By Dominique Arlettaz
The second one Arolla convention on algebraic topology introduced jointly experts protecting quite a lot of homotopy thought and $K$-theory. those complaints replicate either the range of talks given on the convention and the variety of promising learn instructions in homotopy conception. The articles contained during this quantity contain major contributions to classical risky homotopy thought, version type thought, equivariant homotopy thought, and the homotopy concept of fusion structures, in addition to to $K$-theory of either neighborhood fields and $C^*$-algebras
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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 . 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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