Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say we have a finite CW complex with cells only in even degrees. For example a $\mathbb {CP}^n$ or a complex flag variety. If we know the rational cohomology ring, does it also determine the integral cohomology ring?

share|cite|improve this question
up vote 9 down vote accepted

Form two CW-complexes $X_f$ and $X_g$ by choosing attaching maps $f, g: S^3 \to S^2$ with Hopf invariant $H(f) = 1$ and $H(g) = 2$. Then, $H^*(X_f, \mathbb Q)$ and $H^*(X_g, \mathbb Q)$ are isomorphic as graded rings, but $H^*(X_f, \mathbb Z) = \mathbb Z[x_2]/(x_2^3)$ is not isomorphic to $H^*(X_g, \mathbb Z) = \mathbb Z[x_2, y_4]/(x_2^2 - 2y_4, y_4^2, x_2y_4)$.

share|cite|improve this answer
Do these complexes only have cells in even dimensions? – MartianInvader Oct 27 '11 at 19:01
Yes, they both have 1 cell each in dimensions 0, 2 and 4. (Using the appropriate CW structure on $S^2$). – Justin Young Oct 27 '11 at 19:04
Ah, I see, I misunderstood how your construction worked. I now understand you're simply describing an attaching map of a 4-cell onto $S^2$. – MartianInvader Oct 27 '11 at 19:13
The OP asked for a finite CW complex, but then gave examples which are closed manifolds. Do you happen to know the answer if we restrict to closed manifolds? I'd be surprised if the rational cohomology ring determines the integral ring in this case, but I don't know of any counter examples off hand. – Jason DeVito Oct 27 '11 at 19:21
Thanks for the fast answer! – Jan Oct 27 '11 at 19:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.