F(\lambda/\mu)}\ $
$\displaystyle\geq_{\eqref{eq:condition 1 min}}\ C_{d,\varepsilon}\hskip
1.70709pt\left(\frac{\varepsilon}{d}\right)^{\frac{d(d-1)}{2}}\prod_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ {\sim}\
}j\end{subarray}}({\mu_{i}-\mu_{j}+j-i})\prod_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ i\operatorname{\ {\nsim}\
}j\end{subarray}}\frac{\lambda_{i}+d-i}{\lambda_{i}-\lambda_{j}+j-i}$
$\displaystyle=_{\eqref{eq:HB(lambda/mu)}}\ C_{d,\varepsilon}\hskip
1.70709pt\left(\frac{\varepsilon}{d}\right)^{\frac{d(d-1)}{2}}K_{\mathcal{B}}(\lambda/\mu)\
\geq_{\text{Lem~{}\ref{l:G exact estimate}}}\ C_{d,\varepsilon}\hskip
1.70709pt\left(\frac{\varepsilon}{d}\right)^{\frac{d(d-1)}{2}}\Phi(\lambda/\mu)\hskip
1.70709pt,$

as desired. ∎

###  11.4 Proof of Lemma 4.5

Recall the definition of a Thoma pair $(\alpha,\beta)$ in $\S$1.2. Let
$\varepsilon:=\varepsilon(\alpha,\beta)$ be given by

$\varepsilon:=\frac{1}{2(\alpha_{1}+\ldots+\alpha_{d})}\,\min\Big{\\{}\min_{1\leqslant
i<j\leqslant
d}\left\\{{\alpha_{i}-\beta_{i}}{}\right\\},\min_{\begin{subarray}{c}1\leq
i<j\leq d\\\
\alpha_{i}\neq\alpha_{j}\end{subarray}}\\{\alpha_{i}-\alpha_{j}\\}\Big{\\}}.$
(11.15)

We have $\varepsilon>0$ since $\alpha_{i}>\beta_{i}$.

Let $\lambda\simeq\alpha\hskip 0.85355ptn$, $\mu\simeq\beta\hskip 0.85355ptn$
be a TVK $(\alpha,\beta)$-shape. Note that $(\lambda,\mu)$ is
$\varepsilon$-admissible for sufficiently large $n$. Indeed,

$\lambda_{i}\hskip 1.70709pt-\hskip 1.70709pt\mu_{i}\ =\
\lfloor\alpha_{i}n\rfloor\hskip 1.70709pt-\hskip
1.70709pt\lfloor\beta_{i}n\rfloor\ \geq\ (\alpha_{i}\hskip 0.85355pt-\hskip
0.85355pt\beta_{i})\hskip 0.85355ptn\hskip 1.70709pt-\hskip 1.70709pt1\ \geq\
\frac{\alpha_{i}\hskip 0.85355pt-\hskip 0.85355pt\beta_{i}}{2\hskip
0.85355pt(\alpha_{1}\hskip 1.70709pt+\hskip 1.70709pt\ldots\hskip
1.70709pt+\hskip 1.70709pt\alpha_{d})}\hskip 1.70709pt|\lambda|\ \geq\
\varepsilon\hskip 0.85355pt|\lambda|\hskip 0.85355pt,$ (11.16)

for sufficiently large $|\lambda|=|\alpha|\hskip 0.85355ptn+O(1)$, and for all
$1\leq i\leq d$.

Let $\mathcal{B}$ be the interval decomposition of $[d]$ that puts two
integers $i,j\in[d]$ in the same block if and only if $\alpha_{i}=\alpha_{j}$.
Then (11.1) holds for sufficiently large $|\lambda|$, since for all
$i\operatorname{\ \overset{\mathcal{B}}{\nsim}\ }j$, $1\leqslant i<j\leqslant
d$, we have:

