# Spaces which have the same homology groups, the same cohomology groups, but have different cohomology rings?

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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sounds like a homework problem to me. – Ian Agol Mar 2 '11 at 17:59
it isn't. I am teaching myself the material. – rhl Mar 2 '11 at 19:40
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. – Allen Knutson Mar 4 '11 at 4:07

## 2 Answers

A very standard example would be $S^2\vee S^4$ and $\mathbf{C}P^2$.

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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? – rhl Mar 2 '11 at 19:44
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 $\mathbb{C}P^2$. – Sean Tilson Mar 2 '11 at 21:43
oh, and cohomology is easier to compute than homotopy for finite dimensional cell/CW complexes. – Sean Tilson Mar 2 '11 at 21:44

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.

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