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).

  • $\begingroup$ 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
  • 1
    $\begingroup$ I guess this procedure works equally well for Archimedean primes. $\endgroup$ – Keenan Kidwell Oct 30 '12 at 11:09
  • 1
    $\begingroup$ 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).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.