Context: The Futaki Invariant in Kähler Geometry.
Reference: Page 24 of Gang Tian's book, Canonical Metrics in Kähler Geometry.
Let $(M, g, \omega)$ be a compact Kähler manifold with Kähler metric $g$ and Kähler form $\omega$. Let $X$ be a holomorphic vector field on $M$. Since $\omega$ is closed and $X$ is holomorphic, $$\overline{\partial}(i_X \omega) =0.$$ Hence, by the Hodge theorem, there exists a smooth function $\vartheta_X$ and a harmonic 1-form $\alpha$ such that $$i_X \omega = \alpha - \overline{\partial} \vartheta_X.$$ In particular, $$\alpha = i_X \omega + \overline{\partial} \vartheta_X$$ and $\overline{\partial} \alpha =0$.
Claim: $\overline{\partial}^{\ast} \alpha =0$, where $\overline{\partial}^{\ast}$ denotes the formal adjoint of $\overline{\partial}$.
This is clearly an elementary observation, but it is not clicking for me. Any help is appreciated. Thanks in advance.
(Edit): Apologies for asking two questions in the same post.
Claim: If $X^i$ denote the components of the holomorphic vector field described above. How does $$X^i = g^{i\overline{j}} \frac{\partial \alpha}{\partial \overline{z}^j} - g^{i \overline{j}} \frac{\partial \vartheta_X}{\partial \overline{z}^j}?$$
I want to get a really clear understanding of every computation and do not want to simply brush things by.
Question I.
It follows from the Hodge theory on Kahler manifolds, more specifically from commutation identities. You can read more about them in chapter VI of this book, beware though - the author uses $d''$ for $\bar{\partial}$. The specific identity that is being used here is $$ \Delta'' = \frac{1}{2}\Delta, $$ where $\Delta'' = \bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}$. Now if we use that identity on $\alpha$, right-hand is zero by harmonicity and left-hand side reduces to $\bar{\partial}\bar{\partial}^*\alpha = 0$, by the fact that $\bar{\partial}\alpha =0$ and this implies that $\bar{\partial}^*\alpha =0$, since $$ 0 = \int \langle\bar{\partial}\bar{\partial}^*\alpha,\alpha\rangle = \int \langle\bar{\partial}^*\alpha,\bar{\partial}^*\alpha\rangle = \int \|\bar{\partial}^*\alpha\|^2, $$ from the definition of adjoint.
Question II. $$ X^i = X^kg_{k\bar{j}}g^{\bar{j}i} = (X^kg_{k\bar{j}})g^{\bar{j}i} = (i_X\omega)_{\bar{j}}g^{\bar{j}i} = (\alpha - \bar{\partial}\vartheta_X)_{\bar{j}}g^{\bar{j}i}. $$ So as you see it should be the $\bar{j}$-th coefficient of $\alpha$, not the derivative.
Hope it is clear now.