# Does the Casimir depend on normalization of generators?

This question is motivated by something in physics: the area operator in loop quantum gravity is given by the Casimir of $SU(2)$, that is $j(j+1)$ for a dimension $2j+1$ representation of $SU(2)$. I will cross-list if this doesn't work but it's really a question in representation theory.

So our generators $T^a$ of a Lie algebra $\mathfrak{g}$ satisfy

$Tr(T^aT^b)=k_D\delta^{ab}$

for a representation $D$ of $\mathfrak{g}$, where $k_D$ is the index. Then the Casimir $C$ is defined as

$\sum_aT^aT^a=C_D\times 1$

with $1$ being the identity matrix; this thing is a distinguished element of the universal enveloping algebra $\mathcal{U}(\mathfrak{g})$. In physics, we say for $SU(2)$ that $k_D=1/2$, and find that the Casimir for a representation $j$ is $j(j+1)$. It appears to me that this answer depends on the choice for the index, which would depend on how we chose our generators (scaling, for instance). Is that true? Is there any greatly compelling reason to choose $k_D=1/2$ here, except that is gives the right answer on spins?

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## 1 Answer

I believe there's no question of scaling here because the sum is actually over the products of basis elements with elements of the dual basis with respect to the Killing form, so if you scale the basis, you scale the dual basis inversely and the Casimir invariant is, as the name suggests, invariant.

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Great sounds good; I guess we can scale the Killing form though right? I mean, the Killing form is only unique up to scale, so choosing a different inner product would be choosing a different set of Casimir invariants? –  levitopher Mar 19 '12 at 18:08
@cduston: I'm not sure. You still get an invariant symmetric bilinear form if you scale the Killing form, but at least according to the Wikipedia definition the Killing form is uniquely defined in terms of the Lie bracket and the trace, so as far as I can tell it's distinguished from its scalar multiples. –  joriki Mar 19 '12 at 18:55
I think you are right. I was thinking of the discussion from Fulton and Harris, but actually they say "the Killing form can be computed [in terms of the Lie bracket and the trace] or (up to scalars) by using it's invariance under the automorphism group of $\mathfrak{g}$." So, using $B(X,Y)=Tr(ad(X)ad(Y))$ as the Killing form is unique. Thanks for all this! –  levitopher Mar 19 '12 at 20:03
@cduston: You're welcome! –  joriki Mar 20 '12 at 1:04