text:
\section{Introduction}

The study of nuclear masses [1-22] is of great significance not only to nuclear physics, but also to astrophysics.
Early database (AME2003) [19] contained many experimental values of nuclear masses and many predictions of unknown nuclear masses,
the database published in 2017 (AME2016) [20] has some new nuclear masses than those published in 2012(AME2012)${[21]}$ and 2003.
In addition, the new measurement methods and the experimental instruments have made some nuclear masses more accurate.

In recent years, in addition to some global mass relations [1-9], local mass relations have also attracted much attention.
For example: Audi-Wapstra systematics, Garvey-Kelson${[10]}$ local mass relations and the local mass relations of residual proton-neutron interactions [12-15, 22].
In Refs.[12-15, 22], the accuracy of residual interaction is closely related to the accuracy of nuclear masses, many methods and formulas have been developed to study the residual interaction in order to obtain more accurate predictions of nuclear masses.
There are many papers using artificial neural networks in nuclear physics [23-33],
so we study the residual interaction of nuclei with $A\geq100$ based on BP neural network (using MATLAB2015b to build the model),
then obtained the RMSD of odd-A nuclei and even-A nuclei:
comparing with AME2003, RMSDs are 112 keV and 128 keV;
comparing with AME2012, RMSDs are 103 keV and 121 keV;
comparing with AME2016, RMSDs are 106 keV and 122 keV, respectively.
The predicted values based on databases (AME2003 and AME2012) are compared well with the experimental values.

Neural network is applied in many fields [34-37].
The idea that we introduce is enlightened by Refs.[23-33].
In this paper, we use BP neural network and AME databases to study nuclear masses with $A\geq100$.
We describe and predict residual interaction based on the strong descriptive and predictive properties of BP neural network.
It is found that the RMSD of known nuclear mass has been significantly improved, and the accuracy of some nuclear mass predictions has also been improved.
In Sec. 2, the RMSD of known nuclear mass is calculated based on BP neural network and databases.
In Sec. 3, the predicted values of nuclear mass calculated by residual interaction and experimental values are compared with the experimental values. The comparison shows that our predicted values are close to the experimental values.
In Sec. 4, summarizes and discusses this article.
\section{Residual interaction and BP neural network}\label{S:N}

The residual interaction between nucleons is very important.
We assume that there is a nucleus with $Z$ protons and $N$ neutrons,
the relationship between binding energy $B(Z, N)$ and nuclear mass $M(Z, N)$: $B(Z, N) = ZM_p +NM_n - M(Z, N)$,
where the $M_p$ and the $M_n$ are the mass of a free proton and a free neutron,
we then use a definition of the $B(Z,N)$ which yields a positive quantity and obtains a positive $\delta V_{1p-1n}$.
In Ref. [38], the authors studied the effects of electron mass, electron binding energy and Coulomb energy on nuclear masses.
It is shown from residual interaction formula that electron mass, electron binding energy and Coulomb energy have no effect on the residual interaction. Therefore, in this paper these electronic properties have no effect on the nuclear masses.

\subsection{residual interaction}
The proton-neutron interactions between the last $i$ protons and $j$ neutrons is defined as[12]
\begin{eqnarray}
\label{eq:1}
\delta V_{ip-jn}(Z,N)=B(Z,N)+B(Z-i,N-j)-\nonumber\\[1mm]
B(Z,N-j)-B(Z-i,N).
\end{eqnarray}
It is easy to obtain the formula of $\delta V_{1p-1n}$,
\begin{eqnarray}
\label{eq:2}
\delta V_{1p-1n}(Z,N)=B(Z,N)+B(Z-1,N-1)-\nonumber\\[1mm]
B(Z,N-1)-B(Z-1,N)\nonumber\\[1mm]
\ =M(Z,N-1)+M(Z-1,N)-\nonumber\\[1mm]
M(Z,N)-M(Z-1,N-1).
\end{eqnarray}
we then get the the mass equation as follows:
\begin{eqnarray}
\label{eq:3}
\ M(Z,N)=M(Z,N-1)+M(Z-1,N)- \nonumber\\[1mm]
M(Z-1,N-1)-\delta V_{1p-1n}.
\end{eqnarray}
Eq. (3) shows that the accuracy of residual interaction determines the accuracy of nuclear masses.
Ref. [33] uses Bayesian neural networks to construct prediction models to study nuclear mass, while the prior knowledge of Bayesian neural networks has certain limitations.
Inspired by Ref. [23-33], this paper describes and predicts the residual interaction based on BP neural network, thus describing and predicting the nuclear masses.

\subsection{BP neural network}
BP (back-propagation) neural network [39-42] was proposed by Rumelhart and McClelland in 1986. Multi-layer feedforward neural network trained by error back propagation algorithm is the most representative and widely used neural network learning model.
According to the learning rule, the neural network can adjust the weighted value of the link chain to achieve the convergence of the target through training.
BP neural network architecture is composed of several layers of interconnected neurons, usually including input layer, output layer and several hidden layers, each layer contains a number of neurons.
The network structure as shown in Fig. 1.

\begin{figure}[tbp]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig1.eps}
\end{center}
\caption{(Color online) The BP neural network structure.}
\end{figure}

In this paper, we study the known nuclear masses in databases (AME2003, AME2012, AME2016), and obtain the residual interaction of odd-A and even-A nuclei by using Eq. (2).
First, we use the BP neural network to train the samples with odd-A nuclei and even-A nuclei of $\delta V_{1p-1n}$;
second, with the proton number ($Z$), neutron number ($N$) and mass number ($A$) as the inputs of the network, the global minimum error is allowed to be 10$^{-8}$;
third, we use output to describe and predict residual interaction.
The nuclear mass of $A\geq100$ is obtained by the residual interaction calculated values combined with the database, we then compared with the experimental values.
The RMSDs and training times are listed in Table 1.

As is known from Refs. [12, 22], the residual interaction of odd-A nuclear is more statistical than even-A nuclear.
Therefore, the RMSD of odd-A nuclear mass is less than that of even-A nuclear mass based on Eq. (3).
Table 1 also shows that the RMSD of odd-A nuclei is less than that of even-A nuclei when the nuclear mass is calculated by residual interaction trained by BP neural network.

\begin{table}
\centering
\caption{Training times and RMSDs.}\label{tab1}
\begin{tabular}{cccc}
\hline  databases &  odd-even  &   training times  & RMSD (keV)  \\
\hline  AME2003   &odd                      &1562               &$\simeq$112\\
                  &even                     &3540               &$\simeq$128\\
\hline  AME2012   &odd                      &1649               &$\simeq$103\\
                  &even                     &3445               &$\simeq$121\\
\hline  AME2016   &odd                      &2457               &$\simeq$106\\
                  &even                     &2429               &$\simeq$122\\
\hline
\end{tabular}
\end{table}

For nuclei with mass number $A\geq100$ (AME2003),
Ref. [12] uses two empirical formulas to describe the residual interaction of odd-A nuclei and even-A nuclei,
then obtained the RMSD(odd-A and even-A nuclei) between the calculated and experimental values of nuclei mass are 132 keV and 168 keV;
in Ref. [22], the RMSD of odd-A and even-A nuclei is 133 keV and 164 keV;
in our paper, the RMSD of odd-A and even-A nuclei is about 112 keV and 128 keV based on BP neural network and database.
In summary, the accuracy of describing residual proton-neutron interactions based on BP neural network has been improved, which will contribute to the development of nuclide maps.

Fig. 2 shows the deviations between calculated and experimental values.
The solid diamond in Fig. 2(a) represents the difference between the experimental value in the database AME2003 and the calculated value $M^{cal1}$;
in Fig. 2(b) , solid diamonds show that the difference between the experimental value in the database AME2012 and the calculated value $M^{cal2}$;
in Fig. 2(c), solid diamonds are plotted by using the difference between the experimental value in the database AME2016 and the calculated value $M^{cal3}$.
These results demonstrate the feasibility and accuracy of using BP neural network to describe residual interaction.
Therefore, we can predict the residual interaction of unknown nuclear mass on the basis of training model.

\begin{figure}[tbp]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig2.eps}
\end{center}
\caption{(Color online) Deviations between our calculated values and the experimental values.}
\end{figure}

In addition, we study the binding energy and mass excess of nuclei with $A\geq100$ (AME2012) based on the BP neural network, then obtained the RMSD of binding energy and mass excess is about 1210 kev and 908 kev.
When we study the binding energies of even-A nuclei and odd-A nuclei, the RMSD is 1147 keV and 482 keV;
for the mass excess of nuclei, the RMSD is 1087 keV and 577 keV.
It can be seen from the above that the RMSD of binding energy and mass excess of odd-A nuclei is smaller than that of even-A nuclei, which also shows that the systematicness of binding energy and mass excess of odd-A nuclei is better than that of even-A nuclei.
Although the deviation of nuclear mass obtained by nuclear binding energy and residual mass with BP neural network is relatively large,
but the method can be used to estimate the residual interaction, which is helpful to analyze or discover useful information from a large number of experimental data.
Further, it provides reference for finding physical regularity, and provide support for the feasibility of physical experiment.

\section{Prediction of Nuclear Masses}\label{S:specific}

In this section, the residual interaction of unknown nuclear masses are obtained based on BP neural network, and combined with experimental data (AME2003) obtained some unknown nuclear masses.
We get the other mass equation based on Eq. (3):

\begin{eqnarray}
\label{eq:6}
\ M(Z,N)=M(Z+1,N)+M(Z,N+1)- \nonumber\\[1mm]
M(Z+1,N+1)-\delta V_{1p-1n}.
\end{eqnarray}
We obtain the residual interaction of unknown nuclear mass by using BP neural network.
We predict unknown masses with Eqs. (3) and (4), we calculate the average value
if they obtain the same nuclear mass.
In table 2 we present a set of selected data of mass excess of our predicted values and experimental
values in the AME database (AME2012 and AME2016).

\begin{table}
  \centering
  \caption{The difference between the predicted and experimental values (AME2012 and AME2016).}\label{tab2}
  \vskip 2mm
\begin{tabular}{cccccc}
\hline
Nucleus	&	ME2012	&	ME2016	&	ME$^{pred1}$	&	dev1	&	dev2	\\
	&	/keV	&	/keV	&	/keV	&	/keV	&	/keV	\\
\hline
$^{103}Y$	&	-58458	&	-58458	&	-58912	&	454	&	454	\\
$^{119}Pd$	&	-71408	&	-71408	&	-71091	&	-317	&	-317	\\
$^{129}Cd$	&	-63510$^{The.}$	&	-63058	&	-62943	&	Null	&	-115	\\
$^{147}Ba$	&	-60264	&	-60264	&	-60617	&	353	&	353	\\
$^{149}Ba$	&	-53020$^{The.}$	&	-53120	&	-53211	&	Null	&	91	\\
$^{155}Nd$	&	-62284	&	-62284	&	-62444	&	160	&	160	\\
$^{161}Ta$	&	-38701	&	-38779	&	-38885	&	184	&	106	\\
$^{106}Sb$	&	-66473	&	-66473	&	-66541	&	68	&	68	\\
$^{112}I$	&	-67063	&	-67063	&	-66947	&	-116	&	-116	\\
$^{160}Eu$	&	-63480	&	-63480	&	-63613	&	133	&	133	\\
$^{164}Re$	&	-27520	&	-27470	&	-27388	&	-132	&	-82	\\
$^{166}Re$	&	-31890	&	-31890	&	-31737	&	-153	&	-153	\\
$^{176}Au$	&	-18400	&	-18520	&	-18277	&	-123	&	-243	\\
$^{180}Tl$	&	-9260	&	-9390	&	-9309	&	49	&	-81	\\
$^{228}Fr$	&	33369	&	33384	&	33048	&	321	&	336	\\
$^{228}Np$	&	33600	&	33600	&	33879	&	-279	&	-279	\\
\hline
\end{tabular}
\end{table}

In Table 2, we have listed some predicted values in comparison with the experimental values.
Where, dev1 represents the difference between the experimental values (ME2012) of the mass excess in the AME2012 database and our predicted values (ME$^{pred1}$);
we use dev2 to represent the difference between the experimental values (ME2016) of the mass excess in the AME2016 database and our predicted values (ME$^{pred1}$);
These experimental values given in AME2012 and AME2016 are rounded.
From the above table, we can see that the predicted value is close to the experimental value, and some nuclear mass deviations are only tens of keV or even keV.
In addition, two nuclei ($^{129}Cd$ and$^{149}Ba$) have only predicted values in the AME2012 database without experimental values, it is found that our predicted value is closer to the experimental value (AME2016) than the predicted value in AME2012.
Although the  deviation of some nuclear mass is too large, but it has little effect on the overall description and prediction effect of nuclear mass.
The residual interaction based on BP neural network is inaccurate, which leads to the deviation between the calculated value and the experimental value.
Therefore, the smaller the residual interaction deviation is, the more accurate the nuclear mass will be.

