Riemannian Geometry – Lecture 18 Homogeneous spaces
Dr. Emma Carberry
September 21, 2015
Example 18.1. Proposition
The Lie group SO(n + 1) acts smoothly on the unit sphere Sn. For every p ∈ Sn the isotropy subgroup is conjugate to SO(n). Hence Sn is a homogeneous space and is diffeomorphic to the quotient manifold SO(n + 1)/SO(n).
Here we are identifying SO(n) with its image under the embedding
SO(n) → SO(n + 1) ! 1 0 A 7→ 0 A
Example 18.1 (Continued). Proof: The action
n+1 n+1 SO(n + 1) × R → R (g, p) 7→ g · p preserves the Euclidean inner product and hence gives an action
SO(n + 1) × Sn → Sn, which is clearly smooth. Example 18.1 (Continued). To see that the action is transitive, note that it suffices to show that n for any p ∈ S , we may find g ∈ SO(n + 1) such that ge1 = p.
n n But given p ∈ S , we may take v2, . . . , vn+1 ∈ S such that v1 = p, v2, . . . , vn+1 is an orthonor- mal basis of Rn+1 with the same orientation as the standard basis. Then g ∈ SO(n + 1) defined by j vi = gi ej, satisfies p = ge1.
1 Example 18.1 (Continued). The isotropy subgroup SO(n + 1)e1 (i.e. the subgroup fixing e1) clearly contains ! 1 SO(n) = . SO(n)
j Conversely, suppose e1 = g · e1 = g1ej. Then
1 j g1 = 1 and g1 = 0 for j 6= 1.
Since g ∈ SO(n + 1) ⊂ O(n + 1), g tg = I, so
X 1 2 (gj ) = 1 j
1 1 and as g1 = 1 this gives gj = 0 for j 6= 1.
Proposition 18.2. Let H be a closed subgroup of the Lie group G and suppose that H and G/H are connected. Then G is connected.
Proof: Suppose G = U ∪ V for ∅= 6 U, V open subsets of G.
Writing π : G → G/H for the projection then
G/H = π(U) ∪ π(V ) and ∅= 6 π(U), π(V ) open subsets of G/H.
Since G/H is connected, for some g ∈ G we have
gH ∈ π(U) ∩ π(V ).
Since G = U ∪ V then gH = (gH ∩ U) ∪ (gH ∩ V ).
Since gH ∼= H inherits its topology from G, these are open subsets of gH and since H is connected, (gH ∩ U) ∩ (gH ∩ V ) 6= ∅ so U ∩ V 6= ∅ so G is connected.
Corollary 18.3. SO(n) is connected for all n ≥ 1.
2 Proof. We use induction on n. SO(1) is a single point so clearly connected. Then from the above proposition and the fact that Sn is connected, the result follows.
Corollary 18.4. O(n) has two connected components for all n ≥ 1.
Proof. The map
R : SO(n) → O(n) g 7→ Ag where −1 1 A = 1 · 1 is a homeomorphism from SO(n) to the subset of O(n) consisting of matrices of determinant −1.
O(n) = SO(n) ∪ R(SO(n)) then expresses O(n) as the union of two open, nonempty, disjoint connected subsets.
Example 18.5. Proposition The Lie group O(n + 1) acts smoothly on the unit sphere Sn. For every p ∈ Sn the isotropy subgroup is conjugate to O(n). Hence Sn is a homogeneous space and is diffeomorphic to the quotient manifold O(n + 1)/O(n). Proof: Exercise
Active Learning
Question 18.6. Show that the Lie group U(n + 1) acts transitively on the sphere S2n+1, and for each p ∈ S2n+1 the isotropy subgroup is conjugate to U(n). Hence the sphere S2n+1 is a homogeneous space and is diffeomorphic to U(n + 1)/U(n).
Again, we may replace the unitary group by the special unitary group in this example and hence obtain that S2n+1 is diffeomorphic to SU(n + 1)/SU(n). Taking n = 1, note that SU(1) consists of a single point, the identity matrix, and hence S3 is diffeomorphic to the Lie group SU(2). In fact S3 and S1 are the only spheres which admit Lie group structures.
3 Example 18.8 ( Real projective space, RPn). Recall that RPn is the space of lines through the origin in Rn+1. Equivalently, n+1 \{0} n = R , RP ∼ where 0 1 n 0 1 n (X ,X ,...,X ) ∼ λ(X ,X ,...,X ), λ ∈ R \{0}. We write the equivalence class of (X0,X1,...,Xn) as [X0 : X1 : ... : Xn]. Example 18.8 (Continued). The natural projection
n+1 n R \{0} → RP (X0,X1,...,Xn) 7→ [X0 : X1 : ... : Xn] restricts to Sn to exhibit Sn as a double cover of RPn. That is, we have a diffeomorphism
n ∼ n S / ∼ = RP .
Example 18.8 (Continued). We have
Sn ∼= SO(n + 1)/SO(n) ∼= O(n + 1)/O(n)
n n S is a double cover of RP O(n) is a double cover of SO(n).
This suggests trying to show that
n RP is diffeomorphic to SO(n + 1)/O(n).
Example 18.8 (Continued). Identify O(n) with its image under the embedding
O(n) → SO(n + 1) ! det A A 7→ . A
Using the identification n ∼ n S / ∼ = RP we have a smooth transitive action of SO(n + 1) on RPn.
Exercise 18.9. Show that the isotropy group of [1 : 0 ... : 0] is O(n) .
4 Example 18.10 (Complex projective space, CPn). CPn is the space of complex lines through the origin in Cn+1. Equivalently, n+1 \{0} n = C , CP ∼ where 0 1 n 0 1 n (X ,X ,...,X ) ∼ λ(X ,X ,...,X ), λ ∈ C \{0}. We write the equivalence class of (X0,X1,...,Xn) as [X0 : X1 : ... : Xn].
The proof that CPn is a smooth manifold of real dimension 2n is completely analogous to the argument for real projective space.
Exercise 18.11. Show that CPn is a homogeneous manifold diffeomorphic to SU(n+1)/S(U(1)× U(n)).
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