I'm following Gilmore's recipe to compute the abstract Casimir operator of a given algebra (in this example, I refer to algebra su(2)). This recipe bring up a matrix representation of the algebra and my problem is that I obtain different results according to the specific matrix representation that I choose. To be more concrete, look at the difference between these two cases:

Consider the defining commutators: $$ [J_1,J_2]=iJ_3, \qquad [J_2,J_3]=iJ_1, \qquad [J_3,J_1]=iJ_2 $$ The following $3\times 3$ matrices are a representation of this algebra: $$ J_1=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \\ \end{array} \right), \qquad J_2=\left( \begin{array}{ccc} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \\ \end{array} \right), \qquad J_3= \left( \begin{array}{ccc} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right). $$ Applying Gilmore's method (see pag. 140), one can write matrix $$ X= \sum_{i=1}^3a_iJ_i, \qquad a_i\in\mathbb{R} $$ One therefore obtains: $$ X=\left( \begin{array}{ccc} 0 & -i a_3 & i a_2 \\ i a_3 & 0 & -i a_1 \\ -i a_2 & i a_1 & 0 \\ \end{array} \right) $$ At this point one computes the characteristic polynomial $$ P(\lambda)=\mathrm{det}(X-\lambda\mathbb{I})=-\lambda^3+\lambda(a_1^2+a_2^2+a_3^2) $$ Performing the substitution $a_i\to J_i$ in the coefficient of $\lambda$, as prescribed by the algorithm, one obtains that the algebra's Casimir is $$ C=\left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{array} \right) $$ which indeed commutes with $J_i, \quad \forall i$. This scheme therefore gives a

*correct*result.Now consider the following alternative (but equivalent) definition of the algebra su(2):

$$ [J_+,J_-]=2 J_3, \qquad [J_3,J_+]=+J_+, \qquad [J_3,J_-]=-J_-. $$ The following $3\times 3$ matrices are a representation of this algebra: $$ J_+= \left( \begin{array}{ccc} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 2 & 0 \\ \end{array} \right), \qquad J_-=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ -2 & 0 & 0 \\ \end{array} \right), \qquad J_3=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ Introducing three unknown coefficients $a_+,\,a_-,\,a_3$, one can computes matrix $X$ and the characteristic polynomial $P(\lambda)$ as seen before. One obtains that $$ P(\lambda)= -\lambda^3 +\lambda(a_3^2+4a_+a_-) $$ At this point, one has to perform the substitution $a_i\to J_i$. Even if one takes into accunt the need for symmetrization, i.e. $a_+a_-\to (J_+J_-+J_-J_+)/2$, the algorithm retrieves an incorrect result, i.e. the following matrix $$ \bar{C}=\left( \begin{array}{ccc} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \\ \end{array} \right) $$ which does not commute with matrices $J_+$ and $J_-$ and so constitutes a*wrong*result.

My question is: why does the first way give the correct result but the second scheme does not work? Did I miss any hypothesis which is needed by Gilmore's algorithm? I suspect that the problem might be either in the use of $J_\pm$ as basis elements of algebra su(2) or in the substitutions $a^i\to J_i$ (the book uses, in fact, upper and lower indices, a formalism which I am not familiar with). Please, take into account that my target is to understand how to compute the abstract Casimir operator of a certain Lie algebra in an algorithmic way. The matrix representation of linear and quadratic operators is just auxiliary to reach the target but it is not my core business.

notin the Lie algebra, and arenotmatrices of the same size as the model of the Lie algebra, etc. They involve products in the universal enveloping algebra... and these are not matrix products. So the 3x3 matrices you have produced arenotat all Casimir. The commutativity is non-trivial to truly verify... $\endgroup$ – paul garrett Oct 13 '18 at 20:19notmatrix products, but products in the universal enveloping algebra, in any case. So when you map that quadratic expression in the $a_i$'s to $J_i$'s, do not multiply the matrices! $\endgroup$ – paul garrett Oct 13 '18 at 21:09