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My question is regarding the lemma on page 81 of the book by Griffiths and Harris.

The lemma says the following: A $\bar{\partial}$-closed form $\psi\in Z^{p,q}_{\bar{\partial}}(M)$ is of minimal norm in $\psi + \bar{\partial} A^{p,q-1}(M)$ iff $\bar{\partial}^* \psi = 0 $.

Now in the first direction of the equivalence they take $\eta \in A^{p,q-1}(M)$ s.t $\bar{\partial} \eta = 0$; and then they calculate $\| \psi + \bar{\partial}\eta \|^2$; somewhere along the calculations they get because that $\bar{\partial}^* \psi = 0$ that $$\| \psi + \bar{\partial} \eta \|^2 = \| \psi \|^2 + \| \bar{\partial}\eta\|^2 > \|\psi \|^2$$

But obviously if $\bar{\partial} \eta = 0$ then also $\| \bar{\partial} \eta \|^2 = 0 $; Am I missing something here?

I am asking here and not in MSE since its an advanced graduate book, I have slim chances of getting answers over there.

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    $\begingroup$ Yeah, this is a typo in the book, just remove the words "with $\overline{\partial}\eta=0$" from the text, and the argument works. $\endgroup$
    – YangMills
    Commented Mar 15, 2016 at 20:17
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    $\begingroup$ I can assure you this question would have received an answer on MSE. $\endgroup$
    – Michael Albanese
    Commented Mar 15, 2016 at 20:37
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    $\begingroup$ perhaps change the condition "with ∂¯¯¯η=0" , to "with ∂¯¯¯η ≠ 0"? $\endgroup$
    – roy smith
    Commented Mar 16, 2016 at 17:33
  • $\begingroup$ @roy smith: right, since they put a strict inequality $\endgroup$
    – YangMills
    Commented Mar 16, 2016 at 22:52

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