John Stillwell - Naive Lie Theory.pdf - Index of

7.5 SO(n), SU(n), and Sp(n) revisited 151
Then e tX Be −tX B −1 ∈ H as well, so C(t) is in fact a smooth path in H and
C ′ (0)=X − BXB −1 ∈ T 1 (H).
Thus to prove that T 1 (H) ≠ {0} it suffices to show that X − BXB −1 ≠ 0.
Well,
X − BXB −1 = 0 ⇒ BXB −1 = X
⇒ e BXB−1 = e X
⇒ Be X B −1 = e X
⇒ Be X = e X B
⇒ BA = AB,
contrary to our choice of A and B.
This contradiction proves that T 1 (H) ≠ {0}.
□
Corollary. If H is a nontrivial normal subgroup of G under the conditions
above, then T 1 (H) is a nontrivial ideal of T 1 (G).
Proof. We know from Section 6.1 that T 1 (H) is an ideal of T 1 (G), and
T 1 (H) ≠ {0} by the theorem.
If T 1 (H) =T 1 (G) then H fills N δ (1) in G, by the log-exp bijection
between neighborhoods of the identity in G and T 1 (G). But then H = G
because G is path-connected and hence generated by N δ (1). Thus if H ≠ G,
then T 1 (H) ≠ T 1 (G).
□
It follows from the theorem that any nondiscrete normal subgroup H
of G = SO(n),SU(n),Sp(n) gives a nonzero ideal T 1 (H) in T 1 (G). The
corollary says that T 1 (H) is nontrivial, that is, T 1 (H) ≠ T 1 (G) if H ≠ G.
Thus we finally know for sure that the only nontrivial normal subgroups of
SO(n), SU(n), andSp(n) are the subgroups of their centers. (And hence
all the nontrivial normal subgroups are finite cyclic groups.)
SO(3) revisited
In Section 2.3 we showed that SO(3) is simple—the result that launched
our whole investigation of Lie groups—by a somewhat tricky geometric
argument. We can now give a proof based on the easier facts that the center
of SO(3) is trivial, which was proved in Section 3.5 (also in Exercises 3.5.4
and 3.5.5), and that so(3) is simple, which was proved in Section 6.1. The
hard work can be done by general theorems.