In the seminal work of Stanley <cit.>, several conjectures were made on the structure of Littlewood-Richardson coefficients for the multiplication of Jack symmetric functions. Motivated by recent results of Alexandersson and the present author <cit.>, we postulate that a certain `windowing' property holds for all such Jack L-R coefficients. This property provides a vast set of relations between these coefficients and allows for their direct computation in certain novel cases. We demonstrate compatibility between our windowing conjecture and the conjectures of Stanley, with the hope of illuminating the structures within.

§ STRUCTURE OF PAPER

In Section <ref>, we begin by providing a review the titular Stanley Conjecture(s) of <cit.>. Our first novel conjecture (<ref>) on the general structure of Jack LR coefficients provides a modest but straightforward generalization of the strong form of Stanley's conjecture (<ref>). We then provide a summary of some relevant known results, and revisit the recently proven case of a rectangular union <cit.>, which was the primary motivation for this present work.

In Section <ref>, we deliver a wide ranging generalization of the Rectangular Union result in the central conjecture of this paper: the Window Lensing Conjecture (<ref>). We show that this conjecture holds in several known cases of the Stanley Conjecture (where the corresponding Schur LR coefficient is $c=1$), including the Maximal Filling Tableau case (which was proven by Stanley).
The work of Bravi-Gandini <cit.> has substantial overlap with the topics of this section, and we use their work to prove a significant case of the window lensing conjecture. We provide further evidence for this conjecture by showing that it produces a solution to a large 6-parameter family of $c=2$ examples (<ref>).

In Section <ref>, we explore compatibility between the Window Lensing conjecture and a `factorization' result of Bravi-Gandini <cit.>. We use these results to produce alternate formulas for the family of $c=2$ examples of the previous section.

We are optimistic that the conjectures outlined in this paper, if proven, could provide a pathway to a potential proof of the Stanley Conjectures.

§ PRELIMINARIES

§.§ Definitions

We use the same notation and conventions as <cit.>, except where stated otherwise. For a partition $\mu$ of size $n$, let $J_\mu$ denote the associated Jack symmetric function with parameter $\alpha \in \setR$, normalized so that the coefficient of the power sum variable $p_1^{n}$ in $J_\mu$ is $1$. These form a basis of the ring of symmetric functions. The Jack Littlewood-Richardson coefficients $g_{\mu\nu}^{\lambda}$ are defined by the expansion of a product of Jack functions

\begin{equation} J_\mu \cdot J_\nu = \sum_{\lambda} g_{\mu\nu}^{\lambda} J_\lambda. \end{equation}

The Schur Littlewood-Richardson coefficients are recovered by

\begin{equation}\label{eq:schurLRjack}
c_{\mu\nu}^{\lambda} = \left(  \frac{  \lhs{\lambda}{} }{\lhs{\mu}{} \lhs{\nu}{}  }  \jackjLR_{\mu\nu}^\lambda    \right)_{\alpha \to 1}.
\end{equation}

We use the following standard notation for the $\alpha$-Hall norm of the Jacks,

\begin{equation}\label{eq:jacknorm}
j_\lambda :=  \norm{\jackj_\lambda}^2 = \prod_{b\in \lambda} \lh{\lambda}{b} \, \uh{\lambda}{b}.
\end{equation}

where the equality is given by a theorem of Stanley.

§.§ The Stanley Conjectures

In <cit.>, several remarkable conjectures were made regarding the
structure of the Jack Littlewood–Richardson coefficients.

The Jack LR coefficients are ratios of non-negative integer polynomials in $\alpha$. In particular,

\begin{equation*}
\jackjLR_{\mu\nu}^\lambda \cdot j_\lambda  \in \setZ_{\geq0}[\alpha].
\end{equation*}

In general, we use the terminology of a strong form of the Stanley conjecture to refer to any conjecture that proposes an explicit form for $\jackjLR_{\mu\nu}^\lambda$ which manifestly satisfies the positivity of Conjecture (<ref>). Stanley conjectured such a form in the case $c_{\mu\nu}^\lambda =1$.

If $c_{\mu\nu}^\lambda =1$, then the corresponding Jack LR coefficient has the form

\begin{equation*}\label{formula:strongstanley}
\jackjLR_{\mu \nu}^{\lambda} = \frac{
\prod_{b \in \mu} {h}^{\Phi_b}_{\mu}(b) \,\,
\prod_{b \in \nu} {h}^{\Phi_b}_{\nu}(b) }{
\prod_{b \in \lambda} {h}^{\Phi_b}_{\lambda}(b) },
\end{equation*}

where $\Phi_b$ is a choice of either $U$ or $L$ for each box $b$ in $\mu,\nu$ and $\lambda$.
Moreover, one chooses $\lh{\,}{b}$ and $\uh{\,}{b}$ exactly the same number of times in the numerator and denominator.

§.§ Structural Conjecture

In the closing section of <cit.>, Stanley contemplates whether an expression of the type of Conjecture (<ref>) may also hold for cases where $c_{\mu\nu}^\lambda \geq 1$, where one might instead sum over $c_{\mu\nu}^\lambda \in \BZ$ terms of the form given therein. However, a counter example was presented that halted this idea (see Example <ref>). In this paper we pursue that train of thought, and we will build evidence for the following straightforward generalization.

For any collections of boxes $\mu,\nu, \lambda$ (e.g. partitions), define the set of channels $\hookspace_{\mu,\nu}^{\lambda}$ as the set of assignments $\Phi : b \mapsto \{U,L\}$ for every box $b$ in each of $\mu,\nu$ or $\lambda$.

For a hook assignment $A \in \{U,L\}$, we let $\bar A$ denote the opposite assignment.

For each $\mu,\nu,\lambda$, there exist a choice of channel coefficients $c_\Phi \in \BZ$ for $\Phi \in \hookspace_{\mu,\nu}^{\lambda}$ such that

\begin{equation}
\jackjLR_{\mu\nu}^{\lambda} = \sum_{\Phi \in \hookspace_{\mu\nu}^{\lambda}} c_\Phi \cdot \frac{
\prod_{b \in \mu} {h}^{\Phi_b}_{\mu}(b) \,\,
\prod_{b \in \nu} {h}^{\Phi_b}_{\nu}(b) }{
\prod_{b \in \lambda} {h}^{\Phi_b}_{\lambda}(b) }.
\end{equation}

Consequentially, we have

\begin{equation}
c_{\mu\nu}^{\lambda} = \sum_{\Phi \in \hookspace_{\mu\nu}^{\lambda}} c_\Phi.
\end{equation}

We observe that Conjecture (<ref>) is compatible with the Stanley Conjecture (<ref>) in that the poles of $\jackjLR_{\mu\nu}^{\lambda}(\alpha)$ as a meromorphic function of $\alpha$ can only be of the form $1/\ulh{\lambda}{b}$. However, Conjecture (<ref>) is much weaker in the sense that it does not suggest any positivity property of the coefficients $c_\Phi$. We do however make a modest refinement to this positivity conjecture later in (<ref>).

Above, we introduced the notion of channels, which capture the assignment of an upper or lower hook to a particular box in a diagram, which is more general than considering only the hook lengths corresponding to such an assignment. We now formalize this distinction.

For a box $b$ in the tableau associated to a partition $\lambda$, we will consider the assignment of an upper or lower hook to be denoted[In <cit.>, this notation is used instead to denote the hook vector, a closely related concept.] by the hook symbols of either $\uhc{\lambda}{b}$ or $ \lhc{\lambda}{b}$, and the hook lengths (i.e. the $\alpha$-content of that hook) to be denoted $\uh{\lambda}{b}$ or $\lh{\lambda}{b}$. We use bullet $(\bullet)$ to indicate ambiguity between $U$ or $L$, e.g. $\bm{h}^\bullet_{\lambda}(b)$. Inside a tableau, we denote upper hook symbol by $\begin{ytableau}*(boxU) U \end{ytableau}$ and lower hook symbol by $\begin{ytableau}*(boxL) L \end{ytableau}$. For example, for the partition $\lambda = 532$ and the box $b=(1,0)$,

\begin{equation}
\begin{ytableau}
{}& \\
& & \\
& b& & &
\end{ytableau}
\end{equation}

The upper hook symbol is

\begin{equation} \uhc{\lambda}{b} =
\begin{ytableau}
{}& \\
& & \\
& *(boxU) U & & &
\end{ytableau}
\end{equation}

whereas the upper hook length is $\uh{\lambda}{b}= 2+4\alpha$. The $\alpha$-evaluation map $[ \cdot  ]$, is defined by sending hooks symbols to their corresponding hook lengths[One could instead consider the Macdonald evaluation, e.g.

\[ [\uhc{\lambda}{b}]_{q,t} \coloneqq 1-q^{\arm(b)+1} t^{\leg(b)}. \]
]. That is

\[ [  \uhc{\lambda}{b} ] := \uh{\lambda}{b} = \leg(b) + (\arm(b)+1)\alpha.\]

To such a choice, we associate a fraction of products of hook symbols, known as a channel factor

\begin{equation}\label{def:Gchannel}
\bm{G}_\Phi \coloneqq \frac{
\prod_{b \in \mu}  \ahc{\Phi_b}{{\mu}}{b} \,\,
\prod_{b \in \nu}  \ahc{\Phi_b}{\nu}{b} }{
\prod_{b \in \lambda}  \ahc{\Phi_b}{\lambda}{b} }.
\end{equation}

In such a factor, boxes for $\lambda$ appear in the denominator, and boxes for $\mu,\nu$ appear in the numerator. For example for $\lambda = 32$, $\mu = 21$, and $\nu=1$,

\begin{equation}
\bm{G}^{\sU\sL\sU;\sL}_{\sL\sL\sU\sU \sL} \coloneqq \frac{
\begin{ytableau}
*(boxU) U \\
*(boxL) L & *(boxU) U  \\
\end{ytableau}\,\, \begin{ytableau}
\none \\
*(boxL) L  \\
\end{ytableau}}{\begin{ytableau}
*(boxL) L & *(boxL) L  \\
*(boxU) U & *(boxU) U & *(boxL) L  \\
\end{ytableau}} = \frac{\left(\uhc{21}{0,1}\lhc{21}{0,0}\uhc{21}{1,0}\right)\cdot\left(\lhc{1}{0,0}\right)}{\left(\lhc{32}{0,1}\lhc{32}{1,1}\uhc{32}{0,0}\uhc{32}{1,0}\lhc{32}{2,0}\right)} \in \hookspace_{21,1}^{32}.
\end{equation}

A rule is a choice of an element in the $\setZ$ span of such elements, denoted $\hookspace_{\mu,\nu}^{\lambda}(\setZ)$. The evaluation map $[.]$ extends to a map $\hookspace_{\mu,\nu}^{\lambda}(\setZ) \to \setQ(\alpha)$.

With this definition we may restate the General Structure Conjecture (<ref>) as

There exists a rule $\JackjLR_{\mu\nu}^{\lambda}$ that describes Jack Littlewood-Richardson coefficients. That is, there exists

\begin{equation}
\JackjLR_{\mu\nu}^\lambda = \sum_{\Phi \in \hookspace_{\mu\nu}^{\lambda}} c_\Phi\,\bm{G}_\Phi  \in \hookspace_{\mu\nu}^\lambda(\setZ) \quad \text{with  } c_\Phi \in \BZ ,
\end{equation}

with the property that $\jackjLR_{\mu\nu}^{\lambda} = [\JackjLR_{\mu\nu}^{\lambda}]$.

Where appropriate in the remainder of this paper, we freely move between considering $\JackjLR_{\mu\nu}^{\lambda}$ and its evaluation.

§.§ Known Cases

Here we collect some known cases of the Strong Stanley conjecture (<ref>), that is, in all the cases below the corresponding Schur LR coefficient satisfies $c=1$.

Let $\lambda/\mu$ be a horizontal $r$-strip, then the Jack LR coefficient is given by the following expression

\begin{equation*}
\jackjLR_{\mu,(r)}^\lambda = \frac{
\prod_{b \in \mu} A_{\mu,\lambda/\mu}(b) \,\,
\prod_{b \in (r)}  \uh{(r)}{b}  }{
\prod_{b \in \lambda} A_{\lambda,\lambda/\mu}(b) },
\end{equation*}

\begin{equation*}
A_{\sigma,\lambda/\mu}(b) \coloneqq \begin{cases}
\lh{\sigma}{b}  & \text{ if $\lambda/\mu$ does not contain a box in the same column as $b$} \\
\uh{\sigma}{b}  &  \text{ otherwise.}
\end{cases}
\end{equation*}

\begin{equation*}
\JackjLR_{75411,3}^{76431} =
\ytableausetup{boxsize=1.1em}
\frac{
\begin{ytableau}
*(boxL) L\\
*(boxL) L \\
*(boxL) L & *(boxU) U  & *(boxU) U  & *(boxL) L\\
*(boxL) L& *(boxU) U  & *(boxU) U  & *(boxL) L  &*(boxL) L \\
*(boxL) L & *(boxU) U  &*(boxU) U  &*(boxL) L &*(boxL) L  & *(boxU) U & *(boxL) L
\end{ytableau}
\quad
\begin{ytableau}
*(boxU) U & *(boxU) U & *(boxU) U
\end{ytableau}}{
\begin{ytableau}
*(boxL) L\\
*(boxL) L & *(boxU) U  & *(boxU) U \\
*(boxL) L& *(boxU) U  & *(boxU) U  & *(boxL) L\\
*(boxL) L  & *(boxU) U  & *(boxU) U  & *(boxL) L&*(boxL) L  & *(boxU) U \\
*(boxL) L& *(boxU) U  &*(boxU) U  & *(boxL) L &*(boxL) L  & *(boxU) U & *(boxL) L
\end{ytableau}} \\
\end{equation*}

Stanley also generalized the above Pieri rule to the case of Maximal Filling tableaus.

Suppose that $c_{\mu\nu}^{\lambda} =1$, and the unique $LR$-tableau $T$ of shape $\lambda/\mu$ with weight $\nu$ has the maximal filling property, that is, that every column $C$ of $T$ consists of the sequential integers $1,2, \ldots, n_C$.

