# 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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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