Hence we obtain the $4$-dimensional Galois representation $$\rho_E \otimes \text{ind}_{G_F}^{G_{{\mathbb Q}}} \chi_G
\simeq \text{ind}_{G_F}^{G_{{\mathbb Q}}}(\chi_G \otimes
\text{Res}^{G_{{\mathbb Q}}}_{G_F}\rho_E),$$ where $\rho_E$ is the Galois representation associated to $H^1(E)$. The operation of restricting $\rho_E$ to $G_F$ and of twisting by characters have automorphic analogs. Let $\pi_E$ be the automorphic representation associated to $E$. Then $\pi' = \chi\otimes$ Res$^{{\mathbb Q}}_F\pi_E$ makes sense as an automorphic cuspidal irreducible representation of $GL(2,F)$, and we have the characterizing relationship $$L(\pi',s) = L(\text{ind}_F^{{\mathbb Q}}\pi',s) =
L(\pi_E \otimes \text{ind}_F^{{\mathbb Q}}\chi,s) =
L(g_E\otimes g_Y,s).$$

For the last part, write the $p$th Euler factors of $g_E$ and $g_Y$ respectively as $$(1-\alpha_p p^{-s})(1-\alpha'_p p^{-s}) =
1-A(p) p^{-s} + p^{1-2s} \qquad\text{and}$$ $$(1-\beta_p p^{-s})(1-\beta'_p p^{-s}) =
1 - B(p) p^{-s} +\epsilon_Y(p) p^{2-2s}.$$ Then the Euler factor $L_p$ is defined as $$L_p(s) =(1-\alpha_p\beta_p p^{-s})
(1-\alpha'_p\beta_p p^{-s})
          (1-\alpha_p\beta_p' p^{-s})
          (1-\alpha_p'\beta_p' p^{-s}),$$ and the claim follows by a direct calculation.

[**Remark**]{}: [*If a K3 surface has the Picard number $19$ or $18$, the modularity question for the product $Y\times E$ is still open. However, if the Picard number is 19, one knows at least that the rank $3$ motive $T(Y)$ of the transcendental cycles is self dual orthogonal via the cup product. (For explicit examples of 