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Riemannian – Lecture 18 Homogeneous

Dr. Emma Carberry

September 21, 2015

Example 18.1. Proposition

The Lie SO(n + 1) acts smoothly on the unit Sn. For every p ∈ Sn the subgroup is conjugate to SO(n). Hence Sn is a homogeneous and is diffeomorphic to the quotient SO(n + 1)/SO(n).

Here we are identifying SO(n) with its image under the

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 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 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 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 which admit Lie group structures.

3 Example 18.8 ( Real , 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 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

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 (, 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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