Let X be a smooth hypersurface in Projective space Pn of degree d defined by the equation f=0. Given that we have a vector bundle E of rank r≥1 on X such that we have the following exact sequence on Pn:0→O(−1)rd→Ord→E→0.My question is as follows. What is the morphism from O(−1)rd→Ord? A paper indicated that it is given by a rd×rd matrix of linear forms. Why is this? I am not able to see it. If this is so, can we say where in Pn the determinant of that matrix vanishes?

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alagmamGynccip · Accepted Answer

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 ⊕i=1mO(−1)ei→⊕i=1mOvi, where ei,vi are just place holders. A map from O(−1)ei→Ovj is given by a linear form (possibly zero) lij. Thus, you get m2 linear forms lij given by an m×m matrix.For the latter part, notice that determinant of this matrix is a homogeneous polynomial F of degree rd. At any point f≠0,E=0 and thus the matrix must be non-singular there. This says, the only &#039; factor of F is f and by degree considerations, F is fr up to a non-zero constant.

Let X be a smooth hypersurface in Projective space \mathbb{P}^n of degree d defined by the equ

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