By Paolo Francia, Fabrizio Catanese, C. Ciliberto, A. Lanteri, C. Pedrini, Mauro Beltrametti

ISBN-10: 3110171805

ISBN-13: 9783110171808

Eighteen papers, many drawing from shows on the September 2001 convention in Genova, disguise quite a lot of algebraic geometry. specific cognizance is paid to raised dimensional forms, the minimum version software, and surfaces of the final sort. an inventory of Francia's guides is incorporated. members contain mathematicians from Europe, the U.S., Japan, and Brazil

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Extra info for Algebraic Geometry: A Volume in Memory of Paolo Francia

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1. , for A a noetherian subring of R such that A ⊗ Q is a field, an object H of MHS is defined as a triple H = (HA , W, F ) where HA is a finitely generated A-module, W is a finite increasing filtration on HA ⊗ Q and F is a finite decreasing filtration on HA ⊗ C such that W, F and F is a system of opposed filtrations. 1. An ∞-mixed Hodge structure H is a triple (HA , W, F ) where HA is any A-module, W is a finite increasing filtration on HA ⊗Q and F is a finite decreasing filtration on HA ⊗ C such that W, F and F is a system of opposed filtrations.

Exotic (1, 1)-classes Consider X singular. 4 for p = 1. Moreover we show that there are edge maps generalizing the cycle class maps constructed in the previous section. For X a proper irreducible C-scheme, consider the mixed Hodge structure on H 2+i (X, Z) modulo torsion. The extension (5) is the following 0 → H 1+i ((H 1 )• ) → W2 H 2+i (X)/W0 → H i ((H 2 )• ) → 0. (15) Since the complex (H 1 )• is made of level 1 mixed Hodge structures then H 2+i (X)h = H 2+i (X)e in our notation. If X is nonsingular then H 2+i (X) is pure and there are only two cases where this extension is non-trivial.

Let X be a proper smooth C-scheme. Note that we have N 1 H j (X) = ker(H j (X) → H 0 (X, H j )) = { Zariski locally trivial classes in H j (X)}. Thus H j (X, Q) ∩ F 1 H j (X) = gr 0N H j (X) ∩ F 1 ⊆ H 0 (X, H j ) ∩ F 1 . N 1 H j (X) We remark that H j /F 1 is the constant sheaf associated to H j (X, OX ). Thus F 1 ∩ H 0 (X, H j ) ∼ = ker(H 0 (X, H j ) → H j (X, OX )). If j = 1 then H 1 (X) = H 0 (X, H 1 ) from (13) and (8) is trivially an equality. If j = 2 then F 1 ∩ H 0 (X, H 2 ) = 0 from the exponential sequence.

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Algebraic Geometry: A Volume in Memory of Paolo Francia by Paolo Francia, Fabrizio Catanese, C. Ciliberto, A. Lanteri, C. Pedrini, Mauro Beltrametti


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