A bielliptic surface $S$ is a smooth projective complex surface of Kodaira dimension 0 with $h^1(\mathcal O_S)=1$ and $h^2(\mathcal O_S)=0$. It is well known that $S=(A\times B)/G$, where $G$ is a finite group acting on the elliptic curve $A$ by translations and on the elliptic curve $B$ such that $B/G\cong\mathbb{P}^1$. It turns out that $S$ always has torsion canonical bundle (of order 2,3,4, or 6), and there are 7 families of such surfaces based on the 7 different groups that can appear. For 4 of these families, the $G$ is cyclic (of order $2,3,4$ or $6$), so the canonical cover $X$ (i.e. the \'{e}tale cover $X\to S$ with trivial $\omega_X$ and covering degree equal to the degree of torsion of $\omega_S$) is simply $A\times B$. For the other 3 families, the canonical cover is an intermediate abelian $X$.
While Serrano describes $\mathrm{Num}(S)$ explicitly in ("Divisors of bielliptic surfaces and embeddings in $\mathbb{P}^4$," Mathematische Zeitschrift 203 (1990), 527-533) as being generated (over $\mathbb{Q}$) by $A$ and $B$ with $A^2=B^2=0$ and $A.B=|G|$, it is unclear to me how to describe $\mathrm{Num}(X)$, though I've seen it claimed that it has the same rank and generators except that $A.B$ is some divisor of $|G|$. I'm just not sure if this is correct/ or how to show it.
Now, I know that $\mathrm{NS}(A\times B)=\mathbb{Z}A\oplus\mathbb{Z}B\oplus\mathrm{Hom}(A,B)$ with intersection product given by $(a,b,f).(a',b',f')=ab'+a'b-(\deg(f+f')-\deg(f)-\deg(f'))$. I believe there is a theorem of Liebermann that says that $\mathrm{Num}=\mathrm{NS}$ for abelian varieties. So it seems that the above quoted claim about $\mathrm{Num}(X)$ can't be true when $G$ is cyclic, $A$ and $B$ are isogenous with complex multiplication. What about in the other cases? Then $X$ is the quotient of $A\times B$ by a group of translations $H$ (such that $G/H$ is cyclic). Do translations act non-trivially on $\mathrm{Hom}(A,B)$ so that the $H$-invariant part of $\mathrm{NS}(A\times B)$ is really just $\mathbb{Z}A\oplus\mathbb{Z}B$?
