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Oct 25, 2013 at 0:39 comment added Jonathan Beardsley Perhaps checkout Borel's Paper? ncbi.nlm.nih.gov/pmc/articles/PMC1063923/pdf/pnas01596-0040.pdf
Jul 1, 2012 at 13:54 answer added Peter May timeline score: 5
Jun 30, 2012 at 13:27 answer added Peter May timeline score: 2
Jun 29, 2012 at 11:10 comment added Lennart Meier Indeed, I would call the product Vel explained the Pontryagin product (which is, more generally, defined for any H-space).
Jun 29, 2012 at 9:43 answer added Pierre timeline score: 4
Jun 29, 2012 at 9:20 comment added Chris Gerig Also, there is the Pontryagin product in homology (under certain conditions).
Jun 29, 2012 at 6:25 comment added Bruce Westbury The homology and cohomology of a Lie group are Hopf algebras. This example led Hopf to define Hopf algebras.
Jun 29, 2012 at 6:12 comment added Will Sawin It's well-defined, since clearly a cycle-wthout boundary cross a cycle without boundary is a cycle without boundary, so you have a map on the chain complex, and $(\partial A)\times B=\partial (A\times B)$ as long as $\partial B$ is trivial, so it sends boundaries to boundaries. Then obviously the push-forward is also well-defined.
Jun 29, 2012 at 5:57 comment added Vidit Nanda Of course, I have just realized that the cohomology cup product construction requires the Kunneth map also, since the cohomology of the product is not the product of the cohomology. I'm not sure how this translates into the "homology of Lie groups" setting though, so I will leave the question as it stands.
Jun 29, 2012 at 5:48 history asked Vidit Nanda CC BY-SA 3.0