entropy is the same as (\[m34entropy\]). Now, the scalar curvature $R(\tilde{h})$ becomes $$R(\tilde{h})= R(\tilde{q}) +
(D-4)(D-5)\left(\frac{\bar{k}}{r_0^2}\right)\, ,$$ where $R(\tilde{q})$ is the scalar curvature of the metric $\tilde{q}$. Thus the entropy (\[ent\]) has the form $$\mathfrak{S}=4\pi
(r_0^{D-4}\Omega_{D-4})V_2\left[1+2c_2(D-4)(D-5)\left(\frac{\bar{k}}{r_0^2}\right)\right]+
8\pi c_2 (r_0^{D-4}\Omega_{D-4}) \int
\sqrt{\tilde{q}}d^2zR(\tilde{q}) \, .$$ Using the explicit relation among $c_2$, $r_0$ and $D$ in equation (\[c0c2r0Dm4\]), and $\bar{k}=-1$, we see that only the last term remains $$\label{entropym4} \mathfrak{S}= 8\pi c_2 (r_0^{D-4}\Omega_{D-4})
\int \sqrt{\tilde{q}}d^2zR(\tilde{q}) \, .$$ In other words, in this case, the entropy is totally determined by the integration of the Euler characteristic $R(\tilde{q})$ on $\Sigma_2$, in agreement with (\[entropy1\]). Note that the the two-dimensional induced horizon $\tilde{q}$ is a constant curvature space with scalar curvature $2\delta$. The entropy can be further expressed as $$\mathfrak{S}= 64\pi^2 \cdot \delta \cdot c_2 \cdot
(r_0^{D-4}\Omega_{D-4})\, .$$

In fact, in the case of $m=4$, one has $w=1$, and $\ell$ in (\[3eq22\]) can take values $1$ and $2$. We have $A_1=0$ from (\[Alequation\]), while $A_2=c_2$. By the general entropy expression (\[3eq22\]), we arrive at the same result as Eq. (\[entropym4\]). As argued above, the constant entropy does not make sense in black hole thermodynamics, we should drop it and conclude that the physical entropy is zero.

