$(p,q)$-forms and notation I have a couple of questions about a notation I'm not very familiar with.
Let $D$ a bounded domain in $\mathbb{C}^n$, $n \geq 2$, not necessarily with a smooth boundary. Let $C_{(p, q)}^\infty(D)$ denote the $(p, q)$-forms on $D$, where $0 \leq p \leq n$, $0 \leq q \leq n$. 
Let $(z_1, \dots, z_n)$ be the complex coordinates for $\mathbb{C}^n$. Then any $(p, q)$-form $f \in C_{(p, q)}^\infty(D)$ can be expressed as
\begin{equation} 
f = \sum_{I,J}\,'f_{I,J}\,dz^I \wedge d\bar{z}^J \tag{1},
\end{equation}
where $I = (i_1, \dots, i_p)$ and $J = (j_1, \dots, j_q)$ are multiindices and $dz^I = dz_{i_1} \wedge \dots \wedge dz_{i_p}$, $d\bar{z}^J = d\bar{z}_{i_1} \wedge \dots \wedge d\bar{z}_{i_q}$. The notation $\sum \, '$ means the summation over strictly increasing multiindices and the $f_{I,J}$ ’s are defined for arbitrary $I$ and $J$ so that they are antisymmetric. My questions are:
1) If $f$ is expressed by $(1)$ and $g = \sum_{|I|=p,|K|=q-1}^{'}g_{I,K}\,dz^I \wedge d\bar{z}^K$, is it correct to say, in the case where $n = p = q = 2$, that $f$ and $g$ are of this type
\begin{equation}
f = f_{12,12} \, dz_{1} \wedge dz_{2} \wedge d\bar{z}_{1} \wedge d\bar{z}_{2}, \,\,\,\,\,\, f_{12,12} \in C^\infty(D),
\end{equation}
\begin{equation}
g = g_{12,1} \, dz_{1} \wedge dz_{2} \wedge d\bar{z}_{1} + g_{12,2} \, dz_{1} \wedge dz_{2} \wedge d\bar{z}_{2}, \,\,\,\,\,\, g_{12,1}, g_{12,2} \in C^\infty(D)?
\end{equation}
2) My book at some point writes that
\begin{equation}
(f,\bar{\partial}g)=(-1)^p\sum_{I,K}\,' \sum_{k=1}^n (f_{I,kK}, \frac{\partial g_{I,K}}{\partial \bar{z}_k}),
\end{equation}
where $f$ is a $(p,q)$-form and $g$ is a $(p,q-1)$-form.
Here I don't understand what it means by the term $f_{I,kK}$ for $k = 1, \dots, n$. In particular, what do I get if I take $f$ defined in 1)?
Thanks to whoever will answer.
 A: Let $f$ and $g$ be as described in your question. Then, by the definition of the differential, we have
\begin{align*}
\overline\partial g=\sum_{k=1}^n\frac{\partial g_{I,K}}{\partial\overline{z_k}}d\overline{z_k}\wedge d z^I\wedge d\overline{z}^J=(-1)^p\sum_{k=1}^n\frac{\partial g_{I,K}}{\partial\overline{z_k}}dz^I\wedge d\overline{z_k}\wedge d\overline{z}^J
\end{align*}
since $|I|=p$. If we are pairing this result with $f=\sum_{I,K}f_{I,J}dz^I\wedge d\overline{z}^J$ we get the result of 2) by noting that $(dz^I\wedge d\overline{z}^J,dz^K\wedge d\overline{z}^L)\neq 0$ if and only if $I=K$ and $J=L$ as sets. The $f_{I,kK}$ term is just the coefficient of $d z^I\wedge d\overline z_k\wedge d\overline{z}^K$ (so you have to reorder the overlined terms if necessary; maybe getting a minus sign). And yes, you are right with 1). In your example one gets
\begin{align*}
(f,\overline\partial g)=(-1)^2 ((f_{12,12},\frac{\partial g_{12,2}}{\partial \overline z_1})+(-f_{12,12},\frac{\partial g_{12,1}}{\partial\overline z_2}))
=(f_{12,12},\frac{\partial g_{12,2}}{\partial \overline z_1}-\frac{\partial g_{12,1}}{\partial\overline z_2}).
\end{align*}
