Cohomology of vector bundles and syzygies by Jerzy Weyman
By Jerzy Weyman
The imperative subject matter of this ebook is a close exposition of the geometric means of calculating syzygies. whereas this is often a big device in algebraic geometry, Jerzy Weyman has elected to write down from the viewpoint of commutative algebra so as to stay away from being tied to big situations from geometry. No previous wisdom of illustration idea is believed. Chapters on numerous functions are integrated, and various workouts will supply the reader perception into how you can follow this crucial technique.
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Extra info for Cohomology of vector bundles and syzygies
This is done in the same way as in the case of Schur modules. Then it is enough to check two things: (1) The costandard tableaux of shape λ/µ generate the factor Dλ1 −µ1 E ⊗ Dλ2 −µ2 E ⊗ . . ⊗ Dλs −µs E/U (λ/µ, E). (2) The images ψλ/µ (T ) for the costandard tableaux T of shape λ/µ are linearly independent. Facts (1) and (2) allow us to identify the image of ψλ/µ with the factor Dλ1 −µ1 E ⊗ Dλ2 −µ2 E ⊗ . . ⊗ Dλs −µs E/U (λ/µ, E). The proof of (2) is exactly the same as for Schur modules, so we leave it to the reader.
A partition of the form λ = ( p, 1q−1 ). Graphically, ... λ= . , 36 Schur Functors and Schur Complexes with p boxes in the ﬁrst row and q boxes in the ﬁrst column. The relations between two rows of length 1 express the symmetry (cf. example (b)). There is only one type of relations corresponding to the ﬁrst two rows, for the pair u = v = 0. It follows that the Schur functor L ( p,1q−1 ) E is the cokernel of the composition map p+1 ⊗1 E ⊗ Sq−2 E −→ p 1⊗m E ⊗ E ⊗ Sq−2 E −→ p E ⊗ Sq−1 E. 2 that a tableau T is standard if the numbers in each row of T form an increasing sequence and the numbers in each column of T form a nondecreasing sequence.
Since H i (R Hom X (ω X , ω X )) = O X for i = 0 and 0 for i > 0, the left hand side can be identiﬁed with R f ∗ O X . 28), R i f ∗ (ω X ) = 0 for i > 0. Since f is proper, birational, X is smooth, and Y is normal, we have f ∗ O X = OY . Therefore by the assumption of the proposition we have the isomorphism α : R f ∗ (ω X ) → OY , α = f (φ)−1 , and (∗) gives the isomorphism β : R f ∗ (O X ) → ωY . Recall that we have a morphism φ : O X → ω X . It induces the morphism ψ = αφβ −1 : ωY → OY . By assumption of the proposition we know that ψ induces an isomorphism H 0 (ψ) in zero cohomology, since R 0 f ∗ (φ) = f ∗ (φ).