# complex gradient of a function

Let $M$ be a complex $n-$dim manifold and $u : M \rightarrow \mathbb{R}$ be some smooth function. On $M$ assume that we have a Kaehler metric $h$. How is the complex gradient vectorfield defined with respect to this metric ?

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This seems to be a pretty standard definition. The complex gradient vector field of a complex valued smooth function $u$ with respect to $h$ is the dual of $\bar\partial u$ with respect to $h$, namely, it is characterized by $\bar\partial u(X)=h(\mathrm{grad}u,X)$ for every vector field $X$. Since $\bar\partial u$ is of type $(0,1)$, we have that $\mathrm{grad}u$ is of type $(1,0)$, i.e. $\mathrm{grad u}$ is a holomorphic vector field.
Suppose $M$ is compact. Using a bit of Hodge theory, one sees that a holomorphic vector field is a gradient field if and only if it kills every holomorphic one form. In particular, if a holomorphic vector field has a zero then it is a gradient vector field. If $M$ is simply-connected (or more generally $h^{0,1}(M,\mathbf C)=0$) then every holomorphic vector field is a gradient field.