The physics answer can be seen by looking at the action including a cosmological constant
$$
S = \frac{\Lambda c^4}{8\pi G} \int d^4 x \sqrt{-g} + \frac{c^4}{16\pi G}\int d^4 x \sqrt{-g} R + S_{\rm matter} + \cdots
$$
where $c$ is the speed of light, $G$ is Newton's constant, $\Lambda$ is the cosmological constant, $S_{\rm matter}$ contains any matter fields (standard model particles), and the $\cdots$ allows for higher order terms.
The cosmological constant term is the first one, $\sim \Lambda \int d^4 x \sqrt{-g}$. You can see that if the metric $g$ is a dynamical field, meaning that we should compute its equation of motion and solve for it (like we do in general relativity), then there will be a term in the equations of motion proportional to $\Lambda$. Therefore, the "constant" term in the energy contributes to the dynamics of the system through the metric. On the other hand, if we fix the metric (which is what we do in special relativity), then $\Lambda$ will not appear in any of the equations of motion for the matter fields, and so the "constant" term in the energy does not contribute to the dynamics of the system.
I am not sure what you are looking for by a "group theory explanation." I think what you want to see is a smooth limit where the energy transitions from being a "1-d vector" to being a "torsor." I don't know if such a limit makes sense. But, at a looser, less precise level, what you can say is that if you make $G$ smaller and smaller (so gravity becomes less and less important), then it becomes harder and harder to see a cosmological constant (or, in other words, you need a larger and larger cosmological constant before you can observe any effect, for a fixed observational precision). As a crude model, say there is some $\Lambda_{\rm min}$ that is the minimum cosmological constant you can observe. If $\Lambda < \Lambda_{\rm min}$, then the constant term effectively doesn't contribute to any observations, and you can treat energy "like a torsor" (meaning, adding constants to the energy won't matter, so long as you stay below $\Lambda_{\rm min}$. If $\Lambda > \Lambda_{\rm min}$, then energy behaves "like a 1D vector", in that the actual value of the energy does matter (not just the value up to an overall constant). Then in the limit that gravity becomes unimportant ($G$ becomes large), $\Lambda_{\rm min}$ becomes larger and larger until you are always in the "torsor like" regime.
Well, that's actually the limit of a fixed metric, where gravity isn't present. You actually asked about Newtonian gravity, where the gravitational field is present but simply small. This case is actually a bit subtle. Without matter, the normal equation for the Newtonian potential $\phi$, without matter, is Laplace's equation
$$
\nabla^2 \phi = 0
$$
If we take the Newtonian limit $c\rightarrow \infty$, and assume the background spacetime is Minkowski spacetime (eg, we are working in approximately special relativity with small gravity effects), and ignore the matter action, we can identify $g_{00}\approx -c^2 - 2\phi$ (as well as $g_{0i}\approx 0$ and $g_{ij}\approx \delta_{ij}$), and Laplace's equation get's modified.
What's going on is that the cosmological constant introduces a non-trivial gravitational background, and we strictly speaking can't expand around Minkowski space, because Minkowski isn't a solution of the background equations of motion. One way to think of it is that with a cosmological constant, we shouldn't require the normal asymptotic boundary conditions where $\phi$ falls off to $0$ at large $r$, but rather allow it to grow, to allow for the fact that the gravitational field grows at large distances. The most rigorous way to handle this situation to do is to include $\Lambda$ in the background equations of motion, in which case we will expand around de Sitter space (for a positive cosmological constant, or anti-de Sitter space for a negative cosmological constant) instead of Minkowski space, and then $\phi$ will obey an analogue of Laplace's equation suitable for de Sitter space (assuming a positive cosmological constant; anti-de Sitter space if the CC is negative). However, if we think of the cosmological constant as being "small", we can drop the extra cosmological constant term in Laplace's equation, and recover normal Newtonian gravity where the absolute value of the energy doesn't matter -- essentially "zooming in" to the region well within the horizon in de Sitter space.
The above description is a pretty condensed and informal description of some rigorous calculations that you can find in the literature. The punchline is that for a small cosmological constant, if we're only interested in "local" observations (like the solar system), where the cosmological horizon is irrelevant, then at least as far as the cosmological constant goes, the Newtonian limit behaves a lot like the limit of a fixed metric where the gravitational field is kind of like a matter field that simply decouples from the cosmological constant. This decoupling is what allows energy to move from "vector like" to "torsor like".
This post imported from StackExchange Physics at 2025-01-21 21:34 (UTC), posted by SE-user Andrew
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