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I'm looking for a proof of the analytic implicit function theorem (IFT). The only related proof I could find was the holomorphic inverse function theorem (by Henri Cartan). On Wikipedia, the analytic IFT is mentioned casually in the general article "Implicit function theorem", saying that "Similarly, if f is analytic inside U×V, then the same holds true for the explicit function g inside U. This generalization is called the analytic implicit function theorem." Mmmh, that's fast...

A sketch of the proof may be the following :

  • use analytic continuation to transform f into a holomorphic function
  • use the holomorphic inverse function theorem (Cartan) to prove a holomorphic IFT
  • restriction : g is holomorphic on $\mathbb{C}$, therefore analytic on $\mathbb{R}$.

But it seems weird and I don't think it would work (I have no idea whether a so-called holomorphic IFT exists or not). What would be an efficient proof of the theorem ? Thanks a lot by advance.

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  • $\begingroup$ What is the context? Holomorphic functions/maps between: open subsets of complex affine space, or complex manifolds, or complex analytic spaces... $\endgroup$
    – Qfwfq
    Commented Aug 22, 2011 at 11:16
  • $\begingroup$ (I didn't downvote, btw) $\endgroup$
    – Qfwfq
    Commented Aug 22, 2011 at 11:41
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    $\begingroup$ The question seems fine to me. The problem is local, it makes no difference whether one is working with complex manifolds or affine spaces. $\endgroup$
    – Donu Arapura
    Commented Aug 22, 2011 at 11:47
  • $\begingroup$ @Donu: sure. On the other hand I don't know how things work when there are singularities... $\endgroup$
    – Qfwfq
    Commented Aug 22, 2011 at 12:17
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    $\begingroup$ It seems to me that either the real or complex case can be proved in a rather straightforward fashion by finding a recursive formula for the coefficients of the power series and then proving that the series has a positive radius of convergence. $\endgroup$
    – Deane Yang
    Commented Aug 22, 2011 at 15:16

4 Answers 4

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One possible reference is "Holomorphic functions of several variables: an introduction to the fundamental theory" by Ludger Kaup and Burchard Kaup (section 8 of chapter 0).

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  • $\begingroup$ It seems a very good reference. Unfortunately, pages 27 and 28, in which the theorem is proved, are not available on Google Books (at least in my country, France). And I can't spend $ 126 to read two pages once... $\endgroup$
    – Pierre
    Commented Aug 22, 2011 at 10:19
  • $\begingroup$ If you wish, I can send you per mail a numerized copy of the book. $\endgroup$
    – Henri
    Commented Aug 22, 2011 at 10:27
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In one variable, this is a trivial consequence of the standard local inversion theorem. Indeed, holomorphic functions are $C^1$ functions characterized by the fact that their differential is a similitude. And this property is stable by taking the inverse.

to be more precise, if $g$ is holomorphic on some open set $U\subset \mathbb C$, and its differential (as a function $U \to \mathbb R^2$) satisfies that it is invertible everywhere, with differential being a similitude. So the functions is a local diffeomorphism, and the differential of the inverse is the inverse of the differential, so is still a similitude. Therefore, $g$ is a local biholomorphism.

The statement of IFT is a direct consequence of the local inversion theorem then.

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    $\begingroup$ This is also true in higher dimention, at least for manifolds (nonsingular analytic spaces). The $\mathcal{C}^1$ inverse function theorem + the fact that, loosely, if the differential belongs to $\mathrm{GL}(n,\mathbb{C})\subset\mathrm{GL}(2n,\mathbb{R})$, also its inverse does. $\endgroup$
    – Qfwfq
    Commented Aug 22, 2011 at 11:19
  • $\begingroup$ If I don't misremember the analytic implicit function thm should be a relatively 'formal' consequence of the analytic inverse function thm. $\endgroup$
    – Qfwfq
    Commented Aug 22, 2011 at 11:24
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    $\begingroup$ @unknowngoogle: You're perfectly right for the general $n$-dimensional case. As for the second comment, I agree too, this is more or less the meaning of my last sentence in the answer. So, thanks! $\endgroup$
    – Henri
    Commented Aug 22, 2011 at 12:20
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There's a proof of both the analytic inverse function thm and the analytic implicit function thm (where the second is rather "formally" deduced from the first) in the following book:

Fritzsche, Grauert, From Holomorphic Functions to Complex Manifolds

Chapter 7 "Holomorphic maps" (in which both theorems are proved) is freely available online.

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  • $\begingroup$ A bit terse, I agree. And I'm not familiar with all the notations. But I keep it, in case ! $\endgroup$
    – Pierre
    Commented Aug 22, 2011 at 12:24
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The implicit function theorem is deduced from the inverse function theorem in most standard texts, such as Spivak's "Calculus on Manifolds", and Guillemin and Pollack's "Differential Topology". Basically you just add coordinate functions until the hypotheses of the inverse function theorem hold.

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    $\begingroup$ You can ensure that the inverse function is analytic because its derivative is analytic, using the analyticity of matrix inversion and the explicit equation $(f^{-1})'=(f')^{-1}$. $\endgroup$
    – Ben McKay
    Commented Jul 6, 2017 at 14:23

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