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I am reading "Complex geometry" by D. Huybrecht. On p.223 the books says that "If $c_{1}(X)=0$, e.g. if the canonical bundle $K_{X}$ is trivial, and $g$ is Kähler–Einstein metric, then $\operatorname{Ric}(X,g)=0$. Indeed in this case the scalar factor $\lambda$ is necessarily trivial and hence $\operatorname{Ric}(X,g)=\lambda\omega=0$, i.e. the Kähler metric $g$ is Ricci-flat."

Remember that we say a metric $g$ is Kähler–Einstein if $\operatorname{Ric}(X,g)=\lambda \omega$ for some constant $\lambda\in \mathbb{R}$. Here $\omega$ is the Kähler form associated to $g$.

My question is, why does Kähler–Einstein mean $\lambda=0$ in $c_{1}(X)=0$ case? If this is true, any Kähler–Einstein metric on Calabi–Yau manifold is Ricci-flat?

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I now understand why; the scalar factor $\lambda$ can be explicitly computed as $$ \lambda=\frac{2\pi \int_{X}c_{1}(X)\wedge \omega^{n-1}}{\int_{X}\omega^n}. $$ So Kähler–Einstein metric on Calabi–Yau manifold is necessarily Ricci flat metric!

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The answer given by M. K. is nice, but perhaps a little sophisticated. Let me offer a more naive argument (albeit, invoking Chern-Weil theory).

Suppose $X$ is a compact complex manifold with $c_1(X)=0$, i.e., $c_1(-K_X)=0$, where $-K_X$ denotes the dual of the canonical bundle. Suppose $X$ admits a Kähler-Einstein metric $\omega$ with $\text{Ric}(\omega) = \lambda \omega$. We want to show that $\lambda =0$.

By Chern-Weil theory, the de Rham cohomology class represented by the Ricci form of a Kähler metric represents $2\pi c_1(-K_X)$. Hence, $$2\pi c_1(-K_X) = \{ \text{Ric}(\omega) \} = \lambda \{ \omega \},$$ and this implies $\lambda =0$ if $\{ \omega \} \neq 0$. To verify that $\{ \omega \} \neq 0$, proceed by contradiction and suppose that $\{ \omega \} =0$ in $H_{\text{DR}}^2(X, \mathbf{R})$. The $\partial \bar{\partial}$-lemma then implies that $\omega = \sqrt{-1} \partial \bar{\partial} \varphi$ for some $\varphi \in \mathcal{C}^{\infty}(X, \mathbf{R})$. Taking the trace with respect to the metric $\omega$, we see that $\Delta_{\omega} \varphi : = \text{tr}_{\omega}(\sqrt{-1} \partial \bar{\partial} \varphi) = \text{tr}_{\omega}(\omega) = \dim_{\mathbf{C}} X >0$. By the maximum principle, $\varphi =0$ and this contradicts the non-degeneracy of $\omega$.

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