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Is $C^\infty_p$ a finite dimensional vector space or infinite dimensional, for any point $p$ in a manifold $M$?

The notation means the space of smooth germs.

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Sushil
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  • what's $C$? And what "infinite p dimensional" means? – user40276 Oct 16 '15 at 11:01
  • C^(infinity) means we take all C infinity function in some nbd around p and define equivalence relation: two function related if they match on some nbd of p. And then it is collection of equivalence classes – Sushil Oct 16 '15 at 11:12
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    it's infinite dimensional as an algebra over the reals. Just pick Taylor series expansion around $p$ – user40276 Oct 16 '15 at 11:26
  • @user40276 taylor series expansion of what? – Sushil Oct 16 '15 at 11:28
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    Ok. Let me try to clarify myself. There's a surjective morphism from smooth germs to power series, see for instance http://math.stackexchange.com/questions/63050/every-power-series-is-the-taylor-series-of-some-c-infty-function . – user40276 Oct 16 '15 at 11:34
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    Furthermore for germs it does not matter if $M$ is a manifold or the euclidean space, so Borel theorem still holds. – user40276 Oct 16 '15 at 11:35

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For sake of easiness suppose that $M=\mathbb{R}$ and $p=0.$

Let $n$ be a natural number. We want to show that $n<\dim C^\infty_p.$

So consider functions $1,x,x^2,\dots,x^n$ and take their germs $[1],[x],[x^2],\dots,[x^n].$ We will prove that $[1],[x],[x^2],\dots,[x^n]$ are lineary independent.

Suppose that $a_0[1]+a_1[x]+a_2[x^2]+\dots+a_n[x^n]=[0].$ This means that $[a_0+a_1x+a_2x^2+\dots+a_nx^n]=[0].$ And by definition of germ, there exists some open set $U=(-\varepsilon,\varepsilon)$ such that $a_0+a_1x+a_2x^2+\dots+a_nx^n=0$ on $U.$ This means that $$\forall(x\in (-\varepsilon,\varepsilon))\hspace{10pt} a_0+a_1x+a_2x^2+\dots+a_nx^n=0.$$ Lets denote the left hand side by $f:(-\varepsilon,\varepsilon)\rightarrow\mathbb{R}.$ We see that $f(0)=0.$ Hence $a_0=0.$ We see that $f'(0)=0.$ Hence $a_1=0.$ Generally $f^{(k)}(0)=0=a_k$ holds for $k\leqslant n.$ And it means exactly that $[1],[x],[x^2],\dots,[x^n]$ are lineary independent. Hence $n<\dim C^\infty_p.$

Since $n$ was arbitrary, we get that $\dim C^\infty_p=\infty.$

On arbitrary manifold $M$ consider just functions, which around point $p$ takes form $1, \phi_1,\phi_1^2,\dots,\phi_1^n$ where $\phi:M\supset V\rightarrow\phi(V)\subset\mathbb{R}^n=(\phi_1,\dots,\phi_n)$ is some map around $p$ such that $\phi(p)=\theta.$

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