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I suspect this is very elementary, but it is not stated anywhere. A Hodge structure of weight $k$ consists of a finite rank lattice $H_{\mathbb{Z}}$ together with a decomposition of its complexification $H : = H_{\mathbb{Z}} \otimes \mathbb{C}$, $$H = \bigoplus_{p+q=k} H^{p,q},$$ with $H^{p,q} = \overline{H^{p,q}}$. Set $F^p : = \bigoplus_{r \geq p} H^{r,k-r}$.

A blinear form $Q$ on $H$ polarizes $H$ if

$Q(u,v) = (-1)^k Q(v,u)$,

$Q(F^p, F^{k-p+1})=0$,

The Hermitian form $Q((\sqrt{-1})^{p-q} \cdot, \overline{\cdot})$ is positive definite.

Question: Can every Hodge structure be polarized?

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    $\begingroup$ I guess you may be right it’s hard to find, but the answer is no. Hodge structures of type (1,0), (0,1) are equivalent to complex tori, and the polarizable ones correspond to abelian varieties. $\endgroup$
    – Donu Arapura
    Commented Sep 8, 2020 at 2:55
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    $\begingroup$ However the OP does not mention that $Q$ takes integral values on $H$; if you don't put that condition, finding such a $Q$ is just an exercise in linear algebra. $\endgroup$
    – abx
    Commented Sep 8, 2020 at 3:14

1 Answer 1

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Let me summarize comments.

  1. abx points out that $Q$ should be integer valued, otherwise it's not an interesting notion.
  2. (Assuming integrality) the answer is no because the categories of Hodge structures of type $\{(1,0), (0,1)\}$ and complex tori are equivalent. The polarizable ones correspond to abelian varieties, which is a proper subset.
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  • $\begingroup$ Thanks for your answer. The equivalence between Hodge structures of type $\{ (1,0), (0,1) \}$ and complex tori is given as follows? Let $H_{\mathbb{Z}}$ be a lattice of rank $2n$ with $\phi : H_{\mathbb{Z}} \to H_{\mathbb{Z}}$ an endomorphism. Take the eigenvalues $\lambda_1, ..., \lambda_{2n}$ of $\phi$ to be distinct with non-zero imaginary part. Let $H^{1,0}$ be the eigenspace of $\lambda_1, ..., \lambda_n$, such that $\lambda_i \neq \overline{\lambda_j}$ for any $1 \leq j \leq n$. The torus is then $T = H/(H^{1,0} \oplus H_{\mathbb{Z}})$? $\endgroup$
    – AmorFati
    Commented Sep 10, 2020 at 22:44

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