Riemannian Geometry

Home , Sectional curvature

Riemannian Geometry

Dr Emma Carberry

Semester 2, 2015

Lecture 15

[Riemannian Geometry – Lecture 15]Riemannian Geometry – Lecture 15 Riemann Curvature Tensor Dr. Emma

Carberry September 14, 2015

Sectional curvature

Recall that as a (1, 3)-tensor, the Riemann curvature endomorphism of the Levi-Civita connection ∇ is

R(X,Y )Z = ∇X (∇Y Z) − ∇Y (∇X Z) − ∇[X,Y ]Z. It measures the failure of the manifold to be locally isometric to Euclidean space. Starting with the Riemann curvature tensor, there are various simpliﬁcations of this tensor one can deﬁne. An important one is sectional curvature, because it

• is the natural generalisation of Gauss curvature of the surface • completely determines the Riemann curvature tensor

A two-dimensional subspace π of TpM is called a tangent 2-plane to M at p.

Sectional curvature is deﬁned on tangent 2-planes π ⊂ TpM. We will deﬁne the sectional curvature of a tangent 2-plane π in terms of a basis for π. In order to obtain an expression which is independent of this choice of basis, we will need a normalisation factor.

Deﬁnition 15.1. For linearly independent vectors X,Y ∈ TpM, let Q(X,Y ) = (area of the parallelogram with sides X and Y )2 = hX,Xi hY,Y i − hX,Y i2 . Deﬁnition 15.2. Given any point p ∈ M and any tangent 2-plane π to M at p, the sectional curvature of π is the real number hR(X,Y )Y,Xi Rm(X,Y,Y,X) sec(π) := = , Q(X,Y ) Q(X,Y ) where {X,Y } denotes a basis of π. In components Kij := sec (span{∂i, ∂j})

Independence of basis choice Proposition 15.3. Sectional curvature sec(π) is well-deﬁned, independent of the choice of basis {X,Y } for π.

Proof: If {X,˜ Y˜ } is another basis for π, then X˜ = aX + bY, Y˜ = cX + dY,

1 with non-zero determinant ad − bc. An easy computation gives Q(X,˜ Y˜ ) = (ad − bc)2Q(X,Y ).

Using symmetry of Rm, Rm(X,˜ Y,˜ Y,˜ X˜) = Rm(aX + bY, cX + dY, cX + dY, aX + bY ) = (ad − bc) Rm(X, Y, cX + dY, aX + bY ) = (ad − bc)2Rm(X,Y,Y,X) so sec(π) is independent of the choice of basis {X,Y }. We have seen that the sectional curvature of M is a real-valued function deﬁned on the set of all tangent 2-planes of M. Although sectional curvature seems simpler than the curvature tensor Rm, it encapsulates the same information.

Lemma 15.4. Given sec(π) for all tangent 2-planes π ⊂ TpM, we can algebraically determine the curvature tensor Rm at p.

Proof: The idea is to use the symmetry of the Riemann curvature tensor. Set X,Y,Z,W ∈ TpM, and consider the polynomial in t defined by f(t) = Rm(X + tW, Y + tZ, Y + tZ, X + tW ) − t2(Rm(X,Z,Z,X) + Rm(W, Y, Y, W )) Due to the symmetries in each term, we can write f in terms of sectional curvatures (and the function Q which is given by the inner product). The coefficient of t2 in f(t) is Rm(X,Y,Z,W ) + Rm(X,Z,Y,W ) + Rm(W, Z, Y, X) + Rm(W, Y, Z, X). Recall that Rm is skew-symmetric in each of the first and last 2 entries ( Rm(X,Y,Z,W ) = −Rm(Y,X,Z,W ), Rm(X,Y,Z,W ) = −Rm(X, Y, W, Z) and it is symmetric between the first pair and last pair Rm(X,Y,Z,W ) = Rm(Z, W, X, Y ).

Hence the coefﬁcient of t2 in f(t) is 2Rm(X,Y,Z,W ) − 2Rm(Z,X,Y,W ). (1) Interchanging X and Y , we similarly obtain that the expression 2Rm(Y,X,Z,W ) − 2Rm(Z,Y,X,W ), is determined by the sectional curvatures of M at p. Again using symmetry of Rm, this is − 2Rm(X,Y,Z,W ) + 2Rm(Y,Z,X,W ). (2)

Recall also the algebraic Bianchi identity: X Rm(X,Y,Z,W ) = 0. all cyclic permutations of (X,Y,Z) Using this, (1) − (2) = 6Rm(X,Y,Z,W ), so the Riemann curvature tensor is determined by the sectional curvature.

2 Deﬁnition 15.5. A Riemannian manifold M has constant curvature if its sectional curvature function is constant, that is there exists a constant k such that secp(π) = k for all p ∈ M and all tangent planes π ⊂ TpM.

Remark 15.6. Let p ∈ M. If the sectional curvature function at p is zero (that is, secp = 0), then the curvature Rm = 0 at p. Hence, a Riemannian manifold (M, g) is ﬂat if and only if the sectional curvature is identically zero.

Let us consider the special case when our Riemannian manifold is a surface. In that case we had already an intrinsic notion of curvature, namely the Gauss curvature. Proposition 15.7. For a Riemannian surface (M, g), the sectional curvature of M is equal to its Gauss curvature.

Proof: By deﬁnition, for p ∈ M 2,

Rm2112 sec(TpM) = 2 (g11g22 − g12) h(∇ (∇ ∂ ) − ∇ (∇ ∂ )), ∂ i = ∂2 ∂1 1 ∂1 ∂2 1 2 , det g which is a formula for the Gauss curvature. Exercise 15.8. Prove this expression, using

• the deﬁnition of Gauss curvature K(p) as the determinant of the shape operator −dNp, where N is the Gauss map det(II) • the resulting expression K = det I where I and II denote the 1st and 2nd fundamental forms. (You should remind yourself how to obtain this expression.) • the fact (which you should make sure you understand) that k ∇∂i (∂j) = Γij∂k + hijN

where hij denotes the coefﬁcients of the 2nd fundamental form.

We will next study some spaces which have constant sectional curvature, for which the following remark will prove helpful. Remark 15.9. Let (M, g) be a Riemannian manifold and consider the (0, 4)-tensor Q deﬁned by Q(X,Y,Z,W ) = hX,W i hY,Zi − hY,W i hX,Zi for X, Y, W, Z ∈ X (M). Then M has constant sectional curvature equal to K0 if and only if Rm = K0Q, where Rm denotes the Riemann curvature tensor.

