[Math] When are the homology and cohomology Hopf algebras of topological groups equal

at.algebraic-topologycohomologyhomologyhopf-algebrastopological-groups

Suppose we have a topological group $G$, then the multiplication map $\mu$ and the diagonal map $\Delta$ provide the cohomology $H^\ast(G;R)$ (with Pontryagin coproduct and cup coproduct) and homology $H_\ast(G;R)$ (with Pontryagin product and diagonal coproduct) of $G$ with the structure of a Hopf algebra (assuming vanishing of the Tor-terms in the Kunneth theorem). If both satisfy some mild conditions, these are in fact dual. A nice historical explanation of this can be found in Cartier's "Primer on Hopf Algebras" (http://inc.web.ihes.fr/prepub/PREPRINTS/2006/M/M-06-40.pdf). My general question is the following: when we know they are in fact isomorphic as Hopf algebras?

The Samelson theorem says that if $G$ is a Lie group and $R$ is a field of characteristic zero, then $H^\ast(G;R)$ is commutative, associative and coassociative and an exterior algebra on primitive generators of odd degree. I think this implies that $H^\ast(G;R)$ is isomorphic to $H_\ast(G;R)$ as a Hopf algebra. Is this correct? To what extent does Samelson's result extend to non-zero characteristic?

My goal is to limit the amount of calculation when I do calculations of the homology and cohomology of Lie groups.

Best Answer

The mod $2$ cohomology of $\mathrm{SO}_n$ is not an exterior algebra as soon as $n\geq3$ (there is a unique non-zero element in degree $1$ whose square is non-zero, think of the case $\mathrm{SO}_{3}$ which is $3$-dimensional real projective space) while the homology ring is an exterior algebra.

Another example is $K(\mathbb Z,2)$ (aka $\mathbb C\mathbb P^\infty$), its integral cohomology ring is a polynomial ring on a degree $2$-generator while its homology ring is the free divided power algebra on a degree $2$-generator.

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