Fix an algebraic closure, $\overline{\mathbb{Q}}$ for the rationals and consider the set, $B_p$, of all places of $\overline{\mathbb{Q}}$ over a fixed (possibly infinite) prime, $p$, of $\mathbb{Q}$. Let $G_\mathbb{Q}=\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.

1: Let $G_\mathbb{Q}$ act on $B_p$ in the natural way. Then what is the structure of the stabilizer, $H_v$ of $v$, for $v|p$ an element of $B_v$?

Infinite primes are just embeddings of $\overline{\mathbb{Q}}$ into $\mathbb{C}$, so the case $p=\infty$ is really asking for the stabilizer of the action of left multiplication in $G_\mathbb{Q}$, and for finite $p$ it should be the inverse limit--over finite, Galois extensions of $\mathbb{Q}$--of the decomposition groups of primes $\mathfrak{p}|p$. However, actually getting ones hands on the structure of the stabilizers in question seems to be a hard question, at least no one I've talked to so far seems to know much more than the obvious things stated here.

2: Is is perhaps easier to describe the cosets $G/H_v$? If so, what should they look like?


If $p$ is a rational prime, then choosing a prime $v$ of $\overline{\mathbf{Q}}$ lying over $p$ amounts to choosing an embedding $i:\overline{\mathbf{Q}}\hookrightarrow\overline{\mathbf{Q}}_p$. This gives rise to a map $\varphi:G_{\mathbf{Q}_p}\rightarrow G_{\mathbf{Q}}$defined as follows: given $s$ in the source, $\varphi(s)$ is the unique automorphism of $\overline{\mathbf{Q}}$ such that $i\circ\varphi(s)=s\circ i$. This is continuous. Its image is the decomposition group (i.e. the stabilizer of) $v$, $G_v\subseteq G_\mathbf{Q}$. The kernel is the Galois group of $\overline{\mathbf{Q}}_p$ over $\mathbf{Q}_pi(\overline{\mathbf{Q}})$, which is trivial by Krasner's lemma. So you have an isomorphism $G_{\mathbf{Q}_p}\cong G_v$ (it is a homeomorphism because it is bijective with compact source and Hausdorff target).

The case of Archimedean primes is identical. In particular, $G_v$ for $v$ Archimedean has order $2$. When people talk about "choosing a complex conjugation" in $G_{\mathbf{Q}}$, they are referring to the choice of an embedding $\overline{\mathbf{Q}}\rightarrow\mathbf{C}$ which gives rise to an injection $\mathrm{Gal}(\mathbf{C}/\mathbf{R})\hookrightarrow G_{\mathbf{Q}}$, and the image of the unique non-trivial element of the source is the ``complex conjugation."

Whenever you have a, say, $\ell$-adic, Galois representation $\rho:G_\mathbf{Q}\rightarrow\mathrm{GL}_d(\mathbf{Q}_\ell)$, and a theorem talks about the local structure at $p$ of $\rho$, it means the restriction of $\rho$ to a decomposition group for a prime above $p$, which, by the paragraph above, can be identified with $G_{\mathbf{Q}_p}$. So a representation of $G_\mathbf{Q}$ gives rise to representations of $G_{\mathbf{Q}_p}$ for all $p$ by restriction...at least after choosing a decomposition group, which is unique up to conjugacy.

EDIT: This is in response to the question posed in the comments. The reason the map $\varphi$ (which depends on $i$) is well-defined is because $\overline{\mathbf{Q}}$ is a normal extension of $\mathbf{Q}$, so the image of any embedding of $\overline{\mathbf{Q}}$ into $\overline{\mathbf{Q}}_p$ is the same (the subfield of elements algebraic over $\mathbf{Q}$). So, given $s\in G_{\mathbf{Q}_p}$, the embeddings $s\circ i$ and $i$ have the same image, so $i^{-1}\circ s\circ i$ makes sense, and is an element of $G_\mathbf{Q}$. This is $\varphi(s)$.

Since $\varphi$ is a homomorphism, it's enough to check continuity at the identity. Take a finite extension $F$ of $\mathbf{Q}$ in $\overline{\mathbf{Q}}$, so $U=G_F\leq G_\mathbf{Q}$ is a typical neighborhood of the identity. Suppose $\varphi(s)\in G_F$. We want to prove that there is an open subgroup $U^\prime$ of $G_{\mathbf{Q}_p}$ with $s\in U^\prime$ and $\varphi(U^\prime)\subseteq G_F$. Let $F^\prime=i(F)\mathbf{Q}_p\subseteq\overline{\mathbf{Q}}_p$. This is a finite extension of $\mathbf{Q}_p$. Since $\varphi(s)=i^{-1}\circ s\circ i$ fixes $F$, $s$ fixes $i(F)$, and therefore, since $i(F)$ generates $F^\prime$ over $\mathbf{Q}_p$ and $s$ is $\mathbf{Q}_p$-linear, $s$ fixes $F^\prime$. Conversely anything in $U^\prime=G_{F^\prime}$ has image under $\varphi$ in $G_F$, so $s\in U^\prime\leq\varphi^{-1}(U)$, as desired.

