A criteria of being killing vector field on Kahler manifold

Question

On a Riemannian manifold, let $\omega$ be the dual $1$-form of a vector field X, then the condition for X to be a killing vector field is $$ \nabla_{i}\omega_{j}+\nabla_{j}\omega_{i}=0. $$ I am readling a lecture notes by Yum-Tong Siu called Hermitian-Eintein Metrics for Stable Bundles and Kahler-Einstein Metrics. On page 148, he mentioned sort of a complex version of the killing vector condition. On a Kahler manifold (M,g,J). Let $Y^{i}=\uparrow\bar{\partial}\phi$, i.e.$Y$ is a vector field obtianed by raising the index of the $(0,1)$-form $\bar{\partial}\phi$ where $\phi$ is real. Then the imaginary part of Y is $\frac{1}{2\sqrt{-1}}(Y+\bar{Y})=(-\frac{\sqrt{-1}}{2}Y^{i},-\frac{\sqrt{-1}}{2}Y^{\bar{i}})$. Lowering down the index using the Kahler form we get a $(1,1)$-form $\eta=(\eta_{i},\eta_{\bar{i}})$. Then he claimed that if we want to check that $ImY$is killing, we only need to check that $\nabla_{i}\eta_{j}+\nabla_{j}\eta_{i}$,$\nabla_{\bar{i}}\eta_{j}+\nabla_{j}\eta_{\bar{i}}$, $\nabla_{\bar{i}}\eta_{\bar{j}}+\nabla_{\bar{j}}\eta_{\bar{i}}$ all vanish. I guess here he used the Chern connection. I have tried to prove this condition by the equivalence of Chern connection and Levi-Civita connection under the isomorphism of the tangent bundle $TM$ and the holomorphic vector bundle $TM_{\mathbb{C}}^{(1,0)}$. But I find it is rather complicated. Can anyone give me some thoughts on how to prove the claim?

Answer 1

This is true in general, and does not depend on the particular choice of $Y=\mathrm{grad}^{1,0}(\phi)$. Let's call $V$ the imaginary part of $Y$, for brevity. The Killing condition for $V$ is $$g(\nabla_XV,Z)+g(X,\nabla_ZV)=0\quad\forall\, X,Z\mbox{ vector fields}.$$ Choosing in particular $X=\partial_{i}$ and $Z=\partial_{j}$ the condition becomes $$g(\nabla_iV^{1,0}+\nabla_iV^{0,1},\partial_j)+g(\partial_i,\nabla_jV^{1,0}+\nabla_jV^{0,1})=0.$$ Since the metric is Kähler, the Levi-Civita connection preserves the $(1,0)$ and $(0,1)$ decomposition of vector fields. But $g$ is of type $(1,1)$, so that simplifies to $$g(\nabla_iV^{0,1},\partial_j)+g(\partial_i,\nabla_jV^{0,1})=0$$ which in a system of local coordinates becomes $$g_{\bar{k}j}\nabla_iV^{\bar{k}}+g_{i\bar{k}}\nabla_jV^{\bar{k}}=0,$$ which is one of the relations you are seeking: $\nabla_i\eta_j+\nabla_j\eta_i=0$, for $\eta=g(V,\cdot)$.

Similarly, if we choose $X=\partial_{\bar{i}}$ and $Z=\partial_{j}$, we obtain the second condition \begin{equation} \begin{gathered} 0=g(\nabla_{\bar{i}}V^{1,0}+\nabla_{\bar{i}}V^{0,1},\partial_j)+g(\partial_{\bar{i}},\nabla_jV^{1,0}+\nabla_jV^{0,1})=\\ =g(\nabla_{\bar{i}}V^{0,1},\partial_j)+g(\partial_{\bar{i}},\nabla_jV^{1,0})=g_{j\bar{k}}\nabla_{\bar{i}}V^{\bar{k}}+g_{k\bar{i}}\nabla_jV^k=\nabla_{\bar{i}}\eta_j+\nabla_j\eta_{\bar{i}}. \end{gathered} \end{equation} It is also easy to obtain the third condition in the same way. So the three conditions are necessary, but the three together are sufficient, since the vectors $\partial_i$ and $\partial_{\bar{i}}$ generate the complexified tangent bundle of the manifold. Hope this helps!

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By the way, it can be interesting to figure out what that $\eta$ is. In a system of local holomorphic coordinates $(z^1,\dots,z^n)$ we have $$Y=\mathrm{grad}^{1,0}(\phi)=g^{i\bar{j}}\partial_{\bar{z}^j}\phi\,\partial_{z^i}=:\nabla^i\phi\,\partial_{z^i}$$ and $$\mathrm{Im}(Y)=\frac{Y-\bar{Y}}{2\sqrt{-1}}=\frac{1}{2\sqrt{-1}}(\nabla^{1,0}-\nabla^{0,1})\phi.$$ Lowering the indices, we obtain a familiar expression for $\eta$ (by the way, $\eta$ is \emph{not} a $(1,1)$-form, it is just a $1$-form): $$\eta=\frac{1}{2\sqrt{-1}}(\bar{\partial}-\partial)\phi=-\frac{1}{2}\mathrm{d}^c\phi.$$