Character expansion and Casimir

Question

Is there a simple way to extract the quadratic Casimir of a representation from the character? I keep hearing things such as "Chern characters have an expansion that goes like" $$\chi(r) = dim(r) + \text{something}\; C(r)+\dots$$ (with $C(r)$ the Casimir) but I haven't been able to find a reference so far. Any help much appreciated!

Answer 1

Are you thinking of the Chern character generating funtion formula $$ {\rm ch}[F]= {\rm dim}(R)+\frac 1 2 {\rm tr} \left\{\frac{F^2}{4\pi^2}\right\}+\ldots $$ where $$ F=\frac 12 \lambda_a F^a_{\mu\nu} dx^\mu\wedge dx^\nu $$ is the curvature tensor with the $\lambda_a$ in the representation $R$?
Here we are going to have a representation dependent coefficient involving ${\rm tr}\{\lambda_a\lambda_b\}=2x_R \delta_{ab}$ where $x_R$ is the Dynkin index of the representation in which the $\lambda_a$ live. There is a standard formula $$ x_R =\frac{{\rm dim}(R)(\lambda, \lambda+\rho)}{2 {\rm dim} g}= \frac{{\rm dim}(R)C_2(R)}{2 {\rm dim} g} $$ where $\lambda$ is the highest weight in $R$ and $\rho$ is the Weyl vector and $C_2=(\lambda, \lambda+\rho)$ is the quadratic Casimir. I do not know how $x_R$ is encoded in the character table though. Chern characters have nothing to do with group chaacters

Answer 2

I'd doubt you are really talking about Chern characters per se. Your basic QFT book should review elementary Lie Group theory for you.

A group element in representation r of dimension d(r) is an r×r matrix with a trace given by the group character,
$$ \chi (e^{i\theta^a T^a})=\operatorname{tr} ~(e^{i\theta^a T^a})=\operatorname {tr} ~\left (1\!\! 1+ i\theta \cdot T -\frac{\theta^a \theta^b}{2} T^a T^b+...\right )= d(r) - \frac{\theta\cdot \theta}{2} ~ T(r) +... $$ The generators are of course traceless, and have a characteristic index T(r), a real number, for every representation, s.t. $$ \operatorname{tr} ~(T^a T^b)= T(r) \delta^{ab}. $$

The quadratic Casimir invariant characteristic of each representation is also an r×r matrix proportional to the identity, $$ C_r= T^a T^a =c(r) 1\!\! 1, $$ with real eigenvalue c(r).

Tracing it, then, yields $$ \operatorname {tr} C_r= T(r) D =d(r) c(r), $$ where D is the dimension of the algebra, the number of independent generators. Thus, if you have the index of the representation and the most general character of the generic group element, taking $-\frac{\partial^2}{\partial \theta^2}$ of it and evaluating at θ=0, yields $$ T(r) = \frac{d(r)}{D} ~~ c(r). $$

So, for example, knowing that the index of the doublet representation of SU(2), i.e. spin 1/2, the fundamental one, is T(2)=1/2, and recalling D=3, you have $$ \frac{1}{2}=\frac {2}{3} c(2) \qquad \Longrightarrow \qquad c(2) = \frac {3}{4}, $$ which is what you would also find directly from $\frac{\vec \sigma}{2}\cdot \frac{\vec \sigma}{2} $ .

For the triplet, spin 1, representation, the index is 2, so the Casimir eigenvalue is also 2.

Similarly, for SU(3), D=8, knowing that for the sextuplet representation $T(6)=5/2$, yields $c(6)=10/3$.