A question on Deligne's paper "Valeurs de fonctions L et périodes d'intégrales" 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?
 A: 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
