The question is complementary to this question. How to construct pointless real forms of Grassmannians $\operatorname{Gr}(k,n)$? For which $k$ and $n$ do they exist? Any reference will be very helpful.
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1$\begingroup$ They exist when $n$ is even and $k$ is odd. Then the identitity component of the automorphism group is quaternionic. I will type details tomorrow. $\endgroup$Mikhail Borovoi– Mikhail Borovoi2025-04-28 10:39:54 +00:00Commented Apr 28 at 10:39
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3$\begingroup$ I'm sure@MikhailBorovoi will say it far more clearly than I can, but in the mean time: if $n=2m$, consider the space $V_{k,n}$ of left ideals of dim $kn$ of the central simple algebra $A := \mathbb{M}_m(\mathbb{H})$ (so $A\otimes_{\mathbb{R}}\mathbb{C} \cong \mathbb{M}_n(\mathbb{C})$): its complex points are the Grassmannian of $\mathbb{C}^k$ in $\mathbb{C}^n$, and for $k$ odd this has no real points. Equivalently, this is the space of linear subspaces of dimension $k-1$ of the Severi-Brauer variety associated to $A$. (I don't know if there are other twisted forms than these, however.) $\endgroup$Gro-Tsen– Gro-Tsen2025-04-28 13:59:24 +00:00Commented Apr 28 at 13:59
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$\newcommand{\R}{{\mathbb R}} \newcommand{\C}{{\mathbb C}} \newcommand{\cH}{{\mathcal H}} \DeclareMathOperator\IGr{IGr} \DeclareMathOperator\Gr{Gr} $(Too long for a comment.) In addition to the quaternionic Grassmannian described by Gro-Tsen, I will describe the isotropic Grassmannian $\IGr(k,n,p)$ where $n=2k$ and $0\le p\le k$.
Write $c_p=(-1,\dots,-1,1,\dots,1)$ ($p$ times $-1$ and $n-p$ times $+1$). Consider the Hermitian form $\cH_p$ on $V=\C^n$ given by the Hermitian matrix $c_p$.
We consider the isotropic Grassmannian $\IGr(k,n,p)$ whose $\R$-points are the $k$-dimensional subspaces $W\subset V$ that are totally isotropic with respect to $\cH_p$ (that is, the restriction of $\cH_p$ to $W$ is identically 0). I leave to the reader to describe the set of $\C$-points of $\IGr(n,k,p)$. Then $\IGr(k,n,p)$ is a twisted form of the usual Grassmannian $\Gr_\R(k,n)$; see [1, starting from A.16] (but this appendix is not well written…). Clearly, our isotropic Grassmannian $\IGr(k,n,p)$ has $\R$-points if and only if $p=k$. I think that the usual Grassmannian $\Gr_\R(k,n)$, Gro-Tsen's quaternionic Grassmannian (for even $n$), and our isotropic Grassmannian (for $n=2k$) exhaust all twisted forms of $\Gr_\R(k,n)$.
[1] Elizaveta Vishnyakova, with an appendix by Mikhail Borovoi, Automorphisms and real structures for a $\Pi$-symmetric super-Grassmannian, J. Algebra 644 (2024), 232–286.
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$\begingroup$ I didn't edit in in case you know it and don't like it, but, if you don't already, then you might like to know the
\underbraceconstruction: $c_p = (\underbrace{-1, \dotsc, -1}_\text{$p$ times}, \underbrace{1, \dotsc, 1}_\text{$n - p$ times})$c_p = (\underbrace{-1, \dotsc, -1}_\text{$p$ times}, \underbrace{1, \dotsc, 1}_\text{$n - p$ times}). $\endgroup$LSpice– LSpice2025-04-29 10:41:23 +00:00Commented Apr 29 at 10:41 -
$\begingroup$ @LSpice: I did not know it. Anyway, the current version is acceptable; please don't edit it. $\endgroup$Mikhail Borovoi– Mikhail Borovoi2025-04-29 10:50:10 +00:00Commented Apr 29 at 10:50