Strictness of mixed Hodge morphisms

(Weight strictness.) Let $y=f(x)\in f(H)\cap W_{n}K$ and decompose $x={\sum }_{p,q}{x}^{p,q}$ along the splitting of $H_{\mathbb{C}}$; then $y={\sum }_{p,q}{f}_{\mathbb{C}}(x^{p,q})$ with $f_{\mathbb{C}}(x^{p,q})\in I^{p,q}(K)$ by [F1]. Since $y\in W_{n}K={⨁}_{p+q\le n}{I}^{p,q}(K)$ and the sum is direct, all components with $p+q>n$ vanish: $f_{\mathbb{C}}(x^{p,q})=0$ whenever $p+q>n$. Hence $x^{\mathrm{\prime }}:={\sum }_{p+q\le n}{x}^{p,q}$ lies in $W_n{H}_{\mathbb{C}}$ and satisfies $f_{\mathbb{C}}(x^{\mathrm{\prime }})=y$, so $y\in f_{\mathbb{C}}(W_n{H}_{\mathbb{C}})$. The reverse inclusion $f(W_{n}H)\subseteq f(H)\cap W_{n}K$ is the preservation of $W$ by $f$.

(Hodge strictness.) The same computation with the condition $p<a$ instead of $p+q>n$: if $y=f_{\mathbb{C}}(x)\in f(H_{\mathbb{C}})\cap F^a{K}_{\mathbb{C}}$, decompose $x$ and use $F^a{K}_{\mathbb{C}}={⨁}_{p\ge a}{I}^{p,q}(K)$ to conclude that $f_{\mathbb{C}}(x^{p,q})=0$ for $p<a$, so $x^{\mathrm{\prime }}={\sum }_{p\ge a}{x}^{p,q}$ lies in $F^a{H}_{\mathbb{C}}$ and maps to $y$.

(Rational descent.) Step 1.1 gives equality of the complexifications of the rational subspaces $A=f(W_{n}H)$ and $B=f(H)\cap W_{n}K$. Images and intersections commute with extension from $\mathbb{Q}$ to $\mathbb{C}$: in finite rational coordinates they are kernels or images of matrices, whose rank is unchanged on extension since the same nonzero rational minors remain nonzero. To descend $A_{\mathbb{C}}=B_{\mathbb{C}}$, map $A$ into the rational quotient $K\mathrm{/}B$. Its complexification is zero, hence the rational map is zero (a nonzero rational coordinate remains nonzero over $\mathbb{C}$), so $A\subseteq B$; interchanging them gives equality. Together with step 1.2 this proves the stated rational $W$ strictness and complex $F$ strictness. Complex conjugation fixes the real scalar extension of a rational space and is not used to recover rational vectors.

(Exactness of ${\mathrm{Gr}}_n^W$.) Let $0\to A\stackrel{i}{\to }B\stackrel{p}{\to }C\to 0$ be a short exact sequence of mixed Hodge structures. By [F1] the splitting identifies ${\mathrm{Gr}}_n^{W}A$, ${\mathrm{Gr}}_n^{W}B$, ${\mathrm{Gr}}_n^{W}C$ with ${⨁}_{p+q=n}{I}^{p,q}(A)$ and its analogues, and the maps ${\mathrm{Gr}}_n^{W}i$, ${\mathrm{Gr}}_n^{W}p$ with the direct sums of the maps induced by $i$ and $p$ between the pieces $I^{p,q}$. Each piecewise sequence $0\to I^{p,q}(A)\to I^{p,q}(B)\to I^{p,q}(C)\to 0$ is exact: injectivity follows by restriction of $i$; to lift $c\in I^{p,q}(C)$, choose a preimage in $B_{\mathbb{C}}$ and decompose it, whose $(p,q)$ component maps to $c$ by functoriality and uniqueness of the target decomposition. An element $b\in I^{p,q}(B)$ lies in $\ker p$ precisely when $b=i(a)$ for some $a\in A$, and decomposing $a$ into its $I$-pieces shows that all other pieces map to zero; since $i$ is injective, they vanish and $a\in I^{p,q}(A)$. A direct sum of short exact sequences is short exact, so $0\to {\mathrm{Gr}}_n^{W}A\to {\mathrm{Gr}}_n^{W}B\to {\mathrm{Gr}}_n^{W}C\to 0$ is exact after complexification. The rational graded sequence is exact as well, since kernels and images commute with scalar extension and rational ranks descend as in step 2.1.

(Kernel, image and cokernel.) The functorial splitting [F1] makes $f_{\mathbb{C}}$ a direct sum of maps $I^{p,q}(H)\to I^{p,q}(K)$. Thus its kernel, image and cokernel decompose into the respective kernels, images and cokernels of these maps, and their induced or quotient $W$ and $F$ are the partial sums by $p+q\le n$ and $p\ge a$. The projection isomorphisms in [F1] identify their weight-$n$ pieces, with these induced $F$ filtrations, with the kernel, image and cokernel of ${\mathrm{Gr}}_n^{W}f$, computed componentwise in the pure decompositions $H_n^{p,q},K_n^{p,q}$. These identifications are the canonical rational graded maps: for a kernel this follows from $W$ strictness, since if $f(x)\in W_{n-1}K$ with $x\in W_{n}H$, choose $y\in W_{n-1}H$ with $f(y)=f(x)$ and replace $x$ by $x-y$; the image and cokernel follow by the same induced/quotient description. The graded rational map commutes with conjugation and preserves $F$, hence also $\bar{F}$; between equal-weight pure structures it preserves each type $F^p\cap {\bar{F}}^q$ with $p+q=n$. Consequently its pure kernels, images and cokernels have conjugation exchanging their $(p,q)$ and $(q,p)$ pieces. Their $F$ partial sums and their conjugates are therefore $n$-opposed. Every weight-graded piece of the three rational filtered spaces satisfies the mixed Hodge axiom, proving the claim. This argument requires no exact conjugate decomposition on the mixed vector space.

(Pure image.) Now suppose $W_{n-1}H=0$ and $W_{n}K=K$. By step 3.2 the image is a mixed Hodge structure with the induced filtration, and step 1.1 gives $W_{n-1}(\mathrm{im}f)=f(W_{n-1}H)=0$ and $W_n(\mathrm{im}f)=f(H)\cap W_{n}K=f(H)=\mathrm{im}f$; hence the image is pure of weight $n$.