Let $\mathbb S$ be $\mathbb C^{\times}$'s restriction of scalar to $\mathbb R$. To give a real Hodge structure on an $\mathbb Q$ vector space $V$ is to give a real representation of $\mathbb S$ on $V\_{\mathbb R}$. Let $\mathbb G\_m\rightarrow \mathbb S\rightarrow GL(V\_{\mathbb R})$ be the weight homomorphism. If it is actually defined over $\mathbb Q$ we say the hodge structure is rational. But can we say that the weight homomorphism is always algebraic, that is, defined over $\overline{\mathbb Q}$? Every resource claims this, but I can't see why...



The algebraicity of Hodge structure map
It looks false to me. Let $V=\mathbb{Q}^{2}$, and let $V(\mathbb{R})=V^{0}\oplus V^{2}$ where $V^{0}$ is the line defined by $y=ex$ and $V^{2}$
is the line defined by $y=\pi x$. Give $V^{0}$ the unique Hodge structure of
type $(0,0)$ and $V^{2}$ the unique Hodge structure of type $(1,1)$. To say
that $w$ is defined over the subfield $\mathbb{Q}^{\mathrm{al}}$ of
$\mathbb{C}$ means that the gradation $V(\mathbb{R})=V^{0}\oplus V^{2}$
arises from a gradation of $V(\mathbb{Q}{}^{\mathrm{al}})$ by tensoring up,
but this isn't true. Perhaps the all the "resources" have additional
conditions, or perhaps they are all ...


Added: When you are defining a Shimura variety, the weight homomorphism w factors through a Q-subtorus of GL(V), and then it is true that w is defined over the algebraic closure of Q (because, for tori T,T', the group Hom(T,T') doesn't change when you pass from one algebraically closed field to a larger field).


