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Let $M$ be a simply connected complex manifold (of dimension greater than one), $L$ a line bundle, and $\nabla$ a connection on $L$ with possibly singularities along a divisor $D$. We define the curvature as $$ R_{\nabla}(X,Y)=[\nabla_X,\nabla_Y]-\nabla_{[X,Y]} $$

Suppose that a second connection $\nabla'$ on $L$ has the same singularities and the same curvature. Is $\nabla$ equal to $\nabla'$?

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    $\begingroup$ What about flat connections on $L$? These have zero curvature. The parameter space of flat connections on $L=\mathbb{C}\times M$ is $\text{Hom}_{\text{Groups}}(\pi_1(M),\mathbb{C}^\times)$. $\endgroup$
    – Jason Starr
    Commented Aug 27, 2018 at 11:55
  • $\begingroup$ Consider changing your $\Delta$ to $\nabla$ as $\nabla$ is most often used for connection.. $\endgroup$
    – Praphulla Koushik
    Commented Aug 28, 2018 at 14:27
  • $\begingroup$ Thank you both!!! @JasonStarr what about assuming that $M$ is simply connected? Or even contractible. $\endgroup$
    – Giulio
    Commented Sep 17, 2018 at 7:14
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    $\begingroup$ Answer to the edited question: this is already false for $\mathbb{P}^1$, with $D=p_1+\ldots +p_n$, $n>2$ and the points $p_i$ distinct. The connections with poles along $D$ are all flat, they form a homogeneous space under $H^0(\mathbb{P}^1,\Omega ^1(D))$, and two general ones have a pole at each $p_i$. $\endgroup$
    – abx
    Commented Sep 17, 2018 at 7:28

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The connection can be equivalently given by parallel transport along curves. So if two connections are equal (or isomorphic in appropriate sense) then they must have the same holonomy. Now by the Ambrose--Singer theorem there is a close relationship between holonomy and curvature. But full holonomy group is a global object and can see holes ($\pi_1(M)$) in your manifold. As has been already mentioned in the comments, point singularities of connections pretty much behave as punctures in $M$.

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