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Let $K$ be a number field, and $v_0$ a finite place of $K$. Let $\eta$ be a quasicharacter of $K_{v_0}^{\ast}$. Does there exist a character $\chi = \otimes \chi_v$ of the idele class group $\mathbb I_K/K^{\ast}$ such that $\chi_{v_0} = \chi$, and $\chi_v$ is unramified for all finite $v \neq v_0$?

Similarly, if $v_1, ... , v_n$ are places of $K$, is it possible to have an idele class character having specified restrictions to each $K_{v_i}^{\ast}$? What is the best way to think about characters of $\mathbb I_K/K^{\ast}$ to answer questions like these?

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This question, which I only found recently, has some overlap with the phenomenon described in the MO question here that was asked at around the same time as this question.

I'll show how to settle the existence question for a single quasi-character $\eta$ of $K_{v_0}^\times$: find a quasi-character $\chi$ of $\mathbb I_K/K^\times$ with $v_0$-component $\eta$ that is unramified at all other finite places of $K$. In fact, $v_0$ can be an arbitrary place of $K$, including an archimedean one, and $\chi$ can be chosen to be unramified at all places of $K$ other than $v_0$.

Let $v_0$ be a place of $K$ (finite or infinite) and $\eta$ be a quasi-character of $K_{v_0}^\times$. I want to reduce to the setting of characters. There is $\sigma \in \mathbf R$ such that $|\eta(x)| = |x|_{v_0}^\sigma$ for all $x$ in $K_{v_0}^\times$. This means $\eta'(x) := \eta(x)|x|_{v_0}^{-\sigma}$ defines a character ($S^1$-valued) of $K_{v_0}^\times$. If we can find a character $\chi'$ of $\mathbb I_K/K^\times$ that has $v_0$-component $\eta'$ then $\chi(\mathbf x) := \chi'(\mathbf x)|\!|\mathbf x|\!|^\sigma$ (here and below I write $|\!|\mathbf x|\!|$ for the idelic norm on $\mathbb I_K$) is a quasi-character of $\mathbb I_K/K^\times$ that has $v_0$-component $\eta$: for $x \in K_{v_0}^\times$ placed in the $v_0$-component of $\mathbb I_K$ in the calculation below, \begin{eqnarray*} \chi(1,\ldots,1,x,1,\ldots) & = & \chi'(1,\ldots,1,x,1,\ldots)|\!|(1,\ldots,1,x,1,\ldots)|\!|^\sigma \\ & = & \chi'(1,\ldots,1,x,1,\ldots)|x|_{v_0}^\sigma \\ & = & \eta'(x)|x|_{v_0}^\sigma \\ & = & \eta(x). \end{eqnarray*} Thus $\eta$ is the $v_0$-component of $\chi$. Furthermore, $\chi$ is unramified at exactly the places where $\chi'$ is unramified: for a place $v \not= v_0$ of $K$ and $y \in K_{v}^\times$ with $|y|_v = 1$ that is placed into $\mathbb I_K$ in the $v$-component, \begin{eqnarray*} \chi(1,\ldots,1,y,1,\ldots) & = & \chi'(1,\ldots,1,y,1,\ldots)|\!|(1,\ldots,1,y,1,\ldots)|\!|^\sigma \\ & = & \chi'(1,\ldots,1,y,1,\ldots)|y|_{v}^\sigma \\ & = & \chi'(1,\ldots,1,y,1,\ldots), \end{eqnarray*} so for each place $v \not= v_0$ of $K$ the $v$-components of $\chi$ and $\chi'$ are equal on $\{y \in K_v^\times : |y|_v = 1\}$.

Our goal now is to show each character $\eta'$ of $K_{v_0}^\times$ is the $v_0$-component of some character $\chi'$ of $\mathbb I_K$ and make $\chi$' unramified at all places of $K$ other than $v_0$.

