Is anything known about $\Omega_3(BG)$, where $G$ is an arbitrary compact Lie group; i.e., is it possible to describe the structure of $\Omega_3(BG)$ for any compact Lie group? I know that $H_3(BG)$ consists completely of torsion when $G$ is compact and I would like (if possible) a similar type of statement for $\Omega_3(BG)$.
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If you think about oriented bordism, the answer is that $\Omega_3 (BG) \cong H_3 (BG)$. This is true for any space $X$ instead of $BG$, because of the Atiyah-Hirzebruch spectral sequence and because $\Omega_i (pt)=0$ for $i=1,2,3$.
answered Apr 29, 2011 at 21:46
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$\begingroup$ Thank you! Can you say anything as nice for the unoriented case? $\endgroup$ Commented Apr 29, 2011 at 22:30
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$\begingroup$ Is it obvious that Atiyah-Hirzebruch applies to the associated homology of a spectrum in the same way as the associated cohomology? $\endgroup$ Commented Apr 30, 2011 at 2:54
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$\begingroup$ @Dylan: I think the exact couple set up makes it reasonably clear. Have you looked at Adams? $\endgroup$ Commented Apr 30, 2011 at 3:30
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$\begingroup$ The unoriented case is even simpler as unoriented bordism is just homology with coefficients in the (unoriented) bordism ring. $\endgroup$ Commented Apr 30, 2011 at 6:57
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$\begingroup$ Wait, I thought I'd read somewhere that $\Omega_3(BG)$ was equivalent to $H_3(BG)$ up to torsion. $\endgroup$ Commented Apr 30, 2011 at 19:09