I'm looking for a quick way to compute the Pontryagin dual group of the n-dimensional torus $\mathbb{T}^n$ (with $\mathbb{T} := \mathbb{R} / \mathbb{Z}$). The only way I know is from "Dikran Dikranjan – Introduction to Topological Groups" and is very tedious.
I thought so: $\mathbb{T}^n$ is compact then $\widehat{\mathbb{T}^n}$ has the discrete topology. The set $E$ of all the functions
$\mathbb{T}^n \to \mathbb{C} : x \mapsto e^{2 \pi a \cdot x}$
with $a \in \mathbb{Z}^d$ is a subset of $\widehat{\mathbb{T}^d}$ which separates the points of $\mathbb{T}^d$ (I mean: for
any $x,y \in \mathbb{T}^d$ with $x \neq y$ exists some $\chi \in E$ such that $\chi(x) \neq \chi(y)$.
Now I was hoping that Stone-Weierstass theorem could help me to prove that $E$ is dense in $\widehat{\mathbb{T}^d}$ and so that $\widehat{\mathbb{T}^d} = E$, because $\widehat{\mathbb{T}^d}$ is discrete. However I don't know how to apply Stone-Weierstass theorem in this case.
I thank you for any ideas!
Best Answer
The key case is the 1-dimensional torus. We want to show every continuous homomorphism $\chi \colon {\mathbf R}/{\mathbf Z} \rightarrow {\mathbf T}$ has the form $x \bmod {\mathbf Z} \mapsto e^{2\pi inx}$ for some integer $n$.
Any character of ${\mathbf R}/{\mathbf Z}$ can be pulled back to a character of ${\mathbf R}$ by composing with the canonical map ${\mathbf R} \rightarrow {\mathbf R}/{\mathbf Z}$. Let $\chi' \colon {\mathbf R} \rightarrow {\mathbf T}$ be the resulting character of ${\mathbf R}$, i.e., $\chi'(x) = \chi(x \bmod {\mathbf Z})$. All continuous homomorphisms ${\mathbf R} \rightarrow {\mathbf T}$ have the form $x \mapsto e^{2\pi ixy}$ for some real number $y$. Accepting this for a moment, we'd have $\chi'(x) = e^{2\pi ixy}$ for some $y$. Thus $\chi(x \bmod {\mathbf Z}) = e^{2\pi ixy}$. Taking $x = 1$, we get $1 = e^{2\pi iy}$. Therefore $y$ is an integer. Writing $y$ as $n$, for psychological purposes, $\chi(x \bmod {\mathbf Z}) = e^{2\pi inx}$ for an integer $n$, which is what you wanted.
Our task now is to compute the characters of ${\mathbf R}$. For this I like the method from Conway's "A Course in Functional Analysis". For any continuous homomorphism $\gamma \colon {\mathbf R} \rightarrow {\mathbf T}$, we want to show $\gamma(x) = e^{2\pi ixy}$ for some real number $y$. We will use differential equations. Because $\gamma(0) = 1$, by continuity $\int_0^a \gamma(t)dt \approx a$ for small positive $a$, so $\int_0^a\gamma(t)dt \not= 0$ for an appropriate $a$. Fixing such $a$, for all real $x$ we have $$\int_x^{x+a} \gamma(t)dt = \int_0^a \gamma(t+x)dt = \int_0^a \gamma(t)\gamma(x)dt = \gamma(x)\int_0^a \gamma(t)dt.$$ Therefore $$ \gamma(x) = \frac{\int_x^{x+a} \gamma(t)dt}{\int_0^a\gamma(t)dt}. $$ The right side, by the fundamental theorem of calculus, is a differentiable function of $x$, and therefore $$ \gamma'(x) = \frac{\gamma(x+a) - \gamma(x)}{\int_0^a\gamma(t)dt} = \frac{\gamma(a)-1}{\int_0^a\gamma(t)dt}\gamma(x), $$ so $\gamma'(x) = s\gamma(x)$ for a complex number $s$. By the theory of ODE, $\gamma(x) = Ce^{sx}$. Since $\gamma(0) = 1$, we have $C = 1$, so $\gamma(x) = e^{sx}$. Since $|\gamma(x)| = 1$, we have $e^{{\rm Re}(s)x} = 1$ for all real $x$, so ${\rm Re}(s) = 0$. Write $s = 2\pi iy$ for a real number $y$ and we're done.
I don't think you're going to find a 2-line proof of the computation of the character group of ${\mathbf R}/{\mathbf Z}$ or ${\mathbf R}$ because you need to get exponential functions from somewhere. Above we get them from our knowledge of solutions to the most basic first-order ODE, and to bring in an ODE we needed a clever idea to see that the a priori continuous function $\gamma$ is in fact differentiable.