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Define $C^\infty (M)_x := \{ (U,f) | x \in U $ open $ , f \in C^\infty (U) \} / \sim $ where $M$ is a manifold and $(U,f) \sim (V,g)$ if $\exists W$ open, $x \in W$ such that $W \subset V \cap U$ and $f|_W = g|_W$.

This is a ring with the following operations: $[(U,f)] + [(V,g)] := [(U \cap V, f + g)]$ and $[(U,f)]\cdot [(V,g)]:= [(U \cap V, fg)]$.

I'm trying to understand what multiplicative inverses look like, i.e. if I have $[(U, f)]$ then I want to show that there exists an open set $V$ containing $x$ such that $\frac{1}{f}$ is smooth on $V$.

Is this right? And can someone explain to me how I can show this? Many thanks for your help.

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Hint: a multiplicative inverse of $[(U,f)]$ exists iff $f$ does not vanish at $x$. – Sebastian Oct 5 '11 at 8:21
@Sebastian: why is that? $f$ and $\frac{1}{f}$ have to be smooth so in particular continuous. Wouldn't that mean that $f$ has to be non-zero on entire $U$ for $(U,f)$? – Rudy the Reindeer Oct 6 '11 at 7:55
The argument roughly goes like this: Let $[(U,f)]$ s.t. $f$ does not vanish at $x$, then by continuity, there is an open neighbourhood $V$ of $x$, contained in $U$ where this doesn't happen. By definition, we have $[(U,f)] = [(V,f)]$. Now, $[(V,1/f)]$ is an element of our ring, as $V$ is an open neighbourhood of $x$ and $1/f$ is smooth. We clearly have that $[(V,f)] \cdot [(V,1/f)] = [(V,1)] = 1$. – Sebastian Oct 6 '11 at 9:59
Why 1/f is smooth? @Sebastian – Sushil Oct 17 '15 at 9:56
up vote 9 down vote accepted

There are two rings involved in your question:

1) The ring you described, which should actually be denoted $\mathcal C^\infty_{M,x}$
2) The ring $\mathcal C^\infty(M)_{\frak m}$, described as follows. Call $\frak m$ the ideal in $\mathcal C^\infty(M)$ consisting of global functions zero at $x$, that is ${\frak m}=\lbrace f\in \mathcal C^\infty(M): f(x)=0\rbrace$. This is a maximal ideal in $\mathcal C^\infty(M)$, and you can localize $\mathcal C^\infty(M)$ at this ideal to get $\mathcal C^\infty(M)_{\frak m}$. The elements of this localized ring are formal fractions $f/g$ with $f,g\in C^\infty(M)$ and $g(x)\neq 0$ . Beware that these formal fractions are by no stretch of the imagination interpretable as functions on $M$ , since $g$ might very well vanish outside, say, the interval $(x-1,x+1)$ !

There is a canonical ring morphism $\mathcal C^\infty(M)_{\frak m} \to \mathcal C^\infty_{M,x}$ and the strange, perhaps underappreciated, result is that it is an isomorphism.

To answer your question:
1) In the first incarnation an invertible element is a germ $[(U,h)]$ with $h(x)\neq o$ and its inverse is $[(U',1/h)]$, where $U'\subset U$ is a neighbourhood of $x$ on which $h$ has no zero.
2) In the second incarnation an invertible element of $\mathcal C^\infty(M)_{\frak m}$ is a fraction of the form $f/g$ with $f(x)\neq 0$ (in addition to $g(x)\neq 0$, of course), and its inverse is $g/f$.

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Dear Georges, I don't understand what you are hinting at in point 1) because I've never seen the notation $C^{\infty}_{M,x}$. Could you be so kind as to add some elucidation for the uninformed? Best wishes, Theo – t.b. Oct 5 '11 at 12:17
Dear Theo, the notation $\mathcal O_{X,x}$ is standard to denote the local ring at $x\in X$ of the locally ringed space $(X, \mathcal O_X)$. It is systematically used in algebraic geometry and in complex holomorphic geometry . If the locally ringed space is the manifold $(M, \mathcal C^\infty_M)$, then its local ring at $x$ will logically be denoted $\mathcal C^\infty _{M,x}$. However, since a mathematician of your calibre asks the question, it is proof that this point of view is not widespread in differential topology/geometry.(To be continued) . – Georges Elencwajg Oct 5 '11 at 13:00
(Continued) The reason being probably that sheaf theory is only moderately used in this domain of mathematics. Anyway thanks a lot for alerting me to that obscurity in my answer, of which I was completely unaware: a case of professional deformation, or more bluntly of provincialism! – Georges Elencwajg Oct 5 '11 at 13:04
Dear @Pierre-Yves, thanks a lot for alerting t.b.. It was very kind of you to make it your business and I'm very grateful for this friendly action. – Georges Elencwajg Oct 5 '11 at 14:07
Dear @Matt, just take $U'=\lbrace y\in U:h(y)\neq0\rbrace$. This will be an open subset of $U$ (since $h$ is continuous) and will contain $x$, since $h(x)\neq 0$ by hypothesis. – Georges Elencwajg Oct 6 '11 at 9:29

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