Negative temperature of the de-Sitter horizon?

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 , dOmega^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 , dmathcal{V} = T , dS + rho , dmathcal{V}.
end{equation}
Inserting (2) and (4) give the following:
begin{equation}tag{6}
T , dS = -, frac{3 pi}{4 G} , dell_{Lambda}.
end{equation}
If $dell_{Lambda} > 0$ 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 $dS = 2 pi ell_{Lambda} , dell_{Lambda} / G$ 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: $T , dS ; Rightarrow ; -, tau , dmathcal{A}$ (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 $-, tau , dmathcal{A}$ on the horizon instead?

In (4) and (5), is it legit to use the energy inside the horizon only, excluding the exterior part?

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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 $dell_{Lambda} > 0$ (which is the same as $dLambda < 0$) and assume $dS > 0$ (or $S = mathcal{A}/ 4 G$, which may be false for the de-Sitter spacetime).