text:
\section{Introduction}

Orthogonal polynomials appear in many places in mathematics, physics and engineering. In mathematical physics, orthogonal
polynomials arise in the context of eigenfunctions of linear, second order differential operators.
Bochner (see \cite{11-1}, p150) classified those operators which have exactly one polynomial eigenfunction
of each degree.
The classical, one-dimensional examples possess important structures which enable their explicit
construction.  Among these are the $3-$point recurrence relations and the differential ladder
operators, which allow the polynomials to be built in a recursive manner.  Starting from Rodrigues
formulae and generating functions, these structures can themselves be built in a systematic
way.

The classification of orthogonal polynomials in the plane and higher
dimensional spaces is much more complicated than the case of the real line.  However, requiring the
polynomials to be eigenfunctions of linear differential operators makes the classification problem more feasible.
An early and systematic treatment of such polynomials in two variables, satisfying a {\em second order} eigenvalue problem, was
given by Krall and Sheffer \cite{67-2}.  They showed that, up to affine transformation,
there are 9 distinct cases.  Three of these just correspond to products of classical $1-$dimensional polynomials,
but the remaining six are non-trivial extensions.

To generalise Bochner's result for the classical orthogonal polynomials, Krall and Sheffer \cite{67-2} introduced the concept of an admissible operator.  In two dimensions we have to expect the existence of $N+1$ polynomial eigenfunctions of degree $N$.

\bd[Admissible Partial Differential Operator]\label{def:admiss}
The differential operator $L$ is said to be {\em admissible} if, for the eigenvalue problem
$$
L \varphi = \lambda \varphi,
$$
there exists an infinite sequence of eigenvalues $\{\lambda_N\}_{N=0}^\infty$, such that for each
$\lambda=\lambda_N$ there are no non-zero polynomial solutions of degree $d<N$
and there are exactly $N+1$ linearly independent polynomial solutions of degree $N$.
\ed
It is easily seen that a {\em second order} admissible differential operator must have the form
\be                               \label{admiss}
\begin{array}{l}
L \varphi:= (\alpha x^2 + d_1 x + e_1 y+f_1) \varphi_{xx} + (2\alpha x y + d_2 x + e_2 y+f_2) \varphi_{xy}\\[2mm]
 \hspace{2cm} + (\alpha y^2 +d_3 x + e_3 y+f_3) \varphi_{yy} + (\beta x +\kappa_1) \varphi_x + (\beta y + \kappa_2) \varphi_y.
 \end{array}
\ee
The number of parameters is reduced further by making affine transformations on the $x-y$ space, giving $9$ canonical forms (see Section \ref{geom}).
The degree $N$ polynomial eigenfunctions form an $N+1$ dimensional vector space $V_N$ and we may choose a basis of {\em monic} polynomials:
\be  \label{pmn}
P_{m,n} = x^m y^n + \mbox{lower order terms} ,\quad\mbox{for}\;\;\; m+n=N .
\ee
The eigenvalue for $P_{m,n}$ is easily calculated from the leading order term to be
\be\label{lambda}
\lambda_{m+n}=(m+n) ((m+n-1)\alpha+\beta).
\ee
We see that when $\beta$ is not a negative integer, $\lambda_{m_1+n_1}\neq\lambda_{m_2+n_2}$ for $m_1+n_1\neq m_2+n_2$.

This family of polynomial eigenfunctions can be arranged into a triangular array (see Figure \ref{trilatt}), with each horizontal level representing a particular $V_N$.  In place of the {\em three point recurrence relations} for the one dimensional case,  Krall and Sheffer showed that in two-dimensions there are two {\em three {\bf level} recurrence relations} of the form
\bea
P_{m+1,n} &=& x P_{m,n} +\sum_{i=0}^{m+n} a_{i,m,n} P_{i,m+n-i}+\sum_{i=0}^{m+n-1}b_{i,m,n} P_{i,m+n-1-i}, \nn\\[0mm]
&&  \label{ks3level}  \\[0mm]
P_{m,n+1} &=& y P_{m,n} +\sum_{i=0}^{m+n} c_{i,m,n} P_{m+n-i,i}+\sum_{i=0}^{m+n-1}d_{i,m,n} P_{m+n-1-i,i}, \nn
\eea
but gave no specific forms, apart from the case of direct products of classical one dimensional polynomials.
Such 3 level recurrence relations have been discussed more generally in \cite{95-6,01-11,06-4}.  A detailed discussion of orthogonal polynomials in two variables can be found in \cite{99-13}.

The coefficients of the second order terms of the operator (\ref{admiss}) can be considered as the components of an upper index metric tensor, with $L$ being a deformation of the corresponding Laplace-Beltrami operator $L_b$.  In \cite{03-4,01-1} it was observed that each of the 9 metrics was either {\em flat} or {\em constant curvature} and that the additional first order terms correspond to ``trivial'' magnetic field, which can therefore be gauged to a purely potential term.  They also showed that each of these cases (considered as a quantum system) was {\em super-integrable} and presented two commuting operators for each case.

In this paper we first discuss the symmetries of these Laplace-Beltrami operators and use these to group together the 9 canonical forms into 4 classes (see Section \ref{geom}).  Within each class the metrics are related through (non-affine) coordinate transformations.  Each metric is invariant with respect to an involutive transformation, corresponding to an involutive automorphism of the algebra of symmetries. These involutions are very useful in Section \ref{super}, where we discuss super-integrability.  We briefly review separable potentials, defining $L=L_b+U(x,y)$, each of which is characterised by the existence of a commuting, second order operator and each of which depends upon two arbitrary functions of one variable (as in the classical mechanical case; see \cite{76-8}).  Requiring two such commuting operators (for super-integrability) reduces these potentials to simple rational functions depending upon (at most) 3 arbitrary parameters.
Super-integrability (with linear or quadratic commuting operators) is associated with separability in more than one coordinate system (see, for example, \cite{99-11}). It is then possible to gauge transform these super-integrable cases to Krall-Sheffer form (just the reverse calculation of \cite{01-1}), but this sometimes restricts the parameters to a sub-case.

Our main results are in Sections \ref{pols} and \ref{genfuns}.  In the first of these, we discuss the algebraic structure of the polynomial eigenfunctions of the Krall-Sheffer operators.  We first emphasise the connection between {\em super-integrability} and the construction of eigenfunctions.  The commuting operators can be used to build the two-dimensional polynomial eigenfunctions out of one-dimensional polynomials, satisfying a reduced differential operator.  In Section \ref{pols}, we also give {\em explicit formulae} for the 3-level recurrence relations for all the (non-trivial) Krall-Shefffer operators.  We also present {\em explicit formulae} for a pair of differential ladder operators, which satisfy $R_{+,x}P_{m,n}=P_{m+1,n},\; R_{+,y}P_{m,n}=P_{m,n+1}$ ({\em raising operators}).  None of these formulae have previously appeared in the literature.

In Section \ref{genfuns} we present {\em generating functions} for the polynomials of Cases V and VIII (both new) and of Case IX.  The relationship (discussed in Section \ref{geom}) between Cases I and IX, gives us an array of eigenfunctions (not all polynomial) of a restricted form of the Case I operator.

\section{The Geometry of Krall-Sheffer Operators}\label{geom}

For an $n-$dimensional (pseudo-)Riemannian space, with local coordinates
$x^1,\cdots ,x^n$ and metric $g_{ij}$, the Laplace-Beltrami operator is defined
by $L_b f = g^{ij} \nabla_i\nabla_j f$, which has explicit form
\be                \label{lbf}
L_b f = \sum_{i,j=1}^n \frac{1}{\sqrt{g}}
\frac{\pa}{\pa x^j}\left(
               \sqrt{g} g^{ij}\frac{\pa f}{\pa x^i}\right),
\ee
where $g$ is the determinant of the matrix $g_{ij}$.  The coefficients of
leading order terms in the Laplace-Beltrami operator are the coefficients of
the {\em inverse metric} $g^{ij}$.  For a metric with isometries, the
infinitesimal generators (Killing vectors) are just first order differential
operators which commute with the Laplace-Beltrami operator (\ref{lbf}).  When
the space is either flat or constant curvature, it possesses the maximal group
of isometries, which is of dimension $\frac{1}{2}n(n+1)$.  In this case, $L_b$
is proportional to the second order {\em Casimir} function of the symmetry algebra
(see \cite{74-7}).  The Krall-Sheffer metrics fall into this category, with $n=2$.

When $n=2$ we have $3$ independent Killing vectors for flat and constant curvature metrics.  For the particular metrics defined by each of the Krall-Sheffer operators, a basis of Killing vectors is easy to calculate explicitly.  Four Lie algebras occur and the $9$ canonical forms can be grouped under these, with the members of each group being related through {\em nonlinear transformations} of the coordinates (see Table \ref{symtab}).
\begin{table}[hbt]
\begin{center}
\begin{tabular}{|c|c|}
  \hline

   Cases & Symmetry algebra \\ \hline\hline
  I, IX & $so(3)$ \\ \hline
  II, III & $sl(2)$ \\ \hline
  IV, VI, VII & $e(2)$ \\ \hline
  V, VIII & $e(1,1)$ \\
  \hline
\end{tabular}
\end{center}
\caption{The symmetry algebras for the Krall-Sheffer metrics} \label{symtab}
\end{table}
The algebras $so(3)$ and $sl(2)$ generate metrics with constant curvature, while the Euclidean and pseudo-Euclidean algebras generate flat metrics.
For the purposes of this grouping we label the coordinates $(x_i,y_i),\, i=1,\dots ,9$.

\subsection{The Rotation Algebra $so(3)$: Cases I and XI}\label{i-ix}

The abstract algebra, with ${\bf A}, {\bf B}$ and ${\bf C}$ satisfying
$$
[{\bf A},{\bf B}]={\bf C},\;\; [{\bf B},{\bf C}]={\bf A},\;\; [{\bf C},{\bf A}]={\bf B},\quad\mbox{has Casimir function}\quad
{\cal C} = {\bf A}^2+{\bf B}^2+{\bf C}^2.
$$
The concrete realisation
$$
{\bf A}=2 \sqrt{x_1(1-x_1-y_1)}\,\pa_{x_1},\quad {\bf B}=2 \sqrt{y_1(1-x_1-y_1)}\,\pa_{y_1},\quad {\bf C}=2 \sqrt{x_1y_1}\,(\pa_{x_1}-\pa_{y_1}),
$$
gives the Laplace-Beltrami operator
$$
L_b=-\frac{1}{4}{\cal C}=(x_1^2-x_1) \pa_{x_1}^2+2x_1y_1 \pa_{x_1}\pa_{y_1}+(y_1^2-y_1)\pa_{y_1}^2+\frac{1}{2}((3x_1-1)\pa_{x_1}+(3y_1-1)\pa_{y_1}).
$$
The concrete realisation
$$
{\bf A}=\sqrt{1-x_9^2-y_9^2}\,\pa_{x_9},\quad {\bf B}= \sqrt{1-x_9^2-y_9^2}\,\pa_{y_9},\quad {\bf C}=y_9\pa_{x_9}-x_9\pa_{y_9},
$$
gives the Laplace-Beltrami operator
$$
L_b=- {\cal C} =(x_9^2-1) \pa_{x_9}^2+2x_9y_9 \pa_{x_9}\pa_{y_9}+(y_9^2-1)\pa_{y_9}^2+2(x_9\pa_{x_9}+y_9\pa_{y_9}).
$$

In each case, the {\em involution}
$$
(\bar x,\bar y)=(y,x) \quad\mbox{gives the automorphism}\quad {\bf A}\leftrightarrow {\bf B},\;\; {\bf C}\rightarrow -{\bf C},
$$
which leaves the $L_b$ invariant.  The two realisations are related through the change of coordinates $x_1=x_9^2,\quad y_1=y_9^2$.

\subsection{The Special Linear Algebra $sl(2)$: Cases II and III}\label{ii-iii}

The abstract algebra, with ${\bf E}, {\bf F}$ and ${\bf H}$ satisfying
$$
[{\bf H},{\bf E}]=2{\bf E},\quad [{\bf H},{\bf F}]=-2{\bf F},\quad [{\bf E},{\bf F}]={\bf H},\quad\mbox{has Casimir function}\quad
{\cal C} = {\bf H}^2+2{\bf E}{\bf F}+2{\bf F}{\bf E}.
$$
The concrete realisation
$$
{\bf H}=4x_2\pa_{x_2},\quad {\bf E}=2 \sqrt{x_2y_2}\,\pa_{y_2},\quad {\bf F}=4 \sqrt{x_2y_2}\,\pa_{x_2}+2(y_2-1)\sqrt{\frac{y_2}{x_2}}\pa_{y_2},
$$
gives the Laplace-Beltrami operator
$$
L_b=\frac{1}{16} {\cal C} =x_2^2 \pa_{x_2}^2+2x_2y_2 \pa_{x_2}\pa_{y_2}+(y_2^2-y_2)\pa_{y_2}^2+\frac{1}{2}(3x_2\pa_{x_2}+(3y_2-1)\pa_{y_2}),
$$
which is invariant under the {\em involution}
$$
(\bar x_2,\bar y_2)=\left(\frac{(y_2-1)^2}{x_2},y_2\right), \quad\mbox{giving the automorphism}\quad
                                     {\bf E}\leftrightarrow {\bf F},\;\; {\bf H}\rightarrow -{\bf H}.
$$
The concrete realisation
$$
{\bf H}=4x_3\pa_{x_3}+2y_3\pa_{y_3},\quad {\bf E}= x_3\pa_{y_3},\quad {\bf F}=4y_3\pa_{x_3}+\frac{4x_3+3y_3^2}{x_3}\,\pa_{y_3},
$$
gives the Laplace-Beltrami operator
$$
L_b=\frac{1}{16} {\cal C} = x_3^2 \pa_{x_3}^2+2x_3y_3 \pa_{x_3}\pa_{y_3}+(y_3^2+x_3)\pa_{y_3}^2+\frac{3}{2}(x_3\pa_{x_3}+y_3\pa_{y_3}),
$$
which is invariant under the {\em involution}
$$
(\bar x_3,\bar y_3)=\left(\frac{(y_3^2+4x_3)^2}{x_3^3},\frac{(y_3^2+4x_3)y_3}{x_3^2}\right), \quad\mbox{giving the automorphism}\quad
                                     {\bf E}\leftrightarrow {\bf F},\;\; {\bf H}\rightarrow -{\bf H}.
$$
The two realisations are related through the change of coordinates $x_2=-64 x_3,\quad y_2=-y_3^2/(4x_3)$.

