text:
\section{Introduction}
Quasi-exactly solvable (QES) equations refer to a class of spectral
differential equations for which a part of the
spectrum can be obtained by solving
algebraic equations \cite{tur0,ush,shif}.
Linear differential operators preserving a finite dimensional
space of smooth functions constitute in this respect a basic
ingredient in the topic of QES equations.

In the case of operators of one real variable  acting on
a scalar function, the possibilities of finite dimensional
invariant vector spaces are rather limited \cite{tur1}.
Up to a change of the variable and a redefinition of the function,
the vector space can only be the set $P(N)$ of polynomials with degree less
than or equal to $N$.
The relevant operators are the elements of
the enveloping algebra of $sl(2)$ whose generators are
suitably represented by three differential operators \cite{tur0,tur1}.
The QES equations they define therefore possess an $sl(2)$ hidden symmetry.

When the number of variables or (and) the number of components
of the function is (are) larger than one,
the number of possible invariant vector spaces
and  hidden symmetry algebras increases considerably.
The  scalar QES operators in two variables were
classified in \cite{gko1,gko2}. Seven inequivalent
spaces of functions appear to be possible. Correspondingly,
the hidden symmetry algebras can be of several
types, e.g. $sl(2)$, $sl(2) \otimes sl(2)$, $sl(3)$.
A few cases of this classification
generalise easily to operators involving
an arbitrary number of variables.
In particular, the case labelled 2.3 in ref. \cite{gko2}
can be extended  to the space  $P(N,V)$
of polynomials of maximal degree less than or equal to $N$
in their $V$ independent variables.
The related algebra is  $sl(V+1)$.

The construction of the matrix operators in one variable
preserving the direct sum $P(N_1) \oplus \ldots \oplus P(N_k)$
has also been considered \cite{bk,bggk}. These operators
are closely related to graded algebras. As an example,
the case $P(N) \oplus P(N+1)$ is related to the graded Lie
algebra $osp(2,2)$ \cite{shtu}.

The purpose of this paper
is to classify  the operators preserving the vector space
$P(M,V)$ \break
$\oplus P(N,V)$ for arbitrary values of the integers $M,N,V$
and to construct a series of associative algebras
corresponding to the hidden symmetries of these operators.
In Sect. 2 we fix the notations and point out the relevant
representations of the algebra $gl(n)$.
The $2 \times 2$ matrix operators preserving the space
$P(N,V) \oplus P(M,V)$ are contructed in Sect. 3 and are
shown to obey a set of normal ordering rules.
In Sect. 4, these ordering rules are modified into sets
of commutation and anticommutation relations which fulfil
all Jacobi identities. We obtain in this way a series
of  associative abstract algebras which appear to be labelled by
$V$, by  $\Delta=\mid N-M\mid $ and by $\Delta -1$ arbitrary complex
parameters.
The technical details related to the proof of our main result
are given in Sect. 5.

\section{Operators preserving $P(N,V)$}

\par Let $N,V$ be two positive integers. Let $x_i\ (i=1,\ldots,V)$
represent $V$ independent real variables. We define the finite dimensional
vector space $P(N,V)$ of polynomials in the variables $x_i$ and
of maximal total degree  $N$
\begin{equation}
P(N,V) = {\rm span} \ \lbrace x^{n_1}_1, x^{n_2}_2\ldots x^{n_V}_V
\rbrace
\quad , \quad  0 \leq \sum_{j=1}^V n_j \leq N
\end{equation}
\begin{equation}
P(N,1) \equiv P(N) = {\rm span} \ \lbrace 1,x,\ldots,x^N \rbrace \ .
\end{equation}
The dimension of $P(N,V)$ is given by
\begin{equation}
 1 + V + {V(V+1)\over 2} + \ldots +
{V(V+1) \ldots (V+N-1) \over {N!}} = C^{N+V}_V \ .
\end{equation}

The set of linear differential operators preserving $P(N,V)$ can be perceived
as the enveloping algebra generated by the following operators
\begin{eqnarray}
J^0_0(N) &=& D-N \quad , \quad
D\equiv \sum^V_{j=1} x_j {\partial \over {\partial x_j}} \nonumber \\
J^k_0(N) &=& {\partial \over
{\partial x_k}}\quad , \quad k=1,\ldots, V   \nonumber \\
J^0_k(N) &=& -x_k(D-N)\quad , \quad k=1,\ldots, V \nonumber \\
J^l_k(N) &=& - x_k{\partial\over{\partial x_l}}\quad , \quad k,l=1,\ldots, V
\ .
\label{j}
\end{eqnarray}
These $(V+1)^2$ independent operators fulfil the commutation
rules of the Lie algebra $gl(V+1)$.
Acting on the finite dimensional space $P(N,V)$, they
lead to an irreducible representation of this algebra.
The commutation relations are
\begin{equation}
[J^{b}_{a}, J^{d}_{c}] =
 \delta^{d}_{a} J^{b}_{c}
- \delta^{b}_{c}J^{d}_{a}
\quad \ \ , \ \ a,b,c,d = 0,1, \ldots , V \ .
\label{comj}
\end{equation}

Within the representation (\ref{j}), the Casimir operators of $gl(V+1)$
\begin{equation}
C_p \equiv \sum_{a_1,\ldots, a_p = 0}^V
 J^{a_1}_{a_2}  J^{a_2}_{a_3} \ldots  J^{a_p}_{a_1}
\ \ \ , \ \ \
 p=1, \ldots, V+1
\label{cas}
\end{equation}
have the values
$C_p= (-1)^p N(N+V)^{p-1}$.
The operators ${J'}_a^b$ defined by
\begin{equation}
   {J'}_{a}^{b} =   J_{a}^{b}
                    + C \delta_{a}^{b}  \  \ ,
\label{trans}
\end{equation}
where $C$ is any operator which commutes with all $J$'s,
satisfy also the relations (\ref{comj}).
For instance, this is the case for the $(V+1)^2-1$ independent operators
\begin{equation}
\tilde J_{a}^{b}=J_{a}^{b}- {1 \over V+1} C_1 \delta_{a}^{b}
\end{equation}
(since $\tilde C_1 \equiv 0$) which
form  an irreducible
representation of $sl(V+1)$ when acting on $P(N,V)$.
The usual form \cite{tur0} of the of $sl(2)$  generators
\begin{eqnarray}
J_+(N) &=& -\tilde J_1^0 = x(x\partial _x-N) \nonumber \\
J_0(N) &=& - \tilde J^1_1 = (x\partial _x -{N\over 2}) \nonumber \\
J_-(N) &=& \tilde J_0^1 = \partial _x
\end{eqnarray}
is recovered for  $V=1$.
These operators play a major role in the topic of
quasi-exactly solvable equations.

\par More generally, an element, say $A$,
of the enveloping algebra constructed over the
$J^{b}_{a}$ (or the $\tilde J^{b}_{a}$) is a
quasi-exactly solvable operator preserving $P(N,V)$.
That is to say that  the spectral equation
\begin{equation}
A p = \lambda p \quad , \quad p \in P(N,V)
\end{equation}
admits $C^{V+N}_{V}$ solutions.
Recently, the Calogero and Sutherland
quantum hamiltonians were shown to be expressible
in terms of the operators $J^{b}_{a}$ \cite{tur3},
this result reveals the hidden symmetries of these models.

\section{Operators preserving  P(M,V) $\oplus$ P(N,V)}

We now  put the emphasis on the 2$\times$2 matrix
operators which preserve the vector space
\begin{equation}
P(M, V) \oplus P(N,V) \ \ \ , \ \ \ \Delta \equiv N-M \ .
\label{esp}
\end{equation}
Without loss of generality, we assume the
integer $\Delta$ to be non negative.
In order to classify the operators preserving (\ref{esp})
we  define a list of generators.
First the "diagonal" generators that we choose as
\begin{equation}
J^{b}_{a} (N,\Delta) =  \left(\begin{array}{cc}
J^{b}_{a}(N-\Delta) &0\\
0 &J^{b}_{a}(N)
\end{array}\right)
- {1\over 2}
\left(\begin{array}{cc}
1+\Delta &0\\
0        &1-\Delta
\end{array}\right) \delta_{a}^{b}
\label{op1}
\end{equation}
for $0 \leq a , b \leq V$.
They are built as a direct sum of two operators
of the type (4)-(7), translated by (\ref{trans}) in such a way that
that $J^0_0(N,\Delta)$ is proportional to the unit matrix.
The interest for this translation will appear later.

The ``non diagonal'' generators naturally split
into ``$Q$ operators'', proportional to the matrix
$\sigma_-$  (as usual $\sigma_{\pm} = (\sigma_1 \pm i \sigma_2)/2$).
and ``$\overline Q$ operators'' proportional to the matrix
$\sigma_+$. It is convenient to write them by using a multi index
$[A] \equiv a_1,a_2,\ldots,a_{\Delta}$.
For later convenience we also define $[\hat A_i]$
as the set $[A]$ where the index $a_i$ has been removed.
We choose the non diagonal generators respectively as
\begin{equation}
Q_{[A]}
 = (-1)^{\delta} x_{a_1} \ldots x_{a_{\Delta}} \ \sigma_-  \quad ,
 0 \leq a_i \leq V  \quad , \ \ x_0 \equiv 1
\label{op2}
\end{equation}
where $\delta$ represents the overal degree in $x_1,\dots,x_V$
of the monomial
$x_{a_1} \ldots x_{a_{\Delta}}$ and
\begin{equation}
\overline Q^{[B]}  =
 \overline q^{[B]} \ \sigma_+ \quad ,
 0 \leq b_i \leq V  \quad
\label{op3}
\end{equation}
where the scalar operators $\overline q^{[B]}$, fully symmetric
in their $\Delta$ indices  $b_k$, are defined by
\begin{eqnarray}
\overline q^{[B]} &=&\partial_{b_1} \ldots \partial_{b_{\Delta}} \quad \ \ {\rm if}
   \     0< b_1 \leq b_2 \leq b_3 \ldots \leq b_{\Delta}  \nonumber  \\
& &(D-N+\Delta-1)\partial_{b_2} \ldots \partial_{b_{\Delta}} \quad \ \ {\rm if}
   \      0= b_1 < b_2 \leq b_3 \ldots \leq b_{\Delta}   \\
& &(D-N+\Delta-1)(D-N+\Delta-2) \partial_{b_3} \ldots \partial_{b_{\Delta}} \quad \
    {\rm if}
     \    0= b_1 = b_2  < b_3 \ldots \leq b_{\Delta}  \nonumber \\
& &(D-N+\Delta-1)(D-N+\Delta-2) \ldots (D-N)     \quad \
     \ {\rm if}
      \      0= b_1 = b_2  = b_3 = b_{\Delta} \ .  \nonumber
\end{eqnarray}
The operators $Q_{[A]}$ (and similarly the  $\overline Q^{[A]}$)
are fully symmetric in their $\Delta$ indices $a_k$.
Hence there are $C_{\Delta}^{V+\Delta}$ independent operators of both types.
We then have the following proposition.

\vspace{0.5cm}
\noindent {\bf {Proposition  1}}
\vspace{0.5cm}

\noindent {\it The operators preserving the space
$P(N-\Delta, V) \oplus P(N,V)$
are the elements of the enveloping algebra constructed over
the generators (\ref{op1}),(\ref{op2}),(\ref{op3}) }.
\vspace{0.5cm}

This result
(whose demonstration follows the same lines
as in the scalar case \cite{tur1})
allows to write formally
all the operators preserving (\ref{esp}).
However, in order to classify these operators,
it is useful to set up normal ordering rules
between the generators.
In particular, these rules allow to write any product of operators
(the enveloping algebra) in a
canonical form, e.g. as a sum of terms where, in each term,
the $Q$ operators (if any)
are written on the left, the $\overline Q$ operators (if any) on
the right and the $J$ operators in between.
As we show next, such rules exist for the operators
(\ref{op1}),(\ref{op2}),(\ref{op3}).

