Complex Analytic Bijection: Is it a Local Diffeo?

[Bacle]

Hi, Everyone:

Say f(z) defined on a region R , is a complex-analytic bijection. Does it follow

that f:R--->f(R) is a diffeomorphism, i.e., is f^-1 also analytic?

I know this is not true for the real-analytic case, e.g., f(x)=x³ , but complex-

analytic is stronger than real-analytic ; I think this may be some corollary of

the inverse function theorem.

Any Ideas?

 

[micromass]

Yes, you are correct. First, I need to tell you that the word diffeomorphic is not standard in complex analysis. More often, they talk about conformal or biholomorphic functions.

It is indeed true that any analytic bijection is conformal. Here's a nice proof:
Take f an analytic bijection. By the open mapping theorem, it follows immediately that f^-1 is continuous.
By the implicit function theorem (of real analysis), it follows that f^-1 is analytic outside the set of all f(z) where f'(z)=0. This is a discrete set (=has no accumulation points). The function f^-1 is bounded near the points of this discrete set, so by the Riemann removability condition, it follows that f^-1 has an analytic continuation. Thus f^-1 is analytic.

 

[Bacle]

Thanks, Micromass:

I am curious about this: does the implicit function theorem guarantee the

definability of f as a function wherever f'(z) not zero? Or, alternatively, that

f^-1 is defined wherever the Jacobian Jf has non-zero determinant?

Otherwise, I am not too clear on how/where the implicit function theorem applies.

Can we just conclude from f(f^-1(z))=z (since f bijective) , that

f'(f^-1)(z)*f'^-1(z)=1 (since f is analytic in R)

 

[micromass]

The implicit function theorem I'm referring to is:

Let f:D\rightarrow\mathbb{C} be analytic with D open. Assume f is injective and f' does not vanish on D. Then

• The range f(D) is open,
• The inverse function f^{-1}:f(D)\rightarrow \mathbb{C} is analytic,
• The derivative is (f^{-1})^\prime(f(z))=\frac{1}{f^\prime(z)}

The book where I'm getting this from is complex analysis by Freitag and Busam. I actually think that they should have called this the inverse function theorem instead of the implicit function theorem...

 

[lavinia]

Hi, Everyone:

Say f(z) defined on a region R , is a complex-analytic bijection. Does it follow

that f:R--->f(R) is a diffeomorphism, i.e., is f^-1 also analytic?

I know this is not true for the real-analytic case, e.g., f(x)=x³ , but complex-

analytic is stronger than real-analytic ; I think this may be some corollary of

the inverse function theorem.

Any Ideas?

One way to answer this question is to prove that one can always choose coordinates around any point so that the holomorphic function(non-constant) has the form z^m, for some positive integer, m. While this requires using Taylor series, it reveals the structure of the function. Generally,diffeomorphisms can have many forms. But if the function it is analytic its form is restricted.

 

[Bacle]

Your right, lavinia:

If f'(z)=0 , then we can find an expression f(z)-f(zo)= z^kg(z) k>1

and an open set U where g(z) is non-zero. Then zⁿ is not 1-1.

I was having some technical difficulties about showing the differentiability of

f_-1 , but I think using the inverse of the Jacobian Jf evaluated at

some point ( using the fact that Det(Jf)=|f'(z)|^2 , which is only zero for f'(z)=0),

follows that Jf is invertible and (Jf)^-1 gives us the differentiable inverse of f,

but I am looking into the technical details-- I have a thing for detail-and-example-oriented

classical, get-dirty-with -chalk Mathematics (sorry for the rant ).

Anyway, I'll try to clean up above argument and then post it.

P.S: Sorry, the Tex is not compiling; don't know what I could be doing wrong.

 

[Bacle]

Sorry to bother you all , but I think this too will work:

If f is analytic and 1-1 on a region R, we can show that f'(z)=/0 on R (e.g., assume f'(z)=0 by contradiction, then f has a local rep f(z)=z^k.g(z) for g(z) non-zero, and z^k is k-to-1, and it follows that f is k-to-1).

Then the Jacobian matrixJ(f) is anti-symmetric, with non-zero determinant (since f'(z)=/0, and DetJ(f)=u_x^2+ u_y^2 is non-zero, so J(f) is invertible, and it is easy to show that its inverse has the "right form" , i.e., it is also antisymmetric, which shows that
f^-1 satisfies Cauchy-Riemann.

BTW: Def of antisymmetric : my dad's sister; her name is symmetric.