9
$\begingroup$

Notation

Let $S$ be a scheme, proper over a field $k$. Let $\mathrm{SmPr}_{S}$ denote the category of smooth projective $S$-schemes. Let $\mathcal{M}_{S}$ denote the category of relative Chow motives over the base $S$. Let $h_{S} \colon \mathrm{SmPr}_{S}^{\mathrm{op}} \to \mathcal{M}_{S}$ be the functor assigning to $X/S$ its Chow motive, and to a morphism $X \to Y$ the transpose of the (relative) graph in $Y \times_{S} X$.

Motivation

This might pretty long. Scroll down if you want to read the question (-;

If $f \colon S \to S'$ is smooth projective (and a morphism of $k$-schemes), there is a pushforward functor $f_{*} \colon \mathcal{M}_{S} \to \mathcal{M}_{S'}$. It maps $(X, p, m)$ to $(X, j_{*}(p), m)$, where $j$ is the canonical map $X \times_{S} X \to X \times_{S'} X$. One can check that this is well defined (see e.g., [MNP, Cor 8.1.7]).

It is easy to check that the following diagram commutes. $$ \begin{array}{ccc} \mathrm{SmPr}_{S}^{\mathrm{op}} & \stackrel{f \circ \_}{\longrightarrow} & \mathrm{SmPr}_{S'}^{\mathrm{op}} \\ \quad\downarrow h_{S} & & \downarrow h_{S'} \\ \mathcal{M}_{S} & \stackrel{f_{*}}{\longrightarrow} & \mathcal{M}_{S'} \end{array} $$

Suppose $\ell$ is a prime different from $\mathrm{char}(k)$. We can define a cohomology functor $$ \begin{array}{rrll} \mathrm{H}_{S} \colon & \mathcal{M}_{S} & \longrightarrow & \{ \text{$\mathbb{Q}_{\ell}$-sheaves on $S$} \} \\ & ((g \colon X \to S), p, m) & \longmapsto & \mathrm{Im}(p_{*}|\mathrm{R}g_{*} \mathbb{Q}_{\ell}[2m])(m) \end{array} $$ (Obviously, it also makes sense to do this with $\mathbb{Q}$-coefficients if $S$ is a complex scheme.)

To make this more explicit, let $\bar{s} \to S$ be a geometric point. A $\mathbb{Q}_{\ell}$-sheaf on $S$ is the same as a $\pi(S, \bar{s})$-representation (with $\mathbb{Q}_{\ell}$-coefficients). Also $(\mathrm{R}g_{*}\mathbb{Q}_{\ell})_{\bar{s}}$ is $\mathrm{H}^{\bullet}(X_{\bar{s}}, \mathbb{Q}_{\ell})$. Pulling back $p$ along $X_{\bar{s}} \times X_{\bar{s}} \to X \times_{S} X$ gives an autocorrespondence $p_{\bar{s}}$ on $X_{\bar{s}}$. This shows that we can think about $\mathrm{Im}(p_{*}|\mathrm{R}g_{*} \mathbb{Q}_{\ell}[2m])(m)$ as $\mathrm{Im}(p_{\bar{s},*}|\mathrm{H}^{\bullet+2m}(X_{\bar{s}}, \mathbb{Q}_{\ell}))(m)$.

Question

A natural question to ask is whether the following diagram commutes. $$ \begin{array}{ccc} \mathcal{M}_{S} & \stackrel{f_{*}}{\longrightarrow} & \mathcal{M}_{S'} \\ \quad\downarrow h_{S} & & \downarrow h_{S'} \\ \{ \text{$\mathbb{Q}_{\ell}$-sheaves on $S$} \} & \stackrel{\mathrm{R}f_{*}}{\longrightarrow} & \{ \text{$\mathbb{Q}_{\ell}$-sheaves on $S'$} \} \end{array} $$

It commutes on objects, because $\mathrm{R}f_{*}\mathrm{R}g_{*} = \mathrm{R}(f \circ g)_{*}$. However, I have no clue how to show that it commutes on morphisms.

I would be very happy with an answer to:

$\frac{1}{2}$Q: Does the above diagram commute on morphisms that come from $\mathrm{SmPr}_{S}$?

And I would be even more happy if one can show that the diagram actually commutes:

Q: Does the above diagram commute on morphisms in general?

Spelling out $\frac{1}{2}$Q

If $X$ is an $S$-scheme, denote the structure morphism $X \to S$ with $g_{X/S}$. If $\phi/S \colon X/S \to Y/S$ is a morphism of $S$-schemes, we also have a morphism $\phi/S' \colon X/S' \to Y/S'$ of $S'$-schemes.

We have induced morphisms $$ (\phi/S)^{*} \colon \mathrm{R}(g_{Y/S})_{*}\mathbb{Q}_{\ell} \to \mathrm{R}(g_{X/S})_{*}\mathbb{Q}_{\ell} $$ and $$ (\phi/S')^{*} \colon \mathrm{R}(g_{Y/S'})_{*}\mathbb{Q}_{\ell} \to \mathrm{R}(g_{X/S'})_{*}\mathbb{Q}_{\ell}. $$

Applying the $\mathrm{R}f_{*}$-functor to the first morphism, we get a morphism $\mathrm{R}f_{*}(\phi/S)^{*} \colon \mathrm{R}f_{*}\mathrm{R}(g_{Y/S})_{*}\mathbb{Q}_{\ell} \to \mathrm{R}f_{*}\mathrm{R}(g_{X/S})_{*}\mathbb{Q}_{\ell}.$ But, of course we can rewrite the source and target, to get $$ \mathrm{R}f_{*}(\phi/S)^{*} \colon \mathrm{R}(g_{Y/S'})_{*}\mathbb{Q}_{\ell} \to \mathrm{R}(g_{X/S'})_{*}\mathbb{Q}_{\ell}. $$

$\frac{1}{2}$Q': Do we have $\mathrm{R}f_{*}(\phi/S)^{*} = (\phi/S')^{*}$?

References

[MNP] — Murre, Nagel, Peters. Pure Motives. (2013)

$\endgroup$
4
$\begingroup$

Yes it commutes. To prove this you should express the map $(\phi/S)^\ast$ in "six functors" language. All functors below are derived. What's needed is that $\newcommand{\Q}{\mathbf Q_\ell} \renewcommand{\S}{\mathrm{pt}}\Q$ is pulled back from $\S = \mathrm{Spec}(k)$. Then $$ g_{X/S,\ast} \Q= g_{X/S,\ast}g_{X/\S}^\ast \Q = g_{Y/S,\ast}\phi_\ast\phi^\ast g_{Y/\S}^\ast \Q $$ which receives a map from $g_{Y/S,\ast}g_{Y/\S}^\ast \Q$ because of the unit of the adjunction $1 \to \phi_\ast\phi^\ast$. This map is exactly $(\phi/S)^\ast$. If we express $(\phi/S')^\ast$ in the same way and apply $f_\ast$ then we get the equality you want, precisely because $S \to S'$ is a map of $k$-schemes.

$\endgroup$
  • $\begingroup$ Thanks! I had the feeling that it should be such a formality. I had heard a lot about “six functors” but never actually studied them. Now that I am starting to use derived categories, it seems like a good time to look into them. You prove $\frac{1}{2}$Q' (equivalent to $\frac{1}{2}$Q); do you have an idea for Q? My hope is that $\frac{1}{2}$Q can reduce Q to some compatibility between cycle class maps and pullbacks/pushforwards (i.e. some formal argument about Weil cohomologies). $\endgroup$ – jmc Feb 27 '14 at 13:19

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.