Graduate Diffeomorphic manifolds of equal constant curvature

[zinq]

" ... 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."

Yes — I wish I knew more about that.

Let M₂ denote the topological surface of genus = 2. I guess that for the two conformally inequivalent genus-2 curves, since H² is the universal cover of any complex curve of genus ≥ 2, the deck transformation subgroups G, H of

Aut(H²) = PSL(2, R)

corresponding to the two inequivalent curves must each be isomorphic as a group to π₁(M₂).* And G and H would need to act on H² without fixed points, and properly discontinuously, but they would need to be inequivalent under conjugation by any element of PSL(2,R): That is, for all group elements u ∈ Aut(H²), it must be true that G ≠ uHu^-1.

(Hmm ... maybe one way to get G and H explicitly would be to notice how the union of hyperbolic hexagons (four per M₂ surface) tessellate H² as their universal cover. That ought to give generators and relations.)
_____
* A common presentation for π₁(M₂) is

{a,b,c,d | aba^-1b^-1cdc^-1d^-1 = 1}.

 

[mathwonk]

this reference may be of interest:

https://www.acadsci.fi/mathematica/Vol20/kuusalo.pdf

 

[zinq]

Yes, very interesting.

I was wondering about dimensions higher than 2: Do there exist pairs of n-manifolds for n > 2 that are diffeomorphic and have equal constant sectional curvature, but which are not isometric to each other?

There are easy examples by just taking cartesian products of 2-dimensional examples with themselves or each other. For instance, the product of two square tori — and the product of two hexagonal tori. Or likewise the cartesian square of a surface of genus g — and another conformally inequivalent surface of genus g.

Are there other examples as well?

 

[zinq]

Addendum: In my various mentions of "hyperbolic hexagons" beginning with post #26 of this thread, I should have specified that these are all right-angled hexagons.

This enables the glueings of pairs of pants with each other to be varied with arbitrary twists along the closed geodesic boundary circles for which lengths have been assigned. (By the Gauss-Bonnet theorem, all right-angled hyperbolic hexagons have the same area = π.)

 

[lavinia]

Yes, very interesting.

I was wondering about dimensions higher than 2: Do there exist pairs of n-manifolds for n > 2 that are diffeomorphic and have equal constant sectional curvature, but which are not isometric to each other?

There are easy examples by just taking cartesian products of 2-dimensional examples with themselves or each other. For instance, the product of two square tori — and the product of two hexagonal tori. Or likewise the cartesian square of a surface of genus g — and another conformally inequivalent surface of genus g.

Are there other examples as well?

For starters, in higher dimensions there are compact flat Riemannian manifolds that are not tori. As for the flat torus and flat Klein bottle they come in different sizes and shapes.

There are something like 75 non-homeomorphic flat 4 dimensional compact flat Riemannian manifolds. As the dimension goes up so does the number in each dimension. I think there are only 10 three dimensional.

Many fo these manifolds are orientable. Here is one example:

Start with the lattice $L$ in Euclidean 3 space of points with integer coordinates and consider the group of isometries generated by this lattice by translation together with the transformation

$(x,y,z)→ (x+$1/4$, -z,y)$. This transformation defines an isometry of order 4 on the flat torus $R^3/L$.

 

[zinq]

Yes, for Riemannian flat manifolds it's easier to imagine examples that are diffeomorphic, and locally isometric, but which are not globally isometric.

Are there examples of dimension n ≥ 3 with constant sectional curvature K ≠ 0 ?

 

[lavinia]

Yes, for Riemannian flat manifolds it's easier to imagine examples that are diffeomorphic, and locally isometric, but which are not globally isometric.

Are there examples of dimension n ≥ 3 with constant sectional curvature K ≠ 0 ?

As for surfaces, higher dimensional manifolds of constant curvature are quotients of one of three simply connected spaces by the action of a group of isometries. Positive curvature spaces are quotients of the sphere, $S^{n}$,zero curvature are quotients of Euclidean space $R^{n}$, and constant negative curvature spaces are quotients of the hyperbolic half space $H^{n}$.

For negative curvature I came across this amazing theorem called the Mostow Rigity Theorem which says that if two manifolds of dimension greater than two and of constant negative curvature ${^-}1$ have isomorphic fundamental groups and also have finite volume then they are isometric. So the case of Riemann surfaces is unique.

For curvature zero you have flat manifolds and these can vary just as in the case of flat tori.

For constant positive curvature in even dimensions, I believe that even dimensional spheres are the only orientable possibility.

For odd dimensional spheres things get more complicated since finite groups can act properly discontinuously on them. Not sure of the theorems. Finite groups actions on the 3 sphere have been a source of key examples in topology.

 

[zinq]

I think this settles the remaining cases at least for compact manifolds: https://mathoverflow.net/questions/21848/isometry-classification-of-spherical-space-forms, in which it is stated that a 1950 result of de Rham shows that diffeomorphic manifolds covered by a sphere Sⁿ (of constant curvature) are isometric. I *think* this is the paper referred to: http://www.numdam.org/article/AIF_1950__2__51_0.pdf.

 

[lavinia]

@zinq For flat Riemannian manifolds, one can ask a weaker question which is not when are they isometric but when are they affine equivalent. That is: when is there a connection preserving diffeomorphism between them? I think then one gets all flat tori of the same dimension are affinity equivalent and further any two compact flat Riemannian manifolds are affinity equivalent whenever their fundamental groups are isomorphic.

 

[zinq]

I haven't thought about that equivalence relation and I don't have an intuition about it. (Do you?)

I have heard of affine manifolds in the sense that the pseudogroup of the transition maps of its atlas can be reduced to affine maps of Euclidean space. For instance,

M = Rⁿ - {0} / x ~ λx

for some λ > 1. M is clearly diffeomorphic to S^n-1 x S¹.

A famous unsolved problem is: Must a compact affine manifold have Euler characteristic equal to zero?

 

[lavinia]

I haven't thought about that equivalence relation and I don't have an intuition about it. (Do you?)

I have heard of affine manifolds in the sense that the pseudogroup of the transition maps of its atlas can be reduced to affine maps of Euclidean space. For instance,

M = Rⁿ - {0} / x ~ λx

for some λ > 1. M is clearly diffeomorphic to S^n-1 x S¹.

A famous unsolved problem is: Must a compact affine manifold have Euler characteristic equal to zero?

I don't know anything about general affine manifolds. The idea of affine equivalence that I know is that there is a connection preserving diffeomorphism between the two manifolds. For flat manifolds the projections of straight lines in Euclidean space would be mapped onto each other.

There is an unbelievable example - which I don't understand at all - due to Charlap - of two compact flat Riemannian manifolds of the same dimension that are not homotopy equivalent but whose Cartesian products with the circle are not only diffeomorphic but are affinity equivalent. Their holonomy groups are cyclic of prime order.

 

[zinq]

I wonder if that's related to pairs of flat 4-dimensional manifolds with the same spectrum of their Laplacians, I think due to Conway and Sloane. (The first such example was two 16-dimensional manifolds, due to Milnor.)

 

[lavinia]

I wonder if that's related to pairs of flat 4-dimensional manifolds with the same spectrum of their Laplacians, I think due to Conway and Sloane. (The first such example was two 16-dimensional manifolds, due to Milnor.)

It has to do with equivalence classes of integer representations of cyclic groups of prime order. I suppose if one extended Milnor's analysis of the Laplacian to manifolds such as these one might get some number theory out of it.