- #1

Pond Dragon

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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!

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- Thread starter Pond Dragon
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- #1

Pond Dragon

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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

Incnis Mrsi

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- #3

Pond Dragon

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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?half-line, e.g. open interval (0, +∞). You can’t shift it isometrically. Read math.SE answers with more attention, please.

- #4

Incnis Mrsi

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Diffeomorphic ℝ and (0, +∞) are. *Isometric* they are not, even up to (constant) rescaling. This is what you don’t follow, but you ought.

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- #5

WWGD

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## 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.

- #6

Pond Dragon

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Instead of continuing to make snide comments, perhaps you could work to uphold PF's values by actually explaining?Diffeomorphic ℝ and (0, +∞) are.Isometricthey are not, even up to (constant) rescaling.This is what you don’t follow, but you ought.

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

- #7

WWGD

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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 anisometryif 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.

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 ?

- #8

Pond Dragon

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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].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

WWGD

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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].

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.

- #10

Pond Dragon

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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!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?

Not just homeomorphic, though. Remember that an isometry must be a diffeomorphism.

- #11

WWGD

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I think Riemannian curvature tensor and Gaussian curvature are isometric invariants.

- #12

Pond Dragon

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See Theorema Egregium.I think Riemannian curvature tensor and Gaussian curvature are isometric invariants.

How does this relate?

- #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.

- #14

Pond Dragon

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- #15

Incnis Mrsi

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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.

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- #16

Pond Dragon

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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].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.

- #17

Incnis Mrsi

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