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Let $X={\rm Gr}_{n,k,{\Bbb R}}$ denote the Grassmannian of $k$-dimensional subspaces in ${\Bbb R}^n$. We regard $X$ as an ${\Bbb R}$-variety with the set of complex points $X({\Bbb C})={\rm Gr}_{n,k,{\Bbb C}}$. We consider the twisted $\Bbb R$-forms $_c X$ of $X$ where $c\in {\rm Aut\,} X_{\Bbb C}$ is a $1$-cocycle. They have the following property: $({}_c X)\times_{\Bbb R}{\Bbb C}\simeq X\times_{\Bbb R}{\Bbb C}$.

Question 1. What are the twisted $\Bbb R$-forms $_c X$ of $X$ having ${\Bbb R}$-points?

Question 2. What are references for the version of Question 1 over an arbitrary field rather than over ${\Bbb R}$?

Concerning Question 1, I have found the following twisted forms $_c X$, for which I write the set of ${\Bbb R}$-points $({}_c X)({\Bbb R})$.

  1. ${\rm Gr}_{n,k,{\Bbb R}}.$

  2. For $n=2n',\ k=2k'$, the quaternionic Grassmannian ${\rm Gr}_{n',k',{\Bbb H}}$ of $k'$-dimensional subspaces in ${\Bbb H}^{n'}$, where ${\Bbb H}$ denotes the division algebra of Hamilton's quaternions.

  3. For $n=2k$, the isotropic Hermitian Grassmannian whose ${\Bbb R}$-points are the $k$-dimensional subspaces $W\subset{\Bbb C}^n={\Bbb C}^{2k}$ that are totally isotropic with respect to the Hermitian form $\mathcal H$ with matrix ${\rm diag}(-1,\dots,-1,+1,\dots, +1)$ where $-1$ appears $k$ times and also $+1$ appears $k$ times. Here a subspace $W\subset {\Bbb C}^{2k}$ is called totally isotropic if the restriction of $\mathcal H$ to $W$ is zero.

Question 3. Do these exhaust all twisted real forms of $X$ having real points?

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2 Answers 2

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Updated

Let $G:={\rm Aut}(X)^0$. Then it is well-known that $$G_{\mathbb C}= ({\rm Aut}(X)^0)_{\mathbb C}= ({\rm Aut}(X)_{\mathbb C})^0= {\rm Aut}(X_{\mathbb C})^0\cong {\rm PGL}(n,\mathbb C).$$ So $G$ is a real form of ${\rm PGL}(n,\mathbb C)$ which means that $G$ is isomorphic to ${\rm PGL}(n,\mathbb R)$, ${\rm PGL}(n/2,\mathbb H)$, or ${\rm PU}(p,n-p)$. On the other hand, $X$ has a real point $x$ which means $X\cong G/P$ with $P=G_x$. It follows from $G_{\mathbb C}/P_{\mathbb C}\cong X_{\mathbb C}\cong {\rm Gr}_{n,k,\mathbb C}$ that $P_{\mathbb C}$ is a maximal parabolic which is defined over $\mathbb R$. Looking at the Satake diagram of $G$ it corresponds therefore to a non-compact self-conjugate simple root $\alpha_k$. This gives the following possibilities $$ G={\rm PGL}(n,\mathbb R), k=1,\ldots,n-1\quad\Rightarrow\quad\text{your first case} $$ $$ G={\rm PGL}(n/2,\mathbb H), k=2,4\ldots,n-2\quad\Rightarrow\quad\text{your second case} $$ $$ G={\rm PU}(p,n-p), k=p=n/2\quad\Rightarrow\quad\text{your third case} $$ This argument should generalize to arbitrary fields. The first two cases combine to $X=$ space of $k$-dimensional subspaces of $D^n$ where $D$ is a central simple algebra.

