Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $K$ be a field of characteristic zero but not algebraically closed. Let $C$ be a smooth projective curve over $K$. Let $r, d$ be two positive integers that are coprime. Consider the moduli space of stable vector bundles of degree $d$ and rank $r$ over $C$ with fixed determinantal line bundle. Is it fano? If so could someone suggest a reference for this fact.

share|improve this question
1  
This isn't even the case over $\mathbb{C}$. For genus 1 curves Atiyah showed that this is isomorphic to the curve itself (an elliptic curve). For other genus curves you could take $r=1$ so that it is $Pic^d(C)$. –  Matt Sep 20 '13 at 21:29
1  
The OP probably wanted to consider the moduli space where the determinant line bundle is fixed. –  Jim Bryan Sep 21 '13 at 2:32

2 Answers 2

up vote 2 down vote accepted

Actually Drezet and Narasimhan prove that the canonical class is (-2n) times the positive generator of the Picard group, where n = g.c.d (r,d). Hence the moduli space (with fixed determinant) is Fano.

share|improve this answer

Yes, if $g > 1$, it is Fano because its Picard group (over algebraic closure, and hence over K) is isomorphic to $\mathbb{Z}$. You can find the needed references , for example, in Drezet and Narasimhan, Invent. Math. 97 (1989), no. 1, 53–94.

share|improve this answer
1  
I'm confused. Why do you say that Picard number 1 implies it is Fano? A K3 surface can have $\mathbb{Z}$ for a Picard group. –  Jim Bryan Sep 21 '13 at 3:53
1  
@Jim Bryan: It is easy to see that the moduli space (with fixed determinant bundle) is unirational. –  ulrich Sep 21 '13 at 7:12
1  
@ulrich: Does unirational imply Fano? –  Chen Sep 21 '13 at 9:15
2  
@Chen: Unirationality and Picard number 1 does. –  ulrich Sep 21 '13 at 9:41
2  
@Chen:(This assumes that we're in characteristic zero. Otherwise the statement is not true.) If $X$ is unirational, then $K_X$ cannot be trivial or even torsion. If the Picard number is one, then any such divisor is either ample or negative ample. Again if $X$ is unirational, then $K_X$ cannot be ample. So $-K_X$ is ample, and hence $X$ is Fano. (If the Picard group is $\mathbb Z$, then you don't even need to worry about torsion.) –  Sándor Kovács Sep 21 '13 at 23:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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