Let $r_i$ be defined as the largest entry of $T$ in row $i$ of $\lambda/\mu$, and let $c_j$ be the largest entry of $T$ in column $j$ of $\lambda/\mu$. For $b=(j,i)$ define

\begin{equation}\label{eq:maxfillingrule}
C_{\sigma,T}(b)  \coloneqq
\begin{cases}
\uh{\sigma}{b} & \text{ if $r_i\leq c_j$ and $c_j>0$} \\
\lh{\sigma}{b} &  \text{ otherwise}.
\end{cases}
\end{equation}

Furthermore, for $b = (j,i)$, let $b_* \coloneqq (j,i+c_j)$.  The Jack LR coefficient is given by

\begin{equation}\label{eq:maxfilling}
g_{\mu\nu}^{\lambda} = \frac{ \prod_{b \in \mu} C_{\mu,T}(b_*) \cdot \prod_{b\in\nu} \uh{\nu}{b} }{\prod_{b \in \lambda} C_{\lambda,T}( b ) }.
\end{equation}

Take $\lambda = 766654211$, $\mu = 7553322$, $\nu=542$. The unique $LR$ tableau $T$ of shape $\lambda/\mu$ and weight $\nu$ is given by

\[ \ytableausetup{boxsize=1.0em}
T = \begin{ytableau}
2  \\
1 \\
&   \\
&  & 1 & 3 \\
&  &   & 2 & 2 \\
&   &  & 1&1 & 3 \\
&   &   & &&2\\
&    &    &   &   & 1 \\
&  &  &  &   &  &  \\
\end{ytableau}
\]

We see that $T$ has the maximal filling property, with vectors $c = \{2,0,1,3,2,3,0\}$, and $r = \{0,1,2,3,2,3,0,1,2\}$. Using the formula above, we find the following expression for the rule

\begin{equation*}
\JackjLR_{\mu,\nu}^{\lambda} =
\ytableausetup{boxsize=1.0em}
\frac{
\begin{ytableau}
*(boxU) & *(boxL)   \\
*(boxU) & *(boxL)   \\
*(boxU) & *(boxL)  & *(boxL)  \\
*(boxL) & *(boxL)  & *(boxL)   \\
*(boxU) & *(boxL)  & *(boxL)  & *(boxU) & *(boxU)  \\
*(boxL) & *(boxL)  & *(boxL)  & *(boxU) & *(boxL)  \\
*(boxU) & *(boxL)  & *(boxU)  & *(boxU) & *(boxU) & *(boxU) & *(boxL) \\
\end{ytableau} \quad
\begin{ytableau}
*(boxU)  &*(boxU)   \\
*(boxU)  &*(boxU)  &*(boxU)  &*(boxU)     \\
*(boxU)  &*(boxU)  &*(boxU)  &*(boxU)  &*(boxU)
\end{ytableau}}{
\begin{ytableau}
*(boxU)  \\
*(boxU)  \\
*(boxU) & *(boxL)   \\
*(boxL) & *(boxL)  & *(boxL)  & *(boxU) \\
*(boxU) & *(boxL)  & *(boxL)  & *(boxU) & *(boxU)\\
*(boxL) & *(boxL)  & *(boxL)  & *(boxU) & *(boxL) & *(boxU) \\
*(boxU) & *(boxL)  & *(boxL)  & *(boxU) & *(boxU) & *(boxU) \\
*(boxU) & *(boxL)  & *(boxU)  & *(boxU) & *(boxU) & *(boxU) \\
*(boxU) & *(boxL)  & *(boxU)  & *(boxU) & *(boxU) & *(boxU) & *(boxL)\\
\end{ytableau}}  \\
\end{equation*}

Let $\mu \subseteq m^n$, and $\bar\mu$ be the partition that is the reverse[I.e. rotate the shape 180 degrees.] of the shape $m^n/\mu$. Then the strong Stanley conjecture (<ref>) holds for $\jackjLR_{\mu,\bmu}^{m^n}$, in particular

\begin{equation}\label{eq:stanleyrectangular}
\jackjLR_{\mu,\bmu}^{m^n}  = \frac{
\prod_{b \in \mu}  \lh{\mu}{b}  \,\,
\prod_{b \in \bmu}  \uh{\bmu}{b} }{
\prod_{b \in m^n}  D'_{\mu,m^n}(b) },
\end{equation}

where, for $b=(b_1,b_2)\in m^n$, we define

\begin{equation}\label{defn:rectD}
\defin{ D'_{\mu,m^n}(b)} \coloneqq
\begin{cases}
\lh{m^n}{b} & \text{ if $(b_1,n-1-b_2) \in \mu$} \\
\uh{m^n}{b} &  \text{ otherwise}.
\end{cases}
\end{equation}

\begin{equation*}\JackjLR_{211,221}^{333}=
\ytableausetup{boxsize=1.1em}
\frac{
\begin{ytableau}
*(boxL) L   \\
*(boxL) L   \\
*(boxL) L & *(boxL) L
\end{ytableau}\quad
\begin{ytableau}
*(boxU) U  \\
*(boxU) U & *(boxU) U  \\
*(boxU) U & *(boxU) U
\end{ytableau}}{\begin{ytableau}
*(boxL) L & *(boxL) L & *(boxU) U \\
*(boxL) L & *(boxU) U & *(boxU) U  \\
*(boxL) L & *(boxU) U & *(boxU) U
\end{ytableau}}.
\end{equation*}

Note: as with all Jack LR cofficients $\jackjLR_{\mu,\nu}^\lambda$ the evaluations $\jackjLR_{\mu,\bmu}^{m^n}=\jackjLR_{\bmu,\mu}^{m^n}$ are symmetric in the lower arguments[For a demonstration of this property in this case see <cit.>], however in this case the corresponding rules $\JackjLR_{\mu,\bmu}^{m^n}$ are not. A priori, we don't have any reason to expect that rules need be symmetric in these arguments, in fact, the evidence herein suggests them not to be.

The next example is the motivation for the present work. Wherein the surprising simplicity of the answer, and the compatibility with the Rectangular case (<ref>), merited further investigation.

Let $m^n$ be a rectangular partition, and $\mu$ be any partition (i.e. not necessarily $m^n \subset \mu$). Let $\sigma \coloneqq m^n \cap \mu$, and $\bsigma$ be the partition given by the reverse of $m^n / \sigma$. Then, the Strong Stanley Conjecture (<ref>) holds for $\jackjLR_{\mu,\bsigma}^{\mu\cup m^n}$.

In particular, let $R \subset m^n$ be the smallest rectangular shape such that $\mu$ and $\mu\cup m^n$ agree outside $R$. Say $R$ is of shape $k^\ell$. Let $\mu_R$ be the partition giving by shifting the shape $\mu \cap R$ back to the origin, and $\bar \mu_R$ be the reverse of $k^\ell / \mu_R$. We then have the following decomposition for the LR coefficient

\begin{equation}\label{eq:rectunion}
\jackjLR_{\mu,\bsigma}^{\mu\cup m^n} = \raisebox{-0.5\height}{\includegraphics[scale=0.5,page=1]{stanleyConjecturePics}} \times  \jackjLR_{\mu_R, \bar \mu_R}^{k^\ell}
\end{equation}

where the left factor (henceforth denoted $F_R$)  is a products of $L$-hooks for every box to the left of $R$ (denoted by dashed line),
and $U$-hooks for every box below it, for both $\mu$ in the numerator
and $\mu \cup m^n$ in the denominator. For boxes sharing neither a row or column with a box in $R$, we can assign either upper or lower hooks as these cancel between the numerator and denominator, and so we indicate this ambiguity with light grey *(boxG). The second factor $\jackjLR_{\mu_R, \bar \mu_R}^{k^\ell}$ is covered by the rectangular case above.

Thus we see that the rule for the Rectangular union case, $\JackjLR_{\mu,\bsigma}^{\mu\cup m^n}$, is equivalent to starting with the rule for the Rectangular case (<ref>) inside the `window' $R$, given by $\JackjLR_{\mu_R, \bar \mu_R}^{k^\ell}$, and extending outside of this window by $F_R$.

\begin{equation*}
\JackjLR_{42211,211}^{43331} =
\ytableausetup{boxsize=1.1em}
\frac{
\begin{ytableau}
*(boxG)  \\
*(boxL) L   \\
*(boxL) L & *(boxS)  \\
*(boxL) L & *(boxS)  \\
*(boxG)& *(boxU) U & *(boxU) U & *(boxG)  \\
\end{ytableau}\quad
\begin{ytableau}
*(boxS)   \\
*(boxS)   \\
*(boxS)  & *(boxS)
\end{ytableau}}{
\begin{ytableau}
*(boxL) L & *(boxS)  & *(boxS)  \\
*(boxL) L & *(boxS)  & *(boxS)  \\
*(boxL) L & *(boxS)  & *(boxS)  \\
*(boxG) & *(boxU) U & *(boxU) U & *(boxG)  \\
\end{ytableau}} \times (\JackjLR_{11,211}^{222})_R \\
\end{equation*}

The notation $(\JackjLR)_R$ denotes taking hook choices for the boxes inside $R \cap\mu$, $R \cap\lambda$ and $\nu$ to agree with those choices of the rule $\JackjLR$ inside the brackets.

§ WINDOWING CONJECTURE

You rows of houses! you window-pierc'd façades! you roofs!

You porches and entrances! you copings and iron guards!

You windows whose transparent shells might expose so much!

You doors and ascending steps! you arches!

You gray stones of interminable pavements! you trodden crossings!

From all that has touch'd you I believe you have imparted to yourselves, and now would impart the same secretly to me...Walt Whitman
Song of the Open Road

Motivated by the Rectangular Union case (<ref>), we consider the following constructions.

A window is a shape $R$ that is closed under meets and joins of its boxes (hence, necessarily a union of rectangles).

For a partition $\mu$, let $\mu_R$ be the windowed partition obtained by viewing $\mu$ through the window $R$. That is, $\mu_R$ is given by the shape $\mu \cap R$ joined along edges of $R$ and re-centered back to the origin as a partition. We color the boxes of windows with beige

For any shape $\sigma$, define the meet-join of $\sigma$, denoted $\meetjoin(\sigma)$, to be the window that is given by the closure of $\sigma$ under meets and joins of its boxes. In particular, for $\mu\subset \lambda$ we often consider $\meetjoin(\lambda/\mu)$, which is the smallest window such that $\mu$ and $\lambda$ agree outside of it.

For $\lambda= 44322$, and the window $R$ given below, we have $\lambda_R = 3211$.

\begin{equation*}
\ytableausetup{boxsize=0.7em}
\lambda \coloneqq
\begin{ytableau}
{}&   \\
&    \\
&   &  \\
&  &  &   \\
&   &   &\\
\end{ytableau}  \\
\qquad
R \coloneqq
\begin{ytableau}
{}& *(boxS)   & *(boxS)  & *(boxS) \\
& *(boxS)   & *(boxS)  & *(boxS) \\
& *(boxS)  & *(boxS) & *(boxS)\\
&  &  &   \\
& *(boxS)   &*(boxS)   & *(boxS) \\
\end{ytableau}  \\
\qquad
\lambda_R = \begin{ytableau}
{}   \\
\\
&  \\
& & \\
\end{ytableau}
\end{equation*}

For further demonstration, let $\mu \subset \lambda$ be such that $\mu$ and $\lambda$ agree (as shapes) outside of two disjoint[And which can be totally separated horizontally and vertically.] rectangles $R_1, R_2$. Let $R_3 = \bmin{R_1}  {R_2}$ and $R_4 = \bmax{R_1}  {R_2}$.

\begin{equation}\label{fig:twowindowed}
\end{equation}

Viewing through the window $R\coloneqq R_1 + R_2 + R_3 + R_4$, the shapes are joined on their edges, defining the two windowed partitions $\lambda_R$ and $\mu_R$.

\[  {\includegraphics[scale=0.8,page=4]{stanleyConjecturePics}}  \]

Following from Formula (<ref>), we make the following key definitions.

Let $\mu \subset \lambda$ be such that $\mu$ and $\lambda$ agree (as shapes) outside of a window $R$, i.e. $\meetjoin({\lambda/\mu}) \subset R$.

Define the window factor for $R$, denoted $\bm{F}_{\mu,R}^\lambda$, as the ratio of hook symbols given by the following product over all boxes in $\mu$ and $\lambda$ which are outside $R$,

\begin{equation}
\bm{F}_{\mu,R}^\lambda \coloneqq \prod_{b \in \mu/R} \frac{ \ula{\mu}{b} }{ \ula{\lambda}{b}  },
\end{equation}

where for those boxes that:

* share a row with a box in $R$, we assign a lower hook symbol *(boxL).

* share a column with a box in $R$, we assign a upper hook symbol *(boxU).

* share neither a row or column with a box in $R$, then in this case the hook lengths for this box in $\mu$ and $\lambda$ agree, and thus they will cancel out when the hook symbols are evaluated. Thus, we can assign either an upper or lower hook symbol (or neither), an ambiguity indicated by the neutral symbol *(boxG).

For example, if we consider the window given by Figure (<ref>), we have

\begin{equation}
\bm{F}_{\mu,R}^\lambda :=   \raisebox{-0.5\height}{\includegraphics[scale=0.8,page=5]{stanleyConjecturePics}}
\end{equation}

Correspondingly, let $F_{\mu,R}^\lambda \coloneqq [\bm{F}_{\mu,R}^\lambda]$ be the evaluation that sends hook symbols to hook lengths. Often we just write $\bm{F}_{R} \equiv \bm{F}_{\mu,R}^\lambda$ if the context of the partitions is clear.

In the earlier example (<ref>), with $\lambda=43331$ and $\mu=42211$, we have

\begin{equation*}
\ytableausetup{boxsize=0.8em}
R := \begin{ytableau}
{} &   & &  \\
& *(boxS)  & *(boxS) &   \\
& *(boxS)  & *(boxS) &  \\
& *(boxS)  & *(boxS) &\\
&  & &   \\
\end{ytableau}
\quad\quad\quad
\bm{F}_{\mu,R}^\lambda =
\frac{
\begin{ytableau}
*(boxG)  \\
*(boxL)    \\
*(boxL)  & *(boxS)  \\
*(boxL)  & *(boxS)  \\
*(boxG)& *(boxU)  & *(boxU)  & *(boxG)  \\
\end{ytableau}}{
\begin{ytableau}
*(boxL) & *(boxS)  & *(boxS)  \\
*(boxL) & *(boxS)  & *(boxS)  \\
*(boxL) & *(boxS)  & *(boxS)  \\
*(boxG) & *(boxU) & *(boxU) & *(boxG)  \\
\end{ytableau}} \\
\end{equation*}

If $\lambda/\mu \subseteq R' \subset R$, then $\bm{F}_{\mu,R'}^\lambda$ agrees with $\bm{F}_{\mu,R}^\lambda$ outside of $R$. Thus any window containing $\lambda/\mu$ is compatible with the smallest such window $\meetjoin(\lambda/\mu)$.

Any box $b$ outside of $R$ that lies in a row (or column) that intersects $R'$ but not $R$ has the property that $\ulh{\mu}{b}= \ulh{\lambda}{b}$, and hence there is a cancellation between this hook length in the numerator and denominator, and thus we assign this box the neutral symbol.

For a window $R$, define the de-windowing operator $\left( \cdot \right)_R$ by its action on hook symbols as

\begin{equation}
\left( \bm{h}^\bullet_{\lambda_R}(b)\right)_R \coloneqq \bm{h}^\bullet_{\lambda}(b^R),
\end{equation}

where the box $b\in \lambda_R $ corresponds to $b^R \in \lambda \cap R$. This operator is extended by linearity to act on rules.

For ease of notation, we write de-windowing followed by evaluation as

\[ [ \cdot ]_R \coloneqq [\,( \cdot )_R \,]. \]

In the case of Example (<ref>), we have

\begin{equation*}
\ytableausetup{boxsize=1.0em}
\left( \begin{ytableau}
*(boxU) U  \\
*(boxL) L    \\
*(boxL) L  &  *(boxU) U \\
*(boxU) U &  *(boxL) L &  *(boxL) L \\
\end{ytableau} \right)_R =
\begin{ytableau}
{}& *(boxU) U   \\
&  *(boxL) L    \\
& *(boxL) L  &  *(boxU) U  \\
&  &  &   \\
& *(boxU) U  &  *(boxL) L  &  *(boxL) L\\
\end{ytableau}
\end{equation*}

§.§ Main Conjecture

Motivated by the Rectangular Union case (<ref>), we propose that a similar formula of that type holds in general for all Jack LR coefficients. The following is the main conjecture of this paper.

Assume the General Structure Conjecture (<ref>) holds. The Jack Littlewood-Richardson rules $\JackjLR_{\mu\nu}^{\lambda}$ satisfy the window lensing property: for a window $R$ with $\lambda/\mu \subset R$, we have

\begin{equation}
\JackjLR_{\mu,\nu}^{\lambda} ={\bm{F}_{\mu,R}^\lambda} \times \left(\JackjLR_{\mu_R,\nu}^{\lambda_R}\right)_{R}.
\end{equation}

According to this conjecture, to prescribe a rule for a given Jack LR coefficient which is constrained by a window $R$, we fill in the boxes outside of $R$ with the window factor $\bm{F}_R$, and inside $R$ we fill in with the de-windowed rule $\JackjLR_{\mu_R,\nu}^{\lambda_R}$ (extended by linearity over the sum of terms in the rule). We can window over the pair $\mu,\lambda$, or $\nu,\lambda$, and such a formula should hold for each pair. As before, we don't assume the rule $\JackjLR_{\mu,\nu}^{\lambda}$ is symmetric in the two lower arguments.

This conjecture gives a vast set of relations between the various LR coefficients, which we investigate in the next subsection.

§.§ Consequences

In this section, we provide evidence for the Window Lensing conjecture, by considering special cases of it.

Firstly, we consider the case where $\mu \subset \lambda$ are such that $\mu$ and $\lambda$ agree outside of a window $R$ that is a single rectangle. In this case, we have $ [\JackjLR_{\mu_R,\nu}^{\lambda_R}]_R  = \jackjLR_{\mu_R,\nu}^{\lambda_R}$, and the window lensing conjecture reduces to the following result, which we are able to prove as a straightforward consequence of the work of Bravi-Gandini <cit.>.

