Isomorphism classes of line bundles over a scheme $X$ are described by the Picard group $Pic(X)$. Now there is a paper that describes the moduli space of line bundles with connections. This paper is quite technical for me to understand but I do understand that it must be quite different than $Pic(X)$. We can fix our $X$ (I am mainly interested in complex surfaces). Since we have fixed rank as well ($r=1$). I suspect that the moduli space I am interested in is a fibration over the Picard group (maybe of the moduli space of connections?).

Maybe we can take as an example $X = \mathbb{CP}^2$. Then we know that $Pic(X) = \mathbb{Z}$ since the invertible sheaves $=$ line bundles are classified by the degree of the corresponding divisor. Now, let us fix for a moment such a degree say $l$ for the class of line bundle associated to $\mathcal{O}(l)$. Equip this space with a connection $A \in \mathcal{A}^{(l)}$ where the latter is the affine space of connections over $X$. Of course we want to consider connections up to endomorphisms $f \in G^{(l)}$. Therefore, for the line bundle $\mathcal{O}(l)$ the corresponding moduli space of connections on it is $\mathcal{M}^{(l)} := \mathcal{A}^{(l)}/G^{(l}$. I can only assume then that the moduli space of line bundles with connections is a "combination" of $Pic(X)$ and $\sqcup_l \mathcal{M}^{(l)}$. How far off am I?

Of course I might be saying completely crazy things. I would like to get some intuition so anything you might say might be of some help.

P.S. In this question the OP claims that line bundles over $X$ are parametrised by their Chern classes. Is this parametrisation an isomorphism class one? And if so what is the relation with Pic($X$) and the question above?

  • $\begingroup$ You are indeed far off. A vector bundle with a connection has trivial Chern classes. For $X=\mathbb{P}^2$, the only line bundle with a connection is $(\mathcal{O}_{\mathbb{P}^2},d)$. $\endgroup$
    – abx
    Mar 3, 2018 at 11:39
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    $\begingroup$ What does this notation mean (the $d$ part)? But I do not understand why a vector bundle with connection has trivial Chern classes? Do you mean fixed? $\endgroup$
    – Marion
    Mar 3, 2018 at 11:45
  • $\begingroup$ @abx do you have a reference / proof for the fact you mention? (i.e. a vector bundle with connection has trivial chern classes) I'd never heard of it and I'm very interested. $\endgroup$ Mar 3, 2018 at 17:57
  • $\begingroup$ @YosemiteSam I don't think this statement is quite true. Analytically a connection will define a curvature and that is directly related to the $c_1$, the first Chern class. For a line bundle $c_1$ is far from trivial while indeed $c_2=0$. $\endgroup$
    – Marion
    Mar 3, 2018 at 17:59
  • $\begingroup$ @YosemiteSam: the classic refence is Atiyah, Complex analytic connections in fibre bundles, Transactions of the Amer. Math. Soc. 85 (1957), 181-207. MR 19:172c. $\endgroup$
    – Ben McKay
    Apr 2, 2018 at 20:48

1 Answer 1


This is more of a comment. First of all, you have to be clear about which category you're working in. Do you want $C^\infty$ line bundles with $C^\infty$ connections, or holomorphic line bundles with holomorphic connections, or...? If it's the second, as I suspect, then abx is correct, the first Chern class is zero for $X$ smooth projective (cf, Atiyah, Complex analytic connections in fibre bundles, Transactions AMS 1957). Assuming this case, then the a collection of line bundles with connection is well known to be $$LineConn(X)=\mathbb{H}^1(X,dlog: O_X^*\to \Omega_X^1)$$ For an explanation, see Esnault-Viehweg's article on Deligne-Beilinson cohomology, or probably the paper you linked. From the standard exact sequence, we get $$ 0\to H^0(X,\Omega_X^1)\to LineConn(X)\to Pic^\tau(X) \to 0$$ I think this is what you're looking for. $Pic^\tau$ is the group of line bundles with trivial $c_1$ in rational cohomology.

  • $\begingroup$ Sorry, physicist here. I am not sure that I agree that $c_1$ is zero even in the case of holomorphic line bundles with holomorphic connections. Assume $X = \mathbb{CP}^2$ and $L \to E$ a hermitian line bundle over $X$ and equip it with a hermitian connection $A$ (local one-form presented here) s.t. $F = dA$ is the curvature. Then over a cycle $S$ we can have $\int_S F \backsim c_1(L)$ and this does not have to be zero. This is what physicists call magnetic flux. $\endgroup$
    – Marion
    Mar 3, 2018 at 18:24
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    $\begingroup$ You seem to be talking about $C^\infty$ connection on a holomorphic line bundle. And that's fine. But as I said, you have to clear from the beginning about what the category is that you're working in. $\endgroup$ Mar 3, 2018 at 18:36
  • $\begingroup$ But from Kobayashi-Hitchin a $C^{\infty}$ connection corresponds to a hermitian connection. And a hermitian connection is defined on a hermitian line bundle which is holomorphic. $\endgroup$
    – Marion
    Mar 3, 2018 at 18:38
  • $\begingroup$ No. The connection form for holomorphic connection is holomorphic, whereas for a hermitian connection it isn't ( its $\partial \log ||f||$ where $f$ is a $C^\infty$ section.) Anyway, that's all I have to say about this. $\endgroup$ Mar 3, 2018 at 18:45
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    $\begingroup$ @Marion: Arapura is correct of course. Do you want a smooth connection on a holomorphic vector bundle or a holomorphic connection on that bundle? The question is whether the connection has holomorphic coefficients $A^{\mu}_{\nu}$ in a holomorphic trivialization. A Hermitian connection which is not flat cannot be holomorphic, because a holomorphic connection has $(2,0)$ type curvature, while a Hermitian connection has $(1,1)$ type curvature. Look at Huybrechts, Complex Geometry. $\endgroup$
    – Ben McKay
    Apr 2, 2018 at 20:45

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