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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?

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  • $\begingroup$ What is the Levi-Civita connection on $T^{1,0}$? $\endgroup$
    – abx
    Commented Jul 25, 2022 at 18:18
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    $\begingroup$ 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 $$Z(h(X,Y))=g(\nabla_ZX,Y)+g(X,\nabla_Z Y)$$ $\endgroup$
    – James Chiu
    Commented Jul 25, 2022 at 18:33

1 Answer 1

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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.

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