text:
\section{ Introduction}\label{int1}
Recently Tremblay, Turbiner and Winternitz \cite{TTW1, TTW2}  studied a  family
of quantum and classical mechanical systems in two dimensions with classical
Hamiltonian
\begin {equation}\label{KMP1}
{\cal H}=p^2_r+\frac{1}{ r^2}p^2_\theta -\omega^2r^2+ \frac{\alpha}{ r^2\cos ^2k\theta }
+\frac{\beta}{r^2\sin ^2k\theta }.\end {equation}
We call this the TTW system. They conjectured  and gave strong evidence that for  $k$  a rational number
 this system is superintegrable, as is the corresponding quantum analog

\begin {equation}\label{TTWham}
H=\partial ^2_r+\frac{2}{ r}\partial _r+\frac{1}{ r^2}\partial ^2_\theta -\omega^2r^2+
\frac{\alpha}{ r^2\cos ^2k\theta } +\frac{\beta}{ r^2\sin ^2k\theta }.\end {equation}
 In papers \cite{KKM10,KMPog10} the authors and collaborators proved
quantum and classical superintegrability for these systems, the first proofs that covered  all rational $k$. Related results are
 \cite{KKM10a,KKM10b,KKM10c}.
This system has sparked a great deal of interest because it includes several special cases of definite physical interest,
it provides many examples
of superintegrability where the generating constants of the motion are of very high order as polynomials in the momenta or
as differential operators in the quantum case, and because it
gave a hint as to how many other such families of classical and quantum systems could be constructed. The previous classical proofs of
superintegrability for these systems have usually not determined the structure of the symmetry algebras, or even
verified closure at finite order under commutation. For 2D quantum systems, however, the authors and J. M. Kress have recently
introduced a method for verifying superintegrability based on recurrence relations for hypergeometric functions that enables one to
compute the associated  structure relations for the symmetry algebra with relative ease. In \cite{KKM10c} structure relations were obtained for 5
families of superintegrable systems, each indexed by a rational parameter $k$, and including the TTW family.
Here we give a simplified exposition of some of these results.

Each  of these quantum systems  in 2D has an analogous classical counterpart that can be shown to be  classically superintegrable. Natural questions are
\begin{itemize}\item What is the structure of the classical symmetry algebra for each of these systems? Does it close at finite order?
 \item How are the structures of the classical and quantum symmetry algebras, and the symmetries themselves, related.
\end{itemize}
We will present new results in this paper that provide at least partial answers  to these questions for each of the systems
 studied in \cite{KKM10c}, as well as for a family that contains the 2D system on the 2-sphere with generic potential.

\section{Brief resume of definitions}
Suppose we have a Hamiltonian system on a 2D local Riemannian manifold (real or complex) such that the Hamilton-Jacobi equation admits
additive separation in some orthogonal
coordinate system. Then the separable coordinates $x_1,x_2$ can always be chosen such that the Hamiltonian takes the form
\begin {equation}\label{sepconst2} {\cal H}={\cal L}_1=\frac{1}{f_1(x_1)+f_2(x_2)}(p_{x_1}^2+p_{x_2}^2+v_1(x_1)+v_2(x_2),\end {equation}
and the phase space function
$$  {\cal  L}_2 =\frac{f_2(x_2)}{f_1(x_1)+f_2(x_2)}\left(p_{x_1}^2+v_1(x_1)\right)-\frac{f_1(x_1)}{f_1(x_1)+f_2(x_2)}\left(p_{x_2}^2+v_2(x_2)\right).$$
is the second order constant of the motion determining the separation. Here, $\{{\cal  L}_2,{\cal H}\}=0$ where
 $$\{{\cal F}, {\cal G}\}=\sum_{j=1}^2\left(\partial_{x_j}{\cal F}\partial_{p_{x_j}}{\cal G}-\partial_{p_{x_j}}{\cal F}\partial_{x_j}{\cal G}\right)$$
is the Poisson bracket of two phase space functions ${\cal F},{\cal G}$. A constant of the motion is a nonzero function on the phase space
whose Poisson bracket with
 $\cal H$ vanishes. Clearly $\cal H$ and ${\cal L}_2$ are functionally independent constants of the motion, second order in the momenta.
We will present strategies for determining functions $f_1,f_2,v_1,v_1$ such that there exists a 3rd constant of the motion, polynomial in the momenta,
 and such that
the three constants are functionally independent. In this case the system is superintegrable.
By what is essentially an action angle variable construction \cite{KKM10b}, one can compute the integrals $M(x)=\frac12\int\frac{dx}{p_x}$ and
$N(y)=\frac12\int\frac{dy}{p_y}$. Then  ${\tilde {\cal L}}_2=M-N$ is a third constant of the motion, but usually not a polynomial in the momenta,
hence not directly useful in verifying superintegrability. In \cite{KMPog10,KKM10b}, see also \cite{CG, MPT}, we described a procedure for obtaining a
polynomial constant from $M-N$, based on the observation that the integrals
 $$M=\frac12\int \frac{dx_1}{\sqrt{f_1{\cal H}+{\cal L}_2-v_1}},\quad N=\frac12\int \frac{dx_2}{\sqrt{f_2{\cal H}-{\cal L}_2-v_2}},$$
can often be expressed in terms of multiples of the inverse hyperbolic sine or cosine (or the ordinary inverse sine or cosine), and the
hyperbolic sine and cosine satisfy addition formulas. Our method works for  functions $f_j,v_j$ such that $M$ and $N$ possess this property.

For 2D quantum systems such that the Schr\"odinger eigenvalue equation $H\Psi=E\Psi$ admits multiplicative separation in an
orthogonal coordinate system we have
\begin {equation}\label{sepconst2Q} { H}={ L}_1=\frac{1}{f_1(x_1)+f_2(x_2)}(\partial_{x_1}^2+\partial_{x_2}^2+v_1(x_1)+v_2(x_2),\end {equation}
and
$$  {  L}_2 =\frac{f_2(x_2)}{f_1(x_1)+f_2(x_2)}\left(\partial_{x_1}^2+v_1(x_1)\right)-\frac{f_1(x_1)}{f_1(x_1)+f_2(x_2)}\left(\partial_{x_2}^2+v_2(x_2)\right).$$
is the second order operator  determining the separation. Here, $[  L_2, H]=0$ where
 $[ F, G]=FG-GF$ is the commutator
 of two operators $ F,G$. The separable eigenfunctions $\Psi=\Phi(x_1)\Theta(x_2)$ of $H$ are then characterized by the equations
$$ H\Psi=E\Psi,\quad L_2\Psi=\lambda \Psi$$
where $E$ is the energy eigenvalue and $\lambda$ is the separation constant. A symmetry operator is a nonzero differential operator in $x_1,x_2$
whose commutator with
 $ H$ vanishes. Clearly the second order differential operators  $ H$ and $ L_2$ are algebraically independent symmetries. In \cite{KKM10c}
we presented a recurrence relation method  for determining  a 3rd finite order symmetry operator  such that the three symmetries are algebraically
independent, i.e. such that the system is quantum superintegrable.

\section{The quantum TTW system} We use the quantum TTW system to illustrate our recurrence relation approach to proving superintegrability and determining the structure
of the symmetry algebra. For complete details on this construction and on its application to other systems, see  \cite{KKM10c}.
 The quantum Hamiltonian operator is
(\ref{TTWham}).The general solution of the eigenvalue
problem $H\Psi =E\Psi $ is
\begin {equation}\label{sepeigenfunctions}\Psi  = e^{-\frac{\omega }{ 2}r^2}r^{k(2n+a+b+1)}
L^{k(2n+a+b+1)}_m(\omega r^2)
(\sin (k\theta ))^{a+\frac{1}{ 2}}(\cos (k\theta ))^{b+\frac{1}{ 2}}
P^{a,b}_n(\cos (2k\theta ))\end {equation}
where we have taken $\alpha =k^2(\frac{1}{ 4} -a^2)$ and $\beta =k^2(\frac{1}{
4}
-b^2)$. The $L$-functions are associated Laguerre and the $P$-functions are
Jacobi,
not polynomials in general, \cite{AAR}.   We consider the functions
$$\Pi  = e^{-\frac{\omega }{2}r^2}r^{k(2n+a+b+1)}
L^{k(2n+a+b+1)}_m(\omega r^2)P^{a,b}_n(x)=Y^{ A}_m(r)X^{a,b}_n(x)$$
where $x=\cos (2k\theta )$,  $X=P$ or $Q$ and $Y=S$ or $T$. (Here $\Pi$ is obtained from $\Psi$ by a
gauge transformation to remove the angular factors
$(\sin (k\theta ))^{a+\frac{1}{ 2}}(\cos (k\theta ))^{b+\frac{1}{ 2}}$.)
 By this we mean that $P^{a,b}_n(x)$ is a
Jacobi polynomial if $n$ is an integer. Otherwise it is given by its
hypergeometric expression. If $X=Q$ then this denotes the associated second
solution of the Jacobi differential equation. Similar remarks apply to the
choice of $Y=S$.  We  have defined this function to be
$$S^{ A}_m(r)=e^{-\frac{\omega }{ 2}r^2}r^{k(2n+a+b+1)}
L^{k(2n+a+b+1)}_m(\omega r^2)$$
and $T^{ A}_m(r)$ to be a second independent solution. The energy eigenvalue
is given
by
\begin {equation}\label{energyev}E=-2\omega \left[2(m+nk)+1+(a+b+1)k\right]\end {equation}
and $ A=k(2n+a+b+1)$.
The separation equation for $\Theta (\theta )$ is
$${\tilde L}_2\Theta=(\partial ^2_\theta + \frac{\alpha }{ \sin ^2(k\theta )} +
\frac{\beta }{\cos ^2(k\theta )})\Theta (\theta )=-k^2(2n+a+b+1)^2\Theta (
\theta )=-A^2 \Theta (\theta )$$
and ${\tilde L}_2$ is a symmetry operator for  the system. Under the gauge
transformation ${\tilde L}_2$  goes to a symmetry
that we shall  call $L_2$ and which has the same eigenvalues.
We see from the expression for $E$ that in order that an energy eigenvalue to be
unchanged for different values of $m,n$
 we must fix $u=m+nk$. The   two transformations
$$
n\rightarrow n+q,\quad m\rightarrow m-p$$
and
$$n\rightarrow n-q,\quad m\rightarrow m+p$$
will each achieve this.

Consider the functions $X^{a,b}_n(x)$. If we want to raise or lower
the index $n$ we can do so with the operators
$$J^+_n X^{a,b}_n(x)=$$
$$(2n+a+b+2)(1-x^2)\partial
_xX^{a,b}_n(x)+(n+a+b+1)(-(2n+a+b+2)x-(a-b))X^{a,b}_n(x)$$
$$=2(n+1)(n+a+b+1)X^{a,b}_{n+1}(x)$$
and
$$J^-_n X^{a,b}_n(x)=$$
$$-(2n+a+b)(1-x^2)\partial
_xX^{a,b}_n(x)-n((2n+a+b)x-(a-b))X^{a,b}_n(x)=2(n+a)(n+b)X^{a,b}_{n-1}(x),$$
\cite{AAR}.
Similarly, for the  functions ${\cal Y}^A_m(R)=\omega^{A/2}Y^{A}_m(r)$ where
$R= r^2$ we can deduce the
relations
$$K^+_{A,m}{\cal Y}^A_m(R)=\left\{(A+1)\partial _R-\frac{E}{ 4 }-\frac{1}{
2R}A(A+1)\right\}{\cal Y}^A_m(R)
= -\omega{\cal Y}^{A+2}_{m-1}(R),$$
$$K^-_{A,m}{\cal Y}^A_m(R)=\left\{(-A+1)
\partial _R-\frac{E}{ 4}+\frac{1}{ 2R}A(1-A)\right\}{\cal Y}^A_m(R)
=-\omega(m+1)(m+A) {\cal Y}^{A-2}_{m+1}(R),$$ \cite{AAR}.
 We now  construct the two operators
\begin {equation}\label{recurrence3} \Xi _+=K^+_{A+2(p-1),m-(p-1)}\cdots
K^+_{A,m}J^+_{n+q-1}\cdots J^+_n\end {equation}
and
\begin {equation}\label{recurrence4}\Xi _-=K^-_{A-2(p-1),m+p-1}\cdots
K^-_{A,m}J^-_{n-q+1}\cdots J^-_n. \end {equation}
When applied to a basis function $\Psi_n={\cal Y}^A_m(R)X^{a,b}_n(x)$ for fixed
$u=m+kn$, (so $m=u-kn$)
these operators  raise and lower indices according to
\begin {equation}\label{raise1} \Xi _+\Psi_n=2^q(-1)^p \omega^p(n+1)_q(n+a+b+1)_q\Psi_{n+q},\end {equation}
\begin {equation}\label{lower1} \Xi _-\Psi_n=2^q \omega^p
(-n-a)_q(-n-b)_q(u-kn+1)_p(-u-k[n+a+b+1])_p\Psi_{n-q},\end {equation}
where $(\alpha)_q=(\alpha)(\alpha+1)\cdots(\alpha+q-1)$ for nonnegative integer
$q$. It is easy to verify that
under the transformation $n \rightarrow  -n-a-b-1$  we have ${\Xi_+} \rightarrow
\Xi_-$ and ${\Xi_-} \rightarrow \Xi_+$.
Thus $\Xi=\Xi_++\Xi_-$  as a polynomial in $n$ and $u$ is unchanged under this
transformation.
Therefore it is a polynomial in $(2n+a+b+1)^2$ and $u$.  As a consequence of
the relation $\lambda =-k^2(2n+a+b+1)^2$,   in the
expansion of
$\Xi$ in terms of powers of $(2n+a+b+1)^2$ and $E$ we can replace $(2n+a+b+1)^2$
by $L_2/k^2$ and $E$ by $H$ everywhere they occur, and express $\Xi$ as a
pure differential operator, independent of parameters.  Thus, in the
expansion of $\Xi$ in terms of the parameters, a term $ W E$
 is replaced by
$WH$
with the $W$ operator on the left.
Clearly this operator, which we will also call $\Xi$ commutes with $H$ on the formal
eigenspaces of $H$. However, in  \cite{KKM10c} we showed  shows that in fact $\Xi$ commutes with $H$ in general,
thus it is a symmetry operator for $H$.(Indeed it is shown there that any polynomial operator
 identity established on all formal eigenspaces must hold identically.This is a general fact that holds for all
applications of the recurrence operator method.)

