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%%%%% Auteur
%%%1
\author{\firstname{Kazumasa} \lastname{Nomura} }
\address{Tokyo Medical and Dental University\\
Kohnodai, Ichikawa\\
272-0827, Japan}
\email{knomra@pop11.odn.ne.jp}

%%%2
\author{\firstname{Paul} \lastname{Terwilliger}}
\address{University of Wisconsin\\
Dept. of mathematics\\
480 Lincoln Drive\\
Madison, WI 53706 USA}
\email{terwilli@math.wisc.edu}


%%%%% Sujet

\keywords{idempotent system, association scheme, Leonard pair}

\subjclass{17B37, 15A21}

%%%%% Gestion

\DOI{10.5802/alco.159}
\datereceived{2020-05-12}
\daterevised{2020-10-19}
\dateaccepted{2020-10-19}


%%%%% Titre et résumé
\title[Idempotent systems]{Idempotent systems}

\begin{abstract}
In this paper we introduce the notion of an idempotent system.
This linear algebraic object is motivated by the structure of an association scheme.
We focus on a family of idempotent systems, said to be symmetric.
A symmetric idempotent system is an abstraction of the
primary module for the subconstituent algebra of a symmetric association scheme.
We describe the symmetric idempotent systems in detail.
We also consider a class of symmetric idempotent systems,
said to be $P$-polynomial and $Q$-polynomial.
In the topic of orthogonal polynomials there is an object called a Leonard system.
We show that a Leonard system is essentially the same thing as a symmetric
idempotent system that is $P$-polynomial and $Q$-polynomial.
\end{abstract}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}

\maketitle

\section{Introduction}
\label{sec:intro}

In this paper we introduce the notion of an idempotent system.
This linear algebraic object is motivated by the structure of an
association scheme.
We focus on a family of idempotent systems, said to be symmetric. 
As we will see, a symmetric idempotent system is an abstraction of the
primary module for the subconstituent algebra of a symmetric association scheme. 
Before we go into more detail, we
recall the notion of a symmetric association scheme.
A symmetric association scheme is a sequence $(X, \{R_i\}_{i=0}^d)$,
where $X$ is a finite nonempty set,
and $\{R_i\}_{i=0}^d$ is a sequence of nonempty subsets of $X \times X$ 
such that
\begin{enumerate}[label=(\roman*)]
\item\label{intro_i} %[(i)]
$X \times X = R_0 \cup R_1 \cup \cdots \cup R_d \quad \text{(disjoint union)}$;
\item\label{intro_ii} %[(ii)]
$R_0 = \{(x,x) \,|\, x \in X\}$;
\item\label{intro_iii} %[(iii)]
$(x,y) \in R_i$ implies $(y,x) \in R_i$; 
\item\label{intro_iv} %[(iv)]
there exist integers $p^h_{i j}$ $(0 \leq h,i,j \leq d)$ such that
for any $(x,y) \in R_h$ the number of $z \in X$ with $(x,z)\in R_i$ and $(z,y) \in R_j$
is equal to $p^h_{i j}$.
\end{enumerate}
The integers $p^h_{i j}$ are called the intersection numbers.
By~\ref{intro_iii} they satisfy $p^h_{i j} = p^h_{j i}$ for $0 \leq h,i,j \leq d$.
The concept of a symmetric association scheme first arose in design theory 
\cite{BN, BS, BM, RC} and group theory~\cite{Wi}.
A systematic study began with~\cite{Del, Hig}.
A comprehensive treatment is given in~\cite{BI, BCN}.

\looseness-1
Let $(X, \{R_i\}_{i=0}^d)$ denote a symmetric association scheme.
As we study this object, the following concepts and notation will be useful.
Let $\R$ denote the real number field.
Let $\Mat_X(\R)$ denote the $\R$-algebra consisting of the matrices with
rows and columns indexed by $X$, and all entries in $\R$.
Let $I$ (resp.\ $J$) denote the identity matrix (resp.\ all $1$'s matrix) in $\Mat_X(\R)$.
Let $\V$ denote the vector space over $\R$ consisting of the column vectors with coordinates
indexed by $X$, and all entries in $\R$. 
The algebra $\Mat_X(\R)$ acts on $\V$ by left multiplication.
We define a bilinear form $\b{ \; , \; } : \V \times \V \to \R$
such that $\b{u,v} = \sum_{y \in X} u_y v_y$ for $u,v \in \V$.
We have $\b{B u, v} = \b{u, B^{\sf t} v}$ for $B \in \Mat_X(\R)$ and $u,v \in \V$.
Here $B^{\sf t}$ denotes the transpose of $B$.
For $y \in X$ define $\widehat{y} \in \V$ that has $y$-entry $1$ and all other entries $0$.
Note that $\{ \widehat{y} \; | \; y \in X\}$ form an orthonormal basis of $\V$.

We now recall the Bose--Mesner algebra.
For $0 \leq i \leq d$ define $A_i \in \Mat_X(\R)$ that has $(y,z)$-entry $1$
if $(y,z) \in R_i$ and $0$ if $(y,z) \not\in R_i$ $(y,z \in X)$.
The matrix $A_i$ is symmetric.
We have 
\begin{align*}
A_0 &= I, & A_i A_j &= \sum_{h=0}^d p^h_{i j} A_h \qquad (0 \leq i,j \leq d).
\end{align*}
The $\{A_i\}_{i=0}^d$ form a basis for a commutative subalgebra $M$ of $\Mat_X(\R)$.
We call $M$ the Bose--Mesner algebra of the scheme.
Each matrix in $M$ is symmetric.
By~\cite[Section~2.2]{BCN}
there exists a basis $\{E_i\}_{i=0}^d$ for $M$ such that 
\begin{align*}
E_0 &= |X|^{-1} J, &
I &= \sum_{i=0}^d E_i, &
E_i E_j &= \delta_{i,j} E_i \quad (0 \leq i,j \leq d).
\end{align*}
We have
\[
 \V = \sum_{i=0}^d E_i \V \qquad\qquad \text{(orthogonal direct sum)}.
\]
\looseness-1
For $0 \leq i \leq d$, $E_i \V$ is the $i^\text{th}$ common eigenspace for $M$,
and $E_i$ is the orthogonal projection from $\V$ onto $E_i \V$.
There exist real numbers $p_i (j)$, $q_i (j)$ $(0 \leq i,j \leq d)$ such that
\begin{align*}
 A_i &= \sum_{j=0}^d p_i (j) E_j, & 
 E_i &= |X|^{-1} \sum_{j=0}^d q_i (j) A_j
\end{align*}
for $0 \leq i \leq d$.

We now recall the Krein parameters.
Note that $A_i \circ A_j = \delta_{i,j} A_i$ $(0 \leq i,j \leq d)$,
where $\circ$ denotes entry-wise multiplication.
Therefore $M$ is closed under $\circ$.
Consequently there exist real numbers $q^h_{i j}$ $(0 \leq h,i,j \leq d)$ such that 
\begin{align*}
E_i \circ E_j &= |X|^{-1} \sum_{h=0}^d q^h_{i j} E_h && (0 \leq i,j \leq d).
\end{align*}
By~\cite[Theorem~3.8]{BI}, $q^h_{i j} \geq 0$ for $0 \leq h,i,j \leq d$.
The $q^h_{i j}$ are called the Krein parameters of the scheme.

We now recall the dual Bose--Mesner algebra.
For the rest of this section fix $x \in X$.
For $B \in M$ let $B^\rho$ denote the diagonal matrix in $\Mat_X(\R)$
that has $(y,y)$-entry $B_{x,y}$ for $y \in X$.
Roughly speaking, $B^\rho$ is obtained by turning column $x$ of $B$ at a 45 degree angle.
For $0 \leq i \leq d$ define $E^*_i = A_i^\rho$.
For $y \in X$ the $(y,y)$-entry of $E^*_i$ is $1$ if $(x,y) \in R_i$ and $0$
if $(x,y) \not\in R_i$.
Note that $E^*_0$ has $(x,x)$-entry $1$ and all other entries $0$.
The matrices $\{E^*_i\}_{i=0}^d$ satisfy
\begin{align*}
 I &= \sum_{i=0}^d E^*_i, & 
 E^*_i E^*_j &= \delta_{i,j} E^*_i \qquad (0 \leq i,j \leq d).
\end{align*}
Therefore $\{E^*_i\}_{i=0}^d$ form a basis for a commutative subalgebra $M^*$ of $\Mat_X(\R)$.
We call $M^*$ the dual Bose--Mesner algebra with respect to $x$.
We have
\[
 \V = \sum_{i=0}^d E^*_i \V \qquad\qquad \text{(orthogonal direct sum)}.
\]
For $0 \leq i \leq d$, $E^*_i \V$ has basis $\{\widehat{y} \mid y \in X, \; (x,y) \in R_i\}$.
Moreover $E^*_i \V$ is the $i^\text{th}$ common eigenspace for $M^*$,
and $E^*_i$ is the orthogonal projection from $\V$ onto $E^*_i \V$.

The map $\rho : M \to M^*$, $B \mapsto B^\rho$ is $\R$-linear and bijective.
For $0 \leq i \leq d$ define $A^*_i = |X| E_i^\rho$.
The $\{A^*_i\}_{i=0}^d$ form a basis of $M^*$, and
\begin{align*}
 A^*_0 &= I, &
 A^*_i A^*_j &= \sum_{h=0}^d q^h_{i j} A^*_h \qquad (0 \leq i,j \leq d).
\end{align*}
For $0 \leq i \leq d$,
\begin{align*}
 A^*_i &= \sum_{j=0}^d q_i (j) E^*_j,
&
 E^*_i &= |X|^{-1} \sum_{j=0}^d p_i (j) A^*_j.
\end{align*}

We now recall the subconstituent algebra $T$ and the primary $T$-module.
Let $T$ denote the subalgebra of $\Mat_X(\R)$ generated by $M$ and $M^*$.
We call $T$ the subconstituent algebra (or Terwilliger algebra) with respect to $x$.
The algebra $T$ is closed under the transpose map.
By~\cite[Lemma~3.4]{T:subconst1} the algebra $T$ is semisimple.
Moreover by~\cite[Lemma~3.4]{T:subconst1} the $T$-module $\V$ 
decomposes into an orthogonal direct sum of irreducible $T$-modules.
Among these modules there is a distinguished one, said to be primary.
We now describe the primary $T$-module.
Let $\mathbf{1}$ denote the vector in $\V$ that has all entries $1$.
So $\mathbf{1} = \sum_{y \in X} \widehat{y}$.
For $0 \leq i \leq d$,
\begin{align*}
A_i \widehat{x} &= E^*_i \mathbf{1}, & |X|^{-1} A^*_i \mathbf{1} &= E_i \widehat{x}.
\end{align*}
Therefore $M \widehat{x} = M^* \mathbf{1}$; denote this common vector space by $V$.
By construction $V$ is a $T$-module with dimension $d+1$.
By~\cite[Lemma~3.6]{T:subconst1} the $T$-module $V$ is irreducible.
The $T$-module $V$ is said to be primary.
For $0 \leq i \leq d$ define 
\begin{align*}
 \mathbf{1}_i &= A_i \widehat{x} = E^*_i \mathbf{1}.
\end{align*}
The vector $\mathbf{1}_i$ is a basis of $E^*_i V$.
Moreover $\{ \mathbf{1}_i \}_{i=0}^d$ is a basis of $V$.
This basis is orthogonal and 
$|| \mathbf{1}_i ||^2 = k_i$ where $k_i = \text{rank}(E^*_i)$ $(0 \leq i \leq d)$.
The basis $\{ \mathbf{1}_i\}_{i=0}^d$ diagonalizes $M^*$.
For $0 \leq i,j \leq d$,
\begin{align*}
 E^*_i \mathbf{1}_j &= \delta_{i,j} \mathbf{1}_j,
&
 A_i \mathbf{1}_j &= \sum_{h=0}^d p^h_{i j} \mathbf{1}_h.
\end{align*}
For $0 \leq i \leq d$ define 
\begin{align*}
 \mathbf{1}^*_i &= |X|^{-1} A^*_i \mathbf{1} = E_i \widehat{x}.
\end{align*}
The vector $\mathbf{1}^*_i$ is a basis of $E_i V$.
Moreover $\{ \mathbf{1}^*_i \}_{i=0}^d$ is a basis of $V$.
This basis is orthogonal and
$|| \mathbf{1}^*_i ||^2 = k^*_i$ where $k^*_i = \text{rank}(E_i)$ $(0 \leq i \leq d)$.
The basis $\{ \mathbf{1}^*_i \}_{i=0}^d$ diagonalizes $M$.
For $0 \leq i,j \leq d$,
\begin{align*}
 E_i \mathbf{1}^*_j &= \delta_{i,j} \mathbf{1}^*_j,
&
 A^*_i \mathbf{1}^*_j &= \sum_{h=0}^d q^h_{i j} \mathbf{1}^*_h.
\end{align*}
The bases $\{ \mathbf{1}_i \}_{i=0}^d$ and $\{ \mathbf{1}^*_i \}_{i=0}^d$ are related by
\begin{align*}
\mathbf{1}_i &= \sum_{j=0}^d p_i (j) \mathbf{1}^*_j,
&
\mathbf{1}^*_i &= |X|^{-1} \sum_{j=0}^d q_i (j) \mathbf{1}_j
\end{align*}
for $0 \leq i \leq d$.
The following bases for $V$ are of interest:
\begin{align*}
& \textup{\hypertarget{intro2_i}{(i)} }\{ \mathbf{1}_i\}_{i=0}^d,
&& \textup{\hypertarget{intro2_ii}{(ii)} } \{ k_i^{-1} \mathbf{1}_i \}_{i=0}^d,
&& \textup{\hypertarget{intro2_iii}{(iii)} } \{ \mathbf{1}^*_i \}_{i=0}^d,
&& \textup{\hypertarget{intro2_iv}{(iv)} } \{ |X| (k^*_i)^{-1} \mathbf{1}^*_i \}_{i=0}^d.
\end{align*}
The bases~\hyperlink{intro2_i}{\textup{(i)}}, \hyperlink{intro2_ii}{\textup{(ii)}} are dual with respect to $\b{ \; ,\; }$.
Moreover the bases~\hyperlink{intro2_iii}{\textup{(iii)}}, \hyperlink{intro2_iv}{\textup{(iv)}} are dual with respect to $\b{ \; , \; }$.

The algebras $M$ and $M^*$ are related as follows.
For $0 \leq h,i,j \leq d$,
\begin{align*}
 E^*_i A_j E^*_h &= 0 \qquad \text{if and only if} \qquad p^h_{i j}=0,
\\
 E_i A^*_j E_h &= 0 \qquad \text{if and only if} \qquad q^h_{i j} = 0.
\end{align*}
For $0 \leq i \leq d$,
\begin{align*}
A_i E^*_0 E_0 &= E^*_i E_0, &
A^*_i E_0 E^*_0 &= E_i E^*_0.
\end{align*}
For $0 \leq i \leq d$,
\begin{align*}
 E_0 E^*_i &\neq0, &
 E^*_0 E_i & \neq 0, &
 E^*_i E_0 &\neq 0, &
 E_i E^*_0 &\neq 0.
\end{align*}

We summarize the above description with four statements about $V$:
\begin{enumerate}[label=(\roman*)]
\item\label{intro3_i} %[\rm (i)]
the $\{E_i\}_{i=0}^d$ act on $V$ as a system of mutually orthogonal rank $1$ idempotents;
\item\label{intro3_ii} %[\rm (ii)]
the $\{E^*_i\}_{i=0}^d$ act on $V$ as a system of mutually orthogonal rank $1$ idempotents;
\item\label{intro3_iii} %[\rm (iii)]
$E_0 E^*_i E_0$ is nonzero on $V$ for $0 \leq i \leq d$;
\item\label{intro3_iv} %[\rm (iv)]
$E^*_0 E_i E^*_0$ is nonzero on $V$ for $0 \leq i \leq d$.
\end{enumerate}
The above statements~\ref{intro3_i}--\ref{intro3_iv} have the following significance.
We will show that~\ref{intro3_i}--\ref{intro3_iv} together with the symmetry of the matrices 
$\{E_i\}_{i=0}^d$, $\{E^*_i\}_{i=0}^d$
are sufficient to recover the $T$-module $V$ at an algebraic level.

We now turn our attention to idempotent systems.
An idempotent system is defined as follows.
Let $\F$ denote a field.
Let $d$ denote a nonnegative integer,
and let $V$ denote a vector space over $\F$ with dimension $d+1$.
Let $\End (V)$ denote the $\F$-algebra consisting of the $\F$-linear maps from $V$ to $V$.
An idempotent system on $V$ is a sequence 
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ such that
\begin{enumerate}[label=(\roman*)]
\item\label{intro4_i} %[\rm (i)]
$\{E_i\}_{i=0}^d$ is a system of mutually orthogonal rank $1$ idempotents in $\End (V)$;
\item\label{intro4_ii} %[\rm (ii)]
$\{E^*_i\}_{i=0}^d$ is a system of mutually orthogonal rank $1$ idempotents in $\End (V)$;
\item\label{intro4_iii} %[\rm (iii)]
$E_0 E^*_i E_0 \neq 0 \;\; (0 \leq i \leq d)$;
\item\label{intro4_iv} %[\rm (iv)]
$E^*_0 E_i E^*_0 \neq 0 \;\; (0 \leq i \leq d)$.
\end{enumerate}
The above idempotent system $\Phi$ is said to be symmetric whenever there exists an antiautomorphism $\dagger$
of $\End (V)$ that fixes each of $E_i$, $E^*_i$ for $0 \leq i \leq d$.
The map $\dagger$ corresponds to the transpose map.

Let $\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ denote a symmetric idempotent system on $V$.
Using $\Phi$ we will define some elements $\{A_i\}_{i=0}^d$, \{$A^*_i\}_{i=0}^d$ in $\End (V)$ 
and some scalars
\begin{equation}
\nu, \quad 
k_i, \quad 
k^*_i, \quad 
p^h_{i j}, \quad
q^h_{i j}, \quad
p_i (j), \quad
q_i (j) \label{eq:scalars}
\end{equation}
in $\F$.
The scalar $\nu$ corresponds to $|X|$.
We will endow $V$ with a nondegenerate symmetric bilinear form $\b{ \; , \;}$.
We will define four orthogonal bases of $V$ that correspond to the four earlier bases of interest.
We will show that the resulting construction matches the primary $T$-module at an algebraic level.

Our definitions are summarized as follows.
Note that $\{E_i\}_{i=0}^d$ form a basis for a commutative subalgebra $\mathcal M$ of $\End (V)$.
We show that for $0 \leq i \leq d$ there exists a unique $A_i \in {\mathcal M}$ such that 
\[
 A_i E^*_0 E_0 = E^*_i E_0.
\]
We show that $\{A_i\}_{i=0}^d$ is a basis for the vector space $\mathcal M$.
Similarly, the $\{E^*_i\}_{i=0}^d$ form a basis for a commutative subalgebra ${\mathcal M}^*$ of $\End (V)$.
We show that for $0 \leq i \leq d$ there exists a unique $A^*_i \in {\mathcal M}^*$ such that 
\[
 A^*_i E_0 E^*_0 = E_i E^*_0.
\]
We show that $\{A^*_i\}_{i=0}^d$ is a basis for the vector space ${\mathcal M}^*$.

