I was reading Huybrechts complex geometry book,in page 28-29 there is a linear operator defined as follows $\mathbf{I}: \bigwedge^{*} V_{\mathbb{C}} \rightarrow \bigwedge^{*} V_{\mathbb{C}}$ such that:
$$\mathbf{I}=\sum_{p, q} i^{p-q} \cdot \Pi^{p, q}$$
with $\Pi^{p,q}$ is the natural projection on to the form of bidgree $(p,q)$.
Then in Lemma1.2.4,we try to show fundamental form $\omega(x,y) = <I(x),y>$ over Eucliean vector space $V$ with compatible almost complex structure $I$ is type (1,1).Using the arguement as below:
Since $$ (\mathbf{I} \omega)(v, w)=\omega(\mathbf{I}(v), \mathbf{I}(w))=\langle I(I(v)), I(w)\rangle=\omega(v, w) $$ one finds $\mathbf{I}(\omega)=\omega$, i.e. $\omega \in \Lambda^{1,1} V_{\mathbb{C}}^{*}$.
I don't really understant the $\mathbf{I}$,can someone explain this operator a little bit,why it can used to check the type of $\omega$?
I know since $\omega = \omega_{0,2}+\omega_{1,1} + \omega_{2,0}$ after taking $\mathbf{I}\omega = -\omega_{0,1} + \omega_{1,1} - \omega_{2,0}$ which means it only has $(1,1)$ component.
That's clear, since $I \omega = -\omega ^{2,0} - \omega^{0,2}+ \omega^{1,1}$ if $\omega = I \omega$ then $\omega^{2,0}+\omega^{0,2} = 0$ then one is $(2,0)$ type another is $(0,2) $ type, there are disjoint therefore both $\omega^{2,0} = 0 = \omega^{0,2}$.