Global sections of coherent sheaves on determinantal hypersurfaces in $\mathbb{P}^n$

Question

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 $N$?

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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?

Answer 1

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 $$ ^tN^* \stackrel{\rightarrow}{\alpha}=0 $$

Answer 2

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.