Curvature is antisymmetric in last two variables - Revision history

Vipul at 01:52, 24 July 2009

2009-07-24T01:52:11Z

← Older revision		Revision as of 01:52, 24 July 2009
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	<math>\! R(X,Y,Z,W) = -R(X,Y,W,Z)</math>.		<math>\! R(X,Y,Z,W) = -R(X,Y,W,Z)</math>.

			==Related facts==

			* [[Curvature is tensorial]]
			* [[Curvature is antisymmetric in first two variables]]
			* [[Curvature is symmetric in the pairs of first and last two variables]]
	==Proof==		==Proof==

Vipul: /* Proof */

2009-07-24T01:51:24Z

Proof

← Older revision		Revision as of 01:51, 24 July 2009
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	And:		And:

	<math>YXg(Z,W) = Yg(\nabla_X(Z),W) + Yg(Z,nabla_X(W)) = g(\nabla_Y \circ \nabla_X(Z),W) + g(\nabla_X(Z),\nabla_Y(W)) + g(\nabla_Y(Z),\nabla_X(W)) + g(Z,\nabla_Y \circ \nabla_X(W)) \qquad (2)</math>.		<math>YXg(Z,W) = Yg(\nabla_X(Z),W) + Yg(Z,\nabla_X(W)) = g(\nabla_Y \circ \nabla_X(Z),W) + g(\nabla_X(Z),\nabla_Y(W)) + g(\nabla_Y(Z),\nabla_X(W)) + g(Z,\nabla_Y \circ \nabla_X(W)) \qquad (2)</math>.

	Substituting (1) and (2) in <math>(\dagger\dagger)</math> yields <math>(\dagger)</math>.		Substituting (1) and (2) in <math>(\dagger\dagger)</math> yields <math>(\dagger)</math>.

Vipul: /* Proof */

2009-07-24T01:50:47Z

Proof

← Older revision		Revision as of 01:50, 24 July 2009
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	<math>g(\nabla_{[X,Y]}(Z),W) + g(\nabla_{[X,Y]}(W),Z) = [X,Y]g(Z,W) = XYg(Z,W) - YXg(Z,W) \qquad (\dagger\dagger)</math>.		<math>g(\nabla_{[X,Y]}(Z),W) + g(\nabla_{[X,Y]}(W),Z) = [X,Y]g(Z,W) = XYg(Z,W) - YXg(Z,W) \qquad (\dagger\dagger)</math>.

	Simplifying each of the two terms on the right side of <math>~~\tag{~~\dagger\dagger}</math>, we get:		Simplifying each of the two terms on the right side of <math>(\dagger\dagger)</math>, we get:

	<math>XYg(Z,W) = Xg(\nabla_Y(Z),W) + Xg(Z,\nabla_Y(W)) = g(\nabla_X \circ \nabla_Y(Z),W) + g(\nabla_Y(Z),\nabla_X(W)) + g(Z,\nabla_X \circ \nabla_Y(W)) + g(\nabla_X(Z),\nabla_Y(W)) \qquad (1)</math>.		<math>XYg(Z,W) = Xg(\nabla_Y(Z),W) + Xg(Z,\nabla_Y(W)) = g(\nabla_X \circ \nabla_Y(Z),W) + g(\nabla_Y(Z),\nabla_X(W)) + g(Z,\nabla_X \circ \nabla_Y(W)) + g(\nabla_X(Z),\nabla_Y(W)) \qquad (1)</math>.

Vipul at 01:50, 24 July 2009

2009-07-24T01:50:25Z

← Older revision		Revision as of 01:50, 24 July 2009
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	Then:		Then:

	<math>\! R(X,Y,Z,W) = R(X,Y,W,Z)</math>.		<math>\! R(X,Y,Z,W) = -R(X,Y,W,Z)</math>.

