@def mintoclevel=2 @def maxtoclevel=3
The 19th lecture of algebraic curves by Karl Christ
Recall the following proposition from last time:
Proposition: (Base point free pencil trick). If \(V\subseteq H^0(X,L)\) is a pencil with base locus \(B\) then the kernel of \(V\otimes H^0(X,F)\to H^0(X,L\otimes F)\) given by \(s\otimes r \mapsto s\cdot r\) is \(H^0(X,L^{-1}(B))\) for any locally free sheaf \(F\).
We then had the following corollary, and we had some confusion regarding the numerics of the statement:
Cor: If \(L\) is a complete \(g^1_d\) with \(2d < g+2\) then \(\mu_0\) is not injective.
Proof. Let \(\mu_0:H^0(X,L)\otimes H^0(X,K_X-L) \to H^0(X,K_X)\) be the Petri map. Then
\[\begin{aligned}
\operatorname{ker}(\mu_0) = H^0(X,K_X\otimes L^{-1}\otimes L^{-1}(B))
\end{aligned}\]
by the pencil trick. By Riemann-Roch,
\[\begin{aligned}
h^0(X,K_X\otimes L^{-1}\otimes L^{-1}(B)) = 2g - 2 - 2d + b - g + 1 + h^0(L\otimes L(-B)).
\end{aligned}\]
Assume first that \(B = \emptyset\), i.e. that \(L\) is basepoint free. Then \(b = 0\) and we get that the numerics above satisfy
\[\begin{aligned}
... > g - 1 - g - 2 + 3 =0.
\end{aligned}\]
Hence \(K_X\otimes L^{-1}\otimes L^{-1}(B)\) has global sections.
Now suppose that \(L\) is a \(g^1_2\) and \(F\) is a basepoint free \(g^1_3\),
\[\begin{aligned}
\varphi:H^0(X,L)\otimes H^0(X,F) \to H^0(X,L\otimes F)
\end{aligned}\]
and \(\operatorname{ker}\varphi = H^0(X,L\otimes F^{-1}) = 0\), which implies \(L\otimes F\) is a \(g^3_5\). Since \(g\geq 3\), \(5\leq 3g\) which implies that \(r(L\otimes F)\leq d/2\) by Clifford's inequaltiy.
This is a contradiction.
\(\square\)
Remark: (Alternative Proof.) \(L\) and \(F\) would define a bnirational map \(X\to \mathbb P^1 \times \mathbb P^1\) with image of bidegree \((2,3)\) and thus genus 2. Contradiction.
I think the above is missing details – cross reference with Karl's notes.
Motivating Question: Given \(C\subset \mathbb P^r\) of degree \(d\), what is the largest possible value for \(g(C)\)?
Remark: For \(r \geq 3\), \(g(C)\) will not be determined by \(d\).
Example of remark:
Take \(Q_1\cap Q_2\) has genus 1 and degree 4.
\(\mathbb P^1 \to \mathbb P^3\) given by \([x,y] \mapsto [x^4 : x^3y : x^2y^2 : y^4]\), where One of the monomials is omitted. This has genus \(0\) and degree \(4\).
This remark illustrates that the best we can hope for is a bound on the genus. One bound that we've already seen, at least in \(\mathbb P^2\), comes from the genus-degree formula:
\[\begin{aligned}
g = \frac{(g-1)(g-2)}{2}
\end{aligned}\]
which we obtained from the adjunction formula. We also saw that any curve in \(\mathbb P^r\) projects to a singular curve of the same degree in \(\math P^2\). From this it follows that
\[\begin{aligned}
g(C) < \frac{(d-1)(d-2)}{2}
\end{aligned}\]
for \(C\) non-degenerate whenever \(r\geq 3\). We can get a better bound, however.
Here's the plan: estimate the dimension of \(\alpha_\ell = r(\ell D)\) where \(D = C\cap H\) with \(H\) a general hyperplane in \(\mathbb P^r\). We know that \(D\) has degree \(d\) – that's our assumption – and so for high enough value of \(\ell\), \(r(\ell \cdot D)\) will exceed \(d/2\). Clifford's bound says that the rank of a line bundle within the special range is bounded by \(d/2\), so the line bundle corresponding to \(\ell D\) must fall outside the special range. Riemann-Roch then allows us to calculate the rank precisely; it's \(\ell\cdot d - g\). Rearranging will then give us a bound on the genus.
Let \(E_\ell \subset |\ell D|\) given by the restriction of degree \(\ell\) hyperplanes in \(\mathbb P^r\) to \(C\):
\[\begin{aligned}
H^0(\mathbb P^r, \mathcal O_{\mathbb P^r}(\ell)) \to H^0(C,\mathcal O_C(\ell)).
\end{aligned}\]
Set \(\beta_\ell = \operatorname{dim} E_\ell\). Then \(D+E_{\ell - 1}\subseteq E_\ell\) hence \(\beta_{\ell - 1}\leq \operatorname{dim} E_\ell(-D)\). So
\[\begin{aligned}
\beta_\ell - \beta_{\ell - 1} \geq \operatorname{dim} E_\ell - \operatorname{dim} E_\ell (-D).
