Quadratic form of $SO(2,1)$ from the actions of its generators

Question

I have 3 operators which are generators of $SO(2,1)$. I don't know the expressions of these operators as differential operators, but I know their action on a basis $\left|l',m\right>$:

$$J_{\mu\nu}\left|l',m\right>=\sum_{\tilde{l'},\tilde{m}}C_{\mu\nu}^{\left(l',m\right),\left(\tilde{l'},\tilde{m}\right)}\left|\tilde{l'},\tilde{m}\right>$$

These relations are quite long, but the shortest one is of the form:

$$J_{01}\left|l',m\right>=\frac{i}{2}\left(l'+m\right)\left|l',m\right>-\frac{il'}{4|l'|}\sqrt{\left(l'+m+\frac{1}{2}\right)\left(l'+m+\frac{3}{2}\right)}\left|l'+1,m+1\right>$$

So, in my notation, $C_{01}^{\left(l',m\right),\left(l',m\right)}=\frac{i}{2}\left(l'+m\right)$. $J_{02}$ and $J_{12}$ split a state labeled by $(l',m)$ amongst $(l'-1,m-1)$, $(l',m)$, and $(l'+1,m+1)$, if that is of any importance. These generators satisfy the commutation relations:

\begin{array}{c} \left[J_{01},J_{02}\right]=iJ_{12} \\\ \left[J_{01},J_{12}\right]=iJ_{02} \\\ \left[J_{02},J_{12}\right]=-iJ_{01} \end{array}

Which satisfy $[J_{\mu \nu },J_{\rho \sigma }]=i(\eta_{\nu \rho }J_{\mu \sigma }+\eta_{\mu \sigma }J_{\nu \rho }-\eta_{\mu \rho }J_{\nu \sigma }-\eta_{\nu \sigma }J_{\mu \rho})$ for $\eta$ with the metric $(-,+,+)$.

Question: How can I find the invariant quadratic form, $-x_0^2+x_1^2+x_2^2$, in terms of the coordinates $x$, $y$, and $z$ and possibly other known quantities like momentum or the Pauli matrices from just this information?

Additionally, from this information, can I write the operators $J_{\mu\nu}$ as differential operators?

Further, I know the Hamiltonian of this system, the closed forms of all the basis states, $\left|l',m\right>$, and the larger symmetry group $SO(3,2)\supset SO(2,1)$.

Follow up: The $e_{ij}$ that I found are: \begin{array}{ccc} e_{00}=\mathbf{1}-J_{12}J_{12} & e_{01}=-J_{02}J_{12} & e_{02}=J_{01}J_{12}\\\ e_{10}=J_{12}J_{02} & e_{11}=\mathbf{1}+J_{02}J_{02} & e_{12}=-J_{01}{J_02}\\\ e_{20}=-J_{12}J_{01} & e_{21}=-J_{02}J_{01} & e_{22}=\mathbf{1}+J_{01}J_{01} \end{array}

And if each of these is $e_{ij}=Q_iP_j$, I get 9 equations of the form:

$$f(x_0,x_1,x_2)-J_{12}J_{12}f(x_0,x_1,x_2)=x_0\frac{\partial}{\partial x_0}f(x_0,x_1,x_2)$$

Answer 1

So, first, the Lie algebra $\mathfrak{so}(2,1)$ is isomorphic to $\mathfrak{sl}(2)$. For example $$ J_{01} = \frac{i}{2} \begin{pmatrix} 1&0\\0&-1 \end{pmatrix}\quad J_{02} = \frac{i}{2} \begin{pmatrix} 0&1\\1&0 \end{pmatrix}\quad J_{12} = \frac{i}{2} \begin{pmatrix} 0&1\\-1&0 \end{pmatrix}\quad $$ have the same commutation relations and they span $\mathfrak{sl}(2)$ as a linear space.

Second, for any representation $\pi$ of a Lie algebra, the invariant symmetric bilinear form is $$B(X,Y)=\operatorname{trace}(\pi(X)\pi(Y))$$ where $X, Y$ are in the Lie algebra and the trace is computed on linear operators of the space the Lie algebra is acting on.

