Cohomology of relative motives

Question

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)

Answer 1

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.