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I found a contradiction in the Principle of Algebraic Geometry by G&H, section 1.2. I have post this on MSE but it didn't get enough attention. I couldn't sleep or eat or do anything else due to this contradiction. Orz.

Assume $\eta$ is a harmonic form, $L$ is multiplying by Kähler and $\Lambda$ its adjoint. Then we have $$(L\Lambda \eta , \eta) = (\Lambda \eta , \Lambda \eta) \ge 0.$$ But as in the Principles of Algebraic Geometry by Griffiths and Harris, p. 154, in the proof of the Kodaira Vanishing, we have $$ (L \Lambda \eta , \eta) = i/2\pi (\Theta \Lambda \eta , \eta) = i/2\pi (D' \bar{\partial} \Lambda \eta ,\eta) $$ since $\Theta = D'D'' + D''D'$, $\bar{\partial} = D''$, $(\bar{\partial} x , \eta) = (x, \bar{\partial}^* \eta)$ and $\bar{\partial}^*\eta = 0$ since $\eta$ is a harmonic form. By the famous identity $[ \Lambda , \bar{\partial}] = -iD'^*$ we have $$ \begin{split} i/2\pi (D' \bar{\partial} \Lambda \eta ,\eta) & = i/2\pi (D' (\Lambda \bar{\partial}+iD'^*) \eta ,\eta) \\ & = -1/2\pi (D'D'^* \eta ,\eta) \\ & =- 1/2\pi (D'^* \eta , D'^* \eta) \le 0. \end{split} $$

We get two inequality of different directions. It would implies $\Lambda \eta = 0$. What's wrong here?

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    $\begingroup$ @FrancescoPolizzi In this book, the parenthesis product $(a,b)$ means $<a, \bar{b}>$. $\endgroup$ – XT Chen Dec 12 '20 at 8:00
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    $\begingroup$ @FrancescoPolizzi You may download the book from your institute's library. $\endgroup$ – XT Chen Dec 12 '20 at 8:05
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    $\begingroup$ With no disrespect to G or H, their book is known to have many mistakes. So I wouldn't lose sleep over it. I suppose that there must be a sign inconsistency in the parts you were reading. $\endgroup$ – Donu Arapura Dec 12 '20 at 13:12
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    $\begingroup$ @DonuArapura It is not only about G&H. The nightmare is when I refer to the Demailly's note, I get the same conclusion. $\endgroup$ – XT Chen Dec 12 '20 at 13:15
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    $\begingroup$ @DeaneYang If you don't believe in G&H, you should (maybe) believe in Demailly. This identity is the same as the one from the Demailly's note. The first equality in this post is just the definition of adjoint, without any possible confusion. $\endgroup$ – XT Chen Dec 12 '20 at 15:05
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I can give a explanation. Noting that in this question $\eta$ is a harmonic form in $\Omega^{p,q}(L)$ where $L$ is a positive line bundle. And harmony means $\Delta_{\bar{\partial}} \eta =0$.

First of all, we find that $[\Delta , L] = i(\bar{\partial} D' + D' \bar{\partial})$ which is not necessary zero, which is how everything behaves in $\Omega(L)$, unlike in $\Omega$. This means though $\eta$ is harmonic in $\Omega(L)$, $L\eta$ is not necessary harmonic. So there is no way to conclude $L \eta = 0$ for all harmonic $\eta$ from the fact $\Lambda \eta = 0$ for all harmonic $\eta$. So things is not absurd enough. Though $\Lambda \eta = 0$ is already very astonishing.

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