Third bordism group of BG, where G is an arbitrary compact Lie group. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T20:25:57Z http://mathoverflow.net/feeds/question/63470 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63470/third-bordism-group-of-bg-where-g-is-an-arbitrary-compact-lie-group Third bordism group of BG, where G is an arbitrary compact Lie group. Kevin Wray 2011-04-29T20:27:13Z 2011-04-29T21:46:46Z <p>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)\$.</p> http://mathoverflow.net/questions/63470/third-bordism-group-of-bg-where-g-is-an-arbitrary-compact-lie-group/63478#63478 Answer by Johannes Ebert for Third bordism group of BG, where G is an arbitrary compact Lie group. Johannes Ebert 2011-04-29T21:46:46Z 2011-04-29T21:46:46Z <p>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\$.</p>