Let $C$ be a general genus $g$ curve, how can we describe the Neron-Severi group of its $n$-th self product $C^n=C\times \dots \times C$? 


It is a lattice in $H^2(C^n,\mathbb{Z})\cong \mathbb{Z}^{n+2n(n-1)g^2}$, do we know its rank and basis if $C$ is general?



Neron-Severi group for product of curves
Let me give the answer for $n=2$. 


Fix a point $p \in C$ and call $x\_1, \, x\_2, \, \Delta$ the divisor classes


$\begin{equation\*}
\begin{split}
x\_1 & := \{(p, \, x) \, | \, x \in C\} \\
x\_2 & := \{(x, \, p) \, | \, x \in C \} \\ 
\Delta & := \{(x, \, x) \, | \, x \in C \}. 
\end{split}
\end{equation\*}
$ 


Let $C^{(2)}$ be the second symmetric product of $C$, namely the quotient of $C^2$ by the action of $\mathbb{Z}/2 \mathbb{Z}$ exchanging the two factors, and let $x$, $\delta$ be the images of $x\_1$ (or $x\_2$) and $\Delta$ in $C^{(2)}$, respectively. 


It is a classical result that for a general curve $C$ of positive genus the Neron-Severi group of $C^{(2)}$ is generated by $x$ and $\delta$, hence a basis for the Neron-Severi group of $C^2$ is given by $$x\_1, \, x\_2, \, \Delta.$$
The same result holds for the Neron-Severi group of the $n$-fold symmetric product $C^{(n)}$, hence one shows that the Neron Severi group of $C^n$ is generated, for general $C$, by the coordinate divisors and by the classes of the diagonals. 


**References.** 
A. Kouvidakis: Divisors on symmetric product of curves, *Trans. Amer. Math. Soc.* **337** (1993), 117-128.


