What is a realization of this surface in Euclidean space?

  • Thread starter lavinia
  • Start date
  • #1
lavinia
Science Advisor
Gold Member
3,170
575

Main Question or Discussion Point

I am interested to know how to realize this abstract surface as a subset of Euclidean space.

The surface as a point set is the 2 dimenional Euclidean plane minus the origin.
the metric is given by declaring the following 2 vector fields to be an orthonormal frame:

e[itex]_{1}[/itex] = x[itex]\partial[/itex]x - y[itex]\partial[/itex]y

e[itex]_{2}[/itex] = x[itex]^{2}[/itex][itex]\partial[/itex]x + e[itex]^{-xy}[/itex][itex]\partial[/itex]y

I think that this surface has constant negative curvature equal to -1 but I do not think it is a Poincare disk minus a point because the origin is infinitely far away from any point on the x or y axes.

BTW: Here is the curvature calculation but I am not 100% sure of it.

the Lie bracket, [e[itex]_{1}[/itex],e[itex]_{2}[/itex]], equals e[itex]_{2}[/itex] so the covariant derivative can be defined using the following formulas:


[itex]\nabla_{e_{1}}[/itex]e[itex]_{1}[/itex] = 0
[itex]\nabla_{e_{1}}[/itex]e[itex]_{2}[/itex] = 0
[itex]\nabla_{e_{2}}[/itex]e[itex]_{1}[/itex] = -e[itex]_{2}[/itex]
[itex]\nabla_{e_{2}}[/itex]e[itex]_{2}[/itex] = e[itex]_{1}[/itex]

Extend these formulas to arbitrary vector fields by linearity and the Leibniz rule.

The connection 1 form is ω(e[itex]_{2}[/itex]) = -1, ω(e[itex]_{1}[/itex]) = 0.

Since this connection is torsion free the exterior derivative of ω easily computes and is the volume element of the metric. Therefore the Gauss curvature is -1.

It may be helpful to note that the rectangular hyperpolas y = k/x are geodesics as are the rays from the origin along the axes.
 
Last edited:

Answers and Replies

  • #2
120
0
Well I have never before tried to find the coordinates of a manifold from its tangent space but I will propose an idea. Let θ,λ be the coordinates with tangent vectors ∂θ and ∂λ. Then when pulled back to the euclidean spaca they are given as ∂x/∂θ∂x + ∂y/∂θ∂y
and ∂x/∂λ∂x + ∂y/∂λ∂y . Looking at your equations this requires

∂x/∂θ = x(θ,λ), ∂y/∂θ= -y(θ,λ), and
∂x/∂λ = x2(θ,λ), ∂y/∂λ= e-xy(θ,λ).

I think that if your tangent distribution is integrable, you should be able to solve these to get the manifold structure.
 
  • #3
lavinia
Science Advisor
Gold Member
3,170
575
thanks Sina

You are answering another problem.
This problem is to find an embedding of this surface in euclidean space not to find coordinates.
 
  • #4
120
0
I see now what you mean by realization. However for instance I think if you repeat the same procedure for the circle, you get an embedding no? You would get (x,y) = (cos(θ),sin(θ))
 
  • #5
lavinia
Science Advisor
Gold Member
3,170
575
I see now what you mean by realization. However for instance I think if you repeat the same procedure for the circle, you get an embedding no? You would get (x,y) = (cos(θ),sin(θ))
I don't think so. An embedding requires that the normal directions move around the surface in the right way, i.e. in a way that defines the surface's shape. this can not come from a coordinate system. further it is not obvious what dimension Euclidean space the surface can be realized in. Maybe 3 dimensions, maybe 4.
 
  • #6
120
0
Hmm you are right yes. It would only be a luck I suppose if finding the coordinates helped to find the embedding.

I happen to recall that whitney's embedding theorem seemed constructive to me (though I am not sure) but I can not recall the details right now. However it did not make use of concepts such as curvature, connection etc that I am sure about. Maybe that would be also of help.
 
