Diffeomorphic manifolds of equal constant curvature

[C0nstantine]

TL;DR

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?

 

[WWGD]

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]

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]

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]

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?

 

[WWGD]

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]

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]

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?

 

[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]

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.

 

[mathwonk]

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.

 

[C0nstantine]

Thank you for the suggestion!

 

[lavinia]

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.

 

[lavinia]

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]

yes, that is my understanding.

 

[zinq]

A nice counterexample to the original question is found among various symmetric tori T². Any parallelogram in the plane, once its opposite sides are identified with each other, gives a specific Riemannian T² with curvature everywhere equal to 0. Occasionally different parallelograms give isometric tori, but usually not. For a simple example, use a square for the first torus and a non-square rectangle for the second one. A more interesting case is to identify the opposite edges of a regular hexagon to get the second torus. Even if their areas are equal, they won't be isometric. (Exercise: Which parallelograms result in the torus identical to the regular hexagon example?)

Question: Is there an example of constant Gaussian curvature K ≠ 0 where two compact diffeomorphic surfaces both having that constant curvature and also having the same area, are not isometric? (Or a case where two compact diffeomorphic n-manifolds, both with the same constant sectional curvature K ≠ 0, and both with the same n-dimensional volume, are not isometric?) I suspect there are no such examples, but don't have a proof.

 

[lavinia]

Question: Is there an example of constant Gaussian curvature K ≠ 0 where two compact diffeomorphic surfaces both having that constant curvature and also having the same area, are not isometric? (Or a case where two compact diffeomorphic n-manifolds, both with the same constant sectional curvature K ≠ 0, and both with the same n-dimensional volume, are not isometric?) I suspect there are no such examples, but don't have a proof.

This is not an area that I have studied but I believe that every Riemann surface of genus at least 2 admits a metric of constant negative curvature equal to -1. All surfaces with the same genus and curvature -1 have the same area.

A Riemann surface is a smooth 2 dimensional manifold with a conformal structure. There are many conformal structures on any smooth surface of genus 2 or higher. So your question boils down to whether two isometric compact Riemann surfaces must have the same conformal structure.

 

[zinq]

Yes, of course — good point! All compact orientable surfaces can be given a Riemannian structure with constant (Gaussian) curvature equal to +1 (for the sphere), 0 (the torus), or -1 (surfaces of higher genus).

In fact, the Gauss-Bonnet theorem implies that all compact surfaces M with the same topology and the same constant curvature must necessarily have the same area:

∫ K dA = K A = 2π χ(M)

where the integral is taken over the surface M. Here K is the constant curvature, A is the area of the surface M, and χ(M) denotes its Euler characteristic, which depends only on its topology.

[Correction: This is valid only when the constant curvature K is nonzero, for then we have

A = 2π χ(M) / K.

But when K = 0 the topology and curvature do *not* determine the area. As is easily seen with a small square or a large square with opposite sides identified to make two tori with distinct areas (that happen to be conformally equivalent).]

But because all topological surfaces (other than the sphere, the plane, and the projective plane) have infinitely many inequivalent conformal structures, it follows that — despite the fact that the topology, the constant Gaussian curvature, and the area are identical — there are examples with no isometry between them. Maybe the easiest examples are the square torus and the hexagonal torus (defined by identifying opposite edges of a square or regular hexagon, respectively) — and then adjusted to have equal areas.

Now for higher dimensions: It is almost clear that a similar argument works for any-dimensional torus with all sectional curvatures equal to 0. Not sure about other manifolds of constant sectional curvature.

—————
Topologically equivalent surfaces of equal constant Gaussian curvature and the same area that are not isometric also exist for almost all non-orientable and/or non-compact surfaces as well, with the only two exceptions being 1) the projective plane P², because it's covered by the sphere, and 2) the plane. For, the universal covering space must be the complex plane, the hyperbolic plane, or (only in the case of P²) the sphere.

