Is there a technique to derive the groups of rotations of various objects?
Apart from simply memorising them or being able to visualise them on the spot and jot them down is their any way to derive the group of rotations for various shapes in 3-d. (2-d is easy enough to just draw them and derive so never mind those)
group-theory rotations
asked Dec 8 at 23:05
can'tcauchy
968417
add a comment |
Apart from simply memorising them or being able to visualise them on the spot and jot them down is their any way to derive the group of rotations for various shapes in 3-d. (2-d is easy enough to just draw them and derive so never mind those)
group-theory rotations
asked Dec 8 at 23:05
can'tcauchy
968417
add a comment |
Apart from simply memorising them or being able to visualise them on the spot and jot them down is their any way to derive the group of rotations for various shapes in 3-d. (2-d is easy enough to just draw them and derive so never mind those)
group-theory rotations
asked Dec 8 at 23:05
can'tcauchy
968417
Apart from simply memorising them or being able to visualise them on the spot and jot them down is their any way to derive the group of rotations for various shapes in 3-d. (2-d is easy enough to just draw them and derive so never mind those)
group-theory rotations
group-theory rotations
asked Dec 8 at 23:05
can'tcauchy
968417
asked Dec 8 at 23:05
can'tcauchy
968417
asked Dec 8 at 23:05
can'tcauchy
968417
asked Dec 8 at 23:05
can'tcauchy
968417
asked Dec 8 at 23:05
can'tcauchy
968417
968417
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Rotations of a symmetric 3D shape can generally be broken down into:
1) Considering the number of faces
2) The number of vertices per face.
For example, with the icosahedron, the shape has 20 faces, all with three vertices. The rotation group clearly has size 3x20=60. The elements of the group consist of permuting the vertices in ways as to not tear or reflect the shape
answered Dec 9 at 20:58
Amnotwhy
362
That's helpful , thank you :) but if I wanted to understand the structure rather than just knowing the size is their a similar way to work that out ?
– can'tcauchy
Dec 9 at 21:00
2
What do you mean by the structure? You can work out what the different elements are by considering fixing a face/vertex and permuting the other vertices/faces. ie, there are 20 faces, so fix a face and permute the vertices of that face. This is clearly an element of order three and since the group has 20 faces it has 20 elements of order three. You can look for the other elements accordingly
– Amnotwhy
Dec 9 at 21:03
2
Almost. There is a symmetry that you are missing that will result in twice as many elements as you have stated. Because of this, there are 24 elements of order 5
– Amnotwhy
Dec 9 at 21:35
2
For each vertex there is a rotation of 72 degrees, but also a separate rotation of 144 degrees
– Amnotwhy
Dec 9 at 21:42
1
ahh I see ... Thank you very much for your help :)
– can'tcauchy
Dec 9 at 21:47
|
show 2 more comments
I saw this thread earlier when it was overrun by trolls (a worrying trend that I've noticed on this website as of late). A good general method is basically to look at the vertices and faces of the shape, and permute the faces, and see what information this gives you.
answered Dec 9 at 21:30
thegrouptheorist
293
1
Credit to @Amnotwhy, he essentially posted my answer but more in depth.
– thegrouptheorist
Dec 9 at 21:31
Yes trolling does seem to have become an issue it's pretty much the one guy... he always has bob in his name I wrote a post on the meta about it. some people have nothing better to do i suppose. As for your answer could you tell me then if the following is correct say we have a 20 sided die , if we rotate about a vertex we have 5 rotations by $72^0$ and then that means we have 12 elements of order 5 because there are 12 vertices right ?
– can'tcauchy
Dec 9 at 21:36
1
You are close: there are 24 elements of order 5.
– thegrouptheorist
Dec 9 at 21:38
1
Don't forget to upvote the answers :)
– thegrouptheorist
Dec 9 at 21:38
1
No problem, glad I could help! Also, that bob phenomenon does seem pretty strange.
– thegrouptheorist
Dec 9 at 21:43
|
show 1 more comment
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Rotations of a symmetric 3D shape can generally be broken down into:
1) Considering the number of faces
2) The number of vertices per face.
For example, with the icosahedron, the shape has 20 faces, all with three vertices. The rotation group clearly has size 3x20=60. The elements of the group consist of permuting the vertices in ways as to not tear or reflect the shape
answered Dec 9 at 20:58
Amnotwhy
362
That's helpful , thank you :) but if I wanted to understand the structure rather than just knowing the size is their a similar way to work that out ?
– can'tcauchy
Dec 9 at 21:00
2
What do you mean by the structure? You can work out what the different elements are by considering fixing a face/vertex and permuting the other vertices/faces. ie, there are 20 faces, so fix a face and permute the vertices of that face. This is clearly an element of order three and since the group has 20 faces it has 20 elements of order three. You can look for the other elements accordingly
– Amnotwhy
Dec 9 at 21:03
2
Almost. There is a symmetry that you are missing that will result in twice as many elements as you have stated. Because of this, there are 24 elements of order 5
– Amnotwhy
Dec 9 at 21:35
2
For each vertex there is a rotation of 72 degrees, but also a separate rotation of 144 degrees
– Amnotwhy
Dec 9 at 21:42
1
ahh I see ... Thank you very much for your help :)
– can'tcauchy
Dec 9 at 21:47
|
show 2 more comments
Rotations of a symmetric 3D shape can generally be broken down into:
1) Considering the number of faces
2) The number of vertices per face.
