Let $X$ be a smooth projective variety over $\mathbb{Q}$. For simplicity assume that $X$ has rational point. I think there are various algebraic constructions of the Albanese in the 1950's literature, but I have to confess I've never gone through the details. So instead define $A=Alb(X) =Pic^0(Pic^0(X))$. After fiddling with the Poincare sheaf and the given rational point, you should get a Abel-Jacobi morphism $\alpha:X\to A$. I claim that this induces an isomorphism $$H^1(X_{et},\mathbb{Q}_\ell)\cong H^1(A_{et},\mathbb{Q}_\ell)$$$H^1(\overline{X}_{et},\mathbb{Q}_\ell)\cong H^1(\overline{A}_{et},\mathbb{Q}_\ell)$$ (where \overline{X}= X\otimes \overline{\mathbb{Q}}) necessarily compatible with the Galois action. To see this, base change up to \mathbb{C}, then by the comparison theorem, I it suffices to get an isomorphism on the first singular (aka Betti) cohomology with \mathbb{Q} coefficients. But the analytic construction of Alb(X) gives this immediately because$$Alb(X)= H^0(X_{an},\Omega_X^1)^*/H_1(X_{an},\mathbb{Z})$$So the rational first homologies coincide, now dualize. 1 Let me slightly expand my comment from yesterday. Unfortunately, because of various time constraints, this will still be quite sketchy. Note that I'm only addressing the titular question. I have nothing to say about the automorphic aspects, since it is too far from what I know. Let X be a smooth projective variety over \mathbb{Q}. For simplicity assume that X has rational point. I think there are various algebraic constructions of the Albanese in the 1950's literature, but I have to confess I've never gone through the details. So instead define A=Alb(X) =Pic^0(Pic^0(X)). After fiddling with the Poincare sheaf and the given rational point, you should get a Abel-Jacobi morphism \alpha:X\to A. I claim that this induces an isomorphism $$H^1(X_{et},\mathbb{Q}_\ell)\cong H^1(A_{et},\mathbb{Q}_\ell)$$ necessarily compatible with the Galois action. To see this, base change up to \mathbb{C}, then by the comparison theorem, I suffices to get an isomorphism on the first singular (aka Betti) cohomology with \mathbb{Q} coefficients. But the analytic construction of Alb(X) gives this immediately because$$Alb(X)= H^0(X_{an},\Omega_X^1)^*/H_1(X_{an},\mathbb{Z})$\$ So the rational first homologies coincide, now dualize.