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Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015

Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

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Page 1: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

Riemannian Geometry – Lecture 20

Isotropy Continued

Dr. Emma Carberry

October 12, 2015

Page 2: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

Hyperbolic Space: upper half space model

Example 20.1

The upper half plane model for hyperbolic space of radius r isgiven by

Un(r) = {(x1, . . . , xn) ∈ Rn | xn > 0}

together with the Riemannian metric defined in coordinates(x1, . . . , xn)

gUij (x

1, . . . , xn) =r2

(xn)2 δij

where we are taking the standard basis vectors∂i = ei = (0, . . . ,0,1,0, . . . ,0).A computation very similar to that we performed earlier showsthat this has constant sectional curvature −1

r2 .

Page 3: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

Example 20.1 (Continued)

Page 4: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

Hyperbolic Space: ball model

Example 20.2

The ball or disc model for hyperbolic space of radius r is givenby

Bn(r) = {(x1, . . . , xn) ∈ Rn | (x1)2 + · · ·+ (xn)2 < r2}

together with the Riemannian metric defined in coordinates(x1, . . . , xn) by

gBij (x

1, . . . , xn) =4r4δij

(r2 − |x |2)2.

Page 5: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

Example 20.2 (Continued)

Page 6: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

Hyperbolic Space: hyperboloid model

Example 20.3

The hyperboloid model for hyperbolic space of radius r is givenby

Sn−1,1(r) ={(x1, . . . , xn, xn+1) ∈ Rn,1 |

(xn+1)2 − (x1)2 − · · · (xn)2 = r2, xn+1 > 0}

together with the Riemannian metric gH = ι∗ρ induced from theinclusion ι : Sn−1,1(r)→ Rn,1. Here Rn,1 is the manifold Rn+1

together with the Minkowski metric

ρ(x , y) = x1y1 + · · ·+ xnyn − xn+1yn+1.

Note that ρ is not a Riemannian metric so the fact that gH isRiemannian (the hyperboloid is “spacelike”) needs to bechecked.

Page 7: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

x^1

x^2

x^(n+1)

Page 8: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

Proposition 20.4

These three models of hyperbolic space are all isometricRiemannian manifolds.

An isometry from the ball model to the upper half plane modelis given by, with 1 ≤ i ≤ n − 1:

f : Bn(r) → Un(r)

(. . . , x i , . . . , xn) 7→(. . . , 2r2xi

(x1)2+···+(xn−1)2+(xn−r)2, . . . ,

r(r2−(x1)2−...−(xn)2)(x1)2+···+(xn−1)2+(xn−r)2

),

which is a generalisation of the Cayley transform of complexanalysis.

An isometry from the hyperboloid model to the ball model isgiven by hyperbolic stereographic projection.

We omit the check that these are indeed isometries.

Page 9: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

S

p

Page 10: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

We write (Hn(r),g) for any one of the model spaces definedabove and call it hyperbolic space of radius r .

It is also a homogeneous and isotropic Riemannian manifoldwhich is most easily seen with the hyperboloid model.

Let O(n,1) denote the group of invertible linear transformationsRn,1 → Rn,1 which preserve the Minkowski metric ρ. This iscalled the Lorentz group.

Page 11: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

Then the Lorentz group preserves the hyperboloid of twosheets {

(x1, . . . , xn, xn+1) ∈ Rn,1 |

(xn+1)2 − (x1)2 − · · · (xn)2 = r2}

and the subgroup of it preserving the upper sheet

Sn−1,1(r) ={(x1, . . . , xn, xn+1) ∈ Rn,1 |

(xn+1)2 − (x1)2 − · · · (xn)2 = r2, xn+1 > 0}

is denoted by O+(n,1).

Page 12: Riemannian Geometry – Lecture 20Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry October 12, 2015 Hyperbolic Space: upper half space model Example 20.1 The

Proposition 20.5

The positive Lorentz group O+(n,1) acts transitively onSn−1,1(r) and moreover acts transitively on the set oforthonormal bases on Sn−1,1(r). Hence hyperbolic space Hn(r)is an isotropic homogeneous Riemannian manifold.

The proof is similar to that for the sphere.

Corollary 20.6

Hyperbolic space Hn(r) has constant sectional curvature.