Fano-ness of moduli space of stable vector bundles when determinant line bundle is *not* fixed... According to Drezet-Narasimhan, Invent. Math. 97 (1989), no. 1, 53--94, the moduli space $\mathbb M$ of slope-stable holomorphic vector bundles with fixed rank $r$ and fixed determinant line bundle $L$ (such that $\gcd(\deg L,r)=1$) on a smooth, compact, connected Riemann surface $X$ is Fano, which means that the anticanonical line bundle of $\mathbb M$ has positive degree.  By earlier works (I believe), $\mathbb M$ is smooth and compact itself.
What can be said about the structure of the moduli space when the degree of $L$ is fixed but the isomorphism class of $L$ is allowed to vary?
I suspect it is no longer Fano, for any $r$.
Evidence: the $r=1$ version of $\mathbb M$ with $\deg L$ fixed, but with the isomorphism class of $L$ allowed to vary, is the Jacobian of $X$ itself. If the genus of $X$ is $g$, then $\mbox{Jac}(X)$ is a $g$-dimensional complex torus, and hence has trivial anticanonical (and canonical) bundle.  In particular, it is "Calabi-Yau" rather than Fano.
If we take $\mathbb M$ to be the moduli space of slope-stable holomorphic vector bundles with fixed rank $r>1$ and fixed degree $d$ that are coprime (but not fixed determinant) on a smooth, compact, connected Riemann surface $X$, then is it: (a) Fano (suspect not), (b) Calabi-Yau, or (c) something else?
A reference would be helpful, but an intuitive explanation of how going from fixed determinant to non-fixed determinant changes the overall geometry of the moduli space would be great.
 A: I know nothing about this topic, but here is one thing one can say about the overall picture.
If $M$ is your big moduli space, then it will have a morphism $\operatorname{det} : M \rightarrow \operatorname{Pic}^d(X)$ given by taking a vector bundle to its determinant. 
The Drezet–Narasimhan result you quote shows that the fibres of this map are Fano varieties; on the other hand, the target is an abelian variety (hence Calabi– Yau in your sense.) 
Here's where my ignorance really shines: I don't know whether the morphism $\operatorname{det}$ is surjective. (I guess so, but I wouldn't bet the barn on it.) But never mind: in any case, the image is at least a subvariety $Z$ of an abelian variety, hence has "nonpositive curvature".
So the whole space $M$ is of mixed type: it is a fibre space whose base has nonpositive curvature, and whose fibres have positive curvature. So the answer to your final question is (c).
Of course, the more you know about the morphism $\operatorname{det}$ (is it surjective? equidimensional? etc.) the clearer your picture of $M$ will become.
