Does Hermite-Einstein imply Kähler-Einstein?

Question

Let $M$ be a compact Kähler manifold and let $\nabla$ be its Levi-Civita, or equivalently its Chern, connection. Denoting the vector bundle of complexified one forms of $M$ by $\Omega^1_{\mathbb{C}}$, let us assume that $\Omega^1_{\mathbb{C}}$ is an Hermite-Einstein vector bundle. Does it follow that $M$ is a Kähler-Einstein manifold?

Answer 1

If one interprets the OP's question literally, the answer is 'yes', but I imagine that the OP didn't literally mean what the OP wrote.

First, interpret everything literally: Assume that $(M,g,\omega)$ is a Kähler manifold (where $\omega\in\Omega^{1,1}(M)$ is the Kähler form) of complex dimension $n$ and that $\Omega^1_\mathbb{C} = \mathbb{C}\otimes T^*M = \Lambda^{1,0}(M)\oplus\Lambda^{0,1}(M)$ is endowed with its natural connection induced from the Levi-Civita connection of $g$, i.e., just the complexification of the connection induced on $T^*M$. Because of the Kähler assumption, this connection preserves the two complex subbundles $\Lambda^{1,0}(M)$ and $\Lambda^{0,1}(M)$.

When one writes $\omega = \frac{\mathrm{i}}2(\eta_1\wedge\overline{\eta_1}+\cdots +\eta_n\wedge\overline{\eta_n})$ locally, where $\eta_1,\ldots,\eta_n$ is a local basis for the sections of $\Lambda^{1,0}(M)$, the associated connection forms are the $\theta_{i\bar j}=-\overline{\theta_{j\bar{i}}}$ satisfying $\mathrm{d}\eta_j = -\theta_{j\bar{k}}\wedge\eta_k$. The curvature forms $\Theta_{j\bar{k}}= \mathrm{d}\theta_{j\bar{k}}+\theta_{j\bar{\ell}}\wedge\theta_{\ell\bar{k}} $ are of type $(1,1)$. There are functions $R_{j\bar{k}} = \overline{R_{k\bar{j}}}$ for which $\omega^{n-1}\wedge\Theta_{j\bar{k}} = \mathrm{i} R_{j\bar{k}}\,\omega^n$. The connection on $\Lambda^{1,0}(M)$ is Hermite-Einstein if and only if $R_{j\bar{k}} = R^{1,0}\,\delta_{j\bar k}$ for some constant $R^{1,0}$. This latter condition is equivalent to $(M,g,\omega)$ being Kähler-Einstein because $\rho=\mathrm{i} R_{j\bar{k}}\,\eta_k\wedge\overline{\eta_j}$ is, up to a constant multiple, the Ricci form of $(M,g,\omega)$.

However, in this case the constant $R^{0,1}$ that one gets for the induced connection on $\Lambda^{0,1}(M)$ is $R^{0,1}=-R^{1,0}$. This is because the connection forms for the conjugate local coframing $\overline{\eta_1},\ldots,\overline{\eta_n}$ for $\Lambda^{0,1}(M)$ are the conjugates of the connection forms for the original coframe, so that the curvature forms for $\Lambda^{0,1}(M)$ are the conjugates of the curvature forms for $\Lambda^{1,0}(M)$.

What this means is that, if one assumes that the complexified connection on $\Omega^1_\mathbb{C} = \Lambda^{1,0}(M)\oplus \Lambda^{0,1}(M)$ be Hermite-Einstein (instead of each summand separately being Hermite-Einstein), then one will have to have $R^{1,0}=R^{0,1}$, i.e., $R^{1,0}=R^{0,1}=0$, and this stronger condition is equivalent to $(M,g,\omega)$ being Ricci-flat, not just Kähler-Einstein.

Thus, the literal answer to the OP's question is 'yes', but the stronger conclusion is that $(M,g,\omega)$ must be Calabi-Yau, not just Kähler-Einstein. However, if one only asks that, say, $\Lambda^{1,0}(M)$ be Hermite-Einstein (using the connection induced from Levi-Civita), then it suffices for $(M,g,\omega)$ to be Kähler-Einstein, and I suspect that this is what the OP really wanted to know.