We can also easily see that under the transformation $n \rightarrow  -n-a-b-1$
the operator $\Xi_+ -\Xi_-$ changes sign, hence the operator
 ${\tilde \Xi}=(1/(2n+a+b+1))(\Xi_+ -\Xi_-)$ is unchanged under this
transformation. Again, making the replacements $(2n+a+b+1)^2$
by $L_2/k^2$ and $u$ by
$-2 -(H+2\omega k(a+b+1))/4\omega$ we can express ${\tilde \Xi}$ as a pure
differential operator, independent of parameters,
and it is a symmetry operator for $H$. Each of these symmetries has a nonzero
commutator with $L_2$, so each is
algebraically independent of the set $H,L_2$. This proves that the TTW system is
quantum superintegrable for all rational $k$.
We set $L_3= \Xi$, $L_4={\tilde \Xi}$.

Using the explicit relations (\ref{raise1}), (\ref{lower1}) for the action of
the raising and lowering operators $\Xi_\pm$ on a basis
we can obtained very detailed information about the structure of the symmetry
algebra generated by $L_2,L_3,L_4$.
Applying the raising operator to a basis function, followed by the lowering
operator, we obtain the result
\begin {equation}\label{LR1}  \Xi_-\Xi_+\Psi_n\end {equation}
$$=(-1)^p4^q\omega^{2p}(n+1)_q(n+a+1)_q(n+b+1)_q(n+a+b+1)_q(-u+kn)_p(u+k[n+a+b+1]
+1)_p\Psi_n$$
$$=\xi_n\Psi_n,$$
Reversing the order we find
\begin {equation}\label{RL1} \Xi_+\Xi_-\Psi_n\end {equation}
$$=(-1)^p4^q\omega^{2p}(-n)_q(-n-a)_q(-n-b)_q(-n-a-b)_q(u-kn+1)_p(-u-k[n+a+b+1])_p\Psi_n$$
$$ =\eta_n\Psi_n.$$
Thus the action of the operator $\Xi_-\Xi_++ \Xi_+)\Xi_-$ on any
basis function $\Psi_n$ is to multiply it by
$\xi_n+\eta_n$. However, it is easy to check from expressions (\ref{raise1}),
(\ref{lower1} and from  (\ref{RL1}), (\ref{LR1})
that under the transformation $n \rightarrow  -n-a-b-1$  we have
$\Xi_-\Xi_+\leftrightarrow \Xi_+\Xi_-$ and
  $\xi_n\leftrightarrow \eta_n$. Thus $\Xi^{(+)}=\Xi_-\Xi_++
\Xi_+\Xi_-$ is an even polynomial operator in $(2n+a+b+1)$,
polynomial in $u$, and
 $\xi_n+\eta_n$ is an even polynomial function in $(2n+a+b+1)$, polynomial in
$u$. Furthermore, each of $\Xi_-\Xi_+$ and $\Xi_+\Xi_-$ is unchanged under the transformation
$u\longrightarrow -u-(a+b+1)-1$, hence each is a polynomial of order $p$ in
$[2u+(a+b+1)k+1]^2=E^2/4\omega^2$. Due to the multiplicative factor $\omega^{2p}$ in each of these expressions
we conclude that $\Xi^{(+)}$ is a
symmetry operator whose action on a basis is
 given by a polynomial operator $P^{(+)}(H^2,L_2,\omega^2,a,b)$. In fact,
$$ \Xi^{(+)}=P^{(+)}(H^2,L_2,\omega^2,a,b).$$

Similarly, the operator $\Xi^{(-)}=(\Xi_-\Xi_+-
\Xi_+\Xi_-)/(2n+a+b+1)$ is an even polynomial
in $(2n+a+b+1)$, as is $(\xi_n-\eta_n)/(2n+a+b+1)$. Also it is polynomial in $E^2$ and $\omega^2$. Thus
$\Xi^{(-)}=P^{(-)}(H^2,L_2,\omega^2,a,b)$ is a symmetry operator
which is a polynomial function of all of its variables.

Now we can compute the structure relations explicitly by evaluating the
operators on an eigenfunction basis.
These relations must then hold everywhere. The results are:
\begin {equation}\label{[L2,L4]} [L_2,L_4]=-4k^2qL_3-4k^2q^2L_4,\end {equation}
\begin {equation}\label{[L2,L3]} [L_2,L_3]=2q\{L_2,L_4\}+4k^2q^2L_3+8k^2q^3L_4,\end {equation}
\begin {equation}\label{[L3,L4]} [L_3,L_4]=2qL_4^2-2P^{(-)}(H^2,L_2,\omega^2,a,b),\end {equation}
\begin {equation}\label{L3^2Q}
6k^2L_3^2+\{L_2,L_4,L_4\}+6k^2q\{L_3,L_4\}+28k^2q^2L_4^2-4k^2qP^{(-)}(H^2,
L_2,\omega^2,a,b)\end {equation}
$$-12k^2P^{(+)}(H^2,L_2,\omega^2,a,b)=0.$$
Here, $\{A,B\}=AB+BA$ and $\{A,B,C\}$ is the analogous  6-term symmetrizer of 3
operators.
A more transparent realization, for $R=-4k^2qL_3-4k^2q^2L_4$, is
\begin {equation}\label{[L2,L4R]} [L_2,L_4]=R,\end {equation}
\begin {equation}\label{[L2,R]} [L_2,R]=-8k^2q^2\{L_2,L_4\}-16k^4q^4L_4,\end {equation}
\begin {equation}\label{[L4,R]} [L_4,R]=8k^2q^2L_4^2-8k^2qP^{(-)}(H^2,L_2,\omega^2,a,b),\end {equation}
\begin {equation}\label{R^2}
\frac{3}{8k^2q^2}R^2+22k^2q^2L_4^2+\{L_2,L_4,L_4\}-4k^2qP^{(-)}(H^2,L_2,\omega^2,a,b)\end {equation}
$$-12k^2P^{
(+)}(H^2,L_2,\omega^2,a,b)=0.$$
From this result we see that the symmetries $H,L_2,L_4$ generate a closed
symmetry algebra.

For $k=1$ the operator is of order 3, whereas we know that there is a 2nd order symmetry, due to the fact that the TTW Hamiltonian-Jacobi equation  for $k=1$ is multiply separable. Thus we have produced only a proper subalgebra of the maximal symmetry algebra. However, for $k\ne 1$ this multiple separation no longer occurs and we think that the generators we have constructed are of minimum order.

\section{The classical TTW system}
The TTW Hamiltonian  (\ref{KMP1}) admits a second order constant of the motion
corresponding to separation of variables in polar coordinates, viz
\begin {equation}\label{angmom}
{\cal L}_2=p^2_\theta +\frac{\alpha}{ \cos ^2k\theta } +\frac{\beta} {\sin ^2k\theta }.\end {equation}
To demonstrate superintegrability we need to exhibit a third polynomial constant of the motion
that is functionally independent of ${\cal H}, {\cal L}_2$. The method for constructing this symmetry is based on
action angle variables and the existence of an additively separable coordinate system for the Hamilton-Jacobi equation,
 \cite{KMPog10, Evans2008a}. We briefly review the construction.
In terms of the new variable $r=e^R$ the Hamiltonian assumes the form
$${\cal H}=e^{-2R}(p^2_R+p^2_\theta -\omega^2 e^{4R}+\frac{\alpha }{ \cos ^2(k\theta )}+
\frac{\beta }{ \sin ^2(k\theta )}).$$
Applying the method of \cite{KMPog10}
to find  extra invariants we  first need to construct a
function $M(R,p_R)$ which satisfies $\{M,{\cal H}\}=e^{-2R}$, or
$$(4\omega^2 e^{4R}+2{\cal H}e^{2R})\partial _{p_R}M+2p_R\partial_R M=1.$$
This equation has a solution
$$M=\frac{i}{ 4\sqrt{ {\cal L}_2}} B$$
where
$$
\sinh{ B} =i \frac{(2{\cal L}_2e^{-2R}-{\cal H})}{ \sqrt{ {\cal  H}^2+4\omega^2 {\cal  L}_2}},\quad
\cosh{B }= \frac{2\sqrt{ {\cal L}_2}e^{-2R}p_R}{ \sqrt{ {\cal  H}^2+4\omega^2 {\cal L}_2}},$$
and
\begin {equation}\label{L2}{\cal L}_2=p^2_\theta +\frac{\alpha }{ \cos ^2(k\theta )}+
\frac{\beta }{ \sin ^2(k\theta )},\end {equation}
and we also have the relation (which we can use to consider $M$ as a function of $R$ alone):
\begin {equation}\label{LHident}p^2_R+{\cal L}_2-\omega^2 e^{4R}-e^{2R}{\cal H}=0.\end {equation}
We now compute a corresponding function $N(\theta ,p_\theta )$ which
 satisfies $\{N,{\cal H}\}=e^{-2R}$, or $$-(\frac{\alpha }{ \cos ^2(k\theta )}+
\frac{\beta }{ \sin ^2(k\theta )})'\partial _{p_\theta }N+2p_\theta \partial _\theta N=1$$
where the prime denotes differentiation with respect to $\theta$. This equation has a
solution
$N= - \frac{i}{ 4\sqrt{{\cal  L}_2}k} A$
where
$$
\sinh{A }=
i\frac{-\beta +\alpha -{\cal L}_2\cos (2k\theta )}{ \sqrt{({\cal L}_2-\alpha -\beta )^2-4\alpha\beta }},\quad
\cosh{ A }=
\frac{\sqrt{ L_2}\sin (2k\theta )p_\theta }{\sqrt{ ({\cal L}_2-\alpha -\beta )^2-4\alpha \beta }}.$$
According to the action angle variable theory  $M-N$ is a constant of the motion and, since it is
constructed such that $\{M-N,{\cal L}_2\}\ne 0$, it is functionally independent of ${\cal L}_2$, \cite{KKMPog02}.

From these expressions for $M$ and $N$  we see that if $k$ is rational,
$k=\frac{p}{ q}$ ( where $p, q$ are relatively prime integers) then
$$-4ip\sqrt{{\cal  L}_2}[N-M]=qA+pB$$ and any function of $qA+pB$ is a constant of the motion (but not polynomial in the momenta).
Our original proofs of superintegrability for this system were based on the fact that the hyperbolic functions
 $\sinh x$, $\cosh x$ satisfied addition theorems.
Here, however, we will make use of  an observation in \cite{CG} that simplifies the analysis and work with the exponential function.
We have
\begin {equation}\label{expA} e^A=\cosh A +\sinh A= {X}/{U},\quad e^{-A}=\cosh A -\sinh A= {\overline{X}}/{U},\end {equation}
\begin {equation}\label{expB} e^B=\cosh B +\sinh B= {Y}/{S},\quad e^{-B}=\cosh B -\sinh B= {\overline{Y}}/{S},\end {equation}
where
$$ X= \sqrt{{\cal L}_2}\sin (2k\theta) p_\theta+i(-\beta+\alpha-{\cal L}_2\cos(2k\theta)),\
{\overline X}= \sqrt{{\cal L}_2}\sin (2k\theta) p_\theta-i(-\beta+\alpha-{\cal L}_2\cos(2k\theta)),$$
$$Y= 2\sqrt{{\cal L}_2} e^{-2R}p_R+i(2{\cal L}_2e^{-2R}-{\cal H}),\
{\overline Y}= 2\sqrt{{\cal L}_2} e^{-2R}p_R-i(2{\cal L}_2e^{-2R}-{\cal H}),$$
$$ U=\sqrt{({\cal L}_2-\alpha-\beta)^2-4\alpha\beta},\quad S=\sqrt{{\cal H}^2+4\omega^2{\cal L}_2}.$$

Now note that $e^{qA+pB}$ and $e^{-(qA+pB)}$ are constants of the motion, where
\begin {equation}\label{expAB} e^{qA+pB}=(e^A)^q (e^B)^p=\frac{X^qY^p}{U^qS^p},\quad e^{-(qA+pB)}=(e^{-A})^q (e^{-B})^p=\frac{(\overline{X})^q
(\overline{Y})^p}{U^qS^p}.\end {equation} Moreover, the identity $e^{qA+pB}e^{-(qA+pB)}=1$ can be expressed as
\begin {equation}\label{fundident} X^q(\overline{X})^qY^p(\overline{Y})^p=U^{2q}S^{2p}=P({\cal L}_2, {\cal H})\end {equation}
where $P$ is a polynomial in ${\cal L}_2$ and ${\cal H}$.

We make some observations. Let $a,b,c,d$ be nonzero complex numbers and consider the
binomial expansion
\begin {equation}\label{plusbinom}\frac12\left[(\sqrt{{\cal L}_2}a+ib)^q(\sqrt{{\cal L}_2}c+id)^p+(\sqrt{{\cal L}_2}a-ib)^q(\sqrt{{\cal L}_2}c-id)^p\right]
\end {equation}
$$=\sum_{0\le \ell\le q, 0\le s\le p}\left(\begin {array}{cc} q\\ \ell\end {array}\right)
\left(\begin {array}{cc} p\\ s\end {array}\right)b^\ell d^s a^{q-\ell}c^{p-s}\frac{{\cal L}_2^{(q+p-\ell-s)/2}}{2}[i^{\ell+s}+(-i)^{\ell+s}].$$
Suppose $p+q$ is odd. Then it is easy to see that the sum (\ref{plusbinom}) takes the form
 $\sqrt{{\cal L}_2}T_{\rm odd}({\cal L}_2) $ where
$T_{\rm odd}$ is a polynomial in ${\cal L}_2$. On the other hand, if $p+q$ is even then the sum (\ref{plusbinom}) takes the
form $T_{\rm even}({\cal L}_2) $ where
$T_{\rm even}$ is a polynomial in ${\cal L}_2$.

Similarly, consider the
binomial expansion
\begin {equation}\label{minusbinom}\frac{1}{2i}\left[(\sqrt{{\cal L}_2}a+ib)^q(\sqrt{{\cal L}_2}c+id)^p-
(\sqrt{{\cal L}_2}a-ib)^q(\sqrt{{\cal L}_2}c-id)^p\right]
\end {equation}
$$=\sum_{0\le \ell\le q, 0\le s\le p}\left(\begin {array}{cc} q\\ \ell\end {array}\right)
\left(\begin {array}{cc} p\\ s\end {array}\right)b^\ell d^s a^{q-\ell}c^{p-s}\frac{{\cal L}_2^{(q+p-\ell-s)/2}}{2i}[i^{\ell+s}-(-i)^{\ell+s}].$$
Suppose $p+q$ is odd. Then  the sum (\ref{minusbinom}) takes the form
 $V_{\rm odd}({\cal L}_2) $ where
$V_{\rm odd}$ is a polynomial in ${\cal L}_2$. On the other hand, if $p+q$ is even then the sum (\ref{minusbinom}) takes the
form $\sqrt{{\cal L}_2}V_{\rm even}({\cal L}_2) $ where
$V_{\rm even}$ is a polynomial in ${\cal L}_2$.

