Minimal triangulation of Klein bottle
What is minimal triangulation of Klein bottle?
А triangulation is a subdivision of a geometric object into simplices.
Minimal in sense of vertex count.
So, I know that minimal count of vertex in the shortest triangulation must be greater then $7$, because the shortest triangulation of torus consist of $7$ vertex and Euler characteristic is equal to $0$.
I would be cool if you can show me the picture.
triangulation klein-bottle
edited Mar 9 '15 at 22:52
azimut
16.2k1051100
asked Jul 27 '13 at 12:05
Gleb
425210
add a comment |
What is minimal triangulation of Klein bottle?
А triangulation is a subdivision of a geometric object into simplices.
Minimal in sense of vertex count.
So, I know that minimal count of vertex in the shortest triangulation must be greater then $7$, because the shortest triangulation of torus consist of $7$ vertex and Euler characteristic is equal to $0$.
I would be cool if you can show me the picture.
triangulation klein-bottle
edited Mar 9 '15 at 22:52
azimut
16.2k1051100
asked Jul 27 '13 at 12:05
Gleb
425210
I'd be impressed if there was a way to do it in less than $8$.
– Thomas Andrews
Jul 27 '13 at 12:11
10
The minimal number is 8. Section 4 of this paper has a proof of that. Section 5 of same paper derive all six distinct 8-vertex triangulations of the Klein bottle and has picture for them.
– achille hui
Jul 27 '13 at 13:02
Thanks, it`s very good paper.
– Gleb
Jul 27 '13 at 13:21
1
These papers might be interesting for you: arxiv.org/pdf/math/0407008v2.pdf , sciencedirect.com/science/article/pii/S0095895697999998
– taper
Dec 22 '16 at 14:22
add a comment |
What is minimal triangulation of Klein bottle?
А triangulation is a subdivision of a geometric object into simplices.
Minimal in sense of vertex count.
So, I know that minimal count of vertex in the shortest triangulation must be greater then $7$, because the shortest triangulation of torus consist of $7$ vertex and Euler characteristic is equal to $0$.
I would be cool if you can show me the picture.
triangulation klein-bottle
edited Mar 9 '15 at 22:52
azimut
16.2k1051100
asked Jul 27 '13 at 12:05
Gleb
425210
What is minimal triangulation of Klein bottle?
А triangulation is a subdivision of a geometric object into simplices.
Minimal in sense of vertex count.
So, I know that minimal count of vertex in the shortest triangulation must be greater then $7$, because the shortest triangulation of torus consist of $7$ vertex and Euler characteristic is equal to $0$.
I would be cool if you can show me the picture.
triangulation klein-bottle
triangulation klein-bottle
edited Mar 9 '15 at 22:52
azimut
16.2k1051100
asked Jul 27 '13 at 12:05
Gleb
425210
edited Mar 9 '15 at 22:52
azimut
16.2k1051100
asked Jul 27 '13 at 12:05
Gleb
425210
edited Mar 9 '15 at 22:52
azimut
16.2k1051100
edited Mar 9 '15 at 22:52
azimut
16.2k1051100
edited Mar 9 '15 at 22:52
azimut
16.2k1051100
16.2k1051100
asked Jul 27 '13 at 12:05
Gleb
425210
asked Jul 27 '13 at 12:05
Gleb
425210
asked Jul 27 '13 at 12:05
Gleb
425210
425210
I'd be impressed if there was a way to do it in less than $8$.
– Thomas Andrews
Jul 27 '13 at 12:11
10
The minimal number is 8. Section 4 of this paper has a proof of that. Section 5 of same paper derive all six distinct 8-vertex triangulations of the Klein bottle and has picture for them.
– achille hui
Jul 27 '13 at 13:02
Thanks, it`s very good paper.
– Gleb
Jul 27 '13 at 13:21
1
These papers might be interesting for you: arxiv.org/pdf/math/0407008v2.pdf , sciencedirect.com/science/article/pii/S0095895697999998
– taper
Dec 22 '16 at 14:22
add a comment |
I'd be impressed if there was a way to do it in less than $8$.
– Thomas Andrews
Jul 27 '13 at 12:11
10
The minimal number is 8. Section 4 of this paper has a proof of that. Section 5 of same paper derive all six distinct 8-vertex triangulations of the Klein bottle and has picture for them.
– achille hui
Jul 27 '13 at 13:02
Thanks, it`s very good paper.
– Gleb
Jul 27 '13 at 13:21
1
These papers might be interesting for you: arxiv.org/pdf/math/0407008v2.pdf , sciencedirect.com/science/article/pii/S0095895697999998
– taper
Dec 22 '16 at 14:22
I'd be impressed if there was a way to do it in less than $8$.
– Thomas Andrews
Jul 27 '13 at 12:11
I'd be impressed if there was a way to do it in less than $8$.
