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p1 and then the shortest geodesic from p1 to p2 we obtain (after suitable reparametrization) a smooth curve γ : [0, 2] → M such that γ(0) = p, γ(1) = p1, γ(2) = p2, and L(γ\|[0,1]) = d(p, p1) = \|w\|, L(γ\|[1,2]) = d(p1, p2) = \|v\| − \|w\|. Thus L(γ) = \|v\| = d(p, p2). Hence, by Theorem 4.4.4, there is a smooth function β : [0, 2] → [0, 1] satisfying β(0) = 0, β(2) = 1, ˙β(t) ≥ 0, γ(t) = expp(β(t)v) for every t ∈ [0, 2]. This implies expp(w) = p1 = γ(1) = expp(β(1)v), 0 ≤ β(1) ≤ 1. Since w and β(1)v are both elements of Bε(p) and expp is injective on Bε(p), this implies w = β(1)v. Since β(1) ≥ 0 we have β(1) = \|w\| / \|v\|. This proves (5.1.1) and Lemma 5.1.2. 226 CHAPTER 5. CURVATURE Proof of Theorem 5.1.1. That (ii) implies (iii) follows from the deﬁnition of the length of a curve. Namely L(φ ◦ γ dt φ(γ(t)) dt \|dφ(γ(t)) ˙γ(t)\| dt \| ˙γ(t)\| dt a = L(γ). In the third equation we have used (ii). That (iii) implies (i) follows immediately from the deﬁnition of the intrinsic distance functions d and d. We prove that (i) implies (ii). Fix a point p ∈ M and choose ε > 0 so small that ε < min{inj(p; M ), inj(φ(p); M )} and
that the assertion of Lemma 5.1.2 holds for the point p := φ(p) ∈ M. Then there is a unique homeomorphism Φp : Bε(p) → Bε(φ(p)) such that the following diagram commutes. TpM M ⊃ ⊃ Bε(p) Φp Bε(φ(p)) expp exp φ(p) Uε(p) φ Uε(φ(p)) ⊂ ⊂ Tφ(p)M. M Here the vertical maps are diﬀeomorphisms and φ : Uε(p) → Uε(φ(p)) is a homeomorphism by (i). Hence Φp : Bε(p) → Bε(φ(p)) is a homeomorphism. Claim 1. The map Φp satisﬁes the equations exp φ(p)(Φp(v)) = φ(expp(v)), \|Φp(v)\| = \|v\|, Φp(tv) = tΦp(v) (5.1.3) (5.1.4) (5.1.5) for every v ∈ Bε(p) and every t ∈ [0, 1]. Equation (5.1.3) holds by deﬁnition. To prove (5.1.4) we observe that, by Theorem 4.4.4, we have \|Φp(v)\| = d(φ(p), exp φ(p)(Φp(v))) = d(φ(p), φ(expp(v))) = d(p, expp(v)) = \|v\|. 5.1. ISOMETRIES 227 Here the second equation follows from (5.1.3) and the third equation from (i). Equation (5.1.5) holds for t = 0 because Φp(0) = 0 and for t = 1 it is a tautology. Hence assume 0 < t < 1. Then d(exp φ(p)(Φp(tv)), exp φ(p)(Φp(v))) = d(φ(expp(tv)), φ(expp(v
))) = d(expp(tv), expp(v)) = \|v\| − \|tv\| = \|Φp(v)\| − \|Φp(tv)\|. Here the ﬁrst equation follows from (5.1.3), the second equation from (i), the third equation from Theorem 4.4.4 and the fact that \|v\| < inj(p), and the last equation follows from (5.1.4). Since 0 < \|Φp(tv)\| < \|Φp(v)\| < ε we can apply Lemma 5.1.2 and obtain Φp(tv) = \|Φp(tv)\| \|Φp(v)\| Φp(v) = tΦp(v). This proves Claim 1. By Claim 1, Φp extends to a bijective map Φp : TpM → Tφ(p)M via Φp(v) := 1 δ Φp(δv), where δ > 0 is chosen so small that δ \|v\| < ε. The right hand side of this equation is independent of the choice of δ. Hence the extension is well deﬁned. It is bijective because the original map Φp is a bijection from Bε(p) to Bε(φ(p)). The reader may verify that the extended map satisﬁes the conditions (5.1.4) and (5.1.5) for all v ∈ TpM and all t ≥ 0. Claim 2. The extended map Φp : TpM → Tφ(p)M is linear and preserves the inner product. It follows from the equation (4.6.3) in the proof of Lemma 4.6.8 that \|v − w\| = lim t→0 d(expp(tv), expp(tw)) t = lim t→0 = lim t→0 = lim t→0 d(φ(expp(tv)), φ(expp(tw))) t φ(p)(Φp(tv)), exp d(exp φ(p)(Φp(tw))) t d(exp φ(p)(tΦp(v)), exp φ(p)(tΦp(w))) t =
\|Φp(v) − Φp(w)\|. 228 CHAPTER 5. CURVATURE Here the second equation follows from (i), the third from (5.1.3), the fourth from (5.1.4), and the last equation follows again from (4.6.3). By polarization we obtain 2v, w = \|v\|2 + \|w\|2 − \|v − w\|2 = \|Φp(v)\|2 + \|Φp(w)\|2 − \|Φp(v) − Φp(w)\|2 = 2Φp(v), Φp(w). Thus Φp preserves the inner product. Hence, for all v1, v2, w ∈ TpM, we have Φp(v1 + v2), Φp(w) = v1 + v2, w = v1, w + v2, w = Φp(v1), Φp(w) + Φp(v2), Φp(w) = Φp(v1) + Φp(v2), Φp(w). Since Φp is surjective, this implies Φp(v1 + v2) = Φp(v1) + Φp(v2) for all v1, v2 ∈ TpM. With v1 = v and v2 = −v we obtain Φp(−v) = −Φp(v) for every v ∈ TpM and by (5.1.5) this gives Φp(tv) = tΦp(v) for all v ∈ TpM and t ∈ R. This proves Claim 2. Claim 3. φ is smooth and dφ(p) = Φp. By (5.1.3) we have φ = exp φ(p) ◦Φp ◦ exp−1 p : Uε(p) → Uε(φ(p)). Since Φp is linear, this shows that the restriction of φ to the open set Uε(p) is smooth. Moreover, for every v ∈ TpM we have dφ(p)v = d dt t=0 φ(expp(tv)) = d dt
t=0 exp φ(p)(tΦp(v)) = Φp(v). Here we have used equations (5.1.3) and (5.1.5) as well as Lemma 4.3.6. This proves Claim 3 and Theorem 5.1.1. 5.1. ISOMETRIES 229 Exercise 5.1.4. Prove that every isometry ψ : Rn → Rn is an aﬃne map ψ(p) = Ap + b where A ∈ O(n) and b ∈ Rn. Thus ψ is a composition of translation and rotation. Hint: Let e1,..., en be the standard basis of Rn. Prove that any two vectors v, w ∈ Rn that satisfy \|v\| = \|w\| \|v − ei\| = \|w − ei\| for i = 1,..., n and must be equal. Remark 5.1.5. If ψ : Rn → Rn is an isometry of the ambient Euclidean space with ψ(M ) = M, then certainly φ := ψ\|M is an isometry from M onto M. On the other hand, if M is a plane manifold M = {(0, y, z) ∈ R3 \| 0 < y < π/2} and M is the cylindrical manifold M = {(x, y, z) ∈ R3 \| x2 + y2 = 1, x > 0, y > 0}, Then the map φ : M → M deﬁned by φ(0, y, z) := (cos(y), sin(y), z) is an isometry which is not of the form φ = ψ\|M. Indeed, an isometry of the form φ = ψ\|M necessarily preserves the second fundamental form (as well as the ﬁrst) in the sense that dψ(p)hp(v, w) = h ψ(p)(dψ(p)v, dψ(p)w) for v, w ∈ TpM but in the example h vanishes identically while h does not. We may thus distinguish two fundamental question: I. Given M and M when are they extrinsically isomorphic,
i.e. when is there an ambient isometry ψ : Rn → Rn with ψ(M ) = M? II. Given M and M when are they intrinsically isomorphic, i.e. when is there an isometry φ : M → M from M onto M? 230 CHAPTER 5. CURVATURE As we have noted, both the ﬁrst and second fundamental forms are preserved by extrinsic isomorphisms while only the ﬁrst fundamental form need be preserved by an intrinsic isomorphism (i.e. an isometry). A question which occurred to Gauß (who worked for a while as a cartographer) is this: Can one draw a perfectly accurate map of a portion of the earth? (i.e. a map for which the distance between points on the map is proportional to the distance between the corresponding points on the surface of the earth). We can now pose this question as follows: Is there an isometry from an open subset of a sphere to an open subset of a plane? Gauß answered this question negatively by associating an invariant, the Gaußian curvature K : M → R, to a surface M ⊂ R3. According to his Theorema Egregium K ◦ φ = K for an isometry φ : M → M. The sphere has positive curvature; the plane has zero curvature; hence the perfectly accurate map does not exist. Our aim is to explain these ideas. Local Isometries We shall need a concept slightly more general than that of “isometry”. Deﬁnition 5.1.6 (Local isometry). A smooth map φ : M → M is called a local isometry iﬀ its derivative dφ(p) : TpM → Tφ(p)M is an orthogonal linear isomorphism for every p ∈ M. Remark 5.1.7. Let M ⊂ Rn and M ⊂ Rn be manifolds and φ : M → M be a map. The following are equivalent. (i) φ is a local isometry. (ii) For every p ∈ M there are open neighborhoods U ⊂ M and U ⊂ M such that the restriction of φ to U is an isometry from U onto U. That (ii) implies (i) follows immediately from
Theorem 5.1.1. On the other hand (i) implies that dφ(p) is invertible so that (ii) follows from the inverse function theorem. Example 5.1.8. The map R → S1 : θ → eiθ is a local isometry but not an isometry. 5.1. ISOMETRIES 231 Exercise 5.1.9. Let M ⊂ Rn be a compact connected 1-manifold. Prove that M is diﬀeomorphic to the circle S1. Deﬁne the length of a compact connected Riemannian 1-manifold. Prove that two compact connected 1manifolds M, M ⊂ Rn are isometric if and only if they have the same length. Hint: Let γ : R → M be a geodesic with \| ˙γ(t)\| ≡ 1. Show that γ is not injective; otherwise construct an open cover of M without ﬁnite subcover. If t0 < t1 with γ(t0) = γ(t1), show that ˙γ(t0) = ˙γ(t1); otherwise show that γ(t0 + t) = γ(t1 − t) for all t and ﬁnd a contradiction. The next result asserts that two local isometries that have the same value and the same derivative at a single point must agree everywhere, provided that the domain is connected. Lemma 5.1.10. Let M ⊂ Rn and M ⊂ Rn be smooth m-manifolds and assume that M is connected. Let φ : M → M and ψ : M → M be local isometries and let p0 ∈ M such that φ(p0) = ψ(p0) =: p 0, dφ(p0) = dψ(p0) : Tp0M → Tp M. 0 Then φ(p) = ψ(p) for every p ∈ M. Proof. Deﬁne the set M0 := {p ∈ M \| φ(p) = ψ(p), dφ(p) = dψ(p)}. This set is obviously closed. We prove that M0
is open. Let p ∈ M0 and choose U ⊂ M and U ⊂ M as in Remark 5.1.7 (ii). Denote Φp := dφ(p) = dψ(p) : TpM → TpM, p := φ(p) = ψ(p) Then it follows from equation (5.1.3) in the proof of Theorem 5.1.1 that there exists a constant ε > 0 such that Uε(p) ⊂ U and Uε(p) ⊂ U and q ∈ Uε(p) =⇒ φ(q) = exp p ◦Φp ◦ exp−1 p (q) = ψ(q). Hence Uε(p) ⊂ M0. Thus M0 is open, closed, and nonempty. Since M is connected it follows that M0 = M and this proves Lemma 5.1.10. Exercise 5.1.11. (i) If a sequence of local isometries φi : M → M converges uniformly to a local isometry φ : M → M, then it converges in the C∞ topology. Hint: Let p ∈ M. Then every suﬃciently small tangent vector v ∈ TpM satisﬁes the equation dφ(p)v = (exp φ(p))−1(φ(expp(v))). Use this to prove that dφi(p) converges to dφ(p). Deduce that φi converges to φ uniformly with all derviatives in a neighborhood of p. (ii) The C∞ topology on the space of local isometries from M to M agrees with the C0 topology. 232 CHAPTER 5. CURVATURE 5.2 The Riemann Curvature Tensor This section deﬁnes the Riemann curvature tensor and proves the Gauß– Codazzi formula (§5.2.1), introduces the covariant derivative of a global vector ﬁeld (§5.2.2), expresses the curvature tensor in terms of a global formula (§5.2.3), establishes its symmetry properties (§5.2.4),
and examines the curvature for a class of Riemannian metrics on Lie groups (§5.2.5). 5.2.1 Deﬁnition and Gauß–Codazzi Let M ⊂ Rn be a smooth manifold and γ : R2 → M be a smooth map. Denote by (s, t) the coordinates on R2. Let Z ∈ Vect(γ) be a smooth vector ﬁeld along γ, i.e. Z : R2 → Rn is a smooth map such that Z(s, t) ∈ Tγ(s,t)M for all s and t. The covariant partial derivatives of Z with respect to the variables s and t are deﬁned by ∇sZ := Π(γ) ∂Z ∂s, ∇tZ := Π(γ) ∂Z ∂t. In particular ∂sγ = ∂γ/∂s and ∂tγ = ∂γ/∂t are vector ﬁelds along γ and we have ∇s∂tγ − ∇t∂sγ = 0 as both terms on the left are equal to Π(γ)∂s∂tγ. Thus ordinary partial diﬀerentiation and covariant partial diﬀerentiation commute. The analogous formula (which results on replacing ∂ by ∇ and γ by Z) is in general false. Instead we have the following. Deﬁnition 5.2.1. The Riemann curvature tensor assigns to each p ∈ M the bilinear map Rp : TpM × TpM → L(TpM, TpM ) characterized by the equation Rp(u, v)w = ∇s∇tZ − ∇t∇sZ(0, 0) (5.2.1) for u, v, w ∈ TpM where γ : R2 → M is a smooth map and Z ∈ Vect(γ) is a smooth vector ﬁeld along γ such that γ(0, 0) = p, ∂sγ(0, 0) = u, ∂tγ(0, 0) = v, Z(0, 0) = w.
