Lie algebra $su(1,1)$ from Legendre polynomials; question regarding Coherent Orthogonal Polynomials Apologies if this question is a duplicate.
OK, so my question heavily involves the arXiv preprint Coherent Orthogonal Polynomials, which nicely details the Lie algebra $su(1,1)$ coming from the Laguerre $L_n$, and Legendre $P_n$, polynomials. In section 3 which details the Laguerre polynomials the authors build the raising and lowering operators ($K_{\pm}$) from the recurrence formulas involving derivatives for the Laguerre polynomials - specifically Eq.(14) in the text. These raising/lowering operators involve the number operator $N$ which is given by the 2nd order O.D.E. definition of the Laguerre polynomials,
\begin{equation*}
N L_n \equiv -\{x D^2_x +(1-x)D_x \}L_n = n L_n.
\end{equation*}
So far so good.
My problem now comes in section 4 when the equivalent process for the Legendre polynomials is detailed. The corresponding raising/lowering operators - specifically Eqs.(20,21) in the text also involve the corresponding number operator $N$ for the Legendre polynomials. My problem is that the 2nd order O.D.E. definition of the Legendre polynomials,
\begin{equation*}
\{(1-x^2) D^2_x -2 x D_x\}P_n = -n(n+1) P_n,
\end{equation*}
is no longer linear in $n$. How am I to view the corresponding Legendre number operator? Do I simply complete the square of the above equation i.e.
$$
N =? -\frac{1}{2} \pm \sqrt{(x^2-1)D^2_x +2x D_x +\frac{1}{4}},
$$
hence meaning that the resulting number operator is actually a pseudo-differential operator, similar to those detailed in Chapters 2 and 3 of Solitons: Differential Equations, Symmetries and Infinite Dimensional Algebras, T. Miwa, M. Jimbo and E. Date, 2012? Specifically they detail how to construct the square root of a generic Schrödinger operator. The overall concept seems similar here, but I am unclear if the $N$ detailed above, when applied to the
$n$th Legendre polynomial $P_n$, would give $n P_n$.
If anyone can clarify my quandary, thanks in advance. Apologies if this question seems trivial or poorly worded. I am happy to clear up any points which are confusing/inconsistent.

The Laguerre equation is indeed linear in $n$ and the Legendre equation is not, but that is irrelevant; it is the recurrence relations that allow us to define the $J_{+}$ and $J_{-}$ operators.
For the Legendre polynomials the recurrence relations to be used are
$
\left\{
\begin{matrix}
\begin{align}
&\left[-(1-x^2)\frac{d}{dx} + (n+1)x\right] P_n=(n+1)P_{n+1}\\
&\left[(1-x^2)\frac{d}{dx} + nx\right] P_n=nP_{n-1}
\end{align}
\end{matrix}
\right.
$
which allows for the definition of the following operators:
$
\left\{
\begin{matrix}
\begin{align}
&J_+=-(1-X^2)D_x + (N+1)X\\
&J_-=(1-X^2)D_x + NX\\
&J_3=N+1/2
\end{align}
\end{matrix}
\right.
$
where
$
\left\{
\begin{matrix}
\begin{align}
&Xf(x)=xf(x)\\
&NP_n(x)=nP_n(x)\\
&D_xf(x)=f'(x)\\
\end{align}
\end{matrix}
\right.
$
When doing so, the $su(1,1)$ algebra nature of $\left\{J_+,J_-,J_3\right\}$ can be derived.
