# About the 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 $K$ of degree at most $d$, such that $\exists$ $\sigma\in G(K)$ with $\sigma H_1 \sigma^{-1}=H_2$?

The motivation of this question is from A finiteness property for semi-simple algebraic groups. According to the answer of that question, we know that $G$ has only finitely many semi-simple $\mathbb{Q}$-subgroups modulo $G(\mathbb{R})$ conjugation, now what I want to know is that if two semi-simple $\mathbb{Q}$-subgroups are conjugate, can we always choose a conjugation element inside a "small" number field.

To sum up, I would like to know if this $d$ exists and the corresponding proof. Thanks very much!

• Since $GL_n(\mathbf{Z})$ has finite subgroups of bounded order, you already know that there exists $d(k)$ such that every semisimple group of dimension $\le k$ splits over an extension of degree $\le d(k)$. Of course this is weaker than what you ask. – YCor May 16 '16 at 19:13
• The answer seems to be YES. We may pass to a finite extension $k$ of bounded degree. By the comment of @YCor we may assume that $H_1$ and $H_2$ are $k$-isomorphic. After we choose a conjugating element $g\in G( \bar{k})$, we obtain a cohomology class in $H^1(k,N)$, where $N=N(H_1)$ is the normalizer of $H_1$ in $G$. It remains to show that a cohomology class in $H^1(k,N)$ can be killed by a field extension of bounded degree. – Mikhail Borovoi May 16 '16 at 20:06
• If it helps, $N$ is reductive with a bounded number of connected components, and can be supposed to be split. – YCor May 16 '16 at 23:52
• @YCor: You write that $GL_n(\mathbf{Z})$ has finite subgroups of bounded order. Could you please give a reference? – Mikhail Borovoi May 21 '16 at 16:00
• @MikhailBorovoi: see mathoverflow.net/questions/71969, namely Robinson's answer and its comments: some congruence subgroup is torsion-free. So the index is an upper bound on the order of a finite subgroup. – YCor May 21 '16 at 21:39

The answer is YES. It suffices to assume that $H_1$ and $H_2$ are conjugate over $\mathbb{C}$ or, what is the same, that they are conjugate over $\overline{\mathbb{Q}}$.

Theorem 1. Let $G$ be a connected semisimple linear algebraic group defined over a number field $k$. 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 ${\bar{k}}$, then there exists a finite field extension $K=K(H_1,H_2)$ of $k$ of degree $[K:k]\le d$ such that $H_1$ and $H_2$ are conjugate over $K$.

EDIT: This resut is valid over any perfect field, see the end of the answer.

Proof of Theorem 1. Choose an element $g\in G({\bar{k}})$ such that $g\cdot H_1\cdot g^{-1} =H_2$. Let $s\in\mathrm{Gal}({\bar{k}}/k)$, then $^s\!g\cdot H_1\cdot \,^s\!g^{-1}=H_2$. Set $c_s=g^{-1}\cdot \,^s\!g$, then $c_s\in N({\bar{k}})$, where $N$ is the normalizer of $H_1$ in $G$, and $c=(c_s)\in Z^1(k,N)$. Let $[c]\in H^1(k,N)$ denote the cohomology class of $c$.

If $[c]=1$, then there exists $n\in N({\bar{k}})$ such that $c_s=n^{-1}\cdot \,^s\!n$ for all $s\in\mathrm{Gal}({\bar{k}}/k)$. Set $g'=gn^{-1}$, then $g'\in G(k)$ and $g'\cdot H_1\cdot (g')^{-1} =H_2$. We see that if $[c]=1$, then $H_1$ 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, we see that Theorem 1 follows from the next Theorem 2.

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

Proof of Theorem 2. We argue by dévissage. If $N_1$ is a normal subgroup of $N$, and $N_2=N/N_1$, then one sees from the corresponding cohomology exact sequence that it suffices to prove the theorem for $N_1$ and $N_2$.

Let $N^0$ denote the identity component of $N$, then $N/N^0$ is a finite group. For a finite group the theorem is clearly true (over any perfect field, not necessary a number field), so it suffices to prove the theorem for $N^0$.

We see that we may assume that $N$ is connected (and reductive). Let $R=Z(N)^0$ denote the identity component of the center of $N$, then $R$ is a torus. A torus (over any perfect field) splits over a field extension of bounded degree, see YCor's comment, so we may assume that $R$ is split, and then $H^1(k,R)=1$ by Hilbert's Theorem 90. Therefore, it suffices to prove the theorem for the connected semisimple group $N/R$.

Now let $N$ be a connected semisimple group, $S$ the universal covering of $N$, then we have a short exact sequence $$1\to C\to S\to N\to 1$$ where $S$ is a simply connected semisimple group and $C$ is a finite central subgroup of $S$. We obtain a cohomology exact sequence $$H^1(k,S)\to H^1(k,N)\to H^2(k,C).$$

Now we shall use the assumption that $k$ is a number field. We may assume that $k$ has no real embeddings. Then by the Hasse principle for simply connected groups we have $H^1(k,S)=1$. Therefore, Theorem 2 follows from the next Proposition 3.

Proposition 3. Let $C$ be a finite abelian $k$-group over a number field $k$. Then there exists $d=d(C_{\bar{k}})$ such that any cohomology class $\eta\in H^2(k,C)$ can be killed by a finite field extension $K=K(C,\eta)$ of degree $[K:k]\le d$.

Proof of Proposition 3. We may assume that $C$ is constant (i.e, that $C({\bar{k}})=C(k)$. Further, we may assume that $C$ is cyclic. Further, we may assume that $C=\mu_n$ for some natural $n$. Then $H^2(K,\mu_n)=\mathrm{Br}(k)_n$ (the subgroup of elements of the Brauer group of $k$ killed by multiplication by $n$).

Let $b\in \mathrm{Br}(k)_n$. We may assume that $k$ has no real embeddings. We know from the class field theory that $b$ defines local invariants $\mathrm{inv}_v(b)\in\frac{1}{n}\mathbb{Z}/\mathbb{Z}$ for all finite places $v$ of $k$, and these local invariants are nonzero only for a finite set $\Xi$ of places. By weak approximation there exists a Galois extension $K/k$ of degree $n$ such that $K_v:=K\otimes_k k_v$ is a field for $v\in\Xi$, where $k_v$ is the completion of $k$ at $v$. Then the extension $K_v/k_v$ kills $\mathrm{inv}_v(b)$ for all $v\in\Xi$, hence the extension $K/k$ of degree $n$ kills $b$. This completes the proofs of Proposition 3, Theorem 2, and Theorem 1.

EDIT: As Vladimir Chernousov has explained me, theorems 1 and 2 are valid over any perfect field $k$, not necessarily a number field. It suffices to prove that if $N$ is a connected reductive $k$-group, then any cohomology class $\xi\in H^1(k,N)$ can be killed by a field extension of bounded degree. We may assume that $N$ is a split group. Then by Steinberg's theorem (Theorem 11.1 in "Regular elements of semisimple groups", republished in Serre's "Galois Cohomology"), $\xi$ comes from some torus $T\subset N$, and hence, can be killed by an extension of bounded degree.