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Please excuse me if this question is too naive. I know very little about the nonabelian Hodge correspondence but I am trying to understand how the correspondence works in the simplest case of the group $\mathbb{G}_m$ and over a smooth complex projective curve $X$.

If I understand correctly there should be a correspondence between degree zero rank 1 Higgs bundles on $X$ and rank 1 local systems on $X$, by which I mean algebraic line bundles equipped with a flat connection (flatness is in fact automatic since $\dim X = 1$). The former is parametrized by $\operatorname{Jac}(X)\times H^0(X, \Omega^1)$ while the latter is parametrized by some bundle over $\operatorname{Jac}(X)$ of affine spaces modelled over $H^0(X, \Omega^1)$. Assuming that the correspondence preserve the maps to $\operatorname{Jac}(X)$ on both sides and that it preserves the affine space structure on the fibers, the non-abelian Hodge correspondence in this case seems to be equivalent to choosing a section of the affine bundle of local systems, i.e. a preferred connection on every degree zero line bundle on $X$. It is somewhat surprising to me that there is always a canonical connection, though at least I believe that this section is only gonna be $C^\infty$ and not algebraic.

Is my interpretation of the nonabelian Hodge correspondence in this simple case correct? If so, what is the easiest way to describe this canonical connection on every line bundle?

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  • $\begingroup$ The canonical connection will be the unique unitary connection. $\endgroup$
    – Will Sawin
    Commented Nov 13 at 14:30
  • $\begingroup$ I think I see, thanks. I guess the fact that there exists a unique unitary connection on every line bundle follows from the Narasimhan–Seshadri theorem, but is there an easy way to see it for line bundles? $\endgroup$ Commented Nov 13 at 19:07
  • $\begingroup$ Also, am I understanding correctly that the nonabelian Hodge correspondence for $GL_n$ on the locus of stable bundles follows immediately from Narasimhan–Seshadri by the same reasoning, but that you need to do more work to understand the semistable case? $\endgroup$ Commented Nov 13 at 19:10

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