Let $\mu \subset \lambda$ be such that $\lambda/\mu$ in contained inside a rectangular window $R$. We then have

\begin{equation}\label{eq:locality}
\jackjLR_{\mu,\nu}^{\lambda} = F_{\mu,R}^{\lambda} \times \jackjLR_{\mu_R,\nu}^{\lambda_R},
\end{equation}

where, as before,

\[ \windfact_{\mu,R}^{\lambda} \coloneqq \raisebox{-0.5\height}{\includegraphics[scale=0.5,page=2]{stanleyConjecturePics}} \]

We show this follows directly from a result of Bravi-Gandini <cit.>. For $\lambda \subset \mu$, let $J_{\lambda/\mu}$ denote the skew Jack function. For $b\in \mu$, define[In <cit.>, the notation $c_{\mu,\lambda,s} \equiv h^{\sL}_{\lambda,\mu}(s)$, and $c'_{\lambda,\mu,s} \equiv h^{\sU}_{\lambda,\mu}(s) $ is used.]

\begin{equation}
h^{\sL}_{\mu,\lambda}(b) = \arm_{\mu}(b) \alpha + \leg_{\lambda}(b) + 1, \,\,\, h^{\sU}_{\lambda,\mu}(b) \coloneqq (\arm_{\lambda}(b)+1) \alpha + \leg_{\mu}(b).
\end{equation}

For the window $R$ consisting of a single rectangle, let $\lambda_R$ and $\mu_R$ be the corresponding windowed partitions.
In the language of Bravi-Gandini, the skew shapes $\lambda/\mu$ and $\lambda_R/\mu_R$ coincide up to translation, in which case they show the following relation of skew Jack functions

\begin{equation}\label{result:bg}
h^{\sL}_{\mu,\lambda} h^{\sU}_{\lambda,\mu} \cdot J_{\lambda_R/\mu_R} = h^{\sL}_{\mu_R,\lambda_R} h^{\sU}_{\lambda_R,\mu_R} \cdot J_{\lambda/\mu},
\end{equation}

where $h^{\sL}_{\mu,\lambda} \coloneqq \prod_{b \in \mu} h^{\sL}_{\lambda,\mu}(b)$, and $h^{\sU}_{\lambda,\mu} \coloneqq \prod_{b \in \mu} h^{\sU}_{\mu,\lambda}(b)$.
We notice that for a box $b \in \mu \cap R \subseteq \mu$, we have $h^{\sL}_{\lambda,\mu}(b) = h^{\sL}_{ \lambda_R,\mu_R}(b_R)$. Thus, after cancelling factors from all boxes inside $R$, we re-write equation (<ref>) as

\begin{equation}
J_{\lambda/\mu} = \prod_{b \in \mu/ R} h^{\sL}_{\lambda,\mu}(b) h^{\sU}_{\mu,\lambda}(b) \cdot J_{\lambda_R/\mu_R}.
\end{equation}

As we are interested in LR coefficients, which are given by $g_{\mu\nu}^{\lambda} = \langle J_{\lambda/\mu}/j_\lambda ,J_\nu \rangle$, we write the above in terms of the quantities $J_{\lambda/\mu}/j_\lambda$,

\begin{equation}\label{eq:bgbigprod}
\frac{J_{\lambda/\mu}}{j_\lambda} = \prod_{b \in \mu/ R} \frac{h^{\sL}_{\lambda,\mu}(b) h^{\sU}_{\mu,\lambda}(b)}{h^{\sL}_{\lambda}(b) h^{\sU}_{\lambda}(b)} \cdot \frac{J_{\lambda_R/\mu_R}}{j_{\lambda_R}}.
\end{equation}

Now, if $b \in \mu/R$ does not share a row or column with $R$, then $\arm_\mu(b) = \arm_\lambda(b)$ and $\leg_\mu(b) = \leg_\lambda(b)$, and hence $h^{\sL}_{\lambda,\mu}(b) = h^{\sL}_{\lambda}(b)= h^{\sL}_{\mu}(b)$, and similarly $h^{\sU}_{\mu,\lambda}(b) = h^{\sU}_{\lambda}(b)= h^{\sU}_{\mu}(b)$. We see that the contribution from those such boxes $b$ to the product in (<ref>) vanishes. Thus, that same product reduces to one over the two regions $R_{below}$ and $R_{left}$, of those boxes $b \in \mu/R$ either below or left of $R$. For a box $b \in R_{left}$, we have $\leg_\mu(b) = \leg_\lambda(b)$, and thus $h^{\sL}_{\mu,\lambda}(b) = h^{\sL}_{\mu}(b)$, and $h^{\sU}_{\lambda,\mu}(b) = h^{\sU}_{\lambda}(b)$. For a box $b \in R_{below}$, we have $\arm_\mu(b)=\arm_\lambda(b)$, and thus $h^{\sL}_{\mu,\lambda}(b) = h^{\sL}_{\lambda}(b)$ and $h^{U}_{\lambda/\mu}(b) = h^{\sU}_\mu(b)$. Thus we have shown that equation (<ref>) reduces to

\begin{equation}
\frac{J_{\lambda/\mu}}{j_\lambda} = \prod_{b \in R_{left}} \frac{h^{\sL}_{\mu}(b) }{h^{\sL}_{\lambda}(b) }\prod_{b \in R_{below}} \frac{ h^{\sU}_{\mu}(b)}{ h^{\sU}_{\lambda}(b)} \cdot \frac{J_{\lambda_R/\mu_R}}{j_{\lambda_R}}.
\end{equation}

We see this factor is precisely $F^\lambda_{\mu,R}$, that is,

\begin{equation}\label{eq:skeqjackwindow}
\frac{J_{\lambda/\mu}}{j_\lambda} = F^\lambda_{\mu,R} \cdot \frac{J_{\lambda_R/\mu_R}}{j_{\lambda_R}}.
\end{equation}

Taking the inner product with $J_\nu$, we recover the result.

The Rectangular union case (<ref>) follows immediately from the Locality property (<ref>).

In <cit.>, the Rectangular Union LR coefficient was calculated directly using a very different approach. The corollary above demonstrates a vast simplification over that calculation.

By computation, we can verify the following $c=2$ example

\begin{equation*}
\jackjLR_{421,21}^{4321} =
\ytableausetup{boxsize=1.1em}
\frac{
\begin{ytableau}
*(boxS) \\
*(boxS)  & *(boxS)  \\
*(boxU) U & *(boxU) U & *(boxU) U & *(boxG)  \\
\end{ytableau}\quad \begin{ytableau}
*(boxS) \\
*(boxS) &*(boxS) \\
\end{ytableau}
\begin{ytableau}
*(boxS) \\
*(boxS) &*(boxS) \\
*(boxS)  & *(boxS)  &*(boxS)   \\
*(boxU) U & *(boxU) U & *(boxU) U & *(boxG)  \\
\end{ytableau}} \times  \jackjLR_{21,21}^{321} .\\
\end{equation*}

The next result demonstrates the compatibility between the Strong Stanley conjecture (<ref>) and the Window Lensing conjecture (<ref>). First we note compatibillity between Schur LR coefficients and windowing.

For any $\mu \subset \lambda$, and $R$ a window with $\lambda/\mu \subseteq R$, and for all $\nu$, we have the equality of Schur Littlewood-Richardson coeffiecients

\begin{equation}
c_{\mu,\nu}^{\lambda} = c_{\mu_R,\nu}^{\lambda_R}.
\end{equation}

Follows from a simple counting of LR tableau, or from the equality of skew Schur functions $s_{\lambda/\mu} = s_{\lambda_R/\mu_R}$.

Assume $c_{\mu\nu}^{\lambda} = 1$. Let $\lambda/\mu \subset R$ as before. As noted in Lemma (<ref>), we also have $c_{\mu_R\nu}^{\lambda_R}=1$ inside the window $R$.
Assuming the Strong Stanley conjecture (<ref>), there exists a one-term rule $\JackjLR_{\mu_R,\nu}^{\lambda_R}$. Then, by assuming the Window Lensing property (<ref>), a one-term rule for $\JackjLR_{\mu,\nu}^{\lambda}$ is given by

\[ \JackjLR_{\mu,\nu}^{\lambda} = \windfact_{R} \times \left( \JackjLR_{\mu_R,\nu}^{\lambda_R} \right)_R. \]

We show that the Pieri rule (Theorem (<ref>)) is a special case of the Window Lensing conjecture (<ref>), in the form of the Simple Lensing property (<ref>). We consider the earlier example ((<ref>)), however now we note that we can cancel several hooks in the fraction that appears within, to recover a simpler picture:

\begin{equation*}
\JackjLR_{75411,3}^{76431} =
\ytableausetup{boxsize=1.1em}
\frac{
\begin{ytableau}
*(boxL) L \\
*(boxG)& *(boxU) U  & *(boxU) U  & *(boxG) \\
*(boxL) L& *(boxU) U  & *(boxU) U  & *(boxL) L  &*(boxL) L \\
*(boxG) & *(boxU) U  &*(boxU) U  & *(boxG) & *(boxG)  & *(boxU) U & *(boxG)
\end{ytableau}
\quad
\begin{ytableau}
*(boxU) U & *(boxU) U & *(boxU) U
\end{ytableau}}{
\begin{ytableau}
*(boxL) L & *(boxU) U  & *(boxU) U \\
*(boxG)& *(boxU) U  & *(boxU) U  & *(boxG) \\
*(boxL) L  & *(boxU) U  & *(boxU) U  & *(boxL) L&*(boxL) L  & *(boxU) U \\
*(boxG)& *(boxU) U  &*(boxU) U  & *(boxG)  & *(boxG)  & *(boxU) U & *(boxG)
\end{ytableau}} \\
\end{equation*}

We split this up according to the window $R = \meetjoin(\lambda/\mu)$:

\begin{equation*}
\ytableausetup{boxsize=1.1em}
\frac{
\begin{ytableau}
*(boxL) L \\
*(boxG)& *(boxU) U  & *(boxU) U  & *(boxG) \\
*(boxL) L& *(boxS)   & *(boxS)   & *(boxL) L  &*(boxL) L \\
*(boxG) & *(boxU) U  &*(boxU) U  & *(boxG) & *(boxG)  & *(boxU) U & *(boxG)
\end{ytableau}
\quad
\begin{ytableau}
*(boxS)  & *(boxS) & *(boxS)
\end{ytableau}}{
\begin{ytableau}
*(boxL) L & *(boxS)  & *(boxS) \\
*(boxG)& *(boxU) U  & *(boxU) U  & *(boxG) \\
*(boxL) L  & *(boxS)  & *(boxS)   & *(boxL) L&*(boxL) L  & *(boxS)  \\
*(boxG)& *(boxU) U  &*(boxU) U  & *(boxG)  & *(boxG)  & *(boxU) U & *(boxG)
\end{ytableau}} \\
\end{equation*}

We see that this factor is precisely $\windfact_{R}$, as predicted by the conjecture.
The remaining windowed (i.e. beige) boxes are filled in with the de-windowed basic Pieri rule,

\begin{equation}
\JackjLR_{2,3}^{32} = \ytableausetup{boxsize=1.1em}
\frac{
\begin{ytableau}
*(boxU) U & *(boxU) U
\end{ytableau}
\quad
\begin{ytableau}
*(boxU) {U} & *(boxU) U & *(boxU) U
\end{ytableau}}{
\begin{ytableau}
*(boxU) U & *(boxU) U   \\
*(boxU) U & *(boxU) U & *(boxU) U
\end{ytableau}}. \\
\end{equation}

Consider the earlier Maximal Filling tableau example ((<ref>)) of $\lambda = 766654211$, $\mu = 7553322$, $\nu=542$. We will show this is also an example of the Window Lensing conjecture (<ref>). As in the previous example, we cancel out hooks that appear equally in the numerator and denominator and extract the minimal window $R=\meetjoin(\lambda/\mu)$

\begin{equation*}
\ytableausetup{boxsize=0.8em}
\frac{
\begin{ytableau}
*(boxU) & *(boxL)   \\
*(boxU) & *(boxL)   \\
*(boxU) & *(boxL)  & *(boxL)  \\
*(boxL) & *(boxL)  & *(boxL)   \\
*(boxU) & *(boxL)  & *(boxL)  & *(boxU) & *(boxU)  \\
*(boxL) & *(boxL)  & *(boxL)  & *(boxU) & *(boxL)  \\
*(boxU) & *(boxL)  & *(boxU)  & *(boxU) & *(boxU) & *(boxU) & *(boxL) \\
\end{ytableau} \quad
\begin{ytableau}
*(boxU)  &*(boxU)   \\
*(boxU)  &*(boxU)  &*(boxU)  &*(boxU)     \\
*(boxU)  &*(boxU)  &*(boxU)  &*(boxU)  &*(boxU)
\end{ytableau}}{
\begin{ytableau}
*(boxU)  \\
*(boxU)  \\
*(boxU) & *(boxL)   \\
*(boxL) & *(boxL)  & *(boxL)  & *(boxU) \\
*(boxU) & *(boxL)  & *(boxL)  & *(boxU) & *(boxU)\\
*(boxL) & *(boxL)  & *(boxL)  & *(boxU) & *(boxL) & *(boxU) \\
*(boxU) & *(boxL)  & *(boxL)  & *(boxU) & *(boxU) & *(boxU) \\
*(boxU) & *(boxL)  & *(boxU)  & *(boxU) & *(boxU) & *(boxU) \\
*(boxU) & *(boxL)  & *(boxU)  & *(boxU) & *(boxU) & *(boxU) & *(boxL)\\
\end{ytableau}}
\,\,\to\,\,
\frac{
\begin{ytableau}
*(boxU) & *(boxG)    \\
*(boxS) & *(boxL)   \\
*(boxS) & *(boxL)  & *(boxS)  \\
*(boxS) & *(boxL)  & *(boxS)   \\
*(boxS) & *(boxL)  & *(boxS)  & *(boxS) & *(boxS)  \\
*(boxS) & *(boxL)  & *(boxS)  & *(boxS) & *(boxS)  \\
*(boxU) & *(boxG)   & *(boxU)  & *(boxU) & *(boxU) & *(boxU) & *(boxG) \\
\end{ytableau} \quad
\begin{ytableau}
*(boxS)  &*(boxS)   \\
*(boxS)  &*(boxS)  &*(boxS)  &*(boxS)     \\
*(boxS)  &*(boxS)  &*(boxS)  &*(boxS)  &*(boxS)
\end{ytableau}}{
\begin{ytableau}
*(boxS)  \\
*(boxS)  \\
*(boxU) & *(boxG)  \\
*(boxS) & *(boxL)  & *(boxS)  & *(boxS) \\
*(boxS) & *(boxL)  & *(boxS)  & *(boxS) & *(boxS)\\
*(boxS) & *(boxL)  & *(boxS)  & *(boxS) & *(boxS) & *(boxS) \\
*(boxS) & *(boxL)  & *(boxS)  & *(boxS) & *(boxS) & *(boxS) \\
*(boxS) & *(boxL)  & *(boxS)  & *(boxS) & *(boxS) & *(boxS) \\
*(boxU) & *(boxG)  & *(boxU)  & *(boxU) & *(boxU) & *(boxU) & *(boxG) \\
\end{ytableau}}  \\
\end{equation*}

We see the right hand side agrees with the $\windfact_{R}$ factor.
The beige windowed region is then filled in with the following maximal filling rule

\begin{equation}\label{fig:stanelyfactor}
\JackjLR_{44221,542}^{5554311} =
\ytableausetup{boxsize=0.8em}
\frac{
\begin{ytableau}
*(boxU)   \\
*(boxU)  & *(boxL)  \\
*(boxL)  & *(boxL)   \\
*(boxU)   & *(boxL)  & *(boxU) & *(boxU)  \\
*(boxL)   & *(boxL)  & *(boxU) & *(boxL)  \\
\end{ytableau} \quad
\begin{ytableau}
*(boxU)  &*(boxU)   \\
*(boxU)  &*(boxU)  &*(boxU)  &*(boxU)     \\
*(boxU)  &*(boxU)  &*(boxU)  &*(boxU)  &*(boxU)
\end{ytableau}}{
\begin{ytableau}
*(boxU)  \\
*(boxU)  \\
*(boxL)   & *(boxL)  & *(boxU) \\
*(boxU)  & *(boxL)  & *(boxU) & *(boxU)\\
*(boxL)  & *(boxL)  & *(boxU) & *(boxL) & *(boxU) \\
*(boxU)   & *(boxL)  & *(boxU) & *(boxU) & *(boxU) \\
*(boxU)   & *(boxU)  & *(boxU) & *(boxU) & *(boxU) \\
\end{ytableau}}, \\
\end{equation}

where the maximal filling is now

\[ T_R =\ytableausetup{boxsize=1.0em} \begin{ytableau}
2  \\
1 \\
& 1 & 3 \\
&  & 2& 2\\
&    & 1 & 1 & 3 \\
&   & &  & 2 \\
&    &  &  & 1 \\
\end{ytableau}.\]

Indeed, we can show the example of (<ref>) holds in general.

