By Conference on Algebraic Geometry (1988 Sundance Institute), Brian Harbourne, Robert Speiser

This quantity comprises the lawsuits of the NSF-CBMS neighborhood convention on Algebraic Geometry, held in Sundance, Utah, in July 1988. The convention desirous about algebraic curves and comparable kinds. a few of the papers accumulated right here symbolize lectures brought on the convention, a few document on examine performed throughout the convention, whereas others describe similar paintings conducted in other places.

**Read Online or Download Algebraic Geometry: Sundance 1988 : Proceedings of a Conference on Algebraic Geometry Held July 18-23, 1988 With Support from Brigham Young Universi (Contemporary Mathematics) PDF**

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**Extra info for Algebraic Geometry: Sundance 1988 : Proceedings of a Conference on Algebraic Geometry Held July 18-23, 1988 With Support from Brigham Young Universi (Contemporary Mathematics)**

**Example text**

10). This completes the proof. The next entry can be found in Slater’s compendium [262, equation (35)]. 6 (p. 35). 8). Then ∞ (−q; q 2 )n q (n+1)(n+2)/2 qf (−q, −q 7 ) . 2 and Auxiliary Results 33 Proof. 1), set h = 2, a = c = −q 2 , t = −q, and b = 0. Multiplying both sides of the resulting identity by q/(1 + q), we ﬁnd that ∞ ∞ (−q; q 2 )n q (n+1)(n+2)/2 (−q)n q(−q; q 2 )∞ = . 8) three times altogether, we ﬁnd that ∞ ∞ qn 1 q (n−1)/2 = (1 − (−1)n ) (q; q)2n+1 2 n=0 (q; q)n n=0 1 = √ 2 q 1 1 − √ √ ( q; q)∞ (− q; q)∞ 1 (−q 1/2 ; q 2 )∞ (−q 3/2 ; q 2 )∞ (q 2 ; q 2 )∞ = √ 2 q(q; q)∞ −(q 1/2 ; q 2 )∞ (q 3/2 ; q 2 )∞ (q 2 ; q 2 )∞ 1 = √ 2 q(q; q)∞ 1 =√ q(q; q)∞ = = 1 (q; q)∞ ∞ 2 qn −n/2 ∞ − 2 (−1)n q n −n/2 n=−∞ n=−∞ ∞ (2n+1)2 −(2n+1)/2 q n=−∞ ∞ 4n2 +3n q n=−∞ f (q, q 7 ) .

14), we ﬁnd that ∞ ∞ ∞ (−1)n q (n+1)(n+2)/2 (−1)n q (n+1)(n+2)/2+m(2n+1) = 2n+1 (q)n (1 − q ) (q)n n=0 n=0 m=0 ∞ ∞ q m+1 = m=0 ∞ (−1)n q n(n−1)/2+n(2m+2) (q)n n=0 q m+1 (q 2m+2 )∞ = m=0 ∞ = q(q)∞ qm (q)2m+1 m=0 = qf (q, q 7 ), which completes the proof. 8 can be found in Slater’s paper [262, equation (37)]. 8 (p. 35). 8), then ∞ (−q; q 2 )n q n(n+1)/2 f (−q 3 , −q 5 ) . 15) Proof. 1), we set h = 2, a = −q 2 , b = 0, and c = t = −q. Upon using Euler’s identity, we ﬁnd that ∞ ∞ (−q; q 2 )n q n(n+1)/2 (−q)n 2 2 = (−q; q) (−q; q ) .

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.