Recently I read about the dee dee bar lemma ($\partial\bar \partial$-lemma) in Gang Tian's *Canonical metrics in Kähler Geometry*. In the middle of Page 16, the author writes that: "The following equation can be solved for $u$,
$$\bar\partial^\*\theta=\bar\partial^\*\bar\partial u=-g^{i\bar j}u\_{\bar j i}$$
because $\int\_M\bar\partial^\*\theta\omega^n\_g=0$."


However, I can not understand why this condition guarantees this equation to be solvable? Even I know for standard $R^n$, a form $\alpha$ is a $df$ iff $\int\_{c}\alpha=0$... So what is the relationship between them?



Solvable question of dee dee bar lemma
Firstly, I think you meant to write $$\bar\partial^\* \theta = \bar\partial^\*\partial u = -g^{i\bar j}u\_{i\bar j}$$


On Kahler manifolds the R.H.S is just the laplacian of $u$. So basically you want to solve $$-\Delta u = f$$ whenever $\int\limits\_{M}f\omega^n = 0$. Here $f = \bar\partial^\*\theta$ in your question. This is actually quite standard. The integral vanishing is ofcourse necessary by Stokes theorem. One transparent way to see that a solution must exist is via an eigenvalue expansion. Arguing formally (these statements can be made rigorous, consult any book on elliptic PDEs), $-\Delta$ has a discrete spectrum $0=\lambda\_0\leq\lambda\_1\leq\cdots\leq\lambda\_k\cdots$, with corresponding eigenfunctions $\varphi\_0 \equiv 1, \varphi\_k$. Then $f$ has an eigenfunction expansion $$f=\mu\_0 + \sum\_{k=1}^{\infty}\mu\_k\varphi\_k$$ $\int\limits\_{M}f\omega^n = 0$ is equivalent to $\mu\_0 = 0$. So then the function $$u = \sum\_{k=1}^{\infty}\frac{\mu\_k}{\lambda\_k}\varphi\_k$$ is the solution you seek. To make all this rigorous you basically need to prove that $(-\Delta + I)^{-1}$ is a compact self adjoint operator on $L^2(M,\omega)$. An excellent online reference for PDEs in geometry is Jerry Kazdan's book - <http://www.math.upenn.edu/~kazdan/japan/japan.pdf>