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  • $\begingroup$ Dear Friedrich, many thanks! However, it is not quite correct that ${\rm Aut}(X_{\Bbb C})\cong {\rm PGL}(n,{\Bbb C})$. When $n=2k$, the connected group $ {\rm PGL}(n,{\Bbb C})$ is a subgroup of index 2 in ${\rm Aut}(X_{\Bbb C})$. $\endgroup$ Commented Aug 11, 2023 at 13:48
  • $\begingroup$ The other connected component of the automorphism group contains the following automorphism: $\phi_B\colon W\mapsto W^{\bot B}$. Here $W$ is a $k$-dimensional subspace of $V={\Bbb C}^n$, and $W^{\bot B}$ is the annihilator of $W$ in $V$ with respect to a non-degenerate symmetric bilinear form $B$ on $V$. For example, we can take the form $B=B_0$ with matrix $I_n={\rm diag}(1,\dots,1)$. $\endgroup$ Commented Aug 11, 2023 at 14:02
  • $\begingroup$ If instead of $B_0$ we choose the symmetric bilinear form $B_1$ with matrix ${\rm diag}(-1,\dots,-1,+1,\dots, +1)$ where $-1$ appears $k$ times and also $+1$ appears $k$ times, then, I think, $\phi_{B_1}$ is a cocycle giving my third case. $\endgroup$ Commented Aug 11, 2023 at 14:06
  • $\begingroup$ You are right, I overlooked the graph automorphism when $n=2k$. But still, if one defines $G:={\rm Aut}(X)^0$ then the argument should go through because of $G_{\mathbb C}=({\rm Aut}(X)^0)_{\mathbb C}=({\rm Aut}(X)_{\mathbb C})^0={\rm PGL}(n,\mathbb C)$. $\endgroup$ Commented Aug 11, 2023 at 15:45
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Let $X$ be a smooth projective variety over a field $K$ of characteristic zero such that $$ X_{\bar{K}} \cong \mathrm{Gr}(k,n)_{\bar{K}} $$ and $X(K) \ne \varnothing$.

First, consider the case $n \ne 2k$. Let $r = \gcd(k,n)$. Then there is an $r$-torsion Brauer class $\beta \in \mathrm{Br}(K)$ representable by a Severi--Brauer variety of dimension $k - 1$ and a Severi--Brauer variety $Y$ of dimension $n - 1$ with class $\beta$ such that $$ X \cong \mathrm{Hilb}_{\mathrm{P}^{k-1}}(Y) $$ is the component of the Hilbert scheme that parameterizes subschemes with the same Hilbert polynomial as $\mathrm{P}^{k-1}$. Let us call such $X$ a Severi--Brauer Grassmannian of class $\beta$.

Now consider the case $n = 2k$. Then either $X$ is a Severi--Brauer Grassmannian as before, or there is a quadratic extension $K'/K$ such that $X_{K'}$ is a Severi--Brauer Grassmannian.

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  • $\begingroup$ Dear Sasha, I do believe that the Severi-Brauer Grassmannian of class $\beta$ has a $K$-point. However, I don't believe that any variety $X$ over $K$ such that $X_{K'}$ is a Severi-Brauer Grassmannian, has a $K$-point. Could you please comment? $\endgroup$ Commented Aug 10, 2023 at 16:18
  • $\begingroup$ @MikhailBorovoi: you are absolutely right, I forgot the condition that $\beta$ should be the class of a Severi--Brauer variety of dimension $k - 1$. I added the assumption to the answer. $\endgroup$
    – Sasha
    Commented Aug 10, 2023 at 19:31
  • $\begingroup$ Thank you, but I meant the case $n=2k\,$. I don't believe that any variety $X$ over $K$ such that there is a quadratic extension $K'/K$ for which $X_{K'}$ is a Severi-Brauer Grassmannian, has a $K$-point. Here you probably have to mention a Hermitian form of maximal possible index.... $\endgroup$ Commented Aug 10, 2023 at 19:58
  • $\begingroup$ @MikhailBorovoi: Yes, right, the existence of a $K$-point in this case may be tricky. But it is quite plausible that it can be phrased in terms of a Hermitian form; I will think about that. $\endgroup$
    – Sasha
    Commented Aug 11, 2023 at 6:29

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