Third, if $v_1, \ldots v_k$ is a basis of the space, there is a corresponding dual-basis of linear functionals $v_1^*, \dots, v_k^*$ with the property $v_i^*(v_j) = \delta_{ij}$. Using a basis and dual-basis, the trace is computed as $$\operatorname{trace}(\pi(X)\pi(Y)) = \sum_{i=1}^k v_i^*(\pi(X)\pi(Y)v_i)$$

About the differential operators - You have the associative Weyl algebra which is generated by linearly-independent generators $Q_1,\ldots,Q_l,P_1,\ldots,P_l$ with the relations $$[P_i, Q_j] = \delta_{ij}\mathbf{1}\quad [P_i,P_j]=0\quad [Q_i,Q_j]=0\tag{1}$$ where $\mathbf{1}$ is the unit of the algebra. This associative algebra can realized as operators on functions $f(x_1,\ldots,x_l)$ via $$ \begin{align} Q_j f(x_1,\ldots,x_l)&:= x_j f(x_1,\ldots,x_l)\\ P_j f(x_1,\ldots,x_l)&:= \frac{\partial}{\partial x_j} f(x_1,\ldots,x_l)\\ \mathbf{1}f(x_1,\ldots,x_l)&:=f(x_1,\ldots,x_l) \end{align} \tag{2} $$ Now, from the commutation relations $(1)$ you get $$ \begin{align} [Q_i P_j, Q_k P_l] &= [Q_i P_j, Q_k] P_l + Q_k[Q_i P_j, P_l]\\ &= [Q_i,Q_k]P_j P_l + Q_i[P_j,Q_k]P_l + Q_k[Q_i,P_l]P_j + Q_k Q_i[P_j, P_l]\\ &= \delta_{jk} Q_i P_l - \delta_{il} Q_k P_j\tag{3} \end{align} $$

Turning for a moment to the matrix Lie algebra $\mathfrak{gl}(n)$, it has a basis $e_{ij}$ of matrices with $1$ in the ($i$-th row, $j$-th column) and zero elsewhere that satisfy $$[e_{ij}, e_{kl}] = \delta_{ij} e_{il} - \delta_{il} e_{kj}\tag{4}$$

Comparing $(3)$ and $(4)$ we see that the matrix Lie algebra is embedded inside the Weyl algebra via the mapping $e_{ij}\mapsto Q_i P_j$. Furthermore, the $Q_i P_j$ can be realized as differential operators via $(2)$. So if you have any Lie algebra spanned by matrices whatsoever, you can turn it into a Lie algebra of differential operators by mapping linearly the basis $e_{ij}$ to $Q_i P_j$ to the differential operators. If this sounds strange, when carrying out this process on the Lie algebra $\mathfrak{so}(3)$, the Lie algebra of rotations in $3$-space, you get the differential operators of angular momentum acting on functions (up to scaling by $i\hbar$) - and these in turn have spherical harmonics as eigenfunctions.

Edit: For differential operators try $$ \begin{align} J_{01}&= i \left(x_0\frac{\partial}{\partial x_1} + x_1\frac{\partial}{\partial x_0}\right)\\ J_{02}&= i \left(x_0\frac{\partial}{\partial x_2} + x_2\frac{\partial}{\partial x_0}\right)\\ J_{12}&= i \left(x_1\frac{\partial}{\partial x_2} - x_2\frac{\partial}{\partial x_1}\right) \end{align} $$ These have the same commutation relations as the $J$-s in the original post.

For the invariant symmetric bilinear form, in an irreducible representation it's always going to be $$B(J_{01},J_{01})=B(J_{02},J_{02})=-\alpha\quad B(J_{12},J_{12})=\alpha$$ with other pairs zero for some $\alpha>0$ that depends on the dimension of the irreducible representation. For a $2$-dimensional representation, $\alpha=1/2$. For a $3$-dimensional representation, $\alpha=2$.

I can't tell from the notation in the original post what dimension representation is talked about and whether it's irreducible. It looks like it's a representation that's a tensor product of two irreducible ones decomposed according to Clebsch-Gordan.