[proofplan] We prove both directions by linear algebra on the complex [vector space](/page/Vector%20Space) $V_{\mathbb C}$. Starting from a pure [Hodge decomposition](/theorems/2745), the Hodge filtration is the sum of all summands whose first index is at least $p$, while the conjugate shifted filtration is the complementary sum of the remaining summands. Conversely, assuming the opposed direct-sum condition, we define the candidate Hodge summands as intersections of the two opposed filtrations and show that each filtration step splits as one new summand plus the next step. This gives a finite direct-sum decomposition, recovers the original filtration, and conjugation exchanges the summands with reversed indices. [/proofplan]

[step:Derive the opposed direct sum from an existing Hodge decomposition] Assume first that $F^\bullet V_{\mathbb C}$ is the Hodge filtration of a pure $\mathbb Q$-Hodge structure of weight $k$. Thus there is a finite direct-sum decomposition by complex vector subspaces \begin{align*} V_{\mathbb C}=\bigoplus_{a+b=k} V^{a,b} \end{align*} such that $\overline{V^{a,b}}=V^{b,a}$ for all $a,b\in\mathbb Z$, and the filtration is \begin{align*} F^pV_{\mathbb C}=\bigoplus_{\substack{a+b=k, a\ge p}} V^{a,b}. \end{align*} For every $p\in\mathbb Z$, conjugating the filtration step $F^{k-p+1}V_{\mathbb C}$ gives \begin{align*} \overline{F^{k-p+1}V_{\mathbb C}}=\bigoplus_{\substack{a+b=k, a\ge k-p+1}} \overline{V^{a,b}}. \end{align*} Using $\overline{V^{a,b}}=V^{b,a}$ and $a+b=k$, this becomes \begin{align*} \overline{F^{k-p+1}V_{\mathbb C}}=\bigoplus_{\substack{r+s=k, r\le p-1}} V^{r,s}. \end{align*} Therefore $F^pV_{\mathbb C}$ is the direct sum of the Hodge summands with first index at least $p$, while $\overline{F^{k-p+1}V_{\mathbb C}}$ is the direct sum of the Hodge summands with first index at most $p-1$. These two index sets are disjoint and together contain all pairs $(a,b)$ with $a+b=k$. Hence \begin{align*} V_{\mathbb C}=F^pV_{\mathbb C}\oplus \overline{F^{k-p+1}V_{\mathbb C}} \end{align*} for every $p\in\mathbb Z$. [/step]

[step:Construct the candidate Hodge summands from the opposed filtrations]Conversely, assume that \begin{align*} V_{\mathbb C}=F^pV_{\mathbb C}\oplus \overline{F^{k-p+1}V_{\mathbb C}} \end{align*} for every $p\in\mathbb Z$. For each $p\in\mathbb Z$, define the complex subspace \begin{align*} H_p:=F^pV_{\mathbb C}\cap \overline{F^{k-p}V_{\mathbb C}}. \end{align*} We will prove that the subspaces $H_p$ give the Hodge summands of weight $k$, namely $V^{p,k-p}=H_p$. Fix $p\in\mathbb Z$. Applying the opposed condition with index $p+1$ gives \begin{align*} V_{\mathbb C}=F^{p+1}V_{\mathbb C}\oplus \overline{F^{k-p}V_{\mathbb C}}. \end{align*} Since $F^\bullet V_{\mathbb C}$ is decreasing, $F^{p+1}V_{\mathbb C}\subset F^pV_{\mathbb C}$. Therefore every $x\in F^pV_{\mathbb C}$ has a unique decomposition \begin{align*} x=y+z \end{align*} with $y\in F^{p+1}V_{\mathbb C}$ and $z\in \overline{F^{k-p}V_{\mathbb C}}$. Because $x\in F^pV_{\mathbb C}$ and $y\in F^{p+1}V_{\mathbb C}\subset F^pV_{\mathbb C}$, we also have $z=x-y\in F^pV_{\mathbb C}$. Hence $z\in H_p$, and so \begin{align*} F^pV_{\mathbb C}=F^{p+1}V_{\mathbb C}+H_p. \end{align*} The sum is direct because $H_p\subset \overline{F^{k-p}V_{\mathbb C}}$ and the decomposition \begin{align*} V_{\mathbb C}=F^{p+1}V_{\mathbb C}\oplus \overline{F^{k-p}V_{\mathbb C}} \end{align*} has zero intersection. Thus \begin{align*} F^pV_{\mathbb C}=H_p\oplus F^{p+1}V_{\mathbb C}. \end{align*}[/step]

