Exercise 3.1.12 (Homomorphism $r \mapsto e^{2\pi i r}$)

Question

Let $G$ be the additive group of real numbers, let $H$ be the multiplicative group of complex numbers of absolute value $1$ (the unit circle $S^1$ in the complex plane) and let $\varphi : G 	o H$ be the homomorphism $\varphi = r \mapsto e^{2\pi i r}$. Draw the points on a real line which lie in the kernel of $\varphi$. Describe similarly the elements in the fibers of $\varphi$ above the points $-1,i$, and $e^{4\pi i/3}$ of $H$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

\begin{proof} We recall that for all real numbers $x \in \R$,
$$e^{ix} = 1 \iff \exists k \in \Z,\ x = 2k \pi.$$ 
Equivalently, for all real $r \in \R$,
\begin{align}
e^{2\pi i r} = 1 \iff r \in \Z.
\end{align}
If  $\varphi : G 	o H$ is the homomorphism $\varphi = r \mapsto e^{2\pi i r}$, then by (1)
$$\ker(\varphi) = \Z.$$
(I leave it to the patient reader to make nice drawings.)

If $z_0 = e^{2\pi i r_0}$ ($r_0 \in \R$), we obtain 
\begin{align*}
r \in \varphi^{-1}(z_0) &\iff  e^{2\pi i r} = e^{2\pi i r_0}\
&\iff e^{2\pi i\left(r - r_0ight)} = 1\
&\iff r - r_0 \in \Z\
&\iff r \in r_0 + \Z,
\end{align*}
so
$$\varphi^{-1}(\{e^{2i\pi r_0}\}) = r_0 + \Z.$$
Since $-1= e^{i\pi} = e^{2i\pi\cdot(1/2)},\ i = e^{i\pi/2} = e^{2i\pi\cdot(1/4)},\ e^{4i\pi/3} = e^{2i\pi\cdot (2/3)}$, this gives the fibers
\begin{align*}
\varphi^{-1}(\{-1\}) &= \frac{1}{2} + \Z,\
\varphi^{-1}(\{i\}) &= \frac{1}{4} + \Z,\
\varphi^{-1}(\{e^{4i\pi/3} \}) &= \frac{2}{3} + \Z.
\end{align*}
\end{proof}

Exercise 3.1.12 (Homomorphism $r \mapsto e^{2\pi i r}$)

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Exercise 3.1.12 (Homomorphism $r \mapsto e^{2\pi i r}$)

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