4
$\begingroup$

Let $f\colon E\to B$ be a fiber bundle with a connected fiber $F$, $f$ is proper. Let $\underline{\mathbb{C}}_E$ be the constant sheaf on $E$. Let $f_*(\underline{\mathbb{C}}_E)$ denote its direct image in the derived category $D(Sh_B)$ of sheaves of $\mathbb{C}$-vector spaces. Since $F$ is connected, it is clear that $\tau_{\leq 0}(f_*(\underline{\mathbb{C}}_E))=\underline{\mathbb{C}}_B$. Hence there exists an exact triangle in $D(Sh_B)$ $$\underline{\mathbb{C}}_B\to f_*(\underline{\mathbb{C}}_E)\to \mathcal{F},$$ where $\mathcal{F}\in D^{\geq 1}(Sh_B)$.

Question. Is it true that the above exact triangle splits? If not, under what conditions this is true?

Remark. I am aware of a situation when the above triangle splits, but it is too restrictive for my purposes; in fact much more it true in that case. Let $f\colon E\to B$ be a smooth morphism of smooth projective complex algebraic manifolds. Then it is a special case of the decomposition theorem due to Beilinson-Bernstein-Deligne-Gabber that not only the above exact triangle does split, but moreover $f_*(\underline{\mathbb{C}}_E)$ is isomorphic in $D(Sh_B)$ to the direct sum of its cohomology sheaves (which are necessarily shifted local systems).

$\endgroup$
  • 3
    $\begingroup$ Consider the Hopf fibration $S^3 \to S^2$. If your sequence split, then $H^*(S^2)$ would be a summand of $H^*(S^3)$. $\endgroup$ – Vivek Shende Apr 6 '16 at 16:03
3
$\begingroup$

No, it is not true. For example, let $E=\mathbb C^2\backslash \{ 0\}$, $B=\mathbb C\mathbb P^1$ (with the obvious map $f$). Then the pushforward as a complex of sheaves on $\mathbb C\mathbb P^1$ has the following cohomology: constant sheaf in degree 0 and constant sheaf in degree +1. If the triangle you are asking about were split, then the cohomology of the pushforward would be $H^*(\mathbb C\mathbb P^1)\oplus H^*(\mathbb C\mathbb P^1)[-1]$. But it the cohomology of the pushforward is the same as $H^*(\mathbb C^2\backslash \{ 0\})$, which has total dimension 2 and not 4.

$\endgroup$
  • 1
    $\begingroup$ Interesting example, thanks. Here the map is not proper, but homotopically your example is equivalent to that in Vivek Shende' comment, where the map is proper. $\endgroup$ – MKO Apr 6 '16 at 16:17
  • 3
    $\begingroup$ Well, unless you work in an algebraic (or, at least, complex analytic) context, properness is not a sensible condition (almost any nice map is homotopy equivalent to a proper one). $\endgroup$ – Alexander Braverman Apr 6 '16 at 16:19

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.