We make progress towards understanding the structure of Littlewood–Richardson
coefficients $\jackjLR_{\lambda,\mu}^{\nu}$ for products of Jack symmetric functions.
Building on recent results of the second author, we are able to prove new cases of a
conjecture of Stanley in which certain families of these coefficients can
be expressed as a product of upper or lower hook lengths for every box in
each of the partitions. In particular, we prove that conjecture in the case
of a rectangular union, i.e. for $\jackjLR_{\mu,\bsigma}^{\mu \cup m^n}$
where $\bsigma$ is the complementary partition of $\sigma = \mu \cap m^n$
in the rectangular partition $m^n$. We give a formula for these coefficients
through an explicit prescription of such choices of hooks.
Lastly, we conjecture an analogue of this conjecture of Stanley holds
in the case of shifted Jack functions.

§ PRELIMINARIES

§.§ Young diagrams, hooks and the star product

Given an integer partition $\lambda = (\lambda_1,\dotsc,\lambda_\ell)$, we
associate to it the Young diagram as the set of coordinates $(i,j)$ where $0\leq j < \ell$ and
$0 \leq i \leq \lambda_{j-1}$ (French notation).
Each such coordinate $(i,j)$ is called a box, and is pictorially
illustrated by the square with the four corners $\{(i,j),(i+1,j),(i,j+1),(i+1,j+1)\}$, see (<ref>).

The bottom left hand corner of the diagram has coordinate $(0,0)$.
Given any box $s=(x,y)$, we let $\defin{\bps{s}}$ and $\defin{\bms{s}}$ denote the boxes (coordinate) $(x+1,y+1)$
and $(x-1,y-1)$, respectively.
In the diagram below, we have the marked the box $s=(2,1)$.

\begin{equation}\label{eq:yDiagram}
\ytableausetup{boxsize=0.9em}
(7,4,2,2,1) \qquad \longleftrightarrow \qquad
\begin{ytableau}
\, \\
\, & \, \\
\, & \, \\
\, & \, & s & \, \\
\, & \, & \, & \, & \, & \, & \, \\
\end{ytableau}
\end{equation}

We let $\lambda$ also denote the set of boxes in the Young diagram, and $\defin{\lambda^\times}$ denotes
the set of boxes excluding $(0,0)$.

Given a box $s \in \lambda$, let $\defin{\arm(s)}$ be the number of boxes to the right of $s$,
and let $\defin{\leg(s)}$ be the number of boxes above $s$.
For example, in Figure <ref>, we have $\arm(s)=5$, $\leg(s)=3$.

We let $\defin{\addset_\lambda}$ be the set of boxes (called outer corners)
which can be added to $\lambda$ and still have a Young diagram
(denoted $\defin{\lambda \cup s}$), see Figure <ref>.

Similarly, we let $\defin{\remset_\lambda}$ be the set of boxes (called inner corners)
which can be removed from $\lambda$ and still have a Young diagram.
For such $s \in \remset_\lambda$, let $\defin{\lambda \setminus s}$
denote the partition (diagram) obtained by removing $s$ from $\lambda$.

We shall also make use of
$\defin{\remset^+_\lambda} \coloneqq \{ \bps{s} : s \in \remset_\lambda\}$.
This set of coordinates has a natural interpretation, see Figure <ref>.

Given two boxes $a=(a_1,a_2)$ and $b=(b_1,b_2)$
we let the join $\bmin{a}{b}$ be defined as the box at
$(\min(a_1,b_1),\min(a_2,b_2))$. Similarly,
we define the meet $\defin{\bmax{a}{b}}\coloneqq (\max(a_1,b_1),\max(a_2,b_2))$,
see Figure <ref>.

\begin{equation*}
\centering
\includegraphics[scale=1.0,page=1]{addSetRemSet}
\hspace{1cm}
\includegraphics[scale=1.0,page=1]{boxWedge}
\end{equation*}

Left: Consider $\lambda = 74221$ and $s=(1,0)$. For this box,
we have $\arm(s)=5$, $\leg(s)=3$.
All coordinates belonging to $\addset_\lambda$
have been marked with $\bullet$
and all elements in $\remset^+_\lambda$
are marked $\times$.

Right: $c= \bmin{a}{b}$ and $d = \bmax{a}{b}$.

We let $\defin{m^n}$ denote the rectangular partition $(m,m,\dotsc,m)$ with $n$ entries equal to $n$.
For fixed $m$, $n$ and $\mu \subseteq m^n$, we define the complement partition of $\mu$ as

\[
\defin{\bar{\mu}} \coloneqq (m-\mu_n,m-\mu_{n-1},\dotsc,m-\mu_2,m-\mu_1).
\]

For example, with $m=7$, $n=6$ and $\mu = 75421$ we have $\bar{\mu} = 76532$, as
seen below:

\begin{equation*}\label{eq:complement}
\centering
\includegraphics[scale=1.0,page=1]{partitionComplement}
\end{equation*}

Observe that $\addset_\mu$ mostly corresponds to $\remset^+_{\bar{\mu}}$,
except for some boundary cases.

We let $\alpha$ be a fixed indeterminate and introduce the following notation.
For $s = (x,y) \in \setZ^2$, we set $\defin{[(x,y)]} \coloneqq  (x,y)\cdot(\alpha, -1) = \alpha x-y$.
We often simplify the notation and write $[x,y]$ instead of $[(x,y)]$.

Given $s \in \lambda$, we define the upper hook vector and lower hook vector as

\begin{align*}
\defin{\uhv{\lambda}{s}} & \coloneqq (\arm(s)+1,-\leg(s)) \\
\defin{\lhv{\lambda}{s}} & \coloneqq (\arm(s), -\leg(s)-1),
\end{align*}

where we note that

\begin{equation}\label{eq:lrhookshift}
\lhv{\lambda}{s} = \uhv{\lambda}{s} - (1,1),
\end{equation}
and then the upper hook value and lower hook value as

\begin{align*}
\defin{\uh{\lambda}{s}} & \coloneqq [\uhv{\lambda}{s}] = \alpha (\arm(s)+1) + \leg(s),
&& \text{(also known as $h^*_\lambda(s)$)}\\
\defin{\lh{\lambda}{s}} & \coloneqq [\lhv{\lambda}{s}] = \alpha \arm(s) + (\leg(s)+1)
&& \text{(also known as $h_*^\lambda(s)$).}
\end{align*}

We note that the difference is whether the contribution of the box at $s$ should count horizontally,
or vertically—see Figure <ref> for an illustration.

In the left picture, we have that $\uh{\lambda}{s} = 3\alpha +4$, while
$\lh{\lambda}{s} = 2\alpha + 5$.

\begin{equation*}\label{eq:rowcolSets}
\ytableausetup{boxsize=0.9em,aligntableaux=center}
\substack{
\begin{ytableau}
\, \\
\, & \, \\
*(pBlue) \, & *(pBlue)  \, & *(pBlue)  \,  & *(pBlue)  s \\
\, & \, & \, & \,  & \, \\
\, & \, & \, & \, & \, & \, & \, \\
\end{ytableau}
\\
\qquad
\substack{
\begin{ytableau}
\, \\
\, & \, \\
\, & \, & \, & *(pBlue) s \\
\, & \, & \, & *(pBlue) \,  & \, \\
\, & \, & \, & *(pBlue) \, & \, & \, & \, \\
\end{ytableau}
\\
\end{equation*}

Given a box $s$, we let $\defin{R^s_\lambda}$ denote the set of boxes in $\lambda$
in the same row as $s$. Similarly, $\defin{C^s_\lambda}$ is the set of boxes in the same column.
Note that this definition makes sense even when $s \notin \lambda$.

Let $\lambda$ and $\mu$ be Young diagrams.
Since these are sets of boxes, we may define their intersection, $\defin{\lambda \cap \mu}$ and union $\defin{\lambda \cup \mu}$,
and these are the Young diagrams for some integer partitions.

§.§ Jack symmetric functions and structure constants

There are two popular versions of the Jack symmetric functions; the Jack $P$ functions,
and the integral form Jack $J$ functions—the latter has integer coefficients
in the monomial basis. We shall only use the latter functions.
The book <cit.> is recommended as a reference on this topic.
For example, the $\jackj_\lambda(\xvec;\alpha)$ for $\lambda \partition 3$,
expanded in the monomial basis are as follows:

\begin{align*}
\jackj_{3}(\xvec;\alpha) &= (1+3\alpha + 2\alpha^2) \monomial_{3} + (3+3 \alpha) \monomial_{21} + 6\monomial_{111} \\
\jackj_{21}(\xvec;\alpha) &= (2+ \alpha) \monomial_{21} + 6\monomial_{111} \\
\jackj_{111}(\xvec;\alpha) &= 6\monomial_{111}.
\end{align*}

The expansions in the power-sum basis are

\begin{align*}
\jackj_{3}(\xvec;\alpha) &= 2\alpha^2 \powerSum_{3} + 3\alpha \powerSum_{21} + \powerSum_{111} \\
\jackj_{21}(\xvec;\alpha) &= -\alpha \powerSum_{3} + (\alpha-1) \powerSum_{21} + \powerSum_{111} \\
\jackj_{111}(\xvec;\alpha) &= 2\powerSum_{3} -3\powerSum_{21} +\powerSum_{111}.
\end{align*}

Let $\defin{\jackjLR_{\mu\nu}^{\gamma}(\alpha)} \in \setQ(\alpha)$ be the Jack Littlewood–Richardson constants,
determined via the relation

\begin{equation}\label{eq:jackLR}
\jackj_\lambda \jackj_\mu = \sum_\nu \jackjLR_{\lambda\mu}^{\nu}(\alpha) \jackj_\nu.
\end{equation}

We shall often omit the dependence on $\alpha$, and just write $\jackjLR_{\lambda\mu}^{\nu}$.
Observe that these are closely related to the classical Littlewood–Richardson
coefficients for Schur functions, denoted $\defin{c_{\lambda\mu}^{\nu}}$.
The Jack polynomials form an orthogonal basis with respect to
the $\alpha$-deformed Hall inner product, $\langle \,\cdot \, , \,\cdot\, \rangle$.
This is the inner product on symmetric functions with the property that

\[
\langle \powerSum_\lambda, \powerSum_\mu \rangle =
\begin{cases}
z_\lambda \alpha^{\ell(\lambda)} & \text{if $\lambda=\mu$}\\
0 &\text{otherwise,}
\end{cases}
\]

and where $n!/\defin{z_\mu}$ is the number of permutations of size $n$ with cycle type $\mu$.

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

§.§ The Stanley Conjectures

In <cit.>, several remarkable conjectures were stated regarding the
structure of the Jack Littlewood–Richardson coefficients.
These conjectures are stated in terms of the quantities
$\langle \jackj_\mu \jackj_\nu , \jackj_\lambda \rangle = \jackjLR_{\mu\nu}^\lambda \norm{\jackj_\lambda}^2$, herein referred to as Stanley structure coefficients,
but can be easily restated in terms of the Littlewood–Richardson coefficients using (<ref>).

The Stanley structure coefficients are non-negative integer polynomials in $\alpha$, that is,

\begin{equation*}
\langle \jackj_\mu \jackj_\nu , \jackj_\lambda \rangle \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 $\langle \jackj_\mu \jackj_\nu , \jackj_\lambda \rangle$ which is manifestly a non-negative polynomial in $\alpha$. Stanley conjectured such a form in the case $c_{\mu\nu}^\lambda =1$.

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

\begin{equation*}
\langle \jackj_\mu \jackj_\nu , \jackj_\lambda \rangle =
\left( \prod_{b \in \mu} \tilde{h}_{\mu}(b) \right)
\left( \prod_{b \in \nu} \tilde{h}_{\nu}(b) \right)
\left( \prod_{b \in \lambda} \tilde{h}_{\lambda}(b) \right),
\end{equation*}

where $\tilde{h}_{\sigma}(b)$ is a choice of either $\uh{\sigma}{b}$ or $\lh{\sigma}{b}$ for each box $b$.
Moreover, one chooses $\lh{\,}{b}$ and $\uh{\,}{b}$ exactly $|\lambda|$ times each in the above expression.

Stanley proved Conjecture <ref> in an important special case.

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

\begin{equation*}
\langle \jackj_\mu  \jackj_{(r)} , \jackj_\lambda \rangle =
\left( \prod_{b \in \mu} A_{\mu\lambda}(b) \right)
\left( \prod_{b \in (r)}  \uh{(r)}{b}  \right)
\left( \prod_{b \in \lambda} A'_{\mu\lambda}(b) \right),
\end{equation*}

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

and $A'$ is given by the same formula as $A$, except with upper and lower looks exchanged.

In 2013, another case of Conjecture <ref> was
proven using a technique involving vertex operators.

Let $\mu \subseteq m^n$.
The Stanley structure coefficient in the rectangular case has the following expression as a product of lower/upper hook lengths:

\begin{equation}\label{eq:stanleyrectangular}
\langle \jackj_\mu  \jackj_{\bmu} , \jackj_{m^n} \rangle =
\left( \prod_{b \in \mu}  \lh{\mu}{b}   \right)
\left( \prod_{b \in \bmu}  \uh{\bmu}{b}  \right)
\left( \prod_{b \in m^n}  D_{\mu,m^n}(b) \right),
\end{equation}

where, for $b\in m^n$, we have defined

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

For example, $\langle J_{211}J_{221},J_{333} \rangle$ is computed using the following hook assignments

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

In the paper <cit.>, the authors provide formula (<ref>)
and note that the answer must be symmetric under the
exchange of $\mu \leftrightarrow \bmu$, however that symmetry is not explicitly demonstrated in this formula.

In this paper, we provide an explicit demonstration for the symmetry property in formula (<ref>), by re-proving it using a new technique.
Furthermore, we prove a new case of Conjecture <ref>,
the rectangular union case, which generalises the
rectangular case (Thm. <ref>), and provide an
explicit prescription of the required hook choices. In particular, let $\mu$ be a partition and $m^n$ be
any rectangular partition. Let $\sigma = \mu \cap m^n$, and let $\bsigma$
be the complementary partition of $\sigma$ in $m^n$.
Note that this reduces to the rectangular case when $\mu \subset m^n$.

The setup of the rectangular union case.

The strong Stanley conjecture holds for $\langle J_\mu J_{\bsigma} , J_{\mu \cup m^n} \rangle$.
The assignment of lower and upper hooks can be read off from Figure <ref>.

