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By a) H b (X, OX (−B) ⊗ L−1 ) −−→ H b (X, L−1 ) is surjective. One obtains b), since the left hand side is zero. ✷ It is not difficult to modify both parts of this proof to include in b) the case that L is nef and κ(L) = dim X. Moreover, considering very ample divisors on X and using induction on dim(X), one can as well remove the assumption “κ(L) = dim(X)” and obtain the vanishing for b < κ(L). We leave the details to the reader. 12) anyway, when we prove a more general statement. 7. Lemma. For an invertible sheaf L on a projective manifold X the following two conditions are equivalent: a) L is numerically effective.

14. Corollary (I. Bauer, S. Kosarew [4]). Let Z be a projective variety in characteristic zero, U ⊆ Z be an open non-singular subvariety. Then, for k < n − dim (Z − U ) − 1 one has dim IHk (U, Ω•U ) = dim H b (U, ΩaU ). 13). Then we have a natural map of spectral sequences E1ab = H b (U, ΩaU ) ⏐ϕa,b ⏐ =⇒ IHa+b (U, Ω•U ) ⏐ϕ ⏐ E1ab = H b (X, ΩaX (log E)) =⇒ IHa+b (X, Ω•X (log E)). Since E1ab degenerates in E1 and since ϕa,b are isomorphisms for a + b < n − dim (Z − U ) − 1 the second spectral sequence has to degenerate for those a, b.

Let X be a projective manifold and L an invertible sheaf. 11) and that dim X ≤ p. Then, if L is ample, H b (X, L−1 ) = 0 for b < n = dim (X) Proof: Choose N , prime to p = char (k), such that H b (X, L−N −1 ) = H n−b (X, ωX ⊗ LN +1 ) = 0 for b < n, and such that LN is generated by global sections. 1) and find H b (X, L−1 (−D)) −−→ H b (X, L−1 ) to be surjective. Since the group on the left hand side is zero, we are done. ✷ 44 H. Esnault, E. 3) that it is sufficient to assume that X lifts to W2 (k).

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A 1 psi 1 Summation Theorem for Macdonald Polynomials by Kaneko J.

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