Let $X$ and $Y$ be compact Riemann surfaces that are both hyperbolic (i.e. genus > 1). A classical result of de Franchis implies that the space of nonconstant holomorphic maps from $X$ into $Y$ is a finite set. I am investigating the structure of the space of holomorphic mappings from $X$ into $Y^n_\textsf{sym} := Y^n/S_n$ (the $n$th symmetric product of $Y$). I believe that any holomorphic mapping $f:X \to Y^n_\textsf{sym}$ lifts to a holomorphic mapping $\widetilde{f}:X \to Y^n$ such that $f = \Pi \circ f$ where $\Pi:Y^n \to Y^n_\textsf{sym}$ is the quotient map. Is this assertion true?
2 Answers
No, assume that there exists an $n$sheet cover $p:Y\to X$(such certainly exist, take $Y$ to be the Riemann surface with field of meromorphic functions equal to some degree $n$ extension of $\mathbb{C}(X)$). Define $f(x)$ to be the element in symmetric power corresponding to the fiber $p^{1}(x)$. If $f$ factors through $Y^n$, the cover would admit a section(given, for example, by assigning to $x$ the first coordinate of $\tilde{f}(x)$), which is false.
As SashaP already pointed out, this is false. In fact it fails very badly: The space of nonconstant holomorphic maps from $X$ to $\operatorname{Sym}^2 Y$ can already contain infinitely many connected components of arbitrarily large dimension. Indeed, both of these occur when $Y$ is hyperelliptic, so $\operatorname{Sym}^2 Y$ contains a copy of $\mathbb P^1$ parameterizing orbits under the hyperelliptic involution. The dimension of the space of degree $d$ maps from $X$ to $\mathbb P^1$ grows to $\infty$ with $d$, so the dimension of components of the space of maps from $X$ to $\operatorname{Sym}^2 Y$ can be arbitrarily large as well.

$\begingroup$ Thanks! What if I consider a surjective holomorphic map from from $X^n$ to $Y^n_\textsf{sym}$? In this case, the space of surjective holomorphic maps is finite. $\endgroup$ Oct 16, 2016 at 10:38


$\begingroup$ More examples of failure: for $n> 2g2$, $\mathrm{Sym}^n Y$ is a $\mathbb{P}^{ng}$ bundle over $\mathrm{Pic}(Y)$. Any curve $X$ can map into projective space, so we can always map into one of the fibers of this bundle. Actually, we can use a smaller $n$. As soon as $n \geq g/2+1$, there are fibers of $\mathrm{Sym}^n Y \to \mathrm{Pic}(Y)$ which are $\mathbb{P}^1$'s, and any $X$ maps to a $\mathbb{P}^1$. $\endgroup$ Oct 16, 2016 at 11:40

$\begingroup$ @ Will Sawin I will clarify. Let $f: X^n \to Y^n_\textsf{sym}$ be a surjective holomorphic map where $X$ and $Y$ are hyperbolic compact Riemann surfaces. Does $f$ lift to a map into $Y^n$? $\endgroup$ Oct 16, 2016 at 11:52