In addition, we use BP neural network and AME2012 database to predict unknown nuclear mass.
We calculate the average value if Eqs. (3) and (4) obtain the same nuclear mass.
Table 3 shows some mass excess of the experimental values (AME2016) compared with the predicted values.
Results show that the predicted values are in good agreement with the experimental values (AME2016).
Where dev0 represents the measurement deviation in the AME2016 database;
we use dev3 to represent the difference between the experimental values of the mass excess (ME2016) in the AME2016 database and our predicted values (ME$^{pred2}$).

\begin{table}
  \centering
  \caption{ The difference between the predicted and experimental values (AME2016). }\label{tab3}
  \vskip 2mm
\begin{tabular}{ccccc}
\hline
Nucleus	&	ME2016(keV)	&	dev0(keV)	&	ME$^{pred2}$(keV)	&	dev3(keV)	\\
\hline
$^{129}Cd$	&	-63058	&	17	&	-63369	&	311	\\
$^{141}I$	&	-59927	&	16	&	-59824	&	-103	\\
$^{149}Ba$	&	-53120	&	440	&	-52937	&	-183	\\
$^{190}Tl$	&	-24372	&	8	&	-24432	&	60	\\
$^{194}Bi$	&	-16029	&	6	&	-15985	&	-44	\\
$^{198}At$	&	-6715	&	6	&	-6653	&	-62	\\
$^{202}Fr$	&	3096	&	7	&	3164	&	-68	\\
\hline
\end{tabular}
\end{table}

These results show that the paper based on BP neural network to describe and predict the residual interaction is feasible.
The RMSD of the known nuclear masses is reduced,
at the same time, the predicted value based on the residual interaction and the database (AME2003 and AME2012) is also close to the experimental values.

\section{Discussion and Conclusions}

The main work of this paper is to study residual interactions ($\delta V_{1p-1n}$).
Before us there are many people using artificial networks in nuclear physics.
In our paper, we use BP neural network to build description and prediction models based on odd-A nuclei and even-A nuclei residual interaction.
We then describe and predict nuclear masses using residual interaction combined with experimental values.
Although the mass of the deviation is too large, it does not affect the overall description and prediction effect.

In this paper, our calculated values based on BP neural network are in good agreement with the experimental data.
At the same time, the residual interaction of odd-A nuclei is statistically good.
Therefore, the RMSD of odd-A nuclei is less than even-A nuclei when describing the known nuclear mass.
As shown in Table 2 and table 3, the predicted values based on AME2003 (AME2012) are close to the experimental values in AME2012 and AME2016 (AME2016),
and some of the deviations between predicted values and experimental values are only tens of keV.
In summary: BP neural network can better describe and predict the residual interaction, and can better describe the AME database (AME2003, AME2012, AME2016) and predict the nuclear quality.

When using BP neural network for data fitting, as long as we select the appropriate network structure, we can get a better fitting curve and meet the requirements of different users.
In this paper, BP neural network is suitable for medium and long term prediction.
It has the advantages of fast calculation speed, high precision and strong non-linear fitting ability, which reduces the errors in describing and predicting the nuclear mass.
In addition, the number of nuclei involved in describing and predicting nuclear mass formula is 3, and the accuracy of the extrapolation method can be improved with the smaller number of nuclei involved ${[43]}$.
Due to it can not participate in the prediction process, the algorithm is incomplete and so on, so these factors can lead to the deviation.
The more precise residual interactions, the more accurate nuclear masses will be obtained.


text:
\section{Introduction}

Let $PRP_d$ denote the set of holomorphic functions $f:\D^d \to RHP$ such
that $f(0) = 1$.  Here $\D^d$ is the $d$-dimensional unit polydisk in
$\C^d$ and $RHP$ is the right half plane in $\C$.

The set $PRP_d$ is a normal family and a convex set.
Rudin posed the following natural question in his 1970 ICM address \cite{RudinICM}.
\begin{quote}
What are the extreme
points of $PRP_d$?
\end{quote}

A complete answer is known only in the case $d=1$.
 The extreme points
of $PRP_1$ are the functions
\[
z \mapsto \frac{1+\bar{\alpha} z}{1-\bar{\alpha}z}
\]
where $\alpha \in \T$, the unit circle.  Remarkably, this fact
together with the Krein-Milman theorem suffices to prove the Herglotz representation
theorem for elements of $PRP_1$: for every $f\in PRP_1$ there is a unique probability
measure $\mu$ on $\T$ such that
\[
f(z) = \int_{\T} \frac{1+\bar{\zeta}z}{1-\bar{\zeta}z} d\mu(\zeta).
\]
In turn, this fact can be used to prove a variety of spectral theorems.

There is no known characterization of the extreme points of $PRP_d$
for $d>1$ and we would guess there is not a simple characterization.
Every $f\in PRP_d$ has a Poisson type representation
\[
\text{Re} f(z) = \int_{\T^d} P_{z_1} P_{z_2}\cdots P_{z_d} d\mu
\]
where $P_{w}(\zeta) = \frac{1-|w|^2}{|1-\bar{w}\zeta|^2}$ is the Poisson kernel
and $\mu$ is a probability measure.  However, when $d>1$, $\mu$ cannot
be an arbitrary probability measure;
it must satisfy additional moment conditions since $f$ is analytic.
Specifically, we must have
$\text{supp}(\hat{\mu}) \subset \mathbb{Z}_{+}^d \cup \mathbb{Z}_{-}^d$ where $\mathbb{Z}_+$ (resp. $\mathbb{Z}_{-}$) denotes
the set of non-negative (resp. non-positive) integers.
One might expect that extreme points of $PRP_d$ would be represented via
measures with ``small'' support but McDonald has constructed an extreme point
whose boundary measure is absolutely continuous with respect to Lebesgue measure
on $\T^d$ (see \cite{McDonald2}).  Not all of the known results are negative in spirit however.

Forelli has a useful necessary condition for $f \in
PRP_d$ to be an extreme point \cite{Forelli}.
Let $S_d$ denote
the Schur class of the polydisk;
the set of holomorphic functions $F:\D^d \to \cd$.
First, via a Cayley transform
we may write
\begin{equation} \label{fF}
f = \frac{1+F}{1-F}
\end{equation}
where $F \in S_d$ and $F(0) = 0$.  Forelli has proven that if $F$ is
reducible in the sense that $F = GH$ with non-constant $G,H \in S_d$ with $G(0)=0$
(but with no assumption on $H(0)$), then $f$ is \emph{not} an extreme point of
$PRP_d$.  Forelli's approach actually gives an elementary proof of the
characterization of extreme points for the case $d=1$.

Since the general problem of characterizing extreme points
of $PRP_d$ seems difficult, it seems reasonable
to restrict the problem to a more tractable subclass of
functions.  We shall examine the class
\[
PRPR_d = \{ f \in PRP_d: F=\frac{f-1}{f+1} \text{ is a rational inner function} \}.
\]
\emph{Inner} means
 for almost every $z\in \T^d$, $|F(rz)|\to 1$ as $r\nearrow 1$.
This implies $\Re f(rz) \to 0$ at least away from $z\in \T^d$ such that
$F(rz) \to 1$.  This class, while very special, is
also dense in $PRP_d$ using the topology of
local uniform convergence.  (Rudin \cite{FTIP}
proves that rational inner functions are
dense in $S_d$ and it then follows that $PRPR_d$ is
dense in $PRP_d$.) Thus, we shall modify our
opening question from Rudin.

\begin{quote}
Which $f\in PRPR_d$ are extreme points of $PRP_d$?
\end{quote}

It is proven in Rudin \cite{FTIP} that rational inner functions are
of the form
\[
F(z) = \frac{\tilde{p}(z)}{p(z)}
\]
where $p \in \C[z_1,\dots,z_d]$
\begin{enumerate}
\item has no zeros in $\D^d$,
\item multidegree at most $n=(n_1,\dots,n_d)$
(the degree in $z_j$ is at most $n_j$), and
\item the reflection of $p$ given by
\[
\tilde{p}(z) = {z_1}^{n_1} \cdots {z_d}^{n_d} \overline{p(1/\bar{z}_1, \dots, 1/\bar{z}_d)}
\]
has no factors in common with $p$.
\end{enumerate}

We say multidegree ``at most'' above to allow for $\tilde{p}$ to have a monomial factor.
Since $p(0)\ne 0$, we get that $\tilde{p}$ has multidegree exactly $n$.
For this reason it often makes more sense to refer to the multidegree of $\tilde{p}$.

\begin{definition}
We say $p\in \C[z_1,\dots,z_d]$ is \emph{scattering stable} if it satisfies conditions
(1)-(3) above for $n$ equal to the multidegree of $p$.
\end{definition}

 Some simple
examples of rational inner functions are
\begin{equation} \label{schurexamples}
G_1 = \frac{3z_1z_2-z_1-z_2}{3-z_1-z_2} \qquad G_2=\frac{2z_1z_2-z_1-z_1}{2-z_1-z_2}
\end{equation}
and corresponding elements of $PRP_2$ are
\begin{equation} \label{prpexamples}
g_1 = \frac{3-2(z_1+z_2) + 3z_1z_2}{3(1-z_1z_2)} \qquad g_2= \frac{(1-z_1)(1-z_2)}{1-z_1z_2}.
\end{equation}

One way to analyze a convex set is to examine its \emph{faces}.
Given an element $x$ of a convex set $S$,
its face $\mathcal{F}(x)$ is the union of all
line segments $L$ in $S$ with $x$ in the interior of $L$.
We set $\mathcal{F}(x) = \{x\}$ when $x$ is an extreme point.
McDonald has characterized the faces of $f\in PRPR_d$ as
elements of the convex set $PRP_d$.

\begin{theorem}[McDonald \cite{McDonald1}] \label{McDonald}
Let $f\in PRPR_d$ which we write as
$f = \frac{1+F}{1-F} = \frac{p+\tilde{p}}{p-\tilde{p}}$ where $F = \frac{\tilde{p}}{p}$ is
a rational inner function, written as above with $\tilde{p}(0)=0$ and
multidegree $n$.  Then,
every element of the face $\mathcal{F}(f)$ of $f$ in $PRP_d$ is
of the form
\[
\frac{p+\tilde{p}+2v}{p-\tilde{p}}
\]
where $v \in \C[z_1,\dots,z_d]$ has multidegree at most $n$,
satisfies $v=\tilde{v}$, $v(0)=0$, and $p+v$ and $p-a v$ have no zeros
in $\D^d$ for some $a>0$.  The face $\mathcal{F}(f)$ has real dimension at most
$\prod (n_j+1) - 2$.
\end{theorem}

The condition with $a$ is just to make $f$
in the interior of an appropriate line segment.
This theorem does not solve our problem because the
condition that $p+v$ and $p-av$ have no zeros in $\D^d$
is difficult to work with. Nevertheless, the theorem
certainly provides a useful reduction.  Indeed,
it shows that $\mathcal{F}(f)$ does not depend
on the choice of underlying convex set ($PRP_d$ versus $PRPR_d$).

The original proof of McDonald involves one variable slices and the Herglotz representation theorem.
We shall give a very similar proof that uses only complex analysis (i.e. no measure theory).
The number $\prod(n_j+1) - 2$ simply counts the real-linear dimension
of the set of $v$ of multidegree at most $n$ satisfying $v(0)=0$ and $v=\tilde{v}$.
In particular, if $p$ has no zeros in $\cd^d$, then
this dimension is exactly $\prod(n_j+1)-2$
since $p\pm tv$ will have no zeros in $\cd^d$ for $|t|$ small enough.
Thus, if $p$ has no zeros in $\cd^d$ then $f=(p+\tilde{p})/(p-\tilde{p})$
is never an extreme point unless $p$ depends on only one variable.

For example, the function $g_1$ in \eqref{prpexamples} is not extreme and we can perturb
$p_1=3-z_1-z_2$ using any $v=az_1 + \bar{a}z_2$ with small enough $a\in \C$.
In fact, $p_1+az_1+\bar{a} z_2$ has no zeros in $\D^2$ if and only if $|1-a|\leq 3/2$.

