Hi, I have been working on dual-meshing for more than 4 years now. Still found this this lecture series so enlightening. Comprehensively impressed by the way it is presented and it clarified so many of my intuitive confusions. Looking forward to more new materials.
I encounter the same confusion as Jeffrey Zhang working on the coding exercise for Discrete Exterior Calculus. I check, the test program for Hodge 0 does not pass when you use circumcentric dual, so you have to use barycentric dual. The way to check by yourselves is to write circumcentricDualArea program in assignment 2, make sure it is correct using the test program for assignment 2, then use this function for assignment 1 Hodge 0 operator. You will see it does not pass test case while for barycentric it pass. I understand it is a teaching decision to reduce complexity, but it would be nice to note this irregularity somewhere officially. Personally I would add circumcentric vs barycentric dual to the materials, underscore we use circumcentric dual, and keep everything consistent (use circumcentric for hodge 0).
Hi, Prof Crane, thank you for your wonderful lectures! Btw, would you kindly update the lecture 9 slides on brickisland when you're free? The slides there seem to be from 2019 and do not exactly match the slides in the video.
At 13:42 there is an example of a 0-form problem where the solution is the difference between the vertices with orientation considertations. I was wondering how this changes if the edge is curvilinear with multiple points along it. It seems like we are assuming a linear variation along an edge.
Hi, Prof Crane, Is it possible for the DEC to reduce the dimension of mesh used in FEM? What is the typical dimension for the matrix to solve a real-world problem?
Is it possible/reasonable to start with the barycentric dual, but instead of using the barycenters, try to use something close to them, but moved somewhat in order to get orthogonality back? Also, is it necessary that the thing taken as the 0-dimensional dual cell be the same "in each direction"? Like, if this is all in 3d, could it be reasonable to have for each face, the edge dual to it, be perpendicular and somewhere through the face, with the length on each side depending on the dimensions of the simplex in some way, but without requiring that the endpoints of these dual vertices *actually* meet up? (but still treating each of the endpoints inside the same simplex as being "the same vertex") Err, not that these endpoints would actually be computed. And I'm not sure how this would work in other numbers of dimensions.. I guess what I've described just boils down to "can you just choose values for the diagonal in some way that doesn't correspond to the volumes of any actual consistent shapes?" which seems like probably a less interesting question than I initially thought. Maybe I'm just trying to find a free lunch after being informed that there isn't one there to find.
1:03:34 I draw a diagram and it shows that the map from \Omega_1^* to \Omega_2^* should be -d_0^T, not d_0^T. Am I wrong? For example, a vertice v_i in the primal diagram has an outgoing halfedge e, its hodge star *e in the dual diagram rotates e 90 degrees counterclockwise so it contributes to *v_i positively. But an outgoing halfedge means the coefficient of v_i is negetive.
Hi Professor Cane, Thanks for the brilliant lectures. Just wondering, how did you assign the orientation of e_3 at 23:49? If two triangles are with the same orientation, then the edge they share should have two opposite orientations? What did I miss here? Thank you very much.
Yes, this is a common source of confusion. Every element of the mesh (vertex, edge, triangle, ...) has its own orientation, which isn't required to have anything to do with the orientation of any other mesh elements. Then you can ask whether two elements have consistent or inconsistent orientation. In other words, the orientation of a k-simplex need not induce an orientation on any of its faces.
@@keenancrane Thank you very much for this clarification, Professor. Just one more quick question, in the discrete case, if my underlying manifold is a topological manifold, how should I define a mapping (for example linear) as orientation preserving or reserving? I know the orientation for the topological manifold should be defined as a coherent choice of the homology group H_d(M, M/p, Z) around one point or H_d(M, Z) for the d degree fundamental class, does that mean that as long as the mapping can map cycle to cycle (closed d-chain), then it is orientation preserving/reserving? How about piece-wise linear mapping in which each piece maps a part of the manifold into different dimensions? Thanks again.
I've been following along in the coding exercises, and it seems like finding the hodge star of a 1 form is meant to be done with the assumption of a circumcentric dual. But then it also seems encouraged to use the barycentric dual area in the computation of the hodge star for a 0 form - in fact the tests seem to pass when done this way. It seems strange to be assuming a circumcentric dual in one case and a barycentric dual in another case. Is this intended?
