Formula for decomposing a form into $(p,q)$ forms

Question

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.

Answer 1

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.

Answer 2

While @Moishe Kohan's answer is certainly helpful and insightful, it has one detail that I find somewhat confusing. Namely, the need to assume that a $(p,q)$ form is linear in the first $p$ variables and antilinear in the last $q$ variables. So, the following is a variation based on OP's definition of a $(p,q)$ form where all variables of a differential form are treated the same. I only came up with it after reading @Moishe's answer.

Let $\mathbb{C}^*$ act on $k$-forms as in @Sam's answer to this post. That is, $$z \star \omega (v_1, \dots, v_k) = \omega(z\cdot v_1, \dots, z\cdot v_k).$$ Then for a form $\omega$ of type $(p,q)$ and $z \in \mathbb{C}$ with $|z| = 1$, we have $z \star \omega = z^{p-q} \omega$.

Write $\alpha = e^{i \frac{\pi}{2k}}$. Fix $p$ and $q$ with $p + q = k$ and write $\beta = \alpha ^{q - p}$. So, using Moishe's argument with the group $G$ consisting of the $4k$th roots of unity, the $(p, q)$ component of the $k$-form $\omega$ is given by $$\omega^{(p, q)} = \frac{1}{4k} \sum_{j = 1}^{4k} \beta^j \cdot \alpha^j \star \omega.$$

The above group $G$ could be replaced by any other subgroup of the unit circle that has "enough elements" in the sense that it acts differently on each one of the spaces $\Lambda^{(p,q)}$. If, for some purpose, one prefers a continuous integral over a finite sum, one could replace $G$ by the unit circle itself, $S^1$, and get $$\omega^{(p,q)} = \frac{1}{2\pi}\int_{S^1} \zeta^{q-p} \cdot \zeta \star \omega d\zeta.$$