[guided]The main point is to locate the part of $F^pV_{\mathbb C}$ that disappears when we pass to the next filtration step $F^{p+1}V_{\mathbb C}$. The opposed condition tells us exactly which complementary subspace to use. With $p+1$ in place of $p$, it states that \begin{align*} V_{\mathbb C}=F^{p+1}V_{\mathbb C}\oplus \overline{F^{k-p}V_{\mathbb C}}. \end{align*} This direct sum gives a unique splitting of every vector in $V_{\mathbb C}$ into a part in $F^{p+1}V_{\mathbb C}$ and a part in $\overline{F^{k-p}V_{\mathbb C}}$. Now take $x\in F^pV_{\mathbb C}$. By the displayed direct sum, there are unique vectors $y\in F^{p+1}V_{\mathbb C}$ and $z\in \overline{F^{k-p}V_{\mathbb C}}$ such that \begin{align*} x=y+z. \end{align*} Since the filtration is decreasing, $F^{p+1}V_{\mathbb C}\subset F^pV_{\mathbb C}$. Thus both $x$ and $y$ lie in $F^pV_{\mathbb C}$, so their difference \begin{align*} z=x-y \end{align*} also lies in $F^pV_{\mathbb C}$. Combining this with $z\in \overline{F^{k-p}V_{\mathbb C}}$, we get \begin{align*} z\in F^pV_{\mathbb C}\cap \overline{F^{k-p}V_{\mathbb C}}=H_p. \end{align*} Therefore every $x\in F^pV_{\mathbb C}$ belongs to $F^{p+1}V_{\mathbb C}+H_p$. The reverse inclusion follows from the definitions: $F^{p+1}V_{\mathbb C}\subset F^pV_{\mathbb C}$ and $H_p\subset F^pV_{\mathbb C}$. Hence \begin{align*} F^pV_{\mathbb C}=F^{p+1}V_{\mathbb C}+H_p. \end{align*} Finally, this sum is direct because $H_p\subset \overline{F^{k-p}V_{\mathbb C}}$ and the opposed condition at $p+1$ says that \begin{align*} F^{p+1}V_{\mathbb C}\cap \overline{F^{k-p}V_{\mathbb C}}=0. \end{align*} Thus \begin{align*} F^pV_{\mathbb C}=H_p\oplus F^{p+1}V_{\mathbb C}. \end{align*} This is the key splitting: $H_p$ is the new graded piece introduced at filtration level $p$.[/guided]

[step:Iterate the splitting to obtain a finite direct-sum decomposition] Because the filtration is bounded, there exist integers $m$ and $N$ such that \begin{align*} F^mV_{\mathbb C}=V_{\mathbb C} \end{align*} and \begin{align*} F^{N+1}V_{\mathbb C}=0. \end{align*} Iterating the direct splitting \begin{align*} F^pV_{\mathbb C}=H_p\oplus F^{p+1}V_{\mathbb C} \end{align*} from $p=m$ to $p=N$ gives \begin{align*} V_{\mathbb C}=\bigoplus_{p=m}^{N} H_p. \end{align*} For any integer $r$, the same iteration starting at $r$ gives \begin{align*} F^rV_{\mathbb C}=\bigoplus_{p\ge r} H_p, \end{align*} where only finitely many summands are nonzero. Indeed, if $p<m$, then $F^pV_{\mathbb C}=F^{p+1}V_{\mathbb C}=V_{\mathbb C}$, so the direct splitting $F^pV_{\mathbb C}=H_p\oplus F^{p+1}V_{\mathbb C}$ forces $H_p=0$. If $p>N$, then $F^pV_{\mathbb C}=F^{p+1}V_{\mathbb C}=0$, so again $H_p=0$. The directness follows at each stage from the directness of the splitting $F^pV_{\mathbb C}=H_p\oplus F^{p+1}V_{\mathbb C}$. [/step]