 \vspace{0.5cm}
 \noindent {\bf {Normal  ordering  rules}}
 \vspace{0.5cm}

The operators (\ref{op1}) obey the commutation rules (\ref{comj})
and assemble into a reducible representation of $gl(V+1)$
when acting on the vector space (\ref{esp}).
The dimension of it is
\begin{equation}
C^{N+V}_V + C^{N+V-\Delta}_V  \ .
\end{equation}

By construction, the operators $Q_{[A]}$
(resp. $\overline Q^{[A]}$)
transform as an irreducible  multiplet of dimension $C^{V+\Delta}_{\Delta}$
under the adjoint action of the generators $J^{b}_{a}(N,\Delta)$.
More precisely, we have
\begin{eqnarray}
\label{comjq}
[ J^{b}_{a}, Q_{[A]} ] &=&
 k \  \delta^{b}_{a} Q_{[A]}
- \sum_{k=1}^{\Delta}
\delta^{b}_{a_k}
Q_{[\hat A_k, a]}
 \\
\label{comjqb}
[ J^{b}_{a}, \overline Q^{[A]} ] &=&
- k \ \delta^{b}_{a} \overline Q^{[A]}
+ \sum_{k=1}^{\Delta}
\delta_{a}^{a_k}
\overline Q^{[\hat A_k, b]} \ .
\end{eqnarray}
The explicit form of the generators leads to the value
$k = \Delta$.  The first Casimir constructed with (\ref{op1}),
i.e.
\begin{equation}
 T \equiv  \sum_{a=0}^V  J_a^a({N,\Delta}) \ ,
\end{equation}
 plays the role of a grading operator :
\begin{equation}
   [T,J_a^b] = 0 \ \ , \ \
   [T,Q_{[A]}] = \Delta V Q_{[A]} \ \ , \ \
   [T,\overline Q^{[A]}] = -  \Delta V \overline Q^{[A]}  \ .
\label{grading}
\end{equation}

The product of any two  operators $Q$
(and separately of two $\overline Q$'s) vanishes,
hence also their anticommutator
\begin{equation}
       \{ Q_{[A]} ,  Q_{[C]} \} = 0 \ \ \ , \ \ \
       \{ \overline Q^{[B]} , \overline Q^{[D]} \} = 0 \ .
\label{comqq0}
\end{equation}
The evaluation of the anti-commutator $\{ Q, \overline Q \} $ is more
involved. Its form can be guessed
from the covariance under $gl(V+1)$, from the symmetries
of $Q$ and $\overline Q$ in their indices and
from the fact that the anti-commutator involves at most derivatives of
the order $\Delta$. It is therefore likely that
the anticommutator  $\{ Q, \overline Q \} $  should be
expressed as a combination  of the tensors
\begin{equation}
 W_{[A]}^{[B]}(k) \equiv {1 \over (\Delta !)^2}
               S[A] \ S[B]
    ( J_{a_1}^{b_1}  J_{a_2}^{b_2}
      \ldots  J_{a_k}^{b_k}
       \delta_{a_{k+1}}^{b_{k+1}} \ldots
       \delta_{a_{\Delta}}^{b_{\Delta}} )
\end{equation}
where the operator $S[.]$ denotes the sum over all permutations
of all indices entering in the argument $[.]$.
After calculation, we found the following relations between
(\ref{op1}),(\ref{op2}),(\ref{op3}),
\begin{equation}
    \{ Q_{[A]} , \overline Q^{[B]} \} =
    \sum_{k=0}^{\Delta} \alpha_k W_{[A]}^{[B]}(k)
\label{comqqb}
\end{equation}
and the parameters $\alpha_k$ are numbers which
are uniquely determined by the polynomial equation
\begin{equation}
    \prod_{j=0}^{\Delta-1} (y+j) =
    \sum_{k=0}^{\Delta} \alpha_k (y + {\Delta -1 \over 2})^k  \ .
\label{cond}
\end{equation}
As a consequence of (\ref{cond}),
the right hand side of (\ref{comqqb}) is an even (resp. odd) polynomial
in the operators $J$ if $\Delta$ is even (resp. odd).
We would like to stress that this particularly simple expression
is due to the labelling of the generators and to the
translation used in (\ref{op1}).
A priori, the undetermined coefficients
$\alpha_k$  could be $2 \times 2$ diagonal  matrices.

The non vanishing parameters $\alpha_k$ appear only for
 $k={\Delta}, {\Delta-2},{\Delta-4},\ldots$ and read as follows
 for the first few values of $\Delta$~:
\begin{eqnarray}
&\Delta = 1  \ \ , \ \ &\alpha_k : \ 1 \                         \nonumber \\
&\Delta = 2  \ \ , \ \ &\alpha_k : \ 1 \ , \ -{1\over 4}         \nonumber \\
&\Delta = 3  \ \ , \ \ &\alpha_k : \ 1 \ , \ -1                  \nonumber \\
&\Delta = 4  \ \ , \ \ &\alpha_k : \ 1 \ , \ -{5\over 2} \ ,\ {9\over 16} \nonumber \\
&\Delta = 5  \ \ , \ \ &\alpha_k : \ 1 \ , \ - 5 \  , \  4   \nonumber \\
&\Delta = 6  \ \ , \ \ &\alpha_k : \ 1 \ , \  -{35 \over 4} \ , \
{259 \over 16} \  , \  -{225 \over 64} \ .
\end{eqnarray}

\section{Abstract algebras}

\par We now investigate the possibility that the operators
(\ref{op1}),(\ref{op2}),(\ref{op3})
represent  the generators of an abstract associative
algebra. We will see  that there are two
types of such algebras that we note generically  ${\cal A}(V,\Delta)$
and  ${\cal B}(V,\Delta)$.
A few  cases are known  to
coincide with Lie algebras \cite{shtu,bk}
\begin{equation}
{\cal A} (1,0) \simeq sl(2) \otimes sl(2)
\end{equation}
\begin{equation}
{\cal A}(1,1) \simeq osp (2,2) \simeq spl (2,1) \ .
\end{equation}
For $\Delta > 1$,  ${\cal A}(1,\Delta)$ corresponds
to a non linear superalgebra \cite{bk}. The algebra
${\cal A}(1,2)$ was treated in great detail in \cite{bgkn}.
Here we want to move away from the case $V=1$ in order to access
the hidden symmetries of the operators
preserving (\ref{esp}) in general.

With the aim to promote the normal ordering rules
of the previous section into a set of relations
defining an abstract associative algebra, we first note
that the operators $J_{a}^{b}$  (resp. $Q$, $\overline Q$)
should naturally
be interpreted as the bosonic (resp. fermionic) generators
of the algebra (this refers of course to the most natural
choice of the commutator or of the anti-commutator used to
exchange the order between these generators).
Therefore, we expect some graded algebras to come out.
However, it is well known that the knowledge of
a particular representation (here
 (\ref{op1}),(\ref{op2}),(\ref{op3})) is not
sufficient in general to infer the whole algebraic structure :
the Jacobi identities are not automatically fulfilled.
In the present case, the identities which are not obeyed
are those involving a  $\{Q,Q \}$
(or a  $\{\overline Q,\overline Q \}$) anticommutator
(remember that they vanish).
Although we can try to modify the whole set of (anti) commutation
relations, we limit our research of the underlying abstract
algebras in relaxing only the relation (\ref{comqq0}).
In order to present the way to modify it, a few
notations are worth introducing.

Due to its symmetry in the indices $[A]$,
the representation defined by the
 $Q_{[A]}$
(and similarly by the  $\overline Q^{[A]}$)
corresponds to a Young diagram with one line of $\Delta$ boxes.
The products
\begin{equation}
  Q_{a_1 \ldots a_{\Delta} }     Q_{c_1 \ldots c_{\Delta} }
\label{rep}
\end{equation}
assemble into a representation of  $gl(V+1)$
under the adjoint action of the operators $J$.
This representation can be decomposed into irreducible pieces.
The symmetry of $Q$ is such that the irreducible representations
appearing  in the decomposition of (\ref{rep}) correspond to the
Young diagrams consisting of two lines with total number of
$2 \Delta$ boxes.
When applied to the anticommutators
\begin{equation}
    Q_{a_1 \ldots a_{\Delta} }  Q_{c_1 \ldots c_{\Delta} }
 +  Q_{c_1 \ldots c_{\Delta} }  Q_{a_1 \ldots a_{\Delta} } \ ,
\end{equation}
the same decomposition
selects only the representations which are symmetric under the
exchange $[A] \leftrightarrow [C]$. In terms of Young diagrams
they correspond to the diagrams  with two lines and total number of
$2 \Delta$ boxes; the upper line is of length
$2 \Delta-2p$ (with $2p \leq \Delta$)
 and the lower line is of even length $2p$.
One Young tableau, corresponding to this Young diagram with fixed $p$,
is obtained by filling the first (resp. the second)  line with
\begin{equation}
[a_1,a_2,\ldots,a_{\Delta},c_{2p+1},c_{2p+2},\ldots,c_{\Delta}]
 \ \ \ , \ \ \ (\ {\rm resp.\ }
[c_1,c_2,\ldots,c_{2p}] \ ) \ .
\end{equation}
The Young element $S_Y$ corresponding to this Young tableau reads
\begin{equation}
    S_Y =
    S[a_1, \ldots a_{\Delta},c_{2p+1}, \ldots,c_{\Delta}]
    S[c_1, \ldots c_{2p}] E_x \ \ \ \ , \ \ \
\label{young}
\end{equation}
with
\begin{equation}
    E_x \equiv  \prod_{k=1}^{2p} (E - (a_k,c_k))
\label{exchange}
\end{equation}
where the operator $S[.]$ (defined previously) denotes the sum
over all permutations of all indices appearing in the argument $[.]$,
where $(a,b)$ denotes the transposition $a \leftrightarrow b$
and where $E$ is the identity operator.

With these notations, we are ready to describe
the conditions we choose for the anticommutations of two $Q$ operators .
We restrict them by imposing
\begin{equation}
    S_Y  \{ Q_{[A]} ,  Q_{[C]} \} = 0  \ .
\label{comqqy}
\end{equation}
This corresponds to the vanishing of
a particular representation
contained in decomposition of the symmetrized product
of $Q_{[A]}$ with  $Q_{[C]}$ into irreducible representations
of $gl(V+1)$.
The absent  representation is exactly the one
related to the Young diagram defined above,
characterized by  $p$
and by the Young element (\ref{young}).

