Timeline for Twisted forms with real points of a real Grassmannian
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 16, 2023 at 12:20 | vote | accept | Mikhail Borovoi | ||
| Aug 12, 2023 at 12:42 | history | edited | Friedrich Knop | CC BY-SA 4.0 |
added 77 characters in body
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| Aug 11, 2023 at 15:45 | comment | added | Friedrich Knop | 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 $G_{\mathbb C}=({\rm Aut}(X)^0)_{\mathbb C}=({\rm Aut}(X)_{\mathbb C})^0={\rm PGL}(n,\mathbb C)$. | |
| Aug 11, 2023 at 14:06 | comment | added | Mikhail Borovoi | 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. | |
| Aug 11, 2023 at 14:02 | comment | added | Mikhail Borovoi | 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 $I_n={\rm diag}(1,\dots,1)$. | |
| Aug 11, 2023 at 13:48 | comment | added | Mikhail Borovoi | 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 ${\rm Aut}(X_{\Bbb C})$. | |
| Aug 11, 2023 at 12:59 | history | answered | Friedrich Knop | CC BY-SA 4.0 |