Concerning the scalars~\eqref{eq:scalars}, 
we show that $\tr (E_0 E^*_0) \neq 0$.
The scalar $\nu$ is defined by
\begin{align*}
\nu &= \tr (E_0 E^*_0)^{-1}.
\end{align*}
The scalars $k_i$, $k^*_i$ are defined by
\begin{align*}
k_i &= \nu \, \tr (E_0 E^*_i), &
k^*_i &= \nu \, \tr (E^*_0 E_i) && (0 \leq i \leq d).
\end{align*}
We show that $\sum_{i=0}^d k_i = \nu = \sum_{i=0}^d k^*_i$,
and each of $k_i$, $k^*_i$ is nonzero for $0 \leq i \leq d$.
The scalars $p^h_{i j}$, $q^h_{i j}$ are defined by
\begin{align*}
 A_i A_j &= \sum_{h=0}^d p^h_{i j} A_h,
&
 A^*_i A^*_j &= \sum_{h=0}^d q^h_{ i j} A^*_h && (0 \leq i,j \leq d).
\end{align*}
The scalars $p_i (j)$, $q_i (j)$ are defined by
\begin{align*}
 A_i &= \sum_{j=0}^d p_i (j) E_j,
&
 A^*_i &= \sum_{j=0}^d q_i (j) E^*_j && (0 \leq i \leq d).
\end{align*}

We define a bilinear form $\b{ \; , \; }$ on $V$ as follows.
By linear algebra, there exists a nondegenerate bilinear form $\b{ \; , \; }$ on $V$
such that $\b{B u, v} = \b{ u, B^\dagger v}$ for all $B \in \End (V)$ and $u, v \in V$.
The bilinear form $\b{ \; , \;}$ is unique up to multiplication by a nonzero scalar in $\F$.
The bilinear form $\b{ \; , \;}$ is symmetric.

Fix nonzero $\xi, \zeta$ in $E_0 V$ and nonzero $\xi^*, \zeta^*$ in $E^*_0 V$.
We show that each of the following~\hyperlink{intro5_i}{\textup{(i)}}--\hyperlink{intro5_iv}{\textup{(iv)}} is an orthogonal basis for $V$:
\begin{align*}
& \textup{\hypertarget{intro5_i}{(i)} } \{ E^*_i \xi\}_{i=0}^d,
&& \textup{\hypertarget{intro5_ii}{(ii)} } \{ k_i^{-1} E^*_i \zeta \}_{i=0}^d,
&& \textup{\hypertarget{intro5_iii}{(iii)} } \{ E_i \xi^* \}_{i=0}^d,
&& \textup{\hypertarget{intro5_iv}{(iv)} } \{ (k^*_i)^{-1} E_i \zeta^* \}_{i=0}^d.
\end{align*}

The bases~\hyperlink{intro5_i}{\textup{(i)}}, \hyperlink{intro5_ii}{\textup{(ii)}} are dual if and only if $\b{\xi, \zeta} = \nu$,
and the bases~\hyperlink{intro5_iii}{\textup{(iii)}}, \hyperlink{intro5_iv}{\textup{(iv)}} are dual if and only if $\b{\xi^*, \zeta^*} = \nu$.

We just summarized our definitions.
In the main body of the paper, we show that the resulting defined objects are related 
in a manner that matches the primary $T$-module.
To describe this relationship,
we use some equations involving the $\{E_i\}_{i=0}^d$, $\{E^*_i\}_{i=0}^d$,
$\{A_i\}_{i=0}^d$, $\{A^*_i\}_{i=0}^d$ called the reduction rules.

Near the end of the paper we introduce the $P$-polynomial and $Q$-polynomial properties
for symmetric idempotent systems.
We show that a symmetric idempotent system that is $P$-polynomial and $Q$-polynomial
is essentially the same thing as a Leonard system in the sense of~\cite[Definition~4.1]{T:Leonard}.

The paper is organized as follows.
In Section~\ref{sec:pre} we recall some basic results from linear algebra.
In Section~\ref{sec:ips} we introduce the concept of an idempotent system.
In Section~\ref{sec:pi} we introduce the scalar $\nu$ and discuss some related topics.
In Section~\ref{sec:symips} we introduce the symmetric idempotent systems.
In Sections~\ref{sec:rho},~\ref{sec:Ai} we introduce a certain linear bijection $\rho : {\mathcal M} \to {\mathcal M}^*$
and use it to define the elements $A_i$, $A^*_i$.
In Sections~\ref{sec:ki},~\ref{sec:reduction} we introduce the scalars $k_i$, $k^*_i$ and
obtain some reduction rules involving these scalars. 
In Sections~\ref{sec:phij}, \ref{sec:red2} we introduce the scalars $p^h_{i j}$, $q^h_{i j}$
and obtain some reduction rules involving these scalars.
In Sections~\ref{sec:pij}, \ref{sec:red3} we introduce the scalars $p_i (j)$, $q_i (j)$ 
and obtain some reduction rules involving these scalars.
In Section~\ref{sec:matrices} we put some of our earlier results in matrix form.
In Sections~\ref{sec:standard}--\ref{sec:dualstandard} we introduce the four bases of interest
and discuss their properties.
In Section~\ref{sec:4bases} we obtain the transition matrices between these four bases, and the
inner products between these four bases.
We also obtain the matrices
representing $A_i$, $A^*_i$, $E_i$, $E^*_i$ with respect to these four bases.
In Section~\ref{sec:Ppoly} we introduce the $P$-polynomial and $Q$-polynomial properties.
In Section~\ref{sec:LP} we recall the notion of a Leonard pair and a Leonard system.
In Section~\ref{sec:IPSLS} we show that a Leonard system is essentially the same thing
as a symmetric idempotent system that is $P$-polynomial and $Q$-polynomial.

The reader might wonder how the concept of a symmetric idempotent system is
related to the concept of a character algebra~\cite{Kawada}.
Roughly speaking, a symmetric idempotent system is obtained by gluing together
a character algebra and its dual;
we will discuss this in a future paper.



















\section{Preliminaries}
\label{sec:pre}

In this section we fix some notation and recall some basic concepts.
Throughout this paper $\F$ denotes a field.
By a \emph{scalar} we mean an element of $\F$.
All algebras and vector spaces discussed in this paper are over $\F$.
All algebras discussed in this paper are associative and have a multiplicative identity.
For an algebra $\mathcal A$, 
by an \emph{automorphism} of $\mathcal A$ we mean an algebra isomorphism ${\mathcal A} \to {\mathcal A}$,
and by an \emph{antiautomorphism} of $\mathcal A$ we mean an $\F$-linear bijection
$\tau : {\mathcal A} \to {\mathcal A}$ such that $(Y Z)^\tau = Z^\tau Y^\tau$ for $Y, Z \in {\mathcal A}$.
For the rest of this paper, fix an integer $d \geq 0$ and let $V$ denote a vector space with dimension $d+1$.
Let $\End (V)$ denote the algebra consisting of the $\F$-linear
maps from $V$ to $V$. 
Let $\Mat_{d+1}(\F)$ denote the algebra consisting of the $d+1$ by $d+1$
matrices that have all entries in $\F$.
We index the rows and columns by $0,1,\ldots, d$.
The identity of $\End (V)$ or $\Mat_{d+1}(\F)$ is denoted by $I$.
For $A \in \End (V)$,
the dimension of $A V$ is called the \emph{rank of $A$}.
A matrix $M \in \Mat_{d+1}(\F)$ is said to be \emph{tridiagonal} whenever
the $(i,j)$-entry $M_{i,j}=0$ if $|i-j|>1$ $(0 \leq i,j \leq d)$.
Assume for the moment that $M$ is tridiagonal.
Then $M$ is said to be \emph{irreducible}
whenever $M_{i,j} \neq 0$ if $|i-j|=1$ $(0 \leq i,j \leq d)$.
We recall how each basis $\{v_i\}_{i=0}^d$ of $V$ gives an algebra
isomorphism $\End (V) \to \Mat_{d+1}(\F)$.
For $A \in \End (V)$ and $M \in \Mat_{d+1}(\F)$,
we say that \emph{$M$ represents $A$ with respect to $\{v_i\}_{i=0}^d$}
whenever $A v_j = \sum_{i=0}^d M_{i,j} v_i$ for $0 \leq j \leq d$.
The isomorphism sends $A$ to the unique matrix in $\Mat_{d+1}(\F)$ that represents
$A$ with respect to $\{v_i\}_{i=0}^d$.
Next we recall the transition matrix between two bases of $V$.
Let $\{u_i\}_{i=0}^d$ and $\{v_i\}_{i=0}^d$ denote bases of $V$.
By the \emph{transition matrix from $\{u_i\}_{i=0}^d$ to $\{v_i\}_{i=0}^d$}
we mean the matrix $T \in \Mat_{d+1}(\F)$ such that 
$v_j = \sum_{i=0}^d T_{i,j} u_i$ for $0 \leq j \leq d$.
Let $T$ denote the transition matrix from $\{u_i\}_{i=0}^d$ to $\{v_i\}_{i=0}^d$.
Then $T$ is invertible and $T^{-1}$ is the transition matrix
from$\{v_i\}_{i=0}^d$ to $\{u_i\}_{i=0}^d$.
Let $T'$ denote the transition matrix from $\{v_i\}_{i=0}^d$ to a basis $\{w_i\}_{i=0}^d$ of $V$.
Then the transition matrix from $\{u_i\}_{i=0}^d$ to $\{w_i\}_{i=0}^d$ is $T T'$.
For $A \in \End (V)$ let $M$ denote the matrix representing $A$ with respect to
$\{u_i\}_{i=0}^d$.
Then $T^{-1} M T$ represents $A$ with respect to $\{v_i\}_{i=0}^d$.
Let $A \in \End (V)$.
A subspace $W \subseteq V$ is called an \emph{eigenspace} of $A$ whenever
$W \neq 0$ and there exists a scalar $\theta$ such that $W = \{ v \in V \, |\, A v = \theta v\}$;
in this case $\theta$ is the \emph{eigenvalue} of $A$ associated with $W$.
We say that $A$ is \emph{diagonalizable} whenever $V$ is spanned by the eigenspaces of $A$.
We say that $A$ is \emph{multiplicity-free} whenever $A$ is diagonalizable and its eigenspaces
all have dimension one.

\begin{defi} \label{def:decomp} \samepage
By a \emph{decomposition of $V$} we mean a sequence $\{V_i\}_{i=0}^d$
consisting of one-dimensional subspaces of $V$ such that
$V = \sum_{i=0}^d V_i$ (direct sum).
\end{defi}

\begin{defi}[{\cite[Section~6A]{CR}}]
\label{def:orth} \samepage
By a \emph{system of mutually orthogonal rank $1$ idempotents} in $\End (V)$
we mean a sequence $\{E_i\}_{i=0}^d$ of elements in $\End (V)$ such that
\begin{align*}
 E_i E_j &= \delta_{i,j} E_i &&(0 \leq i,j \leq d),
\\
 \rank (E_i) &= 1 &&(0 \leq i \leq d).
\end{align*}
\end{defi}

\begin{lemma} \label{lem:decomp} \samepage
%\ifDRAFT {\rm lem:decomp}. \fi
The following hold.
\begin{enumerate}[label=(\roman*)]
\item\label{lemma2.3_i} %[\rm (i)]
Let $\{V_i\}_{i=0}^d$ denote a decomposition of $V$.
For $0 \leq i \leq d$ define $E_i \in \End (V)$ such that
$(E_i - I)V_i = 0$ and $E_i V_j = 0$ if $j \neq i$ $(0 \leq j \leq d)$.
Then $\{E_i\}_{i=0}^d$ is a system of mutually orthogonal rank $1$ idempotents in $\End (V)$.
\item\label{lemma2.3_ii} %[\rm (ii)]
Let $\{E_i\}_{i=0}^d$ denote a system of mutually orthogonal rank $1$ idempotents in $\End (V)$.
Then $\{E_i V\}_{i=0}^d$ is a decomposition of $V$.
\end{enumerate}
\end{lemma}

\begin{defi} \label{def:primitiveofA} \samepage
Let $A$ denote a multiplicity-free element in $\End (V)$, and let
$\{V_i\}_{i=0}^d$ denote an ordering of the eigenspaces of $A$.
Then $\{V_i\}_{i=0}^d$ is a decomposition of $V$.
Let $\{E_i\}_{i=0}^d$ denote the corresponding system of mutually orthogonal rank $1$ idempotents
from Lemma~\ref{lem:decomp}\ref{lemma2.3_i}.
We call $\{E_i\}_{i=0}^d$ the \emph{primitive idempotents of $A$}.
\end{defi}

For the rest of this section,
let $\{E_i\}_{i=0}^d$ denote a system of mutually orthogonal rank $1$
idempotents in $\End (V)$.
The next two lemmas are routinely verified.

\begin{lemma} \label{lem:trE} \samepage
The following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma2.5_i} %[\rm (i)]
$\tr (E_i) = 1$ $(0 \leq i \leq d)$, where $\tr $ means trace.
\item\label{lemma2.5_ii} %[\rm (ii)]
 $I = \sum_{i=0}^d E_i$;
\item\label{lemma2.5_iii} %[\rm (iii)]
$\{E_i\}_{i=0}^d$ form a basis for a commutative subalgebra of $\End (V)$.
\end{enumerate}
\end{lemma}

\begin{lemma} \label{lem:sumEiAEj} \samepage
For ${\mathcal A} = \End (V)$,
\begin{enumerate}[label=(\roman*)]
\item\label{lemma2.6_i} %[\rm (i)]
the sum ${\mathcal A} = \sum_{i=0}^d \sum_{j=0}^d E_i {\mathcal A} E_j$ is direct;
\item\label{lemma2.6_ii} %[\rm (ii)]
$\dim E_i {\mathcal A} E_j = 1$ for $0 \leq i,j \leq d$.
\end{enumerate}
\end{lemma}


\section{Idempotent systems}
\label{sec:ips}


Recall the vector space $V$ with dimension $d+1$.
In this section we introduce the notion of an idempotent system on $V$.

\begin{defi} \label{def:ips} \samepage
By an \emph{idempotent system} on $V$ we mean a sequence
\begin{equation*}
 ( \{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d) 
\end{equation*}
such that
\begin{enumerate}[label=(\roman*)]
\item\label{defi3.1_i} %[\rm (i)]
$\{E_i\}_{i=0}^d$ is a system of mutually orthogonal rank $1$ idempotents in $\End (V)$;
\item\label{defi3.1_ii} %[\rm (ii)]
$\{E^*_i\}_{i=0}^d$ is a system of mutually orthogonal rank $1$ idempotents in $\End (V)$;
\item\label{defi3.1_iii} %[\rm (iii)]
$E_0 E^*_i E_0 \neq 0 \quad (0 \leq i \leq d)$;
\item\label{defi3.1_iv} %[\rm (iv)]
$E^*_0 E_i E^*_0 \neq 0 \quad (0 \leq i \leq d)$.
\end{enumerate}
\end{defi}

Let $\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ denote an idempotent system on $V$.
Define
\[
 \Phi^* = \left( \{E^*_i\}_{i=0}^d; \{E_i\}_{i=0}^d\right).
\]
Then $\Phi^*$ is an idempotent system on $V$, called the \emph{dual} of $\Phi$.
We have $(\Phi^*)^* = \Phi$.
For an object $f$ attached to $\Phi$, the corresponding object attached to $\Phi^*$
is denoted by $f^*$.

Let $\Phi' = (\{E'_i\}_{i=0}^d; \{E^{* \prime}_i\}_{i=0}^d)$ denote an idempotent system on
a vector space $V'$.
By an \emph{isomorphism of idempotent systems from $\Phi$ to $\Phi'$} we mean
an algebra isomorphism $\End (V) \to \End (V')$ that sends
$E_i \mapsto E'_i$, $E^*_i \mapsto E^{* \prime}_i$ for $0 \leq i \leq d$.
We say that $\Phi$ and $\Phi'$ are \emph{isomorphic} whenever there exists an isomorphism
of idempotent systems from $\Phi$ to $\Phi'$.
By the Skolem--Noether theorem (see~\cite[Corollary~7.125]{Rot}),
a map $\sigma : \End (V) \to \End (V')$ is an algebra isomorphism
if and only if there exists an $\F$-linear bijection $S : V \to V'$ such that $A^\sigma = S A S^{-1}$
for all $A \in \End (V)$.


\begin{defi} \label{def:D}
Let $\mathcal M$ denote the subalgebra of $\End (V)$ generated by $\{E_i\}_{i=0}^d$.
Note that $\mathcal M$ is commutative,
and $\{E_i\}_{i=0}^d$ form a basis of the vector space $\mathcal M$.
\end{defi}


























\section{The scalars \texorpdfstring{$m_i$}{mi}, \texorpdfstring{$\nu$}{nu}}
\label{sec:pi}

Let $\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ denote an idempotent system
on $V$.
In this section we use $\Phi$ to introduce some scalars $\{m_i\}_{i=0}^d$, $\nu$.

\begin{defi} \label{def:mi2} \samepage
For $0 \leq i \leq d$ define 
\begin{align}
 m_i &= \tr (E^*_0 E_i). \label{eq:defmi2}
\end{align}
\end{defi}

\begin{lemma} \label{lem:E0EsiE0} \samepage
For $0 \leq i \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma4.2_i} %[\rm (i)]
$E^*_0 E_i E^*_0 = m_i E^*_0$;
\item\label{lemma4.2_ii} %[\rm (ii)]
$E_0 E^*_i E_0 = m^*_i E_0$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma4.2_i}
Abbreviate ${\mathcal A} = \End (V)$.
By Lemma~\ref{lem:sumEiAEj}\ref{lemma2.6_ii},
$E^*_0$ is a basis for the vector space $E^*_0 {\mathcal A} E^*_0$.
So there exists a scalar $\alpha_i$ such that $E^*_0 E_i E^*_0 =\alpha_i E^*_0$.
In this equation, take the trace of each side and simplify the result
using Lemma~\ref{lem:trE}\ref{lemma2.5_i} and $\tr (M N) = \tr (N M)$
to obtain $\alpha_i = m_i$.
The result follows.

