# Smooth function

## Contents

## 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

- ↑ Introduction to smooth manifolds - John M Lee - Second Edition