Let $C$ be a 1-dimensional complex manifold whose universal covering is provided by the half-plane $\mathcal{H}=\{z \in \mathbb{C} \mid \operatorname{Im}z>0\}$. The symmetric product $C^{(n)} = C^n / S\_n$ is a complex manifold. My question is: is the universal covering of $C^{(n)}$ provided by $\mathcal{H}^{(n)}$? Or is it some different quotient of $\mathcal{H}^n$?



Universal covering of symmetric product
In fact, the universal cover of $C^{(n)}$ will not be $\mathcal H^n$ once $n \gg 0$. Indeed, if $C$ is a compact Riemann surface of genus $g \geq 2$ (so the universal cover is $\mathcal H$) with a base point $x$, then the Riemann–Roch theorem implies that the map
\begin{align\*}
C^{(n)} &\to \operatorname{Jac}\_C = \operatorname{Pic}^0\_C\\
(x\_1,\ldots,x\_n) &\mapsto [x\_1 + \ldots + x\_n - nx]
\end{align\*}
is a $\mathbf P^{n-g}$-bundle if $n > 2g-2$ (see e.g. [CS86, Ch. VII, Rmk. 5.6(c)]). So the universal cover of $C^{(n)}$ is a $\mathbf P^{n-g}$-bundle over $\mathbf C^g$, which is very far from $\mathcal H^n$ (e.g. it is not Kobayashi hyperbolic since $C^{(n)}$ is not Brody hyperbolic).




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**References.**


[CS86] G. Cornell, J. H. Silverman (eds.), [*Arithmetic geometry*](https://doi.org/10.1007/978-1-4613-8655-1). Springer-Verlag, 1986. [ZBL0596.00007](https://zbmath.org/?q=an:0596.00007).


