Character group of the multiplicative rationals

Question

I was reading some stuff on Hecke characters and came across an issue I have not been able to resolve. I posted it here on math stack exchange first.

Let $\mathbb{Q}^{\times}$ be the multiplicative group of the rationals equipped with the discrete topology. Let $\chi : \mathbb{Q}^{\times} \to \mathbb{S}^1$ be a homomorphism, that is $\chi$ is an element of the dual group of the multiplicative rationals. Then $\chi$ is determined by where it maps the primes and $-1$ and conversely any assignment of modulus 1 complex numbers to the primes and of $\pm 1$ to $-1$ gives rise to a multiplicative character.

Now let $J$ denote the ideles and $J^0 = \{x \in J : \lVert x\rVert = 1\}$ be the closed subgroup of $J$. Then $\mathbb{Q}^{\times}$ is a discrete, closed subgroup of $J^0$ and, by Fujisaki's lemma, the quotient $J^0 / \mathbb{Q}^{\times}$ is compact.

Then by general Fourier analysis (i.e. Theorem C.13 in Einsiedler and Ward's book on ergodic theory), we have that $$\widehat{J^0} / (\mathbb{Q}^{\times})^{\perp} \cong \widehat{\mathbb{Q}^{\times}}.$$

But, as explained in these notes on Hecke characters (Wayback Machine) $\widehat{J} / (\mathbb{Q}^{\times})^{\perp} $ is precisely the unitary Hecke characters of $\mathbb{Q}^{\times}$. In light of the first paragraph, it seems as though $\widehat{\mathbb{Q}^{\times}}$ contains much more than that, for instance the Liouville function.

What am I missing?

Answer 1

In Section 8, the notes say that the set of Hecke characters is $\widehat{J/\mathbb{Q}^\times}$ or $\widehat{J^0/\mathbb{Q}^\times}$, which can be identified with $(\mathbb{Q}^\times)^\perp$, but I do not see any reason to think that the Hecke characters should be $\widehat J/(\mathbb{Q}^\times)^\perp$. Indeed, $\widehat{J^0/\mathbb{Q}^\times}$ is discrete, but $\widehat J/(\mathbb{Q}^\times)^\perp \cong \widehat{\mathbb{Q}^\times}$ is compact.