Lectures on D-modules by Ginzburg V.

By Ginzburg V.

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I Let M be a DX -module and let Y → X is a submanifold. 22 For any i ≥ 0, H[Y ] M has a natural structure of a DX -module. Proof. Since this claim is very important we will give two equivalent constructions of the DX -module structure. 23 First construction. We use the isomorphisms H[Y Hi (X \ Y, A). Choose a Cech ] (X, A) • covering for X \Y . All the groups C (Ui1 ∩. ∩Uin , M) have a natural DX -module structure which descends to the cohomology due to combinatorial nature of the Cech differential.

Induce an equivalence of categories between Mcrys (Y ) and Mcrys (Z)Y . The proof is easy, and left to the reader. 21 There exists a natural equivalence between the category of right DY -modules (in the sense of the last subsection) and Mcrys (Y ). Proof. Suppose first that Y is non-singular and F is a D-crystal on Y . 19). 1 FY ∼ = FY ⊗OY DY and adjunction, we may regard τ as a(τ ) ∈ Homp−1 OY (FY ⊗OY DY , FY ). We claim that a(τ ) makes FY into a right DY 2 module. 19) implies that a(τ ) is a ring action.

For example, if one removes a point from C2 the resulting variety is no longer affine, so the functor R1 j∗ will in fact be non-trivial. 5 Example In the algebraic situation described above one has a vector space isomorphism 1 dx1 dxn 1 n ,... , ∧ ... ∧ . H[0] (U, Ωnalg ) C x1 xn x1 xn In fact, Ωn (U1 ∩ . . ∩ Un ) = C x1 , x11 , . . , xn , x1n dx1 ∧ . . ∧ dxn . However, any form that does not have a pole along all the divisors {x1 = 0}, . . , {xn = 0}, represents a Cech coboundary. The quotient of Ωn (U1 ∩ .

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