Second Bianchi identity - Revision history

Vipul at 01:23, 24 July 2009

2009-07-24T01:23:31Z

← Older revision		Revision as of 01:23, 24 July 2009
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	==Statement==		==Statement==

	Let <math>\nabla</math> be a [[linear connection]] and <math>R</math> be the [[Riemann curvature tensor]] of <math>\nabla</math>. Then <math>R</math> can itself be differentiated via <math>\nabla</math>, since <math>R</math> is a <math>(1,3)</math>-tensor and we can define the connection on all <math>(p,q)</math>-tensors. With this meaning, the following cyclic summation is zero:		Let <math>\nabla</math> be a [[fact about::linear connection]] and <math>R</math> be the [[fact about::Riemann curvature tensor]] of <math>\nabla</math>. Then <math>R</math> can itself be differentiated via <math>\nabla</math>, since <math>R</math> is a <math>(1,3)</math>-tensor and we can define the connection on all <math>(p,q)</math>-tensors. With this meaning, the following cyclic summation is zero:

	<math>(\nabla_X R)(Y,Z) + (\nabla_Y R) (Z,X) + (\nabla_Z R)(X,Y) = 0</math>		<math>(\nabla_X R)(Y,Z) + (\nabla_Y R) (Z,X) + (\nabla_Z R)(X,Y) = 0</math>

Vipul: 2 revisions

2008-05-18T20:08:42Z

2 revisions

← Older revision	Revision as of 20:08, 18 May 2008
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Vipul at 14:41, 2 September 2007

2007-09-02T14:41:36Z

← Older revision		Revision as of 14:41, 2 September 2007
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	<math>(\nabla_X R)(Y,Z) + (\nabla_Y R) (Z,X) + (\nabla_Z R)(X,Y) = 0</math>		<math>(\nabla_X R)(Y,Z) + (\nabla_Y R) (Z,X) + (\nabla_Z R)(X,Y) = 0</math>

			This result is known as the '''second Bianchi identity''' and also as the '''differential Bianch identity'''.

	==Proof==		==Proof==
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	We now concentrate on all the other terms.		We now concentrate on all the other terms.

			==Applications==

			The second Bianchi identity is typically used for the studyo of manifolds of dimension three or higher, since we can choose <math>X,Y,Z</math> as linearly independent at a particular point. Some examples of its application:

			* [[Schur's theorem]] which states that if the [[sectional curvature]] of a manifold of dimension at least 3 depends only on the point and not on the choice of tangent plane, then it is in fact constant at all points. The idea here is to show that the scalar function defined by the sectional curvature has zero derivative. To prove this, we employ the second Bianchi identity by choosing <math>X,Y,Z</math> as linearly independent
			* Gradient estimates for [[scalar curvature]], which are used for studying things like the [[Ricci flow]] on higher-dimensional manifolds

Vipul at 11:06, 30 August 2007

2007-08-30T11:06:32Z

New page

==Statement==

Let <math>\nabla</math> be a [[linear connection]] and <math>R</math> be the [[Riemann curvature tensor]] of <math>\nabla</math>. Then <math>R</math> can itself be differentiated via <math>\nabla</math>, since <math>R</math> is a <math>(1,3)</math>-tensor and we can define the connection on all <math>(p,q)</math>-tensors. With this meaning, the following cyclic summation is zero:

<math>(\nabla_X R)(Y,Z) + (\nabla_Y R) (Z,X) + (\nabla_Z R)(X,Y) = 0</math>

==Proof==

To prove this we look more closely at what <math>\nabla_X R</math> means.

<math>\nabla_X R</math> must satisfy the following compatibility condition:

<math>((\nabla_X R) (Y,Z)) (W) + R (\nabla_XY,Z) W + R(Y,\nabla_XZ) W + R(Y,Z) \nabla_X W = \nabla_X (R(Y,Z)W)</math>

We now concentrate on all the other terms.