\subsection{The Euclidean Algebra $e(2)$: Cases IV, VI and VII}\label{iv-vii}

The abstract algebra, with ${\bf A}, {\bf B}$ and ${\bf C}$ satisfying
$$
[{\bf A},{\bf B}]=0,\quad [{\bf C},{\bf A}]={\bf B},\quad [{\bf B},{\bf C}]={\bf A},\quad\mbox{has Casimir function}\quad
{\cal C} = {\bf A}^2+{\bf B}^2.
$$
The concrete realisation
$$
{\bf A}=\sqrt{x_4}\pa_{x_4},\quad {\bf B}= \sqrt{y_4}\pa_{y_4},\quad {\bf C}=2\sqrt{x_4y_4}\,(\pa_{y_4}-\pa_{x_4}),
$$
gives the Laplace-Beltrami operator
$$
L_b= {\cal C} = x_4 \pa_{x_4}^2+y_4\pa_{y_4}^2+\frac{1}{2}(\pa_{x_4}+\pa_{y_4}),
$$
which is invariant under the {\em involution}
$$
(\bar x_4,\bar y_4)=(y_4,x_4), \quad\mbox{giving the automorphism}\quad {\bf A}\leftrightarrow {\bf B},\;\; {\bf C}\rightarrow -{\bf C}.
$$
The concrete realisation
$$
{\bf A}=\sqrt{x_6}\pa_{x_6},\quad {\bf B}= \pa_{y_6},\quad {\bf C}=\sqrt{x_6}\,(2\pa_{y_6}-y_6\pa_{x_6})
$$
gives the Laplace-Beltrami operator
$$
L_b= {\cal C} = x_6 \pa_{x_6}^2+\pa_{y_6}^2+\frac{1}{2}\pa_{x_6},
$$
which is invariant under the {\em involution}
$$
(\bar x_6,\bar y_6)=\left(\frac{y_6^2}{4},2\sqrt{x_6}\right), \quad\mbox{giving the automorphism}\quad {\bf A}\leftrightarrow {\bf B},\;\; {\bf C}\rightarrow -{\bf C}.
$$
The concrete realisation
$$
{\bf A}=\pa_{x_7},\quad {\bf B}=\pa_{y_7},\quad {\bf C}=x_7\pa_{y_7}-y_7\pa_{x_7}
$$
gives the Laplace-Beltrami operator
$$
L_b= {\cal C} = \pa_{x_7}^2+\pa_{y_7}^2,
$$
which is invariant under the {\em involution}
$$
(\bar x_7,\bar y_7)=(y_7,x_7), \quad\mbox{giving the automorphism}\quad {\bf A}\leftrightarrow {\bf B},\;\; {\bf C}\rightarrow -{\bf C}.
$$
These realisations are related through the changes of coordinates: $x_4=x_6,\; y_4=y_6^2$, and $x_6=x_7^2,\; y_6=y_7$.

\subsection{The Pseudo-Euclidean Algebra $e(1,1)$: Cases V and VIII}\label{v-viii}

The abstract algebra, with ${\bf E}, {\bf F}$ and ${\bf H}$ satisfying
$$
[{\bf H},{\bf E}]={\bf E},\quad [{\bf H},{\bf F}]=-{\bf F},\quad [{\bf E},{\bf F}]=0,\quad\mbox{has Casimir function}\quad
{\cal C} = {\bf E}{\bf F}.
$$
The concrete realisation
$$
{\bf H}=2x_5\pa_{x_5},\quad {\bf E}= \sqrt{x_5}\pa_{y_5},\quad {\bf F}=2\sqrt{x_5}\,\pa_{x_5}+\frac{y_5}{\sqrt{x_5}}\, \pa_{y_5},
$$
gives the Laplace-Beltrami operator
$$
L_b= {\cal C} = 2x_5 \pa_{x_5}\pa_{y_5}+y_5\pa_{y_5}^2+\pa_{y_5}.
$$
which is invariant under the {\em involution}
$$
(\bar x_5,\bar y_5)=\left(\frac{y_5^2}{x_5},y_5\right) \quad\mbox{giving the automorphism}\quad
                                     {\bf E}\leftrightarrow {\bf F},\;\; {\bf H}\rightarrow -{\bf H}.
$$
The concrete realisation
$$
{\bf H}=\left(x_8-\frac{3}{4}y_8^2\right)\pa_{x_8}-y_8\pa_{y_8},\quad {\bf E}= y_8\pa_{x_8}+2\pa_{y_8},\quad {\bf F}=\pa_{x_8},
$$
gives the Laplace-Beltrami operator
$$
L_b= {\cal C} = y_8\pa_{x_8}^2+2\pa_{x_8}\pa_{y_8},
$$
which is invariant under the {\em involution}
$$
(\bar x_8,\bar y_8)=\left(\frac{(y_8^2+4x_8)^2}{16}+\frac{y_8}{2},2x_8-\frac{y_8^2}{2}\right) \quad\mbox{giving the automorphism}\quad
                                     {\bf E}\leftrightarrow {\bf F},\;\; {\bf H}\rightarrow -{\bf H}.
$$
These realisations are related through the changes of coordinates: $x_5=\frac{1}{16}\, (4x_8-y_8^2)^2,\; y_5=\frac{y_8}{8}\, (4x_8-y_8^2)$.

\section{Adding Potentials: Separability and Super-Integrability}\label{super}

Second order operators, commuting with $L_b$, are just symmetric quadratic
forms of Killing vectors.  Suppose $\bf K$ is such an operator.  Then we may
seek functions $U$ and $V$, such that
\be  \label{lk=0}
[L_b+U,{\bf K}+V]=0,
\ee
which constitutes a coupled system of partial differential equations for $U$
and $V$.  The solution depends upon a pair of arbitrary functions, each of one
variable, typical of separable systems.  Requiring that there exist two of
these second order commuting operators, $I_j={\bf K}_j+V_j$, strongly
constrains these (formerly) arbitrary functions, which reduce to rational
functions depending upon only a finite number of parameters.  The coefficients
of the second order derivatives in such an operator $\bf K$ define a
contravariant, rank-two Killing tensor.  For brevity, we will refer to these
operators as Killing tensors in what follows.

Such calculations are standard, so we only present (brief) details of Case I and then just some super-integrable cases for the others.  In the calculations it is only necessary to consider one of each of the equivalence classes listed in Table \ref{symtab}.  Within each equivalence class we can (approximately) halve the number of choices of quadratic Killing tensor, by using the corresponding involutive automorphism.

\subsection{Krall-Sheffer Metrics I and IX}

First consider $I_1={\bf A}^2+V$ in (\ref{lk=0}).  The bracket of two second order operators would generally be {\em third} order, but in this case the third and second order terms automatically vanish, since ${\bf A}^2$ {\em commutes} with $L_b$.  For the remaining 3 coefficients to vanish, we require
$$
(x-1)V_x+yV_y=4(1-x-y)U_x,\quad xV_x+(y-1)V_y=0,\quad L_bV={\bf A}^2U.
$$
The integrability condition for the first pair is just
$$
\pa_x (xU_x+(y-1)U_y+U)=0 ,
$$
leading to the general solution of both $U$ and $V$ depending upon 2 arbitrary functions:
\be  \label{pots1a}
U(x,y)=U_1(y)+\frac{U_2\left(\frac{y-1}{x}\right)}{y-1},\quad V(x,y)=4 U_2\left(\frac{y-1}{x}\right).
\ee
The third condition is then automatically satisfied.

\br[Separation of Variables]
Written in terms of the coordinates $\xi=(y-1)/x,\; \eta=y$, the operator $L$ (with the potential (\ref{pots1a})) is separable.
\er

The case of $I_2={\bf B}^2+V$ is obtained from $I_1$ by using the involution.  In this particular case, this is a very elementary operation of switching the roles of $x$ and $y$.

The case of $I_3={\bf C}^2+V$ is calculated independently and leads to another second order PDE for $U(x,y)$:
$(x\pa_x+y\pa_y+2) (U_x-U_y)=0$, giving
\be  \label{pots1c}
U(x,y)=U_3(x+y)+\frac{U_4\left(\frac{y}{x}\right)}{x+y},\quad V(x,y)=-4 U_4\left(\frac{y}{x}\right),
\ee
which is symmetric (in form) under the involution.

\subsubsection{Super-Integrability}

Requiring two of above operators (say $I_1$ and $I_2$) to commute with $L$ reduces the potential to a rational function.
In this case we have {\em three} such commuting operators, $L=L_b+U,\; I_1=A^2+V_1,\; I_2=B^2+V_2,\; I_3=C^2+V_3$, where
\be\label{superpots1}
\begin{array}{ll}
\displaystyle U=\frac{k_1}{(1-x-y)}+\frac{k_2}{x}+\frac{k_3}{y},
    & \displaystyle   V_1=\frac{4k_1x}{(x+y-1)}+\frac{4k_2(y-1)}{x},\\[3mm]
\displaystyle V_2=\frac{4k_1y}{(x+y-1)}+\frac{4k_3(x-1)}{y},
    & \displaystyle    V_3=-\, \frac{4k_2y}{x}-\, \frac{4k_3x}{y}.
\end{array}
\ee
These operators are not independent, satisfying $4L+I_1+I_2+I_3=4k_1$, which reduces to $4L_b+A^2+B^2+C^2=0$, when the parameters are zero.  If we extend the involution to act on the parameters as $k_1\rightarrow k_1,\; k_2\leftrightarrow k_3$, then
$$
L\rightarrow L,\quad I_1 \leftrightarrow I_2,\quad I_3 \rightarrow I_3.
$$

\subsubsection{The Transformation to Case IX}

We have a transformation between the Laplace-Beltrami operators of cases I and IX.  The potential (\ref{pots1a}) is transformed to
$$
U(x,y)=U_1(y^2)+\frac{U_2\left(\frac{y^2-1}{x^2}\right)}{y^2-1},\quad V(x,y)=4 U_2\left(\frac{y^2-1}{x^2}\right).
$$
The potential (\ref{pots1c}) is transformed to
$$
U(x,y)=U_3(x^2+y^2)+\frac{U_4\left(\frac{y^2}{x^2}\right)}{x+y},\quad V(x,y)=-4 U_4\left(\frac{y^2}{x^2}\right).
$$
In particular, the potentials (\ref{superpots1}) give us the corresponding super-integrable case for Case IX.

\subsubsection{The Krall-Sheffer Gauge}

Starting with the operator $L=L_b+U$, with $U$ defined by (\ref{superpots1}), the gauge transformation
$$
L\mapsto G^{-1}LG,\quad\mbox{with}\quad G=(1-x-y)^ax^by^c
$$
leads to $k_1=\frac{1}{2}a(2a-1),\, k_2=\frac{1}{2}b(2b-1),\, k_3=\frac{1}{2}c(2c-1)$ if we require the potential to reduce to a constant.  The operator then reduces to the Krall-Sheffer form with
$$
a=\frac{1}{4}(2(\beta+\kappa_1+\kappa_2)-1),\quad b=-\frac{1}{4}(2\kappa_1+1),\quad c=-\frac{1}{4}(2\kappa_2+1).
$$
The operators $I_1, I_2$ and $I_3$ then (up to an additive constant) take the form
\bea
 I_1 &=& x(1-x-y)\pa_x^2+(\kappa_1(y-1)-(\beta+\kappa_2)x)\pa_x,\nn\\
  I_2 &=& y(1-x-y)\pa_y^2+(\kappa_2(x-1)-(\beta+\kappa_1)y)\pa_y, \nn\\
  I_3 &=& xy(\pa_x-\pa_y)^2+(\kappa_2x-\kappa_1y)(\pa_x-\pa_y).  \nn
\eea

\br
The choice of variables entering the definition of $G$ is dictated by the potential $U$, but the specific form needs to be determined.
\er

If we transform to case IX the second order parts of the two $L$ operators are related in the usual way, but the first order parts take the form
$$
\frac{2 \kappa_1+1+(2\beta-1)x^2}{x}\, \pa_x+ \frac{2 \kappa_2+1+(2\beta-1)y^2}{y}\, \pa_y,
$$
containing non-polynomial parts.  To remove these we must set $\kappa_1=\kappa_2=-\frac{1}{2}$.  The other coefficients are related by $\beta_9=2\beta_1-1$.  It should be noted that this gives the {\em full} parameter version of Case IX from a restricted version of Case I.