\medskip\noindent {\bf Case $p+q$ odd}:  Let
\begin {equation} \label{oddL34} {\cal L}_4=\frac{1}{\sqrt{{\cal L}_2}}({\cal L}^+ +{\cal L}^-),\
{\cal L}_3=\frac{1}{i}({\cal L}^+ -{\cal L}^-),\end {equation}
where
\begin {equation}\label{oddL+-} {\cal L}^+=X^qY^p,\quad {\cal L}^-=(\overline{X})^q(\overline{Y})^p.\end {equation}
Then we see from (\ref{plusbinom}), (\ref{minusbinom}) that ${\cal L}_3,{\cal L}_4$ are constants of the motion,
 polynomial in the momenta. Moreover, the identity (\ref{fundident}) becomes
\begin {equation}\label{fundident1}{\cal L}^+{\cal L}^-=P({\cal L}_2,{\cal H}).\end {equation}
Here ${\cal L}^\pm$ are constants of the motion, but are not polynomial in the momenta. We will see that they are the classical
versions of quantum
raising and lowering operators for angular momentum.

It is straightforward to verify the Poisson bracket computations
\begin {equation}\label{raising}\{{\cal L}_2,{\cal L}^+\}=q\frac{X^{q-1}Y^p}{U^qS^p}\{{\cal L}_2,X\}=-4ip\sqrt{{\cal L}_2}{\cal L}^+,\end {equation}
\begin {equation}\label{lowering}\{{\cal L}_2,{\cal L}^-\}=q\frac{(\overline{X})^{q-1}(\overline{Y})^p}{U^qS^p}\{{\cal L}_2,\overline{X}\}
=4ip\sqrt{{\cal L}_2}{\cal L}^-.\end {equation}
From these results we can obtain
$$\{{\cal L}_2,{\cal L}_4\}=\frac{1}{\sqrt{{\cal L}_2}}(-4ip\sqrt{{\cal L}_2}{\cal L}^++ 4ip\sqrt{{\cal L}_2}{\cal L}^-)$$
$$=-4ip{\cal L}^++4ip{\cal L}^-=4p{\cal L}_3,$$
$$\{{\cal L}_2,{\cal L}_3\}= \frac{1}{i}(-4ip\sqrt{{\cal L}_2}{\cal L}^+-4ip\sqrt{{\cal L}_2}{\cal L}^-)$$
$$=-4p{\cal L}_2{\cal L}_4.$$

To get the remaining structure equations we need some more identities. First of all, note that
\begin {equation}\label{L4L4} {\cal L}_4^2=\frac{1}{{\cal L}_2}\left[ ({\cal L}^+)^2+2{\cal L}^+{\cal L}^-+({\cal L}^-)^2\right],\end {equation}
\begin {equation}\label{L3L3}{\cal L}_3^2=-\left[ ({\cal L}^+)^2-2{\cal L}^+{\cal L}^-+({\cal L}^-)^2\right].\end {equation}
Thus,
$ {\cal L}_3^2+{\cal L}_2{\cal L}_4^2=4{\cal L}^+{\cal L}^-=4P({\cal L}_2,{\cal H})$, so
\begin {equation}\label{L3^2} {\cal L}_3^2=-{\cal L}_2{\cal L}_4^2+4P({\cal L}_2,{\cal H}).\end {equation}

Further,
$$\{{\cal L}^+,{\cal L}^-\} =\{{\cal L}^+, \frac{P}{{\cal L}^+}\}=  ({\cal L}^+)^{-1}\{{\cal L}^+,P\}=({\cal L}^+)^{-1}\left(\frac{\partial P}{\partial {\cal L}_2}\{{\cal L}^+,{\cal L}_2\}+\frac{\partial P}{\partial {\cal H}}\{{\cal L}^+,{\cal H}\}\right),$$
so
\begin {equation}\label{L+L-} \{{\cal L}^+,{\cal L}^-\}=4ip\sqrt{{\cal L}_2}\frac{\partial P}{\partial{\cal L}_2}.\end {equation}
It remains only to evaluate $\{{\cal L}_3,{\cal L}_4\}$. We have
$$\{{\cal L}_3,{\cal L}_4\}=\frac{1}{i}\{{\cal L}^+-{\cal L}^-,\frac{{\cal L}^+}{\sqrt{{\cal L}_2}}+\frac{{\cal L}^-}{\sqrt{{\cal L}_2}}\}=$$
$$\frac{1}{i}\left[-({\cal L}_2)^{-3/2}\{{\cal L}^+,{\cal L}_2\}{\cal L}^+ -({\cal L}_2)^{-3/2}\{{\cal L}^+,{\cal L}_2\}{\cal L}^-
+({\cal L}_2)^{-3/2}\{{\cal L}^-,{\cal L}_2\}{\cal L}^++({\cal L}_2)^{-3/2}\{{\cal L}^-,{\cal L}_2\}{\cal L}^-\right.$$
$$\left. +({\cal L}_2)^{-1/2}\{{\cal L}^+,{\cal L}^-\}-({\cal L}_2)^{-1/2}\{{\cal L}^-,{\cal L}^+\}\right]$$
$$=-\frac{4p}{{\cal L}_2}[({\cal L}^+)^2+2{\cal L}^+{\cal L}^-+({\cal L}^-)^2]+8p\frac{\partial P}{\partial{\cal L}_2}.$$
Then, using (\ref{L4L4}), we conclude that
\begin {equation}\label{L3L4bracket}\{{\cal L}_3,{\cal L}_4\}=-4p{\cal L}_4^2+8p\frac{\partial P}{\partial{\cal L}_2}.\end {equation}
To list the structure equations in standard form we define ${\cal R}=4p{\cal L}_3$. Then the structure equations take the form
\begin {equation}\label{structure+} \{{\cal L}_2,{\cal L}_4\}={\cal R},\ \{{\cal L}_2,{\cal R}\}=-16p^2{\cal L}_2{\cal L}_4,\ \{{\cal L}_4,{\cal R}\}=16p^2{\cal L}_4^2-32p^2\frac{\partial P}{\partial{\cal L}_2},\end {equation}
$${\cal R}^2=-16p^2{\cal L}_2{\cal L}_4^2+64p^2P({\cal L}_2,{\cal H}),$$
where
$$ P({\cal L}_2,{\cal H})=(({\cal L}_2-\alpha-\beta)^2-4\alpha\beta)^q({\cal H}^2+4\omega^2{\cal L}_2)^p.$$
This shows that the symmetry algebra generated by ${\cal H}, {\cal L}_2, {\cal L}_4$ closes at finite order.

\medskip\noindent {\bf Case $p+q$ even}: Let
\begin {equation} \label{evenL34} {\cal L}_4=\frac{1}{i\sqrt{{\cal L}_2}}({\cal L}^- -{\cal L}^+),\
{\cal L}_3={\cal L}^+ +{\cal L}^-,\end {equation}
where ${\cal L}^\pm$ are defined as in (\ref{oddL+-}).
Then we see from (\ref{plusbinom}), (\ref{minusbinom}) that ${\cal L}_3,{\cal L}_4$ are constants of the motion,
 polynomial in the momenta. Moreover, the identity (\ref{fundident1}) is still applicable, as are (\ref{raising}) and (\ref{lowering}).
Now we obtain
$$\{{\cal L}_2,{\cal L}_3\}=-4ip\sqrt{{\cal L}_2}{\cal L}^++4ip\sqrt{{\cal L}_2}{\cal L}^-=-4p{\cal L}_2{\cal L}_4,$$
$$\{{\cal L}_2,{\cal L}_4\}=\frac{1}{i\sqrt{{\cal L}_2}}[4ip\sqrt{{\cal L}_2}{\cal L}^+4ip\sqrt{{\cal L}_2}{\cal L}^-]=4p{\cal L}_3.$$
Similarly, since
\begin {equation}\label{L4L41} {\cal L}_4^2=-\frac{1}{{\cal L}_2}\left[ ({\cal L}^+)^2-2{\cal L}^+{\cal L}^-+({\cal L}^-)^2\right],\end {equation}
\begin {equation}\label{L3L31}{\cal L}_3^2=\left[ ({\cal L}^+)^2+2{\cal L}^+{\cal L}^-+({\cal L}^-)^2\right].\end {equation}
  we can re derive (\ref{L3^2}) again. Using (\ref{L+L-}) and (\ref{L4L41}) we re derive (\ref{L3L4bracket}). Hence the structure equations are exactly
(\ref{structure+}) again, even though the expressions for ${\cal L}_3,{\cal L}_4$ are different.

These structure results for classical TTW correspond to the highest order terms in the structure equations for quantum TTW, with the correspondence
$${\cal H}\sim H, \quad {\cal L}_2\sim L_2,\quad {\cal L}_3\sim -kL_3,\quad {\cal L}_4\sim L_4.$$
Furthermore, modulo multiplication by constants, the raising and lowering operators are associated with the classical  lowering and raising functions
via $\Xi_\pm\sim {\cal L}^\mp$ and the quantum recurrences are associated with the classical factors via $K^-\sim Y$, $K^+\sim \overline{Y}$, $J^-\sim X$, $J^+\sim \overline{X}$. Multiplication or division of the quantum recurrence operators by $2n+a+b+1$ corresponds to multiplication or division of the classical raising and lowering symmetries by $\sqrt{{\cal L}_2}$.

\subsection{The classical constant of the motion ${\cal L}_5$}\label{L5}
As for the quantum TTW system with $k=1$, our method doesn't  give the generator of minimal order. In fact, for all rational $k$,  we can extend our
symmetry algebra by adding a generator of order one less than ${\cal L}_4$. To see this, consider first the case
 $p+q$ odd. Then ${\cal L}_3=\frac{1}{i}({\cal L}^+-{\cal L}^-)$ is a polynomial in ${\cal L}_2$. Setting ${\cal L}_2=0$ in the expressions
for $X, \overline{X}, Y,\overline{Y}$  we see that the constant term in ${\cal L}_3$ is  $2(-1)^{(p-q+1)/2} (\alpha-\beta)^q {\cal H}^p$. Thus
\begin {equation} \label{L51} {\cal L}_5=\frac{{\cal L}_3-2(-1)^{(p-q+1)/2} (\alpha-\beta)^q {\cal H}^p}{{\cal L}_2}\end {equation}
is a constant of the motion, polynomial in the momenta. From (\ref{L51}) we see that
\begin {equation} \label{L52} {\cal L}_2{\cal L}_5={\cal L}_3-2(-1)^{(p-q+1)/2} (\alpha-\beta)^q {\cal H}^p,\end {equation}
so taking the Poisson bracket with ${\cal L}_2$ on both sides of the equation, and using the relation $\{{\cal L}_2,{\cal L}_3\} =-4p{\cal L}_2{\cal L}_4$,
we
find
\begin {equation}\label{L53} \{{\cal L}_2,{\cal L}_5\}= -4p{\cal L}_4.\end {equation}
Then the identities we have already derived show that ${\cal L}_2,{\cal L}_5,{\cal H}$ generate a closed
symmetry algebra that properly contains our original symmetry algebra.

If $p+q$ is even, then ${\cal L}_3={\cal L}^++{\cal L}^-$. Again ${\cal L}_3$ is  a polynomial in ${\cal L}_2$, but now with constant term
$2(-1)^{(q-p)/2} (\alpha-\beta)^q {\cal H}^p$. Thus
\begin {equation}\label{L54} {\cal L}_5=\frac{{\cal L}_3-2(-1)^{(q-p)/2} (\alpha-\beta)^q {\cal H}^p}{{\cal L}_2}\end {equation}
is a constant of the motion, polynomial in the momenta.
For example, if $p=q=1$ then
$${\cal L}_5=\frac{{\cal L}_3-2(\alpha-\beta){\cal H}}{{\cal L}_2},$$
is a 2nd order constant of the motion. Just as before ${\cal L}_2,{\cal L}_5, {\cal H}$ generate a closed symmetry algebra that properly contains our
original algebra.

The construction we have given here is quite general and not at all restricted to the TTW system. All that is
required is that the constant term in ${\cal L}_3$ be a polynomial in ${\cal H}$ alone. In all of the examples we know, this construction
gives the generator of minimum order. However, we don't have a proof that the order is minmal for all $k$.

\section{The classical potential $V=k^2\delta^2 (x+iy)^{k-1}/(x-iy)^{k+1}$}
Consider the classical Hamiltonian
\begin {equation}\label{Vclass} {\cal H} =p_x^2+p_y^2+k^2\delta^2 \frac{(x+iy)^{k-1}}{(x-iy)^{k+1}}\end {equation}
where $k=p/q$.  This system is real in Minkowski space with coordinates $x_1=x,$ $y_1=-iy$. The corresponding quantum system has Hamiltonian
\begin {equation}\label{Vquant}  H =\partial_x^2+\partial_y^2 +k^2\delta^2\frac{(x+iy)^{k-1}}{(x-iy)^{k+1}}\end {equation}
The energy eigenvalue equation is $H\Psi=E\Psi$. In\cite{KMPog10} this system was shown to be both classically and quantum superintegrable, and in
\cite{KKM10c} the recurrence relation method was used to obtain the structure of the quantum symmetry algebra.
Here, we revisit the classical problem and determine
the structure of the classical symmetry algebra.

We use the fact that the Hamilton-Jacobi equation is additively separable in polar coordinates $x=r\cos\theta$, $y=r\sin\theta$ where $r=e^R$.
As in the previous example it is convenient to pass to variables $R$ and
$\theta$. Then
$${\cal H}= {\cal L}_1=
\frac{(p^2_R+p^2_\theta +k^2\delta^2 e^{2ik\theta })}{ e^{2R}},\quad V=k^2\delta^2e^{2ik\theta-2R}$$
and there is is a constant of the motion
$$ {\cal L}_2=p^2_\theta +k^2\delta^2 e^{2ik\theta},$$
responsible for the variable separation.