Proof:

Suppose M has constant sectional curvature K0 and take p ∈ M and linearly independent X,Y ∈ TpM. Then 2 2 2 Rm(X,Y,Y,X) = K0(|X| |Y | − hX,Y i )

= K0Q(X,Y,Y,X). As in the above proof, Lemma 15.4, this then implies that

Rm(X,Y,Z,W ) = K0Q(X,Y,Z,W ) for all X,Y,Z,W ∈ X (M), as required. The converse is trivial.

Corollary 15.10 (Constant Sectional Curvature). Let (M, g) be a Riemannian manifold, p ∈ M and {e1, . . . , en} an orthonormal basis of TpM. Then the following are equivalent:

1. sec(π) = K0 for all 2-planes π ⊂ TpM

2. Rmijkl = K0(δilδjk − δikδjl).

3. Rmijji = −Rmijij = K0 for all i 6= j and Rmijkl = 0 otherwise.

3 Lecture 16

[Riemannian Geometry – Lecture 16]Riemannian Geometry – Lecture 16 Computing Sectional Curvatures Dr. Emma

Carberry September 14, 2015

Stereographic projection of the sphere

Example 16.1. We write Sn or Sn(1) for the unit sphere in Rn+1, and Sn(r) for the sphere of radius r > 0. Stereographic projection φ from the north pole N = (0,..., 0, r) gives local coordinates on Sn(r) \{N}

Example 16.1 (Continued).

Example 16.1 (Continued). Example 16.1 (Continued). We have for each i = 1, . . . , n that

ui = cxi (3) and using the above similar triangles, we see that r c = . (4) r − xn+1

Hence stereographic projection φ : Sn \{N} → Rn is given by rx1 rxn φ(x1, . . . , xn, xn+1) = ,..., r − xn+1 r − xn+1

The inverse φ−1 of stereographic projection is given by (exercise) 2r2ui xi = , i = 1, . . . , n |u|2 + r2 2r3 xn+1 = r − . |u|2 + r2 and so we see that stereographic projection is a diffeomorphism.

4 Stereographic projection is conformal Deﬁnition 16.2. A local coordinate chart (U, φ) on Riemannian manifold (M, g) is conformal if for any p ∈ U and X,Y ∈ TpM, 2 gp(X,Y ) = F (p) hdφp(X), dφp(Y )i where F : U → R is a nowhere vanishing smooth function and on the right-hand side we are using the usual Euclidean inner product. Exercise 16.3. Prove that this is equivalent to:

1. dφp preserving angles, where the angle between X and Y is deﬁned (up to adding multiples of 2π) by g(X,Y ) cos θ = pg(X,X)pg(Y,Y ) and also to

2 2. gij = F δij for some nowhere vanishing smooth function F on U. Example 16.1 (Continued). We now proceed to show that stereographic projection is a conformal map. Now

−1 −1 gij(u) = hdφu (ei), dφu (ej)i and ∂xj ∂ dφ−1(e ) = (u) u i ∂ui ∂xj ! 2r2 4r2(ui)2 ∂ 4r3ui ∂ = − + |u|2 + r2 (|u|2 + r2)2 ∂xi (|u|2 + r2)2 ∂xn+1 | {z } | {z } A(i) B(i) so gij = A(i)A(j)δij + B(i)B(j) and simplifying gives 4r4 gij = δij. (5) (|u|2 + r2)2

Pre-composing with the rotation of the sphere is another coordinate chart which includes the point N = (0,..., 0, 1) and since rotation is an isometry, (5) holds for this coordinate chart also and hence we have covered the sphere with an atlas satisfying 4r4 gij = δij. (|u|2 + r2)2 We next use this to compute the sectional curvature of Sn(r). In fact we will do the computation for a general conformal coordinate chart. Recall Corollary 15.10

Corollary (Constant Sectional Curvature). Let (M, g) be a Riemannian manifold, p ∈ M and {e1, . . . , en} an orthonormal basis of TpM. Then the following are equivalent:

1. sec(π) = K0 for all 2-planes π ⊂ TpM

2. Rmijkl = K0(δilδjk − δikδjl).

3. Rmijji = −Rmijij = K0 for all i 6= j and Rmijkl = 0 otherwise.

Hence if we have a conformal coordinate chart with

Rmijji Kij = = K0 for all i 6= j giigjj and Rmijkl = 0 otherwise, then our manifold of constant sectional curvature K0.

5 Proposition 16.4. Suppose (U, φ) is a conformal coordinate chart on the Riemannian manifold (M, g) and

2 gij = F δij

Writing f = log F , the sectional curvature on U satisﬁes   −2 2 2 X 2 Kij = −F ∂i f + ∂j f + (∂kf)  . k6=i,j

Furthermore Rmijkl = 0 if all four indices are distinct, and if i, j, k are distinct we have

2 i 2 Rmijki = F Rijk = F ((∂kf)(∂jf) − ∂k(∂j(f))) .

The symmetries of Rm then determine all Rmijkl with precisely 3 distinct indices.

Proving this proposition is your 1st assignment question. Example 16.1 (Continued). For the sphere Sn(r) we have 2r2 F = , f = log(2r2) − log(|u|2 + r2) |u|2 + r2 so l l 2 −2u 2 −2 4(u ) ∂lf = , ∂l f = + |u|2 + r2 |u|2 + r2 (|u|2 + r2)2 and 4ulum ∂m(∂lf) = . (|u|2 + r2)2 Example 16.1 (Continued). Hence

Kij =   2 (|u| + r2)2 4 4((ui)2 + 4(uj)2 X 4(ul)2  − −  4r4 2 2 2 2 2 2 2 2 |u| + r (|u| + r ) l6=i,j (|u| + r ) 4r2 1 = = . 4r4 r2 Furthermore

i Rijk = (∂kf)(∂jf) − ∂k(∂j(f)) 1 = 4ukuj − 4ukuj = 0 (|u|2 + r2)2

Example 16.1 (Continued). Similarly

j Rijk = −(∂if)(∂kf) − ∂k(∂if) = 0. Hence by the corollary from last time about constant sectional curvature we have 1 sec(π) = r2

n n for any 2-plane π ⊂ TpS (r) and any p ∈ S (r).

