TITLE: Formula for decomposing a form into $(p,q)$ forms
QUESTION [12 upvotes]: Let $L: \mathbb{C}^n \to \mathbb{C}$ be a real linear map.  In other words, $L(a\vec{v}_1+b\vec{v_2}) = aL(\vec{v}_1)+bL(\vec{v}_2)$ for all $a,b \in \mathbb{R}$.  Then $L$ decomposes uniquely into a complex linear $T$ map and a complex antilinear map $\overline{T}$.  They have the formulas
$$
\begin{align*}
       T(\vec{v}) &= \frac{1}{2}\left( L(\vec{v}) - i L(i\vec{v})\right)
    \\ \overline{T}(\vec{v}) &= \frac{1}{2}\left( L(\vec{v}) + iL(i\vec{v})\right)
    \end{align*}
$$
A real $k-$linear form $\omega:(\mathbb{C}^n)^k \to \mathbb{C}$ can similarly be decomposed into a sum of forms $\omega^{(p,q)}$ for which $\omega^{(p,q)}(z\vec{v}_1,z\vec{v}_2,...,z\vec{v}_k) = z^p\overline{z}^q\omega^{(p,q)}(\vec{v}_1,\vec{v}_2,...,\vec{v}_k)$.
When this construction is applied to differential forms, we obtain the so called $(p,q)-$forms, which are very important in complex geometry.
I have a question of linear algebra, or maybe combinatorics.  What is a formula for $\omega^{(p,q)}$ in terms of $\omega$?  For example, I have figured out that 
$$
\omega^{(1,1)}(\vec{v}_1,\vec{v}_2) = \frac{1}{2}\left(\omega(\vec{v}_1,\vec{v}_2)+\omega(i\vec{v}_1,i\vec{v}_2)\right)
$$
I have a similar kind of formula for $\omega^{(2,0)}$, but it is rather ugly, and relies on getting the above formula first.  
Does anyone have a pretty formula for $\omega^{p,q}$ in terms of $\omega$?
Also, the proof that I know that this decomposition holds is heavily basis dependent.  Does anyone have a clean basis free proof?  Maybe someone with a better handle on tensor algebra can help m out here.
You may also want to look at this question
When is a $k$-form a $(p, q)$-form? for further background.

REPLY [6 votes]: I first describe a general construction of equivariant projections. Let $G$ be a finite abelian group, $V$ a (finite dimensional) complex vector space and $\rho: G\to GL(V)$ a representation. Since $G$ is abelian and $V$ is a complex vector space, the representation $\rho$ is "simultaneously diaginalizable'', i.e., there exist a finite list of (distinct) characters $\chi_1,...,\chi_k: G\to {\mathbb C}^*$ and a $G$-invariant direct sum decomposition
$$
V=\oplus_{j=1}^k V_{\chi_j}
$$ 
such that $G$ acts on each $V_{\chi_j}$ via the character $\chi_j$:
$$
\rho(g)(v)=\chi_j(g)v, \forall v\in V_{\chi_j}, j=1,...,k. 
$$
For each $\ell=1,...,k$, define the following $G$-equivariant projection:
$$
R_\ell: V\to V_{\chi_\ell}, R_\ell(v)= \frac{1}{|G|} \sum_{g\in G} \chi_\ell^{-1}(g)\rho(g)(v). 
$$
This projection fixes $V_{\chi_\ell}$ pointwise. Since $R_\ell$ is $G$-equivariant, it preserves each 
summand $V_{\chi_j}$ and, hence, vanishes on 
$$
\oplus_{j\ne \ell}^k V_{\chi_j}. 
$$
Now, we consider the case when $V$ is the complexification of the vector space $V_{\mathbb R}$ of real exterior forms $\phi: \wedge^n W\to {\mathbb R}$, where $W$ is a complex vector space. (A similar procedure will work for arbitrary tensors, but it will be a bit more complicated and, besides, I think, the motivation comes from differential forms on complex manifolds.) 
Define the group
$$
G=\oplus_{m=1}^n {\mathbb Z}_4.
$$ 
Its order is, of course, $4n$. I will denote the generator of the $m$-th direct summand of $G$ by $g_m$. I define a representation $\rho$ of $G$ on $V_{\mathbb R}$ (and, hence, on $V$), by
$$
\rho(g_m)(\phi(z_1,...z_m,...,z_n))= \phi(z_1,..., i z_m,..., z_n). 
$$
Then $V$ has the $G$-invariant decomposition
$$
\oplus_{p=0}^{n} V^{p,q},
$$
where $p+q=n$ and each $V^{p,q}$ is nothing but $V_{\chi_{p,q}}$, where
$$
\chi_{p,q}(g_m)= i, 1\le m\le p, 
$$
$$
\chi_{p,q}(g_m)= -i, p+1\le m\le p+q. 
$$
Here, because our forms are antisymmetric, I can assume without loss of generality that any $\omega\in V^{p,q}$ is complex-linear in the first $p$ variables and complex antilinear in the last $q$ variables (each variable, of course, is a vector in $W$). 
Thus, for each $(p,q)$, $p+q=n$, as above, we obtain the projection 
$$
R_{p,q}: V\to V^{p,q} ,
$$
whose kernel is
$$
\oplus_{r\ne p} V^{r,q}. 
$$
The projection is defined by the formula: 
$$
R_{p,q}(\omega)= \frac{1}{4n} \sum_{g\in G} \overline{\chi_{p,q}}(g)\rho(g)(\omega). 
$$
Now, if you wish, you can write down this in more details using definitions of $\chi_{p,q}$ and $\rho$ given above, but I find the above formula most transparent. One advantage is that the formula and its proof are independent of the choice of coordinates.
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