Difference between revisions of "Smooth function"

From Maths
Jump to: navigation, search
(Created page with "==Definition== A '''smooth function''' on a manifold}}, {{M|(M,\mathcal{A})}}, is a function<ref>Introduction to smooth manifolds - John M Lee -...")
 
m (→‎See also)
 
(3 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
==Definition==
 
==Definition==
−
A '''smooth function''' on a [[Smooth manifold|smooth {{n|manifold}}]], {{M|(M,\mathcal{A})}}, is a function<ref>Introduction to smooth manifolds - John M Lee - Second Edition</ref> {{M|f:M\rightarrow\mathbb{R}^k}} that satisfies:
+
A '''smooth function''' on a [[Smooth manifold|smooth {{n|manifold}}]], {{M|(M,\mathcal{A})}}, is a function<ref>Introduction to smooth manifolds - John M Lee - Second Edition</ref> {{M|f:M\rightarrow\mathbb{R} }} that satisfies:
  
−
{{M|\forall p\in M\ \exists\ (U,\varphi)\in\mathcal{A} }} such that {{M|f\circ\varphi^{-1}\subseteq\mathbb{R}^n\rightarrow\mathbb{R}^k }} is [[Smooth|smooth]] in the usual sense, of having continuous partial derivatives of all orders.  
+
*{{M|\forall p\in M\ \exists\ (U,\varphi)\in\mathcal{A}[p\in U\wedge f\circ\varphi^{-1}:\varphi(U)\subseteq\mathbb{R}^n\rightarrow\mathbb{R}\in C^\infty]}}
 +
** That is to say {{M|f\circ\varphi^{-1} }} is [[Smooth|smooth]] in the usual sense - of having continuous partial derivatives of all orders.  
  
−
Any smoothly compatible map (so all in the atlas of the smooth manifold) will have a smooth transition function, by composition, the result will be smooth, so {{M|f}} is still smooth.
+
{{Begin Theorem}}
 +
Theorem: Any other chart in {{M|(M,\mathcal{A})}} will also satisfy the definition of {{M|f}} being smooth
 +
{{Begin Proof}}
 +
Let {{M|(M,\mathcal{A})}} be a given [[Smooth manifold|smooth manifold]]
 +
 
 +
Let {{M|(U,\varphi)}} be a chart, on which {{M|f\circ\varphi^{-1}:\varphi(U)\rightarrow\mathbb{R} }} is [[Smooth|smooth]]
 +
 
 +
We wish to show any [[Smoothly compatible charts|smoothly compatible chart]] with {{M|(U,\varphi)}} will support the definition of {{M|f}} being smooth.
 +
 
 +
That is to say all other charts in the [[Smooth atlas|smooth atlas]] {{M|\mathcal{A} }}
 +
 
 +
'''Proof:'''
 +
: Let {{M|(V,\psi)}} be any chart in {{M|\mathcal{A} }} be given.
 +
:: Then {{M|(U,\varphi)}} and {{M|(V,\psi)}} are smoothly compatible
 +
:: this means that either:
 +
::* {{M|U\cap V}} is empty - in which case there is nothing to show OR
 +
::* {{M|\varphi\circ\psi^{-1}:\psi(U\cap V)\rightarrow\varphi(U\cap V)}} is a [[Diffeomorphism|diffeomorphism]]
 +
::: We can compose this with {{M|f\circ\varphi^{-1}:\varphi(U)\rightarrow\mathbb{R} }} as follows
 +
:::: {{M|(f\circ\varphi^{-1})\circ(\varphi\circ\psi^{-1}):\psi(U\cap V)\rightarrow\mathbb{R} }} which is [[Smooth|smooth]] as it is a composition of smooth functions
 +
:::: {{M|1==f\circ\varphi^{-1}\circ\varphi\circ\psi^{-1}:\psi(U\cap V)\rightarrow\mathbb{R} }}
 +
:::: {{M|1==f\circ\psi^{-1}:\psi(U\cap V)\rightarrow\mathbb{R} }} - which we know to be smooth as it is ''equal to'' {{M|(f\circ\varphi^{-1})\circ(\varphi\circ\psi^{-1}):\psi(U\cap V)\rightarrow\mathbb{R} }} - which as we've said is smooth
 +
 
 +
QED
 +
{{End Proof}}
 +
{{End Theorem}}
 +
 
 +
===Extending to vectors===
 +
 
 +
Note that given an {{M|f:M\rightarrow\mathbb{R}^k}} this is actually just a set of functions, {{M|f_1,\cdots,f_k}} where {{M|f_i:M\rightarrow\mathbb{R} }} and {{M|1=f(p)=(f_1(p),\cdots,f_k(p))}}
 +
 
 +
 
 +
We can define {{M|f:M\rightarrow\mathbb{R}^k}} as being smooth {{M|1=\iff\forall i=1,\cdots k}} we have {{M|f_i:M\rightarrow\mathbb{R} }} being smooth
 +
 
