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.

1$\begingroup$ 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 $Pic^d(C)$. $\endgroup$– MattSep 20, 2013 at 21:29

1$\begingroup$ The OP probably wanted to consider the moduli space where the determinant line bundle is fixed. $\endgroup$– Jim BryanSep 21, 2013 at 2:32
2 Answers
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.
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.

1$\begingroup$ 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. $\endgroup$ Sep 21, 2013 at 3:53

1$\begingroup$ @Jim Bryan: It is easy to see that the moduli space (with fixed determinant bundle) is unirational. $\endgroup$– nafSep 21, 2013 at 7:12

1

3$\begingroup$ @Chen: Unirationality and Picard number 1 does. $\endgroup$– nafSep 21, 2013 at 9:41

3$\begingroup$ @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.) $\endgroup$ Sep 21, 2013 at 23:54