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Casimir operator of a Lie Group .. how can i calculate ??'

+ 3 like - 0 dislike
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given a Lie group with generators $ X_{i} $ how can i calculate the Casimir Generator ??

$ H= X_{i}X^{i} $ if possible with two examples please

  1. the Generator of traslation in 2 dimension $ P_{i} $ i=x,y with comutation relations $[P_{i},P_{j} ]=0 $

  2. The generator of the Angular momentum with Commutation relations $ [L_{i} , L_{j}]=\epsilon _{ijk}L_{k} $

here $ X^{i} $ is the 'dual' of $ X_{i} $

This post imported from StackExchange Mathematics at 2014-06-02 10:58 (UCT), posted by SE-user Jose Garcia
asked Oct 20, 2011 in Mathematics by Jose Garcia (15 points) [ no revision ]
I think you are confusing Lie groups with Lie algebras, what you are talking about is the algebras, not the groups (more precisely, Casimir elements are not even elements of the algebra but of the universal enveloping algebra of it). I only know the definition for semi-simple Lie algebras (since it needs the Killing form). If you want the general answer, you need to study the theory of semi-simple Lie algebras. Otherwise the answer will be very confusing if you don't know anything about it.

This post imported from StackExchange Mathematics at 2014-06-02 10:58 (UCT), posted by SE-user Heidar
But for the algebra of translations, I don't think the Casimir element is useful since everything is commuting. For $\mathfrak{so}(3)$ (angular momentum) you have written the answer yourself: $L^2 = L_x^2 + L_y^2 + L_z^2$ and this commutes with the other generators (which is the point of the Casimir element). But you need to understand the Killing form (en.wikipedia.org/wiki/Killing_form), in order to understand what 'dual' means and thereby how to calculate this for more general algebras.

This post imported from StackExchange Mathematics at 2014-06-02 10:58 (UCT), posted by SE-user Heidar
For the first example: since $[P_i,P_j] = 0,$ all structure constants are zero, so the Killing form $g_{ab}$ vanishes. Since $H := g_{ab} X^a X^b$, $H = 0$ in this case. In the second case, you need slightly more work: $C_{ab}^d = \epsilon_{abd},$ so $g_{ab} = C_{ak}^l C_{bl}^k = \epsilon_{akl} \epsilon_{blk} = \delta_{ab},$ which means that $H = \delta_{ab} X^a X^b = L_x^2 + L_y^2 + L_z^2.$

This post imported from StackExchange Mathematics at 2014-06-02 10:58 (UCT), posted by SE-user Gerben
OK Gerben, i think i understand.. so from the commutators i must read the structure constants of the group (see that in wikipedia) and from this estructure constans i find the elements of the metric :) .. the identity $ g_{ab}= C_{ak}^{l}C_{bl}^{k}$ is ALWAYS valid ??

This post imported from StackExchange Mathematics at 2014-06-02 10:58 (UCT), posted by SE-user Jose Garcia
By the way: I made a mistake. I believe it should be $\epsilon_{akl} \epsilon_{blk} = -2 \delta_{ab}.$ However, this comes down to multiplying a Casimir by -2, which isn't a big deal (can you see why?). And yes, that formula can be used as the definition of $g_{ab}$ (it's entirely equivalent to the mathematicians' definition of the Killing form).

This post imported from StackExchange Mathematics at 2014-06-02 10:58 (UCT), posted by SE-user Gerben
however for the traslation operator should'nt be the Casimir operator equal to $ P_{x}^{2}+P_{y}^{2}$ by analogy with the Hamiltonian of a Free particle in 2 dimension.

This post imported from StackExchange Mathematics at 2014-06-02 10:58 (UCT), posted by SE-user Jose Garcia
No! The algebra of translation operators is Abelian, i.e. everything commutes. You can only find a Casimir invariant if an algebra is semisimple; in practice, this means that $g_{ab}$ is non-degenerate - which isn't the case for the translation operator algebra. The main idea is that the Casimir is a special ('distinguished') element of the algebra (technically: the universal enveloping algebra). For $\mathfrak{so}(3)$, $L^2$ is (up to a scale factor) the only nontrivial element that commutes with all $L_i$. For the algebra of the $P_i$, any product of $P_i$ would work...

This post imported from StackExchange Mathematics at 2014-06-02 10:58 (UCT), posted by SE-user Gerben

1 Answer

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First, you are confusing Lie groups with Lie algebras. Casimir elements are objects that can be attached to certain Lie algebras.

Second, Casimir elements do not always exist. For any Lie algebra $\mathfrak{g}$, there is a canonical bilinear form, the Killing form

$$B(x, y) = \text{tr}(\text{ad}_x \text{ad}_y)$$

where $\text{ad}_x(y) = [x, y]$ is the adjoint action of $\mathfrak{g}$ on itself. The Casimir element exists if and only if the Killing form is nondegenerate, which is equivalent to $\mathfrak{g}$ being semisimple (in particular, finite-dimensional). Concretely this means that $B(x, -)$ is a nonzero linear functional for any nonzero $x$. Abstractly this means that the map $$\mathfrak{g} \ni x \mapsto B(x, -) \in \mathfrak{g}^{\ast}$$

(where $\mathfrak{g}^{\ast}$ is the dual space of linear functionals $\mathfrak{g} \to k$ for our base field $k$ of characteristic zero) is an isomorphism. For an abelian Lie algebra, the Killing form is identically zero, and so the Casimir element does not exist in that case.

In the semisimple case, the Killing form itself defines a linear functional $\mathfrak{g} \otimes \mathfrak{g} \to k$ (where $\otimes$ denotes the tensor product), or an element of $\mathfrak{g}^{\ast} \otimes \mathfrak{g}^{\ast}$, and because of the above isomorphism one can equivalently write the Killing form as an element of $\mathfrak{g} \otimes \mathfrak{g}$. This is the Casimir element.

Concretely, we can compute the Casimir element as follows. Given a basis $e_1, ... e_n$ of $\mathfrak{g}$, compute the Killing form $B$ using the structure constants of $\mathfrak{g}$, then compute the dual basis $f_1, ... f_n$, which is the unique basis satisfying $$B(e_i, f_j) = \delta_{ij}.$$

Then the Casimir element is given by $\sum e_i \otimes f_i$. Since you are aware of this definition, perhaps what you're stuck on is either computing the Killing form or computing the dual basis. For fixed $\mathfrak{g}$ both of these are fairly straightforward linear algebra. Which step are you having trouble with?

This post imported from StackExchange Mathematics at 2014-06-02 10:58 (UCT), posted by SE-user Qiaochu Yuan
answered Oct 21, 2011 by Qiaochu Yuan (380 points) [ no revision ]

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