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Now, I am working through a particular case in the book on smooth manifolds by John.M.Lee used in my graduate math class, let's say we have a smooth manifold X which has positive dimension. He then claims that the vector space $C^\infty$ of all smooth functions from $X \rightarrow \mathbb{R} $ will have infinite dimension over $\mathbb{R}$. While I do have a fair idea of the claim, I'm a little bit lost to show that the vector space has infinite dimension.

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3 Answers 3

This follows from the existence of bump functions. Take an open set $U$ in your manifold with a chart $\phi:U\to \mathbb{R}^n$. Let $k$ be a positive integer, and let $x_1,x_2,\ldots,x_k$ be distinct points in $\phi(U)$. For each $j$ from $1$ to $k$, let $f_j:\mathbb{R}^n\to\mathbb{R}$ be a smooth function whose support is a compact subset of $\phi(U)$ such that $f_j(x_j)=1$ and $f_j(x_m)=0$ if $m\neq j$. Define $g_j:X\to\mathbb{R}$ by $g_j(x)=0$ if $x$ is in $X\setminus U$, and $g_j(x)=f_j(\phi(x))$ if $x$ is in $U$. Then $\{g_1,\ldots,g_k\}$ is a linearly independent set of smooth functions on $X$. Since $k$ was arbitrary, the space of smooth functions is infinite dimensional.

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I originally posted the following as a comment but I think it is a particularly simple argument, so worth upgrading to a answer.

Choose any $\theta\in C^\infty(M)$ which is not constant on a connected component of $M$. Then its image is an infinite subset of $\mathbb{R}$ (it includes a nontrivial interval). As the space $\mathbb{R}[X]$ of real polynomials is infinite dimensional, and a polynomial is fully determined by its values on an infinite set, you can conclude that $$ \left\{f\circ\theta\colon f\in\mathbb{R}[X]\right\} $$ is an infinite dimensional subspace of $C^\infty(M)$.

Note that this argument is applicable much more generally. E.g., if a Riemann surface has at least one non-constant meromorphic function then the space of meromorphic functions is infinite dimensional.

If you like, you can modify this method to show that the dimension of $C^\infty(M)$ is (at least) the cardinality of the continuum, by proving the same is true for smooth functions on any nontrivial interval in $\mathbb{R}$. (Hint: the exponentials $x\mapsto e^{ax}$ are linearly independent).

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You still need to justify the existence of non-locally constant smooth functions on $M$. –  Pete L. Clark Mar 14 '11 at 1:12
    
Yes. This is only conditional. You still have show the existence of one such function (and the only way I can think of doing that is by using a bump function in a coordinate chart). But if you can construct one, then you automatically get an infinite dimensional space of such functions. –  George Lowther Mar 14 '11 at 19:40

I'm not sure of the details for smooth manifolds, but certainly the polynomials have infinite dimension and are contained in $C^\infty$.

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You need to know that $X$ smoothly embeds into $\mathbb{R}^n$ to claim that "the polynomials" has any meaning, and this question is easier than that. –  Qiaochu Yuan Mar 13 '11 at 23:16
    
You can use this fact to prove the desired result. Define a polynomial locally in a chart and extend it to the entire manifold by multiplying with a bump function. Then you'll get an infinite dimensional subspace of $C^\infty$ in this way. –  Eric O. Korman Mar 13 '11 at 23:18
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If $\theta\in C^\infty(M)$ is not constant, then $\{f\circ\theta\colon f\in\mathbb{R}[X]\}\subseteq C^\infty(M)$ has infinite dimension. –  George Lowther Mar 13 '11 at 23:20

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