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Showing that a two-dimensional Euclidean CFT ghost action is hermitian

Physics Asked on July 8, 2021

At the end of chapter 6 in Polchinski’s String Theory book he says that the $c$ ghost is anti-hermitian. With that information, I tried to show that the action for the $bc$ system

begin{equation}
S= frac{1}{2pi} int d^2z ; b bar{partial}c
end{equation}

is hermitian. We have

$$S^dagger = int d^2 z ;(b bar{partial}c)^dagger = int d^2 z; – partial c b = int d^2 z ; b partial c, $$
which is not the same as the original action that I started with because $bar{partial}^dagger = partial$.

Question: How do I show that the action for the $bc$ system is hermitian? Do I need to add the antiholomorphic part and define $b^dagger = tilde{b}$ and $c^dagger = – tilde{c}$? But this seems wrong to me because $b$ and $c$ should be functions of $z$ and $bar{z}$ for the action to make sense.

I am very confused and, for sure, missing something basic. I would like a very detailed answer so that I can learn how to take the hermitian conjugate in general for a Euclidean two-dimensional CFT.

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