	==~~Facts used~~==		==Proof==

			We consider the expression <math>R(X,Y,Z,W) + R(X,Y,W,Z)</math>:

			<math>g(\nabla_X \circ \nabla_Y(Z) - \nabla_Y \circ \nabla_X(Z) - \nabla_{[X,Y]}(Z),W) - g(\nabla_X \circ \nabla_Y(W) - \nabla_Y \circ \nabla_X(W) - \nabla_{[X,Y]}(W),Z)</math>

			By the bilinearity of <math>g</math>, this simplifies to:

	# [[~~uses::First Bianchi identity~~]~~]: This states that if <math>R~~</math> ~~is a [[torsion-free linear connection]], then:~~		<math>g(\nabla_X \circ \nabla_Y(Z),W) - g(\nabla_Y\circ \nabla_X(Z),W) - g(\nabla_{[X,Y]}(Z),W) + g(\nabla_X \circ \nabla_Y(W),Z) - g(\nabla_Y\circ \nabla_X(W),Z) - g(\nabla_{[X,Y]}(W),Z)</math>

	~~<math>R(X~~,~~Y)Z + R(Y,Z)X + R(Z,X)Y = 0</math>.~~		To prove that this is zero, it thus suffices to show that:

	==~~Proof~~==		<math>g(\nabla_{[X,Y]}(Z),W) + g(\nabla_{[X,Y]}(W),Z) = g(\nabla_X \circ \nabla_Y(Z),W) + g(\nabla_X \circ \nabla_Y(W),Z) - g(\nabla_Y \circ \nabla_X(Z),W) - g(\nabla_Y \circ \nabla_X(W), Z) \qquad (\dagger)</math>.

			We now show <math>\dagger</math>. Since <math>g</math> is a [[metric connection]], the left side simplifies to:

			<math>g(\nabla_{[X,Y]}(Z),W) + g(\nabla_{[X,Y]}(W),Z) = [X,Y]g(Z,W) = XYg(Z,W) - YXg(Z,W) \qquad (\dagger\dagger)</math>.

			Simplifying each of the two terms on the right side of <math>\tag{\dagger\dagger}</math>, we get:

			<math>XYg(Z,W) = Xg(\nabla_Y(Z),W) + Xg(Z,\nabla_Y(W)) = g(\nabla_X \circ \nabla_Y(Z),W) + g(\nabla_Y(Z),\nabla_X(W)) + g(Z,\nabla_X \circ \nabla_Y(W)) + g(\nabla_X(Z),\nabla_Y(W)) \qquad (1)</math>.

			And:

			<math>YXg(Z,W) = Yg(\nabla_X(Z),W) + Yg(Z,nabla_X(W)) = g(\nabla_Y \circ \nabla_X(Z),W) + g(\nabla_X(Z),\nabla_Y(W)) + g(\nabla_Y(Z),\nabla_X(W)) + g(Z,\nabla_Y \circ \nabla_X(W)) \qquad (2)</math>.

	~~{{fillin}}~~		Substituting (1) and (2) in <math>(\dagger\dagger)</math> yields <math>(\dagger)</math>.

Vipul: Created page with '==Statement== Suppose $M$ is a differential manifold and $g$ is a fact about::Riemannian metric or fact about::pseudo-Riemannian metric and
2009-07-24T01:28:59Z

Created page with '==Statement== Suppose <math>M</math> is a differential manifold and <math>g</math> is a fact about::Riemannian metric or fact about::pseudo-Riemannian metric and <ma…'

New page
==Statement==

Suppose <math>M</math> is a [[differential manifold]] and <math>g</math> is a [[fact about::Riemannian metric]] or [[fact about::pseudo-Riemannian metric]] and <math>\nabla</math> is the [[fact about::Levi-Civita connection]] for <math>g</math>. Consider the [[fact about::Riemann curvature tensor]] <math>R</math> of <math>\nabla</math>. In other words, <math>R</math> is the [[Riemann curvature tensor of Levi-Civita connection|Riemann curvature tensor of the Levi-Civita connection]] for <math>g</math>. We can treat <math>R</math> as a <math>(0,4)</math>-tensor:

<math>\! R(X,Y,Z,W) = g(R(X,Y)Z,W)</math>.

Then:

<math>\! R(X,Y,Z,W) = R(X,Y,W,Z)</math>.

==Facts used==

# [[uses::First Bianchi identity]]: This states that if <math>R</math> is a [[torsion-free linear connection]], then:

<math>R(X,Y)Z + R(Y,Z)X + R(Z,X)Y = 0</math>.

==Proof==

{{fillin}}