(5.2.2) We must prove that R is well deﬁned, i.e. that the right hand side of equation (5.2.1) is independent of the choice of γ and Z. This follows from the Gauß–Codazzi formula which we prove next. Recall that the second fundamental form can be viewed as a linear map hp : TpM → L(TpM, TpM ⊥) and that, for u ∈ TpM, the linear map hp(u) ∈ L(TpM, TpM ⊥) and its dual hp(u)∗ ∈ L(TpM ⊥, TpM ) are given by hp(u)v = dΠ(p)uv, hp(u)∗w = dΠ(p)uw for v ∈ TpM and w ∈ TpM ⊥. 5.2. THE RIEMANN CURVATURE TENSOR 233 Theorem 5.2.2. The Riemann curvature tensor is well deﬁned and given by the Gauß–Codazzi formula Rp(u, v) = hp(u)∗hp(v) − hp(v)∗hp(u) (5.2.3) for u, v ∈ TpM. Proof. Let u, v, w ∈ TpM and choose a smooth map γ : R2 → M and a smooth vector ﬁeld Z along γ such that (5.2.2) holds. Then, by the Gauß– Weingarten formula (3.2.2), we have ∇tZ = ∂tZ − hγ(∂tγ)Z = ∂tZ − dΠ(γ)∂tγZ = ∂tZ − ∂t Π ◦ γ Z. Hence ∂s∇tZ = ∂s∂tZ − ∂s ∂t = ∂s∂tZ − ∂s∂t = ∂s∂tZ − ∂s∂t = ∂s∂tZ − ∂s∂t Π ◦ γ ∂sZ Π ◦ γ Z − d�
�(γ)∂tγ∇sZ + hγ(∂sγ)Z Π ◦ γ Z − hγ(∂tγ)∇sZ − hγ(∂tγ)∗hγ(∂sγ)Z. Interchanging s and t and taking the diﬀerence we obtain ∂s∇tZ − ∂t∇sZ = hγ(∂sγ)∗hγ(∂tγ)Z − hγ(∂tγ)∗hγ(∂sγ)Z + hγ(∂sγ)∇tZ − hγ(∂tγ)∇sZ. Here the ﬁrst two terms on the right are tangent to M and the last two terms on the right are orthogonal to TγM. Hence ∇s∇tZ − ∇t∇sZ = Π(γ)∂s∇tZ − ∂t∇sZ = hγ(∂sγ)∗hγ(∂tγ)Z − hγ(∂tγ)∗hγ(∂sγ)Z. Evaluating the right hand side at s = t = 0 we ﬁnd that ∇s∇tZ − ∇t∇sZ(0, 0) = hp(u)∗hp(v)w − hp(v)∗hp(u)w. This proves the Gauß–Codazzi equation and shows that the left hand side is independent of the choice of γ and Z. This proves Theorem 5.2.2. 234 CHAPTER 5. CURVATURE 5.2.2 Covariant Derivative of a Global Vector Field So far we have only deﬁned the covariant derivatives of vector ﬁelds along curves. The same method can be applied to global vector ﬁelds. This leads to the following deﬁnition. Deﬁnition 5.2.3 (Covariant derivative). Let M ⊂ Rn be an m-dimensional submanifold and X be a vector ﬁeld on M. Fix a point p ∈ M and a tangent vector v
∈ TpM. The covariant derivative of X at p in the direction v is the tangent vector ∇vX(p) := Π(p)dX(p)v ∈ TpM, where Π(p) ∈ Rn×n denotes the orthogonal projection onto TpM. Remark 5.2.4. Let γ : I → M be a smooth curve on an interval I ⊂ R and let X ∈ Vect(M ) be a smooth vector ﬁeld on M. Then X ◦ γ is a smooth vector ﬁeld along γ and the covariant derivative of X ◦ γ is related to the covariant derivative of X by the formula ∇(X ◦ γ)(t) = ∇˙γ(t)X(γ(t)). (5.2.4) Remark 5.2.5 (Gauß–Weingarten formula). Diﬀerentiating the equation X = ΠX (understood as a function from M to Rn) and using the notation ∂vX(p) := dX(p)v for the derivative of X at p in the direction v we obtain the Gauß–Weingarten formula for global vector ﬁelds: ∂vX(p) = ∇vX(p) + hp(v)X(p). (5.2.5) Remark 5.2.6 (Levi-Civita connection). Diﬀerentiating a vector ﬁeld Y on M covariantly in the direction of another vector ﬁeld X we obtain a vector ﬁeld ∇X Y ∈ Vect(M ) deﬁned by (∇X Y )(p) := ∇X(p)Y (p) for p ∈ M. This gives rise to a family of linear operators ∇X : Vect(M ) → Vect(M ), one for each vector ﬁeld X ∈ Vect(M ), and the assignment Vect(M ) → L(Vect(M ), Vect(M )) : X → ∇X (5.2.6) is itself a linear operator. This linear operator is called the Levi-Civita connection on
the tangent bundle T M. 5.2. THE RIEMANN CURVATURE TENSOR 235 The Levi-Civita connection (5.2.6) satisﬁes the conditions ∇f X (Y ) = f ∇X Y, ∇X (f Y ) = f ∇X Y + (LX f ) Y, LX Y, Z = ∇X Y, Z + Y, ∇X Z, ∇Y X − ∇X Y = [X, Y ] (5.2.7) (5.2.8) (5.2.9) (5.2.10) for all X, Y, Z ∈ Vect(M ) and all f ∈ F (M ), where LX f = df ◦ X and [X, Y ] ∈ Vect(M ) denotes the Lie bracket of the vector ﬁelds X and Y. The conditions (5.2.7) and (5.2.8) assert that the linear operator (5.2.6) is a connection on the tangent bundle T M, condition (5.2.9) asserts that the connection (5.2.6) is Riemannian (i.e. it is compatible with the ﬁrst fundamental form), and condition (5.2.10) asserts that it is torsion-free. The next lemma shows that the Levi-Civita connection (5.2.6) is uniquely determined by (5.2.9) and (5.2.10), and hence is the unique torsion-free Riemannian connection on the tangent bundle T M. Lemma 5.2.7 (Uniqueness Lemma). There is a unique linear operator Vect(M ) → L(Vect(M ), Vect(M )) : X → ∇X satisfying equations (5.2.9) and (5.2.10) for all X, Y, Z ∈ Vect(M ). Proof. Existence follows from the properties of the Levi-Civita connection. We prove uniqueness. Let X → DX be any linear operator from Vect(M ) to L(Vect(M ), Vect(M )) that satisﬁes (5.2.9) and (5.2.10). Then
we have LX Y, Z = DX Y, Z + Y, DX Z, LY X, Z = DY X, Z + X, DY Z, −LZX, Y = −DZX, Y − X, DZY. Adding these three equations we ﬁnd LX Y, Z + LY Z, X − LZX, Y = 2DX Y, Z + DY X − DX Y, Z + X, DY Z − DZY + Y, DX Z − DZX = 2DX Y, Z + [X, Y ], Z + X, [Z, Y ] + Y, [Z, X]. The same equation holds for the Levi-Civita connection and hence DX Y, Z = ∇X Y, Z. This implies DX Y = ∇X Y for all X, Y ∈ Vect(M ). 236 CHAPTER 5. CURVATURE Exercise 5.2.8. In the proof of Lemma 5.2.7 we did not actually use the assumption that the operator DX : Vect(M ) → Vect(M ) is linear nor that the operator X → DX is linear. Prove directly that if a map DX : L(M ) → L(M ) satisﬁes (5.2.9) for all Y, Z ∈ Vect(M ), then DX is linear. Prove that every map Vect(M ) → L(Vect(M ), Vect(M )) : X → DX that satisﬁes (5.2.10) is linear. Exercise 5.2.9. Let φt be the ﬂow of a complete vector ﬁeld X on M and let ψt be the ﬂow of a complete vector ﬁeld Y on M. (i) Prove that the formula X(p, v) := (X(p), dX(p)v) deﬁnes a vector ﬁeld on the tangent bundle T M. Hint: Lemma 4.3.1 and equation (5.2.5). (ii) Prove that the ﬂow of X is given by φt(p, v) := (φt(p), dφt(p)v). (iii) Prove that the vector ﬁ
elds t−1((ψt)∗X − X) converge to [X, Y ] in the C1 topology as t tends to zero. Hint: Establish C0 convergence in Lemma 2.4.18 and then use this result for the vector ﬁelds X and Y. Remark 5.2.10 (The Levi-Civita connection in local coordinates). Let φ : U → Ω be a coordinate chart on an open set U ⊂ M with values in an open set Ω ⊂ Rm. In such a coordinate chart a vector ﬁeld X ∈ Vect(M ) is represented by a smooth map ξ = (ξ1,..., ξm) : Ω → Rm deﬁned by ξ(φ(p)) = dφ(p)X(p) for p ∈ U. If Y ∈ Vect(M ) is represented by η, then ∇X Y is represented by the map (∇ξη)k := m i=1 ∂ηk ∂xi ξi + m i,j=1 ijξiηj. Γk (5.2.11) Here the Γk ij : Ω → R are the Christoﬀel symbols deﬁned by ∂gi m gk 1 2 ∂xj + ∂gj ∂xi − ∂gij ∂x Γk ij :=, =1 (5.2.12) where gij is the metric tensor and gij is the inverse matrix so that gijgjk = δk i j (see Lemma 3.6.5). This formula can be used to prove the existence statement in Lemma 5.2.7 and hence deﬁne the Levi-Civita connection in the intrinsic setting. 5.2. THE RIEMANN CURVATURE TENSOR 237 5.2.3 A Global Formula Lemma 5.2.11. For X, Y, Z ∈ Vect(M ) we have R(X, Y )Z = ∇X ∇Y Z − ∇Y ∇X Z + ∇[X,Y ]Z. (5
.2.13) Proof. Fix a point p ∈ M. Then the right hand side of equation (5.2.13) at p remains unchanged if we multiply each of the vector ﬁelds X, Y, Z by a smooth function f : M → [0, 1] that is equal to one near p. Choosing f with compact support we may therefore assume that the vector ﬁelds X and Y are complete. Let φs denote the ﬂow of X and ψt the ﬂow of Y. Deﬁne the map γ : R2 → M by Then γ(s, t) := φs ◦ ψt(p), s, t ∈ R. ∂sγ = X(γ), ∂tγ = (φs ∗Y )(γ). Hence, by Remark 5.2.4 we have ∇s(Z ◦ γ) = (∇X Z) (γ), ∇t(Z ◦ γ) = ∇φs ∗Y Z (γ). Using Remark 5.2.4 again we obtain ∇s∇t(Z ◦ γ) = ∇∂sγ ∇t∇s(Z ◦ γ) = ∇φs ∇φs ∗Y ∇X Z (γ). ∗Y Z (γ) + ∇∂sφs ∗Y Z (γ), Since ∂ ∂s s=0 φs ∗Y = [X, Y ] and ∂sγ = X(γ), it follows that ∇s∇t(Z ◦ γ)(0, 0) = ∇X ∇Y Z + ∇[X,Y ]Z(p), ∇t∇s(Z ◦ γ)(0, 0) = ∇Y ∇X Z(p). Hence Rp(X(p), Y (p))Z(p) = ∇s∇t(Z ◦ γ) − ∇t∇s(Z ◦ γ)(0, 0) = ∇X ∇Y Z − ∇Y ∇X Z + ∇[X,Y ]Z(p). This proves Lemma 5
.2.11. 238 CHAPTER 5. CURVATURE Remark 5.2.12. Equation (5.2.13) can be written succinctly as [∇X, ∇Y ] + ∇[X,Y ] = R(X, Y ). (5.2.14) This can be contrasted with the equation [LX, LY ] + L[X,Y ] = 0 (5.2.15) for the operator LX on the space of real valued functions on M. Remark 5.2.13. Equation (5.2.13) can be used to deﬁne the Riemann curvature tensor. To do this one must again prove that the right hand side of equation (5.2.13) at p depends only on the values X(p), Y (p), Z(p) of the vector ﬁelds X, Y, Z at the point p. For this it suﬃces to prove that the map Vect(M ) × Vect(M ) × Vect(M ) → Vect(M ) : (X, Y, Z) → R(X, Y )Z is linear over the Ring F (M ) of smooth real valued functions on M, i.e. R(f X, Y )Z = R(X, f Y )Z = R(X, Y )f Z = f R(X, Y )Z (5.2.16) for X, Y, Z ∈ Vect(M ) and f ∈ F (M ). The formula (5.2.16) follows from the equations (5.2.7), (5.2.8), (5.2.15), and [X, f Y ] = f [X, Y ] − (LX f )Y. It follows from (5.2.16) that the right hand side of (5.2.13) at p depends only on the vectors X(p), Y (p), Z(p). The proof requires two steps. One ﬁrst shows that if X vanishes near p, then the right hand side of (5.2.13) vanishes at p (and similarly for Y and Z). Just multiply X by a smooth function equal to zero at p and equal to one on the support of X; then f X
= X and hence the vector ﬁeld R(X, Y )Z = R(f X, Y )Z = f R(X, Y )Z vanishes at p. Second, we choose a local frame E1,..., Em ∈ Vect(M ), i.e. vector ﬁelds that form a basis of TpM for each p in some open set U ⊂ M. Then we may write m m X = ξiEi, Y = ηjEj, Z = ζ kEk m i=1 j=1 k=1 in U. Using the ﬁrst step and the F (M )-multilinearity we obtain R(X, Y )Z = m i,j,k=1 ξiηjζ kR(Ei, Ej)Ek in U. If X (p) = X(p), then ξi(p) = ξi(p) so if X(p) = X (p), Y (p) = Y (p), Z(p) = Z(p), then (R(X, Y )Z)(p) = (R(X, Y )Z)(p) as required. 5.2. THE RIEMANN CURVATURE TENSOR 239 5.2.4 Symmetries Theorem 5.2.14. The Riemann curvature tensor satisﬁes R(Y, X) = −R(X, Y ) = R(X, Y )∗, R(X, Y )Z + R(Y, Z)X + R(Z, X)Y = 0, R(X, Y )Z, W = R(Z, W )X, Y (5.2.17) (5.2.18) (5.2.19) for X, Y, Z, W ∈ Vect(M ). Equation (5.2.18) is the ﬁrst Bianchi identity. Proof. The ﬁrst equation in (5.2.17) is obvious from the deﬁnition and the second follows from the Gauß–Codazzi formula (5.2.3). Alternatively, choose a smooth map γ : R2 → M and two
vector ﬁelds Z, W along γ. Then 0 = ∂s∂tZ, W − ∂t∂sZ, W = ∂s∇tZ, W + ∂sZ, ∇tW − ∂t∇sZ, W − ∂tZ, ∇sW = ∇s∇tZ, W + Z, ∇s∇tW − ∇t∇sZ, W − Z, ∇t∇sW = R(∂sγ, ∂tγ)Z, W + Z, R(∂sγ, ∂tγ)W. This proof has the advantage that it carries over to the intrinsic setting. We prove the ﬁrst Bianchi identity using (5.2.10) and (5.2.13): R(X, Y )Z + R(Y, Z)X + R(Z, X)Y = ∇X ∇Y Z − ∇Y ∇X Z + ∇[X,Y ]Z + ∇Y ∇ZX − ∇Z∇Y X + ∇[Y,Z]X + ∇Z∇X Y − ∇X ∇ZY + ∇[Z,X]Y = ∇[Y,Z]X − ∇X [Y, Z] + ∇[Z,X]Y − ∇Y [Z, X] + ∇[X,Y ]Z − ∇Z[X, Y ] = [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]]. The last term vanishes by the Jacobi identity. We prove (5.2.19) by combining the ﬁrst Bianchi identity with (5.2.17): R(X, Y )Z, W − R(Z, W )X, Y = −R(Y, Z)X, W − R(Z, X)Y, W − R(Z, W )X, Y = R(Y, Z)W, X + R(Z, X)W, Y + R(W, Z)X, Y = R(Y, Z)W, X − R(X, W )Z, Y = R(
Y, Z)W, X − R(W, X)Y, Z. Note that the ﬁrst line is related to the last by a cyclic permutation. Repeating this argument we ﬁnd R(Y, Z)W, X − R(W, X)Y, Z = R(Z, W )X, Y − R(X, Y )Z, W. Combining the last two identities we obtain (5.2.19). This proves Theorem 5.2.14. 240 CHAPTER 5. CURVATURE Remark 5.2.15. We may think of a vector ﬁeld X on M as a section of the tangent bundle. This is reﬂected in the alternative notation Ω0(M, T M ) := Vect(M ). A 1-form on M with values in the tangent bundle is a collection of linear maps A(p) : TpM → TpM, one for every p ∈ M, which is smooth in the sense that for every smooth vector ﬁeld X on M the assignment p → A(p)X(p) deﬁnes again a smooth vector ﬁeld on M. We denote by Ω1(M, T M ) the space of smooth 1-forms on M with values in T M. The covariant derivative of a vector ﬁeld Y is such a 1-form with values in the tangent bundle which assigns to every p ∈ M the linear map TpM → TpM : v → ∇vY (p). Thus we can think of the covariant derivative as a linear operator ∇ : Ω0(M, T M ) → Ω1(M, T M ). The equation (5.2.7) asserts that the operators X → ∇X indeed determine a linear operator from Ω0(M, T M ) to Ω1(M, T M ). Equation (5.2.8) asserts that this linear operator ∇ is a connection on the tangent bundle of M. Equation (5.2.9) asserts that ∇ is a Riemannian connection and equation (5.2.10) asserts that ∇ is torsion-free. Thus Lemma 5.2.7 can be rest
ated as asserting that the Levi-Civita connection is the unique torsion-free Riemannian connection on the tangent bundle. Exercise 5.2.16. Extend the notion of a connection to a general vector bundle E, both as a collection of linear operators ∇X : Ω0(M, E) → Ω0(M, E), one for every vector ﬁeld X ∈ Vect(M ), and as a linear operator ∇ : Ω0(M, E) → Ω1(M, E) satisfying the analogue of equation (5.2.8). Interpret this equation as a Leibniz rule for the product of a function on M with a section of E. Show that ∇⊥ is a connection on T M ⊥. Extend the notion of curvature to connections on general vector bundles. Exercise 5.2.17. Show that the ﬁeld which assigns to each p ∈ M the multi-linear map R⊥ p : TpM × TpM → L(TpM ⊥, TpM ⊥) characterized by t ∇⊥ t Y − ∇⊥ R⊥(∂sγ, ∂tγ)Y = ∇⊥ s Y for γ : R2 → M and Y ∈ Vect⊥(γ) satisﬁes the equation p (u, v) = hp(u)hp(v)∗ − hp(v)hp(u)∗ s ∇⊥ R⊥ for p ∈ M and u, v ∈ TpM. 5.2. THE RIEMANN CURVATURE TENSOR 241 5.2.5 Riemannian Metrics on Lie Groups We begin with a calculation of the Riemann curvature tensor on a Lie subgroup of the orthogonal group O(n) with the Riemannian metric inherited from the standard inner product v, w := trace(vTw) (5.2.20) on the ambient space gl(n, R) = Rn×n. This ﬁts into the extrinsic setting used throughout most of this book. Note that every Lie subgroup of O(n) is a closed subset of O(n)
by Theorem 2.5.27 and hence is compact. Example 5.2.18. Let G ⊂ O(n) be a Lie subgroup and let g := Lie(G) = T1lG be the Lie algebra of G. Consider the Riemannian metric on G induced by the inner product (5.2.20) on Rn×n. Then the Riemann curvature tensor on G can be expressed in terms of the Lie bracket (see item (d) below). (a) The maps g → ag, g → ga, g → g−1 are isometries of G for every a ∈ G. (b) Let γ : R → G be a smooth curve and X ∈ Vect(γ) be a smooth vector ﬁeld along γ. Then the covariant derivative of X is given by γ(t)−1∇X(t) = d dt γ(t)−1X(t) + 1 2 γ(t)−1 ˙γ(t), γ(t)−1X(t). (5.2.21) (Exercise: Prove equation (5.2.21). Hint: Since g ⊂ o(n) we have the identity trace((ξη + ηξ)ζ) = 0 for all ξ, η, ζ ∈ g.) (c) A smooth map γ : R → G is a geodesic if and only if there exist matrices g ∈ G and ξ ∈ g such that γ(t) = g exp(tξ). (5.2.22) For G = O(n) we have seen this in Example 4.3.12 and in the general case this follows from equation (5.2.21) with X = ˙γ. Hence the exponential map exp : g → G deﬁned by the exponential matrix (as in §2.5) agrees with the time-1-map of the geodesic ﬂow (as in §4.3). (d) The Riemann curvature tensor on G is given by g−1Rg(u, v)w = − 1 4 [[g−1u, g−1v], g−1w] (5.