 

[mathwonk]

One way to "see" many conformal structures on a surface of genus g, is to cut it up along closed curves that render it planar, thus obtaining a sphere with 2g slits, two slits for each of the g loops that were cut open. Each of these slits has a center on the sphere and a length, thus three real parameters, hence this configuration has 6g parameters. The conformal group of the sphere has only 6 (real) parameters, so the family of all conformal structures on a surface of genus g, i.e. the family of all 2g-slitted spheres modulo conformal isomorphism, has 6g-6 real parameters.

 

[zinq]

mathwonk — that "adjustment" can be done for surfaces of curvature = 0, since in that case the Gauss-Bonnet theorem says only that K A = 0.

Which shows that I neglected to exclude that case where I used the Gauss-Bonnet theorem. (If I can still edit it to correct that mistake, I will.)

 

[mathwonk]

@ zinq: nice point! Having at first misread your post, I have also edited my comment to provide more useful content.

 

[zinq]

"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 {Σt} are of constant curvature, and I need to prove that the flow Φt of u is a conformal map between every two surfaces Σt, Σs."

Consider for example the unit circle C in the xy-plane in 3-space R³. Now imagine all the tori T_r of revolution, with each T_r at a constant distance r from C, for 0 < r < 1. These are nested in each other, and their union forms an open set U. (U does not include C.) Finally, imagine the outward unit vector field V defined on U that is perpendicular to all the T_r.

The tori {T_r} are clearly the result of integrating the differential 1-form dual to the unit vector field V.

Now the flow {φ_t} of V will clearly carry the T_r's into each other. (Specifically, φ_s-r carries T_r into T_s.)

But, just as clearly, the various T_r are not conformally equivalent to each other. (By symmetry, they are all conformally equivalent to rectangular tori, but for r close to 0 they come from a long, skinny rectangle, while for r close to 1 they come from a much fatter rectangle.)

This shows that without additional hypotheses one cannot conclude that the surfaces are conformally equivalent to each other.

 

[lavinia]

@ zinq: nice point! Having at first misread your post, I have also edited my comment to provide more useful content.

@mathwonk Could you explain a couple different Riemann surfaces of genus 2?

 

[zinq]

It can be shown that for any three positive numbers A, B, C, there exists a hexagon in the hyperbolic plane H² with edges that are all geodesics, and such that the lengths of three non-adjacent edges are A, B, C. And that such a hexagon is unique up to isometry.

By glueing two copies of such a hexagon together along the remaining three sides we obtain what Thurston called a "pair of pants". The numbers A, B, C are called its "cuff lengths". A pair of pants is topologically a sphere with three open disks removed, and has constant curvature K = -1. Also, its three boundary circles are each the shortest simple closed curve in its homotopy class.

Finally, by taking two copies of a pair of pants with cuff lengths A B, C these can be glued together to form a surface of genus 2. For each pair of boundary circles of the same length, any angle can be chosen with which to glue them together. The three lengths A, B, C and these three angles form six parameters that determine the conformal equivalence class of the resulting compact surface. As expected 6 = 6g - 6.

This construction generalizes to obtain all conformal classes of all compact surfaces of genus g >= 2.

 

[lavinia]

It can be shown that for any three positive numbers A, B, C, there exists a hexagon in the hyperbolic plane H² with edges that are all geodesics, and such that the lengths of three non-adjacent edges are A, B, C. And that such a hexagon is unique up to isometry.

By glueing two copies of such a hexagon together along the remaining three sides we obtain what Thurston called a "pair of pants". The numbers A, B, C are called its "cuff lengths". A pair of pants is topologically a sphere with three open disks removed, and has constant curvature K = -1. Also, its three boundary circles are each the shortest simple closed curve in its homotopy class.

Finally, by taking two copies of a pair of pants with cuff lengths A B, C these can be glued together to form a surface of genus 2. For each pair of boundary circles of the same length, any angle can be chosen with which to glue them together. The three lengths A, B, C and these three angles form six parameters that determine the conformal equivalence class of the resulting compact surface. As expected 6 = 6g - 6.

This construction generalizes to obtain all conformal classes of all compact surfaces of genus g >= 2.

Can you give two examples and prove directly that they are not conformally equivalent? I guess what I am asking is a picture like the one you gave for flat tori that shows the difference.