Using the usual prescription for obtaining the extra constant we  compute $M(R)=\frac12\int\frac{dR}{p_R}$,
 $N(\theta)=\frac12\int\frac{d\theta}{p_\theta}$. The new constant of the motion (not a polynomial) will be $M-N$.
If $k$ is an integer it is convenient to consider the solution such that  (in terms of the coordinates $R,\theta$)
we have
$2ip\sqrt {{\cal L}_2}(M-N)=qA+pB$ as a constant of the motion
and
$$
\sinh{ A}=\frac{i \sqrt{{\cal L}_2}}{ \delta k  }e^{-ik\theta},\
\cosh{ A}=\frac{ip_\theta}{ \delta k}e^{-ik\theta},$$
$$\sinh{ B}=\frac{i\sqrt{{\cal L}_2}}{ \sqrt{ {\cal H}}} e^{-R},\
\cosh{ B}=\frac{p_R}{ \sqrt{{\cal H}}}e^{-R}.$$
There is the identity $$e^{2R}{\cal H}=p_R^2+{\cal L}_2.$$
Using the same reasoning as for the classical TTW  case,
we have
\begin {equation}\label{expAV} e^A=\cosh A +\sinh A= {X}/{U},\quad e^{-A}=\cosh A -\sinh A= {\overline{X}}/{U},\end {equation}
\begin {equation}\label{expBV} e^B=\cosh B +\sinh B= {Y}/{S},\quad e^{-B}=\cosh B -\sinh B= {\overline{Y}}/{S},\end {equation}
where
$$ X= (ip_\theta+i\sqrt{{\cal L}_2})e^{-ik\theta},\
{\overline X}= (ip_\theta-i\sqrt{{\cal L}_2})e^{-ik\theta},$$
$$Y= (p_R+i\sqrt{{\cal L}_2})e^{-R},\
{\overline Y}= (p_R-i\sqrt{{\cal L}_2})e^{-R},$$
$$ U=k\delta,\quad S=\sqrt{{\cal H}}.$$
Then $e^{qA+pB}$ and $e^{-(qA+pB)}$ are constants of the motion, where
\begin {equation}\label{expABV} e^{qA+pB}=(e^A)^q (e^B)^p=\frac{X^qY^p}{U^qS^p},\quad e^{-(qA+pB)}=(e^{-A})^q (e^{-B})^p=\frac{(\overline{X})^q
(\overline{Y})^p}{U^qS^p}.\end {equation} The identity $e^{qA+pB}e^{-(qA+pB)}=1$ can be expressed as
\begin {equation}\label{fundidentV} X^q(\overline{X})^qY^p(\overline{Y})^p=U^{2q}S^{2p}=(k\delta)^{2q}{\cal H}^p.\end {equation}

  Using an analysis similar to  (\ref{plusbinom}), (\ref{minusbinom}), we see that, independent of $p+q$ being even or odd, we can set
\begin {equation} \label{oddL34V} {\cal L}_4=\frac{1}{\sqrt{{\cal L}_2}}({\cal L}^+ -{\cal L}^-),\
{\cal L}_3=({\cal L}^+ +{\cal L}^-),\end {equation}
where
\begin {equation}\label{oddL+-V} {\cal L}^+=X^qY^p,\quad {\cal L}^-=(\overline{X})^q(\overline{Y})^p.\end {equation}
Then  ${\cal L}_3,{\cal L}_4$ are constants of the motion,
 polynomial in the  momenta. Moreover, the identity (\ref{fundidentV}) becomes
\begin {equation}\label{fundident1V}{\cal L}^+{\cal L}^-=(k\delta)^{2q}{\cal H}^p.\end {equation}

It is straightforward to verify the Poisson bracket computations
\begin {equation}\label{raisingV}\{{\cal L}_2,{\cal L}^+\}=q\frac{X^{q-1}Y^p}{U^qS^p}\{{\cal L}_2,X\}=2ip\sqrt{{\cal L}_2}{\cal L}^+,\end {equation}
\begin {equation}\label{loweringV}\{{\cal L}_2,{\cal L}^-\}=q\frac{(\overline{X})^{q-1}(\overline{Y})^p}{U^qS^p}\{{\cal L}_2,\overline{X}\}
=-2ip\sqrt{{\cal L}_2}{\cal L}^-.\end {equation}
From these results we can obtain
$$\{{\cal L}_2,{\cal L}_4\}=\frac{1}{\sqrt{{\cal L}_2}}(2ip\sqrt{{\cal L}_2}{\cal L}^++ 2ip\sqrt{{\cal L}_2}{\cal L}^-)
=2ip{\cal L}_3,$$
$$\{{\cal L}_2,{\cal L}_3\}= 2ip\sqrt{{\cal L}_2}{\cal L}^+-2ip\sqrt{{\cal L}_2}{\cal L}^-=2ip{\cal L}_2{\cal L}_4.$$

Note that
\begin {equation}\label{L4L4V} {\cal L}_4^2=\frac{1}{{\cal L}_2}\left[ ({\cal L}^+)^2-2{\cal L}^+{\cal L}^-+({\cal L}^-)^2\right],\
{\cal L}_3^2= ({\cal L}^+)^2+2{\cal L}^+{\cal L}^-+({\cal L}^-)^2.\end {equation}
Thus,
\begin {equation}\label{L3^2V} {\cal L}_3^2={\cal L}_2{\cal L}_4^2+4(k\delta)^{2q}{\cal H}^p.\end {equation}

Note that ${\cal L}^-=(k\delta)^{2q}{\cal H}^p/{\cal L}^+$, so
$$\{{\cal L}^+,{\cal L}^-\}=(k\delta)^{2q}\{{\cal L}^+,\frac{{\cal H}^p}{{\cal L}^+}\}=\frac{(k\delta)^{2q}}{{\cal L}^+}\{{\cal L}^+,{\cal H}^p\}=0$$
since ${\cal L}^+$ is a constant of the motion.

We find
$$\{{\cal L}_3,{\cal L}_4\}=\{{\cal L}^++{\cal L}^-,\frac{{\cal L}^+}{\sqrt{{\cal L}_2}}-\frac{{\cal L}^-}{\sqrt{{\cal L}_2}}\}=$$
$$-({\cal L}_2)^{-3/2}\{{\cal L}^+,{\cal L}_2\}{\cal L}^+ +({\cal L}_2)^{-3/2}\{{\cal L}^+,{\cal L}_2\}{\cal L}^-
-({\cal L}_2)^{-3/2}\{{\cal L}^-,{\cal L}_2\}{\cal L}^++({\cal L}_2)^{-3/2}\{{\cal L}^-,{\cal L}_2\}{\cal L}^-$$
$$ +({\cal L}_2)^{-1/2}\{{\cal L}^+,{\cal L}^-\}+({\cal L}_2)^{-1/2}\{{\cal L}^-,{\cal L}^+\}$$
$$=\frac{2pi}{{\cal L}_2}[({\cal L}^+)^2-2{\cal L}^+{\cal L}^-+({\cal L}^-)^2].$$
Then, using (\ref{L4L4V}), we conclude that
\begin {equation}\label{L3L4bracketV}\{{\cal L}_3,{\cal L}_4\}=2pi{\cal L}_4^2.\end {equation}
To list the structure equations in standard form we define ${\cal R}=-2pi{\cal L}_3$. Then the structure equations take the form
\begin {equation}\label{structure+V} \{{\cal L}_2,{\cal L}_4\}={\cal R},\ \{{\cal L}_2,{\cal R}\}=-4p^2{\cal L}_2{\cal L}_4,\ \{{\cal L}_4,{\cal R}\}=4p^2{\cal L}_4^2,\end {equation}
$${\cal R}^2=-4p^2{\cal L}_2{\cal L}_4^2-16p^2(k\delta)^{2q}{\cal H}^p.$$
This shows that the symmetry algebra generated by ${\cal H}, {\cal L}_2, {\cal L}_4$ closes at finite order.
As for the TTW case, these classical structure relations correlate with the quantum structure equations in \cite{KKM10c} (with the sign of the potential reversed) and give the classical version of recurrence relations.
\section{A system on the 2-sphere}
In \cite{KKM10c}  we considered a  quantum Hamiltonian  on the
two-sphere:
\begin {equation} \label{ham1} H=\partial ^2_\theta +\cot\theta \partial _\theta + \frac{1}{
\sin ^2\theta }
\partial ^2_\varphi  + \frac{\alpha }{ \sin ^2\theta \cos^2k\varphi }\end {equation}
where $k=\frac{p}{ q}$ and $p,q$ are relatively prime positive integers, showed it was superintegrable, and determined the closure of its
 symmetry algebra via the recurrence relation approach.  Now we treat the classical analog
\begin {equation} \label{ham2} {\cal  H}=p ^2_\theta  + \frac{1}{
\sin ^2\theta }
p ^2_\varphi  + \frac{\alpha }{ \sin ^2\theta \cos^2k\varphi }.\end {equation}
In this case the variables are not in the standard form for our classical construction and we need to consider  the
 separation variables $x_1,x_2$ as $ \varphi, \psi$ where  $x_1=\varphi$, $x_2=\psi$, $f_1=0$ , $f_2=1/\cosh^2\psi$,
$v_1=\alpha/\cos^2 k\varphi$. Then
\begin {equation} \label{ham3} {\cal  H}=\cosh^2\psi( p ^2_\psi  +
p ^2_\varphi  + \frac{\alpha }{  \cos^2k\varphi })\end {equation}
and
\begin {equation}\label{L2S} {\cal L}_2=p ^2_\varphi  + \frac{\alpha }{  \cos^2k\varphi },\end {equation}
so ${\cal H}=\cosh^2\psi(p_\theta^2+{\cal L}_2)$. The variables $\psi,\theta$ and their corresponding momenta are
related by $\cosh\psi=1/\sin\theta$, $\sinh\psi=\cot\theta$ and $p_\psi=-\sin\theta\  p_\theta$. Then
$$M(\varphi,p_\varphi)=\frac12\int\frac{d\varphi}{p_\varphi}=\frac12\int\frac{d\varphi}{\sqrt{{\cal L}_2 -\frac{\alpha}{\cos^2 k\varphi}}},\
N(\psi ,p_\psi)=\frac12\int\frac{d\psi}{p_\psi}=\frac12\int\frac{d\psi}{\sqrt{\frac{{\cal H}}{\cosh^2\psi}-{\cal L}_2}},$$
so
$$M=\frac{-iA}{2k\sqrt{{\cal L}_2}},\quad \sinh A= \frac{i\sqrt{{\cal L}_2}\sin  k\varphi}{\sqrt{{\cal L}_2-\alpha}},
\quad \cosh A=\frac{\cos k\varphi\ p_\varphi}{\sqrt{{\cal L}_2-\alpha}},$$
and
$$ N=\frac{iB}{2\sqrt{{\cal L}_2}},\quad \sinh B=\frac{-ik\sqrt{{\cal L}_2}\sinh\psi}{\sqrt{{\cal H}-{\cal L}_2}}=
\frac{-ik\sqrt{{\cal L}_2}\cot\theta}{\sqrt{{\cal H}-{\cal L}_2}},\quad
\cosh B=\frac{\cosh \psi\  p_\psi}{\sqrt{{\cal H}-{\cal L}_2}}=\frac{- p_\theta}{\sqrt{{\cal H}-{\cal L}_2}}.$$

From these expressions for $M$ and $N$ we see that
$$2ip\sqrt{{\cal  L}_2}[M-N]=qA+pB$$ is a constant of the motion. As usual,
\begin {equation}\label{expAS} e^A=\cosh A +\sinh A= {X}/{U},\quad e^{-A}=\cosh A -\sinh A= {\overline{X}}/{U},\end {equation}
\begin {equation}\label{expBS} e^B=\cosh B +\sinh B= {Y}/{S},\quad e^{-B}=\cosh B -\sinh B= {\overline{Y}}/{S},\end {equation}
where
$$ X= \cos k\varphi\ p_\varphi+i\sqrt{{\cal L}_2}\sin k\varphi,\
{\overline X}= \cos k\varphi\ p_\varphi-i\sqrt{{\cal L}_2}\sin k\varphi,$$
$$Y=-p_\theta-ik\sqrt{{\cal L}_2}\cot \theta,\
{\overline Y}= -p_\theta+ik\sqrt{{\cal L}_2}\cot \theta,$$
$$ U=\sqrt{{\cal L}_2-\alpha},\quad S=\sqrt{{\cal H}-{\cal L}_2}.$$

Again, $e^{qA+pB}$ and $e^{-(qA+pB)}$ are constants of the motion, where
\begin {equation}\label{expABS} e^{qA+pB}=(e^A)^q (e^B)^p=\frac{X^qY^p}{U^qS^p},\quad e^{-(qA+pB)}=(e^{-A})^q (e^{-B})^p=\frac{(\overline{X})^q
(\overline{Y})^p}{U^qS^p},\end {equation} and the identity $e^{qA+pB}e^{-(qA+pB)}=1$ can be expressed as
\begin {equation}\label{fundidentS} X^q(\overline{X})^qY^p(\overline{Y})^p=U^{2q}S^{2p}=P({\cal L}_2, {\cal H})=({\cal L}_2-\alpha)^q({\cal H}-{\cal L}_2)^p\end {equation}
where $P$ is a polynomial in ${\cal L}_2$ and ${\cal H}$.

