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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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  • I am very new to this subject. The exercise was to see how the block diagonal form would look in a character table. That said, I (naively) assumed that one can construct the table in the same way that one does for point groups. Clearly that is not the case here. Would it be true to say that character tables are not valid in the context of Lie groups and tensor product representations? – user1884367 Apr 20 '23 at 04:50
  • The linked question and Chapter 16 of J Mathews & R Walker, Methods of Mathematical Physics (Addison-Wesley) tell you how to extend to "infinite" character tables, with continuous class parameters (here θ) versus representations (here r) obeying an orthonormality relation integrated over the Haar measure, (in lieu of summing over classes)... Above, $J_3$=diag(1,0,-1). – Cosmas Zachos Apr 20 '23 at 07:30
  • Also, Creutz's book, Ch 8, has al the right formulas in usable form... – Cosmas Zachos Apr 20 '23 at 13:44
  • I haven't yet fully absorbed the math but I think what you are saying is that when one moves to continuous groups, integrals of the aforementioned type become the analog of summations and character tables for discrete groups. Correct? – user1884367 Apr 26 '23 at 00:25
  • Yes. That's it. summing over all equivalence classes corresponds to integrals over the parameters characterizing them, here θ. You may check the above formulas explicitly. – Cosmas Zachos Apr 26 '23 at 00:31