These arguments are totally general. They show that, if $k\hookrightarrow K$ is a map of fields and $i:k_s\hookrightarrow K_s$ is a choice of map of separable closures lifting $k\hookrightarrow K$ , then we get a continuous homomorphism $G_K\rightarrow G_k$. This homomorphism is not always injective though, as the injectivity of $\varphi$ above used a property particular to that setup (Krasner's lemma).

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  • $\begingroup$ Just so we're clear, the way to think of the big prime over p as an embedding is to look at a uniformizer and looking at the power series for the element of $\overline{\mathbb{Q}}$ in $\overline{\mathbb{Q}_p}$, or is there something I'm missing? $\endgroup$ – Adam Hughes Oct 29 '12 at 21:42
  • $\begingroup$ I'm not sure what uniformizer or power series you're referring to here. I can tell you how to go from a prime $v$ of $\overline{\mathbf{Q}}$ to an embedding. I'll assume $v$ is non-Archimedean. Pick an absolute value in $v$. By restriction to each finite subextension, i.e., each number field $F$, you get an absolute value, which I'll also denote by $v$, and you can form the completion $F_v$. If $F\subseteq F^\prime$, then the absolute value on the bigger field restricts to the absolute value on the smaller one, so there is a canonical injection $F_v\rightarrow F_v^\prime$ over $\endgroup$ – Keenan Kidwell Oct 30 '12 at 11:05
  • $\begingroup$ $F\subseteq F^\prime$. Then set $\overline{\mathbf{Q}}_v$ equal to the directed colimit of these completions of finite subextensions. This will be an algebraic extension of $\mathbf{Q}_p$, where $p$ is the residue characteristic of $v$, containing $\overline{\mathbf{Q}}$. Therefore you can uniquely extend the usual absolute value on $\mathbf{Q}_p$ to $\overline{\mathbf{Q}}_v$. You can also choose an embedding of $\overline{\mathbf{Q}}_v$ into $\overline{\mathbf{Q}}_p$. The induced absolute value will have to coincide with the given one, and by restriction to $\overline{\mathbf{Q}}$, you get $\endgroup$ – Keenan Kidwell Oct 30 '12 at 11:08
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    $\begingroup$ I guess this procedure works equally well for Archimedean primes. $\endgroup$ – Keenan Kidwell Oct 30 '12 at 11:09
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    $\begingroup$ Dear @Adam, I edited in an answer to your questions. $\endgroup$ – Keenan Kidwell Dec 2 '13 at 4:06

I do not understand what is $B_v$ in your notation, neither I see what you mean by "stabilizer of left multiplication".

Anyhow you are right in saying that $H_v$ is the inverse limit of decomposition groups and is isomorphic to the absolute Galois group of $\mathbb{Q}_p$. If $p=\infty$ you get a cyclic group with two elements, and if $p\neq \infty$ the structure of this Galois group is relatively well understood, although no explicit description is available. But since the structure of Galois groups of local fields is well-known, you see there is a sequence $$ 1\longrightarrow I_p\longrightarrow H_v\longrightarrow\widehat{\mathbb{Z}}\longrightarrow 1 $$ where $I_p$ is the inertia subgroup at $p$ (and not "at $v$" in the sense that they are all non-canonically isomorphic, the isomorphism being induced from the identification of your $H_v$ with the absolute Galois group of $\mathbb{Q}_p$ a choice of which is equivalent to having fixed your representative $v$). This inertia subgroup is a semi-direct product of the wild inertia subgroup $T_p$ (a pro-$p$ group which is difficult to put your hands on) and a tame inertia subgroup which is actually pro-cyclic of super-order prime to $p$.

Moreover, by local class field theory, you know that the abelianization of $H_v$ is isomorphic to $\widehat{\mathbb{Z}}^\times$. All this can be found in Serre's Corps Locaux, although only small hints are given at the infinite theory, or in Neukirch's Algebraische Zahlentheorie (I guess).

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  • $\begingroup$ Sorry, Filippo, I was a bit incomplete. What I meant was $G$ acts on the $B_p$p--which are homogeneous spaces for $G$ (since the action is continuous and transitive)--so the action is basically just left-multiplication of the cosets $G/H_v$ where $H_v$ is the stabilizer of any $v$ (a place of $\overline{\mathbb{Q}}$ over $p$) since $B_p$ is topologically isomorphic to $G/H_v$ for any $v|p$ of $\overline{\mathbb{Q}}$. $\endgroup$ – Adam Hughes Dec 2 '13 at 4:42

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