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A central advantage of cohomology theory over homology -- at least in terms of richness of structure and strength as an invariant -- is the additional ring structure from the cup product. Recall that this arises from applying the cohomology functor to the following inclusion map of topological spaces $$X \hookrightarrow X \times X$$ where each $x \in X$ is mapped to $(x,x)$ in the product. The "key" insight here is that homology theory lacks an analogous structure precisely because there is no natural candidate for a continuous map $X \times X \to X$. Fair enough.

But Lie groups provide examples of spaces where there is a great candidate for such a map: the group multiplication. I expected that this would make homology of Lie groups interesting by imposing some nice multiplicative structure on homology generators inherited from the group multiplication. On the other hand, the cohomology ring would reveal nothing that you couldn't already learn from the cohomology of the underlying manifold independent of group structure.

Is this wrong? Why is the literature full of material on Lie group cohomology whereas Lie group homology is relatively sparse?

I suspect that maybe this product structure is not even well defined on the level of homology, but I'm not sure how one would prove that.

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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. –  Vidit Nanda Jun 29 '12 at 5:57
    
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. –  Will Sawin Jun 29 '12 at 6:12
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The homology and cohomology of a Lie group are Hopf algebras. This example led Hopf to define Hopf algebras. –  Bruce Westbury Jun 29 '12 at 6:25
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Also, there is the Pontryagin product in homology (under certain conditions). –  Chris Gerig Jun 29 '12 at 9:20
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Indeed, I would call the product Vel explained the Pontryagin product (which is, more generally, defined for any H-space). –  Lennart Meier Jun 29 '12 at 11:10
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With field coefficients, homology and cohomology are dual to each other. The cohomology of a space is an algebra, the homology of a space is a coalgebra. When the space is a group (or loop space or $H$-space...) both its homology and cohomology are Hopf algebras.

"The literature" is full of computations of the cohomology of the classifying space $BG$. In some cases by the way, Hopf algebras help, for example if you compute with coefficients in $\mathbb{Q}$. Then graded Hopf algebras over the rational field are classified, which gives you a strong indication on what the cohomology of $G$ looks like, and a spectral sequence argument tells you that $H^*(BG, \mathbb{Q})$ is a polynomial ring, in the end. Details in McCleary's book A user's guide to spectral sequences.

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I am disturbed by this question. The literature of algebraic topology has abounded in use of the product in homology for more than the half century that I've been working in it. Even a cursory knowledge of the subject makes that clear. Shouldn't some effort be made to know just a tiny bit about a subject before asking a question? Similarly for answers. The rational case is by far the easiest case, and of course the classification of Hopf algebras rationally requires commutativity and is a classification of the underlying algebras, not of the Hopf algebras. I apologize if I seem harsh, but in general I am concerned that the decreasing level in MathOverflow, on average, is leading many experts to cease participating (that is not guesswork). That is a pity.

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Dear Peter, pointing out that a sentence such as "the graded Hopf algebras over the rational field are classified" lacks precision is indeed harsh, if fair. The context being cohomology rings, graded commutativity seems implicit to me, and the fact that the Hopf-Borel theorem does not classify the possible diagonal maps seems irrelevant for the application I'm describing. I would have been more careful if (i) I had not given a reference for the full details and (ii) the original question had been more advanced. –  Pierre Jul 1 '12 at 8:56
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Dear Pierre, I apologize if I seem harsh again, but it is not true that the possible diagonal maps (in cohomology) are irrelevant for the application you are describing. One of the very nicest ways to compute $H^*(BG;k)$ is to use the Rothenberg-Steenrod spectral sequence which converges from $Ext_{H_*(G;k)}(k,k)$ to $H^*(BG;k)$. It is precisely the algebra structure on $H_*(G;k)$ that is relevant, hence the coalgebra structure on $H^*(G;k)$, and there is no reason to restrict to $k=\mathbf{Q}$. One answer to the original question is to point to this spectral sequence, which has been in common use since the early 1960's. From the perspective of Lie groups in particular, what is calculationally easy but perhaps conceptually surprising is how often the algebra $H_*(G;k)$ is (graded) commutative even though the product on $G$ is almost never homotopy commutative. Commutative and cocommutative Hopf algebras over $\mathbf{Q}$ are classified, and in this case the relevant $Ext$ groups are trivial to compute.

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If we want to be precise (and I think we do), this spectral sequence is indeed irrelevant if one wants to prove the result I mentioned, to the effect that $H^*(BG,\mathbb{Q})$ is a polynomial ring (see theorem 6.38 in McCleary's book as well as th 3.27). I never said one had to restrict to $\mathbb{Q}$. As should be apparent now, I was trying to state something as elementary as possible, while... (to be continued) –  Pierre Jul 1 '12 at 15:46
    
while using the Hopf algebra structure a little. It was worth providing complements to what I said. I object to the way it was phrased. PS I should make the following point clearer: the simple (and very nice IMHO) approach in McCleary's book does not work for other fields, unless you throw more hypotheses in. –  Pierre Jul 1 '12 at 15:48
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Surely an alternative proof is not irrelevant! In the Rothenberg-Steenrod spectral sequence, when H_*(G;k) is an exterior algebra, E_2 is a polynomial algebra, there are no differentials for dimensional reasons (if char k is not 2) and it follows immediately that H^*(BG;k) is a polynomial algebra. In char 0, being an exterior algebra is equivalent to being commutative. McCleary's 6.38 is stated for general (perfect) fields, with the evident hypotheses. The universally transgressive hypothesis ensures that H_*(G;k) is an exterior algebra, which is what connects the proofs. Peace. –  Peter May Jul 1 '12 at 16:24
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