\ref{lemma4.2_ii}~Apply~\ref{lemma4.2_i} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:EsiE0Esi} \samepage
For $0 \leq i \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma4.3_i} %[\rm (i)]
$E_i E^*_0 E_i = m_i E_i$;
\item\label{lemma4.3_ii} %[\rm (ii)]
$E^*_i E_0 E^*_i = m^*_i E^*_i$.
\end{enumerate}
\end{lemma}

\begin{proof}
Similar to the proof of Lemma~\ref{lem:E0EsiE0}.
\end{proof}

\begin{lemma} \label{lem:mi3} \samepage
The following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma4.4_i} %[\rm (i)]
$m_i \neq 0$ $\quad (0 \leq i \leq d)$;
\item\label{lemma4.4_ii} %[\rm (ii)]
$\sum_{i=0}^d m_i = 1$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma4.4_i}
Use Definition~\ref{def:ips}\ref{defi3.1_iv} and Lemma~\ref{lem:E0EsiE0}\ref{lemma4.2_i}.

\ref{lemma4.4_i}
By Lemma~\ref{lem:trE}\ref{lemma2.5_ii}, $\sum_{i=0}^d E_i = I$.
In this equation, multiply each side on the left by $E^*_0$ to get
$\sum_{i=0}^d E^*_0 E_i = E^*_0$.
In this equation, take the trace of each side, and evaluate the result
using Lemma~\ref{lem:trE}\ref{lemma2.5_i} and Definition~\ref{def:mi2}.
\end{proof}

\begin{defi} \label{def:nu} \samepage
Setting $i=0$ in~\eqref{eq:defmi2} we find that $m_0 = m^*_0$;
let $\nu$ denote the multiplicative inverse of this common value.
We emphasize $\nu = \nu^*$ and
\begin{equation}
 \tr (E_0 E^*_0) = \nu^{-1}. \label{eq:nu}
\end{equation}
\end{defi}

\begin{lemma} \label{lem:nuE0Es0E0} \samepage
We have
\begin{align}
 \nu E_0 E^*_0 E_0 &= E_0, &
 \nu E^*_0 E_0 E^*_0 &= E^*_0. \label{eq:nu2}
\end{align}
\end{lemma}

\begin{proof}
To get the equation on the left in~\eqref{eq:nu2},
set $i=0$ in Lemma~\ref{lem:E0EsiE0}\ref{lemma4.2_ii} and use Definition~\ref{def:nu}.
Applying this to $\Phi^*$ we get the equation on the right in~\eqref{eq:nu2}.
\end{proof}

\begin{lemma} \label{lem:EsiE0Esj} \samepage
Each of the following is a basis of the vector space $\End (V)$:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma4.7_i} %[\rm (i)]
$\{E_i E^*_0 E_j \,|\, 0 \leq i,j \leq d\}$;
\item\label{lemma4.7_ii} %[\rm (ii)]
$\{E^*_i E_0 E^*_j \,|\, 0 \leq i,j \leq d\}$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma4.7_i}
In view of Lemma~\ref{lem:sumEiAEj},
it suffices to show that $E_i E^*_0 E_j \neq 0$ for $0 \leq i,j \leq d$.
Let $i$, $j$ be given, and suppose $E_i E^*_0 E_j = 0$.
Using Lemmas~\ref{lem:E0EsiE0}\ref{lemma4.2_i} and~\ref{lem:mi3}\ref{lemma4.4_i},
\[
 0 = E^*_0 (E_i E^*_0 E_j) E^*_0 = m_i m_j E^*_0 \neq 0
\]
for a contradiction.
The result follows.

\ref{lemma4.7_ii}
Apply~\ref{lemma4.7_i} to $\Phi^*$.
\end{proof}


\begin{lemma} \label{lem:generate} \samepage
Each of the following is a generating set for the algebra $\End (V)$:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma4.8_i} %[\rm (i)]
$E^*_0$ and $\mathcal M$;
\item\label{lemma4.8_ii} %[\rm (ii)]
$E_0$ and ${\mathcal M}^*$;
\item\label{lemma4.8_iii} %[\rm (iii)]
$\mathcal M$ and ${\mathcal M}^*$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma4.8_i}
By Definition~\ref{def:D} and Lemma~\ref{lem:EsiE0Esj}\ref{lemma4.7_i}.

\ref{lemma4.8_ii}
Apply~\ref{lemma4.8_i} to $\Phi^*$.

\ref{lemma4.8_iii}
By~\ref{lemma4.8_i} above and Definition~\ref{def:D}.
\end{proof}














\section{Symmetric idempotent systems}
\label{sec:symips}

We continue to discuss an idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on $V$.


\begin{defi} \label{def:sym} \samepage
We say that $\Phi$ is \emph{symmetric} whenever there exists an antiautomorphism
$\dagger$ of $\End (V)$ that fixes each of $E_i$, $E^*_i$ for $0 \leq i \leq d$. 
\end{defi}

Recall the algebra $\mathcal M$ from Definition~\ref{def:D}.

\begin{lemma} \label{lem:dagger0} \samepage
Assume that $\Phi$ is symmetric, and let $\dagger$ denote an antiautomorphism of
$\End (V)$ from Definition~\ref{def:sym}.
Then the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma5.2_i} %[\rm (i)]
$\dagger$ is unique;
\item\label{lemma5.2_ii} %[\rm (ii)]
$(A^\dagger)^\dagger = A$ for $A \in \End (V)$;
\item\label{lemma5.2_iii} %[\rm (iii)]
$\dagger$ fixes every element in $\mathcal M$ and every element in ${\mathcal M}^*$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma5.2_iii}
By Definitions~\ref{def:D} and~\ref{def:sym}.

\ref{lemma5.2_ii}
The composition $\dagger \circ \dagger$ is an automorphism of $\End (V)$
that fixes everything in $\mathcal M$ and everything in ${\mathcal M}^*$.
This automorphism is the identity in view of Lemma~\ref{lem:generate}\ref{lemma4.8_iii}.

\ref{lemma5.2_i}
Let $\mu$ denote an antiautomorphism of $\End (V)$ that fixes each of $E_i$, $E^*_i$
for $0 \leq i \leq d$.
We show $\mu = \dagger$.
The composition $\dagger \circ \mu$ is an automorphism of $\End (V)$
that fixes everything in $\mathcal M$ and everything in ${\mathcal M}^*$.
So this automorphism is the identity.
We have $\dagger = \dagger^{-1}$ by~\ref{lemma5.2_ii} above, so $\mu = \dagger$.
\end{proof}


\section{The map \texorpdfstring{$\rho$}{rho}}
\label{sec:rho}

Let $\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ denote a symmetric 
idempotent system on $V$.
Recall the algebra $\mathcal M$ from Definition~\ref{def:D}.
In this section we introduce a certain map $\rho :{\mathcal M} \to {\mathcal M}^*$
that will play an essential role in our theory.
As we will see, $\rho$ is an isomorphism of vector spaces but not algebras.

\begin{lemma} \label{lem:EiEs0} \samepage
For ${\mathcal A} = \End (V)$, 
\begin{enumerate}[label=(\roman*)]
\item\label{lemma6.1_i} %[\rm (i)]
the elements
$\{E_i E^*_0\}_{i=0}^d$ form a basis of ${\mathcal A} E^*_0$;
\item\label{lemma6.1_ii} %[\rm (ii)]
the elements
$\{E^*_i E_0\}_{i=0}^d$ form a basis of ${\mathcal A} E_0$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma6.1_i}
By Lemmas~\ref{lem:trE}\ref{lemma2.5_ii} and~\ref{lem:sumEiAEj}\ref{lemma2.6_i}
the sum ${\mathcal A} E^*_0 = \sum_{i=0}^d E^*_i {\mathcal A} E^*_0$ is direct.
Each summand has dimension one by Lemma~\ref{lem:sumEiAEj}\ref{lemma2.6_ii}, so
${\mathcal A} E^*_0$ has dimension $d+1$.
The elements $\{E_i E^*_0\}_{i=0}^d$ are contained in ${\mathcal A} E^*_0$.
We show that these elements are linearly independent.
For scalars $\{\alpha_i\}_{i=0}^d$ suppose $0 = \sum_{i=0}^d \alpha_i E_i E^*_0$.
For $0 \leq r \leq d$,
multiply each side of this equation on the left by $E_r$ to obtain
$0 = \alpha_r E_r E^*_0$.
We have $E_r E^*_0 \neq 0$ by Definition~\ref{def:ips}\ref{defi3.1_iv}, so $\alpha_r = 0$.
We have shown that $\{E_i E^*_0\}_{i=0}^d$ are linearly independent, and hence
a basis of ${\mathcal A} E^*_0$.

\ref{lemma6.1_ii}
Apply~\ref{lemma6.1_i} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:DtoDEs0} \samepage
For ${\mathcal A} = \End (V)$, 
\begin{enumerate}[label=(\roman*)]
\item\label{lemma6.2_i} %[\rm (i)]
the map ${\mathcal M} \to {\mathcal A} E^*_0$, $Y \mapsto Y E^*_0$ is an $\F$-linear bijection;
\item\label{lemma6.2_ii} %[\rm (ii)]
the map ${\mathcal M}^* \to {\mathcal A} E_0$, $Y \mapsto Y E_0$ is an $\F$-linear bijection.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma6.2_i}
Clearly the map is $\F$-linear.
By Lemma~\ref{lem:EiEs0}\ref{lemma6.1_i},
the map sends the basis $\{E_i\}_{i=0}^d$ of $\mathcal M$ to the basis $\{E_i E^*_0\}_{i=0}^d$
of ${\mathcal A} E^*_0$.
So the map is bijective.

\ref{lemma6.2_ii}
Apply~\ref{lemma6.2_ii} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:rho} \samepage
There exists a unique $\F$-linear map $\rho : {\mathcal M} \to {\mathcal M}^*$
such that for $Y \in {\mathcal M}$,
\begin{align}
 Y E^*_0 E_0 &= Y^\rho E_0. \label{eq:defrho}
\end{align}
\end{lemma}

\begin{proof}
Abbreviate ${\mathcal A} = \End (V)$.
Concerning existence, consider the $\F$-linear map $g : {\mathcal M} \to {\mathcal A} E_0$,
$Y \mapsto Y E^*_0 E_0$.
Let $\mu$ denote the map in Lemma~\ref{lem:DtoDEs0}\ref{lemma6.2_ii}.
The composition
\[
 \rho : {\mathcal M} \xrightarrow{\;\;\; g \;\;\; } {\mathcal A} E_0 \xrightarrow{\;\; \mu^{-1}\;\; } {\mathcal M}^*
\]
satisfies~\eqref{eq:defrho}.
We have shown that $\rho$ exists.
The map $\rho$ is unique by Lemma~\ref{lem:DtoDEs0}\ref{lemma6.2_ii}.
\end{proof}


\begin{lemma} \label{lem:rhorhospre} \samepage
The maps $\rho$ and $\nu \rho^*$ are inverses.
In particular, the maps $\rho$, $\rho^*$ are bijective.
\end{lemma}
 
\begin{proof}
Pick $Y \in {\mathcal M}$.
Using Lemma~\ref{lem:nuE0Es0E0} and applying~\eqref{eq:defrho}
to both $\Phi$ and $\Phi^*$,
\[
 (Y^\rho)^{\rho^*} E^*_0
 = Y^\rho E_0 E^*_0
 = Y E^*_0 E_0 E^*_0
 = \nu^{-1} Y E^*_0.
\]
By this and Lemma~\ref{lem:DtoDEs0}\ref{lemma6.2_i} we get $(Y^\rho)^{\rho^*} = \nu^{-1} Y$.
Applying this to $\Phi^*$,
$(Z^{\rho^*})^\rho = \nu^{-1} Z$ for $Z \in {\mathcal M}^*$.
Thus the maps $\rho$ and $\nu \rho^*$ are inverses.
\end{proof}


\begin{lemma} \label{lem:rhoI} \samepage
The map
$\rho$ sends $I \mapsto E^*_0$ and $E_0 \mapsto \nu^{-1} I$.
\end{lemma}

\begin{proof}
Using Lemma~\ref{lem:rho},
$E^*_0 E_0 = I E^*_0 E_0 = I^\rho E_0$.
This forces $E^*_0 = I^\rho$ by Lemma~\ref{lem:DtoDEs0}\ref{lemma6.2_ii}.
Using Lemmas~\ref{lem:nuE0Es0E0} and~\ref{lem:rho},
$E_0^\rho E_0 = E_0 E^*_0 E_0 = \nu^{-1} E_0$.
This forces $E_0^\rho = \nu^{-1} I$ by Lemma~\ref{lem:DtoDEs0}\ref{lemma6.2_ii}.
\end{proof}














\section{The elements \texorpdfstring{$A_i$}{Ai}}
\label{sec:Ai}


We continue to discuss a symmetric idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on~$V$.


\begin{defi} \label{def:Ai} \samepage
%\ifDRAFT {\rm def:Ai}. \fi
For $0 \leq i \leq d$ define 
\begin{equation}
 A_i = \nu (E^*_i)^{\rho^*}. \label{eq:defAiAsi}
\end{equation}
\end{defi}


\begin{lemma} \label{lem:rhorhos} \samepage
%\ifDRAFT {\rm lem:rhorhos}. \fi
For $0 \leq i \leq d$,
$\rho$ sends $A_i \mapsto E^*_i$ and $E_i \mapsto \nu^{-1} A^*_i$.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:rhorhospre} and Definition~\ref{def:Ai},
$A_i^\rho = \nu ((E^*_i)^{\rho^*})^\rho = E^*_i$.
Applying~\eqref{eq:defAiAsi} to $\Phi^*$, $E_i^\rho = \nu^{-1} A^*_i$.
\end{proof}


\begin{lemma} \label{lem:dagger4} \samepage
%\ifDRAFT {\rm lem:dagger4}. \fi
The antiautomorphism $\dagger$ from Definition~\ref{def:sym} fixes each of $A_i$, $A^*_i$ for $0 \leq i \leq d$.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:dagger0}\ref{lemma5.2_iii} and since $A_i \in {\mathcal M}$, $A^*_i \in {\mathcal M}^*$
for $0 \leq i \leq d$.
\end{proof}


\begin{lemma} \label{lem:AiEs0E0} \samepage
%\ifDRAFT {\rm lem:AiEs0E0}. \fi
For $0 \leq i \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma7.4_i} %[\rm (i)]
$A_i E^*_0 E_0 = E^*_i E_0$;
\item\label{lemma7.4_ii} %[\rm (ii)]
$A^*_i E_0 E^*_0 = E_i E^*_0$;
\item\label{lemma7.4_iii} %[\rm (iii)]
$E_0 E^*_0 A_i = E_0 E^*_i$;
\item\label{lemma7.4_iv} %[\rm (iv)]
$E^*_0 E_0 A^*_i = E^*_0 E_i$.
\end{enumerate}
\end{lemma}


\begin{proof}
\ref{lemma7.4_i}
Use Lemmas~\ref{lem:rho}, \ref{lem:rhorhos}.

\ref{lemma7.4_ii} 
Apply~\ref{lemma7.4_i} to $\Phi^*$.

\ref{lemma7.4_iii}, \ref{lemma7.4_iv}
For the equations in~\ref{lemma7.4_i} and~\ref{lemma7.4_ii}, apply $\dagger$ to each side and use Lemma~\ref{lem:dagger4}.
\end{proof}


\begin{lemma} \label{lem:A0As0} \samepage
%\ifDRAFT {\rm lem:A0As0}. \fi
We have $A_0 = I$.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:rhoI}, $I^\rho = E^*_0$.
In this equation, apply $\rho^*$ to each side and evaluate the result using Lemma~\ref{lem:rhorhospre}
and Definition~\ref{def:Ai}.
\end{proof}


\begin{lemma} \label{lem:sumAi} \samepage
%\ifDRAFT {\rm lem:sumAi}. \fi
We have
$\sum_{i=0}^d A_i = \nu E_0$.
\end{lemma}

\begin{proof}
In the equation $\sum_{i=0}^d E^*_i = I$, apply $\rho^*$ to each side
and evaluate the result using Definition~\ref{def:Ai} along with Lemma~\ref{lem:rhoI}
applied to $\Phi^*$. 
\end{proof}


\begin{lemma} \label{lem:AiAsi} \samepage
The elements $\{A_i\}_{i=0}^d$ form a basis of the vector space $\mathcal M$.
\end{lemma}

\begin{proof}
By Lemmas~\ref{lem:rhorhospre}, \ref{lem:rhorhos}
and since $\{E^*_i\}_{i=0}^d$ form a basis of the vector space ${\mathcal M}^*$.
\end{proof}























\section{The scalars \texorpdfstring{$k_i$}{ki}}
\label{sec:ki}

We continue to discuss a symmetric idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on~$V$.
In this section we use $\Phi$ to define some scalars $k_i$ that will play a role 
in our theory.

\begin{defi} \label{def:ki} \samepage
%\ifDRAFT {\rm def:ki}. \fi
For $0 \leq i \leq d$ let $k_i$ denote the eigenvalue of $A_i$ 
corresponding to~$E_0$.
\end{defi}

\begin{lemma} \label{lem:AiE0} \samepage
%\ifDRAFT {\rm lem:AiE0}. \fi
For $0 \leq i \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma8.2_i} %[\rm (i)]
$A_i E_0 = E_0 A_i = k_i E_0$;
\item\label{lemma8.2_ii} %[\rm (ii)]
$A^*_i E^*_0 = E^*_0 A^*_i = k^*_i E^*_0$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma8.2_i}
By Definition~\ref{def:ki}.

\ref{lemma8.2_ii} 
Apply~\ref{lemma8.2_i} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:kimi} \samepage
%\ifDRAFT {\rm lem:kimi}. \fi
For $0 \leq i \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma8.3_i} %[\rm (i)]
$k_i = \nu m^*_i$;
\item\label{lemma8.3_ii} %[\rm (ii)]
$k^*_i = \nu m_i$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma8.3_i}
By Lemma~\ref{lem:AiEs0E0}\ref{lemma7.4_i},
$E_0 A_i E^*_0 E_0 = E_0 E^*_i E_0$.
In this equation, evaluate the left-hand side using Lemmas~\ref{lem:nuE0Es0E0}, \ref{lem:AiE0}\ref{lemma8.2_i},
and evaluate the right-hand side using Lemma~\ref{lem:E0EsiE0}\ref{lemma4.2_ii}.
This gives $k_i \nu^{-1} E_0 = m^*_i E_0$.
The result follows. 

\ref{lemma8.3_ii}
Apply~\ref{lemma8.3_i} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:ki} \samepage
%\ifDRAFT {\rm lem:ki}. \fi
The following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma8.4_i} %[\rm (i)]
$k_i \neq 0 \qquad (0 \leq i \leq d)$;
\item\label{lemma8.4_ii} %[\rm (ii)]
$\nu = \sum_{i=0}^d k_i$;
\item\label{lemma8.4_iii} %[\rm (iii)]
$k_0 = 1$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma8.4_i}
Apply Lemma~\ref{lem:mi3}\ref{lemma4.4_i} to $\Phi^*$ and use Lemma~\ref{lem:kimi}\ref{lemma8.3_i}.

