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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jan 23, 2010 at 7:26 vote accept algori
Jan 22, 2010 at 4:58 comment added Tom Church Characteristic classes are topological, so all we need from the almost complex structure is a splitting of TM tensor C; we have an isomorphism over R between the tangent bundle and the holomorphic tangent bundle, with isomorphism given by v -> v + Jv. The top Chern class is the Euler class [Morita, Geometry of characteristic classes, Proposition 5.43: e(E) = c_n(E) for E an n-dim complex vector bundle]. Thus c_n(T_holo M) = e(TM) = chi(M).
Jan 22, 2010 at 4:02 comment added algori David, thanks! Yes, the top Chern class of any complex bundle is the Euler class of the realization of that bundle, so I think this works.
Jan 22, 2010 at 3:49 history answered David E Speyer CC BY-SA 2.5