Consider the complex projective space $\mathbb{P}^4$ and the Grassmannian $\mathbb{G}(1:\mathbb{P}^4)$ of lines in it, seen as a projective manifold through plucker embedding. Take $l_1,l_2\subset \mathbb{P}^4$ lines such that $l_x$ and $l_y$ span a projective space $\mathbb{P}^3_{x,y}$. In
https://arxiv.org/pdf/1308.2800.pdf
page $5$, first line, it is said that, for any conic $c\subset \mathbb{G}(1:\mathbb{P}^4)$ that contains the points corresponding to $l_x$ and $l_y$, then $c$ lies in the Plucker quadric $\mathbb{G}(1:\mathbb{P}^3_{1,2})$.
How can I prove it?
I can see from the following lines in the paper that we should use the fact that $c$ and $\mathbb{G}(1:\mathbb{P}^3_{1,2})$ are both quadrics, since by imposing that $c$ lies on some generic $\mathbb{P}^6\subset \mathbb{P}^9$ one eventually find $c=\mathbb{P}^6\cap \mathbb{G}(1:\mathbb{P}^3_{1,2})$ without further utilising the degree of $c$.
First, note that any conic lies on some $Gr(2,4)$. Indeed, let $\mathcal{U} \subset V \otimes \mathcal{O}$ and $\mathcal{U}^\perp \subset V^\vee \otimes \mathcal{O}$ be the rank-2 and rank-3 tautological bundles on the Grassmannian $Gr(2,V)$, where $V$ is 5-dimensional. Restricting to $c$ we obtain an embedding $$ \mathcal{U}^\perp\vert_c \subset V^\vee \otimes \mathcal{O}_c. $$ Since the rank of $\mathcal{U}^\perp\vert_c$ is 3 and the degree is $-2$ (because $c$ is a conic) and it is a subbundle of the trivial bundle, it follows that $$ \mathcal{U}^\perp\vert_c \cong \mathcal{O}_c \oplus \mathcal{O}_c(-1) \oplus \mathcal{O}_c(-1), $$ or $$ \mathcal{U}^\perp\vert_c \cong \mathcal{O}_c \oplus \mathcal{O}_c \oplus \mathcal{O}_c(-2). $$ In particular, $$ 0 \ne H^0(\mathcal{U}^\perp\vert_c) \subset H^0(c,V^\vee \otimes \mathcal{O}) = V^\vee. $$ A non-zero hyperplane section thus corresponds to a function $f \in V^\vee$. Moreover, it follows that the corresponding section of $\mathcal{U}^\vee$ vanishes on $c$, hence $c$ is contained in the zero locus of $f$, which is equal to $Gr(2,V_f) \subset Gr(2,V)$, where $V_f \subset V$ is the hyperplane given by $f$.
Now, note that if $U_1 \subset V$ and $U_2 \subset V$ are the 2-dimensional subspaces corresponding to two points of $c$, then $U_1 \subset V_f$ and $U_2 \subset V_f$, hence $U_1 + U_2 \subset V_f$. If $U_1 \cap U_2 = 0$, it follows that $V_f = U_1 \oplus U_2$.