\ref{lemma8.4_ii}
Apply Lemma~\ref{lem:mi3}\ref{lemma4.4_ii} to $\Phi^*$ and use Lemma~\ref{lem:kimi}\ref{lemma8.3_i}.

\ref{lemma8.4_iii}
By Definition~\ref{def:nu} and Lemma~\ref{lem:kimi}\ref{lemma8.3_i}.
\end{proof}

\section{Some reduction rules}
\label{sec:reduction}

We continue to discuss a symmetric idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on $V$.
In this section we obtain some reduction rules for $\Phi$.
Recall the antiautomorphism~$\dagger$ of $\End (V)$ from Definition~\ref{def:sym}.


\begin{lemma} \label{lem:EiEs0E0} \samepage
%\ifDRAFT {\rm lem:EiEs0E0}. \fi
For $0 \leq i \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma9.1_i} %[\rm (i)]
$E_i E^*_0 E_0 = \nu^{-1} A^*_i E_0$;
\item\label{lemma9.1_ii} %[\rm (ii)]
$E^*_i E_0 E^*_0 = \nu^{-1} A_i E^*_0$;
\item\label{lemma9.1_iii} %[\rm (iii)]
$E_0 E^*_0 E_i = \nu^{-1} E_0 A^*_i$;
\item\label{lemma9.1_iv} %[\rm (iv)]
$E^*_0 E_0 E^*_i = \nu^{-1} E^*_0 A_i$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma9.1_i}
Set $Y=E_i$ in~\eqref{eq:defrho} and use Lemma~\ref{lem:rhorhos}.

\ref{lemma9.1_ii} 
Apply~\ref{lemma9.1_i} to $\Phi^*$.

\ref{lemma9.1_iii}, \ref{lemma9.1_iv}
For the equations in~\ref{lemma9.1_i} and~\ref{lemma9.1_ii}, apply $\dagger$ to each side.
\end{proof}


\begin{lemma} \label{lem:EsjAiEs0} \samepage
%\ifDRAFT {\rm lem:EsjAiEs0}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma9.2_i} %[\rm (i)]
$E^*_j A_i E^*_0 = \delta_{i,j} A_i E^*_0$;
\item\label{lemma9.2_ii} %[\rm (ii)]
$E_j A^*_i E_0 = \delta_{i,j} A^*_i E_0$;
\item\label{lemma9.2_iii} %[\rm (iii)]
$E^*_0 A_i E^*_j = \delta_{i,j} E^*_0 A_i$;
\item\label{lemma9.2_iv} %[\rm (iv)]
$E_0 A^*_i E_j = \delta_{i,j} E_0 A^*_i$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma9.2_i}
For the equation in Lemma~\ref{lem:EiEs0E0}\ref{lemma9.1_ii},
multiply each side on the left by $E^*_j$ to get
$\delta_{i,j} E^*_i E_0 E^*_0 = \nu^{-1} E^*_j A_i E^*_0$.
In this equation, evaluate the left-hand side using Lemma~\ref{lem:EiEs0E0}\ref{lemma9.1_ii}.

\ref{lemma9.2_ii}
Apply~\ref{lemma9.2_i} to $\Phi^*$.


\ref{lemma9.2_iii}, \ref{lemma9.2_iv}
For the equations in~\ref{lemma9.2_i} and~\ref{lemma9.2_ii}, apply $\dagger$ to each side.
\end{proof}

\begin{lemma} \label{lem:E0EsjAiEs0} \samepage
%\ifDRAFT {\rm lem:E0EsjAiEs0}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma9.3_i} %[\rm (i)]
$E_0 E^*_j A_i E^*_0 = \delta_{i,j} k_i E_0 E^*_0$;
\item\label{lemma9.3_ii} %[\rm (ii)]
$E^*_0 E_j A^*_i E_0 = \delta_{i,j} k^*_i E^*_0 E_0$;
\item\label{lemma9.3_iii} %[\rm (iii)]
$E^*_0 A_i E^*_j E_0 = \delta_{i,j} k_i E^*_0 E_0$;
\item\label{lemma9.3_iv} %[\rm (iv)]
$E_0 A^*_i E_j E^*_0 = \delta_{i,j} k^*_i E_0 E^*_0$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma9.3_i}
Using Lemmas~\ref{lem:EsjAiEs0}\ref{lemma9.2_i}, \ref{lem:AiE0}\ref{lemma8.2_i} in order,
\[
 E_0 E^*_j A_i E_0 = \delta_{i,j} E_0 A_i E^*_0 = \delta_{i,j} k_i E_0 E^*_0.
\]

\ref{lemma9.3_ii}
Apply~\ref{lemma9.3_i} to $\Phi^*$.

\ref{lemma9.3_iii}, \ref{lemma9.3_iv}
For the equations in~\ref{lemma9.3_i} and~\ref{lemma9.3_ii}, apply $\dagger$ to each side.
\end{proof}


\begin{lemma} \label{lem:AiEs0Aj} \samepage
%\ifDRAFT {\rm lem:AiEs0Aj}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma9.4_i} %[\rm (i)]
$A_i E^*_0 A_j =\nu E^*_i E_0 E^*_j$;
\item\label{lemma9.4_ii} %[\rm (ii)]
$A^*_i E_0 A^*_j = \nu E_i E^*_0 E_j$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma9.3_i}
Using Lemmas~\ref{lem:EiEs0E0}\ref{lemma9.1_iv}, \ref{lem:AiEs0E0}\ref{lemma7.4_i} in order,
\[
 A_i E^*_0 A_j = \nu A_i E^*_0 E_0 E^*_j
 = \nu E^*_i E_0 E^*_j.
\]

\ref{lemma9.4_ii}
Apply~\ref{lemma9.4_i} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:EiEs0Aj} \samepage
%\ifDRAFT {\rm lem:EiEs0Aj}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma9.5_i} %[\rm (i)]
$E_i E^*_0 A_j = A^*_i E_0 E^*_j$;
\item\label{lemma9.5_ii} %[\rm (ii)]
$E^*_i E_0 A^*_j = A_i E^*_0 E_j$;
\item\label{lemma9.5_iii} %[\rm (iii)]
$A_j E^*_0 E_i = E^*_j E_0 A^*_i$;
\item\label{lemma9.5_iv} %[\rm (iv)]
$A^*_j E_0 E^*_i = E_j E^*_0 A_i$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma9.5_i}
Using Lemmas~\ref{lem:EiEs0E0}\ref{lemma9.1_iv}, \ref{lem:EiEs0E0}\ref{lemma9.1_i} in order,
\[
E_i E^*_0 A_j = \nu E_i E^*_0 E_0 E^*_j
 = A^*_i E_0 E^*_j.
\]

\ref{lemma9.5_ii}
Apply~\ref{lemma9.5_i} to $\Phi^*$.


\ref{lemma9.5_iii}, \ref{lemma9.5_iv}
For the equations in~\ref{lemma9.5_i} and~\ref{lemma9.5_ii}, apply $\dagger$ to each side.
\end{proof}


\section{The scalars \texorpdfstring{$p^h_{i j}$}{phij}, \texorpdfstring{$q^h_{i j}$}{qhij}}
\label{sec:phij}


We continue to discuss a symmetric idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on~$V$.


\begin{lemma} \label{lem:defp} \samepage
%\ifDRAFT {\rm lem:defp}. \fi
There exist scalars $p^h_{i j}$ $(0 \leq h,i,j \leq d)$ such that
\begin{align}
 A_i A_j &= \sum_{h=0}^d p^h_{i j} A_h && (0 \leq i,j \leq d). \label{eq:AiAj}
\end{align}
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:AiAsi}.
\end{proof}


\begin{defi} \label{def:intersectionnumber} \samepage
%\ifDRAFT {\rm def:intersectionnumber}. \fi
Referring to Lemma~\ref{lem:defp}, the scalars $p^h_{i j}$ 
are called the \emph{intersection numbers} of $\Phi$.
\end{defi}

\begin{defi} \label{def:qhij} \samepage
%\ifDRAFT {\rm def:qhij}. \fi
For $0 \leq h,i,j \leq d$ define $q^h_{i j} = (p^h_{i j})^*$.
We call these scalars the \emph{Krein parameters} of $\Phi$.
\end{defi}


\begin{lemma} \label{lem:AsiAsj} \samepage
%\ifDRAFT {\rm lem:AsiAsj}. \fi
For $0 \leq i,j \leq d$, 
\begin{align}
 A^*_i A^*_j &= \sum_{h=0}^d q^h_{i j} A^*_h. \label{eq:AsiAsj}
\end{align}
\end{lemma}

\begin{proof}
Apply Lemma~\ref{lem:defp} to $\Phi^*$ and use Definition~\ref{def:qhij}.
\end{proof}


\begin{lemma} \label{lem:phijphji} \samepage
%\ifDRAFT {\rm lem:phijphji}. \fi
For $0 \leq h,i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma10.5_i} %[\rm (i)]
$p^h_{i j} = p^h_{j i}$;
\item\label{lemma10.5_ii} %[\rm (ii)]
$q^h_{i j} = q^h_{j i}$.
\end{enumerate}
\end{lemma}


\begin{proof}
\ref{lemma10.5_i} 
By~\eqref{eq:AiAj} and since the algebra $\mathcal M$ is commutative.

\ref{lemma10.5_ii} Apply~\ref{lemma10.5_i} to $\Phi^*$.
\end{proof}


\begin{lemma} \label{lem:phi0} \samepage
%\ifDRAFT {\rm lem:phi0}. \fi
For $0 \leq h,i \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma10.6_i} %[\rm (i)]
$p^h_{i 0} = \delta_{h,i}$;
\item\label{lemma10.6_ii} %[\rm (ii)]
$p^h_{0 i} = \delta_{h,i}$;
\item\label{lemma10.6_iii} %[\rm (iii)]
$q^h_{i 0} = \delta_{h,i}$;
\item\label{lemma10.6_iv} %[\rm (iv)]
$q^h_{0 i} = \delta_{h,i}$.
\end{enumerate}
\end{lemma}


\begin{proof}
\ref{lemma10.6_i}
In~\eqref{eq:AiAj} set $j=0$ and use Lemmas~\ref{lem:A0As0}, \ref{lem:AiAsi}.

\ref{lemma10.6_ii} By~\ref{lemma10.6_i} and Lemma~\ref{lem:phijphji}\ref{lemma10.5_i}.

\ref{lemma10.6_iii}, \ref{lemma10.6_iv}
Apply~\ref{lemma10.6_i}, \ref{lemma10.6_ii} to $\Phi^*$.
\end{proof}


\begin{lemma} \label{lem:sumpthr} \samepage
%\ifDRAFT {\rm lem:sumpthr}. \fi
For $0 \leq h,i,j,t \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma10.7_i} %[\rm (i)]
$\sum_{r=0}^d p^t_{h r} p^r_{i j} = \sum_{s=0}^d p^s_{h i} p^t_{s j}$;
\item\label{lemma10.7_ii} %[\rm (ii)]
$\sum_{r=0}^d q^t_{h r} q^r_{i j} = \sum_{s=0}^d q^s_{h i} q^t_{s j}$.
\end{enumerate}
\end{lemma}


\begin{proof}
\ref{lemma10.7_i}
Expand $A_h (A_i A_j) = (A_h A_i) A_j$ in two ways using~\eqref{eq:AiAj},
and compare the coefficients using Lemma~\ref{lem:AiAsi}.

\ref{lemma10.7_ii} Apply~\ref{lemma10.7_i} to $\Phi^*$.
\end{proof}


\begin{lemma} \label{lem:sumphij} \samepage
%\ifDRAFT {\rm lem:sumphij}. \fi
For $0 \leq h,i \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma10.8_i} %[\rm (i)]
$k_i = \sum_{j=0}^d p^h_{i j}$;
\item\label{lemma10.8_ii} %[\rm (ii)]
$k^*_i = \sum_{j=0}^d q^h_{i j}$.
\end{enumerate}
\end{lemma}


\begin{proof}
\ref{lemma10.8_i}
Using Lemmas~\ref{lem:sumAi} and~\ref{lem:AiE0}\ref{lemma8.2_i},
\[
 A_i \sum_{j=0}^d A_j = k_i \sum_{h=0}^d A_h.
\]
By~\eqref{eq:AiAj},
\[
 A_i \sum_{j=0}^d A_j = \sum_{h=0}^d \sum_{j=0}^d p^h_{i j} A_h.
\]
Compare the above two equations using Lemma~\ref{lem:AiAsi}.

\ref{lemma10.8_ii}
Apply~\ref{lemma10.8_i} to $\Phi^*$.
\end{proof}
 

\begin{lemma} \label{lem:kip0ii} \samepage
%\ifDRAFT {\rm lem:kip0ii}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma10.9_i} %[\rm (i)]
$p^0_{i j} = \delta_{i,j} k_i$;
\item\label{lemma10.9_ii} %[\rm (ii)]
$q^0_{i j} = \delta_{i,j} k^*_i$.
\end{enumerate}
\end{lemma}


\begin{proof}
\ref{lemma10.9_i} 
For the equation~\eqref{eq:AiAj}, multiply each side on the left by $E_0 E^*_0$ and on the right
by $E^*_0 E_0$. 
Evaluate the result using Lemma~\ref{lem:AiEs0E0}\ref{lemma7.4_i},\ref{lemma7.4_iii}
along with Lemmas~\ref{lem:E0EsiE0}, \ref{lem:kimi}.

\ref{lemma10.9_ii}
Apply~\ref{lemma10.9_i} to $\Phi^*$.
\end{proof}


\begin{lemma} \label{lem:kikj} \samepage
%\ifDRAFT {\rm lem:kikj}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma10.10_i} %[\rm (i)]
$k_i k_j = \sum_{h=0}^d p^h_{i j} k_h$;
\item\label{lemma10.10_ii} %[\rm (ii)]
$k^*_i k^*_j = \sum_{h=0}^d q^h_{i j} k^*_h$.
\end{enumerate}
\end{lemma}


\begin{proof}
\ref{lemma10.10_i}
In~\eqref{eq:AiAj}, multiply each side by $E_0$, and simplify the result
using Lemma~\ref{lem:AiE0}\ref{lemma8.2_i}.

\ref{lemma10.10_ii}
Apply~\ref{lemma10.10_i} to $\Phi^*$.
\end{proof}


\begin{lemma} \label{lem:khphij} \samepage
%\ifDRAFT {\rm lem:khphij}. \fi
For $0 \leq h,i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma10.11_i} %[\rm (i)]
 $k_h p^h_{i j} = k_i p^i_{j h} = k_j p^j_{h i}$;
\item\label{lemma10.11_ii} %[\rm (ii)]
 $k^*_h q^h_{i j} = k^*_i q^i_{j h} = k^*_j q^j_{h i}$.
\end{enumerate}
\end{lemma}


\begin{proof}
\ref{lemma10.10_i}
In view of Lemma~\ref{lem:phijphji}\ref{lemma10.5_i},
it suffices to show that $k_h p^h_{i j} = k_j p^j_{h i}$.
To obtain this equation,
set $t=0$ in Lemma~\ref{lem:sumpthr}\ref{lemma10.7_i},
and evaluate the result using Lemma~\ref{lem:kip0ii}\ref{lemma10.9_i}.

\ref{lemma10.10_ii}
Apply~\ref{lemma10.10_i} to $\Phi^*$.
\end{proof}

\section{Reduction rules involving \texorpdfstring{$p^h_{i j}$}{phij}, \texorpdfstring{$q^h_{i j}$}{qhij}}
\label{sec:red2}

We continue to discuss a symmetric idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on $V$.
In this section we give some reduction rules for $\Phi$
that involve the intersection numbers and Krein parameters.

\begin{lemma} \label{lem:AjEsiE0} \samepage
%\ifDRAFT {\rm lem:AjEsiE0}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma11.1_i} %[\rm (i)]
$A_j E^*_i E_0 = \sum_{h=0}^d p^h_{i j} E^*_h E_0$;
\item\label{lemma11.1_ii} %[\rm (ii)]
$A^*_j E_i E^*_0 = \sum_{h=0}^d q^h_{i j} E_h E^*_0$;
\item\label{lemma11.1_iii} %[\rm (iii)]
$E_0 E^*_i A_j = \sum_{h=0}^d p^h_{i j} E_0 E^*_h$;
\item\label{lemma11.1_iv} %[\rm (iv)]
$E^*_0 E_i A^*_j = \sum_{h=0}^d q^h_{i j} E^*_0 E_h$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma11.1_i}
Using Lemmas~\ref{lem:AiEs0E0}\ref{lemma7.4_i}, \ref{lem:defp}, \ref{lem:AiEs0E0}\ref{lemma7.4_i} in order,
\[
 A_j E^*_i E_0 = A_j A_i E^*_0 E_0
 = \sum_{h=0}^d p^h_{i j} A_h E^*_0 E_0
 = \sum_{h=0}^d p^h_{i j} E^*_h E_0.
\]

\ref{lemma11.1_ii} 
Apply~\ref{lemma11.1_i} to $\Phi^*$.

\ref{lemma11.1_iii}, \ref{lemma11.1_iv}
For the equations in~\ref{lemma11.1_i} and~\ref{lemma11.1_ii}, apply $\dagger$ to each side.
\end{proof}


\begin{lemma} \label{lem:E0EsiAjEsh} \samepage
%\ifDRAFT {\rm lem:E0EsiAjEsh}. \fi
For $0 \leq h,i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma11.2_i} %[\rm (i)]
$E^*_h A_j E^*_i E_0 = p^h_{i j} E^*_h E_0$;
\item\label{lemma11.2_ii} %[\rm (ii)]
$E_h A^*_j E_i E^*_0 = q^h_{i j} E_h E^*_0$;
\item\label{lemma11.2_iii} %[\rm (iii)]
$E_0 E^*_i A_j E^*_h = p^h_{i j} E_0 E^*_h$;
\item\label{lemma11.2_iv} %[\rm (iv)]
$E^*_0 E_i A^*_j E_h = q^h_{i j} E^*_0 E_h$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma11.2_i}
Using Lemma~\ref{lem:AjEsiE0}\ref{lemma11.1_i},
\[
E^*_h A_j E^*_i E_0
 = \sum_{s=0}^d p^s_{i j} E^*_h E^*_s E_0
 = p^h_{i j} E^*_h E_0.
\]

\ref{lemma11.2_ii}
Apply~\ref{lemma11.2_i} to $\Phi^*$.

\ref{lemma11.2_iii}, \ref{lemma11.2_iv}
For the equations in~\ref{lemma11.2_i} and~\ref{lemma11.2_ii}, apply $\dagger$ to each side.
\end{proof}


\begin{lemma} \label{lem:EsiAjEsh} \samepage
%\ifDRAFT {\rm lem:EsiAjEsh}. \fi
For $0 \leq h,i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma11.3_i} %[\rm (i)]
$E_i A^*_j E_h = m_i^{-1} q^h_{i j} E_i E^*_0 E_h$;
\item\label{lemma11.3_ii} %[\rm (ii)]
$E^*_i A_j E^*_h = (m^*_i)^{-1} p^h_{i j} E^*_i E_0 E^*_h$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma11.3_i}
In Lemma~\ref{lem:E0EsiAjEsh}\ref{lemma11.2_iv}, multiply each side on the left by $E_i$.
Simplify the result using Lemma~\ref{lem:EsiE0Esi}\ref{lemma4.3_i}.