For example, $\langle J_{42211}J_{211}, J_{43331}\rangle $
is computed using the following hook assignments

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

The recent paper by Matveev–Wei <cit.> proves Conjecture <ref> in the special case when
we have $c_{\mu\nu}^\lambda = K_{\mu, \lambda-\mu}=1$, where the $K_{\mu, \lambda-\mu}$
is a Kostka coefficient.
This required equality between Kostka coefficients and Littlewood–Richardson coefficients
holds in the rectangular case, i.e., the setup in Theorem <ref> is covered by their work.
We also understand the proof therein to not be constructive, that is,
no explicit prescription for a choice of upper or lower hooks is presented for each box in the three partitions.

However, our Theorem <ref> is in general not covered by the Matveev–Wei result,
as we sometimes have the strict inequality $c_{\mu\nu}^\lambda < K_{\mu, \lambda-\mu}$ in the rectangular union case.
In the example above, we have $c^{43331}_{42211,211} =1$
but $K_{42211,43331-211} = K_{42211,22231} = 3$, as evidenced by the three semistandard Young tableax

\[
\ytableausetup{boxsize=1.0em}
\ytableaushort{1144,22,33,4,5} \quad \ytableaushort{1134,22,34,4,5} \quad  \ytableaushort{1124,23,34,4,5}.
\]

§.§ A formula involving Jack structure constants

For any multiset[A collection with multiplicities.] of boxes $\Gamma$, we define the rational function[This also depends on $\alpha$.]
$T_{\Gamma}(u) : \setR \to \setR$ by

\begin{equation}\label{def:TGamma}
\defin{ T_{\Gamma}(u) } \coloneqq \prod_{b \in \Gamma} T_\lbox(u-[b])
\text{ where }
\defin{T_\lbox(u)} \coloneqq \frac{(u-[0,0])(u-[1,1])}{(u-[1,0])(u-[0,1])} .
\end{equation}

We give some examples of such functions in Example <ref>.

For any two Young diagrams $\lambda, \mu$, we define the star product $\lambda \* \mu$ as the multiset of boxes

\[
\defin{\lambda \* \mu} \coloneqq \bigsqcup_{\substack{s\in \lambda \\ t \in \mu}} s+t.
\]

For example, $31 \* 221$ is computed as

\begin{equation*}
\ytableausetup{boxsize=1.0em} \ydiagram{1,3}\, \*\, \ydiagram{1,2,2} \,=\,\ytableausetup{boxsize=1.0em}
\begin{ytableau}
\, \\
2 & 2  & \, \\
2 & 3 & 2 & \, \\
\, & 2& 2 & \,
\end{ytableau}
\end{equation*}

where the number inside a box indicates its multiplicity (if $> 1$).
Lastly, we define
$\defin{\jacktop_\mu} \coloneqq \prod_{b \in \mu^\times} [b]$.

With these tools introduced, we can now state the main motivating result that is used in this work.

Let $\mu,\nu$ be partitions and $n= |\mu|+|\nu|$,
then the following `sum-product' identity of rational functions (in a complex variable $u$) holds,
involving the Jack Littlewood–Richardson coefficients defined earlier
in (<ref>):

\begin{equation}\label{eq:sumproductident}
\sum_{\substack{\gamma\partition  n \\ \mu,\nu \subseteq \gamma } }  \jackjLR_{\mu\nu}^{\gamma}(\alpha) \frac{\jacktop_\gamma}{\jacktop_\mu \jacktop_\nu} \left( \sum_{s \in \gamma \setminus (\mu \cup \nu)} \frac{1}{u-[s]} \right)  = T_{\mu\* \nu}(u) - 1.
\end{equation}

For ease of notation, we set

\begin{equation*}
\defin{\hat \jackjLR_{\mu\nu}^{\gamma}} \coloneqq
\jackjLR_{\mu\nu}^{\gamma}(\alpha) \frac{\jacktop_\gamma}{\jacktop_\mu \jacktop_\nu}.
\end{equation*}

§.§ Properties of Tmunu

To make use of (<ref>), we will
need to produce some structural
results about the right hand side.
By definition, we have

\[
T_{\mu\* \nu}(u)  = \prod_{b \in \mu\* \nu } T_\lbox(u-[b])= \prod_{ \substack{s \in \mu \\ t \in \nu }} T_\lbox(u-[s+t]).
\]

Following from Definition <ref>, we may view $T_\lbox(u)$ as the rational
function with zeros at coordinates $(0,0)$ and $(1,1)$, and poles in $(1,0)$
and $(0,1)$. These coordinates are the four vertices of the box $(0,0)$.
This allows us to use Young diagrams to visualize the rational function $T_{\mu}(u)$ or $T_{\mu\* \nu}(u)$,
as we see in the following example.

With $\mu = 331$, we have that

\[
T_{\mu}(u) = \frac{(u-[0,0])(u-[1,3])(u-[3,2])}{(u-[0,3])(u-[1,2])(u-[3,0])}.
\]

We can illustrate such a rational function by marking the zeros and poles in the plane.
Let us also set $\nu = 44311$.
Then $T_{\mu}(u)$, $T_{\nu}(u)$ and $T_{\mu \* \nu}(u)$ can be illustrated as

\[
\includegraphics[scale=1.0,page=1]{starProduct}
\]

where $\bullet$ represents a simple pole,
and $\times$ and $\times^2$ represent a zero of order one and order $2$, respectively.
Note that the number of poles and the number of zeros (counting multiplicity) must be the same in every row (and every column)
of the diagram since any $T_{\Gamma}$
is a product of $T_{\lbox}(u-[b])$ over $b \in \Gamma$.

Note that the poles of $T_{\mu}$ are precisely the elements in $\addset_\mu$,
while the zeros are given by $\remset^+_\mu \cup \{(0,0)\}$.
In other words,

\begin{equation}\label{eq:TmuFormula}
T_{\mu}(u) = u\cdot \frac{\prod_{s \in \remset_\mu^+ }(u-[s])}{\prod_{t
\in\addset_\mu}(u-[t ])}.
\end{equation}

For any partitions $\mu$, $\nu$,
the order of the pole at $u=[s]$ of the function $T_{\mu\* \nu}(u)$
is given by the difference

\begin{equation}\label{eq:Torderformula}
\ord_{u=[s]} T_{\mu\*\nu}(u) = \left| \{  b \in \nu :   s-b \in \addset_\mu \} \right| - \left| \{  b \in \nu : s- b \in ( {\remset^+_\mu} \cup (0,0))\} \right|.
\end{equation}

We note that $T_{\mu \* \nu}(u) = \prod_{b\in \nu} T_{\mu}(u-[b])$.
By (<ref>), the poles of $T_{\mu}(u-[b])$ are at $t+b$ for $t\in \addset_\mu$
and similarly its zeros are at $\bps{t} + b$, for $\bps{t} \in \remset^+_\mu$ and an extra zero at $u=[b]$.
From these observations we can deduce (<ref>).

Left: The contributions to poles at $[s]$ of $T_{\mu\*\nu}$.
Here, $b\in \nu$, $t \in \addset_\mu$ and $s = t+b$.
The contributions to zeroes at $[s]$ of $T_{\mu\*\nu}$.
Here, $b\in \nu$, $t' \in \remset^+_\mu \cup (0,0)$ and $s = \bps{t} + b$.

Fix $\mu$ and some $s \in \setZ^2$.
The orders of the poles of $T_{\mu \* \nu}(u)$ are bounded by

\[
1- \max( |\addset_\mu|,|\addset_\nu|) \leq \mathrm{ord}_{u=[s]} T_{\mu\*\nu}(u) \leq 1 .
\]

Furthermore, $[\bms{s}]$ is a simple pole of $T_{\mu\*\nu}(u)$  if and only
if $\bms{s} \notin \nu$ and $\lambda \subseteq \nu$,
where $\lambda^R$ is the shape between $\mu$ and $s$.

According to Lemma <ref>, in order for $T_{\mu\*\nu}(u)$ to have a
pole $[\bms{s}]$, we must include an outer corner $t\in \addset_\mu$ in $\nu^R \coloneqq \bms{s}-\nu$.
The smallest shape that $\nu^R$ can be so that it contains $t$ and $\bms{s}$ is the rectangle
of boxes between $t$ and $s$, denoted $\Rect(t,s)$.
However, $\Rect(t,s)$ also includes the two (shifted) inner corners $m_1, m_2 \in \remset^+_\mu$ adjacent to $t$,
as seen in Figure <ref>.

Whenever we include an outer corner $t\in \addset_\mu$ in $\bms{s}-\nu$, we must also
include the adjacent inner corners $m_1, m_2 \in \remset^+_\mu$.

Thus according to Lemma <ref>,
$T_{\mu\* \Rect(0,s-t)}(u)$ will have a zero of order $1$ at $u=[\bms{s}]$.
We repeat this inclusion of rectangles for each inner corner of $\mu$ that is inside $\Rect(s)$.
This process yields $\lambda^R \coloneqq \cup_{t \in (\addset_\mu \cap \Rect(s))} \Rect(t,s)$, which is
precisely the shape between $\mu$ and $s$. It is clear from Figure <ref> that $\lambda^R$
contains one more box from $\addset_\mu$ than from $\remset^+_\mu$, and thus $T_{\mu\* \lambda}(u)$
has a pole at $[\bms{s}]$. Thus, if $\nu^R \supset \lambda^R$, we should have a pole of degree $1$.
However, this pole is annihilated if $\nu^R$ also includes the zero coming from $(0,0) \in \mu$, that is, if $\bms{s} \in \nu$.

On the left, the largest degree zero is obtained when $\nu^R \coloneqq \bms{s}-\nu$ contains
all of $\remset^+_\mu$, and avoids $\addset_\mu$.
On the right, a simple pole occurs when $\nu^R$ contains $\lambda^R$ and avoids $(0,0)$.

To describe the corners to the left (or right) of a particular
corner $s=(s_1,s_2)$, we introduce the notation

\[
\defin{\remset_\sigma^{<s}} \coloneqq  \{ t = (t_1,t_2) \in \remset_\sigma : t_1 < s_1 \},
\]

and mutatis mutandis for $\defin{\remset_\sigma^{>s}}$,
$\defin{\addset_\sigma^{<s}}$ and $\defin{\addset_\sigma^{>s}}$.
One can check directly the following reinterpretations
of differences of corners as hook vectors.

For fixed $s \in \remset_\sigma$, the following hold:

For $t \in \remset_\sigma^{<s}$,
we have $\bps{s} - \bps{t} = \uhv{\sigma \setminus s}{\bmin{s}{t}}$.

For $t \in \addset_\sigma^{<s}$, we have $\bps{s} - t = \uhv{\sigma}{\bmin{s}{t}}$.

For $t \in \remset_\sigma^{>s}$, we have $\bps{s} - \bps{t} = -\lhv{\sigma \setminus s}{\bmin{s}{t}}$.

For $t \in \addset_\sigma^{>s}$, we have $\bps{s} - t = -\lhv{\sigma}{\bmin{s}{t}}$.

The proof of Lemma <ref> is straighforward and
we simply refer to Figure <ref>.

The difference between partition corners are hook lengths.
In the first figure, $\bps{s} - \bps{t} = \uhv{\sigma\setminus s}{\bmin{s}{t}}$, while in the second figure we have $\bps{s} - t = \uhv{\sigma}{\bmin{s}{t}}$. The other two cases of Lemma <ref> are verified in a similar manner.

Let $\sigma$ be a partition, and pick $s \in \remset_\sigma$.
Then $T_{\sigma}(u)$ can be expanded[Alternatively, one may shift the argument to write the simpler expression,

\begin{equation*}
T_{\sigma}(u+[\bps{s}]) = u (u+[\bps{s}])
\cdot
\tfrac{
\prod_{b \in R^s_{\sigma \setminus s}} \left( u+  \uh{\sigma  \setminus s}{b}  \right) }{
\prod_{b \in R^s_\sigma } \left( u+ \uh{\sigma}{b} \right)
\cdot
\tfrac{
\prod_{b \in C^s_{\sigma \setminus s}  }\left(u- \lh{\sigma \setminus s}{b} \right) }{
\prod_{b \in C^s_{\sigma}}\left(u-\lh{\sigma}{b} \right)}.
\end{equation*}

] as a row-column product relative to $s$ as

\begin{equation*}
T_{\sigma}(u) = u \cdot \left(u-[\bps{s}]\right)
\cdot
\tfrac{
\displaystyle\prod_{b \in R^s_{\sigma \setminus s}} \left( u-[\bps{s} - \uhv{\sigma  \setminus s}{b} ] \right) }{
\displaystyle\prod_{b \in R^s_\sigma } \left( u-[\bps{s} - \uhv{\sigma}{b} ] \right)
\cdot
\tfrac{
\displaystyle\prod_{b \in C^s_{\sigma \setminus s}  }\left(u-[\bps{s} + \lhv{\sigma \setminus s}{b}] \right) }{
\displaystyle\prod_{b \in C^s_{\sigma}}\left(u-[\bps{s} + \lhv{\sigma}{b}] \right)}.
\end{equation*}

Each of the four factors in the product precisely captures the zeros and poles in each of the four quadrants with a corner in $s$,
see Figure <ref>.

The four quadrants of $T_{\mu}(u)$ expanded around $u=[s]$ for $s\in \remset_\mu$.