1 The sphere n(r) of radius r has constant sectional curvature equal to . S r2

6 Hyperbolic Space Example 16.5. The upper half space model for hyperbolic space is given by

1 n n n M = {(x , . . . , x ) ∈ R | x > 0} together with the Riemannian metric 1 g (x1, . . . , xn) = δ ij (xn)2 ij where we are taking the standard basis vectors ∂i = ei = (0,..., 0, 1, 0,..., 0). 1 n 1 This then is clearly a conformal coordinate system, with F (x , . . . , x ) = xn . Example 16.5 (Continued). Our proposition gives that (for i 6= j)   −2 2 2 X 2 Kij = −F ∂i f + ∂j f + (∂kf)  k6=i,j and then since f(x1, . . . , xn) = − log(xn) we have

0, if i 6= n ∂ f = i −(xn)−1, if i = n

0, if i 6= n ∂2f = i (xn)−2, if i = n so 2 −(xn)2 (−xn)−1 = −1, i, j, n K = for all distinct ij −(xn)2(xn)−2 = −1, for i 6= j = n Example 16.5 (Continued). To conclude that the sectional curvature is −1 for every tangential 2-plane, not just the cooordinate ones, according to the corollary it remains to check that the Riemann coefﬁcients which are not of the form Rmijji or Rmijij vanish. From Proposition 15.10, Rmijkl = 0 if all four indices are distinct, and for i, j, k distinct we have n −2 i n −2 Rmijki = (x ) Rijk = (x ) ((∂kf)(∂jf) − ∂k(∂j(f))) = 0. and the other one is similiar. Hence

hyperbolic space has constant sectional curvature −1.

Lecture 17

[Riemannian Geometry – Lecture 17]Riemannian Geometry – Lecture 17 Lie Groups Dr. Emma Carberry Septem- ber 21, 2015

Lie groups

Deﬁnition 17.1. A Lie group is a manifold which is also a group and is such that the group operations of

1. group multiplication G × G → G :(g, h) 7→ gh and 2. taking inverses G × G : g 7→ g−1 are both smooth.

Example 17.2. Rn is a Lie group under addition.

7 Lie groups

Let Mn(R) denote the space of n × n matrices with real entries. Example 17.3. The general linear group

GL(n, R) = {A ∈ Mn(R) | det A 6= 0}

2 is an open subspace of Rn and hence a manifold of dimension n2. It is easy to verify that the group operations (g, h) 7→ gh and g 7→ g−1 are smooth.

More intrinsically, GL(n, R) is the space of invertible linear transformations Rn → Rn. Example 17.4. The positive general linear group

+ GL (n, R) = {A ∈ Mn(R) | det A > 0} is a Lie subgroup (i.e. a subgroup and a submanifold) of the general linear group and has the same dimension n2 since it is an open subset.

Exercise 17.5. GL+(n, R) is the group of linear transformations Rn → Rn which preserve orientation.

Recall: Deﬁnition 17.6. An orientation of an n-dimensional vector space V is a choice of one of the two possible equivalence classes of ordered basis for V under the equivalence relation ∼ deﬁned by:

(X1,...,Xn) ∼ (Y1,...,Yn) for ordered bases (X1,...,Xn) and (Y1,...,Yn) of V if the n × n matrix A deﬁned by

(Y1,...,Yn) = (X1,...,Xn)A has positive determinant. Example 17.7. The set of symmetric matrices

t Sym(n) = {A ∈ Mn(R) | A = A} is not a Lie group under multiplication since it contains the zero matrix, and so is not a group. It is however a manifold since taking the diagonal and upper triangular entries deﬁnes a global diffeomorphism to Rk, where n(n + 1) k = 1 + 2 + ··· + n = . 2 Exercise 17.8. A matrix is symmetric if and only if

n hAx, yi = hx, Ayi for all x, y ∈ R . Example 17.9. The orthogonal group is deﬁned by

t O(n) = {G ∈ Mn(R) | AA = I}. Note that this forces the determinant of A to be either 1 or −1. In fact the group O(n) consists of two connected components.

Exercise 17.10. Prove that O(n) is the group of linear transformations Rn → Rn which preserve the Euclidean inner product in the sense that

n A ∈ O(n) ⇔ hAx, Ayi = hx, yi for all x, y ∈ R . Example 17.11. Consider the special orthogonal group

t SO(n) = {A ∈ Mn(R) | det A = 1 and AA = I}.

8 Exercise 17.12. Prove that SO(n) is precisely the group of linear transformations Rn → Rn which preserve the Euclidean inner product and orientation. Example 17.11 (Continued). Moreover deﬁning

f : GL+(n, R) → Sym(n) A 7→ tAA − I, we have SO(n) = f −1(0). The map f is smooth, and furthermore it is a submersion.

n+m n Deﬁnition 17.13. A smooth map f : M → N between smooth manifolds is a submersion at p ∈ M if dfp has (maximal) rank n. The map is a submersion if it is is a submersion at every point in its domain.

Equivalently, dfp is surjective. Example 17.11 (Continued). In our example we have

+ n2 n(n+1)/2 f : GL (n, R) ⊂ R → Sym(n) =∼ R open and f being a submersion tells us that it satisﬁes the criteria of the implicit function theorem.

Recall Theorem 17.14 (Implicit Function Theorem). Suppose

F : W ⊂ Rm+n = Rm × Rn → Rn 1 n (x1, . . . , xm, y1, . . . , yn) 7→ (F (x, y),...,F (x, y)) is a smooth map, and that for (a, b) ∈ W ,

 1 1 1  F1 (a, b) F2 (a, b) ··· Fn (a, b) 2 2 2  F1 (a, b) F2 (a, b) ··· Fn (a, b)     . . . .   . . . .  n n n F1 (a, b) F2 (a, b) ··· Fn (a, b)

i is invertible, where F i = ∂F . Write c = F (a, b). Then there are open neighbourhoods U ⊂ m of a and j ∂yj R V ⊂ Rn of b and a smooth map g : U → V so that for (x, y) ∈ U × V , F (x, y) = c ⇔ y = g(x).

Example 17.11 (Continued). The implicit function theorem tells us then that SO(n) = f −1(0) is a manifold of dimension n(n + 1) 2n2 − n2 − n n(n − 1) n2 − = = . 2 2 2 The group operations (multiplication and taking inverses) are smooth and hence SO(n) is a Lie group. Example 17.15. Similarly, applying the same argument to

f : GL(n, R) → Sym(n) A 7→ tAA − I,

−1 n(n−1) we recognise O(n) as f (0) and hence see that O(n) is a Lie group of dimension 2 . As stated above, O(n) has two connected components, one of which is SO(n). Example 17.16. The unitary group U(n) ⊂ GL(n, C) is the subgroup of the complex general linear group consisting of matrices satisfying tAA¯ = I. The special unitary group SU(n) consists of those unitary matrices which additionally satisfy det A = 1.

9 Exercise 17.17. Prove that U(n) is the group of linear transformations Cn → Cn which preserve the Hermitian inner product (z, w) = z1w1 + ··· znwn in the sense that n A ∈ U(n) ⇔ (Az, Aw) = (z, w) for all z, w ∈ C .

The space of Hermitian symmetric matrices

t HermSym(n) = {A ∈ Mn(C) | A = A} consists of matrices of the form   x1 z12 z13 ··· z1n  z12 x2 z23 ··· z2n     .. . .   z13 z23 . . .     . . .   . . . xn−1 zn−1,n    . . z1n . . zn−1,n xn

n(n−1)+n n2 where zij ∈ C, xi ∈ R and hence is a manifold diffeomorphic to R = R . Exercise 17.18. Show that A ∈ HermSym if and only if for all z, w ∈ Cn, (Az, w) = (z, Aw).

