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I'm studying holomorphic vector bundles $(E,h)$ on Kähler manifolds that admit a Hermite-Einstein metrics. Particularly, I'm trying to find the motivation for the definition.

An hermitian structure $h$ on a holomorphic vector bundle $E$ is called Hermite-Einstein if $$i\Lambda_\omega F_\nabla=\lambda\cdot \text{id}_E$$ for some constant scalar $\lambda \in \Bbb R$. Here $\Lambda_\omega F_\nabla$ is the contraction by $\omega$.

What I'm currently thinking is that this is analogous to the definition of an Einstein manifold. We say that a Riemannian manifold $(M,g)$ is Einstein if $\text{Ric} = \lambda g$, i.e. the Ricci tensor is proportional to the Riemannian metric.

To this end I would like to think about $i\Lambda_\omega F_\nabla$ as somehow capturing the same information as the Ricci tensor, but I have been stuck with understanding this expression for a multiple days now.

I found an answer here stating that $\Lambda_\omega \alpha = g(\omega,\alpha)$, whenever $\alpha$ is a $2$-form or a $(1,1)$-form. Since $F_\nabla$ is a $(1,1)$-form this should give $\Lambda_\omega F_\nabla = g(\omega,F_\nabla)$, but now this deviates from what the definition stated that $\Lambda_\omega F_\nabla$ would have been a contraction by $\omega$?

If anyone here can clarify this or has some conceptual understanding about the definition, I would gladly appreciate that! Apologies if this is a question suited more for stackexchange, I did not find much about this there.

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  • $\begingroup$ Because $\omega\in\Lambda^2 T^*M$ is nondegenerate, there is a dual bivector in $\Lambda^2TM$ that you can contract with. Up to some constant factor, this is the same as $g(\omega,-)$. $\endgroup$ Commented Apr 21 at 19:17
  • $\begingroup$ Consider the case that $E$ is the holomorphic tangent bundle and $h$ is a Kähler metric. Then the two notions coincide. $\endgroup$ Commented Apr 23 at 16:20
  • $\begingroup$ I found and old paper by Donaldson "Anti Self-Dual Yang-Mills Connections Over Complex Algebraic Surfaces and Stable Vector Bundles." where the author states that $\Lambda_\omega\left(a_{\alpha\beta}dz^\alpha \wedge d\bar{z}^\beta\right) = -2i\sum a_{\alpha\alpha}$, though at the moment this just expands my confusion. @QuartoBendir $\endgroup$
    – Johannes
    Commented Apr 23 at 19:20
  • $\begingroup$ @SebastianGoette I presume that you mean $\widetilde{\omega} \in \Lambda^2TM$ defined by $\widetilde{\omega}(X) = \omega(X,-)$ and $\Lambda_\omega F_\nabla$ is just abusively using $\Lambda_\omega $ instead of $\Lambda_{\widetilde{\omega}}$? $\endgroup$
    – Johannes
    Commented Apr 30 at 7:39

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