Schur's theorem - Revision history

Vipul: 5 revisions

2008-05-18T20:08:39Z

5 revisions

← Older revision	Revision as of 20:08, 18 May 2008
(No difference)

Vipul: /* Applying the second Bianchi identity */

2007-08-30T10:47:16Z

Applying the second Bianchi identity

← Older revision		Revision as of 10:47, 30 August 2007
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	===Applying the second Bianchi identity===		===Applying the second Bianchi identity===

	The second Bianchi identity states that the cyclic summation:		The [[second Bianchi identity]] states that the cyclic summation:

	<math>\nabla_XR (Y,Z) + \nabla_YR(Z,X) + \nabla_ZR(X,Y) = 0</math>		<math>\nabla_XR (Y,Z) + \nabla_YR(Z,X) + \nabla_ZR(X,Y) = 0</math>

Vipul: /* Proof */

2007-07-18T11:18:37Z

Proof

← Older revision		Revision as of 11:18, 18 July 2007
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	For this, we convert the given condition on sectional curvature to a condition purely in terms of the [[Riemann curvature tensor]].		For this, we convert the given condition on sectional curvature to a condition purely in terms of the [[Riemann curvature tensor]].

	{{~~fillin}~~}		===Obtaining a scalar multiple relationship===

			We start with the assumption that the sectional curvature is a scalar function, or in other words, that at each point, the sectional curvature is constant for all tangent planes.

			Recall that the [[sectional curvature]] of the vector space with basis <math>X</math> and <math>Y</math> is given by:

			<math>\frac{R(X,Y,X,Y)}{\|\|X \times Y\|\|^2}</math>

			where the denominator is the cross product. Rewriting this in terms of the metric tensor, we get:

			<math>R(X,Y,X,Y) = K (g(X,Y)^2 - g(X,X)g(Y,Y)</math>

			By polarization, we can now show that:

			<math>R(X,Y,Z,W) = K(g(X,Y)g(X,W) - g(X,Z)g(Y,W))</math>

			Let <math>s(X,Y,Z,W) = g(X,Y)g(X,W) - g(X,Z)g(Y,W)</math>. Then we have:

			<math>R = Ks</math>

			where, according to our assumptions, <math>K</math> is a function of the point alone, viz <math>K</math> is a scalar function.

			===Applying the second Bianchi identity===

			The second Bianchi identity states that the cyclic summation:

			<math>\nabla_XR (Y,Z) + \nabla_YR(Z,X) + \nabla_ZR(X,Y) = 0</math>

			Applying it to this situation, we get:

			<math>\sum \nabla_X s(Y,Z) + \sum (XK) s(Y,Z) = 0</math>

			Since <math>\nabla g = 0</math>, the product rule tells us that <math>\nabla s \equiv 0</math>. Hence we get:

			<math>\sum (XK) s(Y,Z) = 0</math>

			===Using the high dimension===

			If the tangent space at <math>p \in M</math> has dimension at least 3, then given any vector <math>X</math>, do the following:

			* Pick vectors <math>Y</math> and <math>Z</math> such that <math>X,Y,Z</math> are linearly independent at <math>p</math>
			* Using the definition of <math>s</math>, observe that the linear independence of <math>X,Y,Z</math> at <math>p</math> implies the linear independence of <math>s(Y,Z)</math>, <math>s(Z,X)</math> and <math>s(X,Y)</math> at <math>p</math>.
			* Thus, conclude that <math>XC = 0</math> at <math>p</math>.

Vipul: /* Proof= */

2007-05-22T19:35:11Z

Proof=

← Older revision		Revision as of 19:35, 22 May 2007
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	* The sectional curvature is constant (that is, the metric is a [[constant-curvature metric]])		* The sectional curvature is constant (that is, the metric is a [[constant-curvature metric]])

	==Proof===		==Proof==

	The idea behind the proof of this result is as follows. The second condition clearly implies the first, so what we essentially need to show is that the first implies the second.		The idea behind the proof of this result is as follows. The second condition clearly implies the first, so what we essentially need to show is that the first implies the second.

Vipul at 19:34, 22 May 2007

2007-05-22T19:34:52Z

← Older revision		Revision as of 19:34, 22 May 2007
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	{{sectional curvature result}}		{{sectional curvature result}}

			{{in all dimensions}}
	==Statement==		==Statement==

Vipul: Started the page

2007-03-08T16:48:25Z

Started the page

New page

{{sectional curvature result}}

==Statement==

Let <math>(M,g)</math> be a Riemannian manifold of dimension at least 3. Then the following are equivalent:

* For every point <math>p \in M</math>, the sectional curvature is constant across all tangent planes at <math>p</math>.
* The sectional curvature is constant (that is, the metric is a [[constant-curvature metric]])

==Proof===

The idea behind the proof of this result is as follows. The second condition clearly implies the first, so what we essentially need to show is that the first implies the second.

For this, we convert the given condition on sectional curvature to a condition purely in terms of the [[Riemann curvature tensor]].

{{fillin}}