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I know that a Casimir for a Lie algebra $\mathfrak{g}$ is a central element of the universal enveloping algebra. For example in $\mathfrak{so}(3)$ the generators are the angular momentum operators $J_1,J_2,J_3$ and a quadratic Casimir is $\mathbf{J}^2$.

Consider an irreducible representation $d$ of $\mathfrak{g}$ on $V$. By Schur's Lemma any Casimir $C$ is mapped to a matrix with a single eigenvalue $C_d$. If $d$ and $e$ are equivalent representations then $C_d=C_e$ since change of basis preserves eigenvalues.

Under what circumstances are there sufficiently many Casimirs that we get a converse to this statement? In other words when can we find enough Casimirs that $\{C_d,D_d,\dots F_d\}$ uniquely labels irreducible representations? I can't work out how many we would need in general, or really how to prove this!

Many thanks in advance.

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Same question at Math.SE: math.stackexchange.com/q/288816/11127 –  Qmechanic May 17 '13 at 14:34

2 Answers 2

up vote 9 down vote accepted

Here we will only discuss the case of finite-dimensional irreducible representations (irreps) of a complex semisimple Lie algebra $L$.

Recall that the set $Z$ of Casimir invariants is the center $Z(U(L))$ of the universal enveloping algebra $U(L)$, cf. e.g. this Phys.SE post.

OP's question is answered without proof on p. 253 in Ref. 1:

Theorem 2. For every semisimple Lie algebra $L$ of rank $r$, there exists a set of $r$ invariant polynomial of generator $t_a$, whose eigenvalues characterize the finite-dimensional irreducible representations.

Ref. 2 (which is one the most important books on Lie algebras, at least if one is interested in the proofs) does not bother to mention Theorem 2 explicitly. However, it is possible to string together a set of more fundamental results (and their proofs) from Ref. 2 to get the sought-for result. We outline the proof strategy below.

Recall furthermore that there is associated a root system $\Phi$ to the Lie algebra $L$, and let us imagine that we have picked a base $\Delta$ for $\Phi$. The order $|W|$ of the Weyl group $W$ is equal to the possible choices of (unordered) bases and equal to the possible choices of (fundamental) Weyl chambers.

It is proven in chapters 20-21 of Ref. 2. that a finite-dimensional irrep has a unique highest weight vector (unique up to normalization) with some dominant integral weight $\lambda$. We will from now on denote such irrep $V(\lambda)$. (Ref. 2. also defines a notion of a highest weight irrep $V(\lambda)$ when $\lambda$ is integral but not dominant. Such irreps are necessarily infinite-dimensional, so we will ignore those.) It follows that

Two irreps $V(\lambda)$ and $V(\mu)$ are equivalent (i.e. isomorphic) iff their highest weights are equal $\lambda=\mu$.

As a consequence of Harish-Chandra's theorem, the set $Z$ of Casimirs takes the same value on two highest weight irreps $V(\lambda)$ and $V(\mu)$ iff $\lambda+\delta$ and $\mu+\delta$ belong to the same Weyl orbit,

$$ \sigma(\lambda+\delta)~=~\mu+\delta, \qquad \sigma \in W. $$

Here $\delta$ is half the sum of the positive roots. However if both integral weights $\lambda$ and $\mu$ are dominant, then $\lambda+\delta$ and $\mu+\delta$ must both belong to (the interior of) the fundamental Weyl chamber, so that the Weyl reflection $\sigma={\bf 1}$ must be the identity element. In conclusion, we get that

The set $Z$ of Casimirs takes the same value on two finite-dimensional irreps $V(\lambda)$ and $V(\mu)$ iff their highest weights are equal $\lambda=\mu$.

Harish-Chandra's theorem is proven in chapter 23 of Ref. 2. See also this and this related Math.SE posts.

Example: Consider the Lie algebra $L=sl(3,\mathbb{C})$. The Weyl group is $S_3$. The Lie algebra $L$ has two independent Casimir invariants $C_2$ and $C_3$,

$$C_n ~:=~ {\rm str}({\rm ad} t_{a_1}\circ\ldots\circ{\rm ad} t_{a_n}) t^{a_1} \otimes\ldots\otimes t^{a_n}, \qquad n~\in~ \{2,3\}.$$

Consider the 3-dimensional fundamental representation $F$ and the dual/contragredient representation $\bar{F}$ of $L$, which are non-equivalent irreps. They have highest weights $\lambda=(1,0)$ and $\mu=(0,1)$, respectively. In detail, if $t_a$, $a=1, \ldots, 8$ are generators for $L=sl(3,\mathbb{C})$, then (hattip: Peter Kravchuk)

$$\bar{F}(t_a)~=~ -F(t_a)^t,$$

so that the Casimirs $C_2$ (and $C_3$) take the same (opposite) value on $F$ and $\bar{F}$

$$ {\rm tr}_{\bar{F}}\bar{F}(C_n)~=~(-1)^n{\rm tr}_{F}F(C_n), \qquad n~\in~ \{2,3\}. $$

One may prove that the values are non-zero, so that the Casimirs $C_2$ and $C_3$ distinguish between the two non-equivalent irreps $F$ and $\bar{F}$, as they should.

References:

  1. A. O. Barut and R. Raczka, Theory of group representations and applications, 2nd ed., 1980.

  2. J.E. Humphreys, Introduction to Lie Algebras and Representation Theory, (1980).

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Thanks - that's marvellous. I think I've got a much better intuition for it now! –  Edward Hughes May 17 '13 at 22:05
    
@Qmechanic, do you have any reference for the fact that $C_3$ is the same in $F$ and $\bar{F}$? It is just not obvious for me, since we get the dual by mapping $\bar{F}(g)=-F(g)^T$, and IIRC $C_3$ is cubic in generators as well as diagonal, so $\bar{F}(C_3)=(-1)^3F(C_3)^T=-F(C_3)$. Or is it just zero in both cases? –  Peter Kravchuk May 20 '13 at 18:32
    
@Qmechanic, I have found some formulae for the values of $C_2$ and $C_3$ in representation with weight $(m_1,m_2)$: $c_2=\frac{2}{3}(m_1^2+m_2^2+m_1 m_2+3m_1+3m_2)$ and $c_3=\frac{1}{9}(m_1-m_2)(2m_1+m_2+3)(2m_2+m_1+3)$ which lead to $F(C_3)=\frac{20}{9}$ and $\bar{F}(C_3)=-\frac{20}{9}$. It is from Lyahovsky-Bolokhov, but I doubt that there is an English translation. My claim in the deleted answer is also from this book (well, I was not precise, the statement is that for simple algebras Casimirs can distinguish any two nonequivalent reps). I was yet unable to find their reference though. –  Peter Kravchuk May 20 '13 at 18:57
    
@Peter Kravchuk. Thanks. I accidentally first drew the three Weyl reflection axes of $L=sl(3,\mathbb{C})$ wrong (tilded by $30^{\circ}$), which somehow lead me to believe a string of wrong conclusions. –  Qmechanic May 20 '13 at 21:56
    
@Qmechanic, yeah, I've done the same mistake, but eventually noticed it.) –  Peter Kravchuk May 20 '13 at 21:58

It can be proven (Racah's theorem) that the number of Casimir operators is the same as the rank of the algebra (number of simultaneosly conmuting generators). This is at least true for semi simple algebras.

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