Complex Geometry by Daniel Huybrechts

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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Example text

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.

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