Let $R$ be a ring and $G$ be a finite group. Assume there is a $G$ action on $\operatorname{Spec} R$. For example, the group $\{\pm1\}$ acts on $\mathbb{A}^1_{\mathbb{C}}$ by $(-1) z = -z$. let $\pi : \operatorname{Spec} R \longrightarrow (\operatorname{Spec} R) /G$ be the canonical projection map to the quotient stack.
We can identify $\operatorname{QCoh} (\operatorname{Spec} R)$ with $R$-module and $\operatorname{QCoh} ( (\operatorname{Spec} R) /G)$ with $G$-equivariant $R$-module. Under this identification, the pullback $\pi^\ast$ is just the forgetful functor from $G$-equivariant $R$-module to $R$-module. My question is what is $\pi_\ast$ in this case?
Answers or reference are both welcome. The definition I know for this pushforward is from D. Gaitsgory and N. Rozenblyum's book which is not very computational.
It is the right adjoint to the forgetful functor. You might call this the "co-free representation" or the "co-induced representation from the trivial group" (suitably adjusted to the case where $G$ acts non-trivially on $R$) or something like that.
In case $G$ is finite and you think of "$G$-equivariant $R$-modules" in terms of some sort of twisted (noncommutative) group algebra $R[G]$, the right (respectively, left) adjoint to the forgetful functor should look like $\text{Hom}_R(R[G], -)$ (respectively, $R[G] \otimes_R -$). A more "stacky" description would be to look at the Cartesian square below, and try to establish that the base-change morphism is an isomorphism—identifying the underlying $R$-module with$$r_* p_1^* M \cong \mathcal{O}_{X \times G} \otimes_R M,$$and so forth.
$$\require{AMScd} \begin{CD} X \times G @> r >> X \\ @V p_1 VV @VV \pi V \\ X @>> \pi > X/G \end{CD} $$