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By Bliss G.A.

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Extra resources for A Necessary and Sufficient Condition for the Existence of a Stieltjes Integral (1917)(en)(5s)

Example text

Then [E → X] = [E → X] in Ω∗ (X). m Proof. Write E = i=1 ni Ei with the Ei smooth and irreducible and let Li = OW (Ei ). ,nm . Let FJ = uJ FJ , so that F = FJ . J Let iJ : E J → W be the inclusion. 8, we have iJ∗ [FJ (LJ1 , . . , LJm )] = [FJ (L1 , . . , Lm )]. Thus iJ∗ [FJ (˜ c1 (LJ1 ), . . , c˜1 (LJm )] [E → W ] = J = [FJ (L1 , . . , Lm )] J = [F (L1 , . . , Lm )] = F c˜1 (L1 ), . . , c˜1 (Lm ) (IdW ) 1 m = c˜1 L⊗n ⊗ . . ⊗ L⊗n (IdW ) (cf. 2) m 1 = [OW (E)]. 2. Localization Let X be a finite type k-scheme i : Z → X a closed subscheme, and j : U → X i j∗ ∗ the open complement.

Nm (u1 , . . , um ) ∈ Ω∗ (k)[[u1 , . . ,nm (u1 , . . , um ). ,nm (u1 , . . 4. Assume m = 2 and n1 = n2 = 1. Then F 1,1 (u, v) = F (u, v) = ai,j ui v j i,j ai,j ui v j =u+v+ i≥1,j≥1 ai,j ui−1 v j−1 , =u+v+u·v· i≥1,j≥1 1,1 1,1 1,1 so F(0,0) = 0, F(1,0) (u, v) = 1, F(0,1) (u, v) = 1 and 1,1 F(1,1) (u, v) = ai,j ui−1 v j−1 . ,0) = 0 for all (n1 , . . , nr ). 1. 2. Normal crossing divisors. Let W be in Smk and let E be a strict normal m crossing divisor on W . Write E = j=1 nj Ej , with the Ej irreducible.

Power series. Suppose that A∗ satisfies the axiom (Dim). Let F (u1 , . . , ur ) ∈ A∗ (k)[[u1 , . . , ur ]] be a formal power series in (u1 , . . , ur ) with coefficients in the graded ring A∗ (k). Suppose that F is absolutely homogeneous of degree n. Given line bundles (L1 , . . , Lr ) on X ∈ V, the operations c˜1 (L1 ), . , c˜1 (Lr ) are locally nilpotent on A¯∗ (X) (by axiom (Dim)) and commute with each other. 1) F (˜ c1 (L1 ), . . , c˜1 (Lr )) : A¯∗ (X) → A¯∗−n (X) ⊂ A∗−n (X). If X is a smooth equi-dimensional k-scheme of dimension d, we have the class 1X ∈ A¯d (X) and we set [F (L1 , .

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A Necessary and Sufficient Condition for the Existence of a Stieltjes Integral (1917)(en)(5s) by Bliss G.A.


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