How Does Discretizing a Smooth Manifold Affect Its Geometric Properties?

[lokofer]

"Discrete" Geommetry...

-What can you do if you have a "Smooth" Manifold..but it's very hard to work with?..perhaps you could "discretize" the surface splitting it into "triangles" and take the basic coordinates to be the angles (3) of every triangle..but my question is What would happen with the Metric the "ricci Tensor" (Riemann Tensor contracted ?) and other Differential Geommetry identities?.. How could you recover them or calculate them approximately?..thanks.

- for example an idea applied to Gr you have the Einstein Hamiltonian..if you discretize it using the angles of every "triangle" you would have:

L= \sqrt (-g) R \rightarrow L( \theta_i (t), \dot \theta_i (t),t) for i=1,2,3,4,5,6,7,8,9,... From this Lagrangian i could obtain the Hamiltonian and hence the "Energy levels"...but I'm missing R_{ab} and metric Tensor...How could i achieve the problem.

 

[ObsessiveMathsFreak]

-What can you do if you have a "Smooth" Manifold..but it's very hard to work with?..perhaps you could "discretize" the surface splitting it into "triangles" and take the basic coordinates to be the angles (3) of every triangle.

My understanding of smooth manifolds is that you could express them precisely with overlapping triangles, the triangles being open subsets of the plane. From my expierience with approximating smooth manifolds with discrete polygons, I can tell you that repeated application of any differential operator will make the solution more singular, or jagged. I was under the impression however, that the Ricci tensor is a smoothing operation, so perhaps it's an integral operator.

 

[lokofer]

- The Main problem here is "Obsessive... " that GR can't be managed in such an easy way if you don't use "discrete Polygons" (simplices ?) because then it would be a system with Infinite degrees of freedom..using discrete Polygons and its angles so "Gauss-Bonet" theorem applies \pi - \sum _j \theta _j (t) = K(t)dA where dA= Area of triangle, K is the "Curvature" of the surface (Associated to Riemann Tensor perhaps? ) and t is the time..in that case you can apply usual Quantization process so H( \theta _i (t) ,P_{\theta _i })\Psi ( \theta _i )=E_{n} \Psi(\theta _i ) but i would like to know if given each "angle" depending on time you can recover the usual "Geommetry" elements and Tensor.