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edited Oct 15, 2012 at 16:02

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You have my sympathies for trying come to grips with this stuff. Fortunately the answer, which is both yes and no, can be explained in the simplest case of a point. A polarization on a pure Hodge structure $H$ of weight $w$ is a pairing $$\langle, \rangle: H\otimes H\to \mathbb{Q}(-w)$$ such that $(2\pi i)^n\langle -, C -\rangle$ is positive definite and symmetric, where $C$ is the Weil operator which acts by $i^{p-q}$ on $H^{pq}$. These conditions known as the Hodge-Riemann bilinear relations imply that $H\cong H^*(-w)$ as Hodge structures, so in this sense $H$ self dual. However, just having such an isomorphism would not give everything else.

If $H$ is replaced by a variation of Hodge structures, then a polarization is flat pairing as above satisfying the Hodge-Riemann conditions on the fibres. For Hodge modules the story is similar but more delicate. Since the whole theory is constructed by induction on dimension of support, a polarization is also defined in this manner. Given Hodge module $H$ of weight $w$ on $X$, a polarization is a pairing $H\otimes H\to \mathbb{Q}(\dim X-w)[2\dim X]$ satisfying the inductive conditions (0.8)-(0.10) of Saito's "Modules Hodge Polarizables". Such a pairing should induce an isomorphism $H\cong DH(?)$ up to twist, but it's not equivalent. (Added I don't have a precise reference or proof for this, just a feeling that one could prove it as follows: Buried in Saito's second paper, "Mixed Hodge modules", is a proof that any simple polarizable Hodge module is $j_{!*}V$, where $V$ is a VHS. This effectively reduces the duality statement to the previous case.)

You have my sympathies for trying come to grips with this stuff. Fortunately the answer, which is both yes and no, can be explained in the simplest case of a point. A polarization on a pure Hodge structure $H$ of weight $w$ is a pairing $$\langle, \rangle: H\otimes H\to \mathbb{Q}(-w)$$ such that $(2\pi i)^n\langle -, C -\rangle$ is positive definite and symmetric, where $C$ is the Weil operator which acts by $i^{p-q}$ on $H^{pq}$. These conditions known as the Hodge-Riemann bilinear relations imply that $H\cong H^*(-w)$ as Hodge structures, so in this sense $H$ self dual. However, just having such an isomorphism would not give everything else.

If $H$ is replaced by a variation of Hodge structures, then a polarization is flat pairing as above satisfying the Hodge-Riemann conditions on the fibres. For Hodge modules the story is similar but more delicate. Since the whole theory is constructed by induction on dimension of support, a polarization is also defined in this manner. Given Hodge module $H$ of weight $w$ on $X$, a polarization is a pairing $H\otimes H\to \mathbb{Q}(\dim X-w)[2\dim X]$ satisfying the inductive conditions (0.8)-(0.10) of Saito's "Modules Hodge Polarizables". Such a pairing should induce an isomorphism $H\cong DH(?)$ up to twist, but it's not equivalent.

You have my sympathies for trying come to grips with this stuff. Fortunately the answer, which is both yes and no, can be explained in the simplest case of a point. A polarization on a pure Hodge structure $H$ of weight $w$ is a pairing $$\langle, \rangle: H\otimes H\to \mathbb{Q}(-w)$$ such that $(2\pi i)^n\langle -, C -\rangle$ is positive definite and symmetric, where $C$ is the Weil operator which acts by $i^{p-q}$ on $H^{pq}$. These conditions known as the Hodge-Riemann bilinear relations imply that $H\cong H^*(-w)$ as Hodge structures, so in this sense $H$ self dual. However, just having such an isomorphism would not give everything else.

If $H$ is replaced by a variation of Hodge structures, then a polarization is flat pairing as above satisfying the Hodge-Riemann conditions on the fibres. For Hodge modules the story is similar but more delicate. Since the whole theory is constructed by induction on dimension of support, a polarization is also defined in this manner. Given Hodge module $H$ of weight $w$ on $X$, a polarization is a pairing $H\otimes H\to \mathbb{Q}(\dim X-w)[2\dim X]$ satisfying the inductive conditions (0.8)-(0.10) of Saito's "Modules Hodge Polarizables". Such a pairing should induce an isomorphism $H\cong DH(?)$ up to twist, but it's not equivalent. (Added I don't have a precise reference or proof for this, just a feeling that one could prove it as follows: Buried in Saito's second paper, "Mixed Hodge modules", is a proof that any simple polarizable Hodge module is $j_{!*}V$, where $V$ is a VHS. This effectively reduces the duality statement to the previous case.)

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created Oct 15, 2012 at 13:22

35.2k
2
94
160

You have my sympathies for trying come to grips with this stuff. Fortunately the answer, which is both yes and no, can be explained in the simplest case of a point. A polarization on a pure Hodge structure $H$ of weight $w$ is a pairing $$\langle, \rangle: H\otimes H\to \mathbb{Q}(-w)$$ such that $(2\pi i)^n\langle -, C -\rangle$ is positive definite and symmetric, where $C$ is the Weil operator which acts by $i^{p-q}$ on $H^{pq}$. These conditions known as the Hodge-Riemann bilinear relations imply that $H\cong H^*(-w)$ as Hodge structures, so in this sense $H$ self dual. However, just having such an isomorphism would not give everything else.

If $H$ is replaced by a variation of Hodge structures, then a polarization is flat pairing as above satisfying the Hodge-Riemann conditions on the fibres. For Hodge modules the story is similar but more delicate. Since the whole theory is constructed by induction on dimension of support, a polarization is also defined in this manner. Given Hodge module $H$ of weight $w$ on $X$, a polarization is a pairing $H\otimes H\to \mathbb{Q}(\dim X-w)[2\dim X]$ satisfying the inductive conditions (0.8)-(0.10) of Saito's "Modules Hodge Polarizables". Such a pairing should induce an isomorphism $H\cong DH(?)$ up to twist, but it's not equivalent.