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Let $X^n$ be a compact complex manifold, and $\omega$ be a Hermitian metric on $X$.

Define an operator $P:=i\Lambda_\omega \bar{\partial} \partial$ on the space of the smooth function $C^\infty(X, \mathbb{C})$, where $\Lambda_\omega$ is the dual operator of $L_{\omega} = \omega \wedge \cdot$.

It is clear that $P$ is an elliptic operator and not self-adjoint in general.

In fact, $P^\ast$ (w.r.t $L^2$-inner product) can be expressed by $P^\ast = \frac{i}{(n-1)!} \ast_{\omega} \bar{\partial} \partial L^{n-1}_{\omega}.$

In particular, the second principle symbol of $P$ is equal to the second principle symbol of $\Delta_{\bar{\partial}}$.

Combining this result and the theory of Fredholm operators provides that the $\mathrm{ind}(P)=0$.

Applying maximum principle and calculating locally, $\mathrm{ker}(P)=\mathbb{C}$, and function $f \in \mathrm{im}(P|_{C^\infty(X,\mathbb{R})})$ are not non-positive or not non-negative (ie. not constant sign) other than the zero function.

For the adjoint operator $P^\ast$, the dimension of the kernel can be obtained by $$\mathrm{dim}\, \mathrm{ker} (P^\ast) = \mathrm{dim}\, \mathrm{coker} (P) = \mathrm{dim}\, \mathrm{ker} (P) = 1.$$

I would like to show that every real smooth function $f \in \mathrm{ker} (P^\ast|_{C^\infty(X, \mathrm{R})})$ are always non-positive or non-negative.

It is obvious to see that $\mathrm{ker} (P^\ast)$ is orthogonal complement to $\mathrm{im} (P)$ in the $L^2$-inner product.

In the appendix of the Kobayashi-Hitchin correspondence, Lübke and Teleman said that the orthogonal decomposition of $C^\infty(X, \mathbb{R}) = \mathrm{ker} (P^\ast|_{C^\infty(X, \mathrm{R})}) \oplus \mathrm{im} (P^\ast|_{C^\infty(X, \mathrm{R})})$ would provide that the result I want.

How can I gain that every real smooth function $f \in \mathrm{ker} (P^\ast|_{C^\infty(X, \mathrm{R})})$ are always non-positive or non-negative by the previous statement?

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  • $\begingroup$ This statement is proved in Gauduchon - Le théorème de l'excentricité nulle, C. R. Acad. Sci. Paris 285 (1977), 387-390, Lemme 1. $\endgroup$ – user48958 Jan 21 '18 at 14:59
  • $\begingroup$ Is there any access that provides the article? I have tried any method that I can do... $\endgroup$ – Pan Jan 22 '18 at 2:38
  • $\begingroup$ Gauduchon's article is available here: gallica.bnf.fr/ark:/12148/bpt6k5620896n/f93.item $\endgroup$ – j.c. Jan 22 '18 at 13:38
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Here is the proof in Gauduchon's paper. Denote by $((.,.))$ the scalar product on various $L^2$ spaces.

Remark: If $f\in Ker(P^*)$, $f$ not identically zero, then $((f,1))\neq 0$. Indeed, if $((f,1))=0$, then $1\perp Ker(P^*)$, that is $1\in Im(P)$, therefore there exists $g\in {\mathcal C}^{\infty}(X, {\mathbb R})$ such that $P(g)=1$, and this is impossible since $g$ must have an absolute maximum on $X$, and at that point $P(g)\leq 0$.

If $f\in Ker(P^*)\setminus \{0\}$, we want to show that either $f\leq 0$ or else $f\geq 0$ so suppose that $f$ takes both negative and positive values on $X$. Then it is easy to find $h\in {\mathcal C}^{\infty}(X, {\mathbb R})$, $h>0$ such that $((hf,1))=0$.

If we consider the metric $h\omega$ which is conformal to $\omega$, and do all of the above with respect to $h\omega$, then it is easy to see that $Ker(P^*_h)$ is generated by $h^{1-n}f$ and that $((h^{1-n}f,1))_h=((hf,1))$ (here the index $h$ means with respect to the metric $h\omega$). Since $((hf,1))=0$, it follows that $((h^{1-n}f,1))_h=0$, and this contradicts the Remark above for the metric $h\omega$.

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