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S Apr 17, 2021 at 21:44 history bounty ended CommunityBot
S Apr 17, 2021 at 21:44 history notice removed CommunityBot
Apr 9, 2021 at 21:35 comment added DamienC Would you consider the approach of Alekseev-Meinreinken-Woodward as geometric (and modern)? arxiv.org/pdf/math/0005047.pdf -- though it is for compact groups
S Apr 9, 2021 at 19:08 history bounty started Pulcinella
S Apr 9, 2021 at 19:08 history notice added Pulcinella Authoritative reference needed
Apr 9, 2021 at 12:40 comment added naf The proof in Beauville's paper assumes the factorisation rules: the main point there is to show why this implies the formula actually conjectured/claimed by Verlinde.
Apr 9, 2021 at 8:49 history edited Pulcinella CC BY-SA 4.0
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Apr 7, 2021 at 16:51 answer added Carlo Beenakker timeline score: 4
Apr 7, 2021 at 11:08 history edited Pulcinella CC BY-SA 4.0
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Apr 7, 2021 at 10:26 comment added ssx I don't know what "modern" or "geometric" mean to you, but I would say the proof of Betram-Szenes (type A) is quite geometric. The main idea is that because the Verlinde dimensions are deformation-invariant and the moduli space of curves is irreducible, it suffices to prove it for a hyperelliptic curve, and then use the connection between moduli of bundles on hyperelliptic curves and Fano schemes of intersections of quadrics. The proof by Thaddeus (using stable pairs) I also geometric.
Apr 7, 2021 at 9:48 comment added Pulcinella P.S. There is a chance that Falting's 1994 paper A proof for the Verlinde formula just answers this, but I can't access a copy online.
Apr 7, 2021 at 9:47 history asked Pulcinella CC BY-SA 4.0