Riemannian Geometry – Lecture 19

Isotropy

Dr. Emma Carberry

October 12, 2015

Recap from lecture 17:

Definition 17.21a Lie group G acts on a manifold M if there is a map

G ×M → M(g,p) 7→ g · p

such thate · p = p for all p ∈ M

and

(gh) · p = g · (h · p) for all g,h ∈ G, p ∈ M.

(these force the action of each element to be a bijection)

Definition 17.21 (continued)if for every g ∈ G,

g : M → Mp 7→ g · p

is smooth then we say that G acts smoothlythe isotropy subgroup Gp of G at p is the subgroup fixing pG acts transitively on M if for every p,q ∈ M there existsg ∈ G such that q = g · p

Definition 17.22

A homogeneous space is a manifold M together with a smoothtransitive action by a Lie group G.

Informally, a homogeneous space “looks the same” at everypoint.

Since the action is transitive, the isotropy groups are allconjugate

Gg·p = gGpg−1

and for any p ∈ M we can identify the points of M with thequotient space G/Gp.

Definition 17.23

A Lie subgroup H of a Lie group G is a subset of G such thatthe natural inclusion is an immersion and a grouphomomorphism (i.e. H is simultaneously a subgroup andsubmanifold).

Since the the group action on a homogeneous space is inparticular continuous, the isotropy subgroups are closed in thetopology 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 ofG.

Theorem 17.26

If G is a Lie group and H a Lie subgroup then the quotientspace 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”, pp120–124.

Corollary 17.27

A homogeneous space M acted upon by the Lie group G withisotropy subgroup Gp is diffeomorphic to the quotient manifoldG/Gp where the latter is given the unique smooth structure ofthe previous theorem.

Lecture 18 continued:

Example 18.8 ( Real projective space, RPn)

Recall that RPn is the space of lines through the origin in Rn+1.Equivalently,

RPn =Rn+1 \ {0}∼

,

where

(X 0,X 1, . . . ,X n) ∼ λ(X 0,X 1, . . . ,X n), λ ∈ R \ {0}.

We write the equivalence class of (X 0,X 1, . . . ,X n) as[X 0 : X 1 : . . . : X n].

Example 18.8 (Continued)

The natural projection

Rn+1 \ {0} → RPn

(X 0,X 1, . . . ,X n) 7→ [X 0 : X 1 : . . . : X n]

restricts to Sn to exhibit Sn as a double cover of RPn.That is, we have a diffeomorphism

Sn/ ∼ ∼= RPn.

Example 18.8 (Continued)

We have

Sn ∼= SO(n + 1)/SO(n) ∼= O(n + 1)/O(n)

Sn is a double cover of RPn

O(n) is a double cover of SO(n).

This suggests trying to show that

RPn 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)

A 7→(

det AA

).

Using the identification

Sn/ ∼ ∼= RPn

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) .

Example 18.10 (Complex projective space, CPn)

CPn is the space of complex lines through the origin in Cn+1.Equivalently,

CPn =Cn+1 \ {0}∼

,

where

(X 0,X 1, . . . ,X n) ∼ λ(X 0,X 1, . . . ,X n), λ ∈ C \ {0}.

We write the equivalence class of (X 0,X 1, . . . ,X n) as[X 0 : X 1 : . . . : X n].The proof that CPn is a smooth manifold of real dimension 2n iscompletely analogous to the argument for real projective space.

Exercise 18.11

Show that CPn is a homogeneous manifold diffeomorphic toSU(n + 1)/S(U(1)× U(n)).

Lecture 19:

Isometry group

Definition 19.1

A diffeomorphism ϕ : (M,g)→ (N,h) is called an isometry ifϕ∗(h) = g, equivalently if

h(dϕp(v),dϕp(w)) = g(v ,w)

for all p ∈ M, v ,w ∈ TpM.

Definition 19.2

The set of isometries ϕ : (M,g)→ (M,g) forms a group, calledthe isometry group I of the Riemannian manifold (M,g).

