3
$\begingroup$

Let $f\colon E\to B$ locally trivial bundle of 'nice' topological spaces (say finite CW-complexes) with fiber $F$. Assume also that the base $B$ is simply connected.

Assume that either the cohomology spectral sequence (with coefficients in a field) degenerates in the second term $H^*(B)\otimes H^*(F)$ or that the push-forward in the derived category of the constant sheaf is isomorphic (in derived category) to the direct sum of its cohomology sheaves (I think that the second condition implies the first one, but I am not sure about the converse).

Is it true that there exists an isomoprhism of graded algebras $$H^*(E)\simeq H^*(B)\otimes H^*(F)$$ which is compatible with the pull-back $f^*\colon H^*(B)\to H^*(E)$ and with restriction to fiber $H^*(E)\to H^*(F)$?

$\endgroup$
  • 1
    $\begingroup$ If you have an isomorphism, it will be compatible with the two maps by naturality. However, in general there can be an algebra extension problem, if I am not mistaken. $\endgroup$ – user43326 Dec 24 '19 at 10:46
  • 4
    $\begingroup$ A counterexample is the non-trivial S^2 bundle over S^2 (equivalently the blow-up of CP^2 at one point). $\endgroup$ – Gustavo Granja Dec 25 '19 at 10:00
  • $\begingroup$ @GustavoGranja: What is cohomology ring of that space? $\endgroup$ – makt Dec 25 '19 at 10:40
  • $\begingroup$ @MKO $Z[x]/x^3$ with $deg(x)=2$. $\endgroup$ – user43326 Dec 25 '19 at 11:58
  • $\begingroup$ @user43326: The cohomology should be 4-dimensional. $\endgroup$ – makt Dec 25 '19 at 13:54
5
$\begingroup$

The statement is false, here is a counterexample. First note that for a Lie Group $G$ and its closed subgroup $H$, we have a fibration $G/H\to BH\to BG$. $BG$ and $BH$ are not finite, but they are almost just as good, if you really want finite CW complex, you can restrict the fibration over a finite skelton of $BH$. Now, let's take $G=S^3$, $H=S^1$. Then $G/H$ is homeomorphic to $S^2$. If we restrict this fibration to the $2$-skelton of $BH$, one gets the example mentioned in the comment by @Gustavo Granja, but let's look at the whole thing. We have $BG=CP^{\infty }$, $BH=HP^{\infty }$ so, the $E^2$ term of the spectral sequence is $$P[z]\otimes \Lambda (y)$$ with $y$ in degree 2, $z$ in degree 4, whereas the cohomology of the total space is $P[x]$ with $x$ in degre 2.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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