3
$\begingroup$

Let $G$ be a semi-simple algebraic group over $\mathbb{Q}$, I would like to find an integer $d>0$ only depending on $G$ with the following property. For any two semi-simple $\mathbb{Q}$-subgroups $H_1,H_2\subset G$ that are conjugated by an element $g\in G(\mathbb{R})$, there exists a number field $K\subset \mathbb{R}$ of degree at most $d$, such that $\exists$ $\sigma\in G(K)$ with $\sigma H_1 \sigma^{-1}=H_2$?

Note that we have already known that $K$ can be chosen as a number field of uniformly bounded degree, see: About the conjugation of semi-simple subgroups. Now I would like to figure out if $K$ can be further taken as a real field. Thanks very much!

$\endgroup$
4
$\begingroup$

The answer is YES.

Theorem 1. Let $G$ be a connected semisimple linear algebraic group defined over a number field $k\subset{\mathbb{R}}$. There exists a natural number $d=d(G_{\bar{k}})$ with the following property: let $H_1$ and $H_2$ be two connected semisimple $k$-subgroups of $G$ that are conjugate over ${\mathbb{R}}$, then there exists a finite real field extension $K=K(H_1,H_2)\subset {\mathbb{R}}$ of $k$ of degree $[K:k]\le d$ such that $H_1$ and $H_2$ are conjugate over $K$.

Since there are only finitely many conjugacy classes of connected semisimple subgroups in $G_{\bar{k}}$, see Friedrich Knop's answer to this question, Theorem 1 follows from the next Theorem 2 (as in my previous answer).

Theorem 2. Let $N$ be a linear algebraic group (not necessarily connected or reductive) over a real number field $k\subset{\mathbb{R}}$. The there exists $d=d(N_{\bar{k}})$ such that any cohomology class $\xi\in H^1(k,N)$ that vanishes over ${\mathbb{R}}$ can be killed by a finite real field extension $K=K(N,\xi)\subset{\mathbb{R}}$ of degree $[K:k]\le d$.

Note that the analog of Theorem 2 for abelian varieties is false.

We need three lemmas.

Lemma 1. Let $N$ be a finite $k$-group. Then any ${\mathbb{R}}$-point of $N$ is defined over a real number field of bounded degree.

This is obvious.

Lemma 2. If $N$ is a connected linear algebraic $k$-group, then $N(k)$ is dense in $N({\mathbb{R}})$.

See Sansuc's paper, Cor. 3.5(iii).

Lemma 3. If $N$ is a connected or abelian linear algebraic $k$-group, then the localization map $$ H^1(k,N)\to H^1({\mathbb{R}},N) $$ is surjective.

See Sansuc's paper, Lemma 1.6 and Lemma 1.12.

Note that the analog of Lemma 3 for nonabelian finite $k$-groups is an open question.

With the help of Lemmas 1, 2, and 3, we use dévissage in order to reduce Theorem 2 to the following four propositions.

Proposition 1. Theorem 2 is true when $N$ is a finite $k$-group.

Indeed, then any $k$-torsor $\mathcal{T}$ of $N$ is a finite $k$-scheme, hence any ${\mathbb{R}}$-point of $\mathcal{T}$ is real algebraic of bounded degree.

Now let $N$ be connected, then it splits over an algebraic extension $L\subset \mathbb{C}$ of bounded degree. Set $K=L\cap{\mathbb{R}}$. After passing to $k=K$, we may assume that $N$ splits over $k$ or over a quadratic extension $L$ of $k$. After passing to a real quadratic extension, we may assume that $k$ has only one real embedding.

Proposition 2. Theorem 2 is true when $N$ is a simply connected semisimple $k$-group.

Indeed, under our assumptions the localization map $$ H^1(k,N)\to H^1({\mathbb{R}},N) $$ is bijective (the Hasse principle for simply connected groups), hence $\xi=1$.

Proposition 3. Theorem 2 is true when $N$ is a $k$-torus.

Proof. We assume that $N$ splits over a quadratic extension $L$ of $k$. Then we may assume that $N$ is a nonsplit one-dimensional $k$-torus. Write $L=k(\sqrt{-\alpha})$, where $\alpha\in k$, $\alpha>0$. Then $N$ is given by the equation $$ x^2+\alpha y^2=1,$$ and a $k$-torsor $\mathcal{T}$ of $N$ is given by the equation $$ x^2+\alpha y^2=c,$$ where $c\in k^\times$. By assumption $\mathcal{T}$ has an ${\mathbb{R}}$-point $(x_1,y_1)$, and we may assume that $x_1,y_1>0$. Choose $y_2\in k$ such that $0< y_2<y_1$ and set $\zeta=\sqrt{c-\alpha y_2^2}$, then $K:=k(\zeta)$ is a real quadratic extension of $k$, and $\mathcal{T}$ has a $K$-point $(\zeta, y_2)$. This $\mathcal{T}$ splits over a real quadratic extension, as required.

Proposition 4. Theorem 2 is true for $H^2(k,A)$, where $A$ is a finite abelian $k$-group.

Proof. Set $A'=\mathrm{Hom}(A,\mathbb{G}_m)$, the dual group. We say that $A$ is split over $k$ if the Galois group $G_k$ acts trivially on $A'$. As above, we may assume that $A$ splits over a quadratic extension $L$ of $k$ and that $k$ has only one real embedding. Then, since $A$ splits over a quadratic extension, the Hasse principle for $H^1(k,A)$ and $H^2(k,A)$ holds. The Galois group $G_k$ acts on $A'$ via its quotient group $\Gamma=G_{L/k}$ of order 2. We may assume that $A'$ is a simple $\Gamma$-module, i.e., is has no proper $\Gamma$-submodules. Then $\#A=\ell$ is a prime number, and either $\Gamma$ acts on $A'$ trivially, or $\ell$ is odd and $\Gamma$ acts on $A'$ by multiplication by $-1$. The first case was treated in my previous answer: we can kill a cohomology class $\eta\in H^2(k,A)$ that vanishes over $L$ by a real extension of degree $\ell$. Similarly, in the second case case one can kill a cohomology class $\eta$ that vanishes over $L$ by a real extension of degree $\ell$. (One should use the Tate duality and the Tate-Poitou exact sequence, see Serre's book "Galois cohomology".)

This completes the proofs of Theorem 1 and Theorem 2.

$\endgroup$

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.