# Coincide between Chern-connection and Levi-Civita connection

I am a beginner in complex geometry and I am going to show Levi-Civita connection $$\nabla$$ and the Chern connection $$D$$ are the same on the holomorphic tangent bundle $$T^{1,0}M$$ on Kahler manifold. By definition, obviously $$D$$ satisfy the metric compatibility on hermitian metric on $$T^{1,0}M$$. However I cant show the torsion free s.t $$D_XY-D_YX=[X,Y]$$ where $$X,Y$$ are vector field on the holomorphic tangent bundle at the same point. Then use the uniqueness of Levi-Civita connection.

Is there any hints to show it or another approach?

• What is the Levi-Civita connection on $T^{1,0}$?
– abx
Commented Jul 25, 2022 at 18:18
• The $T^{1,0}M$ is spanned by $\partial_{z_j}$ for $j=1,\cdots ,n$. The Levi-Civita Connection $\nabla:\Gamma^{\infty}(T^{1,0}M)\times \Gamma^{\infty}(T^{1,0}M)\to \Gamma^{\infty}(T^{1,0}M)$ where $X,Y,Z\in \Gamma^{\infty}(T^{1,0}M)$ s.t $$\nabla_XY-\nabla_YX=[X,Y]$$ and $$Z(h(X,Y))=g(\nabla_ZX,Y)+g(X,\nabla_Z Y)$$ Commented Jul 25, 2022 at 18:33

It is easier to prove this result for 1-forms, instead of vector fields. On (1,0)-forms, $$\nabla^{0,1}=\bar\partial$$ because the Levi-Civita connection is torsion-free, hence $$\bigwedge(\nabla(\eta))=d\eta$$, where $$\bigwedge:\; \Lambda^1M \otimes \Lambda^k M \to \Lambda^{k+1}(M)$$ is the exterior product. This is actually true for all torsion-free connection, not necessarily the Levi-Civita. From $$\nabla^{0,1}=\bar\partial$$ we obtain that $$(0,1)$$-part of Levi-Civita connection applied to (1,0)-forms is equal to $$(0,1)$$-part of Chern connection applied to (1,0)-forms. Now, the Chern connection is by definition the only connection on $$\Lambda^{1,0}(M)$$ which has this property and preserves the Hermitian metric. Since the Levi-Civita connection preserves the Hermitian metric, it is equal to Chern connection.