By Daniel Huybrechts

Easily obtainable

Includes contemporary developments

Assumes little or no wisdom of differentiable manifolds and useful analysis

Particular emphasis on issues with regards to reflect symmetry (SUSY, Kaehler-Einstein metrics, Tian-Todorov lemma)

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3), we ﬁnd that ∞ (ax + by n )(ax2 + by n−1 ) · · · (axn + by) (xy; xy)n n=0 40 1 The Heine Transformation ∞ = y −n(n+1)/2 (axy + by n+1 )(ax2 y 2 + by n+1 ) · · · (axn y n + by n+1 ) (xy; xy)n n=0 ∞ = y −n(n+1)/2 (xy; xy)n n=0 ∞ = n n j j=0 (by n+1 )n−j (axy)j (xy)j(j−1)/2 xy j n j(j+1)/2 n(n+1)/2 a b x y (xy; xy) (xy; xy)n j n,j=0 = φ(a, x, y)φ(b, y, x), where in the penultimate line we inverted the order of summation and then replaced n by n + j. 17 (p. 57). If ∞ φ(a) := an q n(n+1)/2 , (q 2 ; q 2 )n n=0 then ∞ φ(a)φ(b) = (a + bq n−1 )(a + bq n−3 ) · · · (a + bq 1−n )q n(n+1)/2 .

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 = . (aq; q 2 )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 ∞ = 2 (−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 ∞ = −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 ∞ = 2 (−aq/b; q)2n b2n q 2n (−b; q)2n+1 n=0 +n 1+ +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.

4) was proved by Berndt in [56, p. 152, Entry 17]. 4), we see that we need to show that (−a/n; k)∞ (−bn; k)∞ (k; k)2∞ (n + 1)f (a/n, bn)f 3 (−k) . 8), this is an easy exercise. 2, but in a diﬀerent notation. 8). Therefore, we have replaced f (x, y) by F (x, y) below. 3 (p. 312). For any complex numbers x and y with |x| < 1, let ∞ 2 xk y k . F (x, y) := k=−∞ Then, for complex numbers a, b, and n with |ab| < 1, √ ∞ F ( ab, n b/a)(ab; ab)3∞ aj 1 bj √ √ √ + = + . 6) Proof. 8). 4). 2, and so the proof is complete.