I'm considering the $4D$ de-Sitter spacetime, in static coordinates (I'm using $c = 1$ and $k_{\text{B}} = 1$): \begin{equation}\tag{1} ds^2 = (1 - \frac{\Lambda}{3} \, r^2) \, dt^2 - \frac{1}{1 - \frac{\Lambda}{3} \, r^2} \, dr^2 - r^2 \, d\Omega^2, \end{equation} where $\Lambda > 0$ is the cosmological constant. This spacetime has an horizon around any static observer, at $r = \ell_{\Lambda} \equiv \sqrt{3 / \Lambda}$. The whole space volume inside that horizon is easily calculated from the metric above (it is not $4 \pi \ell_{\Lambda}^3 / 3$): \begin{equation}\tag{2} \mathcal{V} = \pi^2 \ell_{\Lambda}^3, \end{equation} and the horizon area is $\mathcal{A} = 4 \pi \ell_{\Lambda}^2$. The vacuum has an energy density and pressure: \begin{align}\tag{3} \rho &= \frac{\Lambda}{8 \pi G}, & p &= -\, \rho. \end{align} Thus, the vacuum energy inside the whole volume of the observable de-Sitter universe is \begin{equation}\tag{4} E = \rho \, \mathcal{V} = \frac{3 \pi \ell_{\Lambda}}{8 G}. \end{equation} Note that the enthalpy is trivially 0 (what does that mean?): \begin{equation} H = E + p \mathcal{V} = 0. \end{equation}
I'm now considering the thermodynamic first law, comparing various de-Sitter universes which have slightly different $\Lambda$ (or $\ell_{\Lambda}$): \begin{equation}\tag{5} dE = T \, dS - p \, d\mathcal{V} = T \, dS + \rho \, d\mathcal{V}. \end{equation} Inserting (2) and (4) give the following: \begin{equation}\tag{6} T \, dS = -\, \frac{3 \pi}{4 G} \, d\ell_{\Lambda}. \end{equation} If $d\ell_{\Lambda} > 0$ and $dS > 0$, this implies a negative temperature! If I use the entropy $S = \mathcal{A}/ 4 G$ (take note that this entropy formula is very controversial for $\Lambda > 0$), then $dS = 2 \pi \ell_{\Lambda} \, d\ell_{\Lambda} / G$ and \begin{equation}\tag{7} T = -\, \frac{3}{8 \, \ell_{\Lambda}}. \end{equation} This result is puzzling!
I'm now wondering if the $T \, dS$ term would better be replaced with the work done by the surface tension on the horizon, instead: $T \, dS \; \Rightarrow \; -\, \tau \, d\mathcal{A}$ (I'm not sure of the proper sign in front of $\tau$). In this case, I get the tension of the horizon (I don't know if this makes any sense!): \begin{equation}\tag{8} \tau = \frac{3}{32 G \ell_{\Lambda}}. \end{equation} So is the reasoning above buggy? What is wrong with all this? Any reference that confirms that the de-Sitter Horizon's temperature could be negative, or that the entropy is really undefined there (or that $S = \mathcal{A} / 4 G$ is wrong in this case)? Or should the entropy term $T \, dS$ actually be interpreted as the tension work $-\, \tau \, d\mathcal{A}$ on the horizon instead?
In (4) and (5), is it legit to use the energy inside the horizon only, excluding the exterior part?
EDIT: The energy (4) is the energy of vacuum inside the horizon. It doesn't take into account the gravitational energy. I now believe that it's the Komar energy in the same volume that should be considered. The integration gives the following Komar energy inside the volume (2): \begin{equation}\tag{9} E_K = -\, \frac{\ell_{\Lambda}}{G}. \end{equation} But then, the trouble with the temperature is still the same: temperature is negative if $d\ell_{\Lambda} > 0$ (which is the same as $d\Lambda < 0$) and assume $dS > 0$ (or $S = \mathcal{A}/ 4 G$, which may be false for the de-Sitter spacetime).
