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Context: auxiliary fields

I am new to the field of CFT (conformal field theory), and I am reading two articles, $H^+_3$ WZNW model from Liouville field theory, by Y. Hikida and V. Schomerus; and Free Field Approach to String Theory on AdS_3 by K. Hosomichi, K. Okuyama and Y. Satoh. I will name the former "article 1", and the latter "article 2".

I am specifically interested in the link between the $H_3^+$-WZW model and Liouville quantum gravity. These are two 2-dimensional theories linked at the level of their correlation functions (according to Wikipedia and these articles, among others). I am working on the Riemann sphere $\partial \equiv \mathbb{C}$, and $H_3^+ \equiv \text{SL}(2,\mathbb{C})/\text{SU}(2)$ is the symmetry group of the Wess-Zumino-Witten (WZW) model.

My problem is quite simple to state but longer to justify, so I will trust my calculations and give the results here. As you can see in eq(2.1) of the "article 1" and eq(3.1) of the "article 2", auxiliary fields $\beta$ and $\overline{\beta}$ are introduced. Ultimately, my question is "Why?".

Liouville without auxiliary fields

The WZW action is: \begin{align} &S_\text{WZW}[h]=\frac{k}{2\pi}\left[\int_\partial \text{tr}[\partial_z h \wedge \partial_{\overline{z}} h^{-1}]+ \frac{i}{6}\int_\text{B}\text{tr}[(h^{-1} d h)^{\wedge 3}]\right] \end{align} Upon decomposing the field $h\in H_3^+$ as \begin{equation} h=\left( \begin{matrix} 1 & \gamma \\ 0 & 1 \end{matrix} \right) \left( \begin{matrix} e^{-\phi} & 0 \\ 0 & e^{\phi} \end{matrix} \right) \left( \begin{matrix} 1 & 0 \\ \overline{\gamma} & 1 \end{matrix} \right), \end{equation} and writing $d^2z = dz\wedge d\overline{z}$, I obtain \begin{align} S_\text{WZW}[h] =& \frac{k}{2\pi}\int_\partial d^2z \left( \partial_z \phi \partial_{\overline{z}} \phi + \partial_z \overline{\gamma} \partial_{\overline{z}}\gamma e^{2\phi} \right). \end{align} And this is at this very step that "article 1" and "article 2" introduce their $\beta$, $\overline{\beta}$. But, what I find is that the partition function can be written as follows. \begin{align} Z =& \int \mathcal{D}\phi \mathcal{D}^2\gamma\,e^{-\frac{k}{2\pi}\int_\partial d^2z \left( \partial_z \phi \partial_{\overline{z}} \phi+\partial_z \overline{\gamma} \partial_{\overline{z}}\gamma e^{2\phi} \right)} \nonumber \\ =& \int \mathcal{D}\phi \mathcal{D}^2(e^\phi \gamma)\, e^{2J(g,\phi)} e^{-\frac{k}{2\pi}\int_\partial d^2z \left( \partial_z \phi \partial_{\overline{z}} \phi+\partial_z \overline{\gamma} \partial_{\overline{z}}\gamma e^{2\phi} \right)} \nonumber \\ =& \int \mathcal{D}\phi\, e^{2J(g,\phi)}\det(\Delta)^{-1}e^{-\frac{k}{2\pi}\int_\partial d^2z \, \partial_z \phi \partial_{\overline{z}} \phi} \end{align} With $\Delta = (\nabla_z+\partial_z\phi)(\nabla_{\overline{z}}-\partial_{\overline{z}}\phi)$. The Heat Kernel method gives for $\det(\Delta)^{-1}$: \begin{equation} \det(\Delta)^{-1} = C\,e^{\int_\mathbb{C} d^2z\sqrt{g}\left(\frac{1}{3\pi}\partial_z\phi\partial_{\overline{z}} \phi - \frac{1}{24\pi}\phi\,{}^{(2)}R \right)} \tag{1} \end{equation} (I am not that sure of this calculation). It is simply (3.9) of "article 2", but with a mysterious factor of 6 lacking in the exponent. I can adjust my method to include this factor, but it will be hard to justify. The Fujikawa method used to calculate the anomaly of the change in functional integration measure gives: \begin{equation} J(g,\phi) = -\frac{\Lambda}{8\pi} \int_\mathbb{C} d^2z \sqrt{g} e^{2 \phi}+ \ln(D) - \frac{1}{24\pi}\int_\mathbb{C} d^2z \sqrt{g} [ \phi R_g-8\partial_z \phi \partial_{\overline{z}} \phi] \tag{2} \end{equation} I am more confident in this result because I found it in D'hocker's article 2-D quantum gravity and Liouville theory (see eq(4.10)). Upon ignoring the constants $C$ and $D$ in $(1)$ and $(2)$, I find the partition function: \begin{equation} Z\propto \int \mathcal{D}\phi\,e^{-\frac{1}{2\pi}\int_\partial d^2z \sqrt{g} \left( (k-2)\partial_z \phi \partial_{\overline{z}} \phi + \frac{1}{4} \phi\,{}^{(2)}R + \frac{\Lambda}{2}e^{2\phi} \right)} \end{equation} And so the action is the following, after rescaling of $\phi$: \begin{equation} \frac{1}{2\pi} \int_\partial d^2z \sqrt{g} \left(\partial_z \phi \partial_{\overline{z}} \phi + \frac{Q_\phi}{4} \phi\,{}^{(2)}R + \lambda e^{2b\phi}\right) \end{equation} with $Q_\phi=b$, and $b^{-2}=k-2$. It is essentially (3.14) of "article 2" when integrating over the auxiliary fields.

