Riemannian Geometry – Lecture 19 Isotropy Dr. Emma Carberry October 12, 2015 Recap from lecture 17: Definition 17.21. • a Lie group G acts on a manifold M if there is a map G × M ! M (g; p) 7! g · p such that e · p = p for all p 2 M and (gh) · p = g · (h · p) for all g; h 2 G; p 2 M: (these force the action of each element to be a bijection) Definition 17.21 (continued). • if for every g 2 G, g : M ! M p 7! g · p is smooth then we say that G acts smoothly • the isotropy subgroup Gp of G at p is the subgroup fixing p • G acts transitively on M if for every p; q 2 M there exists g 2 G such that q = g · p Definition 17.22. A homogeneous space is a manifold M together with a smooth transitive action by a Lie group G. 1 Informally, a homogeneous space “looks the same” at every point. Since the action is transitive, the isotropy groups are all conjugate −1 Gg·p = gGpg and for any p 2 M we can identify the points of M with the quotient space G=Gp. Definition 17.23. A Lie subgroup H of a Lie group G is a subset of G such that the natural inclusion is an immersion and a group homomorphism (i.e. H is simultaneously a subgroup and submanifold). Since the the group action on a homogeneous space is in particular continuous, the isotropy subgroups are closed in the topology of G. Theorem 17.24 (Lie-Cartan). A closed subgroup of a Lie group G is a Lie subgroup of G. Corollary 17.25. An isotropy subgroup Gp of a Lie group G is a Lie subgroup of G. Theorem 17.26. If G is a Lie group and H a Lie subgroup then the quotient space G=H has a unique smooth structure such that the map G × G=H ! G=H (g; kH) 7! gkH is smooth. We omit the proof; it is given for example in Warner, “Foundations of Differentiable Manifolds and Lie Groups”, pp 120–124. Corollary 17.27. A homogeneous space M acted upon by the Lie group G with isotropy sub- group Gp is diffeomorphic to the quotient manifold G=Gp where the latter is given the unique smooth structure of the previous theorem. Lecture 18 continued: 2 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) . 3 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)). Lecture 19: Isometry group Definition 19.1. A diffeomorphism ' :(M; g) ! (N; h) is called an isometry if '∗(h) = g, equivalently if h(d'p(v); d'p(w)) = g(v; w) for all p 2 M, v; w 2 TpM. Definition 19.2. The set of isometries ' :(M; g) ! (M; g) forms a group, called the isometry group I of the Riemannian manifold (M; g). The isometry group is always a finite-dimensional Lie group acting smoothly on M (see eg Kobayashi, Transformation Groups in Differential Geometry, Thm II.1.2). 4 Homogeneous and isotropic Riemannian manifolds Definition 19.3. A Riemannian manifold (M; g) is a homogeneous Riemannian manifold if its isometry group I(M) acts transitively on M. A homogeneous space M is just a differentiable manifold, no Riemannian metric. M is diffeo- morphic to the quotient manifold G=Gp. A homogeneous Riemannian manifold also has a Riemannian metric compatible with the group action. A homogeneous Riemannian manifold “looks the same” at every point in terms both of its smooth structure and of the Riemannian metric. Definition 19.4. A Riemannian manifold (M; g) is isotropic at p 2 M if the isotropy subgroup Ip acts transitively on the set of unit vectors of TpM, via Ip × fX 2 TpM j hX; Xi = 1g ! fX 2 TpM j hX; Xi = 1g (g; X) 7! dgp(X): The geometric interpretation is that the Riemannian manifold M near p looks the same in all directions. Definition 19.5. A homogeneous Riemannian manifold which is isotropic at one point must be isotropic at every point. Such a manifold is called a homogeneous and isotropic Riemannian manifold. A homogeneous and isotropic Riemannian manifold then looks the same at every point and in every direction: can you think of any examples? Example 19.6. Proposition The isometry group of Sn is O(n + 1). It acts transitively on Sn. The isotropy group acts transitively on the unit vectors of the tangent space. Hence, Sn is a homogeneous and isotropic Riemannian manifold. Example 19.6 (Continued). Proof Since O(n+1) is by definition the isometry group of Rn+1 and Sn inherits its Riemannian metric from Rn+1, the action of SO(n + 1) on Sn is by isometries. Conversely, any isometry of the sphere can be extended linearly to give an isometry of Rn+1 (we will check this momentarily). Hence I(Sn) = O(n + 1). 5 Example 19.6 (Continued). The proofs of the next two statements are similar. n Choose any point p 2 S . It is unit length, so extend it to an orthonormal basis v1 = n+1 p; v2; ··· ; vn+1 of R . The matrix g with columns v1; ··· ; vn+1 is in O(n + 1) and takes e1 to p. Thus, by going via e1, the action is transitive. n To see it is isotropic, we compute at e1 2 S . We have seen that the isotropy subgroup is just n O(n) acting on the standard basis e2; ··· ; en+1. Choose any unit vector v 2 TpS . We need to find g such that dge1 (e2) = v. But the differential of the linear map is itself, so we may replace dge1 with g. Again, extend v to an orthonormal basis and then we may take the columns of g to be v; v3; ··· ; vn+1, similarly to what we have seen before. Example 19.6 (Continued). To see that every isometry ' of a sphere extends to a linear map, we can write any point p 2 Rn+1 as λs for λ 2 R≥0 and s 2 Sn. Define '~(p) = λϕ(s). This is well defined (check at p = 0). n+1 To see it is linear, take any basis (e0; : : : ; en) of R . By checking inner products, so too is ('(e0);:::;'(en)). We compute X '~(p) = h'~(p);'(ei)i'(ei) X = hλϕ(s);'(ei)i'(ei) X = λh'(s);'(ei)i'(ei) X X = λhs; eii'(ei) = hp; eii'(ei): Note the right-hand side is a linear function of p. It’s an easy check that it is an isometry. Corollary 19.7. The sphere Sn has constant sectional curvature. We could just as easily have argued above with the sphere of radius r > 0, the special orthogonal group similarly acts transitively on it by isometries and furthermore acts transitively on the orthonormal frames of Sn(r). Hence the sphere Sn(r) of radius r > 0 has constant sectional curvature too. 6.