Let $M$ be a fine moduli space of vector bundles on curve which is an algebraic variety as well. The first example of such an object that I have in mind is rank 2, deg 1 VB on a genus 2 curve. This is an intersection of 2 4-dimensional quadrics, and it is Fano. If I recall correctly, all moduli spaces of bundles with odd degree on an algebraic curve are fine.

My question is: are all fine moduli varieties of VB on an algebraic curve Fano? If not, please give counterexamples.

  • $\begingroup$ The moduli space of vector bundles on a curve is fine if the degree is coprime with the rank. So, if you are interested only in rank 2 case, then indeed any odd degree gives a fine moduli space. But if you are interested in other ranks, then this is not true. $\endgroup$ – Sasha Feb 6 '12 at 16:17
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    $\begingroup$ Yes, whenever the moduli space of semistable bundles of rank 2 and fixed, degree $1$ determinant is a fine moduli space, then it is a smooth, proper, geometrically connected variety with ample anticanonical bundle. $\endgroup$ – Jason Starr Feb 6 '12 at 16:57

I assume you are asking about $SU(r,L),$ (semistable rank-$r$ bundles with determinant $L$) rather than $U(r,d)$ (semistable rank-$r$ bundles with determinant of degree d).

Drezet-Narasimhan showed that even when $SU(r,L)$ is not a fine moduli space, it is locally factorial with Gorenstein singularities, and that its dualizing sheaf is isomorphic to $\mathscr{L}^{-2(r,c_{1}(L))}$ where $\mathscr{L}$ is the (ample) determinant bundle; consequently $SU(r,L)$ is Fano.

$U(r,d)$ is then a Fano fibration over the degree-$d$ Picard variety of the underlying curve via the determinant map (the fiber over a degree-$d$ line bundle $L$ is just $SU(r,L)$).

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  • $\begingroup$ Thank you everybody!! Is it smooth also in higher rank? $\endgroup$ – IMeasy Feb 6 '12 at 18:43
  • $\begingroup$ As Sasha pointed out in the comment above, the coprimality of rank and degree is sufficient for smoothness, so the answer is yes. For instance, $SU(r,L)$ is smooth when $r≥2$ and $c_{1}(L)=r-1.$ $\endgroup$ – Yusuf Mustopa Feb 6 '12 at 20:23

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