[step:Verify conjugation symmetry of the constructed summands] For each $p\in\mathbb Z$, conjugating the definition of $H_p$ gives \begin{align*} \overline{H_p}=\overline{F^pV_{\mathbb C}}\cap F^{k-p}V_{\mathbb C}. \end{align*} By commutativity of intersection, this is \begin{align*} \overline{H_p}=F^{k-p}V_{\mathbb C}\cap \overline{F^pV_{\mathbb C}}=H_{k-p}. \end{align*} Therefore conjugation exchanges the summand indexed by $p$ with the summand indexed by $k-p$. [/step]

[step:Define the pure Hodge decomposition and recover the given filtration] Define complex subspaces $V^{a,b}\subset V_{\mathbb C}$ for all $a,b\in\mathbb Z$ as follows: set $V^{a,b}=H_a$ when $a+b=k$, and set $V^{a,b}=0$ when $a+b\ne k$. Then \begin{align*} V_{\mathbb C}=\bigoplus_{a+b=k} V^{a,b} \end{align*} by the decomposition from the previous step. The conjugation symmetry $\overline{H_a}=H_{k-a}$ gives \begin{align*} \overline{V^{a,b}}=V^{b,a} \end{align*} for all $a,b\in\mathbb Z$. Hence the subspaces $V^{a,b}$ define a pure $\mathbb Q$-Hodge structure of weight $k$ on $V_{\mathbb Q}$. It remains to check that its Hodge filtration is the original filtration. For every $r\in\mathbb Z$, the Hodge filtration associated to the constructed decomposition is \begin{align*} \bigoplus_{\substack{a+b=k, a\ge r}} V^{a,b}=\bigoplus_{a\ge r} H_a \end{align*} by the iterated splitting already proved. This equals $F^rV_{\mathbb C}$, so $F^\bullet V_{\mathbb C}$ is exactly the Hodge filtration of the constructed pure Hodge structure. For the uniqueness statement, let $p,q\in\mathbb Z$ with $p+q=k$. Then \begin{align*} V^{p,q}=H_p=F^pV_{\mathbb C}\cap \overline{F^{k-p}V_{\mathbb C}}=F^pV_{\mathbb C}\cap \overline{F^qV_{\mathbb C}}. \end{align*} If $p+q\ne k$, we have $V^{p,q}=0$ by definition. This proves both the characterization and the asserted uniqueness. [/step]

[step:Identify the summands by the stated intersection formula and prove uniqueness] For $p,q\in\mathbb Z$ with $p+q=k$, the definition of $H_p$ gives \begin{align*} V^{p,q}=H_p=F^pV_{\mathbb C}\cap \overline{F^{k-p}V_{\mathbb C}}=F^pV_{\mathbb C}\cap \overline{F^qV_{\mathbb C}}. \end{align*} For $p+q\ne k$, the construction sets $V^{p,q}=0$. Finally, any pure [Hodge decomposition](/theorems/3941) with Hodge filtration $F^\bullet V_{\mathbb C}$ and weight $k$ must satisfy the same intersection formula. Indeed, in the forward-direction computation, $F^pV_{\mathbb C}$ is the direct sum of summands $V^{a,k-a}$ with $a\ge p$, while $\overline{F^{k-p}V_{\mathbb C}}$ is the direct sum of summands $V^{a,k-a}$ with $a\le p$. Their intersection inside the direct sum decomposition is exactly the single summand $V^{p,k-p}$. Hence the decomposition is uniquely determined by \begin{align*} V^{p,q}=F^pV_{\mathbb C}\cap \overline{F^qV_{\mathbb C}} \end{align*} when $p+q=k$, and by $V^{p,q}=0$ otherwise. This proves both the characterization and the asserted uniqueness. [/step]