The base field $k$ is algebraically closed.
Let $C$ be a smooth projective variety and $G$ be a finite group of automorphisms of $C$ (as an algebraic variety). As is well known, up to isomorphism there is only one smooth projective curve $C'$ whose function field is $k(C)^G$. I will denote the quotient of $C$ by $G$ as follows: $C^G:=C'$.
remark: I am denoting the quotient variety by $C^G$ instead of the usual $C/G$ to not creat confustion in what follows.
Let $E$ be an elliptic curve, and $G$ be a finite subgroup of $E$. $G$ can be seen as a group of automorphisms of $E$ (as an algebraic variety, not as a group): $G \times E \rightarrow E$ is $(g,p) \mapsto g+p$. The quotient group $E/G$ clearly is the space of orbits for this action. Does $E/G$ with the quotient topology has a natural structure as a smooth projective variety making it isomorphic to its quotient $E^G$?