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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!

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See mathoverflow.net/questions/47082/… for a very similar question. –  David Speyer May 24 '11 at 19:57
    
@David Speyer: I have edited my question because the one you are talking about told me how to write such forms in local coordinates, thanks. –  Benjamin May 24 '11 at 20:56
4  
Just use induction on the total degree. It's obviously true for total degree 1. –  Lucio Guerberoff May 24 '11 at 21:06
    
@Lucio Guerberoff: Thank you! –  Benjamin May 29 '11 at 4:42

2 Answers 2

up vote 5 down vote accepted

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 $p,q=0,1,\dots,n$.
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Try Chern, Complex Manifolds without Potential Theory, p. 18.

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