\subsection{Super-Integrability for Metrics II and III}

If we consider $L=L_b+U,\; I_1=H^2+V_1,\; I_2=E^2+V_2$, then we find
\be\label{superpots2he}
U=\frac{k_1}{x}+\frac{k_2}{y}+\frac{k_3(y-1)}{x^2},\quad
 V_1=\frac{16k_1(1-y)}{x}-\frac{16k_3(1-y)^2}{x^2},\quad V_2=-4k_2\, \frac{x}{y}-4k_3\, \frac{y}{x}.
\ee

\br
The commutator $[I_1,I_2]$ is a third order operator, so cannot be written as a polynomial expression in $L,\, I_1,\, I_2$.  However, the next commutators, $[I_1,[I_1,I_2]]$ and $[I_2,[I_1,I_2]]$ can, giving us a quadratic algebra (see \cite{f07-1}), but this will not be considered here.
Such operator algebras play an important role in the theory of exactly and quasi-exactly solvable operators \cite{09-5}.
\er

Under the involution, these operators transform to $\bar L=L_b+\bar U,\; \bar I_1=H^2+\bar V_1,\; I_3=F^2+V_3$, where
$$
\bar U=\frac{k_1x}{(y-1)^2}+\frac{k_2}{y}+\frac{k_3x^2}{(y-1)^3},\quad
\bar V_1=\frac{16k_1x}{(1-y)}-\frac{16k_3x^2}{(1-y)^2},\quad V_3=-4k_2\, \frac{(y-1)^2}{xy}-4k_3\, \frac{xy}{(y-1)^2}.
$$

If we consider $L=L_b+U,\; I_2=E^2+V_2,\; I_3=F^2+V_3$, then we find
$$
U=\frac{k_2}{y},\quad V_2= \frac{-4k_2x}{y}, \quad V_3=\frac{-4k_2(y-1)^2}{xy},
$$
and, furthermore, $[H,L]=0$.  These operators all commute with $L$ and form the algebra:
$$
 [H,I_2]=4I_2,\quad  [H,I_3]=-4I_3,\quad [I_2,I_3]=16 HL-H^3+2(8k_2-1) H.
$$
They also satisfy the constraint:
$$
I_2I_3+I_3I_2=32L^2-16(4k_2+1)L-4H^2L+\frac{1}{8}H^4+\frac{1}{2} (5-8k_2)H^2+16k_2(2k_2-1).
$$

\subsubsection{The Transformation to Case III}

All of these formula can be transformed to Case III.  For example, this last algebra is satisfied by
$$
L=L_b-\frac{4k_2x}{y^2},\quad I_2=256\left(E^2-\frac{4k_2x^2}{y^2}\right),\quad I_3=\frac{1}{256}\left(F^2-\frac{4k_2x^2}{y^2}\right), \quad H,
$$
where $L_b,\, E,\, F$ and $H$ are those in Case III.
The numerical factor follows from the fact that $E_2\mapsto -16 E_3,\;\; F_2\mapsto -\frac{1}{16} F_3,\;\; H_2\mapsto H_3$, which satisfy the same commutation relations, since such a scaling is just another isomorphism of the algebra.

\subsubsection{The Krall-Sheffer Gauge}

Starting with the operator $L=L_b+U$, with $U$ defined by (\ref{superpots2he}), the gauge transformation
$$
L\mapsto G^{-1}LG,\quad\mbox{with}\quad G=e^{a(y-1)/x}x^by^c
$$
leads to $k_1=a(1-2b),\, k_2=\frac{1}{2}c(2c-1),\, k_3=a^2$ if we require the potential to reduce to a constant.  The operator then reduces to the Krall-Sheffer form with
$$
a=\frac{\kappa_1}{2},\quad b=\frac{1}{2}(\beta+\kappa_2-1),\quad c=-\frac{1}{4}(2\kappa_2+1).
$$
The operators $I_1$ and $I_2$ then (up to an additive (and an overall multiplicative) constant) take the form
$$
 I_1 = x^2\pa_x^2+((\beta+\kappa_2)x+\kappa_1(1-y))\pa_x,   \quad
  I_2 = xy\pa_y^2+(\kappa_1y-\kappa_2x)\pa_y.
$$

\paragraph{Transforming to case III:} the second order parts of the two $L$ operators are related in the usual way, but the first order parts take the form
$$
\left(\beta x-\frac{\kappa_1}{64}\right)\, \pa_x+ \left(\beta y-\frac{\kappa_1y}{128 x}-(2\kappa_2+1)\frac{x}{y}\right)\, \pa_y,
$$
containing non-polynomial parts.  To remove these we must set $\kappa_1=0, \kappa_2=-\frac{1}{2}$, which leaves us with the reduced operator with first order part given by $\beta x\pa_x+\beta y\pa_y$.

To obtain the most general Krall-Sheffer operator we need to consider the operators $\frac{1}{4}(HE+EH)$ and $E^2$.  We find the following result:
$$
L=L_b+\frac{k_1}{x}+\frac{k_2y}{x^2}+\frac{k_3(x+y^2)}{x^3}
$$
commutes with the two operators
$$
I_1=\frac{1}{4}(HE+EH)-\frac{k_1y}{x}-\frac{k_2(x+y^2)}{x^2}-\frac{k_3y(2x+y^2)}{x^3}\quad\mbox{and}\quad I_2=E^2+\frac{k_2y}{x}+\frac{k_3y^2}{x^2}.
$$
The gauge transformation
$$
L\mapsto G^{-1}LG,\quad\mbox{with}\quad G=x^a e^{b(2x+y^2)/x^2+cy/x}
$$
leads to $k_1=b(4a-3)-c^2,\, k_2=-4bc,\, k_3=-4b^2$ if we require the potential to reduce to a constant.  The operator then reduces to the Krall-Sheffer form with
$$
a=\frac{\kappa_1}{2},\quad b=\frac{1}{2}(\beta+\kappa_2-1),\quad c=-\frac{1}{4}(2\kappa_2+1).
$$
The operators $I_1$ and $I_2$ then take the form
$$
 I_1 = 2x^2\pa_x\pa_y+xy\pa_y^2+(\kappa_2x-\kappa_1y)\pa_x+(\beta x+\kappa_1)\pa_y,   \quad
  I_2 = x^2\pa_y^2+(\kappa_2x-\kappa_1y)\pa_y.
$$

\subsection{The Metrics IV, VI and VII}

These cases are separated in Cartesian coordinates, so the examples of super-integrable cases are very simple and well known.  In the Krall-Sheffer gauge the operators $L$ are just direct sums of $1-$dimensional ones, corresponding to products of classical polynomials, so we omit these examples here.

\subsection{Super-Integrability for Metrics V and VIII}

If we consider $L=L_b+U,\; I_1=H^2+V_1,\; I_2=E^2+V_2$, then we find
\be\label{superpots5he}
U=k_1 y+\frac{k_2}{x}+\frac{k_3 y}{x^2},\quad
 V_1=-\,\frac{k_2 y}{x}-\,\frac{k_3 y^2}{x^2},\quad V_2= k_1 x-\, \frac{k_3}{x}.
\ee
The involution transforms to the case with $L=L_b+\bar{U},\; I_1=H^2+\bar{V}_1,\; I_3=F^2+\bar{V}_3$.

We may also consider the case $L=L_b+U,\; I_2=E^2+V_2,\; I_3=F^2+V_3$, which gives
$$
U=k_1 y+ k_2\sqrt{x}+\frac{k_3 y}{\sqrt{x}},\quad
 V_2= k_1 x +2 k_3 \sqrt{x},\quad V_3= \frac{k_1 y^2}{x}+ \frac{2k_2 y}{\sqrt{x}},
$$
which is invariant under the involution.

\subsubsection{The Krall-Sheffer Gauge}

Starting with the case (\ref{superpots5he}), the gauge transformation
$$
L\mapsto G^{-1}LG,\quad\mbox{with}\quad G=e^{(a x+b)y/x}x^c
$$
leads to $k_1=-a^2,\, k_2=b(1-2c),\, k_3=b^2$ if we require the potential to reduce to a constant.  The operator then reduces to the Krall-Sheffer form with
$$
a=\frac{\beta}{2},\quad b=\frac{\kappa_1}{2},\quad c=-\frac{(\kappa_2-1)}{2}.
$$
The operators $I_1$ and $I_2$ then (up to an additive (and an overall multiplicative) constant) take the form
$$
 I_1 = x^2\pa_x^2+(\kappa_2x-\kappa_1y)\pa_x,   \quad
  I_2 = x\pa_y^2+(\beta x+\kappa_1)\pa_y.
$$

\paragraph{Case VIII:} Transforming between these cases restricts the parameters to the Laplace-Beltrami case.
Here we need the choice $L=L_b+U,\; I_1=F^2+V_1,\; I_2=\frac{1}{4}(E^2-2(HF+FH))+V_2$, where
\bea
&& U= \frac{\beta^2}{4} y(y^2-2x)+\frac{\beta \kappa_2}{2}(y^2-x)+\frac{y}{4}(\kappa_2^2-2\beta \kappa_1),\nn\\
&& V_1= -\frac{1}{4}(\beta y+\kappa_2)^2,\quad V_2= \frac{\beta^2}{4} (y^2-x)x + \frac{x}{4}(\kappa_2^2+2\beta (\kappa_2 y-\kappa_1)).\nn
\eea
Using $G= e^{(6\kappa_2 x+6\kappa_1 y+6\beta xy-3\kappa_2 y^2-2 \beta y^3)/12}$, these gauge transform (up to additive constants) to:
\bea
L &=& y\pa_x^2+2\pa_x\pa_y+(\beta x+\kappa_1)\pa_x+(\beta y+\kappa_2)\pa_y,\nn\\
I_1 &=& \pa_x^2+(\beta y+\kappa_2)\pa_x,  \label{8ks}\\
I_2 &=& (y^2-x)\pa_x^2+2y\pa_x\pa_y+\pa_y^2+(\kappa_1 y-\kappa_2 x)\pa_x+(\beta x+\kappa_1)\pa_y,\nn
\eea

\section{The Polynomial Eigenfunctions}\label{pols}

By Definition \ref{def:admiss}, an {\em admissible operator} possesses a
sequence of eigenvalues $\lambda_N$ (with $\lambda_N\neq\lambda_M$ for $M\neq N$), such that, for each $N\ge 0$, there
exist $N+1$ linearly independent polynomial eigenfunctions of degree $N$, with
eigenvalue $\lambda_N$.  The {\em degeneracy} (for $N>0$) of $\lambda_N$ stems
from the super-integrability in a very explicit way.  We show below how to use
the operators $I_1$ and $I_2$ to build sequences of polynomial eigenfunctions
of $L$.

We may choose a basis of {\em monic} polynomials $P_{m,n}$ of the form (\ref{pmn})
and organise them in a triangular array (see Figure \ref{trilatt}), with the degree $N=m+n$ constant on horizontal layers.
Since the eigenvalue (\ref{lambda}) depends only on the degree, it is constant on each horizontal level.

\br
Cases $IV, VI$ and $VII$ separate into sums of elementary $1-$dimensional operators so their eigenfunctions are just products of classical orthogonal polynomials (respectively Laguerre-Laguerre, Laguerre-Hermite and Hermite-Hermite), so will not be discussed further.
\er

The Krall-Sheffer operators reduce to one dimensional operators on at least one of the edges of this triangle (usually both).  The two exceptional cases are III, which only reduces on the left and VIII, which only reduces on the right.  This means that there exist {\em 3 point recurrence relations} for the eigenfunctions $P_{m,0}$ and/or $P_{0,n}$.  Below, we give explicit {\em 3 {\bf level} recurrence relations} for each case and show how these reduce to the {\em 3 point recurrence relations} on the appropriate edges.  The one dimensional reductions on the left and/or right edges also possess {\em differential ladder operators}.  We also present the ladder operators in directions which are ``parallel'' to the edges for each case and show how these reduce to their one dimensional counterparts on appropriate edges.

\paragraph{What is the role of the commuting operators?} Since
$$
\left. \begin{array}{l}
Lf=\lambda f \\[2mm]
[I,L]=0
\end{array}\right\}\quad\Rightarrow\quad L(If)=\lambda (If),
$$
the function $If$ is also an eigenfunction with the same eigenvalue.  For our Krall-Sheffer operators, we have two such operators $I_1, I_2$ which take us respectively right and left on each horizontal level.  Starting from $P_{N,0}$ or $P_{0,N}$, we use these operators to build the eigenfunctions $P_{m,n}$, with $N=m+n$.
\begin{figure}[hbt]
\begin{center}
\unitlength=0.5mm
\begin{picture}(160,80)
\put(80,80){\makebox(0,0){$P_{0,0}$}}
\put(60,60){\makebox(0,0){$P_{1,0}$}}
\put(100,60){\makebox(0,0){$P_{0,1}$}}
\put(40,40){\makebox(0,0){$P_{2,0}$}}
\put(80,40){\makebox(0,0){{$P_{1,1}$}}}
\put(120,40){\makebox(0,0){$P_{0,2}$}}
\put(20,20){\makebox(0,0){$P_{3,0}$}}
\put(60,20){\makebox(0,0){$P_{2,1}$}}
\put(100,20){\makebox(0,0){$P_{1,2}$}}
\put(140,20){\makebox(0,0){$P_{0,3}$}}
\put(0,0){\makebox(0,0){$P_{4,0}$}}
\put(40,0){\makebox(0,0){{$P_{3,1}$}}}
\put(80,0){\makebox(0,0){$P_{2,2}$}}
\put(120,0){\makebox(0,0){{$P_{1,3}$}}}
\put(160,0){\makebox(0,0){$P_{0,4}$}}
\put(75,75){\vector(-1,-1){10}}
\put(85,75){\vector(1,-1){10}}
\put(55,55){\vector(-1,-1){10}}  \put(105,55){\vector(1,-1){10}}
\put(35,35){\vector(-1,-1){10}}  \put(125,35){\vector(1,-1){10}}
\put(15,15){\vector(-1,-1){10}}  \put(145,15){\vector(1,-1){10}}
\put(80,60){\makebox(0,0){$\rightleftarrows$}}
\put(60,40){\makebox(0,0){$\rightleftarrows$}}
\put(100,40){\makebox(0,0){$\rightleftarrows$}}
\put(40,20){\makebox(0,0){$\rightleftarrows$}}
\put(80,20){\makebox(0,0){$\rightleftarrows$}}
\put(120,20){\makebox(0,0){$\rightleftarrows$}}
\put(20,0){\makebox(0,0){$\rightleftarrows$}}
\put(60,0){\makebox(0,0){$\rightleftarrows$}}
\put(100,0){\makebox(0,0){$\rightleftarrows$}}
\put(140,0){\makebox(0,0){$\rightleftarrows$}}
\put(65,75){\makebox(0,0){${\bf r}_x$}}
\put(95,75){\makebox(0,0){${\bf r}_y$}}
\end{picture}
\end{center}
\caption{The triangular lattice of polynomials $P_{m,n}$
with $P_{0,0}=1$.  Horizontal arrows denote the
action of $I_1$ (right) and $I_2$ (left). The operators ${\bf r}_x$ and ${\bf r}_y$ represent the edge ladder operators.} \label{trilatt}
\end{figure}

