1
$\begingroup$

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 $c$.

$\endgroup$
2
$\begingroup$

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 $V_f = U_1 \oplus U_2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.