For example, with the icosahedron, the shape has 20 faces, all with three vertices. The rotation group clearly has size 3x20=60. The elements of the group consist of permuting the vertices in ways as to not tear or reflect the shape
answered Dec 9 at 20:58
Amnotwhy
362
That's helpful , thank you :) but if I wanted to understand the structure rather than just knowing the size is their a similar way to work that out ?
– can'tcauchy
Dec 9 at 21:00
2
What do you mean by the structure? You can work out what the different elements are by considering fixing a face/vertex and permuting the other vertices/faces. ie, there are 20 faces, so fix a face and permute the vertices of that face. This is clearly an element of order three and since the group has 20 faces it has 20 elements of order three. You can look for the other elements accordingly
– Amnotwhy
Dec 9 at 21:03
2
Almost. There is a symmetry that you are missing that will result in twice as many elements as you have stated. Because of this, there are 24 elements of order 5
– Amnotwhy
Dec 9 at 21:35
2
For each vertex there is a rotation of 72 degrees, but also a separate rotation of 144 degrees
– Amnotwhy
Dec 9 at 21:42
1
ahh I see ... Thank you very much for your help :)
– can'tcauchy
Dec 9 at 21:47
|
show 2 more comments
Rotations of a symmetric 3D shape can generally be broken down into:
1) Considering the number of faces
2) The number of vertices per face.
For example, with the icosahedron, the shape has 20 faces, all with three vertices. The rotation group clearly has size 3x20=60. The elements of the group consist of permuting the vertices in ways as to not tear or reflect the shape
answered Dec 9 at 20:58
Amnotwhy
362
Rotations of a symmetric 3D shape can generally be broken down into:
1) Considering the number of faces
2) The number of vertices per face.
For example, with the icosahedron, the shape has 20 faces, all with three vertices. The rotation group clearly has size 3x20=60. The elements of the group consist of permuting the vertices in ways as to not tear or reflect the shape
answered Dec 9 at 20:58
Amnotwhy
362
answered Dec 9 at 20:58
Amnotwhy
362
answered Dec 9 at 20:58
Amnotwhy
362
answered Dec 9 at 20:58
Amnotwhy
362
362
That's helpful , thank you :) but if I wanted to understand the structure rather than just knowing the size is their a similar way to work that out ?
– can'tcauchy
Dec 9 at 21:00
2
What do you mean by the structure? You can work out what the different elements are by considering fixing a face/vertex and permuting the other vertices/faces. ie, there are 20 faces, so fix a face and permute the vertices of that face. This is clearly an element of order three and since the group has 20 faces it has 20 elements of order three. You can look for the other elements accordingly
– Amnotwhy
Dec 9 at 21:03
2
Almost. There is a symmetry that you are missing that will result in twice as many elements as you have stated. Because of this, there are 24 elements of order 5
– Amnotwhy
Dec 9 at 21:35
2
For each vertex there is a rotation of 72 degrees, but also a separate rotation of 144 degrees
– Amnotwhy
Dec 9 at 21:42
1
ahh I see ... Thank you very much for your help :)
– can'tcauchy
Dec 9 at 21:47
|
show 2 more comments
That's helpful , thank you :) but if I wanted to understand the structure rather than just knowing the size is their a similar way to work that out ?
– can'tcauchy
Dec 9 at 21:00
2
What do you mean by the structure? You can work out what the different elements are by considering fixing a face/vertex and permuting the other vertices/faces. ie, there are 20 faces, so fix a face and permute the vertices of that face. This is clearly an element of order three and since the group has 20 faces it has 20 elements of order three. You can look for the other elements accordingly
– Amnotwhy
Dec 9 at 21:03
2
Almost. There is a symmetry that you are missing that will result in twice as many elements as you have stated. Because of this, there are 24 elements of order 5
– Amnotwhy
Dec 9 at 21:35
2
For each vertex there is a rotation of 72 degrees, but also a separate rotation of 144 degrees
– Amnotwhy
Dec 9 at 21:42
1
ahh I see ... Thank you very much for your help :)
– can'tcauchy
Dec 9 at 21:47
That's helpful , thank you :) but if I wanted to understand the structure rather than just knowing the size is their a similar way to work that out ?
– can'tcauchy
Dec 9 at 21:00
That's helpful , thank you :) but if I wanted to understand the structure rather than just knowing the size is their a similar way to work that out ?
