# Does the holomorphic curvature determine the connection?

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 $\nabla'$?

• What about flat connections on $L$? These have zero curvature. The parameter space of flat connections on $L=\mathbb{C}\times M$ is $\text{Hom}_{\text{Groups}}(\pi_1(M),\mathbb{C}^\times)$. – Jason Starr Aug 27 '18 at 11:55
• Consider changing your $\Delta$ to $\nabla$ as $\nabla$ is most often used for connection.. – Praphulla Koushik Aug 28 '18 at 14:27
• Thank you both!!! @JasonStarr what about assuming that $M$ is simply connected? Or even contractible. – Giulio Sep 17 '18 at 7:14
• 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 $p_i$. – abx Sep 17 '18 at 7:28

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 $M$.