4
$\begingroup$

I have asked this question on Math StackExchange, but have not got any reply.

In section 1.7 of Deligne's paper "Valeurs de fonctions L et périodes d'intégrales", which has an English translation (pdf here), there are two subspaces $F^+$ and $F^-$ appearing in the Hodge filtration of the de Rham realisation $H_{DR}(M)$ of a pure motive $M$. However these two subspaces haven't been defined, could anyone explain the definition?

Also in this section, the author defines subspaces, \begin{equation} H^+_{DR}(M):=H_{DR}(M)/F^-,~~H^-_{DR}(M):=H_{DR}(M)/F^+ \end{equation} and it has said later in the same section that the dual of $H^+_{DR}(M)$ (resp. $H^-_{DR}(M)$) is the subspace $F^+$ (resp. $F^-$) of $H_{DR}(M^\vee)$, where $M^\vee$ is the dual motive of $M$. I don't understand this statement, and could anyone explain it carefully?

$\endgroup$

1 Answer 1

3
$\begingroup$

In the original French version of the article, Deligne actually defines $F^+$ and $F^-$ to be the subspaces of $H_{\mathrm{dR}}(M)$ occurring in its Hodge filtration and having the same dimension as $H_B^+(M)$ and $H_B^-(M)$ respectively. (I agree that the translation is slightly imprecise as it may suggest these subspaces have been defined before.)

Concretely, using Deligne's notations: let $M$ be a pure motive of weight $w$. If $w=2p$ is even, we assume $F_\infty$ acts as $\pm 1$ on $H^{p,p}(M)$. Then via the comparison isomorphism between Betti and algebraic de Rham cohomology, $F^+ \otimes \mathbf{C}$ (resp. $F^- \otimes \mathbf{C}$) corresponds to \begin{equation*} \bigoplus_{\substack{p+q=w \\ p \geq q}} H^{p,q}(M) \qquad \textrm{resp. } \bigoplus_{\substack{p+q=w \\ p > q}} H^{p,q}(M) \end{equation*} unless $w=2p$ and $F_\infty=-1$ on $H^{p,p}(M)$, in which case one should remove the $H^{p,p}$ from $F^+ \otimes \mathbf{C}$ and put it in $F^- \otimes \mathbf{C}$.

Regarding your second question, I will do the case of the motive $M=H^i(X)$ where $X$ is a smooth projective variety over $\mathbf{Q}$ of pure dimension $d$, leaving you other cases like $M=H^i(X)(n)$ as an exercise.

The dual motive of $M=H^i(X)$ is $M^\vee = H^{2d-i}(X)(d)$. Betti cohomology and algebraic de Rham cohomology are examples of Weil cohomology theories, so there exist perfect pairings coming from Poincaré duality

\begin{align*} H_B(M) \otimes H_B(M^\vee) & \to \mathbf{Q}\\ H_{\mathrm{dR}}(M) \otimes H_{\mathrm{dR}}(M^\vee) & \to \mathbf{Q} \end{align*} where for example $H_B(M^\vee)=H^{2d-i}_B(X(\mathbf{C}),\mathbf{Q}(d))$ and so on. The first pairing is given by \begin{equation*} \langle \alpha,\beta \rangle = \frac{1}{(2\pi i)^d} \int_{X(\mathbf{C})} \alpha \wedge \beta \end{equation*} Let $c : X(\mathbf{C}) \to X(\mathbf{C})$ be the complex conjugation. The operator $F_\infty$ in Deligne's article acts on $H_B(M)$ by $c^*$ and on $H_B(M^\vee)$ by $(-1)^d c^*$. Since $c$ multiplies the orientation of $X(\mathbf{C})$ by $(-1)^d$, it is easy to check that $\langle F_\infty \alpha, F_\infty \beta \rangle = \langle \alpha,\beta \rangle$. In particular \begin{equation*} \langle H^+_B(M),H^-_B(M^\vee) \rangle = \langle H^-_B(M),H^+_B(M^\vee) \rangle = 0 \end{equation*} and the induced pairings $H^{\pm}_B(M) \otimes H^{\pm}_B(M^\vee) \to \mathbf{Q}$ are perfect. After extending scalars to $\mathbf{C}$, Poincaré duality also induces \begin{equation*} (*) \qquad H^{p,q}(M) \otimes H^{p',q'}(M^\vee) \to \mathbf{C} \qquad (p+q=i \textrm{ and } p'+q'=2d-i) \end{equation*} which is 0 unless $(p',q')=(d-p,d-q)$, which you can check by considering the types of the differential forms: $(p,q)$ forms on $X(\mathbf{C})$ can only pair with $(d-p,d-q)$ forms, otherwise the wedge product $\alpha \wedge \beta$ is 0. Since the pairing was $F_\infty$-equivariant, we also deduce that $F_\infty = \pm 1$ on $H^{p,p}(M)$ implies $F_\infty = \pm 1$ on $H^{d-p,d-p}(M^\vee)$.

Now the dual of $H^{\pm}_{\mathrm{dR}}(M) = H_{\mathrm{dR}}(M)/F^{\mp}$ is (by definition) the orthogonal of $F^{\mp}$ in $H_{\mathrm{dR}}(M^\vee)$ under Poincaré duality. By considering again the types of the differential forms (or simply using $(*)$), I let you check that \begin{equation*} \langle F^{\mp} H_{\mathrm{dR}}(M) , F^{\pm} H_{\mathrm{dR}}(M^\vee) \rangle =0. \end{equation*} Hence the orthogonal $(F^\mp)^\perp$ of $F^\mp$ contains $F^{\pm} H_{\mathrm{dR}}(M^\vee)$. But its dimension is given by \begin{align*} \dim ((F^\mp)^\perp) & = \dim H_{\mathrm{dR}}(M) - \dim F^\mp = \dim H_B(M) - \dim H^{\mp}_B(M) \\ & = \dim H^{\pm}_B(M) = \dim H^{\pm}_B(M^\vee) = \dim F^{\pm} H_{\mathrm{dR}}(M^\vee), \end{align*} so we have equality of the subspaces. QED

$\endgroup$
1
  • 3
    $\begingroup$ This is so great, thank you very much. I have a pure motive to learn French now! $\endgroup$
    – Wenzhe
    Apr 3, 2018 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.