2
$\begingroup$

Let me assume everything in sight is as nice as possible, probably if the result I want is true then these conditions are too restrictive. All spaces will be smooth algebraic varieties over the complex numbers. We are given maps $j \colon U \to X$, $g \colon X \to S$, $f = g\circ j$. The map $j$ is an open immersion whose complement is a simple normal crossing divisor, $g$ is a smooth projective morphism, and $f$ is topologically a locally trivial fibration.

On one hand, we can restrict cohomology classes on $X$ to $U$ fiberwise, giving us $$ \newcommand{\Q}{\mathbf{Q}}R^qg_\ast \Q \twoheadrightarrow W_qR^qf_\ast\Q \hookrightarrow R^qf_\ast\Q.$$ Here $W_\bullet$ denotes the weight filtration on $R^qf_\ast\Q$, considered as a variation of mixed Hodge structure.

On the other hand, one can also consider $$ H^\bullet(X,\Q) \twoheadrightarrow \mathrm{Im}(j^\ast) \hookrightarrow H^\bullet(U,\Q)$$ given by restricting cohomology classes globally.

Question 1: Is there a "Leray" spectral sequence $H^p(S,W_qR^qf_\ast\Q) \implies \mathrm{Im}(j^\ast)$, compatible with the maps above and the Leray spectral sequences for $f$ and $g$?

Question 2: If so, does it always degenerate at $E_2$, like the Leray spectral sequence for $g$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

You can identify $$W_qR^qf_*\mathbb{Q}=im[R^qg_*\mathbb{Q}\to R^qf_*\mathbb{Q}]$$ It is enough to check this fibrewise, where it's Deligne's Hodge II, cor 3.2.17. Now compare Leray spectral sequences for $g$ and $f$, and take the image $$im ([E_2(g)\Rightarrow H^*(X)]\to [E_2(f)\Rightarrow H^*(U)])$$ This should give your desired answer to Question 1. [Note: there's a subtle strictness question that I overlooked. I'll try to sort it out when I have more time. ]

For Q2, let's first suppose that $S$ is smooth and proper. Then $H^p(S, W_qR^qf_*\mathbb{Q})$ is pure of weight $p+q$, so $d_2,\ldots$ must be zero because it goes between Hodge structures of different weights. This is just the barest outline, but see my paper for some more details. I think the result is true in general, but you would need to use Saito's version of the decomposition theorem in his category of polarizable Hodge modules. I'll see if I can supply some more precise arguments later on.

Added Note: As Dan noted below, Q2 follows easily from the first paragraph, and Deligne's degeneration argument for $g$.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Thanks! Actually, doesn't this prove Q2, even if $S$ is not smooth/proper? Since $[E_2(g) \implies H^\bullet(X)]$ has zero differentials, $\mathrm{Im}([E_2(g) \implies H^\bullet(X)] \to [E_2(f) \implies H^\bullet(U)])$ should also have vanishing differentials. $\endgroup$
    – Dan Petersen
    Commented Jul 13, 2012 at 12:29
  • $\begingroup$ Yes, good point. You've answered your own question, which is always the best way. $\endgroup$
    – Donu Arapura
    Commented Jul 13, 2012 at 12:39
  • $\begingroup$ @Donu. Actually, now I am starting to doubt the argument. In general for a map $f \colon A^\bullet \to B^\bullet$ of complexes, there is no reason to have $H^\bullet(\operatorname{Im} f) \cong \operatorname{Im}(H^\bullet(f))$. So now I don't see why it should be automatic that $H^p(S,W_qR^qf_\ast\mathbf Q)$ converges to $\operatorname{Im}(j^\ast)$. $\endgroup$
    – Dan Petersen
    Commented Jul 13, 2012 at 13:07
  • $\begingroup$ Dan, right again. I was too quick to type my original answer. Unfortunately, I tend not to think deeply about MO questions. So you if it's something that's really important to you, you can send me email (I'm easy to find). Let me give a quick fix with a stronger hypothesis for now. $\endgroup$
    – Donu Arapura
    Commented Jul 13, 2012 at 13:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .