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If $\gamma : S^1 \rightarrow S^1$, $\gamma(t) = e^{2 \pi i t}$ and $\gamma_+$ denotes $\gamma$ restricted to $[0, 0.5]$ i.e. the upper half of $S^1$ and $f:S^1 \rightarrow S^1$ is an odd map, how do I see that the angle swept by $f \circ \gamma_+$ is $\pi + 2 \pi$?

Thanks for your help!

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Do you mean the angle is $\pi$ modulo $2\pi$ ? – PseudoNeo Jul 5 '11 at 11:27
Besides, it is quite confusing to use alternatively two different constructions of $S^1$ : $\mathbb R/\mathbb Z$ and $\{ z\in\mathbb C^* : |z|=1\}$. If you take this second construction from the very beginning, I think your question becomes clearer. – PseudoNeo Jul 5 '11 at 11:29
I am meaning to use $S^1 = [0,1]/\{0, 1\}$! – Rudy the Reindeer Jul 5 '11 at 11:38
My goal is to use this to show that an odd map $f:S^1 \rightarrow S^1$ has odd degree... – Rudy the Reindeer Jul 5 '11 at 11:43
up vote 1 down vote accepted

I consider the map $z \mapsto - z$ from $S^1$ to itself ($S^1$ is hereafter identified to the unit circle of $\mathbb C$). An odd map is a map $\gamma : S^1 \to S^1$ obeying the law $\gamma(- z) = -\gamma(z)$. The restriction $\gamma_+$ of $\gamma$ to the top half-circle is a map $C_+ \to S^1$. Its endpoints are $\gamma(1)$ and $\gamma(-1) = -\gamma(1)$. So this path goes from a point to its antipodal. As such, it makes an odd number of half-turns and it sweeps an angle of $\pi$ modulo $2\pi$.

As any map $C_+ \to S^1$ sending $1$ and $-1$ to antipodal points can be extended to an odd map $S^1 \to S^1$, this result cannot be sharpened: for any angle $\theta$ equal to an odd number of half-turns, there's an odd map $\gamma$ whose $\gamma_+$ sweeps an angle equal to $\theta$.

Please note that in the other canonical model of $S^1$ (the quotient $\mathbb R / \mathbb Z$) you have to replace $z \mapsto -z$ by the transformation $t \mapsto t + .5$ to define an odd map. The transformation $t \mapsto -t$ of $\mathbb R/\mathbb Z$ corresponds to $z \mapsto \overline z$, and it is not true that if $\gamma : S^1 \to S^1$ obeys $\gamma(\overline z) = \overline{\gamma(z)}$, its restriction $\gamma_+$ sweeps an odd number of half-turns (the constant map $\gamma(z) = 1$ is a counterexample).

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thank you! And why does it follow that $f \circ \gamma_+$ also sweeps an angle of $\pi + 2 \pi n$? Could it not be that it sweeps an angle $\Theta + 2 \pi n$? – Rudy the Reindeer Jul 5 '11 at 12:10
oh noes, that's not possible because as you pointed out in your answer: the endpoints are antipodal. Maybe it's time for me to take a coffee break now : ( – Rudy the Reindeer Jul 5 '11 at 12:16

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