# When does a hyperelliptic Riemann surface admit a map of degree 3

Let $X$ be a hyperelliptic curve of genus $g>1$.

For which $g$ does $X$ admit a map $X\to \mathbb P^1$ of degree $3$?

I think a genus two curve $X$ admits a map of degree $3$.

Proof: Pick $P$ and $Q$ such that $P+Q$ is not linearly equivalent to $K_X$. Now compute $h^0(P+Q) = 1$ using Riemann-Roch and the fact that $h^0(K_X-P-Q) =0$. For any point $R$ we have that $h^0(P+Q+R) = 2$, so there exists a function of degree $3$ on $X$ (whose divisor of poles is $P+Q+R$). QED

This computation doesn't work in higher genus making me suspect that for $g>2$ a hyperelliptic curve does not admit a degree 3 map.

• Suppose a curve $C$ admitted maps $f: C\to \mathbb{P}^1$ and $g: C\to \mathbb{P}^1$ of degree $2$ and $3$. Consider the map $(f, g): C\to \mathbb{P}^1\times \mathbb{P}^1$. This exhibits $C$ as birational to a $(2, 3)$-curve in $\mathbb{P}^1\times\mathbb{P}^1$, which has arithmetic genus $2$. Thus the geometric genus of $C$ is at most $2$. – Daniel Litt Jan 25 '15 at 22:06
• @DanielLitt That's a very nice argument! Thank you. – Hybrids Jan 25 '15 at 22:16

Theorem. Assume that there are three compact Riemann surfaces $X_{g}$, $X_{g_1}$, $X_{g_2}$, of respective genera $g, \, g_1, \, g_2$, related by two holomorphic maps $$f_1 \colon X_g \to X_{g_1}, \quad f_2 \colon X_g \to X_{g_2},$$ of respective degree $d_1$ and $d_2$.
If the two maps admit no common, proper factorization, then $$g \leq d_1g_1+d_2g_2 + (d_1-1)(d_2-1).$$
As a consequence, a hyperelliptic curve admitting a morphism of degree $d$ over $\mathbb{P}^1$, non composed with the hyperelliptic involution, has genus at most $d-1$.
As another consequence, a hyperelliptic curve which is also bielliptic has genus at most $3$, in other words the hyperelliptic and the bielliptic locus are disjoint in the moduli space $\mathcal{M}_g$ as soon as $g \geq 4$.