Proof.Recall that if we diagonalize a unitary matrix, the diagonal entries are
the eigenvalues, but their order is undetermined: acting by permutations on
these eigenvalues we get different diagonalizations of the same matrix. In the
case ofSU(2) the matrix
P=(
0 1
−1 0
)
has the property that conjugation by it permutes the diagonal elements, in
particular
P(
eiθ 0
0 e−iθ)
P−^1 =
(
e−iθ 0
0 eiθ)
So
π(P)π((
eiθ 0
0 e−iθ)
)π(P)−^1 =π((
e−iθ 0
0 eiθ)
)
and we see thatπ(P) gives a change of basis ofVsuch that the representation
matrices on theU(1) subgroup are as before, withθ→ −θ. Changingθ→ −θ
in the representation matrices is equivalent to changing the sign of the weights
qj. The elements of the set{qj}are independent of the basis, so the additional
symmetry under sign change implies that for each non-zero element in the set
there is another one with the opposite sign.
Looking at our three examples so far, we see that, restricted toU(1), the
scalar or spin 0 representation of course is one dimensional and of weight 0
(π 0 ,C) =C 0and the spin^12 representation decomposes intoU(1) irreducibles of weights
− 1 ,+1:
(π 1 ,C^2 ) =C− 1 ⊕C+1
For the spin 1 representation, recall (theorem 6.1) that the double cover
homomorphism Φ takes
(
eiθ 0
0 e−iθ)
∈SU(2)→
cos 2θ sin 2θ 0
−sin 2θ cos 2θ 0
0 0 1
∈SO(3)
Acting with theSO(3) matrix above onC^3 will give a unitary transformation
ofC^3 , which therefore is in the groupU(3). One can show that the upper left
diagonal 2 by 2 block acts onC^2 with weights− 2 ,+2, whereas the bottom right
element acts trivially on the remaining part ofC^3 , which is a one dimensional
representation of weight 0. So, restricted toU(1), the spin 1 representation
decomposes as
(π 2 ,C^3 ) =C− 2 ⊕C 0 ⊕C+2
Recall that the spin 1 representation ofSU(2) is often called the “vector” rep-
resentation, since it factors in this way through the representation ofSO(3) by
rotations on three dimensional vectors.