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Let $X$ be a smooth projective curve of genus $g$. Let $\Omega$ be the sheaf of differentials. Mumford (in Abelian Varieties, sec. 2.6, in proving the theorem of the cube) asserts that there is an effective divisor of degree $g$ on $X$ such that $H^0(X, \Omega \otimes L(-D))=0$.

This should be easy, but I'm missing the argument. Could someone explain?

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1 Answer 1

up vote 6 down vote accepted

Let $\mathcal L$ be an invertible sheaf on the curve $X$, and suppose that $H^0(X,\mathcal L)$ has dimension $d$. (In the case when $\mathcal L = \Omega$, we have $d = g$.)

If we choose a closed point $x \in X$, then there is a map $H^0(X,\mathcal L) \to \mathcal L_x/\mathfrak m_x \mathcal L_x$ given by mapping a section to its fibre at $x$.

To make my life easier, let's assume that $k$ (the field over which $X$ is defined) is algebraically closed, so that $x$ is defined over $k$, and we may identify $\mathcal L_x/\mathfrak m_x\mathcal L_x$ with $k$ (since it one dimensional, being the fibre of an invertible sheaf).

Evaluation is then a functional $H^0(X,\mathcal L) \to k.$ Now this functional will be identically zero if and only if every section vanishes at $x$. But a non-zero section has only finitely many zeroes, and $X$ has an infinite number of closed points, so we may certainly choose $x$ so that this functional is not identically zero.

The evaluation map sits in an exact sequence $$0 \to H^0(X,L(-x)) \to H^0(X,L) \to k,$$ and so if we choose $x$ such that evaluation is surjective, we find that $H^0(X,L(-x))$ has dimension $d - 1$. Proceeding by induction, we find points $x_1,\ldots,x_d$ such that $H^0(X,L(-x_1-\cdots - x_d)) = 0$.

In summary: we have shown that we may find an effective degree $d$ divisor $D$ such that $H^0(X,L(-D))$ vanishes. (And in fact, looking at the proof, we see that this vanishing will hold for a generic choice of $D$.)

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Dear Matt E, thank you for the answer! –  Akhil Mathew Jan 8 '11 at 7:05

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