By Jean-Pierre Demailly
This quantity is a selection of lectures given by way of the writer on the Park urban arithmetic Institute (Utah) in 2008, and on different events. the aim of this quantity is to explain analytic concepts worthy within the research of questions relating linear sequence, multiplier beliefs, and vanishing theorems for algebraic vector bundles. the writer goals to be concise in his exposition, assuming that the reader is already a bit accustomed to the elemental ideas of sheaf concept, homological algebra, and complicated differential geometry. within the ultimate chapters, a few very contemporary questions and open difficulties are addressed--such as effects concerning the finiteness of the canonical ring and the abundance conjecture, and effects describing the geometric constitution of Kahler kinds and their confident cones.
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Extra info for Analytic Methods in Algebraic Geometry
7) and letting b → 0. The second line follows from the fact that each of the right-hand entries is equal to 2 2 ∞ am+n q n +m +m+2mn . 4) to (−aq 2n+2 ; q 2 )∞ . 4) to (−aq 2n+1 ; q 2 )∞ and then switch the roles of m and n. M. Somos has observed that if we set ∞ F (a, b; q) := (−bq; q 2 )∞ 2 an q n , 2 2 (q ; q )n (−bq; q 2 )n n=0 then F (a, b; q) = F (b, a; q). 1). 12, set n = 1, and replace q by q 2 , a by a/q, and b by b/q. 1). 2 (p. 42). If a is any complex number, then ∞ ∞ 2 2 a2n q 4n an q n = (aq; q 2 )∞ .
Furthermore, the right-hand side has ellipses after the products forming the numerator and the denominator. If nothing is added, the result is clearly false. However, the following identity has the same left-hand side that Ramanujan gave, and the inﬁnite products from the right-hand side of his proposed identity are isolated in front of our right-hand side. 18 (p. 27, corrected). For any complex numbers a and b with b = 0, ∞ (−a/b; q 2 )n bn q n(n+1)/2 (q; q)n (aq 2 ; q 2 )n n=0 = (−bq; q)∞ (aq; q)∞ (−a/b; q 2 )∞ (aq; q)∞ (aq 2 ; q 2 )∞ ∞ a (−bq 2 ; q 2 )n − 2 2 (q ; q )n (−bq; q)2n b n=0 n .
9). 4 (p. 35). 8). Then ∞ 2 f (−q 6 , −q 10 ) + qf (−q 2 , −q 14 ) (−q 2 ; q 4 )n q n +n = . 4) (q 2 ; q 2 )n (q 2 ; q 4 )n (1 − q 2n+1 ) ϕ(−q 2 ) n=0 Proof. 4 with q replaced by −q. 2). 8), we ﬁnd that S= (iq 2 ; q 2 )∞ (−i1/2 q; q)∞ (−i−1/2 q; q)∞ 2(−q; q 2 )∞ (iq 2 ; q 2 )∞ +(i1/2 q; q)∞ (i−1/2 q; q)∞ = 1 2(q; q)∞ (−q; q 2 )∞ (−i1/2 ; q)∞ (−i−1/2 q; q)∞ (q; q)∞ 1 + i1/2 + = 1 2(q; q)∞ (−q; q 2 )∞ 1 1 + i1/2 + = (i1/2 ; q)∞ (i−1/2 q; q)∞ (q; q)∞ 1 − i1/2 1 1 − i1/2 1 2(1 − i)(q; q)∞ (−q; q 2 )∞ ∞ in/2 q n(n−1)/2 n=−∞ ∞ (−1)n in/2 q n(n−1)/2 n=−∞ ∞ (1 − i1/2 ) in/2 q n(n−1)/2 n=−∞ ∞ +(1 + i1/2 ) (−1)n in/2 q n(n−1)/2 n=−∞ = 1 (1 − i)(q; q)∞ (−q; q 2 )∞ ∞ ∞ in q n(2n−1) − n=−∞ in+1 q n(2n+1) n=−∞ 50 2 The Sears–Thomae Transformation = ∞ 1 (1 − i)(q; q)∞ (−q; q 2 )∞ ∞ in q n(2n−1) − i (−i)n q n(2n−1) n=−∞ , n=−∞ where we replaced n by −n in the latter sum.
Analytic Methods in Algebraic Geometry by Jean-Pierre Demailly