We have studied the associativity conditions, i.e. the
Jacobi identities, compatible with the
relation (\ref{comj}),
the relations (\ref{comjq}),(\ref{comjqb}) for an arbitrary value
of the parameter $k$, the relation
(\ref{comqqb})  for arbitrary  values  of the  parameters
$\alpha_k$ and the relation (\ref{comqqy}) for an arbitrary integer $p$
(in fact $2p \leq \Delta$).
The results of our calculation is summarized by the following
proposition :

\vspace{0.5cm}
\noindent {\bf {Proposition 2}}
\vspace{0.5cm}

\noindent {\it
The set of relations
(\ref{comj}),(\ref{comjq}),(\ref{comjqb}),(\ref{comqqb}),(\ref{comqqy})
are compatible with all the  Jacobi identities
in two cases only :

1. $p=0$ and $k=\Delta$

2. $\Delta$ even, $\Delta = 2p$ and $k=-1$ }
\vspace{0.5cm}

 In both cases, the anticommutation relations
of two $\overline Q$'s have to follow the same symmetry
pattern as the anticommutation relations (\ref{comqqy})
of two $ Q$'s.
Associativity is realized irrespectively
of the values of the parameters $\alpha_k$.
That is to say that we obtained two families of associative
algebras ${\cal{A}},{\cal{B}}$, each indexed by $\Delta + 1$ parameters and
by the integers $V$ and $\Delta$.
By a suitable choose of the normalisation of the  $Q$'s
and/or of the $\overline Q$'s one can set
$\alpha_{\Delta} = 1$ in (\ref{comqqb}).
 One can also set  $\alpha_{\Delta -1}=0$
by using an appropriate translation (\ref{trans})
on the operators $J(N,\Delta)$.
Before presenting the proof of this result, let us
discuss a few properties of the algebras.

\vspace{0.5cm}
\begin{description}
\item [Case 1. The abstract algebra ${\cal A}(V,\Delta,\alpha_k)$]
\ \ \ \

In the case $p=0$, $k = \Delta$,
the constraint (\ref{comqqy}) reads
\begin{equation}
S[A,C]
  \{ Q_{[a_1, a_2 \ldots a_{\Delta}]},
     Q_{[c_1, c_2 \ldots c_{\Delta}]}
  \}     = 0
\label{comqq1}
\end{equation}
and the same relation has to be imposed on the operators $\overline Q$.
Using some combinatoric, one can show that (\ref{comqq1})
encodes a total number of $C^{2 \Delta + V}_V$ independent relations
among the anticommutators of two $Q$'s. Remembering that there are
$C^{V+\Delta}_V$ operators $Q$, we see easily that
the number of constraints is lower than the number of
independent anticommutators,
that is to say that not all anticommutators are constrained.

The  operators (\ref{op1}),(\ref{op2}),(\ref{op3})
constitute a particular representation of the algebras
of this type~: the ones corresponding to the values
$\alpha_k$ determined by (\ref{cond}).
For these operators the conditions (\ref{comqq1})
are  trivially realized.

In the case $\Delta = 1$, the relation
(\ref{comqqy}) just implies that all
anticommutators of two operators $Q$ vanish
(and similarly for  two $\overline Q$).
The algebra ${\cal A}(V,1)$ is linear,
it coincides with the  Lie superalgebra denoted
 $spl(V+1,1)$ in the classification \cite{kac}.
If  $V=1$ one recovers the algebra $osp(2,2)$
(remember the equivalence of $osp(2,2)$ with  $spl(2,1)$).

\vspace{0.5cm}

\item[Case 2. The abstract algebra ${\cal B}(V,\Delta,\alpha_k)$]
\ \ \

If $\Delta$ is even and if $p=\Delta/2$
the constraints (\ref{comqqy}) on the $Q$'s can be set in the form
\begin{equation}
   \lbrace Q_{a_1 a_2 \ldots a_{\Delta}},
           Q_{a_{\Delta+1} a_{\Delta +2} \ldots a_{2\Delta}}  \rbrace
=  \lbrace Q_{\sigma(a_1) \sigma(a_2) \ldots \sigma(a_{\Delta})},
           Q_{\sigma(a_{\Delta+1}) \sigma(a_{\Delta +2})
           \ldots \sigma(a_{2\Delta})}   \rbrace
\end{equation}
for any permutation $\sigma$ of the $2\Delta$ indices.
The total number of independent constraints is not as easy to find
as in the case 1; we obtained it in two particular cases
\begin{equation}
   {\Delta (\Delta -1) \over 2} \ \ \ \ {\rm if} \ \ \  V = 1
\end{equation}
and
\begin{equation}
   {V (V +1)(V^2+9V-4) \over 12}
 \ \ \ \ {\rm if} \ \ \  \Delta = 2 \ .
\end{equation}
\end{description}

\vspace{0.5cm}
\noindent {\bf {Exceptional solutions}}
\vspace{0.5cm}

It should be stressed that associative algebra could
also exist with the same structure as above,
i.e. with (\ref{comj}), (\ref{comjq}), (\ref{comjqb}), (\ref{comqqy})
and (\ref{comqqb}) but where some of
the parameters $\alpha_k$ are $2\times2$ diagonal matrices.
That is to say they depend on the Casimir operators
constructed with the $gl(V+1)$ subalgebra generated by the
operators (\ref{op1}).
We could not solve this problem for  general values
of $\Delta$ but we studied completely the
cases $\Delta = 1,2,3$. We obtained one new solution
in the case $\Delta=2,V=1,p=1$.
The  most general relation for $ \{Q, \overline Q \}$
which is compatible with associativity depends on four parameters.
It is of the form
\begin{equation}
 \{ Q_{a_1 a_2} ,  \overline Q^{b_1 b_2} \}
 =
   \sum_{j=0}^{2} \alpha_j \ W_{a_1 a_2}^{b_1 b_2}(j)
    +
    \beta \Bigl(   C_1\  W_{a_1 a_2}^{b_1 b_2}(1)
            + (4 C_2 - 3 C_1^2) \ W_{a_1 a_2}^{b_1 b_2}(0)
          \Bigr)
\end{equation}
where $\beta$ is the additional  parameter
while $C_1,C_2$ represent the Casimir operators
(\ref{cas}) computed in the representation (\ref{op1}).

\section{Proof of proposition 2}

Let us come to the proof of proposition 2.
The relevant Jacobi identities are
\begin{equation}
\left[\{Q_{[A]},\overline Q^{[B]}\},Q_{[C]}\right]
+\left[\{Q_{[C]},\overline Q^{[B]}\},Q_{[A]}\right]
+\left[\{Q_{[A]},Q_{[C]}\},\overline Q^{[B]}\right]=0 \ .
\end{equation}
The application of $S_Y$ (see (\ref{young}))
to this equation and the use of (\ref{comqqy})
lead to the necessary and sufficient conditions
\begin{equation}
S_Y\left (\left[\{Q_{[A]},\overline Q^{[B]}\},Q_{[C]}\right]
+\left[\{Q_{[C]},\overline Q^{[B]}\},Q_{[A]}\right]\right )=0 \ .
\end{equation}
Moreover $S_Y$ (with an even second line) applied to a tensor
$T_{[A,C]}$, symmetrical in $[A]$ on one side and in $[C]$ on the
other side, selects automatically the piece in $T$ symmetrical
under the exchange $[A]\leftrightarrow [C]$. Hence, the
necessary and sufficient condition becomes simply
\begin{equation}
S_Y\left(\left[\{Q_{[C]},\overline Q^{[B]}\},Q_{[A]}\right]\right )=0
\ .
\label{jacobi}
\end{equation}

Let us first suppose that the anticommutation relations
of $Q$ and  $\overline Q$ take the form
\begin{equation}
\left\{Q_{[A]},\overline Q^{[B]}\right\}=
     S[A]S[B]J^{b_1}_{a_1}J^{b_2}_{a_2}\ldots J^{b_{\Delta}}_{a_{\Delta}}
\label{comqqbs}
\end{equation}
rather than the more general one (\ref{comqqb}).
Using (\ref{comqqbs}) together with (\ref{comjq}),
and separating the terms, say $X'$, which come out without $k$
(through (\ref{comjq}))
from the terms, say  $k Y'$, which come out linear in $k$,
the expression (\ref{jacobi}) becomes
\begin{equation}
X'+ k Y'=0
\end{equation}
where
\begin{eqnarray}
X'&=&-S_Y S[B]\sum_{i=1}^{\Delta}\sum_{j=1}^{\Delta}
    \delta_{a_i}^{b_1}Q_{[c_j,\hat A_i]}S[\hat C_j]
    J^{b_2}_{c_1}J^{b_3}_{c_2}\ldots J^{b_{\Delta}}_{c_{\Delta}}
\ \  - \ \ \ldots
\label{XP} \\
Y'&=& S_Y S[B]\sum_{j=1}^{\Delta}
    \delta_{c_j}^{b_1}Q_{[A]}S[\hat C_j]
    J^{b_2}_{c_1}J^{b_3}_{c_2}\ldots J^{b_{\Delta}}_{c_{\Delta}}
\ \    + \ \  \ldots  \ .
\label{YP}
\end{eqnarray}
In (\ref{XP}) and (\ref{YP}), the $\dots$ refer to the terms where the
$Q$ does not appear as the first operator,
but rather after a $J$ operator. Remark
also that the index $c_j$ is absent in the set
$[\hat C_j]$ and accordingly does not appear as a lower index in the
$J$'s.. It follows that, as it should,
the number ($\Delta -1$) of indices $b_k$ in the product of the $J$'s
matches the number of $c_m$ indices.

Since $X'+ k Y'$ has to be zero identically, every coefficient of
every (independent) operator entering in it has to be zero. This
allows a great simplification in the necessary and sufficient
conditions.
\vskip 0.5 cm
\begin{itemize}
\item The terms labeled $\ldots$ in (\ref{XP}) and (\ref{YP}) can
be forgotten altogether. Indeed, the terms where the $Q$'s are in
the first position are independent of the terms where they are not.
\item The symmetry on the $[B]$ can also be eliminated. Every
term, for every value of the indices $b_k$, has to vanish on its
own.
\item Let us introduce the notations
\begin{equation}
W(a_i)=\delta_{a_i}^{b_1}
\end{equation}
for some arbitrary fixed value of $b_1$ and
\begin{equation}
V[\hat C_k]=J^{b_2}_{c_1}J^{b_3}_{c_2}\ldots J^{b_{\Delta}}_{c_{\Delta}}
\end{equation}
where $c_k$ is absent as a lower index and
$b_2,\ldots,b_{\Delta}$ have also fixed values..
\end{itemize}
\vskip 0.5 cm
With these simplifications, the condition $X'+kY'=0$ reduces to
the necessary and sufficient condition $X+kY=0$ with
\begin{equation}
X=-S_Y \sum_{i=1}^{\Delta}\sum_{j=1}^{\Delta}
    W(a_i) Q_{[c_j,\hat A_i]}S[\hat C_j] V[\hat C_j]
\label{X}
\end{equation}
and
\begin{equation}
Y=  S_Y \sum_{j=1}^{\Delta}
    W(c_j) Q_{[A]}S[\hat C_j]
    V[\hat C_j] \ .
\label{Y}
\end{equation}
The operator $X+k Y$ is composed of  exactly two types of
independent operators. They can be written canonically as
\begin{eqnarray}
O_1&=&W(c_1)Q_{[A]}V[\hat C_1]  \ ,   \\
O_2&=&W(c_{\Delta})Q_{[A]}V[\hat C_{\Delta}]  \ .
\label{OP}
\end{eqnarray}
Indeed~:
\vskip 0.5 cm
\begin{itemize}
\item The indices of the $Q$ operator have to be completely
symmetrical. Hence they must belong to the first line of the
Young tableau and by
symmetry of $S_Y$ can be chosen as the $[A]$ set.
\item If the index in
$W$ is taken in the first line, it can be chosen to be $c_{\Delta}$.
This is due to the fact that any of the indices
(except the those belonging to the set $[A]$ which already pertain
to the $Q$) in the first line is equivalent by symmetry to any
other in the first line. The remaining indices for the
$V$ can the be chosen in any order and for example in the natural order.
\item If the index in
$W$ is taken in the second line, it can be chosen to be $c_{1}$.
Indeed any of the indices in the second line is equivalent by symmetry to any
other in the second line. The remaining indices for the
$V$ can again be chosen in any order and for example in the natural order.
\end{itemize}
\vskip 0.5 cm
The remaining task is to extract in $X$ and in $Y$ the number of times
the operators $O_1$ and $O_2$ occur. This is a rather delicate
operation in terms of the symmetries involved.
Let us call $X_i$ (resp.  $Y_i$) with $i=1,2$ the coefficient of
the operator $O_i$ in $X$ (resp. $Y$)
With these notations
the condition $X+kY=0$ becomes equivalent to
\begin{equation}
    X_1 + k Y_1 = 0 \ \ \ , \ \ \ X_2 + k Y_2 = 0 \ .
\label{xi+yi}
\end{equation}

