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Recall the Grassmanian $G(2,4)$ is cut out in $\mathbb{P}^5$ by the equation $a_{12}a_{34}-a_{13}a_{24}+a_{14}a_{23}=0$.

Let $X$ be the subset of $G(2,4)\times G(2,4)\subset\mathbb{P}^5\times\mathbb{P}^5$ which is the complement of $V((a_{12}b_{34}-a_{13}b_{24}+a_{14}b_{23})+(b_{12}a_{34}-b_{13}a_{24}+b_{14}a_{23}))$, where the $a$ and $b$ are coordinates on the first and second $\mathbb{P}^5$ factors respectively.

How to show $X$ is affine?

My feeling is this can probably done in a coordinate-free way, too?

Thank you.

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  • $\begingroup$ This equation must come from somewhere, right? $\endgroup$
    – Hoot
    Sep 3, 2016 at 9:16
  • $\begingroup$ I'm just thinking about a problem. Ofc this is not a textbook homework or anything because, now you know, it's not hard. $\endgroup$
    – HLC
    Sep 3, 2016 at 15:00
  • $\begingroup$ I just mean that there must be some geometric reason and I am too dim to see it. $\endgroup$
    – Hoot
    Sep 3, 2016 at 20:01

1 Answer 1

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It has nothing to do with the Grassmanian. Consider the Segre embedding

$$\begin{split} \mathbb P^5\times \mathbb P^5 &\to \mathbb P^{35}, \\([X_0, \cdots X_5],[Y_0, \cdots, Y_5])&\mapsto [X_0Y_0, X_0Y_1, \cdots, X_iY_j\cdots X_5Y_5] \end{split}$$

then $(a_{12}b_{34}-a_{13}b_{24}+a_{14}b_{23})+(b_{12}a_{34}-b_{13}a_{24}+b_{14}a_{23})$ is a pullback of a linear function on $\mathbb P^{35}$. Thus under the embedding, $\mathbb P^5\times \mathbb P^5\setminus V$ lies in an affine space of $\mathbb P^{35}$, thus is affine. Thus $G(2, 4)\times G(2, 4)\setminus V$ is also affine.

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