By Goro Shimura

Reciprocity legislation of assorted types play a relevant position in quantity conception. within the simplest case, one obtains a clear formula through roots of solidarity, that are exact values of exponential features. the same concept will be constructed for exact values of elliptic or elliptic modular features, and is named advanced multiplication of such capabilities. In 1900 Hilbert proposed the generalization of those because the 12th of his well-known difficulties. during this ebook, Goro Shimura offers the main finished generalizations of this kind by way of pointing out numerous reciprocity legislation by way of abelian types, theta capabilities, and modular features of a number of variables, together with Siegel modular services.

This topic is heavily attached with the zeta functionality of an abelian style, that's additionally coated as a primary topic within the ebook. The 3rd subject explored through Shimura is many of the algebraic relatives one of the classes of abelian integrals. The research of such algebraicity is comparatively new, yet has attracted the curiosity of more and more many researchers. a few of the issues mentioned during this e-book haven't been lined earlier than. particularly, this can be the 1st publication during which the subjects of varied algebraic family members one of the sessions of abelian integrals, in addition to the detailed values of theta and Siegel modular services, are handled commonly.

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Example text

4. 1 is complete. 2 (p. 5). If |b| < 1 and a is an arbitrary complex number, then ∞ ∞ (−1)n (−q; q)n (−aq/b; q)n bn (−1)n (−aq/b; q)n bn q n(n+1)/2 = . 2 (aq; q )n+1 (−b; q)n+1 n=0 n=0 Proof. 1), set h = 2 and t = q 2 , and replace b, c, and a by −b, aq, and aq, respectively. 1) with h = 1, q replaced by q 2 , t = q 2 , and a, b, and c replaced by −b, −bq, and aq 2 , respectively. 1) with q replaced by q 2 , α = −aq/b, β = −bq 2 , and τ = −bq to deduce that, after multiplying both sides by 1/(1 + b), ∞ (−aq/b; q 2 )n (−bq)n 2) (−b; q n+1 n=0 ∞ = (−aq/b; q 2 )n (−aq 2 /b; q 2 )n (−bq 2 )n (−bq)n q 2n (−b; q 2 )n+1 (−bq; q 2 )n+1 n=0 ∞ 2 −2n (1 − aq 4n+2 ) 2 (−aq/b; q 2 )n (−aq 2 /b; q 2 )n b2n q 2n +n (1 − aq 4n+2 ) = (−b; q 2 )n+1 (−bq; q 2 )n+1 n=0 = = ∞ 2 ∞ 2 (−aq/b; q)2n b2n q 2n +n (1 − aq 4n+2 ) (−b; q)2n+2 n=0 (−aq/b; q)2n b2n q 2n (−b; q)2n+1 n=0 ∞ (−aq/b; q)2n b2n q 2n = (−b; q)2n+1 n=0 +n 1+ 2 +n −bq 2n+1 − aq 4n+2 1 + bq 2n+1 ∞ − (−aq/b; q)2n+1 b2n+1 q (n+1)(2n+1) (−b; q)2n+2 n=0 ∞ = (−1)n (−aq/b; q)n bn q n(n+1)/2 , (−b; q)n+1 n=0 which is the desired result.

11). For any complex number a, ∞ (q 2 ; q 4 )∞ ∞ (aq 2 ; q 2 )n q n(n+1)/2 (aq 2 ; q 4 )n q 4n = (aq 4 ; q 4 )∞ (q; q)n (q 2 ; q 2 )2n n=0 n=0 2 ∞ + (aq 2 ; q 4 )∞ 2 (aq 4 ; q 4 )n q 4n +4n+1 . (q 2 ; q 2 )2n+1 n=0 Proof. 13, replace q by q 2 and set b = −1/q. This yields ∞ 1 an q 2n = 4 ; q 4 ) (−q; q 2 ) 2 ; q 4 ) (−q; q 2 ) (q (aq n n ∞ ∞ n=0 + ∞ (aq 2 ; q 4 )n q 4n (q 2 ; q 2 )2n n=0 1 4 4 (aq ; q )∞ (−q; q 2 )∞ ∞ 2 2 (aq 4 ; q 4 )n q 4n +4n+1 . 1). More precisely, let h = 2, c = −q, and a = 0, and let b tend to 0.

17) which is implicit in the work of Ramanujan in his lost notebook [244]. 2]. 10 involving two additional parameters. 10 (p. 10). ∞ 1 qn = 2 (q)n (q)2∞ n=0 ∞ (−1)n q n(n+1)/2 . 18) n=0 Proof. 1), set h = 1, t = c = q, and a = 0, and then let b → 0. 10 follows immediately. 11 (p. 10). ∞ ∞ 1 q 2n = 2 2 (q) (q) n ∞ n=0 (−1)n q n(n+1)/2 1+2 . 19) n=1 Proof. 1), set h = 1, a = 0, c = q, and t = q 2 . Now let b → 0 to deduce that ∞ ∞ n+1 1 q 2n n1 − q q n(n+1)/2 = (1 − q) (−1) 2 2 (q) (q) 1 − q n ∞ n=0 n=0 1 = (q)2∞ = 1 (q)2∞ ∞ ∞ n n(n+1)/2 (−1) q n=0 (−1)n+1 q (n+1)(n+2)/2 + n=0 ∞ (−1)n q n(n+1)/2 1+2 .

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