\paragraph{The $3$ Level Recurrence and Ladder Operators.} It is possible to explicitly determine the coefficients of the 3 level recurrence relations (\ref{ks3level}).  This is a laborious calculation, so only the resulting recurrence relations are presented.  The general form of (\ref{ks3level}) constitutes a difference equation on $2(m+n+1)$ points, but our formulae contain at most $9$ points, and each equation fits into the following $12$ point relation:
\bea
P_{m+1,n} &=& x P_{m,n} +\sum_{i=0}^3 a_i P_{m-1+i,n+1-i}+\sum_{i=0}^{6}b_i P_{m-3+i,n+2-i}, \nn\\[-1.5mm]
&&  \label{12point}  \\
P_{m,n+1} &=& y P_{m,n} +\sum_{i=0}^{m+n} c_i P_{m+1-i,n-1+i}+\sum_{i=0}^6 d_i P_{m+2-i,n-3+i}, \nn
\eea
{\em Given} the first of these, we operate with
$$
L=A\pa_x^2+2B\pa_x\pa_y+C\pa_y^2+D\pa_x+E\pa_y,
$$
taking into account that $P_{m,n}$, is an eigenfunction, with eigenvalue $\lambda_{m+n}$, we find
\bea
\lambda_{m+n+1}P_{m+1,n} &=& \lambda_{m+n} (x P_{m,n}+\sum_{i=0}^3a_i P_{m-1+i,n+1-i})+(2A\pa_x+2B\pa_y+D)P_{m,n}\nn\\
&& \qquad\qquad +\lambda_{m+n-1} \sum_{i=0}^{6}b_i P_{m-3+i,n+2-i}.  \nn
\eea
Subtracting $\lambda_{m+n-1}$ times the first of (\ref{12point}), leads to a first order {\em two level} differential-difference relation:
\bea
(\lambda_{m+n+1}-\lambda_{m+n-1})P_{m+1,n} &=& (\lambda_{m+n}-\lambda_{m+n-1}) (x P_{m,n}+\sum_{i=0}^3a_i P_{m-1+i,n+1-i}) \nn\\
&& \qquad\qquad +(2A\pa_x+2B\pa_y+D)P_{m,n}.  \label{2leveldiffdel}
\eea
The next step in the calculation is to use the {\em horizontal shift} operators $I_1$ and $I_2$ to write all the polynomials of degree $m+n$ (arising in (\ref{2leveldiffdel})) in terms of the polynomial $P_{m,n}$.  In principle, this could result in a fourth order differential operator, but only second and first order operators occur.  This, together with a similar manipulation of the second of (\ref{12point}), gives two differential operators which {\em raise the degree} of the polynomial eigenfunction:
\be\label{r+}
R_{+,x}P_{m,n}=P_{m+1,n},\quad R_{+,y}P_{m,n}=P_{m,n+1}.
\ee
It is also possible to construct {\em lowering operators}, but these are usually considerably more complicated and
(usually) depend explicitly upon both $m$ and $n$ (not just through the combination $m+n$).

The formulae for $[L,R_{+,x}]$ and $[L,R_{+,y}]$ are non-standard but still imply that $R_{+,x}P_{m,n}$ and $R_{+,y}P_{m,n}$ are eigenfunctions of $L$, with eigenvalue $\lambda_{m+n+1}$.

\subsection{Krall-Sheffer Metric I}\label{pols1}

Here we have
\bea
L &=& (x^2-x) \pa_x^2+2 x y \pa_x\pa_y+ (y^2-y) \pa_y^2+ (\beta x+\kappa_1)\pa_x+ (\beta y+\kappa_2)\pa_y, \nn\\
 I_1 &=& x(1-x-y)\pa_x^2+(\kappa_1(y-1)-(\beta+\kappa_2)x)\pa_x,\nn\\
  I_2 &=& y(1-x-y)\pa_y^2+(\kappa_2(x-1)-(\beta+\kappa_1)y)\pa_y, \nn\\
  I_3 &=& xy(\pa_x-\pa_y)^2+(\kappa_2x-\kappa_1y)(\pa_x-\pa_y).  \nn
\eea
The polynomial eigenfunctions of $L$ satisfy the $6$ point difference equations:
\bea
P_{m+1,n} &=& x P_{m,n}  \nn\\
&& +\frac{1}{\gamma_{2N}\gamma_{2N-2}}\, (((\beta+2n-2)(\kappa_1-2m)-2m(m+1)) P_{m,n} + 2n(n-\kappa_2-1) P_{m+1,n-1}) \nn\\
&& +\frac{1}{\gamma_{2N-1}\gamma_{2N-2}^2\gamma_{2N-3}}\, (m(\beta+m+2n-2)(\kappa_1-m+1)(\beta+\kappa_1+m+2n-1) P_{m-1,n} \nn\\ && +n(\kappa_2-n+1)((\beta+2n-3)(\kappa_1-2m)-2m(m+1)) P_{m,n-1}  \nn\\
&& \hspace{2cm} -n(n-1)(\kappa_2-n+1)(\kappa_2-n+2)P_{m+1,n-2}),  \label{1pmp1n}
\eea
and
\bea
P_{m,n+1} &=& y P_{m,n}  \nn\\
&&   +\frac{1}{\gamma_{2N}\gamma_{2N-2}}\,
   (((\beta+2m-2)(\kappa_2-2n)-2n(n+1))P_{m,n} + 2m(m-\kappa_1-1) P_{m-1,n+1})  \nn\\
&& +\frac{1}{\gamma_{2N-1}\gamma_{2N-2}^2\gamma_{2N-3}}\, (n(\beta+2m+n-2)(\kappa_2-n+1)(\beta+\kappa_2+2m+n-1)P_{m,n-1}  \nn\\
&& +m(\kappa_1-m+1)((\beta+2m-3)(\kappa_2-2n)-2n(n+1))P_{m-1,n}  \nn\\
&&  \hspace{2cm} -m(m-1)(\kappa_1-m+1)(\kappa_1-m+2) P_{m-2,n+1}),  \label{1pmnp1}
\eea
where $N=m+n, \gamma_n=\beta+n$.  The $6$ points of these difference equations are depicted by bullet points in Figure \ref{diff1fig}.
\begin{figure}[htb]
\begin{center}
\unitlength=0.3mm
\begin{picture}(160,60)
\put(60,60){\makebox(0,0){$\circ$}}
\put(50,50){\makebox(0,0){$\circ$}}
\put(70,50){\makebox(0,0){$\circ$}}
\put(40,40){\makebox(0,0){$\circ$}}
\put(60,40){\makebox(0,0){{$\circ$}}}
\put(80,40){\makebox(0,0){$\circ$}}
\put(30,30){\makebox(0,0){$\circ$}}
\put(50,30){\makebox(0,0){$\circ$}}
\put(70,30){\makebox(0,0){$\circ$}}
\put(90,30){\makebox(0,0){$\circ$}}
\put(20,20){\makebox(0,0){$\circ$}}
\put(40,20){\makebox(0,0){$\bullet$}}
\put(60,20){\makebox(0,0){$\bullet$}}
\put(80,20){\makebox(0,0){$\bullet$}}
\put(100,20){\makebox(0,0){$\circ$}}
\put(10,10){\makebox(0,0){{$\circ$}}}
\put(30,10){\makebox(0,0){{$\circ$}}}
\put(50,10){\makebox(0,0){{$\bullet$}}}
\put(70,10){\makebox(0,0){{$\bullet$}}}
\put(90,10){\makebox(0,0){{$\circ$}}}
\put(110,10){\makebox(0,0){{$\circ$}}}
\put(0,0){\makebox(0,0){$\circ$}}
\put(20,0){\makebox(0,0){{$\circ$}}}
\put(40,0){\makebox(0,0){$\circ$}}
\put(60,0){\makebox(0,0){$\bullet$}}
\put(80,0){\makebox(0,0){$\circ$}}
\put(100,0){\makebox(0,0){$\circ$}}
\put(120,0){\makebox(0,0){$\circ$}}
\end{picture}
\end{center}
\caption{The pattern for the difference equations for both $P_{m+1,n}$ and $P_{m,n+1}$ for both cases I and II.}\label{diff1fig}
\end{figure}
Notice that when $n=1$, the point $(m+1,n-2)$ is out of the Figure, but also that in formula (\ref{1pmp1n}) the coefficient of $P_{m+1,n-2}$ vanishes when $n=1$, so the formula is still valid.  When $n=0$, the three points $(m+1,n-2), (m+1,n-1)$ and $(m,n-1)$ fall outside the Figure, but again, the corresponding coefficients vanish, leaving us with the {\em 3 {\bf point} recurrence relation} on the left edge:
\bea
P_{m+1,0} &=& \left(x+\frac{(\beta-2)(\kappa_1-2m)-2m(m+1)}{(\beta+2m)(\beta+2m-2)}\right) P_{m,0}  \nn\\
&& +\left(\frac{m(\beta+m-2)(\kappa_1-m+1)(\beta+\kappa_1+m-1)}{(\beta+2m-1)(\beta+2m-2)^2(\beta+2m-3)}\right) P_{m-1,0}.  \label{1pmp10}
\eea
Similarly, as we approach the right edge, the formula (\ref{1pmnp1}) consistently reduces to the {\em 3 point recurrence relation} on the right edge:
\bea
P_{0,n+1} &=& \left(y+\frac{(\beta-2)(\kappa_2-2n)-2n(n+1)}{(\beta+2n)(\beta+2n-2)}\right) P_{0,n}  \nn\\
&& +\left(\frac{n(\beta+n-2)(\kappa_2-n+1)(\beta+\kappa_2+n-1)}{(\beta+2n-1)(\beta+2n-2)^2(\beta+2n-3)}\right) P_{0,n-1}.  \label{1p0np1}
\eea
The left and right edge polynomials satisfy respective Jacobi equations:
\bea
L^{(x)} P_{m,0}=(x(x-1)\pa_x^2+(\beta x+\kappa_1)\pa_x)P_{m,0} &=& m(m-1+\beta)P_{m,0}, \nn\\
L^{(y)} P_{0,n}=(y(y-1)\pa_y^2+(\beta y+\kappa_2)\pa_y)P_{0,n} &=& n(n-1+\beta)P_{0,n}.  \nn
\eea
To proceed with these {\em second order} difference equations, we expect to need the three lowest polynomials:
\be\label{p00}
P_{0,0}=1,\quad P_{1,0}= x+\frac{\kappa_1}{\beta}, \quad P_{0,1}= y+\frac{\kappa_2}{\beta},
\ee
which are calculated directly as monic, degree one polynomial eigenfunctions. However, when $m=0$, (\ref{1pmp10}) is consistent and can be used to construct $P_{1,0}$ (and similarly for $P_{0,1}$).  We then use the difference equations (\ref{1pmp1n}) and (\ref{1pmnp1}) to calculate the degree 2, degree 3, etc polynomials successively.

Alternatively, we may calculate the polynomials $P_{m,0}$ and $P_{0,n}$, using (\ref{1pmp10}) and (\ref{1p0np1}) and then use the commuting operators to calculate the remaining polynomials on each level.  We know that, since the operators $I_1$ and $I_2$ commute with $L$, the functions $I_kP_{m,n}, k=1,2$, are also polynomial eigenfunctions of the same degree.  By looking at the leading order (monic) part, we obtain:
\bea
I_1P_{m,n}+m(\beta+\kappa_2+m-1)P_{m,n} &=& m(\kappa_1-m+1)P_{m-1,n+1} ,\nn\\
I_2P_{m,n}+n(\beta+\kappa_1+n-1)P_{m,n} &=& n(\kappa_2-n+1)P_{m+1,n-1} ,\nn
\eea
We see that, starting with $P_{m,0}$, $I_1$ moves us to the right, until we reach $P_{0,n}$, which satisfies $I_1P_{0,n}=0$.  $I_2$ similarly moves us from right to left.

Finally, we may construct the operators (\ref{r+}), which are given by
\bea
R_{+,x} &=& \frac{1}{(\beta+2N)(\beta+2N-1)}((\beta+N-1)((\beta+2N)x+\kappa_1-N)+(\beta+2N)x(x-1)\pa_x  \nn\\
  &&  +((\beta+2N)xy+(\beta+\kappa_1)y+\kappa_2(1-x))\pa_y+y(x+y-1)\pa_y^2),  \nn\\

R_{+,y} &=& \frac{1}{(\beta+2N)(\beta+2N-1)}((\beta+N-1)((\beta+2N)y+\kappa_2-N)+(\beta+2N)y(y-1)\pa_y  \nn\\
  &&  +((\beta+2N)xy+(\beta+\kappa_2)x+\kappa_1(1-y))\pa_x+x(x+y-1)\pa_x^2).  \nn
\eea
The formulae interchange under the transformation $x\leftrightarrow y, \kappa_1\leftrightarrow \kappa_2$.
These operators satisfy the commutation relations
\bea
[L,R_{+,x}] &=& \frac{(2x-1)(L-\lambda_N)}{\beta+2N-1} +(\lambda_{N+1}-\lambda_N)R_{+,x}, \nn\\[2mm]
[L,R_{+,y}] &=& \frac{(2y-1)(L-\lambda_N)}{\beta+2N-1} +(\lambda_{N+1}-\lambda_N)R_{+,y}. \nn
\eea

Starting from $P_{0,n}$, $R_{+,x}$ is used to build the functions $P_{m,n}, m\geq 1$, with $n$ fixed.  When $n=0$, $R_{+,x}$ reduces to the $1$ dimensional ladder operator
$$
r_{+,x} = \frac{1}{(\beta+2m)(\beta+2m-1)}((\beta+m-1)((\beta+2m)x+\kappa_1-m)+(\beta+2m)x(x-1)\pa_x).
$$
Similar statements can be made regarding $R_{+,y}$.

This particular case is symmetric in $(x,y)$ and $(\kappa_1,\kappa_2)$, which is reflected in all the formulae.

\subsection{Krall-Sheffer Metric II}\label{pols2}

Here we have
\bea
L &=& x^2 \pa_x^2+2 x y \pa_x\pa_y+ (y^2-y) \pa_y^2+ (\beta x+\kappa_1)\pa_x+ (\beta y+\kappa_2)\pa_y, \nn\\
 I_1 &=& x^2\pa_x^2+((\beta+\kappa_2)x+\kappa_1(1-y))\pa_x,\nn\\
  I_2 &=& xy\pa_y^2+(\kappa_1y-\kappa_2x)\pa_y. \nn
\eea
The polynomial eigenfunctions of $L$ satisfy the $6$ point difference equations:
\bea
P_{m+1,n} &=& x P_{m,n}  \nn\\
&& +\frac{1}{\gamma_{2N}\gamma_{2N-2}}\, (\kappa_1(\beta+2n-2) P_{m,n} + 2n(n-\kappa_2-1) P_{m+1,n-1}) \nn\\
&& +\frac{1}{\gamma_{2N-1}\gamma_{2N-2}^2\gamma_{2N-3}}\, (m\kappa_1^2(\beta+m+2n-2) P_{m-1,n} \nn\\
&& +n\kappa_1(\beta+2n-3)(\kappa_2-n+1) P_{m,n-1}  \nn\\
&& \hspace{2cm} -n(n-1)(\kappa_2-n+1)(\kappa_2-n+2)P_{m+1,n-2}),  \label{2pmp1n}
\eea
and
\bea
P_{m,n+1} &=& y P_{m,n}  \nn\\
&&   +\frac{1}{\gamma_{2N}\gamma_{2N-2}}\,
   (((\beta+2m-2)(\kappa_2-2n)-2n(n+1))P_{m,n} -2m\kappa_1 P_{m-1,n+1})  \nn\\
&& +\frac{1}{\gamma_{2N-1}\gamma_{2N-2}^2\gamma_{2N-3}}\, (n(\beta+2m+n-2)(\kappa_2-n+1)(\beta+\kappa_2+2m+n-1))P_{m,n-1}  \nn\\
&& +m\kappa_1((\beta+2m-3)(\kappa_2-2n)-2n(n+1))P_{m-1,n} -m(m-1)\kappa_1^2 P_{m-2,n+1}),  \label{2pmnp1}
\eea
where $N=m+n, \gamma_n=\beta+n$. The $6$ points of these difference equations are again depicted by the bullet points in Figure \ref{diff1fig}.
Again, as we approach the left edge, the formula (\ref{2pmp1n}) consistently reduces to the {\em 3 point recurrence relation}:
\bea
P_{m+1,0} &=& \left(x+\frac{\kappa_1(\beta-2)}{(\beta+2m)(\beta+2m-2)}\right) P_{m,0} \nn\\
&& +\left(\frac{m\kappa_1^2(\beta+m-2)}{(\beta+2m-1)(\beta+2m-2)^2(\beta+2m-3)}\right) P_{m-1,0}.  \label{2pmp10}
\eea
Similarly, as we approach the right edge, the formula (\ref{2pmnp1}) consistently reduces to the {\em 3 point recurrence relation}:
\bea
P_{0,n+1} &=& \left(y+\frac{(\beta-2)(\kappa_2-2n)-2n(n+1)}{(\beta+2n)(\beta+2n-2)}\right) P_{0,n}  \nn\\
&& +\left(\frac{n(\beta+n-2)(\kappa_2-n+1)(\beta+\kappa_2+n-1)}{(\beta+2n-1)(\beta+2n-2)^2(\beta+2n-3)}\right) P_{0,n-1}.  \label{2p0np1}
\eea
The left and right edge polynomials satisfy respectively:
\bea
L^{(x)} P_{m,0}=(x^2\pa_x^2+(\beta x+\kappa_1)\pa_x)P_{m,0} &=& m(m-1+\beta)P_{m,0}, \nn\\
  &&  \label{LxLy2}  \\
L^{(y)} P_{0,n}=(y(y-1)\pa_y^2+(\beta y+\kappa_2)\pa_y)P_{0,n} &=& n(n-1+\beta)P_{0,n}.  \nn
\eea
Again, we have the polynomials (\ref{p00}), which are the seeds for the difference equations (\ref{2pmp1n}) and (\ref{2pmnp1}) when calculating the degree 2, degree 3, etc polynomials.

Alternatively, we may calculate the polynomials $P_{m,0}$ and $P_{0,n}$, using (\ref{2pmp10}) and (\ref{2p0np1}) and then use the commuting operators to calculate the remaining polynomials on each level.  By looking at the leading order (monic) part of $I_kP_{m,n}$, we obtain:
\bea
I_1P_{m,n}-m(\beta+\kappa_2+m-1)P_{m,n} &=& -m\kappa_1P_{m-1,n+1} ,\nn\\
I_2P_{m,n}-n\kappa_1P_{m,n} &=& -n(\kappa_2-n+1)P_{m+1,n-1} ,\nn
\eea
We see that, starting with $P_{m,0}$, $I_1$ moves us to the right, until we reach $P_{0,n}$, which satisfies $I_1P_{0,n}=0$.  $I_2$ similarly moves us from right to left.

Finally, we may construct the operators (\ref{r+}), which are given by
\bea
R_{+,x} &=& \frac{1}{(\beta+2N)(\beta+2N-1)}((\beta+N-1)((\beta+2N)x+\kappa_1)+(\beta+2N)x^2\pa_x  \nn\\
  &&  +((\beta+2N)xy+\kappa_1y-\kappa_2x)\pa_y+xy\pa_y^2),  \nn\\

R_{+,y} &=& \frac{1}{(\beta+2N)(\beta+2N-1)}((\beta+N-1)((\beta+2N)y+\kappa_2-N)  \nn\\
  &&  +((\beta+2N)xy+(\beta+\kappa_2)x+\kappa_1(1-y))\pa_x+(\beta+2N)y(y-1)\pa_y+x^2\pa_x^2).  \nn
\eea
The formulae no longer interchange under the transformation $x\leftrightarrow y, \kappa_1\leftrightarrow \kappa_2$.
These operators satisfy the commutation relations
\bea
[L,R_{+,x}] &=& \frac{2x(L-\lambda_N)}{\beta+2N-1} +(\lambda_{N+1}-\lambda_N)R_{+,x}, \nn\\[2mm]
[L,R_{+,y}] &=& \frac{(2y-1)(L-\lambda_N)}{\beta+2N-1} +(\lambda_{N+1}-\lambda_N)R_{+,y}. \nn
\eea

Starting from $P_{0,n}$, $R_{+,x}$ is used to build the functions $P_{m,n}, m\geq 1$, with $n$ fixed.  When $n=0$, $R_{+,x}$ reduces to the $1$ dimensional ladder operator
\be\label{r+x2}
r_{+,x} = \frac{1}{(\beta+2m)(\beta+2m-1)}((\beta+m-1)((\beta+2m)x+\kappa_1)+(\beta+2m)x^2\pa_x).
\ee
On the right edge, $R_{+,y}$ reduces to
$$
r_{+,y} = \frac{1}{(\beta+2m)(\beta+2m-1)}((\beta+n-1)((\beta+2n)y+\kappa_2-n) +(\beta+2n)y(y-1)\pa_y).
$$

\subsection{Krall-Sheffer Metric III}\label{pols3}

Here we have
\bea
L &=& x^2 \pa_x^2+2 x y \pa_x\pa_y+ (y^2+x) \pa_y^2+ (\beta x+\kappa_1)\pa_x+ (\beta y+\kappa_2)\pa_y, \nn\\
 I_1 &=& 2x^2\pa_x\pa_y+xy\pa_y^2+(\kappa_2x-\kappa_1y)\pa_x+(\beta x+\kappa_1)\pa_y,\nn\\
  I_2 &=& x^2\pa_y^2+(\kappa_2x-\kappa_1y)\pa_y. \nn
\eea
The polynomial eigenfunctions of $L$ satisfy the $9$ point difference equations:
\bea
P_{m+1,n} &=& x P_{m,n}+\frac{1}{\gamma_{2N}\gamma_{2N-2}}\, (\kappa_1(\beta+2n-2) P_{m,n}-2n\kappa_2P_{m+1,n-1}-2n(n-1)P_{m+2,n-2}) \nn\\
&& +\frac{1}{\gamma_{2N-1}\gamma_{2N-2}^2\gamma_{2N-3}}\, (m\kappa_1^2(\beta+m+2n-2) P_{m-1,n} +n\kappa_1\kappa_2(\beta+2n-3) P_{m,n-1}  \nn\\
&& +n(n-1)(\kappa_1(\beta+2n-4)-\kappa_2^2)P_{m+1,n-2} -2n(n-1)(n-2)\kappa_2 P_{m+2,n-3}  \nn\\
&& \qquad\qquad -n(n-1)(n-2)(n-3)P_{m+3,n-4}),  \label{3pmp1n}
\eea
and
\bea
P_{m,n+1} &=& y P_{m,n}  \nn\\
&&   +\frac{1}{\gamma_{2N}\gamma_{2N-2}}\,
   (\kappa_2(\beta+2m-2)P_{m,n} + 2(n(\beta+2m+n-1) P_{m+1,n-1} -m\kappa_1 P_{m-1,n+1}))  \nn\\
&& +\frac{1}{\gamma_{2N-1}\gamma_{2N-2}^2\gamma_{2N-3}}\, (n(n-1)(n-2)(2\beta+4m+3n-3)P_{m+2,n-3}  \nn\\
&& +n(n-1)(2\beta+6m+4n-5)\kappa_2P_{m+1,n-2} +n((\beta+2m+n-2)(\kappa_2^2-\kappa_1(\beta+3n-3)) \nn\\
&&  +\kappa_1(1-n^2)) P_{m,n-1} +m\kappa_1\kappa_2(\beta+2m-3)P_{m-1,n} -m(m-1)\kappa_1^2P_{m-2,n+1}),  \label{3pmnp1}
\eea
where $N=m+n, \gamma_n=\beta+n$. The $9$ points of these difference equations are depicted by the bullet points in Figure \ref{diff3fig}.
\begin{figure}[htb]
\begin{center}
\unitlength=0.3mm
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\put(40,20){\makebox(0,0){$\bullet$}}
\put(60,20){\makebox(0,0){$\bullet$}}
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\put(100,20){\makebox(0,0){$\bullet$}}
\put(10,10){\makebox(0,0){{$\circ$}}}
\put(30,10){\makebox(0,0){{$\circ$}}}
\put(50,10){\makebox(0,0){{$\bullet$}}}
\put(70,10){\makebox(0,0){{$\bullet$}}}
\put(90,10){\makebox(0,0){{$\bullet$}}}
\put(110,10){\makebox(0,0){{$\circ$}}}
\put(0,0){\makebox(0,0){$\circ$}}
\put(20,0){\makebox(0,0){{$\circ$}}}
\put(40,0){\makebox(0,0){$\circ$}}
\put(60,0){\makebox(0,0){$\circ$}}
\put(80,0){\makebox(0,0){$\bullet$}}
\put(100,0){\makebox(0,0){$\circ$}}
\put(120,0){\makebox(0,0){$\circ$}}
\end{picture}
\end{center}
\caption{The pattern for the difference equations for both $P_{m+1,n}$ and $P_{m,n+1}$ for case III.}\label{diff3fig}
\end{figure}

Note that in this case, the operator $L$ does not reduce to a 1 dimensional operator on the right edge and, similarly, the difference equations cannot be reduced here.  However on the left edge, the operator $L$ reduces to $L^{(x)}$ of (\ref{LxLy2}) and the difference equation (\ref{3pmp1n}) reduces to (\ref{2pmp10}).  Nevertheless, on the right edge, the difference equation (\ref{3pmnp1}) still makes sense, so can be used to construct the (2 variable) polynomials on the right edge.

Starting from the polynomials $P_{m,0}$ we may use the commuting operators to calculate the remaining polynomials on each level.  By looking at the leading order (monic) part of $I_kP_{m,n}$, we obtain:
\bea
I_1P_{m,n}-m\kappa_2P_{m,n} &=& -m\kappa_1P_{m-1,n+1}+n(\beta+2m+n-1)P_{m+1,n-1} ,\nn\\
I_2P_{m,n}+n\kappa_1P_{m,n} &=& n\kappa_2P_{m+1,n-1}+n(n-1)P_{m+2,n-2} ,\nn
\eea
which are each {\bf 3-point} differential, recurrence relations on each level.  At the beginning, when $n=0, m=N$, the first gives a 2-point relation, giving $P_{N-1,1}$.  When $m=0, n=N$, we have $I_1P_{0,N}=N(\beta+N-1)P_{1,N-1}$.  Having constructed the polynomials $P_{m,n}, m+n=N$ in this way, we may operate with $I_2$ to move to the left. In this case $I_2P_{N,0}=0$.

