Minimum valence of a vertex in a simple triangular mesh with no boundaries
Suppose M is a simple (manifold, orientable and connected) triangular mesh with no boundaries (in other words, a mesh homeomorphic to a sphere).
I have the conjecture that the following statement is true:
The valence (the number of connecting edges) of
each vertex is greater or equal than 3.
My current definition of manifoldness is that each edge is incident to only one or two faces and the faces incident to a vertex form a closed or an open fan.
Valence is a natural number by definition, so:
- Valence = 0 breaks connectivity
- Valence = 1 breaks manifoldness because the only edge associated to the vertex would not be able to be incident to any face.
However I am having problems proving that Valence = 2 should break the manifoldness condition.
Can someone help me to better understand those conditions so I can prove if the statement is either false or true?
general-topology manifolds computer-science
asked Dec 26 '18 at 23:11
Albuquerque
214
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Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Suppose M is a simple (manifold, orientable and connected) triangular mesh with no boundaries (in other words, a mesh homeomorphic to a sphere).
I have the conjecture that the following statement is true:
The valence (the number of connecting edges) of
each vertex is greater or equal than 3.
My current definition of manifoldness is that each edge is incident to only one or two faces and the faces incident to a vertex form a closed or an open fan.
Valence is a natural number by definition, so:
- Valence = 0 breaks connectivity
- Valence = 1 breaks manifoldness because the only edge associated to the vertex would not be able to be incident to any face.
However I am having problems proving that Valence = 2 should break the manifoldness condition.
Can someone help me to better understand those conditions so I can prove if the statement is either false or true?
general-topology manifolds computer-science
asked Dec 26 '18 at 23:11
Albuquerque
214
New contributor
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Suppose M is a simple (manifold, orientable and connected) triangular mesh with no boundaries (in other words, a mesh homeomorphic to a sphere).
I have the conjecture that the following statement is true:
The valence (the number of connecting edges) of
each vertex is greater or equal than 3.
My current definition of manifoldness is that each edge is incident to only one or two faces and the faces incident to a vertex form a closed or an open fan.
Valence is a natural number by definition, so:
- Valence = 0 breaks connectivity
- Valence = 1 breaks manifoldness because the only edge associated to the vertex would not be able to be incident to any face.
However I am having problems proving that Valence = 2 should break the manifoldness condition.
Can someone help me to better understand those conditions so I can prove if the statement is either false or true?
general-topology manifolds computer-science
asked Dec 26 '18 at 23:11
Albuquerque
214
New contributor
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Suppose M is a simple (manifold, orientable and connected) triangular mesh with no boundaries (in other words, a mesh homeomorphic to a sphere).
I have the conjecture that the following statement is true:
The valence (the number of connecting edges) of
each vertex is greater or equal than 3.
My current definition of manifoldness is that each edge is incident to only one or two faces and the faces incident to a vertex form a closed or an open fan.
Valence is a natural number by definition, so:
- Valence = 0 breaks connectivity
- Valence = 1 breaks manifoldness because the only edge associated to the vertex would not be able to be incident to any face.
However I am having problems proving that Valence = 2 should break the manifoldness condition.
Can someone help me to better understand those conditions so I can prove if the statement is either false or true?
general-topology manifolds computer-science
general-topology manifolds computer-science
asked Dec 26 '18 at 23:11
Albuquerque
214
New contributor
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 26 '18 at 23:11
Albuquerque
214
New contributor
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Dec 26 '18 at 23:37
asked Dec 26 '18 at 23:11
Albuquerque
214
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Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Dec 26 '18 at 23:11
Albuquerque
214
asked Dec 26 '18 at 23:11
Albuquerque
214
214
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Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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Okay. It took me the entire day to realize. It probably deserves a more formal proof, but at least it is an argument:
If a vertex has valence 2, the 2 neighboring vertices must have 2 edges connecting the same 2 vertices. This would make the edges in the star of v boundaries, as they would be homeomorphic to the closed upper half-plane.
So, in this case, it is guaranteed that each vertex in a 2-manifold mesh with no boundaries has valence greater or equal to 3.
answered Dec 27 '18 at 18:11
Albuquerque
214
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Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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What is the "manifold condition"? Surely if you glue triangles together like in your diagram, the result will be a perfectly good topological manifold.
– Henning Makholm
Dec 27 '18 at 18:41
The "manifold condition" I am referring is that each vertex has a neighborhood homeomorphic to an open disk. But you are right! My writing was poor and perhaps even wrong. The diagram indeed is a good topological manifold, but it has a boundary (which could not exist according to hypothesis).
– Albuquerque
Dec 27 '18 at 19:11
I'm assuming that there are other triangles than the ones in the diagram. But even so, if you take two triangles and glue them together at the edges, you get a perfectly good topological manifold, homeomorphic to the sphere. I don't get why you think the red edges in your diagram would be a boundary -- there's clearly surface to both sides of each of them.
