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The gonality of a compact Riemann surface $\Sigma$ is defined to be the lowest degree $d$ of a non-constant holomorphic map $f\colon \Sigma\to\mathbb CP^1.$ This means the gonality is 1 only for $\mathbb CP^1$, and $2$ only for hyperelliptic (and elliptic) curves.

I would like to know the gonality of the compact Riemann surface $\Sigma_{k,l}$ determined by the algebraic equation $$y^k=\tfrac{z^l-1}{z^l+1},$$ where $k,l\geq 3$ are integers. It would already be good to know the gonality for the case $k=l.$ Note that $\Sigma_{k,l}$ is likewise given by the algebraic equation $$(u^k+i)(v^l+i)=-2,$$ and the gonality is clearly less or equal than $\text{min}(k,l).$ I suspect the gonality is always $\text{min}(k,l),$ but I can only prove this in particular situations so far: $(k,3)$ can't be hyper-elliptic, and for $(k,l)$ for fixed $k$ and all large enough $l$.

So far, I haven't seen many examples in the literature, besides the case of smooth plane algebraic curves, and any idea or pointer to relevant literature would be appreciated as well.

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  • $\begingroup$ Your algebraic equation appears to define your surface đťś®_đť’Ś,đť’Ť as a submanifold of â„‚^2. Is that right? $\endgroup$
    – Daniel Asimov
    Commented Nov 26, 2023 at 18:43
  • $\begingroup$ Of course, the equations define a complex 1-dimensional submanifold in $\mathbb C^2$, but then you add some points ($k$ many over $v=\infty$ and $l$ many over $u=\infty$) to obtain the compact Riemann surface $\Sigma_{k,l}$. $\endgroup$
    – Sebastian
    Commented Nov 27, 2023 at 1:24
  • $\begingroup$ Is that because you want your surfaces to be projective, i.e., complex submanifolds of â„‚â„™^2. or something else? $\endgroup$
    – Daniel Asimov
    Commented Nov 27, 2023 at 5:44
  • $\begingroup$ I want the Riemann surface to be compact (otherwise, the above definition of gonality does not make sense). I therefore add these points, in the analogous way you do add two points for hyperelliptic curve determined by the algebraic equation $y^2=(z-z_1)\dots (z-z_{2g+2})$. Note that these curves are (in general) not smooth in $\mathbb CP^2$, as their genus $g=(k-1)(l-1)$ is quite arbitrary. $\endgroup$
    – Sebastian
    Commented Nov 27, 2023 at 6:47

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I found the answer to my own question for the case $k=l$. There is a theorem due to Lazarsfeld which states that the gonality of a smooth complete intersection $C$ of hypersurfaces $A_1,\dots,A_{n-1}$ of degree $2\leq a_1\leq a_2\dots\leq a_{n-1}$ in $\mathbb CP^n$ is at least $$gon(C)\geq (a_1-1)a_2\dots a_{n-1}.$$ The Riemann surface $\Sigma_{k,k}$ is given as the smooth intersection of the quadric $x_1x_2=x_3x_4$ and the Fermat-type surface $x_1^k+x_2^k+ix_3^k+ix_4^k=0,$ so the gonality is at least $k$. On the other hand, $\Sigma_{k,k}$ clearly admits non-constant holomorphic maps to $\mathbb CP^1$ of degree $k.$

Lazarsfeld, Robert, Lectures on linear series. With the assistance of Guillermo Fernández del Busto, Kollár, János (ed.), Complex algebraic geometry. Lectures of a summer program, Park City, UT, 1993. Providence, RI: American Mathematical Society. IAS/Park City Math. Ser. 3, 163-219 (1997). ZBL0906.14002.

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