Castelnuovo's basepoint free pencil trick

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I am having trouble solving Exercise 4.13 in Eisenbud's "Geometry of Syzygies":

Suppose that $\mathscr{L}$ is a line bundle on a (smooth) curve $X\subset \mathbb{P}^r$ over an infinite field, and suppose that $\mathscr{L}$ is basepoint-free. Show that we may choose 2 sections $\sigma_1, \sigma_2$ of $\mathscr{L}$ which together form a base-point free pencil--that is, $V=\langle \sigma_1,\sigma_2 \rangle$ is a two-dimensional subspace of $H^0(\mathscr{L})$ which generates $\mathscr{L}$ locally everywhere. Show that the Koszul complex on $\sigma_1, \sigma_2$ $$ \mathbb{K}: 0\to \mathscr{L}^{-2}\to \mathscr{L}^{-1}\oplus \mathscr{L}^{-1}\to \mathscr{L}\to 0 $$ is exact, and remains exact when tensored with any sheaf.

We know that $\mathscr{L}$ is generated by a vector space $W\subseteq H^0(\mathscr{L})$ if there is a surjection $W\otimes \mathcal{O}_X\to \mathscr{L}$.

  • How do we show that $W=V$ is two-dimensional?
  • Where do the bundles $\mathscr{L}^{-1}\oplus \mathscr{L}^{-1}$ and $\mathscr{L}^{-2}$ come from?

My ultimate interest is the final assertion of the exercise:

Suppose that $X$ is embedded in $\mathbb{P}^r$ as a curve of degree $d\geq 2g+1$, where $g$ is the genus of $X$. Use the argument above to show that $$ H^0(\mathcal{O}_X(1))\otimes H^0(\mathcal{O}_X(n))\to H^0(\mathcal{O}_X(n+1)) $$ is surjective...

I am hoping that someone can work out this exercise for some so I may understand the "pencil trick". I'm happy to assume that the field is $\mathbb{C}$, as opposed to "infinite field" in the statement of the exercise. This trick is used frequently in the study of syzygies of curves and I cannot find a proof anywhere. I've tried to do this a few times in the past, and there must be something silly that I'm missing. Thanks.

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Let $0\neq s$ be a section of $L$ (calligraphy is too much to type). Then, we have an exact sequence $0\to \mathcal{O}_X\stackrel{s}{\to} L\to T\to 0$. Since $T$ is a coherent sheaf and zero at the generic point, we see that it is a skyscraper sheaf and since it is a quotient of a line bundle, easy to see that $T\cong\mathcal{O}_Z$ for some proper closed subset $Z\subset X$. Since $L$ is globally generated, we have images global sections of $L$ generate $T$ and then (using infinite field), there is a single section $t$ of $L$ such that its image generate $T$. Then, the subspace generated by $s,t$ will do what you want.

Your Koszul complex is not correct. It should be, $0\to L^{-2}\to L^{-1}\oplus L^{-1}\to \mathcal{O}_X\to 0$, and comes from the surjection $\mathcal{O}_X^2\to L$ given by $s,t$. The kernel is $L^{-1}$ by determinant considerations and twist the sequence by $L^{-1}$ to get what you want.

Finally, for your last part, most of what you want can be proved by the vanishing of $H^1$ for all large degree line bundles, inducting on $n$. The case of $n=1$ requires some care.