2.23) for g ∈ G and u, v, w ∈ TgG. Note that the ﬁrst Bianchi identity is equivalent to the Jacobi identity. (Exercise: Prove equation (5.2.23).) 242 CHAPTER 5. CURVATURE Deﬁnition 5.2.19 (Bi-invariant Riemannian metric). Let G be a Lie subgroup of GL(n, R) and let g = Lie(G) = T1lG be its Lie algebra. A Riemannian metric on G is called bi-invariant iﬀ it has the form v, wg := vg−1, wg−1 for g ∈ G and v, w ∈ TgG, where ·, · is an inner product on the Lie algebra g that is invariant under conjugation, i.e. it satisﬁes the equation (5.2.24) ξ, η = gξg−1, gηg−1. (5.2.25) for all ξ, η ∈ g and all g ∈ G. Exercise 5.2.20. Prove that the Riemannian metric induced by (5.2.20) on any Lie subgroup G ⊂ O(n) is bi-invariant. Exercise 5.2.21. Use the Haar measure ([66, Chapter 8]) to prove that every compact Lie group admits a bi-invariant Riemannian metric. Exercise 5.2.22. Prove that all the assertions in Example 5.2.18 carry over verbatim to any Lie group equipped with a bi-invariant Riemann metric. Exercise 5.2.23 (Invariant inner product). Prove that, if an inner product on the Lie algebra g of a Lie group G is invariant under conjugation, then it satisﬁes the equation [ξ, η], ζ = ξ, [η, ζ] (5.2.26) for all ξ, η, ζ ∈ g. If G is connected, prove that, conversely, equation (5.2.26) implies (5.2.25). An inner product on an arbitrary Lie algebra g is called invari
ant iﬀ it satisﬁes equation (5.2.26). Exercise 5.2.24 (Commutant). Let g be a ﬁnite-dimensional Lie algebra. The linear subspace spanned by all vectors of the form [ξ, η] is called the commutant of g and is denoted by [g, g] := span {[ξ, η] \| ξ, η ∈ g}. If g is equipped with an invariant inner product, prove that [g, g]⊥ = Z(g) is the center of g (Exercise 2.5.34) and hence g = [g, g] ⊕ Z(g). Prove that the Heisenberg algebra h in Exercise 2.5.15 satisﬁes [h, h] = Z(h) and hence does not admit an invariant inner product. Example 5.2.25 (Killing form). Every ﬁnite-dimensional Lie algebra g admits a natural symmetric bilinear form κ : g × g → R that satisﬁes equation (5.2.26). It is called the Killing form and is deﬁned by κ(ξ, η) := tracead(ξ)ad(η), ξ, η ∈ g, (5.2.27) where ad : g → Der(g) is the adjoint representation in Example 2.5.23. The Killing form may have a kernel (which always contains the center of g), and even if it is nondegenerate, it may be indeﬁnite. 5.2. THE RIEMANN CURVATURE TENSOR 243 Exercise 5.2.26. Prove that κ([ξ, η], ζ) = κ(ξ, [η, ζ]) for all ξ, η, ζ ∈ g. Exercise 5.2.27. Assume that g admits an invariant inner product. For each ξ ∈ g prove that the derivation ad(ξ) is skew-adjoint with respect to this inner product and deduce that κ(ξ, ξ) = −trace(ad(ξ)∗ad(ξ)) = −\|ad(
ξ)\|2. Deduce that the Killing form is nondegenerate whenever g has a trivial center and admits an invariant inner product. Example 5.2.28 (Right invariant Riemannian metric). Let G be any Lie subgroup of GL(n, R) (not necessarily contained in O(n)), and let g := Lie(G) = T1lG be its Lie algebra. Fix any inner product on the Lie algebra g (not necessarily invariant under conjugation) and consider the Riemannian metric on G deﬁned by v, wg := vg−1, wg−1 (5.2.28) for v, w ∈ TgG. This metric is called right invariant. (a) The map g → ga is an isometry of G for every a ∈ G. (b) Deﬁne the linear map A : g → End(g) by A(ξ)η, ζ = 1 2 ξ, [η, ζ] − η, [ζ, ξ] − ζ, [ξ, η] (5.2.29) for ξ, η, ζ ∈ g. Then A is the unique linear map that satisﬁes A(ξ) + A(ξ)∗ = 0, A(η)ξ − A(ξ)η = [ξ, η] for all ξ, η ∈ g, where A(ξ)∗ is the adjoint operator with respect to the inner product on g. Let γ : R → G be a smooth curve and X ∈ Vect(γ) be a smooth vector ﬁeld along γ. Then the covariant derivative of X is given by ∇X = d dt (Xγ−1) + A( ˙γγ−1)Xγ−1 γ. (5.2.30) (Exercise: Prove this. Moreover, if the inner produt on g is invariant, prove that A(ξ)η = − 1 (c) A smooth curve γ : R → G is a geodesic if and only if it satisﬁes 2 [ξ, η] for all ξ, η �
� g.) d dt ( ˙γγ−1) + A( ˙γγ−1) ˙γγ−1 = 0. (5.2.31) (Exercise: G is complete.) (d) The Riemann curvature tensor on G is given by Rg(ξg, ηg)ζgg−1 = A ([ξ, η]) + [A (ξ), A (η)] ζ (5.2.32) for g ∈ G and ξ, η, ζ ∈ g. (Exercise: Prove this.) 244 CHAPTER 5. CURVATURE 5.3 Generalized Theorema Egregium We will now show that geodesics, covariant diﬀerentiation, parallel transport, and the Riemann curvature tensor are all intrinsic, i.e. they are intertwined by isometries. In the extrinsic setting these results are somewhat surprising since these objects are all deﬁned using the second fundamental form, whereas isometries need not preserve the second fundamental form in any sense but only the ﬁrst fundamental form. Below we shall give a formula expressing the Gaußian curvature of a surface M 2 in R3 in terms of the Riemann curvature tensor and the ﬁrst fundamental form. It follows that the Gaußian curvature is also intrinsic. This fact was called by Gauß the “Theorema Egregium” which explains the title of this section. 5.3.1 Pushforward We assume throughout this section that M ⊂ Rn and M ⊂ Rn are smooth submanifolds of the same dimension m. As in §5.1 we denote objects on M by the same letters as objects in M with primes aﬃxed. In particular, g denotes the ﬁrst fundamental form on M and R denotes the Riemann curvature tensor on M. Let φ : M → M be a diﬀeomorphism. Using φ we can move objects on M to M. For example the pushforward of a smooth curve γ : I → M is the curve φ∗γ := φ ◦ γ : I → M, the pushforward of a smooth function f : M → R
is the function φ∗f := f ◦ φ−1 : M → R, the pushforward of a vector ﬁeld X ∈ Vect(γ) along a curve γ : I → M is the vector ﬁeld φ∗X ∈ Vect(φ∗γ) deﬁned by (φ∗X)(t) := dφ(γ(t))X(t) for t ∈ I, and the pushforward of a global vector ﬁeld X ∈ Vect(M ) is the vector ﬁeld φ∗X ∈ Vect(M ) deﬁned by (φ∗X)(φ(p)) := dφ(p)X(p) for p ∈ M. Recall that the ﬁrst fundamental form on M is the Riemannian metric g deﬁned as the restriction of the Euclidean inner product on the 5.3. GENERALIZED THEOREMA EGREGIUM 245 ambient space to each tangent space of M. It assigns to each p ∈ M the bilinear map gp : TpM × TpM → R given by gp(u, v) = u, v, u, v ∈ TpM. Its pushforward is the Riemannian metric which assigns to each p ∈ M the inner product (φ∗g)p : TpM × TpM → R deﬁned by (φ∗g)φ(p)(dφ(p)u, dφ(p)v) := gp (u, v) for p := φ−1(p) ∈ M and u, v ∈ TpM. The pushforward of the Riemann curvature tensor is the tensor which assigns to each p ∈ M the bilinear map (φ∗R)p : TpM × TpM → L TpM, TpM, deﬁned by (φ∗R)φ(p)(dφ(p)u, dφ(p)v) := dφ(p)Rp (u, v) dφ(p)−1 for p := φ−
1(p) ∈ M and u, v ∈ TpM. 5.3.2 Theorema Egregium Theorem 5.3.1 (Theorema Egregium). The ﬁrst fundamental form, covariant diﬀerentiation, geodesics, parallel transport, and the Riemann curvature tensor are intrinsic. This means that for every isometry φ : M → M the following holds. (i) φ∗g = g. (ii) If X ∈ Vect(γ) is a vector ﬁeld along a smooth curve γ : I → M, then ∇(φ∗X) = φ∗∇X, and if X, Y ∈ Vect(M ) are global vector ﬁelds, then ∇ φ∗X φ∗Y = φ∗(∇X Y ). (5.3.1) (5.3.2) (iii) If γ : I → M is a geodesic, then φ ◦ γ : I → M is a geodesic. (iv) If γ : I → M is a smooth curve, then for all s, t ∈ I, we have Φ φ◦γ(t, s)dφ(γ(s)) = dφ(γ(t))Φγ(t, s). (5.3.3) (v) φ∗R = R. 246 CHAPTER 5. CURVATURE Proof. Assertion (i) is simply a restatement of Theorem 5.1.1. To prove (ii) we choose a local smooth parametrization ψ : Ω → U of an open set U ⊂ M, deﬁned on an open set Ω ⊂ Rm, so that ψ−1 : U → Ω is a coordinate chart. Suppose without loss of generality that γ(t) ∈ U for all t ∈ I and deﬁne c : I → Ω and ξ : I → Rm by γ(t) = ψ(c(t)), X(t) = m i=1 ξi(t) ∂ψ ∂xi (
c(t)). Recall from equations (3.6.6) and (3.6.7) that ∇X(t) =   ˙ξk(t) + m k=1 m i,j=1  ij(c(t)) ˙ci(t)ξj(t) Γk  ∂ψ ∂xk (c(t)), where the Christoﬀel symbols Γk ij : Ω → R are deﬁned by Π(ψ) ∂2ψ ∂xi∂xj = m k=1 Γk ij ∂ψ ∂xk. Now consider the same formula for φ∗X using the parametrization ψ := φ ◦ ψ : Ω → U := φ(U ) ⊂ M. The Christoﬀel symbols Γk U are deﬁned by the same formula as the Γk the metric tensor for ψ agrees with the metric tensor for ψ: ij : Ω → R associated to this parametrization of ij with ψ replaced by ψ. But gij = ∂ψ ∂xi, ∂ψ ∂xj = ∂ψ ∂xi, ∂ψ ∂xj. Hence it follows from Lemma 3.6.5 that Γk ij = Γk that the covariant derivative of φ∗X is given by  m   ˙ξk + m ij(c) ˙ciξj Γk  ∇(φ∗X) = ij for all i, j, k. This implies ∂ψ ∂xk (c)  ij(c) ˙ciξj Γk  ∂ψ ∂xk (c) i,j=1   ˙ξk + m k=1 m i,j=1 k=1 = dφ(ψ(c)) = φ∗∇X. This proves (5.3.1). Equation (
5.3.2) follows immediately from (5.3.1) and Remark 5.2.4. 5.3. GENERALIZED THEOREMA EGREGIUM 247 Here is a second proof of (ii). For every vector ﬁeld X ∈ Vect(M ) we deﬁne the operator DX : Vect(M ) → Vect(M ) by DX Y := φ∗ (∇φ∗X φ∗Y ). Then, for all X, Y ∈ Vect(M ), we have DY X − DX Y = φ∗ (∇φ∗Y φ∗X − ∇φ∗X φ∗Y ) = φ∗[φ∗X, φ∗Y ] = [X, Y ]. Moreover, it follows from (i) that φ∗LX Y, Z = Lφ∗X φ∗Y, φ∗Z = ∇φ∗X φ∗Y, φ∗Z + φ∗Y, ∇φ∗X φ∗Z = φ∗DX Y, φ∗Z + φ∗Y, φ∗DX Z DX Y, Z + Y, DX Z. = φ∗ and hence LX Y, Z = DX Y, Z + Y, DX Z for all X, Y, Z ∈ Vect(M ). Thus the operator X → DX satisﬁes equations (5.2.9) and (5.2.10) and, by Lemma 5.2.7, it follows that DX Y = ∇X Y for all X, Y ∈ Vect(M ). This completes the second proof of (ii). We prove (iii). Since φ preserves the ﬁrst fundamental form it also preserves the energy of curves, namely E(φ ◦ γ) = E(γ) for every smooth map γ : [0, 1] → M. Hence γ is a critical point of the energy functional if and only if φ ◦ γ is a critical point of the energy functional. Alternatively it follows from (ii) that ∇ d dt φ ◦ γ = ∇φ
∗ ˙γ = φ∗∇ ˙γ for every smooth curve γ : I → M. If γ is a geodesic, the last term vanishes and hence φ◦γ is a geodesic as well. As a third proof we can deduce (iii) from the formula φ(expp(v)) = expφ(p)(dφ(p)v) in the proof of Theorem 5.1.1. We prove (iv). For t0 ∈ I and v0 ∈ Tγ(t0)M deﬁne X(t) := Φγ(t, t0)v0, X (t) := Φ φ◦γ(t, t0)dφ(γ(t0))v0. By (ii) the vector ﬁelds X and φ∗X along φ ◦ γ are both parallel and they agree at t = t0. Hence X (t) = φ∗X(t) for all t ∈ I and this proves (5.3.3). 248 CHAPTER 5. CURVATURE We prove (v). Fix a smooth map γ : R2 → M and a smooth vector ﬁeld Z along γ, and deﬁne γ = φ ◦ γ : R2 → M, Z := φ∗Z ∈ Vect(γ). Then it follows from (ii) that R(∂sγ, ∂tγ)Z = ∇ t∇ s∇ sZ tZ − ∇ = φ∗ (∇s∇tZ − ∇t∇sZ) = dφ(γ)R(∂sγ, ∂tγ)Z = (φ∗R)(∂sγ, ∂tγ)Z. This proves (v) and Theorem 5.3.1. The assertions of Theorem 5.3.1 carry over in slightly modiﬁed form to local isometries φ : M → M. In particular, the pushforward of a vector ﬁeld on M under φ is only deﬁned when φ is a diﬀeomorphism while the pushforward
of a vector ﬁeld along a curve is deﬁned for any smooth map φ. Also, the pushforward of the Riemann curvature tensor under a local isometry is only deﬁned locally, and local isometries satisfy the local analogue of the ﬁrst assertion in Theorema Egregium by deﬁnition. Corollary 5.3.2 (Theorema Egregium for Local Isometries). Every local isometry φ : M → M has the following properties. (i) Every vector ﬁeld X along a smooth curve γ : I → M satisﬁes (5.3.1). (ii) If γ : I → M is a geodesic, then so is φ ◦ γ : I → M. (iii) Parallel transport along a smooth curve γ : I → M satisﬁes (5.3.3). (iv) The curvature tensors R of M and R of M are related by the formula R φ(p)(dφ(p)u, dφ(p)v) = dφ(p)Rp(u, v)dφ(p)−1 (5.3.4) for all p ∈ M and all u, v ∈ TpM. Proof. Let p0 ∈ M. Then, by the Inverse Function Theorem 2.2.17, there exists an open neighborhood U ⊂ M of p0 such that U := φ(U ) is an open subset of M and the restriction φ\|U : U → U is a diﬀeomorphism. This restriction is an isometry by Theorem 5.1.1. Hence, by Theorem 5.3.1 the assertions (i) and (ii) hold for the restriction of γ to I0 := γ−1(U ) (a union of subintervals of I) and (iv) holds for all p ∈ U. Since these are local statements and p0 was chosen arbitrary, this proves (i), (ii), and (iv). Part (iii) follows directly from (i) as in the proof of Theorem 5.3.1 and this proves Corollary 5.3.2. 5.3. GENERALIZ
ED THEOREMA EGREGIUM 249 The next corollary spells out a useful consequence of Corollary 5.3.2. For suﬃciently small tangent vectors equation (5.3.5) below already appeared in the proof of Theorem 5.1.1 and was used in Lemma 5.1.10 and Exercise 5.1.11. When M is not complete, recall the notation Vp ⊂ TpM for the domain of the exponential map of M at a point p (Deﬁnition 4.3.5). For p ∈ M denote the domain of the exponential map by V p ⊂ TpM. Corollary 5.3.3. Let φ : M → M be a local isometry and let p ∈ M. Then dφ(p)Vp ⊂ V φ(p) and, for every v ∈ Vp, φ(expp(v)) = exp φ(p)(dφ(p)v). (5.3.5) Proof. Let v ∈ Vp ⊂ TpM and deﬁne γ(t) := expp(tv) for 0 ≤ t ≤ 1. Then γ : [0, 1] → M is a geodesic by Lemma 4.3.6, and hence so is the curve γ := φ ◦ γ : [0, 1] → M by Corollary 5.3.2. Moreover, γ(0) = φ(γ(0)) = φ(p), ˙γ(0) = dφ(γ(0)) ˙γ(0) = dφ(p)v by the chain rule. Hence it follows from the deﬁnition of the exponential map (Deﬁnition 4.3.5) that dφ(p)v ∈ V φ(p) and exp φ(p)(dφ(p)v) = γ(1) = φ(γ(1)) = φ(expp(v)). This proves Corollary 5.3.3. 5.3.3 Gaußian Curvature As a special case we shall now consider a hypersurface M ⊂ Rm+1, i
.e. a smooth submanifold of codimension one. We assume that there exists a smooth map ν : M → Rm+1 such that, for every p ∈ M, we have ν(p) ⊥ TpM, \|ν(p)\| = 1. Such a map always exists locally (see Example 3.1.3). Note that ν(p) is an element of the unit sphere in Rm+1 for every p ∈ M and hence we can regard ν as a map from M to Sm, i.e. ν : M → Sm. Such a map is called a Gauß map for M. Note that if ν : M → Sm is a Gauß map, so is −ν, but this is the only ambiguity when M is connected. Diﬀerentiating ν at p ∈ M we obtain a linear map dν(p) : TpM → Tν(p)Sm = TpM Here we use the fact that Tν(p)Sm = ν(p)⊥ and, by deﬁnition of the Gauß map ν, the tangent space of M at p is also equal to ν(p)⊥. Thus dν(p) is a linear map from the tangent space of M at p to itself. 250 CHAPTER 5. CURVATURE Deﬁnition 5.3.4. The Gaußian curvature of the hypersurface M is the real valued function K : M → R deﬁned by K(p) := detdν(p) : TpM → TpM for p ∈ M. (Replacing ν by −ν has the eﬀect of replacing K by (−1)mK; so K is independent of the choice of the Gauß map when m is even.) Remark 5.3.5. Given a subset B ⊂ M, the set ν(B) ⊂ Sm is often called the spherical image of B. If ν is a diﬀeomorphism on a neighborhood of B, the change of variables formula for an integral gives ν(B) µS = B \|K\|µM. Here µM and µS denote the volume elements on M and Sm, respectively. Introducing the notation AreaM
(B) := B µM we obtain the formula \|K(p)\| = lim B→p AreaS(ν(B)) AreaM (B). This says that the curvature at p is roughly the ratio of the (m-dimensional) area of the spherical image ν(B) to the area of B where B is a very small open neighborhood of p in M. The sign of K(p) is positive when the linear map dν(p) : TpM → TpM preserves orientation and negative when it reverses orientation. Remark 5.3.6. We see that the Gaußian curvature is a natural generalIndeed if M ⊂ R2 is a ization of Euler’s curvature for a plane curve. 1-manifold and p ∈ M, we can choose a curve γ = (x, y) : (−ε, ε) → M such that γ(0) = p and \| ˙γ(s)\| = 1 for every s. This curve parametrizes M by the arclength and the unit normal vector pointing to the right with respect to the orientation of γ is ν(x, y) = ( ˙y, − ˙x). This is a local Gauß map and its derivative (¨y, −¨x) is tangent to the curve. The inner product of the latter with the unit tangent vector ˙γ = ( ˙x, ˙y) is the Gaußian curvature. Thus K := dx ds d2y ds2 − dy ds d2x ds2 = dθ ds where s is the arclength parameter and θ is the angle made by the normal (or the tangent) with some constant line. With this convention K is positive at a left turn and negative at a right turn. 5.3. GENERALIZED THEOREMA EGREGIUM 251 Exercise 5.3.7. The Gaußian curvature of an m-dimensional sphere of radius r is constant and has the value r−m (with respect to an outward pointing Gauß map when m is odd). Exercise 5.3.8. Show that the Gaußian curvature of the surface z = x2 −y2 is −4 at the origin. We now restrict to the case of surfaces, i.e.
