Algebraic Surfaces and Holomorphic Vector Bundles by Robert Friedman

By Robert Friedman

This e-book covers the idea of algebraic surfaces and holomorphic vector bundles in an built-in demeanour. it really is aimed toward graduate scholars who've had an intensive first-year path in algebraic geometry (at the extent of Hartshorne's Algebraic Geometry), in addition to extra complicated graduate scholars and researchers within the parts of algebraic geometry, gauge conception, or 4-manifold topology. the various effects on vector bundles must also be of curiosity to physicists learning string thought. a unique function of the publication is its built-in method of algebraic floor idea and the research of vector package deal idea on either curves and surfaces. whereas the 2 topics stay separate during the first few chapters, and are studied in exchange chapters, they turn into even more tightly interconnected because the booklet progresses. hence vector bundles over curves are studied to appreciate governed surfaces, after which reappear within the evidence of Bogomolov's inequality for solid bundles, that is itself utilized to review canonical embeddings of surfaces through Reider's approach. equally, governed and elliptic surfaces are mentioned intimately, after which the geometry of vector bundles over such surfaces is analyzed. some of the effects on vector bundles seem for the 1st time in ebook shape, compatible for graduate scholars. The ebook additionally has a robust emphasis on examples, either one of surfaces and vector bundles. There are over a hundred routines which shape a vital part of the textual content.

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2 (p. 50). We have (i) 16qf 2 (−q 2 )f 2 (−q 10 ) =(ϕ2 (q) − ϕ2 (q 5 ))(5ϕ2 (q 5 ) − ϕ2 (q)) and (ii) f 2 (−q)f 2 (−q 5 ) =(ψ 2 (q) − qψ 2 (q 5 ))(ψ 2 (q) − 5qψ 2 (q 5 )). Proof. 8). 7), which we state in the following entry. 1 (p. 56). 1), we have (i) (ii) (iii) ϕ(q) + ϕ(q 5 ) = 2q 4/5 f (q, q 9 )R−1 (q 4 ), ϕ(q) − ϕ(q 5 ) = 2q 1/5 f (q 3 , q 7 )R(q 4 ), (iv) ψ(q 2 ) + qψ(q 10 ) = q 1/5 f (q 2 , q 8 )R−1 (q), ψ(q 2 ) − qψ(q 10 ) = q −1/5 f (q 4 , q 6 )R(q), (v) ψ(q 2 ) + qψ(q 10 ) = f (−q 10 ) .

205). Let ω = exp(2πi/3), u = R(q), and v = R(q 3 ). 5) then 4u = − 1 − v3 1 8 + 4R − + 6 v v 1 8 + 4Rω − + 6 v v 1 8 + 4Rω 2 − . 7) then 4v = u3 − u6 + u(8 + 4R) + u6 + u(8 + 4Rω) + u6 + u(8 + 4Rω 2 ). 1. 10) 8(1 − R + 2v 5 ) 1 4R 8 √ − − . 11) √ √ Proof. To prove the first equality, we use the property a ± b = √ a + b ± 2 ab. 9). Thirdly, we can verify (after a long and tedious calculation) that 2 v3 2 8(1 − R + 2v 5 ) 1 4R 8 − − √ − 6 v v v v 1 − 8v 5 − 4Rv 5 ⎛ 2 1 8 1 8 = ⎝2 − + − 6 6 v v v v 4R v ⎞2 16R2 ⎠ + 2 .

The proof of (iv) is similar to that of (ii). 4) to find that √ (q 2 ; q 4 )∞ ψ(q 2 ) − q 5ψ(q 10 ) = f 2 (−q 2 ) 10 20 (q ; q )∞ = f (−q 2 ) = 1 √ ψ(q ) + q 5ψ(q 10 ) 2 (ζq; q 2 )∞ (−ζ 2 q; q 2 )∞ (−ζ 3 q; q 2 )∞ (ζ 4 q; q 2 )∞ (ζq 2 ; q 4 )∞ (ζ 2 q 2 ; q 4 )∞ (ζ 3 q 2 ; q 4 )∞ (ζ 4 q 2 ; q 4 )∞ f (−q 2 ) (−ζq; q 2 )∞ (ζ 2 q; q 2 )∞ (ζ 3 q; q 2 )∞ (−ζ 4 q; q 2 )∞ f (−q 2 ) = (1 − αq n + q 2n ) n odd (1 + βq n + q 2n ) , n odd as desired. 8 Identities Involving the Parameter k = R(q)R2 (q 2 ) Recall again that R(q) denotes the Rogers–Ramanujan continued fraction.

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