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I have problems on how to consider the Kahler curvature operator. I know that one can consider the Riemannian curvature operator $R$ as a linear transformation from $\mathfrak{so}(n,\mathbb{R})$ to $\mathfrak{so}(n,\mathbb{R})$. But how to see that in Kahler case, the (Kahler) curvature operator $K$ is a linear transformation from $\mathfrak{u}(n)$ to $\mathfrak{u}(n)$?

Here $\mathfrak{so}(n,\mathbb{R})$ is the Lie algebra of all skew-symmetric real $n\times n$ matrices and $\mathfrak{u}(n)$ is the Lie algebra of all conjugate skew-symmetric $n\times n$ matrices. I know that one may see $\mathfrak{u}(n)\subset \mathfrak{so}(2n)$ as a Lie sub-algebra.

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The curvature tensor $R_x$ at each point of a Kaehler manifold is a symmetric map $$R:\Lambda^2 T_xM=\Lambda^2\mathbb{R}^{2n}\to \mathfrak{u}(n)\subset\mathfrak{so}(2n)=\Lambda^2\mathbb{R}^{2n},$$ i.e. $$g(R(X\wedge Y),Z\wedge U)=g(R(Z\wedge U),X\wedge Y),$$ consequently it is zero on the orthogonal complement to $\mathfrak{u}(n)\subset\mathfrak{so}(2n)=\Lambda^2\mathbb{R}^{2n}$.

Alternatively, this follows from the equality $R(JX,Y)+R(X,JY)=0$.

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  • $\begingroup$ Thank you very much. I know why one can see the Kaehler curvature operator is from $\mathfrak{u}(n)$ to $\mathfrak{u}(n)$. $\endgroup$ – Koma May 6 '15 at 13:08

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