of 2-dimensional submanifolds of R3. Figure 5.1 illustrates the diﬀerence between positive and negative Gaußian curvature in dimension two. Figure 5.1: Positive and negative Gaußian curvature. Theorem 5.3.9 (Gaußian curvature). Let M ⊂ R3 be a surface and ﬁx a point p ∈ M. If u, v ∈ TpM is a basis, then K(p) = R(u, v)v, u \|u\|2\|v\|2 − u, v2. (5.3.6) Moreover, R(u, v)w = −K(p)ν(p), u × vν(p) × w (5.3.7) for all u, v, w ∈ TpM. Proof. The orthogonal projection of R3 onto the tangent space TpM = ν(p)⊥ is given by the 3 × 3-matrix Π(p) = 1l − ν(p)ν(p)T. Hence dΠ(p)u = −ν(p)(dν(p)u)T − dν(p)uν(p)T. Here the ﬁrst summand is the second fundamental form, which maps TpM to TpM ⊥, and the second summand is its dual, which maps TpM ⊥ to TpM. Thus hp(v) = −ν(p)dν(p)vT : TpM → TpM ⊥, hp(u)∗ = −dν(p)uν(p)T : TpM ⊥ → TpM. K > 0K < 0K = 0 252 CHAPTER 5. CURVATURE By the Gauß–Codazzi formula this implies Rp(u, v)w = hp(u)∗hp(v)w − hp(v)∗hp(u)w = dν(p)udν(p)vTw − dν(p)vdν(p)uTw = dν(p)v, wdν(p)u − dν(p)u, wdν(p)v and hence Rp(u
, v)w, z = dν(p)u, zdν(p)v, w − dν(p)u, wdν(p)v, z. (5.3.8) Now ﬁx four tangent vectors u, v, w, z ∈ TpM and consider the composition R3 A−→ R3 B−→ R3 C−→ R3 of the linear maps Aξ := ξ1ν(p) + ξ2u + ξ3v, dν(p)η, η, Bη := if η ⊥ ν(p), if η ∈ Rν(p),  Cζ :=   ζ, ν(p) ζ, z ζ, w . This composition is represented by the matrix   CBA = 0 1 0 dν(p)u, z 0 dν(p)u, w 0 dν(p)v, z dν(p)v, w  . Hence, by (5.3.8), we have Rp(u, v)w, z = det(CBA) = det(A) det(B) det(C) = ν(p), u × vK(p)ν(p), z × w = −K(p)ν(p), u × vν(p) × w, z. This implies (5.3.7) and Rp(u, v)v, u = K(p)ν(p), u × v2 = K(p) \|u × v\|2 = K(p) \|u\|2 \|v\|2 − u, v2. This proves Theorem 5.3.9. 5.3. GENERALIZED THEOREMA EGREGIUM 253 Remark 5.3.10. Equation (5.3.6) implies Rp(u, v)w, z = K(p) u, zv, w − u, wv, z (5.3.9) for all p ∈ M and all u, v, w, z ∈ TpM. This is proved in Theorem 6.4.8 below. Exercise: Deduce this formula from (
5.3.7). Corollary 5.3.11 (Theorema Egregium of Gauß). The Gaußian curvature is intrinsic, i.e. if is an isometry of surfaces in R3, then. Proof. Theorem 5.3.1 and Theorem 5.3.9. Exercise 5.3.12. For m = 1 the Gaußian curvature is clearly not intrinsic as any two curves are locally isometric (parameterized by arclength). Show that the curvature K(p) is intrinsic for even m while its absolute value \|K(p)\| is intrinsic for odd m ≥ 3. Hint: We still have the equation (5.3.8) which, for z = u and v = w, can be written in the form Rp(u, v)v, u = det dν(p)u, u dν(p)v, u dν(p)u, v dν(p)v, v. Thus, for every orthonormal basis v1,..., vm of TpM, the 2 × 2 minors of the matrix (dν(p)vi, vj)i,j=1,...,m are intrinsic. Hence everything reduces to the following assertion. Lemma. The determinant of an m × m matrix is an expression in its 2 × 2 minors if m is even; the absolute value of the determinant is an expression in the 2 × 2 minors if m is odd and greater than or equal to 3. The lemma is proved by induction on m. For the absolute value, note the formula det(A)m = det(det(A)1lm) = det(AB) = det(A) det(B) for an m × m-matrix A where B is the transposed matrix of cofactors. 254 CHAPTER 5. CURVATURE 5.4 Curvature in Local Coordinates* Riemann Let M ⊂ Rk be an m-dimensional manifold and let φ = ψ−1 : U → Ω be a local coordinate chart on an open set U ⊂ M with values in an open set Ω ⊂ Rm. Deﬁne the vector ﬁelds E1,..., Em along ψ by Ei(x) := ∂�
� ∂xi (x) ∈ Tψ(x)M. These vector ﬁelds form a basis of Tψ(x)M for every x ∈ Ω and the coeﬃcients gij : Ω → R of the ﬁrst fundamental form are gij = Ei, Ej. Recall ij : Ω → R are the coeﬃcients of from Lemma 3.6.5 that the Christoﬀel Γk the Levi-Civita connection, deﬁned by m ∇iEj = Γk ijEk k=1 and that they are given by the formula m Γk ij := gk 1 2 ∂igj + ∂jgi − ∂gij. =1 ijk : Ω → R and Rijk : Ω → R of the Riemann cur- Deﬁne the coeﬃcients R vature tensor by R(Ei, Ej)Ek = m =1 R ijkE, Rijk := R(Ei, Ej)Ek, E = m ν=1 Rν ijkgν. These coeﬃcients are given by R ijk = ∂iΓ jk − ∂jΓ ik + m ν=1 iνΓν Γ jk − Γ jνΓν ik (5.4.1) (5.4.2). (5.4.3) The coeﬃcients of the Riemann curvature tensor have the symmetries Rijk = −Rjik = −Rijk = Rkij (5.4.4) and the ﬁrst Bianchi identity has the form R ijk + R jki + R kij = 0, Rijk + Rjki + Rkij = 0. (5.4.5) Warning: Care must be taken with the ordering of the indices. Some authors use the notation R kij for what we call R ijk and Rkij for what we call Rijk. 5.4. CURVATURE IN LOCAL COORDINATES* 255 Exercise 5.4.1. Prove equations (5
.4.3), (5.4.4), and (5.4.5). Use (5.4.3) to give an alternative proof of Theorem 5.3.1. Gauß If M ⊂ Rn is a 2-manifold (not necessarily embedded in R3), we can use equation (5.3.6) as the deﬁnition of the Gaußian curvature K : M → R. Let ψ : Ω → U be a local parametrization of an open set U ⊂ M deﬁned on an open set Ω ⊂ R2. Denote the coordinates in R2 by (x, y) and deﬁne the functions E, F, G : Ω → R by E := \|∂xψ\|2, F := ∂xψ, ∂yψ, G := \|∂yψ\|2. We abbreviate D := EG − F 2. Then the composition of the Gaußian curvature K : M → R with the parametrization ψ is given by 2 ∂xG ∂yF − 1 1 2 ∂yG y E + ∂x∂yF − 1 2 ∂2 xG   K ◦ ψ =   1 D2 det − 1 D2 det = − 1 √ D 2 ∂ ∂x 1 E F F G 2 ∂xE ∂xF − 1  E F 2 ∂yE 1 ∂xG − ∂yF √ F G 2 ∂xG  1 D 2 ∂yE − 1 2 ∂2  1 2 ∂yE 1 2 ∂xG 0  − 1 4D2 det   E ∂xE ∂yE F ∂xF ∂yF G ∂xG ∂yG − ∂yE − ∂xF √ D ∂ ∂y 1 √ D 2  . This expression simpliﬁes dramatically when F = 0 and we get K ◦ ψ = − 1 √ EG 2 ∂ ∂x ∂xG √ EG + �
� ∂y ∂yE √ EG. (5.4.6) Exercise 5.4.2. Prove that the Riemannian metric E = G = 4 (1 + x2 + y2)2, on R2 has constant constant curvature K = 1 and the Riemannian metric 4 (1 − x2 − y2)2, E = G = F = 0, F = 0, on the open unit disc has constant curvature K = −1. 256 CHAPTER 5. CURVATURE Chapter 6 Geometry and Topology In this chapter we address what might be called the “fundamental problem of intrinsic diﬀerential geometry”: when are two manifolds isometric? The central tool for addressing this question is the Cartan–Ambrose–Hicks Theorem (§6.1). In the subsequent sections we will use this result to examine ﬂat spaces (§6.2), symmetric spaces (§6.3), and constant sectional curvature manifolds (§6.4). The chapter then examines manifolds of nonpositive sectional curvature and includes a proof of the Cartan Fixed Point Theorem (§6.5). The last three sections introduce the Ricci tensor and show that complete manifolds with uniformly positive Ricci tensor are compact (§6.6) and discuss the scalar curvature (§6.7) and the Weyl tensor (§6.8). 6.1 The Cartan–Ambrose–Hicks Theorem The Cartan–Ambrose–Hicks Theorem answers the question (at least locally) when two manifolds are isometric. In general the equivalent conditions given there are probably more diﬃcult to verify in most examples than the condition that there exist an isometry. However, under additional assumptions it has many important consequences. The section starts with some basic observations about homotopy and simple connectivity. 6.1.1 Homotopy Deﬁnition 6.1.1. Let M be a manifold and let I = [a, b] be a compact interval. A (smooth) homotopy of maps from I to M is a smooth map γ : [0, 1] × I → M. We often write γλ(t) = γ(λ, t) for λ ∈ [0, 1] and t �
� I and call γ a (smooth) homotopy between γ0 and γ1. We say the 257 258 CHAPTER 6. GEOMETRY AND TOPOLOGY homotopy has ﬁxed endpoints if γλ(a) = γ0(a) and γλ(b) = γ0(b) for all λ ∈ [0, 1]. (See Figure 6.1.) We remark that a homotopy and a variation are essentially the same thing, namely a curve of maps (curves). The diﬀerence is pedagogical. We used the word “variation” to describe a curve of maps through a given map; when we use this word we are going to diﬀerentiate the curve to ﬁnd a tangent vector (ﬁeld) to the given map. The word “homotopy” is used it is a global rather than a local to describe a curve joining two maps; (inﬁnitesimal) concept. Figure 6.1: A homotopy with ﬁxed endpoints. Deﬁnition 6.1.2. A manifold M is called simply connected iﬀ for any two curves γ0, γ1 : [a, b] → M with γ0(a) = γ1(a) and γ0(b) = γ1(b) there exists a homotopy from γ0 to γ1 with endpoints ﬁxed. (The idea is that the space Ωp,q of curves from p to q is connected.) Remark 6.1.3. Two smooth maps γ0, γ1 : [a, b] → M with the same endpoints can be joined by a continuous homotopy if and only if they can be joined by a smooth homotopy. This follows from the Weierstrass approximation theorem. Remark 6.1.4. Assume M is a connected smooth manifold. Then the topological space Ωp,q of all smooth curves in M with the endpoints p and q is connected for some pair of points p, q ∈ M if and only if it is connected for every pair of points p, q ∈ M. (Prove this!) Example 6.1.
5. The Euclidean space Rm is simply connected; any two curves γ0, γ1 : [a, b] → Rm with the same endpoints can be joined by the homotopy γλ(t) := γ0(t) + λ(γ1(t) − γ0(t)). Example 6.1.6. The punctured plane C \ {0} is not simply connected; two curves of the form γn(t) := e2πint, are not homotopic with ﬁxed endpoints for distinct integers n. Exercise 6.1.7. Prove that the m-sphere Sm is simply connected for m = 1. 0 ≤ t ≤ 1, 1γγ0M 6.1. THE CARTAN–AMBROSE–HICKS THEOREM 259 6.1.2 The Global C-A-H Theorem Theorem 6.1.8 (Global C-A-H Theorem). Let M ⊂ Rn and M ⊂ Rn be nonempty, connected, simply connected, complete m-manifolds. Fix two M be an orthogonal 0 ∈ M and let Φ0 : Tp0M → Tp elements p0 ∈ M and p 0 linear isomorphism. Then the following are equivalent. (i) There exists an isometry φ : M → M satisfying φ(p0) = p 0, dφ(p0) = Φ0. (6.1.1) (ii) If (Φ, γ, γ) is a development satisfying the initial condition γ(0) = p0, γ(0) = p 0, Φ(0) = Φ0, (6.1.2) then γ(1) = p0 =⇒ γ(1) = p 0, Φ(1) = Φ0 (iii) If (Φ0, γ0, γ 0) and (Φ1, γ1, γ 1) are developments satisfying (6.1.2), then γ0(1) = γ1(1) =⇒ 0(1) = γ γ 1(1). (iv) If (Φ, γ, γ) is a development
satisfying (6.1.2), then Φ∗Rγ = R γ. Figure 6.2: Diagram for Example 6.1.9. Example 6.1.9. Before giving the proof let us interpret the conditions in case M and M are two-dimensional spheres of radius r and r respectively in three-dimensional Euclidean space R3. Imagine that the spheres are tangent at p0 = p 0. Clearly the spheres will be isometric exactly when r = r. Condition (ii) says that if the spheres are rolled along one another without sliding or twisting, then the endpoint γ(1) of one curve of contact depends only on the endpoint γ(1) of the other and not on the intervening curve γ(t). This condition is violated in the case r = r (see Figure 6.2). MpM’0 260 CHAPTER 6. GEOMETRY AND TOPOLOGY By Theorem 5.3.9 the Riemann curvature of a 2-manifold at p is determined by the Gaußian curvature K(p); and for spheres we have K(p) = 1/r2. Exercise 6.1.10. Let γ be the closed curve which bounds an octant as shown in the diagram for Example 6.1.9. Find γ. Exercise 6.1.11. Show that in case M is two-dimensional, the condition Φ(1) = Φ0 in Theorem 6.1.8 may be dropped from (ii). Lemma 6.1.12. Let φ : M → M be a local isometry and let γ : I → M be a smooth curve on an interval I. Fix an element t0 ∈ I and deﬁne p0 := γ(t0), q0 := φ(p0), Φ0 := dφ(p0). (6.1.3) Then there exists a unique development (Φ, γ, γ) of M along M on the entire interval I satisfying the initial conditions γ(t0) = q0, Φ(t0) = Φ0. (6.1.4) This development is given by γ(t) = φ(γ(t)), Φ(t) = dφ(γ(t)) (6.1.5) for t
∈ I. Proof. Deﬁne γ(t) := φ(γ(t)), Φ(t) := dφ(γ(t)) for t ∈ I. Then ˙γ(t) = Φ(t) ˙γ(t) for all t ∈ I by the chain rule, and every vector ﬁeld X along γ satisﬁes Φ∇X = ∇(ΦX) by Corollary 5.3.2. Hence (Φ, γ, γ) is a development by Lemma 3.5.19. By (6.1.3) this development satisﬁes the initial condition (6.1.4). Hence the assertion follows from the uniqueness result for developments in Theorem 3.5.21. This proves Lemma 6.1.12. Proof of Theorem 6.1.8. We ﬁrst prove a slightly diﬀerent theorem. Namely, we weaken condition (i) to assert that φ is a local isometry (i.e. not necessarily bijective), and prove that this weaker condition is equivalent to (ii), (iii), and (iv) whenever M is connected and simply connected and M is complete. Thus we drop the hypotheses that M be complete and M be connected and simply connected. We prove that (i) implies (ii). Given a development as in (ii) we have, by Lemma 6.1.12, γ(1) = φ(γ(1)) = φ(p0) = p 0, Φ(1) = dφ(γ(1)) = dφ(p0) = Φ0, as required. 6.1. THE CARTAN–AMBROSE–HICKS THEOREM 261 We prove that (ii) implies (iii) when M is complete. Choose developi) for i = 0, 1 as in (iii). Deﬁne a curve γ : [0, 1] → M by ments (Φi, γi, γ “composition”, i.e. γ(t) := γ0(2t), 0 ≤ t ≤ 1/2, γ1(2 − 2t), 1/2 ≤ t ≤
1, so that γ is continuous and piecewise smooth and γ(1) = p0. By Theorem 3.5.21 there exists a development (Φ, γ, γ) on the interval [0, 1] satisfying (6.1.2) (because M is complete). Since γ(1) = p0 it follows from (ii) that γ(1) = p 0 and Φ(1) = Φ0. By the uniqueness of developments and the invariance under reparametrization, we have (Φ(t), γ(t), γ(t)) = (Φ0(2t), γ0(2t), γ 0(2t)), (Φ1(2 − 2t), γ1(2 − 2t), γ 0 ≤ t ≤ 1/2, 1(2 − 2t)), 1/2 ≤ t ≤ 1. Hence γ 0(1) = γ(1/2) = γ 1(1) as required. We prove that (iii) implies (i) when M is complete and M is connected. Deﬁne the map φ : M → M as follows. Fix an element p ∈ M. Since M is connected, there exists a smooth curve γ : [0, 1] → M such that γ(0) = p0 and γ(1) = p. Since M is complete, there exists a development (Φ, γ, γ) with γ(0) = p 0 and Φ(0) = Φ0 (Theorem 3.5.21). Now deﬁne φ(p) := γ(1). By (iii) the endpoint p := γ(1) is independent of the choice of the curve γ, and so φ is well-deﬁned. We prove that this map φ satisﬁes the following (a) If (Φ, γ, γ) is a development satisfying γ(0) = p0, γ(0) = p then φ(γ(t)) = γ(t) for 0 ≤ t ≤ 1. (b) If p, q ∈ M satisfy 0 < d(p, q) < inj(p; M ) and d(p,
q) < inj(φ(p); M ), then d(φ(p), φ(q)) = d(p, q). That φ satisﬁes (a) follows directly from the deﬁnition and the fact that the triple (Φt, γt, γ t(s) := γ(st) t) deﬁned by Φt(s) := Φ(st), γt(s) := γ(st), γ for 0 ≤ s ≤ 1 is a development. To prove (b), choose v ∈ TpM such that 0, Φ(0) = Φ0, \|v\| = d(p, q) expp(v) = q (Theorem 4.4.4) and let γ : [0, 1] → M be a smooth curve with γ(0) = p0, γ(t) = expp((2t − 1)v) for 1 satisfying γ(0) = p 2 ≤ t ≤ 1. Let (Φ, γ, γ) be the unique development of M along M 0 and Φ(0) = Φ0 (Theorem 3.5.21). Then, by (a), γ( 1 2 ) = φ(p), γ(1) = φ(q). 262 CHAPTER 6. GEOMETRY AND TOPOLOGY Also, by part (ii) of Lemma 3.5.19 with X = ˙γ, the restriction of γ to the interval [ 1 2, 1] is a geodesic. Thus γ(t) = expφ(p)((2t − 1)v) for 1 2 ≤ t ≤ 1, where the tangent vector v ∈ Tφ(p)M is given by v := ˙γ( 1 2 )v and hence satisﬁes \|v\| = \|v\| = d(p, q) < inj(φ(p), M ). Thus it follows from Theorem 4.4.4 that d(φ(p), φ(q)) = d(φ(p), exp φ(p)(v)) = \|v\| = d(p, q) and this proves (b). It follows from (b)
and Theorem 5.1.1 that φ is a local isometry. We prove that (i) implies (iv). Given a development as in (ii) we have 2 ) = Φ( 1 γ(t) = φ(γ(t)), Φ(t) = dφ(γ(t)) for every t, by Lemma 6.1.12. Hence it follows from part (iv) of Corollary 5.3.2 (Theorema Egregium for local isometries) that Φ(t)∗Rγ(t) = (φ∗R)γ(t) = R γ(t) for all t as required. We prove that (iv) implies (iii) when M is complete and M is simply i) for i = 0, 1 as in (iii). Since M connected. Choose developments (Φi, γi, γ is simply connected there exists a homotopy [0, 1] × [0, 1] → M : (λ, t) → γ(λ, t) = γλ(t) from γ0 to γ1 with endpoints ﬁxed. By Theorem 3.5.21 there is, for each λ, a development (Φλ, γλ, γ λ) on the interval [0, 1] with initial conditions λ(0) = p γ 0, Φλ(0) = Φ0 (because M is complete). The proof of Theorem 3.5.21 also shows that γλ(t) and Φλ(t) depend smoothly on both t and λ. We must prove that 1(1) = γ γ 0(1). To see this we will show that, for each ﬁxed t, the curve λ → (Φλ(t), γλ(t), γ λ(t)) is a development; then by the deﬁnition of development we have that the curve λ → γ λ(1) is smooth and ∂λγ λ(1) = Φλ(1)∂λγλ(1) = 0 as required. 6.1. THE CARTAN–AMBROSE–HICKS THEOREM 263 First choose a basis e
1,..., em of Tp0M and extend it to obtain vector ﬁelds Ei ∈ Vect(γ) along the homotopy γ by imposing the conditions that the vector ﬁelds t → Ei(λ, t) be parallel, i.e. ∇tEi(λ, t) = 0, Ei(λ, 0) = ei. (6.1.6) Then the vectors E1(λ, t),... Em(λ, t) form a basis of Tγλ(t)M for all λ and t. Second, deﬁne the vector ﬁelds E i along γ by i(λ, t) := Φλ(t)Ei(λ, t) E (6.1.7) so that ∇ tE i = 0. Third, deﬁne the functions ξ1,..., ξm : [0, 1]2 → R by ∂tγ =: m i=1 ξiEi, ∂tγ = m i=1 ξiE i. (6.1.8) Here the second equation follows from (6.1.7) and the fact that Φλ∂tγ = ∂tγ. Now consider the vector ﬁelds X := ∂λγ, along γ. They satisfy the equations ∇ tX = ∇ t∂λγ = ∇ λ∂tγ = ∇ λ and i := ∇ Y λE i (6.1.9) ξiE i = m i=1 m i=1 ∂λξiE i + ξiY i i = ∇ tY ∇ To sum up we have X (λ, 0) = Y m λE t∇ ∂λξiE i + ξiY i, ∇ tX = i = R(∂tγ, ∂λγ)E i. i − ∇ λ∇ tE i (λ, 0) = 0 and ∇ tY i = R(∂tγ, X )E i. (6.1.10) i=1 On the other hand, the vector �
�elds X := Φλ∂λγ, Y i := Φλ∇λEi (6.1.11) along γ satisfy the same equations, namely ∇ tX = Φλ∇t∂λγ = Φλ∇λ∂tγ = Φλ∇λ ξiEi m i=1 m = Φλ ∂λξiEi + ξi∇λEi = m ∂λξiE i + ξiY i, i=1 i=1 ∇ tY i = Φλ ∇t∇λEi − ∇λ∇tEi = ΦλR(∂tγ, ∂λγ)Ei = R(Φλ∂tγ, Φλ∂λγ)ΦλEi = R(∂tγ, X )E i. Here the last but one equation follows from (iv). 264 CHAPTER 6. GEOMETRY AND TOPOLOGY Since the tuples (6.1.9) and (6.1.11) satisfy the same diﬀerential equa- tion (6.1.10) and vanish at t = 0 they must agree. Hence ∂λγ = Φλ∂λγ, ∇ λE i = Φλ∇λEi for i = 1,..., m. This says that λ → (Φλ(t), γλ(t), γ For t = 1 we obtain ∂λγ(λ, 1) = 0 as required. λ(t)) is a development. Now the modiﬁed theorem (where φ is a local isometry) is proved. The original theorem follows immediately. Condition (iv) is symmetric in M and M. Thus, if we assume (iv), there are local isometries φ : M → M and ψ : M → M satisfying φ(p0) = p 0) = p0, dψ(p 0. But then ψ ◦ φ is a local isometry with ψ ◦ φ(p0) = p0 and d(ψ ◦ φ)(p0) = id. Hence ψ ◦ φ is the