We have been able to prove the following add-on to McDonald's theorem, which
also says that the dimension of $\mathcal{F}(f)$ is maximal
if and only if $p$ has no zeros in $\cd^d$.

\begin{theorem}\label{addon}
Assume the setup of Theorem \ref{McDonald}.
If $p$ vanishes at $\zeta \in \T^d$ to order $M$, then
$v$ vanishes at least to order $M$ at $\zeta$.
\end{theorem}

We can also say something about the homogeneous expansion
of $v$.  See Theorem \ref{vhomo} for details.

Unfortunately, Theorem \ref{addon} has limitations
because order of vanishing is a crude way
to count zeros in this setting.
Consider the example $g_2$ above.
Since $p_2=2-z_1-z_2$ has a zero at $(1,1)$,
so must any valid perturbation $v=az_1+\bar{a}z_2$.
This implies $a=ir$ for some $r\in \mathbb{R}$.
It turns out $g_2$ is actually
an extreme point, but it takes additional
work to prove this (i.e. to show $r=0$).

Our main goal is to prove a theorem which describes a number of extreme points
of $PRP_d$ coming from $PRPR_d$
 in the case $d=2$.
The denominators of the $F$'s appearing in these extreme points have a special property.

\begin{definition}
Let $p\in \C[z_1,z_2]$ be scattering stable and of bidegree $(n_1,n_2)$.
We say  $p$ is \emph{$\T^2$-saturated} if $p$ and $\tilde{p}$ have $2n_1 n_2$ common zeros on $\T^2$.
\end{definition}

The common zeros of $p$ and $\tilde{p}$ are
counted with multiplicities as in B\'ezout's theorem.
The number $2n_1n_2$ counts all of the common zeros
of $p$ and $\tilde{p}$ on $\C_{\infty}\times \C_{\infty}$.
See \cite{Fischer, Fulton} for more details about intersection
multiplicities.

The class of $\T^2$-saturated polynomials is interesting in its own right.
In the paper \cite{GK}, we gave two related characterizations of $\T^2$-saturated polynomials.
One relates to a sums-of-squares formula for
scattering stable polynomials.
If $p\in \C[z_1,z_2]$
is scattering stable of bidegree $(n_1,n_2)$ then
\begin{equation} \label{sos}
|p|^2-|\tilde{p}|^2 = (1-|z_1|^2)SOS_1 + (1-|z_2|^2)SOS_2
\end{equation}
where $SOS_j$ is sum of squared moduli of $n_j$ polynomials.
A scattering stable polynomial $p$ is $\T^2$-saturated if and only if the
sums of squares terms $SOS_1,SOS_2$ are unique (see Corollary 13.6 of \cite{GK}).

A second characterization says that
a scattering stable polynomial $p$ is $\T^2$-saturated if and only if
\[
\int_{\T^2} \left|\frac{f}{p}\right|^2 d\sigma =\infty
\]
for all nonzero $f\in \C[z_1,z_2]$ satisfying $\deg_j f < \deg_j p$ for  $j=1,2$ (see
Corollary 6.5 of \cite{GK}).
Thus, $\T^2$-saturated polynomials have so many boundary zeros
that lower degree polynomials cannot match them (in the above sense).

A simple example of a $\T^2$-saturated polynomial is $p=2-z_1-z_2$.
Note $p$ and $\tilde{p}$ share a single zero $(1,1)$ on $\T^2$ but
it has multiplicity $2$. This can be computed using the resultant
of $p$ and $\tilde{p}$ with respect to $z_2$.

Our main theorem \emph{suggests} a third characterization of $\T^2$-saturated polynomials
and constructs a family of extreme points in $PRP_2$.

\begin{theorem} \label{thmsaturated}
Let $p \in \C[z_1,z_2]$ be scattering stable and $\tilde{p}(0,0)=0$.
Let $f=\frac{p+\tilde{p}}{p-\tilde{p}}$.
If $p$ is $\T^2$-saturated and $p-\tilde{p}$ is irreducible,
then $f$ is an extreme point of $PRP_2$.
\end{theorem}

For example, one can conclude the example $g_2$ is an extreme point
simply by noticing $p=2-z_1-z_2$ is $\T^2$-saturated
and $p-\tilde{p} = 2(1-z_1z_2)$ is irreducible.
On the other hand, $p$ being $\T^2$-saturated is not
sufficient for $f$ to be extreme as the example
\[
\frac{1-G_2}{1+G_2} = \frac{(1-z_1z_2)}{(1-z_1)(1-z_2)}
 = \frac{1}{2}\left(\frac{1+z_1}{1-z_1} + \frac{1+z_2}{1-z_2}\right)
\]
shows.

Along the path to proving Theorem \ref{thmsaturated}
we have been able to prove a more refined version
of Theorem \ref{addon}.
Given two polynomials $p,q\in \C[z_1,z_2]$ with a common
zero $\zeta$, we let $I_{\zeta}(p,q)$ denote the intersection
multiplicity of $p,q$ at $\zeta$.

\begin{theorem} \label{thmmults}
Let $f=\frac{p+\tilde{p}}{p-\tilde{p}} \in PRPR_2$ where
$p\in \C[z_1,z_2]$ is scattering stable.
Suppose
\[
g = f+ 2\frac{v}{p-\tilde{p}}
\]
belongs to the face of $f$ as in Theorem \ref{McDonald}.
Suppose $p(\zeta)=0$ for some $\zeta\in \T^2$.  Then,
\[
I_{\zeta}(p,\tilde{p}) \leq I_{\zeta} (p+v,\tilde{p}+v).
\]
\end{theorem}

We have been unable to prove the following.

\begin{conjecture} \label{conjecture}
If $p\in \C[z_1,z_2]$ is scattering stable, $\tilde{p}(0,0)=0$, and $f=\frac{p+\tilde{p}}{p-\tilde{p}}$ is
an extreme point of $PRP_2$, then $p$ is $\T^2$-saturated and $p-\tilde{p}$ is irreducible.
\end{conjecture}

A very modest piece of evidence is the following.  See section \ref{modest}.

\begin{theorem}\label{oneone}
Conjecture \ref{conjecture} holds if $p$ has bidegree $(1,1)$.
\end{theorem}

The reason Theorem \ref{oneone} works out is that if $f\in PRPR_2$
is extreme and depends on both variables, then the associated $p$ must have at least one
zero on $\T^2$.  In the case of degree $(1,1)$ polynomials, a single zero implies
saturation.  In Section \ref{nonsaturated} we present an example of
a non-saturated scattering stable polynomial with a zero on $\T^2$ and
we show that the associated $f$ is not extreme.  This at least
shows that $\T^2$-saturated should not be replaced with $p$
having at least one zero on $\T^2$.

One final piece of evidence in favor of
Conjecture \ref{conjecture} is the following.

\begin{theorem} \label{intthm}
Suppose $p\in \C[z_1,z_2]$ is scattering stable with $\tilde{p}(0,0)=0$.

If there exists $v\in\C[z_1,z_2]$ such that
$\deg v \leq \deg \tilde{p}$, $v(0,0)=0$, $v=\tilde{v}$, and $p\pm v$ have
no zeros in $\D^2$, then $v/p \in L^2(\T^2)$.

Going in the other direction, if $p$ is not $\T^2$-saturated
then there exists nonzero $v\in \C[z_1,z_2]$ such that $v/p\in L^2(\T^2)$,
$\deg v \leq \deg \tilde{p}$, $v=\tilde{v}$, and $v(0,0)=0$.
\end{theorem}

Of course the key thing missing in the last statement is
$p\pm t v$ have no zeros in $\D^2$ for some small $t>0$.  This theorem
requires some machinery from \cite{GK}.

In Section \ref{secdets}
we discuss some connections to determinantal representations
for stable polynomials
and transfer function representations for analytic functions.
In particular, we point out that saturated polynomials $p$
possess symmetric, contractive determinantal representations,
and the corresponding $F=\tilde{p}/p$ possess symmetric unitary
transfer function realizations.  These notions are all defined in
Section \ref{secdets}.

\section{Proof of McDonald's characterization of faces}

In this section we will give a slightly more elementary proof
of McDonald's characterization of faces (Theorem \ref{McDonald})
for $f=\frac{1+F}{1-F}$ where
$F$ is rational inner.  It is essentially the same proof except we avoid
measure theory and gain a little more
information and the face of $f$.

We first tackle the one variable case where $F$ is a Blaschke product
which we may write as $F = \frac{\tilde{p}}{p}$ for a polynomial $p\in \C[z]$
having no zeros in $\cd$ (because zeros on $\T$ get cancelled out)
and with $\tilde{p}(0)=0$.
Then, $f = \frac{p+\tilde{p}}{p-\tilde{p}}$.

Let $f_{1}$ be in the face $\mathcal{F}(f)$
corresponding to $f$.  This means there exist $a>0$ and analytic $H$
such that $f_t= f+tH \in PRP_1$ for
$t \in [-a,1]$.  It will be convenient later to write $H = \frac{2v}{p-\tilde{p}}$
for some holomorphic function $v$.
Division by $p-\tilde{p}$ is no loss of generality
since this function is non-vanishing in $\D$ (indeed, by the maximum principle $F\ne 1$ in $\D$).
Note that
\[
f_{1} = \frac{p+\tilde{p}}{p-\tilde{p}} + \frac{2 v}{p-\tilde{p}}
=\frac{1+\frac{\tilde{p}+v}{p+v}}{1-\frac{\tilde{p}+v}{p+v}}.
\]

Let $A=\{a_1,\dots, a_n\}$ be the set of zeros of $p-\tilde{p}$ on $\T$.
Since $\Re f = \frac{a}{a+1} \Re f_{1} + \frac{1}{a+1} \Re f_{-a} \geq 0$ and
since $\Re f = \frac{|p|^2-|\tilde{p}|^2}{|p-\tilde{p}|^2} = 0$
for $z \in \T\setminus A$ we see that for all $t\in [-a,1]$
\[
\Re f_{t}(z) \to 0
\]
as $z \to \T\setminus A$.

By the Schwarz reflection principle, the $f_{t}$ extend analytically
across $\T\setminus A$ via the formula $f_{t} (z) = -\overline{f_{t}(1/\bar{z})}$
which even extends analytically to $\infty$.  The singularities at
the $a_j$ are either removable or simple poles.
(Without going into too much detail we can give an elementary explanation.
The derivative of $f_{t}$ cannot vanish
on $\T\setminus A$ because of local mapping properties.  Thus, $f_{t}(e^{i\theta})$
is imaginary-valued and monotone off singularities and this is enough to prove
$f_{t}$ omits an imaginary line segment near singularities.
This implies $f_{t}$ cannot have an essential singularity at any $a_j$
by either the Big Picard theorem or by conformal mapping and the Casorati-Weierstrass theorem.
Any poles on $\T$ must be simple because of local mapping properties of $1/f_{t}$.)
Therefore, the $f_{t}$ are rational and by their symmetry properties we get

\[
\begin{aligned}
f_1(z) &= \frac{p(z)+\tilde{p}(z)}{p(z)-\tilde{p}(z)} + \frac{v(z)}{p(z)-\tilde{p}(z)} \\
&= -\overline{f_1(1/\bar{z})} = - \frac{\overline{p(1/\bar{z})}+\overline{\tilde{p}(1/\bar{z})}}{
\overline{p(1/\bar{z})}-\overline{\tilde{p}(1/\bar{z})}} -\frac{\overline{v(1/\bar{z})}}{
\overline{p(1/\bar{z})}-\overline{\tilde{p}(1/\bar{z})}}
\end{aligned}
\]
and after some simplification we get
\[
\frac{v(z)}{p(z)-\tilde{p}(z)} = \frac{z^n \overline{v(1/\bar{z})}}{p(z)-\tilde{p}(z)}
\]
for $z\notin A$.  Necessarily, $v$ has no poles (because
simple poles in the above function are accounted for by the
denominator) and hence $v$ must be a polynomial.
Also, $v(z) = z^n\overline{v(1/\bar{z})}$ and $v(0)=0$.

Thus, every element of the face associated to $f$ is of the form
\[
\frac{p+\tilde{p}}{p-\tilde{p}} + \frac{2v}{p-\tilde{p}}
\]
where $v$ is a polynomial such that $v=\tilde{v}$, $v(0)=0$, $p+v$ has no
zeros in $\D$, and there exists $a>0$ such that $p-av$ has no zeros in $\D$.

Conversely, given $v$ satisfying all of the above conditions,
$\frac{\tilde{p}+v}{p+v}, \frac{\tilde{p}-av}{p-av}$ are finite Blaschke products,
and $f=\frac{p+\tilde{p}}{p-\tilde{p}}$ is a convex combination of $f_{1},f_{-a}$ as
defined above.

