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If one localizes the ring of $\mathcal{C}^\infty$ functions on $\mathbb{R}$ at the maximal ideal of those vanishing at the origin, one obtains the ring of germs of $\mathcal{C}^{\infty}$ functions. Does the same thing work if one replaces $\mathcal{C}^{\infty}$ by "analytic" ? That is, if one localizes the the ring of power series with infinite radius at the maximal ideal of those vanishing at the origin, does one obtain the ring of power series with positive radius? I think one obtains just a proper subset but any confirmation from you guys would help me sleep better.

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    Follow-up question. Preliminaries. Suppose that we have a function f which is analytic in some open, simply connected neighborhood, U, of C. Suppose there are two paths, P and P', which are simply connected, though piecewise so. For p on P, p' on P', we have f(p) = f(p') $T$hen we can define a function X(z,z') = f(z) - f(z') and extend it to U [Q?]. X(z,z') = 0 then implicitly defines a function $g$, where g(p) = p' is analytic on U. If f - f(g) = h, f(p) - f(g(p)) = 0 on P implies h is zero where it is defined. Conclusion: f and f(g) are in the same equivalence class as germs. $T$rue?2012-09-27

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Unfortunately no. Roughly speaking, the reason is that analytic functions are very rigid: if two analytic functions are defined on some connected domain and they agree in some neighborhood of a point, then they are equal everywhere. Put differently, each germ is determined by at most one function, and as such, we will not distinguish between analytic functions and their germs. This is in sharp contrast to smooth functions, where thanks to bump functions and partitions of unity, if a smooth function is defined on $\overline{U}$ for some open set $U$, then $f$ can be extended to all of $\mathbb{R}^n$, even when $U$ is not connected.

Moreover, the local behavior of an analytic function at a point is nice: if $f(z)$ is analytic, we can locally (near $0$) write it as $z^m g(z)$, where $g(0)\neq 0$ and $g$ is analytic. If $f$ and $g$ are analytic with $g$ nonvanishing (on some open set), then $f/g$ is also analytic. Combining these two observations, we see that the quotient of two entire functions is at worst meromorphic, requiring no branch points or branch cuts to define.

Thus, the localization of the analytic functions at $0$ is a subset of the restriction of "entire" meromorphic functions.

What sorts of things must we be missing?

  • Functions which are multivalued, like $\log (1+z)$ or $\sqrt{1+z}$
  • The restrictions of functions which contain singularities other than poles. As Jiangwei mentioned, we will not have the restriction of $e^{1/(z-1)}$, as this has an essential singularity at $1$
  • Possibly more (every meromorphic function is locally the quotient of two analytic functions, but is this true globally?)

This is why it is necessary to use sheafs when dealing with Riemann surfaces (or more generally, when dealing with varieties). The local functions are not at all determined by global functions, and so we really do need to store the local information as additional data. For an extreme example, there are no non-constant globally defined functions in projective space.


Edit I just realized that you asked your question for functions on $\mathbb{R}$, not $\mathbb{C}$. However, because real analytic functions extend to complex analytic functions in a neighborhood of the real line, essentially everything written above still holds. In particular, the examples still work.

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    Every meromorphic function $f$ is a ratio of entire functions. Let $z_1$, $z_2$, ... be the sequence of poles of $f$ listed with multiplicity, so $|z_n| \to \infty$. By the Weierstrass theorem http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem#Existence_of_entire_function_with_specified_zeroes there is an entire function $g$ with zeroes at the $z_i$ (of matching multiplicity). So $h:=fg$ is also entire, and $f=h/g$.2012-09-27