The isometry group is always a finite-dimensional Lie groupacting smoothly on M (see eg Kobayashi, TransformationGroups in Differential Geometry, Thm II.1.2).

Homogeneous and isotropic Riemannian manifolds

Definition 19.3

A Riemannian manifold (M,g) is a homogeneous Riemannianmanifold if its isometry group I(M) acts transitively on M.

A homogeneous space M is just a differentiable manifold, noRiemannian metric. M is diffeomorphic to the quotient manifoldG/Gp.

A homogeneous Riemannian manifold also has a Riemannianmetric compatible with the group action.

A homogeneous Riemannian manifold “looks the same” atevery point in terms both of its smooth structure and of theRiemannian metric.

Definition 19.4

A Riemannian manifold (M,g) is isotropic at p ∈ M if theisotropy subgroup Ip acts transitively on the set of unit vectorsof TpM, via

Ip × {X ∈ TpM | 〈X ,X 〉 = 1} → {X ∈ TpM | 〈X ,X 〉 = 1}(g,X ) 7→ dgp(X ).

The geometric interpretation is that the Riemannian manifold Mnear p looks the same in all directions.

Definition 19.5

A homogeneous Riemannian manifold which is isotropic at onepoint must be isotropic at every point. Such a manifold is calleda homogeneous and isotropic Riemannian manifold.

A homogeneous and isotropic Riemannian manifold then looksthe same at every point and in every direction: can you think ofany examples?

Example 19.6

PropositionThe isometry group of Sn is O(n + 1). It acts transitively on Sn.The isotropy group acts transitively on the unit vectors of thetangent space. Hence, Sn is a homogeneous and isotropicRiemannian manifold.

Example 19.6 (Continued)

ProofSince O(n + 1) is by definition the isometry group of Rn+1 andSn inherits its Riemannian metric from Rn+1, the action ofSO(n + 1) on Sn is by isometries. Conversely, any isometry ofthe sphere can be extended linearly to give an isometry of Rn+1

(we will check this momentarily). Hence I(Sn) = O(n + 1).

Example 19.6 (Continued)

The proofs of the next two statements are similar.Choose any point p ∈ Sn. It is unit length, so extend it to anorthonormal basis v1 = p, v2, · · · , vn+1 of Rn+1.The matrix g with columns v1, · · · , vn+1 is in O(n + 1) and takese1 to p. Thus, by going via e1, the action is transitive.To see it is isotropic, we compute at e1 ∈ Sn. We have seenthat the isotropy subgroup is just O(n) acting on the standardbasis e2, · · · ,en+1. Choose any unit vector v ∈ TpSn. We needto find g such that dge1(e2) = v .But the differential of the linear map is itself, so we may replacedge1 with g.Again, extend v to an orthonormal basis and then we may takethe columns of g to be v , v3, · · · , vn+1, similarly to what we haveseen before.

Example 19.6 (Continued)

To see that every isometry ϕ of a sphere extends to a linearmap, we can write any point p ∈ Rn+1 as λs for λ ∈ R≥0 ands ∈ Sn.Define ϕ̃(p) = λϕ(s). This is well defined (check at p = 0).To see it is linear, take any basis (e0, . . . ,en) of Rn+1. Bychecking inner products, so too is (ϕ(e0), . . . , ϕ(en)). Wecompute

ϕ̃(p) =∑〈ϕ̃(p), ϕ(ei)〉ϕ(ei)

=∑〈λϕ(s), ϕ(ei)〉ϕ(ei)

=∑

λ〈ϕ(s), ϕ(ei)〉ϕ(ei)

=∑

λ〈s,ei〉ϕ(ei) =∑〈p,ei〉ϕ(ei).

Note the right-hand side is a linear function of p. It’s an easycheck 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 ofradius r > 0, the special orthogonal group similarly actstransitively on it by isometries and furthermore acts transitivelyon the orthonormal frames of Sn(r).

Hence the sphere Sn(r) of radius r > 0 has constant sectionalcurvature too.