Algebraic Curves Lecture 23

We return to the case that \(d > 2r\). Assume that \(C\subseteq \mathbb P^r\) is extremal, i.e.

1. non-degenerate
2. \(g(C) = \pi(d,r)\), where \(\pi(d,r)\) is the bound from Castelnuova’s bound.

Set \(H\) to be a general hyperplane and let \(D = C\cap H\) as always. Then

\begin{align*} \alpha_1 = h^0(C, \mathcal O_C(D)) = r + 1 \end{align*}

and on the level of quadrics

\begin{align*} \alpha_2 = h^0(C, \mathcal O_C(2D)) = (r+1) + (2r - 1) = 3r. \end{align*}

So for an extremal curve, I exactly know the dimension of the restriction of quadrics to \(C\): we’ve got the map

\begin{align*} \varphi:H^0(\mathbb P^r, \mathcal O_{\mathbb P^r}(2)) \twoheadrightarrow H^0(C, \mathcal O_C(2)) \end{align*}

where the domain has dimension \(\binom{r+2}{2}\) and the codomain has dimension \(3r\), so

\begin{align*} \dim(\ker\varphi) = \frac{(r+2)(r+1)}{2}-3r = \frac{r^2 - 3r + 2}{2} = \binom{r-1}{2}. \end{align*}

This is the number of quadrics containing a rational normal scroll.

New questions: What is \(\rho = \bigcap Q_i\), where the \(Q_i\) are the \(\binom{r-1}{2}\) linearly independent quadrics containing \(C\)?

Pick a general hyperplane \(H\) and consider \(D = C\cap H\). Then \(|D| = d \geq 2r + 1\). The points in \(D\) are in linearly general position and impose \(2r - 1\) conditions on quadrics. By Castelnuova’s \(2r + 3\) lemma, these is a rational normal curve \(C_D \subseteq H\) containing \(D\).

Claim: \(C_D\subseteq S\). Suppose there is \(Q_i\) such that \(C_D \not\subseteq Q_i\). Then

\begin{align*} C_D . Q_i = 2\cdot (r - 1) \geq d = \deg (D) \end{align*}

because \(D\subseteq C_D\cap Q_i\). The count above is simply Bezout’s theorem. This is a contradiction however, since \(d > 2r\), so we can’t have \(2\cdot (r - 1) \geq d\). Thus \(C_D\not\subseteq Q_i\) implies that \(C_d\cap Q_i\) has dimension \(0\).

Conversely: \(C_D\) is cut out by \(\binom{r - 1}{2}\) quadrics in \(H\cong \mathbb P^{r-1}\) which implies \(S\cap H = C_D\). A rational normal curve in \(\mathbb P^r\) is cut out by the \(2\times 2\) minors of a matrix of the form

\begin{align*} \begin{pmatrix} x_0 & x_1 & x_2 & \dots & x_{r-1} \\ x_1 & x_2 & x_3 & \dots & x_r \end{pmatrix}, \end{align*}

of which there are \(\binom{r}{2}\). Because \(C_D\) is a rational normal curve in \(H\cong \mathbb P^{r-1}\), it is therefore cut out by \(\binom{r-1}{2}\) quadrics. We know already that \(C_D\) vanishes on the \(Q_i\), of which there are \(\binom{r-1}{2}\), so there are no degrees of freedom left and hence \(S\cap H \supset C_D\). This gives us both inclusions and hence \(S\cap H = C_D\).

Summary: We get that \(\bigcap Q_i = S\) gives a rational normal curve of degree \(r-1\) when we intersect with a general hyperplane:

\begin{align*} C_D := S\cap H. \end{align*}

This implies \(S\) is a surface of degree \(r-1\), and hence that \(C\) lies on a rational normal scroll or on \(\mathbb P^2 \to \mathbb P^5\) (the Veronese embedding \([x:y:z] \mapsto [x^2:y^2:z^2:xy:xz:yz:]\)).

Suppose \(S = X_{a_1, a_2}\) with \(a_1\neq 0\) is a smooth, rational normal scroll. We saw (or claimed without proof? Can’t remember) last time that \(\operatorname{Pic}(S) = \langle L, H\rangle\) with \(L.L = 0\), \(L.H = 1\) and \(H.H = r-1\).

Recall if you have a curve \(C\) with bidegree \((d_1,d_2)\) then \(2g - 2 = K_S. C + C.C\) and \(C = \alpha L + \beta H\). Then

\begin{align*} K_S = K_1\cdot L + K_2\cdot H. \end{align*}

From \(L\) you have

\begin{align*} -2 = K_S\cdot L + L^2 = K_2 \end{align*}

and for \(H\)

\begin{align*} -2 = K_S H + H^2 = K_1 + K_2(r-1) + (r-1) = K_1 - (r-1). \end{align*}

This implies \(K_1 = r-3\). Hence \(K_S = (r-3)L - 2H\), \(2g(C) - 2 = K_S.C + C.C\) and so

\begin{align*} g(C) = \frac{(\alpha - 1)(2m - \alpha)}{2}(r-1) + \epsilon\cdot (\alpha - 1), \end{align*}

hence \(d-1 = m(r-1) + \epsilon\). So \(g\) is maximal for \(\alpha = m+1\), \(\beta = \epsilon + 2 - r\) for any \(\epsilon\) or \(\alpha = m\) and \(\beta = 1\) and \(\epsilon = 0\). For these values, equality in Castelneuovo’s bound is achieved.