De Rham derivative satisfies Leibniz rule - Revision history

Vipul: 3 revisions

2008-05-18T19:37:35Z

3 revisions

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Vipul at 22:13, 11 April 2008

2008-04-11T22:13:04Z

← Older revision		Revision as of 22:13, 11 April 2008
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	==Statement==		==Statement==

	Let <math>M</math> be a [[differential manifold]]. Consider the [[de Rham derivative ~~for functions~~]] on <math>M</math>, defined as a map:		Let <math>M</math> be a [[differential manifold]]. Consider the [[de Rham derivative of a function\|de Rham derivative]] on <math>M</math>, defined as a map:

	<math>d: C^\infty(M) \to \Gamma(T^*M)</math>		<math>d: C^\infty(M) \to \Gamma(T^*M)</math>

Vipul at 22:12, 11 April 2008

2008-04-11T22:12:40Z

← Older revision		Revision as of 22:12, 11 April 2008
Line 1:		Line 1:
	==Statement==		==Statement==

	Let <math>M</math> be a [[differential manifold]]. Consider the [[de Rham derivative]] on <math>M</math>, defined as a map:		Let <math>M</math> be a [[differential manifold]]. Consider the [[de Rham derivative for functions]] on <math>M</math>, defined as a map:

	<math>d: C^\infty(M) \to \Gamma(T^*M)</math>		<math>d: C^\infty(M) \to \Gamma(T^*M)</math>

Vipul: New page: ==Statement== Let $M$ be a differential manifold. Consider the de Rham derivative on $M$ , defined as a map: $d: C^\infty(M) \to \Gamma(T^*M)$ Th...

2008-04-11T22:12:21Z

New page: ==Statement== Let <math>M</math> be a differential manifold. Consider the de Rham derivative on <math>M</math>, defined as a map: <math>d: C^\infty(M) \to \Gamma(T^*M)</math> Th...

New page

==Statement==

Let <math>M</math> be a [[differential manifold]]. Consider the [[de Rham derivative]] on <math>M</math>, defined as a map:

<math>d: C^\infty(M) \to \Gamma(T^*M)</math>

Then <math>d</math> satisfies the Leibniz rule:

<math>d(fg) = g(df) + f(dg)</math>

In other words, <math>d</math> is a [[derivation to a module|derivation]] from the algebra <math>C^\infty(M)</math> to the module <math>\Gamma(T^*M)</math>.

==Proof==

To verify equality of the two sides, we need to verify that for any [[vector field]] <math>X</math>, the two sides evaluate to the same thing for <math>X</math>. The left side, evaluated on <math>X</math>, yields:

<math>X(fg)</math>

The right side, evaluated on <math>X</math>, yields:

<math>g(Xf) + f(Xg)</math>

The equality of these two sides follows from the Leibniz rule for derivations.

De Rham derivative satisfies Leibniz rule - Revision history

Vipul: 3 revisions

Vipul at 22:13, 11 April 2008

Vipul at 22:12, 11 April 2008

Vipul: New page: ==Statement== Let M be a differential manifold. Consider the de Rham derivative on M, defined as a map: d: C^\infty(M) \to \Gamma(T^*M) Th...

Vipul: New page: ==Statement== Let $M$ be a differential manifold. Consider the de Rham derivative on $M$ , defined as a map: $d: C^\infty(M) \to \Gamma(T^*M)$ Th...