The composite group homomorphism $K_{v_0}^\times \rightarrow \mathbb I_K \rightarrow \mathbb I_K/K^\times$ is continuous and injective. This embedding is closed and the quotient $(\mathbb I_K/K^\times)/K_{v_0}^\times$ is compact; see the MO page here (the question, answers, and comments) for a discussion of these properties. Inside $\mathbb I_K$ let $$ U_{v_0}' = \prod_{v \not= v_0} \{y \in K_v^\times : |y|_v = 1\} \times \{1\}, $$ where the $\{1\}$ at the end is the $v_0$ component. The continuous group homomorphism $\mathbb I_K \rightarrow \mathbb I_K/K^\times$ is an embedding on $U_{v_0}'$, and since $U_{v_0}'$ is compact its image in $\mathbb I_K/K^\times$ is compact. From this we will show the image of $$ U_{v_0}K_{v_0}^\times = \prod_{v \not= v_0} \{y \in K_v^\times : |y|_v = 1\} \times K_{v_0}^\times = U_{v_0}' \times K_{v_0}^\times $$ in $\mathbb I_K/K^\times$ is closed.

When $G$ is an abelian topological group with closed subgroups $C$ and $H$ such that $C$ is compact and $G/H$ is compact, the subgroup $CH = \{hc : h \in H, c \in C\}$ in $G$ is closed: the continuous mapping $G \rightarrow G/H$ sends $C$ onto $CH/H$, so $CH/H$ is closed and thus its inverse image $CH$ in $G$ is closed. Apply this with $G = \mathbb I_K/K^\times$, $C = U_{v_0}'$ (as an embedded subgroup of $G$), and $H = K_{v_0}^\times$ (as an embedded subgroup of $G$) to see that the image of $U_{v_0}' \times K_{v_0}^\times$ under $\mathbb I_K \rightarrow \mathbb I_K/K^\times$ is closed.

The character $\eta'$ on $K_{v_0}^\times$ extends to a character of $U_{v_0}' \times K_{v_0}^\times$ by declaring it to be trivial on $U_{v_0}'$. That is, use the composite $U_{v_0}' \times K_{v_0}^\times \rightarrow K_{v_0}^\times \rightarrow S^1$ where the second function is $\eta'$. Now view $U_{v_0}' \times K_{v_0}^\times$ as an embedded closed subgroup of $\mathbb I_K/K^\times$ by the process described above. (The groups $K_{v_0}^\times$ and $U_{v_0}'$ intersect trivially in $\mathbb I_K/K^\times$ since no element of $K^\times$ is in $U_{v_0}'$ inside $\mathbb I_K$ other than $1$.) From Pontryagin duality (for characters, not quasi-characters!) every character of a closed subgroup of a locally compact abelian group can be lifted to a character of the whole group. Therefore the character we gave on $U_{v_0}' \times K_{v_0}^\times$ that is trivial on $U_{v_0}'$ and is $\eta'$ on $K_{v_0}^\times$ can be lifted to a character of $\mathbb I_K/K^\times$ (in many ways). Let $\chi'$ be any such lifting. Then $\chi'$ is a character of $\mathbb I_K/K^\times$ that equals $\eta'$ on the embedded copy of $K_{v_0}^\times$ and is trivial on the embedded copy of $U_{v_0}'$. This means, in terms of local components of a character on $\mathbb I_K/K^\times$, that the $v_0$-component of $\chi'$ is $\eta'$ and, for each place $v$ of $K$ other than $v_0$, the $v$-component of $\chi'$ is trivial on $\{y \in K_v^\times : |y|_v = 1\}$, so $\chi'$ is unramified at all places of $K$ other than $v_0$, finite or infinite, and therefore the quasi-character $\chi$ constructed above that has $v_0$-component $\eta$ is unramified at every place of $K$ other than $v_0$.

Note: If $\eta$ is a finite-order character of $K_{v_0}^\times$ then we may ask for $\chi$ to be a finite-order character of $\mathbb I_K/K^\times$ with the same order as $\eta$ and ramification related to that of $\eta$. This can be achieved except in some rare situations described by the Grunwald-Wang theorem; see the paper by Song Wang (not the "Wang" from Grunwald-Wang) on the arXiv here, especially paragraph 4 of Section 1 and Theorem 2.2.