 +
==Notations==
 +
===The set of all smooth functions===
 +
Without knowledge of [[Smooth manifold|smooth manifolds]] we may already define {{M|C^\infty(\mathbb{R}^n)}} - the set of all functions with continuous partial derivatives of all orders.
 +
 
 +
However with this definition of a smooth function we may go further:
 +
===The set of all smooth functions on a manifold===
 +
Given a [[Smooth manifold|smooth {{n|manifold}}]], {{M|M}}, we now know what it means for a function to be smooth on it, so:
 +
 
 +
Let <math>f\in C^\infty(M)\iff f:M\rightarrow\mathbb{R}</math> is smooth
  
 
==See also==
 
==See also==
 +
* [[Smooth map]]
 
* [[Smooth manifold]]
 
* [[Smooth manifold]]
 
* [[Smoothly compatible charts]]
 
* [[Smoothly compatible charts]]

Latest revision as of 16:16, 14 April 2015

Definition

A smooth function on a smooth [ilmath]n[/ilmath]-manifold, [ilmath](M,\mathcal{A})[/ilmath], is a function[1] [ilmath]f:M\rightarrow\mathbb{R} [/ilmath] that satisfies:

  • [ilmath]\forall p\in M\ \exists\ (U,\varphi)\in\mathcal{A}[p\in U\wedge f\circ\varphi^{-1}:\varphi(U)\subseteq\mathbb{R}^n\rightarrow\mathbb{R}\in C^\infty][/ilmath]
    • That is to say [ilmath]f\circ\varphi^{-1} [/ilmath] is smooth in the usual sense - of having continuous partial derivatives of all orders.

Theorem: Any other chart in [ilmath](M,\mathcal{A})[/ilmath] will also satisfy the definition of [ilmath]f[/ilmath] being smooth

Let [ilmath](M,\mathcal{A})[/ilmath] be a given smooth manifold

Let [ilmath](U,\varphi)[/ilmath] be a chart, on which [ilmath]f\circ\varphi^{-1}:\varphi(U)\rightarrow\mathbb{R} [/ilmath] is smooth

We wish to show any smoothly compatible chart with [ilmath](U,\varphi)[/ilmath] will support the definition of [ilmath]f[/ilmath] being smooth.

That is to say all other charts in the smooth atlas [ilmath]\mathcal{A} [/ilmath]

Proof:

Let [ilmath](V,\psi)[/ilmath] be any chart in [ilmath]\mathcal{A} [/ilmath] be given.
Then [ilmath](U,\varphi)[/ilmath] and [ilmath](V,\psi)[/ilmath] are smoothly compatible
this means that either:
  • [ilmath]U\cap V[/ilmath] is empty - in which case there is nothing to show OR
  • [ilmath]\varphi\circ\psi^{-1}:\psi(U\cap V)\rightarrow\varphi(U\cap V)[/ilmath] is a diffeomorphism
We can compose this with [ilmath]f\circ\varphi^{-1}:\varphi(U)\rightarrow\mathbb{R} [/ilmath] as follows
[ilmath](f\circ\varphi^{-1})\circ(\varphi\circ\psi^{-1}):\psi(U\cap V)\rightarrow\mathbb{R} [/ilmath] which is smooth as it is a composition of smooth functions
[ilmath]=f\circ\varphi^{-1}\circ\varphi\circ\psi^{-1}:\psi(U\cap V)\rightarrow\mathbb{R}[/ilmath]
[ilmath]=f\circ\psi^{-1}:\psi(U\cap V)\rightarrow\mathbb{R}[/ilmath] - which we know to be smooth as it is equal to [ilmath](f\circ\varphi^{-1})\circ(\varphi\circ\psi^{-1}):\psi(U\cap V)\rightarrow\mathbb{R} [/ilmath] - which as we've said is smooth

QED


Extending to vectors

Note that given an [ilmath]f:M\rightarrow\mathbb{R}^k[/ilmath] this is actually just a set of functions, [ilmath]f_1,\cdots,f_k[/ilmath] where [ilmath]f_i:M\rightarrow\mathbb{R} [/ilmath] and [ilmath]f(p)=(f_1(p),\cdots,f_k(p))[/ilmath]


We can define [ilmath]f:M\rightarrow\mathbb{R}^k[/ilmath] as being smooth [ilmath]\iff\forall i=1,\cdots k[/ilmath] we have [ilmath]f_i:M\rightarrow\mathbb{R} [/ilmath] being smooth

Notations

The set of all smooth functions

Without knowledge of smooth manifolds we may already define [ilmath]C^\infty(\mathbb{R}^n)[/ilmath] - the set of all functions with continuous partial derivatives of all orders.

However with this definition of a smooth function we may go further:

The set of all smooth functions on a manifold

Given a smooth [ilmath]n[/ilmath]-manifold, [ilmath]M[/ilmath], we now know what it means for a function to be smooth on it, so:

Let [math]f\in C^\infty(M)\iff f:M\rightarrow\mathbb{R}[/math] is smooth

See also

References

  1. ↑ Introduction to smooth manifolds - John M Lee - Second Edition
Retrieved from "https://www.maths.kisogo.com/index.php?title=Smooth_function&oldid=611"