# Conics in Grassmannians

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 $$V_f = U_1 \oplus U_2$$.
• 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 $\mathcal{O}_c(1)$? – Nutella Warrior Jan 7 at 18:29
• 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. – Sasha Jan 7 at 18:44
• I see. Last thing: why do you have $\mathcal{U}^\perp_c\subset \mathcal{O} _c$? – Nutella Warrior Jan 7 at 21:40
• @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). – Sasha Jan 8 at 7:59