Does anyone have an example of two spaces which have the same homology groups, the same cohomology groups, but have different cohomology rings?
is it possible?
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2$\begingroup$ sounds like a homework problem to me. $\endgroup$– Ian AgolCommented Mar 2, 2011 at 17:59
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$\begingroup$ it isn't. I am teaching myself the material. $\endgroup$– rhlCommented Mar 2, 2011 at 19:40
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$\begingroup$ Slightly different example I quite like: the two partial flag manifolds for $SO(5)$ have the same cohomology groups (free abelian, in even degrees), the same rational cohomology rings, and different integer cohomology rings. $\endgroup$– Allen KnutsonCommented Mar 4, 2011 at 4:07
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2 Answers
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A very standard example would be $S^2\vee S^4$ and $\mathbf{C}P^2$.
answered Mar 2, 2011 at 15:59
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$\begingroup$ hm, OK. but these have different homotopy groups, Looks like \pi_5(S^2 \wedge s^4) = Z_2 x Z_2 but \pi_5(CP^2) = Z correct? It's not clear to me that the cohomology ring is actually a finer invariant than cohomology. When I read, all sources say 'the structure is richer than homology,' but is it useful? maybe mathoverflow is not an appropriate forum for this question? $\endgroup$– rhlCommented Mar 2, 2011 at 19:44
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$\begingroup$ math.stackexchange.com might be better for you. Also, this is the first example I learned. It satisfies all of the requirements listed in your question. The ring structure on $S^2 \vee S^4$ is trivial, but not on $\mathbb{C}P^2$. $\endgroup$ Commented Mar 2, 2011 at 21:43
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$\begingroup$ oh, and cohomology is easier to compute than homotopy for finite dimensional cell/CW complexes. $\endgroup$ Commented Mar 2, 2011 at 21:44
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There are also standard examples in which both spaces are compact manifolds. For instance, if $n \geq 1$ is an integer and $Q_n \subset \mathbb{P}^{2n+2}$ is a non-singular quadric, then $Q_n$ has the same integral homology and cohomology groups as $\mathbb{P}^{2n+1}$, but the cohomology rings are different.