The Window Lensing conjecture (<ref>) holds in the Maximal Filling Tableau case.

Firstly, we show that outside $R= \meetjoin(\lambda/\mu)$ the assignment rule (<ref>) yields a factor of $\windfact_{R}$. We note that $R$ can be described as the union of all columns satisfying $c_j=0$ with all rows satisfying $r_i = 0$. For each box in a column with $c_j=0$, the rule (<ref>) assigns all lower hooks. Of the boxes $b$ in such columns that are in rows with $r_i=0$ have $\lh{\lambda}{b} = \lh{\mu}{b_*} = \lh{\mu}{b} $ since for such boxes we have $b_*=b$. Since the partitions are identical in that row-column, these hooks factors cancel between the numerator and denominator. Thus each such column agrees with $\windfact_{R}$. For the remaining hook choices outside $R$, that is, those in rows with $r_i=0$ and in positions with $c_j>0$, we can just look at the transposed problem (as it it known that this prescription of hooks is invariant under transposition of all data), the solution to which is given by flipping all hooks, and thus the previous argument for columns with $c_j=0$ must apply now to rows with $r_i=0$, however with upper hooks instead of lower hooks. Thus, outside of R we assign hooks according to $\windfact_{R}$.

Next, we need to show that the assignment of hooks inside $R$ agrees with the same rule applied to $\lambda_R, \mu_R$. First we note that the maximal filling LR tableau agrees with its windowed version $T_R$. Hence the vectors of windowed row-column maximums $c^R, r^R$ , for the windowed LR tableau $T_R$ agree with the un-windowed $c,r$ however will all zeros removed. Thus, for each box $b' \coloneqq (j',i')$ in the windowed partitions $\lambda_R$, corresponding to the unwindowed box $b=(j,i)$ in $R$, we have the equivalent conditions $c^R_{j'} \leq r_{i'}^R \Leftrightarrow c_j \leq r_i$. Hence the windowed choice of hooks inside $\lambda_R$ agrees with the un-windowed choices inside $R \cap \lambda$, i.e. $d_{\lambda_R,T_R}(b')=d_{\lambda,T}(b)$, where $d_{\lambda,T} \in \{U,L\}$ represents the choice of hook for $C_{\lambda,T}$ in formula (<ref>), and $D_{\mu,T}(b) \coloneqq d_{\lambda,T}(b_*)$ be the corresponding choices for $\mu$.

So, to complete the proof we have to show that the hook choices all boxes $b'=(j',i') \in \mu_R$ also agree with those for their un-windowed counterparts $b=(j,i) \in \mu$, that is, we need to show that

\begin{equation}\label{eq:Dmu}
\end{equation}

or equivalently $d_{\lambda_R,T_R}((b')_*) = d_{\lambda,T}(b_*)$, where as before $b_* \coloneqq (j,i+c_j)$.

The complication is that we may have removed certain rows from $\mu$ that have $r_i=0$ (i.e, those in $R$), and so we have to check that these hook choices still agree after taking into consideration the vertical shift by $c_j$. In other words, it is not the case that $(b')_*=(b_*)' $ for all $b$. If indeed this was so, equation (<ref>) would follow immediately, since then we would have $d_{\lambda_R,T_R}((b')_*) =d_{\lambda_R,T_R}((b_*)') = d_{\lambda,T}(b_*)$. This complication is demonstrated in the following example of $\mu=44441, \nu = 431, \lambda = 5554321$.

\[ T := \begin{ytableau}
1& 2\\
& 1 &1 \\
&  & & \\
&  & & & 3 \\
&  & & & 2 \\
&  & &  &1 \\
\end{ytableau}, \quad T_R := \begin{ytableau}
1& 2\\
& 1 &1 \\
&  & &  3 \\
&  & &  2 \\
&  & &  1 \\
\end{ytableau} \]

\begin{equation*}
\ytableausetup{boxsize=1.0em}
\frac{
\begin{ytableau}
*(boxU) a\\
*(boxU) b& *(boxU) &*(boxU) &*(boxL)\\
*(boxU) c & *(boxU) &*(boxU)& *(boxL)\\
*(boxU) d& *(boxU) &*(boxL) &*(boxL) & \none \\
*(boxL) e& *(boxL) &*(boxL) &*(boxL) & \none \\
\end{ytableau}
\begin{ytableau}
*(boxU) b_*&*(boxU)  \\
*(boxU) c_*&  *(boxU) & *(boxU)  \\
*(boxU) d_*& *(boxU) &*(boxU) & *(boxL) \\
*(boxL) e_*& *(boxL) &*(boxL) &*(boxL) & *(boxU)  \\
*(boxU)&*(boxU)  & *(boxL)& *(boxL) &*(boxU)   \\
*(boxU)&*(boxU)  & *(boxU)& *(boxL) &*(boxU)   \\
\end{ytableau}} \leftrightarrow
\ytableausetup{boxsize=1.0em}
\frac{
\begin{ytableau}
*(boxU)c'&*(boxU)  &*(boxU)  \\
*(boxU) d'& *(boxU) & *(boxL)& \none \\
*(boxL) e'& *(boxL) & *(boxL)& \none \\
\end{ytableau} }{
\begin{ytableau}
\none \\
*(boxU)a'_*  \\
*(boxU)f & *(boxU) \\
*(boxU)g& *(boxU)  &*(boxU) \\
*(boxL)e'_*&*(boxL) &*(boxL) &  *(boxU) \\
*(boxU)& *(boxU) & *(boxL)& *(boxU)  \\
*(boxU)& *(boxU) & *(boxU) & *(boxU)  \\
\end{ytableau}}. \\
\end{equation*}

In this example, we find the following box correspondences,

\[ (a')_*=(a_*)', \quad  f= (b_*)'=(c')^*, \quad g=(c_*)' = (d')_* \quad (e')_*=(e_*)'. \]

In $\mu$ the row containing $b$ is removed, in $\lambda$ the row containing $d_*$ is removed. Here we see that the two boxes $c,d$ in $\mu$, ie, those that are in rows at up to $c_j$ below the row that is being removed (the one containing $b$), need to be shifted up by 1 row inside $\lambda$ for the correspondence to work out.

We will show that indeed the hook choices are consistent when one removes only a single row, and then by iteration the result will hold for all $R$. So let $R$ be the complement of a single row, $\tau$. All the considerations below are for boxes in a column with $c_j >0$.

We have shown that for boxes $x \in \mu$ at most $c_j$ rows beneath the row $\tau$ in $\mu$, which we say are close to $\tau$, we have

\[ (x_* + (0,1))' = (x')_*, \]

whereas for all other boxes one can easily show $(x_*)'=(x')_*$.
We have to check now that the rule (<ref>) assigns consistent hooks for $x$ close to $\tau$ across the four boxes $x, x', (x')_*$ and $x_* + (0,1)$.

For any $x$ that is between $1$ and $c(x)$ rows below $\tau$ in $\mu$, $x_* = x+(0,c(x))$ is at most $c(x)-1$ rows above the row $\tau$ in $\lambda$. Then let $y_*=x_*+(0,1)$ be the box that satisfies $(y_*)' = (x')_*$. We then know that $y_*$ is at most $c(x)$ rows above the row $\tau$ in $\lambda$.

Next, consider $z_* \in \lambda$ in the row $\tau$ in $\lambda$, for which we have $c(z_*)>0$ and $r(z_*) = 0$. We will show that $d_{\lambda,T}(z_*) = U = d_{\lambda,T}(z_*+(0,k) )$ for all $k \leq c(z_*)$. In particular this applies to $y_*$ from the previous paragraph. We recall that the maximal filling condition forces the the entries of the LR tableau to be increasing incrementally as we move up columns, and so because we know that $r(z_*)=0$ we must have $r(z_* + (0,k))\leq k$. Thus, for $k \leq c(z_*)$, we find $r(z_* + (0,k)) \leq c(z_*) = c(z_* + (0,k)) > 0$, and so $d_{\lambda,T}(z_*+(0,k) ) = U$.

Thus for $x$ close to $\tau$, we have shown the following agreement of hook assignments:

\begin{eqnarray*}
D_{\mu,T}(x) &=& d_{\lambda, T}(x_*) \\&=& d_{\lambda, T}(x_*+(0,1)) \\&=& d_{\lambda_R, T_R}((y_*)') \\&=& d_{\lambda_R, T_R}((x')_*) \\&=&  D_{\mu_R,T_R}(x')
\end{eqnarray*}

For $x$ not close to $\tau$, we have $(x')_* = (x_*)'$ and so the same agreement follows directly. Thus the theorem is proven.

§.§ Positivity Conjecture

With the constructions of the previous sections, we can propose a sharpening of the Stanley Conjecture (<ref>).

Let $\mu \subset \lambda$ be such that $\lambda/\nu \subset R$, for a window $R$. In terms of the positivity of Conjecture (<ref>), the factor $F_{R}$ accounts for all boxes outside $R$ in the following precise sense:

\begin{equation} \jackjLR_{\mu\nu}^\lambda \cdot \left( F_{\mu,R}^\lambda\right)^{-1} \cdot j_{\lambda,R} \in \setZ_{\geq0}[\alpha], \end{equation}

where $j_{\lambda,R} \coloneqq \prod_{b \in \lambda \cap R} \uh{\lambda}{b} \lh{\lambda}{b} $.

Note that $F_{R}$ is a rational function in $\alpha$ with numerator and denominator each a positive polynomial of equal degrees, and $j_{\lambda,R}$ is a positive polynomial of degree $2|\lambda \cap R|$, and so conjecture (<ref>) claims a stronger positivity than that of Conjecture (<ref>).

To demonstrate conjecture (<ref>), we take $\mu = 654321$, $\nu=211$, $\lambda=664432$. We find

\[ \jackjLR_{\mu,\nu}^{\lambda} = \frac{\alpha^2 (2+3\alpha) f(\alpha)}{2700 (1+\alpha)^{12} (3+2\alpha)(4+3\alpha)} \]

where $f(\alpha)$ is an irreducible degree 8 positive polynomial.
We take the minimal window $R$,

\begin{equation*}
\ytableausetup{boxsize=0.7em}R \coloneqq
\begin{ytableau}
{}& *(boxS)   & *(boxS)  & *(boxS)& & *(boxS) \\
& *(boxS)   & *(boxS)  & *(boxS)& & *(boxS) \\
& *(boxS)  & *(boxS) & *(boxS)& & *(boxS)\\
&  &  &&  &    \\
& *(boxS)   &*(boxS)   & *(boxS) & & *(boxS) \\
&   &  & & & \\
\end{ytableau}  \\
\end{equation*}

for which we find

\begin{equation*}
\windfact_{\mu,R}^\lambda = \frac{\begin{ytableau}
*(boxL)  \\
*(boxL)& *(boxS)   \\
*(boxL) & *(boxS)  & *(boxS) \\
*(boxG)  & *(boxU) & *(boxU)  & *(boxU)  \\
*(boxL) & *(boxS)   &*(boxS)   & *(boxS) & *(boxL) \\
*(boxG)  & *(boxU)  & *(boxU) & *(boxU)& *(boxG) & *(boxU)  \\
\end{ytableau}}{
\begin{ytableau}
*(boxL) & *(boxS)     \\
*(boxL)& *(boxS)   & *(boxS)   \\
*(boxL) & *(boxS)  & *(boxS) & *(boxS)\\
*(boxG)  & *(boxU) & *(boxU)  & *(boxU)  \\
*(boxL) & *(boxS)   &*(boxS)   & *(boxS) & *(boxL) & *(boxS) \\
*(boxG)  & *(boxU)  & *(boxU) & *(boxU)& *(boxG) & *(boxU)  \\
\end{ytableau}}
}, \quad F_{\mu,R}^\lambda = \frac{\hkl{1}{0}\hkl{2}{1}\hkl{3}{2}\hkl{5}{4}\hkl{2}{3}\hkl{1}{2}\hkl{0}{1}\hkl{4}{5}\hkl{3}{4}\hkl{2}{3}\hkl{1}{0}\hkl{0}{1} }{ \hkl{1}{1}\hkl{2}{2}\hkl{3}{3}\hkl{5}{5}\hkl{3}{3}\hkl{2}{2}\hkl{1}{1}\hkl{5}{5}\hkl{4}{4}\hkl{3}{3}\hkl{1}{1}\hkl{1}{1} }
\end{equation*}

where $\hkl{a}{b} \coloneqq a+b\alpha$. We also have

\[ j_{\lambda,R} = (\!\hkl{0}{1}\hkl{1}{0})^4(\!\hkl{1}{2}\hkl{2}{1})^2\hkl{2}{3}\hkl{3}{2}\hkl{4}{5}\hkl{5}{4}\hkl{3}{4}\hkl{4}{3}\hkl{2}{3}\hkl{3}{2}. \]

We check the window positivity conjecture

\[ \jackjLR_{\mu,\nu}^{\lambda} \cdot \left(F_{\mu,R}^\lambda\right)^{-1} \cdot j_{\lambda,R} = 4\alpha^4(2+\alpha)(1+2\alpha)(2+3\alpha) f(\alpha) \in \setZ_{\geq0}[\alpha]. \]

Note that the Stanley conjecture (<ref>) gives the weaker positivity property

\[ \jackjLR_{\mu,\nu}^{\lambda}\cdot  j_{\lambda} = (\text{  degree 32 polynomial in } \alpha \, ) f(\alpha) \in \setZ_{\geq0}[\alpha]. \]

[P. Hanlon's example <cit.>]
Take $\mu = 31$, $\nu=21$, $\lambda=421$. Here we have $c=2$, and the LR coefficient is given by

\[ \jackjLR_{31,21}^{421} = \frac{\alpha( 9+97\alpha+294\alpha^2 + 321\alpha^3 + 131\alpha^4+12\alpha^5 )}{3(1+\alpha)^3(2+\alpha)(1+2\alpha)^2(1+3\alpha)} \]

We take the minimal $R= \meetjoin(\lambda/\mu)$, which gives

\[ F_{R} = \frac{
\ytableausetup{boxsize=1.1em}
\begin{ytableau}
*(boxS)  \\
*(boxS) & *(boxS) & *(boxL) L    \\
\end{ytableau}
\begin{ytableau}
*(boxS) \\
*(boxS) & *(boxS)\\
*(boxS) & *(boxS) & *(boxL) L & *(boxS)   \\
\end{ytableau}} = \frac{1}{1+\alpha}.
\]

To verify conjecture (<ref>), we check

\[  \jackjLR_{31,21}^{421} \cdot F_{R}^{-1}\cdot j_{\lambda,R}=  4\alpha^4( 9+97\alpha+294\alpha^2 + 321\alpha^3 + 131\alpha^4+12\alpha^5 ), \]

\begin{eqnarray*}
j_{\lambda,R} &=& \hkl{0}{1}^3\hkl{1}{2}\hkl{2}{1}\hkl{1}{3}\hkl{2}{2}\hkl{2}{3}\hkl{3}{3} \\
&=& 12 \,\alpha^3(1+2\alpha)^2(2+\alpha)(1+3\alpha)(1+\alpha)^2.
\end{eqnarray*}

§.§ Further Consequences

In this section, we demonstrate more examples that provide evidence for the Window Lensing conjecture (<ref>).
We'll be using the base case Pieri rules

\begin{equation}\label{eqn:basecase}
\JackjLR_{1,1}^{1^2} = \ytableausetup{boxsize=1.1em}
\frac{
\begin{ytableau}
\end{ytableau}
\,\,
\begin{ytableau}
*(boxU) {}
\end{ytableau}}{
\begin{ytableau}
*(boxU)     \\
\end{ytableau}}, \qquad
\JackjLR_{1,1}^{2} = \ytableausetup{boxsize=1.1em}
\frac{
\begin{ytableau}
\end{ytableau}
\,\,
\begin{ytableau}
*(boxL) {}
\end{ytableau}}{
\begin{ytableau}
*(boxL)  &*(boxL)
\end{ytableau}}.
\end{equation}

We begin with two very simple examples.

