3
$\begingroup$

There has already been a question on this exact Proposition in Hartshorne (Proof of Halphen's Theorem), but there was not a satisfactory answer, and I'll try to make more precise the source of my confusion.

The harder half of the Proposition is to show that, if $X$ is a genus $g\geq 2$ curve of degree $d$, then there exists a nonspecial ample divisor $D$ of degree $d$ if $d\geq g+3$. Here, nonspecial means $h^{0}(\mathscr{O}(K-D))=0$, where $K$ is the canonical divisor, and but remembering this definition and the exact bounds on $d$ and $g$ aren't necessary to understand my question.

Hartshorne attempts to prove this using a dimension counting argument, where he lets the points of $X^d$ parametrize the space of effective divisors of degree $d$ and he shows that the set of divisors $D$ that we don't want has dimension strictly less than $d$. To be clear, a point $(P_1,\ldots,P_d)$ in $X^d$ represents the divisor $P_1+\cdots+P_d$, so the correspondence of the points of $X^d$ and divisors of degree $d$ is only up to permutation.

There are two important steps that are confusing to me:

1) Hartshorne claims that, since ${\rm dim}|K|:=h^{0}(\mathscr{O}(K))-1=g-1$, the space of degree $d-2$ divisors in $X^{d-2}$ that are linearly equivalent to a subsystem of $K$ is at most $g-1$ in dimension.

2) Since the set $S\subset X^{d-2}$ of effective divisors that are linearly equivalent to a subsystem $K$ has dimension at most $g-1$, the set $T$ in $X^d$ consisting of divisors of the form $E+P+Q$, where $P$ and $Q$ are points and $E$ is linearly equivalent to $K$ has dimension at most $g+1$.

The claims about dimension are plausible, but it's not so clear to me how, for example, the effective divisors linearly equivalent to a subsystem $K$ sit inside $X^d$ (it could be some nasty set). Why, for example, does the dimension ${\rm dim}|K|$ correspond to the dimension of these effective divisors in $X^d$?

$\endgroup$
4
$\begingroup$

It's not linearly equivalent to $K$; it's linearly equivalent to a subsystem of $K$, which can be quantified as saying that $K-D$ has a global section, in other words $H^0(K-D) > 0$. $K$ gives a divisor on $X^d$, and the property that $H^0(K-D) > 0$ is a closed property by semicontinuity. That the divisors of degree $d$ with this property have dimension $\le g-1$ is simply because the set of all full canonical divisors have dimension less than this.

$\endgroup$
1
  • $\begingroup$ Sorry, you're right in that we're looking at divisors that are linearly equivalent to a subsystem of $K$, not $K$. I will fix that. I'm still confused about why $h^0(K-D)>0$ in $X^d$ is a closed property or why, for example, the canonical divisors in $X^g$ has dimension at most $g-1$, though. Since you mentioned semicontinuity, I can only wildly guess that you might be considering a flat sheaf $\mathscr{F}$ over $X^d$ with the fiber over each divisor $D$ being $H^0(K-D)$. If such an $\mathscr{F}$ exists, I can believe both statements. $\endgroup$ – DCT Mar 8 '15 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.