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Let $\mathbf{G}$ be the image of the natural embedding of $\operatorname{SL}_4(\mathbb{R})$ inside $\operatorname{SL}_4(\mathbb{C})\subset \operatorname{SL}_8(\mathbb{R})$. Then $\mathbf{G}$ is an algebraic group defined over $\mathbb{Q}$ with $\mathbf{G}(\mathbb{Q})$ being $\mathbb{Q}$-isotropic.

My question: is it possible to find $g\in \operatorname{SL}_8(\mathbb{R})$ such that $g\mathbf{G}g^{-1}$ is $\mathbb{Q}$-algebraic and $\mathbb{Q}$-anisotropic?

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    $\begingroup$ Do you know of an example of this happening, e.g. for $SL_2$? $\endgroup$
    – Vincent
    Commented Oct 23, 2023 at 16:03
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    $\begingroup$ The "natural embedding" is $A\mapsto \begin{pmatrix} A & 0\\ 0 & I_4\end{pmatrix}$ or $A\mapsto \begin{pmatrix} A & 0\\ 0 & A\end{pmatrix}$? $\endgroup$
    – YCor
    Commented Oct 23, 2023 at 16:15
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    $\begingroup$ @Vincent Yes for $\operatorname{SL}_2$ this happens. It can be obtained using a quaternionic central division algebra over $\mathbb{Q}$. $\endgroup$
    – user515519
    Commented Oct 23, 2023 at 16:28
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    $\begingroup$ @Vincent I think that the difference is that we allow conjugacy by $\operatorname{SL}_8(\mathbb{R})$, instead of just $\operatorname{SL}_8(\mathbb{Q})$, such that the resulting group is $\mathbb{Q}$-algebraic. This yields, in principle, different isomorphism classes over $\mathbb{Q}$. $\endgroup$
    – user515519
    Commented Oct 23, 2023 at 16:46
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    $\begingroup$ @Vincent Consider rational quadratic forms on $\mathbb{R}^3$ of signature $2+1$. Their orthogonal groups are $\mathbb{Q}$-algebraic and conjugate over $\mathbb{R}$ but not over $\mathbb{Q}$ in general as some of them are $\mathbb{Q}$-isotropic and some are not. This also happens for $\operatorname{SL}_2$ acting on $2\times 2$ matrices i.e there are $\mathbb{Q}$-isotropic $\mathbb{Q}$-forms conjugate in $\operatorname{SL}_4(\mathbb{R})$ to anisotropic ones. $\endgroup$
    – user515519
    Commented Oct 24, 2023 at 10:18

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$\newcommand{\nN}{{\mathcal N}} \newcommand{\SL}{{\rm SL}} \newcommand{\G}{{\bf G}} \newcommand{\Q}{{\Bbb Q}} \newcommand{\R}{{\Bbb R}} \newcommand{\C}{{\Bbb C}} $The answer is No.

Write $\nN$ for the normalizer of $\G$ in $\SL_{8,\Q}$. We wish to find an anisotropic $\Q$-subgroup $\G'\in\SL_{8,\Q}$ such that $\G'_\R$ is conjugate to $\G_\R$ under $\SL(8,\R)$. Then $\G'\simeq{} _c \G$ for some 1-cocycle $c\in Z^1(\Q,\nN)$ such that $c$ becomes a coboundary in $ Z^1(\R,\nN)$. Since $\nN$ acts on $\G$ by inner automorphisms (see a comment of @YCor above), we see that $\G'$ must be an anisotropic inner form of $\G$ with an 8-dimensional representation defined over $\Q$.

Any inner form $\G'$ of $\G=\SL_{4,\Q}$ is isomorphic to $\SL(1,D)$ where $D$ is a central simple algebra of degree 4 over $\Q$. Since $\G'$ is anisotropic, $D$ is a division algebra. However, a representation $\rho$ of $\G'=\SL(1,D)$ over $\Q$ such that $\rho\otimes_\Q\C$ contains the standard 4-dimensional representation of $\G'_\C\simeq\SL_{4,\C}$ in $\C^4$, has dimension at least 16. Thus there is no anisotropic $\Q$-subgroup $\G'\subset\SL_{8,\Q}$ conjugate to $\G$ over $\R$ or over $\C$.

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