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This question comes from an old complex analysis qual. First denote $\mathbb{C}^{\times} = \mathbb{C} \backslash \{ 0 \}$, $u = \{ e^{it} : 0 \leq t < 2 \pi \}$, and let $f : \mathbb{C}^{\times} \to \mathbb{C}^{\times}$ be some holomorphic function. Then the winding number $$ \frac{1}{2 \pi i} \int_{f(u)} \frac{dz}{z} $$ can take any integer value depending on $f$ (by letting $f(z) = z^n$, for example). However, if we change the domain and codomain of $f$ to something like $\{ z \in \mathbb{C} : \frac{1}{4} < |z| < 4 \}$, then we can't use those $f$'s we used before. The winding number can still be $1$ of course by using the identity. I see that it can also be $0$ by letting $f(z) = 1$. We can also get the winding number to be $-1$ by setting $f(z) = z^{-1}$. Can we get the winding number to be $2$? I suspect not but am having trouble showing it. Thoughts?

  • I don't understand your problem. Why is there a problem with $f(z)=z^n$ on the unit circle in your new domain? What is the exact question on the qualifying exam? Are they trying to get you to understand that the winding number of $f(\Gamma)$ gives the number of roots of $f$ inside $\Gamma$? – Ted Shifrin Apr 24 '14 at 00:26
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    @Ted Using $z^2$, for example, would expand the range from ${\frac14<|z|<4}$ to ${\frac1{16}<|z|<16}$. In his problem, both the domain and range are restricted to ${\frac14<|z|<4}$. – Mario Carneiro Apr 24 '14 at 00:29
  • I don't see how to solve this, yet, but I think the Hadamard three circle theorem will come into play. – Mario Carneiro Apr 24 '14 at 00:30
  • But we're looking only at the winding number of $f$ restricted to the unit circle!! I'm confused. Is the question about winding numbers or is it about invariant domains? – Ted Shifrin Apr 24 '14 at 00:31
  • @Ted We are looking at the winding numbers of functions with invariant domains. – Mario Carneiro Apr 24 '14 at 00:32
  • @Mario: Ah. It would be nice if there had been a clearly stated problem. I certainly hope no university gave such a vague and confusing qualifying exam question! :P I would be shot if I wrote such a question. – Ted Shifrin Apr 24 '14 at 00:33
  • The question is about winding numbers. The first example in the OP was an unbounded domain, while the second is a bounded domain. I suppose that the possibilities for the winding number changes but am not sure how to show it. – user131576 Apr 24 '14 at 00:33
  • @TedShifrin The restrictions placed on $f$ are a bit artificial - I was confused by that at first too. The integral only ever evaluates $f$ on the unit circle, but quite independently of that it is also required that $f$ maps the annulus $\frac{1}{4} < |z| < 4$ into itself. Which prevents $z\mapsto z^2$ from being used, because even though it maps the unit circle to itself, the annuluas isn't invariant under it. – fgp Apr 24 '14 at 00:34
  • @TedShifrin The problem is "can we get a winding number of 2", or any other winding number for that matter. Also, I mentioned that this problem "came from" (meaning was inspired by) a qual, it wasn't a question in itself. Please edit the question if it is that confusing. – user131576 Apr 24 '14 at 00:35
  • @user131576 The annulus (call it $A$) is bounded. For continuous $f\colon A \to A$, the winding number of the image of $u$ is the degree of $f$, $\deg f$. We have $\deg (f\circ g) = \deg f \cdot \deg g$. Now suppose $f\colon A \to A$ is holomorphic. What does the normality of $\mathscr{F} = { f^n : n \in \mathbb{N}}$ tell you about the degree of $f$? – Daniel Fischer Apr 30 '14 at 10:43
  • @DanielFischer What is the degree of $f$ (is $f$ nec. a polynomial?)? The normality of $\mathcal{F}$ says that the degree of $f$ is bounded, which can't happen if the abs. value of the winding number is greater than 1. Is this what you mean? – user131576 Apr 30 '14 at 16:49
  • Oh, sorry, degree as in topology, doesn't need to mean anything to you yet, you can interpret it as just a name for that number, and use the multiplicative property. If/when you learn homology or homotopy, you'll formally meet the degree of a map. Yes, precisely, the degree of the maps in $\mathscr{F}$ must be bounded, and since $\deg (f^n) = (\deg f)^n$, that implies $\lvert \deg f\rvert \leqslant 1$. Can you say why the degree must be bounded on a normal family $\mathscr{F}$? – Daniel Fischer Apr 30 '14 at 17:54
  • @DanielFischer I can't right away (I thought you meant degree as in polynomial). Does it have something to do with the fact that $f(u)$ must intersect itself at least one time if the winding number of $f(u)$ is greater than 1? – user131576 Apr 30 '14 at 21:56
  • Take a sequence $f^{n_k}$ that converges locally uniformly to $g$. What can you say about $\deg g$? – Daniel Fischer Apr 30 '14 at 21:58
  • By the multiplicative property you mentioned earlier, is the degree of $g$ infinite? I'd wager a guess that the degree of a holomorphic map can't be infinite, but I can't explain why. – user131576 Apr 30 '14 at 22:23

