$X$ = bi-elliptic surface (smooth and over $\mathbb{C}$),
Aut($X$) = the group of automorphisms of $X$,
Aut$^0(X)$ = connected component of the identity in Aut($X$).


Is Aut$^0(X)$ always an affine algebraic group?



Automorphism group of bi-elliptic surface
The answer is always "no". By classification, a bielliptic surface over $\mathbb C$ has the form $(E\times F)/G$ where $E,F$ are elliptic curves, $G=\subset Aut(E,0)$ is an abelian group acting by complex multiplications on $E$ and by translations on $F$. ($G$ is not necessarily cyclic as Tuan correctly points out.)


($X$ maps to an elliptic curve $F/G$ and every fiber is isomorphic to an elliptic curve $E$, hence the name *bielliptic*.)


Then $F$ acts on $E\times F$ by $(x,y)\mapsto (x,y+f)$, and this action commutes with the $G$-action. Thus, $F\subset Aut^0(X)$. As $F$ is a projective variety, $Aut^0(X)$ is not affine.