As in the example immediately preceding, we need consider only one case, independent of $p+q$ even or odd:
\begin {equation} \label{L34S} {\cal L}_4=\frac{1}{\sqrt{{\cal L}_2}}({\cal L}^+ -{\cal L}^-),\
{\cal L}_3={\cal L}^+ +{\cal L}^-,\end {equation}
where
\begin {equation}\label{L+-S} {\cal L}^+=X^qY^p,\quad {\cal L}^-=(\overline{X})^q(\overline{Y})^p.\end {equation}
Then  ${\cal L}_3,{\cal L}_4$ are constants of the motion,
 polynomial in the momenta. and
\begin {equation}\label{fundident1S}{\cal L}^+{\cal L}^-=P({\cal L}_2,{\cal H})=({\cal L}_2-\alpha)^q({\cal H}-{\cal L}_2)^p.\end {equation}

Now,
\begin {equation}\label{raisingS}\{{\cal L}_2,{\cal L}^+\}=q\frac{X^{q-1}Y^p}{U^qS^p}\{{\cal L}_2,X\}=-2ip\sqrt{{\cal L}_2}{\cal L}^+,\
\{{\cal L}_2,{\cal L}^-\}=q\frac{(\overline{X})^{q-1}(\overline{Y})^p}{U^qS^p}\{{\cal L}_2,\overline{X}\}
=2ip\sqrt{{\cal L}_2}{\cal L}^-,\end {equation}
so
$$\{{\cal L}_2,{\cal L}_4\}=\frac{1}{\sqrt{{\cal L}_2}}(-2ip\sqrt{{\cal L}_2}{\cal L}^+- 2ip\sqrt{{\cal L}_2}{\cal L}^-)
=-2ip{\cal L}_3,$$
$$\{{\cal L}_2,{\cal L}_3\}= (-2ip\sqrt{{\cal L}_2}{\cal L}^++2ip\sqrt{{\cal L}_2}{\cal L}^-)
=-2ip{\cal L}_2{\cal L}_4.$$

Since
\begin {equation}\label{L4L4S} {\cal L}_4^2=\frac{1}{{\cal L}_2}\left[ ({\cal L}^+)^2-2{\cal L}^+{\cal L}^-+({\cal L}^-)^2\right],\
{\cal L}_3^2= ({\cal L}^+)^2+2{\cal L}^+{\cal L}^-+({\cal L}^-)^2,\end {equation}
we have
$ {\cal L}_3^2-{\cal L}_2{\cal L}_4^2=4{\cal L}^+{\cal L}^-=4P({\cal L}_2,{\cal H})$, so
\begin {equation}\label{L3^2S} {\cal L}_3^2={\cal L}_2{\cal L}_4^2+4P({\cal L}_2,{\cal H}).\end {equation}

Further,
$$\{{\cal L}^+,{\cal L}^-\} =\{{\cal L}^+, \frac{P}{{\cal L}^+}\}=  ({\cal L}^+)^{-1}\{{\cal L}^+,P\}=({\cal L}^+)^{-1}\left(\frac{\partial P}{\partial {\cal L}_2}\{{\cal L}^+,{\cal L}_2\}+\frac{\partial P}{\partial {\cal H}}\{{\cal L}^+,{\cal H}\}\right),$$
so
\begin {equation}\label{L+L-S} \{{\cal L}^+,{\cal L}^-\}=2ip\sqrt{{\cal L}_2}\frac{\partial P}{\partial{\cal L}_2}.\end {equation}
Lastly,
$$\{{\cal L}_3,{\cal L}_4\}=\{{\cal L}^++{\cal L}^-,\frac{{\cal L}^+}{\sqrt{{\cal L}_2}}-\frac{{\cal L}^-}{\sqrt{{\cal L}_2}}\}=$$
$$-({\cal L}_2)^{-3/2}\{{\cal L}^+,{\cal L}_2\}{\cal L}^+ +({\cal L}_2)^{-3/2}\{{\cal L}^+,{\cal L}_2\}{\cal L}^-
-({\cal L}_2)^{-3/2}\{{\cal L}^-,{\cal L}_2\}{\cal L}^++({\cal L}_2)^{-3/2}\{{\cal L}^-,{\cal L}_2\}{\cal L}^-$$
$$ -({\cal L}_2)^{-1/2}\{{\cal L}^+,{\cal L}^-\}+({\cal L}_2)^{-1/2}\{{\cal L}^-,{\cal L}^+\}$$
\begin {equation}\label{L3L4bracketS}=2ip{\cal L}_4^2-4ip\frac{\partial P}{\partial{\cal L}_2}.\end {equation}
For standard form we set ${\cal R}=-2ip{\cal L}_3$. Then the structure equations take the form
\begin {equation}\label{structure+S} \{{\cal L}_2,{\cal L}_4\}={\cal R},\ \{{\cal L}_2,{\cal R}\}=-4p^2{\cal L}_2{\cal L}_4,\ \{{\cal L}_4,{\cal R}\}=-4p^2{\cal L}_4^2+8p^2\frac{\partial P}{\partial{\cal L}_2},\end {equation}
$${\cal R}^2=-4p^2{\cal L}_2{\cal L}_4^2-16p^2P({\cal L}_2,{\cal H}),$$
where
$$ P({\cal L}_2,{\cal H})=({\cal L}_2-\alpha)^q({\cal H}-{\cal L}_2)^p.$$
Thus the symmetry algebra generated by ${\cal H}, {\cal L}_2, {\cal L}_4$ closes at finite order.

\section{The caged anisotropic oscillator revisited} The quantum caged anisotropic oscillator is defined by the Hamiltonian operator
\begin {equation}\label{cagedoscillatorQ} H=\partial ^2_x+\partial ^2_y-\omega_1 ^2x^2-\omega_2
^2y^2+
\frac{\alpha}{x^2} + \frac{\beta}{ y^2},\end {equation}
where $\omega _1=p\omega $ and $\omega _2=q\omega $ and $p$, $q$ are positive integers that
we assume are relatively prime. The classical version is
 \begin {equation}\label{cagedoscillatorC} H=p^2_x+p ^2_y-\omega_1 ^2x^2-\omega_2
^2y^2+
\frac{\alpha}{x^2} + \frac{\beta}{ y^2},\end {equation}
Marquette \cite{Marquette20101, Marquette20103}, using a ladder operator method,  was the first to work out the quantum structure
relations for this system. We treated it again, from the recurrence relation point of
 view, in \cite{KKM10c} and will revisit it here to demonstrate the correspondence between the quantum recurrence relations and our
classical construction. We do not claim novelty for the final results for this very simple but very important system, only the approach.
This system separates in the Cartesian coordinates $x,y$ and either of the classical constants of the motion
$${\cal L}_1=p_x^2-\omega_1^2x^2+\frac{\alpha}{x^2},\quad
{\cal L}_2=p_y^2-\omega_2^2y^2+\frac{\beta}{y^2},$$
characterizes the separation, where  ${\cal H}={\cal L}_1+{\cal L}_2$.

Constructing $M(x,p_x)$ which satisfies $M=\frac12\int dx/p_x$, we find a solution
$$ M=\frac{A}{4\omega_1},\ \sinh A= \frac{2\omega_1 x p_x}{ \sqrt{ {\cal L} ^2_1+4\omega ^2_1\alpha }}, \
\cosh A = \frac{2x^2\omega ^2_1+{\cal L}_1}{ \sqrt {L^2_1+4\omega ^2_1\alpha}}.$$
Similarly, $N(y,p_y)$ can be found as
$$ N=\frac{-B}{ 4\omega _2},\ \sinh B= \frac{-2\omega _2yp_y}{ \sqrt {L^2_2+4\omega ^2_2\beta}},\
\cosh B= \frac{2y^2\omega ^2_2+{\cal L}_2}{ \sqrt {L^2_2+4\omega ^2_2\beta}}.$$
Thus, $4pq\omega (M-N)=qA+pB$ is a constant of the motion, as are
\begin {equation}\label{expABO} e^{qA+pB}=(e^A)^q (e^B)^p=\frac{X^qY^p}{U^qS^p},\quad e^{-(qA+pB)}=(e^{-A})^q (e^{-B})^p=\frac{(\overline{X})^q
(\overline{Y})^p}{U^qS^p},\end {equation} and the identity $e^{qA+pB}e^{-(qA+pB)}=1$ can be expressed as
\begin {equation}\label{fundidentO} X^q(\overline{X})^qY^p(\overline{Y})^p=U^{2q}S^{2p}=P({\cal L}_1,
 {\cal L}_2)=({\cal L} ^2_1+4\omega ^2_1\alpha )^q({\cal L} ^2_2+4\omega ^2_2\alpha )^p\end {equation}
where $P$ is a polynomial in ${\cal L}_1$ and ${\cal L}_2$. In  particular,

$$ X= 2x^2\omega_1^2+{\cal L}_1+2\omega_1x p_x,\
{\overline X}= 2x^2\omega_1^2+{\cal L}_1-2\omega_1x p_x,$$
$$Y=2y^2\omega_2^2+{\cal L}_2-2\omega_2y p_y,\
{\overline Y}= 2y^2\omega_2^2+{\cal L}_2+2\omega_2y p_y,$$
$$ U=\sqrt {L^2_1+4\omega ^2_1\alpha},\quad S=\sqrt {L^2_2+4\omega ^2_2\beta} .$$

As usual, we set
\begin {equation}\label{L+-O} {\cal L}^+=X^qY^p,\quad {\cal L}^-=(\overline{X})^q(\overline{Y})^p.\end {equation}
In this degenerate case, ${\cal L}^+,{\cal L}^-$ are each constants of the motion, polynomial in the momenta, We set
$${\cal L}_3= {\cal L}^++{\cal L}^-,\quad {\cal L}_4= {\cal L}^+-{\cal L}^-.$$
Now
$$\{{\cal L}_1, {\cal L}^+\}=qX^{q-1}Y^p\{{\cal L}_1,X\}=-4q\omega_1{\cal L}^+,\ \{{\cal L}_1,{\cal L}^-\}=4q\omega_1{\cal L}^-,$$
$$\{{\cal L}_2,{\cal L}^+\}=pX^qY^{p-1}\{{\cal L}_2,Y\}=4p\omega_2{\cal L}^+,\{{\cal L}_2,{\cal L}^-\}=-4p\omega_2{\cal L}^-,$$
and, since ${\cal L}^+{\cal L}^-=P$,
$$\{{\cal L}^+,{\cal L}^-\}=\{{\cal L}^+,\frac{P}{{\cal L}^+}\}=({\cal L}^+)^{-1}\{{\cal L}^+,P\}$$
$$=({\cal L}^+)^{-1}\left(\frac{\partial P}{\partial {\cal L}_1}\{{\cal L}^+,{\cal L}_1\}+\frac{\partial P}{\partial {\cal L}_2}\{{\cal L}^+,{\cal L}_2\}\right) =4q\omega_1\frac{\partial P}{\partial {\cal L}_1}-4p\omega_2\frac{\partial P}{\partial {\cal L}_2}.$$
Thus,
$$\{{\cal L}_1,{\cal L}_3\}=-4q\omega_1{\cal L}_4,\ \{{\cal L}_1,{\cal L}_4\}=-4q\omega_1{\cal L}_3,\ \{{\cal L}_2,{\cal L}_3\}=4p\omega_2{\cal L}_4,\ \{{\cal L}_2,{\cal L}_4\}=4p\omega_2{\cal L}_3,$$
$$\{{\cal L}_1,{\cal L}_2\}=0,\ \{{\cal L}_3,{\cal L}_4\}=8q\omega_1\frac{\partial P}{\partial {\cal L}_1}-8p\omega_2\frac{\partial P}{\partial {\cal L}_2}.$$
$$ {\cal L}_3^2-{\cal L}_4^2=4P,\quad P({\cal L}_1,{\cal L}_2)=({\cal L} ^2_1+4\omega ^2_1\alpha )^q({\cal L} ^2_2+4\omega ^2_2\alpha )^p.$$

\section{Extension of the generic potential on the 2-sphere}
A general Hamiltonian on the 2-sphere, separable in spherical coordinates can be taken in the standard form
$${\cal H}=\cosh^2\psi(p_\psi^2+p_\varphi^2+v_1(\varphi)+v_2(\psi)),\quad x=\varphi,\ y=\psi,\ f_1=0,\ f_2=\frac{1}{\cosh^2\psi}.$$
 Embedded in complex Euclidean 3-space with Cartesian coordinates, such 2-sphere systems can be written as
$$ H={\cal J}_1^2+{\cal J}_2^2+{\cal J}_3^2 +V({\bf s}),$$
where
$${\cal J}_3=s_1p_{s_2}-s_2p_{s_1},\ {\cal J}_1=s_2p_{s_3}-s_3p_{s_2},\ {\cal J}_2=s_3p_{s_1}-s_1p_{s_3},\  s_1^2+s_2^2+s_3^3=1.$$
where $s_1=\sin\theta\cos\varphi$, $s_2=\sin\theta\sin\varphi$, $s_3=\cos\theta$ are the usual spherical coordinates.
 The variables $\psi,\theta$ and their corresponding momenta are related by $\cosh\psi=1/\sin\theta$, $\sinh\psi=\cot\theta$ and $p_\psi=-\sin\theta\  p_\theta$.
Now we consider the superintegrable Hamiltonian system
\begin {equation} \label{hamS9} {\cal H}=\cosh^2\psi [p^2_\psi +p^2_\varphi + \frac{\alpha }{\cos ^2k\varphi } +
\frac{\beta }{\sin ^2k\varphi } + \frac{\gamma }{ \sinh^2\psi } ]\end {equation}
$${\cal L}_2=p^2_\varphi + \frac{\alpha }{ \cos ^2k\varphi } + \frac{\beta }{ \sin ^2k\varphi }$$
Here $k=p/q$ where $p,q$ are relatively prime integers.
As shown in \cite{KKM10c},
 the functions that determine the extra constant of the motion are
$$M(\varphi ,p_\varphi )= \frac{i}{4k\sqrt{ {\cal L}_2}} {\cal A},\quad \sinh{\cal  A }=
\frac{i({\cal L}_2\cos (2k\varphi )-\alpha +\beta )}{ \sqrt {({\cal L}_2-\alpha -\beta )^2-4\alpha \beta }},\
\cosh{\cal A}=\frac{\sqrt{{\cal L}_2}\sin (2k\varphi )p_\varphi }{ \sqrt {({\cal L}_2-\alpha -\beta )^2-4\alpha \beta }},$$
$$N(\psi ,p_\psi )= \frac{i}{ 4\sqrt {{\cal L}_2}}{\cal B},\quad \sinh{\cal B }=
\frac{i({\cal L}_2\cosh(2\psi )+\gamma -{\cal H})}{ \sqrt {({\cal H}-{\cal L}_2-\gamma )^2-4{\cal L}_2\gamma }},\
\cosh{\cal B }= \frac{\sqrt{{\cal L}_2}\sinh(2\psi )p_\psi }
{ \sqrt{({\cal H}-{\cal L}_2-\gamma )^2-4{\cal L}_2\gamma }}.$$
(Here, we have corrected some misprints in the expressions listed there.) If $k=1$ this is the generic potential on the 2-sphere,
 a 2nd order superintegrable system.