– Henning Makholm
Dec 27 '18 at 19:18
If you glue two (flat) triangles at the edges, wouldn't you have a set that is itself a (flat) triangle, hence being homeomorphic to a disk, not a sphere?
– Albuquerque
Dec 28 '18 at 3:15
x @Albu: Um, no. Only of you glue the interiors of the triangles together rather than just the edges.
– Henning Makholm
Dec 28 '18 at 14:42
add a comment |
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Okay. It took me the entire day to realize. It probably deserves a more formal proof, but at least it is an argument:
If a vertex has valence 2, the 2 neighboring vertices must have 2 edges connecting the same 2 vertices. This would make the edges in the star of v boundaries, as they would be homeomorphic to the closed upper half-plane.
So, in this case, it is guaranteed that each vertex in a 2-manifold mesh with no boundaries has valence greater or equal to 3.
answered Dec 27 '18 at 18:11
Albuquerque
214
New contributor
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is the "manifold condition"? Surely if you glue triangles together like in your diagram, the result will be a perfectly good topological manifold.
– Henning Makholm
Dec 27 '18 at 18:41
The "manifold condition" I am referring is that each vertex has a neighborhood homeomorphic to an open disk. But you are right! My writing was poor and perhaps even wrong. The diagram indeed is a good topological manifold, but it has a boundary (which could not exist according to hypothesis).
– Albuquerque
Dec 27 '18 at 19:11
I'm assuming that there are other triangles than the ones in the diagram. But even so, if you take two triangles and glue them together at the edges, you get a perfectly good topological manifold, homeomorphic to the sphere. I don't get why you think the red edges in your diagram would be a boundary -- there's clearly surface to both sides of each of them.
– Henning Makholm
Dec 27 '18 at 19:18
If you glue two (flat) triangles at the edges, wouldn't you have a set that is itself a (flat) triangle, hence being homeomorphic to a disk, not a sphere?
– Albuquerque
Dec 28 '18 at 3:15
x @Albu: Um, no. Only of you glue the interiors of the triangles together rather than just the edges.
– Henning Makholm
Dec 28 '18 at 14:42
add a comment |
Okay. It took me the entire day to realize. It probably deserves a more formal proof, but at least it is an argument:
If a vertex has valence 2, the 2 neighboring vertices must have 2 edges connecting the same 2 vertices. This would make the edges in the star of v boundaries, as they would be homeomorphic to the closed upper half-plane.
So, in this case, it is guaranteed that each vertex in a 2-manifold mesh with no boundaries has valence greater or equal to 3.
answered Dec 27 '18 at 18:11
Albuquerque
214
New contributor
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is the "manifold condition"? Surely if you glue triangles together like in your diagram, the result will be a perfectly good topological manifold.
– Henning Makholm
Dec 27 '18 at 18:41
The "manifold condition" I am referring is that each vertex has a neighborhood homeomorphic to an open disk. But you are right! My writing was poor and perhaps even wrong. The diagram indeed is a good topological manifold, but it has a boundary (which could not exist according to hypothesis).
– Albuquerque
Dec 27 '18 at 19:11
I'm assuming that there are other triangles than the ones in the diagram. But even so, if you take two triangles and glue them together at the edges, you get a perfectly good topological manifold, homeomorphic to the sphere. I don't get why you think the red edges in your diagram would be a boundary -- there's clearly surface to both sides of each of them.
– Henning Makholm
Dec 27 '18 at 19:18
If you glue two (flat) triangles at the edges, wouldn't you have a set that is itself a (flat) triangle, hence being homeomorphic to a disk, not a sphere?
– Albuquerque
Dec 28 '18 at 3:15
x @Albu: Um, no. Only of you glue the interiors of the triangles together rather than just the edges.
– Henning Makholm
Dec 28 '18 at 14:42
add a comment |
Okay. It took me the entire day to realize. It probably deserves a more formal proof, but at least it is an argument:
If a vertex has valence 2, the 2 neighboring vertices must have 2 edges connecting the same 2 vertices. This would make the edges in the star of v boundaries, as they would be homeomorphic to the closed upper half-plane.
So, in this case, it is guaranteed that each vertex in a 2-manifold mesh with no boundaries has valence greater or equal to 3.
answered Dec 27 '18 at 18:11
Albuquerque
214
New contributor
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Okay. It took me the entire day to realize. It probably deserves a more formal proof, but at least it is an argument:
If a vertex has valence 2, the 2 neighboring vertices must have 2 edges connecting the same 2 vertices. This would make the edges in the star of v boundaries, as they would be homeomorphic to the closed upper half-plane.
So, in this case, it is guaranteed that each vertex in a 2-manifold mesh with no boundaries has valence greater or equal to 3.
answered Dec 27 '18 at 18:11
Albuquerque
214
New contributor
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 27 '18 at 19:14
answered Dec 27 '18 at 18:11
Albuquerque
214
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Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered Dec 27 '18 at 18:11
Albuquerque
214
answered Dec 27 '18 at 18:11
Albuquerque
214
214
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Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Albuquerque is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is the "manifold condition"? Surely if you glue triangles together like in your diagram, the result will be a perfectly good topological manifold.
