# A more precise description of conjugation of semi-simple subgroups

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!

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.