BTW: I find it remarkable that different conformal structures determine non-isometric metrics of constant curvature $^{-}1$ on the same smooth surface.

 

[zinq]

I won't try to prove each step, but:

1) Conformally equivalent compact surfaces of the same constant curvature K ≠ 0 are isometric.

2) On a compact surface of constant negative curvature, each non-trivial free homotopy class of closed loops contains a unique shortest representative, and that is a simple closed geodesic.

3) The set of lengths of the simple closed geodesics of a compact Riemannian surface form an isometry invariant.

This means that for any A, B, C > 0 , glueing together four A, B, C, hexagons to make two pairs of pants, and then glueing together the two pairs of pants — as described in my previous post — will result in a surface of genus 2 with three closed geodesics of lengths 2A, 2B, 2C. Using three different numbers D, E, F will result in a surface of genus 2 with closed geodesics of different lengths.

I omitted showing that there don't exist other geodesics in the D, E, F case of lengths 2A, 2B, 2C, or vice versa. But in almost all cases, that is true.

 

[lavinia]

I won't try to prove each step, but:

1) Conformally equivalent compact surfaces of the same constant curvature K ≠ 0 are isometric.

2) On a compact surface of constant negative curvature, each non-trivial free homotopy class of closed loops contains a unique shortest representative, and that is a simple closed geodesic.

3) The set of lengths of the simple closed geodesics of a compact Riemannian surface form an isometry invariant.

This means that for any A, B, C > 0 , glueing together four A, B, C, hexagons to make two pairs of pants, and then glueing together the two pairs of pants — as described in my previous post — will result in a surface of genus 2 with three closed geodesics of lengths 2A, 2B, 2C. Using three different numbers D, E, F will result in a surface of genus 2 with closed geodesics of different lengths.

I omitted showing that there don't exist other geodesics in the D, E, F case of lengths 2A, 2B, 2C, or vice versa. But in almost all cases, that is true.

Nice

 

[mathwonk]

Roughly speaking, all smooth Riemann surfaces of genus 2 are double covers of the (complex) projective "line" P^1, branched at 6 distinct points. If they are complex holomorphically isomorphic, then it seems there must be an isomorphism of P^1, i.e. a Mobius transformation, taking the branch points of one to those of another. But the Mobius group has only 3 complex parameters, so we could produce two non isomorphic Riemann surfaces of genus 2, by letting them both have branch points at {0,1,infinity} as well as at three more points, such that the two choices of the extra three points are not carried to one another by the finite group of Mobius transforms that permute the three points...Hmmmm, I guess it is not quite that simple.

I guess I would have to identify the finite group that permutes the given 6 points, including {0,1,infinity} and then choose another point not in any of those orbits.OK here is a link (below) to a paper distinguishing surfaces of genus 2 by their different automorphism groups. the point is that all such surfaces have equations of form y^2 = f(x), where f has degree 5 (or 6 if you wish), according to whether or not you put a branch point (the zeroes of f) at infinity. All such surfaces admit the automorphism (x,y)-->(x,-y), and some have more. The quotient by that distinguished automorphism is P^1, and is called therefore a genus zero involution. All other involutions have quotient an elliptic curve, (and 2 branch, i.e. fixed points) and are called genus one involutions.

It follows that such special genus 2 curves are also branched double covers of elliptic curves, with 2 branch points, so you might construct examples of distinct special genus 2 curves by looking at special elliptic curves with known automorphisms, and choosing the branch points to be zero and one other. The possible automorphisms of elliptic curves (with a chosen fixed point, hence group isomorphisms) are well known, and the special cases correspond to certain special symmetric plane lattices, such as repetitions of a square or a hexagon.

Anyway, the following genus 2 curves seem to have distinct automorphism groups, hence are not isomorphic:

y^2 = X^5 - X, y^2 = X^6-X, y^2 = X^6-1.

see this paper, esp.pages 9,10,17?
https://arxiv.org/pdf/math/0107142.pdfsomewhat more in the spirit of zinq's discussion, one could view a genus 2 curve as a quotient of the disc, or upper half plane, by a hyperbolic group, and describe two inequivalent hyperbolic tessellations of the disc, by distinct hyperbolic groups.