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).

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