Pushforward under Segre Embedding

Question

This question originated from an answer to this post. Take $\mathcal{O}(1,0)$ on $\mathbf{P^1}\times \mathbf{P}^1$ and consider its push-forward $F$ on the quadric $\subset \mathbf{P}^3$ under the Segre embedding. Then there should be a resolution $0\rightarrow \mathcal{O}(-1)^2 \rightarrow \mathcal{O}^2\rightarrow F \rightarrow 0$. There are powerful tools like the Tate resolution that allow us to get to this sequence, but I think this should be possible with more modest methods? Glad for any help. I thought about $\mathcal{O}(-1)^2 \xrightarrow{\begin{pmatrix} x &y \\ z & w\end{pmatrix}} \mathcal{O}^2$ which has I think a co-kernel where $\det=xw-yz$ is zero which is the equation of the Segre quadric...?

Answer 1

Yes, your guess is correct. To show this note that the kernel of the morphism is zero at the general point of $\mathbb{P}^3$, hence zero everywhere, and the cokernel is annihilated by $xw - yz$, hence is supported (scheme-theoretically) on this quadric (let me call it $Q$). This means that we have an exact sequence $$ 0 \to \mathcal{O}(-1)^{\oplus 2} \to \mathcal{O}^{\oplus 2} \to i_*F \to 0 $$ for some sheaf $F$. To understand what $F$ is, pull back the sequence to the quadric. Then you get a right-exact sequence $$ \mathcal{O}_Q(-1,-1)^{\oplus 2} \to \mathcal{O}_Q^{\oplus 2} \to F \to 0. $$ Now, recalling that the isomorphism $Q \cong \mathbb{P}^1 \times \mathbb{P}^1$ is given by $$ x = u_1u_2,\qquad y = u_1v_2,\qquad z = v_1u_2,\qquad w = v_1v_2, $$ we see that in the second sequence the first arrow is given by the matrix $$ \begin{pmatrix} u_1u_2 & u_1v_2 \\ v_1u_2 & v_1v_2 \end{pmatrix} = \begin{pmatrix} u_1 \\ v_1 \end{pmatrix} \cdot \begin{pmatrix} u_2 & v_2 \end{pmatrix}, $$ therefore the corresponding morphism is a composition $$ \mathcal{O}_Q(-1,-1)^{\oplus 2} \to \mathcal{O}_Q(-1,0) \to \mathcal{O}_Q^{\oplus 2}, $$ where the first arrow is given by $(u_2,v_2)$, hence surjective, and the second arrow is given by $(u_1,v_1)$ (transposed), hence injective, and the cokernel of the composition equal to the cokernel of the second arrow, hence to $\mathcal{O}_Q(1,0)$.