Finally, we may construct the operators (\ref{r+}), which are given by
\bea
R_{+,x} &=& \frac{1}{\gamma_{2N}\gamma_{2N-1}}\,(\gamma_{N-1}(\gamma_{2N}x+\kappa_1)+\gamma_{2N}x^2\pa_x   +(\gamma_{2N}xy+\kappa_1y-\kappa_2x)\pa_y-x^2\pa_y^2),  \nn\\

R_{+,y} &=& \frac{1}{\gamma_{2N}\gamma_{2N-1}}\,(\gamma_{N-1}\,(\gamma_{2N}y+\kappa_2) +(\gamma_{2N}xy+\kappa_2x-\kappa_1y)\pa_x  \nn\\
  && \hspace{3cm} +(\gamma_{2N}y^2+2\gamma_{N}x+\kappa_1)\pa_y+2x^2\pa_x\pa_y+xy\pa_y^2).  \nn
\eea
The formulae no longer interchange under the transformation $x\leftrightarrow y, \kappa_1\leftrightarrow \kappa_2$.
These operators satisfy the commutation relations
\bea
[L,R_{+,x}] &=& \frac{2x(L-\lambda_N)}{\gamma_{2N-1}} +(\lambda_{N+1}-\lambda_N)R_{+,x}, \nn\\[2mm]
[L,R_{+,y}] &=& \frac{2y(L-\lambda_N)}{\gamma_{2N-1}} +(\lambda_{N+1}-\lambda_N)R_{+,y}. \nn
\eea
On the left edge, $R_{+,x}$ reduces to the $1$ dimensional operator (\ref{r+x2}), which generates the polynomials $P_{m,0}$.  On the right edge, $R_{+,y}$ {\em does not} reduce to a $1$ dimensional operator, but still generates the polynomials $P_{0,n}$.

\subsection{Krall-Sheffer Metric V}\label{pols5}

Here we have
\bea
L &=& 2 x \pa_x\pa_y+ y \pa_y^2+ (\beta x+\kappa_1)\pa_x+ (\beta y+\kappa_2)\pa_y, \nn\\
 I_1 &=& x^2\pa_x^2+(\kappa_2x-\kappa_1y)\pa_x,\nn\\
  I_2 &=& x\pa_y^2+(\beta x+\kappa_1)\pa_y. \nn
\eea
The polynomial eigenfunctions of $L$ satisfy the $5$ and $4$ point difference equations:
\be
P_{m+1,n} =  \left(x+\frac{\kappa_1}{\beta}\right)P_{m,n}+\frac{2n}{\beta}P_{m+1,n-1}-\frac{n}{\beta^2}((n-1)P_{m+1,n-2}
         +\kappa_1P_{m,n-1}),  \label{5pmp1n}
\ee
and
\be
P_{m,n+1}=\left(y+\frac{\kappa_2+2(m+n)}{\beta}\right)P_{m,n}-\frac{1}{\beta^2}(n(\kappa_2+2m+n-1)P_{m,n-1}+m\kappa_1P_{m-1,n}).  \label{5pmnp1}
\ee
The points of these difference equations are depicted by the bullet points in Figure \ref{diff5fig}.
\begin{figure}[htb]
\begin{center}
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\put(100,0){\makebox(0,0){$\circ$}}
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\end{picture}
}
\end{center}
\caption{The pattern for the difference equations for (a)\, $P_{m+1,n}$, (b)\, $P_{m,n+1}$.}\label{diff5fig}
\end{figure}

On the left edge (\ref{5pmp1n}) reduces to
$$
P_{m+1,0} =  \left(x+\frac{\kappa_1}{\beta}\right)P_{m,0}\quad\Rightarrow\quad P_{m,0}= \left(x+\frac{\kappa_1}{\beta}\right)^m.
$$
On the right edge, (\ref{5pmnp1}) reduces to
$$
P_{0,n+1}=\left(y+\frac{\kappa_2+2n}{\beta}\right)P_{0,n}-\frac{1}{\beta^2}(n(\kappa_2+n-1)P_{0,n-1}),
$$
which is the {\em three point recurrence relation} for the Laguerre polynomials.

Notice that the operator $L$ reduces to
$$
(\beta x+\kappa_1)\frac{\pa P_{m,0}}{\pa x}=m\beta P_{m,0},
$$
on the left edge, giving the simple solution above for $P_{m,0}$.  On the right edge, the operator reduces to that of Laguerre.

Starting from the polynomials $P_{m,0}$ we may use the commuting operators to calculate the remaining polynomials on each level.  By looking at the leading order (monic) part of $I_kP_{m,n}$, we obtain:
$$
I_1P_{m,n}-m(\kappa_2+m-1)P_{m,n} = -m\kappa_1P_{m-1,n+1}, \quad I_2P_{m,n}+n\kappa_1P_{m,n} = n\beta P_{m+1,n-1},
$$
with $I_1P_{0,n}=0$ and $I_2P_{m,0}=0$.

Finally, we may construct the operators (\ref{r+}), which are given by
\bea
R_{+,x} &=& \frac{1}{\beta^2}\,(x\pa_y^2+(2 \beta x+\kappa_1)\pa_y +\beta (\beta x+\kappa_1)),  \nn\\
R_{+,y} &=& \frac{1}{\beta}\,(x\pa_x + y\pa_y+\beta y+m+n+\kappa_2).  \nn
\eea
These operators satisfy the commutation relations
$$
[L,R_{+,x}] = \beta R_{+,x},\quad
[L,R_{+,y}] = \frac{(L-\lambda_N)}{\beta} +\beta R_{+,y}.
$$

On the left edge, $R_{+,x}$ reduces to the {\em multiplicative} operator
$$
r_{+,x} = x+\frac{\kappa_1}{\beta},
$$
giving the above simple formula for $P_{m,0}$.  On the right edge, $R_{+,y}$ reduces to
$$
r_{+,y} = \frac{1}{\beta}\,(y\pa_y+\beta y+n+\kappa_2).
$$

\subsection{Krall-Sheffer Metric VIII}\label{pols8}

Here we have
\bea
L &=& y \pa_x^2+2 \pa_x\pa_y+  (\beta x+\kappa_1)\pa_x+ (\beta y+\kappa_2)\pa_y, \nn\\
 I_1 &=& \pa_x^2+(\beta y+\kappa_2)\pa_x,\nn\\
  I_2 &=& (y^2-x)\pa_x^2+2y\pa_x\pa_y+\pa_y^2+(\kappa_1 y-\kappa_2 x)\pa_x+(\beta x+\kappa_1)\pa_y. \nn
\eea
The polynomial eigenfunctions of $L$ satisfy the $5$ and $3$ point difference equations:
\be
P_{m+1,n} =  \left(x+\frac{\kappa_1}{\beta}\right)P_{m,n}+\frac{1}{\beta}\, (2m P_{m-1,n+1}+ n P_{m,n-1})
    -\frac{m \kappa_2}{\beta^2}\, P_{m-1,n},  \label{8pmp1n}
\ee
and
\be
P_{m,n+1}=\left(y+\frac{\kappa_2}{\beta}\right)P_{m,n}+\frac{m}{\beta} P_{m-1,n}.  \label{8pmnp1}
\ee
The points of these difference equations are depicted by the bullet points in Figure \ref{diff8fig}.
\begin{figure}[htb]
\begin{center}
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\caption{The pattern for the difference equations for (a)\, $P_{m+1,n}$, (b)\, $P_{m,n+1}$.}\label{diff8fig}
\end{figure}

On the right edge, (\ref{8pmnp1}) reduces to
$$
P_{0,n+1}=\left(y+\frac{\kappa_2}{\beta}\right)P_{0,n}\quad\Rightarrow\quad P_{0,n}=\left(y+\frac{\kappa_2}{\beta}\right)^n.
$$
On the left edge (\ref{8pmp1n}) reduces to
$$
P_{m+1,0} =   \left(x+\frac{\kappa_1}{\beta}\right)P_{m,0}+\frac{2m}{\beta}\, P_{m-1,1}-\frac{m \kappa_2}{\beta^2}\, P_{m-1,0},
$$
which is no longer a $3$ point recursion on the edge, since the left edge polynomials are not functions of just $x$.

Starting from the polynomials $P_{0,n}$ we may use the commuting operators to calculate the remaining polynomials on each level.  By looking at the leading order (monic) part of $I_kP_{m,n}$, we obtain:
\bea
I_1P_{m,n} &=& m\beta P_{m-1,n+1}, \nn\\
I_2P_{m,n}+m\kappa_2P_{m,n} &=& n\beta P_{m+1,n-1}+m\kappa_1 P_{m-1,n+1}+m(m-1) P_{m-2,n+2},\nn
\eea
with $I_1P_{0,n}=0$.

Finally, we may construct the operators (\ref{r+}), which are given by
\bea
R_{+,x} &=& x+\frac{\kappa_1}{\beta} + \frac{1}{\beta}\, \pa_y+ \frac{1}{\beta^2}\, ((2 \beta y+\kappa_2)\pa_x+\pa_x^2),  \nn\\
R_{+,y} &=& y+\frac{\kappa_2}{\beta} + \frac{1}{\beta}\,\pa_x.  \nn
\eea
These operators satisfy the commutation relations
$$
[L,R_{+,x}] = \beta R_{+,x},\quad  [L,R_{+,y}] = \beta R_{+,y}.
$$

On the right edge, $R_{+,y}$ reduces to the {\em multiplicative} operator
$$
r_{+,y} = y+\frac{\kappa_2}{\beta},
$$
giving the above simple formula for $P_{0,n}$.  On the left edge, $R_{+,x}$ does not reduce to an operator in just $x$, but still generates the functions $P_{m,0}$ from $P_{0,0}=1$.

\subsection{Krall-Sheffer Metric IX}\label{pols9}

In this case it is convenient to consider the 4 operators, which commute with $L$:
\bea
&& L= (x^2-1) \pa_x^2+2 x y \pa_x\pa_y+ (y^2-1) \pa_y^2+ \beta x\pa_x+ \beta y\pa_y,  \nn\\
&& I_1=(1-x^2-y^2)\pa_x^2+(1-\beta) x \pa_x,\quad  I_2=(1-x^2-y^2)\pa_y^2+(1-\beta) y \pa_y,  \label{case9-4ops} \\
&&  I_3= x \pa_y - y \pa_x,\quad  I_4= 2(1-x^2-y^2)\pa_x\pa_y+(1-\beta) (x \pa_y + y \pa_x).  \nn
\eea
Under the commutator, these form a closed, quadratic algebra (see \cite{scott}), which we don't need here. They are not, of course, independent.
They satisfy the quadratic relations:
$$
I_1+I_2+I_3^2+L=0,\quad 2(I_1I_2+I_2I_1)-(\beta^2-4\beta-1)(I_1+I_2)=(\beta-1)(\beta-5)L+I_4^2.
$$
These also satisfy
$$
\begin{array}{l}
I_1P_{m,n} = (2-\beta-m)m P_{m,n}-m(m-1)P_{m-2,n+2}, \\[1mm]
I_2P_{m,n} = (2-\beta-n)n P_{m,n}-n(n-1)P_{m+2,n-2},\\[1mm]
I_3P_{m,n} = n P_{m+1,n-1}-m P_{m-1,n+1}, \\[1mm]
I_4P_{m,n} = (1-\beta-2m)n P_{m+1,n-1}+(1-\beta-2n) m P_{m-1,n+1},
\end{array}
$$
so we use $I_3$ and $I_4$ (relating ``nearest neighbours'') as our horizontal ladders.

The polynomial eigenfunctions of $L$ satisfy the $4$ point difference equations:
$$
P_{m+1,n} =  x P_{m,n}+\frac{1}{\gamma_{2N-1}\gamma_{2N-3}}\, (n(n-1) P_{m+1,n-2}-m(\beta+m+2n-2) P_{m-1,n}),  \label{9pmp1n}
$$
and
\be
P_{m,n+1}= y P_{m,n}+\frac{1}{\gamma_{2N-1}\gamma_{2N-3}}\, (m(m-1) P_{m-2,n+1}-n(\beta+2m+n-2) P_{m,n-1}),  \label{9pmnp1}
\ee
where $N=m+n, \gamma_n=\beta+n$.
The points of these difference equations are depicted by the bullet points in Figure \ref{diff9fig}.  On the left edge, the first reduces to
$$
P_{m+1,0} =  x P_{m,0}-\frac{1}{\gamma_{2m-1}\gamma_{2m-3}}\, m(\beta+m-2) P_{m-1,0},
$$
which is the 3 point recurrence relation for Gegenbauer's equation
$$
(1-x^2)\frac{d^2P_{m,0}}{dx^2}-\beta x \frac{dP_{m,0}}{dx}=m(m+\beta-1)P_{m,0}.
$$
The second relation (\ref{9pmnp1}) similarly reduces on the right edge.

\begin{figure}[htb]
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\end{center}
\caption{The pattern for the difference equations for (a)\, $P_{m+1,n}$, (b)\, $P_{m,n+1}$.}\label{diff9fig}
\end{figure}

Finally, we may construct the operators (\ref{r+}), which are given by
\bea
R_{+,x} &=& \frac{1}{\gamma_{2N-1}}\,(x y\pa_y+(x^2-1)\pa_x + (\beta +N-1)x),  \nn\\
R_{+,y} &=& \frac{1}{\gamma_{2N-1}}\,(x y\pa_x+(y^2-1)\pa_y + (\beta +N-1)y).  \nn
\eea
These operators satisfy the commutation relations
\bea
[L,R_{+,x}] &=& \frac{2x(L-\lambda_N)}{\gamma_{2N-1}} +(\lambda_{N+1}-\lambda_N)R_{+,x}, \nn\\[2mm]
[L,R_{+,y}] &=& \frac{2y(L-\lambda_N)}{\gamma_{2N-1}} +(\lambda_{N+1}-\lambda_N)R_{+,y}. \nn
\eea

\section{Generating Functions}\label{genfuns}

Generating functions are known for the special cases IV, VI and VII, since these are just products of the known classical generating functions.
The generating function for Case IX is just a direct generalisation of that for Gegenbauer polynomials (in 1 dimension).  A multi-dimensional generalisation can be found in \cite{01-11}.  We believe that the generating functions given below for Cases V and VIII are new.

