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!

  • 1
    $\begingroup$ Where did you hear that? ''The Casimir operator'' does depend on the choice of the scalar product of the Lie Algebra, while the ''Chern Character'' should be some topological invariant of a bundle. $\endgroup$ – Creo Nov 2 '18 at 17:22
  • $\begingroup$ Related. $\endgroup$ – Cosmas Zachos Nov 4 '18 at 14:42

I'd doubt you are talking about Chern characters. 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 character
$$ \chi (e^{i\theta^a T^a})=\operatorname {tr} (e^{i\theta^a T^a})=\operatorname {tr} (1\!\! 1+ i\theta \cdot T -\frac{\theta^a \theta^b}{2} T^a T^b+...)= d(r) - \frac{\theta\cdot \theta}{2} ~ T(r) +... $$ 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 minion 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 $c(6)=10/3$.


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