identity. Similarly φ ◦ ψ is the identity so φ is bijective (and ψ = φ−1) as required. This proves Theorem 6.1.8. 0, dφ(p0) = Φ0 and ψ(p 0) = Φ−1 Remark 6.1.13. The proof of Theorem 6.1.8 shows that the various implications in the weak version of the theorem (where φ is only a local isometry) require the following conditions on M and M : (i) always implies (ii), (iii), and (iv); (ii) implies (iii) whenever M is complete; (iii) implies (i) whenever M is complete and M is connected; (iv) implies (iii) whenever M is complete and M is simply connected. Remark 6.1.14. The proof that (iii) implies (i) in Theorem 6.1.8 can be slightly shortened by using the following observation. Let φ : M → M be a map between smooth manifolds. Assume that φ ◦ γ is smooth for every smooth curve γ : [0, 1] → M. Then φ is smooth. Corollary 6.1.15. Let M and M be nonempty, connected, simply connected, complete Riemannian manifolds and let φ : M → M be a local isometry. Then φ is bijective and hence is an isometry. Proof. This follows by combining the weak and strong versions of the global C-A-H Theorem 6.1.8. Let p0 ∈ M and deﬁne p 0 := φ(p0) and Φ0 := dφ(p0). Then the tuple M, M, p0, p 0, Φ0 satisﬁes condition (i) of the weak version of Theorem 6.1.8. Hence this tuple also satisﬁes condition (iv) of Theorem 6.1.8. Since M and M are connected, simply connected, and complete we may apply the strong version of Theorem 6.1.8 to obtain an isometry ψ : M → M satisfying ψ(p0) = p 0 and dψ(p0) = Φ0. Since every isometry is also a local isometry and M
is connected it follows from Lemma 5.1.10 that φ(p) = ψ(p) for all p ∈ M. Hence φ is an isometry, as required. 6.1. THE CARTAN–AMBROSE–HICKS THEOREM 265 Remark 6.1.16. Reﬁning the argument in the proof of Corollary 6.1.15 one can show that a local isometry φ : M → M must be surjective whenever M is complete and M is connected. None of these assumptions can be removed. (Take an isometric embedding of a disc in the plane or an embedding of a complete space M into a space with two components, one of which is isometric to M.) Likewise, one can show that a local isometry φ : M → M must be injective whenever M is complete and connected and M is simply connected. Again none of these asumptions can be removed. (Take a covering R → S1, or a covering of a disjoint union of two isometric complete simply connected spaces onto one copy of this space, or some noninjective immersion of a disc into the plane and choose the pullback metric on the disc.) 6.1.3 The Local C-A-H Theorem Theorem 6.1.17 (Local C-A-H Theorem). Let M and M be smooth m-manifolds, let p0 ∈ M and p M be an orthogonal linear isomorphism. Let r > 0 be smaller than the injectvity radii of M at p0 and of M at p 0 ∈ M, and let Φ0 : Tp0M → Tp 0 and deﬁne 0 Ur := {p ∈ M \| d(p0, p) < r}, r := p ∈ M \| d(p U 0, p) < r. Then the following are equivalent. (i) There exists an isometry φ : Ur → U (ii) If (Φ, γ, γ) is a development on an interval I ⊂ R with 0 ∈ I, satisfying the initial condition (6.1.2) as well as γ(I) ⊂ Ur and γ(I) ⊂ U r satisfying (6.1.1). r, then γ(1
) = p0 =⇒ γ(1) = p 0, Φ(1) = Φ0. (iii) If (Φ0, γ0, γ 0) and (Φ1, γ1, γ γ0(1) = γ1(1) 1) are developments as in (ii), then 0(1) = γ γ 1(1). =⇒ (iv) If v ∈ Tp0M with \|v\| < r and γ(t) := exp p 0 γ(t) := expp0(tv), then Φ(t)∗Rγ(t) = R If these equivalent conditions are satisﬁed, then γ(t) for 0 ≤ t ≤ 1. (tΦ0v), Φ(t) := Φ γ(t, 0)Φ0Φγ(0, t), φ(expp0(v)) = exp p 0 (Φ0v) for all v ∈ Tp0M with \|v\| < r. The proof is based on the following lemma. 266 CHAPTER 6. GEOMETRY AND TOPOLOGY Lemma 6.1.18. Let p ∈ M and v, w ∈ TpM. For 0 ≤ t ≤ 1 deﬁne γ(t) := exp(tv), X(t) := ∂ ∂λ λ=0 expp t(v + λw) ∈ Tγ(t)M. Then ∇t∇tX + R(X, ˙γ) ˙γ = 0, X(0) = 0, ∇tX(0) = w. (6.1.12) A vector ﬁeld along γ satisfying the ﬁrst equation in (6.1.12) is called a Jacobi ﬁeld along γ. Proof. Deﬁne γ(λ, t) := expp(t(v + λw)), X(λ, t) := ∂λγ(λ, t) for all λ and t. Since γ(λ, 0) = p for all λ we have X(λ, 0) = 0 and ∇tX(λ
, 0) = ∇t∂λγ(λ, 0) = ∇λ∂tγ(λ, 0) = v + λw = w. d dλ Moreover, ∇t∂tγ = 0 and hence ∇t∇tX = ∇t∇t∂λγ = ∇t∇λ∂tγ − ∇λ∇t∂tγ = R(∂tγ, ∂λγ)∂tγ = R(∂tγ, X)∂tγ. This proves Lemma 6.1.18. Proof of Theorem 6.1.17. The proofs (i) =⇒ (ii) =⇒ (iii) =⇒ (i) =⇒ (iv) are as before; the reader might note that when L(γ) ≤ r we also have L(γ) ≤ r for any development so that there are plenty of developments with γ : [0, 1] → Ur and γ : [0, 1] → U r. The proof that (iv) implies (i) is a little diﬀerent since (iv) here is somewhat weaker than (iv) of the global theorem: the equation Φ∗R = R is only assumed for certain developments. Hence assume (iv) and deﬁne φ : Ur → U r by φ := exp p 0 ◦Φ0 ◦ exp−1 p0 : Ur → U r. We must prove that φ is an isometry. Thus we ﬁx a point q ∈ Ur and a tangent vector u ∈ TqM and choose v, w ∈ TpM with \|v\| < r such that expp0(v) = q, d expp0(v)w = u. (6.1.13) 6.1. THE CARTAN–AMBROSE–HICKS THEOREM 267 Deﬁne γ : [0, 1] → Ur, γ : [0, 1] → U r, X ∈ Vect(γ), and X ∈ Vect(γ) by γ(t) = expp0(tv), X(t) := γ(t) = exp p 0 (tΦ0v), X
(t) := Then, by deﬁnition of φ, we have ∂ ∂λ ∂ ∂λ λ=0 λ=0 expp0(t(v + λw)), exp p 0 (t(Φ0v + λΦ0w)). γ := φ ◦ γ, dφ(γ)X = X. (6.1.14) By Lemma 6.1.18, X is a solution of (6.1.12) and X is a solution of ∇t∇tX = R(∂tγ, X )∂tγ, X (λ, 0) = 0, ∇ tX (λ, 0) = Φ0w. (6.1.15) Now deﬁne Φ(t) : Tγ(t)M → Tγ(t)M by Φ(t) := Φ γ(t, 0)Φ0Φγ(0, t). Then Φ intertwines covariant diﬀerentiation. Since ˙γ and ˙γ are parallel vector ﬁelds with ˙γ(0) = Φ0v = Φ(0) ˙γ(0), we have Φ(t) ˙γ(t) = ˙γ(t) for every t. Moreover, it follows from (iv) that Φ∗Rγ = R with (6.1.12) we obtain γ. Combining this ∇ t∇ t(ΦX) = Φ∇t∇tX = R(Φ ˙γ, ΦX)Φ ˙γ = R( ˙γ, ΦX) ˙γ. Hence the vector ﬁeld ΦX along γ also satisﬁes the initial value problem (6.1.15) and thus ΦX = X = dφ(γ)X. Here we have also used (6.1.14). Using (6.1.13) we ﬁnd γ(1) = expp0(v) = q, X(1) = d expp0(v)w = u, and so dφ
(q)u = dφ(γ(1))X(1) = X (1) = Φ(1)u. Since Φ(1) : Tγ(1)M → Tγ(1)M is an orthogonal transformation this gives \|dφ(q)u\| = \|Φ(1)u\| = \|u\|. Hence φ is an isometry as claimed. This proves Theorem 6.1.17. 268 CHAPTER 6. GEOMETRY AND TOPOLOGY 6.2 Flat Spaces Our aim in the next few sections is to give applications of the CartanAmbrose-Hicks Theorem. It is clear that the hypothesis Φ∗R = R for all developments will be diﬃcult to verify without drastic hypotheses on the curvature. The most drastic such hypothesis is that the curvature vanishes identically. Deﬁnition 6.2.1. A Riemannian manifold M is called ﬂat iﬀ the Riemann curvature tensor R vanishes identically. Theorem 6.2.2. Let M ⊂ Rn be a smooth m-manifold. (i) M is ﬂat if and only if every point has a neighborhood which is isometric to an open subset of Rm, i.e. at each point p ∈ M there exist local coordinates x1,..., xm such that the coordinate vectorﬁelds Ei = ∂/∂xi are orthonormal. (ii) Assume M is connected, simply connected, and complete. Then M is ﬂat if and only if there is an isometry φ : M → Rm onto Euclidean space. Proof. Assertion (i) follows immediately from Theorem 6.1.17 and (ii) follows immediately from Theorem 6.1.8. Exercise 6.2.3. Carry over the Cartan–Ambrose–Hicks theorem and Theorem 6.2.2 to the intrinsic setting. Exercise 6.2.4. A one-dimensional manifold is always ﬂat. Exercise 6.2.5. If M1 and M2 are ﬂat, so is M = M1 × M2. Example 6.2.6. By Exercises 6.2.4 and 6.
2.5 the standard torus Tm = z = (z1,..., zm) ∈ Cm \|z1\| = · · · = \|zm\| = 1 is ﬂat. Exercise 6.2.7. For a, b > 0 and c ≥ 0 deﬁne M ⊂ C3 by M := M (a, b, c) := (u, v, w) ∈ C3 \|u\| = a, \|v\| = b, w = c. u a v b Then M is diﬀeomorphic to a torus (a product of two circles) and M is ﬂat. If a, b > 0 and c ≥ 0, prove that there is an isometry φ from M = M (a, b, c) to M = M (a, b, c) if and only if the triples (a, b, c) and (a, b, c) are related by a permutation. 6.2. FLAT SPACES 269 Hint: Show ﬁrst that an isometry φ : M → M that satisﬁes the condi- tion φ(a, b, c) = (a, b, c) must have the form φ(u, v, w+γ v b β+δ for integers α, β, γ, δ that satisfy αδ − βγ = ±1. Show that this map φ is an isometry if and only if a2 + c2 = α2a2 + γ2b2 + (α + γ)2c2, c2 = αβa2 + γδb2 + (α + γ)(β + δ)c2, b2 + c2 = β2a2 + δ2b2 + (β + δ)2c2. Exercise 6.2.8 (Developable manifolds). Let n = m + 1 and let E(t) be a one-parameter family of hyperplanes in Rn. Then there exists a smooth map u : R → Rn such that E(t) = u(t)⊥, \|u(t)\| = 1, (6.2.1) for every t. We assume that ˙u(t) =
0 for every t so that u(t) and ˙u(t) are linearly independent. Show that L(t) := u(t)⊥ ∩ ˙u(t)⊥ = lim s→t E(t) ∩ E(s). (6.2.2) Thus L(t) is a linear subspace of dimension m − 1. Now let γ : R → Rn be a smooth curve such that ˙γ(t), u(t) = 0, ˙γ(t), ˙u(t) = 0 (6.2.3) for all t. This means that ˙γ(t) ∈ E(t) and ˙γ(t) /∈ L(t); thus E(t) is spanned by L(t) and ˙γ(t). For t ∈ R and ε > 0 deﬁne L(t)ε := {v ∈ L(t) \| \|v\| < ε}. Let I ⊂ R be a bounded open interval such that the restriction of γ to the closure of I is injective. Prove that, for ε > 0 suﬃciently small, the set M0 := γ(t) + L(t)ε t∈I is a smooth manifold of dimension m = n − 1. A manifold which arises this way is called developable. Show that the tangent spaces of M0 are the original subspaces E(t), i.e. TpM0 = E(t) for p ∈ γ(t) + L(t)ε. 270 CHAPTER 6. GEOMETRY AND TOPOLOGY (One therefore calls M0 the “envelope” of the hyperplanes γ(t) + E(t).) If (Φ, γ, γ) is a deShow that M0 is ﬂat. (Hint: use Gauß–Codazzi.) velopment of M0 along Rm, show that the map φ : M0 → Rm, deﬁned by φ(γ(t) + v) := γ(t) + Φ(t)v 0 ⊂ Rm. Thus a development for v ∈
L(t)ε, is an isometry onto an open set M “unrolls” M0 onto the Euclidean space Rm. When n = 3 and m = 2 one can visualize M0 as a twisted sheet of paper (see Figure 6.3). Figure 6.3: Developable surfaces. Remark 6.2.9. Given a codimension-1 submanifold M ⊂ Rm+1 and a curve γ : R → M we may form the osculating developable M0 to M along γ by taking E(t) := Tγ(t)M. This developable has common aﬃne tangent spaces with M along γ as Tγ(t)M0 = E(t) = Tγ(t)M for every t. This gives a nice interpretation of parallel transport: M0 may be unrolled onto a hyperplane where parallel transport has an obvious meaning and the identiﬁcation of the tangent spaces thereby deﬁnes parallel transport in M. (See Remark 3.5.16.) Exercise 6.2.10. Each of the following is a developable surface in R3. (i) A cone on a plane curve Γ ⊂ H, i.e. M = {tp + (1 − t)q \| t > 0, q ∈ Γ} where H ⊂ R3 is an aﬃne hyperplane, p ∈ R3\H, and Γ ⊂ H is a 1-manifold. 6.2. FLAT SPACES 271 (ii) A cylinder on a plane curve Γ, i.e. M = {q + tv \| q ∈ Γ, t ∈ R} where H and Γ are as in (i) and v is a ﬁxed vector not parallel to H. (This is a cone with the cone point p at inﬁnity.) (iii) The tangent developable to a space curve γ : R → R3, i.e. M = {γ(t) + s ˙γ(t) \| \|t − t0\| < ε, 0 < s < ε}, where ˙γ(t0) and ¨γ(t0) are linearly independent and ε > 0
is suﬃciently small. (iv) The paper model of a M¨obius strip (see Figure 6.3). Figure 6.4: A circular one-sheeted hyperboloid. Remark 6.2.11. A 2-dimensional submanifold M ⊂ R3 is called a ruled surface iﬀ there is a straight line in M through every point. Every developable surface is ruled, however, there are ruled surfaces that are not developable. An example is the manifold M = {γ(t) + s¨γ(t) \| \|t − t0\| < ε, \|s\| < ε} where γ : R → R3 is a smooth curve with \| ˙γ\| ≡ 1 and ¨γ(t0) = 0, and ε > 0 is suﬃciently small; this surface is not developable in general. Other examples are the elliptic hyperboloid of one sheet M := (x, y, z) ∈ R3 x2 a2 + y2 b2 − z2 c2 = 1 depicted in Figure 6.4, the hyperbolic paraboloid M := (x, y, z) ∈ R3 z = x2 a2 − y2 b2. (6.2.4) (6.2.5) (both with two straight lines through every point in M ), Pl¨ucker’s conoid M := (x, y, z) ∈ R3 x2 + y2 = 0, z = 2xy x2 + y2, (6.2.6) 272 CHAPTER 6. GEOMETRY AND TOPOLOGY the helicoid M := (x, y, z) ∈ R3 x + iy x2 + y2 = eiαz, (6.2.7) and the M¨obius strip   M :=   +    cos(s) sin(s) 0   t 2 cos(s/2) cos(s) cos(s/2) sin(s) sin(s/2)   s ∈ R and −6.2.8) These ﬁve surfaces have negative Gaußian curvature. The M¨obius strip
in (6.2.8) is not developable, while the paper model of the M¨obius strip is. The helicoid in (6.2.7) is a minimal surface, i.e. its mean curvature (the trace of the second fundamental form) vanishes. A minimal surface which is not ruled is the catenoid M := {(x, y, z) ∈ R3 \| x2 + y2 = c2 cosh (z/c)}. (Exercise: Prove all this.) 6.3 Symmetric Spaces In the last section we applied the Cartan-Ambrose-Hicks Theorem in the ﬂat case; the hypothesis Φ∗R = R was easy to verify since both sides vanish. To ﬁnd more general situations where we can verify this hypothesis note that for any development (Φ, γ, γ) satisfying the initial conditions γ(0) = p0, γ(0) = p 0, Φ(0) = Φ0, we have Φ(t) = Φ γ(t, 0)Φ0Φγ(0, t) so that the hypothesis Φ∗R = R is certainly implied by the three hypotheses Φγ(t, 0)∗Rp0 = Rγ(t) Φ γ(t, 0)∗R = R p 0 (Φ0)∗Rp0 = R p 0 γ(t). The last hypothesis is a condition on the initial linear isomorphism Φ0 : Tp0M → Tp 0 M while the former hypotheses are conditions on M and M respectively, namely, that the Riemann curvature tensor is invariant by parallel transport. It is rather amazing that this condition is equivalent to a simple geometric condition as we now show. 6.3. SYMMETRIC SPACES 273 6.3.1 Symmetric Spaces Deﬁnition 6.3.1. A Riemannian manifold M is called symmetric about the point p ∈ M iﬀ there exists a (necessarily unique) isometry φ : M → M satisfying φ(p) = p, dφ(p) = −id; (6.3.1) M is called a symmetric space iﬀ
it is symmetric about each of its points. A Riemannian manifold M is called locally symmetric about the point p ∈ M iﬀ, for r > 0 suﬃciently small, there exists an isometry φ : Ur(p, M ) → Ur(p, M ), Ur(p, M ) := {q ∈ M \| d(p, q) < r}, satisfying (6.3.1); M is called a locally symmetric space iﬀ it is locally symmetric about each of its points. Remark 6.3.2. The proof of Theorem 6.3.4 below will show that, if M is locally symmetric, the isometry φ : Ur(p, M ) → Ur(p, M ) with φ(p) = p and dφ(p) = −id exists whenever 0 < r ≤ inj(p). Exercise 6.3.3. Every symmetric space is complete. Hint: If γ : I → M is a geodesic and φ : M → M is a symmetry about the point γ(t0) for t0 ∈ I, then φ(γ(t0 + t)) = γ(t0 − t) for all t ∈ R with t0 + t, t0 − t ∈ I. Theorem 6.3.4. Let M ⊂ Rn be an m-dimensional submanifold. Then the following are equivalent. (i) M is locally symmetric. (ii) The covariant derivative ∇R (deﬁned below) vanishes identically, i.e. (∇vR)p(v1, v2)w = 0 for all p ∈ M and v, v1, v2, w ∈ TpM. (iii) The curvature tensor R is invariant under parallel transport, i.e. Φγ(t, s)∗Rγ(s) = Rγ(t) (6.3.2) for every smooth curve γ : R → M and all s, t ∈ R. Proof. See §6.3.2. 274 CHAPTER 6. GEOMETRY AND TOPOLOGY Corollary 6.3.5. Let M and M be locally symmetric spaces and ﬁx
two M be an orthogonal points p0 ∈ M and p 0 ∈ M, and let Φ0 : Tp0M → Tp linear isomorphism. Let r > 0 be less than the injectivity radius of M at p0 and the injectivity radius of M at p (i) There exists an isometry φ : Ur(p0, M ) → Ur(p and dφ(p0) = Φ0 if and only if Φ0 intertwines R and R, i.e. 0. Then the following holds. 0, M ) with φ(p0) = p 0 0 (Φ0)∗Rp0 = R p 0. (6.3.3) (ii) Assume M and M are connected, simply connected, and complete. Then there exists an isometry φ : M → M with φ(p0) = p 0 and dφ(p0) = Φ0 if and only if Φ0 satisﬁes (6.3.3). Proof. In (i) and (ii) the “only if” statement follows from Theorem 5.3.1 (Theorema Egregium) with Φ0 := dφ(p0). To prove the “if” statement, let (Φ, γ, γ) be a development satisfying γ(0) = p0, γ(0) = p 0, and Φ(0) = Φ0. Since R and R are invariant under parallel transport, by Theorem 6.3.4, it follows from the discussion in the beginning of this section that Φ∗R = R. Hence assertion (i) follows from the local C-A-H Theorem 6.1.17 and (ii) follows from the global C-A-H Theorem 6.1.8. Corollary 6.3.6. A connected, simply connected, complete, locally symmetric space is symmetric. Proof. Corollary 6.3.5 (ii) with M = M, p 0 = p0, and Φ0 = −id. Corollary 6.3.7. A connected symmetric space M is homogeneous; i.e. given p, q ∈ M there exists an isometry φ : M → M with
φ(p) = q. Proof. If M is simply connected, the assertion follows from part (ii) of Corollary 6.3.5 with M = M, p0 = p, p 0 = q, and Φ0 = Φγ(1, 0) : TpM → TqM, where γ : [0, 1] → M is a curve from p to q. If M is not simply connected, we can argue as follows. There is an equivalence relation on M deﬁned by p ∼ q : ⇐⇒ ∃ isometry φ : M → M φ(p) = q. Let p, q ∈ M and suppose that d(p, q) < inj(p). By Theorem 4.4.4 there is a unique shortest geodesic γ : [0, 1] → M connecting p to q. Since M is symmetric there is an isometry φ : M → M such that φ(γ(1/2)) = γ(1/2) and dφ(γ(1/2)) = −id. This isometry satisﬁes φ(γ(t)) = γ(1 − t) and hence φ(p) = q. Thus p ∼ q whenever d(p, q) < inj(p). This shows that each equivalence class is open, hence each equivalence class is also closed, and hence there is only one equivalence class because M is connected. This proves Corollary 6.3.7. 6.3. SYMMETRIC SPACES 275 6.3.2 Covariant Derivative of the Curvature For two vector spaces V, W and an integer k ≥ 1 we denote by Lk(V, W ) the vector space of all multi-linear maps from to W. Thus L1(V, W ) = L(V, W ) is the space of all linear maps from V to W. Deﬁnition 6.3.8. The covariant derivative of the Riemann curvature tensor assigns to every p ∈ M a linear map (∇R)p : TpM → L2(TpM, L(TpM, TpM )) such that (∇R)(X)(X1, X2)Y = �
�X R(X1, X2)Y − R(∇X X1, X2)Y − R(X1, ∇X X2)Y − R(X1, X2)∇X Y (6.3.4) for all X, X1, X2, Y ∈ Vect(M ). We also use the notation (∇vR)p := (∇R)p(v) for p ∈ M and v ∈ TpM so that (∇X R)(X1, X2)Y := (∇R)(X)(X1, X2)Y for all X, X1, X2, Y ∈ Vect(M ). Remark 6.3.9. One veriﬁes easily that the map Vect(M )4 → Vect(M ) : (X, X1, X2, Y ) → (∇X R)(X1, X2)Y, deﬁned by the right hand side of equation (6.3.4), is multi-linear over the ring of functions F (M ). Hence it follows as in Remark 5.2.13 that ∇R is well deﬁned, i.e. that the right hand side of (6.3.4) at p ∈ M depends only on the tangent vectors X(p), X1(p), X2(p), Y (p). Remark 6.3.10. Let γ : I → M be a smooth curve on an interval I ⊂ R and X1, X2, Y ∈ Vect(γ) be smooth vector ﬁelds along γ. Then equation (6.3.4) continues to hold with X replaced by ˙γ and each ∇X on the right hand side replaced by the covariant derivative of the respective vector ﬁeld along γ: (∇˙γR)(X1, X2)Y = ∇(R(X1, X2)Y ) − R(∇X1, X2)Y − R(X1, ∇X2)Y − R(X1, X2)∇Y. (6.3.5) 276 CHAPTER 6. GEOMETRY AND TOPOLOGY Theorem 6.