To now compute these four coefficients, we will make
use of the
fundamental theorem of group theory which states that,
if $P$ is any permutation of the elements in $[A]$
\begin{equation}
S[A]=PS[A]=S[A]P
\label{fonda}
\end{equation}
\vskip 0.5 cm
\begin{description}
\item[Computation of $X_1$]
\ \ \

Let us rewrite $X$ as
\begin{eqnarray}
X=&-&
   \sum_{i=1}^{\Delta}\sum_{j=1}^{\Delta}
   S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
   E_x   \nonumber \\
   && W(a_i) Q_{[c_j,\hat A_i]}S[\hat C_j] V[\hat C_j]
\label{X1a}
\end{eqnarray}
where we have interchanged the finite sommation on $i$ and $j$
with the symmetry operations.
First, the $a_i$ in $W(a_i)$ which belongs to the first line has
to be replaced by a $c$ belonging to the second line. This can
be done at the intervention of the operator $E_x$ only. At the
same time none of the other $a_j,j \neq i$ in $Q$ should be
replaced by an element of the second line. Hence, from the
$2^{2p}$ terms in $E_x$ we can restrict ourselves
to the transposition $(a_i,c_i)$ which comes
with a minus sign. At the same time $i$ can be restricted to the range $1,2p$.
The sommation on $j$ then has one term with $j=i$. For the terms
with $j \neq i$, the $c_j$ in $Q$ has to belong to the set
$j=2p+1,\ldots,\Delta$ in order to be able to replace it by an
$a$ by the first symmetry operator $S$ in (\ref{X1a}). Hence
the restricted part of $X$, say $\hat X$, is composed of two
pieces, say $\hat X_{\alpha}$ and $\hat X_{\beta}$,
\begin{eqnarray}
\hat X_{\alpha}=&&
   \sum_{i=1}^{2p}
   S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
   (a_i,c_i)
   W(a_i) Q_{[c_i,\hat A_i]}S[\hat C_i] V[\hat C_i]
         \nonumber  \\
          =&&
   \sum_{i=1}^{2p}
   S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
   W(c_i) Q_{[A]}S[\hat C_i] V[\hat C_i]
         \nonumber  \\
          =&&
   \sum_{i=1}^{2p}
   S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
   (c1,c_i)
    W(c_i) Q_{[A]}S[\hat C_i] V[\hat C_i]
         \nonumber  \\
          =&&
   S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
   \sum_{i=1}^{2p}
     W(c_1) Q_{[A]}S[\hat C_1] V[\hat C_1]
\label{X1alpha}
\end{eqnarray}
and
\begin{eqnarray}
\hat X_{\beta}=&&
   \sum_{i=1}^{2p}\sum_{j=2p+1}^{\Delta}
   S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
   (a_i,c_i)
    \nonumber  \\
   && W(a_i) Q_{[c_j,\hat A_i]}S[\hat C_j] V[\hat C_j]  \ .
\label{X1beta}
\end{eqnarray}

Using (\ref{fonda}), we
easily conclude that in $X_{\alpha}$ the following coefficient appears
\begin{equation}
(\Delta)!(2p)!(\Delta-2p)!  \ .
\label{xbeta}
\end{equation}
The first factor $(\Delta)!$ comes from the permutation of the $[A]$ set
which always contributes to an equal factor due to the symmetry
of the $Q$. The second term $(2p)!$ comes from the product of
 of the sommation over $i$ (a factor $2p$)
and of a factor $(2p-1)!$ coming from the repetition
of the symmetries in $[c_2,\ldots,c_{2p}]$ contained in the
second and in the third $S$ factors. The last term $(\Delta-2p)!$
comes from from the repetition
of the symmetries in $[c_{2p+1},\ldots,c_{\Delta}]$ contained in the
first and in the third $S$ factors.

Let us now focuss our attention on the $X_{\beta}$ term. Using
(\ref{fonda}) we can factor out of
$S[A,c_{2p+1},\ldots,c_{\Delta}]$, at no cost, a transposition
factor $(a_i,c_j)$ and from $S[c_1,\ldots,c_{2p}]$ a factor
$(c_1,c_i)$. The product of these two transpositions together
with the transposition in $X_{\beta}$ leads to the cyclic
permutation $(a_i,c_1,c_i,c_j)$ and the relevant part $\hat X_{\beta}$
becomes
\begin{eqnarray}
\hat X_{\beta}=&&
   \sum_{i=1}^{2p}\sum_{j=2p+1}^{\Delta}
   S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
     \nonumber  \\
    &&W(c_1) Q_{[A]}S[\hat C_1] V[\hat C_1] \ .
\label{X1beta2}
\end{eqnarray}

The following coefficient then appears
\begin{equation}
(\Delta-2p)(\Delta)!(2p)!(\Delta-2p)!  \ .
\label{xalpha}
\end{equation}
The extra factor as compared to the coefficient coming out of
$X_{\alpha}$ is due to the extra sommation over $j$.

Summing up the results (\ref{xalpha},\ref{xbeta}), we find
\begin{equation}
X_1=(1+\Delta-2p)(\Delta)!(2p)!(\Delta-2p)! \ \ .
\label{X_1}
\end{equation}
\vskip 0.5 cm
\item[Computation of $Y_1$]
\ \ \

The same technique applied to $Y_1$ is much simpler as the
relevant term in $E_x$ is simply the identity. Hence
\begin{eqnarray}
\hat Y=&&
     S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
     \sum_{j=1}^{2p}
     W(c_j) Q_{[A]}S[\hat C_j]
    V[\hat C_j]   \nonumber  \\
=&&
     S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
     \sum_{j=1}^{2p}
     W(c_1) Q_{[A]}S[\hat C_1]
    V[\hat C_1]  \ .
\label{Y1}
\end{eqnarray}
To pass from the first to the second line we have factored out of
$S[c_1,\ldots,c_{2p}]$ the transposition $(c_1,c_j)$.

Collecting again the factors, we find
\begin{equation}
Y_1=(\Delta)!(2p)!(\Delta-2p)! \ \ .
\label{Y_1}
\end{equation}
\vskip 0.5 cm
\item[Computation of $X_2$]
\ \ \

The relevant term in $E_x$ is again the identity and
the reduced part of $X$ which can lead to a term of the form $O_2$
(\ref{OP}) is
\begin{equation}
\hat X=-   \sum_{i=1}^{\Delta}\sum_{j=1}^{\Delta}
     S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
    W(a_i) Q_{[c_j,\hat A_i]}S[\hat C_j] V[\hat C_j] \ .
\label{X2a}
\end{equation}
The sommation on $j$ has to be restricted to those values in the
first line of the Young diagram. A transposition $(a_i,c_j)$ can
then be factored out of $S[A,c_{2p+1},\ldots,c_{\Delta}]$ as
well as a
transposition $(a_i,c_{\Delta})$, i.e. in total a cyclic permutation
$(a_i,c_{\Delta},c_j)$ . We find
\begin{eqnarray}
\hat X=&-& \sum_{i=1}^{\Delta}\sum_{j=2p+1}^{\Delta}
     S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
     (a_i,c_{\Delta},c_j)
    \nonumber  \\
    && W(a_i) Q_{[c_j,\hat A_i]}S[\hat C_j] V[\hat C_j]
     \nonumber  \\
     =&-& \sum_{i=1}^{\Delta}
     S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
     \sum_{j=2p+1}^{\Delta}
    \nonumber  \\
    && W(c_{\Delta}) Q_{[A]}S[\hat C_{\Delta}] V[\hat C_{\Delta}] \ .
\label{X2b}
\end{eqnarray}
Collecting the factors as usual, we find
\begin{equation}
X_2=-\Delta(\Delta)!(2p)!(\Delta-2p)! \ .
\label{X_2}
\end{equation}
The extra factor ($\Delta$) comes from the sommation over $i$.
\vskip 0.5 cm
\item[Computation of $Y_2$]
\ \ \

In this last case the relevant term in $E_x$ is again
the identity and
the reduced part of $Y$ which can lead to a term of the form $O_2$
(\ref{OP}) is
\begin{eqnarray}
\hat Y&=&
     S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
     \sum_{j=2p+1}^{\Delta}
    W(c_j) Q_{[A]}S[\hat C_j]
    V[\hat C_j]  \nonumber        \\
    &=&
     S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
     \sum_{j=2p+1}^{\Delta} (c_j,c_{\Delta})
    W(c_j) Q_{[A]}S[\hat C_j]
    V[\hat C_j]    \nonumber   \\
    &=&
     S[A,c_{2p+1},\ldots,c_{\Delta}] S[c_1,\ldots,c_{2p}]
     \sum_{j=2p+1}^{\Delta}
    W(c_{\Delta}) Q_{[A]}S[\hat C_{\Delta}]
    V[\hat C_{\Delta}]        \ .
\label{Y2}
\end{eqnarray}
Collecting the terms, we find
\begin{equation}
Y_2=(\Delta)!(2p)!(\Delta-2p)!  \ .
\label{Y_2}
\end{equation}
\end{description}
\vskip 0.5 cm
We can now summarize the conditions coming from (\ref{xi+yi}) :
\begin{enumerate}
\item The conditions coming from the Jacobi identities are thus two in
number if $p \neq 0$ (the condition for the operator $O_1$ to be
defined) and if $\Delta \neq 2p$ (the condition for $0_2$ to be
defined). These conditions
\begin{eqnarray}
k&=&-(\Delta+1-2p)  \\
k&=&\Delta
\label{gencase}
\end{eqnarray}
are incompatible.
\item More generally, the anticommutators of two $Q$'s cannot
vanish for more than one representation.
\item If $p=0$ the only condition comes from the $0_2$ operator.
It is
\begin{equation}
k=\Delta
\end{equation}
which is a solution to our problem. The corresponding Young
diagram has only one line of length $2\Delta$.
\item If $\Delta$ is even and $\Delta=2p$ the only condition
comes from the $0_1$ operator.
It is
\begin{equation}
k=-1
\end{equation}
which is a second solution to our problem.
The corresponding Young diagram has two lines of equal length $\Delta$.
\end{enumerate}
\vskip 0.5 cm
This achieves the proof of the proposition when the
anticommutator $\{ Q , \overline Q \}$ is restricted to
(\ref{comqqbs})
It is easy to see that the other allowed terms in the
anticommutator of the $Q$'s with the $\overline Q$'s, i.e. those
which do not involve $J$'s only but the products of $J$'s and
$\delta$'s as in (\ref{comqqb}) lead to exactly the same restrictions.
Hence they can all be present at the same time
leaving us with the form (\ref{comqqb}) with the  $\Delta+1$
arbitrary coefficients.

The conditions coming from the Jacobi identities involving two
$\overline Q$ and one $Q$ also lead to exactly the same conditions.
Hence the anticommutators which are chosen to be zero for the
anticommutations
of two $Q$'s on one side or of two $\overline Q$'s on the other
side must be identical.