We start with a simple example that is a case of the Pieri rule.
Let $\nu = 1$, $\mu = (1+s)1^p$, $\lambda = (1+s)1^{p+1}$. Here we have $c_{\mu\nu}^{\lambda}=1$. We write

\begin{equation} \lambda =\begin{ytableau}
\\
p \\
&  s \\
\end{ytableau},\quad \mu = \begin{ytableau}
p \\
& s   \\
\end{ytableau}, \quad \nu = \begin{ytableau}
{}   \\
\end{ytableau}
\end{equation}

Here, we can window $\lambda$ and $\mu$  with

\begin{equation}
R = \begin{ytableau}
*(boxS) &  \\
p  &\\
*(boxS) & s \\
\end{ytableau}
\end{equation}

This gives $\lambda_{R}=(1^2), \mu_{R}=(1)$. In this case, the window lensing conjecture claims

\begin{equation}
\JackjLR_{(1+s)1^p,1}^{(1+s)1^{p+1}} = \bm{F}_{R} \times (\JackjLR_{1,1}^{1^2})_R
\end{equation}

Evaluating this, using the base case (<ref>), we find

\begin{equation}\label{fig:2paramuuchannel}
\JackjLR_{(1+s)1^p,1}^{(1+s)1^{p+1}} = \frac{
\begin{ytableau}
*(boxU) p \\
*(boxS) & *(boxL) s   \\
\end{ytableau}}{
\begin{ytableau}
*(boxS) \\
*(boxU) p  \\
*(boxS) & *(boxL) s \\
\end{ytableau}
\times
\frac{
\begin{ytableau}
p \\
*(boxU) &  s   \\
\end{ytableau}\,\,
\begin{ytableau}
\none \\
*(boxU)    \\
\end{ytableau}
\begin{ytableau}
*(boxU)  \\
p  \\
*(boxU)  &  s \\
\end{ytableau}
} =
\frac{
\begin{ytableau}
*(boxU) p \\
*(boxU) & *(boxL) s   \\
\end{ytableau}\,\,
\begin{ytableau}
\none \\
*(boxU)    \\
\end{ytableau}
\begin{ytableau}
*(boxU)  \\
*(boxU) p  \\
*(boxU)  & *(boxL) s \\
\end{ytableau}
}. \end{equation}

As expected, this answer agrees with the known result from the Pieri rule (<ref>).

Let $\mu = (1+N)1^M$, $\nu = (1+s)1^p$, $\lambda = (1+N+s)1^{M+p+1}$. Here we have $c_{\mu\nu}^{\lambda}=1$.

\begin{equation}
\lambda = \begin{ytableau}
\\
p \\
M \\
& N &  s \\
\end{ytableau}, \quad \mu=
\begin{ytableau}
M \\
& N   \\
\end{ytableau} , \quad \nu=
\begin{ytableau}
p \\
& s   \\
\end{ytableau}.
\end{equation}

We window $\mu$ and $\lambda$ with

\begin{equation}
R_1 = \begin{ytableau}
*(boxS) &  & *(boxS)\\
*(boxS) p &  & *(boxS)\\
M & & \\
*(boxS) & N &  *(boxS) s \\
\end{ytableau}.
\end{equation}

We have $\mu_{R_1} = (1)$, $\lambda_{R_1} = (1+s)1^{p+1}$, and Window Lensing says that

\begin{equation} \JackjLR_{\mu,\nu}^{\lambda} = \windfact_{R_1} \left( \JackjLR_{\mu_{R_1},\nu}^{\lambda_{R_1}}
\right)_{R_1}.
\end{equation}

Now for the inner argument, $\JackjLR_{\mu_{R_1},\nu}^{\lambda_{R_1}}$, we use the results from Example (<ref>), and window on the arguments $\nu, \lambda_{R_1}$,

\begin{equation}\label{fig:2paramuuchannel}
\JackjLR_{(1+N)1^M, (1+s)1^p}^{(1+N+s)1^{M+p+1}} =\frac{
\begin{ytableau}
*(boxU) M \\
*(boxS) & *(boxL) N  & \none \\
\end{ytableau}
\begin{ytableau}
*(boxS)  \\
*(boxS) p \\
*(boxU) M \\
*(boxS) & *(boxL) N & *(boxS) s \\
\end{ytableau}
}\times \frac{
\begin{ytableau}
M \\
*(boxU) &  N   \\
\end{ytableau} \,
\begin{ytableau}
*(boxU) p \\
*(boxU) & *(boxL) s   \\
\end{ytableau}
\begin{ytableau}
*(boxU)  \\
*(boxU) p \\
M \\
*(boxU) & N & *(boxL) s \\
\end{ytableau}
\frac{
\begin{ytableau}
*(boxU) M \\
*(boxU) & *(boxL) N   \\
\end{ytableau} \,
\begin{ytableau}
*(boxU) p \\
*(boxU) & *(boxL) s   \\
\end{ytableau}
\begin{ytableau}
*(boxU)  \\
*(boxU) p \\
*(boxU) M \\
*(boxU) & *(boxL) N & *(boxL) s \\
\end{ytableau}
} \end{equation}

This answer agrees with known results.

In this previous example, we see the composition of multiple window lensing

\begin{equation} \JackjLR_{\mu\nu}^{\lambda} = \windfact_{R_1} \left( \windfact_{R_2} \left(
\JackjLR_{\mu_{R_1},\nu_{R_2}}^{\lambda_{R_1\circ R_2}} \right)_{R_2}\right)_{R_1}.
\end{equation}

When composing multiple windows $R=R_1\circ R_2$, we define the composite window factor

\begin{equation}
\windfact_{R} \coloneqq \windfact_{R_1} \left( \windfact_{R_2} \right)_{R_1}.
\end{equation}

We similarly define a composite de-windowing operation

\[ (\JackjLR)_{R} := \left((\JackjLR)_{R_2}\right)_{R_1} . \]

The examples (<ref>) and (<ref>) combine together to give the composite de-windowing

\begin{equation}
\JackjLR_{(1+N)1^M, (1+s)1^p}^{(1+N+s)1^{M+p+1}}   = \frac{
\begin{ytableau}
*(boxU) M \\
*(boxS) & *(boxL) N   \\
\end{ytableau} \,
\begin{ytableau}
*(boxU) p \\
*(boxS) & *(boxL) s   \\
\end{ytableau}
\begin{ytableau}
*(boxS)  \\
*(boxU) p \\
*(boxU) M \\
*(boxS) & *(boxL) N & *(boxL) s \\
\end{ytableau}
}  \times \left(\JackjLR_{1,1}^{1^2} \right)_R.
\end{equation}

These previous examples are a warm up for the following non-trivial consequence of the Window Lensing conjecture. This example will be the focus of the rest of this article.

Consider the 6 parameter $(K,M,N,T,p,s)$ family of problems given by $\lambda = (2+K+N+s+1)(K+2)^{M+1}(1)^{T+p+1}$, $\mu = (K+2+N)(K+2)^M(K+1)(1)^T$. and $\nu = (s+2)(1)^{p+1}$.

\begin{equation}
\lambda =
\begin{ytableau}
\\
p \\
&  & \\
M &  & \\
& K & & N & s &  \\
\end{ytableau},\quad
\mu = \begin{ytableau}
&  &\none\\
M &  & \\
& K &  &  N  \\
\end{ytableau}, \quad
\nu = \begin{ytableau}
\none \\
p \\
\\
&  &  s  \\
\end{ytableau}.
\end{equation}

This family has $c_{\mu\nu}^{\lambda}=2$.
We first apply the window $R_1$ to $\lambda$ and $\mu$, and then the second window $R_2$ to $\lambda_{R_1}$ and $\nu$, where the windows are given by

\begin{equation}\label{eq:6paramwindows}
*(boxS) &  & *(boxS)&  & *(boxS) & *(boxS)\\
*(boxS) p &  & *(boxS)&  & *(boxS) & *(boxS)\\
T &&&&&\\
*(boxS) &  & *(boxS)&  & *(boxS) & *(boxS)\\
M &  & &&& \\
*(boxS) &  K &*(boxS)  &   N & *(boxS)  s  & *(boxS) \\
\end{ytableau}, \quad
*(boxS) & *(boxS)&  & *(boxS)\\
p &&& \\
*(boxS) & *(boxS)  &  & *(boxS)\\
*(boxS)  & *(boxS)  &   s  & *(boxS) \\
\end{ytableau}.
\end{equation}

For the composite window $R=R_1\circ R_2$ we get $\mu_R = 21$, $\nu_R=21$, and $\lambda_R=321$, and the composite window factor and the composite de-windowing are given by

\begin{equation}\label{eq:6paramwindow}
\windfact_{R} = \frac{\begin{ytableau}
*(boxU)  T\\
*(boxS) &*(boxL)  &\none\\
*(boxU)M & *(boxG) &*(boxU) \\
*(boxS) &*(boxL) K &*(boxS)  & *(boxL) N  \\
\end{ytableau} \,\, \begin{ytableau}
\none \\
*(boxU) p \\
*(boxS)  \\
*(boxS) & *(boxS) & *(boxL)  s  \\
\end{ytableau}
*(boxS)  \\
*(boxU) p \\
*(boxU)  T\\
*(boxS) &*(boxL)  & *(boxS)\\
*(boxU)M & *(boxG) &*(boxU) \\
*(boxS) &*(boxL) K &*(boxS)  & *(boxL) N & *(boxL)  s & *(boxS)  \\
\end{ytableau}},
\qquad
\left(\frac{\begin{ytableau}
A  \\
B & C
\end{ytableau}\,\, \begin{ytableau}
D \\
E & F
\end{ytableau}
j  \\
G &  k \\
H & I &  \ell
\end{ytableau}}\right)_{R} = \frac{\begin{ytableau}
*(boxS) A& &\none\\
M & & \\
*(boxS) B& K &*(boxS) C  & N  \\
\end{ytableau} \,\, \begin{ytableau}
\none \\
p \\
*(boxS) D \\
*(boxS) E& *(boxS)F & s  \\
\end{ytableau}
*(boxS) j \\
p \\
*(boxS)G & & *(boxS)k\\
M &  & \\
*(boxS)H & K &*(boxS) I & N &  s & *(boxS)\ell  \\
\end{ytableau}}.
\end{equation}

If we assume the Window Lensing Conjecture (<ref>), then there exists a rule $\JackjLR_{21,21}^{321} \in \hookspace_{21,21}^{321}(\BZ)$ that simultaneously solves the 6 parameter family (<ref>) of Jack LR coefficients using the composite window $R$ in formula (<ref>). That is,

\begin{equation}\label{eqn:6parammainformula}
\JackjLR_{\mu\nu}^{\lambda} = \windfact_{R} \times (\JackjLR_{21,21}^{321})_{R}.
\end{equation}

So far in this article, we have discussed windowing as a way of reducing to the calculation of LR coefficients to simpler problems. Now, we reverse that previous perspective and view the windowed partitions as fixed, and consider varying the de-windowing via $R$. As we vary the 6 parameters $(K,M,N,T,p,s)$ characterizing $R$, we can use other methods (such as directly computing the Jacks, or by using the recursive technique of <cit.>) to compute the values of $\jackjLR_{\mu\nu}^{\lambda}(K,M,N,T,p,s)$. Then the system (<ref>) gives an overdetermined system of linear equations for the rule $\JackjLR_{21,21}^{321}$, that is, for the integer coefficients $c_\Phi$ within, according to the General Structure Conjecture (<ref>). In particular, we looking for a rule of the form

\begin{equation}
\JackjLR_{21,21}^{321} = \sum_{\Phi \in \hookspace_{21,21}^{321}} c_\Phi\,\bm{G}_\Phi \, \in \hookspace_{21,21}^{321}(\BZ) , \quad \text{ with } c_\Phi \in \BZ.
\end{equation}

The set $\hookspace_{21,21}^{321}$ has $2^{12}$ elements, the channels $\Phi := {}^{\sA\sB\sC;\sD\sE\sF}_{\sG\sH\sI jk\ell}  \in \hookspace_{21,21}^{321} \cong \{U,L\}^{12}$. The corresponding channel factors (c.f (<ref>))

\begin{equation}
\bm{G}_\Phi \equiv \bm{G}^{\sA\sB\sC;\sD\sE\sF}_{\sG\sH\sI jk\ell} \coloneqq \frac{\begin{ytableau}
A  \\
B & C
\end{ytableau}\,\, \begin{ytableau}
D \\
E & F
\end{ytableau}
j  \\
G &  k \\
H & I &  \ell
\end{ytableau}}.
\end{equation}

When such a rule is de-windowed and evaluated, the hook lengths of the symbols at $j,k,\ell$  are indistinguishable, as they all either $1$ or $\alpha$. Hence, we split off these three, and write the de-windowed rule over the $2^9$ channels $\phi := {}^{\sA\sB\sC;\sD\sE\sF}_{\sG\sH\sI} \in \hookspace_{21,21}^{21} \cong \{U,L\}^{9}$, as

\begin{equation}\label{eq:c2rule}
[\JackjLR_{21,21}^{321}]_R = \sum_{\phi \in \hookspace_{21,21}^{21}} G_{\phi,R}\, \frac{d_\phi(\alpha)}{\alpha^3}, \,\, \text{ with } d_\phi (\alpha)=\sum_{i=0}^3 d_\phi^i\alpha^i \,\, \text{ and }  d_\phi^i \in \BZ,
\end{equation}

That is, only $G_{\phi,R} := [\bm{G}_\phi]_R$ depends on the de-windowed variables $(K,M,N,T, p,s)$, and the variables $d_\phi^i$ track the number of $L$-hooks out of $j,k,\ell$, e.g. $d_\phi^1$ counts choices where exactly two of $j,k,\ell$ are upper hooks.

We also consider two operations on the set of channels. For $\phi = {}^{\sA\sB\sC;\sD\sE\sF}_{\sG\sH\sI}$, define

\begin{equation}\label{eq:channelops}
\phi^T \coloneqq {}^{\sC\sB\sA;\sF\sE\sD}_{\sI\sH\sG},\qquad \bar\phi \coloneqq {}^{\bar \sA\bar \sB\bar \sC;\bar \sD\bar \sE\bar \sF}_{\bar \sG\bar \sH\bar \sI}.
\end{equation}

We can check computationally[For these computations we used P. Alexandersson's Mathematica packages <cit.> to solve for a large subset of the 6-parameter space of the Jack LR coefficients. Then the overdetermined system could be solved exactly. The supporting Mathematica code can be supplied upon request.] that the equation (<ref>) has a solution, a result that is our main supporting evidence for the Window Lensing conjecture.

The following coefficients $d_\phi = d^{\sA\sB\sC;\sD\sE\sF}_{\sG\sH\sI}$ for the de-windwoed rule (<ref>) give a solution to equation (<ref>). These coefficients satisfy $d_\phi = d_{ \phi^T}$.
There are only 20 non-zero coefficients, given by

$\phi \quad\quad \quad \,\, \phi^T$     $d_\phi $

${}^{\sL\sU\sU;\sU\sU\sL}_{\sU\sL\sU} \,,\, {}^{\sU\sU\sL;\sL\sU\sU}_{\sU\sL\sU}$      $\alpha$

${}^{\sL\sU\sU;\sL\sU\sU}_{\sU\sL\sU} \,,\, {}^{\sU\sU\sL;\sU\sU\sL}_{\sU\sL\sU} $      $(1-2\alpha)\alpha$

${}^{\sL\sU\sU;\sL\sU\sL}_{\sU\sL\sU} \,,\, {}^{\sU\sU\sL;\sL\sU\sL}_{\sU\sL\sU}$      $(-2+\alpha)\alpha$

${}^{\sL\sU\sU;\sL\sL\sL}_{\sU\sL\sU} \,,\, {}^{\sU\sU\sL;\sL\sL\sL}_{\sU\sL\sU} $     $\alpha$

${}^{\sL\sU\sU;\sL\sU\sU}_{\sU\sL\sL} \,,\, {}^{\sU\sU\sL;\sU\sU\sL}_{\sL\sL\sU}$     $ \alpha^2$

${}^{\sL\sU\sL;\sU\sU\sU}_{\sU\sU\sU}$      $(-2+\alpha)\alpha^2$

${}^{\sL\sU\sL;\sU\sL\sU}_{\sU\sU\sU}$      $(1-2\alpha)\alpha^2$

${}^{\sL\sU\sL;\sU\sL\sL}_{\sU\sU\sU} \,,\, {}^{\sL\sU\sL;\sL\sL\sU}_{\sU\sU\sU}$     $\alpha^2$

${}^{\sL\sU\sL;\sU\sU\sU}_{\sU\sL\sU}$     $\alpha$

${}^{\sL\sU\sL;\sU\sU\sL}_{\sU\sL\sU} \,,\, {}^{\sL\sU\sL;\sU\sU\sL}_{\sU\sL\sU}$     $\alpha(-2+\alpha)$

${}^{\sL\sU\sL;\sL\sU\sL}_{\sU\sL\sU}$     $-2\alpha(-2+\alpha)$

${}^{\sL\sU\sL;\sL\sL\sL}_{\sU\sL\sU}$     $-2\alpha$

${}^{\sL\sL\sL;\sU\sU\sU}_{\sU\sU\sU}$     $\alpha$

That is, the above solution solves

\begin{equation}\label{eq:6paramfull}
g_{\mu\nu}^\lambda = F_{R} \times \sum_{\phi \in \hookspace_{21,21}^{21}} G_{\phi,R} \,\frac{d_\phi(\alpha)}{\alpha^3}.
\end{equation}

for all values of the 6 parameters. Note that, as expected, we have

\begin{equation}
c_{\mu\nu}^\lambda = \sum_\phi d_\phi(1) = 2.
\end{equation}

§.§ Redundancy
Although we claim to have found a solution (<ref>) to the equation (<ref>), each such solution comes with a large degree of redundancy. First, we propose a discrete $\setZ_2$ symmetry corresponding to channel conjugation $\Phi \to \bar \Phi$.

