If you find any mistakes, please make a comment! Thank you.

Represent the real numbers as complex numbers of modulus 1


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.12

Let $G$ be the additive group of real numbers and $H$ the multiplicative group of complex numbers with absolute value 1. Let $\varphi : G \rightarrow H$ be the homomorphism $r \mapsto e^{2 \pi i r}$. Find the kernel of $\varphi$ and the fibers over $-1$, $i$, and $e^{4 \pi i / 3}$.


Solution: By Euler’s formula we have $$e^{2 \pi i r} = \cos(2 \pi r) + i \sin(2 \pi r).$$ Thus for each integer $n$, we have $\varphi(r + n) = \varphi(r)$.

(1) If $\varphi(r) = 1$, we have $\sin(2 \pi r) = 0$. Thus $r$ is an integer or a half integer. But also $\cos(2 \pi r) = 1$, so that $r$ must be an integer. Thus $\mathsf{ker}\ \varphi = \mathbb{Z}$.

(2) If $\varphi(r) = -1$, we have $\sin(2 \pi r) = 0$. Thus $r$ is an integer or a half integer. But also $\cos(2 \pi r) = -1$, so that $r$ must be a half integer. Thus $\varphi^\star(-1) = \{ n + 1/2 \ |\ n \in \mathbb{Z} \}$.

(3) If $\varphi(r) = i$, we have $\cos(2 \pi r) = 0$. Thus $r$ is either a quarter integer or a three-quarter integer (i.e. $n + 1/4$ or $n + 3/4$ for some integer $n$.) Since $\sin(2 \pi r) = 1$, we see that $r$ must be a quarter integer. Thus $\varphi^\star(i) = \{ n + 1/4 \ |\ n \in \mathbb{Z} \}$.

(4) If $\varphi(r) = e^{4 \pi i / 3} = \cos(4 \pi / 3) + i \sin(4 \pi / 3)$, we have $\cos(2 \pi r) = \cos(4 \pi / 3)$. Thus we have $2 \pi n \pm 4 \pi / 3 = 2 \pi r$ for some integer $n$, and thus $r = n \pm 2/3$. Since $\sin$ is an odd function, the condition $\sin(4 \pi / 3) = \sin(2 \pi r)$ eliminates the negative solution. So $$\varphi^\star(e^{4 \pi / 3}) = \{ n + 2/3 \ |\ n \in \mathbb{Z} \}.$$

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This Post Has One Comment

  1. Hi. I did (2) and the fiber of phi above -1 gave me the union of the set you wrote and {n+3/4 | for n in Z}. Is this correct?

Leave a Reply

Close Menu