A topological group is a set G on which a group structure and a topological structure coexist: this means that the product
G× G → G, (g, h) → gh,
and inversion
G→ G, g → g−1
are continuous operations. For any given h ∈ G, the map
Rh: G → G defined as Rh(g) = gh
is the composite of the product in G with the inclusion G → G × G, g → (g, h). Hence Rh, called right multiplication by h, is continuous as composite of continuous
functions. Furthermore, Rhis bijective with inverse Rh−1, hence a homeomorphism.
Lh: G → G defined by Lh(g) = hg
is a homeomorphism with inverse Lh−1.
Lemma 4.53 For anyg and h in a topological group G there exists a homeomor-
phismϕ from G to G such that ϕ(g) = h.
Proof Consider ϕ= Rg−1h, or ϕ= Lhg−1.
Example 4.54 Every group endowed with the discrete topology becomes a topolog-
ical group.
Example 4.55 The additive groups(Rn, +) and (Cn, +), with the Euclidean topol-
ogy, are topological groups.
Example 4.56 The group GL(n, R) is open in the space Rn2and as such it admits a
natural topological structure, with which it becomes a topological group. The entries of the product of two matrices A, B depend in a continuous way on the entries of A and B, and the entries of A−1depend continuously on those of A.
Lemma 4.57 Let G be a topological group with neutral element e. Then G is Haus-
dorff if and only if{e} is closed.
Proof One implication is clear, since points are closed in Hausdorff spaces.
Now suppose{e} is closed in G and consider the map φ: G × G → G, φ(g, h) =
gh−1, composite of continuous maps hence itself continuous. Since φ(g, h) = e iff
g = h, the pre-image of {e} is precisely the diagonal, which is closed. To finish we
invoke Theorem3.69.
Let’s now overview some features of the main linear groups: GL(n, R) = {A ∈ Mn,n(R) | det(A) = 0}. SL(n, R) = {A ∈ Mn,n(R) | det(A) = 1}. SO(n, R) = {A ∈ Mn,n(R) | AAT = I, det(A) = 1}. GL(n, C) = {A ∈ Mn,n(C) | det(A) = 0}. SL(n, C) = {A ∈ Mn,n(C) | det(A) = 1}. U(n, C) = {A ∈ Mn,n(C) | ATA= I }. SU(n, C) = U(n, C) ∩ SL(n, C).
They’re all subspaces ofRn2 orCn2, in particular they are metrisable and so Haus- dorff.
Proposition 4.58 The groups GL+(n, R) = {A ∈ GL(n, R) | det(A) > 0} and
4.6 Topological Groups 81
Proof We shall prove only that GL+(n, R) is connected, for GL(n, C) is completely
analogous.
The group GL+(1, R) coincides with the interval ]0, +∞[ and is connected; by induction we can suppose n > 1 and GL+(n − 1, R) connected. Consider the map
p: Mn,n(R) → Rnsending a matrix to its first column vector. If we isolate the first
column from the others we may write Mn,n(R) = Rn × Mn,n−1(R), and then p
is the projection onto the first factor of the product. This makes p continuous and open, and so its restriction to the open set GL+(n, R) ⊂ Mn,n(R) is open; moreover,
p(GL+(n, R)) = Rn− {0}.
We claim that the fibres of p: GL+(n, R) → Rn− {0} are connected: from this plus Lemma 4.18 the connectedness of GL+(n, R) will follow. The fibre
p−1(1, 0, . . . , 0) is the product Rn−1×GL+(n −1, R), so connected. Given any y ∈
Rn−{0} the fibre p−1(y) is non-empty, so let A ∈ GL+(n, R) be such that p(A) = y.
Left-multiplication by A, LA(B) = AB, is a homeomorphism of GL+(n, R). From the relation p(AB) = Ap(B) follows LA
p−1(1, 0, . . . , 0) = p−1(y), so the fibres
of p are all homeomorphic.
Corollary 4.59 The topological groups SL(n, R) and SL(n, C) are connected,
while GL(n, R) has two connected components.