Generating functions are important for a number of reasons and can be considered as fundamental.  Suppose a given family of polynomials $P_{m,n}$ are coefficients in a generating function expansion
$$
G(x,y,s,t)=\sum_{m,n=0}^\infty A_{m,n} P_{m,n} \frac{s^m t^n}{m!n!},
$$
where the additional coefficients $A_{m,n}$ are included to allow our polynomials to be {\em monic}.
Then, by considering $\frac{\pa G}{\pa s}$ and $\frac{\pa G}{\pa t}$, we can construct explicit forms of difference equations, relating polynomials of degrees $m+n+1, m+n$ and $m+n-1$.
By considering $\frac{\pa G}{\pa x}$ and $\frac{\pa G}{\pa y}$, we can construct explicit forms of differential ladder operators.  Manipulating these, we can construct the eigenvalue problem for the polynomials, thus proving that they really are eigenfunctions of a particular differential operator.

\subsection{Cases I and IX: Generalised Gegenbauer Polynomials}

Since the operator $L$ of (\ref{case9-4ops}), for Case IX, is symmetric in $x$ and $y$ it is easy to generalise the standard Gegenbauer generating function to
$$
G(x,y,s,t)=(1-2sx-2ty+s^2+t^2)^{(1-\beta)/2}=
  \sum_{m,n=0}^\infty \frac{2^{m+n} \Gamma\left(\frac{\beta+2(m+n)-1}{2}\right) P_{m,n} s^m t^n}{\Gamma\left(\frac{\beta-1}{2}\right)m!n!}.
$$
Notice that this generating function has a pair of symmetries $(s,x)\mapsto (-s,-x)$ and $(t,y)\mapsto (-t,-y)$, which must be reflected in each term $P_{m,n}s^mt^n$. When $m$ is even (odd), $P_{m,n}$ is an even (odd) function of $x$ (and similarly for $n$ and $y$).  This means there will be an even polynomial of $x$, $\rho_{2m,n}$, such that $P_{2m+1,n}=x\rho_{2m,n}$ (with a similar statement for $P_{m,2n+1}$).  Indeed, the first few polynomials are
\bea
&& P_{0,0}=1, \;\; P_{1,0}=x,\;\; P_{0,1}=y,\;\; P_{2,0}=x^2-\frac{1}{1+\beta},\;\; P_{1,1}=x y,\;\; P_{0,2}=y^2-\frac{1}{1+\beta},\nn\\
&&  P_{3,0}=x\left(x^2-\frac{3}{3+\beta}\right),\;\; P_{2,1}=y\left(x^2-\frac{1}{3+\beta}\right),\nn\\
&&  P_{1,2}=x\left(y^2-\frac{1}{3+\beta}\right),\;\; P_{0,3}=y\left(y^2-\frac{3}{3+\beta}\right).  \nn
\eea
In Figure \ref{pols9fig} below, polynomials which are even in both $x$ {\bf and} $y$ are depicted by black dots.  They occur only when $N=m+n$ is even.
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\end{picture}
\end{center}
\caption{The pattern of polynomials for case IX. The black dots are when {\bf both} $m$ and $n$ are even.}\label{pols9fig}
\end{figure}
This configuration is unimportant in the context of case IX, but when the transformation $(x_9,y_9)=(\sqrt{x_1},\sqrt{y_1})$ (see Section \ref{i-ix}) is used, the black dots represent the polynomial eigenfunctions in Case I (with the restriction $\kappa_1=\kappa_2=-\frac{1}{2}$).  Generally, $P_{m,n}$ (of Case IX) maps to $P_{\frac{m}{2},\frac{n}{2}}$ (of Case I), giving us ``half-integer'' indices, corresponding to non-polynomial eigenfunctions.  For example, $P_{2,1}\mapsto P_{1,\frac{1}{2}}=\sqrt{y}\left(x-\frac{1}{2(\beta+1)}\right)$.

In Section \ref{pols9} we chose to use $I_3, I_4$ to take us right and left on horizontal levels, since they just related ``nearest neighbours''.  The operators $I_1, I_2$ jump two steps at a time, connecting (in particular) the black dots on the horizontal levels.  These transform to the $I_1,I_2$ given in Section \ref{pols1} (with $\kappa_1=\kappa_2=-\frac{1}{2}$ and the replacement $\beta_9\mapsto 2\beta_1-1$).  With these values of $\kappa_i$, we have
$$
I_1P_{m,n}+m\left(\beta+m-\frac{3}{2}\right)P_{m,n} = m\left(\frac{1}{2}-m\right)P_{m-1,n+1},
$$
so, when $m=\frac{1}{2}$, the right hand side vanishes and $P_{\frac{1}{2},n}$ is just an eigenfunction of $I_1$.  This means that starting on one of the half-integer functions, $I_1$ moves us to the right until we reach the last one on the given level.  The operator $I_2$ acts in a similar way.  The operators $I_3, I_4$ can also be transformed to the reduced Case I, but these operators no longer have polynomial coefficients.

\subsection{Cases V and VIII}

For Case V, the generating function is
$$
G(x,y,s,t)=\exp\left(\frac{\beta^2 s x+(\kappa_1s+\beta t y)(\beta-t)}{(\beta-t)^2}\right)\,\left(1-\frac{t}{\beta}\right)^{-\kappa_2}
= \sum_{m,n=0}^\infty P_{m,n}(x,y) \frac{s^m t^n}{m! n!}.
$$
When $t=0$ this reduces to $G_l(x,s)=\exp\left(s\left(x+\frac{\kappa_1}{\beta}\right)\right)$ and generates the polynomials $P_{m,0}=\left(x+\frac{\kappa_1}{\beta}\right)^m$, located on the left edge.  With $s=0$, the generating function reduces to
$$
G_r(y,t)=\exp\left(\frac{\beta t y}{(\beta-t)}\right)\,\left(1-\frac{t}{\beta}\right)^{-\kappa_2}
$$
and generates Laguerre polynomials $P_{0,n}$ on the right edge.  In the interior of the triangle the polynomials depend on both $x$ and $y$ in a non-trivial way.  For example
$$
P_{1,1}=\left(x+\frac{\kappa_1}{\beta}\right)\left(y+\frac{\kappa_2}{\beta}\right)+\frac{2}{\beta}\left(x+\frac{\kappa_1}{2\beta}\right).
$$

By differentiating $G(x,y,s,t)$ with respect to $s$ we find
$$
(\beta-t)^2\frac{\pa G}{\pa s}=(\beta^2 x+\kappa_1(\beta-t))G,
$$
which leads to the {\em 3 level recurrence relation}
$$
P_{m+1,n}=\left(x+\frac{\kappa_1}{\beta}\right)P_{m,n}+\frac{2n}{\beta}P_{m+1,n-1}-\frac{n}{\beta^2}((n-1)P_{m+1,n-2}+\kappa_1P_{m,n-1}),
$$
which we already saw in Section \ref{pols5}.  When $n=0$, this reduces to $P_{m+1,0}=\left(x+\frac{\kappa_1}{\beta}\right)P_{m,0}$, which gives the above $P_{m,0}$.

Differentiating $G(x,y,s,t)$ with respect to $t$ leads to a more complicated formula, but can be combined with the $s-$derivative to give
$$
(\beta-t)^2\frac{\pa G}{\pa t}-2s(\beta-t)\frac{\pa G}{\pa s}=(\beta^2 y+\kappa_2(\beta-t)-\kappa_1s)G,
$$
which leads to the other {\em 3 level recurrence relation}
$$
P_{m,n+1}=\left(y+\frac{\kappa_2+2(m+n)}{\beta}\right)P_{m,n}-\frac{1}{\beta^2}(n(\kappa_2+2m+n-1)P_{m,n-1}+m\kappa_1P_{m-1,n}),
$$
also given in Section \ref{pols5}.

For Case VIII, the generating function is
\bea
G(x,y,s,t) &=& \exp\left(\frac{4s^3+3(2\beta y+\kappa_2)s^2+6\beta s t+6\beta ((\beta x+\kappa_1)s+(\beta y+\kappa_2)t)}{6\beta^2}\right)\nn\\
&=& \sum_{m,n=0}^\infty P_{m,n}(x,y) \frac{s^m t^n}{m! n!}.\nn
\eea
Similar manipulations as described above lead to the difference equations (\ref{8pmp1n}) and (\ref{8pmnp1}).

The transformation between the Laplace-Beltrami operators of Cases V and VIII (see Section \ref{v-viii}) do not extend to the Krall Sheffer operators, so the polynomials generated for either case cannot be transformed to eigenfunctions of the other.

\section{Conclusions}

In this paper we presented a number of recursive algebraic structures which allow the explicit construction of polynomial eigenfunctions of the Krall-Sheffer differential operators.  By arranging these polynomials in the triangular array of Figure \ref{trilatt}, we showed that the $2-$dimensional Krall-Sheffer eigenvalue problem reduced to a standard $1-$dimensional problem on at least one of the edges, where we could use the standard {\em $3-$point recurrence relation} to build polynomial eigenfunctions of one variable.  We then used the commuting operators (coming from super-integrability) to build the $2-$dimensional polynomials on the horizontal level of the triangle.

We then gave {\em explicit formulae} for the {\em $3-$level recurrence relations} for each of the (non-trivial) Krall-Sheffer cases.  These reduce to the corresponding $3-$point relations on appropriate edges.  The $3-$level recurrence relations give a very effective way of building the polynomials. We also gave {\em explicit formulae} for the differential raising operators, which shift us to the next eigenvalue in directions parallel to the edges of the triangular array.  Lowering operators also exist (see \cite{scott}) and sometimes have simple formulae.  However, for some of the cases, these operators are very complicated, so we didn't include any of them here.

Finally, we discussed the existence of generating functions for the Krall-Sheffer polynomials.  We presented new generating functions for cases V and VIII.  It was shown how the well known generalised Gegenbauer generating function of Case IX can be used to build eigenfunctions of Case I with special parameter values.  It would be interesting to generalise this to a generating function for the {\em full} Case I.  Generating functions for Cases II and III are also currently unknown.

The discussion of Sections \ref{geom} and \ref{super} give us a direction to follow if we wish to build Krall-Sheffer like operators in higher dimensions.  Some calculations in this direction can be found in \cite{scott}, but it is an important subject for future research.

\subsubsection*{Acknowledgments:}
The authors would like to thank Y. Xu and A.V. Turbiner for bringing the papers \cite{95-6} and \cite{09-5} to their attention.

redpajama_set_name: RedPajamaArXiv


My wife and I picked up copies of your publication last night while waiting for a pizza to be made, and we both had the same reaction – your paper is inflammatory in the extreme. While the situation of world climate change certainly is a biggie, there is a large element of 'preaching to the choir' in your paper that turns off people who have already done as much as they reasonably can do to limit their impact on the environment. The fact is, the environment is already damaged in ways unrelated to climate change, but strongly related to human population growth, and THAT is the PROBLEM. Since I perused your front page yesterday and now, the world population has grown by approximately 100,000 people. If all the people in Vermont dropped off the face of the earth today, their numbers would be replenished in less than 7 days time. Understanding that fact casts a blurry haze on the importance of me buying a light bulb that saves 25 watts an hour. We cannot fix this problem with technology that we have today. When the technology is available, it will be deployed by the public service companies, not individuals in their back yards, individuals who have an extra $40,000 laying around looking for a project. Thank you for considering my comments.
Thank you for your letter. I must say I was a bit taken aback by the idea that Green Energy Times is "inflammatory in the extreme." Reviewing the issue you were reading, I could see why you might feel this way. I think you need to balance views on the dangers of climate change with what is understood about current technology to deal with the problem.
To begin with your last point, it is not necessary to have anything remotely close to $40,000 to get a solar system for a household. We have pointed out repeatedly that for Vermont residents it is not necessary to have much more than the ability to pay a monthly electric bill. Green Mountain Power and NeighborWorks of Western Vermont are working together to make it possible for many people to purchase a system on credit and have the payments on the same electric bill with credits for electricity fed into the system. The immediate net effect is a reduction in payments, which will eventually go to nearly zero. There are a number of other organizations that offer similar financial packages both in Vermont and elsewhere.
Next, the consensus is that we actually do have all the technology we need to deal with climate change. And we do not have to wait for utility companies to act. You need not take this on the word of the Green Energy Times editors. Such financial institutions as UBS, Barclays, and Morgan Stanley have given warnings on US utilities because the market is changing so fast that they may not be able to stop "grid defections" as people find it less expensive to go off grid. This is a situation they see developing in the next three to four years. Other organizations with similar views include Lawrence Berkley National Laboratories, the University of Delaware, Lazard, and Navigant. We have seen projections from such organizations that our civilization could be 100% powered by renewables as early as 2040, based on current technology alone.
We agree that the population is too large. That is, however, not an area of our expertise, and we know that climate change and energy are too important to ignore. We do what we can.


