Solvable question of dee dee bar lemma 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?
 A: 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
