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 \, 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).
 A: The future cosmic Event horizon is the source of de Sitter (aka cosmic Hawking) radiation, also characterised by a specific temperature, the de Sitter temperature $T$ (as per the OP). It is the minimum possible temperature of the universe.
To an observer in our universe, a de Sitter Universe is in their infinite future, i.e. when the Hubble sphere and Event horizon are coincident.
Now, we can assign the de Sitter minimum length as $l_Λ=2$ and de Sitter $Λ=3/4$ in natural units. If you don’t like this, no matter, just stick with the symbolic equations.
Unlike a Schwarzschild black hole solution, the de Sitter solution has a non-zero pressure. So, the following by the OP are correct:

*

*having the PV term in equation (5)

*the entropy expression, i.e. $S=A/4G=π.l_Λ^2=4π$

*energy density and pressure in (3)

However, because (4) is an expression of the horizon energy $E_H$ the  relevant volume is not (2) rather it is the so-called areal volume (page 6) which is $V=4πl_Λ^3/3$. Then, the energy is:
$$E_H=U= ρV=(l_Λ^3/6).Λ= (4/3).Λ=1 (Eqn.4)$$
The energy of the horizon equals the energy in the bulk, as per the holographic principle so:
$$TS= ρV=1 (Eqn.4b)$$
$$T.4π=  (l_Λ^3/6).Λ$$
$$T=  (l_Λ^3/24π).Λ=1/4π=1/(2π.l_Λ )$$
Giving the de Sitter temperature $T$ as expected (Page 3, i.e. Gibbons and Hawking, 1977). Or equivalently:
$$T=  (1/2π).√(Λ/3)=  H_o/2π$$
The thermodynamic first law:
$$TS-E=pV (Eqn.5)$$
$$E= TS- pV$$
$$E=2TS=2$$
This is the maximum mass-energy of the de Sitter observable universe, and we have also found the universal relation $E=2TS$ as per Padmanabhan (page 42).  This result also corresponds with Boehmer & Harko (page 3) mass-energy of an observable universe (natural units):
$$m_P.E.c^2=(c^4/G) √(3/Λ)=E=2 (Eqn.5b)$$
Finally, yes, the enthalpy $H$ is indeed zero for a de Sitter universe. This means de Sitter space is unstable, as is known, and so spontaneously (no magician needed) created a rabbit (our Universe).
Free energy $G=H-TS= -TS=-1$
