Math 419H, Spring 2017 -- Hints for HW 7A, Artin Ch 10.9, p 312 
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Ex 10.9.2. Question: The SU(2) operation on a polynomial f(u,v) is defined by substituting the matrix product (u,v)P for the input vector (u,v). But then Q[Pf] substitutes (u,v)PQ, whereas [(QP)f] substitutes (u,v)(QP), and these fail to be equal unless P and Q commute.

Answer: Actually, it does work correctly as an action on _functions_ f(u,v) rather than on vectors (u,v).  Let (Pf)(u,v) = f((u,v)P), matrix multiplying (u,v)P. Then:

   ((QP)f)(u,v) = f((u,v)QP)

   (Q(Pf))(u,v) = (Pf)((u,v)Q) = f(((u,v)Q)P) = f((u,v)QP).

You have to be extra careful with the definitions.

The other way to make this work: define (Pf)(u,v) = f(P^(-1)(u,v)).
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Ex 10.9.3. By dimension, the 3-dim rep of SU(2) must be 3(chi_0) or ch_0 + chi_1, or chi_2. To refute the former two, we need to show it is irreducible in GL_3(C), not just in GL_3(R), since the classification theorem 10.9.7 is for irreducible complex reps.

This is easily done by computing the character of the rep. Since the diagonal matrices give representatives for all conjugacy classes, you only need the trace of each diagonal SU(2) matrix (i.e. x_2 = x_3 = 0) acting on C^3, and see that it matches the desired irreducible.