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Let $M$ be a manifold of $L \to M$ a line bundle (say over $\mathbb{C}$, ie complex line bundle).

Is it true & why that for every such line bundle there exist a flat connection $\nabla_L : \Gamma(X,E)\to \Gamma(X, \Omega_X^1\otimes L)$, i.e. a connection which curvature $\nabla_L^2= \Omega_L \in \Omega ^{2}({\mathrm {End}}\,L)=\Gamma ({\mathrm {End}}\,L\otimes \Lambda ^{2}T^{*}M)$ is zero. Thus an existence problem. Sure, I not see any reason why all connections on $L$ should be flat, nevertheless I'm asking if on the other hand there always exist a flat one. If yes, is the claim independ of the field (so we can replace $\mathbb{C}$ by any other)?

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  • $\begingroup$ By manifold, you mean a smooth manifold? $\endgroup$
    – Haohao Liu
    Nov 23 at 10:04

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No, this is false. Over $\mathbb{C}$ we have the exact sequence of sheaves of abelian groups on $X$ given by $$ 1 \to \mathbb{C}^\times \to \mathcal{O}_X^\times \stackrel{d \log}{\longrightarrow} \Omega_X \to 0. $$ The group $H^1(X, \mathbb{C}^\times)$ parametrizes line bundles with flat connection, and under this correspondence, the map to $H^1(X, \mathcal{O}_X^\times)$ (which parametrizes line bundles) forgets the connection. The obstruction to the surjectivity is $H^1(X, \Omega_X)$; a line bundle given by a transition function such that $d \log$ of the transition function is non-trivial in $H^1(X, \Omega_X)$ then gives a counterexample.

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  • $\begingroup$ Why $H^1(X, \mathbb{C}^x)$ parametrizes line bundles with flat connection? $\endgroup$
    – user738741
    Feb 21, 2020 at 4:53
  • $\begingroup$ @TimGrosskreutz given a line bundle with flat connection, pick an open cover of $X$ which is small enough to trivialize the connection (i.e. over each open your bundle + connection should be isomorphic to $\mathcal{O}, d$). Then on each overlap you get an automorphism of the bundle-with-flat-connection $\mathcal{O}, d$, but that's multiplication by a constant. $\endgroup$
    – hunter
    Feb 21, 2020 at 4:55
  • $\begingroup$ Conversely, given a cech one-cocycle representing a cohomology class (say there are just two opens U and V so that this fits in a comment), you get a sheaf $L$ whose value on U is $\mathbb{C}$ and value on V is $\mathbb{C}$ and restriction map to U \cap V is determined by your one-cocycle. You can tensor this up over the constant sheaf $\mathbb{C}$ with $\mathcal{O}_X$ to get a line bundle $\mathcal{L}$, and this line bundle admits the flat connection $\mathcal{L} = L \otimes _\C\mathcal{O}_X \to L \otimes \Omega$ induced by $d$. $\endgroup$
    – hunter
    Feb 21, 2020 at 5:05
  • $\begingroup$ One point confuses me a bit. If we shrink all members $U_i$ of the covering of $X$ small enough then you claim that over every member $U_i$ (say $=U$) the bundle -with-flat-connection has the shape $\mathcal{O}, d$ and an automorphism of such trivial bundle-with-flat-connection is a multiplication by a constant. I not completetly understand it. Firstly, I know that every connection over a trivial bundle $U \times \mathbb{R}^n$ has the shape $d + \omega$, where $d$ extends the common differential to the $End(\Omega^{\bullet})$ and $\omega$ is a $n \times n$-matrix, $\endgroup$
    – user738741
    Feb 24, 2020 at 22:11
  • $\begingroup$ i.e. a linear map, right? Ok, so flatness of the bundle imply $\omega=0$, right? So the connection is the De-Rham operator. The only remaining question is why every automorphism on such trivial bundle-with connection =De Rham-op, coinsides with multiplication by a constant $\neq 0$? Equivalently, is De-Rham-operator determined up to multiplication by a constant? $\endgroup$
    – user738741
    Feb 24, 2020 at 22:11

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