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:

  1. 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.

  2. 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.

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    $\begingroup$ In fact, both these two computations are not correct, because the quadratic algebraic expressions for the Casimir operator are not in the Lie algebra, and are not matrices 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 are not at all Casimir. The commutativity is non-trivial to truly verify... $\endgroup$ – paul garrett Oct 13 '18 at 20:19
  • $\begingroup$ Thanks a lot for your comment and let me apologize for my naive question but please notice that I'm a Physicist (so, not really an expert of this field). Actually, I am just trying to follow Gilmore's examples and, as far as I understand, computation number 1 is just the copy/paste of the example developed in the book by Gilmore (see eq 9.8 at pag 141 and eq 9.16 at pag 143). Basically, my target is to find the Casimir of a certain algebra starting from a certain matrix representation thereof. Do you think that I'am interpreting this algorithmic technique in a wrong way? $\endgroup$ – AndreaPaco Oct 13 '18 at 20:42
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    $\begingroup$ Ok, but/and the appropriate products $J_iJ_j$ are not matrix 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
  • $\begingroup$ Again, thanks for your help. So, "practically" which is, for example, the correct mapping $a_1^2 \to J_1^2$? In other words, how can I move to the universal enveloping algebra? In addition: is the fact that the first computation gives matrix $C=2\mathbb{I}$ (which indeed does commute with all the algebra generators) just an accident? $\endgroup$ – AndreaPaco Oct 13 '18 at 21:14
  • $\begingroup$ And, on top of that... let's forget about the matrix representation of the casimir operator. Let's focus just on abstract operators, which is the core of what I am interested in, by the way. Clearly you see that the first quadratic expression, (after the mapping $a_i\to J_i$) is correct, but the second one is wrong. In fact, it is well known that $J_1^2+J_2^2+J_3^2$ is the Casimir quadratic invariant of su(2). On the contrary, operator $J_3^2+(J_+J_-+J_-J_+)$ is not invariant. This is my main matter of concern. (I've brought up the matrix representations just because of Gilmore's algorithm) $\endgroup$ – AndreaPaco Oct 13 '18 at 21:30

I think that the reason for this discrepancy is that in the first basis all three vectors have the same norm (with respect to the Killing form) while in the second they do not. You should see the $a_i$s as elements in the dual of your Lie algebra and so the last step should be mapping these elements to the duals of your basis elements. This is where norm discrepancies play a role.

To be a little more specific: For any semisimple Lie algebra $\mathfrak{g}$ there exists a nondegenerate invariant symmetric form $B$ called Killing form. (Which in your case is just a multiple of the trace form $B(X,Y) = \mathrm{Tr}\, XY.$)

With such a form $B$ we have a concrete isomorphism $\mathfrak{g} \to \mathfrak{g}^*$ given by $X \mapsto B(\,\_\,,X).$ For a basis $X_i$ of $\mathfrak{g}$, the dual basis $\alpha_i^*$ is canonically defined by the condition $\alpha_i(X_j) = \delta_{ij}.$ So starting with an orthogonal basis $X_i$ we see that the dual basis can be identified using the previous isomorphism with $X_i / B(X_i, X_i)$. For a general basis $X_i$ some elments $Y_i$ are/represent the dual basis to $X_i$ iff $B(X_i, Y_j) = \delta_{ij}$. If you write this in coordinates you get $$ [X_i]^T [B] [Y_j] = \delta_{ij} \mathrm{I}_n, $$ where $[X_i] = e_i$ and $[B] = (B(X_i, X_j))_{i,j=1}^n$. But this equations shows that $[Y_j] = [B]^{-1}[X_j].$

Let me also clear the confusion from comments. The Casimir operator is an element of the universal enveloping algebra and as such it cannot be represented by a finite dimensional matrix. What you are effectively doing here is you are mapping this element into an endomorphism algebra of some representation (see the universal property of enveloping algebras) and thus you should obtain multiple of the identity matrix. This multiple can be sometimes obtained directly by different means. E.g. for a quadratic Casimir of a complex simple Lie algebra and a highest weight representation there is a formula involving $\lambda, \rho$ and the Killing form.

