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Well, some days ago I've asked here how do we describe functions on manifolds. My idea was that it could be done using the coordinate functions of a chart: if $(x,U)$ is a chart for a manifold $M$ then we can define one function $f : U \to \Bbb R$ as a combination of the $x^i$ functions. Now I have one doubt that seems very silly (the answer is probably obvious, and I'm failing to see it).

Now here comes my doubt: let $C^{\infty}(U\subset M,\Bbb R)$ be the set of all smooth functions defined in the subset $U$ of a manifold $M$ of dimension $n$. I've defined a $k$-combination to be a map

$$c:\prod_{i=1}^k C^{\infty}(U,\Bbb R) \to C^{\infty}(U,\Bbb R)$$

so for instance, for $k=2$ the map $c(f,g)=\lambda f + \sin \circ g$ would be a $2$-combination of $f$ and $g$. Now, let $k = n$, then trivially by the definition we have that:

$$c(x^1,\dots,x^n)\in C^\infty(U,\Bbb R)$$

My question is: do we have that for any $f \in C^\infty(U,\Bbb R)$ there exists a unique $n$-combination of the functions $x^i$ such that $f = c(x^1,\dots, x^n)$? In other words, do we have that any function defined on $U\subset M$ is a suitable combination of the coordinate functions?

Thanks very much in advance!

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Gold
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  • The question is not really clear to me. Can you precise what is a k-combination? Up to now, a function is a 1-combination of itself, no?

    Maybe you should try to answer this question for $M={\mathbb R}^n$ first.

     – Taladris Jul 14 '13 at 01:04
  • Yes, a function is a 1-combination of itself. The question is wether every function can be written as an $n$-combination of the coordinate functions of the manifold. In $\Bbb R^n$ this is trivially true because the natural coordinate functions are the components of the identity. – Gold Jul 14 '13 at 01:19
  • @user1620696 and in a manifold is is also locally true since the identity function can always be locally recovered from a coordinate map. Does this answer your question? – James S. Cook Jul 15 '13 at 12:41
  • @JamesS.Cook now I understood why you composed $f$ with the identity using $f \circ x^{-1} \circ x$: that way we can use the same reasoning we do in $\Bbb R^n$ and write $f$ in terms of the $x^i$ functions. Thanks very much for your help! – Gold Jul 15 '13 at 13:38

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I believe the answer to your question is yes. Suppose $f: U \subseteq M \rightarrow \mathbb{R}$ where $M$ is a manifold. Let $p \in U$ then, by the definition of a manifold, there exists (at least one) $(V,x)$ a coordinate chart with $V \subseteq U$. If $V$ is too large we can construct a new chart by intersection and shrink the domain as needed. Note $$ f|_V = f \circ x^{-1} \circ x $$ which means the formula you desire exists locally at $p$. Now, I may not be able to write a formula for $f$ in terms of coordinates on all of $U$ since it is conceivable that $U$ needs to be covered by several charts.

But, perhaps, the real question you are asking, is how can we define a function in terms of something besides a coordinate chart on a manifold. The answer there is usually given in terms of the explicit structure of the set as a point set. For example, $x+y+z=1$ gives a plane hence $f(x,y,z) = 2x+2y+z$ is some function on the plane not (yet) given in coordinate chart on the plane( which I have not stated). However, it is a simple exercise to choose parameters for the plane as in so doing construct charts which then could be used to formulate $f$. I know this is possible by the general argument I give at the outset of this post.

James S. Cook
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