Page 140 - Matrices theory and applications

7.5. The Orthogonal Groups O(p, q)
Proposition 7.4.1 The unitary group U(p, q) is homeomorphic to U p ×
2pq
.In particular, U(p, q) is connected.
U q × IR
There remains to show connectivity. It is a straightforward consequence of
the following lemma.
Lemma 7.4.1 The unitary group U n is connected. 123
Since GL n (CC) is homeomorphic to U n ×HPD n (via polar decomposition),
hence to U n × H n (via the exponential), it is equivalent to the following
statement.
Lemma 7.4.2 The linear group GL n (CC) is connected.
Proof
Let M ∈ GL n (CC) be given. Define A := CC \{(1 − λ) −1 |λ ∈ Sp(M)}.
The arcwise-connected set A does not contain the origin, nor the point
z =1, since 0 ∈ Sp(M). Therethusexistsa path γ joining0to1in A:
γ ∈C([0, 1]; A), γ(0) = 0 and γ(1) = 1. Let us define M(t):= γ(t)M +(1−
γ(t))I n .By construction, M(t) is invertible for every t,and M(0) = I n ,
M(1) = M. The connected component of I n is thus all of GL n (CC).

7.5 The Orthogonal Groups O(p, q)

The analysis of the maximal compact subgroup and of G∩H n for the group
O(p, q) is identical to that in the previous paragraph. On the one hand,
O(p, q) ∩ O n is isomorphic to O p × O q . On the other hand, G∩ H n is
isomorphic to M p×q (IR), which is of dimension d = pq.

Proposition 7.5.1 Let n ≥ 1. The group O(p, q) is homeomorphic to O p ×
pq
O q ×IR . The number of its connected components is two if p or q is zero,
four otherwise.

Proof
We must show that O n has two connected components. However, O n is
the disjoint union of SO n (matrices of determinant +1) and of O (matri-
−
n
−
−
ces of determinant −1). Since O = M · SO n for any matrix M ∈ O (for
n
n
example a hyperplane symmetry), there remains to show that the special
orthogonal group SO n is connected, in fact arcwise connected. We use the
following property: