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Oct 30, 2021 at 9:08 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Sep 30, 2021 at 9:00 history edited YCor CC BY-SA 4.0
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Sep 30, 2021 at 5:46 answer added AG learner timeline score: 2
Aug 28, 2021 at 19:31 comment added AG learner What you essentially need is that there is a non-degenerate pairing between the Dolbeault cohomology and the space of holomorphic forms: $H^{0,1}_{\bar{\partial}}(X)\times H^{1,0}(X)\to \mathbb C$. This is a version of Serre's duality. It can be proved analytically without using Riemann-Roch. For example, see section 9 of this book.
Aug 27, 2021 at 20:31 comment added Sándor Kovács Actually, in my opinion, the natural direction is to prove Serre duality without Riemann-Roch. Riemann-Roch is essentially a direct consequence of two facts: 1) that holomorphic Euler characteristic is additive on short exact sequences and 2) Serre duality. Also, I don't think the general form of Serre duality would follow from Riemann-Roch....
Aug 24, 2021 at 10:50 history asked Serge the Toaster CC BY-SA 4.0