Complete proof for the shape of quasicharacters of $\mathbb{R}^{\times}, \mathbb{C}^{\times}$

Question

Quasicharacters (:=continuous group homomorphism to $\mathbb{C}^{\times}$) of $\mathbb{R}^{\times}, \mathbb{C}^{\times}$ seems to be known to be following forms (the following is quoted from [Raghuram, p.8]):

(1) Quasicharacters of $\mathbb{R}^{\times}$.

Any continuous homomorphism $\chi: \mathbb{R}^{\times} \rightarrow \mathbb{C}^{\times}$is of the form $$ \chi(x)=\operatorname{sgn}(x)^n|x|^w=\left(\frac{x}{|x|}\right)^n|x|^w, $$ with $n \in\{0,1\}$ and $w \in \mathbb{C}$. Such a character $\chi$ is unitary if and only if $w=i \varphi \in i \mathbb{R}$.

(2) Quasicharacters of $\mathbb{C}^{\times}$.

For $z=x+i y \in \mathbb{C}$, define $|z|:=|z|_{\mathbb{R}}:=\sqrt{x^2+y^2}$, and the normalised valuation on $\mathbb{C}$ as $|z|_{\mathbb{C}}:=|z|_{\mathbb{R}}^2=x^2+y^2$. A continuous homomorphism $\chi$ : $\mathbb{C}^{\times} \rightarrow \mathbb{C}^{\times}$is of the form $$ \chi(z)=\left(\frac{z}{|z|}\right)^n|z|_{\mathbb{C}}^w $$ with $n \in \mathbb{Z}$ and $w \in \mathbb{C}$. If $z=r e^{\mathrm{i} \theta}$ with $r \in \mathbb{R}_{+}^{\times}$and $\theta \in \mathbb{R}$, then $\chi(z)=r^{2 w} e^{in \theta}$. Such a character $\chi$ is unitary if and only if $w=i \varphi \in i \mathbb{R}$.

I want to know the complete proof of these.

Answer 1

$\def\A{\mathbb{A}}$ $\def\B{\mathbb{B}}$ $\def\C{\mathbb{C}}$ $\def\N{\mathbb{N}}$ $\def\O{\mathcal{O}}$ $\def\P{\mathbb{P}}$ $\def\Q{\mathbb{Q}}$ $\def\R{\mathbb{R}}$ $\def\Z{\mathbb{Z}}$ $\def\mf{\mathfrak}$ $\def\mc{\mathcal}$ $\def\x{^{\times}}$ $\newcommand{\abs}[1]{\left \lvert #1 \right \rvert}$ $\def\innertimes{\dot{\times}}$ $\DeclareMathOperator{\sgn}{sgn}$

The group of complex numbers with absolute value 1 is denoted by $\C^1$.

First, let's review the following fundamental result:

Proposition. For a real number $a$, define $\psi_a:\R\to\C^1$ by $\psi_a(x):=e^{2\pi i ax}$. For any continuous homomorphism $\psi:\R\to\C^1$, there exists a unique real number $a\in\R$ such that $\psi=\psi_a$.

Proof. See KConrad's answer in this thread for example. $\square$

(This can be rephrased as follows: The map $\R\xrightarrow{\sim}\R^{\vee};a\mapsto \psi_a$ is an isomorphism between $\R$ and its Pontryagin dual $\R^{\vee}$. In other words, $\R$ is self-dual with respect to the Pontryagin duality.)

Lemma. (1) A continuous homomorphism $\chi:\R_{>0}\to\C\x$ can be written as $\chi(x)=x^w$ for some $w\in\C$.

(2) A continuous homomorphism $\chi:\C^1\to\C\x$ can be written as $\chi(z)=z^N$ for some $N\in\Z$.

Proof. (1) First, decompose the target as $\C\x=\R_{>0}\innertimes \C^1$ (where $z=\abs{z}\cdot \frac{z}{\abs{z}}$ and $\innertimes$ denotes internal direct product). Also, the map $\log:\R_{>0}\xrightarrow{\sim} \R$ is an isomorphism of topological groups. Therefore, we need to consider two continuous homomorphisms $\R\to\R$ and $\R\to\C^1$ separately. For the former, it is known from Cauchy's functional equation that it has the form $x\mapsto rx\ (r\in\R)$. For the latter, from the above proposition, it has the form $x\mapsto e^{isx}\ (s\in\R)$.

Thus, $\chi(x)$ is the product of the values obtained by sending $x$ through $\R_{>0}\xrightarrow{\log} \R\xrightarrow{x\mapsto rx}\R\xrightarrow{\exp} \R$ and $\R_{>0}\xrightarrow{\log} \R\xrightarrow{x\mapsto e^{isx}}\C^1$, resulting in $x^r\cdot x^{is}=x^{w}\ (w:=r+is)$.

(2) Considering the isomorphism $\R\to\C^1;x\mapsto e^{2\pi ix}$ which induces $\R/\Z\cong \C^1$, what we seek is essentially the same as continuous homomorphisms $\R\to\C\x$ that are trivial on $\Z$.

From (1), continuous homomorphisms $\R\to\C\x$ can be written in the form $x\mapsto e^{wx}\ (w\in\C)$, and it is easy to see that being always equal to $1$ on $\Z$ is equivalent to $w\in 2\pi i\Z$.

Therefore, continuous homomorphisms $\R\to\C^1$ that are trivial on $\Z$ are of the form $x\mapsto e^{2\pi i Nx}\ (N\in\Z)$.

(Note: This expresses the fact that $(\R/\Z)^\vee\cong \Z$.)

Hence, $\chi$ equals $x\mapsto x^N$ through $\C^1\xrightarrow{\frac{1}{2\pi i}\log} \R/\Z\xrightarrow{[x]\mapsto e^{2\pi i Nx}}\C^1$. $\square$

(For another proof using integration, see this answer for (1), this answer for(2) but these are essentially same as the above KConrad's answer)

Proof of the statement of original question. The multiplicative groups of Archimedean local fields can be decomposed as topological groups as follows: \begin{align} \R\x&=\{\pm 1\}\innertimes \R_{>0}\ (;a=\sgn(a)\cdot \abs{a}),\\ \C\x&=\R_{>0}\innertimes \C^1\ (;z=\abs{z}\cdot \frac{z}{\abs{z}}) \end{align}

Therefore, considering these element correspondences and the above lemma, the conclusions follow. $\square$