# 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.

• You may want to try cross-posting this question on Math Overflow. – msteve Jan 6 '15 at 4:26
• Thanks for the tip. I'll see what results over here, and if nothing much happens, then I'll migrate it over there. :-) – MathsByTheSea Jan 6 '15 at 4:29
• By the way, let me add one little nitpick. You say "...is Fano, which means that the anticanonical line bundle of $M$ has positive degree". There is no such thing as the degree of a line bundle on a variety of dimension at least 2. – user64687 Jan 6 '15 at 10:43
• I guess the word I am looking for is "ample". Can I replace "degree positive" with saying that the anticanonical line bundle has at least two linearly-independent global holomorphic sections? (I say at least two, to force it to be non-trivial.) – MathsByTheSea Jan 6 '15 at 15:25
• Yes, "ample" is the key word. Annoyingly, no number of linearly independent global sections are enough to guarantee that a bundle is ample! – user64687 Jan 6 '15 at 15:53

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.

• This is great, thanks! It is the moduli space of stable bundles, so in the end, it is stability that will determine whether a particular determinant line bundle is possible or not. (If it contained all vector bundles of a particular rank and degree, stable or not, then the map would be obviously surjective, because we could always build an appropriate bundle by taking a direct sum of the determinant line bundle and as many copies as we need of the trivial line bundle.) – MathsByTheSea Jan 6 '15 at 15:33
• If two holomorphic bundles have the same rank and degree but different determinants, then do they always differ by tensoring with a line bundle? In other words, do two line bundles of the same degree that are not isomorphic always differ by tensoring by a degree $0$ line bundle? If this is the case, this suggests the map from $\mathbb M$ to $\mbox{Pic}^d(X)$ is surjective, since tensoring by a line bundle does not affect slope stability (I think). – MathsByTheSea Jan 6 '15 at 15:36
• @MathsByTheSea: you're welcome. To answer your question here, unfortunately no, you can't always go between bundles of equal rank and degree by tensoring by a line bundle. For example, on $\mathbf P^1$ consider $O \oplus O$ and $O(-1) \oplus O(1)$. – user64687 Jan 6 '15 at 15:56
• That's true -- I might be asking something slightly different, though. The bundles $\mathcal O^{\oplus 2}$ and $\mathcal O(-1)\oplus\mathcal O(1)$ already have the same determinant line bundle, which is just the trivial line bundle. (On $\mathbb{CP}^1$, two vector bundles of the same degree and rank must have the same determinant, because $\mbox{Pic}(\mathbb{CP}^1)=\mathbb Z$.) – MathsByTheSea Jan 6 '15 at 20:35
• @MathsByTheSea Sorry I didn't get back to you on this. And I see what you mean about my earlier comment: I was misreading yours. Indeed, I think you're completely right that we can move between any two points of Pic^d like this: if $L_1$ and $L_2$ both have degree $d$, then $L_1^{-1} \otimes L_2$ is a line bundle $L_0$ of degree 0. Since Pic^0 is an abelian variety, it is divisible, so there is some line bundle $\Lambda$ such that $\Lambda^r=L_0$. Then for a stable bundle $M$ we have $det(M \otimes \Lambda) = det(M) \otimes L_0$. So we can move from $det(M)$ to any other point in this way. – user64687 Jan 8 '15 at 12:22