$\lambda_{i}\hskip 1.70709pt-\hskip 1.70709pt\lambda_{j}\ =\
\lfloor\alpha_{i}n\rfloor\hskip 1.70709pt-\hskip
1.70709pt\lfloor\alpha_{j}n\rfloor\ \geq\ (\alpha_{i}\hskip 0.85355pt-\hskip
0.85355pt\alpha_{j})\hskip 0.85355ptn\hskip 1.70709pt-\hskip 1.70709pt1\ \geq\
\frac{\alpha_{i}\hskip 1.70709pt-\hskip 1.70709pt\alpha_{j}}{2\hskip
0.85355pt(\alpha_{1}\hskip 1.70709pt+\hskip 1.70709pt\ldots\hskip
1.70709pt+\hskip 1.70709pt\alpha_{d})}\hskip 1.70709pt|\lambda|\hskip
0.85355pt,\ \geq\ \varepsilon\hskip 0.85355pt|\lambda|\hskip 0.85355pt,$
(11.17)

Similarly, (11.2) holds, since for all $i\operatorname{\
\overset{\mathcal{B}}{\sim}\ }j$, $1\leqslant i<j\leqslant d$, we have:

$\lambda_{i}\hskip 1.70709pt-\hskip 1.70709pt\lambda_{j}\ =\
\lfloor\alpha_{i}n\rfloor\hskip 1.70709pt-\hskip
1.70709pt\lfloor\alpha_{j}n\rfloor\,=\,0\hskip 0.85355pt.$ (11.18)

By Lemma 11.5 and Lemma 11.9, this implies that there exists an absolute
constant $C_{\alpha,\beta}>0$, such that

$\displaystyle\frac{1}{C_{\alpha,\beta}}\,\Phi(\lambda/\mu)\ \leq\
\frac{f(\lambda/\mu)}{F(\lambda/\mu)}\ \leq\
{C_{\alpha,\beta}}\,\Phi(\lambda/\mu)\,,$

for sufficiently large $n$. This implies the result. $\square$

###  11.5 Proof of Theorem 1.4

Let $\varepsilon:=\varepsilon(\alpha,\beta)$ be as in (11.15). By Lemma 4.3,
it suffices to check that for every $\varepsilon$-admissible triplet
$(\lambda,\gamma,\mu)$, we have:

$\frac{f(\gamma/\mu)}{F(\gamma/\mu)}\ \leq\
{C_{\alpha,\beta}}\,\Phi(\gamma/\mu)\hskip
1.70709pt,\quad\frac{f(\lambda/\gamma)}{F(\lambda/\gamma)}\ \leq\
{{C_{\alpha,\beta}}}\,\Phi(\lambda/\gamma)\hskip 1.70709pt,\quad\text{ and
}\quad\frac{f(\lambda/\mu)}{F(\lambda/\mu)}\ \geq\
\frac{1}{{C_{\alpha,\beta}}}\,\Phi(\lambda/\mu),$ (11.19)

for some absolute constant ${C_{\alpha,\beta}}>0$. Note that the third
inequality in (11.19) is proved in Lemma 4.5.

For the second inequality in (11.19), let $\mathcal{B}$ be the interval
decomposition of $[d]$ that puts two integers $i,j\in[d]$ in the same block if
and only if $\alpha_{i}=\alpha_{j}$. By the same argument as in (11.17) and
(11.18), we have (11.1) and (11.2) hold for the pair $(\lambda,\gamma)$ and
$\mathcal{B}$, and for sufficiently large $n$. By Lemma 11.5, we get the
second inequality for sufficiently large $n$.

For the first inequality in (11.19), let $\mathcal{B}^{\prime}$ be the
interval decomposition of $[d]$ that puts $i,j\in[d]$ in the same block if and
only if $\beta_{i}=\beta_{j}$. Let
$\varepsilon^{\prime}:=\varepsilon^{\prime}(\alpha,\beta)$ be the constant
defined by

$\varepsilon^{\prime}\,:=\,\frac{d\hskip 0.85355pt\varepsilon^{3}}{8\hskip
0.85355pt(\alpha_{1}+\ldots+\alpha_{d})}\,\min_{\begin{subarray}{c}1\leqslant
i<j\leqslant d\\\ \beta_{i}\neq\beta_{j}\end{subarray}}\hskip
1.70709pt\\{\beta_{i}-\beta_{j}\\}\hskip 0.85355pt.$

For all $i\overset{\mathcal{B}^{\prime}}{\nsim}j$, $1\leqslant i<j\leqslant
d$, we have:

$\gamma_{i}\hskip 0.85355pt-\hskip
0.85355pt\gamma_{j}\,\geq_{\eqref{eq:progressive-def}}\,p\hskip
0.85355pt(\lambda_{i}-\lambda_{j})\hskip 0.85355pt+\hskip 0.85355pt(1-p)\hskip
0.85355pt(\mu_{i}-\mu_{j})\hskip 0.85355pt-\hskip 0.85355pt2\hskip
0.85355pt(|\lambda|-|\mu|)^{\frac{3}{4}}\,\geq\,(1-p)\hskip
0.85355pt(\mu_{i}-\mu_{j})\hskip 0.85355pt-\hskip 0.85355pt2\hskip
0.85355ptn^{\frac{3}{4}}.$ (11.20)

Note that

$\mu_{i}-\mu_{j}\,=\,\lfloor\beta_{i}n\rfloor\hskip 1.70709pt-\hskip
1.70709pt\lfloor\beta_{j}n\rfloor\,\geq\,(\beta_{i}-\beta_{j})\hskip
0.85355pt|\lambda|\hskip 1.70709pt-\hskip 1.70709pt1\,\geq\,\hskip
1.70709pt\frac{\beta_{i}-\beta_{j}}{2\hskip
0.85355pt(\alpha_{1}+\ldots+\alpha_{d})}\hskip 1.70709pt|\lambda|\hskip
0.85355pt,$ (11.21)

for sufficiently large $n$. Note also that

$1-p\,=_{\eqref{eq:definition p}}\
\frac{|\lambda|-|\gamma|}{|\lambda|-|\mu|}\,=\,\sum_{i=1}^{d}\frac{\lambda_{i}-\gamma_{i}}{|\lambda|-|\mu|}\,\geq_{\eqref{eq:separated-
def}}\,\frac{d\varepsilon^{3}}{2}\hskip
1.70709pt\frac{|\lambda|}{|\lambda|-|\mu|}\,\geq\,\frac{d\varepsilon^{3}}{2}\hskip
1.70709pt.$ (11.22)

Substituting (11.21) and (11.22) into (11.20), we get

$\gamma_{i}-\gamma_{j}\,\geq\,\frac{d\varepsilon^{3}}{4}\hskip
1.70709pt\frac{\beta_{i}-\beta_{j}}{\alpha_{1}+\ldots+\alpha_{d}}\hskip
1.70709pt|\lambda|\hskip 1.70709pt-\hskip 1.70709pt2\hskip
0.85355ptn^{\frac{3}{4}}\,\geq\hskip
1.70709pt\frac{d\varepsilon^{3}}{8}\frac{\beta_{i}-\beta_{j}}{\alpha_{1}+\ldots+\alpha_{d}}\hskip
1.70709pt|\lambda|\,\geq\,\varepsilon^{\prime}\hskip
0.85355pt|\lambda|\,\geq\,\varepsilon^{\prime}\hskip 0.85355pt|\gamma|,$
(11.23)

for $|\lambda|=\Theta(n)$ large enough. On the other hand, for all
$i\overset{B^{\prime}}{\sim}j$, $1\leqslant i<j\leqslant d$, we have:

$\mu_{i}\hskip 1.70709pt-\hskip 1.70709pt\mu_{j}\ =\
\lfloor\beta_{i}n\rfloor\hskip 1.70709pt-\hskip
1.70709pt\lfloor\beta_{j}n\rfloor\,=\,0\hskip 0.85355pt.$ (11.24)

It follows from (11.23) and (11.24), that (11.1) and (11.3) hold for this case
when $n$ is sufficiently large. Thus, the first inequality in (11.19) follows
by Lemma 11.5. This completes the proof of the theorem. $\square$

##  12 Conjectures and open problems

We believe our results can be further strengthened in several directions, and
would like to mention a few possibilities.