If a solution $\{d_\phi(\alpha)\}$ is given to equation (<ref>), then the set of channel solutions $\{d'_{\phi}(\alpha)\}$ given by

\begin{equation}
d'_{\phi}(\alpha) \coloneqq \alpha^{3} d_{\bar \phi}(\alpha^{-1}),
\end{equation}

is another solution. Here we have used the operation (<ref>).

Next, we show there is a large kernel $K_\phi$ solving the homogeneous equation

\begin{equation} \label{eq:kernel}
0 = \sum_{\phi \in \hookspace_{21,21}^{21}} [\bm{G}_{\phi} ]_R\, K_\phi.\end{equation}

Recall $\hh{\sigma}{a}{L} = \hh{\sigma}{a}{U}+\beta$, where $\beta \coloneqq 1-\alpha$.

For any partition $\sigma$, and boxes $a,b,c \in \sigma$, define

\begin{equation} k_{\sigma,abc}^{\sA\sB\sC} \coloneqq \hh{\sigma}{a}{A}-\hh{\sigma}{b}{B}+\hh{\sigma}{c}{C}.
\end{equation}

Note we have the symmetry $k_{\sigma,abc}^{\sA\sB\sC} = k_{\sigma,abc}^{\sC\sB\sA}$. For the three boxes

b     c
of $\sigma = 21$,

we define $k^{\sA\sB\sC}:=k_{21,abc}^{\sA\sB\sC}$, which is

\begin{equation}\label{eq:kvector}
\begin{split}
k^{\sL\sL\sL}= \alpha, \, k^{\sL\sL\sU}=-1, \, &\, k^{\sL\sU\sL} = 1-2\alpha,\,  k^{\sL\sU\sU}= \alpha, \\
k^{\sU\sU\sU}=-1, \, k^{\sU\sU\sL}=\alpha,\, & \, k^{\sU\sL\sU}=2-\alpha, \,  k^{\sU\sL\sL}=-1.
\end{split}
\end{equation}

Note that we have the symmetry  $-\alpha\, k^{\sA\sB\sC} (\alpha^{-1}) = k^{\bar \sA\bar \sB\bar \sC}(\alpha)$.

The following identity involving these $k$ vectors be useful for describing the kernel.

For any partition $\sigma$, and boxes $a,b,c \in \sigma$, we have

\begin{equation}\label{eq:kident}
\sum_{A,B,C} \hh{\sigma}{a}{A}\hh{\sigma}{b}{B}\hh{\sigma}{c}{C} \times k^{\bar \sA\bar \sB \bar \sC} = \left( -k_{\sigma,abc}^{\sU\sU\sU}+k^{\sU\sU\sU}\right)\beta^3 .
\end{equation}

We expand $\hh{\sigma}{a}{L} = \hh{\sigma}{a}{U}+\beta$, to get

\begin{eqnarray*}
\sum_{A,B,C} \hh{\sigma}{a}{A}\hh{\sigma}{b}{B}\hh{\sigma}{c}{C} \times k^{\bar \sA\bar \sB \bar \sC} &=&  \hh{\sigma}{a}{U}\hh{\sigma}{b}{U}\hh{\sigma}{c}{U}(\sum_{A,B,C}  k^{\bar \sA\bar \sB \bar \sC}) \\
+ \hh{\sigma}{a}{U}\hh{\sigma}{b}{U} (\sum_{A,B}  k^{\bar \sA\bar \sB \sU}) \beta &+& \hh{\sigma}{a}{U}\hh{\sigma}{c}{U} (\sum_{A,C}  k^{\bar \sA\sU\bar \sC })\beta+ \hh{\sigma}{b}{U}\hh{\sigma}{c}{U} (\sum_{B,C}  k^{\sU\bar \sB\bar \sC})  \beta \\
+ \hh{\sigma}{a}{U}  (\sum_{A}  k^{\bar \sA \sU\sU})\beta^2  &+& \hh{\sigma}{b}{U}(\sum_{B}  k^{\sU\bar \sB\sU})\beta^2 + \hh{\sigma}{c}{U} (\sum_{C}  k^{\sU\sU\bar \sC }) \beta^2\\
& +& k^{\sU\sU\sU}\beta^3.
\end{eqnarray*}

We then notice that

\[ \hh{\sigma}{a}{U}  (\sum_{A}  k^{\bar \sA \sU\sU})\beta^2  + \hh{\sigma}{b}{U}(\sum_{B}  k^{\sU\bar \sB\sU})\beta^2 + \hh{\sigma}{c}{U} (\sum_{C}  k^{\sU\sU\bar \sC }) \beta^2 = -k_{\sigma,abc}^{\sU\sU\sU}\beta^3\]

and all the othe sums vanish.

We now express a large subspace of the kernel (<ref>) in terms of the vector $k$ given by (<ref>).

The kernel equation (<ref>) has solutions of the following form

\begin{equation}\label{eq:secondkernel}
\sum_{A,B,C \in \{U,L\}^3} \left[ \bm{G}_{\phi^{\sA\sB\sC}}\right]_R \times k^{\bar \sA \bar \sB \bar \sC}(\alpha) = 0,
\end{equation}

where ${\phi}^{\sA\sB\sC} \in \hookspace_{21,21}^{21}$ equal to any one of the following 6 sets of 8 channels

\begin{equation}\label{eq:kernchannelsets}
\frac{ \begin{ytableau}
\sA  \\
\sB  &  \sC
\end{ytableau}\,\,\begin{ytableau}
\\
\end{ytableau} }{ \begin{ytableau}
&*(boxGr) \\
&   &*(boxGr)
\end{ytableau} },
\frac{ \begin{ytableau}
\\
\end{ytableau}\,\,\begin{ytableau}
\sA  \\
\sB  &  \sC
\end{ytableau} }{ \begin{ytableau}
&*(boxGr) \\
&   &*(boxGr)
\end{ytableau} } ,
\frac{ \begin{ytableau}
\\
\end{ytableau}\,\,\begin{ytableau}
\\
\end{ytableau} }{ \begin{ytableau}
\sA & *(boxGr)\\
\sB  &  \sC   &*(boxGr)
\end{ytableau} },
\frac{ \begin{ytableau}
\sA  \\
\end{ytableau}\,\,\begin{ytableau}
\sC   \\
\end{ytableau} }{ \begin{ytableau}
\bar \sB  &*(boxGr) \\
&   &*(boxGr)
\end{ytableau} },
\frac{ \begin{ytableau}
\\
\sA  &
\end{ytableau}\,\,\begin{ytableau}
\\
\sC  &
\end{ytableau} }{ \begin{ytableau}
& *(boxGr)\\
\bar \sB  &   &*(boxGr)
\end{ytableau} },
\frac{ \begin{ytableau}
\\
&  \sA
\end{ytableau}\,\,\begin{ytableau}
\\
&  \sC
\end{ytableau} }{ \begin{ytableau}
&*(boxGr) \\
& \bar \sB   &*(boxGr)
\end{ytableau} }.
\ytableausetup{boxsize=1.1em}
\end{equation}

where the unfilled, ungreyed boxes are assigned any fixed hook choices, and the greyed boxes are ignored.

We only show this for the first and third sets of the 6 channel sets of (<ref>), and the rest follow similarly. For the first, we need to show

\begin{equation}\label{eq:firstkernel}
\ytableausetup{boxsize=1.0em}
\sum_{A,B,C} \left[ \begin{ytableau}
\sA  \\
\sB & \sC
\end{ytableau}\right]_R \times k^{\bar \sA\bar \sB \bar \sC} = 0.
\end{equation}

We notice that (for $\mu=21$)

\[ \left[ \begin{ytableau}
\sA  \\
\sB & \sC
\end{ytableau}\right]_R = k_{\mu,abc}^{\sA\sB\sC}.\]

After de-windowing we find $\uh{\mu}{b}=\uh{\mu}{a}+\uh{\mu}{c}+1$, and hence $k_{\mu,abc}^{\sU\sU\sU }=-1$.
Thus $-k_{\mu,abc}^{\sU\sU\sU }+k^{\sU\sU\sU} = 0$, and so using (<ref>) we find that equation (<ref>) holds.

Similarly, for the third channel set of (<ref>), we need to show

\begin{equation}\label{eq:lambdaker}
\sum_{A,B,C}\frac{1}{ \left[ \begin{ytableau}
*(boxGr) \\
\sA & *(boxGr) \\
\sB & \sC & *(boxGr)
\end{ytableau}\right]_R  } \times k^{\bar \sA\bar \sB \bar \sC} = 0.
\end{equation}

We find that for $\lambda=321$ (with ungreyed boxes $a,b,c \in \lambda$), this gives

\[\uh{\lambda}{b}=\uh{\lambda}{a}+\uh{\lambda}{c}-\alpha, \]

and thus $k_{\lambda,abc}^{\sL\sL\sL } = -1$.

We then write equation (<ref>) as

\begin{eqnarray*}
\sum_{A,B,C} \frac{1}{\hh{\lambda}{a}{A}\hh{\lambda}{b}{B}\hh{\lambda}{c}{C}} \times k^{\bar \sA\bar \sB \bar \sC} &=& \frac{\sum_{A,B,C} \hh{\lambda}{a}{\bar A}\hh{\lambda}{b}{\bar B}\hh{\lambda}{c}{\bar C} \times k^{\bar \sA\bar \sB \bar \sC}}{\hh{\lambda}{a}{U}\hh{\mu}{a}{L}\hh{\lambda}{b}{U}\hh{\lambda}{b}{L}\hh{\lambda}{c}{U}\hh{\lambda}{c}{L}} \\
&=&\frac{\left( -k_{\lambda,abc}^{\sL\sL\sL}+k^{\sU\sU\sU}\right)\beta^3}{\hh{\lambda}{a}{U}\hh{\lambda}{a}{L}\hh{\lambda}{b}{U}\hh{\lambda}{b}{L}\hh{\lambda}{c}{U}\hh{\lambda}{c}{L}} .
\end{eqnarray*}

to see that (<ref>) holds. The rest follow using the same technique.

We now count the dimension of this subspace of the kernel. For each of the six sets of channels, we can make $2^6$ choices on how to fill the remaining 6 ungreyed boxes. However, each of the $k$ vectors (formula (<ref>)) is a linear polynomial in $\alpha$, so at most we can multiply these kernel solutions by quadratic integer polynomials in $\alpha$ (of dimension 3) to remain in the space of integer polynomials in $\alpha$ of degree of most 3, where the $d_\phi(\alpha)$ reside. Thus, this kernel subspace is $3 \times 6 \times 2^6$ dimensional over $\setZ$. On the other hand, there are $4 \times 2^9=2^{11}$ integer variables $d_\phi^i$.

The relations of Lemma (<ref>) exhaust the kernel (<ref>).

With this large kernel, we expect that there exists a more meaningful solution than the one given in (<ref>), one whose coefficients $d_\phi$ have a combinatorial interpretation. It is this path investigration that we propose might lead to insight into the Stanley conjecture. We will follow up on this in a future work.

§ FACTORIZATION

In this section, we consider the situation where the window $R$  splits the skew shape $\lambda/\mu$ into two components $\lambda_1/\mu_1$, and $\lambda_2/\mu_2$. Bravi-Gandini <cit.>, begin the investigation of how the skew Jack functions $J_{\lambda/\mu}$ (and hence the Jack LR coefficient $\jackjLR_{\mu\nu}^{\lambda}$) factorize under such a splitting, which serves as the motivation for what is to come.

In the Schur case, the situation is quite simple.

\begin{equation}\label{eq:schurfacttwo}
c_{\mu\nu}^{\lambda} = \sum_{\nu_1,\nu_2} c_{\mu_1,\nu_1}^{\lambda_1} c_{\mu_2,\nu_2}^{\lambda_2} \cdot c_{\nu_1,\nu_2}^{\nu}.
\end{equation}

Follows from the factorization of skew Schur functions $s_{\lambda/\mu} = s_{\lambda_1/\mu_2} s_{\lambda_2/\mu_2}$.

Bravi-Gandini give an extension of this result to skew Jack functions in a special case.

Let $\mu$ be a rectangular partition, with $\mu \subset \lambda$. Suppose $\lambda/\mu$ has two connected components, whose shapes are those of partitions $\lambda_1$ and $\lambda_2$. Suppose $\nu$ satisifes $c_{\mu, \nu}^{\lambda} = 1$ and $c_{\lambda_1,\lambda_2}^{\nu}=1$, then we have[We have left in the two factors of $\jackjLR_{\emptyset,\lambda_i}^{\lambda_i}=1$ to express similarity to equation (<ref>)]

\begin{equation}\label{eq:basicfact}
\jackjLR_{\mu, \nu}^{\lambda} =\prod_{b\in \mu} \frac{  h^\bullet_\mu(b) }{  h^\bullet_\lambda(b)} \times  \jackjLR_{\emptyset,\lambda_1}^{\lambda_1}\,\jackjLR_{\emptyset,\lambda_2}^{\lambda_2} \cdot f^{\lambda_1,\lambda_2}_{\nu},
\end{equation}

where we have defined

\begin{equation}
\defin{ f^{\mu,\nu}_{\lambda}} \coloneqq \jackjLR_{\mu,\nu}^{\lambda} \frac{j_\lambda}{j_\mu j_\nu},
\end{equation}

and where the fraction of hooks in (<ref>) has the following property: upper (and hence lower) hooks appear the same number of times in the numerator and denominator - a property we call balanced.

\begin{equation*}
\jackjLR_{2,211}^{321} =
\ytableausetup{boxsize=1.1em}
\frac{
\begin{ytableau}
*(boxU) U & *(boxL) L  \\
\end{ytableau}
\begin{ytableau}
\\
& \\
*(boxL) L & *(boxU) U &   \\
\end{ytableau}} \times f^{1,21}_{211} \\
\end{equation*}

Note here that the hook choices for the numerator and denominator do not agree.

\begin{equation*}
\jackjLR_{222,2211}^{43221} =
\ytableausetup{boxsize=1.1em}
\frac{
\begin{ytableau}
*(boxU) U & *(boxU) U \\
*(boxU) U & *(boxL) L  \\
*(boxL) L &*(boxL) L  \\
\end{ytableau}
\begin{ytableau}
\\
& \\
*(boxU) U & *(boxU) U  \\
*(boxL) L & *(boxU) U &   \\
*(boxL) L & *(boxL) L &  &  \\
\end{ytableau}} \times f^{21,21}_{2211} \\
\end{equation*}

Note that this result is compatible with windowing.