\end{aligned}\]
We have a short exact sequence
\[\begin{aligned}
0 \to I_C(\ell)\to \mathcal O_{\mathbb P^r}(\ell) \to \mathcal O_C(\ell)\to 0,
\end{aligned}\]
where \(I_C(\ell)\) is the sheaf of ideals defining \(C\) twisted by \(\ell\). Passing to the long exact sequence of cohomology gives us a portion saying that
\[\begin{aligned}
H^0(\mathcal O_{\mathbb P^r}(\ell)) \to H^0(\mathcal O_C(\ell)) \to H^1(I_C(\ell)).
\end{aligned}\]
The rightmost map above need not by surjective, so we get
\[\begin{aligned}
\operatorname{dim} h^0(\mathcal O_C(\ell)) \geq h^0(\mathcal O_{\mathbb P^r}(\ell)) - h^0(I_C(\ell)).
\end{aligned}\]
However \(E_\ell\) is precisely the image of \(H^0(\mathcal O_{\mathbb P^r}(\ell))\) in \(H^0(\mathcal O_C(\ell))\), so
\[\begin{aligned}
\beta_\ell = \operatorname{dim} E_\ell = h^0(\mathcal O_{\mathbb P^r}(\ell)) - h^0(I_C(\ell)).
\end{aligned}\]
Now, thinking of \(D\) as a subset of points in \(C\), we have a short exact sequence
\[\begin{aligned}
0 \to I_D(\ell) \to \mathcal O_{\mathbb P^r}(\ell) \xrightarrow{\psi} \mathcal O_D(\ell) \to 0
\end{aligned}\]
where the map \(\psi\) fits into a commutative triangle with
\[\begin{aligned}
\mathcal O_{\mathbb P^r}(\ell) \xrightarrow{\varphi_1} \mathcal O_C(\ell)\xrightarrow{\varphi_2} \mathcal O_D(\ell).
\end{aligned}\]
Then
\[\begin{aligned}
\operatorname{dim}(E_{\ell}(-D)) &= \operatorname{dim} (\operatorname{ker}(\varphi_2|_{E_\ell})) \\
&= \operatorname{dim}(\operatorname{ker} \psi) - \operatorname{dim}(\operatorname{ker} \varphi_1) \\
&= h^0(I_D(\ell)) - h^0(I_C(\ell)).
\end{aligned}\]
This implies
\[\begin{aligned}
\beta_\ell - \beta_{\ell - 1} &= h^0(\mathcal O_{\mathbb P^r}(\ell)) - h^0(I_C(\ell)) - h^0(I_D(\ell)) + h^0(I_C(\ell)) \\
&= h^0(\mathcal O_{\mathbb P^r}(\ell)) -h^0(I_D(\ell)).
\end{aligned}\]
Geometrically, this is the number of conditions imposed by the points of \(D\) on the hypersurfaces in \(\mathbb P^r\), so the intersection points of \(C\) with a general degree \(\ell\) hyperplane of \(\mathbb P^r\). Note that when the points of \(D\) aren't sufficiently general, fewer conditions are imposed than expected. We're going to use the above difference as a bound, so we're looking for the worst configuration of points which impose the least number of conditions.
Remark: \(h^0(\mathcal O_{\mathbb P^r}(\ell)) - h^0(I_D(\ell))\) is the number \(\rho_\ell\) of conditions imposed by the points in \(D\) on hypersurfaces in \(\mathbb P^r\) of degree \(\ell\).
Lemma: Suppose \(p_1,...,p_d \in \mathbb P^{r-1}\) are points chosen such that any \(r\) among them are linearly independent. Then the \(p_i\) impose at least \(\min\{d, \ell(r- 1)+1\}\) conditions on hypersurfaces of deegree \(\ell\).
Before we prove this lemma, let's expalin a little more.
If \(\ell = 2\), then the points impose at least \(2r - 1\) conditions on quadrics.
If \(\ell = 3\), then the points impose at least \(5\) conditions on quadrics in \(\mathbb P^2\) for any \(d\geq 5\) points that are linearly general ( no three points are colinear, no four points are coplanar, etc).
Now let's prove the lemma.
Proof. Assume the interesting case of the lemma, that the minimum is achieved by \(\ell(r - 1) + 1\) so \(\ell(r-1)+1 \leq d\). Order the points:
\[\begin{aligned}
p_1, ..., p_{r - 1}, p_{r}, ..., p_{2(r - 1)}, ... , p_{\ell\cdot (r-1)}, ..., p_d.
\end{aligned}\]
Let \(H_i\) be the collection of points \(p_{(i-1)r},...,p_{i\cdot (r-1)}\) (this should be (r-1) points) and \(L_i\) be the linear form vanishing on \(\operatorname{span}(H_i)\). Thus \(L_i\) does not vanish on \(p_d\) since the points \(p_1,...,p_d\) are lie in linearly general position.
Now consider \(F = \prod^ell_{i = 1}L_i\) which has degree \(\ell\) and vanishes on \(p_1,...,p_{\ell(r-1)}\) but not on \(p_d\).
We're left with the case that \(d < \ell(r - 1) + 1\)...
\(\square\)