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> | ===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.