I encounter the same confusion as Jeffrey Zhang. I check, the test program for Hodge 0 does not pass when you use circumcentric dual, so you have to use barycentric dual. The way to check by yourselves is to write circumcentricDualArea program in assignment 2, make sure it is correct using the test program for assignment 2, then use this function for assignment 1 Hodge 0 operator. You will see it does not pass test case while for barycentric it pass. I understand it is a teaching decision to reduce complexity, but it would be nice to note this irregularity somewhere officially. Personally I would add circumcentric vs barycentric dual to the materials and keep everything consistent
I am confused @15:52 you are taking the exterior derivative by integrating over the boundaries (adding up to sign). Yet in the example to the left you were actually looking at differences between vertices to find the values at edges, which makes perfect sense: the differential in the discrete case becomes the difference as you go from 0-forms (vertices) to 1-forms (edges). But how does it make sense that when you go from 1-forms (edges) to 2-forms (triangles) you are actually integrating? Shouldn't you also be differencing in some fashion?
@@keenancrane Surely that can't possibly be always true. Consider a perfectly symmetric pyramid with 100 faces. The dual 0-cell is then perfectly horizontal. Now move two primal 0-cells downwards. Then we have a bunch of barycenters on the horizontal plane and three barycenters off that plane
@@columbus8myhw I'm not sure I understand your example, but the circumcentric dual corresponds to the Voronoi diagram of the vertices, and each Voronoi cell is an intersection of halfspaces.
Hi, I have been working on dual-meshing for more than 4 years now. Still found this this lecture series so enlightening. Comprehensively impressed by the way it is presented and it clarified so many of my intuitive confusions. Looking forward to more new materials.
I encounter the same confusion as Jeffrey Zhang working on the coding exercise for Discrete Exterior Calculus. I check, the test program for Hodge 0 does not pass when you use circumcentric dual, so you have to use barycentric dual. The way to check by yourselves is to write circumcentricDualArea program in assignment 2, make sure it is correct using the test program for assignment 2, then use this function for assignment 1 Hodge 0 operator. You will see it does not pass test case while for barycentric it pass.
I understand it is a teaching decision to reduce complexity, but it would be nice to note this irregularity somewhere officially. Personally I would add circumcentric vs barycentric dual to the materials, underscore we use circumcentric dual, and keep everything consistent (use circumcentric for hodge 0).
Hi, Prof Crane, thank you for your wonderful lectures!
Btw, would you kindly update the lecture 9 slides on brickisland when you're free? The slides there seem to be from 2019 and do not exactly match the slides in the video.
Good catch-now updated!
00:20 Review-Discrete Differential Forms
Discrete Exterior Derivative
03:32 Reminder: Exterior Derivative
05:33 Discrete Exterior Derivative (0-Forms)
08:40 Discrete Exterior Derivative (1-Forms)
12:06 Discrete Exterior Derivative-Examples
17:32 Discrete Exterior Derivative-Matrix Representation
19:56 Discrete Exterior Derivative d_0-Example
23:04 Discrete Exterior Derivative d_1-Example
25:32 Exterior Derivative Commutes w/ Discretization
28:26 Exactness of Discrete Exterior Derivative
Dual Forms
29:38 Reminder: Poincaré Duality
30:23 Dual Discrete Differential k-Form
31:35 Primal vs. Dual Discrete Differential k-Forms
34:49 Dual Exterior Derivative
36:22 Dual Forms: Interpolation & Discretization
Discrete Hodge Star
40:54 Reminder: Hodge Star
41:50 Geometry of Dual Complex
45:09 Discrete Hodge Star-Basic Idea
48:40 Discrete Hodge Star-1-forms in 2D
50:05 Discrete Hodge Star-2-forms in 3D
51:37 Diagonal Hodge Star
53:01 Matrix Representation of Diagonal Hodge Star
54:10 Computing Volumes
57:26 Possible Choices for Discrete Hodge Star
Summary
1:01:04 Discrete Exterior Calculus-Basic Operators
1:04:12 Composition of Operators
1:05:38 Applications
1:05:53 Other Discrete Operators
1:07:07 Discrete Exterior Calculus - Summary
At 13:42 there is an example of a 0-form problem where the solution is the difference between the vertices with orientation considertations. I was wondering how this changes if the edge is curvilinear with multiple points along it. It seems like we are assuming a linear variation along an edge.
Hi, Prof Crane, Is it possible for the DEC to reduce the dimension of mesh used in FEM? What is the typical dimension for the matrix to solve a real-world problem?
Is it possible/reasonable to start with the barycentric dual, but instead of using the barycenters, try to use something close to them, but moved somewhat in order to get orthogonality back?
