From a physicist: How do I show certain superelliptic curves are also hyperelliptic?

Question

As the title suggests, I am a physicist and have a question about how to show certain superelliptic curves are also hyperelliptic. The superelliptic Riemann surfaces in question has the form $$w^n = \prod_{\alpha=1}^N(z-u_\alpha)(z-v_\alpha)^{n-1},\quad u_1<v_1<\dotsb<u_N<v_N$$ and is of interest for certain questions 1+1D CFTs (see Coser, Tagliacozzo, and Tonni - On Rényi entropies of disjoint intervals in conformal field theory) that I am working on. For N = 2, there is a change of variables that can be done (see section 6 of Enolski and Grava - Singular $Z_N$-curves and the Riemann-Hilbert problem) to bring this into the form $\nu^2 = f(\rho)$ where $f(\rho)$ is a polynomial of degree $2n$. That is, the $N=2$ case is a hyperelliptic curve and there is a change of variables that makes that obvious. My question is: is there a way to prove that for some or all values of $N$, the above curve is also hyperelliptic? While having a way to construct the appropriate change of variables would be nice, if there is a way say whether it is or is not hyperelliptic without constructing such a transformation would be very helpful.

I hope this question isn't too basic for around here.

Answer 1

This curve is not hyperelliptic unless $n=2$ or $N=2$.

First, note that it is more convenient to write the curve as

$$ w^n = \frac{ \prod_{\alpha=1}^N (z- u_\alpha )} {\prod_{\alpha=1}^N (z- v_\alpha)}$$ after a change of variables dividing $w$ by $\prod_{\alpha=1}^N (z- v_\alpha)$.

A straightforward calculation shows that this equation defines a smooth curve in $\mathbb P^1 \times \mathbb P^1$, of bidegree $(n,N$), thus of genus $(n-1)(N-1)$. So, if the curve is hyperelliptic, the hyperelliptic involution must have $2g+2 = 2(n-1)(N-1) +2$ fixed points.

Since hyperelliptic curves have a unique hyperelliptic involution, this involution commutes with all other automorphisms, so it commutes with $w \to e^{ 2\pi i /n} w$.

Thus it sends orbits of $w \to e^{ 2\pi i /n} w$ to orbits. Since these orbits are determined by the value of $z$, the hyperelliptic involution must express the new value of $z$ as a function of the old value of $z$. This must be invertible, so the new $z$ is a rational linear transformation $f(z)$ of the old $z$.

Assume $f$ is not the identity. Then $f$ fixes at most $2$ values of $z$. Each value of $z$ corresponds to at most $n$ values of $w$, so the hyperelliptic involution has at most $2n$ fixed points, and thus $$2n \leq 2(n-1)(N-1) +2$$ or $$ (n-1) \leq (n-1) (N-1)$$ which implies $N \leq 2$ or $n \leq 1$. The $n \leq 1$ case is clearly genus $0$ and can be discarded, so if $N>2$ then we must have $f(z)=z$.

If $f(z)=z$, then the hyperelliptic involution sends $w$ to another $n$th root of $w^n$. This must be obtained by multiplying $w$ by an $n$th root of unity. Since the hyperelliptic involution is an involution, this must be multiplying $w$ by $-1$, so $n$ must be even and the quotient by the hyperelliptic involution is given by

$$ w^{n/2}= \frac{ \prod_{\alpha=1}^N(z- u_\alpha )} {\prod_{\alpha=1}^N (z- v_\alpha)}.$$

This is a smooth curve of bidegree $(n/2,N)$ in $\mathbb P^1 \times \mathbb P^1$, and must have genus $0$, so we must have $n/2 \leq 1$ or $N \leq 1$. Again the case $N \leq 1$ is trivial, so we must have $n=2$.

So the only hyperelliptic cases are $n=2$ and $N=2$.