\ref{lemma11.3_ii}
Apply~\ref{lemma11.3_i} to $\Phi^*$.
\end{proof}


\begin{lemma} \label{lem:EsiAjEsh2} \samepage
%\ifDRAFT {\rm lem:EsiAjEsh2}. \fi
For $0 \leq h,i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma11.4_i} %[\rm (i)]
$E^*_i A_j E^*_h = 0 \;\;$ if and only if $\;\; p^h_{i j} = 0$;
\item\label{lemma11.4_ii} %[\rm (ii)]
$E_i A^*_j E_h = 0 \;\;$ if and only if $\;\; q^h_{i j} = 0$.
\end{enumerate}
\end{lemma}

\begin{proof}
By Lemmas~\ref{lem:EsiE0Esj} and~\ref{lem:EsiAjEsh}.
\end{proof}



\begin{lemma} \label{lem:phij} \samepage
%\ifDRAFT {\rm lem:phij}. \fi
For $0 \leq h,i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma11.5_i} %[\rm (i)]
$p^h_{i j} = (m^*_h)^{-1} \tr (E_0 E^*_i A_j E^*_h)$;
\item\label{lemma11.5_ii} %[\rm (ii)]
$q^h_{i j} = m_h^{-1} \tr (E^*_0 E_i A^*_j E_h)$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma11.5_i}
In Lemma~\ref{lem:E0EsiAjEsh}\ref{lemma11.2_iii}, take the trace of each side,
and simplify the result using Definition~\ref{def:mi2}.

\ref{lemma11.5_ii}
Apply~\ref{lemma11.5_i} to $\Phi^*$.
\end{proof}


















\section{The scalars \texorpdfstring{$p_i(j)$}{pi(j)}, \texorpdfstring{$q_i(j)$}{qi(j)}}
\label{sec:pij}


We continue to discuss a symmetric idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on $V$.
In this section we use $\Phi$ to define some scalars $p_i (j)$, $q_i (j)$
that will play a role in our theory.
Recall the algebra $\mathcal M$ from Definition~\ref{def:D}.


\begin{lemma} \label{lem:pij} \samepage
%\ifDRAFT {\rm lem:pij}. \fi
There exist scalars $p_i(j)$ $(0 \leq i,j \leq d)$ such that
\begin{align}
A_i &= \sum_{j=0}^d p_i(j) E_j && (0 \leq i \leq d). \label{eq:Ai}
\end{align}
\end{lemma}


\begin{proof}
By Definition~\ref{def:D} the elements
$\{E_i\}_{i=0}^d$ form a basis of $\mathcal M$.
By Definition~\ref{def:Ai}, $A_i \in {\mathcal M}$ for $0 \leq i \leq d$.
The result follows.
\end{proof}




\begin{defi} \label{def:qij} \samepage
%\ifDRAFT {\rm def:qij}. \fi
For $0 \leq i,j \leq d$ define $q_i(j) = (p_i(j))^*$.
\end{defi}

\begin{lemma} \label{lem:Asi} \samepage
%\ifDRAFT {\rm lem:Asi}. \fi
For $0 \leq i,j \leq d$,
\begin{align}
 A^*_i &= \sum_{j=0}^d q_i (j) E^*_j && (0 \leq i \leq d). \label{eq:Asi}
\end{align}
\end{lemma}

\begin{proof}
Apply Lemma~\ref{lem:pij} to $\Phi^*$ and use Definition~\ref{def:qij}.
\end{proof}

\begin{lemma} \label{lem:AiEj} \samepage
%\ifDRAFT {\rm lem:AiEj}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma12.4_i} %[\rm (i)]
$A_i E_j = E_j A_i = p_i (j) E_j$;
\item\label{lemma12.4_ii} %[\rm (ii)]
$A^*_i E^*_j = E^*_j A^*_i = q_i(j) E^*_j$.
\end{enumerate}
In other words, $p_i (j)$ (resp.\ $q_i (j)$) is the eigenvalue of $A_i$
(resp.\ $A^*_i$) associated with $E_j V$ (resp.\ $E^*_j V$).
\end{lemma}

\begin{proof}
\ref{lemma12.4_i}
Use~\eqref{eq:Ai}.

\ref{lemma12.4_ii}
Apply~\ref{lemma12.4_i} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:EiEsi} \samepage
%\ifDRAFT {\rm lem:EiEsi}. \fi
For $0 \leq i \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma12.5_i} %[\rm (i)]
$E^*_i = \nu^{-1} \sum_{j=0}^d p_i (j) A^*_j$;
\item\label{lemma12.5_ii} %[\rm (ii)]
$E_i = \nu^{-1} \sum_{j=0}^d q_i(j) A_j$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma12.5_i}
In~\eqref{eq:Ai}, apply $\rho$ to each side and use
Lemma~\ref{lem:rhorhos}.

\ref{lemma12.5_ii}
Apply~\ref{lemma12.5_i} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:pihqhj} \samepage
%\ifDRAFT {\rm lem:pihqhj}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma12.6_i} %[\rm (i)]
$\sum_{h=0}^d p_i (h) q_h (j) = \delta_{i,j} \nu$;
\item\label{lemma12.6_ii} %[\rm (ii)]
$\sum_{h=0}^d q_i (h) p_h (j) = \delta_{i,j} \nu$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma12.6_i}
By~\eqref{eq:Ai}, $A_i = \sum_{h=0}^d p_i (h) E_h$.
In this equation, eliminate $E_h$ using Lemma~\ref{lem:EiEsi}\ref{lemma12.5_ii},
and compare the coefficients of each side.

\ref{lemma12.6_ii}
Apply~\ref{lemma12.6_i} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:p0j} \samepage
%\ifDRAFT {\rm lem:p0j}. \fi
For $0 \leq j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma12.7_i} %[\rm (i)]
$p_0 (j) = 1$;
\item\label{lemma12.7_ii} %[\rm (ii)]
$q_0 (j) = 1$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma12.7_i}
Set $i=0$ in~\eqref{eq:Ai} and recall that $A_0 = I$.

\ref{lemma12.7_ii} Apply~\ref{lemma12.7_i} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:pi0} \samepage
%\ifDRAFT {\rm lem:pi0}. \fi
For $0 \leq i \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma12.8_i} %[\rm (i)]
$p_i (0) = k_i$;
\item\label{lemma12.8_ii} %[\rm (ii)]
$q_i (0) = k^*_i$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma12.8_i}
Set $j=0$ in Lemma~\ref{lem:AiEj}\ref{lemma12.4_i} 
and compare the result with Lemma~\ref{lem:AiE0}\ref{lemma8.2_i}.

\ref{lemma12.8_ii}
Apply~\ref{lemma12.8_i} to $\Phi^*$.
\end{proof}



\begin{lemma} \label{lem:sumphj} \samepage
%\ifDRAFT {\rm lem:sumphj}. \fi
For $0 \leq j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma12.9_i} %[\rm (i)]
$\sum_{h=0}^d p_h (j) = \delta_{0, j} \nu$;
\item\label{lemma12.9_ii} %[\rm (ii)]
$\sum_{h=0}^d q_h (j) = \delta_{0,j} \nu$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma12.9_i}
Set $i=0$ in Lemma~\ref{lem:pihqhj}\ref{lemma12.6_ii}, and evaluate the result
using Lemma~\ref{lem:p0j}\ref{lemma12.7_ii}.

\ref{lemma12.9_ii}
Apply~\ref{lemma12.9_i} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:kshpih} \samepage
%\ifDRAFT {\rm lem:kshpih}. \fi
For $0 \leq i \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma12.10_i} %[\rm (i)]
$\sum_{h=0}^d m_h p_i (h) = \delta_{i,0}$;
\item\label{lemma12.10_ii} %[\rm (ii)]
$\sum_{h=0}^d m^*_h q_i (h) = \delta_{i,0}$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma12.10_i} 
Set $j=0$ in Lemma~\ref{lem:pihqhj}\ref{lemma12.6_i}, and evaluate the result
using Lemmas~\ref{lem:kimi}\ref{lemma8.3_ii}, \ref{lem:pi0}\ref{lemma12.8_ii}.

\ref{lemma12.6_ii}
Apply~\ref{lemma12.6_i} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:pirpjr} \samepage
%\ifDRAFT {\rm lem:pirpjr}. \fi
For $0 \leq i,j,r \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma12.11_i} %[\rm (i)]
$p_i (r) p_j (r) = \sum_{h=0}^d p^h_{i j} p_h (r)$;
\item\label{lemma12.11_ii} %[\rm (ii)]
$q_i (r) q_j (r) = \sum_{h=0}^d q^h_{i j} q_h (r)$.
\end{enumerate}
\end{lemma}

\begin{proof}
$\Mk$\ref{lemma12.11_i}
In~\eqref{eq:AiAj}, multiply each side by $E_r$,
and simplify the result using Lemma~\ref{lem:AiEj}\ref{lemma12.4_i}.

\ref{lemma12.11_ii}
Apply~\ref{lemma12.11_i} to $\Phi^*$.
\end{proof}



\begin{lemma} \label{lem:phijsum} \samepage
%\ifDRAFT {\rm lem:phijsum}. \fi
For $0 \leq h,i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma12.12_i} %[\rm (i)]
$p^h_{i j} = \nu^{-1} \sum_{r=0}^d p_i (r) p_j (r) q_r (h)$;
\item\label{lemma12.12_ii} %[\rm (ii)]
$q^h_{i j} = \nu^{-1} \sum_{r=0}^d q_i (r) q_j (r) p_r (h)$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma12.12_i}
Expand the sum $\sum_{r=0}^d p_i (r) p_j (r) q_r (h)$
using Lemma~\ref{lem:pirpjr}\ref{lemma12.11_i},
and simplify the result using Lemma~\ref{lem:pihqhj}\ref{lemma12.6_i}.

\ref{lemma12.12_ii}
Apply~\ref{lemma12.12_i} to $\Phi^*$.
\end{proof}

\section{Reduction rules involving \texorpdfstring{$p_i (j)$}{pi(j)}, \texorpdfstring{$q_i (j)$}{qi(j)}}
\label{sec:red3}

We continue to discuss a symmetric idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on~$V$.


\begin{lemma} \label{lem:E0AsiAj} \samepage
%\ifDRAFT {\rm lem:E0AsiAj}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma13.1_i} %[\rm (i)]
$E_0 A^*_i A_j = p_j (i) E_0 A^*_i$;
\item\label{lemma13.1_ii} %[\rm (ii)]
$E^*_0 A_i A^*_j = q_j (i) E^*_0 A_i$;
\item\label{lemma13.1_iii} %[\rm (iii)]
$A_j A^*_i E_0 = p_j (i) A^*_i E_0$;
\item\label{lemma13.1_iv} %[\rm (iv)]
$A^*_j A_i E^*_0 = q_j (i) A_i E^*_0$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma13.1_i}
Using Lemmas~\ref{lem:pij} and~\ref{lem:EsjAiEs0}\ref{lemma9.2_iv} in order,
\[
E_0 A^*_i A_j
= \sum_{h=0}^d p_j (h) E_0 A^*_i E_h
= \sum_{h=0}^d p_j (h) \delta_{i,h} E_0 A^*_i
= p_j (i) E_0 A^*_i.
\]

\ref{lemma13.1_ii}
Apply~\ref{lemma13.1_i} to $\Phi^*$.

\ref{lemma13.1_iii}, \ref{lemma13.1_iv}
For the equations in~\ref{lemma13.1_i} and~\ref{lemma13.1_ii}, apply $\dagger$ to each side.
\end{proof}

\begin{lemma} \label{lem:E0EsiEj} \samepage
%\ifDRAFT {\rm lem:E0EsiEj}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma13.2_i} %[\rm (i)]
$E_0 E^*_i E_j = \nu^{-1} p_i (j) E_0 A^*_j$;
\item\label{lemma13.2_ii} %[\rm (ii)]
$E^*_0 E_i E^*_j = \nu^{-1} q_i (j) E^*_0 A_j$;
\item\label{lemma13.2_iii} %[\rm (iii)]
$E_j E^*_i E_0 = \nu^{-1} p_i (j) A^*_j E_0$;
\item\label{lemma13.2_iv} %[\rm (iv)]
$E^*_j E_i E^*_0 = \nu^{-1} q_i (j) A_j E^*_0$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma13.2_i}
Using Lemmas~\ref{lem:EiEsi}\ref{lemma12.5_i} and~\ref{lem:EsjAiEs0}\ref{lemma9.2_iv} in order,
\[
E_0 E^*_i E_j
 = E_0 \left( \nu^{-1} \sum_{h=0}^d p_i (h) A^*_h \right) E_j
 = \nu^{-1} \sum_{h=0}^d p_i (h) \delta_{h,j} E_0 A^*_h
 = \nu^{-1} p_i (j) E_0 A^*_j.
\]

\ref{lemma13.2_ii}
Apply~\ref{lemma13.2_i} to $\Phi^*$.

\ref{lemma13.2_iii}, \ref{lemma13.2_iv}
For the equations in~\ref{lemma13.2_i} and~\ref{lemma13.2_ii}, apply $\dagger$ to each side.
\end{proof}

\begin{lemma} \label{lem:E0AsiAjEs0} \samepage
%\ifDRAFT {\rm lem:E0AsiAjEs0}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma13.3_i} %[\rm (i)]
$E_0 A^*_i A_j E^*_0 = p_j (i) k^*_i E_0 E^*_0$;
\item\label{lemma13.3_ii} %[\rm (ii)]
$E^*_0 A_i A^*_j E_0 = q_j (i) k_i E^*_0 E_0$;
\item\label{lemma13.3_iii} %[\rm (iii)]
$E^*_0 A_i A^*_j E_0 = p_i (j) k^*_j E^*_0 E_0$;
\item\label{lemma13.3_iv} %[\rm (iv)]
$E_0 A^*_i A_j E^*_0 = q_i (j) k_j E_0 E^*_0$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma13.3_i}
Using Lemmas~\ref{lem:E0AsiAj}\ref{lemma13.1_i}, \ref{lem:AiEj}\ref{lemma12.4_ii}, \ref{lem:pi0}\ref{lemma12.8_ii} in order,
\[
E_0 A^*_i A_j E^*_0
= p_j (i) E_0 A^*_i E^*_0
= p_j (i) q_i (0) E_0 E^*_0
= p_j (i) k^*_i E_0 E^*_0.
\]

\ref{lemma13.3_ii}
Apply~\ref{lemma13.3_i} to $\Phi^*$.

\ref{lemma13.3_iii}, \ref{lemma13.3_iv}
For the equations in~\ref{lemma13.3_i} and~\ref{lemma13.2_ii}, apply $\dagger$ to each side.
\end{proof}

\begin{lemma} \label{lem:E0EsiEjEs0} \samepage
%\ifDRAFT {\rm lem:E0EsiEjEs0}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma13.4_i} %[\rm (i)]
$E_0 E^*_i E_j E^*_0 = p_i (j) m_j E_0 E^*_0$;
\item\label{lemma13.4_ii} %[\rm (ii)]
$E^*_0 E_i E^*_j E_0 = q_i (j) m^*_j E^*_0 E_0$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma13.4_i}
Using Lemmas~\ref{lem:E0EsiEj}\ref{lemma13.2_i}, \ref{lem:AiEj}\ref{lemma12.4_ii}, \ref{lem:pi0}\ref{lemma12.8_ii} in order,
\[
E_0 E^*_i E_j E^*_0
= \nu^{-1} p_i (j) E_0 A^*_j E^*_0
= \nu^{-1} p_i (j) q_j (0) E_0 E^*_0
= \nu^{-1} p_i (j) k^*_j E_0 E^*_0.
\]
Now use Lemma~\ref{lem:kimi}\ref{lemma8.3_ii}.

\ref{lemma13.4_ii}
Apply~\ref{lemma13.4_i} to $\Phi^*$.
\end{proof}

\goodbreak

\begin{lemma} \label{lem:pijksj} \samepage
%\ifDRAFT {\rm lem:pijksj}. \fi
For $0 \leq i,j \leq d$,
\begin{equation}
 \frac{ p_i (j) } { k_i } = \frac{ q_j (i) } {k^*_j }. \label{eq:pijksj}
\end{equation}
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:E0AsiAjEs0}\ref{lemma13.3_ii},\ref{lemma13.3_iii},
$p_i (j) k^*_j E^*_0 E_0 = q_j (i) k_i E^*_0 E_0$.
The result follows since $E^*_0 E_0 \neq 0$ by Definition~\ref{def:ips}\ref{defi3.1_iii}.
\end{proof}

\begin{lemma} \label{lem:pij2} \samepage
%\ifDRAFT {\rm lem:pij2}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma13.6_i} %[\rm (i)]
$p_i (j) = \nu m_j^{-1} \, \tr (E_0 E^*_i E_j E^*_0)$;
\item\label{lemma13.6_ii} %[\rm (ii)]
$p_i (j) = \nu m_j^{-1} \, \tr (E^*_0 E_j E^*_i E_0)$;
\item\label{lemma13.6_iii} %[\rm (iii)]
$q_i (j) = \nu (m^*_j)^{-1} \, \tr (E^*_0 E_i E^*_j E_0)$;
\item\label{lemma13.6_iv} %[\rm (iv)]
$q_i (j) = \nu (m^*_j)^{-1} \, \tr (E_0 E^*_j E_i E^*_0)$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma13.6_i}
Using Lemma~\ref{lem:E0EsiEjEs0}\ref{lemma13.4_i} and Definition~\ref{def:nu},
\[
 \tr (E_0 E^*_i E_j E^*_0)
 = p_i (j) m_j \tr (E_0 E^*_0)
 = \nu^{-1} p_i (j) m_j.
\]

\ref{lemma13.6_iii}
Apply~\ref{lemma13.6_i} to $\Phi^*$.

\ref{lemma13.6_ii}
In~\ref{lemma13.6_iii}, exchange $i$, $j$, and use Lemmas~\ref{lem:kimi}, \ref{lem:pijksj}.

