Basic Riemannian Geometry - Department of Mathematical ...

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Let ∂ t and ∂ s denote the coordinate vector fields on I × (−ɛ, ɛ) and setD = h −1 ∇. Since each γ s is a geodesic, we havewhenceD ∂t∂h∂t = 0D ∂s D ∂t∂h∂t = 0.The definition of the curvature tensor, along with the fact that [∂ s , ∂ t ] = 0,allows us to write0 = D ∂s D ∂t∂h∂t = D ∂ tD ∂s∂h∂t + R(∂h ∂t , ∂h∂s )∂h ∂t .Moreover, it follows from the fact that ∇ is torsion-free thatso thatSetting s = 0, this last becomes∂hD ∂s∂t = D ∂h∂ t∂s0 = (D ∂t ) 2 ∂h∂s + R(∂h ∂t , ∂h∂s )∂h ∂t .(∇ t ) 2 Y + R(γ ′ , Y )γ ′ = 0.Fact. All Jacobi fields arise this way.Let us contemplate an example which will compute for us the (constant)value of K for hyperbolic space: let (D n , g) be hyperbolic space and considera path ξ : (−ɛ, ɛ) → S n−1 ⊂ D 0 with ξ ′ (0) = η ⊥ ξ(0).We set h(t, s) = γ ξ(s) (t) = (2 tanh t/2)ξ(s)—a variation of geodesics through0. We then have a Jacobi field Y along γ = γ ξ(0) :Y (t) = ∂h∂s ∣ = 2(tanh t/2)ηs=0= sinh t ( η/ cosh 2 t/2 )= sinh tU(t)where U is a unit length parallel vector field along γ.We therefore have:(∇ t ) 2 Y = sinh ′′ tU(t) = sinh tU(t)20
whenceTake an inner product with U to getU + R(γ ′ , U)γ ′ = 0.K(γ ′ ∧ U) = −1and so conclude that (D n , g) has constant curvature −1.The same argument (that is, differentiate the image under exp m of a familyof straight lines through the origin) computes Jacobi fields in normalcoordinates:Theorem. For ξ ∈ M m , the Jacobi field Y along γ ξ withis given byY (0) = 0(∇ t Y )(0) = η ∈ M mY (t) = (d exp) tξ tη.3.6 Conjugate points and the Cartan–Hadamard theoremLet ξ ∈ M p and let γ = γ ξ : I ξ → R. We say that q = γ(t 1 ) is conjugate top along γ if there is a non-zero Jacobi field Y withY (0) = Y (t 1 ) = 0.In view of the theorem just stated, this happens exactly when (d exp p ) t1 ξ issingular.Theorem (Cartan–Hadamard). If (M, g) is complete and K ≤ 0 thenno p ∈ M has conjugate points.Proof. Suppose that Y is a Jacobi field along some geodesic γ with Y (0) =Y (t 1 ) = 0. Then0 == −∫ t10∫ t10〈∇ 2 t Y + R(γ ′ , Y )γ ′ , Y 〉 dt∫ t1|∇ t T | 2 dt + K(γ ′ ∧ Y )|Y | 2 dt0where we have integrated by parts and used Y (0) = Y (t 1 ) = 0 to kill theboundary term. Now both summands in this last equation are non-negativeand so must vanish. In particular,∇ t Y = 0so that Y is parallel whence |Y | is constant giving eventually Y ≡ 0.21
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Basic Riemannian Geometry - Department of Mathematical ...

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