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Let $A_n$ denote the vector space of $n\times n$ antisymmetric matrices over ${\mathbb{Q}}$, where $n$ is even. Let $A_n^*\subset A_n$ denote the affine ${\mathbb{Q}}$-subvariety of invertible antisymmetric matrices. Since $n$ is even, $A_n^*$ is non-empty (indeed, if $a_2\in A_2(\mathbb{C})$, $a_2\neq 0$, then $\mathrm{diag}(a_2,a_2,\dots,a_2)\in A_n(\mathbb{C})$ is invertible). For $a\in A_n^*({\mathbb{C}})$ we have $a^{-1}\in A_n^*({\mathbb{C}})$. We obtain an involutive automorphism defined over $\mathbb{Q}$ $$ s\colon A_n^*\to A_n^*,\quad a\mapsto a^{-1}.$$ Let $X_n=A_n^*/\langle s\rangle$ denote the quotient variety, it has dimension $n(n-1)/2$.

It is known that the $\mathbb{Q}$-variety $X_n$ is ${\mathbb{C}}$-rational. It is known also that $X_n$ is stably ${\mathbb{Q}}$-rational, hence it is stably ${\mathbb{R}}$-rational.

Question. Is $X_n$ ${\mathbb{R}}$-rational?

(The case $n=2$ is trivial. It is known that $X_4$ is ${\mathbb{R}}$-rational.)

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  • $\begingroup$ Could you explain why $X_n$ is $\Bbb{C}$-rational (or give a reference)? $\endgroup$
    – abx
    Commented May 2, 2016 at 12:08
  • $\begingroup$ For the ignorant: what is known about the similar problem with symmetric matrices (resp. all square matrices)? $\endgroup$
    – Laurent Moret-Bailly
    Commented May 2, 2016 at 12:19
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    $\begingroup$ @EhudMeir: You are right, it seems that I don't need $n$ to be even to define $X_n$. However, when $n$ is even, $X_n$ is $\mathbb{Q}$-birationally isomorphic to $\mathrm{PSO}(n)$, and therefore I know that $X_n$ is $\mathbb{C}$-rational, that it is stably $\mathbb{Q}$-rational (by a theorem of Chernousov, 1994), and that conjecturally it is $\mathbb{R}$-rational (by a conjecture of Platonov). When $n$ is odd, I know nothing.... $\endgroup$
    – Mikhail Borovoi
    Commented May 2, 2016 at 15:55
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    $\begingroup$ For odd $n$, isn't $A_n^*$ empty? $\endgroup$
    – Laurent Moret-Bailly
    Commented May 2, 2016 at 16:29
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    $\begingroup$ @LaurentMoret-Bailly: Yes, it is empty! Indeed, $\mathrm{det}(a^t)=\mathrm{det}(a)$, but $\mathrm{det}(-a)=(-1)^n\mathrm{det}(a)$. Therefore, if $n$ is odd and $a^t=-a$, then $\mathrm{det}(a)=0$ and $a$ cannot be invertible. $\endgroup$
    – Mikhail Borovoi
    Commented May 2, 2016 at 16:54

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