Active Learning Question 17.19. By an analogous argument to that used for SO(n), prove that U(n) is a real smooth manifold of dimension n2.

Note that a unitary matrix satisﬁes det A det A = 1 and hence det A ∈ S1. We may characterise SU(n) as the pre-image of 1 under the map

det : U(n) → S1.

Again det is a submersion and so the implicit function theorem tells us that SU(n) is a smooth real manifold of dimension n2 − 1. 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 ∈ 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 → 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

10 • G acts transitively on M if for every p, q ∈ M there exists g ∈ 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.

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 ∈ 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 subgroup Gp is diffeomorphic to the quotient manifold G/Gp where the latter is given the unique smooth structure of the previous theorem.

Lecture 18

[Riemannian Geometry – Lecture 18]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 ∈ 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.

11 Example 18.1 (Continued). To see that the action is transitive, note that it suffices to show that for any p ∈ Sn, 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 orthonormal 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.

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 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 ∈ G we have

gH ∈ 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.

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.

12 Proof. The map R : SO(n) → O(n) g 7→ Ag where  −1   1    A =  1     ·  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 ∈ 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 spheres which admit Lie group structures. Example 18.8 ( Real projective space, 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 equivalence class 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 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).

13 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) .

Example 18.10 (Complex projective space, 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)).

Lecture 19

[Riemannian Geometry – Lecture 19]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

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 ∈ M there exists g ∈ 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.

14 Informally, a homogeneous space “looks the same” at every point. Since the action is transitive, the isotropy groups are all conjugate

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.

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, 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 equivalence class 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 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).

15 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) .

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 ϕ :(M, g) → (N, h) is called an isometry if ϕ∗(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, 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, Transfor- mation Groups in Differential Geometry, Thm II.1.2).

16 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 diffeomorphic 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 ∈ M if the isotropy subgroup Ip acts transitively on the set of unit vectors of TpM, via

Ip × {X ∈ TpM | hX,Xi = 1} → {X ∈ TpM | hX,Xi = 1}

(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). Example 19.6 (Continued). The proofs of the next two statements are similar. n n+1 Choose any point p ∈ S . It is unit length, so extend it to an orthonormal basis v1 = 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 ∈ S . We have seen that the isotropy subgroup is just O(n) acting on the n standard basis e2, ··· , en+1. Choose any unit vector v ∈ 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 ϕ of a sphere extends to a linear map, we can write any point p ∈ Rn+1 as λs for λ ∈ R≥0 and s ∈ Sn. Deﬁne ϕ˜(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 (ϕ(e0), . . . , ϕ(en)). We compute X ϕ˜(p) = hϕ˜(p), ϕ(ei)iϕ(ei) X = hλϕ(s), ϕ(ei)iϕ(ei) X = λhϕ(s), ϕ(ei)iϕ(ei) X X = λhs, eiiϕ(ei) = hp, eiiϕ(ei).

Note the right-hand side is a linear function of p. It’s an easy check that it is an isometry.

17 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.

Lecture 20

[Riemannian Geometry – Lecture 20]Riemannian Geometry – Lecture 20 Isotropy Continued Dr. Emma Carberry

October 12, 2015

Hyperbolic Space: upper half space model Example 20.1. The upper half plane model for hyperbolic space of radius r is given by

n 1 n n n U (r) = {(x , . . . , x ) ∈ R | x > 0} together with the Riemannian metric deﬁned in coordinates (x1, . . . , xn)

r2 gU (x1, . . . , xn) = δ ij (xn)2 ij where we are taking the standard basis vectors ∂i = ei = (0,..., 0, 1, 0,..., 0). −1 A computation very similar to that we performed earlier shows that this has constant sectional curvature r2 .

Example 20.1 (Continued).

Hyperbolic Space: ball model Example 20.2. The ball or disc model for hyperbolic space of radius r is given by

n 1 n n 1 2 n 2 2 B (r) = {(x , . . . , x ) ∈ R | (x ) + ··· + (x ) < r } together with the Riemannian metric deﬁned in coordinates (x1, . . . , xn) by

4 B 1 n 4r δij gij (x , . . . , x ) = . (r2 − |x|2)2

18 Example 20.2 (Continued).

Hyperbolic Space: hyperboloid model Example 20.3. The hyperboloid model for hyperbolic space of radius r is given by

n−1,1 1 n n+1 n,1 S (r) = (x , . . . , x , x ) ∈ R | (xn+1)2 − (x1)2 − · · · (xn)2 = r2, xn+1 > 0 together with the Riemannian metric gH = ι∗ρ induced from the inclusion 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 is Riemannian (the hyperboloid is “spacelike”) needs to be checked.

x^(n+1)

x^2

x^1

Proposition 20.4. These three models of hyperbolic space are all isometric Riemannian manifolds.

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

f : Bn(r) → Un(r) ! 2 i r(r2−(x1)2−...−(xn)2) (. . . , xi, . . . , xn) 7→ ..., 2r x ,..., , (x1)2+···+(xn−1)2+(xn−r)2 (x1)2+···+(xn−1)2+(xn−r)2 which is a generalisation of the Cayley transform of complex analysis.

19 An isometry from the hyperboloid model to the ball model is given by hyperbolic stereographic projection. We omit the check that these are indeed isometries.

We write (Hn(r), g) for any one of the model spaces deﬁned above and call it hyperbolic space of radius r. It is also a homogeneous and isotropic Riemannian manifold which is most easily seen with the hyperboloid model.

Let O(n, 1) denote the group of invertible linear transformations Rn,1 → Rn,1 which preserve the Minkowski metric ρ. This is called the Lorentz group. Then the Lorentz group preserves the hyperboloid of two sheets 1 n n+1 n,1 (x , . . . , x , x ) ∈ R | (xn+1)2 − (x1)2 − · · · (xn)2 = r2 and the subgroup of it preserving the upper sheet n−1,1 1 n n+1 n,1 S (r) = (x , . . . , x , x ) ∈ R | (xn+1)2 − (x1)2 − · · · (xn)2 = r2, xn+1 > 0 is denoted by O+(n, 1). n−1,1 Proposition 20.5. The positive Lorentz group O+(n, 1) acts transitively on S (r) and moreover acts transitively on the set of orthonormal 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.

Lecture 21

[Riemannian Geometry – Lecture 21]Riemannian Geometry – Lecture 21 Geodesics Dr. Emma Carberry October

19, 2015

Given a smooth curve γ : I → M and t0, t1 ∈ I, recall that we deﬁned a map

Pt0t1 : Tγ(t0)M → Tγ(t1)M by deﬁning Pt0t1 (X0) to be the parallel translate of X0 to γ(t1) along γ.