Recall from (<ref>) that

\begin{align*}
T_{\sigma}(u) &= u\frac{\prod_{t \in \remset_\sigma }(u-[\bps{t}])}{\prod_{t
\in\addset_\sigma }(u-[t ])}  \\
&= u\frac{\prod_{t \in  \remset_\sigma } (u-[\bps{s} - (\bps{s}-\bps{t})])}{\prod_{t
\in\addset_\sigma }(u-[\bps{s} - (\bps{s}-t)])}.
\end{align*}

Using Lemma <ref>, we split this according to $\remset_\sigma = \remset_\sigma^{<s} \cup \{ s\}\cup \remset_\sigma^{>s}$, as

\[
\frac{
\prod_{t \in \remset_\sigma^{<s}}
\left(u-[\bps{s} - \uhv{\sigma \setminus s}{ \bmin{s}{t}  }]\right)
\prod_{t \in \addset_\sigma^{<s}}
\left(u-[\bps{s}- \uhv{\sigma}{  \bmin{s}{t} }]\right)
\frac{
\prod_{t \in \remset_\sigma^{>s}}
\left(u-[\bps{s} + \lhv{\sigma \setminus s}{ \bmin{s}{t} }]\right)
\prod_{t \in \addset_\sigma^{>s}}
\left(u-[\bps{s}+ \lhv{\sigma}{ \bmin{s}{t} } ]\right)
\]

Note that in the first fraction, $\bmin{s}{t}$
is always a box in the same row as $s$,
while in the second fraction, $\bmin{s}{t}$
is a box in the same column as $s$.
Next, we claim the that following holds for the first ratio of products:

\begin{equation}\label{eq:cornersumtorowsum}
\frac{
\prod_{t \in \remset_\sigma^{<s}}
(u-\uh{\sigma \setminus t}{\bmin{s}{t} })}
\prod_{t \in \addset_\sigma^{<s}}
(u-\uh{\sigma}{\bmin{s}{t} })} = \frac{\prod_{b \in R^s_{\sigma \setminus s}  }(u-\uh{\sigma \setminus s}{b})}{
\prod_{b \in R^s_\sigma  }(u-\uh{\sigma}{b})} .
\end{equation}

That is, the ratio of products which only receives contributions from
corners can be extended to include all boxes in the row of $s$.
This holds because for two adjacent boxes $b, b+(1,0)$ in $R_\sigma^s$, with $b$ not in the same column as an inner corner, we have

\[
\uhv{\sigma\setminus s}{b} =  \uhv{\sigma}{b+(1,0)}.
\]

Thus there is pairwise cancellation between pairs of boxes in the right
hand side of (<ref>),
unless $b$ shares a column with an inner corner, in which case $b+(1,0)$ shares a column with the adjacent outer corner, see Figure <ref>.

Relation between $\uhv{\sigma\setminus s}{b}$ and $\uhv{\sigma}{b+(1,0)}$.
In the first diagram, we have that $\uhv{\sigma\setminus s}{b} = \uhv{\sigma}{b + (1,0)}$,
but this does not hold in the second diagram as $b$ shares a column with an inner corner.

We find a similar result for the left hand ratio of products,
this time with products over columns:

\begin{equation*}\label{eq:cornersumtocolsum}
\frac{
\prod_{t \in \remset_\sigma^{>s}}
(u-\lh{\sigma \setminus t}{\bmin{s}{t} })}
\prod_{t \in \addset_\sigma^{>s}}
(u-\lh{\sigma}{\bmin{s}{t} })} = \frac{\prod_{b \in C^s_{\sigma \setminus s}  }(u-\lh{\sigma \setminus s}{b})}{
\prod_{b \in C^s_\sigma  }(u-\lh{\sigma}{b})}.
\end{equation*}

Piecing these two results together this becomes

\[
= u(u-[\bps{s}]) \frac{\prod_{b \in R^s_{\sigma \setminus s}  }(u-[\bps{s} - \uhv{\sigma \setminus s}{b}])}{\prod_{b \in R^s_\sigma  }(u-[\bps{s} - \uhv{\sigma}{b}])}\frac{\prod_{b \in C^s_{\sigma \setminus s}  }(u-[\bps{s} + \lhv{\sigma \setminus s }{b}])}{\prod_{b \in C^s_\sigma  }(u-[\bps{s}+ \lhv{\sigma}{b}])}.
\]

§ RECTANGULAR CASE

Let $c_{\mu\nu}^{\lambda}$  denote the Schur Littlewood–Richardson coefficients. One can easily show the following result.

If $\lambda=m^n$ is a rectangular partition, then
the (Schur) Littlewood–Richardson coefficient
satisfies $c_{\mu\nu}^{\lambda} \in \{0,1\}$,
and takes the value $1$ if and only $\mu \subset m^n$ and $\nu = \bmu$.

Among all integer partitions $\lambda \vdash mn$,
the rectangular partition $m^n$ is the unique partition
containing the box with coordinate $\bms{v}=(m-1,n-1)$.
This simple fact allows us to extract the
rectangular Jack Littlewood–Richardson coefficients
from the formula (<ref>).

Let  $\mu \subseteq m^n$. Then

\begin{equation*}
\langle \jackj_{\mu}\jackj_{\bmu},\jackj_{m^n} \rangle =
\left( \Res_{u=[\bms{v}]}  T_{\mu\*\bmu}(u) \right) \cdot
\norm{ \jackj_{m^n} }^2 \cdot \frac{ \jacktop_\mu \jacktop_{\bmu}} {\jacktop_{m^n}}.
\end{equation*}

To evaluate the residue in Corollary <ref>,
we will need to build up a sequence of structural results about $T_{\mu\*\bmu}(u)$.
We work always in reference to the fixed rectangle $m^n$.
For any box $b$, we define the complementary box $\defin{\bb} \coloneqq \bms{v} - b$. For every $\sigma \subset m^n$, the map $t \to \bar t$, maps $\remset_\sigma$ into $\addset_{\bsigma}$, and the image misses only a single outer corner of $\bsigma$ at the edge.

Let $\sigma \subset m^n$ and recall that $\bsigma$
represents the complementary partition of $\sigma$
inside the rectangle.
Fix the inner corner $t \in \remset_\sigma$ and let
$\bt \in \addset_{\bsigma}$ be the
corresponding outer corner of $\bar{\sigma}$.

* For $s \in \addset_{\sigma}$,
we have $\eh{\sigma}{\bmin{s}{t}} = \eh{\bsigma}{ \bmin{\bs}{\bt} }$;

* for $w \in \remset_\sigma$,
we have $\eh{\sigma\setminus t}{ \bmin{w}{t} } = \eh{\bsigma\cup\bt}{ \bmin{\bw}{\bt} }$,

where $\defin{\eh{\sigma}{b}}$ is either $\uh{\sigma}{b}$ or $\lh{\sigma}{b}$.
See Figure <ref>.

The hook complement for boxes to the left of $t$ and to the right of $t$, respectively.
In the left figure, we see for example that
$\uh{\sigma\setminus t}{\bmin{w}{t}} = \uh{\bsigma \cup \bt}{\bmin{\bw}{\bt}}$ and in the right figure we have
for example $\uh{\sigma}{\bmin{s}{t}} = \uh{\bsigma}{\bmin{\bs}{\bt}}$.

The next lemma conveys how we can `flip' row/column box
products in $\sigma$ to column/row box products in $\bsigma$.

For $t \in \remset_\sigma$, we have

\begin{align}\label{eq:fliprules}
\frac{\prod_{b \in C^t_{\sigma \setminus t}  }(u-\uh{\sigma \setminus t}{b})}{
\prod_{b \in C^t_\sigma  }(u-\uh{\sigma}{b})} &= \frac{1}{(u-[(m,0)-t])}
\frac
{\prod_{b \in R^{\bt}_{\bsigma}  }(u- \uh{\bsigma \cup \bt}{b})}
{\prod_{b \in R^{\bt}_{\bsigma} }(u-\uh{\bsigma}{b} )}, \\
\frac{
\prod_{b \in R^t_{\sigma \setminus  t}  }(u- \lh{\sigma \setminus  t}{b}   )}{
\prod_{b \in R^t_\sigma  }(u-\lh{\sigma}{b} )} &= \frac{1}{(u-[t-(0,n)]) }
\frac
{\prod_{b \in C^{\bt}_{\bsigma} }(u- \lh{\bsigma \cup \bt}{b} )}
{\prod_{b \in C^{\bt}_{\bsigma} }(u- \lh{\bsigma}{b} )}.
\end{align}

From the proof of the expansion Lemma <ref>, we know that the ratio of column products can be expressed as

\begin{equation}\label{eq:preflipped}
\frac{\prod_{b \in C^t_{\sigma \setminus t}  }(u-\uh{\sigma \setminus t}{b})}{
\prod_{b \in C^t_\sigma  }(u-\uh{\sigma}{b})} = \frac{\prod_{s \in \remset_{\sigma}^{>t}}(u-\uh{\sigma \setminus t}{\bmin{s}{t} })}{
\prod_{s \in \addset_{\sigma}^{>t} }(u-\uh{\sigma}{\bmin{s}{t} })} .
\end{equation}

We now use Lemma <ref> to flip each of the terms in this product. All the inner corners to the right of $t$ can be flipped this way, however if $(m,0) \in \addset_\sigma^{>t}$, that is, if $\sigma$ runs to the right edge of $m^n$, then there is no corresponding element in $\remset_{\bsigma}$. We treat this case first, and then consider case when $(m,0) \notin \addset_\sigma^{>t}$. We claim that in both cases the following holds:

\begin{equation}\label{eq:flippoles}
\frac{\prod_{s \in \remset_{\sigma}^{>t}}(u-\uh{\sigma \setminus t}{\bmin{s}{t}})}{
\prod_{s \in \addset_{\sigma}^{>t} }(u-\uh{\sigma}{\bmin{s}{t}})} = \frac{1}{(u-[(m,0)-t])}\frac
{\prod_{\bs \in \addset_{\bsigma}^{< \bt} }(u- \uh{\bsigma \cup  \bt}{\bmin{\bs}{\bt}})}
{\prod_{ \bs \in \remset_{\bsigma}^{<\bt} }(u-\uh{\bsigma}{\bmin{\bs}{\bt}} )}.
\end{equation}

In the case $(m,0) \in \addset_\sigma^{>t}$ we have contribution in the LHS denominator coming from $\bmin{t}{(m,0)}= (t_1,0) $.
As this term cannot be flipped, it contributes an extra pole on the RHS at

\[
\uhv{\sigma}{(t_1,0)} = (m-t_1,-t_2) = (m,0)-t,
\]

which appears in (<ref>).

In the case where $(m,0) \notin \addset_\sigma^{>t}$, we must now have $(0,n) \in \addset_{\bsigma}^{<\bt}$ and that this corner of $\bsigma$ has no corresponding corner in $\sigma$. In the RHS numerator of (<ref>) we find a contribution from $\bmin{\bt}{(0,n)} = (0,\bt_2)$, for which

\[\uhv{\bsigma+\bt}{(0,\bt_2)} = (\bt_1 +1, -(n-1-\bt_2)) =  \bps{\bt}-(0,n) = (m,0)-t.\]

As this zero in the numerator of the RHS of (<ref>) does not appear as a flipped hook from the left hand side, this extra zero will have to be cancelled by the same pole factor that appeared in first case. Thus (<ref>) holds in both cases, and we can then extend the products in the right hand side out over the full row $R_{\bsigma}^{\bt}$ (using a formula similar to equation (<ref>)) to recover equation (<ref>). By a similar technique, we find the other flip rule for row products.

The extra pole factors entering into the formulas of the
flip Lemma <ref> have an alternate interpretation
in terms of hooks in the rectangular partition $m^n$.

Recall that we have a fixed rectangular partition $m^n$ in mind.
For $b=(b_1,b_2)$ we denote $\defin{\hflip{b}} \coloneqq (m-1-b_1,b_2)$
and $\defin{\vflip{b}} \coloneqq (b_1,n-1-b_2)$.

We have that

\[
(m,0) - b = \uhv{m^n}{\vflip{b}}
\quad \text{and} \quad
b - (0,n) = \lhv{m^n}{\hflip{b}}.
\]

See Figure <ref>.

One can easily see that $(m,0) - b = \uhv{m^n}{\vflip{b}}$
and $b - (0,n) = \lhv{m^n}{\hflip{b}}$.

We now arrive at the main structural result, which expresses the rational
function $T_{\mu\*\bmu}(u)$ as a product involving hook lengths.

The function $T_{\mu\*\bmu}(u)$ is given by the following expression[As before, one may shift the argument to write a simplified expression,

\begin{equation*}\label{mumudecompalt}
T_{\mu\*\bmu}(u+[\bms{v}]) = \prod_{b \in \mu} \frac{u+[\bb]}{u+[b] }
\prod_{b \in \bmu} \frac{u+ \lh{\bmu}{b} }{u+\lh{m^n}{{\vflip{ b}}}}
\prod_{b \in \mu} \frac{u-\uh{\mu}{ b}}{u-\uh{m^n}{\hflip{b}}}.
\end{equation*}

\begin{equation}\label{eq:mumudecomp}
T_{\mu\*\bmu}(u) = \prod_{b \in \mu} \frac{u-[b]}{u-[\bb] }
\prod_{b \in \bmu} \frac{u-[\bms{v} - \lhv{\bmu}{b} ]}{u-[\bms{v}-\lhv{m^n}{{\vflip{ b}}}]}
\prod_{b \in \mu} \frac{u-[\bms{v}+\uhv{\mu}{ b}]}{u-[\bms{v}+\uhv{m^n}{\hflip{b}}]},
\end{equation}

Furthermore, each of the three products in the expression (<ref>) is the component of the LHS explicitly decomposed with respect to quadrants around $\bms{v}$, as in Fig. <ref>. Thus, each of these factors, as a function of a partition $\mu \subset m^n$, is manifestly invariant under $\mu \to \bmu$.

