The algebraicity of Hodge structure map 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...
 A: 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).
A: Who is every source? If you fix a weight $k$, then to give a real Hodge structure is the same as giving a homomorphism 
$h:{\mathbb G}_m\to GL(V)$ which is defined over $\mathbb R$. That does not mean it is defined over $\overline{\mathbb Q}$. Indeed, $h$ doesn't know anything about the $\overline{\mathbb Q}$-structure, so why should it respect it?
As an example, let's take $V={\mathbb Q}^2$ and $h$ defined by
$$
h(a)=A(^a{}_1)A^{-1},\ \ \ A=(^1_0{}^{\pi}_1).
$$
Then $h$ is not defined over $\overline{\mathbb Q}$.
But it is conjugate to a homomorphism which is defined over $\mathbb Q$. Maybe that's what you need?