20 Exercise 21.1. 1. Pt0t1 is a vector space isomorphism 2. −1 Pt0t X(t) − X(t0) ∇γ˙ X(t0) = lim . t→t0 t − t0 Proposition 21.2. Let ∇ be a linear connection on a Riemannian manifold (M, g). The following conditions are equivalent:

(a) ∇ is a metric connection. (b) ∇g ≡ 0. (c) If X,Y are vector ﬁelds along any curve γ : I → M, d hX,Y i = h∇ X,Y i + hX, ∇ Y i. dt γ˙ γ˙

(d) If X,Y are parallel vector ﬁelds along a curve γ : I → M, then hX,Y i is constant.

(e) Parallel translation Pt0t1 : Tγ(t0)M → Tγ(t1)M is an isometry for each t0, t1.

Recall that ∇ is a metric connection means that

∇Z hX,Y i = h∇Z X,Y i + hX, ∇Z Y i for all X,Y,Z ∈ X (M).

Proof: (a) ⇔ (b) Immediate from the product rule:

(∇Z g)(X,Y ) = ∇Z hX,Y i − h∇Z X,Y i − hX, ∇Z Y i.

Question 21.3. Prove (a) ⇔ (c)

Answer 21.4. (c) ⇒ (a): Take X,Y,Z ∈ X (M), and p ∈ M. Choose γ : I → M so that γ(t0) = p, γ˙ (t0) = Z(p).

Then (writing Y (t0) for the restriction of Y to the curve, etc)

h(∇Z X)(p),Y (p)i + hX(p), (∇Z Y )(p)i

= h(∇γ˙ X)(t0),Y (t0)i + hX(t0), (∇γ˙ Y )(t0)i

d = hX,Y i dt t=t0 = Z(p)hX,Y i(p)

Answer 21.5. (a) ⇒ (c)

Take t0 ∈ I, and an orthonormal basis Ei(t0) for Tγ(t0)M.

Use parallel translation to extend this to a basis Ei(t) of each Tγ(t)M, t ∈ I. By (a), the Ei(t) are orthonormal. Writing i i V (t) = V (t)Ei(t),W (t) = W (t)Ei(t), we have d d hV,W i = V iW jhE ,E i dt dt i j d = V iW jδ dt ij i j i j = V˙ W δij + V W˙ δij i j i j = hV˙ Ei,W Eji + hV Ei, W˙ Eji

= h∇γ˙ V,W i + hV, ∇γ˙ W i.

21 (c) ⇒ (d): Obvious. (d) ⇒ (e): Parallel translation produces parallel vector ﬁelds, so this is also obvious. (e) ⇒ (c): Similar to (a) ⇒ (c); parallel transport being an isometry again means that an orthonormal basis extended by parallel translation remains orthonormal.

Deﬁnition 21.6. A smooth curve γ : I → M is a geodesic if ∇γ˙ γ˙ (t) = 0 for all t ∈ I.

Note that when the connection ∇ is metric that a geodesic satisﬁes d hγ,˙ γ˙ i = 2h∇ γ,˙ γ˙ i = 0, dt γ˙ and so has constant speed. We have existence and uniqueness of geodesics, in the following sense: Theorem 21.7. Let M be a smooth manifold and ∇ a linear connection on M. Then for each p ∈ M and X ∈ TpM there exists > 0 and a geodesic γ :(−, ) → M such that γ(0) = p and γ˙ (0) = X. Furthermore any two such geodesics agree wherever both are deﬁned.

To prove this we will use the existence and uniqueness of ODEs: Theorem 21.8 (Picard-Lindelöf). Given any system

X˙ k(t) = f(t, X1,...Xn), 1 ≤ k ≤ n of ordinary ﬁrst order differential equations on (−δ, δ) × U where U ⊂ Rn is open and f is smooth, with initial condition 1 n 1 n (X (0),...,X (0))) = (X0 ,...,X0 ) ∈ U there exists > 0 and X = (X1,...Xn):(−, ) → U satisfying the system of ODEs. If f is smooth then so is X . Furthermore any two solutions agree whenever both are deﬁned.

There is a stronger version of the above ODE theorem which guarantees smooth dependence on initial conditions. Namely Theorem 21.9. Given X˙ k(t) = f(t, X1,...Xn), 1 ≤ k ≤ n n on (−δ, δ) × U where U ⊂ R is open and f is smooth, and X0 ∈ U there exists > 0 and open V ⊂ U containing X0 and a unique smooth mapping

χ :(−, ) × V → U

1 n such that for each X0 = (X0 ,...,X0 ) ∈ V , t 7→ χ(t, X0) is the unique solution to the above system of ODEs with initial condition X(0) = X0.

We won’t need this smooth dependence on initial conditions for the existence and uniqueness of a single geodesic, but we will use it in our deﬁnition of the exponential map. Firstly though, let us give the manifold form of the above ODE theorem (the stronger one). Theorem 21.10 (ODE Solutions on Manifold). Given a smooth vector ﬁeld Y on an open subset U of a smooth manifold M and p ∈ U, there exists > 0 and open V ⊂ U containing p and a smooth map

χ :(−, ) × V → U such that for each q ∈ V , t 7→ χ(t, q) is the unique smooth curve in M which passes through q when t = 0 and whose velocity vector ﬁeld is the restriction of Y to the curve.

22 Proof of uniqueness and existence of geodesics (Theorem 21.7)

The geodesic condition ∇γ˙ γ˙ (t) = 0 in local coordinates may be written as

k i j k x¨ +x ˙ x˙ Γij = 0, a system of second order ordinary differential equations. Writing Xk(t) =x ˙ k(t) converts this into a system ˙ k i j k X + X X Γij = 0 of first order ordinary differential equations. The manifold that we apply Theorem 21.10 to in order to conclude the existence and uniqueness of geodesics is not M, but rather TM. The system of equations k k ˙ k i j k x˙ (t) = X (t), X = −X X Γij defines a vector field ∂ ∂ G = Xk − XiXjΓk ∂xk ij ∂Xk on an open subset TU ⊂ TM, where U is a coordinate neighbourhood. Let us call a curve γ :(−, ) → M such that γ˙ (t) = X(γ(t)) and γ(0) = p a trajectory of the vector field X passing through p when t = 0. Lemma 21.11. Given a smooth manifold M and a linear connection ∇ on M, there exists a unique vector field G on TM whose trajectories are the form t 7→ (γ(t), γ˙ (t)) for geodesics γ.