§.§ Conjectures

In this section, we assume the earlier conjectures regarding windowing and we speculate about a window-compatible generalization of the basic factorization result (<ref>). One might expect such a generalization to take the form:

\begin{equation}
\jackjLR_{\mu\nu}^{\lambda}  = F_{R} \times \sum_{\nu_1,\nu_2} \left[ \prod_{b\in \mu_R} \frac{  \bm{h}^\bullet_{\mu_R}(b) }{  \bm{h}^\bullet_{\lambda_R}(b)} \right]_R \times \jackjLR_{\mu_1,\nu_1}^{\lambda_1}\jackjLR_{\mu_2,\nu_2}^{\lambda_2} f_{\nu}^{\nu_1,\nu_2},
\end{equation}

mimicing the form of (<ref>) and (<ref>). We conjecture that a formula similar to this holds.

Let $R$ be a window (potentially composite).
Let $\kappa$ be a factorization region, that is, a partition satisfying $\kappa \subseteq \mu_R \subset \lambda_R$, with the property that $\lambda_R/\kappa$ has multiple components $\{\lambda_i\}$, and similarly let $\{\mu_j\}$ be the components of $\mu_R/\kappa$ (if they exist). We conjecture that the Jack LR coefficient can be expanded in the following form

\begin{equation}
\jackjLR_{\mu\nu}^{\lambda}  = F_{R} \sum_{\Phi \in \hookspace_\kappa} G_{\Phi,R}^{\kappa} \times H_{\Phi}(\vec{v}),
\end{equation}

* $\Phi \in \hookspace_\kappa \coloneqq \hookspace_{\kappa,\emptyset}^{\kappa}$ are a set of factorization channels that are independent of the window $R$.

* To each channel $\Phi$ is associated a channel factor $\bm{G}^\kappa_\Phi$ (compare to (<ref>)), which is the product of hooks for each box $b \in \kappa$ inside both $\lambda_R$ and $\mu_R$.

\begin{equation}
\bm{G}^\kappa_{\Phi} = \prod_{b \in \kappa} \frac{\ahc{\Phi_b}{{\mu_R}}{b}}{\ahc{\Phi_b}{{\lambda_R}}{b}}.
\end{equation}

As before, we let $G_{\Phi,R}^{\kappa} \coloneqq [\bm{G}^\kappa_{\Phi}]_R$.

* The channel solutions $H_\Phi(\vec{v}) \coloneqq [\bm{H}_\Phi ]_R$ are rational functions in $\alpha$ that are `local' to $\lambda_i,\mu_j,\nu_R$. That is, that depend only on the vector of local parameters $\vec{v}$ characterising the de-windowed shapes of $\lambda_i, \mu_j$ and $\nu_R$. Typically, these parameters are of the form

\[ v= [ \bm{h}^{\sA}_\sigma(b)]_R\]
for $\sigma = \lambda_R$, $\mu_R$ or $\nu_R$ and $b\in \sigma/\kappa$, or

\begin{equation}\label{defn:ehook}
v= [\bm{e}_{\kappa}(b) ]_R \coloneqq  \left[\bm{h}^{\bullet}_{\lambda_R}(b) -\bm{h}^{\bullet}_{\mu_R}(b) \right]_R
\end{equation}

for $b\in \kappa$.

In short, the window factor $F_R$ and the channel factors $G_{\Phi,R}^{\kappa}$ depend on all the parameters of the window $R$, but the channel solutions depend on only a subset of these parameters.

One is tempted to think of the channel solutions as capturing the factorized product $H_\Phi \cong \jackjLR_{\mu_1,\nu_1}^{\lambda_1}\jackjLR_{\mu_2,\nu_2}^{\lambda_2} f_{\nu}^{\nu_1,\nu_2}$, for $\Phi$ corresponding to the $\nu_1,\nu_2$ splitting, but we find that such an interpretation does not hold in examples with $c_{\mu\nu}^{\lambda}>1$.

The claims of the above conjecture are best illustrated with examples.

The Bravi-Gandini factorization result (<ref>) reduces to the case where $\mu$ is a rectangular partition and $\kappa=\mu$ (hence $\mu_i = \emptyset$). In this case, the Jack LR coefficient factors over a single channel $\Phi$, with $H_\Phi = f^{\lambda_1,\lambda_2}_{\nu}$. In the example (<ref>), we use the window

\begin{equation}R=
\begin{ytableau}
*(boxS) & *(boxS) \\
*(boxS) & *(boxS) \\
&    \\
*(boxS) &  *(boxS) &  *(boxS) &  *(boxS) \\
*(boxS) &  *(boxS) &  *(boxS) & *(boxS)  \\
\end{ytableau},
\end{equation}

which gives $\mu_R = 2^2 = \kappa$. We sum over the single channel $\Phi := {}^{\left[{}^{\sU\sL}_{\sL\sL}\right]}_{\left[{}^{\sL\sU}_{\sL\sL}\right]}$, with

\begin{equation} g^{43221}_{222,2211} = F_R \times G_{\Phi,R}^{\kappa} \times \jackjLR_{\emptyset,21}^{21}\,\jackjLR_{\emptyset,21}^{21}\, f^{21,21}_{2211}.\end{equation}

In this case the general factorization conjecture (<ref>) correctly predicts that the channel coefficient $H_\Phi = f_{21,21}^{2211}= f_{\lambda_1,\lambda_2}^{\nu}$ is only a function of the the shapes $\lambda_1, \lambda_2$ and $\nu$.

Returning to the maximal filling example (<ref>), we can factorize the windowed problem over the shape $\kappa=1^5$, in which case we see that the maximal filling (MF) condition is local, that is, splits up into two sub-shapes in which both satisfy the maximal filling condition,

\[ T_R =\ytableausetup{boxsize=1.0em} \begin{ytableau}
2  \\
1 \\
& 1 & 3 \\
&  & 2& 2\\
&    & 1 & 1 & 3 \\
&   & &  & 2 \\
&    &  &  & 1 \\
\end{ytableau} = \begin{ytableau} 2  \\
1 \\
\none \\
&   \none & 1 & 3 \\
& \none & & 2& 2\\
&  \none &  & 1 & 1 & 3 \\
&  \none & & &  & 2 \\
& \none &   &  &  & 1 \\
\end{ytableau}\]

In the first factored windowed region $\lambda_1$, we have the solution to the Maximal Filling (MF) problem $g_{\emptyset,11}^{11}$, and in the second region $\lambda_2$ we have the solution to the MF problem $g_{3311,432}^{44432}$. This corresponds to splitting $\nu$ as the horizontal sum $\nu = 542 = 432+11$. We find the example factorizes over $\kappa$ with a single channel $\Phi = {}^{[\sL\sU\sL\sU\sU]'}_{[\sU\sU\sL\sU\sL]'}$ (as read off from figure  (<ref>)), where the following holds

\begin{equation}
g_{\mu\nu}^{\lambda} = F_{R} \times G_{\Phi,R}^\kappa \times H_\Phi,
\end{equation}

where the channel solution splits accordingly

\begin{equation}
H_\Phi = g_{3311,432}^{44432} \, g_{\emptyset,11}^{11} \, f^{432,11}_{542}.
\end{equation}

These previous examples were warm up for the following non-trivial example of factorization. Consider once more the $c=2$ family of Example (<ref>). We continue to use the composite window $R$ given in (<ref>), with its 6 parameters $(K,M,N,T,p,s)$.
We consider factorization over $\kappa = 21 \subset \mu_R \subset \lambda_R$ (indicated with black boxes).

\begin{equation}
\lambda_{R}/\kappa =\begin{ytableau}
*(boxS)  \\
*(boxB)  & *(boxS)\\
*(boxB)  &*(boxB)  & *(boxS)  \\
\end{ytableau}, \quad
\mu_{R}/\kappa =\begin{ytableau}
*(boxB)  \\
*(boxB)  & *(boxB)  \\
\end{ytableau},\quad
\nu_R =\begin{ytableau}
*(boxS)  \\
*(boxS)  & *(boxS)  \\
\end{ytableau}.
\end{equation}

The factorization conjecture (<ref>) claims that

\begin{equation}\label{eq:6paramfact6}
g_{\mu,\nu}^{\lambda} = F_R \times \sum_{\Phi\in \hookspace_\kappa} G_{\Phi,R} \times H_\Phi(p,s)
\end{equation}

That is, the channel solutions $H_\Phi$ only depend on two parameters $(p, s)$ that characterize the shapes of the de-windowing of $\mu_R/\kappa$, $\lambda_R/\kappa$ and $\nu_R$, whereas the window factor $F_R$ and the channel solutions $G_{\Phi,R}$ depend on all six parameters.

To set notation, for the factorization region $\kappa = \begin{ytableau}
x \\
y &  z
\end{ytableau}$, let the channel $\Phi = {}^{\sA\sB\sC}_{\sD\sE\sF}$ correspond to the channel factor

\[ \bm{G}^{\sA\sB\sC}_{\sD\sE\sF} := \frac{\bm{h}^{\sA}_{\mu_R}(x)\bm{h}^{\sB}_{\mu_R}(y)\bm{h}^{\sC}_{\mu_R}(z)}{\bm{h}^{\sD}_{\lambda_R}(x)\bm{h}^{\sE}_{\lambda_R}(y)\bm{h}^{\sF}_{\lambda_R}(z)}.\]

For $\nu_R =\begin{ytableau}
*(boxS) a \\
*(boxS) b & *(boxS)  c
\end{ytableau}$,

Let $\bm{k}^{\sA\sB\sC}_{\nu_R} =-\bm{h}^\sA_{\nu_R}(a)+ \bm{h}^\sB_{\nu_R}(b)-\bm{h}^\sC_{\nu_R}(c)$.

A solution to equation (<ref>) is given as follows.
Define the expression

\[ {\bf{Z}}= \frac{ \bm{j}_{\nu_R} }{\bm{k}^{\sU\sU\sU}_{\nu_R}\bm{k}^{\sL\sL\sL}_{\nu_R}\prod_{b\in \kappa}\bm{e}_{\kappa}(b)}, \]

where we have used the notation (<ref>),
with de-windowed evaluation

\[ Z(p,s) :=\left[ \bf{Z} \right]_R  =  \frac{(\hkl{p}{1}\hkl{p+1}{0})(\hkl{1}{s}\hkl{0}{s+1})(\hkl{p+2}{s+1}\hkl{p+1}{s+2})}{\hkl{0}{1}\hkl{1}{0}\cdot \hkl{p+1}{1}\hkl{1}{s+1}\hkl{p+1}{s+1}}. \]

The channel solutions $H_\Phi(p,s) \coloneqq [\bm{H}_\Phi]_R$ defined below over the 8 channels solve the equation (<ref>),

$\Phi$     $\bm{H}_\Phi / \bm{Z}$      $\Phi$     $\bm{H}_\Phi / \bm{Z}$

${}^{\sU\sU\sU}_{\sU\sU\sU}$      $\frac{-1}{\bm{h}^\sU_{\nu_R}(b)} - \frac{1}{\bm{\beta}} $     ${}^{\sL\sL\sL}_{\sL\sL\sL}$     $ \frac{-1}{\bm{h}^\sL_{\nu_R}(b)} + \frac{1}{\bm{\beta}} $

${}^{\sU\sU\sL}_{\sU\sU\sL}$      $ \frac{1}{\bm{h}^\sU_{\nu_R}(a)}+ \frac{1}{\bm{\beta}}  $     ${}^{\sL\sL\sU}_{\sL\sL\sU}$      $  \frac{1}{\bm{h}^\sL_{\nu_R}(a)}- \frac{1}{\bm{\beta}} $

${}^{\sL\sU\sU}_{\sL\sU\sU}$      $ \frac{1}{\bm{h}^\sU_{\nu_R}(c)}+ \frac{1}{\bm{\beta}}  $    ${}^{\sU\sL\sL}_{\sU\sL\sL}$      $  \frac{1}{\bm{h}^\sL_{\nu_R}(c)}- \frac{1}{\bm{\beta}}  $

${}^{\sL\sU\sL}_{\sL\sU\sL}$     $ \frac{1}{\bm{k}^{\sU\sU\sU}_{\nu_R}}- \frac{1}{\bm{\beta}}  $     ${}^{\sU\sL\sU}_{\sU\sL\sU}$     $  \frac{1}{\bm{k}^{\sL\sL\sL}_{\nu_R}}+ \frac{1}{\bm{\beta}}  $

where $\bm{\beta} = \begin{ytableau}*(boxL)L \end{ytableau}-\begin{ytableau}*(boxU)U \end{ytableau}$ satisfies $[\bm{\beta}]_R=\beta$.
Note that this solution is only over channels of the form $G^X_X$.
Furthermore this solution has the two symmetries

\begin{equation}\label{eq:6symsym}
\bm{H}_{\Phi^T} = (\bm{H}_{\Phi})^T, \qquad \bm{H}_{\bar \Phi} = \overline{\bm{H}_{\Phi}},
\end{equation}

where $\bar{\bm{\beta}}=-\bm{\beta}$ and $\bm{\beta}^T = \bm{\beta}$, and $\bm{Z}$ is invariant under both. This the unique solution with 8 or less non-vanishing channel solutions.

We demonstrate the decomposition of

\[ \jackjLR_{21,21}^{321} = \frac{6\alpha(2+11\alpha+2\alpha^2)}{(2+\alpha)(1+2\alpha)(3+2\alpha)(2+3\alpha)} \]

as $\jackjLR_{21,21}^{321} = \sum_\Phi G_\Phi \times H_\Phi$, with

$\Phi$     $G_\Phi \times H_\Phi$      $\Phi$     $G_\Phi \times H_\Phi$

${}^{\sU\sU\sU}_{\sU\sU\sU}$      $\frac{- 3 \alpha^4 (2+\alpha)}{(\alpha-1)(1+\alpha)^3(1+2\alpha)(2+3\alpha)}$     ${}^{\sL\sL\sL}_{\sL\sL\sL}$      $\frac{3\alpha(1+2\alpha)}{(\alpha-1)(1+\alpha)^3(2+\alpha)(3+2\alpha)} $

${}^{\sU\sU\sL}_{\sU\sU\sL}$      $\frac{\alpha(-1+2\alpha)(1+2\alpha)}{(\alpha-1)(1+\alpha)^3(2+3\alpha)} $     ${}^{\sL\sL\sU}_{\sL\sL\sU}$      $\frac{\alpha(-2+\alpha)(2+\alpha)}{(\alpha-1)(1+\alpha)^3(3+2\alpha)}$

${}^{\sL\sU\sU}_{\sL\sU\sU}$      $\frac{\alpha(-1+2\alpha)(1+2\alpha)}{(\alpha-1)(1+\alpha)^3(2+3\alpha)} $    ${}^{\sU\sL\sL}_{\sU\sL\sL}$      $\frac{\alpha(-2+\alpha)(2+\alpha)}{(\alpha-1)(1+\alpha)^3(3+2\alpha)} $

${}^{\sL\sU\sL}_{\sL\sU\sL}$     $ \frac{\alpha(-2+\alpha)(1+2\alpha)^2}{(\alpha-1)(1+\alpha)^3(2+\alpha)(2+3\alpha)} $     ${}^{\sU\sL\sU}_{\sU\sL\sU}$     $ \frac{\alpha^2(-1+2\alpha)(2+\alpha)^2}{(\alpha-1)(1+\alpha)^3(1+2\alpha)(3+2\alpha)} $

§.§ Channel Identities

Note that in the formula (<ref>), each of the channel solutions $H_\Phi$ has a pole as $\alpha \to 1$, i.e. as $\beta \to 0$. However, the combined residue at this pole vanishes, as shown by the following result.

\[ \sum_{X\in \{U,L\}^3} (-1)^{\#U(X)}  G^{X}_{X}   = -\beta^3 \frac{\prod_{b\in \kappa}e_{\kappa}(b)}{j_{\lambda,\kappa}}  \]

where $\#U(X)$ counts the number of upper hooks $(U)$ in $X$, and $e_\kappa(b) := [\bm{e}_{\kappa}(b)]_R$ as in (<ref>).

\begin{eqnarray}
\sum_{X\in \{U,L\}^3} (-1)^{\#U(X)}  G^{X}_{X}  &=& \prod_{b\in \kappa} \left( \tfrac{h^{\sL}_\mu(b)}{h^{\sL}_\lambda(b)} - \tfrac{h^{\sU}_\mu(b)}{h^{\sU}_\lambda(b)} \right)  \\
&=& \prod_{b\in \kappa} \left( \tfrac{h^{\sL}_\mu(b) - h^{\sL}_\lambda(b)}{h^{\sL}_\lambda(b)} - \tfrac{h^{\sU}_\mu(b)-h^{\sU}_\lambda(b)}{h^{\sU}_\lambda(b)} \right)  \\
&=& \prod_{b\in \kappa}(h^{\sL}_\mu(b) - h^{\sL}_\lambda(b)) \left( \tfrac{h^{\sL}_\lambda(b)-h^{\sU}_\lambda(b)}{h^{\sL}_\lambda(b)h^{\sU}_\lambda(b)}  \right)  \\
&=& \beta^3 \tfrac{\prod_{b\in \kappa}(h^{\sL}_\mu(b) - h^{\sL}_\lambda(b))}{j_{\lambda,\kappa}}.
\end{eqnarray}

The following results show that we always have redundancies between factorization channels, that is, there always exists linear relations between channel factors for $\kappa$ with coefficients that are valued in the local parameters, e.g. $e_\kappa(b)$.

