The Deligne splitting of a mixed Hodge structure

(Two-filtration lemma, spanning and partial sums.) Let $V$ be a finite-dimensional $\mathbb{C}$-vector space with finite decreasing filtrations $F$ and $\bar{F}$ such that $V={⨁}_{p+q=n}(F^{p}V\cap {\bar{F}}^{q}V)$ for an integer $n$. Then (i) $F^{a}V+{\bar{F}}^{n+1-a}V=V$ for every $a$, and (ii) $F^{a}V={⨁}_{p\ge a}(F^{p}V\cap {\bar{F}}^{n-p}V)$, with the conjugate formula ${\bar{F}}^{b}V={⨁}_{q\ge b}(F^{n-q}V\cap {\bar{F}}^{q}V)$. For (i), write $v={\sum }_{p+q=n}{v}_{p,q}$; the terms with $p\ge a$ sum into $F^{a}V$, and each remaining term has $p<a$, hence $q=n-p>n-a$, that is $q\ge n+1-a$, so their sum lies in ${\bar{F}}^{n+1-a}V$. For (ii), prove $F^{a}V\subseteq {\sum }_{p\ge a}(F^{p}V\cap {\bar{F}}^{n-p}V)$ by descending induction on $a$: the claim is trivial when $F^{a}V=0$, and otherwise (i) with $a+1$ gives $V=F^{a+1}V+{\bar{F}}^{n-a}V$, so the modular law applied to $F^{a+1}V\subseteq F^{a}V\subseteq V$ yields $F^{a}V=F^{a+1}V+(F^{a}V\cap {\bar{F}}^{n-a}V)$, in which the induction hypothesis places $F^{a+1}V$ inside ${\sum }_{p\ge a+1}(F^{p}V\cap {\bar{F}}^{n-p}V)$ while $F^{a}V\cap {\bar{F}}^{n-a}V$ is the term with $p=a$; the reverse inclusion is clear, and the sum is direct because the pieces $F^{p}V\cap {\bar{F}}^{n-p}V$ are independent by the decomposition hypothesis, which is symmetric in the two filtrations, so the same argument with $F$ and $\bar{F}$ interchanged gives the formula for ${\bar{F}}^{b}V$.

(Two-filtration lemma, vanishing.) Under the hypothesis of step 1.1, $F^{a}V\cap {\bar{F}}^{b}V=0$ whenever $a+b>n$. Indeed, let $x\in F^{a}V\cap {\bar{F}}^{b}V$ and let $x={\sum }_{p+q=n}{x}_{p,q}$ be its unique decomposition with $x_{p,q}\in F^{p}V\cap {\bar{F}}^{q}V$. Since $x\in F^{a}V$, its decomposition lies in ${⨁}_{p\ge a}(F^{p}V\cap {\bar{F}}^{n-p}V)$ by step 1.1 (ii), so $x_{p,n-p}=0$ for every $p<a$; since $x\in {\bar{F}}^{b}V$, the conjugate half of step 1.1 (ii) gives $x_{n-q,q}=0$ for every $q<b$. If $x\mathrm{\ne }0$, some $x_{p,n-p}$ is nonzero, and it satisfies $p\ge a$ and $n-p\ge b$, whence $n=p+(n-p)\ge a+b>n$, a contradiction; so $x=0$.

(Lifting.) Let $m=p+q$. Every $d\in W_{m-1}$ can be written as $d=z+g$ with $z\in F^p\cap W_{m-1}$ and $g\in ({\bar{F}}^q\cap W_{m-1})+{\sum }_{j\ge 1}({\bar{F}}^{q-j}\cap W_{m-j-1})$. To see this, construct elements $d_1,d_2,\dots$ and vectors $f_j,g_j$ as follows. Put $d_1=d$, and suppose $d_j\in W_{m-j}$ has been constructed for some $j\ge 1$. By [F1] the induced filtrations on ${\mathrm{Gr}}_{m-j}^W$ are $(m-j)$-opposed, and step 1.1 (i) applied there with the pair $(p,q-j+1)$, whose indices sum to $m-j+1$, shows that the images of $F^p\cap W_{m-j}$ and ${\bar{F}}^{q-j+1}\cap W_{m-j}$ generate ${\mathrm{Gr}}_{m-j}^W$; choose $f_j\in F^p\cap W_{m-j}$ and $g_j\in {\bar{F}}^{q-j+1}\cap W_{m-j}$ whose classes sum to the class of $d_j$, and put $d_{j+1}:=d_j-f_j-g_j\in W_{m-j-1}$. By [F2] there is $J$ with $W_{m-J}=0$; the recursion stops there, and summing the identities $d_{j+1}=d_j-f_j-g_j$ gives $d={\sum }_{j=1}^{J-1}(f_j+g_j)$. Each $f_j$ lies in $F^p\cap W_{m-j}\subseteq F^p\cap W_{m-1}$; $g_1$ lies in ${\bar{F}}^q\cap W_{m-1}$ and, for $j\ge 2$, $g_j$ lies in ${\bar{F}}^{q-(j-1)}\cap W_{m-(j-1)-1}$. Hence $z:={\sum }_j{f}_j$ and $g:={\sum }_j{g}_j$ have the required properties.

