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The quintic del Pezzo $3$-fold $V(5)$ of degree $5$ is defined as the intersection of $Grass(2,5)$ and a codimension $3$ linear subspace. I would like to show the following:

For every point $p$ in $V(5)$ there is a line contained in $V(5)$ not passing through $p$.

I know that $\operatorname{Gr}(2,5)$ is defined by $5$ quadrics in $\mathbb{P}^9$, thus we can view $V(5)$ as defined by $5$ quadrics in $\mathbb{P}^6$. Now given a point $p=[1,0,\dots,0]$, write the five quadrics as $$Q_i= q_{i,2}(x_1,\dots, x_n)+ x_0 q_{i,1}(x_1,\dots, x_n),$$ where $q_{i,k}$ is homogeneous of degree $k$. There is a line contained in $V(5)$ passing through $p$ if and only if the $\{q_{i,2},q_{i,1}: i=1,2,3,4,5\}$ have a common zero in $\mathbb{P}^5$. Indeed, for a common zero $[a_0,a_1,\dots,a_n]$, the line joining $p$ and $[a_0,a_1,\dots,a_n]$ is contained in $V(5)$. Since the quadrics are linearly independent, the dimension of the subset defined by these $10$ equations is $\leq0$ which means there are only finitely many lines in $V(5)$ passing through a given point. I'm thinking whether the number of lines on $V(5)$ is finite or infinite so that I can compare the number of lines to get a line not passing through $p$. How to determine the number of lines on $V(5)$? Or is there a better approach to this question?

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    $\begingroup$ Just an extra comment: the five quadrics you are mentioning are way non general by obvious reasons. They are given by the 4-Pfaffians of the 5x5 skew matrix with the linear form $x_{i,j}$ at the $(i,j)$ entry. This is not directly connected to the question (which Sasha already solved), but might be relevant if you keep working with this particular variety. $\endgroup$ – Enrico Mar 8 at 22:13
  • $\begingroup$ @Enrico $V(5)$ can be obtained as the intersection of $Gr(2,5)$ with $3$ hyperplanes in general position. Which hyperplanes should they be? $\endgroup$ – L. Xie Mar 14 at 13:21
  • $\begingroup$ @L.X in this case you should be safe by picking three of $x_{i,j}+x_{k,l}$ (with $i \neq j \neq k \neq l$, all five indexes involved and appropriate different coefficients). $\endgroup$ – Enrico Mar 14 at 13:43
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    $\begingroup$ @L.X by the way, this might help mathoverflow.net/questions/323095/… $\endgroup$ – Enrico Mar 14 at 14:31
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    $\begingroup$ @Enrico Thanks, if I take $x_{12}=x_{34}$, $x_{13}=x_{45}$, $x_{23}=x_{15}$, then by a calculation I get two nonintersecting lines: one joining $( x_{14}=1,x_{ij}=0)$ and $( x_{12}=1,x_{ij}=0)$, the other joining $( x_{24}=1,x_{ij}=0)$ and $( x_{25}=1,x_{ij}=0)$. This should solve the question. $\endgroup$ – L. Xie Mar 14 at 16:32
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Let $X$ be the quintic del Pezzo 3-fold. The Hilbert scheme of lines on $X$ is 2-dimensional (this follows easily from deformation theory). If there is a point $p \in X$ such that every line passes through $p$, then $X$ is a cone with vertex $p$, hence either $X$ is singular at $p$, or $X \cong \mathbb{P}^3$. But, of course, neither of these is true.

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  • $\begingroup$ I don't know what a Hilbert scheme is. Would you mind giving me some references? $\endgroup$ – anon Mar 8 at 23:46
  • $\begingroup$ Is there a way to know the number of lines on $V(5)$? $\endgroup$ – anon Mar 8 at 23:47
  • $\begingroup$ What do you mean? It is infinite of course, read Sasha's answer. $\endgroup$ – abx Mar 9 at 5:24
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    $\begingroup$ Yes. For instance a general hyperplane section contains some lines. $\endgroup$ – abx Mar 9 at 10:39
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    $\begingroup$ A general hyperplane section is a quintic del Pezzo surface. It is well known that it contains 10 lines. $\endgroup$ – abx Mar 9 at 16:16
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I give an answer here using only elementary algebraic geometry.

We know that the Grassmannian $\operatorname{Gr}(2,5)$ is defined by $5$ linearly independent quadrics in $\mathbb{P}^9$ and $V(5)$ can be viewed as defined by $5$ linearly independent quadrics in $\mathbb{P}^6$.