The three quadrants of $T_{\mu\*\bmu}(u)$ around $u=[\bms{v}]$. Note that each region contains an equal number of poles and zeros

We prove this by induction on the number of boxes in $\mu$.
We start with the base case $\mu_0 = \{1\}$ and so $\bar\mu_0=m^n\setminus\bms{v}$, where we find

\begin{eqnarray*}
T_{\mu_0\*\bmu_0}(u) &=& T_{1\*(m^n \setminus \bms{v} )}(u)\\
&=& \frac{u-[0,0]}{u-[m-1,n-1]} \frac{u-[m-1,n]}{u-[0,n]} \frac{u-[m,n-1]}{u-[m,0]}.
\end{eqnarray*}

Of the three terms in this product, the first and last agree with the statement of the theorem, with the product over the the only box $b=(0,0)$ of $\mu_0 = \{1\}$. For the middle factor, we need to check

\begin{equation*}
\prod_{b \in \bmu_0} \frac{u-[\bms{v} - \lhv{\bmu_0}{b} ]}{u-[\bms{v}-\lhv{m^n}{{\vflip{ b}}}]} =  \frac{u-[m-1,n]}{u-[0,n]}.
\end{equation*}

To show this holds, we first fill out the bottom product from $\bmu_0 \to m^n$

\[ \prod_{b \in \bmu_0} \tfrac{u-[\bms{v} - \lhv{\bmu_0}{b} ]}{u-[\bms{v}-\lhv{m^n}{{\vflip{ b}}}]} =(u-[\bms{v}-\lhv{m^n}{{\widehat{\bms{v}}}}])\tfrac{\prod_{b \in \bmu_0} u-[\bms{v} - \lhv{\bmu_0}{b} ]}{\prod_{b \in m^n} u-[\bms{v}-\lhv{m^n}{{b}}]}.  \]

Next, we similarly fill out the numerator product by carefully expanding the row/columns containing $\bms{v}$, which yields

\[ \prod_{b \in \bmu_0} u-[\bms{v} - \lhv{\bmu_0}{b} ] = \frac{(u-[\bms{v} - \lhv{m^n}{\bms{v}}]) \prod_{b \in m^n} (u-[\bms{v} - \lhv{m^n}{b} ])}{(u-[\bms{v} - \lhv{m^n}{(0,n-1)} ])(u-[\bms{v} - \lhv{m^n}{(m-1,0)} ])} . \]

Combining these previous two results, we find

\[ \prod_{b \in \bmu_0} \frac{u-[\bms{v} - \lhv{\bmu}{b} ]}{u-[\bms{v}-\lhv{m^n}{{\vflip{ b}}}]}  = \frac{(u-[\bms{v} - \lhv{m^n}{\bms{v}} ])(u-[\bms{v}-\lhv{m^n}{{\widehat{\bms{v}}}}])}{(u-[\bms{v}- \lhv{m^n}{(0,n-1)} ])(u-[\bms{v} - \lhv{m^n}{(m-1,0)} ])}.  \]

Now using $\widehat{\bms{v}} = (m-1,0)$, $\lhv{m^n}{\bms{v}} = (0,-1)$, and $\lhv{m^n}{(0,n-1)} = (m-1,-1)$, we arrive at

\[\prod_{b \in \bmu_0} \frac{u-[\bms{v} - \lhv{\bmu}{b} ]}{u-[\bms{v}-\lhv{m^n}{{\vflip{ b}}}]} =  \frac{u-[\bms{v} -(0,-1)  ]}{u-[\bms{v} - (m-1,-1) ]} =  \frac{u-[m-1,n]}{u-[0,n]}. \]

Thus we have shown the base case.
Next, for the inductive step, we assume the theorem holds for $\mu$, and we show it holds for $\mu+s$. Here, we have $\overline{\mu+s} = \bmu - \bs$ where  $s+\bs = \bms{v}$.
We then find

\begin{eqnarray*}
T_{\mu+s\*\overline{\mu+s}}(u) &=& T_{\mu+s\*(\bmu - {\bs})}(u)\\
&=&  \frac{ T_{s\* \bmu}(u)}{ T_{(\mu+s)\*{\bs}}(u)}T_{\mu\*\overline{\mu}}(u)\\
&=& \frac{ T_{\bmu}(u-[s])}{ T_{\mu+{s}}(u-[\bs])}  T_{\mu\*\overline{\mu}}(u).
\end{eqnarray*}

We now simplify this factor in the last line by using the
expansion in Lemma <ref> to expand
the numerator at $\bs \in \remset_{\bmu}$
and the denominator at $s \in \remset_{\mu+s}$:

\[
T_{\bmu}(u-[s]) = (u-[s])(u-[v])
\tfrac{\prod_{b \in R^{\bs}_{\bmu-\bs}  }(u-[v- \uhv{\bmu-\bs}{b} ])}{\prod_{b \in R^{\bs}_{\bmu}  }(u-[v-\uhv{\bmu}{b}])}
\textcolor{blue}{ \tfrac{\prod_{b \in C^{\bs}_{\bmu-\bs}  }(u-[v+\lhv{\bmu-\bs}{b}])
}{\prod_{b \in C^{\bs}_{\bmu} }(u-[v+\lhv{\bmu}{b}])}}.    \]

\[ T_{ \mu+s}(u-[\bs]) = (u-[\bs])(u-[v])
\textcolor{blue}{ \tfrac{\prod_{b \in R^s_{\mu} }(u-[v-\uhv{\mu}{b}])}{
\prod_{b \in R^s_{\mu+s}  }(u-[v-\uhv{\mu+s}{b}])}}
\tfrac{\prod_{b \in C^s_\mu  }(u-[v+ \lhv{\mu}{b}])}{\prod_{b \in C^s_{\mu+s}  }(u-[v+\lhv{\mu+s}{b}])}.
\]

We use Lemma <ref> to flip the blue factors
to the following red factors,
and use Equation <ref>,
to arrive at the expressions

\[
T_{\bmu}(u-[s]) = \tfrac{(u-[s])(u-[v])}{
\textcolor{red}{(u-[v+(m,0)-\bps{\bs}])}}
\tfrac{\prod_{b \in R^{\bs}_{\bmu-\bs}  }(u-[v-\uhv{\bmu-\bs}{b}])}{
\prod_{b \in R^{\bs}_{\bmu}  }(u-[v-\uhv{\bmu}{b}])}
\textcolor{red}{ \tfrac{\prod_{b \in R^s_{\mu} }(u-[v+\lhv{\mu+s}{b}])}{
\prod_{b \in R^s_{\mu} }(u-[v+\lhv{\mu}{b}])} },
\]

\[
T_{ \mu+s}(u-[\bs]) = \tfrac{(u-[\bs])(u-[v])}{\textcolor{red}{(u-[v-(\bps{s}-(0,n))])}}
\textcolor{red}{  \tfrac{\prod_{b \in C^{\bs}_{\bmu-\bs}  }(u-[v-\uhv{\bmu}{b}])}{
\prod_{b \in C^{\bs}_{\bmu-\bs}  }(u-[v-\uhv{\bmu-\bs}{b}])} }
\tfrac{\prod_{b \in C^s_\mu  }(u-[v+\lhv{\mu}{b}])}{\prod_{b \in C^s_{\mu+s}  }(u-[v+\lhv{\mu+s}{b}])}.
\]

The ratio of these terms then fills out to products over the entire partitions $\mu$, $\bmu$, as the $s$-row-column products are the only ones that are changed by adding in $s$, and so we arrive at the following expression,

\begin{eqnarray}\label{eq:bigprod}
\tfrac{T_{\bmu}(u-[s])}{T_{ \mu+s}(u-[\bs])} &=&
\tfrac{ (u-[s]) }{ (u-[\bs]) } \tfrac{ (u-[\bms{v}- (s-(0,n))]) }{ (u-[\bms{v}+(m,0)-\bs]) } \\
\times \tfrac{\prod_{b \in \bmu-\bs  }(u-[v-\uhv{\bmu-\bs}{b}])}{\prod_{b \in {\bmu} }(u-[v-\uhv{\bmu}{b}])}\tfrac{\prod_{b \in \mu+s  }(u-[v+\lhv{\mu+s}{b}])}{\prod_{b \in \mu  }(u-[v+\lhv{\mu}{b}])}.
\end{eqnarray}

We consider each of these four terms separately, and note their effect on the formula (<ref>).
For the first term of formula <ref>, it correctly updates the contribution from $s$ in the first term in formula (<ref>), i.e.

\begin{equation*}
\frac{ (u-[s])}{(u-[\bs])} \prod_{b \in \mu} \frac{u-[b]}{u-[\bb] } = \prod_{b \in \mu \cup s} \frac{u-[b]}{u-[\bb] }.
\end{equation*}

The second term of formula (<ref>) may be rewritten using Lemma <ref>, as

\begin{equation*}\frac{ u-[\bms{v}-\lhv{m^n}{\hflip{s}}] }{ u-[\bms{v}+\uhv{m^n}{\vflip{\bs}}] }. \end{equation*}

If we note that $\vflip{\bs} = \hflip{s}$, we see that this term is precisely what is needed to include the $s$ contribution (resp. remove the $\bs$ contribution) in the denominator of the third (resp. second) term in formula (<ref>).
The last two terms in formula (<ref>)  update the numerators of the last two terms in formula (<ref>) respectively.
Thus we see that the factor of $\tfrac{T_{\bmu}(u-[s])}{T_{ \mu+s}(u-[\bs])}$ precisely is what is required for the inductive step to work.

Note: As the function $T_{\mu\*\bmu}$ is manifestly symmetric under $\mu \leftrightarrow \bmu$, it clear that formula (<ref>) is also symmetric under this exchange. As a consequence, each of the three factors in formula (<ref>) must also be symmetric, as they are expressed explicitly as products of poles/zeros (of the form $u-[\bms{v} \pm \lhv{}{b}]$) of the function $T_{\mu\*\bmu}(u)$ that lie in a distinct quadrant of the $u$-plane relative to $[\bms{v}]$.

With this we can provide a new proof of a result first shown by <cit.>.

The strong Stanley conjecture holds in the rectangular case.
In particular, the rectangular Littlewood–Richardson coefficient is given by

\begin{equation}\label{eqn:rectangularLR}
\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}(b) },
\end{equation}

where[Note here that the function $D'$ differs from $D$ given in formula (<ref>) by flipping the hooks $U \leftrightarrow L$, since we are writing an expression for the LR coefficients rather than the Stanley inner products.]

\begin{equation*}
D'_{\mu,m^n}(b) \coloneqq
\begin{cases}
\lh{m^n}{b} & \text{ if $\vflip{b} \in \mu$} \\
\uh{m^n}{b} &  \text{ otherwise}.
\end{cases}
\end{equation*}

The rectangular case of the Littlewood–Richardson coefficient, $\jackjLR_{\mu, \bmu}^{m^n}$

As stated in Corollary <ref>, we have

\begin{equation*}
\jackjLR_{\mu,\bmu}^{m^n} =\frac{\jacktop_\mu \jacktop_{\bmu}} {\jacktop_{m^n}}  \Res_{u=[\bms{v}]} T_{\mu\*\bmu}(u).
\end{equation*}

Using formula <ref>, we find

\begin{equation}\label{eq:resexpression}
\Res_{u=[\bms{v}]}  T_{\mu\*\bmu}(u) =
\frac{\prod_{b \in \mu} [\bb]}{\prod_{b \in \mu^\times} [b] }
\prod_{b \in \bmu} \frac{\lh{\bmu}{b} }{\lh{m^n}{\vflip{b}}} \prod_{b \in \mu} \frac{\uh{\mu}{b}}{\uh{m^n}{\hflip{b}}}.
\end{equation}

As noted earlier, we know that each of the three factors in (<ref>) remains
invariant under the switch $\mu \leftrightarrow \bmu$, so we are free to
switch each of the last two factors.
This provides a direct proof that this result is symmetric in $\mu$ and $\bmu$, and gives an alternate choices of hooks,

\begin{equation*}
\Res_{u=[\bms{v}]}  T_{\mu\*\bmu}(u) =
\frac{\prod_{b \in \mu} [\bb]}{\prod_{b \in \mu^\times} [b] }
\prod_{b \in \mu} \frac{\lh{\mu}{b} }{\lh{m^n}{\vflip{b}}} \prod_{b \in \bmu} \frac{\uh{\bmu}{b}}{\uh{m^n}{\hflip{b}}}.
\end{equation*}

We group these terms as follows,

\begin{eqnarray*}
&=& \frac{ \jacktop_{m^n}/\jacktop_{\bmu} }{ \jacktop_{\mu} }   \frac{ \prod_{b \in \mu}\lh{\mu}{b} \prod_{b \in \bmu}  \uh{\bmu}{b} }{\left( \prod_{b \in \mu} \lh{m^n}{\vflip{b}}   \prod_{b \in \bmu} \uh{m^n}{\hflip{b}} \right) }.
\end{eqnarray*}

The last thing we need to check is the following expression for the denominator

\begin{equation*}
\prod_{b \in \mu} \lh{m^n}{\vflip{b}}   \prod_{b \in \bmu} \uh{m^n}{\hflip{b}} = \prod_{b \in \mu} \lh{m^n}{\vflip{b}}   \prod_{b \in m^n/\mu} \uh{m^n}{\hflip{\bb}}  = \prod_{b \in m^n} D'_{\mu,m^n}(b),
\end{equation*}

where we have used $\hflip{\bb} = \vflip{b}$.

Note, that by swapping $\mu \to \bmu$ in formula (<ref>)
we have the two representations of this Littlewood–Richardson coefficient (see Figure <ref>):

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

The two choices of hooks for the Littlewood–Richardson coefficients in the rectangular case. Each of the two fractions on either side of the equation are equal to the corresponding fraction on the other side. We color in blue the boxes that don't have hooks assigned to them.

§ RECTANGULAR UNION CASE

In this section we prove a generalization of the previous results.
Fix $\mu$ and $m^n$, without necessarily assuming that $\mu \subset m^n$ as in the rectangular case. Let $v = (n,m)$, and $\bms{v} = (n-1,m-1)$.
Let $\sigma = \mu \cap m^n$.
Let $\bsigma$ be the conjugation of $\sigma$ taken with respect to $m^n$, as illustrated in Figure <ref>.

The setup of the rectangular union case.

We have

\begin{equation}\label{eq:claim1}
\jackjLR_{\mu,\bsigma}^{\mu\cup m^n} =
\jackjLR_{\sigma,\bsigma}^{ m^n} \cdot
\prod_{b \in \mu/\sigma} T_{\bsigma} \left([\bb] \right)  .
\end{equation}

First, by Lemma <ref>, we know $ T_{\mu,\bsigma}(u)$ has a pole at $u=[\bms{v}]$. By formula <ref>, we know that

\begin{equation*}
\sum_{\gamma} \hat  \jackjLR_{\mu,\bsigma}^\gamma \sum_{b \in \gamma} \frac{1}{u-[b]} = T_{\mu\*\bsigma}(u)-1.
\end{equation*}

The residue of this equality at $u = [\bms{v}]$ is

\begin{equation*}
\sum_{\gamma: \mu\cup \bsigma \subseteq \gamma, \bms{v} \in \gamma} \hat \jackjLR_{\mu,\bsigma}^\gamma  =\Res_{u=[\bms{v}]} T_{\mu\*\bsigma}(u).
\end{equation*}

Because $\mu\cup m^n$ is the only partition of its size that contains $\mu$ and the box at $\bms{v}$, the sum on the LHS contains only the $\gamma=\mu\cup m^n$ term and we conclude

\[
\hat \jackjLR_{\mu,\bsigma}^{\mu\cup m^n}   =\Res_{u=[\bms{v}]} T_{\mu\*\bsigma}(u).
\]

Next, we note that

\[
T_{\mu\*\bsigma}(u) =  T_{(\mu/\sigma)\*\bsigma}(u) T_{\sigma\*\bsigma}(u),
\]

and as the second factor has a pole $u=[\bms{v}]$, the first factor is regular at $u=[\bms{v}]$. So,

\[
\Res_{u=[\bms{v}]} T_{\mu\*\bsigma}(u) = T_{(\mu/\sigma)\*\bsigma}([\bms{v}]) \Res_{u=[\bms{v}]} T_{\sigma\*\bsigma}(u).
\]

We then use

\[
T_{\{s\} *\bsigma}([\bms{v}])=T_{\bsigma}([\bms{v}-s]),
\]

to arrive at

\begin{equation}\label{eq:withhats}
\hat \jackjLR_{\mu,\bsigma}^{\mu\cup m^n} = \left( \prod_{b \in \mu/\sigma} T_{\bsigma}([\bms{v}-b]) \right) \cdot  \hat\jackjLR_{\sigma,\bsigma}^{ m^n} .
\end{equation}

Finally, we notice that

\begin{equation*}
\frac{\varpi_\mu \varpi_{\bsigma}}{\varpi_{\mu\cup m^n}} = \frac{\varpi_\sigma \varpi_{\bsigma}}{\varpi_{m^n}},
\end{equation*}

so we can drop the hats in equation <ref>.