3.11. (i) If γ : R → M is a smooth curve such that γ(0) = p and ˙γ(0) = v, then (∇vR)p = d dt t=0 Φγ(0, t)∗Rγ(t) (6.3.6) (ii) The covariant derivative of the Riemann curvature tensor satisﬁes the second Bianchi identity (∇X R)(Y, Z) + (∇Y R)(Z, X) + (∇ZR)(X, Y ) = 0. (6.3.7) Proof. We prove (i). Let v1, v2, w ∈ TpM and choose parallel vector ﬁelds X1, X2, Y ∈ Vect(γ) along γ satisfying the initial conditions X1(0) = v1, X2(0) = v2, Y (0) = w. Thus X1(t) = Φγ(t, 0)v1, X2(t) = Φγ(t, 0)v2, Y (t) = Φγ(t, 0)w. Then the last three terms on the right vanish in equation (6.3.5) and hence = (∇vR)(v1, v2)w = ∇(R(X1, X2)Y )(0) d dt d dt d dt t=0 t=0 t=0 = = Φγ(0,t)Rγ(t)(X1(t), X2(t))Y (t) Φγ(0,t)Rγ(t)(Φγ(t, 0)v1, Φγ(t, 0)v2)Φγ(t, 0)w Φγ(0, t)∗Rγ(t) (v1, v2)w. Here the second equation follows from Theorem 3.3.6. This proves (i). We prove (ii). Choose a smooth function γ : R3 → M and denote by (r, s, t) the coordinates on R3. If Y is a vector ﬁeld along γ, we have (∇∂rγR
)(∂sγ, ∂tγ)Y = ∇r R(∂sγ, ∂tγ)Y − R(∂sγ, ∂tγ)∇rY − R(∇r∂sγ, ∂tγ)Y − R(∂sγ, ∇r∂tγ)Y = ∇r (∇s∇tY − ∇t∇sY ) − (∇s∇t − ∇t∇s) ∇rY + R(∂tγ, ∇r∂sγ)Y − R(∂sγ, ∇t∂rγ)Y. Permuting the variables r, s, t cyclically and taking the sum of the resulting three equations we obtain (∇∂rγR)(∂sγ, ∂tγ)Y + (∇∂sγR)(∂tγ, ∂rγ)Y + (∇∂tγR)(∂rγ, ∂sγ)Y = ∇r (∇s∇tY − ∇t∇sY ) − (∇s∇t − ∇t∇s) ∇rY + ∇s (∇t∇rY − ∇r∇tY ) − (∇t∇r − ∇r∇t) ∇sY + ∇t (∇r∇sY − ∇s∇rY ) − (∇r∇s − ∇s∇r) ∇tY. The terms on the right cancel out. This proves Theorem 6.3.11. 6.3. SYMMETRIC SPACES 277 Proof of Theorem 6.3.4. We prove that (iii) implies (i). This follows from the local Cartan–Ambrose–Hicks Theorem 6.1.17 with p 0 = p0 = p, Φ0 = −id : TpM → TpM. This isomorphism satisﬁes (Φ0)∗Rp = Rp. Hence it follows from the discussion in the beginning of this section that Φ∗R = R for every development (Φ, γ
, γ) of M along itself satisfying γ(0) = γ(0) = p, Φ(0) = −id. Hence, by the local C-A-H Theorem 6.1.17, there is an isometry satisfying φ : Ur(p, M ) → Ur(p, M ) φ(p) = p, dφ(p) = −id whenever 0 < r < inj(p; M ). We prove that (i) implies (ii). By Theorem 5.3.1 (Theorema Egregium), every isometry φ : M → M preserves the Riemann curvature tensor and covariant diﬀerentiation, and hence also the covariant derivative of the Riemann curvature tensor, i.e. φ∗(∇R) = ∇R. Applying this to the local isometry φ : Ur(p, M ) → Ur(p, M ) we obtain ∇dφ(p)vR φ(p) (dφ(p)v1, dφ(p)v2) = dφ(p) (∇vR) (v1, v2)dφ(p)−1 for all v, v1, v2 ∈ TpM. Since dφ(p) = −id this shows that ∇R vanishes at p. We prove that (ii) imlies (iii). If ∇R vanishes, then equation (6.3.6) in Theorem 6.3.11 shows that the function s → Φγ(t, s)∗Rγ(s) = Φγ(t, 0)∗Φγ(0, s)∗Rγ(s) is constant and hence is everywhere equal to Rγ(t). This implies (6.3.2) and completes the proof of Theorem 6.3.4. 278 CHAPTER 6. GEOMETRY AND TOPOLOGY Covariant Derivative of the Curvature in Local Coordinates Let φ : U → Ω be a local coordinate chart on M with values in an open set Ω ⊂ Rm, denote its inverse by ψ := φ−1 : Ω → U, and let Ei(
x) := ∂ψ ∂xi (x) ∈ Tψ(x)M, x ∈ Ω, i = 1,..., m, be the local frame of the tangent bundle determined by this coordinate ijk : Ω → R the ij : Ω → R denote the Christoﬀel symbols and R chart. Let Γk coeﬃcients of the Riemann curvature tensor so that ∇iEj = Γk ijEk, k R(Ei, Ej)Ek = R ijkE. Given i, j, k, ∈ {1,..., m} we can express the vector ﬁeld (∇EiR)(Ej, Ek)E along ψ for each x ∈ Ω as a linear combination of the basis vectors Ei(x). This gives rise to functions ∇iRν jk : Ω → R deﬁned by (∇EiR)(Ej, Ek)E =: ∇iRν jkEν. ν These functions are given by ∇iRν jk = ∂iRν jk + iµRµ Γν jk µ Γµ ijRν µk − − µ Γµ ikRν jµ − µ Γµ iRν jkµ. µ The second Bianchi identity has the form ∇iRν jk + ∇jRν ki + ∇kRν ij = 0. (6.3.8) (6.3.9) (6.3.10) Exercise: Prove equations (6.3.9) and (6.3.10). Warning: As in §5.4, care must be taken with the ordering of the indices. Some authors use the notation ∇iRν jk. jk for what we call ∇iRν 6.3.3 Examples and Exercises Example 6.3.12. Every ﬂat manifold is locally symmetric. Example 6.3.13. If M1 and M2 are (locally) symmetric, so is M1 × M2. 6.3. SYMMETRIC SPACES
279 Example 6.3.14. M = Rm with the standard metric is a symmetric space. Recall that the isometry group I(Rm) consists of all aﬃne transformations of the form φ(x) = Ax + b, A ∈ O(m), b ∈ Rm. (See Exercise 5.1.4.) The isometry with ﬁxed point p ∈ Rm and dφ(p) = −id is given by φ(x) = 2p − x for x ∈ Rm. Example 6.3.15. The ﬂat tori of Exercise 6.2.7 in the previous section are symmetric (but not simply connected). This shows that the hypothesis of simple connectivity cannot be dropped in part (ii) of Corollary 6.3.5. Example 6.3.16. Below we deﬁne manifolds of constant curvature and show that they are locally symmetric. The simplest example, after a ﬂat space, is the unit sphere Sm = x ∈ Rm+1 \| \|x\| = 1. The symmetry φ of the sphere about a point p ∈ M is given by φ(x) := −x + 2p, xp for x ∈ Sm. This extends to an orthogonal linear transformation of the ambient space. In fact the group of isometries of Sm is the group O(m + 1) of orthogonal linear transformations of Rm+1 (see Example 6.4.16 below). In accordance with Corollary 6.3.7 this group acts transitively on Sm. Example 6.3.17. A compact two-dimensional manifold of constant negative curvature is locally symmetric (as its universal cover is symmetric) but not homogeneous (as closed geodesics of a given period are isolated). Hence it is not symmetric. This shows that the hypothesis that M be simply connected cannot be dropped in Corollary 6.3.6. Example 6.3.18. The real projective space RPn with the metric inherited from Sn is a symmetric space and the orthogonal group O(n+1) acts on it by isometries. The complex projective space CPn with the Fubini–Study metric in Example 3.7.5 is a symmetric space and the unitary group U(
n + 1) acts on it by isometries. The complex Grassmannian Gk(Cn) in Example 3.7.6 is a symmetric space and the unitary group U(n) acts on it by isometries. (Exercise: Prove this.) Example 6.3.19. The simplest example of a symmetric space which is not of constant curvature is the orthogonal group O(n) = g ∈ Rn×n \| gTg = 1l with the Riemannian metric (5.2.20) of Example 5.2.18. The symmetry φ about the point a ∈ O(n) is given by φ(g) = ag−1a. This discussion extends to every Lie subgroup G ⊂ O(n). (Exercise: Prove this.) 280 CHAPTER 6. GEOMETRY AND TOPOLOGY 6.4 Constant Curvature In the §5.3 we saw that the Gaußian curvature of a two-dimensional surface is intrinsic: we gave a formula for it in terms of the Riemann curvature tensor and the ﬁrst fundamental form. We may use this formula to deﬁne the Gaußian curvature for any two-dimensional manifold (even if its codimension is greater than one). We make a slightly more general deﬁnition. 6.4.1 Sectional Curvature Deﬁnition 6.4.1. Let M ⊂ Rn be a smooth m-dimensional submanifold. Let p ∈ M and let E ⊂ TpM be a 2-dimensional linear subspace of the tangent space. The sectional curvature of M at (p, E) is the number K(p, E) = Rp(u, v)v, u \|u\|2\|v\|2 − u, v2, (6.4.1) where u, v ∈ E are linearly independent (and hence form a basis of E). The right hand side of (6.4.1) remains unchanged if we multiply u or v by a nonzero real number or add to one of the vectors a real multiple of the other; hence it depends only on the linear subspace spanned by u ad v. Example 6.4.2. If M ⊂ R3
is a 2-manifold, then by Theorem 5.3.9 the sectional curvature K(p, TpM ) = K(p) is the Gaußian curvature of M at p. More generally, for any 2-manifold M ⊂ Rn (whether or not it has codimension one) we deﬁne the Gaußian curvature of M at p by K(p) := K(p, TpM ). (6.4.2) Example 6.4.3. If M ⊂ Rm+1 is a submanifold of codimension one and ν : M → Sm is a Gauß map, then the sectional curvature of a 2-dimensional subspace E ⊂ TpM spanned by two linearly independent tangent vectors u, v ∈ TpM is given by K(p, E) = u, dν(p)uv, dν(p)v − u, dν(p)v2 \|u\|2\|v\|2 − u, v2. (6.4.3) This follows from equation (5.3.8) in the proof of Theorem 5.3.9 which holds in all dimensions. In particular, when M = Sm, we have ν(p) = p and hence K(p, E) = 1 for all p and E. For a sphere of radius r we have ν(p) = p/r and hence K(p, E) = 1/r2. 6.4. CONSTANT CURVATURE 281 Example 6.4.4. Let G ⊂ O(n) be a Lie subgroup equipped with the Riemannian metric v, w := trace(vTw) for v, w ∈ TgG ⊂ Rn×n. Then, by Example 5.2.18, the sectional curvature of G at the identity matrix 1l is given by K(1l, E) = 1 4 \|[ξ, η]\|2 for every 2-dimensional linear subspace E ⊂ g = Lie(G) = T1lG with an orthonormal basis ξ, η. Exercise 6.4.5. Let E ⊂ TpM be a 2-dimensional linear subspace, let r
> 0 be smaller than the injectivity radius of M at p, and let N ⊂ M be the 2dimensional submanifold given by N := expp ({v ∈ E \| \|v\| < r}). Show that the sectional curvature K(p, E) of M at (p, E) agrees with the Gaußian curvature of N at p. Exercise 6.4.6. Let p ∈ M ⊂ Rn and let E ⊂ TpM be a 2-dimensional linear subspace. For r > 0 let L denote the ball of radius r in the (n − m + 2)dimensional aﬃne subspace of Rn through p and parallel to the vector subspace E + TpM ⊥: p + v + w \| v ∈ E, w ∈ TpM ⊥, \|v\|2 + \|w\|2 < r2 L =. Show that, for r suﬃciently small, L ∩ M is a 2-dimensional manifold with Gaußian curvature KL∩M (p) at p given by KL∩M (p) = K(p, E). 6.4.2 Constant Sectional Curvature Deﬁnition 6.4.7. Let k ∈ R and m ≥ 2 be an integer. An m-manifold M ⊂ Rn is said to have constant sectional curvature k iﬀ K(p, E) = k for every p ∈ M and every 2-dimensional linear subspace E ⊂ TpM. Theorem 6.4.8. Let M ⊂ Rn be an m-manifold and ﬁx an element p ∈ M and a real number k. Then the following are equivalent. (i) K(p, E) = k for every 2-dimensional linear subspace E ⊂ TpM. (ii) The Riemann curvature tensor of M at p is given by Rp(v1, v2)v3, v4 = k v1, v4v2, v3 − v1, v3v2, v4 (6.4.4) for all v1, v2, v3, v4 ∈ TpM. 282 CHAPTER 6. GEOMETRY
AND TOPOLOGY Proof. That (ii) implies (i) follows directly from the deﬁnition of the sectional curvature in (6.4.1) by taking v1 = v4 = u and v2 = v3 = v in (6.4.4). Conversely, assume (i) and deﬁne the multi-linear map Q : TpM 4 → R by Q(v1, v2, v3, v4) := Rp(v1, v2)v3, v4 − k v1, v4v2, v3 − v1, v3v2, v4. Then, for all u, v, v1, v2, v3, v4 ∈ TpM, the map Q satisﬁes the equations Q(v1, v2, v3, v4) + Q(v2, v1, v3, v4) = 0, Q(v1, v2, v3, v4) + Q(v2, v3, v1, v4) + Q(v3, v1, v2, v4) = 0, Q(v1, v2, v3, v4) − Q(v3, v4, v1, v2) = 0, Q(u, v, u, v) = 0. (6.4.5) (6.4.6) (6.4.7) (6.4.8) Here the ﬁrst three equations follow from Theorem 5.2.14 and the last follows from the deﬁnition of Q and the hypothesis that the sectional curvature is K(p, E) = k for every 2-dimensional linear subspace E ⊂ TpM. We must prove that Q vanishes. Using (6.4.7) and (6.4.8) we ﬁnd 0 = Q(u, v1 + v2, u, v1 + v2) = Q(u, v1, u, v2) + Q(u, v2, u, v1) = 2Q(u, v1, u, v2) for all u, v1, v2 ∈ TpM. This implies 0 = Q(u1 + u2,
v1, u1 + u2, v2) = Q(u1, v1, u2, v2) + Q(u2, v1, u1, v2) for all u1, u2, v1, v2 ∈ TpM. Hence Q(v1, v2, v3, v4) = −Q(v3, v2, v1, v4) = Q(v2, v3, v1, v4) = −Q(v3, v1, v2, v4) − Q(v1, v2, v3, v4). Here the second equation follows from (6.4.5) and the last from (6.4.6). Thus Q(v1, v2, v3, v4) = − 1 2 Q(v3, v1, v2, v4) = 1 2 Q(v1, v3, v2, v4) for all v1, v2, v3, v4 ∈ TpM and, repeating this argument, Q(v1, v2, v3, v4) = 1 4 Q(v1, v2, v3, v4). Hence Q ≡ 0 as claimed. This proves Theorem 6.4.8. 6.4. CONSTANT CURVATURE 283 Remark 6.4.9. The symmetric group S4 on four symbols acts naturally on the space L4(TpM, R) of multi-linear maps from TpM 4 to R. The conditions (6.4.5), (6.4.6), (6.4.7), and (6.4.8) say that the four elements a = id + (12), c = id + (123) + (132), b = id − (34), d = id + (13) + (24) + (13)(24) of the group ring of S4 annihilate Q. This suggests an alternate proof of Theorem 6.4.8. A representation of a ﬁnite group is completely reducible so one can prove that Q = 0 by showing that any vector in any irreducible representation of S4 which is annihilated by the four elements a, b, c and d must necessarily be zero. This can be checked case by case for each irred
ucible representation. (The group S4 has 5 irreducible representations: two of dimension 1, two of dimension 3, and one of dimension 2.) If M and M are two m-dimensional manifolds with constant curvature k, then every orthogonal isomorphism Φ : TpM → TpM intertwines the Riemann curvature tensors by Theorem 6.4.8. Hence by the appropriate version (local or global) of the C-A-H Theorem we have the following corollaries. Corollary 6.4.10. Every Riemannian manifold with constant sectional curvature is locally symmetric. Proof. Theorem 6.3.4 and Theorem 6.4.8. Corollary 6.4.11. Let M and M be m-dimensional Riemannian manifolds with constant curvature k and let p ∈ M and p ∈ M. If r > 0 is smaller than the injectivity radii of M at p and of M at p, then for every orthogonal isomorphism Φ : TpM → TpM there exists an isometry such that φ : Ur(p, M ) → Ur(p, M ) φ(p) = p, dφ(p) = Φ. Proof. This follows from Corollary 6.3.5 and Corollary 6.4.10. Alternatively one can use Theorem 6.4.8 and the local C-A-H Theorem 6.1.17. 284 CHAPTER 6. GEOMETRY AND TOPOLOGY Corollary 6.4.12. Any two connected, simply connected, complete Riemannian manifolds with the same constant sectional curvature and the same dimension are isometric. Proof. Theorem 6.4.8 and the global C-A-H Theorem 6.1.8. Corollary 6.4.13. Let M ⊂ Rn be a connected, simply connected, complete manifold. Then the following are equivalent. (i) M has constant sectional curvature. (ii) For every pair of points p, q ∈ M and every orthogonal linear isomorphism Φ : TpM → TqM there exists an isometry φ : M → M such that φ(p) = q, dφ(p) = �
�. Proof. That (i) implies (ii) follows immediately from Theorem 6.4.8 and the global C-A-H Theorem 6.1.8. Conversely assume (ii). Then, for every pair of points p, q ∈ M and every orthogonal linear isomorphism Φ : TpM → TqM, it follows from Theorem 5.3.1 (Theorema Egregium) that and so Φ∗Rp = Rq K(p, E) = K(q, ΦE) for every 2-dimensional linear subspace E ⊂ TpM. Since, for every pair of points p, q ∈ M and of 2-dimensional linear subspaces E ⊂ TpM, F ⊂ TqM, we can ﬁnd an orthogonal linear isomorphism Φ : TpM → TqM such that this implies (i). ΦE = F, Corollary 6.4.13 asserts that a connected, simply connected, complete Riemannian m-manifold M has constant sectional curvature if and only if the isometry group I(M ) acts transitively on its orthonormal frame bundle O(M ). Note that, by Lemma 5.1.10, this group action is also free. 6.4. CONSTANT CURVATURE 285 Examples and Exercises Example 6.4.14. Any ﬂat Riemannian manifold has constant sectional curvature k = 0. Example 6.4.15. The manifold M = Rm with its standard metric is, up to isometry, the unique connected, simply connected, complete Riemannian m-manifold with constant sectional curvature k = 0. Example 6.4.16. For m ≥ 2 the unit sphere M = Sm with its standard metric is, up to isometry, the unique connected, simply connected, complete Riemannian m-manifold with constant sectional curvature k = 1. Hence, by Corollary 6.4.12, every connected simply connected, complete Riemannian manifold with positive sectional curvature k = 1 is compact. Moreover, by Corollary 6.4.13, the isometry group I(Sm) is isomorphic to the group O(m + 1) of orthog
