There's a connection (the Chern connection) on the Tangent Bundle of a Kahler Manifold which is compatible with both the hermitan metric, and the holomorphic structure. In general, I guess there's no reason for the connection to be holomorphic (i.e. there's no reason for holomorphic sections to go to holomorphic sections and one possible obstruction is that the tangent bundle needn't be flat?).

However, if my Kahler Manifold is actually a Hermitan symmetric domain, then that particular obstruction doesn't mess things up because the tangent bundle is trivial. My question is: is the Chern connection on a Hermitian symmetric domain holomorphic?

  • $\begingroup$ Even for the unique disk, the Chern connection is not holomorphic. $\endgroup$
    – Xin Nie
    Jul 12, 2014 at 7:55

1 Answer 1


If I understood your question correctly, you define a holomorphic connection as one which satisfies $\bar\partial (\nabla_x y)=0$ for any holomorphic vector fields $x, y$. Then the holomorphicity is equivalent to curvature of the connection being a (2,0)+(0,2)-form. Indeed, the (1,1)-part of the curvature applied to $z\in T^{0,1}$ and $x\in T^{1,0}$ is $[\bar\partial_z \nabla_x] y$. This commutator vanishes for any holomorphic $x, y$. However, these $x,y$ generate $T^{1,0}$, hence $[\bar\partial_z \nabla_x]=0$.

On the other hand, the curvature of Chern connection is a (1,1)-form. Therefore, a holomorphic Chern connection is always flat.


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