Hello!


Let $M$ be an almost complex manifold. Let $TM$ denote its tangent bundle. Then we have the decomposition $TM\otimes\mathbb{C}=T^{1,0}M\oplus T^{0,1}M$ corresponding to the eigenvalues of the almost complex structure. This decomposition yields the decomposition:
$$
\Lambda^r(T^\star M\otimes\mathbb{C})=\Lambda^r(T^{1,0}M^\star\oplus T^{0,1}M^\star)=\bigoplus\_{p+q=r}\Lambda^p(T^{1,0}M^\star)\otimes\Lambda^q(\overline{T^{0,1}M}^\star)
$$
Now take a section $\omega$ of the complex vector bundle 
$$
\Lambda^{p,q}:=\Lambda^p(T^{1,0}M^\star)\otimes\Lambda^q(\overline{T^{0,1}M}^\star)
$$
$\omega$ is called a complex differential form of type $(p,q)$. Consider a complex $(p,q)$-form $\omega$ and take its differential. Its differential $\mathrm{d}\omega$ is a section of:
$$
\Lambda^{p+q+1}(T^\star M\otimes\mathbb{C})=\bigoplus\_{m+n=p+q+1}\Lambda^{m,n}
$$
Therefore $\mathrm{d}\omega$ can be decomposed in a sum of complex differential forms of type $(m,n)$ with $m+n=p+q+1$. However I have read that there are only four terms. My second question is:


*How do we prove that in fact $\mathrm{d}\omega$ is a section of: $$\Lambda^{p+2,q-1}\oplus\Lambda^{p+1,q}\oplus\Lambda^{p,q+1}\oplus\Lambda^{p-1,q+2}$$ only?*


I am aware that in the case where the almost complex structure is integrable we get only two terms such that finally we have $\mathrm{d}=\partial+\bar{\partial}$. But in fact it seems that in the almost complex case already we do not have so many terms (namely we have only 4 as above). I think this has something to do with the graduation of the algebra of differential forms and the nilpotence of the differential itself but I am not able to prove it.


At last, since I am interesting in the same kind of question concerning Lie and Courant algebroids, I was wondering if this fact could be recast in the language of homotopical algebras (by which I vaguely mean that usual identities on brackets hold up to something else)? This is because the algebra of differential forms is a supercommutative algebra and that we can reformulate $\mathrm{d}^2=0$ by $[\mathrm{d},\mathrm{d}]=0$. Could somebody point me toward an article?


Thank you very much!



Differential forms on an almost complex manifold
Call $C^{\infty}\_{p,q}(M)$ the space of smooth complex sections of the bundle $\Lambda^{p,q}T^\*\_M$ and let $2n$ be the real dimension of $M$. 


The fact that
$$
dC^{\infty}\_{p,q}(M)\subset C^{\infty}\_{p+2,q-1}(M)+C^{\infty}\_{p+1,q}(M)+C^{\infty}\_{p,q+1}(M)+C^{\infty}\_{p-1,q+2}(M)
$$
follows immediately from the two following facts:


1. The (bigraded) algebra $C^{\infty}\_{\bullet,\bullet}(M)=\bigoplus\_{p,q=0}^n C^{\infty}\_{p,q}(M)$ is locally generated by $C^{\infty}\_{0,0}(M)$, $C^{\infty}\_{1,0}(M)$ and $C^{\infty}\_{0,1}(M)$;
2. There are (obvious) inclusions
$$
dC^{\infty}\_{0,0}(M)\subset C^{\infty}\_{1,0}(M)+C^{\infty}\_{0,1}(M),
$$
$$
dC^{\infty}\_{1,0}(M)\subset C^{\infty}\_{2,0}(M)+C^{\infty}\_{1,1}(M)+C^{\infty}\_{0,2}(M),
$$
$$
dC^{\infty}\_{0,1}(M)\subset C^{\infty}\_{2,0}(M)+C^{\infty}\_{1,1}(M)+C^{\infty}\_{0,2}(M).
$$


Moreover, for an almost complex manifold $M$ with complex structure $J$, the following facts are equivalent:


* $J$ has no torsion (and thus, by Newlander-Nirenberg theorem $J$ is a true complex structure and $M$ a complex analytic manifold);
* $dC^{\infty}\_{1,0}(M)\subset C^{\infty}\_{2,0}(M)+C^{\infty}\_{1,1}(M)$
and $dC^{\infty}\_{0,1}(M)\subset C^{\infty}\_{1,1}(M)+C^{\infty}\_{0,2}(M)$;
* $dC^{\infty}\_{p,q}(M)\subset C^{\infty}\_{p+1,q}(M)+C^{\infty}\_{p,q+1}(M)$ for all $p,q=0,1,\dots,n$.


