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 $n>n'$...


**added** 


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? 



On morphisms of pure Hodge structures of decreasing weight
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 $n > n'$.


