I am following this set of lecture notes. On page 3, the author is considering a metric on a complex Kahler manifold of dimension $n$, which is denoted by $g_{\alpha, \bar{\beta}} = \partial_{\alpha}\partial_{\bar{\beta}} K(\vec{z}, \vec{\bar{z}})$. Immediately afterwards, the author writes down a (1,1)-form given by $J = g_{\alpha, \bar{\beta}} dz^{\alpha} \wedge d\bar{z}^{\bar{\beta}}$, which he shows is a $(1,1)$ form belonging to $H^{(1,1)}$ cohomology class, and subsequently defines a Kahler class.
My question is this, the metric is symmetric in its indices, so then how can it be used to construct a (1,1)-form, which is antisymmetric in its indices? What am I missing here? Is the antisymmetrization independently in the $z$'s and the $\bar{z}$'s and hence this is alright?
Best Answer
Keep in mind that your $J$ is a product of the symmetric metric by the antisymmetric (1,1)-forms $dz^\alpha\wedge d\bar{z}^\beta$. A symmetric thing times an antisymmetric thing is antisymmetric. So, for example, \begin{align*} J(X,Y)=\left(g_{\alpha\bar{\beta}}\,dz^\alpha\wedge d\bar{z}^\beta\right)(X,Y) = g_{\alpha\bar{\beta}}\left( dz^\alpha(X)d\bar{z}^\beta(Y) - dz^\alpha(Y)d\bar{z}^\beta(X)\right) \end{align*} is clearly antisymmetric in $(X,Y)$. To see the indices of $J$ explicitly, compute: \begin{align*} J_{i\bar{j}} = J(\partial_i,\partial_{\bar{j}}) = g_{k\bar{\ell}}(\delta_{ik}\delta_{j\ell}-\delta_{i\ell}\delta_{jk}), \end{align*} where $\delta$ denotes the Kronecker delta. Now it's again explicit that $J_{i\bar{j}}=-J_{j\bar{i}}$.
Just to note, your $J$ is usually called the Kahler form and denoted $\omega$. People usually reserve $J$ for the (almost-) complex structure.