Carrying out our usual construction, we have
\begin {equation}\label{expAG} e^A=\cosh A +\sinh A= {X}/{U},\quad e^{-A}=\cosh A -\sinh A= {\overline{X}}/{U},\end {equation}
\begin {equation}\label{expBG} e^B=\cosh B +\sinh B= {Y}/{S},\quad e^{-B}=\cosh B -\sinh B= {\overline{Y}}/{S},\end {equation}
where
$$ X= \sqrt{{\cal L}_2}\sin (2k\varphi) p_\varphi+i(\beta-\alpha+{\cal L}_2\cos(2k\varphi)),\
{\overline X}= \sqrt{{\cal L}_2}\sin (2k\varphi) p_\varphi-i(\beta-\alpha+{\cal L}_2\cos(2k\varphi)),$$
$$Y=\sqrt{{\cal L}_2}\sinh (2\psi) p_\psi+i(\gamma-{\cal H}+{\cal L}_2\cosh(2\psi)),\
{\overline Y}=\sqrt{{\cal L}_2}\sinh (2\psi) p_\psi-i(\gamma-{\cal H}+{\cal L}_2\cosh(2\psi)),$$
$$ U=\sqrt{({\cal L}_2-\alpha-\beta)^2-4\alpha\beta},\quad S=\sqrt{({\cal H}-{\cal L}_2-\gamma)^2-4\gamma{\cal L}_2}.$$
Then
\begin {equation}\label{expABG} e^{qA+pB}=(e^A)^q (e^B)^p=\frac{X^qY^p}{U^qS^p},\quad e^{-(qA+pB)}=(e^{-A})^q (e^{-B})^p=\frac{(\overline{X})^q
(\overline{Y})^p}{U^qS^p},\end {equation}  and  we have the identity
\begin {equation}\label{fundidentG} X^q(\overline{X})^qY^p(\overline{Y})^p=U^{2q}S^{2p}=P({\cal L}_2, {\cal H})=[({\cal L}_2-\alpha-\beta)^2-4\alpha\beta]^q
[({\cal H}-{\cal L}_2-\gamma)^2-4\gamma{\cal L}_2]^p\end {equation}
where $P$ is a polynomial in ${\cal L}_2$ and ${\cal H}$.

\medskip\noindent {\bf Case $p+q$ odd}:  Let
\begin {equation} \label{oddL34G} {\cal L}_4=\frac{1}{\sqrt{{\cal L}_2}}({\cal L}^+ +{\cal L}^-),\
{\cal L}_3=\frac{1}{i}({\cal L}^+ -{\cal L}^-),\end {equation}
where
\begin {equation}\label{oddL+-G} {\cal L}^+=X^qY^p,\quad {\cal L}^-=(\overline{X})^q(\overline{Y})^p.\end {equation}
Then  ${\cal L}_3,{\cal L}_4$ are constants of the motion,
 polynomial in the momenta and we have
\begin {equation}\label{fundident1G}{\cal L}^+{\cal L}^-=P({\cal L}_2,{\cal H}).\end {equation}

We find
\begin {equation}\label{raisingG}\{{\cal L}_2,{\cal L}^+\}=q\frac{X^{q-1}Y^p}{U^qS^p}\{{\cal L}_2,X\}=4ip\sqrt{{\cal L}_2}{\cal L}^+,\end {equation}
\begin {equation}\label{loweringG}\{{\cal L}_2,{\cal L}^-\}=q\frac{(\overline{X})^{q-1}(\overline{Y})^p}{U^qS^p}\{{\cal L}_2,\overline{X}\}
=-4ip\sqrt{{\cal L}_2}{\cal L}^-,\end {equation}
so
$$\{{\cal L}_2,{\cal L}_4\}=-4p{\cal L}_3,\quad
\{{\cal L}_2,{\cal L}_3\}=4p{\cal L}_2{\cal L}_4.$$
Using (\ref{L4L4}), (\ref{L3L3}) we obtain
\begin {equation}\label{L3^2G} {\cal L}_3^2=-{\cal L}_2{\cal L}_4^2+4P({\cal L}_2,{\cal H}).\end {equation}
Further,
$$\{{\cal L}^+,{\cal L}^-\} =\{{\cal L}^+, \frac{P}{{\cal L}^+}\}=  ({\cal L}^+)^{-1}\{{\cal L}^+,P\}=({\cal L}^+)^{-1}\left(\frac{\partial P}{\partial {\cal L}_2}\{{\cal L}^+,{\cal L}_2\}+\frac{\partial P}{\partial {\cal H}}\{{\cal L}^+,{\cal H}\}\right),$$
so
\begin {equation}\label{L+L-G} \{{\cal L}^+,{\cal L}^-\}=-4ip\sqrt{{\cal L}_2}\frac{\partial P}{\partial{\cal L}_2},\end {equation}
and
\begin {equation}\label{L3L4bracketG}\{{\cal L}_3,{\cal L}_4\}=4p{\cal L}_4^2-8p\frac{\partial P}{\partial{\cal L}_2}.\end {equation}
To list the structure equations in standard form we define ${\cal R}=-4p{\cal L}_3$. Then the structure equations take the form
\begin {equation}\label{structure+G} \{{\cal L}_2,{\cal L}_4\}={\cal R},\ \{{\cal L}_2,{\cal R}\}=-16p^2{\cal L}_2{\cal L}_4,\ \{{\cal L}_4,{\cal R}\}=16p^2{\cal L}_4^2-32p^2\frac{\partial P}{\partial{\cal L}_2},\end {equation}
$${\cal R}^2=-16p^2{\cal L}_2{\cal L}_4^2+64p^2P({\cal L}_2,{\cal H}),$$
where
$$ P({\cal L}_2,{\cal H})=(({\cal L}_2-\alpha-\beta)^2-4\alpha\beta)^q(({\cal H}-{\cal L}_2-\gamma)^2-4\gamma{\cal L}_2)^p.$$
This shows that the symmetry algebra generated by ${\cal H}, {\cal L}_2, {\cal L}_4$ closes at finite order.

\medskip\noindent {\bf Case $p+q$ even}: Let
\begin {equation} \label{evenL34G} {\cal L}_4=\frac{1}{i\sqrt{{\cal L}_2}}({\cal L}^- -{\cal L}^+),\
{\cal L}_3={\cal L}^+ +{\cal L}^-,\end {equation}
where ${\cal L}^\pm$ are defined as in (\ref{oddL+-G}). Then ${\cal L}_3,\ {\cal L}_4$ are polynomial constants of the motion.
It can be verified that the structure equations are exactly the same as for $p+q$ odd.

Just as for the TTW system, in the case $k=1$ our method gives a 3rd order generator, whereas we know that, due to the
 multiseparability of the Hamilton-Jacobi equation, there is a 2nd order generator ${\cal L}_5$ independent of ${\cal L}_2, {\cal H}$.
 The method for finding this symmetry ${\cal L}_5$ for all rational $k$
and showing that it determines a larger symmetry algebra that closes is exactly analogous
 to that in Subsection \ref{L5}. The message is that the symmetries ${\cal L}^\pm$ determine everything.

\section{Discussion and conclusions}
We have found the classical analog of the  recurrence relation approach for verifying quantum superintegrability for 2D Hamiltonian systems that
 admit a separation of variables, and have demonstrated that this approach allows us give explicit expressions for the symmetries and to compute
 structure equations for them. We have not proved that this method always yields the symmetry generators of lowest order, although that is the
case for all of the examples we have checked. A major surprise for us was
the simplicity of the structure equations. Indeed all of the structures in our various examples are essentially the same, except for the
 function $P({\cal L}_2,{\cal H})$ and the polynomial in ${\cal H}$ that occurs in the expression for ${\cal L}_5$.
It is clear that this method can be generalized to Hamiltonian systems in $>2$ dimensions that admit separation in subgroup type coordinates
 \cite{KKM10}. It is not clear what to do if the Hamiltonian system does not admit orthogonal variable separation.

redpajama_set_name: RedPajamaArXiv


The following statement is issued by various progressive individuals and organizations from many parts of the world, but mostly in Asia-Pacific. A copy of the statement has already been sent to Syriza's international department.
Political crises in Ukraine and Greece expose the paralysis and lack of strong leadership from Merkel in Berlin.


Overt and Relational Aggression and Victimization: Multiple Perspectives within the School Setting.
The current study involved a comprehensive comparative examination of overt and relational aggression and victimization across multiple perspectives in the school setting (peers, teachers, observers in the lunchroom, self-report). Patterns of results involving sociometic status, ethnicity and gender were explored among 4(th) graders, with particular emphasis on girls. Controversial and rejected children were perceived as higher on both forms of aggression than other status groups, but only rejected children were reported as victims. Both European American and African American girls showed a greater tendency toward relational aggression and victimization than overt aggression or victimization. Results indicated negative outcomes associated with both relational and overt victimization and especially overt aggression for the target girl sample. Poorer adjustment and a socially unskillful behavioral profile were found to be associated with these three behaviors. However, relational aggression did not evidence a similar negative relation to adjustment nor was it related to many of the behaviors examined in the current study. Implications of these results are discussed.


"It's a dog's life," a term that originated in the 16th century, used to allude to the miserable subservient existence of dogs during this era. Not so in today's world. And, certainly not in dog-friendly Aspen Snowmass. Many lodges in Aspen and Snowmass Village are pet-friendly. Four-legged friends checking in with their human companions will be greeted with much more than just a pat on the head. They are offered complimentary bowls, special dog beds and sometimes even their own room service menu. If the hotel doesn't offer dog walking, the concierge can connect guests with a local dog walking service.
In keeping with the Limelight tradition, the Limelight Snowmass is also welcoming to dogs. Hank the Black Labrador Retriever receives the same hospitality as Jake – a complimentary bowl and placemat. The Lounge and bar are off limits to Hank, due to food service. However, outside play couldn't be better with hiking trails throughout Snowmass, summertime rides on the Elk Camp gondola and walks on Snowmass Mountain. Leashes are required since there are biking trails on the mountain.
Pet owners of overnight canine guests, are required to sign an agreement upon arrival. A fee of $125 is charged for the first night and $25 per night thereafter.
Other pet-friendly lodges in Aspen Snowmass include Aspen Meadows (lots of room to roam), Hotel Jerome (formerly known for its on-site Mountain Bernese), the St. Regis Aspen Resort, Aspen Square and the Viceroy Snowmass.
Susan Linden has been wordsmithing as an advertising copywriter, feature writer, and storyteller since her college days. She left the whirlwind corporate world to live in the mountains and spends as much time as she can hiking, biking, and skiing in them. Working, playing, and raising a family in the Roaring Fork Valley for the last 29 years gives Susan a unique insider's perspective to share.


These shafts are engineered with speed and kinetic energy, designed to cause the maximum zombie face damage. The Zombie Slayer arrows are 100% carbon and manufactured with the tightest .001" or .003" straightness or straighter tolerances, to ensure you never miss that critical head shot.
Don't entrust your family's survival with some second rate shaft, unless you like getting eaten alive.


text:
Q: How to use sourcemaps to restore the original file? I have a file that has been transpiled from ES6 to ES5 using Babel. I have sourcemap. I am assuming I can restore the original file (the way it looked when written in ES6) using these resources.
How is it done?
Is there a CLI tool to do this?

A: Very simple NodeJS implementation which I wrote for my needs.
const fs = require('fs');
const { SourceMapConsumer } = require("source-map");

fs.readFile('./example.js', 'utf8' , (err, data) => {
  if (err) return console.error(err);

  const sourceMapData = data.split('//# sourceMappingURL=data:application/json;base64,')[1];
  let buff = new Buffer.from(sourceMapData, 'base64');
  let rawSourceMap = buff.toString('ascii');

  const parsed = SourceMapConsumer(rawSourceMap);

  fs.writeFile('example.ts', parsed.sourcesContent, function (err) {
    if (err) return console.log(err);
  });
});

A: There's a nice web service to do that: http://sokra.github.io/source-map-visualization/

*

*Click on "Custom"

*Select your *.js file

*Select the corresponding *.js.map file

*Browse the non-minified code on the right

There might be a lot of noise from webpack:///node_modules/ modules that you're probably not interested in. You can try to find the I found that simply searching for webpack:///src usually brings up the actual source code within the whole snippet.

A: Open up the source map in a text editor, and you'll see that it's mostly just a simple JSON object. The "sources" field contains an array of URLs/paths to all source files, which can help you to find them. There is also an optional "sourcesContent" field, which also is an array, where each item contains the contents of the file at the same index in the "sources" array; if so, you could find the original code right in the source map.
A CLI tool? Well, there's source-map-visualize which tries to find all original sources and pre-loads them into this online source map visualization.

A: i wrote a script as well that came handy to me. I'll share it here it might come handy to someone
this script assumes that the .map.(js|css) has a structure like
{
    "version": 3,
    "file": "static/js/15.80b7f5e4.chunk.js",
    "mappings": "iOAkCA,UAzBqB",
    "sources": [
        "path/to/afile.js"
    ],
    "sourcesContent": [
        "content of the file"
    ],
    "names": [
        "post",
        "useState"
    ],
    "sourceRoot": ""
}

thou i didn't make use of names and mapping entry on the map file i just mapped sources with sourceContent and recreated the file structure.
const fs = require("fs");
const path = require("path");

function parseIfMap(filePath, content) {
  console.log("working on:", filePath);
  if (filePath.includes(".map")) {
    const jsonContent = JSON.parse(content);
    jsonContent.sources.forEach(function (value, index) {
      const parsed = path.parse(value);
      if (parsed.dir) {
        fs.mkdirSync(parsed.dir, { recursive: true });
      }
      fs.writeFile(
        value,
        jsonContent.sourcesContent[index],
        (path) => path && console.log(value, path)
      );
    });
  }
}

function readFiles(dirname, onError) {
  fs.readdir(dirname, function (err, filenames) {
    if (err) {
      onError(dirname, err);
      return;
    }
    filenames.forEach(function (filename) {
      const filePath = dirname + path.sep + filename;
      if (fs.lstatSync(filePath).isFile()) {
        fs.readFile(filePath, "utf-8", function (err, content) {
          if (err) {
            onError(err);
            return;
          }
          parseIfMap(filePath, content);
        });
      } else {
        readFiles(filePath, parseIfMap, onError);
      }
    });
  });
}
// change with you own file path
readFiles("static", console.log);

redpajama_set_name: RedPajamaStackExchange


Dear readers, I am wishing you a Happy New Year from Gstaad! Again, we are gifted with 365 days to play with, to fill them up with whatever our hearts desire. A new year is the best time to make fresh new beginnings. May the year 2015 be like a blank book that is waiting for our imagination to fill up its 365 pages with love, emotions, unexpected incidents, adventures and magic moments so that we will have no regrets at the end of the annual cycle.
The beauty of life is that it is full of surprises. Who would have thought that I could ride a camel in Gstaad?! When you get a chance, visit Leonard at the Bellevue Gstaad, he is an institution.
To an amazing New Year's Eve!


