Absolute Hodge implies Galois invariant?

Question

Let $X$ be an Abelian variety defined over a number field $K$, suppose that it has a good reduction over a fine place $\mathfrak{p}$ of $K$. Let $G_{\mathfrak{p}}$ be the local Galois group for $K_{\mathfrak{p}}$. Let $s$ be a Hodge cycle in $H^n_{B}(X^{an}, \mathbb{Q})$, the Betti cohomology. By comparison theorem there is a canonical isomorphism $H^n_{B}(X^{an}, \mathbb{Q})\otimes \mathbb{Q}_p\cong H^n_{et}(X,\mathbb{Q}_p)$, where the right hand side of the equation is the etale cohomology. So via this isomorphism, $s$ corresponds to a etale cycle $s_{et}$. Then Does Deligne's absolute Hodge theory implies that $s_{et}$ is fixed by a finite index subgroup of $G_{\mathfrak{p}}$.

Answer 1

This is Proposition 2.9(b) of Deligne's 1982 notes.

Answer 2

I guess $X = A$, and for the étale cohomology you change base to $\bar{K}$. The answer is then: yes, by definition. (At first, Deligne–Milne define absolute Hodge cycles over algebraically closed fields. Then, for general fields, they ask the étale component to be Galois invariant.)

Let $\mathrm{MT}$ denote the Mumford–Tate group of $H_{B}^{n}(X^{an},\mathbb{Q})$. Then $s$ is invariant under $\mathrm{MT}$. Let $G_{p}^{\circ}$ be the connected component of the Zariski closure of the image of the Galois group under the representation on $H_{et}^{n}(X_{\bar{K}}, \mathbb{Q}_{p})$.

Using the fact that the étale component of absolute Hodge cycles is Galois invariant, we get the following corollary: Deligne's theorem shows (for abelian varieties) that Hodge = absolute Hodge, hence $$G_{\mathfrak{p}} \subset \mathrm{MT} \otimes_{\mathbb{Q}} \mathbb{Q}_{p}.$$ This is precisely because the group $G_{p}^{\circ}$ must fix at least all the étale components of the absolute Hodge cycles (which, by the theorem are precisely the $\mathrm{MT}$-invariants).