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Let $X$ be a smooth projective curve over $\mathbb C$ and $M_{dR}$ denote the moduli space of stable $\Lambda$-connections of fixed rank and degree $0$. Is it known whether $M_{dR}$ is a smooth variety? If yes, then is there a simple argument like in the case of moduli of stable Higgs bundles?

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    $\begingroup$ The degree of a bundle with connection is zero, so it cannot be coprime with the rank. Or am I missing something? $\endgroup$ Jan 26, 2021 at 10:16
  • $\begingroup$ @Piotr Achinger Thanks for the correction $\endgroup$
    – user131608
    Jan 26, 2021 at 10:30
  • $\begingroup$ @Piotr Achinger Although there is a concept of special bundles over a Riemann surface which comes from certain unitary representations of the punctured Riemann surface. These bundles will not have degree $0$. Similarly those representations in GL(n, C) give rise to Higgs bundles with non-zero degree. I am not sure about bundles with connections $\endgroup$
    – user131608
    Jan 26, 2021 at 11:36
  • $\begingroup$ @user131608 I assume you are familiar with Simpson's theorems relating Betti, de Rham, and Dolbeaut moduli spaces. Do none of his comparison results allow you to deduce smoothness of de Rham from smoothness of Dolbeaut? $\endgroup$ Jan 27, 2021 at 0:50
  • $\begingroup$ @Tabes Bridges As i understand the comparison between de Rham and Dolbeault uses the fact the both these varieties are smooth varieties. I am considering here only the stable parts of the moduli. The semistable points can be singular. $\endgroup$
    – user131608
    Jan 28, 2021 at 9:02

1 Answer 1

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For $\lambda \neq 0$, $\lambda$-connections are just $\lambda$ times a holomorphic connection. On a Riemann surface, every holomorphic connection is flat and thus corresponds to a unique local system, whose monodromy gives a linear representation of the fundamental group $\pi_1(X)$ modulo conjugation.

Therefore, the moduli space of semistable pairs $(E,\nabla)$ where $E$ is a holomorphic vector bundle of rank $r$ and $\nabla$ is a $\lambda$-connection on $E$ is isomorphic to the rank $r$ character variety of $\pi_1(X)$, i.e. the GIT quotient of the affine variety $\mathrm{Hom}(\pi_1(X), \mathrm{GL}(r,\mathbb C))$ under the conjugation action of $\mathrm{PGL}(r,\mathbb C)$, and the moduli space of stable pairs is the Zariski open subset consisting of conjugacy classes of irreducible representations.

There are several ways to prove that irreducible representations are smooth points of the character variety. One way would to first show that they are smooth points of $\mathrm{Hom}(\pi_1(X), \mathrm{GL}(r,\mathbb C))$ by an explicit computation, and then remark that $\mathrm{PGL(r,\mathbb C)}$ acts freely on the set of irreducible representations by Schur's lemma. Another way would be to compute the dimension of the tangent space to the character variety at $[\rho]$, which is isomorphic to the twisted cohomology group $H^1(\pi_1(X),\rho)$. When replacing $\mathrm{GL}(r,\mathbb C)$ by some other complex reductive group $G$, one has to be careful that the stable part of the character variety may have orbifold points, corresponding to irreducible representations whose centralizer in $G/Z(G)$ is finite but not trivial. More details can be found for instance in Labourie's Lectures on representations of surface groups: https://math.unice.fr/~labourie/preprints/pdf/surfaces.pdf

Note that the non-Abelian Hodge correspondance of Hitchin, Corlette, Simpson... gives a real analytic isomorphism between the moduli space of stable Higgs bundles of rank $r$ and degree $0$ and the moduli space of rank $r$ $\lambda$-connections, so whatever simple argument giving you the smoothness of the former gives you the smoothness of the latter. The structure of complex algebraic varieties do not agree though.

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  • $\begingroup$ When we say that there is a real analytic isomorphism between the moduli space of stable Higgs bundles of rank r and degree 0 and the moduli space of rank r λ-connections, we are already using the fact that both these spaces fit into a smooth family. So you have actually assumed what you want to argue. $\endgroup$
    – user131608
    Jan 28, 2021 at 8:54
  • $\begingroup$ Also I dont understand how one can use the Riemann-Hilbert correspondence to argue that the moduli of stable lambda connections is a smooth algebraic variety. At least the arguement given above is not complete and probably your following statement is incorrect "and the moduli space of stable pairs is the Zariski open subset consisting of conjugacy classes of irreducible representations. ". Thank you for your answer. $\endgroup$
    – user131608
    Jan 28, 2021 at 8:59
  • $\begingroup$ About your first comment: I confess I never read the proof, but I think you can show the real analyticity of the non abelian Hodge correspondence by simply showing that solutions of Hitchin's equations vary analytically with the parameters. $\endgroup$
    – Nicolast
    Jan 28, 2021 at 11:48
  • $\begingroup$ About the second one: maybe I misunderstood your question: do you want the smoothness of the space of $\Lambda$-connections for all $\lambda$ simultaneously? My answer was only about the smoothness for a fixed $\lambda\neq 0$. $\endgroup$
    – Nicolast
    Jan 28, 2021 at 11:53

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