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Nov 17, 2017 at 16:14 vote accept Vamsi
Nov 17, 2017 at 13:50 answer added Jason Starr timeline score: 9
Nov 17, 2017 at 13:34 comment added Jason Starr That is not true. I will write an explanation as an answer below.
Nov 17, 2017 at 9:42 comment added js21 Oh, ok, I just saw the "up to isomorphism" condition.
Nov 17, 2017 at 9:41 comment added js21 A trivial bundle of rank $>1$ will satisfy your semistability condition with $\mu =0$ while having uncountably many trivial subbundles (all semistable of slope $0$).
Nov 17, 2017 at 8:28 history asked Vamsi CC BY-SA 3.0