This gives the desired characterization of faces in the one variable case.

Next we look at several variables.
Write $f=\frac{p+\tilde{p}}{p-\tilde{p}}\in PRPR_d$ where $p\in \C[z_1,\dots,z_d]$ is
scattering stable and has multidegree at most $n=(n_1,\dots, n_d)$.

Now, if $f+tH \in PRP_d$ for $t \in [-a,1]$ and $H = \frac{2v}{p-\tilde{p}}$ for some
analytic $v:\D^d\to \C$,
then we claim $v$ is a polynomial of degree at most $n$ with $v=\tilde{v}$.

We examine slices $f_{\zeta}(z):= f(z \zeta)$
where $z\in \D, \zeta \in \T^d$. We will use this notation
for other functions, not just $f$.

If we verify the hypotheses of the following lemma, then we are finished.
The lemma will be proven at the end.

\begin{lemma}\label{slices} Let $h:\D^d\to \C$ be analytic.
If for each $\zeta \in \T^d$, $h_{\zeta}(z)$ is a
(one variable) polynomial
of degree at most $|n|=n_1+\dots+n_d$
satisfying $h_{\zeta}(z) = \zeta^n z^{|n|} \overline{h_{\zeta}(1/\bar{z})}$,
then
$h$ is a ($d$ variable) polynomial of multidegree
at most $n$ and $h=\tilde{h}$.
\end{lemma}

Fix $\zeta \in \T^d$.
Since $p$ has multidegree at most $n$, $p_{\zeta}$ has
degree at most $|n|$ and we can calculate that
\[
\zeta^n \widetilde{p_{\zeta}}(z) = \tilde{p}_{\zeta}(z).
\]
Then, for a choice of $\mu= \sqrt{\zeta^n}$
\[
f_{\zeta} = \frac{p_{\zeta} +\zeta^n \widetilde{p_{\zeta}}}{p_{\zeta} -\zeta^n \widetilde{p_{\zeta}}}
=\frac{\bar{\mu} p_{\zeta} +\mu \widetilde{p_{\zeta}}}{\bar{\mu}p_{\zeta} -\mu \widetilde{p_{\zeta}}}.
\]
The point is that we are matching the earlier formulation of the Cayley transform.
However, in this case it is possible for $p_{\zeta}$ and $\widetilde{p_{\zeta}}$
to have common zeros, necessarily on $\T$.  Let $q$ be a greatest common divisor
of $p_{\zeta}$ and $\widetilde{p_{\zeta}}$. We can arrange for $q=\tilde{q}$ by
multiplying by an appropriate constant since
all of the roots of $q$ will be on $\T$.
Let $g=\bar{\mu} p_{\zeta}/q$ which has degree at most $|n|-\deg q$.

Then, $f_{\zeta} = \frac{g+\tilde{g}}{g-\tilde{g}}$ and $H_{\zeta} = \frac{2\bar{\mu}v_{\zeta}/q}{g-\tilde{g}}$.
We can now apply the one variable result
to see that
$\bar{\mu} v_{\zeta}/q$ is a polynomial of degree
$|n|-\deg q$ and is equal to its reflection which means
\[
\bar{\mu} v_{\zeta}(z)/q(z) = \mu z^{|n|} \overline{v_{\zeta}(1/\bar{z})}/q(z)
\]
and this implies the hypotheses of Lemma \ref{slices}, proven below.
This lemma can essentially by found in McDonald \cite{McDonald1} but
we include a proof for convenience.

\begin{proof}[Proof of Lemma \ref{slices}]
We write out the power series
$h(w) = \sum_{\alpha} a_{\alpha} w^{\alpha}$, so that
\[
\begin{aligned}
h_{\zeta}(z) &= \sum_{\alpha} a_{\alpha} z^{|\alpha|} \zeta^{\alpha} \\
&= \sum_{j\geq 0} z^j(\sum_{\alpha: |\alpha|=j}a_{\alpha} \zeta^\alpha)
\end{aligned}
\]
and we see that for $j> |n|$, we have $0\equiv \sum_{|\alpha|=j}a_{\alpha} \zeta^{\alpha}$
and therefore $a_{\alpha} = 0$ for $|\alpha|>|n|$ since this is
an identically zero trigonometric polynomial.

The reflection condition on $h_{\zeta}$ implies for $0\leq j \leq |n|$
\[
\sum_{\alpha: |\alpha|=j}a_{\alpha} \zeta^\alpha =
\sum_{\alpha:|\alpha|=|n|-j} \overline{a_{\alpha}} \zeta^{n-\alpha}.
\]
By matching Fourier coefficients, we see that $a_{\alpha}=0$ if for some $k$ we have $\alpha_k >n_k$.
Thus, $h$ has multidegree at most $n$ and the sum on the right can be reindexed as
\[
\sum_{\alpha: |\alpha| = j} \overline{a_{n-\alpha}} \zeta^{\alpha}
\]
(where $a_{\beta}=0$ if undefined) so that $a_{\alpha} = \overline{a_{n-\alpha}}$.
This exactly means $h=\tilde{h}$ if we examine coefficients.
\end{proof}

This proves McDonald's characterization of faces with the additional information
that $v_{\zeta}$ vanishes to at least the same order at a point of $\T$ as $p_{\zeta}$.
Using work of \cite{GK} it is possible to show
$v$ vanishes to at least the same order
as $p$ at a point of $\T^d$.

\begin{proof}[Proof of Theorem \ref{addon}]

We may assume $\zeta = \mathbf{1}:=(1,\dots,1)$.
Theorem 14.1 of \cite{GK} says that if $p\in \C[z_1,\dots, z_d]$
has no zeros in $\D^d$ and $p$ vanishes at $\mathbf{1}$ to order $M$, meaning
\[
p(\mathbf{1} - \zeta) = \sum_{j=M}^{N} P_M(\zeta)
\]
where the $P_j$ are homogeneous of total degree $j$ and $P_M\ne 0$, then
$P_M$ has no zeros in $RHP^d$.
Then, the order of vanishing of $p_{\mathbf{1}}(z) = p(z\mathbf{1})$ at $z=1$
equals $M$ because $P_M((1-z)\mathbf{1}) = (1-z)^M P_M(\mathbf{1})$ and $P_M(\mathbf{1}) \ne 0$.
(Geometrically, the variety $p=0$ has no radial tangents on $\T^d$.)

Note both $p$ and $p+v$ have no zeros in $\D^d$.
Thus, if $p$ vanishes to order $M$ at $\mathbf{1}$, then $p_{\mathbf{1}}$
vanishes to order $M$ at $1$.   In turn, $v_{\mathbf{1}}$ vanishes to
at least order $M$, and then so does $(p+v)_{\mathbf{1}}$, and we then
conclude
the order of vanishing of $p+v$ at $\mathbf{1}$ is at least $M$.
\end{proof}

If we use an additional result from \cite{GK}
we can say something about the bottom homogeneous term
of $v$ at $\mathbf{1}$.

Proposition 14.5 of \cite{GK} says that if $p \in \C[z_1,\dots,z_d]$
has no zeros in $\D^d$ and vanishes to order $M$ at $\mathbf{1}$
then writing homogeneous expansions:
\[
p(\mathbf{1} - \zeta) = \sum_{j=M}^N P_j(\zeta) \qquad
\tilde{p}(\mathbf{1} - \zeta) = \sum_{j=M}^{N} Q_j(\zeta)
\]
we have that $Q_M$ is a unimodular multiple of $P_M$, say $Q_M = \mu^2 P_M$ for
some $\mu \in \T$.  (It will be convenient to take a square root of $\mu^2$ later.)
Set $F=\tilde{p}/p$.
It is not hard to prove $F(\mathbf{1}-\zeta) \to \mu^2$ as $\zeta \to 0$ non-tangentially in
$RHP^d$.  The details are in Proposition 14.3 of \cite{GK}.

We also need to write out the homogeneous expansion of $h$
\[
v(\mathbf{1}-\zeta) = \sum_{j=M}^N V_j(\zeta).
\]
By the above work, $v$ vanishes at least to order $M$ at $\mathbf{1}$.
It is possible that $v$ vanishes to higher order, so we allow for $V_M=0$.

We assume for simplicity that $p\pm v$ have no zeros in $\D^d$; this
is true after rescaling $v$.  The following inequalities hold
\[
|p\pm v|^2 - |\tilde{p} \pm v|^2 \geq 0 \text{ on } \D^d
\]
(see Lemma 14.4 of \cite{GK}).
We can rewrite this as
\begin{equation} \label{rewrite}
|p(z)|^2-|\tilde{p}(z)|^2 \geq 2|\text{Re}\left((\overline{p(z)}-\overline{\tilde{p}(z)})v(z)\right)|.
\end{equation}
Let $\zeta \in RHP^d$ and set $z=z(t)=\mathbf{1} - t\zeta$.
 For $t>0$ small enough $z(t) \in \D^d$ and since
$P_M$ is a unimodular multiple of $Q_M$, $|p(z(t))|^2-|\tilde{p}(z(t))|^2$ vanishes
to order at least $M(M+1)$ in $t$.  This must also hold
for the right hand side of \eqref{rewrite} which implies
\[
\text{Re}\left( (1-\bar{\mu}^2)\overline{P_M(\zeta)} V_M(\zeta)\right) = 0
\]
which implies $(1-\bar{\mu}^2)\frac{V_M}{P_M} = ic$ for some $c \in \mathbb{R}$.

If $\mu^2\ne 1$, then
the constant $\frac{ic}{1-\bar{\mu}^2}$ is of the form $r\mu$ for
$r\in \mathbb{R}$.  We summarize the above.

\begin{theorem}\label{vhomo}
 Assume the setup of Theorem \ref{McDonald}.
Suppose $p$ vanishes at $\mathbf{1}$ to order $M$.
Let $\mu^2 = F(\mathbf{1})$ in the sense of a
non-tangential limit.  Assume $\mu^2\ne 1$.
Then the
bottom homogeneous terms of $p(\mathbf{1}-\zeta), v(\mathbf{1}-\zeta)$ at
$\zeta=0$, say $P_M, V_M$, satisfy
\[
V_M = r \mu P_M
\]
for some $r \in \mathbb{R}$.
\end{theorem}

This result is strong enough
to prove $g_2$ from the introduction
is extreme.
Note that for $p(z_1,z_2) = 2-z_1-z_2$ we have
\[
p(1-\zeta_1,1-\zeta_2) = \zeta_1+\zeta_2
\]
\[
\tilde{p}(1-\zeta_1,1-\zeta_2) =-\zeta_1-\zeta_2+2\zeta_1\zeta_2
\]
which means $\mu^2=-1$.  Then, any perturbation of $p$ must be
of the form $v(z_1,z_2)=ir z_1-irz_2$ where $r\in \mathbb{R}$ and
\[
v(1-\zeta_1,1-\zeta_2) = ir(\zeta_2-\zeta_1)
\]
which is not a multiple of $\zeta_1+\zeta_2$ unless $r=0$.

The theorem says nothing when $\mu^2= 1$
and this is for good reason.
Consider
\[
q = i(2-z_1-z_2)
\]
For $G = \frac{\tilde{q}}{q}$ we have $\lim_{r\nearrow 1} G(r,r) = 1$ and
$g=(1+G)/(1-G)$ is not extreme:
\[
g = \frac{1-z_1z_2}{(1-z_1)(1-z_2)} = \frac{1}{2}\left(\frac{1+z_1}{1-z_1} + \frac{1+z_2}{1-z_2}\right).
\]

\section{Preliminaries for Theorems \ref{thmsaturated} and \ref{thmmults}}\label{prelims}

McDonald's theorem reduces the possibilities for the face of
a given $f$.  If $f$ is built out of a $\T^2$-saturated polynomial,
we can limit $f$'s face further.

First, let us recall the Schur-Cohn test.

\begin{theorem}[Schur-Cohn test]
Let $p\in \C[z]$ and write $p(z)=\sum_{j=0}^{m}p_j z^j$.  Define
\[
P= \begin{pmatrix} p_0 & p_1 & \dots & p_{m-1} \\
 0 & p_0 & \dots & p_{m-2}\\
 \vdots & 0 & \ddots & \vdots \\
 0 & 0 & 0 & p_{0}
 \end{pmatrix} \quad
 Q = \begin{pmatrix} \overline{p_{m}} & \overline{p_{m-1}} & \dots & \overline{p_1} \\
 0 & \overline{p_m} & \dots & \overline{p_2} \\
 \vdots & 0 & \ddots & \vdots \\
 0 & 0 & 0 & \overline{p_m}
 \end{pmatrix}
\]
Then, $p$ has no zeros in $\cd$ if and only if $P^*P-Q^*Q >0$.
\end{theorem}

Actually this result is due to Schur while Cohn proved a
generalization (see \cite{PtakYoung}).