###  12.1 Sorting probability

The bound
$\delta\bigl{(}P_{\lambda/\mu}\bigr{)}=O\bigl{(}\frac{1}{\sqrt{n}}\bigr{)}$
that we obtain in Theorems 1.2–1.5 is likely not tight. In fact,
$\Omega\bigl{(}\frac{1}{n}\bigr{)}$ is the only lower bound that we know in
some cases (see $\S$1.5). The results in Corollary 3.5 and [CPP21] also seem
to suggest that $O\left(\frac{1}{n}\right)$ is perhaps the best one can aim
for in full generality. We believe the TVK shapes are likely the easiest case
to make progress as they are most similar to the Catalan poset case:

######  Conjecture 12.1.

There is a universal constant $C>0$, such that for all $d\geq 2$, and for
every Thoma sequence $\alpha\in\mathbb{R}^{d}_{>0}$, we have:

$\delta\bigl{(}P_{\lambda}\bigr{)}\hskip 1.70709pt\leq\hskip
1.70709pt\frac{C}{n^{5/4}}\,,$

where $\lambda\simeq\alpha\hskip 0.85355ptn$ is a TVK $\alpha$-shape.

We believe the same bound holds for more general cases. To understand our
reasoning, note that we take $a=\lfloor\lambda_{1}/2\rfloor$ in this case to
minimize the sorting probability. Even if the bound we obtain is tight, by
varying $a$ one is likely to obtain lower global minimum in the definition of
the sorting probability. In fact, we believe the following general claim with
a weaker bound:

######  Conjecture 12.2.

There is a universal constant $C>0$, such that for every $\lambda\vdash n$,
$\lambda\neq(n),(1^{n})$, we have:

$\delta\bigl{(}P_{\lambda}\bigr{)}\hskip 1.70709pt\leq\hskip
1.70709pt\frac{C}{\sqrt{n}}\hskip 1.70709pt.$

This conjecture is suggesting that the constants $C_{d,\varepsilon}$ in
Theorem 1.3 and Theorem 1.5 can be made independent of parameters $d$ and
$\varepsilon$, even though the proofs give dependence that is relatively wild.
See, e.g. the last line of the proof of Theorem 8.4. At the moment, we cannot
even prove that $\delta\bigl{(}P_{\lambda}\bigr{)}\to 0$ for general
partitions $\lambda$, with $n=|\lambda|\to\infty$.

In a different direction, suppose $\lambda$ is a $3$-dimensional diagram
defined as lower ideals in $\mathbb{N}^{3}$. The tools of this paper are
heavily based on the HLF (2.7), NHLF (2.8), asymptotics of Schur functions and
other Algebraic Combinatorics results. None of these are available for
$3$-dimensional diagrams, even for the boxes (products of three chains).
Finding new tools to establish such bounds would be a major breakthrough.

######  Conjecture 12.3.

Fix $d,r\geq 2$. Denote by $P_{d,r,m}$ the $3$-dimensional poset given by a
$[d\times r\times m]\subset\mathbb{N}^{3}$ box (product of chains on size $d$,
and $m$, respectively). Then:

$\delta\bigl{(}P_{d,r,m}\bigr{)}\hskip 1.70709pt=\hskip
1.70709ptO\Bigl{(}\frac{1}{m}\Bigr{)}\,,\ \ \text{as}\ \hskip
1.70709ptm\to\infty.$

A more general problem would be to find conditions on the poset $P=(X,\prec)$
of bounded width, which would guarantee that the sorting probability
$\delta(P)\to 0$ as the size $|X|\to\infty$.

###  12.2 Technical estimates

The tools of this paper are based on bounds for
$f(\lambda/\mu)=|\operatorname{{\rm SYT}}(\lambda/\mu)|$, which are of
independent interest. Recall the definition of $F(\lambda/\mu)$ in (2.9) and
the bound in Theorem 2.2. Recall also the balance function $\Phi(\lambda/\mu)$
defined in (4.1) and the bounds in Lemmas 4.4 and 4.5. The following
conjecture is a natural generalization.

######  Conjecture 12.4.

Fix $d\geq 2$. Let $\lambda/\mu\vdash n$, $\ell(\lambda)\leq d$. Then:

$\frac{1}{C_{d}}\,\Phi(\lambda/\mu)\ \leq\
\frac{f(\lambda/\mu)}{F(\lambda/\mu)}\ \leq\ {C_{d}}\,\Phi(\lambda/\mu)\hskip
0.85355pt,$ (12.1)

for an absolute constant $C_{d}>0$.

