8
$\begingroup$

Let $X$ be a smooth projective variety over $\mathbf{C}$, and $p : X_{\rm an}\to X_{\rm Zar}$ the obvious map of sites.

The Leray spectral sequence

$$H^r(X_{\rm Zar}, R^sp_*\mathbf{C})\Rightarrow H^{r+s}(X_{\rm an},\mathbf{C})$$ induces a filtration on $H^*(X_{\rm an},\mathbf{C})$ that agrees with the coniveau filtration, as shown by Bloch and Ogus in their paper on homology of schemes.

We have an exact sequence:

$$H^n(X_{\rm Zar}, R^np_*\mathbf{C})\to H^{2n}(X_{\rm an},\mathbf{C})\to H^1(X_{\rm Zar}, R^{2n-1}p_*\mathbf{C})$$ if I'm not wrong (can anyone confirm this, first of all?)

My question is: how can one describe $H^1(X_{\rm Zar}, R^{2n-1}p_*\mathbf{C})$? For example, $H^n(X_{\rm Zar}, R^np_*\mathbf{C})$ is $A^n(X)_{\rm alg}\otimes\mathbf{C}$, the Chow group of cycles up to algebraic equivalence.

Is there a Hodge decomposition for $H^1(X_{\rm Zar}, R^{2n-1}p_*\mathbf{C})$, induced by that of the cohomology of $\mathbf{C}$? Can one use the comparison with algebraic de Rham cohomology?

For instance one could say

$$Rp_*\mathbf{C} = Rp_*\Omega^{\bullet}_{X_{\rm an}}=Rp_*p^*\Omega^{\bullet}_X$$

where $\Omega^{\bullet}_X$ is the algebraic de Rham complex. Is the natural map $$\Omega_X^{\bullet}\to Rp_*p^*\Omega^{\bullet}_X$$ a quasi-isomorphism of Zariski sheaves on $X$?

$\endgroup$
3
  • 3
    $\begingroup$ I almost read that as "bogus spectral..." 😊 $\endgroup$
    – Suvrit
    Commented Apr 2, 2018 at 1:04
  • $\begingroup$ You're not the first one to point that out to me :) $\endgroup$
    – user92332
    Commented Apr 2, 2018 at 1:06
  • 2
    $\begingroup$ Can you explain where your sequence comes from? Usually the relationship between the terms of a spectral sequence and its abutment is a bit more complicated, so I don't see what you're using. $\endgroup$ Commented Apr 20, 2018 at 0:49

0

You must log in to answer this question.