Proposition

A rotation of $S^1$ is a map $\rho: S^1 \rightarrow S^1$ of the form $\rho(z) = e^{i\theta}z$ for some fixed $e^{i\theta} \in S^1$. Show that if $\rho$ is a rotation, then $N(\rho \circ f) = N(f)$ for every loop $f$ in $S^1$.

Solution

Let $\rho(z) = e^{i\theta_0}z$ be a rotation in the format specified in the Proposition. Let $f$ be a loop in $S^1$. Let $\tilde{f}$ be any lift of $f$.

Then $N(f) = \tilde{f}(1) - \tilde{f}(0)$.

Let $g: S^1 \rightarrow S^1$ be defined such that

\[\begin{align*} g(e^{i\theta}) = \frac{\theta_0}{2\pi} + \tilde{f}(e^{i\theta}). \end{align*}\]

Then

\[\begin{align*} \epsilon(g(e^{i\theta})) &= \epsilon(\frac{\theta_0}{2\pi} + \tilde{f}(e^{i\theta})) \\ &= \exp(2\pi i(\frac{\theta_0}{2\pi} + \tilde{f}(e^{i\theta}))) \\ &= \exp(\theta_0 i + 2\pi_i\tilde{f}(e^{i\theta})) \\ &= \exp(\theta_0 i)\exp(2\pi_i\tilde{f}(e^{i\theta})) \\ &= \exp(\theta_0 i)\epsilon(\tilde{f}(e^{i\theta})) \\ &= \rho(\epsilon(\tilde{f}(e^{i\theta}))) \\ &= \rho(f(e^{i\theta})) \\ &= (\rho \circ f)(e^{i\theta}). \end{align*}\]

Therefore, $g$ is a lift of $\rho \circ f$. Then

\[\begin{align*} N(\rho \circ f) &= g(1) - g(0) \\ &= (\frac{\theta_0}{2\pi} + \tilde{f}(1)) - (\frac{\theta_0}{2\pi} + \tilde{f}(0)) \\ &= \tilde{f}(1) - \tilde{f}(0) \\ &= N(f). \end{align*}\]

Therefore, $N(\rho \circ f) = N(f)$ for every loop in $S^1$.