The cosmological constant as a Lagrange multiplier? The cosmological constant $\Lambda$ can be introduced into the gravitational action like this :
\begin{equation}
S = \frac{1}{2 \kappa} \int_{\Omega} (R - 2 \Lambda) \sqrt{-g} \; d^4 x + \text{matter terms}.
\end{equation}
The spacetime region $\Omega$ is arbitrary here.  Now, what amaze me is that we can also write this :
\begin{equation}\tag{1}
-\: \frac{\Lambda}{8 \pi G} \int_{\Omega} \sqrt{-g} \; d^4 x = -\: \frac{\Lambda \, \mathcal{V}_4}{8 \pi G},
\end{equation}
where $\mathcal{V}_4$ is the 4-volume of the spacetime region $\Omega$.  So, the cosmological constant $\Lambda$ may be interpreted as the conjugate "variable" to $\mathcal{V}_4$, and as a Lagrange multiplier associated to a 4-volume constraint.  We could suppose that since $\mathcal{V}_4$ should be very large and the action $S$ "reasonable", then $\Lambda$ should be small.  My intuition tells me that there should be an equation like this one :
\begin{equation}\tag{2}
\Lambda \, \mathcal{V}_4 \sim \Lambda_{\text{max}} \mathcal{V}_{\text{min}},
\end{equation}
where $\mathcal{V}_{\text{min}}$ is the smallest 4-volume that is physically meaningfull ; $\mathcal{V}_{\text{min}} \approx \ell_{\text{P}}^4$ ($\ell_{\text{P}}$ is the Planck length), and $\Lambda_{\text{max}} \approx \ell_{\text{P}}^{-2}$ is the "natural" value associated to the quantum vacuum.  We then get
\begin{equation}\tag{3}
\Lambda \sim \frac{\ell_{\text{P}}^2}{\mathcal{V}_4},
\end{equation}
which is thus very small.  The relation $\Lambda \, \mathcal{V}_4 \approx \ell_{\text{P}}^2 \propto \hbar$ is also similar to an Heisenberg uncertainty relation ; $\Delta t \, \Delta E \ge \hbar$, which isn't surprising since the cosmological constant is introduced at the level of the action !
Can we make the previous idea more "rigorous" ?  Does it make sense to interpret $\Lambda$ as a Lagrange multiplier associated to a constrained 4-volume introduced into the action ?
If the universe is spatially closed ($k = 1$) and also closed in time (especially if $\Lambda$ was negative), then the 4-volume of the whole universe would be finite.
Any opinion on this ?
 A: Just a partial and naive "answer", using the uncertainty principle.
An observer makes an energy mesurement in some empty volume $V_3$ during a time intervall $\Delta t$.  According to Heisenberg uncertainty principle, he wil get an uncertainty $\Delta E$ on the energy mesurement :
\begin{align}
\Delta t \; \Delta E &\approx \Delta t \; \rho_{\text{vac}} \, \Delta V_3 \\[12pt]
&= \Delta t \; \frac{\Lambda \, c^4}{8 \pi G} \; \Delta V_3 \\[12pt]
&= \frac{\Lambda \; \Delta\mathcal{V}_4 \; c^3}{8 \pi G} \ge \frac{\hbar}{2},
\end{align}
where I have inserted the uncertainty on 4-volume of the spacetime region the observer is studying ;  $\Delta\mathcal{V}_4 = c \, \Delta t \; \Delta V_3$.  Thus
\begin{equation}\tag{1}
\Lambda \; \Delta \mathcal{V}_4 \ge 4 \pi \, \ell_{\text{P}}^2.
\end{equation}
Apparently, the cosmological constant $\Lambda$ depends on the size of the spacetime region that is sampled.  This is weird !
We could also invert the result, saying that given some experimental value of $\Lambda$, then the accesible portion of spacetime to any observer would have an uncertainty constrained by
\begin{equation}\tag{2}
\Delta\mathcal{V}_4 \ge \frac{4 \pi \, \ell_{\text{P}}^2}{\Lambda_{\text{exp}}}.
\end{equation}
The Planck length is $\ell_{\text{P}} \approx 1.6 \times 10^{-35} \text{m}$.  The current value of the cosmological constant is $\Lambda_{\text{exp}} \sim 10^{-52} \text{m}^{-2}$, which gives the smallest uncertainty on the 4-volume :
\begin{equation}\tag{3}
\Delta \mathcal{V}_{4 \, \text{min}} \sim 3 \times 10^{-17} \text{m}^4 \sim (0.0757 \text{mm})^4.
\end{equation}
(Note : the 4-volume of our observable universe is $\mathcal{V}_4 = c \, \Delta t \; \mathcal{V}_3 \sim (5 \times 10^{26} \text{m})^4$)
The problem with this "answer" is that it doesn't say why $\Lambda$ could be interpreted as a Lagrange multiplier to be added to the total action of the universe.
