19
$\begingroup$

There is a standard theorem that says if $f$ is analytic in the whole complex plane and never zero, then it has the form $e^g$ where $g$ is analytic in the whole plane; i.e. we can define a function $\log f$ that is analytic in the whole plane and satisfies $f=e^{\log f}$. The argument is easy: $f$ is entire and never zero, so $f'/f$ is entire, so $g(z)=\int_0^z f'(\zeta)d\zeta/f(\zeta)$ is well-defined and entire, and then computation shows that $fe^{-g}$ has derivative zero, so is constant, so $f$ is a constant multiple of $e^g$, and the constant can be absorbed into $g$.

The argument does not mention the topology of the situation, and this bugs me a little. By the Little Picard theorem, the image of $f$ is the whole complex plane minus zero. In general, of course, a continuous single-valued branch of $\log$ cannot be defined on this set. That's why the proof involved the integral of the logarithmic derivative of $f$, rather than merely the composition of $f$ with the right branch of $\log$. That said, if we obtained a different function $h$ by picking some branch of $\log$ defined on $\mathbb{C}$ minus some ray from the origin (call this set $S$), and then defining $h(z)=\log f(z)$ on $f^{-1}(S)$, then of course $h$ would satisfy $f=e^h$ where it is defined, and then $1=e^{h-g}$, so that on each connected component of $f^{-1}(S)$, since $h,g$ are defined and continuous they must differ by a constant multiple of $2\pi i$. If I allow myself to use a different branch of $\log$ for each connected component of $f^{-1}(S)$, I could end up with $h$ exactly equaling $g$ and therefore being extendable by continuity to all of $\mathbb{C}$. What this adds up to is that I have the strong feeling that the theorem (in spite of the easy calculation-based proof) has some nontrivial topological content and I am trying to isolate this content.

At this point, I was originally going to ask for help isolating the topological content. But over the course of composing the question, I think I may have answered it. So, following is what I believe to be an alternative proof of the original result that makes the topological content explicit. My question is now, is the following proof correct? I would also be interested in your thoughts about (soft-question) where is the topology hiding in the original proof?

Theorem: If $f$ is entire and never zero it has the form $f=e^g$ for $g$ entire.

Proof: Let $f$ be an entire function that is never zero. Then $f:\mathbb{C}\rightarrow\mathbb{C}\setminus\{0\}$ is continuous, so induces a homomorphism $f_*:\pi_1(\mathbb{C})\rightarrow\pi_1(\mathbb{C}\setminus\{0\})$, which is trivial because $\mathbb{C}$ is simply connected. Therefore, $f$ lifts to a continuous function $\hat{f}$ from $\mathbb{C}$ to $\mathbb{C}\setminus{0}$'s universal cover $U$. But $U$ is the Riemann surface of the $\log$ function; in particular, the logarithm is well-defined and single-valued on $U$; so let $g=\log \hat{f}$; and then $g$ is such that $f=e^g$. $\hat{f}$ is actually holomorphic to $U$, so the $g$ thus defined is entire.

EDIT: When I wrote this proof I was a little uncomfortable with speaking of $\hat{f}$ being holomorphic to $U$ and the log being holomorphic on $U$, since my understanding of covering space theory is strictly topological. Subsequently I realized that we can actually take $U$ to be $\mathbb{C}$, with the covering map given by $z\mapsto e^z$, and then the "logarithm on $U$" is just the identity on $\mathbb{C}$. I think the argument goes through, but can be made even shorter:

By the lifting theorem, any function $f$ from a simply connected domain to $\mathbb{C}\setminus\{0\}$ has a lift $\hat{f}$ to $\mathbb{C}\setminus{0}$'s universal cover, $\mathbb{C}$, that factors through the covering map $z\mapsto e^z$. Then $f=e^{\hat{f}}$ and $\hat{f}$ is the desired $g$. Since the covering map is holomorphic, $f$ and $\hat{f}$ are simultaneously holomorphic. Done!

Thanks to Jonas Meyer for highlighting that the key feature of the original argument is also the simple-connectedness of the domain, on which rests the invocation of Cauchy's theorem to show the integral is well-defined.

$\endgroup$
0

1 Answer 1

14
$\begingroup$

Where is the topology hiding in the original proof?

It is hiding in the phrase "is well-defined," or, equivalently, in Cauchy's theorem. In a domain that is not simply connected, this "definition" of $g$ in general depends on the path taken from your base point to $z$. (For example, if there are isolated singularities, then different paths could wind around the singularities differently and pick up residues.) The well-definedness depends on Cauchy's theorem for simply connected domains.

$\endgroup$
3
  • $\begingroup$ Ah. Your point is that as soon as $f$ is analytic and nonvanishing (forget entire) in a simply-connected domain $D$, it can be written as $e^g$ on $D$, and the original argument goes through to prove this? (I.e. that really the content of "$f$ is entire" was "f is analytic in a simply connected domain," and the scope of the theorem applies to whatever this domain is?) $\endgroup$ Dec 6, 2011 at 23:03
  • $\begingroup$ Yes. And this characterizes simply connected domains in $\mathbb C$. The direction we're discussing follows from Cauchy's theorem for simply connected domains. Conversely, if $G$ is a connected open subset of $\mathbb C$ such that for every nonvanishing analytic $f:G\to\mathbb C$ there exists an analytic $g:G\to\mathbb C$ such that $f=e^g$, then $G$ is simply connected. This is made clear in Conway's Functions of one complex variable, where the only property about a simply connected domain used to prove the Riemann Mapping Theorem is exstnce of analytic sqrts of nonvnshng analytic fnctns. $\endgroup$ Dec 6, 2011 at 23:21
  • $\begingroup$ Good comment. Excellent. $\endgroup$
    – mick
    Jul 13, 2013 at 23:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .