# Analytic implicit function theorem

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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What is the context? Holomorphic functions/maps between: open subsets of complex affine space, or complex manifolds, or complex analytic spaces... –  Qfwfq Aug 22 '11 at 11:16
(I didn't downvote, btw) –  Qfwfq Aug 22 '11 at 11:41
The question seems fine to me. The problem is local, it makes no difference whether one is working with complex manifolds or affine spaces. –  Donu Arapura Aug 22 '11 at 11:47
@Donu: sure. On the other hand I don't know how things work when there are singularities... –  Qfwfq Aug 22 '11 at 12:17
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. –  Deane Yang Aug 22 '11 at 15:16

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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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 $126 to read two pages once... – Pierre Aug 22 '11 at 10:19 If you wish, I can send you per mail a numerized copy of the book. – Henri Aug 22 '11 at 10:27 add comment In one variable, this is a trivial consequence of the standard local inversion theorem. Indeed, holomorphic functions are$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$is a local biholomorphism. The statement of IFT is a direct consequence of the local inversion theorem then. - This is also true in higher dimention, at least for manifolds (nonsingular analytic spaces). The$\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})$, also its inverse does. – Qfwfq Aug 22 '11 at 11:19 If I don't misremember the analytic implicit function thm should be a relatively 'formal' consequence of the analytic inverse function thm. – Qfwfq Aug 22 '11 at 11:24 @unknowngoogle: You're perfectly right for the general$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! –  Henri Aug 22 '11 at 12:20

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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A bit terse, I agree. And I'm not familiar with all the notations. But I keep it, in case ! –  Pierre Aug 22 '11 at 12:24