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Jan 29, 2012 at 6:06 comment added Tom Church Isn't this formula essentially tautological, once you believe that this triangulation exists? i.e. a similar formula exists for any triangulated manifold (with no need for the 1/|Aut| factor for an honest triangulation, of course).
Jan 29, 2012 at 2:27 comment added Steve Right. By ``similar formula of arbitrary degree'' I'm asking for the integral over a cycle of the appropriate dimension.
Jan 28, 2012 at 20:01 answer added Igor Rivin timeline score: 3
Jan 28, 2012 at 18:56 comment added John Pardon (of course, neither side makes sense if $\omega$ is not top dimensional)
Jan 28, 2012 at 18:06 history asked Steve CC BY-SA 3.0