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Let $X$ be a smooth hypersurface in Projective space $\mathbb{P}^n$ of degree $ d$ defined by the equation $f=0$. Given that we have a vector bundle $E$ of rank $r\geq1$ on $X$ such that we have the following exact sequence on $\mathbb{P}^n$:

$$0\rightarrow O(-1)^{rd}\rightarrow O^{rd}\rightarrow E\rightarrow 0.$$ My question is as follows. What is the morphism from $O(-1)^{rd}\rightarrow O^{rd}$? A paper indicated that it is given by a $rd\times rd$ matrix of linear forms. Why is this? I am not able to see it. If this is so, can we say where in $\mathbb{P}^n$ the determinant of that matrix vanishes?

gradstudent
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  • A map from $O(-1)\to O$ is given by multiplication by a linear form. So, you get a matrix of linear forms as you claim. Its determinant is precisely $f^r$ up to a non-zero constant. – Mohan Jun 29 '18 at 18:14
  • @Mohan, for instance a map from $O(-1)\oplus O(-1)\rightarrow O\oplus O$ Could be given by multiplication by $2\times 1$ matrix of linear forms. The question says it is given by a $2\times 2$ matrix. I think the rank of $E$ and degree of $X$ determine the order of th matrix. I am not able to see that. – gradstudent Jun 29 '18 at 18:24
  • Also how can we say that the determinant is $f^r$ upto a constant? – gradstudent Jun 29 '18 at 18:26
  • No, the map you describe is always given by a $2\times 2$ matrix of linear forms (zero is a form of any degree). I will do this in an answer, since it takes a bit too long to explain in a comment if you still have difficulties. – Mohan Jun 29 '18 at 18:36
  • @Mohan, it will be great if you can make it an answer. I am not able to understand why it has to be a $2\times 2$ matrix. Thanks in advance! – gradstudent Jun 29 '18 at 19:09

1 Answers1

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I do not understand your confusion about the size of the matrix. If you fix a basis (to avoid confusion), the map is given as $\oplus_{i=1}^m O(-1)e_i\to \oplus_{i=1}^m Ov_i$, where $e_i, v_i$ are just place holders. A map from $O(-1)e_i\to Ov_j$ is given by a linear form (possibly zero) $l_{ij}$. Thus, you get $m^2$ linear forms $l_{ij}$ given by an $m\times m$ matrix.

For the latter part, notice that determinant of this matrix is a homogeneous polynomial $F$ of degree $rd$. At any point $f\neq 0$, $E=0$ and thus the matrix must be non-singular there. This says, the only prime factor of $F$ is $f$ and by degree considerations, $F$ is $f^r$ up to a non-zero constant.

Mohan
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