In order to expand the product term in (<ref>), we produce a formula for the factors appearing therein.

For $\mu$ as before, and $s \in \addset_{\mu}$ with $s$ outside of $m^n$, then

\begin{equation}\label{eq:tboxfactor}
T_{\bsigma}([\bar s]) = \frac{
\frac{\displaystyle\prod_{b \in R^s_{\mu \cup m^n}  } \uh{\mu \cup m^n}{b} }{
\displaystyle\prod_{b \in R^s_{(\mu+s) \cup m^n} } \uh{(\mu+s) \cup m^n}{b} } \times
\frac{\displaystyle\prod_{b \in C^s_{\mu \cup m^n}  } \lh{\mu \cup m^n}{b}}{
\displaystyle\prod_{b \in C^s_{(\mu+s) \cup m^n}  } \lh{(\mu+s) \cup m^n}{b}}}{
\frac{\displaystyle\prod_{b \in R^s_{\mu} } \uh{\mu }{b} }{
\displaystyle\prod_{b \in R^s_{\mu+s}   } \uh{\mu+s}{b}} \times  \frac{
\displaystyle\prod_{b \in C^s_\mu  } \lh{\mu}{b}}{
\displaystyle\prod_{b \in C^s_{\mu+s}  } \lh{\mu+s}{b} }}.
\end{equation}

We start by noting that for any $\mu$ such that $\mu \cap m^n = \sigma$, we have

\[
T_{\bsigma}([n,m]-u)  = \frac{T_{\mu \cup m^n}(u)}{T_{\mu}(u)}.
\]

\[
T_{\bsigma}([\bar s])=T_{\bsigma}([(n,m)-(\bps{s})])=
\frac{T'_{(\mu+s) \cup m^n}([\bps{s}])}{T'_{(\mu+s)}([\bps{s}])},
\]

where $T'$ indicates that we have dropped the zero at $u=[\bps{s}]$.
We then use the expansion Lemma <ref> to write the terms in this ratio as

\[
T'_{(\mu+s) \cup m^n}([\bps{s}]) =- [\bps{s}] \times
\frac{\prod_{b \in R^s_{\mu \cup m^n}  } \uh{\mu \cup m^n}{b} }{
\prod_{b \in R^s_{(\mu+s) \cup m^n} } \uh{(\mu+s) \cup m^n}{b} } \times
\frac{\prod_{b \in C^s_{\mu \cup m^n}  } \lh{\mu \cup m^n}{b}}{
\prod_{b \in C^s_{(\mu+s) \cup m^n}  } \lh{(\mu+s) \cup m^n}{b}}.
\]

\[ T'_{(\mu+s)}([\bps{s}]) = - [\bps{s}] \times  \frac{\prod_{b \in R^s_{\mu} } \uh{\mu }{b} }{
\prod_{b \in R^s_{\mu+s}   } \uh{\mu+s}{b}} \times  \frac{\prod_{b \in C^s_\mu  } \lh{\mu}{b}}{
\prod_{b \in C^s_{\mu+s}  } \lh{\mu+s}{b} }.
\]

Taking the ratio of these these two equalities, we arrive at the result.

To state the following results, we need the following
decomposition of $\mu = (\mu_0,\mu_1,\mu_2,\mu_3)$ w.r.t a
rectangle $m^n$ as in Figure <ref>.

The decomposition $\mu =: \mu_0 \cup \mu_1 \cup \mu_2 \cup \mu_3 $ corresponding to the intersection with the rectangular partition $m^n$, and the definition of $\bsigma$ as the (reverse of the) boxes that fill the remainder of the rectangle. Note that $\mu_0$ is defined as the union of all boxes under $\mu_1$ with those left of $\mu_3$.

The Jack Littlewood–Richardson coefficient
$\jackjLR_{\mu,\bsigma}^{\mu\cup m^n }$
in the rectangular union case is given by the fraction
of hooks in Figure <ref>.
That is, the figure gives a prescription for upper/lower hooks for each box
in each of the three partitions.

The coefficient $\jackjLR_{\mu,\bsigma}^{m^n \cup \mu}$ expressed as a fraction involving the assignment of upper or lower hooks for every box in each of the three partitions.

We refer to the decomposition in Fig. <ref>, according to which the the result (<ref>) becomes

\begin{equation}\label{eqn:gmusigma}
\hat \jackjLR_{\mu,\bsigma}^{\mu\cup m^n} =\left( \prod_{ b \in \mu_3} T_{\bsigma}([\bb]) \right)  \left( \prod_{b \in \mu_1} T_{\bsigma}([\bb]) \right) \cdot \hat \jackjLR_{\sigma,\bsigma}^{ m^n}.
\end{equation}

We will proceed by beginning with $\sigma= \mu \cap m^n = \mu_0 \cup \mu_2$,
and then first extend to include $\mu_1$ by adding one box at a time, and then extending to $\mu_3$
in a similar manner.

One of these two extensions will be simpler, depending one which choice of hooks we take for $\jackjLR_{\sigma, \bsigma}^{m^n}$.

We start with by choosing the hooks as given by Equation <ref>, represented by
Fig. <ref>, noting that the technique works similarly if we begin with the alternate choice. Importantly, with this choice, every box in $\mu_0$ that is beneath $\mu_1$ has been assigned lower hooks, in both the $\sigma$ factor in the numerator and the $\sigma \cup m^n$ factor in the denominator. This will make including the boxes of $\mu_1$ rather straightforward.

The formula for $\jackjLR_{\sigma, \bsigma}^{m^n}$, indicating the location of the flipped component of $\mu_2$

We now proceed to add in each of the boxes in $s \in \mu_1$.
Since $\mu_1$ is in the top-left region above $m^n$,
for $b \in \mu_1$ we have

\[
\ulh{\mu}{b} = \ulh{\mu\cup m^n}{b}.
\]

Because of this, and because $s \in \mu_1$ implies $R_{\mu_1}^s \subset \mu_1$,
the factor (formula (<ref>)) according to equation (<ref>) coming from adding of the box $s\in\mu_1$ simplifies to

\[
T_{\bsigma}([\bar s])=
\frac{\prod_{b \in C^s_{\mu \cup m^n}  } \lh{\mu \cup m^n}{b} }{
\prod_{b \in C^s_{(\mu \cup s) \cup m^n}  } \lh{(\mu  \cup  s) \cup m^n}{b} } /
\frac{\prod_{b \in C^s_\mu  } \lh{\mu}{b} }{ \prod_{b \in C^s_{\mu  \cup  s}  } \lh{\mu \cup s}{b}}.
\]

We represent this `new factor' pictorially in the right hand side of Fig. <ref>. We color in orange those boxes that we are tasked with assigning hooks to, and color in blue those which aren't assigned hooks.
Of the terms appearing in this new factor, two will remove the $L$-hooks for all $b\in C_\sigma^s$ and $b\in C_{\sigma \cup m^n}^s$, and the other two terms will add these hooks back in for the larger
partitions $\sigma\cup s$ and $(\sigma \cup m^n) \cup s$, and thus extending the choice of a $L$-hook to the new box in the column, as seen in Fig. <ref>.
By continuing in this way for each box $b \in \mu_1$, each new box will always have all lower hooks beneath it, and thus these factors provide exactly for $L$-hooks for all the boxes in $\mu_1$.

The new orange box $s \in \mu_1$ being added, and on the left we have the factor $T_{\bsigma}([\bs])$. The task is to assign hooks to the added orange box(es), using the hooks coming from the `new factor' on the right, where blue indicates boxes that don't have assigned hooks.

We notice that these factors precisely extend the column of $L$-hooks on the far left up one box.

Next, we proceed to adding the first box $s$ of $\mu_3$.
Unlike the boxes we added in $\mu_1$, $s \in \mu_3$ does not have all $U$-hooks to the right of it,
and so a more involved process will be required to add in such a box.
We observe that for $b \in \mu_3$ we have

\[
\ulh{\mu}{b} = \ulh{\mu\cup m^n}{b}.
\]

and thus, similar to before, we find that a box $s \in \mu_3$ implies that $C_{\mu_3}^s \in \mu_3$, and so the new factor reduces to

\[
T_{\bsigma}([\bar s])=
\frac{\prod_{b \in R^s_{\mu \cup m^n}  } \uh{\mu \cup m^n}{b} }{
\prod_{b \in R^s_{(\mu \cup s) \cup m^n}  } \uh{(\mu  \cup  s) \cup m^n}{b} } /
\frac{\prod_{b \in R^s_\mu  } \uh{\mu}{b} }{ \prod_{b \in R^s_{\mu  \cup  s}  } \uh{\mu \cup s}{b}}.
\]

This new factor is represented in Figure <ref>.

On the left, we show the new factor $T_{\bsigma}([\bs])$ corresponding to adding the box $s \in \mu_3$ ($s$ drawn in orange).

The next step is to swap the row $R_\mu^s$ of $L$-hooks in $\mu$ that contains $s \in \mu_3$,
with the row of $U$-hooks in $\mu \cup s$ from the new factor, as indicated in Figure <ref>.

The result of swapping the first term in the numerator of the new factor with the bottom row of the $\mu$ factor.

Then, we split up the two rows of hooks in the numerator of the new factor according
to the staggered row profile of $L$-hooks in the denominator of the right hand side, as in Figure <ref>.

Split up the two rows in the numerator of the new factor according to the profile of lower hooks in the denominator of the left hand side.

We now show that this numerator of the new factor gives precisely the hooks required to
flip the indicated hooks in the denominator of the right hand side from $L \to U$.
Let $\ell_{\lambda}(b)$ be shorthand for $\leg_\lambda(b)$.
Let $b \in \mu$ be such that the row $R_\mu^b$ extends all the way to
the right boundary of $m^n$ (for example, in the case at hand $b$ is in the bottom row of $\mu$).
The claim is

\begin{equation*}
\lh{\mu}{b} = \lh{\mu \cup m^n}{b+(0,\ell_{\mu \cup m^n}(b)-\ell_{\mu}(b))}.
\end{equation*}

One can easily verify this by direct calculation or pictorially.
Thus, we use the hooks on the numerator of the new factor to flip all the hooks on the
staggered row in the denominator. After this, we are left with the
two terms in the denominator of the new factor, as in Figure <ref>.

With the staggered row now flipped to upper hooks, the bottom row of the denominator can now easily be extended.

Now that the entire bottom row of the denominator of the RHS is all upper hooks,
we can see that these two remaining rows in the denominator of the new factor are
precisely what is required to extend this row of upper hooks to include $s \in \mu_3$.
With this, all hooks in the new factor have been used, and all boxes in the
triple of partitions have been assigned a choice of hooks, as in Figure <ref>.

After extending the bottom row, we have assigned hooks for all boxes including $s$.

Now, we can use the above steps to include the the rest of the boxes inside $\mu_3$.
If the newly added box is to the right of a box added previously, the row just extends by $U$-hooks.
If the new box in $\mu_3$ is added to a new row, then repeat all of the above steps, starting in this new row.
Proceeding in this way, we see that all boxes to the left of $\mu_3$ will be
assigned U-hooks, and the remaining rectangular region will
be assigned hooks just as in Figure <ref>.

As we mentioned at the start of Section <ref>, there is a direct
correspondence between the Littlewood–Richardson coefficients and Stanley structure coefficients.
Thus, our Theorem <ref> immediately gives the following main result of the paper.

The strong Stanley conjecture holds in the rectangular union case.

We find that the Stanley structure coefficients are given by Figure <ref>,
except with bringing the denominator up to the numerator with its hooks flipped $L \leftrightarrow U$.

We have

\begin{equation*}
\langle J_{42211}J_{211}, J_{43331}\rangle
\begin{ytableau}
*(boxL) L   \\
*(boxL) L   \\
*(boxL) L & *(boxL) L \\
*(boxL) L & *(boxL) L \\
*(boxU) U & *(boxU) U & *(boxU) U & *(boxU) U \\
\end{ytableau}\quad
\begin{ytableau}
*(boxU) U  \\
*(boxU) U  \\
*(boxU) U & *(boxU) U
\end{ytableau}\quad
\begin{ytableau}
*(boxU) U \\
*(boxU) U & *(boxU) U & *(boxL) L \\
*(boxU) U & *(boxU) U & *(boxL) L \\
*(boxU) U & *(boxL) L & *(boxL) L \\
*(boxL) L & *(boxL) L & *(boxL) L & *(boxL) L \\
\end{ytableau}. \\
\end{equation*}

From the diagrams, we then get (by reading columns from left to right)

\begin{align*}
&\left([1,0][2,0][3,-1][4,-1][4,-4] \cdot
[1,0][2,0][2,-3]\cdot[0,-2]\cdot[0,-1]\right) \\
& \times \left([0,-1][1,-1][2,-2]\cdot[0,-1]\right) \\
& \times \left([0,-1][1,-3][2,-3][3,-3][5,-3] \right. \\
& \phantom{\times(}\cdot [0,-2][1,-2][3,-1][4,-2]\\
& \phantom{\times(}\cdot \left.[1,0][2,0][3,0][4,-1] \cdot [1,0] \right).
\end{align*}

Using $[x,y]=\alpha x - y$, we find that the above expression

\begin{align*}
2^9\cdot 3^2\cdot\alpha^6(1+\alpha)^4(2+\alpha)(3+\alpha)^2(4+\alpha)^2
\end{align*}

Finally, we summarize the two possible solutions to the rectangular union Littlewood–Richardson coefficient, based on the starting choice for $g_{\sigma,\bsigma}^{m^n}$, and by noticing the cancellation of the L/U hooks between the numerator and denominator of Fig <ref> in the $\mu_1$, $\mu_3$ regions and a corner rectangle.
We see that the rectangular union Littlewood–Richardson coefficient can be expressed in terms of a
rectangular Littlewood–Richardson coefficient of a sub-diagram.