Also, is it necessary that the thing taken as the 0-dimensional dual cell be the same "in each direction"? Like, if this is all in 3d, could it be reasonable to have for each face, the edge dual to it, be perpendicular and somewhere through the face, with the length on each side depending on the dimensions of the simplex in some way, but without requiring that the endpoints of these dual vertices *actually* meet up? (but still treating each of the endpoints inside the same simplex as being "the same vertex")
Err, not that these endpoints would actually be computed. And I'm not sure how this would work in other numbers of dimensions.. I guess what I've described just boils down to "can you just choose values for the diagonal in some way that doesn't correspond to the volumes of any actual consistent shapes?" which seems like probably a less interesting question than I initially thought.
Maybe I'm just trying to find a free lunch after being informed that there isn't one there to find.
1:03:34 I draw a diagram and it shows that the map from \Omega_1^* to \Omega_2^* should be -d_0^T, not d_0^T. Am I wrong? For example, a vertice v_i in the primal diagram has an outgoing halfedge e, its hodge star *e in the dual diagram rotates e 90 degrees counterclockwise so it contributes to *v_i positively. But an outgoing halfedge means the coefficient of v_i is negetive.
Possible typo in slide 28 at 56:40. alpha^jk_k doesn't make much sense.
Hi Professor Cane, Thanks for the brilliant lectures. Just wondering, how did you assign the orientation of e_3 at 23:49? If two triangles are with the same orientation, then the edge they share should have two opposite orientations? What did I miss here? Thank you very much.
Yes, this is a common source of confusion. Every element of the mesh (vertex, edge, triangle, ...) has its own orientation, which isn't required to have anything to do with the orientation of any other mesh elements. Then you can ask whether two elements have consistent or inconsistent orientation. In other words, the orientation of a k-simplex need not induce an orientation on any of its faces.
@@keenancrane Thank you very much for this clarification, Professor. Just one more quick question, in the discrete case, if my underlying manifold is a topological manifold, how should I define a mapping (for example linear) as orientation preserving or reserving? I know the orientation for the topological manifold should be defined as a coherent choice of the homology group H_d(M, M/p, Z) around one point or H_d(M, Z) for the d degree fundamental class, does that mean that as long as the mapping can map cycle to cycle (closed d-chain), then it is orientation preserving/reserving? How about piece-wise linear mapping in which each piece maps a part of the manifold into different dimensions?
Thanks again.
I've been following along in the coding exercises, and it seems like finding the hodge star of a 1 form is meant to be done with the assumption of a circumcentric dual. But then it also seems encouraged to use the barycentric dual area in the computation of the hodge star for a 0 form - in fact the tests seem to pass when done this way. It seems strange to be assuming a circumcentric dual in one case and a barycentric dual in another case. Is this intended?
I believe it's just for practice's sake. If you really need closure, Assignment 2 will require us to calculate circumcentricDualArea for 0 forms.
I encounter the same confusion as Jeffrey Zhang. I check, the test program for Hodge 0 does not pass when you use circumcentric dual, so you have to use barycentric dual. The way to check by yourselves is to write circumcentricDualArea program in assignment 2, make sure it is correct using the test program for assignment 2, then use this function for assignment 1 Hodge 0 operator. You will see it does not pass test case while for barycentric it pass.
I understand it is a teaching decision to reduce complexity, but it would be nice to note this irregularity somewhere officially. Personally I would add circumcentric vs barycentric dual to the materials and keep everything consistent
I am confused @15:52 you are taking the exterior derivative by integrating over the boundaries (adding up to sign). Yet in the example to the left you were actually looking at differences between vertices to find the values at edges, which makes perfect sense: the differential in the discrete case becomes the difference as you go from 0-forms (vertices) to 1-forms (edges). But how does it make sense that when you go from 1-forms (edges) to 2-forms (triangles) you are actually integrating? Shouldn't you also be differencing in some fashion?
Answer: "When going from vertices to edges, taking an oriented sum is indistinguishable from taking a difference."
If dual-0 cells aren't always planar, how are we measuring their area?
For the circumcentric dual they are, but in general they may not be (e.g., if you use the barycentric dual). Good question.
@@keenancrane That's super surprising that it always becomes planar.
@@keenancrane Surely that can't possibly be always true. Consider a perfectly symmetric pyramid with 100 faces. The dual 0-cell is then perfectly horizontal. Now move two primal 0-cells downwards. Then we have a bunch of barycenters on the horizontal plane and three barycenters off that plane
@@columbus8myhw I'm not sure I understand your example, but the circumcentric dual corresponds to the Voronoi diagram of the vertices, and each Voronoi cell is an intersection of halfspaces.