One can generalize the definition of $\Phi(\lambda/\mu)$ to continuous
setting:

$\Phi(\textbf{{x}}/\mu)\,:=\,\prod_{1\leqslant i<j\leqslant d}\hskip
1.70709pt\min\left\\{\mu_{i}-\mu_{j}+j-i,\frac{x_{i}}{x_{i}-x_{j}}\right\\},$

where $\textbf{{x}}=(x_{1},\ldots,x_{d})\in\mathbb{R}^{d}$, and
$\mu=(\mu_{1},\ldots,\mu_{d})$ is an integer partition.

######  Conjecture 12.5.

Fix $d\geq 2$ and $\varepsilon>0$. Then, for every
$\mu=(\mu_{1},\ldots,\mu_{d})$ and
$\textbf{{x}}=(x_{1},\ldots,x_{d})\in\mathbb{R}^{d}$, such that
$x_{1}>\ldots>x_{d}>\varepsilon\hskip 0.85355ptx_{1}>0$, we have:

$\frac{s_{\mu}(x_{1},\ldots,x_{d})}{x_{1}^{\mu_{1}}\hskip
1.70709pt\cdots\hskip 1.70709ptx_{d}^{\mu_{d}}}\ \leq\
{C_{d,\varepsilon}}\,\Phi(\textbf{{x}}/\mu)\hskip 0.85355pt,$ (12.2)

for an absolute constant $C_{d,\varepsilon}>0$.

We obtain partial results in favor of this conjecture: a lower bound in Lemma
8.3 and an upper bound in Theorem 8.4. Let us present the former with
simplified notation, as it also gives connection between Conjectures 12.4 and
12.5.

######  Theorem 12.6 ($=$ Lemma 8.3).

Let $\lambda/\mu$ be a skew shape, where
$\lambda=(\lambda_{1},\ldots,\lambda_{d})$, $\mu=(\mu_{1},\ldots,\mu_{d})$.
Then:

$1\ \leq\ \frac{f(\lambda/\mu)}{F(\lambda/\mu)}\ \leq\
\frac{s_{\mu}(\lambda_{1}+d-1,\hskip 0.85355pt\lambda_{2}+d-2,\hskip
1.70709pt\ldots\hskip
1.70709pt,\lambda_{d})}{(\lambda_{1}+d-1)^{\mu_{1}}\hskip
0.85355pt(\lambda_{2}+d-2)^{\mu_{2}}\hskip 1.70709pt\cdots\hskip
1.70709pt\lambda_{d}^{\mu_{d}}}\,.$

######  Remark 12.7.

Conjecture 12.5 in the earlier version of the paper had a matching lower
bound:

$\frac{1}{C_{d,\varepsilon}}\ \Phi(\textbf{{x}}/\mu)\ \leq\
\frac{s_{\mu}(x_{1},\ldots,x_{d})}{x_{1}^{\mu_{1}}\hskip 1.70709pt\cdots\hskip
1.70709ptx_{d}^{\mu_{d}}}\,.$

Unfortunately, this bound fails for the substitution $x_{i}\leftarrow q^{i}$
for $q\in\bigl{[}\varepsilon^{1/(d-1)},1\bigr{)}$, by the _hook-content
formula_ , see [Sta99, Thm 7.21.2]. On the other hand, the upper bound (12.2)
is easy to check in this case.

##  13 Final remarks

### 13.1

Although much of the paper is motivated by the work surrounding the
$\frac{1}{3}$–$\frac{1}{3}$ Conjecture 1.1, we do not resolve the conjecture
in any new cases. As mentioned in the introduction, for all skew Young
diagrams the conjecture was already established in [OS18]. In fact, when
compared with the Kahn–Saks Conjecture 1.6, our results are counterintuitive
since we obtain the conclusion of the conjecture in a strong form, while the
width of our posets remains bounded. Clearly, much of the subject remains
misunderstood and open to further exploration.

### 13.2

The technical assumption in Theorem 1.3, that $\lambda$ is $\varepsilon$-thick
is likely unnecessary, but at the moment we do not know how to avoid it. The
same applies for the $\varepsilon$-smooth assumption, and the Main Theorem 1.5
most likely holds under much weaker assumptions. Let us remark though, that in
some formal sense these two assumptions are equivalent. Indeed, let
$\lambda=(\lambda_{1},\ldots,\lambda_{d})$ and
$\mu=(\lambda_{i+1},\ldots,\lambda_{i+1},\lambda_{i+2},\ldots,\lambda_{d})$.
The skew shape
$\nu:=\lambda/\mu=(\lambda_{1}-\lambda_{i+1},\ldots,\lambda_{i}-\lambda_{i+1})$
is then the straight shape, so the $\varepsilon$-smooth condition
$\lambda_{i}-\lambda_{i+1}\geq\varepsilon\hskip 0.85355ptn$ becomes the
$\varepsilon$-thick condition for $\nu$.

