Formal Adjoint of $\overline{\partial}$ operator 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. 
 A: 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.
