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Let $H_{\mathbf{Q}}$ and $H_{\mathbf{Q}}'$ be two pure Hodge structures of weight $n$ and $n'$ respectively. How do you prove the following simple fact:

fact: If $n>n'$ and $f:H_{\mathbf{Q}}\rightarrow H_{\mathbf{Q}}'$ is a morphism which respects the filtrations over $\mathbf{C}$, then $f=0$.

I don't quite see how to use the assumption $n>n'$...

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My original question was related to the fact that asking the morphism to be compatible with the torus action seems to be a stronger condition that only asking for the filtration to be preserved. And I guess that in general this is all you can say. This reflexion was motivated by Deligne Scholie 5.1 in Hodge 1 which says the following:

Scholie 5.1 Soit $H$ et $H'$ des structures de Hodge de poids $n$ et $n'$ avec $n>n'$.Soit $f:H_{\mathbf{Q}}\rightarrow H_{\mathbf{Q}}'$ un morphisme tel que $f:H_C\rightarrow H_C'$ respecte $F$. Alors $f=0$.

Q: So how should I interpret Scholie 5.1?

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    $\begingroup$ One only needs to assume that $n \neq n'$... $\endgroup$
    – naf
    Commented Aug 12, 2011 at 14:30
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    $\begingroup$ It's easier if you think that a weight HS has a bigrading $H\otimes \C$ with $\bar H^{pq}=H^{qp}$ and $p+q=n$. Since morphisms must preserve this, they would vanish for different weights. There is a lemma to be proved about the equivalence of this notion with representations of Deligne's torus, but it's not difficult. $\endgroup$
    – Donu Arapura
    Commented Aug 12, 2011 at 15:03
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    $\begingroup$ I claim that there are no nonzero morphisms (= $\mathbb{Q}$-linear filtered maps) from $\mathbb{Q}$ and the Tate twist $\mathbb{Q}(-1)$ (your example). But I'm afraid I'm going to have to leave this to you. $\endgroup$
    – Donu Arapura
    Commented Aug 12, 2011 at 16:06
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    $\begingroup$ I hadn't read that before. Scholie 5.1 is baffling to me as well, since a few lines up, Deligne makes a stronger statement "si $f:H\to H'$ un morphisme de structures de Hodge mixte pures de poids différents, alors $f$ torsion." If you figure it out, let us know. $\endgroup$
    – Donu Arapura
    Commented Aug 12, 2011 at 18:49
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    $\begingroup$ I think your interpretation of the Scholie is correct: $f$ is NOT a morphism of Hodge structures but just a linear map of the underlying rational vector spaces which, when based changed to $\mathbb{C}$, preserves $F$. (For pure Hodge structures of a fixed weight $n$ this is a morphism of Hodge structures.) $\endgroup$
    – naf
    Commented Aug 13, 2011 at 9:44

1 Answer 1

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Here is a proof of the Scholie:

For a pure Hodge structure $H$ of weight $n$ we have $H_{\mathbb{C}} = \oplus_{p+q=n} H^{p,q}$ where we have $H^{p,q} := F^p \cap \bar{F}^q$. The key point is that $H^{p,q}$ as defined above is $0$ if $p+q > n$ (but not if $p+q < n$). This is because for a pure Hodge structure of weight $n$ we have $F^p = \oplus_{p' \geq p, p' +q' = n} H^{p',q'}$ and $\bar{F}^q = \oplus_{q' \geq q, p' + q' = n} H^{p',q'}$.

The Scholie now follows easily: It suffices to prove that $f(H^{p,q}) = 0$ for all $p+q = n$. Since $f$ is defined over $\mathbb{Q}$, $f(H^{p,q}) \subset H'^{p,q}$ and the latter is $0$ if $n > n'$.

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  • $\begingroup$ Yes, that's it. I somehow mislead myself into believing that filtered $\mathbb{Q}$-linear maps between pure Hodge structures are morphisms, but this only true if the weights are the same. Sometimes the obvious interpretation is the hardest. $\endgroup$
    – Donu Arapura
    Commented Aug 13, 2011 at 12:08
  • $\begingroup$ .... to accept. $\endgroup$
    – Donu Arapura
    Commented Aug 13, 2011 at 12:10
  • $\begingroup$ But Ulrich I'm confused, you only seem to use the fact that $n\neq n'$ and moreover what about my counterexample $\endgroup$
    – Hugo Chapdelaine
    Commented Aug 13, 2011 at 14:06
  • $\begingroup$ You have $f(F^P H)\subseteq F^p H'$ but I don't quite see how you get from that $f(H^{p,q})\subseteq H'^{p,q}$. In fact my counterexample shows that it is wrong, isn't? $\endgroup$
    – Hugo Chapdelaine
    Commented Aug 13, 2011 at 14:08
  • $\begingroup$ Since $f$ is defined over $\mathbb{Q}$ (so commutes with complex conjugation) one also gets $f(\bar{F}^qH) \subset \bar{F}^q H'$. Your counterexample doesn't apply here since the conclusion uses $n > n'$. $\endgroup$
    – naf
    Commented Aug 13, 2011 at 14:23

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