John Stillwell - Naive Lie Theory.pdf - Index of

68 3 Generalized rotation groups
Case (2): A ∈ Z(SO(2m + 1)).
The argument is very similar to that for Case (1), except for the last
step. The (2m + 1) × (2m + 1) matrix −1 does not belong to SO(2m + 1),
because its determinant equals −1. Hence Z(SO(2m + 1)) = {1}.
Case (3): A ∈ Z(U(n)).
In this case A = Z θ1 ,θ 2 ,...,θ n
for some θ 1 ,θ 2 ,...,θ n and A commutes with
allelementsofU(n). Ifn = 1thenU(n) is isomorphic to the abelian group
S 1 = {e iθ : θ ∈ R}, soU(1) is its own center. If n ≥ 2 we take advantage
of the fact that
( e
iθ 1
0
0 e iθ 2
)
does not commute with
( ) 0 1
1 0
unless e iθ 1
= e iθ 2
. It follows, by building a matrix with ( 01
10)
somewhere
on the diagonal and otherwise only 1s on the diagonal, that A = Z θ1 ,θ 2 ,...,θ n
must have e iθ 1
= e iθ 2
= ···= e iθ n
.
In other words, elements of Z(U(n)) have the form e iθ 1. Conversely,
all matrices of this form are in U(n), and they commute with all other
matrices. Hence
Z(U(n)) = {e iθ 1 : θ ∈ R} = {ω1 : |ω| = 1}.
Case (4): A ∈ Z(SU(n)).
The argument for U(n) shows that A must have the form ω1, where
|ω| = 1. But in SU(n) we must also have
1 = det(A)=ω n .
This means that ω is one of the n “roots of unity”
e 2iπ/n , e 4iπ/n , ..., e 2(n−1)π/n , 1.
All such matrices ω1 clearly belong to SU(n) and commute with everything,
hence Z(SU(n)) = {ω1 : ω n = 1}.
Case (5): A ∈ Z(Sp(n)).
In this case A = Q θ1 ,θ 2 ,...,θ n
for some θ 1 ,θ 2 ,...,θ n and A commutes
with all elements of Sp(n). The argument used for U(n) applies, up to the
point of showing that all matrices in Z(Sp(n)) have the form q1, where
|q| = 1. But now we must bear in mind that quaternions q do not generally
commute. Indeed, only the real quaternions commute with all the others,
and the only real quaternions q with |q| = 1areq = 1andq = −1. Thus
Z(Sp(n)) = {±1}.
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