Lecture 8: Mesh Representations and Geometry Processing (17)
jacklishufan
Why do we consider this concept of the manifold as a "topological concept" when it is geometric invariant? It seems that so long as the geometry is fixed, no matter how we change the placement of lines and edges, we will not change the behavior when we cut the surface with a sphere S^2. A manifold stays a manifold regardless of topology (triangulation)
lwg0320
I'm having trouble visualizing the difference between manifold with border and without. Is the reason that the manifold with border still a manifold despite having a cut disk is the fact that its edges are all connected?
jacklishufan
I believe so. More precisely there is a continuous bijection from the shaded area to a two dimension disk
Staffjamesfobrien
If your object is defined by a mesh, then you can see a connection between the object's topology and the mesh's. Some changes to the mesh's topology would not change the topology of the object, but some will. Changes to the object's topology will always involve a change in the mesh's topology.
Regarding the connection between a non-manifold edge/vertex in an mesh and a non-manifold curve/point on a surface, they are the same concept. If you have a 2D surface (flat plane, surface of sphere, etc) then you can take any point on that surface and have two perpendicular directions of movement. If some combinations of those directions only work going in the + or - direction but not both, then you have a border. If something happens and you can no longer consider a point on your surface to be homeomorphic to a disk, or equivalently if there is no way to embed that point and it's neighborhood in 2D without a discontinuity or overlap, then that was a discrete topological change, just like adding or removing a handle. So while it might seem like you could take a disk and mush it about until it looked like the non-manifold point in the slide, you can only do that if the disk was a 3D shape with finite thickness to start with, and even then you will not get the shape shown in the slide that has zero thickness.
Why do we consider this concept of the manifold as a "topological concept" when it is geometric invariant? It seems that so long as the geometry is fixed, no matter how we change the placement of lines and edges, we will not change the behavior when we cut the surface with a sphere S^2. A manifold stays a manifold regardless of topology (triangulation)
I'm having trouble visualizing the difference between manifold with border and without. Is the reason that the manifold with border still a manifold despite having a cut disk is the fact that its edges are all connected?
I believe so. More precisely there is a continuous bijection from the shaded area to a two dimension disk
If your object is defined by a mesh, then you can see a connection between the object's topology and the mesh's. Some changes to the mesh's topology would not change the topology of the object, but some will. Changes to the object's topology will always involve a change in the mesh's topology.
Regarding the connection between a non-manifold edge/vertex in an mesh and a non-manifold curve/point on a surface, they are the same concept. If you have a 2D surface (flat plane, surface of sphere, etc) then you can take any point on that surface and have two perpendicular directions of movement. If some combinations of those directions only work going in the + or - direction but not both, then you have a border. If something happens and you can no longer consider a point on your surface to be homeomorphic to a disk, or equivalently if there is no way to embed that point and it's neighborhood in 2D without a discontinuity or overlap, then that was a discrete topological change, just like adding or removing a handle. So while it might seem like you could take a disk and mush it about until it looked like the non-manifold point in the slide, you can only do that if the disk was a 3D shape with finite thickness to start with, and even then you will not get the shape shown in the slide that has zero thickness.