We can apply this to bivariate polynomials $p\in \C[z,w]$ of
bidegree $(n,m)$ as follows.
Write $p(z,w) = \sum_{j=0}^{m} p_j(z)w^j$.
Then, $\tilde{p}(z,w) = \sum_{j=0}^{m} \tilde{p}_{m-j}(z) w^j$
where $\tilde{p}_j(z) = z^{n} \overline{p_j(1/\bar{z})}$.
Note that we use $(z,w)$ for the coordinates in $\C^2$ instead
of $(z_1,z_2)$.

Define
\begin{equation}\label{PQ}
P(z)= \begin{pmatrix} p_0(z) & p_1(z) & \dots & p_{m-1}(z) \\
 0 & p_0(z) & \dots & p_{m-2}(z)\\
 \vdots & 0 & \ddots & \vdots \\
 0 & 0 & 0 & p_{0}(z)
 \end{pmatrix} \quad
 Q(z) = \begin{pmatrix} \tilde{p}_{m}(z) & \tilde{p}_{m-1}(z) & \dots & \tilde{p}_1(z) \\
 0 & \tilde{p}_m(z) & \dots & \tilde{p}_2(z) \\
 \vdots & 0 & \ddots & \vdots \\
 0 & 0 & 0 & \tilde{p}_m(z)
 \end{pmatrix}.
\end{equation}
Note that for $z\in \T$, $Q(z)$ is almost a direct analogue of $Q$ above;
it differs by a factor of $z^n$.  This gets
cancelled out in the Schur-Cohn matrix calculation.

Indeed, $p$ has no zeros in $\T\times \cd$
if and only if $T_p(z)= P(z)^*P(z)-Q(z)^*Q(z)>0$ for
all $z\in \T$.

The following formula holds for $z\in \T, w,\eta \in \C$
\begin{equation}\label{scformula}
\frac{\overline{p(z,\eta)}p(z,w)-\overline{\tilde{p}(z,\eta)}\tilde{p}(z,w)}{1-\bar{\eta} w} = (1,\bar{\eta},\dots,\bar{\eta}^{m-1})T_p(z) (1,w,\dots, w^{m-1})^t.
\end{equation}

\begin{lemma} \label{nozonface}
If $p$ has no zeros in $\T\times \D$ or no zeros in $\D^2$, then $P(z)^*P(z)-Q(z)^*Q(z)\geq 0$ for all $z\in \T$.
\end{lemma}
\begin{proof}
If $p$ has no zeros in $\T\times \D$,
then the polynomial $p_r(z,w) = p(z,rw)$ has no zeros in $\T\times \cd$ for $r\in (0,1)$.
We can then form analogues of $P,Q$ corresponding to $p_r, \tilde{p_r}$ which we label $P_r, Q_r$.  The dependence on $r$ is evidently continuous.  Then, $P_r^*P_r-Q_r^*Q_r>0$ on $\T$. If we send $r\nearrow 1$, then $P^*P-Q^*Q\geq 0$ by continuity.

If $p$ has no zeros in $\D^2$, then $p(rz,w)$ has no zeros in $\T\times \D$ for $r\in (0,1)$.
We can again form analogues, say $P^r,Q^r$, of $P,Q$ depending on $r$ for which $(P^r)^*P^r-(Q^r)^*Q^r \geq 0$ on $\T$.
Then, send $r\nearrow 1$ to see $P^*P-Q^*Q \geq 0$ on $\T$.
\end{proof}

\begin{lemma} \label{scatnozonface}
If $p$ is scattering stable, then $p$ has no zeros in $(\T\times \D)\cup(\D\times\T)$.
\end{lemma}
\begin{proof}
For fixed $z_0\in \D$, $p(z_0,\cdot)$ has no zeros in $\D$.  By Hurwitz's theorem, if we send $z_0$ to $\T$, we see that $p(z_0,\cdot)$ either has no zeros in $\D$ or is identically zero.  In the latter case, $z-z_0$ divides $p$ and hence also $\tilde{p}$ since $z_0\in \T$.  Thus, $p$ is non-vanishing on $\T\times \D$ and by symmetry on $\D\times \T$.
\end{proof}

\begin{lemma}\label{possc}
Let $p\in \C[z,w]$ and form $P(z),Q(z)$ as above.
Then,
$P(z)^*P(z) - Q(z)^*Q(z) >0$ for all but finitely many $z\in \T$
if and only if
$p$ has no zeros in $(\T\times \cd)\setminus S$ where
$S$ consists of finitely many points in $\T^2$ and
finitely many ``vertical'' lines $z=z_0$ with $z_0\in \T$.

We can rule out vertical lines if we assume $p$ and $\tilde{p}$
have no common factors.
\end{lemma}

We note that the condition $P^*P-Q^*Q>0$ for all but finitely many $z\in \T$
is equivalent to saying $P^*P-Q^*Q\geq 0$ and $\det(P^*P-Q^*Q)$ is not
identically zero.

\begin{proof}
By the Schur-Cohn test, $P(z)^*P(z)-Q(z)^*Q(z)>0$
for all but finitely many $z\in \T$ if and only if
 $w\mapsto p(z,w)$ has no zeros in $\cd$ for all but finitely many $z\in \T$.
The latter condition means there exists a finite set $S_1 \subset \T$
such that $w\mapsto p(z,w)$ has no zeros in $\cd$ except for when $z\in S_1$.
By Hurwitz's theorem, if $z_0\in S_1$ then $w\mapsto p(z_0,w)$
either has no zeros in $\D$ or is identically zero.
In the latter case, $z-z_0$ divides $p$ which means $p$ vanishes
on a vertical line.
In the former case, $p(z_0,\cdot)$ has no zeros in $\D$ (although it could have zeros in $\T$).
 Thus, the condition $P^*P-Q^*Q>0$ for all
but finitely many $z$ is equivalent to $p$ having
 no zeros in $(\T\times \cd) \setminus S$ where $S$
consists of finitely many vertical lines and finitely many points of $\T^2$.
\end{proof}

\begin{lemma} \label{commonfactor}
Assume $p\in\C[z,w]$ has bidegree $(n,m)$.
Let $r(z)$ be the resultant of $p$ and $\tilde{p}$ with respect
to the variable $w$.  Then, on $\T$
\[
r(z) = z^{nm} \det(P(z)^*P(z) - Q(z)^*Q(z))
\]
where $P,Q$ are defined in \eqref{PQ}.
In particular,
the polynomials $p$ and $\tilde{p}$ have a common factor involving $w$
 if and only if $\det(P(z)^*P(z)-Q(z)^*Q(z))=0$ for all $z\in \T$.
\end{lemma}

\begin{proof}
This proof is from \cite{GW} (specifically Lemma 2.1.3 of that paper,
which says their proof is inspired by similar
arguments for Bezoutians in \cite{LT}, Theorem 1 Section 13.3).

Let $\tilde{P}(z) = z^n P(1/\bar{z})^*$ and $\tilde{Q}(z) = z^n Q(1/\bar{z})^*$.  Then,
\[
\begin{aligned}
R(z) &= \begin{pmatrix} P(z) & \tilde{Q}(z) \\ Q(z) & \tilde{P}(z) \end{pmatrix}\\
&= \begin{pmatrix} p_0(z) & \cdots & p_{m-1}(z) & p_{m}(z) & \ & \bigcirc \\
\ & \ddots & \vdots & \vdots & \ddots & \ \\
\bigcirc & \ & p_0(z) & p_1(z) & \cdots & p_m(z) \\
\tilde{p}_{m}(z) & \cdots & \tilde{p}_{1}(z) & \tilde{p}_{0}(z) & \ & \bigcirc \\
\ & \ddots & \vdots & \vdots & \ddots & \ \\
\bigcirc & \ & \tilde{p}_{m}(z) & \tilde{p}_{m-1}(z) & \cdots & \tilde{p}_0(z) \
\end{pmatrix}
\end{aligned}
\]
is the resultant or Sylvester matrix of $p$ and $\tilde{p}$ with respect to $w$
and its determinant is the resultant $r$ of $p$ and $\tilde{p}$ with respect to $w$.
Since $P$ and $Q$ commute, we can compute
\[
r(z) = \det(\tilde{P}(z)P(z)-\tilde{Q}(z)Q(z))
\]
and on $\T$ this agrees with $z^{nm}\det(P(z)^*P(z)-Q(z)^*Q(z))$.
Since $\T$ is a set of uniqueness, this determinant is identically zero on $\T$
if and only if the resultant $r\equiv 0$
which by standard properties of the resultant holds
if and only if
$p$ and $\tilde{p}$ have a common factor depending on $w$.
\end{proof}

We shall begin the proofs of Theorem \ref{thmsaturated} and
Theorem \ref{thmmults} simultaneously
and then diverge at a certain point.

Let $f=\frac{p+\tilde{p}}{p-\tilde{p}}$ be in $PRPR_2$
with associated scattering stable polynomial
$p\in \C[z,w]$.

 By McDonald's characterization,
points of the faces of
 $f$ are associated to
 $v\in \C[z,w]$ satisfying:
$v(0)=0$, $\deg v = \deg \tilde{p}$, $v=\tilde{v}$, and
there is an interval $I=[a,b]$ with $a<0<b$ such that
for $t \in I$,
$p+ t v$ has no zeros in $\D^2$.
For our purposes, we can shrink the interval $I$
to be symmetric about $0$ and rescale $v$ so that
$p\pm v$ have no zeros in $\D^2$ and in fact $p\pm v$
are scattering stable.
This is because the resultants associated to
$p,\tilde{p}$ will not be identically zero
and thus so will those of $p+tv$, $\tilde{p}+tv$
for $|t|$ small.

Write $v = \sum_{j=0}^{m} v_j(z) w^j$ and form
\[
V(z) = \begin{pmatrix} v_0(z) & v_1(z) & \dots & v_{m-1}(z) \\
 0 & v_0(z) & \dots & v_{m-2}(z)\\
 \vdots & 0 & \ddots & \vdots \\
 0 & 0 & 0 & v_{0}(z)
 \end{pmatrix}.
 \]

All of the matrix functions in the rest of this
proof will be functions of $z$, so we will
omit the evaluations such as $V(z)$ and simply
write $V$.
Since $p\pm v$ are scattering stable,
 $(P\pm V)^*(P\pm V)-(Q\pm V)^*(Q\pm V) > 0$ for
all but finitely many $z\in \T$.

By the matrix Fej\'er-Riesz lemma, we can factor
 \begin{equation} \label{factor}
 (P\pm V)^*(P\pm V)-(Q\pm V)^*(Q\pm V) = E_{\pm}^*E_{\pm}
 \end{equation}
 where $E_{+}, E_{-}$ are matrix polynomials of degree at most $n$
with $\det E_{\pm}(z)$ non-zero for $z\in \D$.
 We can also factor
\begin{equation}\label{Epoly}
P^*P-Q^*Q = E^*E
\end{equation}
 where $E$ is a matrix polynomial of degree at most $n$
and is
non-singular whenever $z\in \D$.

Let $r(z)$ be the resultant
of $p$ and $\tilde{p}$ with respect to $w$,
and let $r_{\pm}(z)$ be
the resultants of $p\pm v, \tilde{p}\pm v$
with respect to $w$ (for the choices of $+$ and $-$).
The number of roots of the
resultant $r(z)$ on $\T$
equals the number of common roots of $p$
and $\tilde{p}$ on $\T^2$
since $p$ has no zeros on $\D\times \T$
by Lemma \ref{scatnozonface}.
More precisely, the multiplicity of a given
root of $r(z)$, say $z_0$, counts the number
of common roots of $p$ and $\tilde{p}$ on the line $z=z_0$.
Note that $r(z) = z^{mn} |\det(E)|^2$ and
$r_{\pm}(z) = z^{mn} |\det(E_{\pm})|^2$ on $\T$.

The left side of \eqref{factor} is
 \[
 P^*P-Q^*Q -(\pm 1)2\text{Re}((P-Q)^*V)
 \]
and averaging over $+$ and $-$ yields
\[
P^*P-Q^*Q = E^*E = \frac{1}{2}(E_{+}^*E_{+} + E_{-}^*E_{-})
\]
on $\T$.  Setting $\Phi_{\pm} = E_{\pm} E^{-1}$ we have
$2I = \Phi_{+}^*\Phi_{+} + \Phi_{-}^{*}\Phi_{-}$ on $\T$
away from poles of $E^{-1}$.
But, this equation shows $\Phi_{\pm}$ are bounded
on $\T$ and so cannot have any poles.  So, $\Phi_{\pm}$
are analytic on $\cd$ since we already know
$E^{-1}$ has no poles in $\D$.

