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According this answer, the recent WMAP experiment has only shown that if our Universe has a spherical geometry, then it should have at least a $3\cdot 10^{11}$ light year big radius.

Now consider the possibility, if our Universe is a 4-sphere, thus it has a small, constant positive curvature.

It means, that we have a new symmetry. Translating any point of any system with $2{\pi}r$, we get the same system back. Note, it is a different thing as the common space translation symmetry (which results the impulse preservation):

  1. it is valid only for $2{\pi}r$ translations
  2. but, it is valid for any point of any system, not only for the whole system.

On Noether's theorem, every differentiable symmetry of an action has a corresponding conservation law.

What conservation law would correspond to this symmetry?

Extension/Fix:

As I understand @conifold 's answer, this is a discrete and not a continuous symmetry, because the translation is possible here only with $n\cdot 2\pi r (n \in \mathbb{Z})$, thus Noether's theorem here doesn't apply in its original form. But, according to this question, yes, there is something similar to Noether's theorem also on discrete symmetries. On the accepted (and bountied) answer, "For infinite symmetries like lattice translations the conserved quantity is continuous, albeit a periodic one." How does this apply in our case?

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  • $\begingroup$ Considering that translating anything with $10^{12}$ light years is impossible, this preservation law has probably no practical importance, but it still could exist. $\endgroup$
    – peterh
    Commented Mar 19, 2017 at 21:01
  • $\begingroup$ This wouldn't be an example of a differentiable symmetry so Noether's theorem would not be applicable. $\endgroup$
    – lemon
    Commented Mar 19, 2017 at 21:20
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    $\begingroup$ Comments to the post (v3): 1. Noether's theorem doesn't work for discrete symmetries, cf. e.g. this Phys.SE post. 2. For continuous/Killing symmetries, see physics.stackexchange.com/q/317946/2451 (with the Schwarzschild metric replaced by the FLRW metric). $\endgroup$
    – Qmechanic ♦
    Commented Mar 19, 2017 at 21:22
  • $\begingroup$ @Qmechanic The similarity is not clear to me, furthermore I can't see the answer, the corresponding law. $\endgroup$
    – peterh
    Commented Mar 19, 2017 at 21:37
  • $\begingroup$ @lemon This symmetry maps the $L(\textbf{q}, \dot{\textbf{q}}, t)$ to $L(\textbf{q} + (2\pi r, 0, 0, ...), \dot{\textbf{q}}, t)$. It looks to me very differentiable. $\endgroup$
    – peterh
    Commented Mar 20, 2017 at 0:28

3 Answers 3

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Your "translation of any point of any system with 2πr" can not be done for all points of the sphere simultaneously. Therefore it is not a symmetry in the sense of the Noether's theorem. I am guessing it refers to something like the full rotation of a 2-sphere around an axis, and you can already see from this example that you can not perform such a rotation on all points at once. Some move the full circle, others less, some not at all (poles). For the 3-sphere there may not be poles, but then there will be invariant circles, for the 4-sphere there will be poles again (this follows from existence of 1D or 2D invariant subspaces in real linear algebra).

But "the shape of the universe" being a sphere refers to a spacetime slice, not the whole spacetime, so it is a 3-sphere. It would be problematic for a 4D spacetime to be spherical even on the cyclic cosmology theories. They are also probing WMAP data for detecting other finite 3D space-forms, the quotients of the sphere by finite groups, see The Poincaré Dodecahedral Space and the Mystery of the Missing Fluctuations by Weeks.

Even if it did work globally "translation by 2πr" has no continuous parameter in it (r is fixed), so the Noether's theorem would still not apply. However, there is a shadow of it for discrete symmetries, involving conserved topological charges, which impose selection rules on various processes, see Is there something similar to Noether's theorem for discrete symmetries?

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  • $\begingroup$ 4-sphere means the surface of a 4D sphere. If its radius is enough big (in our case, at least $3\cdot{10}^{11}$ ly), then it is indistinguishable for us from a planar global geometry, according to the refered answer in the question. I.e. our 3D space can be the surface of a 4D sphere. It is all about the spacelike part. Your first sentence is unclear to me, why should be the whole system translatable? I think, any point of any system would be translatable with $2\pi r$ into any direction, it would be applying the same symmetry multiple times. $\endgroup$
    – peterh
    Commented Mar 20, 2017 at 0:13
  • $\begingroup$ Note: also the space mirror symmetry doesn't have any continuous parameter. It maps $L(\textbf{q}, \dot{\textbf{q}}, t)$ to $L(-\textbf{q}, -\dot{\textbf{q}}, t)$. The only parameter what I can see here is $-1$. $\endgroup$
    – peterh
    Commented Mar 20, 2017 at 0:48
  • $\begingroup$ As I can understand, Noether's theorem prescribes differentiability, i.e. the $L(\textbf{q}, \dot{\textbf{q}}, t) \rightarrow L'(\textbf{q}, \dot{\textbf{q}}, t)$ should be a differentiable function. There is nothing about continuous parameters. And this translation symmetry is obviously differentiable. $\endgroup$
    – peterh
    Commented Mar 20, 2017 at 1:01
  • $\begingroup$ @peterh Standard terminology is to give the dimension of the manifold, not of what it is the boundary of, see n-sphere, so 4-sphere is 4D. Symmetry has to be continuous because otherwise Noether's generators are not even locally defined, and differentiability only makes sense for a continuous parameter, not discrete. It has to apply to the whole manifold because otherwise they are not defined globally. Classical Noether's theorem does not apply to mirror symmetry. $\endgroup$
    – Conifold
    Commented Mar 20, 2017 at 18:04
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I am truly not an expert on this, but my superficial impression is that preservation theorems correspond to local (infinitesimal) symmetries, while you are talking about a (slightly doubtful) global symmetry? (Doubtful insofar as you would assume a perfectly constant curvature, which seems a rather unphysical assumption?)

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  • $\begingroup$ Also I am not sure if it is a differentiable symmetry. What makes a symmetry differentiable? Translating any point of any system with any small - differentiable - length would preserve this symmetry. Constant curvature and spherical geometry can be the result of homogenity and isotropy, what we experience everywhere. Maybe the question would look better as "What additional conservation law would we have in a static, 4-sphere Universe?" $\endgroup$
    – peterh
    Commented Mar 19, 2017 at 21:31
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In short, this symmetry isn't a differentiable symmetry in the sense as it would be affected by the Noether Theorem.

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