For a box $b\in \kappa$, define the single channel factor

\[ G^{U}_{L} := \frac{ \uh{\mu}{b} }{ \lh{\lambda}{b} }.\]

The following linear relation holds between channel factors

\begin{equation}\label{eq:innerparameterredund}
G^L_L-G^L_U-G^U_L+G^U_U =\frac{-\beta}{e_\kappa(b)} \cdot (G^L_L-G^U_U) .
\end{equation}

We start with

\[ G^U_U-G^L_U = \frac{-\beta}{\uh{\lambda}{b}} \]

\begin{eqnarray*}
\frac{1}{-\beta} ( G^U_U-G^L_U-G^U_L+G^L_L ) &=& \frac{1}{\uh{\lambda}{b}} -  \frac{1}{\lh{\lambda}{b}} \\
&=&\frac{1}{e_\kappa(b)}\left( \frac{\uh{\lambda}{b}-\uh{\mu}{b}}{\uh{\lambda}{b}} -  \frac{\lh{\lambda}{b}-\lh{\mu}{b}}{\lh{\lambda}{b}} \right)\\
&=&\frac{1}{e_\kappa(b)}\left( \frac{-\uh{\mu}{b}}{\uh{\lambda}{b}} -  \frac{-\lh{\mu}{b}}{\lh{\lambda}{b}} \right).
\end{eqnarray*}

We can easily generalize Lemma (<ref>) to any $\kappa$.

\begin{equation}
\sum_{X,Y \in \{U,L\}^{|\kappa|}} (-1)^{\# U(X)+\#U(Y)} G^{X}_{Y} = \frac{(-\beta)^{|\kappa|}}{\prod_{b\in\kappa} {e_\kappa(b)}} \sum_{X\in \{U,L\}^{|\kappa|}} (-1)^{\# U(X)} G^{X}_{X}.
\end{equation}

§.§ Regularity

The individual channel solutions $H_\Phi$ in Conjecture (<ref>) have poles in the Schur limit, $\alpha\to1$. This is likely a consequence of the highly factorized nature of that particular solution.

Corresponding to each factorization channel, we define

\begin{equation}
\tilde C_\Phi(\vec{v}) \coloneqq \frac{ \lhs{\lambda}{} }{ \lhs{\mu}{} \lhs{\nu}{}  }\times  F_R \times G_\Phi \times  H_\Phi(\vec{v}).
\end{equation}

If the limit $\left( \tilde C_\Phi(\vec{v}) \right)_{\alpha\to1} $ exists for all $\Phi \in \Gamma$ and  $\vec{v} \in \CT$, then we say the channel solutions $\{H_{\Phi}\}$ are regular.
In that case we can associate a local channel coefficient,

\begin{equation}
C_\Phi(\vec{v}) \coloneqq \left( \tilde C_\Phi(\vec{v}) \right)_{\alpha\to1}.
\end{equation}

Note, one must take the $\alpha$ limit after assigning values to $\vec{v}$. We will often see factors such as $\frac{(-1+\alpha)}{(t-1+\alpha)}$, where the order of these limits matters when $t=0$.

In the case of a regular factorization, we have the following decomposition of Schur LR coefficients (c.f. equation (<ref>))

\[ c_{\mu\nu}^{\lambda} = \sum_{\Phi\in \hookspace_\kappa}  C_\Phi(\vec{v}).  \]

For certain factorizations $\kappa$, there exists factorization solutions such that $C_\Phi(\vec{v}) \in \setZ$, and are independent of $\vec{v}$.

§.§ Regular Solution

In the previous example we factorized over $\kappa = 21$, here we instead factorize over the smaller $\kappa = (2)$.

\begin{equation}
\lambda_{R}/\kappa =\begin{ytableau}
*(boxS)  \\
*(boxS)  & *(boxS)\\
*(boxB)  &*(boxB)  & *(boxS)  \\
\end{ytableau}, \quad
\mu_{R}/\kappa =\begin{ytableau}
*(boxS)  \\
*(boxB)  & *(boxB)  \\
\end{ytableau},\quad
\nu_R =\begin{ytableau}
*(boxS)  \\
*(boxS)  & *(boxS)  \\
\end{ytableau}.
\end{equation}

With this, the channel solutions will be independent of $M,N$.
The difference here is now we expect that the channel solutions will have is a dependence on $t$.

The factorization conjecture (<ref>) claims there exists a solution of the form

\begin{equation}\label{eq:4paramfact4}
g_{\mu,\nu}^{\lambda} = F_R \times \sum_{\Phi\in \hookspace_\kappa} G_{\Phi,R} \times H_\Phi(p,s,T,K)
\end{equation}

That is, the channel factors $H_\Phi$ only depend on the parameters $p, s,T,K$ that characterize the shapes of the de-windowing of $\mu_R/\kappa$, $\lambda_R/\kappa$ and $\nu_R$.

For the factorization region $\kappa = \begin{ytableau}
*(boxS) y & *(boxS) z
\end{ytableau}$, let the channel $\Phi = \frac{\sB\sC}{\sE\sF}$ correspond to the channel solution

\[ \bm{G}^{\sB\sC}_{\sE\sF} := \frac{\bm{h}^{\sB}_{\mu_R}(y)\bm{h}^{\sC}_{\mu_R}(z)}{\bm{h}^{\sE}_{\lambda_R}(y)\bm{h}^{\sF}_{\lambda_R}(z)}.\]

The following[We shift the variables here for simplicity of the formula] regular channel solutions $H_\Phi$ over the four channels $\Phi = \{\frac{\sU\sU}{\sU\sU}, \frac{\sU\sL}{\sL\sU},\frac{\sU\sL}{\sU\sL},\frac{\sL\sU}{\sL\sU}\}$ solve the equation (<ref>), with constant local channel coefficients $C_\Phi \in \BZ$,

$\Phi$     $H_\Phi(p,s,t\coloneqq T+K\alpha) $     $C_\Phi(\vec{v})$

$\frac{\sU\sU}{\sU\sU}$     $ \frac{(1-\alpha)(p+1)(t+1)(p+\alpha)(t+1+p+(s+3)\alpha)(p+2+2(s+1)\alpha)}{\alpha(t+p+1+2\alpha)(t+p+2+(s+2)\alpha)(p+1+(s+1)\alpha)(1+(s+1)\alpha)}  $     $0$

$\frac{\sU\sL}{\sL\sU}$     $ \frac{-(1-\alpha) (p+1)((s+1)\alpha) (p+\alpha) (2t+1+p+2\alpha)(t+p+3+(s+1)\alpha)(p+1+(s+2)\alpha) }{\alpha(t+1+\alpha)(t+1+p+\alpha)(t+p+2+\alpha)(t+1+p+2\alpha)(t+p+2+(s+2)\alpha) } $     $0$

$\frac{\sU\sL}{\sU\sL}$     $ \frac{(t+1)(p+1)(s\alpha)(2+s\alpha)(p+1+(s+2)\alpha)(t+p+2\alpha)}{\alpha(t+1+p+\alpha)(t+1+p+2\alpha)(1+(s+1)\alpha)} $     $1$

$\frac{\sL\sU}{\sL\sU}$     $  \frac{(t+1)((s+1)\alpha)(p+\alpha)(p+2+s\alpha)(p+1+(s+2)\alpha)(t+2\alpha)}{\alpha^2(t+1+\alpha)(p+1+(s+1)\alpha)(t+1+p+2\alpha)}$     $1$

This solution has redundancies, as was noted before in equation (<ref>). There is also a two parameter set of redundancies between the 6 balanced channels for $\kappa$, that is, channels with the same number of upper hooks in the numerator as denominator. Indeed, there exists solutions over just three balanced channels, however these solutions are irregular. The solution (<ref>) is preferable as it is over the fewest channels (i.e, four) while consisting of regular channel solutions.

§.§ Examples

We conclude with several important cases of this 6-parameter family.

[Return to P. Hanlon's example <cit.>]
Take $\mu = 31$, $\nu=21$, $\lambda=421$.

It is known that

\[ \jackjLR_{31,21}^{421} = \frac{\alpha( 9+97\alpha+294\alpha^2 + 321\alpha^3 + 131\alpha^4+12\alpha^5 )}{3(1+\alpha)^3(2+\alpha)(1+2\alpha)^2(1+3\alpha)}. \]

This is the $N=1, M=0,t=0, p=0, s=0$ case of the windowing $R$. The solution (<ref>) gives the following expression for this LR coefficient

\begin{eqnarray*}
g_{31,21}^{421}&=& F_{R}\times \sum_\Phi G_\Phi \times H_\Phi \\
&=&\frac{1}{(1+ \alpha )}\times  \\
&&\left(  \frac{(1+3\alpha)(2\alpha)}{(2+4\alpha)(1+3\alpha)} \times \frac{(1-\alpha)(1+3\alpha)  }{ (1+\alpha )^2(1+2\alpha)} \right.  \\ %(t-qa)(t+(2+q)a) \\
&&+\,\, \frac{(1+3\alpha)(1+\alpha)}{(3+3\alpha)(1+3\alpha)} \times\frac{ (1-\alpha )(-\alpha) (3+\alpha )}{ (1+\alpha )^2 (2+\alpha)} \\
&&+\,\, \frac{(1+3\alpha)(1+\alpha)}{(2+4\alpha)(2+2\alpha)} \times \frac{    2 (2 \alpha )}{ (1+\alpha ) ^2 }\\
&&+\, \left.  \frac{(2+2\alpha)(2\alpha)}{(3+3\alpha)(1+3\alpha)} \times\frac{   2  (2 \alpha )}{ (1+\alpha ) ^2  } \right). \\
\end{eqnarray*}

[Question by A. Ram]
We seek a formula for  $g_{(a_1,a_2),21}^{(a_1+1,a_2+1,1)}$.

\begin{equation}
\lambda =\begin{ytableau}
a_1  &  a_2 \\
&   &   \\
\end{ytableau}, \quad
\mu =\begin{ytableau}
a_1  & a_2 \\
\end{ytableau},\quad
\nu =\begin{ytableau}
\\
&   \\
\end{ytableau}.
\end{equation}

This is corresponds to the windowing $R$ with $t= a_1-a_2-1$, $p=0$, $s=0$, $N=0$, $M=a_2-1$. Using the solution of Conjecture (<ref>)

\[ g_{(a_1,a_2),21}^{(a_1+1,a_2+1,1)} = F_{R}\times \sum_\Phi G_{\Phi,R} \times H_\Phi(p,s,t). \]

we find the LR coefficient is given be the formula

\begin{eqnarray*}
%&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! g_{(a_1,a_2),21}^{(a_1+1,a_2+1,1)} \\
&=& \frac{(t+2\alpha)}{(t-1+\alpha)}\frac{(\alpha)^2}{(a_1-1+2\alpha)(a_2-1+\alpha)}\times  \\
&&\left(  \frac{(a_1-1+2\alpha)(a_2-1+\alpha)}{(a_1+3\alpha)(a_2+2\alpha)} \times\frac{2  (t) (1-\alpha) (t+  3\alpha)}{ (1+\alpha )  (t+2 \alpha ) (t+1+2\alpha  )} \right.  \\ %(t-qa)(t+(2+q)a) \\
&&+\,\,\frac{(a_1-1+2\alpha)(a_2)}{(a_1+1+2\alpha)(a_2+2\alpha)} \times\frac{ -2(1-\alpha )\alpha (1 +  2\alpha) (t+2+\alpha )}{ (t+\alpha ) (t+1+\alpha) (t+2 \alpha ) (t+1 +2\alpha  )} \\
&&+\,\, \frac{(a_1-1+2\alpha)(a_2)}{(a_1+3\alpha)(a_2+1+1\alpha)} \times \frac{    2(t) (1 +2\alpha  ) (t-1+2 \alpha )}{ (1+\alpha ) (t+\alpha ) (t+2 \alpha )}\\
&&+\,\, \left.  \frac{(a_1+\alpha)(a_2-1+\alpha)}{(a_1+1+2\alpha)(a_2+2\alpha)} \times \frac{   2(t) (1 +2\alpha  ) (t-1+2 \alpha )}{ (1+\alpha )  (t+\alpha ) (t+2 \alpha )} \right). \\
\end{eqnarray*}

This factorization results in the Schur LR sum

\begin{equation}
c_{a_1a_2,21}^{(a_1+1,a_2+1,1)} := C_{\frac{\sU\sU}{\sU\sU}}+C_{\frac{\sU\sL}{\sL\sU}} +C_{\frac{\sU\sL}{\sU\sL}}+ C_{\frac{\sL\sU}{\sL\sU}}= \begin{cases}
0+0+1+1&  a_1 > a_2  \\
0+1+0+0 &  a_1= a_2.
\end{cases}
\end{equation}

Here we see the regularity of the solutions continues formally as $a_1 \to a_2$ (i.e. $t\to-1$ and the window $R$ requires modification), and there is a discrete discontinuity in the channel factors $C_\Phi$ at this `boundary' of the parameter space where the problem degenerates to $c=1$.

We find for  $a_1=13, a_2=5$, this gives

\begin{equation}
g_{(13,5),(2,1)}^{(14,6,1)} = \frac{3\alpha^2(772+335\alpha+ 13\alpha^2)}{(6+\alpha)(7+\alpha)(9+\alpha)(5+2\alpha)(13+3\alpha)}.
\end{equation}

§ CONCLUSIONS

We have formulated the Window Lensing Conjecture (<ref>) and presented evidence for its validity. This conjecture may help improve the understanding of Jack Littlewood-Richardson coefficients, and can hopefully lay the groundwork for an eventual proof of the Stanley conjectures. This Window Lensing property is essentially the minimal property that covers both the Maximal Filling case (which we've proved it holds for) and the Rectangular Union case (for which it holds by inspection). However, beyond those $c=1$ examples, we have presented three formulas for the $c=2$ family with windowed rule $\JackjLR_{21,21}^{321}$, each of which displays particular curious properties.

Firstly, Formula (<ref>) gives a conjectural 20-term solution $\{d_\phi(\alpha)\}$ in the most precise form, demonstrating the validity of the General Structure conjecture. This example exhibits a single symmetry, and has a large degree of redundancy through the kernel (<ref>).

Secondly, Formula (<ref>) on the factorized set of 8 channels $\{H_\Phi(p,s)\}$, gives our most symmetric form of an answer. This formula is the most combinatorical in nature, with the Channel Solutions $H_\Phi$ being given explicitly in terms of hooks and the two symmetries (<ref>) are manifest. This formula is has no redundancies within the 8 symmetric channels, but each channel solution is singular in the $\alpha\to1$ limit.

Lastly, Formula (<ref>) over the factorized 4 channels $\{H_\Phi(p,s,t)\}$, represents our most economical formula. This solution is regular, and can easily be used to compute Jack LR coefficients in some large examples.

These three solutions present the same windowed content, as expressed in the following equalities

\begin{eqnarray*}
&& \sum_{\phi \in \hookspace_{21,21}^{21}} G_{\phi,R} \,\frac{d_\phi(\alpha)}{\alpha^3}.\\
&=& \sum_{\Phi\in \hookspace_{\kappa=21}} G_{\Phi,R} \times H_\Phi(p,s) \\
&=& \sum_{\Phi\in \hookspace_{\kappa=2}} G_{\Phi,R} \times H_\Phi(p,s,t).
\end{eqnarray*}

The work continues in the hope for a combinatorical description of any one of these three solutions.

§ ACKNOWLEDGEMENTS

The author would like to thank Per Alexandersson for helpful conversations on to the present work, and Darij Grinberg for explaining key properties of skew Schur functions.