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I am confused about the following question.

Consider $\mathbb C^4$ endowed with nondegenerate symmetric bilinear form $J:=\left(\begin{matrix}0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{matrix}\right)$. Let $X:=OG(2,4)$ denote the orthogonal grassmannian of isotropic 2-planes in $\mathbb C^4$. We have a matrix representation for it by nondegenerate $2\times 4$ matrix $A:=\left(\begin{matrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\end{matrix}\right)$, which satisfies $AJA^t=0.$ If $\left(\begin{matrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{matrix}\right)$ is nondegenerate then we can take a zariski open set of $X$ as $\left(\begin{matrix}1&0&a&0\\0&1&0&-a\end{matrix}\right)$. If $\left(\begin{matrix}a_{11}&a_{13}\\a_{21}&a_{23}\end{matrix}\right)$ is nondegenerate then we have another Zariski open set of $X$ as $\left(\begin{matrix}1&a&0&0\\0&0&1&-a\end{matrix}\right)$.
But then this two sets are disjoint, which is impossible. I think I must make a mistake but where it is?

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  • $\begingroup$ The sets of matrices may be disjoint, but the map from matrices to the Grassmannian is not injective. $\endgroup$
    – Qiaochu Yuan
    Commented Apr 14, 2016 at 20:21
  • $\begingroup$ Could you show me why? for instance which two matrices are mapped to a same point in Grassmannian? In my opinion the equivalence classes are taken under left multiplication, then their image in Grassmanian are different. thanks! $\endgroup$
    – user42804
    Commented Apr 14, 2016 at 20:28

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The orthogonal Grassmannian $OG(2,4)$ has two connected components, and the two Zariski opens you describe are in different components. The same happens to $OG(n,2n)$ for any $n$.

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  • $\begingroup$ But how to tell which component it is in? for example can we check the positivity of the determinants? $\endgroup$
    – user42804
    Commented Apr 14, 2016 at 20:33
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    $\begingroup$ No (over $\mathbb{C}$ positivity makes no sense). The trick is to fix one isotropic space, say $A_0$, then the components are distinguished by the parity of the dimension of intersection with $A_0$. $\endgroup$
    – Sasha
    Commented Apr 14, 2016 at 20:35

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