onal linear transformations of Rm+1. Thus, by Corollary 6.4.13, the orthonormal frame bundle O(Sm) is diﬀeomorphic to O(m + 1). This follows also from the fact that, if is an orthonormal basis of TpSm = p⊥ then v1,..., vm p, v1,..., vm is an orthonormal basis of Rm+1. Example 6.4.17. A product of spheres is not a space of constant sectional curvature, but it is a symmetric space. Exercise: Prove this. Example 6.4.18. For n ≥ 4 the orthogonal group O(n) is not a space of constant sectional curvature, but it is a symmetric space and has nonnegative sectional curvature (see Example 6.4.4). 286 CHAPTER 6. GEOMETRY AND TOPOLOGY 6.4.3 Hyperbolic Space Fix an integer m ≥ 2. The hyperbolic space Hm is, up to isometry, the unique connected, simply connected, complete Riemannian m-manifold with constant sectional curvature k = −1. A model for Hm can be constructed as follows. A point in Rm+1 will be denoted by p = (x0, x), x0 ∈ R, x = (x1,..., xm) ∈ Rm. Let Q : Rm+1 × Rm+1 → R denote the symmetric bilinear form given by Q(p, q) := −x0y0 + x1y1 + · · · + xmym (6.4.9) for p = (x0, x), q = (y0, y) ∈ Rm+1. Since Q is nondegenerate the space Hm := p = (x0, x) ∈ Rm+1 \| Q(p, p) = −1, x0 > 0 is a smooth m-dimensional submanifold of Rm+1 and the tangent space of Hm at p is given by TpHm = v ∈ Rm+1 \| Q(p, v) = 0. For p = (x0, x) ∈ Rm+1 and v = (ξ0, �
�) ∈ Rm+1 we have p ∈ Hm ⇐⇒ x0 = v ∈ TpHm ⇐⇒ ξ0 = 1 + \|x\|2, ξ, x 1 + \|x\|2. Now let us deﬁne a Riemannian metric on Hm by gp(v, w) := Q(v, w) = ξ, η − ξ0η0 = ξ, η − ξ, xη, x 1 + \|x\|2 (6.4.10) for v = (ξ0, ξ) ∈ TpHm and w = (η0, η) ∈ TpHm. Theorem 6.4.19. Hm is a connected, simply connected, complete Riemannian m-manifold with constant sectional curvature k = −1. 6.4. CONSTANT CURVATURE 287 We remark that the manifold Hm does not quite ﬁt into the extrinsic framework of most of this book as it is not exhibited as a submanifold of Euclidean space but rather of “pseudo-Euclidean space”: the positive deﬁnite inner product v, w of the ambient space Rm+1 is replaced by a nondegenerate symmetric bilinear form Q(v, w). However, all the theory developed thus far goes through (reading Q(v, w) for v, w) provided we impose the additional hypothesis (true in the example M = Hm) that the ﬁrst fundamental form gp = Q\|TpM is positive deﬁnite. For then Q\|TpM is nondegenerate and we may deﬁne the orthogonal projection Π(p) onto TpM as before. The next lemma summarizes the basic observations; the proof is an exercise in linear algebra. Lemma 6.4.20. Let Q be a symmetric bilinear form on a vector space V and for each subspace E of V deﬁne its orthogonal complement by E⊥Q := {u ∈ V \| Q(u, v) = 0 ∀v ∈ E}. Assume Q is nondegenerate, i
.e. V ⊥Q = {0}. Then, for every linear subspace E ⊂ V, we have V = E ⊕ E⊥Q ⇐⇒ E ∩ E⊥Q = {0}, i.e. E⊥Q is a vector space complement of E if and only if the restriction of Q to E is nondegenerate. Proof of Theorem 6.4.19. The proofs of the various properties of Hm are entirely analogous to the corresponding proofs for Sm. Thus the unit normal ﬁeld to Hm is given by for p ∈ Hm although the “square of its length” is ν(p) = p Q(p, p) = −1. For p ∈ Hm we introduce the Q-orthogonal projection Π(p) of Rm+1 onto the subspace TpHm. It is characterized by the conditions Π(p)2 = Π(p), ker Π(p) ⊥Q imΠ(p), imΠ(p) = TpHm, and is given by the explicit formula Π(p)v = v + Q(v, p)p for v ∈ Rm+1. 288 CHAPTER 6. GEOMETRY AND TOPOLOGY The covariant derivative of a vector ﬁeld X ∈ Vect(γ) along a smooth curve γ : R → Hm is given by ∇X(t) = Π(γ(t)) ˙X(t) = ˙X(t) + Q( ˙X(t), γ(t))γ(t) = ˙X(t) − Q(X(t), ˙γ(t))γ(t). The last identity follows by diﬀerentiating the equation Q(X, γ) ≡ 0. This can be interpreted as the hyperbolic Gauß–Weingarten formula as follows. For p ∈ Hm and u ∈ TpHm we introduce, as before, the second fundamental form via hp(u) : TpHm → (TpHm)⊥Q hp(u)v := dΠ(p)uv and denote its Q-adjoint by hp(u)∗
: (TpHm)⊥Q → TpHm. For all p ∈ Hm, u ∈ TpHm, and v ∈ Rm+1 we have dΠ(p)u v = d dt t=0 v + Q(v, p + tu)(p + tu) = Q(v, p)u + Q(v, u)p, where the ﬁrst summand on the right is tangent to Hm and the second summand is Q-orthogonal to TpHm. Hence hp(u)v = Q(v, u)p, hp(u)∗w = Q(w, p)u (6.4.11) for v ∈ TpHm and w ∈ (TpHm)⊥Q. With this understood, the Gauß-Weingarten formula ˙X = ∇X + hγ( ˙γ)X extends to the present setting. The reader may verify that the operators ∇ : Vect(γ) → Vect(γ) thus deﬁned satisfy the axioms of Theorem 3.7.8 and hence deﬁne the LeviCivita connection on Hm. 6.4. CONSTANT CURVATURE 289 Now a smooth curve γ : I → Hm is a geodesic if and only if it satisﬁes the equivalent conditions ∇ ˙γ ≡ 0 ⇐⇒ ¨γ(t) ⊥Q Tγ(t)Hm ∀ t ∈ I ⇐⇒ ¨γ = Q( ˙γ, ˙γ)γ. A geodesic must satisfy the equation d dt Q( ˙γ, ˙γ) = 2Q(¨γ, ˙γ) = 0 because ¨γ is a scalar multiple of γ, and hence Q( ˙γ, ˙γ) is constant. Fix an element p ∈ Hm and a tangent vector v ∈ TpHm such that Q(v, v) = 1. Then the geodesic γ : R → Hm with γ(0) = p and ˙γ(0) = v is given by γ
(t) = cosh(t)p + sinh(t)v, (6.4.12) where cosh(t) := et + e−t 2, sinh(t) := et − e−t 2. In fact we have ¨γ(t) = γ(t) ⊥Q Tγ(t)Hm. It follows that the geodesics exist for all time and hence Hm is geodesically complete. Moreover, being diﬀeomorphic to Euclidean space, Hm is connected and simply connected. It remains to prove that Hm has constant sectional curvature k = −1. To see this we use the Gauß–Codazzi formula in the hyperbolic setting, i.e. Rp(u, v) = hp(u)∗hp(v) − hp(v)∗hp(u). (6.4.13) By equation (6.4.11), this gives Rp(u, v)v, u = Q(hp(u)u, hp(v)v) − Q(hp(v)u, hp(u)v) = Q(Q(u, u)p, Q(v, v)p) − Q(Q(u, v)p, Q(u, v)p) = −Q(u, u)Q(v, v) + Q(u, v)2 = −gp(u, u)gp(v, v) + gp(u, v)2 for all u, v ∈ TpHm. Hence, for every p ∈ M and every 2-dimensional linear subspace E ⊂ TpM with a basis u, v ∈ E we have K(p, E) = Rp(u, v)v, u gp(u, u)gp(v, v) − gp(u, v)2 = −1. This proves Theorem 6.4.19. 290 CHAPTER 6. GEOMETRY AND TOPOLOGY Exercise 6.4.21. Prove that the pullback of the metric on Hm under the diﬀeomorphism Rm → Hm : x → 1 + \|x\|2, x is given by \|x\|x = \|x\|2 − x, ξ2 1 + \|x\|
2 for x ∈ Rm and x ∈ Rm = TxRm. Thus the metric tensor is given by gij(x) = δij − (6.4.14) xixj 1 + \|x\|2 for x = (x1,..., xm) ∈ Rm. Exercise 6.4.22. The Poincar´e model of hyperbolic space is the open unit disc Dm ⊂ Rm equipped with the Poincar´e metric \|y\|y = 2 \|y\| 1 − \|y\|2 for y ∈ Dm and y ∈ Rm = TyDm. Thus the metric tensor is given by 4δij y ∈ Dm. (6.4.15) gij(y) = 1 − \|y\|22, Prove that the diﬀeomorphism Dm → Hm : y → 1 + \|y\|2 1 − \|y\|2, 2y 1 − \|y\|2 is an isometry with the inverse Hm → Dm : (x0, x) → x 1 + x0. Interpret this map as a stereographic projection from the south pole (−1, 0). Exercise 6.4.23. The composition of the isometries in Exercise 6.4.21 and Exercise 6.4.22 is the diﬀeomorphism Rm → Dm : x → y given by y = x 1 + \|x\|2 + 1, x = 2y 1 − \|y\|2, 1 + \|x\|2 = 1 + \|y\|2 1 − \|y\|2. Prove that this is an isometry intertwining the Riemannian metrics (6.4.14) and (6.4.15). 6.4. CONSTANT CURVATURE 291 Exercise 6.4.24. This exercise shows that every nonconstant geodesic in the Poincar´e model Dm of hyperbolic space in Exercise 6.4.22 converges to two points on the boundary Sm−1 = ∂Dm in forward and backward time, and that any two distinct points on the boundary are the asymptotic limits of a unique geodesic in Dm up to reparametrization. Fix an element y ∈ Dm and a tangent vector y ∈ Ty
Dm = Rm of norm one in the hyperbolic metric, i.e. λ \|y\| = 1, λ := 2 1 − \|y\|2. (6.4.16) Let γ : R → Dm be the unique geodesic satisfying γ(0) = y and ˙γ(0) = y. Prove the following. (a) The geodesic γ is given by the explicit formula γ(t) = cosh(t)λy + sinh(t)λy + λy, λyy 1 + cosh(t)λ − 1 + sinh(t)λy, λy (6.4.17) for t ∈ R. Hint: Use (6.4.12) and the isometries in Exercise 6.4.22. (b) The limits y± := lim t→±∞ γ(t) ∈ Sm−1 exist and are given by y+ = λy + λy + λy, λyy λ − 1 + λy, λy, y− = λy − λy − λy, λyy λ − 1 − λy, λy. (6.4.18) (c) Assume y /∈ Ry. Then there is a unique circle in Rm through y− and y+ that is orthogonal to Sm−1 at y±. The center c ∈ Rm and the radius r of this circle are given by c = r2 = y+ + y− 1 + y+, y− 1 − y+, y− 1 + y+, y− = = λ2 − λ − λy, λy2y − λy, λyλy λ2 − 2λ − λy, λy2 1 λ2 − 2λ − λy, λy2., (6.4.19) (d) Let c and r be as in (c). Then the geodesic γ in (a) satisﬁes \|γ(t) − c\| = r for all t. Hint: It suﬃces to verify this equation for t = 0. (e) If y ∈ Ry, then y− +
y+ = 0 and the geodesic γ traverses a segment of a straight line through the origin. (f ) Fix two distinct points y− and y+ on the unit sphere Sm−1. Then there exists a geodesic γ : R → Dm such that limt→±∞ γ(t) = y±. If γ : R → Dm is any other geodesic satisfying limt→±∞ γ(t) = y±, then there exist real numbers a, b such that a > 0 and γ(t) = γ(at + b) for all t ∈ R. 292 CHAPTER 6. GEOMETRY AND TOPOLOGY Exercise 6.4.25. Prove that the isometry group of Hm is the pseudo-orthogonal group I(Hm) = O(m, 1) := g ∈ GL(m + 1) Q(gv, gw) = Q(v, w) for all v, w ∈ Rm+1. Thus, by Corollary 6.4.13, the orthonormal frame bundle O(Hm) is diﬀeomorphic to O(m, 1). Exercise 6.4.26. Prove that the exponential map expp : TpHm → Hm is given by expp(v) = cosh Q(v, v) p + sinh Q(v, v) Q(v, v) v (6.4.20) for v ∈ TpHm = p⊥Q. Prove that this map is a diﬀeomorphism for every p ∈ Hm. Thus any two points in Hm are connected by a unique geodesic. Prove that the intrinsic distance function on hyperbolic space is given by d(p, q) = cosh−1 (Q(p, q)) for p, q ∈ Hm. Compare this with Example 4.3.11. (6.4.21) Exercise 6.4.27. In the case m = 2 the Poincar´e model of hyperbolic space in Exercise 6.4.22 is the open unit disc D ⊂ C in the complex plane. It can be identiﬁed with the upper half plane via the diﬀeomorphism H = {
z ∈ C \| Im(z) > 0. Show that the pullback of the Poincar´e metric on D under this diﬀeomorphism is the Riemannian metric on H given by \|z\|z = for z = x + iy ∈ H and z ∈ TzH = C. Show that the isometries of H (in the identity component) have the form \|z\| y φ(z) = az + b cz + d, a b c d ∈ SL(2, R), and deduce that the Lie group PSL(2, R) := SL(2, R)/{±1l} is isomorphic to the identity component of O(2, 1). Prove that every nonconstant geodesic in H traverses either a vertical half line or a semicircle centered at a point on the boundary ∂H = R. 6.5. NONPOSITIVE SECTIONAL CURVATURE 293 6.5 Nonpositive Sectional Curvature In the previous section we have seen that any two points in a connected, simply connected, complete manifold M of constant negative curvature are joined by a unique geodesic (Exercise 6.4.26). Thus the entire manifold M is geodesically convex and its injectivity radius is inﬁnity. This continues to hold in much greater generality for manifolds with nonpositive sectional curvature. It is convenient, at this point, to extend the discussion to Riemannian manifolds in the intrinsic setting. In particular, at some point in the proof of the main theorem of this section and in our main example, we shall work with a Riemannian metric that does not arise (in any obvious way) from an embedding. Deﬁnition 6.5.1. A Riemannian manifold M is said to have nonpositive sectional curvature iﬀ K(p, E) ≤ 0 for every p ∈ M and every 2dimensional linear subspace E ⊂ TpM or, equivalently, Rp(u, v)v, u ≤ 0 for all p ∈ M and all u, v ∈ TpM. A nonempty, connected, simply connected, complete Riemannan manifold with nonpositive sectional curvature is called a Hadamard manifold. 6.5.1 The Cartan
–Hadamard Theorem The next theorem shows that every Hadamard manifold is diﬀeomorphic to Euclidean space and has inﬁnite injectivity radius. This is in sharp contrast to positive curvature manifolds as the example M = Sm shows. Theorem 6.5.2 (Cartan–Hadamard). Let M be a connected, simply connected, complete Riemannan manifold. Then the following are equivalent. (i) M has nonpositive sectional curvature. (ii) The derivative of each exponential map is length increasing, i.e. d expp(v)v ≥ \|v\| for all p ∈ M and all v, v ∈ TpM. (iii) Each exponential map is distance increasing, i.e. d(expp(v0), expp(v1)) ≥ \|v0 − v1\| for all p ∈ M and all v0, v1 ∈ TpM. Moreover, if these equivalent conditions are satisﬁed, then the exponential map expp : TpM → M is a diﬀeomorphism for every p ∈ M. Thus any two points in M are joined by a unique geodesic. The proof makes use of the following two exercises. 294 CHAPTER 6. GEOMETRY AND TOPOLOGY Exercise 6.5.3. Let ξ : [0, ∞) → Rn be a smooth function such that ξ(0) = 0, ˙ξ(0) = 0, ξ(t) = 0 ∀ t > 0. Prove that the function f : [0, ∞) → R given by f (t) := \|ξ(t)\| is smooth. Hint: The function η : [0, ∞) → Rn deﬁned by η(t) := t−1ξ(t), ˙ξ(0), for t > 0, for t = 0, is smooth. Show that f is diﬀerentiable and ˙f = \|η\|−1 η, ˙ξ. Exercise 6.5.4. Let ξ : R → Rn be a smooth function such that ξ(0) = 0, ¨ξ(0) = 0. Prove that there exist constants �
� > 0 and c > 0 such that, for all t ∈ R, \|t\| < ε =⇒ \|ξ(t)\|2 \| ˙ξ(t)\|2 − ξ(t), ˙ξ(t)2 ≤ c \|t\|6. Hint: Write ξ(t) = tv + η(t), ˙ξ(t) = v + ˙η(t) with η(t) = O(t3) and ˙η(t) = O(t2). Show that the terms of order 2 and 4 cancel in the Taylor expansion at t = 0. Proof of Theorem 6.5.2. We prove that (i) implies (ii). Fix a point p ∈ M and two tangent vectors v, v ∈ TpM. Assume without loss of generality that v = 0 and deﬁne the curve γ : R → M and the vector ﬁeld X ∈ Vect(γ) along γ by γ(t) := expp(tv), X(t) := ∂ ∂s s=0 expp(t(v + sv)) ∈ Tγ(t)M (6.5.1) for t ∈ R. Then X(0) = 0, X(t) = d expp(tv)tv, ∇X(0) = v = 0. (6.5.2) To see this, deﬁne the map β : R2 → M by β(s, t) := expp(t(v + sv)). It satisﬁes β(0, t) = γ(t), ∂sβ(0, t) = X(t), β(s, 0) = p, and ∂tβ(s, 0) = v + sv for all s, t ∈ R. Hence ∇X(0) = ∇t∂sβ(0, 0) = ∇s∂tβ(0, 0) = v. Moreover, the curve β(s, ·) is a geodesic for every s, and hence Lemma 6.1.18 asserts that X = ∂sβ(0, ·) is a Jacobi ﬁe
ld along γ, i.e. ∇∇X + R(X, ˙γ) ˙γ = 0. (6.5.3) 6.5. NONPOSITIVE SECTIONAL CURVATURE 295 It follows from Exercise 6.5.3 with ξ(t) := Φγ(0, t)X(t) that the function [0, ∞) → R : t → \|X(t)\| is smooth and t=0 Moreover, for t > 0, we have d dt \|X(t)\| = \|∇X(0)\| = \|v\|. d2 dt2 \|X\| = = = X, ∇X \|X\| d dt \|∇X\|2 + X, ∇∇X \|X\| − \|X\|2\|∇X\|2 − X, ∇X2 \|X\|3 X, ∇X2 \|X\|3 X, R( ˙γ, X) ˙γ \|X\| + (6.5.4) ≥ 0. Here the third equality follows from the fact that X is a Jacobi ﬁeld along γ, and the inequality follows from the nonpositive sectional curvature condition in (i) and from the Cauchy–Schwarz inequality. Thus the second derivative of the function [0, ∞) → R : t → \|X(t)\| − t \|v\| is nonnegative; so its ﬁrst derivative is nondecreasing and it vanishes at t = 0; thus \|X(t)\| − t \|v\| ≥ 0 for every t ≥ 0. In particular, for t = 1 we obtain d expp(v)v = \|X(1)\| ≥ \|v\|. as claimed. Thus we have proved that (i) implies (ii). We prove that (ii) implies (i). Assume, by contradiction, that (ii) holds but there exists a point p ∈ M and a pair of vectors v, v ∈ TpM such that (6.5.5) Rp(v, v)v, v < 0. Deﬁne γ : R → M and X ∈ Vect(γ) by (6.5.1) so that (6.5.2) and (6.5.