– Henning Makholm
Dec 27 '18 at 18:41
The "manifold condition" I am referring is that each vertex has a neighborhood homeomorphic to an open disk. But you are right! My writing was poor and perhaps even wrong. The diagram indeed is a good topological manifold, but it has a boundary (which could not exist according to hypothesis).
– Albuquerque
Dec 27 '18 at 19:11
I'm assuming that there are other triangles than the ones in the diagram. But even so, if you take two triangles and glue them together at the edges, you get a perfectly good topological manifold, homeomorphic to the sphere. I don't get why you think the red edges in your diagram would be a boundary -- there's clearly surface to both sides of each of them.
– Henning Makholm
Dec 27 '18 at 19:18
If you glue two (flat) triangles at the edges, wouldn't you have a set that is itself a (flat) triangle, hence being homeomorphic to a disk, not a sphere?
– Albuquerque
Dec 28 '18 at 3:15
x @Albu: Um, no. Only of you glue the interiors of the triangles together rather than just the edges.
– Henning Makholm
Dec 28 '18 at 14:42
add a comment |
What is the "manifold condition"? Surely if you glue triangles together like in your diagram, the result will be a perfectly good topological manifold.
– Henning Makholm
Dec 27 '18 at 18:41
The "manifold condition" I am referring is that each vertex has a neighborhood homeomorphic to an open disk. But you are right! My writing was poor and perhaps even wrong. The diagram indeed is a good topological manifold, but it has a boundary (which could not exist according to hypothesis).
– Albuquerque
Dec 27 '18 at 19:11
I'm assuming that there are other triangles than the ones in the diagram. But even so, if you take two triangles and glue them together at the edges, you get a perfectly good topological manifold, homeomorphic to the sphere. I don't get why you think the red edges in your diagram would be a boundary -- there's clearly surface to both sides of each of them.
– Henning Makholm
Dec 27 '18 at 19:18
If you glue two (flat) triangles at the edges, wouldn't you have a set that is itself a (flat) triangle, hence being homeomorphic to a disk, not a sphere?
– Albuquerque
Dec 28 '18 at 3:15
x @Albu: Um, no. Only of you glue the interiors of the triangles together rather than just the edges.
– Henning Makholm
Dec 28 '18 at 14:42
What is the "manifold condition"? Surely if you glue triangles together like in your diagram, the result will be a perfectly good topological manifold.
– Henning Makholm
Dec 27 '18 at 18:41
What is the "manifold condition"? Surely if you glue triangles together like in your diagram, the result will be a perfectly good topological manifold.
– Henning Makholm
Dec 27 '18 at 18:41
The "manifold condition" I am referring is that each vertex has a neighborhood homeomorphic to an open disk. But you are right! My writing was poor and perhaps even wrong. The diagram indeed is a good topological manifold, but it has a boundary (which could not exist according to hypothesis).
– Albuquerque
Dec 27 '18 at 19:11
The "manifold condition" I am referring is that each vertex has a neighborhood homeomorphic to an open disk. But you are right! My writing was poor and perhaps even wrong. The diagram indeed is a good topological manifold, but it has a boundary (which could not exist according to hypothesis).
– Albuquerque
Dec 27 '18 at 19:11
I'm assuming that there are other triangles than the ones in the diagram. But even so, if you take two triangles and glue them together at the edges, you get a perfectly good topological manifold, homeomorphic to the sphere. I don't get why you think the red edges in your diagram would be a boundary -- there's clearly surface to both sides of each of them.
– Henning Makholm
Dec 27 '18 at 19:18
I'm assuming that there are other triangles than the ones in the diagram. But even so, if you take two triangles and glue them together at the edges, you get a perfectly good topological manifold, homeomorphic to the sphere. I don't get why you think the red edges in your diagram would be a boundary -- there's clearly surface to both sides of each of them.
– Henning Makholm
Dec 27 '18 at 19:18
If you glue two (flat) triangles at the edges, wouldn't you have a set that is itself a (flat) triangle, hence being homeomorphic to a disk, not a sphere?
– Albuquerque
Dec 28 '18 at 3:15
If you glue two (flat) triangles at the edges, wouldn't you have a set that is itself a (flat) triangle, hence being homeomorphic to a disk, not a sphere?
– Albuquerque
Dec 28 '18 at 3:15
x @Albu: Um, no. Only of you glue the interiors of the triangles together rather than just the edges.
– Henning Makholm
Dec 28 '18 at 14:42
x @Albu: Um, no. Only of you glue the interiors of the triangles together rather than just the edges.
– Henning Makholm
Dec 28 '18 at 14:42
add a comment |
Albuquerque is a new contributor. Be nice, and check out our Code of Conduct.
Albuquerque is a new contributor. Be nice, and check out our Code of Conduct.
Albuquerque is a new contributor. Be nice, and check out our Code of Conduct.
Albuquerque is a new contributor. Be nice, and check out our Code of Conduct.
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