We conclude that on $\T$, $r_{\pm}(z) = r(z) |\det(\Phi_{\pm})|^2$.
Let $I_{\{z=z_0\}}(p,q)$ denote the number
of common zeros of two polynomials on the line $z=z_0$ counting
multiplicities.  We have just proved the following.

\begin{prop} \label{proplines}
Suppose $p\in \C[z,w]$ is scattering stable,
and suppose $v\in \C[z,w]$ satisfies $v(0)=0, \deg v= \deg \tilde{p}, v=\tilde{v}$,
as well as $p\pm v$ are scattering stable.
Then, for $z_0\in \T$
\[
I_{\{z=z_0\}}(p,\tilde{p}) \leq I_{\{z=z_0\}} (p\pm v, \tilde{p}\pm v).
\]
Moreover, both intersection multiplicities are even.
\end{prop}

The last statement follows from the formula $r(z) = z^{mn} |\det E |^2$
which implies that zeros on $\T$ occur with even multiplicity.

If the common zeros of $p$ and $\tilde{p}$ on $\T^2$ had distinct
$z$-coordinates (or $w$-coordinates)
then we would be finished with Theorem \ref{thmmults}.

At this stage the reader can jump to the next section to
finish the proof of Theorem \ref{thmsaturated} or Section \ref{secmults}
to finish the proof of Theorem \ref{thmmults}.

\section{Completion of the proof of Theorem \ref{thmsaturated}}

Assume now that $p$ is $\T^2$-saturated and $p-\tilde{p}$
is irreducible.
Then, the resultant of $p$ and $\tilde{p}$, $r(z)$,
has $2nm$ roots on $\T$ and these must be
all of its roots since $r(z)$
has degree at most $2nm$.
This implies $\det(P^*P - Q^*Q) = \bar{z}^{nm}r(z)$ has $2nm$ roots on $\T$
and therefore $\det E$ has $nm$ roots on $\T$.
 These must be all of the roots of $\det E$
since $E$ is $m\times m$ and of degree at most $n$.
  Thus, $E$ is invertible in $\D$ and $\C\setminus \cd$.
Also, $z^nE(1/z)$ is an $m\times m$ matrix polynomial of degree at most $n$,
which implies $\det (z^n E(1/z))$ has degree at most $nm$.
 Since $\det E$ has $nm$ roots on $\T$,
we can conclude that $\det (z^n E(1/z))$ has all of its roots on $\T$.
Thus, $z^nE(1/z)$ is invertible at $0$.

At the same time,
\[
E_{\pm}(1/z)E^{-1}(1/z)=(z^n E_{\pm}(1/z))(z^n E(1/z))^{-1}
\]
is analytic on $\D$ since $z^n E(1/z)$ is invertible in $\D$.  This implies $\Phi_{\pm}$ are analytic on the Riemann sphere, which implies $\Phi_{\pm}$ are constant $m\times m$ matrices.

Subtract equation \eqref{factor} with $+$ from $-$ to obtain
\begin{equation} \label{subtract}
4\text{Re}((P-Q)^*V) = E_{-}^*E_{-}-E_{+}^*E_{+}= E^*(\Phi_{-}^*\Phi_{-}-\Phi_{+}^*\Phi_{+}) E.
\end{equation}
Now we consider the matrices obtained from $p+t v$ where $t$ is a real parameter.  The matrices we get are
\[
(P+t V)^*(P+t V)-(Q+t V)^*(Q+t V)= E^*(I - t\frac{1}{2}\Phi_{-}^*\Phi_{-}+t\frac{1}{2}\Phi_{+}^*\Phi_{+})E.
\]
Unless $\Phi_{-}^*\Phi_{-} = \Phi_{+}^*\Phi_{+}$, there will be a value of $t$ such that
\[
(I - t(\frac{1}{2}\Phi_{-}^*\Phi_{-}-\frac{1}{2}\Phi_{+}^*\Phi_{+}))
\]
is singular since all matrices involved are self-adjoint.
For such $t$, $\det((P+t V)^*(P+t V)-(Q+t V)^*(Q+t V))$ is identically zero on $\T$
and by Lemma \ref{commonfactor}
this implies $p+tv$ and $\tilde{p}+tv$ have a common factor.
Such a common factor must be proper (meaning it has strictly
lower degree in one of the variables).

Indeed, if it is
not proper, then $p+tv$ divides $\tilde{p}+tv$ or vice versa.
The latter possibility is excluded by the fact $p(0)\ne 0=\tilde{p}(0)=v(0)$.
In the former case, $\tilde{p}+tv = (p+tv) g$
for some non-constant polynomial $g$ with $g(0)=0$.
(Note we could not immediately rule this case out based on
degrees because $p+tv$ only has degree at most $(n,m)$.)
Then, $p-\tilde{p} = (p+tv)(1-g)$ is reducible contrary to our assumption.
However, a proper common factor of $p+tv$ and $\tilde{p}+tv$
implies a proper factor of $p-\tilde{p}$, which again contradicts
irreducibility of $p-\tilde{p}$.

We are left with the possibility $\Phi_{-}^*\Phi_{-}=\Phi_{+}^*\Phi_{+}$.
By \eqref{subtract} we have
$\text{Re}((P-Q)^*V) \equiv 0$, which implies
\[
(P+t V)^*(P+t V)-(Q+t V)^*(Q+t V) = P^*P-Q^*Q
\]
 for all $t\in \mathbb{R}$.  By \eqref{scformula}, this implies
for $z\in \T$, $w,\eta\in \C$ that the following expression does
not depend on $t$
\[
\frac{\overline{(p+tv)(z,\eta)}(p+tv)(z,w) - \overline{(\tilde{p}+tv)(z,\eta)}(\tilde{p}+tv)(z,w)}{1-\bar{\eta}w}.
\]
The coefficient of $t$ is therefore identically
zero yielding
 \[
 \overline{(p-\tilde{p})(z,\eta)} v(z,w)+\overline{v(z,\eta)}(p-\tilde{p})(z,w)=0
 \]
 for $w,\eta\in \C, z\in \T$.
Since $z\in \T$ and $v=\tilde{v}$,
if we replace $\eta$ with $1/\bar{\eta}$ and multiply through by $z^n\eta^m$ we get
 \[
 (\tilde{p}-p)(z,\eta) v(z,w)+ v(z,\eta)(p-\tilde{p})(z,w)=0
 \]
 for $w,\eta \in \C$.
As this is a polynomial in $z$ the identity holds for all $z\in \C$, not just $\T$.
 Since $p-\tilde{p}$ is assumed to be irreducible, either:
\begin{enumerate}
 \item $(p-\tilde{p})(z,w)$ divides
 $(\tilde{p}-p)(z,\eta)$ or
\item  $(p-\tilde{p})(z,w)$ divides $v(z,w)$.
\end{enumerate}

In addressing these two cases it is useful to point out that $p-\tilde{p}$ has
bidegree exactly $(n,m)$.  This is
because $p(0)\ne 0$ and the reflection of $p-\tilde{p}$ at the degree $(n,m)$
is $\tilde{p}-p$ which implies the coefficient of $z^nw^m$ in $(p-\tilde{p})(z,w)$ is
$-\overline{p(0)}$.

Case (1) implies $(p-\tilde{p})(z,w)$ does not
depend on $w$ which is only possible if $m=0$.
But,  $p-\tilde{p}$ being irreducible means $n=0$ or $n=1$ by
the fundamental theorem of algebra.  Having $n=0$ is not allowed and
when $n=1$ we automatically have $v=0$.

Case (2) implies $v$ is a multiple of $p-\tilde{p}$ since $v$ has degree
at most $(n,m)$ and $p-\tilde{p}$ has degree exactly $(n,m)$. Since $p(0)\ne 0$
this is impossible unless $v=0$.

In all cases $v=0$.  This
means the face of $f$ is the singleton $\{f\}$ and
proves $f$ is an extreme point.

\section{Completion of the proof of Theorem \ref{thmmults}}\label{secmults}
We pick up where we left off in Section \ref{prelims}.
Proposition \ref{proplines} shows
the number of common zeros of $p$ and $\tilde{p}$
on $\T^2$
is less than or equal to the number of
common zeros of $p\pm v$ and $\tilde{p}\pm v$ on $\T^2$.
However, since the resultant is a global object
we have only proven that the number of zeros on ``vertical'' lines
in $\T^2$ increases.
To fix this we cannot do a linear change of variables
as is often done when using the resultant to count
common zeros (as in \cite{Fischer} for instance).
Instead we perform a more complicated change of variables
and use a different (but equivalent) definition of
intersection multiplicity.

First, recall that the intersection
multiplicity of $p,q$ at a common zero $x\in \C^2$
is equal to
\[
\dim \mathcal{O}_x/(p,q)\mathcal{O}_x.
\]
Here $\mathcal{O}_x$ is the localization
of $\C[z,w]$ at $x$, namely the ring of rational functions
with denominators non-vanishing at $x$.
Also, $(p,q)\mathcal{O}_x$ denotes the ideal generated by
$p,q$ within the local ring $\mathcal{O}_x$.
This definition of intersection multiplicity is used
in \cite{Fulton}.

Now, suppose $p, \tilde{p}$ have a common zero on $\T^2$
which we assume without loss of generality is $\mathbf{1}=(1,1)$.
The map
\[
\phi_k: F(z,w) \mapsto F(w^k z,w)
\]
is an isomorphism of the local ring $\mathcal{O}_{\mathbf{1}}$
onto itself
since it has inverse $G(z,w)\mapsto G(w^{-k} z, w)$.
This implies that
\begin{equation} \label{phik}
I_{\mathbf{1}}(p,\tilde{p}) = I_{\mathbf{1}}(\phi_k(p), \phi_k(\tilde{p})).
\end{equation}
This applies to $p\pm v$ as well.

Since $p$ is scattering stable, so is $\phi_k(p)$.  Indeed,
$\phi_k(p)$ has no zeros in $\D^2$ since $(z,w) \mapsto (w^k z, w)$
maps the bidisk into itself.
Also,
\[
\widetilde{\phi_k(p)} = z^{n} w^{kn+m} \overline{p(1/\bar{w}^k1/\bar{z}, 1/\bar{w})} = \phi_k(\tilde{p})
\]
If $\phi_k(p)$ and $\phi_k(\tilde{p})$ had
a nontrivial common factor say $\phi_k(p)=g h_1$, $\phi_k(\tilde{p}) = g h_2$ then
\[
p(z,w) = g(w^{-k} z,w) h_1(w^{-k} z,w)
\qquad
\tilde{p}(z,w)= g(w^{-k} z, w) h_2(w^{-k} z,w).
\]
One can then argue that $g(w^{-k} z,w)$ is either already a polynomial
or there exists $N$ such that $w^N g(w^{-k} z,w)$, $w^{-N}h_1(w^{-k} z,w)$, $w^{-N}h_2(w^{-k} z,w)$
are polynomials.  Either case implies $p$ and $\tilde{p}$ have a non-trivial common
factor contrary to assumption.

The same arguments apply to $p\pm v$.  Therefore,
by Proposition \ref{proplines}
\[
I_{\{z=1\}}(\phi_k(p), \phi_k(\tilde{p})) \leq I_{\{z=1\}}(\phi_k(p)\pm \phi_k(v), \phi_k(\tilde{p})\pm \phi_k(v)).
\]
Finally, we can choose $k$ so that $\phi_k(p), \phi_k(\tilde{p})$
and $\phi_k(p)\pm \phi_k(v), \phi_k(\tilde{p})\pm \phi_k(v)$ have a single
common zero on the line $\{z=1\}$ namely $\mathbf{1}=(1,1)$.
Then, the above intersection multiplicities
count the multiplicity at just $\mathbf{1}$ and these agree
with the multiplicities associated to $p,\tilde{p}$ and $p\pm v, \tilde{p}\pm v$
as mentioned in equation \eqref{phik}.

This proves
\[
I_{\mathbf{1}}(p,\tilde{p}) \leq I_{\mathbf{1}} (p\pm v, \tilde{p} \pm v)
\]
which proves Theorem \ref{thmmults}.

