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I am terrifically confused by the notation in Polchinski's string theory book from chapter 3 to chapter 4. The ghost action of the bosonic string in conformal gauge is (3.3.24) $$S = \frac{1}{2 \pi} \int d^2 z (b_{zz} \partial_{\overline{z}}c^z + b_{\overline{z}\overline{z}}\partial_zc^{\overline{z}}).$$ In chapter 4, he drops the indices and starts to refer to $b$, $\tilde{b}$, $c$, and $\tilde{c}$. Am I supposed to identify: \begin{align} b(z) &= b_{zz} \\ \tilde{b}(\overline{z})&=b_{\overline{z}\overline{z}}\\ c(z) &= c^{\overline{z}} \\ \tilde{c}(\overline{z}) &= c^z? \end{align} Then, with this, do the transformations such as (4.3.1c) become \begin{align} \delta_B c &= i \epsilon(c \partial + \tilde{c}\overline{\partial})c &= i \epsilon c \partial c \\ \delta_B \tilde{c} &= i \epsilon(c \partial + \tilde{c}\overline{\partial})\tilde{c} &= i \epsilon \tilde{c} \overline{\partial} \tilde{c}? \end{align} If this is not right, what is the precise functional dependence of $b$, $\tilde{b}$, $c$, and $\tilde{c}$, and how do they relate to the fields $b_{zz}, \, c^z \, b_{\overline{z}\overline{z}}, \, c^{\overline{z}}$?

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    $\begingroup$ You have the correct identification up to normalizations which are irrelevant anyway. $\endgroup$
    – Prahar
    Jan 11, 2023 at 20:24
  • $\begingroup$ Thank you for the response @Prahar . I have to ask then, if c is holomorphic and $\overline{\partial}c = 0$ , was it redundant for Joe to include them in (4.3.1c) ? $\endgroup$
    – Diffycue
    Jan 11, 2023 at 20:51
  • $\begingroup$ @Prahar wait a second. If my identification is right then the first term in the lagrangian would be $b(z) \overline{\partial}\tilde{c}(\overline{z})$. But I don't like that because we should have a holomorphic $bc$ system and an antiholomorphic $\tilde{b}\tilde{c}$ system, (section 2.7, bc CFT Polchinski)? My choice seems to mix these systems. But then, I'm still in trouble by making the other choice, because if I let $c^z = c(z)$ then we have $\overline{\partial}c(z) = 0$ and the lagrangian becomes trivially 0. Sorry, I am really confused. $\endgroup$
    – Diffycue
    Jan 11, 2023 at 21:45
  • $\begingroup$ I'm sorry, I was too fast in my comment. Didn't read your identifications carefully. @ɪdɪətstrəʊlə's answer is absolutely correct. $\endgroup$
    – Prahar
    Jan 12, 2023 at 9:59

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You got your $c$'s mixed up. \begin{align} b(z,\bar{z})&:=b_{zz} \qquad &\bar{b}(z,\bar{z}) :=b_{\bar{z}\bar{z}}\\ c(z,\bar{z})&:=c^{z\phantom{z}} \qquad &\bar{c}(z,\bar{z}) :=c^{\bar{z}\phantom{\bar{z}}}\\ \partial &:=\partial_z \qquad &\bar{\partial}:=\partial_{\bar{z}}. \end{align} The action is $$S = \frac{1}{2\pi}\int\mathrm{d}^2 z\ \big(b\,\bar{\partial} c + \bar{b}\,\partial\,\bar{c}\big).$$ This is not trivially zero (in response to your comment in the original question), since $\bar{\partial}c=0$ is just the equation of motion for $c$; i.e. it is zero on-shell. This is common to all homogeneous actions (see e.g. this phys.SE answer).

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  • $\begingroup$ Thanks! One more point to get through my skull, re: "it is zero on-shell." I was worried that if $c$ is only a function of $z$ then $\overline{\partial}c(z) = 0$ trivially. But I had it the wrong way around-- it is the equation of motion that allows us to conclude that $c$ is holomorphic. (?) $\endgroup$
    – Diffycue
    Jan 12, 2023 at 1:13
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    $\begingroup$ @Diffycue indeed, a priori there is nothing to force the fields to be (anti-)holomorphic. It is the equations of motion that do so. $\endgroup$ Jan 12, 2023 at 7:36

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