# Line bundles have flat connections

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)?

• By manifold, you mean a smooth manifold? Nov 23 at 10:04

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.

• Why $H^1(X, \mathbb{C}^x)$ parametrizes line bundles with flat connection?
– user738741
Feb 21, 2020 at 4:53
• @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. Feb 21, 2020 at 4:55
• 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 $d$. Feb 21, 2020 at 5:05
• 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,
– user738741
Feb 24, 2020 at 22:11
• 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?
– user738741
Feb 24, 2020 at 22:11