– Thomas Andrews
Jul 27 '13 at 12:11
10
10
The minimal number is 8. Section 4 of this paper has a proof of that. Section 5 of same paper derive all six distinct 8-vertex triangulations of the Klein bottle and has picture for them.
– achille hui
Jul 27 '13 at 13:02
The minimal number is 8. Section 4 of this paper has a proof of that. Section 5 of same paper derive all six distinct 8-vertex triangulations of the Klein bottle and has picture for them.
– achille hui
Jul 27 '13 at 13:02
Thanks, it`s very good paper.
– Gleb
Jul 27 '13 at 13:21
Thanks, it`s very good paper.
– Gleb
Jul 27 '13 at 13:21
1
1
These papers might be interesting for you: arxiv.org/pdf/math/0407008v2.pdf , sciencedirect.com/science/article/pii/S0095895697999998
– taper
Dec 22 '16 at 14:22
These papers might be interesting for you: arxiv.org/pdf/math/0407008v2.pdf , sciencedirect.com/science/article/pii/S0095895697999998
– taper
Dec 22 '16 at 14:22
add a comment |
1 Answer
1
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votes
The Klein bottle can be seen as the square $I^2$ with the boundaries identified in a specific way. Thus some triangulations of the square induces a triangulation of the Klein bottle. In particular you have a triangulation with exactly two faces, three edges and one vertex induced by the triangulation of the square obtained by cutting along the diagonal.
answered Jul 27 '13 at 12:14
Daniel Robert-Nicoud
20.3k33596
I think that for a triangulation it is required that the closures of all the triangles are mapped injectively to the triangulated space (here the Klein bottle). Your simple idea fails, because all the vertices of the two triangles are mapped on the same point on the Klein bottle.
– Jyrki Lahtonen
Jul 27 '13 at 12:32
Unfortunately, this partition isnt triangulation, because we have a edge, which incidence only one vertex. So, it isnt simlices.
– Gleb
Jul 27 '13 at 12:35
@JyrkiLahtonen Yes, if you require this of a triangulation then my triangulation doesn't work. I still think that working on the square is the best way to obtain something, though.
– Daniel Robert-Nicoud
Jul 27 '13 at 12:35
1
IOW, this is a fine structure as a CW-complex but not as a simplicial complex.
– Jyrki Lahtonen
Jul 27 '13 at 12:41
add a comment |
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The Klein bottle can be seen as the square $I^2$ with the boundaries identified in a specific way. Thus some triangulations of the square induces a triangulation of the Klein bottle. In particular you have a triangulation with exactly two faces, three edges and one vertex induced by the triangulation of the square obtained by cutting along the diagonal.
answered Jul 27 '13 at 12:14
Daniel Robert-Nicoud
20.3k33596
I think that for a triangulation it is required that the closures of all the triangles are mapped injectively to the triangulated space (here the Klein bottle). Your simple idea fails, because all the vertices of the two triangles are mapped on the same point on the Klein bottle.
– Jyrki Lahtonen
Jul 27 '13 at 12:32
Unfortunately, this partition isnt triangulation, because we have a edge, which incidence only one vertex. So, it isnt simlices.
– Gleb
Jul 27 '13 at 12:35
@JyrkiLahtonen Yes, if you require this of a triangulation then my triangulation doesn't work. I still think that working on the square is the best way to obtain something, though.
– Daniel Robert-Nicoud
Jul 27 '13 at 12:35
1
IOW, this is a fine structure as a CW-complex but not as a simplicial complex.
– Jyrki Lahtonen
Jul 27 '13 at 12:41
add a comment |
The Klein bottle can be seen as the square $I^2$ with the boundaries identified in a specific way. Thus some triangulations of the square induces a triangulation of the Klein bottle. In particular you have a triangulation with exactly two faces, three edges and one vertex induced by the triangulation of the square obtained by cutting along the diagonal.
answered Jul 27 '13 at 12:14
Daniel Robert-Nicoud
20.3k33596
I think that for a triangulation it is required that the closures of all the triangles are mapped injectively to the triangulated space (here the Klein bottle). Your simple idea fails, because all the vertices of the two triangles are mapped on the same point on the Klein bottle.
– Jyrki Lahtonen
Jul 27 '13 at 12:32
Unfortunately, this partition isnt triangulation, because we have a edge, which incidence only one vertex. So, it isnt simlices.
– Gleb
Jul 27 '13 at 12:35
@JyrkiLahtonen Yes, if you require this of a triangulation then my triangulation doesn't work. I still think that working on the square is the best way to obtain something, though.
– Daniel Robert-Nicoud
Jul 27 '13 at 12:35
1
IOW, this is a fine structure as a CW-complex but not as a simplicial complex.