\section{Summary and Conclusions}

The operators preserving globally a system of two polynomials in
$V$ variables ($V\geq 1$) and of degrees $N$ and $N-\Delta$
($\Delta\geq 0$) respectively can
be constructed as the elements of the enveloping algebra of certain
superalgebras.

In this paper, we have constructed a family
of such associative, non linear superalgebras. Any of these
algebras is specified by $V$, by $\Delta$ and by a set of
$\Delta+1$ complex numbers noted $\alpha_k$ with
$k=0,1,\ldots,\Delta$. They are generated by $(V+1)^2$ (bosonic) operators
\begin{equation}
J_a^b\ \ \ , \ \ \ a,b=0,1,\ldots,\Delta
\end{equation}
and by two sets of $C_{\Delta}^{V+\Delta}$ (fermionic) operators
\begin{equation}
Q_{[a_1,\ldots,a_{\Delta}]}\ \ \ , \ \ \
    {\overline Q}^{[a_1,\ldots,a_{\Delta}]} \ \ \ ,\ \ \
        a_k=0,1,\ldots,\Delta
\end{equation}
symmetric in their $\Delta$ indices.

The bosonic generators obey the commutation relations of the Lie
algebra $gl(V+1)$. The operators $Q$ (and separately the
$\overline Q$) assemble into a specific representation of
$gl(V+1)$ under the adjoint action of $J_a^b$
(see (\ref{comjq}),(\ref{comjqb})).
The anticommutators $\{Q,\overline Q\}$ are polynomials of
degree at most $\Delta$ in the bosonic operators. The
arbitrariness of the polynomials is coded in the
$\Delta+1$~parameters $\alpha_k$ (\ref{comqqb}).

All the supplementary conditions on the products of the
operators $Q$ (and of the operators $\overline Q$) necessary to
guarantee associativity (equivalent to the generalised Jacobi
identities) are given by our proposition~2.

For all fixed values of the integers $V$ and $\Delta$ and of the
complex parameters $\alpha_k$ we denote
${\cal{A}}(V,\Delta,\alpha_k)$ the algebra corresponding to
case~1 of proposition~2. If $\Delta$ is even, a supplementary
algebraic structure, that we denote
${\cal{B}}(V,\Delta,\alpha_k)$ is also possible, as predicted by
case~2 of proposition~2.

Refering to the general definition of a $W$-algebra given
recently in \cite{sorba}, it is natural to classify
${\cal{A}}(V,\Delta,\alpha_k)$ and ${\cal{B}}(V,\Delta,\alpha_k)$
as ``finite $W_{\Delta+1}$-superalgebras''.

An analysis of the representations of
${\cal{A}}(1,2,\alpha_0, \alpha_1, \alpha_2)$, performed
recently \cite{bgkn}, leads to a rather rich set of inequivalent
irreducible, finite dimensional representations .

Let us stress again that the operators in the enveloping
algebras that we have constructed are directly relevant for the
study of quasi-exactly solvable systems of equations.

\newpage

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"Legomandala" combines the Tibetan Buddhist tradition of creating and ritualistically dismantling ornate sand mandalas with the contemporary medium of LEGO. A meticulous work of art, it is taken apart just after completion by a group of adults and children invited to transform it into their own creation.
for a timelapse video of the first large Legomandala I completed with some LEGO I had on loan last year. It was an experiment and it worked! Now I need a collection of LEGO dedicated to this work to bring more Legomandalas to the public at large.
Using LEGO, a contemporary toy and a plaything of wealthy nations, in an artistic, political and social context has proved a powerful way of communicating with people today. The LEGO makes accessible to people who might otherwise be unconcerned with spiritual, philosophical or political issues, uncommonly profound concepts in an environment focused on one thing: collaborative change.
The deconstruction of the piece represents the acceptance of material impermanence; calling us to live more presently and to value all life in its precious fragile reality. It is also a statement as to the insanity of valuing material goods over the good of other living beings. This is an ancient Tibetan Buddhist message and has helped the Tibetan people resist the Chinese occupation of their country peacefully and for decades. Their cause is no longer in "fashion" but it is still very much alive and as relevant to the world's crises as ever.
The reconstruction of the LEGO represents the importance of imaginative co-creation and joy in transformative work. When participants are invited to dismantle this "sacred" universe and encouraged to make it their own, they are being called to play an active role in to co-creation of our world. The LEGO represents the building blocks of our universe and how when we are involved in transformation and re-creation collaboratively and joyfully – we are at our best in the world.
It is both a spiritual, political and social piece. The beauty of the mandala, letting it go and participating as a group in creating the new - is an excellent medium for raising awareness. Not only is a Legomandala a profound agent for social impact, it is also fun.
I am currently involved in a number of awesome projects. Another one of note is a street art initiative involving children's drawings, The Monster Project.


Do I need surgery on my kneecap?
my kneecap started hurting last week and makes crunching / popping sounds.
It wont stop it hurts to bend or move.
Did you do anything in particular to hurt your knee?
Is it red and/or hot?
Your problem is common however if there is no reason for the onset of pain it would be best to get a thorough physical assessment by an osteopath/physiotherapist or chiropractor. Your knee pain can be a result of a problem elsewhere i.e. your low back, hip or foot.
You "may" eventually need surgery but surgery should be something to look into only if the symptoms can't me managed through non-surgical means first.
See your preferred physio/osteo/chiro first.
See your GP to discuss combining this with some simple painkiller and antiinflammatories.
If still not settling, you would then be advised get scans and to see a surgeon.
80% of the population will suffer from what is called retro patella pain at some time in their life. Of these most will get better unaided but some will need help. If there has been no history of injury then patellofemoral pain is the most common suspect. My suggestion is to see a Physiotherapist ASAP towards the correct diagnosis and tretment/management. If its is something else underlying that is causing the problem then the Physiotherapist should be able to detect and refer. If it only the patellofemora ljoint that is involved then the Physiotherapist should be able to treat sucsessfully.
What could the pain on the inside of my upper theighs be?


If they're thinking of re-emphasizing teetotalism to keep their members unspotted by the world, they really are Some. Kinda. Stupid. Better to re-institute the ban on movies and to throw in TV sitcoms and dramas, too.


This study was completed following extensive Language Development Specialist (LDS) Training in Stockton Unified School District (SUSD) between 1992 and 1994. A post training survey designed by this researcher was mailed to all (303) teachers and administrators/elementary and secondary who participated in LDS training during this time. This survey asked participants to identify training information applicable to daily work tasks. One hundred seven teachers and twenty-four administrators completed this survey. The survey was divided into four sections. Sections I and II were based on the Nine LDS competencies. A 2 x 2 Chi Square Test of Association was completed for this data. Significant differences (p. $<$ 05) were obtained for teacher and administrator responses for changes made/observed in the classroom for planning lessons, teaching strategies, selecting materials, testing, and certifying limited/non-English (LEP/NEP) speaking students. Both groups noted some to significant understanding for background/culture/language of LEP/NEP students and differences between bilingual education and English as a second language. Section III was a sentence completion task. These items were: (a) most valuable information received from LDS training, (b) information of little or no value, (c) items to include/exclude in future LDS type training programs, and (d) additional comments. Improved understanding of second language acquisition was most valuable information, history and court cases was of little or/no value. Future LDS type training should include practical application to daily work tasks and exclude statistics and theory. Additional comments emphasized pressure from SUSD to complete LDS training and pass the state exam. Section IV asked participants to identify their grade level and date(s) of LDS training. Data beyond grade level was not tabulated since less than 2% identified date(s) of LDS training. Post-training feedback was the basis of this investigation. This promotes reflective thought and allows participants to more accurately identify information applicable to daily work tasks. Implications for future research include increased practical information and employee input in staff training design, content, and presentation.


2 fried eggs, tomatoes, jalapeno, onions and fire roasted salsa. Served with refried beans, Latin-fried potatoes, 2 flour tortillas and salsa verde.
2 scrambled eggs, chorizo, pico de gallo and tortilla strips. Served with refried beans, Latin-fried potatoes, 2 flour tortillas and salsa verde.
Choose a dressing: Buttermilk Ranch Dressing, House Vinaigrette, Avocado Ranch Dressing, Creamy Sriracha Dressing, Fire Roasted Salsa Dressing.
A fresh lettuce blend topped with tomato, onion, shredded cheese, feta, cilantro, tortilla strips and choice of 1 salad dressing.


Museum Development and Cultural Representation critically examines the development of a museum and cultural heritage centre in the indigenous Kelabit Highlands in Sarawak, Malaysia.
Building on their direct involvement in the development of the project, the authors appraise the process in retrospect through a thematic analysis. Themes covered include the project's local and international contexts, community involvement and agency, the balance of tourism and authenticity, and the role of non-local partners. Through their analysis, the authors unpack the complexities of cultural representation and identity in heritage design practice, and investigates the relationship between capacity building and agency in cultural heritage management.
Situating the project within international trends in museology, Museum Development and Cultural Representation offers a valuable case example of a heritage-making process in an indigenous community. It will be of interest to scholars and students studying cultural representation, as well as communities and museum professionals looking to develop similar projects.


