From the $N=1$ superconformal algebra OPEs (A.44), (A.55), (A.56) one obtains the (anti-)commutation algebra relations,

$$[L_m, L_n] = (m-n) L_{m+n} + \frac{c}{12} m(m^2-1) \delta_{m+n,0}, \qquad (A.60)$$

$$\{G_r, G_s\} = 2L_{r+s} + \frac{c}{12}(4r^2-1)\delta_{r+s,0}, \qquad (A.61)$$

$$[L_m, G_r] = \frac{m-2r}{2} G_{m+r}. \qquad (A.62)$$

Expanding the total Kac-Moody currents as$^{50}$

$$j^M = \sum_n \frac{j_n^M}{z^{n+1}}, \qquad j^w = \sum_n \frac{j_n^w}{z^{n+1}}, \qquad (A.63)$$

and the fermions as

$$\psi^M = \sum_{r \in \mathbb{Z}+\frac{1}{2}} \frac{\psi_r^M}{z^{r+\frac{1}{2}}}, \qquad \psi^w = \sum_{r \in \mathbb{Z}+\frac{1}{2}} \frac{\psi_r^w}{z^{r+\frac{1}{2}}}, \qquad (A.64)$$

where $M = a, a', i$ labels the SL(2, $\mathbb{R}$) polarizations $a = 1, 2, 3$, SU(2) polarizations $a' = 1, 2, 3$, and $\mathbb{T}^4$ polarizations $z^i, i = 1, 2, 3, 4$, one obtains from the OPEs (A.50), (A.54)

$$\begin{aligned} [L_m, j_n^M] &= -n j_{m+n}^M, & [L_m, j_n^w] &= -n j_{m+n}^w, \\ [L_m, \psi_r^M] &= -\frac{m+r}{2} \psi_{m+r}^M, & [L_m, \psi_r^w] &= -\frac{m+r}{2} j_{m+r}^w, \end{aligned} \qquad (A.65)$$

and from the OPEs (A.57) we obtain

$$\begin{aligned} \{G_r, j_n^M\} &= -n \psi_{n+r}^M, & \{G_r, j_n^w\} &= -n \psi_{n+r}^w, \\ \{G_r, \psi_s^M\} &= j_{r+s}^M, & \{G_r, \psi_s^w\} &= j_{r+s}^w. \end{aligned} \qquad (A.66)$$

Using the OPE expressions (3.45), (3.50), (6.24), we obtain (here $\delta^{+-} = \delta^{-+} = 2$)

$$\begin{aligned} \{\psi_r^a, \psi_s^b\} &= \frac{k}{2} \delta^{ab} \delta_{r+s,0}, & \{\psi_r^{\prime a}, \psi_s^{\prime b}\} &= \frac{k}{2} \delta^{ab} \delta_{r+s,0}, \\ \{\psi_r^x, \psi_s^x\} &= \frac{1}{2} \delta_{r+s,0}, & \{\psi_r^w, \psi_s^w\} &= -\frac{1}{2} \delta_{r+s,0}, \end{aligned} \qquad (A.67)$$

and similarly for the anti-holomorphic sector.

## B GHY term in string frame

In this appendix, we are going to derive the GHY boundary term in string frame, that needs to be added to the action (3.60). To derive it, we first re-write the action (3.60) in Einstein frame, and identify the boundary terms necessary to allow for a well-defined variational principal. We will identify two contributing boundary terms: the usual GHY action in Einstein frame, and the extra term necessary to offset the boundary contributions due to

<footnote>$^{50}$For abelian currents the total Kac-Moody current coincides with its purely bosonic theory version, $j^x = j^x$, $j^w = j^w$, $j^i = j^i$, $i = 1, 2, 3, 4$.</footnote>
the dilaton total derivative term. Finally, we will combine these two terms and re-write them in terms of metric in string frame.

