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Apr 30, 2011 at 19:09 comment added Kevin Wray Wait, I thought I'd read somewhere that $\Omega_3(BG)$ was equivalent to $H_3(BG)$ up to torsion.
Apr 30, 2011 at 6:57 comment added Torsten Ekedahl The unoriented case is even simpler as unoriented bordism is just homology with coefficients in the (unoriented) bordism ring.
Apr 30, 2011 at 3:30 comment added Sean Tilson @Dylan: I think the exact couple set up makes it reasonably clear. Have you looked at Adams?
Apr 30, 2011 at 2:54 comment added Dylan Wilson Is it obvious that Atiyah-Hirzebruch applies to the associated homology of a spectrum in the same way as the associated cohomology?
Apr 29, 2011 at 22:30 comment added Kevin Wray Thank you! Can you say anything as nice for the unoriented case?
Apr 29, 2011 at 21:46 history answered Johannes Ebert CC BY-SA 3.0