Skip to main content
8 events
when toggle format what by license comment
Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Apr 15, 2013 at 23:17 vote accept Martin Brandenburg
Apr 6, 2010 at 7:33 comment added Martin Brandenburg thanks. I've edited the description of the $K$ ring in my question.
Apr 6, 2010 at 7:32 history edited Martin Brandenburg CC BY-SA 2.5
deleted 1 characters in body; added 37 characters in body
Apr 6, 2010 at 7:05 comment added Torsten Ekedahl Note that just adding additive inverses to the semiring does <em>not</em> give the usual Grothendieck group. In fact the semiring is, by the Krull-Remak-Schmidt theorem, the free abeliam semigroup on the (isomorphism classes of) indecomposable vector bundles. Hence its group completion is just the free abelian group on the same vector bundles. Hence it is, as Angelo pointed out for the semiring, truly huge. Except in the $1$-dimensional case when the only indecomposable vector bundles are the line bundles.
Apr 6, 2010 at 3:33 answer added Angelo timeline score: 10
Apr 6, 2010 at 0:04 history edited Martin Brandenburg CC BY-SA 2.5
added 50 characters in body; added 17 characters in body; deleted 36 characters in body
Apr 5, 2010 at 23:58 history asked Martin Brandenburg CC BY-SA 2.5