Let us consider the short exact sequence of coherent sheaves on $\mathbb{P}^n$ $$0 \to \mathcal{O}_{\mathbb P^n}(-1)^{r} \stackrel{N}{\longrightarrow} \mathcal{O}_{\mathbb P^n}^{r} \longrightarrow \mathcal F \to 0,$$ where $r \geq 2$ and $N=\{L_{ij}\}$ is a $r \times r$ matrix of linear forms in the variables $x_0, \ldots, x_n$ acting on the left (i.e., $\mathbf{x} \to N \mathbf{x}$).

Then the sheaf $\mathcal{F}$ is supported on the determinantal hypersurface $Y:=\{\det N=0 \} \subset \mathbb{P}^n$, and it is locally free of rank $1$ (i.e., a line bundle) on the open dense locus of $Y$ given by $\textrm{rank } N = r-1$.

Passing to cohomology, we obtain an isomorphism $$\sigma \colon H^0(\mathbb{P}^n, \, \mathcal{O}_{\mathbb P^n}^{r}) \stackrel{\cong}{\longrightarrow} H^0(Y, \, \mathcal F),$$ in particular $h^0(Y, \, \mathcal F)=r.$ In other words, this is a bijective correspondence between $n$-ples $(\alpha_1, \ldots, \alpha_r)$ of complex numbers and global sections of $\mathcal F$.

Question. How can we describe the zero locus in $Y$ of the section $\sigma(\alpha_1, \ldots, \alpha_n)$ in terms of the $\alpha_i$ and the matrix $N$?

Toy model. Consider the case $n=3$, $r=2$ and $$N = \pmatrix{x_0 & x_1 \\ x_2 & x_3}.$$ In this situation, $Y$ is the smooth quadric $\{x_0x_3-x_1x_2=0 \} \subset \mathbb{P}^3$, and $\mathcal{F}$ is a line bundle with $2$ global sections, hence a pencil corrisponding to one of the two rulings of $Y$. I made some (unelegant) local computation and it seems to me that in this case $\sigma(\alpha_1, \, \alpha_2)$ should be something like $\alpha_1 \sigma_{02} + \alpha_2 \sigma_{13}$, where $\sigma_{02}$ is the section whose zero locus is the line $\{x_0=x_2=0 \}$ and $\sigma_{13}$ is the section whose zero locus is the line (in the same ruling) $\{x_1=x_3=0 \}.$ In other words, the global sections of $\mathcal{F}$ are expressed in terms of the columns of $N$.

On the other hand, dualizing our exact sequence and applying Grothendieck duality we get $$0 \to \mathcal{O}_{\mathbb P^3}(-1)^{2} \stackrel{ { }^t N}{\longrightarrow} \mathcal{O}_{\mathbb P^3}^{2} \longrightarrow \mathcal F^*(1) \to 0,$$ where $\mathcal{F}^*:=\mathscr{Hom}(\mathcal{F}, \, \mathcal{O}_{\mathbb P^3})$.

Now the sections of $\mathcal{F}^*(1)$ should correspond the other ruling of the quadric, because $\mathcal{F} \otimes \mathcal{F}^*(1)= \mathcal{O}_Y(1)$. Therefore the global sections of $\mathcal{F}^*(1)$ are given in terms of the rows of $N$, that are the columns of ${}^t N$.

Is this correct? If so, is there any intrinsec or general way to see this?


This is meant to be an integration to Yusuf answer.

Consider a section $$ \stackrel{\rightarrow}{\alpha}= \begin{pmatrix} \alpha_1\\ \vdots\\ \alpha_r \end{pmatrix} \in {\mathbb C}^r\cong H^0({\mathbb P}^n,{\mathcal O}^r_{{\mathbb P}^n}). $$ Its image on $H^0(Y,{\mathcal F})$ vanishes in a point $p \in Y$, by definition of quotient, if and only it is in the image of the map induced by $N$ on the stalks over $p$. Then, by the Kronecker-Rouché-Capelli's Theorem, its zero locus equals the set of points where the rank of the matrix $N$ equals the rank of the $r \times (r+1)$ matrix $(N \stackrel{\rightarrow}{\alpha})$.

In your toy model, in all points of $Y$ the rank of $N$ is $1$, so we are looking to the locus where $$ rank \begin{pmatrix} x_0&x_1&\alpha_1\\ x_2&x_3&\alpha_2\\ \end{pmatrix}=1 $$ that is $\{\alpha_1x_2-\alpha_2x_0=\alpha_1x_3-\alpha_2x_1=0\}$.

This is as in your local computation indeed one of the rulings (but the other one, if my computations are not wrong).

A similar computation in the general case, as pointed out by Yusuf, produces equations involving the $(r-1) \times (r-1)$ minors of $N$.

More precisely, we need the adjoint or cofactor matrix $N^*$ whose $(i,j)$ entry is $(-1)^{i+j}$ times the minor of $N_{i,j}$, the determinant of the $(r-1) \times (r-1)$ matrix obtained by $N$ deleting the $i^{th}$ row and the $j^{th}$ column. In your toy model $$ N^*= \begin{pmatrix} x_3&-x_2\\ -x_1&x_0\\ \end{pmatrix} $$

Then the zero locus of the image of $\stackrel{\rightarrow}{\alpha}$ is $$ ^tN^* \stackrel{\rightarrow}{\alpha}=0 $$


The quadric surface example you mention extends to the general case as follows: to each nonzero vector $\vec{\alpha} \in \mathbb{C}^{r} \cong H^{0}(\mathcal{O}_{\mathbb{P}^{n}}^{r})$ we can associate a hyperplane $H_{\vec{\alpha}} \subset H^{0}(\mathcal{O}_{\mathbb{P}^{n}}^{r})^{\ast}$, and we can look at the map from $H_{\vec{\alpha}}$ to $H^{0}(\mathcal{O}_{\mathbb{P}^{n}}(1))^{\oplus r}$ induced by $N^{T}.$ The zero set of the associated section $\sigma(\vec{\alpha}) \in H^{0}(Y,\mathcal{F})$ is the zero set of maximal minors of an $r \times (r-1)$ matrix with coefficients in $H^{0}(\mathcal{O}_{\mathbb{P}^{n}}(1))$ representing this map.


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.