4
$\begingroup$

Suppose $f: X \to Y$ is an $n$-dimensional affine space fibration of schemes: by this I mean that there exists an étale cover $\{U_i \to Y\}_{i \in I}$ such that the map $X \times_Y U_i \to U_i$ is isomorphic to $U_i \times \mathbb{A}^n \to U_i$ for all $i \in I$.

Suppose $\mathcal{F}$ is an object in the bounded derived category of $\ell$-adic sheaves on $Y$. Is it then true that $$ R\Gamma_c(Y,\mathcal{F}) = R\Gamma_c(X,Rf^*\mathcal{F})[2n]?$$ I'm struggling to cite the right base change theorem to make this work.

$\endgroup$

1 Answer 1

3
$\begingroup$

The answer is yes, suitably interpreted (and with an added Tate twist): Let me assume that $Y$ is separated and of finite type over a field $\operatorname{Spec} k$ with structure map $\pi$. Then there is an isomorphism of (complexes of) sheaves on $Y$ inducing your desired isomorphism after applying $R \pi_{!}$.

Let $\omega_Y$ (respectively $\omega_X$) be the dualizing sheaf on $Y$ (respectively $X$) and write $\mathbb{D}_Y$ (respectively $\mathbb{D}_X$) for the Verdier duality functor on $Y$ (respectively $X$). There is a canonical adjunction map $\mathbb{D}_Y(\mathcal{F}) \to R f_{\ast} f^{\ast}\mathbb{D}_Y(\mathcal{F})$ with Verdier dual $$\mathbb{D}_Y\left(R f_{\ast} f^{\ast}\mathbb{D}_Y(\mathcal{F})\right) \to \mathbb{D}_Y\left(\mathbb{D}_Y(\mathcal{F}) \right)=\mathcal{F}.$$ Using relative Verdier duality for $f$, we find that the left hand side simplifies to $$\mathbb{D}_Y\left(R f_{\ast} f^{\ast}\mathbb{D}_Y(\mathcal{F})\right) \simeq Rf_{!} \mathbb{D}_X(f^{\ast}\mathbb{D}_Y(\mathcal{F})) \simeq R f_{!} f^{!} \mathbb{D}_Y\left(\mathbb{D}_Y(\mathcal{F})\right) \simeq R f_{!} f^{!} \mathcal{F}.$$

So we have produced a map $R f_{!} f^{!} \mathcal{F} \to \mathcal{F}$. To identify the left hand side, we note that it follows from your assumption that $f$ is smooth of relative dimension $n$, so that $f^{!} \mathcal{F} \simeq f^{\ast}\mathcal{F}[2n](n)$ (where the square brackets denotes a shift and the round brackets a Tate twist). We deduce that there is a canonical map $$ R f_{!} f^{\ast} \mathcal{F}[2n](n) \to \mathcal{F}. $$ We can check that this is an isomorphism on stalks, and by proper base change this reduces to the situation that $Y$ is an algebraically closed field and $X$ is affine space. But then $\mathcal{F}$ is just a direct sum of (shifts) of the constant sheaf $\mathbb{Q}_{\ell}$, and the result is well known here.

Finally, applying $R \pi_{!}$ we get a canonical isomorphism $$ R \Gamma_c(Y, f^{\ast} \mathcal{F})[2n](n) \to R \Gamma_c(Y, \mathcal{F}), $$ as desired.

$\endgroup$
3
  • $\begingroup$ Sorry, would you elaborate on the "well-known" result? I'm slightly unfamiliar $\endgroup$
    – W. Zhan
    Commented Jun 24 at 20:12
  • 1
    $\begingroup$ The well known result is that $H^{2n}_{c}(\mathbb{A}_{K}^n, \mathbb{Q}_{\ell}) = \mathbb{Q}_{\ell}(-n)$ when $K$ is an algebraically closed field. $\endgroup$ Commented Jun 27 at 6:40
  • $\begingroup$ I got confused by affine = spec of a com. ring. I see you argument now- nice! $\endgroup$
    – W. Zhan
    Commented Jun 27 at 13:32

You must log in to answer this question.

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