Proof of Lemma:

We ﬁrst prove uniqueness. Any such G has trajectories t 7→ (γ(t), γ˙ (t)) such that

k k ˙ k i j k 1 n x˙ (t) = X (t), X = −X X Γij, where γ(t) = (x (t), . . . , x (t)).

But then by the uniqueness of trajectories to such a system, the vector ﬁeld G must be unique. For existence, we can deﬁne G Locally using coordinates as above.

Uniqueness then means that we can define G locally on TUp for a neighbourhood Up of each p ∈ M and it is well-defined on overlaps TUp ∩ TUq. Then applying Theorem 21.10 to the vector field G, we obtain the existence and uniqueness of geodesics, and also that the geodesic with velocity vector X at the point p ∈ M depends smoothly on the initial data (p, X).

In particular there is for any given initial conditions p ∈ M and X ∈ TpM a unique maximal geodesic γ : (−, ) → M with γ(0) = p and γ˙ (0) = X. Here by maximal we mean that this geodesic cannot be extended to any larger interval.

We denote this geodesic by γX . Henceforth we shall consider only the case where the connection ∇ is the Levi-Civita connection associated to the Riemannian metric g on M.

Lecture 22

[Riemannian Geometry – Lecture 22]Riemannian Geometry – Lecture 22 Exponential Map Dr. Emma Carberry

October 26, 2015

23 Lemma 22.1 (Rescaling lemma). For X ∈ TpM and c ∈ R \{0}, if the geodesic γX is deﬁned on the interval (−, ) then the geodesic γcX is deﬁned on interval − c , c and

γcX (t) = γX (ct).

Proof: Deﬁne γ˜ : − c , c → M by γ˜(t) = γX (ct). ˙ ˙ Then writing ∇e γ˜˙ for covariant differentiation along γ˜, since γ˜(t) = cγ˙ (ct) and γ˜(0) = cX

˙ ¨k k ˙ i ˙ j ∇e γ˜˙ γ˜(t) = γ˜ (t) + Γij(˜γ(t))γ˜ (t)γ˜ (t) ∂k 2 = c ∇γ˙ γ˙ X (ct) = 0.

Deﬁne E ⊂ TM by

E = {(p, X) ∈ TM | γX is deﬁned on an interval containing [0, 1]}, and denote by Ep the intersection of E with the set TpM. Then the exponential map is

exp : E → M X 7→ γX (1).

Proposition 22.2. 1. For each p ∈ M and X ∈ TpM, γX (t) = exp(tX) whenever these are deﬁned

2. E is an open subset of TM and for each p ∈ M, we have 0 ∈ Ep.

3. For each p ∈ M, Ep is a star-shaped region about 0. That is, whenever X ∈ Ep then also cX ∈ Ep for all c ∈ [0, 1]. 4. In fact for each p ∈ M there exists a neighbourhood V of p in M and ˜ > 0 such that for every (q, X) ∈ TV with |X| < ˜, we have (q, X) ∈ E. 5. exp is a smooth map. 6. For each p ∈ M there exists > 0 such that

expp : {X ∈ TpM | |X| < } → M is a diffeomorphism onto its image.

Proof:

(1) is simply the rescaling lemma with t = 1:

exp(cX) = γcX (1) = γX (c).

(2) Clearly E contains the 0 section of TM.

24 To prove that E is open, we will use the stronger version of the existence and uniqueness theorem. From Theo- rem 21.10, we have: For each p ∈ M there exists an open neighbourhood U of p in M and an open neighbourhood U of (p, 0) in TU together with > 0 and a smooth map γ :(−, ) × U → TU such that t 7→ γ(t, q, X) is the unique trajectory of the geodesic vector ﬁeld G which for t = 0 passes through the point q with velocity vector X ∈ TqM. (3) This follows immediately from the rescaling lemma; since

γcX (t) = γX (ct) is deﬁned whenever ct ∈ [0, 1] and c ∈ [0, 1], it is certainly deﬁned whenever t ∈ [0, 1]. (4) Taking a compact subset of the open neighbourhood U of p and then an open subset V of this compact set, then there exists δ > 0 such that

{(q, X) ∈ TV | X ∈ TqM satisﬁes |X| < δ} ⊂ U.

Then for q ∈ V , the geodesic γX (t) is deﬁned whenever |t| < and |X| < δ. δ X By the rescaling lemma then the geodesic γY is deﬁned whenever |t| < 2 and |Y | < 2 , where Y = 2 . (5): The exponential map (p, X) 7→ exp(p, X) is the evaluation of the smooth map γ(t, p, X) at t = 1, and hence is smooth.

(6): The differential of expp at 0 ∈ TpM is just the identity map, since

d d(expp)0(X) = (expp(tX)) dt t=0

d = (γ(1, p, tX)) dt t=0

d = (γ(t, p, X)) dt t=0 = X. Then by the inverse function theorem, the exponential map is locally a diffeomorphism.

Deﬁnition 22.3. If V is a neighbourhood of 0 in TpM such that expp restricted to V is a diffeomorphism onto its image, then we call expp(V ) =: U a normal neighbourhood of p.

A choice of orthonormal basis E1,...,En for TpM is equivalent to an isomorphism

n E : TpM → R i 1 n , X Ei 7→ (X ,...,X ) and so on a normal neighbourhood V of p ∈ M we have local coordinates

−1 n E ◦ expp : V → R . Such coordinates are called normal coordinates centered at p. Normal coordinates are extremely useful, in particular they yield a very simple representation for geodesics. Lemma 22.4. Let (M, g) be a Riemannian manifold and (V, {φi}) be a normal coordinate chart centred at p ∈ M. i For any X = X ∂i ∈ TpM, the geodesic γX is given in normal coordinates by −1 1 n γX (t) = φ (tX , . . . , tX ). Question 22.5. Prove this.

25 Lecture 23

[Riemannian Geometry – Lecture 23]Riemannian Geometry – Lecture 23 Length-minimising curves Dr. Emma

Carberry October 26, 2015

Open versus closed intervals

Our convention is that the domain I of a curve is an open interval unless specified otherwise, because we initially defined smoothness of functions only on open sets. Definition 23.1. A map γ :[a, b] → M is smooth if it is the restriction to [a, b] of a smooth map γ˜ : I → M where I is an open interval containing [a, b].

This deﬁnition also allows us to work with curves deﬁned on closed intervals. The symmetry of the Levi-Civita connection is crucial in proving that curves which are length-minimising are geodesics. Lemma 23.2 (Symmetry). Suppose (M, g) is a Riemannian manifold and σ :[a, b] × (−, ) → M (t, s) 7→ σ(t, s) is smooth. Then ∂σ ∂σ ∇ ∂σ = ∇ ∂σ , ∂t ∂s ∂s ∂t where

•∇ ∂σ is the covariant derivative along the curve t 7→ σ(t, s0) for s0 a ﬁxed element of (−, ) ∂t

•∇ ∂σ is the covariant derivative along the curve s 7→ σ(t0, s) for t0 a ﬁxed element of [a, b]. ∂s

Proof:

It sufﬁces to show this locally, so we compute in local coordinates about an arbitrary p0 ∈ Σ = σ ((a, b) × (−, )): φ : p 7→ (x1(p), . . . , xn(p)).

