# A contradiction caused by the Kähler identity and the formal adjoint relation

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?

• @FrancescoPolizzi In this book, the parenthesis product $(a,b)$ means $<a, \bar{b}>$. Dec 12, 2020 at 8:00
• 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. Dec 12, 2020 at 14:48
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.