Expressing the Ricci scalar $\tilde{R}$ in the Einstein frame in terms of its counterpart $R$ in the string frame for the general transformation $G_{\mu\nu} = e^{2\Omega} \tilde{G}_{\mu\nu}$ we obtain in D-dimensional space-time

$$R = e^{-2\Omega} (\tilde{R} - 2(D-1)\nabla^2\Omega - (D-2)(D-1)(\partial\Omega)^2). \qquad (B.1)$$

Fixing $D=10$, $\Omega = (\Phi - \Phi_0)/4$, we obtain

$$S = \frac{1}{2\kappa^2} \int d^6x \tilde{V}_{\mathbb{T}^4} \sqrt{-\det \tilde{G}} \left( \tilde{R} - \frac{9}{2} (\partial\Phi)^2 - \frac{9}{2} \tilde{\nabla}^2 \Phi - \frac{1}{12} e^{\Phi - \Phi_0} \mathcal{H}_{\mu\nu\lambda} \mathcal{H}^{\mu\nu\lambda} \right), \qquad (B.2)$$

where now contraction of indices is done with the Einstein metric tensor, and tilde on top of the covariant derivative indicates that it is calculated in Einstein frame. Here we can extract the pure boundary term, at $\chi = \Lambda$, using the Stokes's theorem,

$$S \supset -\frac{9}{4\kappa^2} \int d^5x \sqrt{-\det \tilde{\gamma}} \tilde{n}_M \partial^M \Phi|_{\chi=\Lambda}, \qquad (B.3)$$

where $\tilde{\gamma}_{\mu\nu}$ is the boundary metric in Einstein frame, and

$$\tilde{n}_M = \left(0, \sqrt{\tilde{G}_{\chi\chi}}, 0, 0, 0, 0, 0, 0, 0\right) \qquad (B.4)$$

is an outward-pointing vector normal to the boundary. The gravity action in string frame is to be supplemented with the GHY boundary term,

$$S_{\text{GHY}} = \frac{1}{\kappa^2} \int d^5x \tilde{V}_{\mathbb{T}^4} \sqrt{-\det \tilde{\gamma}} \tilde{\nabla}_M \tilde{n}^M, \qquad (B.5)$$

The GHY term in Einstein frame needs to be supplemented with an extra contribution, to compensate for (B.3). In total, we obtain the boundary term in Einstein frame

$$S_b = \frac{1}{\kappa^2} \int d^5x \sqrt{-\det \tilde{\gamma}} \tilde{V}_{\mathbb{T}^4} \left( \tilde{\nabla}_M \tilde{n}^M + \frac{9}{4} \tilde{n}^\chi \partial_\chi \Phi \right) |_{\chi=\Lambda}. \qquad (B.6)$$

On the other hand, in string frame we the outward-point unit normal to the boundary

$$n^M = \left(0, \sqrt{\tilde{G}^{\chi\chi}}, 0, 0, 0, 0, 0, 0, 0\right) = e^{\frac{\Phi_0 - \Phi}{4}} \tilde{n}^M. \qquad (B.7)$$

Using the relation between Christoffel symbols,

$$\Gamma^M_{M\chi} = \tilde{\Gamma}^M_{M\chi} + \frac{5}{2} \partial_\chi \Phi, \qquad (B.8)$$

we obtain

$$\nabla_M n^M = e^{\frac{\Phi_0 - \Phi}{4}} \left( \tilde{\nabla}_M \tilde{n}^M + \frac{9}{4} \tilde{n}^\chi \partial_\chi \Phi \right) \qquad (B.9)$$

Introducing the boundary metric in string frame, $\gamma_{\mu\nu} = e^{\frac{\Phi_0 - \Phi}{2}} \tilde{\gamma}_{\mu\nu}$, and combining everything together, we can re-write the total boundary term (B.6) as

$$S_b = \frac{1}{\kappa_0^2} \int d^5x e^{-2\Phi} \sqrt{-\det \gamma} V_{\mathbb{T}^4} \nabla_M n^M. \qquad (B.10)$$

