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, 2020 at 8:00
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    $\begingroup$ @FrancescoPolizzi You may download the book from your institute's library. $\endgroup$
    – XT Chen
    Dec 12, 2020 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. $\endgroup$ Dec 12, 2020 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, 2020 at 13:15
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    $\begingroup$ You have to be REALLY careful about conjugation, the powers of $i$ and signs of terms. I suggest you write everything out with the real and imaginary terms shown explicitly. If you do this a few times, you’ll learn how to use the briefer notation correctly or at least become better at checking your calculations. Unfortunately, as @DonuArapura said, this book is infamous for errors like this. $\endgroup$
    – Deane Yang
    Dec 12, 2020 at 14:48

1 Answer 1


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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