  • $\begingroup$ Thanks a lot for the very nice answer. Can you please expand the first part of your explanation? More specifically, can you please explain how to build the duals of the basis elements? $\endgroup$ – AndreaPaco Oct 14 '18 at 10:43
  • $\begingroup$ Thanks again for the editing and for your time. Nevertheless, I am not sure that I've correctly understood how to apply your explanation. Suppose you want to compute the dual of $J_+$. If I understood correctly, you start computing $B(J_+,J_+)=\mathrm{Tr}(J_+^2)$ which gives... zero. Then the dual of $J_+$ is supposed to be $J_+/B(J_+,J_+)$... which of course is non-sense. What did I do wrong? $\endgroup$ – AndreaPaco Oct 14 '18 at 12:13
  • $\begingroup$ @AndreaPaco Ah yes. I forgot there are isotropic vectors. The short answer is that you just take the inverse matrix to $B$ and it's columns are coefficients of the dual basis elements with respect to your chosen basis. For so called Chevalley basis you will just get the opposite basis elements (i.e. some permutation of basis elements) and norm corrections. $\endgroup$ – Vít Tuček Oct 14 '18 at 12:55
  • $\begingroup$ Thanks again for this clarification. Just two clarifications more, please. Clarification 1: If I've well understood, I have to compute matrix $B$, whose entries are $B_{i,j}=\mathrm{Tr}(J_i,J_j)$, where $i,j=\{+,−,3\}$. Then I have to compute its inverse, i.e. $B^{−1}$. And then? Can I say that the dual of $J+$ is $B^{-1}J_+$ ? $\endgroup$ – AndreaPaco Oct 14 '18 at 13:30
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    $\begingroup$ The Killing form matrix with respect to the basis $(J_+, J_-, J_3)$ is $$\left(\begin{array}{rrr} 0 & 4 & 0 \\ 4 & 0 & 0 \\ 0 & 0 & 2 \end{array}\right)$$ and its inverse is $$\left(\begin{array}{rrr} 0 & \frac{1}{4} & 0 \\ \frac{1}{4} & 0 & 0 \\ 0 & 0 & \frac{1}{2}. \end{array}\right)$$ Thus $a_+ \mapsto 1/4J_-$, $a_- \mapsto 1/4 J_+$ and $a_3 \mapsto 1/2 J_3$. This gives the action of $C$ as $1/2 \mathrm{I}_n.$ $\endgroup$ – Vít Tuček Oct 14 '18 at 14:24

It might be worthwhile to also give the standard description (I think I saw it in Serre's notes long ago) of Casimir in intrinsic, coordinate-free terms... which proves that it is certainly independendent of coordinates! First, given a simple Lie algebra $\mathfrak g$, the Killing form $B(,)$ is non-degenerate, so gives an isomorphism of $\mathfrak g$ and its dual by $x \to B(-,x)$. Also, recall the isomorphism of $\mathrm{End}(\mathfrak g)$ with $\mathfrak g \otimes \mathfrak g^*$ (also for any finite-dimensional vector space). Then we have a chain of $G$-equivariant maps $$\mathrm{End}_{\mathbb R}(\mathfrak g) \to\mathfrak g\otimes \mathfrak g^* \to\mathfrak g\otimes \mathfrak g \subset \bigotimes^\bullet \mathfrak g\to U \mathfrak g.$$ The identity map in $\mathrm{End}(\mathfrak g)$ is $G$-invariant, so its image through these maps is also $G$-invariant in the universal enveloping algebra $U\mathfrak g$. That image is the Casimir element (which needs to be proven non-zero, either by Poincare-Birkhoff-Witt, or by giving a repn in which it does not map to $0$).

In any case, this gives an (Ad) $G$-invariant (or ad-$\mathfrak g$-invariant) element in the universal enveloping algebra.

Examination of the description of it does confirm that for any basis $\{x_i\}$ of $\mathfrak g$, with dual basis $\{x_i'\}$, Casimir can be written as $\sum_i x_i x'_i$.

  • $\begingroup$ Thanks a lot for this great explanation! $\endgroup$ – AndreaPaco Oct 16 '18 at 15:12

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