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By Alessandrini L.

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5) 0 ~ A*(n) ......... ~ A*-- P > A*--* 0 ~A(n) --~E x t rig 0 ~ ~ A*--* 0 is commutative. The right hand square of the morphism A* ~ Extrig(A,Gm). of the left hand square (c) by definition To check the commutatlvity let the extension O ~ Gm ~ E ~ A ~ O represent an element homomorphism in Cartier A*(n). 4). 6) A(n)¢ Thus going A*(n) are Gm~A ¢IA(n)l: A(n)l ~ ~ * Gm around the left hand square: ~ 7 ~A(n) extension ! ~ E~A A ~ Extr~g(A'Gm) together A(n)l ~ G m assigns (e) with the rigidification and the canonical the trivial whose components inclusion A .

Subtracting we find = the result. l = s*(~ ), /2 = ~ ( ~ V, V on V 2 = ~I(~) ® ~ ( ~ ) , ~7~ = v ~ ) e v ~ ( ~ ) . s*(¢). ) is primitive and hence V~ : s * ~ ) , = CeC(A,Q I) But because &(~) = ~ ) . is independent of the connection put on the Z , we can Just as in 4) above define a morphism Ext(A'Gm) ~ ~AwA" As the trivial connection on the trivial extension is compatible with the group structure, any connection placed on any extension is similarly compatible since the morphism is constantly zero.

But/because abelian scheme2this mapping 4) To show the connection trick which will be repeated F(A,~/S) Q' ~ F(S,~) E is an element of hence all global is an is an isomorphism. below in showing that E . (~2A/S)). corresponds to a line bundle this line bundle. 1), ~' to a comn~ction on is an abelian scheme, and 1-forms are closed, the curvature c(~ is 41 actually independent of the connection on allows us to define a morphism if S' is an (absolutely) 0 + Gms,~ E' ~ AS. (QA/S). Namely S -scheme and is an extension, we can take any structure of rigidified extension on it, then by put a connection on E • Notice this E' the above procedure and hence finally obtain the curvature ~'~s,/~) tensor which lies in r(S',~As,.

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1-Convex Manifolds are p-Kahler by Alessandrini L.

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