A corollary of the proof is the following fundamental fact.

\begin{corollary}
If $p\in \C[z,w]$ is scattering stable then for $\zeta \in \T^2$,
$I_{\zeta}(p,\tilde{p})$
is even.
\end{corollary}

This follows because the intersection multiplicities in Proposition \ref{proplines}
are even.  We gave a detailed proof of this corollary in Appendix C of \cite{GK}.

\section{Degree $(1,1)$ polynomials} \label{modest}

\begin{proof}[Proof of Theorem \ref{oneone}]
Suppose $p$ is scattering stable, has degree $(1,1)$,
and $f=(p+\tilde{p})/(p-\tilde{p})$ is
an extreme point.

We can normalize so that $|p(0)|=1$.
(We cannot normalize so that $p(0)=1$
because $p/p(0)$ has reflection $\tilde{p}/\overline{p(0)}$
not $\tilde{p}/p(0)$.)

Since $\tilde{p}(0)=0$,
$p$ has the form $p(z,w) = \mu+az+bw$; $\mu\in \T, a,b,\in \C$.
We can
apply a rotation in each of the variables and
replace $p$ with the polynomial $\mu(1-|a|z-|b|w)$;
we will drop the
absolute values on $a,b$ and assume $a,b >0$.
This change of variables does not change extremality.

For $p$ to be stable we must have $1\geq a+b$.
Since $f$ is extreme, $p$ must have a zero on $\T^2$.
This forces $1=a+b$ and the zero is $(z,w)=(1,1)$.
Zeros on $\T^2$ occur with even multiplicity
and this automatically implies $p$ is $\T^2$-saturated.
(Thus, in this low degree situation a single zero forces
the polynomial to be saturated.)

We need to show $p-\tilde{p}$ is irreducible.  We have
\[
p-\tilde{p} = \mu+(\bar{\mu} b-\mu a)z+(\bar{\mu}a - \mu b)w - \bar{\mu}zw.
\]
If this were reducible it would factor into
$\mu(1-cz)(1-dw)$.  We will omit the
details but the only way this can happen is if $c=d=1$ and
$\mu = \pm i$.  We can normalize so $\mu = i$.  Then,
\[
p-\tilde{p} = i(1 - (z+w) +zw)
\]
while
\[
p+\tilde{p} = i(1+t(z-w) - zw)
\]
where $t=b-a$.
Then, one can directly check
\[
f = b \frac{1+z}{1-z} +a \frac{1+w}{1-w}
\]
contradicting extremality of $f$.
\end{proof}

The next theorem is an amusing consequence of the
above proof.

\begin{theorem}\label{thm:mobius}
Let $p\in \C[z_1,z_2]$ be scattering stable, $\tilde{p}(0,0)=0$, and $\deg p = (1,1)$.
Set $F = \tilde{p}/p$ and for $\nu \in \T$,
\[
f_{\nu} = \frac{1+\nu^2 F}{1-\nu^2 F} = \frac{\bar{\nu} p + \nu \tilde{p}}{\bar{\nu} p - \nu \tilde{p}}.
\]
Then, there exists $\nu \in \T$ such that $f_{\nu}$ is not an extreme point of $PRP_2$.
\end{theorem}

I. Klep and J. Pascoe have found an example
which shows this theorem does not hold more generally.

\begin{proof}
Let $p\in \C[z,w]$ be scattering stable, $\deg p = (1,1)$, and $\tilde{p}(0,0)=0$.
We claim some $f_{\nu}$ as in the theorem statement is not extreme.
By Theorem \ref{oneone} if $p$ is not $\T^2$-saturated, then $f_{\mu}$ is
never extreme.  So, we assume $p$ is $\T^2$-saturated.

If we follow the proof of Theorem \ref{oneone} above, we
see that if we set $\nu = i\mu$ then $\bar{\nu} p - \nu \tilde{p} = i(1-z)(1-w)$
which means $f_{\nu}$ is not extreme.
\end{proof}

\section{Example of a non-saturated polynomial}\label{nonsaturated}

Let $p(z,w) = 3-z-z^2-w$ which is scattering stable, has degree $(2,1)$,
and has a single zero on $\T^2$ at $(z,w)=(1,1)$.
The multiplicity of this zero is $2$, so $p$ is not saturated (which
would require $4$ boundary zeros counting multiplicity).
One can compute that this multiplicity is $2$ using the resultant (with
respect to $w$) of
$p$ and $\tilde{p} = -z^2 -w - zw +3z^2w$ which is
\[
\det\begin{pmatrix} 3-z-z^2 & -1 \\ -z^2 & -1-z+3z^2 \end{pmatrix} = -(z-1)^2(3z^2+8z+3).
\]

Next, we claim that $f=\frac{p+\tilde{p}}{p-\tilde{p}}$ is not extreme.
To do this we must perturb $p$ with $v$ satisfying $v=\tilde{v}$, $v(0,0)=0$,
and stay stable.  Let $v(z,w) = (1-z)(z-w)$.  We claim $p+tv$ has no
zeros in $\D^2$ for $t\in [-7/4,1/2]$.  By the Schur-Cohn test
we check when the following expression is non-negative
on $\T$
\[
|3-z-z^2+t(1-z)z|^2 - |-z^2 + t(1-z)z|^2.
\]
This can be rewritten as
\[
|1-z|^2(2t+8 + 6(t+1)\text{Re} z)
\]
which is non-negative on $\T$ for $t\in [-7/4,1/2]$.
Thus, for such $t$, $p+tv$ has no zeros in $\T\times \cd$
except for finitely many points on $\T^2$ and finitely many
vertical lines by Lemma \ref{possc}.
We can rule out vertical lines since
\[
p-\tilde{p} = 3-z + (3z-z^2)w
\]
is irreducible (and if $p+tv$ had a factor $z-z_0$ with $z_0\in \T$
then so would $\tilde{p}+tv$ and so would $p-\tilde{p}$).
By Lemma \ref{scatnozonface} $p+tv$ has no zeros in $\T\times \D$.
For $w\in \D$, the number of zeros
of $z\mapsto (p+tv)(z,w)$ in $\D$ depends continuously
on $w$ by the argument principle and hence must be constant.
We can show $(p+tv)(z,0) = 3+(1-t)z-(1+t)z^2$ has no zeros
in $\D$ directly by noting $|t-1|+|t+1|\leq 3$ for $t\in [-7/4,1/2]$.
Therefore, $p+tv$ has no zeros in $\D^2$ for $t\in [-7/4,1/2]$.
This proves $f$ is not extreme.

The endpoints of the line segment
$q_{-}=p-7v/4$ and $q_+=p+v/2$ are
\[
q_{-} = 3-(11/4)z+(3/4)w -(3/4)z^2 -(7/4)zw
\]
\[
q_+ = 3-(1/2)z - (3/2)w-(3/2)z^2+(1/2)zw.
\]
Now, $q_{-}$ has two zeros on $\T^2$; $(1,1),(-1,1)$.  Each
has multiplicity 2 which implies $q_{-}$ is $\T^2$-saturated.
Since $p-\tilde{p}= q_{-}-\tilde{q_{-}}$ is irreducible we conclude
that $f_{-} = \frac{q_{-} + \tilde{q}_{-}}{q_{-} - \tilde{q}_-}$
is an extreme point by Theorem \ref{thmsaturated}.

Next, $q_{+}$ has a single zero on $\T^2$; $(1,1)$.  It occurs
with multiplicity $4$ which implies $q_{+}$ is $\T^2$-saturated.
As with $q_{-}$, we see that the analogue of $f_{-}$, namely $f_+$,
is also an extreme point.

Then,
\[
f = (2/9) f_{-} + (7/9) f_{+}
\]
expresses $f$ as a convex combination of extreme points.

\section{Integrability of perturbations}

In this section we prove Theorem \ref{intthm}.
We need to use several definitions and results from \cite{GK}.

\begin{definition}
If $p\in \C[z,w]$ is scattering stable and $\vec{A}_1\in\C^{N}[z,w],\vec{A}_2\in \C^M[z,w]$
are vector polynomials satisfying
\[
|p|^2-|\tilde{p}|^2 = (1-|z|^2)|\vec{A}_1|^2 + (1-|w|^2)|\vec{A}_2|^2
\]
then we call $(\vec{A}_1,\vec{A}_2)$ an \emph{Agler pair}.
\end{definition}

For this definition, $N,M$ need not have any relation
to the bidegree $(n,m)$ of $p$.  One can in fact show $N\geq n, M\geq m$,
but this is not important for now.  Later we do look at Agler pairs with minimal
dimensions.

It is proven in \cite{CW, GKpnoz} that $p$ has at least
one Agler pair. (These vector polynomials are
closely related to so-called Agler kernels for
more general bounded analytic functions.  See \cite{BK}.)

Lemma 7.3 of \cite{GK} proves that the values of $(|\vec{A}_1|, |\vec{A}_2|)$
on $\T^2$ are the same for all Agler pairs.
Therefore, the following proposition does
not depend on the particular Agler pair.

\begin{prop}\label{lprop}
Suppose $p\in \C[z,w]$ is scattering stable
and $(\vec{A}_1,\vec{A}_2)$ is an Agler pair for $p$.

Given $q \in \C[z,w]$, $q/p \in L^2(\T^2)$
if and only if there is a constant $C$ such that
\[
|q| \leq C(|\vec{A}_1| + |\vec{A}_2|)
\]
holds on $\T^2$.
\end{prop}
\begin{proof}
Lemma 7.3 of \cite{GK} proves explicit formulas
for $|\vec{A}_1|, |\vec{A}_2|$ on $\T^2$ in terms of $p$.
Corollary 7.4 of \cite{GK} proves that the given inequality
is equivalent to $q/p\in L^2(\T^2)$ however it is stated using
said explicit formulas from Lemma 7.3 (so that the corollary
does not need to reference Agler pairs).
\end{proof}

The first part of Theorem \ref{intthm} is contained in the following
proposition.

\begin{prop}\label{propvp}
Suppose $p$ is scattering stable, $v=\tilde{v}$, and $p\pm v$ are scattering stable.
Let $q \in \C[z,w]$. If $\frac{q}{p+ v}\in L^2(\T^2)$ then $\frac{q}{p} \in L^2(\T^2)$.
In particular, $v/p \in L^2(\T^2)$.
\end{prop}

\begin{proof}
We can rescale $v$ so that $p\pm v$ are scattering stable.
Let $(E_{\pm}, F_{\pm})$ be Agler pairs for $p\pm v$.
Since
\[
\begin{aligned}
|p\pm v|^2 - |\tilde{p} \pm v|^2 &= |p|^2-|\tilde{p}|^2 \pm 2\text{Re}(\overline{p-\tilde{p}}v)\\
&=(1-|z|^2)|E_{\pm}|^2 + (1-|w|^2)|F_{\pm}|^2
\end{aligned}
\]
it follows that for $E= \frac{1}{\sqrt{2}} \begin{pmatrix}  E_{+}\\ E_{-}\end{pmatrix}$
and $F=\begin{pmatrix} F_{+} \\ F_{-} \end{pmatrix}$, the pair $(E,F)$ is an Agler pair for $p$.
This comes from averaging the above formula over $+$ and $-$.

Note that $|E| \geq \frac{1}{\sqrt{2}} |E_{\pm}|$, $|F| \geq \frac{1}{\sqrt{2}}|F_{\pm}|$
on $\T^2$.  These bounds along with Prop \ref{lprop} automatically
give that $q/(p+v) \in L^2(\T^2)$ implies $q/p \in L^2(\T^2)$ for $q\in \C[z,w]$.

Since $1 = (p+v)/(p+v) \in L^2(\T^2)$ we conclude that $(p+v)/p \in L^2$ and hence
$v/p\in L^2$.
\end{proof}

For any scattering stable polynomial $p\in \C[z,w]$ with
degree at most $(n,m)$ define the subspace
\[
K_p = \{q\in\C[z,w]: \deg q\leq (n-1,m-1) \text{ and } q/p\in L^2(\T^2)\}.
\]
Theorem B of \cite{GK} states that
\[
\dim K_p = nm- \frac{1}{2} I_{\T^2}(p,\tilde{p})
\]
where $I_{\T^2}(p,\tilde{p})$ is the number
of common zeros of $p$ and $\tilde{p}$ on $\T^2$ counted with
multiplicities.  We can then conclude the following.

\begin{corollary}
Under the assumptions of Proposition \ref{propvp} we have
\[
\dim K_{p} \geq \dim K_{p+v}.
\]
\end{corollary}

This also follows from Theorem \ref{thmmults}.
As mentioned in the introduction $p$ is
$\T^2$-saturated if and only if $K_p = \{0\}$ (see Corollary 6.5 of \cite{GK}).
We use this
fact in the proof of the second half of Theorem \ref{intthm}
which is stated in the following proposition.

\begin{prop}
Suppose $p \in \C[z,w]$ is scattering stable and
not $\T^2$-saturated.  Then, there exists nonzero $v\in \C[z,w]$
with $\deg v \leq (n,m), v=\tilde{v}, v(0,0)=0$ and $v/p \in L^2(\T^2)$.
\end{prop}

\begin{proof}
Since $p$ is not $\T^2$-saturated, there
exists $f \in K_p$, $f\ne 0$.  We can
assume $z$ does not divide $f$ because otherwise $f/z \in K_p$
and we could divide out all of the factors of $z$ if necessary.