1 Answers1

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This proof uses some basic facts about Extremal length quoted from the wikipedia article.

The extremal length $EL(\Gamma)$ of a collection of curves $\Gamma$ within some domain $D$ is defined to be

$$EL(\Gamma)=\sup_{\rho\in [0,\infty]^D}\frac{\left(\inf_{\gamma\in\Gamma}\int_\gamma\rho\,|dz|\right)^2}{\int_D\rho^2\,dx\,dy},$$

and the most important property of this quantity is that it is invariant under conformal maps, which is to say, if $\Gamma^*=\{f\circ\gamma:\gamma\in\Gamma\}$ for some bijective holomorphic function $f$, then $EL(\Gamma^*)=EL(\Gamma)$.

The wikipedia article has already gone to the trouble to calculate $$EL(\Gamma)=\frac{2\pi}{\log(r_2/r_1)}$$ for the collection $\Gamma$ of curves that wind once around the annulus of radii $r_1,r_2$, and $EL(\Gamma)=w/h$ for the collection of curves in a rectangle $w\times h$ that travel from one edge to the opposite edge (traveling a distance $w$).

Being an infimum over $\Gamma$, it is clear that $\Gamma\subseteq\Gamma'$ implies $EL(\Gamma)\ge EL(\Gamma')$. So now let us suppose that there is some holomorphic $f$ whose domain is the annulus $r_1,r_2$ and with codomain in this same annulus, and with a winding number $N>1$. We can "unwind" this function to $g=\log\circ f$, where $g$ analytically continues $\log$ so that the result is a bijective holomorphic function with codomain $(\log r_1,\log r_2)\times [0,2\pi N]$.

Every path $\gamma\in\Gamma_1$ is homotopic to $e^{it}:t\in[0,2\pi]$, and we know that the image of this path winds $N$ times around the annulus, so $g\circ\gamma\in\Gamma''$, the set of paths that got from $(\log r_1,\log r_2)\times\{0\}$ to $(\log r_1,\log r_2)\times\{2\pi N\}$, while staying within $(\log r_1,\log r_2)\times (-\infty,\infty)$, because of the codomain restriction on $f$.

Furthermore, for each curve $\gamma''\in\Gamma''$, some subset $\gamma'$ of it (with smaller path length) will be strictly contained in $(\log r_1,\log r_2)\times [0,2\pi N]$. Thus $\gamma'\in\Gamma'$, where $\Gamma'$ is the set of curves that cross from the bottom to the top of $(\log r_1,\log r_2)\times [0,2\pi N]$. so $EL(\Gamma'')\ge EL(\Gamma')$. This implies, from the above calculations:

$$EL(\Gamma_1)=EL(g\circ\Gamma)\ge EL(\Gamma'')\ge EL(\Gamma')\implies \frac{2\pi}{\log(r_2/r_1)}\ge \frac{2\pi N}{\log r_2-\log r_1},$$

which is a contradiction. (Also, given a holomorphic function $f$ with winding number $N<-1$, the composition with $z^{-1}$ satisfies the same conditions as above and so this is also impossible.)

  • The extremal length is conserved by conformal maps, but a holomorphic map $f\colon A \to A$ (where $A$ is the annulus) for which $f\circ u$ has a winding number $\neq \pm 1$ cannot be conformal. Thus you don't have $EL(\Gamma_1) = EL(f\circ\Gamma_1)$ for such a map. – Daniel Fischer Apr 28 '14 at 15:42
  • @Daniel Ah, you're right. I missed the part about $f$ needing to be bijective for $EL(\Gamma)=EL(f\circ\Gamma)$ to hold. I think I can patch the construction by mapping to a rectangle instead of an annulus, though. – Mario Carneiro Apr 28 '14 at 16:55
  • @Daniel I've updated the answer. – Mario Carneiro Apr 28 '14 at 17:35