∂σ ∂σ The vector ﬁelds ∂t and ∂s are deﬁned on a neighbourhood of p0 in Σ and 2 k i j ∂σ ∂ x ∂x ∂x k ∇ ∂σ = + Γij ∂k ∂t ∂s ∂t∂s ∂t ∂s ∂2xk ∂xj ∂xi = + Γk ∂ ∂s∂t ∂s ∂t ji k ∂σ = ∇ ∂σ . ∂s ∂t

Deﬁnition 23.3. A geodesic ball about p ∈ M is a set of the form

expp(B(0)) where B(0) ⊂ TpM denotes the ball of radius about 0 and > 0 is chosen sufﬁciently small that the restriction of expp to B(0) is a diffeomorphism. Deﬁnition 23.4. A geodesic sphere about p ∈ M is a set of the form

expp(S(0)) where S(0) = ∂B(0) and > 0 is as above.

26 Lemma 23.5. Let (M, g) be a Riemannian manifold and take p ∈ M and X ∈ TpM ∩E. Then in radial directions, (d expp)X preserves lengths on TpM = TX (TpM). That is

(d expp)X (λX) = λ|X| for all λ ∈ R. Question 23.6. Prove this.

Lemma 23.8 (Gauss). Let (M, g) be a Riemannian manifold and take p ∈ M. Then the map expp is a radial isometry. This means that for any X ∈ TpM ∩ E and Y ∈ TpM, using the canonical isomorphism

TX (TpM) = TpM, we have g((d expp)X (X), (d expp)X (Y )) = g(X,Y ).

In particular, within a normal neighbourhood U = expp(V ) of p, the geodesics emanating from p are orthogonal to the geodesic spheres.

Proof:

To see that the second statement follows from the ﬁrst, recall that with respect to normal coordinates the geodesics are exactly given by radial rays −1 γX (t) = φ (tX) which are clearly perpendicular to the spheres S(0).

We now prove that expp is a radial isometry. The lemma tells us that

(d expp)X (λX) = λ|X| for all λ ∈ R.

For Y ∈ TpM, we may decompose Y as Y = YT + YN , where YT is parallel to X and YN is perpendicular to X.

Then (d expp)X preserves the lengths of both X and on YT and so certainly

g((d expp)X (X), (d expp)X (YT )) = ±|X||YT | = g(X,YT ).

Assume w.l.o.g that YN 6= 0. Since X ∈ Ep, and Ep is open and star shaped about 0, there exists > 0 such that

tX + stYN ∈ Ep for (t, s) ∈ [0, 1] × (−, ). Consider then the smooth map σ : [0, 1] × (−, ) → M (t, s) 7→ expp(tX + stYN ), and note that the curves t 7→ σ(t, s0) are geodesic rays. Moreover ∂σ ∂σ (1, 0) = (d exp ) (X), (1, 0) = (d exp ) (Y ) ∂t p X ∂s p X N whilst ∂σ (0, 0) = 0. ∂s Hence it sufﬁces to show that ∂ ∂σ ∂σ g , = 0. ∂t ∂s ∂t

But ∂ ∂σ ∂σ ∂σ ∂σ ∂σ ∂σ g , =g ∇ ∂σ , + g , ∇ ∂σ ∂t ∂s ∂t ∂t ∂s ∂t ∂s ∂t ∂t

27 The second term is zero because the tangent vector of a geodesic is parallel along that geodesic.

∂σ ∂σ =g ∇ ∂σ , ∂s ∂t ∂t 1 ∂ ∂σ ∂σ = g , 2 ∂s ∂t ∂t

The inner product is constant in t, again because it is a geodesic, so we may take t = 1

1 ∂ = g(X + sY ,X + sY ) 2 ∂s N N =g(X,YN ) = 0.

Definition 23.9. A piecewise smooth curve is a continuous map γ :[a, b] → M such that for some a = t0 < t1 < ··· < tk = b, the restriction of γ to each (ti, ti+1) is smooth. The length of such γ is defined to be Z b l(γ) = |γ˙ (t)| dt. a Exercise 23.10. Check that the length of a piecewise smooth curve is is independent of its parameterisation. Definition 23.11. A piecewise smooth curve γ :[a, b] → M is length-minimising if for every piecewise smooth curve α from γ(a) to γ(b) we have l(γ) ≤ l(α).

Proposition 23.12. Let (M, g) be a Riemannian manifold and take p ∈ M and a geodesic ball B = expp(B(0)) about p. Then any geodesic γ : [0, 1] → B with γ(0) = p is length-minimising and if another piecewise smooth curve in M from p to γ(1) has the same length then their traces coincide.

Proof: Let α :[a, b] → M be a curve with the same endpoints as γ. Write a˜ for the largest value in [a, b] such that α(˜a) = p, and ˜b for the smallest value in (˜a, b] for which α(˜b) ∈ ∂B or b if α never leaves the ball.

Then l(α) ≥ l(α|[˜a,˜b]), so we may restrict our attention to the latter segment. ˜ Since the restriction of expp to B(0) is a diffeomorphism, throughout [˜a, b], we may uniquely determine a curve + t 7→ X(t) in TpM with |X(t)| = 1 and a piecewise smooth function r : (0, 1] → R by the equation

α(t) = expp(r(t)X(t)).

Writing α(t) = expp(r(t)X(t)) = f(r(t), t) then ∂f ∂f α˙ (t) = r˙(t) + ∂r ∂t and from the Gauss lemma, ∂f ∂f g , = 0. ∂r ∂t

By assumption

∂f = 1 ∂r and hence 2 2 2 ∂f |α˙ (t)| = |r˙(t)| + ∂t ≥ |r˙(t)|2 .

28 Then

Moreover we see that if l(α) = l(γ) then [˜a, ˜b] = [a, b] and moreover the inequalities above must be equalities. Hence throughout this interval we must have

∂f = 0 ∂t which forces the curve X(t) to be constant and furthermore we must have r˙(t) > 0 throughout. Hence α is simply a monotonic reparametrisation of γ.

Lecture 24

[Riemannian Geometry – Lecture 24]Riemannian Geometry – Lecture 24 Sn, Hn and their geodesics Dr. Emma

Carberry October 26, 2015 Recall

Deﬁnition 24.1. If V is a neighbourhood of 0 in TpM such that expp restricted to V is a diffeomorphism onto its image, then we call expp(V ) =: U a normal neighbourhood of p.

A stronger condition is Deﬁnition 24.2. An open set U ⊂ M is said to be uniformly normal if there exists δ > 0 such that for every p ∈ U, expp is a diffeomorphism on Bδ(0) and U ⊂ expp(Bδ(0)). Proposition 24.3. For each p ∈ M there exists a uniformly normal neighbourhood U of p.

Proof: We showed earlier that for each p ∈ M there exists a neighbourhood V of p in M and > 0 such that

V = {(q, X) ∈ TV | |X| < } ⊂ E.

Deﬁne F : V → M × M (q, X) 7→ (q, expq X).

Shrink V if necessary so that it is a normal neighbourhood of p, and take normal coordinates (x1, . . . , xn) about p. i i i For q ∈ V , writing X ∈ TqM as X = a ∂i gives coordinates (x , a ) on TV . With respect to these coordinate systems, we may write

∂xi ∂xi ! ∂xj ∂aj I 0 dF(p,0) = ∂ expi ∂ expi = p p ∗ I ∂xj ∂aj

29 which is clearly invertible. By the inverse function theorem there is a neighbourhood V ⊂ V of (p, 0) such that F | is a diffeomorphism onto e Ve its image. There exists δ > 0 and a neighbourhood W ⊂ V of p such that the open set

Wδ = {(q, X) ∈ TW such that |X| < δ} is contained in Ve. For a proof of this fact, see Lee, Riemannian geometry, Lemma 5.12 (p 80) or try it yourself.

Take a neighbourhood U ⊂ W of p so that U × U ⊂ F (Wδ). We claim that for every q ∈ U,

(1) expq : Bδ(0) → M is a diffeomorphism onto its image

(2) U ⊂ expq(Bδ(0)) and so U is a uniformly normal neighbourhood of p.

(1) By deﬁnition of δ, we have Bδ(0) ⊂ Eq and moreover since

F (q, X) = (q, expq X) the inverse map has the form

F −1(q, m) = (q, f(q, m)) for some smooth f.

Writing fq = f(q, m), then clearly fq is smooth and

fq ◦ exp = id| , exp ◦fq = id| q Bδ(0) q expq (Bδ (0))

so the indeed the restriction of expq to Bδ(0) is a diffeomorphism onto its image.

(2) Since F (Wδ) ⊃ U × U, and U ⊂ W for any q ∈ U we have

{q} × U ⊂ F ({q} × Bδ(0))

and since

F (q, X) = (q, expq X) this gives U ⊂ expq(Bδ(0)).

Corollary 24.4. Let p ∈ M and take a uniformly normal neighbourhood U of p. Then for any q, m ∈ U there is a geodesic from q to m, it is the curve between these two points of minimal length, and any other curve between these two points of the same length has the same trace.

Proof: The open set U is also a uniformly normal neighbourhood of q, so we may take the radial geodesic from q to m and then the property that this is the unique (up to reparameterisation) curve of minimal length we proved last lecture. Corollary 24.5. Suppose γ :[a, b] → M is a piecewise smooth curve, satisfying

1. γ is parameterised proportional to arc length 2. for any other piecewise smooth curve α with endpoints γ(a) and γ(b) we have

l(γ) ≤ l(α).

30 Then γ is a geodesic (so in particular is regular).

Proof: Take t ∈ [a, b] and a uniformly normal neighbourhood U of γ(t). Then take an open interval I ⊂ [a, b] containing t and such that γ(I¯) ⊂ U.

From the unique minimising property of geodesics, there is a unique geodesic whose trace agrees with that of γ|I . But then setting X =γ ˙ (t), we have γ = γX on I¯. Thus γ is locally a geodesic and hence a geodesic. Solving the geodesic equations directly is a lot of work. Fortunately for many fundamental examples, there are alternative ways of finding the geodesics. Example 24.6 (Geodesics on Sn(r).). We have proven that the sphere is a homogeneous and isotropic Riemannian manifold O(n + 1) Sn(r) =∼ . O(n) Hence it suffices to calculate the geodesic at one point with one specific velocity vector and then using the isometries given by the action of O(n + 1) we obtain all geodesics. (Scaling the velocity vector simply re-parameterises the geodesic at a different speed) We will take our point to be the North Pole (0,..., 0, r) and our velocity vector a non-zero multiple of (1, 0,..., 0).

31 Example 24.6 (Continued). The corresponding geodesic γ must have trace the great circle lying in the plane

x2 = x3 = ··· = xn = 0.

Indeed, suppose not: then for some parameter value t0 the point γ(t0) does not lie on this plane, and then writing R ∈ O(n + 1) for the reﬂection in this plane we have

R(γ(t0)) 6= γ(t0).

But the isometry R ﬁxes both the North Pole and the velocity vector X, so it must ﬁx the geodesic γ, a contradic- tion.

32 Since Sn(r) is a homogeneous and isotropic Riemannian manifold, this now determines all geodesics on the sphere. Example 24.7 (geodesics on Hn(r)). Using the hyperboloid model

n ∼ O+(n, 1) S1 (r) = O+(n − 1, 1) we previously realised hyperbolic space as a homogeneous and isotropic Riemannian manifold. By the analogous symmetry argument, the geodesics passing through (0,..., 0, r) with velocity vector parallel to (1, 0,..., 0) are the constant speed parameterisations of the hyperbola given by the intersection of the x0–xn plane with the hyperboloid.

Then the action of O+(n, 1) yields all geodesics on the hyperboloid model

33 Hyperbolic Space: upper half space model Example 24.8. The upper half plane model for hyperbolic space of radius r is given by n 1 n n n U (r) = {(x , . . . , x ) ∈ R | x > 0} together with the Riemannian metric deﬁned in coordinates (x1, . . . , xn) r2 gU (x1, . . . , xn) = δ ij (xn)2 ij where we are taking the standard basis vectors ∂i = ei = (0,..., 0, 1, 0,..., 0). Example 24.8 (Continued). Fix x = (x1, . . . , xn−1) ∈ Rn−1 and consider, for a > 0, γ :[a, b] → U n(r) t 7→ (x1, . . . , xn−1, t). Then for any other curve α :[a, b] → U n(r) with the same endpoints as γ we have Z b q i j l(α) = gijα˙ (t)α ˙ (t) dt a v b un−1 Z r uX = t (x ˙ i(t))2 + (x ˙ n(t))2 dt xn(t) a i=1 Z b r n ≥ n |x˙ (t)| dt a x (t) = l(γ). Example 24.8 (Continued). Then we can use our explicit isometry identifying the hyperboloid and upper half plane n models to obtain a transitive action of O+(n, 1) on U (r) by isometries and hence calculate all the geodesics for the upper half plane model. They are the vertical lines and the semicircles with centre on the xn-axis .