Cutting Edge Carbon Button PSD ideal for Hi-Tech Designs! This PSD file is layered file of the carbon button style desgins which help in your making your design more unique and attractive. Download This Carbon Button and use it in your designs.


text:
Engle, Joyner Come Together To Promote The Call
By Brian Tashman | August 29, 2011 10:55 am
Lou Engle and Rick Joyner will be working together in September to promote The Call: Detroit, Engle's prayer rally on November 11. This shouldn't come as a surprise, since both are closely involved in the New Apostolic Reformation, or the movement which believes that God is ordaining a new generation of prophets and apostles. Both have prophesied that America is turning into Nazi Germany and have claimed to know the reasons behind natural disasters, with Joyner calling Hurricane Katrina God's judgment for homosexuality and Engle asserting that the Joplin, Missouri tornado was God's punishment for legal abortion. Engle and Joyner, who leads MorningStar Ministries and The Oak Initiative, are also strong proponents of Seven Mountains Dominionism, which posits that fundamentalist Christians should take control over the seven most influential sectors of society: government, family, media, business, education, arts and entertainment, and the church.
Engle and Joyner are hosting the 'Mobilization Conference for The Call Detroit' on September 9-10, where the two "will be sharing a message that evening on the seasons of time we are in and the importance of the upcoming gathering at Ford Field in Detroit."
The Call has previously won endorsements from Republican and Religious Right leaders including Mike Huckabee, Sam Brownback, Pat Robertson, James Dobson, Tony Perkins, Tim Wildmon, Harry Jackson and Mat Staver. The rally is intended to use "spiritual warfare" to make abortion illegal and ban same-sex marriage. Engle has said that The Call: Detroit will also focus on converting Muslims to Christianity along with the usual anti-choice and anti-gay efforts. Engle also released a new video publicizing the prayer rally:
Another figure running the 'Mobilizaiton Conference for The Call' with Engle and Joyner is self-proclaimed apostle Barbara Yoder. Yoder says she is a follower of Peter Wagner, who is seen as the architect of the New Apostolic Reformation, and her biography lists her as a: "Member of Eagles' Vision Apostolic Team led by Drs. Peter and Doris Wagner; Michigan State Coordinator Global Apostolic Prayer Network under Dr. Peter Wagner and Dr. Chuck Pierce; Member of the core faculty for Wagner Leadership Institute; Member of the International Coalition of Apostles under the direction of Dr. Peter Wagner."
In her book Taking on Goliath: How to Stand Against The Spiritual Enemies in Your Life and Win, Yoder warns of the loss of America's white majority and the rampant bisexuality unleashed after Britney Spears and Madonna kissed at the MTV Video Music Awards:
In terms of spiritual beliefs, pluralism means that there is no single path to God. Buddhism, Hinduism, New Age, or whatever someone thinks is 'the way' will work as long as we end up with 'god'—and ill-defined god at that. Pluralism rejects the Bible as the source of absolute truth. Many Americans prefer pluralism as the most plausible belief system.
In terms of human cultural mix, American culture is increasingly formed from multiple cultures. Hispanics or Latinos have emerged as the leading minority group, Latinizing the national culture. The Caucasian majority, who came from European roots, is aging and cannot compete with the fertility rates of the minorities. Asian and Middle Eastern nations are sending a flood of immigrants to American shores. This has a noticeable effect on the Judeo-Christian underpinnings of our national character.
It seems as if there are no boundaries anymore. Adultery and fornication are rampant, not only in the society at large but in the church as well. Television and Internet ads are sexualized. Previous taboos have been eliminated, and television programs include sex acts.
It is now 'normal' to be homosexual or bisexual. When Madonna and Britney Spears were photographed kissing each other, a forbidden boundary was removed. Now it is the rage among junior high and high school girls to take pictures of themselves kissing each other and post them on their Myspace.com pages. Anything goes.
Yoder's warning of the deleterious consequences of Madonna and Britney Spears kissing has already been addressed at The Call, when self-proclaimed Prophet Cindy Jacobs preached against it at The Call: Sacramento last year:
Tags: Chuck Pierce Cindy Jacobs Harry Jackson Lou Engle Pat Robertson Rick Joyner Abortion Anti-Choice Anti-Gay choice Education endorsements Legal marriage Media Muslims Religious Right Spiritual Warfare Video Leadership Institute Religious Right
Huckabee: If Someone Breaks Into My House, 'I Call 911 To Tell Them Where To Come Pick Up The Body'
Bobby Jindal's Clueless Defense Of His AFA Prayer Rally
President Gingrich's First Order Of Business Would Be To Call The Nation To Prayer
Steve King: DREAMers Come From Another Planet

redpajama_set_name: RedPajamaCommonCrawl


text:
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Herbert, R.J.H. & Muxagata, E. (2009).
Barnacles (Crustacea: Cirripedia) of the Solent and Isle of Wight. Proceedings of the Isle of Wight Natural History & Archaeological Society, 24:42-56
Herbert, R.J.H., Southward, A.J., Sheader, M. & Hawkins,S.J. (2009).
Persistent Border. An analysis of the geographic boundary of an intertidal species. Marine Ecology Progress Series, 379:135-150
Herbert, R.J.H., Southward, A.J., Sheader, M. & Hawkins,S.J (2007).
Influence of recruitment and temperature on distribution of intertidal barnacles in the English Channel. Journal of the Marine Biological Association of the UK, 87:487-489.
Herbert, R.J.H. & Hawkins, S.J. (2006).
Effect of rock type on the recruitment and early mortality of the barnacle Chthamalus montagui. Journal of Experimental Marine Biology & Ecology, 334: 96-108.
Herbert, R.J.H (2004).
Solent & South Coast Mantis Shrimp Survey. A partnership project with the 'conserving marine sand and gravel biotopes in SE England initiative . Report to Hampshire & IOW Wildlife Trust. Januray 2007
Herbert, R.J.H & Bray, S. (2005)
Ventnor Haven Maintenance Dredging. Monitoring Littoral Rock Biotopes . Initial Survey Report to Royal Haskoning Ltd. Medina Valley Centre. (April 2005).
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Monitoring intertidal rocky reefs below shingle re-nourishment/recycling site at Bembridge , IOW. Report to Royal Haskoning Ltd. Medina Valley Centre. (June 2005).
Herbert, R.J.H., Hawkins, S.J., Sheader, M. & Southward, A.J. (2003)
Range extension and reproduction of the barnacle Balanus perforatus in the eastern English Channel. Journal of the Marine Biological Association of the UK, 83: 73-82.
Herbert, R.J.H. (2001)
Recent observations of the Mantis Shrimp Rissoides desmaresti from the Solent. Porcupine Newsletter9 :12-15.
Intertidal Biotope Survey of Rocky Shores along the SW coast of the Isle of Wight . Report to English Nature.
Testing hypotheses related to changes in abundance and distribution of warm-temperate invertebrates on rocky shores along the south coast of England. Ph.D Thesis. School of Ocean & Earth Sciences, University of Southampton.
Herbert, R., Crowe,T. & Sheader, M. (2000)
Benthic Survey to determine the effect of swinging moorings in the Folly Reach, Medina Estuary, IOW. Report to the Environment Agency 25pp Medina Valley Centre.
Herbert, R.J.H., Bray, S. & Hawkins, S.J.H. (2000)
Use of the Dog-whelk Nucella lapillus as a Bio-Indicator of Tributyltin (TBT) Contamination in the Solent and around the Isle of Wight:307-310. In: Collins, M. & Ansell, K. Solent Science – a review. Proceedings in Marine Science (1).Elsevier.
Bray. S. and Herbert, R.J.H. (1998)
A reassessment of populations of the dog-whelk (Nucella lapillus) on the Isle of Wight following legislation restricting the use of TBT antifouling paints. Proc. Isle. Wight. nat. Hist. archaeol Soc. 14 : 23-40.
Herbert, R.J.H. & Bray, S. (1997)
Ten Years on – the current status of the dog-whelk Nucella lapillus (L.) on the Isle of Wight following the 1987 legislation on the use of TBT antifouling paints . Report to the Environment Agency. Medina Valley Centre, Newport, IW.
Herbert, R.J. (1995).
A supplement to the Marine Flora & Fauna of the Medina Estuary. Medina Valley Wildlife 1994 Report. 14-21. Medina Valley Centre, Newport. IOW.
Herbert, R.J.H. (1995).
The Marine Flora & Fauna of Horse Ledge Shanklin, Isle of Wight. Interim Report & Species List, February 1995 . Isle of Wight Natural History & Archaeological Society, Shanklin.
Herbert, R. (1991).
Autumn Field Meeting at Osborne Bay, Isle of Wight. Porcupine Newsletter.5 (3): 53-58.
Collins, K.C., Herbert, R.J.H., & Mallinson, J.J. (1990).
The Marine Flora & Fauna of Bembridge and St. Helens, Isle of Wight. Proc. Isle. Wight. nat. Hist. archaeol. Soc. 9 : 41-85.
A survey of the Dog-whelk Nucella lapillus (L.) around the coast of the Isle of Wight. Proc. Isle. Wight. nat. Hist. archaeol. Soc. 8 (3): 15-21.
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Sports Cycling
This guy got all in my face because I "touched" his bike. He was an oaf?
I really don't understand if you cause a problem, surely you deserve what happens if you don't take responsibility. I believe its a true "cycle" *ironic* - what goes around comes around. So I try to enter a shop, and there's a bike just inside lying all over the door way. I try and hop over... show more I really don't understand if you cause a problem, surely you deserve what happens if you don't take responsibility. I believe its a true "cycle" *ironic* - what goes around comes around. So I try to enter a shop, and there's a bike just inside lying all over the door way. I try and hop over but can't and trip sort of on the wheel. So I pick up the bike and sort of throw it on to the street as best I can. Its heavy. Well I guess the noise of it scraping along the road alerts the owner I assume and comes from somewhere inside and yelling at me and is in my face I tell him to back off or I will scream rape, and then the store worker comes and gets us both to leave. Did he deserve all (or more ) that he got ?? Is there an authority that could deal with this oaf?
Update: Definitely happened MouldHippie. It was at my local shop, and he was balding, tall in late 20s so NOT a kid and quite intimidating to me a woman. So thanks for your further MISOGYNY!!, adding to my anxiety for confronting a bully, now you are further bullying me by not believing understanding when a woman is... show more Definitely happened MouldHippie. It was at my local shop, and he was balding, tall in late 20s so NOT a kid and quite intimidating to me a woman. So thanks for your further MISOGYNY!!, adding to my anxiety for confronting a bully, now you are further bullying me by not believing understanding when a woman is subjected to male bullying. I'm anonymous because of this very reason.
Best Answer: You're getting all bent out of shape over this? OK...let me run these scenarios past you. Did you ever have to stop & wait for a bus to drop off and/or pick up passengers? Have you ever seen a car double parked on the street & you had to wait for a clear opening to go around? Have you ever had that one fat person in the grocery store leave their cart on one side of the isle while they look for an item on the other side?
Entrance-ways & exits get temporarily blocked from time to time. People sometimes do stupid stuff. Get over it. What if you owned that car double parked on the street? And while you ducked inside for just a moment a tow truck came along & latched onto it? That's exactly what you did to that guy's bike.
Take a chill pill. You're making a mountain out of a molehill.
RaleighBob · 11 months ago
I am not too sure what you mean by OAF,
and what kind of bike it was or how valuable it is, but you sound like a douche.
The whole society is not safe.
Mattman · 11 months ago
What on Earth made you think throwing it into the street was an appropriate response?
If you see a bag on a seat you want, do you pick that up and throw it out too?
OK, so the guy was a bit inconsiderate, but YOU were deliberately vindictive.
David · 11 months ago
The courteous thing to have done would've been to ask if someone could move the bike so you could enter the store. You need to grow up. You certainly had no right to throw it into the street.
However the worst thing you did was threatening to scream rape on the person. It's people like you who cry wolf that make it hard for actual rape victims to be believed.
Like I said you need to grow up. This has nothing to do with a misogynistic view (an ingrained prejudice against women). You may want to be seen as a woman who stands up to male bullying. However what you're actually displaying is you're a woman who has anger issues against men. Which probably stem from you having a severe case of penis envy.
Source(s): Motorized bicycle owner and builder.
LandRiderJerry · 11 months ago
Awww...
David D. · 11 months ago
Doesn't matter if you're a male or a female. Just move the darn thing out of the way & get on with your life. If that's the worst thing that happened to me in a day, I'd call it a good day.
Why did you feel the need to 'throw it on to the street'? I suppose you've never inadvertently left a shopping cart in the middle of an isle while looking for a product. Then you'd be the one blocking the way for other people.
Jib Jab · 11 months ago
I smell a TROLL. A true "cyclist" would NEVER leave his or her bike lying in the entrance to a store. Maybe...and I mean MAYBE a kid with a BMX bike would do this. It's happened to me when a bunch of kids were playing around & one of them left his bike lying in my driveway. I picked it up & rolled it out of the way.
Why is that so friggin' hard? Just another anonymous post from a member of the "Blue Man Group" SMDH
OldHippie · 11 months ago
the oaf seems to be the one who tripped over a bike and then, in a fit of rage, threw it in the street
I really don't understand if you cause a problem, surely you deserve what happens if you don't take responsibility. I believe its a true "cycle" *ironic* - what goes around comes around. So I try to enter a shop, and there's a bike just inside lying all over the door way. I try and hop over but can't and trip sort of on the wheel. So I pick up the bike and sort of throw it on to the street as best I can. Its heavy. Well I guess the noise of it scraping along the road alerts the owner I assume and comes from somewhere inside and yelling at me and is in my face I tell him to back off or I will scream rape, and then the store worker comes and gets us both to leave. Did he deserve all (or more ) that he got ?? Is there an authority that could deal with this oaf?
Can I fit on a 20 inch bike?
What is a good triathlon bike for beginners that won't break the bank?
Are penny farthings worth anything?
What is better a gas powered bicycle or a electric bicycle?
Is my friend fit to ride a bike with Complex PTSD and Epilepsy?
Is it possible to build a bicycle powered by a hampster?
Should the touring the France cycle race be?
How often do you ride a bicycle?


text:
Q: Need to create a dataset on news using python I need to create the dataset on news. I need to extract all the news of given news website which have ever been posted on that website. I have write this code
import requests
from bs4 import BeautifulSoup
import pandas
import csv
from datetime import datetime

records=[]

def cnbc(base_url):

    r = requests.get(base_url)
    c = r.content
    soup = BeautifulSoup(c,"html.parser")
    Title=soup.find("h1","class":"title"}).text.replace("\r","").replace("\n","")
    content=' '
    for content_tag in soup.find_all("p"):
        content = content+content_tag.text.replace("\r","").replace("\n","")
    content= content[18:-458]
    Country ='United States'
    website='https://www.cnbc.com/'
    comments=''
    genre='Political'
    date= soup.find("time",{"class":"datestamp"}).text[35:-2].replace("\r","").replace("\n","")
    d = datetime.strptime(date, "%d %b %Y")
    date = d.strftime("%d-%m-%Y")
    records.append((Title,content,Country,website,comments,genre,date))

cnbc("https://www.cnbc.com/2018/11/02/here-are-the-three-things-pulling-down-the-stock-market-again.html")

but this is only allowing me to extract one news.
Can anyone tell me that how can I extract all the news url from the root directory of website.

