Let $C$ be a smooth irreducible (complex) curve of genus $g\geq2$.

The gonality of $C$ is defined as the minimum degree of surjective morphisms $C\rightarrow\Bbb{P}^1$. So $C$ has gonality $d$ if it has a $g^1_d$ but no $g^1_{d-1}$, where $g^r_d$ denotes a linear system of dimension $r$ and degree $d$ on $C$.

Now my question is very basic, I don't understand the following remark:

If $C$ has gonality $d$ then each $g^1_d$ on $C$ is base point free and complete.

First, why are they complete? But even if they are complete, so we have the system $|A|$ with $A$ divisor on $C$, with $\deg A=d\geq2$, and $h^0(A)=2$. Then why is it base point free?

Let $P\in C$. We need to show $h^0(A-P)=h^0(A)-1=1$. Riemann-Roch gives $$h^0(A-P)\geq d-g.$$

Now I'm stuck here. From Hurwitz's formula we get $d=\frac{\deg R}{2}+1-g$, where $R$ is the ramification divisor of the morphism, but I don't see how this may help.



A rational map $C\dashrightarrow V\subset\mathbb P^n$ from a curve $C$ to a projective variety $V$ is defined at all the smooth points. In particular, taking $V=\mathbb P^n$, any linear series $f:C\dashrightarrow\mathbb P^n$ on a smooth projective curve is in fact base-point-free. Recall that the locus where $f$ is not defined coincides with the base locus of the corresponding linear series; but the latter is defined as the intersection of all the divisors in the system, namely $$\textrm{Bl}(f)=\bigcap_{H\subset\mathbb P^{n}}f^{-1}(H),\,\,\,\,\,\,\,\,\,\,\,\,\,\,H\,\,\textrm{all the hyperplanes in}\,\,\mathbb P^n,$$ and when $n=1$, a point $P\in C$ cannot satisfy $f(P)=x$ for all $x\in \mathbb P^1$, so in the case you are interested in the base locus is empty for this very concrete reason.


Let $\textrm{Pic}^{e}(C)$ denote the group of degree $e$ line bundles on $C$.

As you said, the gonality being $d$ means that any $L\in \textrm{Pic}^{d-1}(C)$ is such that any vector subspace $U\subseteq H^0(C,L)$ has $\dim U\leq 1$. But we have a $g^1_d$, namely a couple $(\mathscr L,V)$ where $\mathscr L\in\textrm{Pic}^{d}(C)$ and $V\subseteq H^0(C,\mathscr L)$ has dimension 2. We want to show that in fact $h^0(C,\mathscr L)=2$.

Let us take any point $P\in C$. Then, $L=\mathscr L(-P)\in \textrm{Pic}^{d-1}(C)$, so by what we said we must have $h^0(C,L)\leq 1$. But the inclusion $H^0(C,L)\subset H^0(C,\mathscr L)$ produces, in general, either no dimension jump, or a jump of one dimension. This means that $h^0(C,\mathscr L)\leq 2$, hence it is in fact $=2$.

  • 3
    $\begingroup$ I'm not sure that I like the explanation for basepoint-freeness: in particular, it isn't true that any linear series on a smooth curve is basepoint-free. What is true is that the base locus is removable, and that's why the associated rational map is actually a morphism. An alternative argument for why a $g^1_d$ must be basepoint-free is the following: if not, remove the base locus, to get a $g^1_k$ for some $k<d$. This contradicts $d$-gonality of $C$. $\endgroup$ – user64687 Oct 15 '13 at 15:05
  • $\begingroup$ On the other hand, the argument for completeness is very nice, +1. $\endgroup$ – user64687 Oct 15 '13 at 15:06
  • $\begingroup$ @AsalBeagDubh: thanks for your comment. But could you explain where I am wrong? I know that a rational map from a smooth curve to a projective variety is a morphism. Next step: there is a correspondence between rational maps to projective space and linear series; conclusion: every linear series is bpf. What is wrong? $\endgroup$ – Brenin Oct 15 '13 at 15:23
  • $\begingroup$ @Brenin There is a correspondence between rational maps to projective space and linear series without fixed locus. $\endgroup$ – Heitor Fontana Oct 15 '13 at 15:34
  • $\begingroup$ The point is that there is a correspondence between rational maps and linear series, but it's not quite one-to-one, exactly because of the possibility of components in the base locus. As a simple example: take the linear series $|p|$ where $p$ is a point on an elliptic curve. It defines a rational map to $\mathbf{P}^0$; a priori that map is not defined at $p$ itself, but of course it extends to $p$. But the extension is exactly the map corresponding to the trivial line bundle (or linear series) --- that is, the map we get by removing the base locus of $|p|$. I hope that clarifies things... $\endgroup$ – user64687 Oct 15 '13 at 15:35

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