decomposition of differential from in (1,1) tpye Suppose $X$ a complex manifold with hermitian metric $g$ and complex structure $J$ over its tangent bundle.
We define a $(1,1)$ from $\omega$ as $\omega(v_1, v_2)=g(v_1,Jv_2)$.
My question is, how to prove that $\Omega^{1,1}=\Omega^{1,1}_0\oplus \Omega^0\omega$. Here $\Omega^{1,1}_0$ consist of all forms with $(1,1)-$type orthogonal to $\omega$ pointwisely. $\Omega^0$ means all smooth function on $X$.
 A: This is more or less just saying that a symmetric matrix can be decomposed as a sum of a trace-free matrix and a multiple of the identity matrix - you can prove this at the level of stalks, where you can take $g$ to be the identity matrix.
To give some more detail, assume that the complex dimension is $n$, fix a basis $(e_a)_{1\leq a\leq n}$ for $T^{1,0}M$ with dual basis $(\varepsilon^a)_{1\leq a\leq n}$. For any $(1,1)$-form $\alpha\in\Omega^{1,1}$, consider the decomposition
$$\alpha=\alpha-\frac{1}{n}\Lambda_\omega\alpha\,\omega+\frac{1}{n}\Lambda_\omega\alpha\,\omega,$$
where $\Lambda_\omega\alpha=-\sqrt{-1}g^{a\bar{b}}\alpha_{a\bar{b}}$ is the contraction with $\omega$ (a sum over repeated indices is implied). It is clear that $\Lambda_\omega\alpha\,\omega$ is a multiple of $\omega$, so we should check that $\alpha-\frac{1}{n}\Lambda_\omega\alpha\,\omega$ is orthogonal to $\omega$:
\begin{equation*}
\begin{split}
g\left(\omega\,,\,\alpha-\frac{1}{n}\Lambda_\omega\alpha\,\omega\right)=&\sqrt{-1}\,g_{a\bar{b}}\alpha_{c\bar{d}}g^{a\bar{d}}g^{c\bar{b}}+\frac{1}{n}\Lambda_\omega\alpha\,g_{a\bar{b}}g_{c\bar{d}}g^{a\bar{d}}g^{c\bar{b}}=\\
=&-\Lambda_\omega\alpha+\Lambda_\omega\alpha.
\end{split}
\end{equation*}
If instead of $(1,1)$-forms you would like to work with symmetric matrices, just consider instead of $\alpha$ the symmetric $2$-tensor defined by $A(v,w):=\alpha(v,Jw)$ for every $v,w\in TM$. Then the decomposition is precisely what I mentioned earlier - trace-free part plus constant multiple of the identity matrix.
