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Trivial Isometry Group for the Reals

  1. Sep 4, 2014 #1
    In the following stackexchange thread, the answerer says that there is a Riemannian metric on [itex]\mathbb{R}[/itex] such that the isometry group is trivial.

    http://math.stackexchange.com/questions/492892/isometry-group-of-a-manifold

    This does not seem correct to me, and I cannot follow what he is saying. Could someone please explain this to me?

    Thank you!
     
  2. jcsd
  3. Sep 5, 2014 #2
    No, studiosus spoke about half-line, e.g. open interval (0, +∞). You can’t shift it isometrically. Read math.SE answers with more attention, please.
     
  4. Sep 5, 2014 #3
    This is precisely what I don't follow. [itex](0,+\infty)\cong\mathbb{R}[/itex] as smooth manifolds and, using a diffeomorphism such as [itex]\varphi:\mathbb{R}\to(0,+\infty),~t\mapsto e^t[/itex], [itex](\mathbb{R},\phi^*\mathrm{g})\cong((0,+\infty),\mathrm{g})[/itex] as Riemannian manifolds. Don't isometric manifolds have isomorphic isometry groups?
     
  5. Sep 5, 2014 #4
    Diffeomorphic ℝ and (0, +∞) are. Isometric they are not, even up to (constant) rescaling. This is what you don’t follow, but you ought.
     
    Last edited: Sep 5, 2014
  6. Sep 5, 2014 #5

    WWGD

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    Like Incnis said (which refreshed my memory), for an isometry here, using ##\phi(t)=e^t ##, and the standard inner-product in ##\mathbb R ## as plain multiplication, we would need , for all pairs of Reals x,y,
    ## xy = e^x e^y ##. It may be interesting to show that there is no map ##\phi(t) ## which would give us an isometry; I guess for that we would need to use some invariants for Riemannian metrics, and show that one space satisfies the invariants but the other does not. I can't think of any at this point.

    Just wanted to refresh my knowledge; I know this is basic for all.
     
  7. Sep 5, 2014 #6
    Instead of continuing to make snide comments, perhaps you could work to uphold PF's values by actually explaining?

    Note that we are talking about Riemannian manifolds with the manifold being [itex]\mathbb{R}[/itex] and the metric being nonstandard. A map [itex]\varphi:(M,\mathrm{g})\to (N,\mathrm{g'})[/itex] is an isometry if and only if [itex]\varphi[/itex] is a diffeomorphism and [itex]\mathrm{g}=\varphi^*\mathrm{g}'[/itex]. If we have a diffeomorphism and define the Riemannian metric to be the pullback, then we automatically have an isometry.
     
  8. Sep 5, 2014 #7

    WWGD

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    But doesn't this imply that given any two diffeomorphic manifolds M,N with diffeomorphism ψ and g is the metric on M, then M,N are isometric by giving N the metric ψ*g ?
     
  9. Sep 5, 2014 #8
    Not quite. We're giving [itex]M[/itex] the pullback metric, so the original metric that we pull back from is the one on [itex]N[/itex].
     
  10. Sep 5, 2014 #9

    WWGD

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    But doesn't the same conclusion then follow? We assign to M the pullback metric ψ*g' given (N,g') and a homeomorphism ψ , then (M,ψ*g') and (N,g') are isometric if they are homeomorphic? Maybe I didn't follow you well, but this seems to imply that any two homeomorphic manifolds are isometric.
     
  11. Sep 5, 2014 #10
    First, yes. Sorry I misinterpreted; that was my fault. It does go both ways, you'd just have to take the inverse of the diffeomorphism. Being isometric is an equivalence relation, after all!

    Not just homeomorphic, though. Remember that an isometry must be a diffeomorphism.
     
  12. Sep 5, 2014 #11

    WWGD

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    I think Riemannian curvature tensor and Gaussian curvature are isometric invariants.
     
  13. Sep 5, 2014 #12
    See Theorema Egregium.

    How does this relate?
     
  14. Sep 5, 2014 #13

    WWGD

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    Sorry, just trying to see how to construct diffeomorphisms that are not isometries, by seeing if these diffeomorphisms preserve either the R. Curvature Tensor, and the Gaussian Curvature. I think every diffeomorphism is isotopic ( but not "diffeotopic", i.e., isotopic thru paths of diffeomorphisms ) to an isometry.
     
  15. Sep 5, 2014 #14
    I have figured it out. The only Riemannian automorphism (or endo-isometry or...better terminology) of [itex](\mathbb{R},(v_p,w_p)\mapsto e^{2p}v_pw_p)[/itex] is the identity. Thus, the isometry group is trivial.
     
  16. Sep 6, 2014 #15
    When a mathematician speaks about real half-line, or (0, +∞), another mathematician understands which metric (and Riemannian metric) are assumed: those inherited for a submanifold of ℝ. A non-crackpot mathematician won’t speak about “ℝ with non-standard metric”, except making special exercises. It defeats the purpose of thinking about such thing as about ℝ. A non-crackpot mathematician will speak as Ī said: ℝ and (0, +∞) are diffeomorphic, but not isometric. In other words: (non-Riemannian) differential geometry doesn’t see any difference between the two, whereas Riemannian geometry does.
     
    Last edited: Sep 6, 2014
  17. Sep 6, 2014 #16
    Yes, but I clearly spoke about looking for a Riemannian metric on [itex]\mathbb{R}[/itex] such that the isometry group is trivial. Thus, I am clearly not talking about the usual metric on [itex]\mathbb{R}[/itex].
     
  18. Sep 6, 2014 #17
    So, Ī failed to compose initial question and the text at math.SE into one picture (in other words, read the question poorly). IMHO now there is no more miscommunication.
     
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