De Rham derivative of a function: Difference between revisions

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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.

Latest revision as of 19:37, 18 May 2008

Definition

Definition with symbols

Let be a differential manifold. The de Rham derivative is an operator that takes as input an infinitely differentiable function on , and outputs a differential 1-form on . It is defined as follows:

In other words, the de Rham derivative of a function sends a vector field to the function .


Facts

Where it lives

The de Rham derivative can be viewed as:

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:

Not tensorial

However, the de Rham derivative is not tensorial. In other words, at a point cannot be determined from .

Leibniz rule

Further information: de Rham derivative satisfies Leibniz rule

The de Rham derivative satisfies the Leibniz rule, namely:

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

Related notions

Exact 1-form

Further information: 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 information: de Rham derivative of a differential 1-form

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