Can we simply comment that the $\text{Tr}[T^a_rT^b_r]\equiv C(r) \delta^{ab}$ depends on the representation. For the case of SU(2) and SO(3), we can relate this to the spin-S representation. By the manner that for SU(2) group is in a spin 1/2 representation and SO(3) group is in a spin 1 representation. One can write down the relation of spin operators as:
$$
S_x^2+S_y^2+S_z^2=S(S+1) \;\mathbb{I}_{2s+1}.
$$
($\hbar=1$).
And
$$
\sum_a(S^a)^2=S_x^2+S_y^2+S_z^2=\sum_{a=x,y,z}(T^a)^2=3 (T^b)^2
$$
here $b$ can be $x,y,z$.
So, combine the two relations above:
$$
\frac{1}{2}\text{Tr}[T^a_rT^b_r]=\frac{1}{2}\frac{S(S+1)}{3}\text{Tr}[\mathbb{I}_{2s+1}]=\frac{S(S+1)(2S+1)}{6}
$$
For SU(2), spin-1/2 representation, we have:
$$
\text{Tr}[T^a_rT^b_r]=2\frac{S(S+1)(2S+1)}{6}|_{S=1/2}=1/2
$$
For SO(3), spin-1 representation, we have:
$$
\text{Tr}[T^a_rT^b_r]=2\frac{S(S+1)(2S+1)}{6}|_{S=1}=2
$$.

For spin-3/2 representation, we have:
$$
\text{Tr}[T^a_rT^b_r]=2\frac{S(S+1)(2S+1)}{6}|_{S=3/2}=5,
$$
etc.
Shall we say the level k quantization of SU(2) C-S and SO(3) C-S are thus related by a factor of: $$(1/2)/2=1/4.$$

And this quantization value presumably is a measurable quantized value for the **spin-Hall conductance**. See for example, the discussion in this paper: Symmetry-protected topological phases with charge and spin symmetries: response theory and dynamical gauge theory in 2D, 3D and the surface of 3D: arXiv-1306.3695v2, in Eq(26) and its p.7 right column and in p.8 left column. See also this Phys Rev B paper.

This way of interpretation simplifies Dijkgraaf-Witten or Moore-Seiberg's mathematical argument to a very physical level of the spin $S$ property. Would you agree?

Any further thoughts/comments?

This post imported from StackExchange Physics at 2014-04-04 16:26 (UCT), posted by SE-user Idear