text:
The Republican Ticket: Trump and Pence
By Lesley Stahl
July 17, 2016 / 8:12 PM / CBS News
The Republican Ticket
The Republican Ticket 21:36
The following is a script from "The Republican Ticket" which aired on July 17, 2016. Lesley Stahl is the correspondent. Richard Bonin and Ruth Streeter, producers.
The Republican National Convention that begins tomorrow in Cleveland will star Donald Trump and his chosen running mate the governor of Indiana, Mike Pence.
The Republican leadership has praised Pence as a good choice to unite the party. He's known as a reliable conservative with close ties to the religious right and he also has good relations with Republicans in Congress, having served six terms in the House of Representatives.
What kind of VP would Pence be? 01:53
But there are significant areas in terms of values and policy where he has differed with Mr. Trump in the past.
We had the chance to ask the two of them about that in their first and only interview together, yesterday afternoon in New York, in Trump's three-story penthouse apartment in the Trump Tower.
Lesley Stahl: First of all, Governor Pence, congratulations.
Mike Pence: Thank you.
"It's very, very humbling and I couldn't be more honored to have the opportunity to run with, and serve with, the next president of the United States."
Lesley Stahl: This has probably come as a huge, life-changing moment for you.
Mike Pence: It has. It's very, very humbling and I couldn't be more honored to have the opportunity to run with, and serve with, the next president of the United States.
Lesley Stahl: Before we actually talk about the politics, you know, there've been so many major world events very recently, in the last week. I don't know if you can remember the last time we have seen a world this much in chaos. You even said, "It's spinning apart." Are you ready for this world that we are facing today?
Waging war against radical Islam 01:16
Donald Trump: We're both ready. I've no doubt. We need toughness. We need strength. Obama's weak, Hillary's weak. And part of it is that, a big part of it. We need law and order. We need strong borders.
Lesley Stahl: But all reactions to what's been going on aren't muscular. For example, look what happened in Turkey. There was a military coup in a democratic country; a NATO ally. How would you respond to that?
Donald Trump: Well, as a president, I'm going to be-- you know, they've been an ally and I stay with our allies. They have been an ally. But that was a quick coup. I was actually surprised to see how well it was handled. And you know who really handled it? The people. So, I mean, we can say what we want, but the people handled it. When they surrounded the army tanks and without the people, you would've never had it. The military would've taken over.
"We're both ready. I've no doubt. We need toughness. We need strength. Obama's weak, Hillary's weak."
Mike Pence: But I truly do believe that the larger issue here is declining American power in the world. I truly do believe that history teaches that weakness arouses evil and whether it be the horrific attack in France, the inspired attacks here in the United States, the instability in Turkey that led to a coup. I think that is all a result of a foreign policy of Hillary Clinton and Barack Obama that has led from behind and that has sent an inexact, unclear message about American resolve. One of the reasons why I said yes in a heartbeat to run with this man, is because he embodies American strength, and I know that he will provide that kind of broad-shouldered American strength on the global stage as well.
Lesley Stahl: Let's talk about what happened in Nice, horrendous, carnage, horrible--
Donald Trump: Horrible.
Lesley Stahl: Horrible. You said you would declare war against ISIS. What exactly do you have--
Donald Trump: It is war. By the way, it is war.
Lesley Stahl: No, but does that-- when you say, "Declare war," do you want to send American troops in there? Is that what you mean?
Donald Trump: Look, we have people that hate us. We have people that want to wipe us out. We're gonna declare war against ISIS. We have to wipe out ISIS. These are people that--
Ivanka as VP? 00:51
Lesley Stahl: With troops on the ground?
Donald Trump: I am going to have very few troops on the ground. We're going to have unbelievable intelligence, which we need; which, right now, we don't have. We don't have the people over there. We are going to use--
Lesley Stahl: You want to send Americans--
Donald Trump: Excuse me-- and we're going to have surrounding states and, very importantly, get NATO involved because we support NATO far more than we should, frankly, because you have a lot of countries that aren't doing what they're supposed to be doing. And we have to wipe out ISIS. And speaking of Turkey, Turkey is an ally. Turkey can do it by themselves. But they have to be incentivized. For whatever reason, they're not. So we have no choice.
Lesley Stahl: But I still don't know if you're going to send troops over--
Donald Trump: Very little. I'm gonna--
Lesley Stahl: But declare war--
Donald Trump: --get neighboring states and I'm going to get-- we are going to get NATO; we're going to wipe 'em out. We're gonna--
Lesley Stahl: But declare war?
Mike Pence: Lesley--
Lesley Stahl: What does that mean--
Mike Pence: This is-- this is the kind-- this is the kind of leadership that America needs and it--
Lesley Stahl: But what--
Mike Pence: -and it begins with deciding to destroy the enemies of our freedom.
Lesley Stahl: How?
Mike Pence: And how we do that? I have every confidence. You-- you remember I served on the Foreign Affairs Committee. And I'm very confident that when Donald Trump becomes president of the United States, he'll give a directive to our military commanders, bring together other nations, and we will use the enormous resources of the United States to destroy that enemy.
Trump and Pence's shared values 01:33
Donald Trump: Now look, we are going to get rid of ISIS, big league. And we're going to get rid of 'em fast. And we're going to use surrounding states. We're going to use NATO, probably. And we're going to declare war. It is war. When the World Trade Center comes tumbling down, with thousands of people being killed, people are still-- I have friends that are still--
Lesley Stahl: But we did go to war, if you remember. We went to Iraq.
Donald Trump: Yeah, you went to Iraq, but that was handled so badly. And that was a war-- by the way, that was a war that we shouldn't have entered because Iraq did not knock down--excuse me
Lesley Stahl: Your running mate--
Donald Trump: Iraq did not--
Lesley Stahl: --voted for it.
Donald Trump: I don't care.
Lesley Stahl: What do you mean you don't care that he voted for?
Donald Trump: It's a long time ago. And he voted that way and they were also misled. A lot of information was given to people.
Lesley Stahl: But you've harped on this.
Donald Trump: But I was against the war in Iraq from the beginning.
Lesley Stahl: Yeah, but you've used that vote of Hillary's that was the same as Governor Pence as the example of her bad judgment.
Trump OK with Pence vote for Iraq war 00:42
Donald Trump: Many people have, and frankly, I'm one of the few that was right on Iraq.
Lesley Stahl: Yeah, but what about he--
Donald Trump: He's entitled to make a mistake every once in a while.
Lesley Stahl: But she's not? OK, come on--
Donald Trump: But she's not--
Lesley Stahl: She's not?
Donald Trump: No. She's not.
Lesley Stahl: Got it.
Lesley Stahl: I have to move on or we're never gonna find out why he chose you. Why did you pick him? You had other candidates--
Donald Trump: OK, I did. I had a lot of people that wanted it; a lot more people than anybody--
Lesley Stahl: That came to you and begged you for it?
Donald Trump: That called me and came to me and wanted it badly. And you know, the press didn't report that. The press said, "Well, maybe he's having hard time picking--"
Lesley Stahl: Well, what about the governor? Did he want it like that?
Donald Trump: I actually brought it up to him.
Lesley Stahl: OK.
Donald Trump: I got to know him during the-- when I was in Indiana during the primaries, and I did very well in Indiana, like I did just about everywhere else in all fairness. But-- I got to know him very well and I gained great respect for him. And I looked at the numbers, meaning the financials, which we would say in business. But I looked at the numbers. Unemployment? What a great job he did. Jobs? What a great job he did. Triple-A rating on his bonds.
Lesley Stahl: --but you went to him and said--
Donald Trump: I did--
Lesley Stahl: Would you want to be considered--
Donald Trump: I broached it.
Lesley Stahl: He didn't -- And then he said, "I-- I really want it." So why did you pick him?
Donald Trump: I would say that he thought about it a little bit. And about two seconds later, he called me, with his incredible wife, and said--
Lesley Stahl: That I'd like to be-- but what--
Donald Trump: Like many others.
Lesley Stahl: How does he help you? How does he help you win in terms of groups of people? And what is your weakness that he compensates for and so forth--
Donald Trump: Well, I went for the quality individual rather than I'm gonna win a state, because I'm doing very well in Indiana, and I guess I'm a lot up. And I think I'm gonna win Indiana. I have a great relationship and Bobby Knight helped me so much with Indiana. Indiana's a great place; great state.
Lesley Stahl: Why didn't you pick him? No, I'm joking -- moving on.
Donald Trump: He would've been very good. But he's a terrific guy. But-- but I really like him as an individual.
Lesley Stahl: So you must've considered, obviously, by the reaction to your choice, a lot of the conservatives are very happy.
Donald Trump: Very happy.
Lesley Stahl: Was that part of the--
Donald Trump: Yes, it was party unity. I'm an outsider. I am a person that used to be establishment when I'd give them hundreds of thousands of dollars. But when I decided to run, I became very anti-establishment because I understand the system--
Lesley Stahl: Is he establishment?
Donald Trump: --than anybody else. He's very establishment, in many ways, and that's not a bad thing. But I will tell you--
Lesley Stahl: That's kind of interesting --
Donald Trump: --I have seen more people that, frankly, did not like me so much, and now they're saying, "What a great pick." You see the kinda reaction. He has helped bring the party together. I understand. Look, I got more votes than anybody, but I also understand there's a faction--
Lesley Stahl: Is it already unified, do you think?
Donald Trump: I think it's very close to unified. And I will say--
Lesley Stahl: Just because of this pick?
Donald Trump: No. I think it's be-- I think it was much more unified than people thought. You saw that with the recent vote where we won in a landslide. You saw that with the big vote, the primary vote. I think it's far more unified than the press lets on. But having Governor Mike Pence has really-- people that I wasn't necessarily liking or getting along are loving this pick, because they have such respect for him.
Lesley Stahl: And that was--
Donald Trump: And the party unity is OK. You know, I think it's OK to say I picked somebody, because I-- as one of the things. But I really believe the main reason I picked him is the incredible job he's done. Just look at the economics of Indiana--
Lesley Stahl: Indiana--
Donald Trump: --and what's going on.
Lesley Stahl: But what about the chemistry between you two? You don't really know each other that well. You're -- at least I've read, a very low-key, very religious, you're a brash New Yorker--
Donald Trump: Religious.
Lesley Stahl: Religious?
Donald Trump: Religious--
Lesley Stahl: Are you?
Donald Trump: Yea, religious.
Lesley Stahl: --you wouldn't--
Donald Trump: Hey, I won the evangelicals. The evangelicals--
Lesley Stahl: That doesn't--
Mike Pence: You know, nobody thought--
Donald Trump: --well, I think it means a lot. I don't think they think I'm perfect, and they would get up and they would say, "You know, he's not perfect," but--
Lesley Stahl: They'd point to the --
Donald Trump: --they like me--
Lesley Stahl: --divorces--
Donald Trump: --but I won-- I won states with evangelicals that nobody thought I'd even come close to--
Lesley Stahl: Well, that's true--
Donald Trump:-and I won--
Lesley Stahl: --so you didn't (UNINTEL)--
Donald Trump: --with landslides--
Lesley Stahl: --need him for the evangelicals?
Donald Trump: I think it helps. But I don't think I needed him, no, because--I won with evangelicals.
Mike Pence: But I think we have more in common--
Lesley Stahl: Yeah, tell me--
Mike Pence: --than--
Lesley Stahl: --what you think you have in common.
Donald Trump: --what might be immediately obvious.
Lesley Stahl: Besides issues. Values and things like that.
Donald Trump: I think we will have very, very good chemistry. I feel that. And I can feel that pretty early on. I don't think you need to be with somebody for two years to find that out. My feeling is-
Lesley Stahl: Your gut feeling.
Donald Trump: I knew him during the primaries, during many trips to Indiana, I'd be with him. I think we have a great chemistry.
Lesley Stahl: I want to ask you though about something you've said about negative campaigning.
Donald Trump: Yeah.
Lesley Stahl: You said negative campaigning is wrong, and a campaign ought to demonstrate the basic decency of the candidate.
Mike Pence: Right.
Lesley Stahl: With that in mind, what do you think about your running mate's campaign and the tone and the negativity of it?
Mike Pence: I think this is a good man who's been talking about the issues the American people care about.
Lesley Stahl: But name-calling?
Mike Pence: In that--
Lesley Stahl: "Lyin' Ted?"
Mike Pence: --in the essay that I wrote a long time ago, I said campaigns oughta be about something more important than just one candidate's election. And-- and this campaign and Donald Trump's candidacy has been about the issues the American people care about.
Lesley Stahl: --but what about--
Donald Trump: Lesley, Lesley--
Lesley Stahl: --the negative side? He apologized for being a negative--
Donald Trump: We're different people. I understand that. I'll give you an example. Hillary Clinton is a liar. Hillary Clinton-- that was just proven--
Lesley Stahl: That's--
Donald Trump: --last week.
Lesley Stahl: --that's negative--
Donald Trump: Hillary Clinton--
Lesley Stahl: By the way --
Donald Trump: --you better believe it. Hillary Clinton is a crook.
Lesley Stahl: That's negative--
Donald Trump: I call her "Crooked Hillary." She's crooked Hillary. He won't-- I-- I don't-- I didn't ask him to do it, but I don't think he should do it because it's different for him.
Lesley Stahl: But-- he--
Donald Trump: He's not that kind of a--
Lesley Stahl: He probably--
Donald Trump: --person. We're different people--
Lesley Stahl: --don't you think he--
Donald Trump: --to me she's--
Lesley Stahl: --thinks that's--
Donald Trump: --Crooked Hillary.
Lesley Stahl: --wrong?
Donald Trump: I don't think he should use that term. I've never said one way or the other. But to him, I don't think it would sound right, but he will say how dishonest she is by going over the facts.
Lesley Stahl: But would you go to him-- let's say you won, you're the vice president, your office is, I assume, down the hall, and you go in and you say, "You know, you shouldn't be saying-- name-calling." Would you do that? Would you go in and say, "You crossed the line, I think you should apologize?" Would you do something like that?
Mike Pence: Look, I--
Lesley Stahl: He's laughing.
Mike Pence: --it's probably-- it's-- it's probably--
Lesley Stahl: It's OK.
Mike Pence: --obvious to people that our styles are different. But I promise you, our vision is exactly the same.
Lesley Stahl: No, but would you--
Mike Pence: And let me be--
Lesley Stahl: --will--
Mike Pence: --clear--
Lesley Stahl: --will you answer that? Would you go in --
Mike Pence: Well the-- one of the things I found out about this man is he appreciates candor. And--
Lesley Stahl: So you would go in--
Mike Pence: --I-- I--
Donald Trump: I'd like him to if he thinks I was doing something wrong--
Lesley Stahl: Would you listen--
Donald Trump: --just say--
Lesley Stahl: --to him--
Mike Pence: Yeah.
Lesley Stahl: --if he said you--
Donald Trump: Absolutely--
Lesley Stahl: --crossed the line.
Donald Trump: --absolutely. I might not apologize. You know, you said apologize, but--
Lesley Stahl: I did say apologize.
Donald Trump: But I might not do that, but I would absolutely want him to come in-- if he thinks I'm doing something wrong, Mike, I would want him to come in and say, really, you're doing, you gotta-- and that's OK. I accept that from my consultants and my people and if Mike came in and told me, you know, "I think you should do this or that--
Lesley Stahl: Back off that.
Donald Trump: I would listen and very likely listen to him.
Lesley Stahl: Do you think John McCain is not a hero because he was captured?
Mike Pence: I have a great deal of respect for John McCain, and--
Lesley Stahl: Do you think he went too far?
Donald Trump: You could say yes. I-- that's OK.
Donald Trump: That one, you could say yes, I mean, you're not--it's fine - hey, look, I like John McCain. But we have to take care of our vets.
Lesley Stahl: No, but I wanna know if--
Donald Trump: OK, but I'm just--
Lesley Stahl:-Mr. Pence would go in and--
Donald Trump:-I'm just saying, that's--
Lesley Stahl: --say to you--
Donald Trump: --why not that many people are that upset--
Lesley Stahl: What are you -- what did you say? You know, would you do something like that?
Mike Pence: I promise you that when the circumstances arise where I have a difference on policy or on presentation, I have-- I can tell you in my heart, I know-- I would have no hesitation, were I privileged to be vice president, to walk into the president's office, close the door, and share my heart. And I also know this good man would listen, and has the leadership qualities to draw from the people around him.
Lesley Stahl: Let's talk about-- some of the issues. Because there seems to be some daylight between you two, and we can just tick-- go quickly through these. Immigration. Mr. Trump, you have called for a temporary ban on Muslims entering the United States. Do you agree with that?
Mike Pence: I do. In fact, in Indiana we suspended the Syrian refugee program in the wake of the terrorist attack. We have no higher priority than the safety and security of the people of this country, and Donald Trump--
Lesley Stahl: Now--
Mike Pence: --is right to--
Lesley Stahl: --in December--
Mike Pence: --articulate that view.
Lesley Stahl: --in December you tweeted, and I quote you, "Calls to ban Muslims from entering the U.S. are offensive and unconstitutional."
Donald Trump: So you call it territories. OK? We're gonna do territories. We're gonna not let people come in from Syria that nobody knows who they are. Hillary Clinton wants 550 percent more people to come in than Obama--
Lesley Stahl: So you--
Donald Trump: --who doesn't know what he's--
Lesley Stahl: --so you're changing--
Donald Trump: --so we're going to--
Lesley Stahl: --your position.
Donald Trump: --no, I-- call it whatever you want. We'll call it territories, OK?
Lesley Stahl: So not Muslims?
Donald Trump: You know-- the Constitution -- there's nothing like it. But it doesn't necessarily give us the right to commit suicide, as a country, OK? And I'll tell you this. Call it whatever you want, change territories, but there are territories and terror states and terror nations that we're not gonna allow the people to come into our country. And we're gonna have a thing called "Extreme vetting." And if people wanna come in, there's gonna be extreme vetting. We're gonna have extreme vetting. They're gonna come in and we're gonna know where they came from and who they are.
Mike Pence: You just asked me--if I'm comfortable with that--
Mike Pence: --and I am. What-- what Donald--
Lesley Stahl: You're on the same--
Mike Pence: Which--
Lesley Stahl: --page on that?
Mike Pence: --clearly-- clearly this man is not a politician. He doesn't speak like a politician--
Lesley Stahl: He's done pretty well.
Mike Pence: --he-- he speaks from his--
Mike Pence: --heart--
Donald Trump: Is that a good thing? I think that's a good thing.
Mike Pence: --he speaks from his heart. And--
Lesley Stahl: Well, I--
Donald Trump: Well, I-- I speak from my heart and my brain. Just so we understand.
Donald Trump: This is (points to head) maybe more important.
Lesley Stahl: Let's go to trade. You have voted for every trade agreement when you were in Congress--
Mike Pence: I have.
Lesley Stahl: --that came before you. You're supporting the Trans-Pacific partnership that Mr. Trump says would rape this country. Now, are you gonna be able to go out and campaign in support of his protectionist positions?
Mike Pence: I support free trade, and so does Donald Trump.
Lesley Stahl: Not really--
Donald Trump: I do. I'm free trade, but I wanna make good deals. No, no, I'm all for free--
Lesley Stahl: You've talked
Donald Trump: --trade. You--
Lesley Stahl: --about--
Donald Trump: --know I'm not an isolationist. A lot of people think because I wanna make good deals--
Lesley Stahl: You wanna undo--
Donald Trump: --these are stupid people--
Lesley Stahl: --these--
Donald Trump: --wait a minute Lesley, these are stupid people that think that. I wanna make great deals for our country. We have deals like the deal signed by Bill Clinton, NAFTA, one of the worst things that ever happened to this country in terms of trade, in terms of economics.
Lesley Stahl: What do you think about NAFTA?
Mike Pence: You're absolutely right. I've supported free trade throughout my career. But--
Mike Pence: --the truth of the matter is NAFTA has provisions in that law that call for it to be reviewed, that have never been-- never been-- initiated. What-- what I hear Donald Trump saying is let's-- let's look at these trade agreements and reconsider them and renegotiate them. And--
Lesley Stahl: And you're OK with--
Mike Pence: --with regard to--
Lesley Stahl: --that?
Mike Pence: --and with regard to other trade agreements, we've talked about this. I-- I really do believe when the American people elect one of the best negotiators in the world as president of the United States, we would do well--
Donald Trump: We're gonna bring back jobs--
Mike Pence: --to negotiate individually with countries.
Donald Trump: We're gonna bring back our jobs, we're gonna bring back our wealth, we're gonna take care of our people. Very simple.
Lesley Stahl: OK. More issues. Waterboarding. Mr. Trump wants to bring back waterboarding, and quote, "A hell of a lot more." Are you comfortable with bringing back waterboarding?
Mike Pence: I don't think we should ever tell our enemy what our tactics are.
Lesley Stahl: But what about that? What--
Mike Pence: I don't--
Lesley Stahl: --about-- he's publicly--
Donald Trump: I like that answer.
Lesley Stahl: --said that--
Mike Pence: I don't think we should-- I-- I think--
Lesley Stahl: But are you OK with the idea of--
Mike Pence: --I think--
Lesley Stahl: --waterboarding?
Mike Pence: --I think enhanced interrogation saved lives.
Mike Pence: I--
Mike Pence: --what I'm OK with-- what I'm OK with is protecting the American people. What I'm OK with is when people have the intent to come to this country and take American lives, that-- that we are-- that we are prepared to do what's necessary to gain the information to protect the people of this country--
Donald Trump: But Lesley, let's step further. We have an enemy, ISIS and others, who chop off heads, who drown people in steel cages and we can't do waterboarding--
Lesley Stahl: OK, but, but why--
Donald Trump: OK, they're not playing--
Lesley Stahl: --would you use their--
Donald Trump: --under -- because you know--
Lesley Stahl: --techniques?
Donald Trump: --what, those techniques get information. I don't care what anyone says.
Lesley Stahl: Are you agreeing with him?
Mike Pence: I am--
Donald Trump: And get information--
Mike Pence: --what I--
Donald Trump: --using those things.
Mike Pence: --what I can tell you is enhanced information gleaned information that saved American lives and, I was informed, prevented incoming terrorist attacks on this country from being successful. The American people expect the president of the United States to be prepared to support action to protect the people of this nation, and I know Donald Trump will.
Lesley Stahl: Have you answered me?
Lesley Stahl: Let's talk about the convention. You're a showman. What are you going to do to keep it from being a snooze-o-rama, as some have happened.
Donald Trump: Well, I think we're gonna have an exciting time. We've got some wonderful speakers. We have some very talented people. My family's gonna speak.
Lesley Stahl: Worried about violence outside? This is an open-carry state. People can carry guns. There'll be demonstrators. They've already said they're going to carry assault rifles. Are you worried? And would you call on people not to carry their guns?
Donald Trump: I have great faith in law enforcement. If they don't want to take their guns, I think that's fantastic. But I have great confidence in law enforcement. The police like Donald Trump. It's law and order. And I have great confidence that they will do a great job.
Lesley Stahl: There's no question in anybody's mind that you want to win this election. I don't think anyone would doubt that. But what about being president? Do you really want to be president of the United States?
Donald Trump: I want to make America great again. Honestly. I want to make Am-- I'm not doing this because-- I'm sacrificing tremendous things. I could be doing other things. It's lovely to sit down with you and be grilled. That's okay. But I could be doing other things right now. And I have some of the greatest properties in the world. I could be out there--
Lesley Stahl: Would you rather be out there?
Donald Trump: I tell you what. I've really enjoyed this process. I've gotten to know the people of this country. I've gotten to know places that I didn't know, that I read about, but I didn't know. I've also gotten to see the problems. And it's a movement.
Donald Trump: Now, when you ask me the question, do I wanna be?
Lesley Stahl: Yeah.
Donald Trump: I wanna be for one reason. I wanna make America safe again, and I wanna make America great again. That's why I'm doing this. And I love it.
Lesley Stahl: And you wanna govern? I mean it--
Donald Trump: I do wanna govern.
Lesley Stahl: It is different--
Donald Trump: I do--
Lesley Stahl: --building a movement--
Donald Trump: I govern my-- I--
Lesley Stahl: -and than-- going in there with the nitty gritty and--
Lesley Stahl: --all that tough decision making.
Donald Trump: Sure. No, no I wanna govern
Lesley Stahl: OK, this is my absolute final question.
Donald Trump: OK.
Lesley Stahl: You're not known to be a humble man. But I wonder--
Donald Trump: I think I am, actually humble. I think I'm much more humble than you would understand.
Lesley Stahl: As you think about-- prospect of running this country in these tough times where the world is spinning apart-- are you awed? Are you intimidated? Are you humbled by the enormity of this?
Donald Trump: You just said it best.
Mike Pence: Mmm.
Donald Trump: In a world that's spinning apart. That's what I'm thinking of. I'm not thinking of, "Oh gee, isn't this wonderful? Isn't this great what I've done?" I've had people that said, "It doesn't matter if you win or lose, what you've done has never been done before. You're gonna go down in the history books." You know what I say to 'em?" I say, "You're wrong." I will consider it, 'cause I funded my own primaries, I'm funding now a lot of this campaign. I'm putting in, you know-- I've spent $55 million in the primaries. I'm spending a fortunate now. I'll tell you, it is spinning. Our world is spinning out of control. Our country's spinning out of control. That's what I think about. And I'll stop that.
Lesley Stahl: Not-- humbled or-- awe.
Mike Pence: I can say to 'ya-- what--
Lesley Stahl: Go ahead.
Mike Pence: Talking with him in private settings, I love the words you used because this man is awed with the American people, and he is not intimidated by the world. And Donald Trump, this good man, I believe, will be a great president of the United States.
Donald Trump: I love what he just said.
Lesley Stahl
One of America's most recognized and experienced broadcast journalists, Lesley Stahl has been a 60 Minutes correspondent since 1991.
First published on July 17, 2016 / 8:12 PM