– Jyrki Lahtonen
Jul 27 '13 at 12:41
add a comment |
The Klein bottle can be seen as the square $I^2$ with the boundaries identified in a specific way. Thus some triangulations of the square induces a triangulation of the Klein bottle. In particular you have a triangulation with exactly two faces, three edges and one vertex induced by the triangulation of the square obtained by cutting along the diagonal.
answered Jul 27 '13 at 12:14
Daniel Robert-Nicoud
20.3k33596
The Klein bottle can be seen as the square $I^2$ with the boundaries identified in a specific way. Thus some triangulations of the square induces a triangulation of the Klein bottle. In particular you have a triangulation with exactly two faces, three edges and one vertex induced by the triangulation of the square obtained by cutting along the diagonal.
answered Jul 27 '13 at 12:14
Daniel Robert-Nicoud
20.3k33596
answered Jul 27 '13 at 12:14
Daniel Robert-Nicoud
20.3k33596
answered Jul 27 '13 at 12:14
Daniel Robert-Nicoud
20.3k33596
answered Jul 27 '13 at 12:14
Daniel Robert-Nicoud
20.3k33596
20.3k33596
I think that for a triangulation it is required that the closures of all the triangles are mapped injectively to the triangulated space (here the Klein bottle). Your simple idea fails, because all the vertices of the two triangles are mapped on the same point on the Klein bottle.
– Jyrki Lahtonen
Jul 27 '13 at 12:32
Unfortunately, this partition isnt triangulation, because we have a edge, which incidence only one vertex. So, it isnt simlices.
– Gleb
Jul 27 '13 at 12:35
@JyrkiLahtonen Yes, if you require this of a triangulation then my triangulation doesn't work. I still think that working on the square is the best way to obtain something, though.
– Daniel Robert-Nicoud
Jul 27 '13 at 12:35
1
IOW, this is a fine structure as a CW-complex but not as a simplicial complex.
– Jyrki Lahtonen
Jul 27 '13 at 12:41
add a comment |
I think that for a triangulation it is required that the closures of all the triangles are mapped injectively to the triangulated space (here the Klein bottle). Your simple idea fails, because all the vertices of the two triangles are mapped on the same point on the Klein bottle.
– Jyrki Lahtonen
Jul 27 '13 at 12:32
Unfortunately, this partition isnt triangulation, because we have a edge, which incidence only one vertex. So, it isnt simlices.
– Gleb
Jul 27 '13 at 12:35
@JyrkiLahtonen Yes, if you require this of a triangulation then my triangulation doesn't work. I still think that working on the square is the best way to obtain something, though.
– Daniel Robert-Nicoud
Jul 27 '13 at 12:35
1
IOW, this is a fine structure as a CW-complex but not as a simplicial complex.
– Jyrki Lahtonen
Jul 27 '13 at 12:41
I think that for a triangulation it is required that the closures of all the triangles are mapped injectively to the triangulated space (here the Klein bottle). Your simple idea fails, because all the vertices of the two triangles are mapped on the same point on the Klein bottle.
– Jyrki Lahtonen
Jul 27 '13 at 12:32
I think that for a triangulation it is required that the closures of all the triangles are mapped injectively to the triangulated space (here the Klein bottle). Your simple idea fails, because all the vertices of the two triangles are mapped on the same point on the Klein bottle.
– Jyrki Lahtonen
Jul 27 '13 at 12:32
Unfortunately, this partition isn
t triangulation, because we have a edge, which incidence only one vertex. So, it isnt simlices.– Gleb
Jul 27 '13 at 12:35
Unfortunately, this partition isn
t triangulation, because we have a edge, which incidence only one vertex. So, it isnt simlices.– Gleb
Jul 27 '13 at 12:35
@JyrkiLahtonen Yes, if you require this of a triangulation then my triangulation doesn't work. I still think that working on the square is the best way to obtain something, though.
– Daniel Robert-Nicoud
Jul 27 '13 at 12:35
@JyrkiLahtonen Yes, if you require this of a triangulation then my triangulation doesn't work. I still think that working on the square is the best way to obtain something, though.
– Daniel Robert-Nicoud
Jul 27 '13 at 12:35
1
1
IOW, this is a fine structure as a CW-complex but not as a simplicial complex.
– Jyrki Lahtonen
Jul 27 '13 at 12:41
IOW, this is a fine structure as a CW-complex but not as a simplicial complex.
– Jyrki Lahtonen
Jul 27 '13 at 12:41
add a comment |
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I'd be impressed if there was a way to do it in less than $8$.
– Thomas Andrews
Jul 27 '13 at 12:11
10
The minimal number is 8. Section 4 of this paper has a proof of that. Section 5 of same paper derive all six distinct 8-vertex triangulations of the Klein bottle and has picture for them.
– achille hui
Jul 27 '13 at 13:02
Thanks, it`s very good paper.
– Gleb
Jul 27 '13 at 13:21
1
These papers might be interesting for you: arxiv.org/pdf/math/0407008v2.pdf , sciencedirect.com/science/article/pii/S0095895697999998
– taper
Dec 22 '16 at 14:22