# How to show that Hodge filtration of CM type is algebraic?

I'm asking for a proof or references of the following claim:

Let $$V$$ be a rational Hodge structure having CM in the sense that its Mumford-Tate group is abelian. Then there is a filtration $$F^{\bullet}_{\overline{\mathbb{Q}}}$$ on $$V_{\overline{\mathbb{Q}}}$$ such that the Hodge filtration $$F^{\bullet}$$ of $$V_{\mathbb{C}}$$ is given by $$F^{\bullet}=F^{\bullet}_{\overline{\mathbb{Q}}}\otimes_{\overline{\mathbb{Q}}}\mathbb{C}.$$

A CM Hodge structure is a polarizable rational Hodge structure whose Mumford-Tate group is commutative, hence a torus $$T$$. The Hodge filtration is split by a cocharacter of $$T$$ over $$\mathbb{C}$$, which is automatically defined over the algebraic closure of $$\mathbb{Q}$$ in $$\mathbb{C}$$.
Added: it's the cocharacter $$\mu$$ such that $$\mu(z)$$ acts on $$V^{p,q}$$ as $$z^{-p}$$. A cocharacter over $$\mathbb{C}$$ of a torus defined over $$\mathbb{Q}^{al}$$ is automatically defined over $$\mathbb{Q}^{al}$$. Since the cocharacter determines the filtration, it also is defined over $$\mathbb{Q}^{al}$$.
• By "a cocharacter", do you mean the one $\mathbb{G}_m\to M_\phi$ induced by $\phi:\mathbb{S}(\mathbb{R})\to GL(V)(\mathbb{R})$? How to use it? On the other hand, shouldn't a representation of a torus automatically decomposes into characters? Why I need a cocharacter? May 22, 2019 at 4:38