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My problem

Let $Z$ be a projective variety. I have to prove that $\Lambda = \{ (\text{Pl}(L), p) \in \text{Pl}(\text{Gr}(k,n))) \times Z \ | \ p \in L\}$ is a projective variety. Here $\text{Gr}(k,n)$ is the Grassmannian and $\text{Pl}: \text{Gr}(k,n) \rightarrow \mathbb{P}^{\binom{n}{k} -1}$ the plucker embedding.

My (flawed) attempt:

I know that $\text{PL}(\text{Gr}(k,n))$ is a variety. So it is described by finitely many polynomials $\{ f_i \}$. Since $Z$ is also a projective variety, it is described by $\{ g_j \}$. We can think of the Grassmanian set of all $(k-1)$ -dimensional linear subvarieties of the projective space $\mathbb{P}^{n-1}$. Thus the Plucker coordinates correspond to the minors of a $k \times n$ matrix. Take some vector space $X \in \text{Gr}(k,n)$, spanned by vectors $x_1, \ldots, x_k$. Now $y \in X$ iff it can be written as linear combination of $x_1, \ldots, x_k$. Then all $k+1$- minors of the matrix with rows $x_1, \ldots, x_k,y$ are zero.

I got some inspiration from this other question, however it's hard to see what applies to my problem and what doens't. Incidence correspondence of Grassmannian is a projective variety

My question

Any hints on how to continue are welcome. If I made some mistakes, please let me know. Thanks!

Oopsilon
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  • What do you mean by $p \in L$? The way you presented the problem there is no link between $Z$ and the Grassmannian. – Matthias Jan 23 '24 at 11:38
  • Notation should imply that $p$ also lies in $Z$ – Oopsilon Jan 23 '24 at 13:13
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    Something is missing here, you say $p$ is a point of a projective variety $Z$, so how am I supposed to interpret $p$ as an element of $L$? – Matthias Jan 23 '24 at 13:22
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    I think it's just bad notation. I'm interpreting this as $$\Lambda = {(\operatorname{Pl}(L),p)\in\operatorname{Pl}(\operatorname{Gr}(k,n)))\times Z \mid p \in L \cap Z }$$ – soggycornflakes Jan 24 '24 at 12:29
  • @Matthias the grassmanian $\operatorname{Gr}(k,n)$ can be considered as a set of linear subspaces of $\Bbb{P}^n$ – soggycornflakes Jan 24 '24 at 12:30
  • @soggycornflakes the issue is that as stated, $Z$ is not even a sub-variety of $\mathbb P^n$, so $L\cap Z$ does not make sense. – Kenta S Jan 25 '24 at 07:56
  • I think it's safe to assume that "$Z$ be a projective variety" implies "$Z \subset \Bbb{P}^n$" – soggycornflakes Jan 25 '24 at 13:08
  • @soggycornflakes Thanks! I was also confused with the notation, that's one of the reason why I posted it. Can I use the other problem I linked to solve this excercise or not? If so, could you help me get further? – Oopsilon Jan 26 '24 at 12:23

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