  • #7
lavinia
Science Advisor
Gold Member
3,170
575
Hmm you are right yes. It would only be a luck I suppose if finding the coordinates helped to find the embedding.

I happen to recall that whitney's embedding theorem seemed constructive to me (though I am not sure) but I can not recall the details right now. However it did not make use of concepts such as curvature, connection etc that I am sure about. Maybe that would be also of help.
Right. The Whitney embedding theorem is for smooth embeddings not for isometric embeddings. That is the Nash embedding theorem but I do not know how it is proved. I have heard that is is monster hard.
 
  • #8
489
0
Hi Lavinia. This is of course just a guess, but why could it not be realized as isometric to a one-sheeted hyperbola?
 
  • #9
lavinia
Science Advisor
Gold Member
3,170
575
Hi Lavinia. This is of course just a guess, but why could it not be realized as isometric to a one-sheeted hyperbola?
I don't know what a one sheeted hyperbola is.
 
  • #10
489
0
The Poincare disk model for the hyperbolic plane also has a realization as the upper sheet of z^2=x^2+y^2+1. This is of course simply connected, so one might try modifying the equation z^2=x^2+y^2-1. Think of fixing r=1/2 on the plane z=0 in R^3, and then blowing the boundary of the circle to z=+infinity, and at the same time blowing the origin out to z=-infinity, where the traces would be hyperbolas in two directions and circles in the other direction.

Now, I can't see that this would work in your example because of the lack of symmetry, but it does give an example of constant negative curvature with the homotopy type of the circle. So my intuition would be that your surface has an isometric embedding in R^3, but the idea of solving the equations gives me indigestion :)
 
  • #11
lavinia
Science Advisor
Gold Member
3,170
575
The Poincare disk model for the hyperbolic plane also has a realization as the upper sheet of z^2=x^2+y^2+1. This is of course simply connected, so one might try modifying the equation z^2=x^2+y^2-1. Think of fixing r=1/2 on the plane z=0 in R^3, and then blowing the boundary of the circle to z=+infinity, and at the same time blowing the origin out to z=-infinity, where the traces would be hyperbolas in two directions and circles in the other direction.

Now, I can't see that this would work in your example because of the lack of symmetry, but it does give an example of constant negative curvature with the homotopy type of the circle. So my intuition would be that your surface has an isometric embedding in R^3, but the idea of solving the equations gives me indigestion :)
I just realized that the two vector fields are dependent on the y axis so the metric can only be defined on one of the open half spaces x>0 or x<0
 
  • #12
431
0
You have a 2 dimensional surface which is simply connected and contractible with constant negative curvature. Surely that implies that it is isometric to the hyperbolic plane, no?
 
  • #13
lavinia
Science Advisor
Gold Member
3,170
575
You have a 2 dimensional surface which is simply connected and contractible with constant negative curvature. Surely that implies that it is isometric to the hyperbolic plane, no?
Yes Jamma. You are right.

I originally thought that the metric extended to the plane minus the origin in which case the manifold would not have been contractible. But I realized that it does not extend to the y-axis so it is valid only on either open half plane.

The problem was trying to see if a vector field of index negative 1 could be tangent to geodesics for some metric in a region of the plane. But this attempt fails. Any ideas?
 
  • #14
431
0
What do you mean by a vector field of index -1? Do you mean the index at a fixed point (I know about the index of a vector field at some fixed point-take the degree of the induced map into the sphere)?
 
  • #15
lavinia
Science Advisor
Gold Member
3,170
575
What do you mean by a vector field of index -1? Do you mean the index at a fixed point (I know about the index of a vector field at some fixed point-take the degree of the induced map into the sphere)?
I mean exactly what you said. The vector field x[itex]\partial[/itex]x -y[itex]\partial[/itex]y has index -1 around the origin.
 

Related Threads on What is a realization of this surface in Euclidean space?

  • Last Post
Replies
11
Views
5K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
2K
Replies
8
Views
6K
  • Last Post
Replies
11
Views
4K
Replies
13
Views
5K
Replies
0
Views
2K
Replies
35
Views
3K
Top