This is sort of a follow-up to: Gauge theory construction of moduli of vector bundles

If I have a complex compact algebraic curve with at worst nodal singularities, is there an analytic description of holomorphic structures on the trivial bundle in terms of (0,1) forms satisfying some constraint? presumably one wants the forms to pick up singular behavior of some kind at the node. Also, is the description well-behaved in families as we smooth the curve? for instance, in the sense that whatever infinite-dimensional vector space is "locally constant" as we smooth the curve.

Other mild singularity types would be interesting too!


1 Answer 1


There are subtleties even in the simplest case - $C$ a compact, irreducible complex curve with one node, $Pic_0(C)$ the Picard variety of line bundles of degree $0$ - so why not start there?

Pulling back line bundles via the normalisation map $\nu\colon \tilde{C}\to C$ defines a map $Pic_0(C)\to Pic_0(\tilde{C})$. The latter space is a complex torus of dimension $g(\tilde{C})$. To recover a line bundle $L$ from $\nu^*L$ we also need an isomorphism $\nu^\ast L_p\to \nu^\ast L_q$, where $p$ and $q$ are the two points of $\tilde{C}$ that lie over the node of $C$. In this way, one sees that $Pic_0(C)$ is a $\mathbb{C}^{\ast}$-bundle over $Pic_0(\tilde{C})$.

So far as I can see, it's straightforward to give a gauge-theoretic description of $Pic_0(C)$: it consists of pairs consisting of a flat, unitary connection (or equivalently a Cauchy-Riemann operator, or equivalently a holomorphic structure) in a complex line bundle of degree zero over $\tilde{C}$, with an isomorphism $I$ of the fibres over $x$ and $y$, modulo gauge transformations respecting $I$.

One can compactify $Pic_0(C)$ to a $\mathbb{C}P^1$ bundle over $Pic_0(\tilde{C})$ in a natural way, which also makes good sense gauge-theoretically. By gluing the zero-section to the infinity-section, covering the map on the base given by translation by the divisor $q-p$, one constructs a variety with normal crossing singularities which is isomorphic to the complex points of the compactified Picard scheme, parametrizing torsion-free sheaves of rank 1 on $C$. You could think about what this means in gauge-theory terms. There's a large literature on such compactified Picard (and Jacobian) varieties and their behaviour in families; see e.g. L. Caporaso, A compactification of the universal Picard variety over the moduli space of stable curves, J. Amer. Math. Soc. 7 (1994), no. 3, 589--660; MR1254134.

For higher rank stable bundles, there's a bewildering array of papers on assorted compactifications (some of which are moduli spaces, some not). I'll refer you only to an extraordinary paper by Donaldson that might give you some clues as to what to expect: Gluing techniques in the cohomology of moduli spaces, Topological methods in modern mathematics (Stony Brook, NY, 1991), 137--170, Publish or Perish, Houston, TX, 1993; MR1215963.

  • $\begingroup$ Thanks Tim! Are things any easier if we're in a situation with no vanishing cycles? so for instance, the Picard variety stays compact now. $\endgroup$
    – Martin
    Mar 30, 2010 at 5:07
  • $\begingroup$ If by "no vanishing cycles" you mean non-singular, then yes - a compact complex curve has a compact degree 0 Picard variety, a g-dimensional torus H^1(C,O_C)/H^1(C,Z) - and everything goes nicely in families. $\endgroup$
    – Tim Perutz
    Mar 30, 2010 at 13:27
  • $\begingroup$ "g-dimensional" meaning as a complex manifold. $\endgroup$
    – Tim Perutz
    Mar 30, 2010 at 13:28
  • $\begingroup$ No, not nonsingular - but where the node is disconnecting - the Picard variety is still a torus, so we're all good but I wasn't sure if the higher rank bundles could still be studied via gauge theory... thanks again for the answers! $\endgroup$
    – Martin
    Mar 31, 2010 at 3:51
  • $\begingroup$ Ah I see. Yes, I think you're probably right. Locally near the node, everything should work the same way as before, but in this case the C^* gluing parameters are really all "the same" for global reasons, so one just gets the product of the Picards of the components of the normalisation. $\endgroup$
    – Tim Perutz
    Mar 31, 2010 at 13:17

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