text:
The Story Behind the Statistics
By Caroline Hendrie
Behind this year's forecast of record-breaking school enrollment are some statistical quirks that have even veteran education-watchers scratching their heads.
Chief among them is the fact that the total number of children ages 5 to 17, the range that demographers traditionally consider school-age, is actually expected to be smaller this year than when enrollment hit its previous peak 25 years ago.
This year, that traditional school-age group is expected to be slightly less than 50 million, compared with 52.6 million in 1971, said Gregory Spencer, the chief of the population-projections branch of the U.S. Bureau of the Census.
So why this year's record enrollment?
"You have a bunch of little contributors," said Wendy Bruno, the bureau's education specialist. "Nothing's going to explain the whole thing."
Schools Extend Reach
Among the factors are:
An increase in the number of children in kindergarten. In 1971, fewer than 3.3 million 5-year-olds were in kindergarten, compared with nearly 3.9 million in 1994, Ms. Bruno said.
Improvement in the dropout rate. In 1993, 12.7 percent of Americans ages 18 to 24 were dropouts, down from 17 percent in 1971.
An increase in the number of students beyond the 5-to-17 age bracket who are showing up on school rosters.
Although the U.S. Department of Education includes only a relatively small number of preschoolers in its count--measuring only those in public and private schools that offer at least some primary grades--the number is thought to be higher than it was 25 years ago.
Similarly, a greater proportion of 18-year-olds are still on high school rosters--26.9 percent in 1993, compared with 17.8 percent in 1971.
Forecasters Surprised
Because of these factors, the department's National Center for Education Statistics predicts that total public and private enrollment will climb to 51.7 million this school year, exceeding the previous high by about 400,000 students.
Of that total, 45.9 million students are expected to attend public schools, with 5.8 million in private institutions.
Those projections are considerably higher than those of five years ago. At that time, demographers predicted that public school enrollment would not climb to 44 million until 1999. Now, they say it hit that level in 1994--five years ahead of schedule.
That surprise has led to complaints that the imperfect projections have made it harder for educators to cope with this decade's enrollment surge.
"The barometers that were there didn't work for us," said Nelson Canton, a spokesman for the National Education Association. "This would not be a problem if we'd been ready, but it surprised people."
In retrospect, demographers say a key reason for the fast growth was the unexpected trend among baby boomers toward later marriage and delayed child-bearing. While birthrates for virtually all age groups have been rising in the past decade, they have climbed more quickly for women in their 30s, explained Thomas D. Snyder, an NCES demographer.
Such changing and unforeseen circumstances make department estimates less accurate the further in the future they go, education officials say.
"It's like trying to predict exactly when the stock market is going to go up or down," Mr. Snyder said. "That's very hard to do."


