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Riemannian Geometry – Lecture 18 Homogeneous Spaces

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Riemannian Geometry – Lecture 18 Homogeneous Spaces 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 2 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 2 S , we may find g 2 SO(n + 1) such that ge1 = p. n n But given p 2 S , we may take v2; : : : ; vn+1 2 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 2 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 2 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 2 G we have gH 2 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 0 1 −1 B C B 1 C B C B C A = B 1 C B C B · C @ A 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 2 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 2 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 n f0g n = R ; RP ∼ where 0 1 n 0 1 n (X ;X ;:::;X ) ∼ λ(X ;X ;:::;X ); λ 2 R n f0g: We write the equivalence class of (X0;X1;:::;Xn) as [X0 : X1 : ::: : Xn]. Example 18.8 (Continued). The natural projection n+1 n R n f0g ! 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 n f0g n = C ; CP ∼ where 0 1 n 0 1 n (X ;X ;:::;X ) ∼ λ(X ;X ;:::;X ); λ 2 C n f0g: 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)). 5.
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