(Surjectivity onto the graded summand.) Let $\xi \in H_m^{p,q}$. Since $H_m^{p,q}\subseteq F^p{\mathrm{Gr}}_m^W$ is the image of $F^p\cap W_m$, choose $x_0\in F^p\cap W_m$ representing $\xi$; since $H_m^{p,q}\subseteq {\bar{F}}^q{\mathrm{Gr}}_m^W$ is the image of ${\bar{F}}^q\cap W_m$, choose $y_0\in {\bar{F}}^q\cap W_m$ also representing $\xi$. Then $d:=x_0-y_0$ lies in $W_{m-1}$, so step 2.2 gives $d=z+g$ with $z\in F^p\cap W_{m-1}$ and $g$ in the displayed sum. Put $x:=x_0-z=y_0+g$. Then $x\in F^p\cap W_m$, and $x\in ({\bar{F}}^q\cap W_m)+{\sum }_{j\ge 1}({\bar{F}}^{q-j}\cap W_{m-j-1})$ because $y_0$ lies in the first summand and $g$ in the sum; so $x\in I^{p,q}$. Finally $x-x_0=-z\in W_{m-1}$, so $x$ and $x_0$ have the same class $\xi$ in ${\mathrm{Gr}}_m^W$. Hence the projection maps $I^{p,q}$ onto $H_m^{p,q}$, and $I^{p,q}\subseteq W_m\cap F^p$ holds by definition.

(Injectivity modulo the lower weight.) Suppose $0\mathrm{\ne }x\in I^{p,q}\cap W_{m-1}$, and choose its least weight index $r$, so $x\in W_r\setminus W_{r-1}$ and $r<m$. Write $x=y+{\sum }_{j\ge 1}{g}_j$ as in the defining formula, with $y\in {\bar{F}}^q\cap W_m$ and $g_j\in {\bar{F}}^{q-j}\cap W_{m-j-1}$; only finitely many terms are needed because $W$ is finite. Put $J=m-r-1\ge 0$. The sum ${\sum }_{j>J}{g}_j$ lies in $W_{r-1}$, while $h:=y+{\sum }_{1\le j\le J}{g}_j$ lies in ${\bar{F}}^{q-J}$ and also in $W_r$, because $h=x-{\sum }_{j>J}{g}_j$. Thus the nonzero class of $x$ in ${\mathrm{Gr}}_r^W$ belongs to $F^p{\mathrm{Gr}}_r^W\cap {\bar{F}}^{q-J}{\mathrm{Gr}}_r^W$. But $p+q-J=r+1>r$, so step 2.1 makes that intersection zero, a contradiction. Together with step 3.1 this proves the projection isomorphism in (a).