3) are satisﬁed. Thus X is a Jacobi ﬁeld with X(0) = 0, ∇X(0) = v = 0. Hence it follows from Exercise 6.5.4 with ξ(t) := Φγ(0, t)X(t) that there is a constant c > 0 such that, for t > 0 suﬃciently small, we have the inequality \|X(t)\|2 \|∇X(t)\|2 − X(t), ∇X(t)2 ≤ ct6. 296 CHAPTER 6. GEOMETRY AND TOPOLOGY Moreover, by (6.5.1) and (6.5.2), limt0 ˙γ(t) = v and limt0 t−1X(t) = v. Hence, by (6.5.5) there exist constants δ > 0 and ε > 0 such that \|X(t)\| ≥ δt, X(t), R( ˙γ(t), X(t)) ˙γ(t) ≤ −εt2, for t > 0 suﬃciently small. By (6.5.4) this implies d2 dt2 \|X\| = \|X\|2\|∇X\|2 − X, ∇X2 \|X\|3 + X, R( ˙γ, X) ˙γ \|X\| ≤ ct3 δ3 − εt δ. Integrate this inequality over an interval [0, t] with ct2 < εδ2 to obtain d dt \|X(t)\| < d dt t=0 \|X(t)\| = \|∇X(0)\| Integrating this inequality again gives \|X(t)\| < t \|∇X(0)\| for small positive t. Hence it follows from (6.5.2) that d expp(tv)tv for t > 0 suﬃciently small. This contradicts (ii). = \|X(t)\| < t \|∇X(0)\| = t \|v\| We prove that (ii) implies that the exponential map expp : TpM → M is a diﬀeomorphism for every p ∈ M. By (ii) expp is a local di�
��eomorphism, i.e. its derivative d expp(v) : TpM → Texpp(v)M is bijective for every v ∈ TpM. Hence we can deﬁne a Riemannian metric on M := TpM by pulling back the metric on M under the exponential map. To make this more explicit we choose a basis e1,..., em of TpM and deﬁne the map ψ : Rm → M by ψ(x) := expp xiei m i=1 for x = (x1,..., xm) ∈ Rm. Deﬁne the metric tensor by, ∂ψ gij(x) := ∂xi (x), i, j = 1,..., m. ∂ψ ∂xj (x) Then (Rm, g) is a Riemannian manifold (covered by a single coordinate chart) and ψ : (Rm, g) → M is a local isometry, by deﬁnition of g. The manifold (Rm, g) is clearly connected and simply connected. Moreover, for every ξ = (ξ1,..., ξn) ∈ Rm = T0Rm, the curve R → Rm : t → tξ is a geodesic with respect to g (because ψ is a local isometry and the image of the curve under ψ is a geodesic in M ). Hence it follows from Theorem 4.6.5 that (Rm, g) is complete. 6.5. NONPOSITIVE SECTIONAL CURVATURE 297 Since both (Rm, g) and M are connected, simply connected, and complete, it follows from Corollary 6.1.15 that the local isometry ψ is bijective. Thus the exponential map expp : TpM → M is a diﬀeomorphism as claimed. It follows that any two points in M are joined by a unique geodesic. We prove that (ii) implies (iii). Fix a point p ∈ M and two tangent vectors v0, v1 ∈ TpM. Let γ : [0, 1] → M be the geodesic with the endpoints γ
(0) = expp(v0), γ(1) = expp(v1) and let v : [0, 1] → TpM be the unique curve satisfying expp(v(t)) = γ(t) for all t. Then v(0) = v0, v(1) = v1, and d(expp(v0), expp(v1)) = L(γ) 1 0 1 d expp(v(t)) ˙v(t) dt \| ˙v(t)\| dt = ≥ ≥ 0 1 ˙v(t) dt 0 = \|v1 − v0\|. Here the third inequality follows from (ii). This shows that (ii) implies (iii). We prove that (iii) implies (ii). Fix a point p ∈ M and a tangent vector v ∈ TpM and denote q := expp(v). By (iii) the exponential map expq : TqM → M is injective and, since M is complete, it is bijective (see Theorem 4.6.6). Hence there exists a unique geodesic from q to any other point in M and therefore, by Theorem 4.4.4, we have \|w\| = d(q, expq(w)) for every w ∈ TqM. Now deﬁne φ := exp−1 q ◦ expp : TpM → TqM. This map satisﬁes φ(v) = 0. Moreover, it is diﬀerentiable in a neighborhood of v and, by the chain rule, dφ(v) = d expp(v) : TpM → TqM. Now choose w := φ(v + v) with v ∈ TpM. Then expq(w) = expq(φ(v + v)) = expp(v + v) and hence it follows from (6.5.6) and part (iii) that (6.5.6) \|φ(v + v))\| = \|w\| = d(q, expq(w)) = d(expp(v), expp(v + v)) ≥ \|v\|. This gives d expp(v)v = \|dφ(v)
v\| = lim t→0 \|φ(v + tv)\| t ≥ lim t→0 \|tv\| t = \|v\|. Thus we have proved that (iii) implies (ii) and this proves Theorem 6.5.2. 298 CHAPTER 6. GEOMETRY AND TOPOLOGY The next lemma establishes a useful inequality for Hadamard manifolds that ampliﬁes the expanding property of the exponential map. Lemma 6.5.5. Let M be a Hadamard manifold. Fix an element p ∈ M and two tangent vectors v0, v1 ∈ TpM. Then, for 0 < t ≤ T, \|v0 − v1\| ≤ d(expp(tv0), expp(tv1)) t ≤ d(expp(T v0), expp(T v1)) T. (6.5.7) Proof. The ﬁrst inequality in (6.5.7) is part (iii) of Theorem 6.5.2. To prove the second inequality, assume v0 = v1 and deﬁne γ0(t) := expp(tv0), γ1(t) := expp(tv1) for t ∈ R. For each t ∈ R let the curve [0, 1] → M : s → γ(s, t) be the unique geodesic with the endpoints γ(0, t) = γ0(t) and γ(1, t) = γ1(t). Then ρ(t) := d(γ0(t), γ1(t)) = 1 0 \|∂sγ\| ds = \|∂sγ(s, t)\| for all s and t and hence 1 ˙ρ(t) = ∂sγ, ∇t∂sγ \|∂sγ\| ds 0 ∂sγ(1, t), ∂tγ(1, t) − ∂sγ(0, t), ∂tγ(0, t) ρ(t) = (6.5.8). Since d dt (ρ ˙ρ) = ρ¨ρ + ˙ρ2 and γ0 and γ1 are geodesics, this implies ρ(t)
¨ρ(t) + ˙ρ(t)2 = d dt ∂sγ(1, t), ∂tγ(1, t) − ∂sγ(0, t), ∂tγ(0, t) = ∇t∂sγ(1, t), ∂tγ(1, t) − ∇t∂sγ(0, t), ∂tγ(0, t) 1 0 1 ∂ ∂s = = = ∇t∂sγ, ∂tγ ds \|∇t∂sγ\|2 + ∇s∇t∂sγ, ∂tγ ds 0 1 \|∇t∂sγ\|2 + R(∂sγ, ∂tγ)∂sγ, ∂tγ ds 0 ≥ ˙ρ(t)2. Here the last step follows from (6.5.8), the Cauchy–Schwarz inequality, and the nonpositive sectional curvature assumption. Thus ρ : [0, T ] → R is a convex function satisfying ρ(0) = 0 and hence ρ(t) ≤ tρ(T )/T for 0 ≤ t ≤ T. This proves Lemma 6.5.5. 6.5. NONPOSITIVE SECTIONAL CURVATURE 299 6.5.2 Cartan’s Fixed Point Theorem Recall from Deﬁnition 6.5.1 that a Hadamard manifold is a nonempty, connected, simply connected, complete Riemannian manifold of nonpositive sectional curvature. Theorem 6.5.6 (Cartan). Let M be a Hadamard manifold and let G be a compact topological group that acts on M by isometries. Then there exists a point p ∈ M such that gp = p for every g ∈ G. The proof follows the argument given by Bill Casselmann in [16] and requires the following two lemmas. The ﬁrst lemma asserts that every complete, connected, simply connected Riemannian manifold of nonpositive sectional curvature is a semi-hyperbolic space in the sense of Alexandrov [3]. Figure 6.5: Alexandrov semi-hyperbolic space. Lemma 6.5.7 (Alex
androv). Let M be a Hadamard manifold, let m ∈ M and v ∈ TmM, and deﬁne p0 := expm(−v), p1 := expm(v). Then 2d(m, q)2 + d(p0, p1)2 2 ≤ d(p0, q)2 + d(p1, q)2 (6.5.9) for every q ∈ M (see Figure 6.5). vmq01pp 300 CHAPTER 6. GEOMETRY AND TOPOLOGY Proof. By Theorem 6.5.2 the exponential map expm : TmM → M is a diﬀeomorphism. Hence d(p0, p1) = 2\|v\|. Now let q ∈ M. Then there is a unique tangent vector w ∈ TmM such that q = expm(w), d(m, q) = \|w\|. Since the exponential map is expanding, by Theorem 6.5.2, we have d(p0, q) ≥ \|w + v\|, d(p1, q) ≥ \|w − v\|. Hence d(m, q)2 = \|w\|2 = ≤ \|w + v\|2 + \|w − v\|2 2 d(p0, q)2 + d(p1, q)2 2 − \|v\|2 − d(p0, p1)2 4. This proves Lemma 6.5.7. Exercise 6.5.8. Equality holds in (6.5.9) whenever M is ﬂat. The next lemma is Serre’s Uniqueness Theorem for the circumcentre of a bounded set in a semi-hyperbolic space. Figure 6.6: The circumcenter of a bounded set. Lemma 6.5.9 (Serre). Let M be a Hadamard manifold and, for p ∈ M and r ≥ 0, denote by B(p, r) ⊂ M the closed ball of radius r centered at p. Let Ω ⊂ M be a nonempty bounded set and deﬁne rΩ := inf {r > 0 \| there exists a p ∈ M such that Ω ⊂ B(p, r)}. Then there exists a unique point
pΩ ∈ M such that Ω ⊂ B(pΩ, rΩ) (see Figure 6.6). rΩΩΩp 6.5. NONPOSITIVE SECTIONAL CURVATURE 301 Proof. We prove existence. Choose sequences ri > rΩ and pi ∈ M such that Ω ⊂ B(pi, ri), lim i→∞ ri = rΩ. Choose q ∈ Ω. Then d(q, pi) ≤ ri for every i. Since the sequence ri is bounded and M is complete, it follows that pi has a convergent subsequence, still denoted by pi. Its limit pΩ := lim i→∞ pi satisﬁes Ω ⊂ B(pΩ, rΩ). We prove uniqueness. Let p0, p1 ∈ M such that Ω ⊂ B(p0, rΩ) ∩ B(p1, rΩ). Since the exponential map expp : TpM → M is a diﬀeomorphism (by Theorem 6.5.2), there exists a unique vector v0 ∈ Tp0M such that p1 = expp0(v0). Denote the midpoint between p0 and p1 by m := expp0 1 2 v0. Then it follows from Lemma 6.5.7 that d(m, q)2 ≤ d(p0, q)2 + d(p1, q)2 2 d(p0, p1)2 4 Ω − ≤ r2 − d(p0, p1)2 4 for every q ∈ Ω. Since supq∈Ω d(m, q) ≥ rΩ (by deﬁnition of rΩ), it follows that d(p0, p1) = 0 and hence p0 = p1. This proves Lemma 6.5.9. Proof of Theorem 6.5.6. Let q ∈ M and consider the group orbit Ω := {gq \| g ∈ G}. Since G is compact, this set is bounded. Let rΩ and pΩ be
as in Lemma 6.5.9. Then Ω ⊂ B(pΩ, rΩ). Since G acts on M by isometries, this implies Ω = gΩ ⊂ B(gpΩ, rΩ) for all g ∈ G. Hence it follows from the uniqueness statement in Lemma 6.5.9 that gpΩ = pΩ for every g ∈ G. This proves Theorem 6.5.6. 302 CHAPTER 6. GEOMETRY AND TOPOLOGY 6.5.3 Positive Deﬁnite Symmetric Matrices We close this section with an example of a nonpositive sectional curvature manifold which plays a key role in Donaldson’s approach to Lie algebra theory [17] (see §7.5.2). Let m be a positive integer and consider the space P ∈ Rm×m P T = P > 0 P := P(Rm) := (6.5.10) of positive deﬁnite symmetric m × m-matrices. (Here the notation “P > 0” means x, P x > 0 for every nonzero vector x ∈ Rm.) Thus P is an open subset of the vector space S := {S ∈ Rm×m \| ST = S} of symmetric matrices and hence the tangent space of P is TP P = S for every P ∈ P. However, we do not use the metric inherited from the inclusion into S but deﬁne a Riemannian metric by P1, P2P := trace P1P −1 P2P −1 (6.5.11) for P ∈ P and P1, P2 ∈ TP P = S. Theorem 6.5.10. The space P with the Riemannian metric (6.5.11) is a connected, simply connected, complete Riemannian manifold with nonpositive sectional curvature, and the distance function on P is given by d(P, Q) = trace log(P −1/2QP −1/2)2 (6.5.12) for P, Q ∈ P. Moreover, P is a symmetric space and the group GL(m, R) of nonsingular m × m-matrices acts
on P by isometries via P → gP gT for g ∈ GL(m, R). Proof. See below. Remark 6.5.11. Let V be an m-dimensional vector space and H ⊂ S2V ∗ be the set of inner products on V. Deﬁne a Riemannian metric on H by h1, h2h := trace(S1S2), (6.5.13) for h ∈ H and h1, h2 ∈ ThH = S2V ∗. Then every vector space isomorphism α : Rm → V determines a diﬀeomorphism φα : H → P via h(·, Si·) := hi, φα(h) = P ⇐⇒ h(αξ, αη) = ξ, P −1ηRm. (6.5.14) The derivative of φα at h in the direction h ∈ ThH is given by dφα(h)h = P ⇐⇒ P P −1 = −α−1Sα, h(·, S·) := h. (6.5.15) Thus φα is an isometry with respect to the Riemannian metrics (6.5.13) on H and (6.5.11) on P. The φα form an atlas on H with the transition βα, where gβα := β−1α ∈ GL(m, R). maps φβα(P ) := φβ ◦ φ−1 α (P ) = gβαP gT 6.5. NONPOSITIVE SECTIONAL CURVATURE 303 Remark 6.5.12. The submanifold P0 := P0(Rm) := {P ∈ P \| det(P ) = 1} (6.5.16) of positive deﬁnite symmetric m × m-matrices with determinant one is totally geodesic (see Remark 6.5.13 below). Hence all the assertions of Theorem 6.5.10 (with GL(m, R) replaced by SL(m, R)) remain valid for P0. Remark 6.5.13. Let M be a Riemannian manifold and
L ⊂ M be a submanifold. Then the following are equivalent. (i) If γ : I → M is a geodesic on an open interval I such that 0 ∈ I and γ(0) ∈ L, ˙γ(0) ∈ Tγ(0)L, then there is a constant ε > 0 such that γ(t) ∈ L for \|t\| < ε. (ii) If γ : I → L is a smooth curve on an open interval I and Φγ denotes parallel transport along γ in M, then Φγ(t, s)Tγ(s)L = Tγ(t)L ∀ s, t ∈ I. (iii) If γ : I → L is a smooth curve on an open interval I and X ∈ Vect(γ) is a vector ﬁeld along γ (with values in T M ), then X(t) ∈ Tγ(t)L ∀ t ∈ I =⇒ ∇X(t) ∈ Tγ(t)L ∀ t ∈ I. A submanifold that satisﬁes these equivalent conditions is called totally geodesic. Exercise 6.5.14. Prove the equivalence of (i), (ii), (iii) in Remark 6.5.13. Hint: Choose suitable coordinates and translate each of the three assertions into conditions on the Christoﬀel symbols. Exercise 6.5.15. Prove that P0 is a totally geodesic submanifold of P. Prove that P is diﬀeomorphic to the quotient GL(m, R)/O(m) via polar decomposition and that P0 is diﬀeomorphic to the quotient SL(m, R)/SO(m). Hint: Consider the map GL(m, R) → P : g → ggT. Exercise 6.5.16. In the case m = 2 prove that P0 is isometric to the hyperbolic space H2. The proof of Theorem 6.5.10 is based on the calculation of the Levi-Civita connection and the formulas for geodesics and the Riemann curvature tensor in the following three lemmas. 304 CHAPTER 6. GEOM
ETRY AND TOPOLOGY Lemma 6.5.17. Let I → P : t → P (t) be a smooth path in P on an interval I ⊂ R and let I → S : t → S(t) be a vector ﬁeld along P. Then the covariant derivative of S is given by ∇S = ˙S − SP −1 ˙P − 1 2 ˙P P −1S. 1 2 (6.5.17) Proof. The formula (6.5.17) determines a family of linear operators on the spaces of vector ﬁelds along paths that satisfy the torsion-free condition ∇s∂tP = ∇t∂sP for every smooth map R2 → P : (s, t) → P (s, t) and the Leibniz rule ∇ S1, S2P = ∇S1, S2P + S1, ∇S2P for any two vector ﬁelds S1 and S2 along P. These two conditions determine the covariant derivative uniquely (see Lemma 3.6.5 and Theorem 3.7.8). This proves Lemma 6.5.17. Lemma 6.5.18. The geodesics in P are given by γ(t) = P exptP −1 P = expt P P −1P = P 1/2 exptP −1/2 P P −1/2P 1/2 (6.5.18) for P ∈ P, P ∈ TP P = S, and t ∈ R. In particular, P is complete. Proof. The curve γ : R → P deﬁned by (6.5.18) satisﬁes ˙γ(t) = P exptP −1 P = P P −1γ(t). Hence it follows from Lemma 6.5.17 that ∇ ˙γ(t) = ¨γ(t) − ˙γ(t)γ(t)−1 ˙γ(t) = ¨γ(t) − P P −1 ˙γ(t) = 0 for every t ∈ R. Hence γ is a geodesic. Since the curve γ : R

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