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I have derived the four 4 x 4 block diagonal matrices (e,A,B,C) in accordance with the transformation TST-1 where T is the C-G matrix for the 1/2 ⊗ 1/2 = 1 ⊕ 0 representation. This part is described here. I now want to construct the character table for this. So far I get :

    e  A  B  C
Γ1  3  0  0  0  (3 x 3)
Γ2  1  0  0  0  (1 x 1)

But this doesn't seem to be correct since it doesn't satisfy the rules for characters (Great Orthogonality Theorem, etc.). What am I missing?

user1884367
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1 Answers1

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1/2 ⊗ 1/2 = 1 ⊕ 0 is visibly a reducible representation. I'm not sure how you got your character table, nor what it would mean. You can check the obvious directly.

Since any rotation in any rep can be rotated to $J_3$, there is only one family of classes in (Lie!) SU(2), characterized by the azimuth angles θ . The generic group element of your rep is $$ e^{i\theta ~\hat n \cdot \vec J} \oplus 1, $$ so, inside the trace, you may merely consider $$ e^{i\theta \operatorname{diag}(1,0,-1) } \oplus 1, $$ tracing which yields the character $_0+ _1$,
$$ \chi_0=1, \qquad \chi_1= 1+2\cos\theta = {\sin 3\theta/2 \over \sin \theta/2}, \implies \\ _0+ _1= 2(1+\cos\theta)\\ =(2\cos\theta/2)^2 =\chi_{1/2}^2~, $$ as expected.

Addendum in response to comment

See this question.

The character table for finite groups generalizes to an "infinite matrix" characters, $\chi_r (U)$, where r denotes the irrep, and U, the group element, as betokened by its class parameters, here merely θ. The sum over classes is replaced by the group invariant integral with the Haar measure, in this parameterization $\int [dU]=\int^{4\pi} _ 0 \!\! d\theta ~~{\sin^2\theta/2 \over 2\pi}$.

The character of a Kronecker product amounts to the product of the characters of the tensor factors, whilst the character of a direct sum to the sum of the respective characters. You might think of the Fourier expansion as you manipulate and analyze expressions, project, etc, using orthonormality, $$ \int [dU] \chi_r^*(U) \chi_s (U)= \delta_{rs}. $$

Check this for the above formulas in your particular case! If all goes well, also verify $$ \chi_{1/2}(\theta) \cdot \chi_{1}(\theta) = \chi_{3/2}(\theta) +\chi_{1/2}(\theta) . $$ Always check the trivial dimensionalities, $\chi_{r}(0) = 2r+1$, the first row (or column) of the "character table".

Cosmas Zachos
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