Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
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

1 Answer 1

up vote 8 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 $g/f$.

share|improve this answer
1  
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
3  
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
1  
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.