# Degree of curve where matrix of polynomials has rank 1

My question is about a step in Exercise 12.8 on page 442 of 3264 & All That by Eisenbud and Harris. Chapter 12 is about Porteous' formula.

The exercise reads: Let $A=(P_{i,j})$ be a $2 \times 3$ matrix whose entries $P_{i,j}$ are general polynomials of degree $a_{i,j}$ on $\mathbb{P}^3$. Assuming that $a_{1,j}+a_{2,k}=a_{1,k}+a_{2,j}$ for all $j$ and $k$--so that the minors of $A$ are homogeneous--what is the degree of the curve where $A$ has rank $1$?

Shuai Wang posted a solution on page 21 of this document https://www.math.columbia.edu/~tedd2013/intersectiontheory.pdf which seems correct, but I'm struggling to understand one step of it.

He claims that the curve where $A$ has rank $1$ is the degeneracy locus of the following bundle map:

$\mathcal{O}_{\mathbb{P}^3}(a_{22}) \oplus \mathcal{O}_{\mathbb{P}^3}(a_{12}) \to \mathcal{O}_{\mathbb{P}^3}(a_{11}+a_{22}) \oplus \mathcal{O}_{\mathbb{P}^3}(a_{12}+a_{22}) \oplus \mathcal{O}_{\mathbb{P}^3}(a_{12}+a_{23})$.

\begin{align*} \begin{bmatrix} a_{22} & a_{12} \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix} \end{align*}

Why is $\mathcal{O}_{\mathbb{P}^3}(a_{11}+a_{22}) \oplus \mathcal{O}_{\mathbb{P}^3}(a_{12}+a_{22}) \oplus \mathcal{O}_{\mathbb{P}^3}(a_{12}+a_{23})$ the target space of this map?

Why is the degeneracy locus of this map equal the locus of the maximal minors of $A$? The maximal minors of $A$ are $\{a_{11}a_{22}-a_{21}a_{12}, a_{11}a_{23}-a_{13}a_{21}, a_{12}a_{23}-a_{22}a_{13} \}$ while the result of the matrix multiplication is

\begin{align*} \begin{bmatrix} a_{22} & a_{12} \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix} = \begin{bmatrix} a_{11}a_{22}+a_{21}a_{12} & a_{12}a_{22}+a_{12}a_{22} & a_{22}a_{13}+a_{12}a_{23} \end{bmatrix}. \end{align*}

In general, how does one find a map of vector bundles such that its degeneracy locus is the locus of maximal minors of a given matrix? (If possible, it would be nice for the vector bundles to be written as sums of line bundles.) This seems like an important tool for applying Porteus' formula.

I believe you got confused with the map. Let's define a bundle map by right multiplication with $$A = \begin{bmatrix} P_{11} & P_{12} & P_{13} \\ P_{21} & P_{22} & P_{23} \end{bmatrix}$$ whose degrees are respectively$$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$$ Start taking $\begin{bmatrix} R & S \end{bmatrix}$ a section of $\mathcal{O}_{\mathbb{P}^3}(a_{22}) \oplus \mathcal{O}_{\mathbb{P}^3}(a_{12})$. Then $$\begin{bmatrix} R & S \end{bmatrix} A = \begin{bmatrix} RP_{11}+SP_{21} & RP_{12}+SP_{22} & RP_{13}+SP_{23} \end{bmatrix} =: C$$ The identity $a_{1,j}+a_{2,k}=a_{1,k}+a_{2,j}$ says that this map is well defined and the target is $\mathcal{O}_{\mathbb{P}^3}(r) \oplus \mathcal{O}_{\mathbb{P}^3}(s)\oplus \mathcal{O}_{\mathbb{P}^3}(t)$, for some integers $r,s,t$.
These numbers are the corresponding degrees of the entries of $C$, so $r= a_{22}+a_{11}$, $s= a_{22}+a_{12}$ and $t= a_{22}+a_{13} = a_{12}+a_{23}$.
Now you want to calculate the points $p\in \mathbb{P}^3$ such that the map $(x,y) \mapsto \begin{bmatrix} x & y \end{bmatrix}A(p)$ is not injective, the so called degenracy locus. This coincides with the vanishing of the minors of $A(p)$, hence the degeneracy locus of this bundle map is given by vanishing of the minors of $A$.