(The weight decomposition and the direct-sum identity.) For every $m$ the projection restricts to an isomorphism ${⨁}_{p+q=m}{I}^{p,q}\to {\mathrm{Gr}}_m^W{H}_{\mathbb{C}}$: by steps 3.1 and 4.1 each $I^{p,q}$ maps isomorphically onto $H_m^{p,q}$, and by [F1] the summands $H_m^{p,q}$ are independent with sum ${\mathrm{Gr}}_m^W$. In particular $\dim I^{p,q}=\dim H_m^{p,q}$, so ${\sum }_{p+q=m}\dim I^{p,q}=\dim {\mathrm{Gr}}_m^W$ for every $m$. The sum map ${\psi }_m\text{\hspace{0.17em}⁣}:{⨁}_{p+q\le m}{I}^{p,q}\to W_m$ is injective: if a finite sum ${\sum }_{m^{\mathrm{\prime }}\le m}{\sum }_{p+q=m^{\mathrm{\prime }}}{x}_{p,q}$ vanishes, its class in ${\mathrm{Gr}}_m^W$ is the sum of the classes of the top pieces $x_{p,m-p}$, which vanishes only if each of those pieces is zero because the restricted projection is an isomorphism; subtracting them leaves a vanishing sum over $m^{\mathrm{\prime }}\le m-1$, and induction on $m$ (with $W_m=0$ for all sufficiently negative $m$) finishes the argument. Its image therefore has dimension ${\sum }_{m^{\mathrm{\prime }}\le m}\dim {\mathrm{Gr}}_{m^{\mathrm{\prime }}}^W=\dim W_m$, the finite telescoping identity for the finite filtration $W$; an injective map between spaces of equal dimension is an isomorphism, so $W_m{H}_{\mathbb{C}}={⨁}_{p+q\le m}{I}^{p,q}$. Taking $m$ large gives $H_{\mathbb{C}}={⨁}_{p,q}{I}^{p,q}$.

(Conjugation modulo lower weights.) Work in $U=W_m\mathrm{/}{W}_{m-2}$ with the images of $F^p\cap W_m$ and ${\bar{F}}^q\cap W_m$. Their intersection $T$ maps injectively to ${\mathrm{Gr}}_m^W$: its intersection with $W_{m-1}\mathrm{/}{W}_{m-2}$ is $F^p{\mathrm{Gr}}_{m-1}^W\cap {\bar{F}}^q{\mathrm{Gr}}_{m-1}^W=0$ by step 2.1, since $p+q=m>m-1$. The formula for $I^{p,q}$ shows its image in $U$ lies in $T$, since every tail term has weight at most $m-2$. Step 3.1 shows this image maps onto $H_m^{p,q}$; the image of $T$ also lies in $H_m^{p,q}$, so injectivity makes the image of $I^{p,q}$ exactly $T$. Conjugating the same argument for $I^{q,p}$ shows that $\sigma (I^{q,p})$ has this same image $T$ in $U$. Therefore $I^{p,q}+W_{m-2}=\sigma (I^{q,p})+W_{m-2}$.

(The Hodge decomposition.) For all $a,m$ one has $F^a\cap W_m={⨁}_{p\ge a,p+q\le m}{I}^{p,q}$. Indeed, decompose $x\in F^a\cap W_m$ by step 5.1. Its class in ${\mathrm{Gr}}_m^W$ lies in ${⨁}_{p\ge a}{H}_m^{p,m-p}$ by step 1.1, so its top components with $p<a$ vanish by the projection isomorphism. Subtracting the remaining top components leaves an element of $F^a\cap W_{m-1}$, and induction through the finite $W$ proves the assertion. Taking $m$ large proves $F^a={⨁}_{p\ge a}{I}^{p,q}$. Conjugating this identity gives ${\bar{F}}^b={⨁}_{q\ge b}\sigma (I^{q,p})$; it does not assert a decomposition of $\bar{F}$ by the original $I^{p,q}$.

(Functoriality.) Let $f\text{\hspace{0.17em}⁣}:H\to K$ be a morphism of rational mixed Hodge structures. Since $f$ is defined over $\mathbb{Q}$, $f_{\mathbb{C}}$ commutes with the conjugations, so $f_{\mathbb{C}}({\bar{F}}^q{H}_{\mathbb{C}})=\overline{f_{\mathbb{C}}(F^q{H}_{\mathbb{C}})}\subseteq {\bar{F}}^q{K}_{\mathbb{C}}$; it also satisfies $f_{\mathbb{C}}(F^p{H}_{\mathbb{C}})\subseteq F^p{K}_{\mathbb{C}}$ and $f_{\mathbb{C}}(W_a{H}_{\mathbb{C}})\subseteq W_a{K}_{\mathbb{C}}$ by definition. Applying $f_{\mathbb{C}}$ to the two intersections and the sum defining $I^{p,q}(H)$ therefore produces an element of the corresponding intersection of $F^p{K}_{\mathbb{C}}$ with the sum of the $\bar{F}$-terms for $K$, and $W$-membership is preserved term by term; hence $f_{\mathbb{C}}(I^{p,q}(H))\subseteq I^{p,q}(K)$.