Given a point $p=[1,0,\dots,0]\in X$ for $X\subset \mathbb{P}^6$ defined by $Q_i, 1\leq i\leq 5$, where $Q_i$ are homogeneous of degree $2$. Write $$Q_i= q_{i,2}(x_1,\dots, x_6)+ x_0 q_{i,1}(x_1,\dots, x_6)$$ with $q_{i,k}$ homogeneous of degree $k$.

I claim that there there is a line contained in $X$ passing through $p$ if and only if $$q_{i,k}=0, \quad 1\leq i \leq 5, k=1,2 \quad (*)$$ have a common zero in $\mathbb{P}^{5}=\mathbb{P}(k[x_1,\dots x_6])$. To see this, note that given a common zero $[a_1,a_2,\dots,a_6]\in\mathbb{P}^{5}$, the line joining $p$ to $[0, a_1,a_2,\dots,a_6]$ is contained in $X$ and the line joining $[1,0,\dots,0]$ and $[1,-a_1,\dots,-a_n]$ is the same as the line joining $[1,0,\dots,0]$ and $[0, a_1,\dots,a_n]$.

Since $Q_i$ are linearly independent, we see that the linear part $q_{i,1}, 1\leq i\leq 5$ are linearly independent. By the geometric version of Krull’s Principal Ideal Theorem[11.3.2, Vakil’s notes], the dimension of the subvariety in $\mathbb{P}^5$ defined by $(*)$ is $\leq 0$. This means $(*)$ has finitely many common zeros, that is, there are at most finitely many lines passing through a given point.

Now back to $X=V(5)$, as suggested by Enrico in the comment. We take $3$ hyperplanes ''general enough'' to get explicit defining equations of $V(5)$. Consider the Plucker coordinate $w_{ij}$ in $\mathbb{P}^9$, take $w_{12}=w_{34}$, $w_{13}=w_{45}$, $w_{23}=w_{15}$, the defining equations for $V(5)$ are

\begin{cases} w_{34}^2-w_{45}w_{24}+w_{14}w_{15}=0\\ w_{45}^2-w_{14}w_{35}+w_{15}w_{34}=0\\ w_{15}^2-w_{45}w_{25}+w_{34}w_{35}=0\\ w_{15}w_{45}-w_{24}w_{35}+w_{25}w_{34}=0\\ w_{34}w_{45}-w_{14}w_{25}+w_{15}w_{24}=0 \end{cases}

By the above argument, choose $p$ to be the point with the only nonzero entry $w_{ij}$ for $\{i,j\}\neq \{3,4\},\{4,5\},\{1,5\}$ we can have some lines. For example, we can find two nonintersecting lines: one joining $[ x_{14}=1,x_{ij}=0]$ and $[x_{12}=1,x_{ij}=0]$, the other joining $[ x_{24}=1,x_{ij}=0]$ and $[ x_{25}=1,x_{ij}=0]$. This has solved the question.


Remark. Suppose we are given the fact that for every point, there is at least one line passing through it, we can show that the subvariety in $\operatorname{Gr}(2,7)$ characterising lines on $V(5)$ has dimension $ 2$. Consider the following incidence correspondence, as we have discussed above, for any $p\in V(5)$ the fibre $\operatorname{pr}_1^{-1} (p)$ has dimension $0$.

enter image description here

By Upper Semicontinuity of Fibre Dimension, we have $\operatorname{dim} \Phi= \operatorname{dim} V(5)=3$. Take a line $l\in V(5)$, $\operatorname{dim}pr_2^{-1}([l])=1$, by Upper Semicontinuity again, $\operatorname{dim}\Phi -\operatorname{dim} pr_2\Phi = 1$, $\operatorname{dim} pr_2\Phi = 2$. We conclude that the subvariety of $\operatorname{Grass}(2,7)$ characterising lines on $V(5)$ has dimension $ 2$.

A stronger result than $\operatorname{dim }pr_2\Phi= 2$ is that the Hilbert scheme of lines on $V(5)$ is isomorphic to $\mathbb{P}^2$ as in Sasha's answer and the reference is [Theorem 1, 2]. And the number of lines passing through a point is $3$ when counted with multiplicity [P.26, 1].

$[1]$ Takagi, Hiromichi; Zucconi, Francesco, Geometries of lines and conics on the quintic del Pezzo 3-fold and its application to varieties of power sums, Mich. Math. J. 61, No. 1, 19-62 (2012). ZBL1262.14048.

$[2]$Furushima, Mikio; Nakayama, Noboru, The family of lines on the Fano threefold (V_ 5), Nagoya Math. J. 116, 111-122 (1989). ZBL0731.14025.

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