Integrate a 2-form on Kaehler manifold Let $X$ be a Kaehler 3-fold, with associated Kaehler form $\omega$ and metric $g_{i\bar{j}}$,
$$
\omega = \omega_{i\bar{j}} \, dz^{i} \wedge d\bar{z}^j 
= \frac{i}{2} g_{i\bar{j}} \, dz^{i} \wedge d\bar{z}^j \,.
$$
Suppose there is also a 2-form $B = B_{i \bar{j}} \, dz^i \wedge d\bar{z}^j$.
I want to calculate the integral
$$
\int_X B \wedge \omega \wedge \omega
$$
and show that it yields, up to a constant
$$
\int_X \sqrt{\mathrm{det}(g)} \, g^{i \bar{j}}B_{i\bar{j}} \,
dz^1\wedge d\bar{z}^1 \wedge dz^2\wedge d\bar{z}^2 \wedge dz^3\wedge d\bar{z}^3
$$
(I asked a similar question about $\int_X \omega^3$ here ).
$$
\epsilon^{ikm} \epsilon^{\bar{j}\bar{l}\bar{n}} \,
B_{i\bar{j}} \, \omega_{k\bar{l}} \, \omega_{m \bar{n}} \,
dz^1\wedge d\bar{z}^1 \wedge dz^2\wedge d\bar{z}^2 \wedge dz^3\wedge d\bar{z}^3 
$$
but have become stuck ...
I also think that the relation
$$
B_{i \bar{j}} = g_{i \bar{b}} g_{a \bar{j}} B^{a \bar{b}}
$$
will be necessary to get three factors of the metric in order to get $\mathrm{det}(g)$, but the square root is throwing me off as well.
 A: Brutal force: For general $n$,
\begin{align}
B\wedge \omega^{n-1}&= \left( \frac i2\right)^{n-1}\left( B_{i_1\bar j_1} dz^{i_1} \wedge d\bar z^{j_1} \right)\wedge \left(g_{i_2\bar j_2} dz^{i_2} \wedge d\bar z^{j_2} \right) \wedge \cdots \wedge \left( g_{i_n\bar j_n} dz^{i_n} \wedge d\bar z^{j_n} \right) \\
&= \left( \frac i2\right)^{n-1}B_{i_1\bar j_1} g_{i_2\bar j_2}\cdots g_{i_n\bar j_n} dz^{i_1} \wedge d\bar z^{j_1} \wedge \cdots \wedge dz^{i_n} \wedge d\bar z^{j_n} \\
&= (-1)^M \left( \frac i2\right)^{n-1}B_{i_1\bar j_1} g_{i_2\bar j_2}\cdots g_{i_n\bar j_n} dz^{i_1} \wedge \cdots \wedge dz^{i_n}  \wedge d\bar z^{j_1} \wedge \cdots \wedge d\bar z^{j_n}
\end{align}
It is clear that we can assume $\{i_1, \cdots, i_n\} = \{j_1, \cdots, j_n\} = \{1, \cdots, n\}$. Fix $i_1,j_1$. Let $i : \{2, \cdots, n\} \to \{1, \cdots, n\}\setminus \{i_1\}$ which send $k\to i_k$. Similar for $j$. Then
\begin{align}
B_{i_1\bar j_1} &g_{i_2\bar j_2}\cdots g_{i_n\bar j_n} dz^{i_1} \wedge \cdots \wedge dz^{i_n}  \wedge d\bar z^{j_1} \wedge \cdots \wedge d\bar z^{j_n}\\
&= (-1)^{i_1+j_1}B_{i_1\bar j_1}\ \  (-1)^{\operatorname{sgn} i + \operatorname{sgn} j} g_{i_2\bar j_2}\cdots g_{i_n\bar j_n} dZ \wedge d\bar Z
\end{align}
where $dZ = dz^1 \wedge \cdots \wedge dz^n$. For each fixed $i$,
\begin{align}
(-1)^{\operatorname{sgn} i} \sum_j(-1)^{ \operatorname{sgn} j} g_{i_2\bar j_2}\cdots g_{i_n\bar j_n} =(-1)^{\operatorname{sgn} i} \det (M_{i_1j_1})_i = \det M_{i_1j_1}.
\end{align}
Here $M_{i_1j_1}$ is the $(i_1, j_1)$ minor of $(g_{i\bar j}$ and $(M_{i_1j_1})_i$ is the permutation of the rows of $M_{i_1j_1}$ by $i$. Since there are $(n-1)!$ such $i$,
\begin{align} (-1)^{i_1+j_1}B_{i_1\bar j_1}\ \  (-1)^{\operatorname{sgn} i + \operatorname{sgn} j} g_{i_2\bar j_2}\cdots g_{i_n\bar j_n} &= (n-1)! (-1)^{i_1+ j_1} M_{i_1j_1} B_{i_1\bar j_1} \\
&= (n-1)! \det g g^{i_1\bar j_1} B_{i_1\bar j_1}
\end{align}
since
$$g^{i_1\bar j_1} = \frac{ (-1)^{i_1 + j_1} M_{i_1 j_1}}{\det (g_{i\bar j})}.$$
So
$$ B \wedge \omega^{n-1} = (n-1)! (-1)^M \left( \frac i2 \right)^{n-1} g^{i\bar j} B_{i\bar j} \det (g_{i\bar j}) dZ \wedge d\bar Z.$$
We do not have $\sqrt{\det g}$ here, unless $g$ is the real metric and in this case $\sqrt{\det g} = \det (g_{i\bar j})$.