A: This is python3 script and it is not flawless but I hope it can serve as starting point so you can achieve what you are trying to. I am not sure does this site from which you are trying to scrape data allows such operation so I will not place their web address for constants WEB_SITE_BASE_URL and WEB_SITE_REGION_URL. It is your choice what you are going to put there.
import requests
from bs4 import BeautifulSoup
from datetime import datetime

# https://www.xxxx.com"
WEB_SITE_BASE_URL= ""
# https://www.xxxx.com/?region=us
WEB_SITE_REGION_URL = ""

def get_categories(web_site_base_url):
    r = requests.get(web_site_base_url)
    c = r.content
    soup = BeautifulSoup(c,"html.parser")
    spans = soup.find_all(attrs={"nav-menu-buttonText"})
    categories = [category.text for category in spans]
    return categories

def get_links(category_url):
    r = requests.get(category_url)
    c = r.content
    soup = BeautifulSoup(c,"html.parser")
    links = [a.get('href') for a in soup.find_all('a', href=True)]
    filtered_links = list(set([k for k in links if '/2018/11/' in k]))
    return filtered_links

def news(link):
    r = requests.get(link)
    c = r.content
    soup = BeautifulSoup(c,"html.parser")
    Title=soup.find("h1",{"class":"title"}).text.replace("\r","").replace("\n","")
    content=' '
    for content_tag in soup.find_all("p"):
        content = content+content_tag.text.replace("\r","").replace("\n","")
    content= content[18:-458]
    Country ='United States'
    website='WEB_SITE_BASE_URL'
    comments=''
    date= soup.find("time",{"class":"datestamp"}).text[35:-2].replace("\r","").replace("\n","")
    d = datetime.strptime(date, "%d %b %Y")
    date = d.strftime("%d-%m-%Y")
    spans = soup.find_all(attrs={"header_title last breadcrumb"})
    categories = [category.text for category in spans]
    genre = categories
    return(Title,content,Country,website,comments,genre,date)

categories = get_categories(WEB_SITE_REGION_URL)
list_of_link_lists = []
for category in categories:
    list_of_link_lists.append(get_links(WEB_SITE_BASE_URL + "/" + category.replace(" ", "20%")))
flat_link_list = list(set([item for sublist in list_of_link_lists for item in sublist]))
articles_list = []
for link in flat_link_list:
    try:
        articles_list.append(news(WEB_SITE_BASE_URL + link))
    except:
        print("Something was wrong")
    continue

print(articles_list)

A: There is a rough method to extract part of all news, the method showed as my code. First, extract all the news which div class is headline news_headline = soup.find_all("div",class_="headline"). Then check if element is what we want.
new = []
for div in news_headline:
    each = ()
    if div.a:
        each[0] = url + div.a.get("href")
        if div.a.text:
            # use split to remove \t \n blankspace
            each[1] = " ".join(div.a.text.split())
        else:
            each[1] = " ".join(div.a.get("title").split())
        new.append(each)
    else:
        continue

It is the full code but i wrote this as short as i can.
import requests
from bs4 import BeautifulSoup

def index(url="https://www.cnbc.com/world/"):
    with requests.Session() as se:
        se.encoding = "UTF-8"
        res = se.get(url)
        text = res.text
    soup = BeautifulSoup(text,"lxml")
    news_headline = soup.find_all("div",class_="headline")
    news_ = [(url + div.a.get("href"), " ".join(div.a.text.split()) if div.a.text else  "".join(div.a.get("title").split()) ) for div in news_headline if div.a]
    print(news_)

index()


text:
× Upcoming Concert Welcome Artistic Leadership Fanfare Magazine Tickets + Events Watch + Listen Donate Board of Directors & Administration Staff Past Concerts
The Planets, Op. 32
Born: September 21, 1874, Cheltenham, United Kingdom
Died: May 25, 1934, London, United Kingdom
Composed: 1914–16; the orchestration was completed in 1917
Premiere: The private premiere took place September 29, 1918, Queen's Hall, London, Adrian Boult conducting; the public premiere of the complete work was November 15, 1920, Albert Coates conducting.
Instrumentation: 4 flutes (incl. 2 piccolos, alto flute), 3 oboes (incl. bass oboe), English horn, 3 clarinets, bass clarinet, 3 bassoons, contrabassoon, 6 horns, 4 trumpets, 3 trombones, 2 tubas, 2 timpani, bass drum, chimes, crash cymbals, glockenspiel, snare drum, suspended cymbals, tam-tam, tambour de Basque, triangle, xylophone, 2 harps, celeste, organ, strings
CSO notable performances: First Performance: November 1934, Eugene Goossens conducting. Most Recent: February 2015, Peter Oundjian conducting. Other: Paavo Järvi and the CSO recorded the work for their 2009 Holst: The Planets CD.
Duration: approx. 51 minutes
Holst's interest in writing a piece of music on the attributes of the astrological signs was apparently spurred by his visit in the spring of 1913 with the writer and avid star-gazer Clifford Bax, who noted that Holst was himself "a skilled reader of horoscopes." (Imogen Holst suggested that one reason her father may have been attracted to composing such a work was because he was having difficulty at the time formulating structural plans for large-scale pieces, and a suite for orchestra seemed appropriate to his compositional needs.) Of the music's inspiration, Holst noted, "As a rule I only study things which suggest music to me. That's why I worried at Sanskrit. Then recently the character of each planet suggested lots to me, and I have been studying astrology fairly closely." Despite his immediate attraction to the planets as the subject for a musical work, however, he took some time before beginning actual composition. He once wrote to William Gillies Whittaker, "Never compose anything unless the not composing of it becomes a positive nuisance to you," and it was not until the summer of 1914, more than a full year after he had conceived the piece, that he could no longer resist the lure of The Planets.
"Once he had taken the underlying idea from astrology, he let the music have its way with him," reported Imogen of her father's writing The Planets. The composition of the work occupied him for over three years. Jupiter, Venus and Mars were written in 1914 (prophetically, Mars, the Bringer of War was completed only weeks before the assassination at Sarajevo precipitated the start of the First World War); Saturn, Uranus and Neptune followed in 1915, and Mercury a year after that. Except for Neptune, all the movements were originally written for two pianos rather than directly into orchestral score, probably because Holst was then having painful problems with his writing hand due to severe arthritis, and he needed to concentrate the physical effort of composition as much as possible. For the mystical Neptune movement, he considered the percussive sounds of the piano too harsh and wrote it first as an organ piece. All seven movements were orchestrated in 1917 with the help of Nora Day and Vally Lasker, two of the composer's fellow faculty members at St. Paul's School in London, who wrote out the full score from Holst's keyboard notations under his guidance. The finished work is superb testimony to Holst's skill as an orchestrator, much of which was gained from his practical experience as a teacher and conductor, and also as a professional trombonist in several British orchestras, a vocation from which he was forced to retire in 1903 because of his arthritis.
The Planets had a complicated performance history when it was new. Holst was judged unfit for military service in World War I because of his health, but he did manage to obtain a post with the YMCA as a music organizer among troops in the Near East. His overseas departure date was set too quickly to allow the new piece to be scheduled for performance in one of London's regular concert series, so his friend Balfour Gardiner, realizing how eager Holst was to hear the work, underwrote a special private performance as a farewell gift. Holst and an invited audience attended the concert, conducted by Adrian Boult on September 29, 1918 in Queen's Hall, London, only days before the composer sailed for Salonica and points beyond. On February 27, 1919, Boult led the Royal Philharmonic Orchestra in a public performance of The Planets, but omitted Venus and Neptune, so the honor of the work's first complete public performance fell to Albert Coates, who conducted the score in London on November 15, 1920. Interest in the work ran so high in America that the premiere in this country was given simultaneously in New York (Albert Coates) and Chicago (Frederick Stock) during the 1920–21 season. The Planets has remained Holst's most popular composition.
Holst gave the following explanation of The Planets for its first performances: "These pieces were suggested by the astrological significance of the planets. There is no program music in them, neither have they any connection with the deities of classical mythology bearing the same names. If any guide to the music is required, the subtitle to each piece will be found sufficient, especially if it is used in a broad sense. For instance, Jupiter brings jollity in the normal sense, and also the more ceremonial kind of rejoicing associated with religious or national festivities. Saturn brings not only physical decay, but also a vision of fulfillment."
The individual movements of The Planets employ a wide spectrum of musical styles in which the influences of Stravinsky, Dukas, Debussy and even Schoenberg may be discerned, but, according to Imogen, "The Planets is written in Holst's own language." It is a language of spectacular variety—a greater contrast than that between the first two movements is hard to imagine. The staggering hammer blows of Mars, the Bringer of War are followed by the sweet luminosity of Venus, the Bringer of Peace. Each of the remaining movements cuts as distinctive a figure as the first two. Mercury, the Winged Messenger is a nimble scherzo that seems, like the fast movements of Baroque music, to be a stream of notes spinning infinitely through the cosmos of which the composer has revealed only a small segment. Within Jupiter, the Bringer of Jollity co-exist a boisterous Bacchanalian dance ("the most joyous jangle imaginable," according to Richard Capell) and a striding hymn tune to which Elgar stood godfather. Hard upon Jupiter, which reportedly inspired the charwomen cleaning the hall during rehearsals for the premiere to toss away their mops and dance a little jig, follow the lugubrious solemnities of Saturn, the Bringer of Old Age, the movement Holst declared to be his favorite piece in the suite. This music is invested with a weighty, Mahlerian seriousness that recalls Das Lied von der Erde. Uranus, the Magician is shown as a rather portly prestidigitator who includes perhaps more broad humor than baffling legerdemain in his act. The haunting finale, Neptune, the Mystic, springs from the misty domain of Debussy's Nocturnes, but possesses an even wispier, more diaphanous orchestral sonority, with the disembodied siren song of the female chorus floating away to inaudibility among the spheres at its close.
—Dr. Richard E. Rodda


text:
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    "use strict";

    app.module.directive('defaultGrid', function (uiGridHelper, broadcastEvents) {
        return {
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            scope: {
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                columns: '='
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            templateUrl: 'appServices/directives/defaultGrid.html',
            link: function (scope, element, attrs) {
                var pagingInformation = {
                    itemsPerPage: 10,
                    currentPage: 1,
                    totalItems: 0
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                scope.paging = pagingInformation;

                function refresh() {
                    var onRefreshHandler = scope.onRefresh();

                    if (angular.isDefined(onRefreshHandler)) {
                        onRefreshHandler(scope.paging)
                            .then(function (result) {
                                scope.gridOptions.data = result.items;
                                scope.paging.totalItems = result.totalCount;
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                scope.refresh = refresh;

                scope.gridOptions = {
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                    useExternalSorting: true,
                    columnDefs: scope.columns,
                    minRowsToShow: scope.paging.itemsPerPage,
                    onRegisterApi: function (gridApi) {
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                        scope.gridApi.core.on.sortChanged(scope, function (grid, sortColumns) {
                            scope.paging.sortColumns = uiGridHelper.createSortInformation(sortColumns);
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                scope.$on(broadcastEvents.DATA_REFRESH, refresh);

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redpajama_set_name: RedPajamaGithub


text:
AIB sets aside extra €40m to cover tracker mortgage scandal
Economic recovery helps bring bank to strong performance in 2017
Thu, Mar 1, 2018, 07:40 Updated: Thu, Mar 1, 2018, 08:44
Ciara O'Brien, Ciarán Hancock
AIB's pretax profit was €1.3 billion, down from almost €1.7 billion a year earlier, with overall profit at €1.1 billion for 2017. Photograph: Paul McErlane/Reuters
AIB has set aside another €40 million to cover the cost of its tracker mortgage redress scheme and expects to have made payments to 95 per cent of impacted customers by the end of this month.
The majority State-owned bank made a provision of €190 million in 2015 to meet the costs of the programme, which was mandated of all lenders in the market by the Central Bank of Ireland.
This comprised €105 million in customer redress payments and €85 million for "other costs".
The bank has now set aside another €30 million to cover the cost of customer redress, and €10 million for other costs. Some 500 AIB employees are dealing with the issue.
Customer redress
On a conference call with media on Thursday, AIB chief executive Bernard Byrne said this additional provision was the bank's best estimate of the total cost of the tracker issue, but was subject to sign off by the Central Bank.
He said 95 per cent of customers would have received their redress by the end of this month with the balance to get theirs in the second quarter of this year.
"We are nearing the end of the programme," Mr Byrne said.
AIB picks Galvin as finance chief as hike in dividend planned
Bank bonus reform could bring down the Government
Workers and businesses pay 'record' €51bn in tax
The bank's annual report shows that €95 million had been paid in redress to 4,100 customers by the end of 2017. The balance of €40 million will be paid this year.
Late last year, AIB identified another 4,900 customers who were impacted by the issue and would be paid compensation.
AIB is expected to be the subject of enforcement proceedings by the Central Bank in relation to the tracker issue. Mr Byrne said it had yet to receive any notice of this from the regulator.
Falling profits
Earlier on Thursday morning, the bank reported profits fell slightly in 2017, but net interest margin rose as the ongoing economic recovery helped boost its figures.
Pretax profit was €1.3 billion, down from almost €1.7 billion a year earlier, with overall profit at €1.1 billion for 2017.
Net interest margin was 2.58 per cent, a rise of 35 basis points year on year.
The bank described the year as "pivotal", as it undertook an IPO, and said its full-year dividend for 2017 would be €326 million, a rise of 30 per cent.
Its share price closed down just more than 2 per cent in Dublin on a truncated day of trading due to heavy snowfall and the expected arrival of Storm Emma.
New lending
New lending rose by 13 per cent to €9.4 billion for the year, with new loans spread across all sectors. Mortgage lending rose 17 per cent, while personal lending climbed 16 per cent and business lending was up 15 per cent.
Impaired loans, meanwhile, fell €2.8 billion to €6.3 billion, bringing the total reduction since 2013 to €23 billion, or 78 per cent.
Net interest income rose 8 per cent to €2.2 billion.
Total operating expenses for the year were €1.4 billion. Credit provision writebacks were €113 million.
"Against the backdrop of an improving Irish economy, these strong results confirm that we are generating significant amounts of capital and on track to meet our medium-term targets on a sustainable basis," said Mr Byrne.
Bernard Byrne
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