A Diffeomorphic manifolds of equal constant curvature

C0nstantine

Summary
Every two semi-Riemannian manifolds of the same dimension, index and constant curvature are locally isometry. If they are diffeomorphic, are they also isometric?
Every two semi-Riemannian manifolds of the same dimension, index and constant curvature are locally isometric. If they are also diffeomorphic, are they also isometric?

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WWGD

Gold Member
I think the answer for the first is no, using an open interval (a,b) on the Real line and the entire Real line . Both same dimension, both zero curvature, diffeomorphic through a variant of arctan, but not isometric.

lavinia

Gold Member
Summary: Every two semi-Riemannian manifolds of the same dimension, index and constant curvature are locally isometry. If they are diffeomorphic, are they also isometric?

Every two semi-Riemannian manifolds of the same dimension, index and constant curvature are locally isometric. If they are also diffeomorphic, are they also isometric?
No. For instance, they could have different volumes.

WWGD

Gold Member
Just take any two equal manifolds ,deform one of them smoothly,diffeomorphically.

C0nstantine

Thank you both for your quick answers. Well, the problem I've come up with is that I have a semi-Riemannian manifold $(M,g)$ equipped with a unit vector field $u$ that satisfies the Frobenius integrability condition. The orthogonal hypersurfaces $\{\Sigma_t\}$ are of constant curvature, and I need to prove that the flow $\Phi_t$ of $u$ is a conformal map between every two surfaces $\Sigma_t$, $\Sigma_s$.

lavinia

Gold Member
Thank you both for your quick answers. Well, the problem I've come up with is that I have a semi-Riemannian manifold $(M,g)$ equipped with a unit vector field $u$ that satisfies the Frobenius integrability condition. The orthogonal hypersurfaces $\{\Sigma_t\}$ are of constant curvature, and I need to prove that the flow $\Phi_t$ of $u$ is a conformal map between every two surfaces $\Sigma_t$, $\Sigma_s$.
Can you prove that the orthogonal manifolds to the flow are even mapped into each other?

• C0nstantine

WWGD

Gold Member
Isnt the flow a path of diffeomorphisms? Isnt that enough?

C0nstantine

Can you prove that the orthogonal manifolds to the flow are even mapped into each other?
I did a calculation and it yields

$\mathcal{L}_u g\rvert_\tau = f(\tau) g\rvert_\tau$

where $g\rvert_\tau$ is the metric of $\Sigma_\tau$. I think that concludes the proof.

lavinia

Gold Member
I did a calculation and it yields

$\mathcal{L}_u g\rvert_\tau = f(\tau) g\rvert_\tau$

where $g\rvert_\tau$ is the metric of $\Sigma_\tau$. I think that concludes the proof.
A couple of things I don't understand:

- How do you show that the Lie derivative of the metric is a multiplemof the metric?

- Even if this is true, how do you know that the flow carries the foliation into itself?

- What does constant curvature of the manifolds in the foliation have to do with anything?

lavinia

Gold Member
Isnt the flow a path of diffeomorphisms? Isnt that enough?
It is true that the flow is by local diffeomorphisms. But the orthogonal planes are tangent to a foliation of the manifold. Why would the flow of the unit vector field move one manifold in the foliation onto another?

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C0nstantine

A couple of things I don't understand:

- How do you show that the Lie derivative of the metric is a multiplemof the metric?

- Even if this is true, how do you know that the flow carries the foliation into itself?

- What does constant curvature of the manifolds in the foliation have to do with anything?
It's not a "general" theorem I'm working on, it's actually a derivation of the FLRW metric. So, you have a lot of additional stuff simplifying the manifold structure. There's the assumption that $\Sigma_\tau$ is the set of points $p\in M$ such $t(p) = \tau$, and $u = \partial / \partial t$, so the flow of $u$ carries the foliation into itself. No...?

lavinia

Gold Member
It's not a "general" theorem I'm working on, it's actually a derivation of the FLRW metric. So, you have a lot of additional stuff simplifying the manifold structure. There's the assumption that $\Sigma_\tau$ is the set of points $p\in M$ such $t(p) = \tau$, and $u = \partial / \partial t$, so the flow of $u$ carries the foliation into itself. No...?
not sure as I am unfamiliar with this.

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mathwonk

Homework Helper
This is not my area, but my impression is that the various metrics that can be put on, say, a real torus, i.e. compact 2 manifold of genus one, correspond to the various complex structures that can be put on it. There are infinitely many of these, parametrized by conjugacy classes of lattice subgroups of the plane, which groups give rise to these tori via quotient action. They are represented by points of a certain famous region in the upper half plane, a sort of infinite triangle, i.e. with vertex at infinity, and base apparently a semi circle with ends on the real axis. This is discussed beautifully and in detail in the lovely book Geometries and Groups, by Nikulin and Shafarevich. I believe it can be shown then that there is exactly one geometry on the torus for each point of the complex plane, (essentially by means of the " j-invariant"). I think this is probably essentially correct, but I am not an expert. The conclusion is that there can be infinitely many different metric geometries on manifolds that are all diffeomorphic.

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• C0nstantine

C0nstantine

Thank you for the suggestion!

lavinia

Gold Member
It seems that you have a unit speed geodesic flow. I think a generalization of Gauss's theorem on geodesic normal coordinates says that such a flow carries the foliation into itself at least if the Space-Time is time orientable.

I think, if the flow maps $M_0$ into $M_1$ then the differential of the mapping $df(v)$ at a point $p$ is the value of a Jacobi field at $f(p)$ obtained from a variation through the geodesics emanating from any curve in $M_0$ whose derivative at $p$ is the tangent vector $v$.

I am unsure so far how to relate this to the sectional curvature of the leaves of the foliation. Maybe one can compute the differential in Fermi coordinates.

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lavinia

Gold Member
This is not my area, but my impression is that the various metrics that can be put on, say, a real torus, i.e. compact 2 manifold of genus one, correspond to the various complex structures that can be put on it. There are infinitely many of these, parametrized by conjugacy classes of lattice subgroups of the plane, which groups give rise to these tori via quotient action. They are represented by points of a certain famous region in the upper half plane, a sort of infinite triangle, i.e. with vertex at infinity, and base apparently a semi circle with ends on the real axis. This is discussed beautifully and in detail in the lovely book Geometries and Groups, by Nikulin and Shafarevich. I believe it can be shown then that there is exactly one geometry on the torus for each point of the complex plane, (essentially by means of the " j-invariant"). I think this is probably essentially correct, but I am not an expert. The conclusion is that there can be infinitely many different metric geometries on manifolds that are all diffeomorphic.
It seems like one can lift the metric on the torus to the plane then apply the Uniformization Theorem to show that the metric is conformally equvalent to a flat metric. One is left with conjugacy classes of lattices in the plane to distinguish the various non-isometric flat tori. Nu?

• mathwonk

mathwonk

Homework Helper
yes, that is my understanding.

"Diffeomorphic manifolds of equal constant curvature"

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