The action (B.10) therefore gives the GHY term in string frame. In the main text we will denote (B.10) as $S_{\text{GHY}}$, having in mind that it is the full boundary term in the string frame.
## C Thermodynamics of SL(2, $\mathbb{R}$) × U(1)/U(1)

In this appendix, we are going to demonstrate that our new results for thermodynamics of the coset geometry (3.1) reproduce the known results for the

$$ \frac{\mathrm{SL}(2, \mathbb{R}) \times \mathrm{U}(1)}{\mathrm{U}(1)} \times \mathrm{SU}(2) \times \mathbb{T}^4 \qquad (C.1) $$

background, that corresponds to the limit of un-gauged SU(2) sub-group.<fnref n="51" /> According to (3.12), this limit can be obtained by setting $a_2 = b_2 = 0$. The anomaly-free conditions (3.24) then impose two constraints $a_1^2 + a_3^2 = 1$, $b_1^2 + b_3^2 = 1$ on the four parameters of the model, $a_{1,3}$, $b_{1,3}$. The remaining two independent parameters can be expressed in terms of the angles $\psi \in [0, \pi/2]$, $\chi \in [0, \pi/2]$, as follows [51]

$$ a_1 = \cos(\chi - \psi), \quad b_1 = \cos \chi, \quad a_3 = \sin(\chi - \psi), \quad b_3 = \sin \chi. \qquad (C.2) $$

The same parameters $\psi$, $\chi$ are also used in the conventions of [32]. In the particular case of [30, 31, 50] one sets $\chi = 0$.<fnref n="52" />

We begin by writing down expression for the temperature (4.4) for the choice of parameters (C.2),

$$ T = \frac{1}{\pi \sqrt{k} \ell_s} \frac{\cos \chi \cos(\chi - \psi)}{\cos \chi + \cos(\chi - \psi)}. \qquad (C.3) $$

In particular, setting $\chi = 0$, (C.3) is seen to agree with the known result for the coset (C.1), that can be found, e.g., in eq. (2.26) of [50].

The entropy of (C.1) can be found in eq. (3.31) of [51]<fnref n="53" />

$$ S = \pi \sqrt{k} \ell_s \left( \sqrt{M^2 - Q_L^2} + \sqrt{M^2 - Q_R^2} \right), \qquad (C.4) $$

where $M$ is the ADM mass, and the $U(1)_{L,R}$ charges are given by (C.16) of [51],<fnref n="54" />

$$ Q_L = M \sin(\psi - \chi), \quad Q_R = M \sin \chi. \qquad (C.5) $$

Combining (C.3), (C.4), (C.5), we then obtain

$$ T S = M \cos \chi \cos(\chi - \psi). \qquad (C.6) $$

This agrees with (4.10) for the choice of parameters (C.2), provided $M = \frac{4\pi^2}{\kappa^2} k^2 \ell_s^2 V_{S1} V_{T4}$. The latter indeed matches our result for the energy, (3.111).

<footnote n="51">⁵¹See [38, 75] for discussion of related thermodynamic systems with microscopic description of NS5-F1 system and single-trace $T\bar{T}$ deformation. See also the seven-parameter black hole solution in [76] with $\lambda_{\pm} = 0$.</footnote>

<footnote n="52">⁵²In conventions of [50] one also sets $\alpha \to \psi$.</footnote>

<footnote n="53">⁵³The extra factor of $1/\sqrt{2}$ relative to that equation is necessary once we relax the convention of $\alpha' = \ell_s^2 = 2$.</footnote>

