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If $f:X\to\text{Spec}(k)$ is a smooth projective variety over a separably closed field, with $i:Z\subset X$ a closed smooth sub variety, with complement $j: U\to X$, and $f':X'\to \text{Spec}(k)$ is the blow-up of $X$ along $Z$, we have, in $\ell$-adic cohomology:

$$(*)\ \ H^a(X',\mathbf{Z}_{\ell}(n)) = H^a(X,\mathbf{Z}_{\ell}(n))\oplus\bigoplus_{j=1}^{c-1}H^{a-2j}(Z,\mathbf{Z}_{\ell}(n-j)).$$

Is there, in addition, a splitting of the complex $Rf'_*\mathbf{Z}_{\ell}(n)$ in the derived category of $\ell$-adic sheaves? What is this splitting?

It should be something like

$$Rf'_*\mathbf{Z}_{\ell}(n) = Rf_*\mathbf{Z}_{\ell}(n)\oplus\bigoplus_j R^j(f\circ i)_*\mathbf{Z}_{\ell}(n-c)[-2j+2c]\ ?$$

I'd like to ask for a reference for the proof of the derived version of formula $(*)$, or an argument, with correct weights and shifts.

Thanks a lot!

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    $\begingroup$ SGA5 Exposé VII is a good source for such computations. Theorem 8.1 treats general blow-ups. Combined with the projective bundle formula (Theorem 2.2.1), it gives the desired splitting in the derived category of l-adic sheaves on X. It's like your first formula with cohomology replaced by pushforwards to X. $\endgroup$
    – Marc Hoyois
    Commented Feb 27, 2018 at 17:13
  • $\begingroup$ @MarcHoyois A quick followup. Where is the $\ell$-adic cohomology of the exceptional divisor in the left side in the above formula supposed to be? $\endgroup$
    – user92332
    Commented Feb 28, 2018 at 3:27
  • $\begingroup$ I'm not sure I understand the question. The exceptional divisor is the big direct sum on the right-hand side of your first formula (minus the summand corresponding to Z). That's the projective bundle formula. $\endgroup$
    – Marc Hoyois
    Commented Feb 28, 2018 at 4:29
  • $\begingroup$ Oh I see, there's an $H^a(Z,\mathbf{Z}_{\ell}(n))$ on both sides that splits off and can be removed, to get the first formula. $\endgroup$
    – user92332
    Commented Feb 28, 2018 at 4:38

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