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2008-05-18T19:37:30Z

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Vipul at 21:51, 11 April 2008

2008-04-11T21:51:13Z

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	==Definition==		==Definition==

	Let <math>M</math> be a [[differential manifold]]. The '''de Rham derivative''' is an operator that takes as input an [[infinitely differentiable function]] on <math>N</math>, and outputs a differential 1-form.		===Definition with symbols===

			Let <math>M</math> be a [[differential manifold]]. The '''de Rham derivative''' is an operator that takes as input an [[infinitely differentiable function]] on <math>M</math>, and outputs a [[differential 1-form]] on <math>M</math>. It is defined as follows:

			<math>df := X \mapsto Xf</math>

			In other words, the de Rham derivative of a function <math>f</math> sends a vector field <math>X</math> to the function <math>Xf</math>.


			==Facts==

			===Where it lives===
			The de Rham derivative can be viewed as:

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

			i.e. it is a map from the algebra of infinitely differentiable functions, to the space of differential 1-forms.

			It can also be viewed as an element:
			<math>d \in C^\infty(M)^* \otimes \Gamma(T^*M)</math>

			===Not tensorial===

			However, the de Rham derivative is ''not'' [[tensorial map\|tensorial]]. In other words, <math>df</math> at a point <math>p \in M</math> cannot be determined from <math>f(p)</math>.

			===Leibniz rule===

			{{further\|[[de Rham derivative satisfies Leibniz rule]]}}

			The de Rham derivative satisfies the Leibniz rule, namely:

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

			This is a direct consequence of the [[Leibniz rule for derivations]].

			==Related notions==

			===Exact 1-form===

			{{further\|[[Exact 1-form]]}}

			An exact 1-form is a [[differential 1-form]], or a section of the [[cotangent bundle]], that can be expressed as the de Rham derivative of a function.

			===de Rham derivative of a differential 1-form===

			{{further\|[[de Rham derivative of a differential 1-form]]}}

			There is a closely related notion called the de Rham derivative of a differential 1-form.

Vipul: New page: ==Definition== Let $M$ be a differential manifold. The '''de Rham derivative''' is an operator that takes as input an infinitely differentiable function on $N$
2008-04-11T20:29:14Z

New page: ==Definition== Let <math>M</math> be a differential manifold. The '''de Rham derivative''' is an operator that takes as input an infinitely differentiable function on <math>N</mat...

New page
==Definition==

Let <math>M</math> be a [[differential manifold]]. The '''de Rham derivative''' is an operator that takes as input an [[infinitely differentiable function]] on <math>N</math>, and outputs a differential 1-form.

De Rham derivative of a function - Revision history

Vipul: 2 revisions

Vipul at 21:51, 11 April 2008