13

So I remember reading once that the only element of $G=Gal(\overline{\Bbb Q} / \Bbb Q)$ that we understand is complex conjugation. Suppose we fix an embedding of $\overline{\Bbb Q}$ into $\Bbb C$. Then complex conjugation in $\Bbb C$ restricts to a "complex conjugation", $\sigma \in G$. Clearly $\sigma$ has order $2$, and any conjugate of $\sigma$ also has order $2$.

Question: Is it true that any element of order $2$ in $G$ is conjugate to $\sigma$?

I've thought about this for a while, but my knowledge of elements of $G$ isn't excellent. For general groups, not all elements of order $2$ are necessarily conjugate. For example, $(12)$ and $(12)(34)$ are both of order $2$ but not conjugate in $S_n$ for any $n\geq 4$. It is unclear how to proceed in the case of $G$.

My reason for asking is that the element $\sigma$ I defined above should only be thought of as being defined up to conjugacy, since a different embedding gives a new complex conjugation which is conjugate to $\sigma$. Thus, if all elements of order $2$ in $G$ were conjugate then I believe I could conclude that there is really only a single honest complex conjugation.

I apologize if any of this is unclear, I'd be glad to elaborate if necessary. Also, I'd appreciate any intuition on how to think about this in terms of Galois representations, since those seem to be the only way to get a handle on $G$.

Dylan Yott
  • 7,189
  • 3
    Dear Dylan, If we look not at all of $\overline{\mathbb Q}$, but at the maximal subfield unramified ouside some set of prime $S$, and consider the corresponding Galois group $G_S$, then there are other explicit conjugacy classes of $G_S$, namely the Frobenius elements Frob${}_p$ for primes $p \not\in S$. Cebotarev density shows that the union of these conjugacy classes is dense in $G_S$, and so these are the main tool for "accessing" $G_S$. Complex conjugation is important (and is important in the theory of Galois representations), but it is only a small part of $G$ or $G_S$. Regards, – Matt E Dec 31 '13 at 05:27

1 Answers1

14

It is a theorem of E. Artin that any non-trivial finite order element of $\mathbb Q$ is a complex conjugation. This follows from the theory of real closed fields; see here for example. (If $H \subset G$ is a non-trivial finite subgroup, then $F:=\overline{\mathbb Q}^H$ is a field whose algebraic closure is a proper finite exension; thus $F$ satisfies property 4 from the linked page. Property 5 then shows that $\overline{\mathbb Q} = F(\sqrt{-1})$, from which one sees that $H$ has order $2$; a little more argument shows that the non-trivial element of $H$ acts as a complex conjugation. Any treatment of real closed fields that is more detailed than the wikipedia page should give the details.)

Matt E
  • 127,227