\ref{lemma13.6_iv}
Apply~\ref{lemma13.6_ii} to $\Phi^*$.
\end{proof}

\section{Some matrices}
\label{sec:matrices}


We continue to discuss a symmetric idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on $V$.
In the previous sections we used $\Phi$ to define several kinds of scalars,
and we described how these scalars are related.
In this section we express these relationships in matrix form.

\begin{defi} \label{def:matrices} \samepage
%\ifDRAFT {\rm def:matrices}. \fi
Let $K$ (resp. $K^*$) denote the diagonal matrix in $\Mat_{d+1}(\F)$
that has $(i,i)$-entry $k_i$ (resp.\ $k^*_i$) for $0 \leq i \leq d$.
Let $P$ (resp.\ $Q$) denote the matrix in $\Mat_{d+1}(\F)$ that has
$(i,j)$-entry $p_j (i)$ (resp.\ $q_j (i)$) for $0 \leq i,j \leq d$.
\end{defi}

\begin{lemma} \label{lem:PtKs} \samepage
%\ifDRAFT {\rm lem:PtKs}. \fi
The following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma14.2_i} %[\rm (i)]
$P Q = Q P = \nu I$;
\item\label{lemma14.2_ii} %[\rm (ii)]
$P^{\sf t} K^* = K Q$;
\item\label{lemma14.2_iii} %[\rm (iii)]
$K^* P = Q^{\sf t} K$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma14.2_i}
By Lemma~\ref{lem:pihqhj}.

\ref{lemma14.2_ii} 
By Lemma~\ref{lem:pijksj}.

\ref{lemma14.2_iii}
In~\ref{lemma14.2_ii}, take the transpose of each side. 
\end{proof}

\begin{defi} \label{def:U} \samepage
%\ifDRAFT {\rm def:U}. \fi
Note by Lemma~\ref{lem:PtKs} that
$K^{-1} P^{\sf t} = Q (K^*)^{-1}$ and
$(K^*)^{-1} Q^{\sf t} = P K^{-1}$;
we define
\begin{align}
 U &= K^{-1} P^{\sf t} = Q (K^*)^{-1}, &
 U^* &= (K^*)^{-1} Q^{\sf t} = P K^{-1}. \label{eq:defU}
\end{align}
\end{defi}

\begin{lemma} \label{lem:U3} \samepage
%\ifDRAFT {\rm lem:U3}. \fi
The following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma14.4_i} %[\rm (i)]
$P = U^* K$;
\item\label{lemma14.4_ii} %[\rm (ii)]
$P^{\sf t} = K U$;
\item\label{lemma14.4_iii} %[\rm (iii)]
$Q = U K^*$;
\item\label{lemma14.4_iv} %[\rm (iv)]
$Q^{\sf t} = K^* U^*$.
\end{enumerate}
\end{lemma}

\begin{proof}
Immediate from Definition~\ref{def:U}.
\end{proof}

\goodbreak
\begin{lemma} \label{lem:U0j} \samepage
%\ifDRAFT {\rm lem:U0j}. \fi
We have $U_{i,0}=1$ and $U^*_{i,0} =1$ for $0 \leq i \leq d$.
Moreover $U_{0,j} = 1$ and $U^*_{0,j} = 1$ for $0 \leq j \leq d$.
\end{lemma}

\begin{proof}
Use Lemmas~\ref{lem:ki}\ref{lemma8.4_iii}, \ref{lem:p0j}, \ref{lem:pi0}.
\end{proof}


\begin{lemma} \label{lem:U2} \samepage
%\ifDRAFT {\rm lem:U2}. \fi
The following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma14.6_i} %[\rm (i)]
$U^{\sf t} = U^*$;
\item\label{lemma14.6_ii} %[\rm (ii)]
$U K^* U^* K = \nu I$;
\item\label{lemma14.6_iii} %[\rm (iii)]
$U^* K U K^* = \nu I$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma14.6_i}
By Definition~\ref{def:U}.

\ref{lemma14.6_ii}, \ref{lemma14.6_iii}
By Lemma~\ref{lem:U3}\ref{lemma14.4_i},\ref{lemma14.4_iii} and Lemma~\ref{lem:PtKs}\ref{lemma14.2_i}.
\end{proof}

\begin{defi} \label{def:matrices2} \samepage
%\ifDRAFT {\rm def:matrices2}. \fi
For $0 \leq i \leq d$ let $B_i$ and $B^*_i$ denote the matrices in $\Mat_{d+1}(\F)$
that have entries
\begin{align*}
 (B_i)_{h,j} &= p^h_{i j}, &
 (B^*_i)_{h,j} &= q^h_{i j} && (0 \leq h,j \leq d).
\end{align*}
We call $B_i$ (resp.\ $B^*_i$) the \emph{$i^\text{th}$ intersection matrix}
(resp.\ \emph{$i^\text{th}$ dual intersection matrix}) of $\Phi$.
\end{defi}

\begin{defi} \label{def:matrices3} \samepage
%\ifDRAFT {\rm def:matrices3}. \fi
For $0 \leq i \leq d$ let $H_i$ and $H^*_i$ denote the diagonal matrices in $\Mat_{d+1}(\F)$ 
that have diagonal entries
\begin{align*}
 (H_i)_{j,j} &= p_i (j), &
 (H^*_i)_{j,j} &= q_i (j) && (0 \leq j \leq d).
\end{align*}
\end{defi}


\begin{lemma} \label{lem:BDPQ} \samepage
%\ifDRAFT {\rm lem:BDPQ}. \fi
For $0 \leq r \leq d$,
\begin{align}
H_r P &= P B_r, & H^*_r Q &= Q B^*_r, \label{eq:DrP}
\\
Q H_r &= B_r Q, & P H^*_r &= B^*_r P, \label{eq:QDr}
\\
K B_r &= (B_r)^{\sf t} K, & K^* B^*_r &= (B^*_r)^{\sf t} K^*, \label{eq:KBr}
\\
U H_r &= B_r U, & U^* H^*_r &= B^*_r U^*. \label{eq:UDr}
\end{align}
\end{lemma}

\begin{proof}
To get the equation on the left in~\eqref{eq:DrP},
compare the entries of each side using Lemma~\ref{lem:pirpjr}\ref{lemma12.11_i}.
In the equation on the left in~\eqref{eq:DrP}, multiply each side on the left and on the right by $Q$
and simplify the result using Lemma~\ref{lem:PtKs}\ref{lemma14.2_i}. 
This gives the equation on the left in~\eqref{eq:QDr}.
To obtain the equation on the left in~\eqref{eq:KBr}, 
compare the entries of each side using Lemma~\ref{lem:khphij}\ref{lemma10.11_i}.
The equation on the left in~\eqref{eq:UDr} follows from $Q H_r = B_r Q$ and
Lemma~\ref{lem:U3}\ref{lemma14.4_iii} together with the fact that $H_r$, $K^*$ commute
since they are both diagonal.
To get the equations on the right in~\eqref{eq:DrP}--\eqref{eq:UDr},
apply the equations on the left in~\eqref{eq:DrP}--\eqref{eq:UDr} to $\Phi^*$.
\end{proof}

\begin{lemma} \label{lem:BrBs} \samepage
%\ifDRAFT {\rm lem:BrBs}. \fi
For $0 \leq i,j \leq d$ the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma14.10_i} %[\rm (i)]
$B_i B_j = \sum_{h=0}^d p^h_{i j} B_h$;
\item\label{lemma14.10_ii} %[\rm (ii)]
$B^*_i B^*_j = \sum_{h=0}^d q^h_{i j} B^*_h$;
\item\label{lemma14.10_iii} %[\rm (iii)]
$H_i H_j = \sum_{h=0}^d p^h_{i j} H_h$;
\item\label{lemma14.10_iv} %[\rm (iv)]
$H^*_i H^*_j = \sum_{h=0}^d q^h_{i j} H^*_h$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma14.10_i}, \ref{lemma14.10_ii}
By Lemma~\ref{lem:sumpthr}.

\ref{lemma14.10_iii}, \ref{lemma14.10_iv}
By Lemma~\ref{lem:pirpjr}.
\end{proof}

\section{The \texorpdfstring{$\Phi$}{Phi}-standard basis}
\label{sec:standard}

We continue to discuss a symmetric idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on~$V$.
In this section we introduce the notion of a $\Phi$-standard basis.


\goodbreak
\begin{lemma} \label{lem:EsiE0V} \samepage
%\ifDRAFT {\rm lem:EsiE0V}. \fi
For $0 \leq i \leq d$, $E^*_i V = E^*_i E_0 V$.
\end{lemma}

\begin{proof}
The vector space $E^*_i V$ has dimension $1$ and contains $E^*_i E_0 V$.
By Definition~\ref{def:ips}\ref{defi3.1_iii}, $E^*_i E_0 V \neq 0$.
The result follows.
\end{proof}

\begin{lemma} \label{lem:Esiu} \samepage
%\ifDRAFT {\rm lem:Esiu}. \fi
\looseness-1
Let $\xi$ denote a nonzero vector in $E_0 V$.
Then for $0 \leq i \leq d$ the vector $E^*_i \xi$ is nonzero and hence
a basis of $E^*_i V$.
Moreover the vectors $\{E^*_i \xi\}_{i=0}^d$ form a basis of $V$.
\end{lemma}

\begin{proof}
Let the integer $i$ be given.
We show $E^*_i \xi \neq 0$.
The vector space $E_0 V$ has dimension $1$ and $\xi$ is a nonzero vector in $E_0 V$, so
$\xi$ spans $E_0 V$.
Therefore $E^*_i \xi$ spans $E^*_i E_0 V$.
The vector space $E^*_i E_0 V$ has dimension $1$ by Lemma~\ref{lem:EsiE0V}
so $E^*_i \xi$ is nonzero.
The remaining assertions are clear.
\end{proof}

\begin{defi} \label{def:standard} \samepage
%\ifDRAFT {\rm def:standard}. \fi
By a \emph{$\Phi$-standard basis} of $V$ we mean a sequence
$\{E^*_i \xi\}_{i=0}^d$, where~$\xi$ is a nonzero vector in $E_0 V$.
\end{defi}

We give a characterization of a $\Phi$-standard basis.

\begin{lemma} \label{lem:standard} \samepage
%\ifDRAFT {\rm lem:standard}. \fi
Let $\{u_i\}_{i=0}^d$ denote a sequence of vectors in $V$, not all $0$.
Then this sequence is a $\Phi$-standard basis if and only if both~\ref{lemma15.4_i}, \ref{lemma15.4_ii}~hold below:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma15.4_i} %[\rm (i)]
$u_i \in E^*_i V$ for $0 \leq i \leq d$;
\item\label{lemma15.4_ii} %[\rm (ii)]
$\sum_{i=0}^d u_i \in E_0 V$.
\end{enumerate}
\end{lemma}

\begin{proof}
To prove the lemma in one direction,
assume that $\{u_i\}_{i=0}^d$ is a $\Phi$-standard basis of $V$.
By Definition~\ref{def:standard} there exists a nonzero $\xi \in E_0 V$
such that $u_i = E^*_i \xi$ for $0 \leq i \leq d$.
By construction $u_i \in E^*_i V$ for $0 \leq i \leq d$, so~\ref{lemma15.4_i} holds.
Recall $I = \sum_{i=0}^d E^*_i$.
In this equation we apply each side to $\xi$, to find that $\xi = \sum_{i=0}^d u_i$,
and~\ref{lemma15.4_ii} follows.
We have now proved the lemma in one direction.
To prove the lemma in the other direction,
assume that $\{u_i\}_{i=0}^d$ satisfy~\ref{lemma15.4_i} and~\ref{lemma15.4_ii}.
Define $\xi = \sum_{i=0}^d u_i$ and observe $\xi \in E_0 V$.
Using~\ref{lemma15.4_i} we find that $E^*_i u_j = \delta_{i,j} u_i$ for $0 \leq i,j \leq d$.
It follows $u_i = E^*_i \xi$ for $0 \leq i \leq d$.
Observe $\xi \neq 0$ since at least one of $\{u_i\}_{i=0}^d$ is nonzero.
Now $\{u_i\}_{i=0}^d$ is a $\Phi$-standard basis of $V$ by Definition~\ref{def:standard}.
\end{proof}

\section{Bilinear forms} 
\label{sec:bilin}

In this section we recall some basic facts concerning bilinear forms on $V$.
See~\cite[Section~8.5]{Rot} for more information.
By a \emph{bilinear form on $V$} we mean a map $\b{\; , \; } : V \times V \to \F$
that satisfies the following four conditions for $u,v,w \in V$ and $\alpha \in \F$:
(i)~$\b{u+v,w} = \b{u,w} + \b{v,w}$;
(ii)~$\b{\alpha u, v} = \alpha \b{u,v}$;
(iii)~$\b{u,v+w} = \b{u,v} + \b{u,w}$;
(iv)~$\b{u,\alpha v} = \alpha \b{u,v}$.
Let $\b{ \; , \; }$ denote a bilinear form on $V$.
We abbreviate $||v||^2 = \b{ v,v }$ for $v \in V$.
The following are equivalent:
(i)~there exists a nonzero $u \in V$ such that
$\b{ u, v} = 0$ for all $v \in V$;
(ii)~there exists a nonzero $v \in V$ such that
$\b{ u,v } = 0$ for all $u \in V$.
The form $\b{ \; , \; }$ is said to be \emph{degenerate}
whenever (i), (ii) hold and \emph{nondegenerate} otherwise.

We recall from~\cite[Theorem~1.1]{F} or~\cite[Ch.~1, Theorem.~4.2]{KMRT}
how bilinear forms on $V$ are related to antiautomorphisms of $\End (V)$.
Let $\gamma$ denote an antiautomorphism of $\End (V)$.
Then there exists a nonzero bilinear form $\b{\; , \;}$ on $V$ such that
$\bbig{A u, v} = \bbig{u, A^\gamma v}$ for $u,v \in V$ and $A \in \End (V)$.
The form is unique up to multiplication by a nonzero scalar.
The form is nondegenerate.
We refer to this form as a \emph{bilinear form on $V$ associated with $\gamma$}.


For the rest of this section let $\b{ \; , \; }$ denote a nondegenerate bilinear form on $V$.

\begin{defi} \label{def:innermatrix} \samepage
%\ifDRAFT {\rm def:innermatrix}. \fi
For bases $\{u_i\}_{i=0}^d$ and $\{v_i\}_{i=0}^d$ of $V$,
the \emph{inner product matrix from $\{u_i\}_{i=0}^d$ to $\{v_i\}_{i=0}^d$} 
is the matrix in $\Mat_{d+1}(\F)$ that has $(i,j)$-entry $\b{ u_i, v_j }$
for $0 \leq i,j \leq d$.
\end{defi}

\goodbreak
Referring to Definition~\ref{def:innermatrix}, the inner product matrix from $\{u_i\}_{i=0}^d$
to $\{v_i\}_{i=0}^d$ is invertible.

\begin{defi} \label{def:symmetric} \samepage
%\ifDRAFT {\rm def:symmetric}. \fi
The form $\b{\; , \;}$ is said to be \emph{symmetric} whenever
$\b{u,v} = \b{v,u}$ for $u,v\in V$.
\end{defi}

\begin{defi} \label{def:dualbasis} \samepage
%\ifDRAFT {\rm def:dualbasis}. \fi
Assume that $\b{ \; , \; }$ is symmetric.
Then two bases $\{u_i\}_{i=0}^d$, $\{v_i\}_{i=0}^d$ of $V$ are said to be 
\emph{dual with respect to $\b{ \; , \; }$}
whenever $\b{u_i, v_j} = \delta_{i,j}$ for $0 \leq i,j \leq d$.
\end{defi}

\begin{lemma} \label{lem:dualbasis} \samepage
%\ifDRAFT {\rm lem:dualbasis}. \fi
Assume that $\b{ \; , \; }$ is symmetric.
Then each basis of $V$ has a unique dual with respect to $\b{ \; , \; }$.
\end{lemma}

\begin{lemma} \label{lem:trans} \samepage
%\ifDRAFT {\rm lem:trnas}. \fi
Assume that $\b{ \; , \; }$ is symmetric.
Let $\{u_i\}_{i=0}^d$ and $\{v_i\}_{i=0}^d$ denote bases of $V$.
Then the following are the same:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma16.5_i} %[\rm (i)]
the inner product matrix from $\{u_i\}_{i=0}^d$ to $\{v_i\}_{i=0}^d$;
\item\label{lemma16.5_ii} %[\rm (ii)]
the inner product matrix from $\{u_i\}_{i=0}^d$ to $\{u_i\}_{i=0}^d$,
times the transition matrix from $\{u_i\}_{i=0}^d$ to $\{v_i\}_{i=0}^d$.
\end{enumerate}
\end{lemma}

\begin{proof}
Routine linear algebra.
\end{proof}


\section{The dual \texorpdfstring{$\Phi$}{Phi}-standard basis}
\label{sec:dualstandard}

We return our attention to a symmetric idempotent system
 $\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on $V$.
In this section we introduce the notion of a dual $\Phi$-standard basis of $V$.
Recall the antiautomorphism $\dagger$ of $\End (V)$ from Definition~\ref{def:sym}.
For the rest of the paper $\b{\; , \;}$ denotes a bilinear form on $V$
associated with $\dagger$.
By the construction, for $A \in \End (V)$ we have
\begin{align}
 \bbig{ A u, v} &= \bbig{u, A^\dagger v} && (u,v \in V). \label{eq:bXuv}
\end{align}
Recall the algebra $\mathcal M$ from Definition~\ref{def:D}.


\begin{lemma} \label{lem:bilin1} \samepage
%\ifDRAFT {\rm lem:bilin1}. \fi
For $A \in {\mathcal M} \cup {\mathcal M}^*$,
\begin{align}
 \bbig{ A u, v} &= \bbig{u, A v} && (u,v \in V). \label{eq:bXuv2}
\end{align}
\end{lemma}

\begin{proof}
By Definition~\ref{def:sym} and~\eqref{eq:bXuv}.
\end{proof}

\begin{lemma} \label{lem:EsiwEsjw} \samepage
%\ifDRAFT {\rm lem:EsiwEsjw}. \fi
For $\xi \in E_0 V$,
\begin{align}
 \bbig{E^*_i \xi, E^*_j \xi} &= \delta_{i,j} \nu^{-1} k_i ||\xi||^2
 && (0 \leq i,j \leq d). \label{eq:EsiwEsjw}
\end{align}
\end{lemma}

\begin{proof}
Using~\eqref{eq:bXuv2} and $E_0 \xi = \xi$,
\[
\bbig{ E^*_i \xi, E^*_j \xi } 
 = \bbig{ E^*_i E_0 \xi, E^*_j E_0 \xi }
 = \bbig{ \xi, E_0 E^*_i E^*_j E_0 \xi }
 = \delta_{i,j} \bbig{ \xi, E_0 E^*_i E_0 \xi}.
\]
By this and Lemmas~\ref{lem:E0EsiE0}\ref{lemma4.2_ii}, \ref{lem:kimi}\ref{lemma8.3_i} we get the result.
\end{proof}

\begin{lemma} \label{lem:bsym} \samepage
%\ifDRAFT {\rm lem:sym}. \fi
The bilinear form $\b{ \; , \;}$ is symmetric.
\end{lemma}

\begin{proof}
Consider a $\Phi$-standard basis $\{E^*_i \xi\}_{i=0}^d$ of $V$, where $0 \neq \xi \in E_0 V$.
By Lemma~\ref{lem:EsiwEsjw}, 
$\bbig{E^*_i \xi, E^*_j \xi} = \bbig{ E^*_j \xi, E^*_i \xi}$ for $0 \leq i,j \leq d$.
Therefore $\b{u,v} = \b{v,u}$ for $u$, $v \in V$.
\end{proof}


\goodbreak
\begin{lemma} \label{lem:nonzero1} \samepage
%\ifDRAFT {\rm lem:nonzero1}. \fi
The following hold for $0 \neq \xi \in E_0 V$ and
$0 \neq \xi^* \in E^*_0 V$:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma17.4_i} %[\rm (i)]
each of $||\xi||^2$, $||\xi^*||^2$, $\bbig{\xi, \xi^*}$ is nonzero;
\item\label{lemma17.4_ii} %[\rm (ii)]
$E^*_0 \xi = \frac{ \b{\xi, \xi^*} } { || \xi^* ||^2 } \, \xi^*$;
\item\label{lemma17.4_iii} %[\rm (iii)]
$E_0 \xi^* = \frac{ \b{\xi, \xi^*} } { || \xi ||^2 } \, \xi$;
\item\label{lemma17.4_iv} %[\rm (iv)]
$ || \xi ||^2 || \xi^* ||^2 = \nu \, \bbig{\xi, \xi^*}^2$.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{lemma17.4_i}
Observe $||\xi||^2 \neq 0$ by Lemma~\ref{lem:EsiwEsjw} and since $\b{ \; , \; }$ is nonzero.
Applying this to $\Phi^*$ we get $||\xi^*||^2 \neq 0$.
To see that $\bbig{\xi, \xi^*} \neq 0$,
observe that $\xi^*$ is a basis of $E^*_0 V$ so there exists a scalar $\alpha$ such that
$E^*_0 \xi = \alpha \xi^*$.
Recall $E^*_0 \xi \neq 0$ by Lemma~\ref{lem:Esiu} so $\alpha \neq 0$.
Using~\eqref{eq:bXuv2} and $E^*_0 \xi^* = \xi^*$ we routinely find that 
$\bbig{\xi, \xi^*} = \alpha ||\xi^*||^2$ and it follows $\bbig{\xi, \xi^*} \neq 0$.

