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>Manifolds are just graphs with many vertices. Okay, I'll bite. How? What is the definition of the tangent space? Dimension?
by woopwoop 9y ago
>Manifolds are just graphs with many vertices.
Okay, I'll bite. How? What is the definition of the tangent space? Dimension?
- contravariant 9y agoYeah you'll want hypergraphs (or even better simplicial complexes) in general but you can make sense of manifolds as discrete objects. In fact I'm fairly sure you can triangulate any (smooth?) manifold. The tangent space can be defined in terms of derivations, as soon as you define what a smooth function on the 'discrete' manifold should look like (you may have to define the derivative at a face, rather than a vertex, or have the function take values on faces rather than vertices).
- woopwoop 9y agoSure. I think the notion of a manifold wouldn't be very interesting if it didn't have some kind of discrete analogue. But, for me at least, smooth manifolds are much easier to think about than simplicial complexes or PL manifolds.
- enriquto 9y agothe tangent space is the set of edges vector fields (sections of the tangent bundle) correspond to real-valued functions defined on edges dimension is always 2 ;) if you want higher dimension you have to consider higher-dimensional cliques beyond edges (that are 2-cliques): triangles, tetrahedra, and so on. But very often this is not necessary, even when discretizing 3d stuff. For example, for Poisson equation, and the associated classical pde, you only need the laplacian, which acts on functions, regardless of the dimension.
- woopwoop 9y agoIs a triangle a circle or a disk?
- contravariant 9y agoThat seems a weird choice of tangent space, surely a circle and a disk should have different tangent spaces?
- enriquto 9y agoAnd they do. When you discretize a circle as a graph, all the edges are along the boundary. When you discretize the disk, you fill its whole interior with edges in all directions.