Assume that $C$ is an elliptic curve and $C_p$ is the $p$-fold symmetric product. Let $\beta:C_p\to C$ be defined by the addition on the elliptic curve. Let $u\in C$ be the zero in the additive group of $C$. Then $\beta^*u=:F$ is a divisor on $C_p$. I want to calculate $H^0(C_p,nF)$. The paper Symmetric products of elliptic curves and surfaces of general type with $p_g = q = 1$ by Cantanese and Ciliberto(Link) indicates that $C_p$ is a projective bundle over $C$ and its Theorem 1.17 gives an explicit answer, but I don't understand how the leading coefficient is determined. Another confusion is that it indicates that all divisors on $C_p$ are algebraically equivalent to $mD+nF$, where $D=u+C_{p-1}$, but we know that the Picard group of a projective bundle is the direct sum of the picard group of the base and the integer $\mathbb{Z}$. We have $Pic(C)=C\oplus\mathbb{Z}$. Therefore we should have $Pic(C_p)=Pic(C)\oplus \mathbb{Z}$. I think the symbol $\mathbb{Z}$ in $Pic(C)$ stands for $\mathbb{Z}F$, and the symbol $\mathbb{Z}$ in $Pic(C_p)$ stands for $\mathbb{Z}D$. Why the Picard group of $C_p$ can be generated by only two elements? If there was any misunderstanding, please indicate it to me.

## 1 Answer

For the second part, as what Francesco Polizzi indicated, the group generated by $F$ and $D$ is not the Picard group of $C_p$. For the calculation of the cohomology, we have $$nF=n\beta^*u=\beta^*(nu).$$ Therefore $$H^0(C_p,nF)=H^0(C_p,\beta^*\mathcal{O}_C(nu))=H^0(C,\beta_*\beta^*\mathcal{O}_C(nu)).$$ By projection formula, $$\beta_*\beta^*\mathcal{O}_C(nu)=\beta_*\mathcal{O}_{C_p}\otimes\mathcal{O}_C(nu).$$ We are left to show that $$\beta_*\mathcal{O}_{C_p}=\mathcal{O}_{C}.$$ As indicated by the paper I mentioned in the question, $C_p$ is in the form $\mathbb{P}(E)$ for some vector bundle $E$ on $C$. For any affine open subset $U\cong\mathrm{Spec}(A)\subset C$, $$\beta_*\mathcal{O}_{C_p}(U)\cong\mathcal{O}_{\mathbb{P}(E)}\bigl(\operatorname{Proj}\operatorname{Sym}E(U)\bigr)\cong E^0(U)=\mathcal{O}_C(U).$$