<footnote n="54">⁵⁴In particular, in case of $\chi = 0$, we recover (2.30) of [50], with $Q_L = Q$, $Q_R = 0$. In the latter, we substitute the mass $M$ and charge $Q$ of the black hole, determined by eqs. (34), (36) of [30]. Specifically, one plugs $M = m e^{-2\phi_0}$, $Q = q e^{-2\phi_0}$ into $S = \pi \sqrt{k} \ell_s (M + \sqrt{M^2 - Q^2})$, where $m = \frac{1}{2} \frac{1+p^2}{1-p^2}$, $q = \frac{p}{1-p^2}$, $p = \tan(\psi/2)$, and $\phi_0 = \Phi_0 + \frac{1}{2} \log \frac{1+p^2}{2(1-p^2)}$, expressed in terms of asymptotic value of the flat space dilaton $\Phi_0 = \log(g_s)$ via (29), (35) of [30].</footnote>
The free energy is determined by

$$ \mathcal{F} = M - TS - \mu_L Q_L - \mu_R Q_R, \qquad (C.7) $$

where $\mu_{L,R}$ are the left- and right-moving components of the chemical potential. As pointed out in [50], the free energy of the theory (C.1) vanishes, which is in agreement with our result (4.24) for (3.1) obtained independently of the choice of the anomaly-free gauging parameters $a_i, b_i$. We can in fact match expressions for the free energy term by term. We have already matched $M$ and $TS$ with our results in the limit of un-gauged SU(2). Now let us match $\mu_L Q_L + \mu_R Q_R$. The chemical potential can be found from the asymptotic behavior of the gauge field components$^{55}$

$$ A_t^{G,B} = \pm \sqrt{2} \mu^{G,B} + \dots, \qquad (C.8) $$

where the gauge fields can be found in eq. (C.9) of [51]. Using eq. (C.9) of [51], we first determine the relation $y = \frac{t\sqrt{2}}{\sec\chi+\sec(\chi-\psi)}$ between the time coordinate $y$, used in that equation, and the time coordinate $t$, featuring asymptotic behavior of the metric $g_{tt} = -1$. We then obtain $\mu_G = \sec(\chi - \frac{\psi}{2})\sin(\frac{\psi}{2})$, $\mu_B = \cos\chi\tan(\frac{\psi}{2}) - \sin\chi$. Consequently

$$ \begin{aligned} \mu_L &= \frac{\mu_G + \mu_B}{2} = -\frac{1}{2} \cos(\chi) \sec\left(\frac{\psi}{2}\right) \sin(\chi - \psi) \sec\left(\chi - \frac{\psi}{2}\right), \\ \mu_R &= \frac{\mu_G - \mu_B}{2} = \frac{1}{2} \sin(\chi) \sec\left(\frac{\psi}{2}\right) \cos(\chi - \psi) \sec\left(\chi - \frac{\psi}{2}\right). \end{aligned} \qquad (C.9) $$

We then obtain

$$ \mu_L Q_L + \mu_R Q_R = \frac{M}{2} (2 - \cos(2\chi - \psi) - \cos\psi). \qquad (C.10) $$

At the same time, from (4.27), using (3.111), (C.2) we obtain

$$ \Omega \cdot \mathcal{J} = \mathcal{E}(1 - \cos(\chi - \psi) \cos \chi), \qquad (C.11) $$

that indeed agrees with (C.10) for $\mathcal{E} = M$. This completes the term-by-term match of our result for the free energy for the choice of parameters (C.2) with the known result for the (C.1).

**Open Access.** This article is distributed under the terms of the Creative Commons Attribution License (CC-BY 4.0), which permits any use, distribution and reproduction in any medium, provided the original author(s) and source are credited. SCOAP$^3$ supports the goals of the International Year of Basic Sciences for Sustainable Development.

<footnote>$^{55}$Again, the factor of $1/\sqrt{2}$ appears when we relax the convention of $\alpha' = \ell_s^2 = 2 \rightarrow \ell_s = \sqrt{2}$ used in [51]. This in turn rescales the time coordinate measured in units of $\ell_s$ by the factor of $1/\sqrt{2}$.</footnote>
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