Let $\tilde{f}$ be the reflection of $f$ at degree $(n-1,m-1)$.
Set $v = w f + z \tilde{f}$.  Note $v$ is nonzero because $z$ does not
divide $f$.  Since $|f|=|\tilde{f}|$ on $\T^2$, we have $\tilde{f}/p \in L^2(\T^2)$.
Therefore, $v/p \in L^2(\T^2)$.  The other properties of $v$ are straightforward.
\end{proof}

\section{Determinantal representations and transfer function realizations}\label{secdets}
In this section we discuss a number of formulas which
are known to hold for scattering stable $p$ and corresponding $F=\tilde{p}/p$
in the case of two variables.  When $p$ is saturated it turns out
the formulas have a special form.  The techniques of this section are well-known
and this section is more of a supplement to the paper.  See \cite{Drexel, GKdv}
for related results.

It is a well-established consequence
of the sums of squares formula\eqref{sos} that if $p \in \C[z_1,z_2]$ is
scattering stable of bidegree $n=(n_1,n_2)$ then
there exists a $(1+|n|)\times (1+|n|)$ unitary $U = \begin{pmatrix} A & B \\ C& D\end{pmatrix}$
where $A$ is $1\times 1$ such that
\begin{equation} \label{tfr}
F(z) = A+ B P(z) (1-D P(z))^{-1} C.
\end{equation}
Here
\begin{equation} \label{Pmatrix}
P(z) = \begin{pmatrix} z_1 I_{n_1} & 0 \\ 0 & z_2 I_{n_2} \end{pmatrix}.
\end{equation}

We call the formula \eqref{tfr} a \emph{unitary transfer function realization} of $F$.

Since $F$ has denominator $p$, we see that $p$ divides $\det(I-D P(z))$.
But, $\det(I-D P(z))$ has bidegree at most $n$
we must have $p(z) = p(0) \det(I-D P(z))$.
Thus, $p$ has a \emph{contractive determinantal representation};
so-called because $D$ is a contractive matrix.
The matrix $D$ is not just contractive;
it is also a rank one perturbation of a unitary matrix.
Indeed, the matrices
\[
V_{\alpha} = D+\frac{\alpha}{1-\alpha A} CB
\]
are unitary for $\alpha \in \T$.
This follows from
\[
U \begin{pmatrix} \frac{\alpha}{1-\alpha A} B \\ I_{|n|} \end{pmatrix} =
 \begin{pmatrix} \frac{1}{1-\alpha A} B \\ V_{\alpha} \end{pmatrix}
\]
which implies
\[
I-V_{\alpha}^*V_{\alpha} = \frac{1-|\alpha|^2}{|1-\alpha A|^2}B^*B
\]
and this is zero when $|\alpha|=1$.

Notice now that the numerator of $F(z)$ is
\[
\tilde{p}(z) = p(0)\det(I-DP(z))F(z) =p(0)( A\det(I-DP(z)) +BP(z) \text{adj}(1-DP(z)) C)
\]
where $\text{adj}$ denotes the classical adjoint or adjugate matrix.
Observe that
\[
\begin{aligned}
p-\alpha \tilde{p} &= p(0)((1-\alpha A)\det(I-DP(z)) - \alpha BP(z) \text{adj}(I-DP(z))C)\\
&= p(0)(1-\alpha A)\det( I-DP(z) -\frac{\alpha}{1-\alpha A} CB P(z))\\
&= p(0)(1-\alpha A) \det( I - V_{\alpha} P(z)).
\end{aligned}
\]
The second line used the ``rank one matrix update formula''
\[
\det(M + xy^*) = \det M + y^*\text{adj}(M) x.
\]
We conclude that $p-\alpha \tilde{p}$ has a \emph{unitary determinantal representation}
whenever $\alpha \in \T$.  This gives a general way to describe the denominators
of elements of $PRPR_2$.

The matrix $U$ can be chosen with special structure when $p$ is saturated.

\begin{theorem}
If $p\in \C[z_1,z_2]$ is $\T^2$-saturated, then $F=\frac{\tilde{p}}{p}$ possesses
a symmetric unitary transfer function realization; i.e.
$U$ in \eqref{tfr} can be chosen to be a unitary and symmetric: $U^*=U^{-1}$ and $U=U^t$.
In particular, $p$ has a determinantal representation
using a symmetric contraction (which is a rank one perturbation of a symmetric unitary)
and $p-\tilde{p}$ has a symmetric unitary determinantal representation.
\end{theorem}

This theorem can be subdivided
into two parts.

\begin{prop}
If $p\in \C[z_1,z_2]$ is $\T^2$-saturated with $\deg p = (n_1,n_2)$
then
there exist $E_1 \in \C^{n_1}[z_1,z_2], E_2 \in \C^{n_2}[z_1,z_2]$
of bidegrees at most $(n_1-1,n_2), (n_1,n_2-1)$
such that
\begin{equation} \label{sos2}
|p(z)|^2 -|\tilde{p}(z)|^2 = (1-|z_1|^2)|E_1(z)|^2 + (1-|z_2|^2)|E_2(z)|^2
\end{equation}
and
\begin{equation} \label{symmetry}
E_1(z) = \underset{:=\tilde{E}_1}{\underbrace{z_1^{n_1-1} z_2^{n_2} \overline{E_1(1/\bar{z}_1, 1/\bar{z}_2)}}}
\qquad E_2(z) = \underset{:=\tilde{E}_2}{\underbrace{z_1^{n_1} z_2^{n_2-1} \overline{E_2(1/\bar{z}_1,1/\bar{z}_2)}}}.
\end{equation}
\end{prop}

This follows directly from
Theorem 1.1.5 of \cite{GKpnoz} and Corollary 13.6 of \cite{GK}.
The condition in \eqref{symmetry} can be replaced with
the weaker conditions $|E_1|=|\tilde{E}_1|$, $|E_2| = |\tilde{E}_2|$.
(An argument is given in
the proof of Theorem 1.1.5 of \cite{GKpnoz}.
It uses the fact that any symmetric unitary $U$ can be
factored as $U=V^tV$ where $V$ is unitary.)

\begin{prop} \label{symmetryprop}
If $p$ is scattering stable and possesses a sums of squares decomposition
as in \eqref{sos2} which in addition satisfies
\eqref{symmetry}, then $F=\frac{\tilde{p}}{p}$ satisfies \eqref{tfr}
where $U$ may be chosen to be symmetric.
\end{prop}

Symmetry in the
sums of squares formulas does not
characterize saturated polynomials.

\begin{example}
Let $p = (3-z_1-z_2)^2$---a non-saturated polynomial.  Then, $\tilde{p} = (3z_1 z_2-z_1-z_2)^2$
and \eqref{sos2} holds with
\[
E_1 = \begin{pmatrix} q(z_2)p(z) \\ \tilde{q}(z_2) \tilde{p}(z) \end{pmatrix}
\qquad E_2 = \begin{pmatrix}  \tilde{q}(z_1) p(z) \\ q(z_1) \tilde{p}(z) \end{pmatrix}
\]
where $q(\zeta) = \sqrt{3}\left( \frac{1-\sqrt{5}}{2} + \frac{1+\sqrt{5}}{2} \zeta\right)$
and $\tilde{q}$ is the reflection at degree $1$ of this one variable polynomial.
Evidently, $|E_j|=|\tilde{E}_j|$.
\end{example}

It may be possible to use
the characterization of all sums of squares
decompositions \eqref{sos2} given in \cite{GK}
to characterize which $p$ have decompositions satisfying
\eqref{symmetry}.

We now prove Proposition \ref{symmetryprop}.
Unfortunately, the only way we see to
give a comprehensible proof is to rehash
a number of standard arguments (including
the proof of \eqref{tfr}).

\begin{proof}[Proof of Proposition \ref{symmetryprop}]
Initially we do not need \eqref{symmetry}.

Equation \eqref{sos2} can be polarized to the following
form
\begin{equation} \label{polarized}
\overline{p(w)}p(z) - \overline{\tilde{p}(w)}\tilde{p}(z) =
(1-\bar{w}_1 z_1) E_1(w)^* E_1(z) + (1-\bar{w}_2 z_2) E_2(w)^* E_2(z)
\end{equation}
by the polarization theorem for holomorphic functions.
We now use what is called a lurking isometry argument.
Let $N=1+n_1+n_2$.
Define
\[
X(z):= \begin{pmatrix} p(z) \\ z_1 E_1(z) \\ z_2 E_2(z) \end{pmatrix} \qquad
Y(z) := \begin{pmatrix} \tilde{p}(z) \\ E_1(z) \\ E_2(z) \end{pmatrix}.
\]
Then, \eqref{polarized} rearranges into $X(w)^*X(z) = Y(w)^*Y(z)$.
The map
\begin{equation} \label{mapsto}
X(z) \mapsto
Y(z)
\end{equation}
extends to a well-defined linear isometry from the span of the elements on
the left to the span of the elements on the right (as $z$ varies over $\C^2$).
It turns out that $\text{span}\{X(z): z\in \C^2\} = \C^{N}$ and this
implies the map above actually extends uniquely to a unitary.
This follows from the construction of $E_1,E_2$ in \cite{GKpnoz}.
However, as this fact is difficult to retrieve from \cite{GKpnoz} without
introducing too much machinery we proceed without it.
The map \eqref{mapsto} can always be extended to some unitary; we would like
to show that if $E_1,E_2$ satisfy \eqref{symmetry} then we can extend to
a symmetric unitary.

So, suppose \eqref{symmetry} holds and
notice that $Y = \tilde{X}$, with the reflection performed at the
degree $(n_1,n_2)$.
We introduce the orthogonal
complements of the spans of the left and right sides in \eqref{mapsto}
\[
\begin{aligned}
S_1 &=\text{span}\{X(z):z\in \C^2\}^{\perp} \\%= \{v \in \C^N: X(z)^* v = 0 \text{ for all } z \in \C^2\}\\
S_2 &=\text{span}\{\tilde{X}(z): z \in \C^2\}^{\perp}
\end{aligned}
\]
Notice that $v \in S_1$ iff $\overline{v} \in S_2$ because we can reflect
the equation $X^* v= 0$ to obtain $\tilde{X}^* \overline{v} = 0$.
Let $v_1,\dots, v_k$ be an orthonormal basis for $S_1$.  Then,
$\bar{v}_1,\dots, \bar{v}_k$ is an orthonormal basis for $S_2$.
We can then extend the map \eqref{mapsto} to all of $\C^N$
by mapping $v_j \mapsto \bar{v}_j$
and extending linearly.  Note that we are not saying $v\mapsto \bar{v}$ for
a general $v\in S_1$.  We get an $N\times N$ unitary matrix $U$ with the property
\[
U X(z) = \tilde{X}(z) \text{ and } U v_j = \bar{v}_j.
\]
Applying the reflection operation at the degree $(n_1,n_2)$ and conjugation reveals
\[
U^t X(z) = \tilde{X}(z) \text{ and } U^t v_j = \bar{v}_j.
\]
Therefore, $U=U^t$.

Write $U = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ where $A$ is $1\times 1$;
the sizes of $B,C,D$ are then determined.
Letting $E = \begin{pmatrix} E_1 \\ E_2 \end{pmatrix}$
we have
\[
\begin{aligned}
A p(z) + B P(z) E(z) &= \tilde{p} \\
C p(z) + D P(z) E(z) &= E(z)
\end{aligned}
\]
where $P(z)$ is given in \eqref{Pmatrix}.
The second line implies $p(z)(I-D P(z))^{-1}C = E(z)$ which implies
$A+BP(z)(I-DP(z))^{-1}C = \tilde{p}/p$ via the first line.
\end{proof}

\section{Acknowledgments}
I would like to thank Mike Jury, John McCarthy, and James Pascoe for useful
discussions.  Thanks to Pascoe and Klep for disproving
a conjectured generalization of Theorem \ref{thm:mobius}.

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\end{document}

redpajama_set_name: RedPajamaArXiv