For $n$ quasi-characters $\eta_1, \ldots \eta_n$ of $K_{v_1}^\times, \ldots, K_{v_n}^\times$ with $n > 1$, to have a quasi-charater $\chi$ on $\mathbb I_K/K^\times$ with $v_i$-component $\eta_i$ for $i = 1, \ldots, n$ definitely imposes a restriction: the exponents of all the $\eta_i$ must be the same. That is, there has to be a $\sigma \in \mathbf R$ such that for $i = 1, \ldots, n$ we have $|\eta_i(x)| = |x|_{v_i}^\sigma$ for all $x \in K_{v_i}^\times$ because the exponent of a quasi-character on $\mathbb I_K/K^\times$ equals the exponent of each of its local components. Using a common exponent $\sigma$ for the $\eta_i$, the $\sigma$-power of the absolute value on each $K_{v_i}$ and the $\sigma$-power of the idelic norm on $\mathbb I_K$ can be used to show the existence question is equivalent to the case of characters (when $\sigma = 0$).

The necessary condition that all $\eta_i$ have the same exponent is not sufficient, and I will explain how to give counterexamples in the case $K = \mathbf Q$. For a character $\chi$ of $\mathbb I_\mathbf Q$ (don't assume yet that $\chi$ is trivial on $\mathbf Q^\times$), let $\chi_v \colon \mathbf Q_v^\times \hookrightarrow \mathbb I_\mathbf Q \rightarrow S^1$ be its $v$-component, so by continuity of $\chi$ all but finitely many $\chi_p$ are trivial on $\mathbf Z_p^\times$. For all $\mathbf x \in \mathbb I_\mathbf Q$, with $v$-component $x_v$, we have $\chi(\mathbf x) = \prod_v \chi_v(x_v)$.

We will show that having $\chi$ be trivial on $\mathbf Q^\times$ imposes a constraint on its components $\chi_v$ that we can use to give an example, for each pair of places $v$ and $w$ of $\mathbf Q$, of characters $\eta_v$ of $\mathbf Q_v^\times$ and $\eta_w$ of $\mathbf Q_w^\times$ that are not the components of a character of $\mathbb I_\mathbf Q/\mathbf Q^\times$. The idea that this should be possible (for a general number field, but I stick to $\mathbf Q$ for concreteness) comes from the main question on the MO post here, which concerns the composite map $K_v^\times \times K_w^\times \hookrightarrow \mathbb I_K \twoheadrightarrow \mathbb I_K/K^\times$ not having a closed image for many choices of $v$ and $w$ (e.g., when $v$ and $w$ are non-archimedean not lying over the same prime of $\mathbf Q$). If the image were closed then my argument above showing that a single local character of $K_{v_0}^\times$ is the $v_0$-component of a character of $\mathbb I_K/K^\times$ could be adapted to a pair of characters of $K_{v_0}^\times$ and $K_{w_0}^\times$.

The characters of $\mathbf Q_v^\times$ have standard formulas. For $v = \infty$, we can write for $x \in \mathbf R^\times$ $$ \chi_\infty(x) = ({\rm sgn } \, x)^\delta|x|^{ic_\infty}, $$ where $\delta$ is 0 or 1 and $c_\infty \in \mathbf R$. The number ${\rm sgn} \, x$ is $\pm 1$. For $v = p$ we can write for $x \in \mathbf Q_p^\times$ $$ \chi_p(x) = \chi_p(u_x)|x|_p^{ic_p} $$ where $u_x = x/p^{{\rm ord}_p(x)}$ is the unit part of a nonzero $p$-adic number $x$ (we use on $\mathbf Q_p^\times$ the standard uniformizer $p$) and $c_p \in \mathbf R$. The number $\chi_p(u_x)$ is a root of unity (of order dividing some $p^k(p-1)$). We'll use these formulas in the following claim.

Claim: If a character $\chi$ of $\mathbb I_\mathbf Q$ is trivial on $\mathbf Q^\times$ then for each prime $p$, $c_p - c_\infty \in (\pi/\log p)\mathbf Q$.

