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284 Lemma 42]. 29. The statement that △f ≤ n−1 f in the sense of distributions (or equivalently in the weak sense) means that for any non-negative test function φ, that is to say for any compactly supported C ∞ -function φ, we have n−1 f △φdvol ≤ φdvol. f M M Since the triangle inequality implies that |f (x) − f (y)| ≤ d(x, y), it follows that f is Lipschitz, and hence that the restriction of ∇f to any compact subset of M is an L2 one-form. Integration by parts then shows that M f △φdvol = − M ∇f, ∇φ dvol.

There are many important related curvatures. 6. The sectional curvature of a 2-plane P ⊂ Tp M is defined as K(P ) = R(X, Y, X, Y ), where {X, Y } is an orthonormal basis of P . , K(P ) < 0) for every 2-plane P . There are analogous notions of non-negative and non-positive sectional curvature. In local coordinates, suppose that X = X i ∂i and Y = Y i ∂i . Then we have K(P ) = Rijkl X i Y j X k Y l . A Riemannian manifold is said to have constant sectional curvature if K(P ) is the same for all p ∈ M and all 2-planes P ⊂ Tp M .

Then d2 E(γu ) du2 u=0 =0 if and only if Y is a Jacobi field. Proof. Suppose that γ(u, t) is a one-parameter family of curves from γ(0) to γ(1) with γ0 = γ and Y is the first-order variation of this family along γ. Since γ is a minimal geodesic we have 1 − rmJac(Y ), Y dt = 0 d2 E(γu ) du2 u=0 ≥ 0. The associated symmetric bilinear form is Bγ (Y1 , Y2 ) = − rmJac(Y1 ), Y2 dt γ 3. GEODESICS AND THE EXPONENTIAL MAP 45 is symmetric when Y1 and Y2 are constrained to vanish at both endpoints. Since the associated quadratic form is non-negative, we see by the usual argument for symmetric bilinear forms that Bγ (Y, Y ) = 0 if and only if Bγ (Y, ·) = 0 as a linear functional on the space of vector fields along γ vanishing at point endpoints.

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