pullback of pushforward In the case where $k$ is a field and $L$ a finite Galois extension of $k$ we can consider the projection map $\pi: X\otimes_k L\rightarrow X$, where $X$ is a smooth projective variety over $k$. If we take a coherent sheaf $\mathcal{F}$ (or locally free sheaf $\mathcal{E}$. I guess it is true that $\pi^*\pi_*\mathcal{E}\simeq \mathcal{E}\otimes_k L[G]\simeq \mathcal{E}^{\oplus r}$, where $G$ is the Galois group and $r=ord(G)$.
My question:
What happens for infinite Galois groups or in the absolute case i.e. $k\subset \bar{k}$ and $G$ the absolute Galois group. What is $\pi^*\pi_*\mathcal{E}$ for $\pi:X\otimes_k\bar{k}\rightarrow X$ ? 
 A: That's not true. Consider a coherent sheaf $\mathcal E$ on $X \otimes_k L$ which is not Galois-invariant, like a line bundle on an elliptic curve correspoinding to a point of the Jacobian which is not defined over $k$, Then $\pi^* \pi_* \mathcal E$ is conjugation-invariant, but $\mathcal E^{\oplus r}$ usually isn't.
In the finite case, this will be the sum over all the Galois conjugates of $\mathcal E$. In the infinite case, $\pi^* \pi _ * \mathcal E$ is the subsheaf of the infinite product sheaf $\Pi_{\sigma \in G} \mathcal E^\sigma$ consisting of functions that are locally constant in the profinite topology on $G$.
The map $\mathcal E \otimes_k L \to \Pi_{\sigma \in G} \mathcal E^\sigma$ just comes from evaluating the product of a function in $\mathcal E$ with an element in $L$ by twisting with the automorphism. The key fact being that the action of $L$ on $\mathcal E \otimes_k L$ in this construction comes from the right $L$, not the left $\mathcal E$, which is why it is $\mathcal E_\sigma$. The functions we get are locally constant because any element in the tensor product depends on only finitely many elements of $L$. The map to locally constant functions is an isomrophism because we can check it's an isomorphism for each finite-degree subfield of $L$ / submodule of functions constant on an open index subgroup of the Galois group.
(What you said is true for $\pi_* \pi^*$, though. ) 
