Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

The following is probably well-known, but I couldn't find anything in the literature. Any reference would be nice.

Let $\Omega$ be a domain in the complex plane, and let $f$ be holomorphic and one-to-one in $\Omega$. Then is it true that $f'$ has a holomorphic logarithm in $\Omega$, i.e. there exists a function $g$ holomorphic in $\Omega$ such that $f'=e^g$ ?

EDIT This is false in general, as seen by David Cohen's answer. However, I would be very interested to know under which conditions is the above true?

Thank you,

Best regards, Malik

share|improve this question
2  
It seems to me that a minor modification of David Cohen's answer provides counterexamples whenever $\Omega$ is not simply connected. Of course, whenever $\Omega$ is simply connected then every non-vanishing function admits a logarithm. –  Andreas Blass May 31 '13 at 19:17
1  
I agree. If $\Omega$ is not simply connected then $\mathbb C\setminus\Omega$ has at least one bounded component, say $K$. If $a$ is any point of $K$, then $z-a$ has no holomorphic logarithm on $\Omega$, and it follows that $f(z)=(z-a)^2$ is a counterexample. –  Etienne May 31 '13 at 21:31
    
@Etienne Matheron : In general, $f(z)=(z-a)^2$ is not necessarily one-to-one on $\Omega$. You probably mean that $f(z)=1/(z-a)$ is a counterexample. –  Malik Younsi May 31 '13 at 23:47
    
@Andreas Blass : Yes indeed, but I meant under which conditions on $\Omega$ and $f$ does the above holds. In particular, I think it is true if $f$ does not reverse orientation of curves inside $\Omega$. –  Malik Younsi May 31 '13 at 23:48
    
@Malik. Yes, $f(z)=\frac1{z-a}$! –  Etienne Jun 1 '13 at 4:12

2 Answers 2

up vote 5 down vote accepted

I believe that your suggestion is correct: $\newcommand{\C}{\mathbb{C}}$


THEOREM. Suppose that $U,V\subset\mathbb{C}$ are domains, and that $f:U\to V$ is a conformal isomorphism. Then $f'$ has a (single-valued) logarithm if and only if $f$ "maps the outer boundary of $U$ to the outer boundary of $V$". (I.e., if $\gamma\subset U$ is a simply closed curve parameterized in positive orientation, then $f\circ\gamma$ also has positive orientation.)


A special case is, of course, that the logarithm always exists when $U$ is simply-connected.

To prove the Theorem, note that the existence of the logarithm of $f'$ means precisely that, on any essential curve in $U$, the derivative does not wind around zero.

Claim. If $\gamma\subset\C$ is a simple closed curve, and $f$ is holomorphic and injective in a neighbourhood of $\gamma$, then the winding number of $f'\circ\gamma$ around zero is either $0$ or $-2$, according to whether $f\circ\gamma$ has the same or the opposite orientation to $\gamma$.

This proves the theorem.


The claim should be intuitively plausible. To prove it, consider the case where $\gamma$ and $f(\gamma)$ are both the unit circle. Then clearly, for $z\in\gamma$, $$ \arg f'(z) = \arg f(z) - \arg(z) $$ if $f$ is orientation-preserving, while $$ \arg f'(z) = -\arg f(z) - \arg(z)$$ if $f$ is orientation-reversing.

To reduce the general case to this observation, e.g. assume w.l.o.g. that the curve $\gamma$ is analytic, and change coordinates using the Riemann mapping theorem.

(I think there should be a direct analytic proof that does not require this change of variable, but I find the above argument rather intuitive.)

share|improve this answer
    
Yes, this is the proof I had in mind, thank you. However, I was wondering : do you have any idea where I could look to find a reference for this result? It is quite natural, so I am pretty sure it appears somewhere in the literature, but I couldn't find anything. –  Malik Younsi Jun 5 '13 at 14:57
    
@Malik: For the reference, try MR1343250 Fulton, William Algebraic topology. A first course. Graduate Texts in Mathematics, 153. Springer-Verlag, New York, 1995. xviii+430 pp. ISBN: 0-387-94326-9; 0-387-94327-7. Although I am not sure if you can find this precise argument there, Chapter 3 treats winding numbers. There are applications given in Chapter 4, and Chapter 6 has a section on winding numbers and homology. –  Margaret Friedland Jun 5 '13 at 15:27
    
@Margaret Friedland : Thank you for the reference. It seems like a good book, but I didn't find what I'm looking for. –  Malik Younsi Jun 6 '13 at 12:35

If $f(z)=\frac{1}{z}$, then $f$ is holomorphic (and one-to-one) in $\mathbb{C}\setminus 0$. But $f^{\prime}(z)=\frac{-1}{z^{2}}$ does not have a logarithm. (If $e^{g}=\frac{-1}{z^{2}}$, then $e^{(i\pi-g)/2}=z$, i.e., $(i\pi-g)/2$ would be a logarithm for $z$, which is impossible.)

share|improve this answer
    
We have $e^g=-1/z^2$, and thus $e^{-g}=-z^2$. How do you obtain $e^{-2g}=z$? But I believe this gives a counter-example because -z^2 does not have a holomorphic logarithm on $\mathbb{C} \setminus \{0\}$. Clearly I overlooked something this morning... Thank you –  Malik Younsi May 31 '13 at 18:01

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.