#TOUR2 tag is a stock design on our 1.25" dia. Brass circle. Tag includes a jump ring to help tag hang straight.
This tag will be engraved with 2007 for this year. The season can be modified to fall, summer or winter in the additional instructions.
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text:
Share this Story: It's getting scary, Chatham renter says of housing crunch
It's getting scary, Chatham renter says of housing crunch
Ellwood Shreve
Kit Britton, left, and her sister Tammy Gorman are among those Chatham-Kent residents facing challenges in finding affordable housing. The sisters were among many who attended a housing support services fair held at Chatham public library on Oct. 19. Britton and Gorman display a picture taken when the Chatham-Kent Fire & Emergency Services responded to a fire at Gorman's apartment this summer. Ellwood Shreve/Postmedia Network
Kit Britton says she's lucky to have $50 left for food after she's paid her monthly rent and utilities.
The Chatham woman was one of those who attended a housing support services fair. Held for the first time, the Oct. 18 event attracted a large crowd, with many hoping to find help to solve their housing challenges.
It's getting scary, Chatham renter says of housing crunch Back to video
Organizers were surprised with the turnout, but Britton wasn't.
"It's getting scary," she said. "Look at all these people here looking for the same thing."
Her sister, Tammy Gorman, is facing a more immediate need since she had to leave her apartment unit in late August because of a fire.
Gorman is staying in an agency-provided accommodation, but must move by the end of the month.
She would also like to find something more affordable. The single mother of two boys said financial benefits received through Ontario Works don't go far.
Gorman said it's also another challenge that she must continue paying rent for an apartment she's not living in.
Britton is facing her own particular housing challenge. She lives in a fourplex apartment but's it's for sale. She wants to leave because she fears a new owner will raise the rent.
"They're all coming from Toronto, they're buying all the units and the housing here and they're increasing the rates to Toronto (prices)," she said. "Our incomes are not Toronto rates, but our housing is Toronto rates."
Dominque McKee, a community navigator with Chatham-Kent, who helps people access services, organized last week's housing support services fair.
She said the services available to those who need housing is dependent upon what their situation is.
If that situation is homelessness, McKee said there is a homeless response line as well as rapid rehousing workers.
She said she was surprised the response at the housing fair.
"Now that we really see the interest and the traction, we're looking at hopefully making it a bit bigger and getting more partners out," McKee said.
She said a number of community service providers came by and said they want to be part of future events.
McKee said the wait list for social housing and the affordability of rent at market rates are issues facing many people.
She noted that if a person can't afford their rent, there are municipal programs to access funding.
If people need help knowing what their tenancy rights are, McKee said the Salvation Army provides assistance and RentSmart provides information on how to be a good tenant.
Jeff Wilkins, a lawyer specializing in housing law who works for the Chatham-Kent Legal Clinic, has been inundated with cases as he tries to help low-income residents and those on social supports who are struggling to pay rent.
"That's unfortunately what we're seeing a lot of, people falling behind on their rent," he said. "Once they get behind a little bit, all of a sudden that's a hole that's pretty difficult to climb out of."
Wilkins has been spending a lot of time representing people are going before the landlord and tenant board because of issues paying their rent.
"Because there's a lack of housing out there, there's a really big need for people to be represented at the board," he said.
Wilkins said the goal is to try to stop the eviction process before it starts by finding solutions, such as repayment plans, that are acceptable to the landlord and tenant.
With winter coming and the added expenses of higher heating bills and Christmas, "This is the time of year we start telling tenants if they are in trouble, let us know before that notice comes in the mail or before you're served," Wilkins said.
"Once they get caught up in the system, there's costs for the filing that the landlord passes onto the tenant and it's just a vicious spiral."

redpajama_set_name: RedPajamaCommonCrawl


Rockaway Recovery & Collection can be found at Union St 140. The following is offered: Business Billing & Collection. The entry is present with us since Sep 9, 2010 and was last updated on Nov 14, 2013. In Lynn there are 1 other Business Billing & Collection. An overview can be found here.


text:
Home Crypto News Japan's High Court Rejects Former Mt Gox CEO's Conviction Appeal
Japan's High Court Rejects Former Mt Gox CEO's Conviction Appeal
The former CEO of Mt. Gox – the bitcoin exchange at the center of one of the most notorious hacks in crypto – has had his conviction upheld by Japan's high court.
On Thursday, Tokyo's the court threw out an appeal from Mark Karpeles, who had been found guilty last year for manipulating electronic data, upholding the original conviction and sentence of two and a half years in prison, suspended for four years.
Karpeles was originally arrested in August 2015, roughly 18 months after an unknown hacker made off with more than 850,000 bitcoins from Mt. Gox. At the time the amount was valued in the hundreds of millions, but would be worth more than $8 billion, today. Around 15% was later recovered and now constitutes the Mt. Gox estate, which is still being fought over by the exchange's creditors.
Prosecutors had originally accused Karpeles of embezzling user funds, breach of trust and manipulating electronic data, and demanded he serve 10 years in prison. The Tokyo District Court found him innocent on the first two counts, but guilty on manipulating electronic data in order to harm his clients.
The legal team representing Karpeles, which has always claimed he is innocent of all charges, contend that prosecutors didn't know how crypto exchanges operated and were merely using him as a fall guy.
"Today's verdict was unfortunate, and I am reviewing its contents alongside my lawyers and will decide how to proceed from there in the coming days," Karpeles said following the ruling, in a report from the Associated Press.
Previous articleTezos and Algorand Latest to Integrate Tech for Anti-Money Laundering Compliance
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text:
För arkitekten, se Adrian C. Peterson

Adrian Lewis Peterson, född 21 mars 1985 i Palestine, Texas, med smeknamnet "AD" (för "All Day") och "Purple Jesus" är en amerikansk fotbollsspelare som tidigare spelade RB (Running Back) för Washington Commanders i National Football League. Han spelade College Football i University of Oklahoma.

I sin debutsäsong i Oklahoma satte Peterson NCAA freshman rushing record med 1.925 yards. Han satte också ett rekord genom att sluta som tvåa i omröstningen till Heisman Trophy (USC:s University of Southern California Matt Leinart vann priset) som förstaårsstudent. Peterson blev utnämnd till skolans tredje bästa rusher någonsin.

Under sin första stjärnsäsong där han satte ett NFL-rekord för flest sprungna yards i en enda match (296), blev Peterson NFL Offensiv Rookie of the Year. Han blev då tilldelad priset som MVP för sin prestation i Pro Bowl och blev den femte spelaren i NFL:s historia att ha sprungit mer än 3.000 yards på sina två första säsonger.

Källor

Amerikanska utövare av amerikansk fotboll
Spelare i Minnesota Vikings
Amerikanska idrottare under 2000-talet
Idrottare från Texas
Personer från Palestine, Texas
Födda 1985
Levande personer
Män

redpajama_set_name: RedPajamaWikipedia


text:
Jamie Kehoe's murderer walks free
This is revolting. Jamie Kehoe's murder was big news in Surrey. He was an innocent bystander sitting on a bus and a guy runs up to him and slashes his throat abruptly ending his life before it really got started. A tragedy that matches the Jesse Cadman murder.
Yesterday the news broke that the crown will not be proceeding with charges against the accused because of the unlikelihood of conviction. Not because he was innocent, but because the court couldn't prove beyond a reasonable doubt that the accused didn't murder him in self defense. That is absolutely outrageous.
Jamie was a small kid. The accused isn't. Jamie was a nonviolent skateboarder. He's sitting on the bus with two friends, a guy and a girl. Another guy gets on the bus with another girl who becomes belligerent and relentlessly antagonizes the girl who is with Jamie and his other friend. The whole bus ride she keeps yelling at the girl, challenging her to a fight while the girl is silent.
As the belligerent girl gets off the bus the girl who was taunted the whole trip tells her to f off after she gets off the bus. The antagonistic girl freaks. She runs back onto the buss and starts kicking the sh*t out of the girl she had been taunting. The guy with the antagonistic girl then runs up to Jamie as he's sitting in his seat and slashes his throat for no reason. That is just plain evil.
After he slashes Jamie's throat, Jamie's friend jumps in and starts to fight the assailant and ends up getting stabbed. That's what happened. Denying the family a day in court because you don't think you can prove it wasn't in self defense is even more evil that the horrific murder itself. Just because two girls are fighting doesn't necessarily mean the guys are going to jump in. They might help separate them or pull one off if the other starts getting seriously beat up but there is no reason this case can't go to trial. Preventing the case from going to trial is a travesty of justice. It's time to mobilize and march for justice. Being silent on this issue is wrong.
Update: A similar case in Chilliwack involving Lori Stevens has shown that particular crown counsel to be negligent.
The Organized Youth of Canada May 11, 2012 at 4:05 PM
THIS IS WHY THERE ARE NO NEED FOR THE CROWN OR THE USELESS LIMPDICK POLICE/RCMP JUST ARM AND DEFEND YOUR SELF
Funny, the chain of events released by the crown is completely different.
THE CROWN WAS NOT PRESENT NO SURPRISE THEY GOT IT WRONG AS ALWAYS MAYBE TIME TO RETURN TO VIGILANTE JUSTICE!
Dennis Watson August 19, 2014 at 5:31 AM
There's no such thing as vigilante justice. Mob law is anarchy. The crown made a mistake and that needs to be examined just like in the Lori Stevens case:
http://gangstersout.blogspot.ca/2014/08/chilliwack-shooting-suspect-arrested.html
Dennis Watson May 12, 2012 at 10:58 AM
That isn't what the crown said and it most certainly isn't funny. The media reported that the crown claimed they couldn't prove beyond the shadow of a doubt that Jamie's friend didn't pull out the baton first. They didn't say that's what happened. The witnesses are traumatized and terrified. They saw a woman terrorize a girl and kick the crap out of her then they saw her boyfriend slit an innocent bystanders' throat. No doubt they are terrified to testify against them. He's free to reoffend. I was at the candlelight vigil. I spoke with a friend of the family that knew and had spoken with Jamie's' two friends on the bus. Yes, it is hearsay but it passes the test of believability. The other claim does not.
trailrunner78 May 15, 2012 at 11:37 AM
TOYOC beat me to it.
But, then won't that make you like 'those Americans'? Wouldn't it be worth any price not to be accused of that? Because of course America and Americans are the source of all the world's problems..... :rolleyes:
Dennis Watson May 15, 2012 at 3:55 PM
You never hear me saying that America and Americans are the source of all the worlds problems. It's just when corrupt agencies and corrupt politicians violate the US constitution that we see problems arise. A criminal is a criminal regardless of what country he's from.
trailrunner78 May 17, 2012 at 1:04 PM
I really didn't mean you K. I was speaking to all the American bashing fucking limp dick hippie losers I knew in the 10 years I lived in the lower mainland, the enlightened "progressive" sheep who believe violence is always wrong. Just so I'm real clear about where I'm coming from here, Jamie Kehoe's murderer should not have survived the offense, he should have been shot down like the trash he is by some lawfully armed citizen on that bus.
freddy October 23, 2012 at 12:17 PM
"Fears of a violent gang are keeping some northern B.C. residents from sharing information about the torture and decapitation of a young man following a party last winter, CBC News has learned."
"The severed head of Fribjon Bjornson, 28, was found in a vacant house on the Nak'azdli reserve near Fort St. James on Feb. 3, three weeks after he was last seen about 60 kilometres away at a 7-Eleven in Vanderhoof, B.C."
"The rest of his body was never found and no one has been charged in the case."
http://tinyurl.com/92jd4u8 (CBC)
Another un-solved murder; but this one & another, recent, same-same, disappearance of yet another young girl on this "Highway Of Tears"...led to a personal revelation. About all this murder up there.
It's someone amongst the Native tribes; living along that Highway; wot's done the murders. That is how they have been, incredibly, hidden and totally opaque for all these years. The chief admits at least twenty persons on his Res., know of this current murder, and dead silence!
There is just a wall of omerta between the tribes and the Law; I've seen it, living up-country, all these years. Even when a native was herself killed, by a Quesnel tribe member, the same wall-of-silence descended and her killer never caught.
It makes so much sense, that a Native, not white, killer/s work along the "Highway Of Tears." Maybe, they aren't seen as threatening by potential victims, than one single white guy. They know the country like no one; and would know when, & how, to locally dispose of a body so it would not be found.
There is no reason to count out someone from the Native population, as we see now there are armed and dangerous Native gangs up North, as in this case.
Dennis Watson October 23, 2012 at 4:46 PM
"Fears of a violent gang are keeping some northern B.C. residents from sharing information about the torture and decapitation of a young man following a party last winter." Wow, that is brutal and extreme. We had heard Fribjon had been tortured before his murder but that news report confirms that that he was indeed decapitated as some reports had claimed. That is outrageous. Now it appears that a gang was involved. I don't think there has been any racial profiling of the suspect yet. He could be white, native or anything else. The fact that his body, or at least part of his body was found on a reservation does imply someone from the reservation was in on it. Many chiefs on several reservations have had trouble with gangs disrupting the peace on Native reservations. I believe the Red Alert around Kamloops were affiliated with the Hells Angels.
We know that the media recently reposted that the DNA of at least one of the murders along the highway of tears belonged to a white man who died in a US prison. Clearly, he was not responsible for all the murders as his dna had been ruled out in several cases. The decapitation reminds me of two disturbing cases in Surrey. Mihaly Illes was a caucasian from Hungary who was convicted of parading around with his victims decapitated head in a bucket. Another case was the murder of Paul Percy Soluk. The hit man was caucasian and so was the person who helped him dismember and dispose of the body. The accused claimed the murder was for the East Vancouver chapter of the Hells Angels. But you will be happy to know that the accused in Jamie Kehoe's murder was non white.
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text:
from django.conf import settings  # noqa
from django.conf.urls.defaults import include  # noqa
from django.conf.urls.defaults import patterns  # noqa
from django.conf.urls.defaults import url  # noqa
from django.views.generic import TemplateView  # noqa

urlpatterns = patterns('horizon.views',
    url(r'^home/$', 'user_home', name='user_home')
)

# Client-side i18n URLconf.
urlpatterns += patterns('',
    url(r'^i18n/js/(?P<packages>\S+?)/$',
        'django.views.i18n.javascript_catalog',
        name='jsi18n'),
    url(r'^i18n/setlang/$',
        'django.views.i18n.set_language',
        name="set_language"),
    url(r'^i18n/', include('django.conf.urls.i18n'))
)

if settings.DEBUG:
    urlpatterns += patterns('',
        url(r'^qunit/$',
            TemplateView.as_view(template_name="horizon/qunit.html"),
            name='qunit_tests'))

redpajama_set_name: RedPajamaGithub