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In complex analysis, the winding number (around the origin) of a continuous loop $\gamma: [0,1] \to \mathbb{C} \setminus \{0\}$ is the number of times the loops "winds" around zero, which is given by the integral

$$\frac{1}{2 \pi i}\int_\gamma \frac{dz}{z}$$

One of the basic results of algebraic topology is that loops with the same winding numbers are homotopic.

I think it is pretty clear that any continuous map $f: \mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\}$ should also carry a similar notion of 'winding number'. It should be defined by the integral $\frac{1}{2 \pi i} \int_\gamma \frac{dz}{z}$ (where $ \gamma: [0,1] \to \mathbb{C} \setminus \{0\}$ is given by $\gamma(t) = f(e^{2 \pi it})$).

In this scenario, does the result above still hold, i.e. that continuous maps $\mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\}$ with the same winding number are homotopic (through continuous maps $\mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\})$? How can I see this? Is it possible to use the result above (the equivalent one for loops) to construct this homotopy?

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  • $\begingroup$ I seriously doubt so. Your "winding number for a continuous map" only sees the values of $f$ on the unit circle. $\endgroup$ – Giuseppe Negro Oct 23 '18 at 7:07
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    $\begingroup$ For continuous maps? I very much doubt it. For holomorphic maps? Possibly. $\endgroup$ – Theo Bendit Oct 23 '18 at 7:18
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    $\begingroup$ The answer is yes because $\mathbb{C}\setminus\{0\}$ deformation retracts onto $S^1$, and the property is true for maps $S^1\to S^1$ $\endgroup$ – Max Oct 23 '18 at 9:24
  • $\begingroup$ @Max Why not an official answer? $\endgroup$ – Paul Frost Nov 11 '18 at 17:56
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Suppose $f : \mathbb C \setminus \{ 0 \} \to \mathbb C \setminus \{ 0 \}$ is any continuous map such that the loop $$\gamma : S^1 \to \mathbb C \setminus \{ 0 \}, \ \ \ \gamma (t) = f(e^{2\pi it})$$ has winding number $n$.

I claim that $f$ is homotopic (via maps $\mathbb C \setminus \{ 0 \} \to \mathbb C \setminus \{ 0 \}$) to the map $$ h_n : \mathbb C \setminus \{ 0 \} \to \mathbb C \setminus \{ 0 \} , \ \ \ h_n (re^{2\pi it})=e^{2\pi int}.$$ If I can prove this, then this would be sufficient for me to conclude that any two maps $f_1, f_2 : \mathbb C \setminus \{ 0 \} \to \mathbb C \setminus \{ 0 \}$ such that $t \mapsto f_1(e^{2\pi i t})$ and $t \mapsto f_2 (e^{2\pi i t})$ have the same winding number must be homotopic to each other.

To construct the homotopy between $f$ and $h_n$, I proceed in two steps.

First, I define the map $$ g : \mathbb C \setminus \{ 0 \} \to \mathbb C \setminus \{ 0 \} , \ \ \ g (re^{2\pi it})=f(e^{2\pi it}),$$ and note that $f$ is homotopic to $g$ via the homotopy, $$ F : \mathbb C \setminus \{ 0 \} \times [0,1] \to \mathbb C \setminus \{ 0 \} , \ \ \ F(re^{2\pi it}, s) = f(r^{1-s} e^{2\pi it}).$$

The next step is to use the fact that, since $\gamma (t) = f(e^{2\pi it})$ has winding number $n$, $\gamma$ must be homotopic to the map $$w_n : S^1 \to \mathbb C \setminus \{ 0 \}, \ \ \ w_n (e^{2\pi it}) = e^{2\pi int}.$$

Suppose that $Z : S^1 \times [0, 1] \to \mathbb C \setminus \{ 0 \}$ is a homotopy between $\gamma$ and $w_n$. Then

$$ G : \mathbb C \setminus \{ 0 \} \times [ 0, 1] \to \mathbb C \setminus \{0 \}, \ \ \ G(re^{2\pi it}, s) = Z(e^{2\pi it}, s)$$ is a homotopy between $g$ and $h_n$.

Having shown that $f$ is homotopic to $g$ and $g$ is homotopic to $h_n$, we conclude that $f$ is homotopic to $h_n$, and we're done.

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  • $\begingroup$ Yes you're probably right but that was actually clear from the fact that $\mathbb{C}\setminus\{0\}$ deformation retracts onto $S^1$ $\endgroup$ – Max Oct 23 '18 at 9:23
  • $\begingroup$ @Max Yes, that's how I came up with this - I defined my $F$ by thinking about the deformation retraction of $\mathbb C \setminus \{ 0 \} $ onto $S^1$. $\endgroup$ – Kenny Wong Oct 23 '18 at 10:43
  • $\begingroup$ Also: Kenny's answer is probably actually useful to OP, who is learning the beginnings of complex analysis, and therefore has probably not encountered the notion of deformation retracts, even though to someone with more experience, statements about such things are "very basic results". Perhaps an improved comment might say "Your conjecture about continuous maps is correct, and when/if you study topology, the proof that Kenny gave can be generalized to make similar claims about maps between similar types of spaces. Watch for the words "deformation retract" as you study..." $\endgroup$ – John Hughes Oct 23 '18 at 12:10
  • $\begingroup$ @JohnHughes : yes, I realized that too long after writing my comment unfortunately. I still gave my own answer but of course Kenny's answer is useful and instructive for someone with less topological background. $\endgroup$ – Max Oct 23 '18 at 13:44
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Kenny Wong's answer is instructive because it does it "by hand", but here's a much more condensed version:

A very basic result is that the winding number of $\gamma$ only depends on the path homotopy class of $\gamma$.

Let $r:\mathbb{C}\setminus\{0\} \to S^1, z\mapsto \frac{z}{|z|}$, and $i:S^1\to \mathbb{C}\setminus\{0\}$. It is then a classical result that $r$ is a retraction by deformation onto $S^1$.

Let $f,g:\mathbb{C}\setminus\{0\}\to \mathbb{C}\setminus\{0\}$ be two maps with the same winding number. Then compare $i\circ r\circ f\circ i$ and $i\circ r\circ g\circ i$: they are two maps $S^1\to \mathbb{C}\setminus\{0\}$ and since $i\circ r \simeq id_{\mathbb{C}\setminus\{0\}}$, $i\circ r \circ f\circ i \simeq f\circ i$ and similarly with $g$, thus they have respectively the winding number of $f$ and $g$. Thus they are homotopic (using the result for $S^1$). Thus $f\circ i$ and $g\circ i$ are. Thus $f\circ i \circ r$ and $g\circ i \circ r$ are, and thus $f$ and $g$ are.

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  • $\begingroup$ This is a very nice way of writing it! $\endgroup$ – Kenny Wong Oct 23 '18 at 12:19

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