Questions

Why use these $\beta$ and $\overline{\beta}$ in the first place if we can directly relate the two theories like so? And where does (2.1) of "article 1" come from? It refers to "article 2" and another one and none of these contain this action... I am quite confused as these two articles seem to be entirely based on their method.

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2 Answers 2

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Your calculation of the integral over $\gamma$ looks correct, but it works only for the partition function. For an $N$-point function you have additional dependence on $\gamma$ due to the vertex operators (2.2) of article 1. You could maybe adapt your method to this case. But Hikida and Schomerus found it simpler and more elegant to introduce the field $\beta$, and integrate over $\gamma$ then $\beta$. This $\beta$ is just what you need to make the Lagrangian linear in $\gamma$, when it was originally quadratic. It is natural to make the Lagrangian linear in $\gamma$ because the vertex operators are exponential of $\gamma$s.

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  • $\begingroup$ Thank you for your answer! So if I understand correctly, the purpose of introducing these auxiliary fields is to simplify the calculations of the correlations of the vertex operators. I have found a way to justify the action (2.1) of Article 1, I will write an answer about it and then accept yours. $\endgroup$ Dec 14, 2023 at 10:15
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About action (2.1) of the article 1:

I was troubled by the difference between the action I found and the action (2.1) of "article 1", but I managed to find a way to justify it by reverse-reasoning:

First, was have to use the measure $\mathcal{D}\phi\mathcal{D}^2(e^{-b\phi}\gamma)$ when taking the partition function. So, using action (2.1), we should re-derive the action of a free scalar field, after integrating out $\gamma$. This action (2.1) is: \begin{equation} S=\frac{1}{2\pi} \int_\mathbb{C} d^2z \sqrt{g} \left( \partial_z \phi \partial_{\overline{z}} \phi - \beta \partial_{\overline{z}} \gamma - \overline{\beta}\partial_z \overline{\gamma} + \frac{Q_\phi}{4} R\phi -b^2 \beta \overline{\beta} e^{2b \phi} \right) \end{equation} Integrating $\beta$ and $\overline{\beta}$ we find: \begin{equation} S=\frac{1}{2\pi} \int_\mathbb{C} d^2z \sqrt{g} \left( \partial_z \phi \partial_{\overline{z}} \phi + \frac{Q_\phi}{4} R \phi + b^{-2} \partial_z \overline{\gamma} \partial_{\overline{z}} \gamma e^{-2b \phi} \right) \end{equation} Then, integrating $e^{-b\phi}\gamma$ we obtain: \begin{equation} S=\frac{1}{2\pi} \int_\mathbb{C} d^2z \sqrt{g} \left( \partial_z \phi \partial_{\overline{z}} \phi + \frac{Q_\phi}{4} R \phi \right)+ \ln\det(\Delta_\alpha) \end{equation} With $\Delta_\alpha$ the same operator as $\Delta$ in my question, but with an additional overall factor $\alpha$ coming from a different choice in the energetic scale used in the zeta regularization of the Heat Kernel. Specifically, we have $\ln\det(\Delta) = -\alpha \int_\mathbb{C} d^2z \sqrt{g} \left( \frac{b}{3\pi} \partial_z \phi \partial_{\overline{z}} \phi-\frac{1}{24\pi}R\phi \right)$. We can fix $\alpha$ by constraining our result to be the action of a free scalar field. So, using the choice $\alpha=-3b$, we obtain: \begin{align} S =& \frac{1}{2\pi} \int_\mathbb{C} d^2z \sqrt{g}\left[ \left( \partial_z \phi \partial_{\overline{z}} \phi + \frac{Q_\phi}{4} R \phi \right)+\left(2b^2 \partial_z \phi \partial_{\overline{z}} \phi - \frac{b}{4} R \phi \right)\right] \nonumber \\ \stackrel{Q_\phi=b}{=}& \frac{1}{2\pi} \int_\mathbb{C} d^2z \sqrt{g} (1+2b^2) \partial_z \phi \partial_{\overline{z}} \phi \nonumber \\ \stackrel{\phi \leadsto \phi b^{-1}}{=}& \frac{1}{2\pi} \int_\mathbb{C} d^2z \sqrt{g} (b^{-2}+2) \partial_z \phi \partial_{\overline{z}} \phi \end{align} But we know that $b^{-2}=k-2$. So the result is consistent with a free scalar field.

So, using this reasoning backward and gluing it after my derivation of $Z$, we obtain a "derivation" of the action (2.1) of the article 1.

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