Let $\mu_2$ be as before, and note that $\mu_2 \cup \bsigma^R$ is a rectangular region $K$ of shape $k^\ell$. We then have

\begin{equation*}
g_{\mu,\bsigma}^{\mu\cup m^n} = F \times g_{\mu_2, \bar \mu_2}^{k^\ell}
\end{equation*}

where $F$ is a products of $L$-hooks for every box to the left of $K$,
and $U$-hooks for every box below it, for both $\mu$ in the numerator
and $\mu \cup m^n$ in the denominator. See Figure <ref>.

We can re-write the result for rectangular union in schematic form.
The blue sections are not assigned any hooks, and for the beige regions
we can choose one of the two solutions to the rectangular case for that subregion.

§ CONJECTURES INVOLVING SHIFTED JACK FUNCTIONS

Is there an analog of (<ref>) for shifted Jack structure constants?

In <cit.>, it is conjectured that for shifted Jack functions, we have that
$\langle \jacks{\mu} \jacks{\nu} , \jacks{\lambda} \rangle$ are Laurent polynomials in $\alpha$ with non-negative
integer coefficients.
One can then ask if a shifted analog of Stanley's conjecture holds.
Let us first set

\[
\defin{H^U_\lambda} \coloneqq  \prod_{b \in \lambda} h^U_\lambda(b), \quad
\defin{H^L_\lambda} \coloneqq  \prod_{b \in \lambda} h^L_\lambda(b).
\]

The definition of the Jack Littlewood–Richardson coefficients
$\defin{ \jackjLR_{\mu\nu}^{\lambda}(\alpha)}$ are then defined via

\begin{equation*}
\jacks{\mu} \jacks{\nu} = \sum_{\lambda}  \jackjLR_{\mu\nu}^{\lambda}(\alpha) \jacks{\lambda} \text{ or equivalently }
\jacksp{\mu} \jacksp{\nu} = \sum_{\lambda}  \jackjLR_{\mu\nu}^{\lambda}(\alpha)
\frac{H^L_\lambda}{H^L_\mu \cdot H^L_\mu}
\jacksp{\lambda},
\end{equation*}

where $\jacks{\mu} = H^L_\mu \jacksp{\mu}$.
When $|\mu|+|\nu|=|\lambda|$, we recover the “non-shifted” coefficients in (<ref>),
but we may have non-zero coefficients also when $|\mu|+|\nu| > |\lambda|$.
The quantity $c_{\mu\nu}^\lambda(\alpha) \coloneqq \jackjLR_{\mu\nu}^{\lambda}(\alpha) \frac{H^L_\lambda}{H^L_\mu \cdot H^L_\mu}$
is then a deformation of the classical Littlewood–Richardson coefficients for Schur functions,
and their shifted generalization.
The inner product for the shifted Jack polynomials are then given by

\[
\langle \jacks{\mu} \jacks{\nu} , \jacks{\lambda} \rangle =
H^L_\lambda \cdot H^U_\lambda \cdot \jackjLR_{\mu\nu}^{\lambda}(\alpha),
\]

as in Subsection <ref>.

If the shifted Littlewood–Richardson coefficient $c_{\mu\nu}^\lambda(1)$ is equal to $1$,
then the (shifted) structure coefficient has the following form:

\begin{equation*}
\langle \jacks{\mu} \jacks{\nu} , \jacks{\lambda} \rangle =
\left( \prod_{b \in \mu} \tilde{h}_{\mu}(b) \right)
\left( \prod_{b \in \nu} \tilde{h}_{\nu}(b) \right)
\left( \prod_{b \in \lambda} \tilde{h}_{\lambda}(b) \right),
\end{equation*}

where $\tilde{h}_{\sigma}(b)$ is a choice of either $\uh{\sigma}{b}$ or $\lh{\sigma}{b}$ for each box $b$.
Moreover, one chooses $\uh{\sigma}{b}$ exactly $|\lambda|$
times each in the above expression
and $\lh{\sigma}{b}$ is chosen $|\mu|+|\nu|$ times.

For $\lambda = 32211$, $\mu = 31$ and $\nu=321$, we have that

\[
c_{\mu\nu}^{\lambda}(\alpha) =
\frac{\alpha ^2 (\alpha +3)^2 (\alpha +4) (2 \alpha +1)^2 (2 \alpha +5) (3 \alpha +1) (3 \alpha +2)}{(\alpha +1)^6 (\alpha +2)^2 (2 \alpha +3)^2 (3 \alpha +4)}
\]

(which evaluates to $1$ at $\alpha=1$) and

\[
\langle \jacks{\mu} \jacks{\nu} , \jacks{\lambda} \rangle
8 \alpha ^5 (\alpha +3)^2 (\alpha +4) (2 \alpha +1)^2 (2 \alpha +5) (3
\alpha +1) (3 \alpha +2).
\]

We can see that this product is obtained from the assignments below:

\begin{equation*}
\langle  \jacks{321} \jacks{31}, \jacks{32211}\rangle
\begin{ytableau}
*(boxU) U  \\
*(boxU) U & *(boxU) U  \\
*(boxU) U & *(boxU) U &  *(boxL) L
\end{ytableau}\,\,
\begin{ytableau}
*(boxU) U \\
*(boxU) U & *(boxU) U & *(boxU) U \\
\end{ytableau}
\,\,
\begin{ytableau}
*(boxL) L   \\
*(boxL) L   \\
*(boxL) L & *(boxL) L \\
*(boxL) L & *(boxL) L \\
*(boxL) L & *(boxL) L & *(boxL) L  \\
\end{ytableau}.
\end{equation*}

Note that here the choices for all boxes that aren't inner corners are forced.

§ ACKNOWLEDGEMENTS

The authors would like to thank the organizers of the
conference Open Problems in Algebraic Combinatorics 2022 (OPAC),
the occasion of which led to this collaboration. RM would like thank Alexander Moll for first suggesting that there might be a path to make progress on the Stanley conjectures starting from a deeper understanding of the Nazarov-Skylanin Lax operator, and this paper represents the culmination of that idea.


# A generalized Routh-Hurwitz criterion for the stability analysis of
polynomials with complex coefficients: application to the PI-control of
vibrating structures

Anthony Hastir<EMAIL_ADDRESS>Riccardo Muolo  Department of
Mathematics and naXys, Namur Institute for Complex Systems, University of
Namur, Rue Grafé 2, 5000 Namur, Belgium

###### Abstract

The classical Routh-Hurwitz criterion is one of the most popular methods to
study the stability of polynomials with real coefficients, given its
simplicity and ductility. However, when moving to polynomials with complex
coefficients, a generalization exists but it is rather cumbersome and not as
easy to apply. In this paper, we make such generalization clear and
understandable for a wider public and develop an algorithm to apply it. After
having explained the method, we demonstrate its use to determine the external
stability of a system consisting of the interconnection between a rotating
shaft and a PI-regulator. The extended Routh-Hurwitz criterion gives then
necessary and sufficient conditions on the gains of the PI-regulator to
achieve stabilization of the system together with regulation of the output.
This illustrative example makes our formulation of the extended Routh-Hurwitz
criterion ready to be used in several other applications.

###### keywords:

Routh-Hurwitz criterion, complex coefficients polynomials, vibrating
structures, PI-control

††journal: Elsevier

##  1 Introduction

Let us consider the following $n$-th order polynomial

$q(s)=s^{n}+\sum_{j=1}^{n}\left(a_{j}+ib_{j}\right)s^{n-j},$ (1)

where $i$ denotes the imaginary unit throughout this note. We want to study
its stability, i.e., all its roots need to have negative real part. If all the
coefficients $b_{j}=0$ $\forall~{}j\in\\{1,...,n\\}$, meaning that we would be
dealing with real coefficients, we would rely on the well-known Routh-Hurwitz
criterion Routh (1877); Hurwitz (1895), which provides a simple algorithm to
verify the stability conditions. However, to study the stability of the
polynomial (1) is not trivial and the method to obtain the stability
conditions is not as straightforward as its real analogous. In the literature
there are some available tools, but they are often developed for specific
cases and their applicability is not immediately clear for a general public.
For instance, one finds the so-called Kharitonov’s theorem, first introduced
in Kharitonov (1978a) for polynomials with real coefficients and then extended
in Kharitonov (1978b) in the complex case. This theorem consists in
determining the region where the roots of a polynomial are located based on
the same conclusion obtained for several upper- and lower-polynomials. That
is, polynomials with coefficients encapsulating the coefficients of the
nominal polynomial, sometimes called interval polynomials. One needs $4$
polynomials in the real case, while $8$ polynomials need to be used when the
coefficients are complex. Few years later, these results have been revisited
in Minnichelli et al. (1989) and Callier and Desoer (1991) by taking an
engineering oriented point of view, notably. The major drawback of such an
approach is that the coefficients of the original polynomial are not used
directly, making this method not systematic.
We found that the most general method is the one developed in Frank (1946) and
then recalled in Xie (1985), which is the one we will discuss pedagogically in
the following. To the best of our knowledge, it is the most natural and direct
extension of the classical Routh-Hurwitz criterion in the complex case.
However, as a simple counter-example on a polynomial of degree $3$ may
highlight, the main result that is presented in Xie (1985) is wrong when the
degree of the considered polynomial is odd. The same mistake has been further
repeated in Yu et al. (2020) on a particular example in network theory. This
constitutes one additional reason for this note to describe the method in a
constructive and a rigorous way. With an eye on applications, such a criterion
turns out to be useful for determining for instance the stability of a
dynamical system whose dynamics exhibit complex coefficients. Such cases arise
often in rotordynamics to describe the behavior of rotating shafts, as
highlighted in e.g. Loewy et al. (1970) and Barnett (1983). Complex
coefficients appear also in the dynamics of electrical networks as is
described in Varricchio and Gomes (2018). Applications to the analysis of
spontaneous self-excitation in induction generators are also found in Bodson
and Kiselychnyk (2013) where the authors recall the generalized criterion for
polynomials with complex coefficients. This criterion has been applied to
fractional systems with applications to population dynamic models in Bourafa
et al. (2020). The criterion discussed in this note should then also be useful
for analyzing the stability or developing control methods for such systems and
beyond.

The paper is organized as follows: the extension of the classical Routh-
Hurwitz criterion is highlighted in Section 2 as an algorithm, in which we
make the distinction between $n$ odd and $n$ even. The case of $n=4$ is
developed in Section 3. An example build from rotordynamics is then considered
in Section 4: a Proportional-Integral (PI) action is applied to the system and
the stability properties of the closed-loop system are analyzed using the
results described in the previous Sections. Some conclusions are addressed in
Section 5. A lot of attention has been paid to describing the method in a
pedagogical way in the form of an algorithm, which then could be
straightforwardly implemented numerically.

##  2 General description of the method

Let us again consider the $n$-th order polynomial given in (1). As a matter of
notation, let $\mathbb{C}^{-}_{\xi}$ (resp. $\mathbb{C}^{+}_{\xi}$) denote the
open subset $\\{s\in\mathbb{C},\mathfrak{Re}(s)<\xi\\}$ (resp.
$\\{s\in\mathbb{C},\mathfrak{Re}(s)>\xi\\}$), $\xi\in\mathbb{R}$. We use also
the notation $|A|$ for the determinant of the matrix $A$. The general
algorithm that determines whether the roots of $q(s)$ in (1) are in
$\mathbb{C}_{0}^{-}$ is presented in Algorithm 1.

Data: The complex coefficients of the polynomial given in (1).

Output: Necessary and sufficient conditions for the roots of (1) to be in
$\mathbb{C}^{-}_{0}$.

Initialization:  In the case where $n$ is even, construct a $2\times n$ matrix
as follows:

$a_{1}^{(1)}=a_{1}$ |  $b_{2}^{(1)}=b_{2}$ |  $\dots$ |  $a_{n-1}^{(1)}=a_{n-1}$ |  $b_{n}^{(1)}=b_{n}$
---|---|---|---|---
$b_{1}^{(1)}=a_{1}^{(1)}b_{1}-b_{2}^{(1)}$ |  $a_{2}^{(1)}=a_{1}^{(1)}a_{2}-a_{3}^{(1)}$ |  $\dots$ |  $b_{n-1}^{(1)}=a_{1}^{(1)}b_{n-1}-b_{n}$ |  $a_{n}^{(1)}=a_{1}^{(1)}a_{n}$
When $n$ is odd, construct the following $2\times n$ matrix:  $a_{1}^{(1)}=a_{1}$ |  $b_{2}^{(1)}=b_{2}$ |  $\dots$ |  $b_{n-1}^{(1)}=b_{n-1}$ |  $a_{n}^{(1)}=a_{n}$
---|---|---|---|---
$b_{1}^{(1)}=a_{1}^{(1)}b_{1}-b_{2}^{(1)}$ |  $a_{2}^{(1)}=a_{1}^{(1)}a_{2}-a_{3}^{(1)}$ |  $\dots$ |  $a_{n-1}^{(1)}=a_{1}^{(1)}a_{n-1}-a_{n}$ |  $b_{n}^{(1)}=a_{1}^{(1)}b_{n}$

for $p=2,\dots,n-1$ do

if ($p$ is even and $n$ is even) or ($p$ is odd and $n$ is odd) then

Construct the following $2\times(n-(p-1))$ matrix:

$a_{p}^{(p)}$ |  $b_{p+1}^{(p)}$ |  $\dots$ |  $b_{n-1}^{(p)}$ |  $a_{n}^{(p)}$
---|---|---|---|---
$b_{p}^{(p)}$ |  $a_{p+1}^{(p)}$ |  $\dots$ |  $a_{n-1}^{(p)}$ |  $b_{n}^{(p)}$
where

  * 1.