Students learn disaster risk management at school.
We believe that addressing existing vulnerabilities, mitigating hazards and risks, and adapting to the impact of climate change requires whole-system capacity-building and transformation. At the foundation of these efforts are women and men in poverty, who must have access to the relevant information and resources so that they are enabled to address their situation, improve their level of preparedness and their capacity to respond to disaster events, innovate in their livelihoods and reduce their vulnerability.
Deprived people, especially women, living in disaster-prone areas are less affected by shocks, stresses and uncertainties, and thrive in the face of them.
To implement, with partners, high-quality and timely humanitarian responses that meet disaster-affected people's (especially women) immediate needs and recovery needs, whilst supporting their future preparedness and resilience.
To support deprived women and men in rural and urban communities to build their adaptive capacity and resilience to climate change, as well as natural and man-made disasters.
To influence the GOL, local governments, INGOs, the corporate sector, and international organisations towards more strategic disaster response that addresses root causes of vulnerability and builds poor people's resilience.
Ensuring high-quality humanitarian response by building the capacity of partners, local government and other actors to undertake coordinated emergency responses that adhere to SPHERE standards, with particular emphasis on the promotion of women's participation and leadership.
Supporting the development of resilient communities through appropriate and adapted disaster risk reduction planning and collaborative approaches linked to the government Socio-Economic Development Plan (SEDP).
Supporting efforts towards greater, more effective learning and coordination across the UN system, disaster management working groups, corporate sector, and government networks towards better information and resource sharing, programming, and humanitarian response.
National ministries and sub-national agencies (Ministry of Labour and Social Welfare, Ministry of Natural Resources and Environment).
OCHA, and AADMER partners in the region.


As young leaders and Youth Ministers meet in Antigua and Barbuda for the Caribbean Region Commonwealth Youth Ministers Meeting (CR-CYMM), Samantha Khan, 22, a Commonwealth Correspondent from Trinidad and Tobago, reflects on a major challenge facing young people: the scourge of unemployment and lack of opportunity.
With the completion of my undergraduate degree looming nearer, I have begun the daunting search for employment. I had hoped to find a job that could help support my postgraduate education while also providing me with some experience in my chosen field of study. As a young graduate, I expected to encounter some difficulty finding a suitable job, but this was complicated by the fact that my chosen field is within the cultural and creative industries.
Youth underemployment is on the rise as young people are being put into jobs for which we are overqualified and unsuitable and, unfortunately, this is especially true for the creative industries in which new artists, writers and cultural workers are often expected to work without pay while they build their career. The Caribbean nations, each with its own vibrant cultural identity and powerful presence in the creative world, should be seeking to support youth involvement in the cultural and creative industries as we start business and seek gainful employment.
With regard to launching projects, perhaps the most obstinate limitation is a lack of funding. Many visionary individuals in the creative and cultural sector are unable to bring their ideas to life because of the lack of available funding. After some research, I found that in Trinidad and Tobago, the Ministry of Arts and Multiculturalism has a programme by which individuals or organisations can apply for grants toward the development of skills or projects in the cultural and creative industries. This is a considerable initiative and is definitely something to be commended. It is my hope that, in future, more institutions will follow this example and establish programmes that lend financial support to young people trying to build careers in the creative sector.
In contrast, some young people may merely want to work in the creative industries rather than start their own ventures, but even this has proven to be extremely difficult. Young people looking for jobs in this sector often find themselves having to undertake unpaid internships in order to gain the experience necessary to move up the career ladder. While this is beneficial to employers and often provides new cultural/creative workers with crucial training, it is highly impractical and some might argue even exploitative. It would probably be helpful for new workers in this industry if there were more paid internships or apprenticeships available. While it may seem impractical from an employer's point of view, even a small stipend to cover living and travel expenses would be invaluable.
In order to combat the growing popularity of unpaid internships, involvement on a governmental level might be necessary. This exploitative practice is not only prevalent in the cultural and creative industries but flourishes in most sectors. It is magnified, however, by the unfortunate fact that we live in a society that values art but is often unwilling to pay the artists for their efforts. It is a sad truth that only a few creative and cultural artisans ever achieve an esteemed enough status that affords them compensation for their work. If policy-makers were to bear this in mind and put measures in place to reduce the number of unpaid internships on offer while also increasing the opportunities for skill-building in the cultural and creative industries through paid apprenticeships, training sessions or workshops, perhaps there would be more young people able to explore careers in the sector.
All in all, while it may seem to be a slightly frivolous topic of discussion when there are undeniably more pressing concerns in today's world such as child poverty and climate change, I believe that it would be remiss to ignore it completely.
Throughout history, the artists of a generation have been the ones who capture the zeitgeist and explore questions of humanity. In order to keep the effervescent, dynamic cultural and creative spirit of the Caribbean nations alive, we must encourage the young artists as they try to find footing in the sector. We must support the young person who dreams of being the Derek Walcott or Anya Ayoung-Chee of tomorrow.


text:
Call Her Ganda
Grassroots activists in the Philippines are spurred into action when a local transgender woman is found dead in a motel room with a 19-year-old U.S. marine as the leading suspect. As they demand answers and a just trial, hidden histories of U.S. colonization come bubbling to the surface.
Director: PJ Raval
Actors: Jennifer Laude, Julita Laude, Meredith Talusan, Naomi Fontanos, Virginia Lacsa Suarez
Country: Philippines, USA
The Long Shadow
When filmmaker and investigative journalist Frances Causey, a daughter of the South, set out to explore the continuing racial divisions in the US, what she discovered was that the politics…
The Man from Mo'Wax
James Lavelle played his first DJ set at 14, launched pioneering record label Mo'Wax at 18 and released the genre defining UNKLE album Psyence Fiction at 22. His phenomenally rapid…
Genre: Documentary, Music
The Truth About Killer Robots
Exploring provocative viewpoints from engineers, factory workers, journalists, philosophers and Asimov himself, The Truth About Killer Robots is a cautionary tale about a world automating beyond control.
Bombshell: The Hedy Lamarr Story
The world's most beautiful woman was also the secret inventor of secure wifi, bluetooth and GPS communications, but her arresting looks stood in the way of her being given the…
The New American Century
The New American Century presents the historical, philosophical, economical and political background — some of which is totally unknown to the public — that lend support to the accusation that…
Nerd Prom: Inside Washington's Wildest Week
From it's humble beginnings as a one night affair, the White House Correspondents' Association's annual dinner has grown into a nearly week long celebration of press, power players, celebrities and…
Genre: Documentary, News
Cobra Gypsies
A beautifully colorful documentary about the Kalbeliya of Northern India.
When America's last standing roller rinks are threatened with closure, a community of thousands battle in a racially charged environment to save an underground subculture.
WWE: You Think You Know Me? The Story of Edge
You think you know him-he's the Rated-R Superstar, the Ultimate Opportunist, and one of the most decorated champions in WWE History. Edge has won more than 30 championships, including 11…
Genre: Action, Documentary
TT3D: Closer to the Edge
By vividly recounting the TT's legendary rivalries and the Isle of Man's unique road racing history, this 3D feature documentary discovers why modern TT riders still risk their lives to…
Private Violence
One in four women experience violence in their homes. Have you ever asked, "Why doesn't she just leave?" Private Violence shatters the brutality of our logic and intimately reveals the…
Psychonautics: A Comic's Exploration of Psychedelics
Comedian Shane Mauss goes on a series of adventures to deepen his understanding of psychedelics. He describes the indescribable and takes us through some of his most intense experiences, while…
Trailer: Call Her Ganda


The unique opportunity to reach the Eurasian aviation and airport markets.
Although nations in the Region – extending from the Balkans to the Turkish Republics of Central Asia – have their own peculiarities, the similarities in the long time accumulated and newly emerging needs in view of the aviation and airport industries, far more exceed those differences.
Having entered the era of restructuring and investment earlier than most nations in the Region, Turkey sets a perfect benchmark for the said markets and thus, plays an important pioneering role both for the aviation and airport industries in Eurasia and the Middle East.


text: Q: Why do we use a complex (Schrodinger) equation in quantum mechanics, instead of two interconnected real equations of real functions? Is it possible to write down two separate differential equations that are interconnected and describe the real and imaginary parts of the wave function, but these equations would still be equivalent to the Schrödinger's complex equation? Then why do we use the Schrödinger's complex partial differential equation?

redpajama_set_name: RedPajamaStackExchange


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