Nonpositively curved implies conjugate-free - Revision history

Vipul: 5 revisions

2008-05-18T19:50:09Z

5 revisions

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Vipul at 14:39, 30 August 2007

2007-08-30T14:39:11Z

← Older revision		Revision as of 14:39, 30 August 2007
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	This tells us that <math><\frac{DJ}{dt},J></math> is nondecreasing, and hence, that <math><J,J></math> is nondecreasing. Thus <math>J</math> cannot be zero at both endpoints unless it is identically zero.		This tells us that <math><\frac{DJ}{dt},J></math> is nondecreasing, and hence, that <math><J,J></math> is nondecreasing. Thus <math>J</math> cannot be zero at both endpoints unless it is identically zero.

			==Application==

			This observation is crucial to the proof of the [[Cartan-Hadamard theorem]].

Vipul: Negatively curved implies conjugate-free moved to Nonpositively curved implies conjugate-free

2007-08-05T12:13:09Z

Negatively curved implies conjugate-free moved to Nonpositively curved implies conjugate-free

← Older revision	Revision as of 12:13, 5 August 2007
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Vipul at 12:12, 5 August 2007

2007-08-05T12:12:42Z

← Older revision		Revision as of 12:12, 5 August 2007
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	===Verbal statement===		===Verbal statement===

	If the [[sectional curvature]] of a [[Riemannian manifold]] is everywhere nonpositive, then the Riemannian manifold os [[conjugate-free Riemannian manifold\|conjugate-free]], viz it does not contain a pair of conjugate points.		If the [[sectional curvature]] of a [[Riemannian manifold]] is everywhere nonpositive, then the Riemannian manifold is [[conjugate-free Riemannian manifold\|conjugate-free]], viz it does not contain a pair of conjugate points.

	==Proof==		==Proof==
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	<math>\frac{D^2J}{dt^2} + R(J,V)V = 0</math>		<math>\frac{D^2J}{dt^2} + R(J,V)V = 0</math>

	Taking the inner product of both these with <math>J</math>, the second term becomes the [[sectional curvature]] of the tangent plane spanned by <math>J</math> and <math>V</math>, which is ~~nonpositive~~. Hence we conclude that:		Taking the inner product of both these with <math>J</math>, the second term becomes the [[sectional curvature]] of the tangent plane spanned by <math>J</math> and <math>V</math>, which is negative. Hence we conclude that:

	<math><\frac{D^2J}{dt^2},J> \ge 0</math>		<math><\frac{D^2J}{dt^2},J> \ge 0</math>

	This tells us that		This tells us that <math><\frac{DJ}{dt},J></math> is nondecreasing, and hence, that <math><J,J></math> is nondecreasing. Thus <math>J</math> cannot be zero at both endpoints unless it is identically zero.

Vipul: Nonpositively curved implies conjugate-free moved to Negatively curved implies conjugate-free: More correct.

2007-08-05T12:10:13Z

Nonpositively curved implies conjugate-free moved to Negatively curved implies conjugate-free: More correct.

← Older revision	Revision as of 12:10, 5 August 2007
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Vipul at 12:09, 5 August 2007

2007-08-05T12:09:41Z

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==Statement==

===Verbal statement===

If the [[sectional curvature]] of a [[Riemannian manifold]] is everywhere nonpositive, then the Riemannian manifold os [[conjugate-free Riemannian manifold|conjugate-free]], viz it does not contain a pair of conjugate points.

==Proof==

Let <math>M</math> be the Riemannian manifold, <math>\gamma</math> a geodesic in <math>M</math> and <math>J</math> a [[Jacobi field]] about <math>\gamma</math> that vanishes at the endpoints. We must show that <math>J</math> is everywhere zero.

Consider the [[Jacobi equation]]:

<math>\frac{D^2J}{dt^2} + R(J,V)V = 0</math>

Taking the inner product of both these with <math>J</math>, the second term becomes the [[sectional curvature]] of the tangent plane spanned by <math>J</math> and <math>V</math>, which is nonpositive. Hence we conclude that:

<math><\frac{D^2J}{dt^2},J> \ge 0</math>

This tells us that