Proof: We have $1 = \chi(p,p,p,\ldots) = \chi_\infty(p)\chi_p(p)\prod_{v \not= p,\infty} \chi_v(p)$. From the formulas for each $\chi_v$, for $v$ not $p$ or $\infty$ the number $\chi_v(p)$ is a root of unity (and most $\chi_v(p) = 1$), so $\chi_\infty(p)\chi_p(p)$ is a root of unity. Feeding into this condition the formulas for $\chi_\infty(p)$ and $\chi_p(p)$, we see that $p^{ic_\infty} (1/p)^{ic_p}$ is a root of unity. Thus $p^{i(c_\infty - c_p)} = 1$, so $(c_\infty - c_p)(\log p)/(2\pi) \in \mathbf Q$, and that completes the proof of the claim.

Example 1: There are characters of $\mathbf R^\times$ and $\mathbf Q_p^\times$ that are not components of a character of $\mathbb I_\mathbf Q/\mathbf Q^\times$.

Pick a real number $c_\infty$. The set $c_\infty + (\pi/\log p)\mathbf Q$ is countable, so there is a real number $c_p$ not lying in $c_\infty + (\pi/\log p)\mathbf Q$. By the claim, the characters $\chi_\infty(x) = ({\rm sgn} \, x)^\delta|x|^{ic_\infty}$ (for $\delta = 0$ or $1$) and $\chi_p(x) = \varepsilon_p(u_x)|x|_p^{ic_p}$ (for any continuous character $\varepsilon_p$ of $\mathbf Z_p^\times$) are not the archimedean and $p$-adic components of a character of $\mathbb I_\mathbf Q/\mathbf Q^\times$.

Note: The number $\pi/\log p$ is almost certainly transcendental, but I believe even a proof of its irrationality is still an open question for every prime $p$. If we accept that $\pi/\log p$ is transcendental or just irrational then we can use for $c_\infty$ and $c_p$ in Example 1 two different integers.

Example 2: For primes $p$ and $q$, there are characters of $\mathbf Q_p^\times$ and $\mathbf Q_q^\times$ that are not components of a character of $\mathbb I_\mathbf Q/\mathbf Q^\times$.

For each character $\chi$ of $\mathbb I_\mathbf Q/\mathbf Q^\times$ we have $c_p - c_\infty \in (\pi/\log p)\mathbf Q$ and $c_q - c_\infty \in (\pi/\log q)\mathbf Q$ by the claim above, so $$ c_\infty \in \left(c_p + \frac{\pi}{\log p}\mathbf Q\right) \cap \left(c_q + \frac{\pi}{\log q}\mathbf Q\right). $$ If we pick $c_p$ and $c_q$ in $\mathbf R$ to make the above intersection empty then there is no choice for $c_\infty$, so a character $\chi_p(x) = \varepsilon_p(u_x)|x|_p^{ic_p}$ on $\mathbf Q_p^\times$ and $\chi_q(x) = \varepsilon_q(u_x)|x|_p^{ic_q}$ on $\mathbf Q_q^\times$ using the exponents $c_p$ and $c_q$ (and arbitrary characters $\varepsilon_p$ on $\mathbf Z_p^\times$ and $\varepsilon_q$ on $\mathbf Z_q^\times$) are not components of a character of $\mathbb I_\mathbf Q/\mathbf Q^\times$.

Pick $c_p \in \mathbf R$. The set $c_p + (\pi/\log p)\mathbf Q + (\pi/\log q)\mathbf Q$ is countable, so there is a real number $c_q$ not belonging to that set, and with that choice of $c_q$ the sets $c_p + (\pi/\log p)\mathbf Q$ and $c_q + (\pi/\log q)\mathbf Q$ are disjoint.

Probably a concrete choice is $c_p = 0$ and $c_q = \pi$, or more generally $c_p$ and $c_q$ are two different rational multiples of $\pi$, since $\{1, 1/\log p, 1/\log q\}$ is almost certainly a linearly independent set over $\mathbf Q$, but I believe this is still an open problem. If there were a linear dependence relation, the coefficient of $1$ would have to be nonzero since $1/\log p$ and $1/\log q$ are linearly independent over $\mathbf Q$, so we'd have $p^{\log q}$ equal to $p^rq^s$ for some rational $r$ and $s$, making $p^{\log q}$ algebraic, but almost certainly $p^{\log q}$ is transcendental and I think this transcendence is still an open question.

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