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


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 $V_f = U_1 \oplus U_2$.

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  • $\begingroup$ Thank you, your answer is very interesting! Just some questions: can you explain why the degree of $\mathcal{U}^\perp|_c$ is $-2$? I feel like I need some reference, I'm not an expert of this kind of things. And why the decomposition of $\mathcal{U}^\perp|_c$ cannot contains, for example, some $\mathcal{O}_c(1)$? $\endgroup$ – Nutella Warrior Jan 7 at 18:29
  • $\begingroup$ First, $\deg(\mathcal{U}^\perp\vert_c) = c_1(\mathcal{U}^\perp) \cdot [c] = -H \cdot [c] = -2$, where $H$ is the Pl\"ucker class. Second, $\mathrm{Hom}(\mathcal{O}_c(1),\mathcal{O}_c) = 0$, so $\mathcal{O}(1)_c$ cannot be a summand of a subsheaf of a trivial vector bundle. $\endgroup$ – Sasha Jan 7 at 18:44
  • $\begingroup$ I see. Last thing: why do you have $\mathcal{U}^\perp_c\subset \mathcal{O} _c$? $\endgroup$ – Nutella Warrior Jan 7 at 21:40
  • $\begingroup$ @NutellaWarrior: That was a typo, sorry, the correct inclusion is $\mathcal{U}^\perp\vert_c \subset V^\vee \otimes \mathcal{O}_c$ (the restriction of the tautological embedding). $\endgroup$ – Sasha Jan 8 at 7:59

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