12
$\begingroup$

I have already asked similar questions before, but now I realized that there a nice general way to ask what I want. Namely let $X$ be a normal affine variety over a field $k$. Assume first that $k$ is finite. Then is it true that

1) IF $X$ is smooth, then the set of isomorphism classes of vector bundles of given rank on $X$ is finite?

2) More generally, is it true that for any $X$ the set of isomorphism classes of Cohen-Macaulay torsion free sheaves of fixed generic rank is finite?

When the field $k$ is arbitrary then I would expect that there exists a finite-dimensional (over $k$) family of vector bundles or Cohen-Macaulay sheaves which contains every isomorphism class. Is this true?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
10
$\begingroup$

Suppose that $X = \mathbb{P}^1 \times \mathbb{P}^1 \setminus \Delta$, where $\Delta$ is the diagonal. The Picard group of $\mathbb{P}^1 \times \mathbb{P}^1$ is $\mathbb{Z}^2$. Yanking out $\Delta$ should just kill the generator $(1,1)$ in $\mathbb{Z}^2$, so $\mathrm{Pic}(X)$ should be $\mathbb{Z}$. So there is an infinite discrete parameter describing line bundles on $X$. Unless I am missing something...

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks, I have somehow missed that kind of example. I wonder whether requiring that all the Chern classes of the vector bundle in question are equal to 0 will change anything... $\endgroup$
    – Alexander Braverman
    Commented Apr 22, 2011 at 14:36
5
$\begingroup$

As shown by David, the answer is no for line bundles.

More generally, the answer in no even for higher dimensional vector bundles. In fact, in his paper "Vector bundle over affine surfaces birationally equivalent to ruled surfaces" Murthy proves the following results:

(1) Let $V$ be an irreducible affine non-singular surface defined over an algebraically closed field $k$ and such that $V$ is birational to $C \times \mathbb{P}^1$, vhere $C$ is a curve. Then any vector bundle over $V$ is a direct sum of a trivial bundle and a line bundle. In general, there are infinitely many equivalence classes of line bundles (as shown by David for $V= \mathbb{P}^1 \times \mathbb{P}^1 \setminus \Delta$).

(2) There exist an affine, nonsingular rational variety of dimension $3$ over which there are infinitely many non-isomorphic indecomposable vector bundles of rank $>1$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ If you already have infinite number of line bundles you can twist any given bundle of higher rank to produce infinite number of vector bundles of that rank, can't you? $\endgroup$
    – Sasha
    Commented Apr 20, 2011 at 21:47
  • $\begingroup$ Of course you can. The point of $(2)$ is that in dimension $3$ you find indecomposable vector bundles of any rank, in contrast with the case of affine ruled surfaces, where every vector bundle splits as a sum of line bundles. $\endgroup$
    – Francesco Polizzi
    Commented Apr 21, 2011 at 8:11
  • 1
    $\begingroup$ Keeping in mind, of course, that you cannot twist beyond the top rank = dimension of your variety, since any vector bundle whose rank exceeds the dimension necessarily splits a trivial line bundle. $\endgroup$
    – Andrew Parker
    Commented Apr 21, 2011 at 15:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .