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Let $H$ be a finite dimensional $\mathbb{Q}$-vector space. I've heard that to give an effectif Hodge structure of weight 1 on $H$ is the same as to give a complex structure on $H_\mathbb{R}$. Why is this true?

If I have a complex structure, that is an endomorphism $I \in \mathrm{End}(H_\mathbb{R})$ such that $I^2=-Id$, then I can define $H^{1, 0}$ (resp. $H^{0, 1}$) as the eigenspaces of the eigenvalues $i$ and $-i$.

How to prove the converse. Here is my idea:

consider the inclusion $j: H_\mathbb{R} \hookrightarrow H_\mathbb{C}=H^{1, 0} \oplus H^{0, 1}$. For any $u \in H_\mathbb{R}$, $j(u)$ decomposes as

$j(u)=v+w, \quad v \in H^{1, 0}, w\in H^{0, 1}$

Define $I(u)=i(v-w)$ for the corresponding $v$ and $w$. One has to verify that $I(u) \in H_\mathbb{R}$. But:

$\overline{I(u)}=-i(\bar{v}-\bar{w})=-i(w-v)=i(v-w)=I(u)$

where I use the Hodge symmetry to say that $\bar{v}=w$ and $\bar{w}=v$.

Is that correct? Is it the standard argument? Is there a more conceptual way of understanding things?

Thanks

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I suppose you mean weight $1$ Hodge structure? –  dhagbert Apr 11 '13 at 21:54
    
You can use the equivalence between Hodge structures and representations of the group scheme $\mathbf{S}$. In weight 1, this information can be encoded by the eigenvalues of your operator $I$, but in higher weights, you lose some information. –  S. Carnahan Apr 12 '13 at 2:54
    
@dhagbert: Yes I mean weight 1, I edited my question. –  SH1 Apr 12 '13 at 7:25
    
@Carnahan: Could you develop a bit more your answer? Anyway, what you are saying should be the generalization to higher weight of the equivalence "Effective HS of weight 1 --- complex structures" –  SH1 Apr 12 '13 at 7:27
    
Help please!!!! –  SH1 Apr 16 '13 at 16:41

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