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...
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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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$\begingroup$ Thanks! So when we define a Shimura datum $(G,X)$, the map $\mathbb S\rightarrow G_{\mathbb R}$ is actually defined over $\mathbb Q$, not only over $\mathbb R$? If it is only defined over $\mathbb R$, I still can't assure myself that $\mathbb G_m\rightarrow \mathbb S\rightarrow G_{\mathbb R}$ factors through a $\mathbb Q$ subtorus. $\endgroup$– TJCMCommented Mar 16, 2010 at 14:15
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$\begingroup$ The maps $\mathbb{S}\to G_{\mathbb{R}}$ are defined over $\mathbb{R}$,not $\mathbb{Q}$. However, one of the axioms (2.1.1.1 in Deligne's Corvallis articles; SV1 in my articles) for Shimura varieties implies that the weight homomorphism $w$ of each $h$ factors through the centre of $G$ (hence through connected centre, which is a torus). As the $h$'s are all conjugate, this implies that the $w$'s are all equal, and that the $w$ is defined over the algebraic closure of $\mathbb{Q}$ (it is a homomorphism between tori defined over $\mathbb{Q}$). $\endgroup$– JS MilneCommented Mar 16, 2010 at 19:14
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$\begingroup$ I see, thank you. Is it in general true for a homomorphism from a reductive group to a linear group, if it is defined over a bigger algebraic closed field, then we can always descent it to a smaller algebraic closed field? (for unipotent groups one may have a lot of non algebraic homomorphisms, e.g. $x\mapsto \pi x$) $\endgroup$– TJCMCommented Mar 17, 2010 at 0:59
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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?