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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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up vote 2 down vote accepted

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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