The elements $a_{k}^{(p)},k=p,p+2,p+4,\dots,n$ and
$b_{l}^{(p)},l=p+1,p+3,\dots,n-1$ of the first row are:
$a_{k}^{(p)}=\left|\begin{matrix}a_{p-1}^{(p-1)}&-b_{k}^{(p-1)}\\\
b_{p-1}^{(p-1)}&a_{k}^{(p-1)}\end{matrix}\right|$ and
$b_{l}^{(p)}=\left|\begin{matrix}a_{p-1}^{(p-1)}&a_{l}^{(p-1)}\\\
b_{p-1}^{(p-1)}&b_{l}^{(p-1)}\end{matrix}\right|$

  * 2.

The elements
$a_{k}^{(p)},k=p+1,p+3,\dots,n-1,b_{l}^{(p)},l=p,p+2,p+4,\dots,n-2$ of the
second row are:
$a_{k}^{(p)}=-\left|\begin{matrix}a_{p-1}^{(p-1)}&a_{k}^{(p-1)}\\\
a_{p}^{(p)}&a_{k+1}^{(p)}\end{matrix}\right|,b_{l}^{(p)}=-\left|\begin{matrix}a_{p-1}^{(p-1)}&b_{l}^{(p-1)}\\\
a_{p}^{(p)}&b_{l+1}^{(p)}\end{matrix}\right|$ and
$b_{n}^{(p)}=-\left|\begin{matrix}a_{p-1}^{(p-1)}&b_{n}^{(p-1)}\\\
a_{p}^{(p)}&0\end{matrix}\right|$

end if

if ($p$ is odd and $n$ is even) or ($p$ is even and $n$ is odd) then

Construct the following $2\times(n-(p-1))$ matrix:  $a_{p}^{(p)}$ |  $b_{p+1}^{(p)}$ |  $\dots$ |  $a_{n-1}^{(p)}$ |  $b_{n}^{(p)}$
---|---|---|---|---
$b_{p}^{(p)}$ |  $a_{p+1}^{(p)}$ |  $\dots$ |  $b_{n-1}^{(p)}$ |  $a_{n}^{(p)}$
where

  * 1.

The elements $a_{k}^{(p)},k=p,p+2,p+4,\dots,n-1$ and
$b_{l}^{(p)},l=p+1,p+3,\dots,n$ of the first row are:
$a_{k}^{(p)}=\left|\begin{matrix}a_{p-1}^{(p-1)}&-b_{k}^{(p-1)}\\\
b_{p-1}^{(p-1)}&a_{k}^{(p-1)}\end{matrix}\right|$ and
$b_{l}^{(p)}=\left|\begin{matrix}a_{p-1}^{(p-1)}&a_{l}^{(p-1)}\\\
b_{p-1}^{(p-1)}&b_{l}^{(p-1)}\end{matrix}\right|$

  * 2.

The elements $a_{k}^{(p)},k=p+1,p+3,\dots,n-2$ and
$b_{l}^{(p)},l=p,p+2,p+4,\dots,n-1$ of the second row are:
$a_{k}^{(p)}=-\left|\begin{matrix}a_{p-1}^{(p-1)}&a_{k}^{(p-1)}\\\
a_{p}^{(p)}&a_{k+1}^{(p)}\end{matrix}\right|,b_{l}^{(p)}=-\left|\begin{matrix}a_{p-1}^{(p-1)}&b_{l}^{(p-1)}\\\
a_{p}^{(p)}&b_{l+1}^{(p)}\end{matrix}\right|$ and
$a_{n}^{(p)}=-\left|\begin{matrix}a_{p-1}^{(p-1)}&a_{n}^{(p-1)}\\\
a_{p}^{(p)}&0\end{matrix}\right|$

end if

end for

Compute the following coefficient  $\displaystyle
a_{n}^{(n)}=\left|\begin{matrix}a_{n-1}^{(n-1)}&-b_{n}^{(n-1)}\\\
b_{n-1}^{(n-1)}&a_{n}^{(n-1)}\end{matrix}\right|$

Necessary and sufficient conditions:  the roots of the polynomial (1) are in
$\mathbb{C}_{0}^{-}$ if and only if $a_{k}^{(k)}>0$ for all $k=1,\dots,n$.

Algorithm 1 Generalized Routh-Hurwitz criterion

##  3 Example for a $4$-th degree polynomial

Let us now show, as a pedagogical example, the table of coefficients for a
$4$-th order polynomial of the form (1) with $n=4$ and explicitly find the
stability conditions.

The coefficients obtained with the method described in the previous section
are given in Table 1.

$a_{1}$ |  $b_{2}$ |  $a_{3}$ |  $b_{4}$
---|---|---|---
$b_{1}^{(1)}=a_{1}b_{1}-b_{2}$ |  $a_{2}^{(1)}=a_{1}a_{2}-a_{3}$ |  $b_{3}^{(1)}=a_{1}b_{3}-b_{4}$ |  $a_{4}^{(1)}=a_{1}a_{4}$
$a_{2}^{(2)}=a_{1}a_{2}^{(1)}+b_{1}^{(1)}b_{2}$ |  $b_{3}^{(2)}=a_{1}b_{3}^{(1)}-b_{1}^{(1)}a_{3}$ |  $a_{4}^{(2)}=a_{1}a_{4}^{(1)}+b_{1}^{(1)}b_{4}$ |
$b_{2}^{(2)}=a_{2}^{(2)}b_{2}-a_{1}b_{3}^{(2)}$ |  $a_{3}^{(2)}=a_{2}^{(2)}a_{3}-a_{1}a_{4}^{(2)}$ |  $b_{4}^{(2)}=a_{2}^{(2)}b_{4}$ |
$a_{3}^{(3)}=a_{2}^{(2)}a_{3}^{(2)}+b_{2}^{(2)}b_{3}^{(2)}$ |  $b_{4}^{(3)}=a_{2}^{(2)}b_{4}^{(2)}-b_{2}^{(2)}a_{4}^{(2)}$ |  |
$b_{3}^{(3)}=a_{3}^{(3)}b_{3}^{(2)}-a_{2}^{(2)}b_{4}^{(3)}$ |  $a_{4}^{(3)}=a_{3}^{(3)}a_{4}^{(2)}$ |  |
$a_{4}^{(4)}=a_{3}^{(3)}a_{4}^{(3)}+b_{3}^{(3)}b_{4}^{(3)}$ |  |  |

Table 1: Coefficients of the generalized Routh-Hurwitz criterion

The necessary and sufficient conditions for the stability of the polynomial
(1) with $n=4$ are then given by

$\begin{cases}a_{1}>0\\\ a_{2}^{(2)}=a_{1}a_{2}^{(1)}+b_{1}^{(1)}b_{2}>0\\\
a_{3}^{(3)}=a_{2}^{(2)}a_{3}^{(2)}+b_{2}^{(2)}b_{3}^{(2)}>0\\\
a_{4}^{(4)}=a_{3}^{(3)}a_{4}^{(3)}+b_{3}^{(3)}b_{4}^{(3)}>0\end{cases}$ (2)

A cumbersome but straightforward computation, which can be found in AppendixA,
gives the expression for the generalized Routh-Hurwitz conditions

$\begin{cases}a_{1}>0\\\
a_{1}^{2}a_{2}-a_{1}a_{3}-a_{1}b_{1}b_{2}+b_{2}^{2}:=\beta>0\\\
\beta^{2}a_{3}-\beta
a_{1}^{3}a_{4}+\beta(a_{1}b_{4}+a_{3}b_{2})(a_{1}b_{1}-b_{2})\\\ \hskip
14.22636pt-\beta a_{1}b_{2}(a_{1}b_{3}-b_{4})+a_{1}[a_{1}(a_{1}b_{3}-b_{4})\\\
\hskip 14.22636pt-a_{3}(a_{1}b_{1}-b_{2})]^{2}:=\gamma>0\\\
\gamma^{2}[a_{1}^{2}a_{4}-b_{4}(a_{1}b_{1}-b_{2})]\\\ \hskip
14.22636pt-\eta[\gamma a_{1}(a_{1}b_{3}-b_{4})\gamma
a_{3}(a_{1}b_{1}-b_{2})-\beta\eta]>0\end{cases}$ (3)

where

$\displaystyle\eta$ $\displaystyle=\beta^{2}b_{3}-[\beta
b_{2}-a_{1}(a_{1}(a_{1}b_{3}-b_{4})-a_{3}(a_{1}b_{1}-b_{2}))]\varepsilon$
$\displaystyle\varepsilon$
$\displaystyle=[a_{1}^{2}a_{4}-b_{4}(a_{1}b_{1}-b_{2})]$

From the above conditions (3), one can recover the classical Routh-Hurwitz
conditions when the polynomial has real coefficients, i.e.,
$b_{j}=0,~{}\forall j\in\\{1,2,3,4\\}$, as we show in AppendixB.

##  4 Application to the PI-regulation of a rotating shaft

An application arising from the theory of rotating shafts is considered in
this section. We emphasize that this application has been chosen in order to
highlight the efficiency of the generalized Routh-Hurwitz criterion, reason
why some physical concepts are omitted. The dynamics of rotating shafts has
triggered the interest of scholars since the very first studies carried out by
William John Macquorn Rankine in 1869, with many notable results. For an
overview on the topic, the reader may refer to Loewy et al. (1970). Therein,
it is shown that such systems may be modeled by the following ordinary
differential equation

$\ddot{x}(t)+(2k\omega+i2\Omega)\dot{x}(t)+(\omega^{2}-\Omega^{2})x(t)=f(t),$
(4)

where $k,\Omega$ and $\omega$ indicate a damping coefficient, an angular
velocity and the frequency of undamped oscillations, respectively. The
quantity $f$ acts on the system as an external force. As a control objective
for that system, let us consider the regulation of the position $x(t)$ to a
constant prescribed reference position denoted by $x_{ref}$. To achieve such
an objective, one will rely on the well-established Proportional Integral (PI)
control, see Borase et al. (2020) and references therein. In that way, the
forcing term will take the following form

$f(t)=k_{p}(x(t)-x_{ref})+k_{I}\ell(t),$ (5)

where the proportional and the integral gains denoted by $k_{p}$ and $k_{I}$
need to be determined for the closed-loop system to be stable. The quantity
$\ell(t)$ is updated adaptively as

$\dot{\ell}(t)=x(t)-x_{ref}.$ (6)

By introducing the auxiliary variables $x_{1}=x$ and $x_{2}=\dot{x}$, the
closed-loop system composed of (4), (5) and (6) reads as

$\displaystyle\left(\begin{smallmatrix}\dot{x}_{1}(t)\\\ \dot{x}_{2}(t)\\\
\dot{\ell}(t)\end{smallmatrix}\right)$
$\displaystyle=A_{(k_{I},k_{p})}\left(\begin{smallmatrix}x_{1}(t)\\\
x_{2}(t)\\\ \ell(t)\end{smallmatrix}\right)-\left(\begin{smallmatrix}0\\\
k_{p}x_{ref}\\\ x_{ref}\end{smallmatrix}\right)$ (7)

where $A_{(k_{I},k_{p})}=\left(\begin{smallmatrix}0&1&0\\\
k_{p}+\Omega^{2}-\omega^{2}&-(2k\omega+i2\Omega)&k_{I}\\\
1&0&0\end{smallmatrix}\right)$. From (7), if the gains $k_{p}$ and $k_{I}$ are
chosen such that the matrix $A$ is stable, then the control objective will be
satisfied. In particular, the quantity $x_{1}$ will reach the equilibrium
$x_{ref}$. One needs then to determine in which cases the matrix $A$ is a
stable matrix. First observe that the characteristic polynomial of that matrix
is given by

$\displaystyle q(s)$ $\displaystyle=|sI-A_{(k_{I},k_{p})}|$
$\displaystyle=s^{3}+(2k\omega+i2\Omega)s^{2}+(\omega^{2}-\Omega^{2}-k_{p})s-k_{I}.$
(8)

We shall therefore rely on the generalized Routh-Hurwitz criterion detailed in
Algorithm 1. The consecutive arrays of numbers generated by this algorithm are
given in Table 2.

$a_{1}^{(1)}=2k\omega$ |  $b_{2}^{(1)}=0$ |  $a_{3}^{(1)}=-k_{I}$
---|---|---
$b_{1}^{(1)}=4k\omega\Omega$ |  $a_{2}^{(1)}=2k\omega(\omega^{2}-\Omega^{2}-k_{p})+k_{I}$ |  $b_{3}^{(1)}=0$
$a_{2}^{(2)}=2k\omega(2k\omega(\omega^{2}-\Omega^{2}-k_{p})+k_{I})$ |  $b_{3}^{(2)}=k_{I}(4k\omega\Omega)$ |
$b_{2}^{(2)}=-8k_{I}k^{2}\omega^{2}\Omega$ |  $a_{3}^{(2)}=-k_{I}a_{2}^{(2)}$ |
$a_{3}^{(3)}=-k_{I}(a_{2}^{(2)})^{2}-32k_{I}^{2}k^{3}\omega^{3}\Omega^{2}$ |  |

Table 2: Generalized Routh-Hurwitz table for the polynomial (8)

According to Algorithm 1, the matrix $A$ is stable if and only if the
following three conditions are satisfied

$\displaystyle 2k\omega$ $\displaystyle>0$ (9) $\displaystyle
2k\omega\left[2k\omega(\omega^{2}-\Omega^{2}-k_{p})+k_{I}\right]$
$\displaystyle>0$ (10)
$\displaystyle-8k_{I}^{2}k\omega\Omega^{2}-k_{I}\left[2k\omega(\omega^{2}-\Omega^{2}-k_{p})+k_{I}\right]^{2}$
$\displaystyle>0.$ (11)

In order to illustrate the feasibility of Conditions (9)–(11), a grid has been
made with different values of $k_{I}$ and $k_{p}$. At each point of the grid,
Conditions (9)–(11) have been tested. If these conditions are all satisfied,
the value $1$ (yellow part) has been placed on the grid and $0$ (blue part)
otherwise. The resulting picture is depicted in Figure 1. As a matter of
comparison, another test has been performed. For each point of the grid
$(k_{I},k_{p})$, the value of the largest real part of the eigenvalues of the
matrix $A$ has been encoded in the grid. A contour plot has then been
performed and it is shown in Figure 2. From the latter, it is clear that the
stability region is the same as the one highlighted with Figure 1, obtained
from the stability conditions (9)–(11). In the above derivation, the
parameters have been set to $k=1,\omega=2$ and $\Omega=2$.

