witheiθgoing aroundU(1) once asθgoes from 0 to 2π, this means we can
choose a basis ofV so that
π(ei^2 θS^3 ) =
eiθq^10 ··· 0
0 eiθq^2 ··· 0
··· ···
0 0 ··· eiθqm
Taking the derivative of this representation to get a Lie algebra representation,
using
π′(X) =
d
dθπ(eθX)|θ=0we find forX=i 2 S 3
π′(i 2 S 3 ) =d
dθ
eiθq^10 ··· 0
0 eiθq^2 ··· 0
··· ···
0 0 ··· eiθqm
|θ=0=
iq 1 0 ··· 0
0 iq 2 ··· 0
··· ···
0 0 ··· iqm
Recall thatπ′is a real-linear map from a real vector space (su(2) =R^3 ) to
another real vector space (u(n), the skew-Hermitianmbymcomplex matrices).
As discussed in section 5.5, we can use complex linearity to extend any such
map to a complex-linear map fromsu(2)C (the complexification ofsu(2)) to
u(m)C(the complexification ofu(m)). Sincesu(2)∩isu(2) = 0 and any element
ofsl(2,C) be written as a complex number times an element ofsu(2), we have
su(2)C=su(2) +isu(2) =sl(2,C)Similarly
u(m)C=u(m) +iu(m) =M(m,C) =gl(m,C)
As an example, multiplyingX=i 2 S 3 ∈su(2) by− 2 i, we haveS 3 ∈sl(2,C)
and the diagonal elements in the matrixπ′(i 2 S 3 ) get also multiplied by− 2 i
(sinceπ′is now a complex-linear map), giving
π′(S 3 ) =
q 1
2 0 ···^0
0 q 22 ··· 0
··· ···
0 0 ··· q 2 m
We see thatπ′(S 3 ) will have half-integral eigenvalues, and make the following
definitions:
Definition(Weights and weight spaces).Ifπ′(S 3 )has an eigenvaluek 2 , we say
thatkis a weight of the representation(π,V).
The subspaceVk⊂V of the representationVsatisfying
v∈Vk =⇒ π′(S 3 )v=k
2v