4
$\begingroup$

Is the Hilbert scheme of points on a Fano variety a Fano variety itself?

$\endgroup$

2 Answers 2

10
$\begingroup$

The Hilbert scheme $S^{[n]}$ of $n$ points on a smooth surface $S$ is smooth for any $n$, but it is never Fano if $n>1$ since the canonical divisor is trivial on the fibers of the morphism to $S^{(n)}$, the $n$-th symmetric product. Indeed, the morphism from $S^{[n]}$ to $S^{(n)}$ is a crepant resolution of singularities. So neither the canonical nor the anticanonical line bundle can be ample on $S^{[n]}$.

$\endgroup$
8
$\begingroup$

The Hilbert scheme of points on a Fano variety of dimension at least 3 is not even a variety if the number of points is sufficiently large. Also it has bad singularities where components of different dimensions meet, which makes it hard to define the anticanonical sheaf so it is not even clear what "Fano" might mean for such a scheme.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.