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I'm not sure which is the exact definition of a Casimir operator.

In some texts it is defined as the product of generators of the form: $$X^2=\sum X_iX^i$$

But in other parts it is defined as an operator that conmutes with every generator of the Lie group.

Are these definitions equivalent? If the answer is yes, how could I prove it (I'm thinking in using Jacobi's identity)?

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Do you know about the Killing form? –  Vibert May 5 '13 at 16:13
    
I've heard of it. Isn't it related to the adjoint representation? –  jinawee May 5 '13 at 16:15

2 Answers 2

up vote 12 down vote accepted

I'll give you enough hints to complete the proof yourself. If you're desperate, I'm following the notes by Zuber, which are available online, IIRC.

Let's start with some notation: pick some basis $\{t_a\}$ of your Lie algebra, then $$ [t_a,t_b] = C_{ab}{}^c t_c$$ defines the structure constants. If you define $$ g_{ab} = C_{ad}{}^e C_{be}{}^d,$$ then this gives you an inner product $$(X,Y) := g_{ab} x^a y^b, \quad X = x^a t_a \text{ and } Y = y^b t_b.$$ Indeed this "Killing form" is related to the adjoint representation, as $$(X,Y) = \text{tr}(\text{ad } X \text{ ad} Y)$$ (exercise!). Similarly, $$g_{ab} =\text{tr}(\text{ad } t_a \text{ ad } t_b).$$ In this language, the Casimir $c_2$ is given by $$ c_2 = g^{ab} t_a t_b, \qquad \text{ so}$$ $$[c_2,t_e] = g^{ab} [t_a t_b,t_e].$$ Now you need to do some basic work (expand the first factor of the commutator, work out the resulting brackets) and you'll see that this gives you $$ \ldots = g^{ab} g^{dk} C_{bek} \{ t_a,t_d \}.$$ This vanishes (why?), so you're done!

Edit (regarding Peter Kravchuk's remark): when you write $c_2 \sim t_a t_b$, it's not really part of the Lie algebra. The only multiplication that "works" in Lie algebras is the commutator $[t_a,t_b]$. So these guys live in some richer structure, which is called the "universal enveloping algebra." Indeed you often hear that "the Casimir is a multiple of the identity matrix," but the identity matrix is seldom part of the Lie algebra (the identity in a Lie algebra is 0). In practice everything is self-evident, because you do calculations in some vector space.

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I guess it is not of major importance here, but you should say some words about either the universal enveloping algebra, or picking some faithfull representation. The latter (if expected to be finite-dimensional) does not exist in some cases, afaik. I mean that there is no non-Lie product in a Lie algebra beforehand. –  Peter Kravchuk May 5 '13 at 18:53
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Since, as Qmechanic notes, Casimir operators need not be quadratic, it appears that the two definitions in the question are not in fact equivalent. –  Art Brown May 5 '13 at 19:08
    
@ArtBrown: you are correct; in general, higher-order Casimirs can exist, but you need to construct them on a case-by-case basis. The quadratic Casimir is "the" Casimir operator physicists talk about usually. (Formally you can write down some huge expressions involving the $g_{ab}$ and $C_{ab}{}^d$ to define higher Casimirs, but that's not what you do in practice, right?) –  Vibert May 5 '13 at 19:24
    
@Vibert, thanks, but we are talking about "exact definitions". –  Art Brown May 5 '13 at 19:28
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@jinawee: sorry, I should have been more explicit. The index is lowered used the Killing form, i.e. $C_{abe} = C_{ab}{}^d g_{de}$ where $g_{ab}$ is the "metric" defined in my post. –  Vibert May 10 '13 at 11:10

I) The Casimir invariants of a Lie algebra $L$ over a field $\mathbb{F}$ are the central elements of the universal enveloping algebra $U(L)$.

Example: The angular momentum square $\vec{J}^2$ is a quadratic Casimir invariant of the Lie algebra $L=sl(2,\mathbb{C})$.

II) Given a bilinear associative/invariant form $B:L\times L\to \mathbb{F}$, one can create a quadratic Casimir invariant, as explained on this Wikipedia page.

A simple Lie algebra has a unique bilinear associative/invariant form (up to an overall normalization factor), namely the Killing form. As a consequence, a simple Lie algebra has a unique quadratic Casimir invariant (up to an overall normalization factor)

$$C_2 ~:=~ t_a \otimes t^a~\in~ U(L), \qquad t^a~:=~(\kappa^{-1})^{ab} t_b, \qquad \kappa_{ab}~:=~{\rm tr}({\rm ad} t_a\circ{\rm ad} t_b). $$

III) More generally, a semisimple Lie algebra that is built from $m$ simple Lie algebras has a basis a $m$ quadratic Casimir invariant.

Example: The linear combination

$$\alpha_L \vec{J}_{\!L}^2+\alpha_R \vec{J}_{\!R}^2$$

is a quadratic Casimir invariant of the Lie algebra $L=sl(2,\mathbb{C})_L\oplus sl(2,\mathbb{C})_R$ for arbitrary constants $\alpha_L,\alpha_R\in\mathbb{C}$.

IV) There also exist cubic and higher-order Casimir invariants. For a semisimple Lie algebra $L$, e.g.,

$$C_n ~:=~ {\rm str}({\rm ad} t_{a_1}\circ\ldots\circ{\rm ad} t_{a_n}) t^{a_1} \otimes\ldots\otimes t^{a_n}~\in~ U(L),$$

where ${\rm str}$ denotes symmetrized trace. They are not all independent, though.

V) Finally, in response to Art Brown's comment: Racah's theorem states that the number of independent Casimir invariants for a complex semisimple Lie algebra $L$ is equal to the rank of the Lie algebra $L$.

There exist generalizations of Racah's theorem to non-semisimple Lie algebras, see e.g. B.G. Wybourne, Classical Groups for Physicists, 1974, p. 142.

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Nice, concise answer, as usual. (Personally, I'd like more explanation, but...) My venerable Schiff asserts that the number of Casimir invariants is equal to the maximum number of mutually commuting generators, which seems like a useful fact. Can you say how it emerges from your analysis (assuming it's true)? –  Art Brown May 5 '13 at 19:14
    
@ArtBrown: I think there is an intuitive proof for Racah's theorem, but I don't have the time right now to work it out. –  Qmechanic May 5 '13 at 22:36
    
No problem, I was not expecting a proof, intuitive or otherwise. Your edit more than answered the mail and is much appreciated. –  Art Brown May 6 '13 at 0:04
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@jinawee. The (complexified) Lorentz algebra $so(4,\mathbb{C})\cong sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})$ is semisimple; the Poincare algebra is not semisimple. –  Qmechanic May 6 '13 at 15:40
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@jinawee. A Lie algebra $L$ has always a Killing form $\kappa$. The following theorem holds: [$L$ is semisimple] $~\Leftrightarrow~$ [$\kappa$ is a non-degenerate]. –  Qmechanic May 6 '13 at 21:43

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