Proof We just saw that GL(n, C) and GL+(n, R) are connected. Dividing the
first column vector by the determinant yields maps GL(n, C) → SL(n, C) and GL+(n, R) → SL(n, R), both continuous and onto.
The group GL(n, R) is the disjoint union of the open sets GL+(n, R) and GL−(n, R) = det−1(] − ∞, 0[), and multiplying by any matrix with negative deter- minant induces a homeomorphism from GL+(n, R) to GL−(n, R). For the study of the orthogonal and unitary groups we’ll need a preliminary result.
Lemma 4.60 Let f: X → Y be an onto, continuous map from X compact to Y
connected and Hausdorff. If all fibres f−1(y) are connected, y ∈ Y , then X is connected.
Proof The map f is closed so we can apply Lemma4.18.
Proposition 4.61 The topological groups SO(n, R), U(n, C), SU(n, C) are com-
pact and connected.
Proof We will only discuss SO(n, R), because the cases U(n, C) and SU(n, C) are
entirely similar. Consider the continuous mapping
Mn,n(R) → Mn,n(R) × R, A→ (AAT, det(A));
SO(n, R) is the pre-image of (I, 1) and hence closed in Mn,n(R). Since the column vectors of an orthogonal matrix have norm 1, SO(n, R) is contained in the bounded set{(ai j) |
The map p: SO(n, R) → Sn−1 sending a matrix to its first column vec- tor is continuous. Moreover, p(AB) = Ap(B) for any A, B ∈ SO(n, R), and
LA(p−1(v)) = p−1(Av), which implies that the fibres are all homeomorphic. The
fibre over(1, 0, . . . , 0) is SO(n −1, R). Since SO(n, R) is compact and Sn−1is con-
nected Hausdorff, connectedness is proven by induction on n, keeping into account Lemma4.60and the fact that SO(1, R) = {1}.
Exercises 4.41 Prove that
SU(2, C) × U(1, C) → U(2, C), (A, z) → A ·
z 0
0 1
is a homeomorphism.
4.42 Prove that SL(n, R) and SL(n, C) are not compact, for any n > 1. 4.43 (♥) Prove that for any n, m ≥ 2 the space
X = {A ∈ Mn,n(R) | Am = I }
is disconnected.
4.44 Let A, B ⊂ Rnbe closed sets and call
W(A, B) = {g ∈ GL(n, R) | g(A) ∩ B = ∅}.
Suppose A is compact and prove that W(A, B) is closed in GL(n, R). Show— by finding an example—that W(A, B) might not be closed if neither A nor B are compact.
4.45 Let G be a topological group and H ⊂ G a subgroup. Prove that H is open and
closed in G provided the interior H◦is not empty. (Hint: write H and its complement as union of open sets of the form Rg(H◦) for suitable g ∈ G.)
4.46 Let G be a topological group with neutral element e. Prove the following facts:
1. if H⊂ G is a subgroup, then H is a subgroup;
2. the connected component of e in G is a closed subgroup.
4.47 Let H be a closed, connected subgroup in(Rn, +). Prove H is a vector subspace
ofRn. (Hint: show H= ∩nHn, where Hnis the subspace generated by B(0, 1/n) ∩ H .)
4.48 Let G be a topological group. For subsets A, B ⊂ G define
A−1= {a−1| a ∈ A}, A B= {ab | a ∈ A, b ∈ B}.
Prove that A= ∩AU = ∩AU−1, where U varies among all neighbourhoods of the neutral element.
4.6 Topological Groups 83 4.49 Let G be a Hausdorff topological group and H ⊂ G a discrete subgroup (the
points in H are open in the subspace topology). Prove that H is closed in G and compare this fact with Exercise3.43. (Hint: use Exercise4.48; show that there’s a neighbourhood U of the neutral element e such that U U−1∩ H = {e}, and also that for every x∈ HU there’s a unique h ∈ H such that x ∈ hU.)
4.50 Prove that:
1. any discrete subgroup of the real additive group(R, +) has the form Za for some
a∈ R;
2. any discrete subgroup of U(1, C) is finite cyclic.