\ref{lemma17.4_ii}
In the proof of part~\ref{lemma17.4_i} we found $E^*_0 \xi = \alpha \xi^*$ where $\bbig{\xi, \xi^*} = \alpha ||\xi^*||^2$.
The result follows.

\ref{lemma17.4_iii}
Apply~\ref{lemma17.4_ii} to $\Phi^*$.

\ref{lemma17.4_iv}
Using $\xi = E_0 \xi$ and Lemma~\ref{lem:nuE0Es0E0} one finds that $\nu^{-1} \xi = E_0 E^*_0 \xi$.
To finish the proof, evaluate $E_0 E^*_0 \xi$ using~\ref{lemma17.4_ii}, \ref{lemma17.4_iii}.
\end{proof}


\begin{defi} \label{def:dualstandard} \samepage
%\ifDRAFT {\rm def:dualstandard}. \fi
By a \emph{dual $\Phi$-standard basis} of $V$ we mean the dual of a
$\Phi$-standard basis with respect to $\b{ \; , \; }$.
\end{defi}

Shortly we will describe the dual $\Phi$-standard bases.
We will use the following definition.


\begin{defi} \label{def:partner} \samepage
%\ifDRAFT {\rm def:partner}. \fi
Note that for nonzero $\xi$, $\zeta \in E_0V$ the following are equivalent:
\begin{align*}
& \textup{\hypertarget{defi17.6_i}{(i)} } \bbig{\xi, \zeta} = \nu;
&& \textup{\hypertarget{defi17.6_ii}{(ii)} } \zeta = \nu \xi / ||\xi||^2;
&& \textup{\hypertarget{defi17.6_iii}{(iii)} } \xi = \nu \zeta / ||\zeta||^2.
\end{align*}
We say that $\xi$, $\zeta$ are \emph{partners} whenever they satisfy \hyperlink{defi17.6_i}{\textup{(i)}}--\hyperlink{defi17.6_iii}{\textup{(iii)}}.
\end{defi}

\begin{lemma} \label{lem:dualstandard3} \samepage
%\ifDRAFT {\rm lem:dualstandard3}. \fi
For nonzero $\xi$, $\zeta$ in $E_0 V$
the following are equivalent:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma17.7_i} %[\rm (i)]
the bases $\{E^*_i \xi\}_{i=0}^d$ and $\{ k_i^{-1} E^*_i \zeta\}_{i=0}^d$ are dual
with respect to $\b{ \; , \; }$;
\item\label{lemma17.7_ii} %[\rm (ii)]
$\xi$, $\zeta$ are partners.
\end{enumerate}
\end{lemma}

\begin{proof}
The vector space $E_0 V$ has dimension $1$, so there exists a scalar $\alpha$
such that $\zeta = \alpha \xi$.
By this and Lemma~\ref{lem:EsiwEsjw},
\[
 \bbig{ E^*_i \xi, k_j^{-1} E^*_j \zeta } = \delta_{i,j} \alpha \nu^{-1} ||\xi||^2.
\]
So~\ref{lemma17.7_i} holds if and only if $\alpha ||\xi||^2 = \nu$.
By this and Definition~\ref{def:partner} we obtain the result.
\end{proof}


\begin{lemma} \label{lem:dualstandard} \samepage
%\ifDRAFT {\rm lem:dualstandard}. \fi
A given basis of $V$ is dual $\Phi$-standard if and only if it has the form
$\{ k_i^{-1} E^*_i \zeta\}_{i=0}^d$ 
for some nonzero $\zeta \in E_0 V$.
\end{lemma}

\begin{proof}
Use Lemma~\ref{lem:dualstandard3}.
\end{proof}

We mention a result for later use.

\begin{lemma} \label{lem:EsixiEjxis} \samepage
%\ifDRAFT {\rm lem:EsixiEjxis}. \fi
For $0 \neq \xi \in E_0V$ and $0 \neq \xi^* \in E^*_0 V$,
\begin{align*}
 \bbig{ E^*_i \xi, E_j \xi^* } &= \nu^{-1} p_i (j) k^*_j \bbig{ \xi, \xi^* } && (0 \leq i,j \leq d).
\end{align*}
\end{lemma}

\goodbreak
\begin{proof}
Using $E_0 \xi = \xi$, $E^*_0 \xi^* = \xi^*$ and 
Lemma~\ref{lem:E0EsiEjEs0}\ref{lemma13.4_i},
\[
\bbig{E^*_i \xi, E_j \xi^* }
 = \bbig{ \xi , E_0 E^*_i E_j E^*_0 \xi^* }
 = p_i (j) m_j \bbig{\xi, \xi^*}.
\]
By this and Lemma~\ref{lem:kimi}\ref{lemma8.3_ii} we obtain the result.
\end{proof}

\section{Four bases of \texorpdfstring{$V$}{V}}
\label{sec:4bases}

We continue to discuss a symmetric idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on $V$.
Recall the elements $A_i$ from Definition~\ref{def:Ai}.
Recall the matrices $K$, $K^*$, $U$, $U^*$ from Definitions~\ref{def:matrices}, \ref{def:U},
and the matrices $B_i$, $B^*_i$, $H_i$, $H^*_i$ from Definitions~\ref{def:matrices2}, \ref{def:matrices3}.
Recall the bilinear form $\b{\; , \;}$ from above Lemma~\ref{lem:bilin1}.

Throughout this section, we fix nonzero vectors $\xi, \zeta \in E_0 V$ and $\xi^*, \zeta^* \in E^*_0 V$,
and consider the following four bases of $V$.%:
\begin{equation}
\renewcommand{\arraystretch}{1.3}
\begin{array}{c@{\qquad}|@{\qquad}c}
\text{basis type} & \text{basis}
\\ \hline
\text{$\Phi$-standard} & \{ E^*_i \xi \}_{i=0}^d \rule{0mm}{3ex}
\\
\text{dual $\Phi$-standard} & \{ k_i^{-1} E^*_i \zeta\}_{i=0}^d \rule{0mm}{3ex}
\\
\text{$\Phi^*$-standard} & \{E_i \xi^*\}_{i=0}^d \rule{0mm}{3ex}
\\
\text{dual $\Phi^*$-standard} & \{ (k^*_i)^{-1} E_i \zeta^* \}_{i=0}^d \rule{0mm}{3ex}
\end{array} 
 \label{eq:4bases}
\end{equation}
In this section we display the matrices that represent $\{A_r\}_{r=0}^d$, $\{A^*_r\}_{r=0}^d$, 
$\{E_r\}_{r=0}^d$, $\{E^*_r\}_{r=0}^d$ with respect to these bases.
We display the inner product matrices between these bases.
We display the transition matrices between these bases.

We introduce some notation.
For $0 \leq i,j \leq d$ define $\Delta_{i,j} \in \Mat_{d+1}(\F)$
that has $(i,j)$-entry $1$ and all other entries $0$.

\begin{prop} \label{prop:matrixAr} \samepage
%\ifDRAFT {\rm prop:matrixAr}. \fi
In the table below we give some matrix representations.
For $0 \leq r \leq d$, each entry in the table is the matrix that represents the map
in the given column with respect to the basis in the given row.%:
\[
\renewcommand{\arraystretch}{1.3}
\begin{array}{c@{\qquad}|@{\qquad}ccccc}
\text{basis} & & A_r & A^*_r & E_r & E^*_r
\\ \hline
 \{ E^*_i \xi \}_{i=0}^d
 & & B_r & H^*_r & \nu^{-1} U K^* \Delta_{r,r} U^* K & \Delta_{r,r} \rule{0mm}{3ex}
\\
\{ k_i^{-1} E^*_i \zeta \}_{i=0}^d
 & & B_r^{\sf t} & H^*_r & (U^*)^{-1} \Delta_{r,r} U^* & \Delta_{r,r} \rule{0mm}{3ex}
\\
\{ E_i \xi^* \}_{i=0}^d
 & & H_r & B^*_r & \Delta_{r,r} & \nu^{-1} U^* K \Delta_{r,r} U K^* \rule{0mm}{3ex}
\\
\{ (k^*_i)^{-1} E_i \zeta^* \}_{i=0}^d
& & H_r & (B^*_r)^{\sf t} & \Delta_{r,r} & U^{-1} \Delta_{r,r} U \rule{0mm}{3ex}
\end{array}
\]
\end{prop}

\begin{proof}
We first consider the matrices representing $A_r$.
The matrix representing $A_r$ with respect to $\{E^*_i \xi\}_{i=0}^d$ is obtained using
Lemma~\ref{lem:AjEsiE0}\ref{lemma11.1_i} and Definition~\ref{def:matrices2}.
The matrix representing $A_r$ with respect to $\{k_i^{-1} E^*_i \zeta\}_{i=0}^d$ is obtained
using Lemmas~\ref{lem:khphij}\ref{lemma10.11_i} and~\ref{lem:AjEsiE0}\ref{lemma11.1_i}.
The matrices representing $A_r$ with respect to $\{E_i \xi^*\}_{i=0}^d$ and $\{(k^*_i)^{-1} E_i \zeta^*\}_{i=0}^d$ 
are obtained using Lemma~\ref{lem:AiEj}\ref{lemma12.4_i} and Definition~\ref{def:matrices3}.
Applying these results to $\Phi^*$ we obtain the matrices representing $A^*_r$.
Next we consider the matrices representing $E_r$.
The matrix representing $E_r$ with respect to $\{E^*_i \xi\}_{i=0}^d$ is obtained using
Lemmas~\ref{lem:E0EsiEj}\ref{lemma13.2_iii}, \ref{lem:Asi}, \ref{lem:U3}\ref{lemma14.4_i},\ref{lemma14.4_iii}.
Multiply this matrix on the left (resp.\ right) by $K$ (resp.\ $K^{-1}$)
and use Lemma~\ref{lem:U2}\ref{lemma14.6_iii} to obtain the matrix representing $E_r$
with respect to $\{k_i^{-1} E^*_i \zeta\}_{i=0}^d$.
The matrices representing $E_r$ with respect to $\{E_i \xi^*\}_{i=0}^d$ and $\{ (k^*_i)^{-1} E_i \zeta^* \}_{i=0}^d$
are routinely obtained.
Applying these results to $\Phi^*$ we obtain the matrices representing $E^*_r$.
\end{proof}

\begin{prop} \label{prop:innermatrix0} \samepage
%\ifDRAFT {\rm prop:innermatrix0}. \fi
In the table below we give the inner product matrices between the bases in~\eqref{eq:4bases}.
Each entry of the table is the inner product matrix from the basis in the given row
to the basis in the given column.%:
\[
\renewcommand{\arraystretch}{1.3}
\begin{array}{c@{\qquad}|@{\qquad}cccc}
 & \{ E^*_i \xi \}_{i=0}^d & \{ k_i^{-1} E^*_i \zeta \}_{i=0}^d 
 & \{ E_i \xi^* \}_{i=0}^d & \{ (k^*_i)^{-1} E_i \zeta^* \}_{i=0}^d
\\ \hline
\{ E^*_i \xi \}_{i=0}^d 
 & \frac{ ||\xi||^2 } { \nu } K 
 & \frac{ \b{ \xi, \zeta} } { \nu } I 
 & \frac{ \b{\xi, \xi^*} } { \nu } K U K^* 
 & \frac{ \b{\xi, \zeta^*} } { \nu } K U \rule{0mm}{3.5ex}
\\
\{ k_i^{-1} E^*_i \zeta \}_{i=0}^d 
 & \frac{ \b{ \zeta, \xi} } { \nu } I 
 & \frac{ || \zeta ||^2 } { \nu } K^{-1} 
 & \frac{ \b{\zeta, \xi^*} } { \nu } U K^* 
 & \frac{ \b{ \zeta, \zeta^*} } { \nu } U \rule{0mm}{3ex}
\\
\{ E_i \xi^* \}_{i=0}^d 
 & \frac{ \b{ \xi^*, \xi } } { \nu } K^* U^* K
 & \frac{ \b{ \xi^* , \zeta } } { \nu } K^* U^*
 & \frac{ ||\xi^*||^2 } { \nu } K^* 
 & \frac{ \b{ \xi^*, \zeta^* } } { \nu } I \rule{0mm}{3.5ex}
\\
\{ (k^*_i)^{-1} E_i \zeta^* \}_{i=0}^d 
 & \frac{ \b{ \zeta^*, \xi } } { \nu } U^* K 
 & \frac{ \b{ \zeta^*, \zeta} } { \nu } U^*
 & \frac{ \b{ \zeta^*, \xi^* } } { \nu } I 
 & \frac{ ||\zeta^*||^2 } { \nu } (K^*)^{-1} \rule{0mm}{3.5ex}
\end{array}
\]
\end{prop}

\begin{proof}
Note that $\zeta$ (resp.\ $\zeta^*$) is a nonzero scalar multiple of $\xi$
(resp.\ $\xi^*$).
Using this and Lemmas~\ref{lem:EsiwEsjw}, \ref{lem:EsixiEjxis}
we represent the inner products in terms of $P$, $Q$, $K$, $K^*$.
Now eliminate $P$, $Q$ using Lemma~\ref{lem:U3} to get the result.
\end{proof}

In the diagram below we display the inner product matrices between
the four bases in~\eqref{eq:4bases}:

\begin{center}
\includegraphics{Figures/fig1}
\end{center}


\begin{prop} \label{prop:transmatrix0} \samepage
%\ifDRAFT {\rm prop:transmatrix0}. \fi
In the table below we give the transition matrices between the
four bases in~\eqref{eq:4bases}.
Each entry of the table is the transition matrix from the
basis in the given row to the basis in the given column.%:
\[
\renewcommand{\arraystretch}{1.3}
\begin{array}{c@{\qquad}|@{\qquad}cccc}
 & \{ E^*_i \xi \}_{i=0}^d & \{ k_i^{-1} E^*_i \zeta \}_{i=0}^d 
 & \{ E_i \xi^* \}_{i=0}^d & \{ (k^*_i)^{-1} E_i \zeta^* \}_{i=0}^d
\\ \hline
\{ E^*_i \xi \}_{i=0}^d 
& I 
& \frac{ \b{\xi, \zeta} } { ||\xi||^2 } K^{-1} 
& \frac{ \b{\xi, \xi^* } } { ||\xi||^2 } U K^* 
& \frac{ \b{\xi, \zeta^*} } { ||\xi||^2 } U \rule{0mm}{3ex}
\\
\{ k_i^{-1} E^*_i \zeta \}_{i=0}^d 
& \frac{ \b{ \zeta, \xi} } { ||\zeta||^2 } K 
& I 
& \frac{ \b{ \zeta, \xi^* } } { ||\zeta||^2 } K U K^* 
& \frac{ \b{ \zeta, \zeta^* } } { ||\zeta||^2 } K U \rule{0mm}{3.5ex}
\\
\{ E_i \xi^* \}_{i=0}^d 
& \frac{ \b{ \xi^*, \xi } } { ||\xi^*||^2 } U^* K 
& \frac{ \b{ \xi^*, \zeta } } { ||\xi^*||^2 } U^*
& I 
& \frac{ \b{ \xi^*, \zeta^*} } { ||\xi^*||^2 } (K^*)^{-1} \rule{0mm}{3.5ex}
\\
\{ (k^*_i)^{-1} E_i \zeta^* \}_{i=0}^d 
& \frac{ \b{ \zeta^*, \xi } } { ||\zeta^*||^2 } K^* U^* K
& \frac{ \b{ \zeta^*, \zeta } } { ||\zeta^*||^2 } K^* U^*
& \frac{ \b{ \zeta^*, \xi^* } } { ||\zeta^*||^2 } K^* 
& I \rule{0mm}{3.5ex}
\end{array}
\]
\end{prop}

\begin{proof}
Use Lemma~\ref{lem:trans} and Proposition~\ref{prop:innermatrix0}.
\end{proof}

\looseness-1
\looseness-1
In the diagram below we display the transition matrices between
the four bases in~\eqref{eq:4bases}.%:

\begin{center}
\includegraphics{Figures/fig2}
\end{center}

\section{\texorpdfstring{$P$}{P}-polynomial and \texorpdfstring{$Q$}{Q}-polynomial idempotent systems}
\label{sec:Ppoly}

We continue to discuss a symmetric idempotent system
$\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ on~$V$.

\begin{defi} \label{def:Ppoly} \samepage
%\ifDRAFT {\rm def:Ppoly}. \fi
We say that $\Phi$ is \emph{$P$-polynomial}
whenever $p^h_{i j}$ is zero (resp.\ nonzero) if one of $h,i,j$ is greater
than (resp.\ equal to) the sum of the other two $(0 \leq h,i,j \leq d)$.
\end{defi}

For the moment, assume that $d \geq 1$ and $\Phi$ is $P$-polynomial.
Then the first intersection matrix $B_1$ has the form
\[
 B_1 =
 \begin{pmatrix}
 a_0 & b_0 & & & & \mathbf{0} \\
 c_1 & a_1 & b_1 \\
 & c_2 & \cdot & \cdot \\
 & & \cdot & \cdot & \cdot \\
 & & & \cdot & \cdot & b_{d-1} \\
 \mathbf{0} & & & & c_d & a_d \\
 \end{pmatrix},
\]
where 
\begin{align*}
c_i &= p^{i}_{1, i-1} \quad (1 \leq i \leq d),
&
a_i &= p^i_{1, i} \quad (0 \leq i \leq d),
&
b_i &= p^{i}_{1, i+1} \quad (0 \leq i \leq d-1). 
\end{align*}
Moreover $c_i \neq 0$ for $1 \leq i \leq d$ and
$b_i \neq 0$ for $0 \leq i \leq d-1$.
So $B_1$ is irreducible tridiagonal.
Shortly we will show that this feature of $B_1$ characterizes the $P$-polynomial property.

\begin{lemma} \label{lem:A1Ai} \samepage
%\ifDRAFT {\rm lem:A1Ai}. \fi
Assume that $d \geq 1$ and $\Phi$ is $P$-polynomial.
Then
\begin{align*}
 A_1 A_0 &= a_0 A_0 + c_1 A_1,
\\
 A_1 A_i &= b_{i-1} A_{i-1} + a_i A_i + c_{i+1} A_{i+1} && (1 \leq i \leq d-1), 
\\
 A_1 A_d &= b_{d-1} A_{d-1} + a_d A_d.
\end{align*}
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:defp} and the comments below Definition~\ref{def:Ppoly}.
\end{proof}

For elements $A$, $B$ in any algebra,
we say that $B$ is an \emph{affine transformation} of $A$ whenever
there exist scalars $\alpha$, $\beta$ such that $\alpha \neq 0$
and $B = \alpha A + \beta I$.

\begin{prop} \label{prop:Ppoly} \samepage
%\ifDRAFT {\rm prop:Ppoly}. \fi
Assume that $d \geq 1$.
Then for $A \in \End (V)$ the following are equivalent:
\begin{enumerate}[label=(\roman*)]
\item\label{prop19.3_i} %[\rm (i)]
$\Phi$ is $P$-polynomial and $A$ is an affine transformation of $A_1$;
\item\label{prop19.3_ii} %[\rm (ii)]
for $0 \leq i \leq d$ there exists $f_i \in \F[x]$ such that $\deg (f_i)=i$
and $A_i = f_i (A)$.
\end{enumerate}
\end{prop}

\begin{proof}
\ref{prop19.3_i} $\Rightarrow$~\ref{prop19.3_ii}
By Lemma~\ref{lem:A1Ai} and since $A_0 = I$.

\ref{prop19.3_ii} $\Rightarrow$~\ref{prop19.3_i}
The elements $\{A_i\}_{i=0}^d$ are linearly independent by Lemma~\ref{lem:AiAsi},
so the elements $\{A^i\}_{i=0}^d$ are linearly independent.
Pick integers $i$, $j$ $(0 \leq i,j \leq d)$ such that $i+j \leq d$.
We show that
\begin{align}
 f_i f_j &= \sum_{h=0}^d p^h_{i j} f_h. \label{eq:fifj}
\end{align}
Define a polynomial $g = f_i f_j - \sum_{h=0}^d p^h_{i j} f_h$.
The degree of $g$ is at most $d$, and $g(A)=0$.
Therefore $g=0$.
We have shown~\eqref{eq:fifj}.
In~\eqref{eq:fifj} we examine the degrees to find
\[
 i + j = \max \{h \,|\, 0 \leq h \leq d, \; p^h_{i j} \neq 0 \}.
\]
By this and Lemma~\ref{lem:khphij}\ref{lemma10.11_i},
we find that $\Phi$ is $P$-polynomial.
Since $A_1 = f_1 (A)$ and $\deg (f_1)= 1$,
$A$ is an affine transformation of $A_1$.
\end{proof}

\begin{prop} \label{prop:Ppoly2} \samepage
%\ifDRAFT {\rm prop:Ppoly2}. \fi
Assume that $d \geq 1$ and $\Phi$ is $P$-polynomial.
Then the following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{prop19.4_i} %[\rm (i)]
$\{A_1^i\}_{i=0}^d$ form a basis for the vector space $\mathcal M$, 
where $\mathcal M$ is from Definition~\ref{def:D};
\item\label{prop19.4_ii} %[\rm (ii)]
$\{p_1 (j)\}_{j=0}^d$ are mutually distinct;
\item\label{prop19.4_iii} %[\rm (iii)]
$\{E_i V\}_{i=0}^d$ are the eigenspaces of $A_1$;
\item\label{prop19.4_iv} %[\rm (iv)]
$A_1$ is multiplicity-free;
\item\label{prop19.4_v} %[\rm (v)]
$\{E_i\}_{i=0}^d$ are the primitive idempotents of $A_1$.
\end{enumerate}
\end{prop}

\begin{proof}
\ref{prop19.4_i}
By Lemma~\ref{lem:AiAsi} and Proposition~\ref{prop:Ppoly}\ref{prop19.3_ii}.

\ref{prop19.4_ii}
By Lemma~\ref{lem:AiEj}, $p_1 (j)$ is the eigenvalue of $A_1$ 
corresponding to $E_j V$ for $0 \leq j \leq d$.
So the characteristic polynomial of $A_1$ is $\prod_{j=0}^d (x - p_1 (j))$.
By~\ref{prop19.4_i} the minimal polynomial of $A_1$ has degree $d+1$.
By these comments, the minimal polynomial of $A_1$ is $\prod_{j=0}^d (x - p_1 (j))$.
The result follows.

\ref{prop19.4_iii}
By Lemma~\ref{lem:AiEj}\ref{lemma12.4_i} and~\ref{prop19.4_ii} above.

\ref{prop19.4_iv}
By~\ref{prop19.4_iii} above and since $E_i V$ has dimension one for $0 \leq i \leq d$.


\ref{prop19.4_v}
By~\ref{prop19.4_iii}, \ref{prop19.4_iv} above. 
\end{proof}

\begin{prop} \label{prop:PpolyB1} \samepage
%\ifDRAFT {\rm prop:PpolyB1}. \fi
For $d \geq 1$ the following are equivalent:
\begin{enumerate}[label=(\roman*)]
\item\label{prop19.5_i} %[\rm (i)]
$\Phi$ is $P$-polynomial;
\item\label{prop19.5_ii} %[\rm (ii)]
the first intersection matrix $B_1$ is irreducible tridiagonal.
\end{enumerate}
\end{prop}

\begin{proof}
\ref{prop19.5_i} $\Rightarrow$~\ref{prop19.5_ii}
We saw this above Lemma~\ref{lem:A1Ai}.

\ref{prop19.5_ii} $\Rightarrow$~\ref{prop19.5_i}
Since $B_1$ is irreducible tridiagonal, we have the equations in Lemma~\ref{lem:A1Ai}.
So for $0 \leq i \leq d$ there exists $f_i \in \F[x]$ such that $\deg (f_i) = i$ 
and $A_i = f_i (A_1)$.
By Proposition~\ref{prop:Ppoly} (with $A = A_1$) we see that $\Phi$ is $P$-polynomial.
\end{proof}

\begin{defi} \label{def:Qpoly} \samepage
%\ifDRAFT {\rm def:Qpoly}. \fi
We say that $\Phi$ is \emph{$Q$-polynomial}
whenever $q^h_{i j}$ is zero (resp.\ nonzero) if one of $h,i,j$ is greater
than (resp.\ equal to) the sum of the other two $(0 \leq h,i,j \leq d)$.
\end{defi}

\begin{lemma} \label{lem:Qpoly} \samepage
%\ifDRAFT {\rm lem:Qpoly}. \fi
$\Phi$ is $Q$-polynomial if and only if $\Phi^*$ is $P$-polynomial.
\end{lemma}

\begin{proof}
Immediate from Definitions~\ref{def:qhij},
\ref{def:Ppoly}, \ref{def:Qpoly}.
\end{proof}




































\section{Leonard pairs and Leonard systems}
\label{sec:LP}

In this section we recall the notion of a Leonard pair and a Leonard system.


\begin{defi}[{\cite[Definition~1.1]{T:Leonard}}]
\label{def:LP} \samepage
%\ifDRAFT {\rm def:LP}. \fi
By a \emph{Leonard pair} on $V$ we mean an ordered pair
$A,A^*$ of elements in $\End (V)$
that satisfy the following~\ref{defi20.1_i}, \ref{defi20.1_ii}.
\begin{enumerate}[label=(\roman*)]
\item\label{defi20.1_i} %[\rm (i)]
There exists a basis of $V$ with respect to which the matrix representing $A$
is irreducible tridiagonal and the matrix representing $A^*$ is diagonal.
\item\label{defi20.1_ii} %[\rm (ii)]
There exists a basis of $V$ with respect to which the matrix representing $A^*$
is irreducible tridiagonal and the matrix representing $A$ is diagonal.
\end{enumerate}
\end{defi}

Let $A,A^*$ denote a Leonard pair on $V$.
By~\cite[Lemma~1.3]{T:Leonard} each of $A$, $A^*$ is multiplicity-free.
Let $\{E_i\}_{i=0}^d$ denote an ordering of the primitive idempotents of $A$.
For $0 \leq i \leq d$ pick a nonzero $v_i \in E_i V$.
Then $\{v_i\}_{i=0}^d$ form a basis of $V$.
We say that the ordering $\{E_i\}_{i=0}^d$ is \emph{standard} whenever
$\{v_i\}_{i=0}^d$ satisfies Definition~\ref{def:LP}\ref{defi20.1_ii}.
In this case, the ordering $\{E_{d-i} \}_{i=0}^d$ is standard and no further
ordering is standard.
A standard ordering of the primitive idempotents of $A^*$ is similarly defined.

\begin{defi}[{\cite[Definition~1.4]{T:Leonard}}] 
 \label{def:LS} \samepage
%\ifDRAFT {\rm def:LS}. \fi
By a \emph{Leonard system} on $V$ we mean a sequence
\begin{equation}
 (A; \{E_i\}_{i=0}^d; A^*; \{E^*_i\}_{i=0}^d) \label{eq:Phi}
\end{equation}
of elements in $\End (V)$
that satisfy the following~\ref{defi20.2_i}--\ref{defi20.2_iii}:
\begin{enumerate}[label=(\roman*)]
\item\label{defi20.2_i} %[\rm (i)]
$A,A^*$ is a Leonard pair on $V$;
\item\label{defi20.2_ii} %[\rm (ii)]
$\{E_i\}_{i=0}^d$ is a standard ordering of the primitive idempotents of $A$;
\item\label{defi20.2_iii} %[\rm (iii)]
$\{E^*_i\}_{i=0}^d$ is a standard ordering of the primitive idempotents of $A^*$.
\end{enumerate}
\end{defi}

For the rest of this section
let $(A; \{E_i\}_{i=0}^d; A^*; \{E^*_i\}_{i=0}^d)$ denote a Leonard system on $V$.
Note that $(A^*; \{E^*_i\}_{i=0}^d; A; \{E_i\}_{i=0}^d)$ is a Leonard system on $V$.
 
\begin{lemma}[{\cite[Lemma~9.2]{T:qRacah}}]
 \label{lem:E0EsiE02} \samepage
%\ifDRAFT {\rm lem:E0EsiE02}. \fi
The following hold:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma20.3_i} %[\rm (i)]
$E_0 E^*_i E_0 \neq 0 \quad (0 \leq i \leq d)$;
\item\label{lemma20.3_ii} %[\rm (ii)]
$E^*_0 E_i E^*_0 \neq 0 \quad (0 \leq i \leq d)$.
\end{enumerate}
\end{lemma}

\begin{lemma}[{\cite[Theorem~6.1 and Lemma~6.3]{T:qRacah}}]
\label{lem:anti} \samepage
%\ifDRAFT {\rm lem::anti}. \fi
There exists a unique antiautomorphism $\dagger$ of $\End (V)$
that fixes each of $A$, $A^*$.
Moreover $\dagger$ fixes each of $E_i$, $E^*_i$ for $0 \leq i \leq d$.
\end{lemma}

\begin{lemma}[{\cite[Theorem~13.4]{T:qRacah}}]
\label{lem:vi} \samepage
%\ifDRAFT {\rm lem:vi}. \fi
There exist polynomials $\{f_i\}_{i=0}^d$ in $\F[x]$ such that
$\deg (f_i) = i$ and $f_i (A) E^*_0 E_0 = E^*_i E_0$
for $0 \leq i \leq d$.
\end{lemma}

\begin{lemma}[{\cite[Theorem~4.2]{NT:split}}]
\label{lem:NTsplit} \samepage
%\ifDRAFT {\rm lem:NTsplit}. \fi
For elements $B$, $B^*$ in $\End (V)$ the following are equivalent:
\begin{enumerate}[label=(\roman*)]
\item\label{lemma20.6_i} %[\rm (i)]
$(B; \{E_i\}_{i=0}^d; B^*; \{E^*_i\}_{i=0}^d)$ is a Leonard system;
\item\label{lemma20.6_ii} %[\rm (ii)]
$B$ (resp.\ $B^*$) is an affine transformation of $A$ (resp.\ $A^*$).
\end{enumerate}
\end{lemma}























\section{Idempotent systems and Leonard systems}
\label{sec:IPSLS}

In this section we show that a Leonard system is essentially the same thing
as a symmetric idempotent system that is $P$-polynomial and $Q$-polynomial.

\begin{theorem} \label{thm:LP} \samepage
%\ifDRAFT {\rm thm:LP}. \fi
Let $\Phi = (\{E_i\}_{i=0}^d; \{E^*_i\}_{i=0}^d)$ denote a sequence
of elements in $\End (V)$.
Then the following are equivalent:
\begin{enumerate}[label=(\roman*)]
\item\label{theo21.1_i} %[\rm (i)]
$\Phi$ is a symmetric idempotent system that is $P$-polynomial and $Q$-polynomial;
\item\label{theo21.1_ii} %[\rm (ii)]
there exist $A$, $A^*$ in $\End (V)$ such that
$(A; \{E_i\}_{i=0}^d; A^*; \{E^*_i\}_{i=0}^d)$ is a Leonard system.
\end{enumerate}
\end{theorem}

\begin{proof}
We assume $d \geq 1$; otherwise the assertion is obvious.

\ref{theo21.1_i} $\Rightarrow$~\ref{theo21.1_ii}
We show that $(A_1; \{E_i\}_{i=0}^d; A^*_1; \{E^*_i\}_{i=0}^d)$ is
a Leonard system on $V$, where $A_1$, $A^*_1$ are from
Definition~\ref{def:Ai}.
By Proposition~\ref{prop:matrixAr}, with respect to a $\Phi$-standard basis of $V$
the matrix representing $A_1$ is $B_1$ 
and the matrix representing $A^*_1$ is $H^*_1$.
By Definition~\ref{def:matrices2} the matrix $H^*_1$ is diagonal,
and by Proposition~\ref{prop:PpolyB1} the matrix $B_1$ is irreducible tridiagonal.
Thus with respect to a $\Phi$-standard basis the matrix representing $A_1$ is irreducible tridiagonal
and the matrix representing $A^*_1$ is diagonal.
Applying this to $\Phi^*$, with respect to a $\Phi^*$-standard basis the matrix representing $A^*_1$
is irreducible tridiagonal and the matrix representing $A_1$ is diagonal.
By these comments $A_1, A^*_1$ is a Leonard pair on $V$.
By Proposition~\ref{prop:Ppoly2}(v) and the construction,
$\{E_i\}_{i=0}^d$ (resp.\ $\{ E^*_i\}_{i=0}^d$) is a standard ordering of the primitive idempotents
of $A_1$ (resp.\ $A^*_1$).
We have shown that $(A_1; \{E_i\}_{i=0}^d; A^*_1; \{E^*_i\}_{i=0}^d)$ is a Leonard system on $V$.

\ref{theo21.1_ii} $\Rightarrow$~\ref{theo21.1_i}
By Lemmas~\ref{lem:E0EsiE02} and~\ref{lem:anti}, $\Phi$ is a symmetric idempotent system on $V$.
By Lemma~\ref{lem:vi} there exist polynomials $\{f_i\}_{i=0}^d$ in $\F[x]$ such that
$\deg (f_i) = i$ and 
$f_i (A) E^*_0 E_0 = E^*_i E_0$ for $0 \leq i \leq d$.
By Lemmas~\ref{lem:rho}, \ref{lem:rhorhospre}, \ref{lem:AiEs0E0}\ref{lemma7.4_i},
$A_i$ is the unique element in $\mathcal M$ such that $A_i E^*_0 E_0 = E^*_i E_0$
$(0 \leq i \leq d)$.
By these comments $f_i (A) = A_i$ for $0 \leq i \leq d$.
By this and Proposition~\ref{prop:Ppoly}, $\Phi$ is $P$-polynomial.
Apply this to the Leonard system $(A^*; \{E^*_i\}_{i=0}^d; A; \{E_i\}_{i=0}^d)$
to find that $\Phi$ is $Q$-polynomial.
\end{proof}

\begin{lemma} \label{lem:LP2} \samepage
%\ifDRAFT {\rm lem:LP2}. \fi
Assume that $d \geq 1$ and the equivalent conditions~\ref{theo21.2_i}, \ref{theo21.2_ii}~hold in Theorem~\ref{thm:LP}.
Then for $A$, $A^*$ in $\End (V)$ the following are equivalent:
\begin{enumerate}[label=(\roman*)]
\item\label{theo21.2_i} %[\rm (i)]
$(A; \{E_i\}_{i=0}^d; A^*; \{E^*_i\}_{i=0}^d)$ is a Leonard system on $V$;
\item\label{theo21.2_ii} %[\rm (ii)]
$A$ (resp.\ $A^*$) is an affine transformation of $A_1$ (resp.\ $A^*_1$),
where $A_1$, $A^*_1$ are from Definition~\ref{def:Ai}.
\end{enumerate}
\end{lemma}

\begin{proof}
\ref{theo21.2_i} $\Rightarrow$~\ref{theo21.2_ii}
By the proof of Theorem~\ref{thm:LP}, 
$(A_1; \{E_i\}_{i=0}^d; A^*_1; \{E^*_i\}_{i=0}^d)$ is a Leonard system on $V$.
By this and Lemma~\ref{lem:NTsplit}, $A$ (resp.\ $A^*$) is an
affine transformation of $A_1$ (resp.\ $A^*_1$).

\ref{theo21.2_ii} $\Rightarrow$~\ref{theo21.2_i}
By Lemma~\ref{lem:NTsplit}.
\end{proof}


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