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Dr. Katz DEq Homework Solutions 12

# Dr. Katz DEq Homework Solutions 12 - ~-~ ~ According...

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12 N.M. Katz: (1.2.2.4) Let us reinterpret the interior product mapping I as a pairing (1.2.2.4.1) I: Homex(O~, ~)|174 From this pairing we deduce a cup-product pairing for all p, q > 0 (1.2.2,4.2) HP(X, Hom(~, ~))| H~(X, A~ ~)-, HP+q(X, ~)| A~- ' ~). (1.2.2.4.3) Taking p=l, consider the element ~eHI(X, Hom(~,~)) which corresponds via (1.2.1.8) to the class of the extension (1.2.1.2), (1.2.2.4.4) 0-, ~r -~ our-+ .~-, 0. (1.2.2.4.5) Proposition. The mapping "cup-product with ~" ~: Hq(X, A~)-, Hq+X(X, [email protected] ~-1 ~) deduced from (1.2.2.4.2) is equal to the coboundary in the long exact cohomology sequence arising from the short exact sequence (1.2.1.5) (1.2.2.4.6) 0-~ ~ | A ~- 1 ~_~ KO/K z (A ~ ~,uf) _~ A ~ ~_~ O. Proof Indeed by definition this map is none other than the cup- product with the image of ~ ~ H 1 (X, Hom(~; if)) under I: Hi(X, Hom(~, ~)) -~ Hi(X, Hom(a~,, ~|
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Unformatted text preview: ~-~ ~)). According to (1.2.2.3), the image 1(0 is none other than the class of the extension (1.2.2.4.6) (via the isomorphism (1.2.1.8 bis)). (1.2.2.4.7) But it is a general fact that the coboundary mapping H~(X, C)-, H q+l (X, A) associated to any short exact sequence 0-~ A-, B-+ C--) 0 of @x-modules is none other than the cup-product with the element of Ext~x(C, A) (this last operation has a sense, thanks to the isomorphism of functors H q (X, -) ~Ext~, ((gx, -)). Q.E.D. 1.3. Application to Hodge Cohomology (1.3.1) Proposition. In the geometric situation of 1.0, consider the short exact sequence (1.1.1) (1.3.1.1) 0 -)f* (O~/T) ---) O~C/T (Iog O) --, O~C/S (log D) -, 0 which gives rise, via the A p construction (1.2.1.6), to a short exact sequence 0 -,f* ((2~/r) | (2w (log D) --~ K~ z (Y2w (log D)) (1.3.1.2) --+ O]/s (log D) -, 0....
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