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By Alexander Polishchuk

This ebook is a contemporary therapy of the idea of theta services within the context of algebraic geometry. the newness of its procedure lies within the systematic use of the Fourier-Mukai remodel. Alexander Polishchuk begins through discussing the classical thought of theta capabilities from the perspective of the illustration conception of the Heisenberg team (in which the standard Fourier rework performs the renowned role). He then indicates that during the algebraic method of this idea (originally as a result of Mumford) the Fourier-Mukai rework can usually be used to simplify the present proofs or to supply thoroughly new proofs of many vital theorems. This incisive quantity is for graduate scholars and researchers with powerful curiosity in algebraic geometry.

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2). Let T be a complex torus, e1 , . . , e2n be the basis of the lattice H 1 (T, Z), ∗ e1∗ , . . , e2n be the dual basis of H 1 (T ∨ , Z), where T ∨ is the dual torus. Show that the first Chern class of the Poincar´e bundle on T × T ∨ is given by 2n c1 (P) = i=1 ei ∧ ei∗ . 2 Representations of Heisenberg Groups I This chapter in an introduction to the representation theory of Heisenberg groups. This theory will be our principal tool in the study of theta functions. Throughout this chapter V is a real vector space with a fixed symplectic form E.

With every non-degenerate quadratic function q on a finite abelian group A we associate the Gauss sum γ (q) = |A|− 2 1 q(a). a∈A Here are some simple properties of γ (q). 6. (i) Let A = A1 ⊕ A2 , q = q1 ⊕ q2 , where qi is a nondegenerate quadratic function on Ai (i = 1, 2). By the definition, this means that q(a1 , a2 ) = q1 (a1 )q2 (a2 ). Then γ (q1 ⊕ q2 ) = γ (q1 ) · γ (q2 ). (ii) Assume that there is a subgroup I ⊂ A such that q| I = 1 and |I | = 1 |A| 2 . Then γ (q) = 1. (iii) One has γ (q −1 ) = γ (q)−1 .

2) has only one potentially nontrivial cohomology group A(L 1 , L 2 , L 3 ) = ker(d3 )/ im(d2 ). One has A(L 1 , L 2 , L 3 ) (L 1 + L 2 ) ∩ L 3 /(L 1 ∩ L 3 + L 2 ∩ L 3 ). 3. Quadratic Function Associated with an Admissible Triple 45 The proof is left to the reader. Let us define a function q : ker(d3 ) → U (1) by setting q(x1 , x2 , x3 ) = σ (x1 )σ (x2 )σ (x3 ), where σ : L i → H , i = 1, 2, 3, are lifting homomorphisms. Since x1 + x2 + x3 = 0, the expression in the RHS belongs to U (1) ⊂ H . 3.

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