Figure 1: A grid $(k_{I},k_{p})$ at which each point is either $1$ or $0$,
based on conditions (9)–(11). Figure 2: Contour plot of the function
$\lambda_{s}(k_{I},k_{p}):=\max\\{\lambda\in\sigma(A_{(k_{I},k_{p})})\\}$

##  5 Conclusion

In this note, we have clarified and explained in a constructive and
pedagogical way an extension of the classical Routh-Hurwitz criterion to
polynomials with complex coefficients. The general algorithm to determine
whether the roots of such a polynomial are located in $\mathbb{C}_{0}^{-}$ or
not is given in Section 2. Then, this method is extensively developed for a
$4$-th order polynomial in Section 3. Finally, an application to the PI
regulation of a rotating shaft whose own dynamics exhibit complex coefficients
is detailed in Section 4, giving rise to the study of the stability of a
$3$-rd order polynomial. Our presentation of the algorithm and the given
examples make the method understandable and ready to use also for scholars and
students outside the control community, paving the way for further
advancements in applications where complex polynomials appear.

## Acknowledgements

The authors are grateful to Alice Bellière for useful discussions and
feedback. This research was conducted with the financial support of
F.R.S-FNRS. A.H. is supported by a FNRS Postdoctoral Fellowship, Grant CR
40010909. R.M. is supported by a FRIA Fellowship, funded by the Walloon
Region, Grant FC 33443.

##  Appendix A Explicit derivation of the Generalized Routh-Hurwitz conditions

Let us now explicitly derive the conditions (3).

Derivation of the second condition:

$\displaystyle a_{2}^{(2)}$
$\displaystyle=a_{1}a_{2}^{(2)}+b_{1}^{(1)}b_{2}=a_{1}(a_{1}a_{2}-a_{3})+b_{2}(a_{1}b_{1}-b_{2})$
$\displaystyle=a_{1}^{2}a_{2}-a_{1}a_{3}+a_{1}b_{1}b_{2}-b_{2}^{2}:=\beta$

Derivation of the third condition:

$\displaystyle a_{3}^{(3)}$
$\displaystyle=a_{2}^{(2)}a_{3}^{(2)}+b_{2}^{(2)}b_{3}^{(2)}$
$\displaystyle=a_{2}^{(2)}(a_{2}^{(2)}a_{3}-a_{1}a_{4}^{(2)})$
$\displaystyle+(a_{2}^{(2)}b_{2}-a_{1}b_{3}^{(2)})(a_{1}b_{3}^{(1)}-b_{1}^{(1)}a_{3})$
$\displaystyle=(a_{2}^{(2)})^{2}a_{3}-a_{1}a_{2}^{(2)}a_{4}^{(2)}+a_{1}b_{2}a_{2}^{(2)}b_{3}^{(1)}-a_{3}b_{2}a_{2}^{(2)}b_{1}^{(1)}$
$\displaystyle-
a_{1}^{2}b_{3}^{(1)}b_{3}^{(2)}+a_{1}a_{3}b_{1}^{(1)}b_{3}^{(2)}$

$\displaystyle=\beta^{2}a_{3}-\beta
a_{1}(a_{1}a_{4}^{(1)}+b_{1}^{(1)}b_{4})+\beta a_{1}b_{2}(a_{1}b_{3}-b_{4})$
$\displaystyle-\beta
a_{3}b_{2}(a_{1}b_{1}-b_{2})-a_{1}^{2}\left(a_{1}b_{3}-b_{4}\right)\left(a_{1}b_{3}^{(1)}-a_{3}b_{1}^{(1)}\right)$
$\displaystyle+a_{1}a_{3}\left(a_{1}b_{1}-b_{2}\right)\left(a_{1}b_{3}^{(1)}-b_{1}^{(1)}a_{3}\right)$
$\displaystyle=\beta^{2}a_{3}-\beta
a_{1}\left[a_{1}^{2}a_{4}+b_{4}\left(a_{1}b_{1}-b_{2}\right)\right]$
$\displaystyle+\beta a_{1}b_{2}\left(a_{1}b_{3}-b_{4}\right)-\beta
a_{3}b_{2}\left(a_{1}b_{1}-b_{2}\right)$ $\displaystyle-
a_{1}^{2}\left(a_{1}b_{3}-b_{4}\right)\left[a_{1}\left(a_{1}b_{3}-b_{4}\right)-a_{3}\left(a_{1}b_{1}-b_{2}\right)\right]$
$\displaystyle+a_{1}a_{3}\left(a_{1}b_{1}-b_{2}\right)\left[a_{1}\left(a_{1}b_{3}-b_{4}\right)-a_{3}\left(a_{1}b_{1}-b_{2}\right)\right]$
$\displaystyle=\beta^{2}a_{3}-\beta
a_{1}\left(a_{1}^{2}a_{4}+b_{4}\left(a_{1}b_{1}-b_{2}\right)\right)$
$\displaystyle+\beta a_{1}b_{2}\left(a_{1}b_{3}-b_{4}\right)-\beta
a_{3}b_{2}\left(a_{1}b_{1}-b_{2}\right)$
$\displaystyle-\left[a_{1}^{2}\left(a_{1}b_{3}-b_{4}\right)-a_{1}a_{3}\left(a_{1}b_{1}-b_{2}\right)\right]\dots$
$\displaystyle\dots\left[a_{1}\left(a_{1}b_{3}-b_{4}\right)-a_{3}\left(a_{1}b_{1}-b_{2}\right)\right]$

$\displaystyle=\beta^{2}a_{3}-\beta a_{1}^{3}a_{4}-\beta
a_{1}b_{4}\left(a_{1}b_{1}-b_{2}\right)+\beta
a_{1}b_{2}\left(a_{1}b_{3}-b_{4}\right)$ $\displaystyle-\beta
a_{3}b_{2}\left(a_{1}b_{1}-b_{2}\right)-a_{1}\left[a_{1}\left(a_{1}b_{3}-b_{4}\right)-a_{3}\left(a_{1}b_{1}-b_{2}\right)\right]^{2}$
$\displaystyle=\beta^{2}a_{3}-\beta
a_{1}^{3}a_{4}-\beta(a_{1}b_{4}+a_{3}b_{2})(a_{1}b_{1}-b_{2})$
$\displaystyle+\beta
a_{1}b_{2}(a_{1}b_{3}-b_{4})-a_{1}[a_{1}(a_{1}b_{3}-b_{4})-a_{3}(a_{1}b_{1}-b_{2})]^{2}$
$\displaystyle:=\gamma$

Derivation of the fourth condition:

$\displaystyle a_{4}^{(4)}$ $\displaystyle=\gamma
a_{4}^{(3)}+b_{3}^{(3)}b_{4}^{(3)}$ $\displaystyle=\gamma\left(\gamma
a_{4}^{(2)}\right)+\left(\gamma b_{3}^{(2)}-\beta
b_{4}^{(3)}\right)b_{4}^{(3)}$
$\displaystyle=\gamma^{2}\left(a_{1}a_{4}^{(1)}+b_{1}^{(1)}b_{4}\right)$
$\displaystyle+\gamma\left(a_{1}b_{3}^{(1)}-a_{3}b_{1}^{(1)}\right)\left(\beta
b_{4}^{(2)}-b_{2}^{(2)}a_{4}^{(2)}\right)-\beta\left(b_{4}^{(3)}\right)^{2}$
$\displaystyle=\gamma^{2}\left[a_{1}^{2}a_{4}+b_{4}\left(a_{1}b_{1}-b_{2}\right)\right]$
$\displaystyle+\gamma\left[a_{1}\left(a_{1}b_{3}-b_{4}\right)-a_{3}\left(a_{1}b_{1}-b_{2}\right)\right]\eta-\beta(b_{4}^{(3)})^{2}$

Where we have defined

$\displaystyle\eta$ $\displaystyle:=\beta^{2}b_{4}-\left(\beta
b_{2}-a_{1}b_{3}^{(2)}\right)\left(a_{1}a_{4}^{(1)}+b_{1}^{(1)}b_{4}\right)$
$\displaystyle=\beta^{2}b_{4}-\left[\beta
b_{2}-a_{1}\left(a_{1}b_{3}^{(1)}-a_{3}b_{1}^{(1)}\right)\right]\varepsilon,$

with
$\varepsilon:=\left[a_{1}^{2}a_{4}+b_{4}\left(a_{1}b_{1}-b_{2}\right)\right]$.
Moreover, we can rewrite the coefficient $b_{4}^{(3)}$ in terms of $\eta$,
obtaining

$\displaystyle b_{4}^{(3)}$
$\displaystyle=\beta^{2}b_{4}-b_{2}^{(2)}a_{4}^{(2)}$
$\displaystyle=\beta^{2}b_{4}-\left[\beta
b_{2}-a_{1}\left(a_{1}b_{3}^{(1)}-a_{3}b_{1}^{(1)}\right)\right]\varepsilon$
$\displaystyle=\eta$

Hence, the explicit expression of the fourth condition becomes

$\displaystyle a_{4}^{(4)}$ $\displaystyle=\gamma^{2}\varepsilon+\eta[\gamma
a_{1}(a_{1}b_{3}-b_{4})-\gamma a_{3}(a_{1}b_{1}-b_{2})-\beta\eta]$

Lastly, let us derive the explicit expression for $\eta$:

$\displaystyle\eta$ $\displaystyle:=\beta^{2}b_{4}-[\beta
b_{2}-a_{1}(a_{1}(a_{1}b_{3}-b_{4})-a_{3}(a_{1}b_{1}-b_{2}))]\varepsilon.$

##  Appendix B Attainment of Classical Routh-Hurwitz criterion in case of real
coefficient

Let us now show that the stability conditions are the same as the classical
Routh-Hurwitz criterion in case of real coefficients, namely $b_{j}=0,\forall
j\in\\{0,..,4\\}$. For simplicity, let us again consider a $4$-order
polynomial

$p(s)=s^{4}+a_{1}s^{3}+a_{2}s^{2}+a_{3}s+a_{4}$

The table of the coefficients is given by

$s^{4}$ |  $1$ |  $a_{2}$ |  $a_{4}$ |  $0$
---|---|---|---|---
$s^{3}$ |  $a_{1}$ |  $a_{3}$ |  $0$ |  $0$
$s^{2}$ |  $\displaystyle\frac{a_{1}a_{2}-a_{3}}{a_{1}}$ |  $a_{4}$ |  $0$ |  $0$
$s^{1}$ |  $\displaystyle\frac{(a_{1}a_{2}-a_{3})a_{3}-a_{1}^{2}a_{4}}{a_{1}a_{2}-a_{3}}$ |  $0$ |  $0$ |  $0$
$s^{0}$ |  $a_{4}$ |  $0$ |  $0$ |  $0$

Table 3: Coefficients of the classical Routh-Hurwitz criterion

and the necessary and sufficient stability conditions are given by

$\begin{cases}a_{1}>0\\\ a_{1}a_{2}-a_{3}>0\\\ a_{4}>0\\\
\left(a_{1}a_{2}-a_{3}\right)a_{3}-a_{1}^{2}a_{4}>0\end{cases}$ (12)

When $b_{j}=0~{}\forall j\in\\{0,..,4\\}$, table 1 becomes the table that is
given in 4.

$a_{1}$ |  $0$ |  $a_{3}$ |  $0$
---|---|---|---
$b_{1}^{(1)}=0$ |  $a_{2}^{(1)}=a_{1}a_{2}-a_{3}$ |  $b_{3}^{(1)}=0$ |  $a_{4}^{(1)}=a_{1}a_{4}$
$a_{2}^{(2)}=a_{1}a_{2}^{(1)}$ |  $b_{3}^{(2)}=0$ |  $a_{4}^{(2)}=a_{1}a_{4}^{(1)}$ |
$b_{2}^{(2)}=0$ |  $a_{3}^{(2)}=a_{2}^{(2)}a_{3}-a_{1}a_{4}^{(2)}$ |  $b_{4}^{(2)}=0$ |
$a_{3}^{(3)}=a_{2}^{(2)}a_{3}^{(2)}$ |  $b_{4}^{(3)}=0$ |  |
$b_{3}^{(3)}=0$ |  $a_{4}^{(3)}=a_{3}^{(3)}a_{4}^{(2)}$ |  |
$a_{4}^{(4)}=a_{3}^{(3)}a_{4}^{(3)}$ |  |  |

Table 4: Coefficients of the generalized Routh-Hurwitz criterion for $b_{j}=0$

and the necessary and sufficient conditions for stability are given by Eq.
(2), which for real coefficients become

$\begin{cases}a_{1}>0\\\ a_{2}^{(2)}=a_{1}a_{2}^{(1)}>0\\\
a_{3}^{(3)}=a_{2}^{(2)}a_{3}^{(2)}>0\\\
a_{4}^{(4)}=a_{3}^{(3)}a_{4}^{(3)}>0\end{cases}$ (13)

Let us remember that they all have to stand simultaneously, in order for the
system to be stable. The equivalence between the first condition of Eq. (13)
and the first of (12) is trivial. The second condition of Eq. (13) gives us

$a_{2}^{(2)}=a_{1}a_{2}^{(1)}=a_{1}(a_{1}a_{2}-a_{3})>0$

Since $a_{1}>0$, we have the second condition of Eq. (12). The third condition
of Eq. (13) gives us

$a_{3}^{(3)}=a_{1}\left(a_{1}a_{2}-a_{3}\right)\left[a_{1}\left(a_{1}a_{2}-a_{3}\right)a_{3}-a_{1}^{3}a_{4}\right]>0$

Given the first two conditions, namely $a_{1}>0$ and $a_{1}a_{2}-a_{3}>0$, we
obtain $\left(a_{1}a_{2}-a_{3}\right)a_{3}-a_{1}^{3}a_{4}>0$, which is exactly
the fourth condition of Eq. (12). Lastly, the fourth condition of Eq. (13)
gives us

$a_{4}^{(4)}=(a_{3}^{(3)})^{2}a_{1}^{2}a_{4}>0$

which reduces to $a_{4}>0$, i.e., the third condition of Eq. (12).
Hence, from the generalized Routh-Hurwitz conditions for the case of real
coefficients, we attained the classical Routh-Hurwitz conditions, proving the
equivalence.

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