How is the Kähler form decomposed in terms of the metric? In countless textbooks and lecture notes (e.g. eqn 4.9 of Lectures on Riemannian Geometry, Part II:
Complex Manifolds by Stefan Vandoren), the Kähler (1,1)-form, $\omega$, is written in terms of the Riemannian metric:
\begin{equation} 
\omega = 2 i g_{a\bar{b}}dz^a \wedge d\bar{z}^{\bar{b}}\,.
\end{equation}
where  $dz^a$ and $d\bar{z}^a$ are holomorphic and antiholomorphic coordinates, respectively. My question is simple - how does this make sense? The metric is symmetric $g_{a\bar{b}}=g_{\bar{b}a}$ and the wedge product is of course antisymmetric suggesting to me that $\omega$ should be zero. Also my understanding was that a differential p-form, $C_p$, was defined:
\begin{equation}
C_p=\frac{1}{p!}C_{m_1\dots m_p}dx^{m_1}\wedge \dots \wedge dx^{m_p}
\end{equation}
where the components, $C_{m_1\dots m_p}$, are completely antisymmetric in the indices. Since $i g_{a\bar{b}}$ is clearly not antisymmetric, how is this possible? Do the components of a (1,1)-form not have to be totally antisymmetric?
I realise there must just be a simple misunderstanding here on my part so I apologise for the simplicity of the question but I would very much appreciate any help. Thank you.
 A: The summation is only through holomorphic indices. So $g_{\bar b a}$ term does not appear in the sum and hence does not cancel out. Its best to write out the definition $w(.,.)=g(J.,.)$ and figure out the coefficients by plugging in co-ordinate vector fields.
If $w=w_{A\bar{B}}dz^{A}\wedge d\bar z^{\bar B}$ then by plugging in the vectors $\dfrac{\partial}{\partial z^{a}},\;\dfrac{\partial}{\partial \bar z^{\bar b}}$ in the above definition, we get that 
$$w_{a\bar b}-w_{\bar b {a}}=ig_{a\bar b}$$
where $A,B$ goes through all indices, but $a,b$ takes only holomorphic indices (without bar). In the author's notation, they used antisymmetry of wedge product to combine $w_{a\bar b}dz^a\wedge d\bar z^{\bar b}+w_{\bar b {a}} d\bar z^{\bar b}\wedge dz^a$ into $(w_{a\bar b}-w_{\bar b {a}})dz^a\wedge d\bar z^{\bar b}$ and then used the formula above.
Also, what you have written about anti-symmetry of coefficients of forms is not true. It's true only if you have tensor products instead of wedges in that equation.
