8
$\begingroup$

Let $F\to E\to B$ be a fibration with $B$ simply-connected. Suppose all differentials in the cohomology Serre spectral sequence (corresponding to the above fibration) are zero maps. Then as a graded module, $$ H^*(E)\cong H^*(F)\otimes H^*(B). $$

Question 1: as cohomology rings with cup products, do we still have the isomorphism $$ H^*(E)\cong H^*(F)\otimes H^*(B)? $$

My idea of Question 1: there is a product in the $E_r$-page $$ E^{p,q}_r\times E^{s,t}_r\to E^{p+q,s+t}_r $$ which induces a product in the $E_{r+1}$-page $$ E^{p,q}_{r+1}\times E^{s,t}_{r+1}\to E^{p+q,s+t}_{r+1}. $$ Since all differentials are zero, the products on $E_r$-pages are same for each $r$. Hence there is a product on $E_\infty$-page which is the same as the product on $E_2$-page, given by the product of the tensor algebra $H^*(F)\otimes H^*(B)$. However, these products are in the pages of the Serre spectral sequence, but not in $H^*(E)$.

Question 2: suppose the cohomology coefficient is taken in $\mathbb{Z}_2$ and the Steenrod operations on $H^*(F;\mathbb{Z}_2)$, $H^*(B;\mathbb{Z}_2)$ are already known. Is it possible to determine the Steenrod operations on $H^*(E;\mathbb{Z}_2)$?

$\endgroup$
  • 1
    $\begingroup$ Note that, even if all differentials are zero, there may still be nontrivial extensions. In particular, $H^*(E)\neq H^*(B)\otimes H^*(F)$ in general. As an example, consider the Serre SS for the fibration $K(\mathbb{Z}/2,1)\rightarrow K(\mathbb{Z},2)\rightarrow K(\mathbb{Z},2)$, where the second map is multiplication by $2$. $\endgroup$ – Achim Krause Aug 29 '15 at 9:18
  • 1
    $\begingroup$ If all differentials are zero, you only get an isomorphism with field coefficients, in general, and even then it is not natural. Things like the cup product and steenrod operations are only determined "modulo lower filtration", so you can't get a complete description. $\endgroup$ – Achim Krause Aug 29 '15 at 9:20
12
$\begingroup$

1) No in general. A counterexample is the projective space bundle associated to a vector bundle. For a rank $n$ vector bundle, the fiber, $\mathbb C \mathbb P^{n-1}$, has cohomology ring $\mathbb Z[x]/(x^n)$ with $x$ in degree $2$. But in the total space, the usual lift of $x$ satisfies the equation

$$\sum_{i=0}^n (-1)^i c_i (V) x^{n-i}=0$$

Since this equation is not the same as $x^n=0$, you don't get an isomorphism unless the Chern classes all vanish. (You can get an isomorphism by changing $x$ if the Chern classes satisfy a very specific equation, but for a general vector bundle it will fail.)

I think an explicit explicit example where it fails is the fibrartion of the complete flag variety in $\mathbb C^3$ over $\mathbb P^2$ by $\mathbb P^1$.

2) I don't know - maybe real vector bundles give a similar counterexample?

$\endgroup$
  • 7
    $\begingroup$ What one does have is an algebra isomorphism of the gr of the limit with the E^2 page. In this example one sees exactly what gets lost when passing to the associated graded algebra. $\endgroup$ – Mariano Suárez-Álvarez Aug 29 '15 at 5:13
12
$\begingroup$

The behavior of the Steenrod squaring operations in the Serre spectral sequence was determined by Araki and independently by Vázquez (whose article I cannot locate online). However, it's a little work to extract the statement from Araki's paper. Roughly, there are $Sq^n$ operations, they are compatible with the differentials (though this is a little sensitive), and on the $E_\infty$ page they give the Steenrod operations on the associated graded of the cohomology of the total space (i.e. they give about the same amount of information as the cup product in the Serre spectral sequence).

One of the big differences is that the target of the Steenrod operations $Sq^n$ moves. For example, we have: $$ \begin{align*} Sq^n:&E_\infty^{p,q} \to E_\infty^{p,q+n} & \text{if }&n \leq q,\\ Sq^n:&E_\infty^{p,q} \to E_\infty^{p+(n-q),2q} & \text{if }&p+q \geq n \geq q.\\ \end{align*} $$ There's actually a good reason for this. Roughly, if you imagine that your terms in the spectral sequence are $\alpha \otimes \beta$ for $\alpha$ in $H^p(B)$ and $\beta \in H^q(F)$, this corresponds to applying the Cartan formula for $Sq^n(\alpha \otimes \beta)$, eliminating the terms $Sq^a(\alpha) \otimes Sq^b(\beta)$ which are forced zero for degree reasons (if $a > p$ or $b > q$), and picking the term with maximal filtration degree $q+b$.

$\endgroup$

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.