### 13.3

For a fixed number of rows $d=\ell(\lambda)$, Corollary 3.5 shows that
$\delta(P_{\lambda})=O\big{(}\frac{1}{n}\big{)}$ for all $\lambda\vdash n$,
such that $\lambda_{2}=O(1)$. This is the opposite extreme of
$\varepsilon$-thick diagrams $\lambda$, suggesting that the
$\varepsilon$-thick assumption in Theorem 1.3 might be unnecessary indeed.

### 13.4

The upper bound in (11.2) and (11.3) can be replaced with an arbitrary
constant $K$ at the cost of changing the positive constant $C_{d,\varepsilon}$
in our results into the positive constant $C_{d,\varepsilon,K}$, which now
also depends on $K$. The rest of the proof follows verbatim and gives a slight
extension of Theorem 1.4 under weaker conditions
$\bigl{|}\lambda_{i}-\alpha_{i}\hskip 0.85355ptn\bigr{|}\leq K$, and the same
for the $\mu$. We omit the details.

### 13.5

The Naruse’s hook-length formula (2.8) works well when $|\lambda/\mu|$ is
relatively small compared to $|\lambda|$. On the other hand, when $|\mu|$ is
very small, there is another positive formula due to Okounkov and Olshanski
[OO98], which was observed in [OO98, Sta99] to give sharp estimates in that
regime. In [MPP1, $\S$9.4], the authors suggested that this rule is equivalent
to the Knutson–Tao “equivariant puzzles” rule. This was proved in [MZ+], which
reworked the Okounkov–Olshanski formula in the NHLF-style. It would be
interesting to see if this formula can be used in place of NHLF to obtain
sharper bounds on the sorting probability of skew Young diagrams, at least in
some cases.

### 13.6

When $\lambda=(m^{d})$ is a rectangle, one can estimate $\delta(P_{\lambda})$
without the NHLF, since $f(\lambda/\mu)$ can be computed by the hook-length
formula (2.7). This greatly simplifies the calculations, and is an approach
take in [CPP21] for the Catalan numbers example
$\lambda=\bigl{(}\frac{n}{2},\frac{n}{2}\bigr{)}$, see $\S$1.5.

### 13.7

As we mentioned in the previous section, there are several places where our
bounds are likely not sharp. First, the argument in $\S$5.2, is a quantitative
version of Linial’s pigeonhole principle argument, which we also employ in
$\S$3.1. But the real obstacle to improving the
$O\bigl{(}\frac{1}{\sqrt{n}}\bigr{)}$ bound is not apparent until Section 8,
where the interval decompositions are introduced and a different pigeonhole
argument is used.

### 13.8

Most recently, Conjecture 1.1 was generalized to all Coxeter groups [GG20]. It
would be interesting to see if our results extend to this setting.

### Acknowledgements

We are grateful to Han Lyu, Alejandro Morales and Fedya Petrov for many
interesting discussions on the subject. We are thankful to Vadim Gorin, Jeff
Kahn, Martin Kassabov, Richard Stanley and Tom Trotter for useful comments.
Dan Romik kindly provided us with Figure 1.3. We thank the anonymous referees
for the careful reading of the paper, especially the suggestion which led us
to Remark 12.7. The first author was partially supported by the Simons
Foundation. The last two authors were partially supported by the NSF.

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[swee] Swee Hong Chan
University of California Los Angeles
Los Angeles, USA
sweehongmathuclaedu
https://www.math.ucla.edu/~sweehong/ [igor] Igor Pak
University of California Los Angeles
Los Angeles, USA
pakmathuclaedu
https://www.math.ucla.edu/~pak/ [greta] Greta Panova
University of Southern California
Los Angeles, USA
gpanovauscedu
https://sites.google.com/usc.edu/gpanova/home