– can'tcauchy
Dec 9 at 21:00
2
2
What do you mean by the structure? You can work out what the different elements are by considering fixing a face/vertex and permuting the other vertices/faces. ie, there are 20 faces, so fix a face and permute the vertices of that face. This is clearly an element of order three and since the group has 20 faces it has 20 elements of order three. You can look for the other elements accordingly
– Amnotwhy
Dec 9 at 21:03
What do you mean by the structure? You can work out what the different elements are by considering fixing a face/vertex and permuting the other vertices/faces. ie, there are 20 faces, so fix a face and permute the vertices of that face. This is clearly an element of order three and since the group has 20 faces it has 20 elements of order three. You can look for the other elements accordingly
– Amnotwhy
Dec 9 at 21:03
2
2
Almost. There is a symmetry that you are missing that will result in twice as many elements as you have stated. Because of this, there are 24 elements of order 5
– Amnotwhy
Dec 9 at 21:35
Almost. There is a symmetry that you are missing that will result in twice as many elements as you have stated. Because of this, there are 24 elements of order 5
– Amnotwhy
Dec 9 at 21:35
2
2
For each vertex there is a rotation of 72 degrees, but also a separate rotation of 144 degrees
– Amnotwhy
Dec 9 at 21:42
For each vertex there is a rotation of 72 degrees, but also a separate rotation of 144 degrees
– Amnotwhy
Dec 9 at 21:42
1
1
ahh I see ... Thank you very much for your help :)
– can'tcauchy
Dec 9 at 21:47
ahh I see ... Thank you very much for your help :)
– can'tcauchy
Dec 9 at 21:47
|
show 2 more comments
I saw this thread earlier when it was overrun by trolls (a worrying trend that I've noticed on this website as of late). A good general method is basically to look at the vertices and faces of the shape, and permute the faces, and see what information this gives you.
answered Dec 9 at 21:30
thegrouptheorist
293
1
Credit to @Amnotwhy, he essentially posted my answer but more in depth.
– thegrouptheorist
Dec 9 at 21:31
Yes trolling does seem to have become an issue it's pretty much the one guy... he always has bob in his name I wrote a post on the meta about it. some people have nothing better to do i suppose. As for your answer could you tell me then if the following is correct say we have a 20 sided die , if we rotate about a vertex we have 5 rotations by $72^0$ and then that means we have 12 elements of order 5 because there are 12 vertices right ?
– can'tcauchy
Dec 9 at 21:36
1
You are close: there are 24 elements of order 5.
– thegrouptheorist
Dec 9 at 21:38
1
Don't forget to upvote the answers :)
– thegrouptheorist
Dec 9 at 21:38
1
No problem, glad I could help! Also, that bob phenomenon does seem pretty strange.
– thegrouptheorist
Dec 9 at 21:43
|
show 1 more comment
I saw this thread earlier when it was overrun by trolls (a worrying trend that I've noticed on this website as of late). A good general method is basically to look at the vertices and faces of the shape, and permute the faces, and see what information this gives you.
answered Dec 9 at 21:30
thegrouptheorist
293
1
Credit to @Amnotwhy, he essentially posted my answer but more in depth.
– thegrouptheorist
Dec 9 at 21:31
Yes trolling does seem to have become an issue it's pretty much the one guy... he always has bob in his name I wrote a post on the meta about it. some people have nothing better to do i suppose. As for your answer could you tell me then if the following is correct say we have a 20 sided die , if we rotate about a vertex we have 5 rotations by $72^0$ and then that means we have 12 elements of order 5 because there are 12 vertices right ?
– can'tcauchy
Dec 9 at 21:36
1
You are close: there are 24 elements of order 5.
– thegrouptheorist
Dec 9 at 21:38
1
Don't forget to upvote the answers :)
– thegrouptheorist
Dec 9 at 21:38
1
No problem, glad I could help! Also, that bob phenomenon does seem pretty strange.
– thegrouptheorist
Dec 9 at 21:43
|
show 1 more comment
I saw this thread earlier when it was overrun by trolls (a worrying trend that I've noticed on this website as of late). A good general method is basically to look at the vertices and faces of the shape, and permute the faces, and see what information this gives you.
answered Dec 9 at 21:30
thegrouptheorist
293
I saw this thread earlier when it was overrun by trolls (a worrying trend that I've noticed on this website as of late). A good general method is basically to look at the vertices and faces of the shape, and permute the faces, and see what information this gives you.
answered Dec 9 at 21:30
thegrouptheorist
293
answered Dec 9 at 21:30
thegrouptheorist
293
answered Dec 9 at 21:30
thegrouptheorist
293
answered Dec 9 at 21:30
thegrouptheorist
293
293
1
Credit to @Amnotwhy, he essentially posted my answer but more in depth.
– thegrouptheorist
Dec 9 at 21:31
Yes trolling does seem to have become an issue it's pretty much the one guy... he always has bob in his name I wrote a post on the meta about it. some people have nothing better to do i suppose. As for your answer could you tell me then if the following is correct say we have a 20 sided die , if we rotate about a vertex we have 5 rotations by $72^0$ and then that means we have 12 elements of order 5 because there are 12 vertices right ?
– can'tcauchy
Dec 9 at 21:36
1
You are close: there are 24 elements of order 5.
– thegrouptheorist
Dec 9 at 21:38
1
Don't forget to upvote the answers :)
– thegrouptheorist
Dec 9 at 21:38
1
No problem, glad I could help! Also, that bob phenomenon does seem pretty strange.
– thegrouptheorist
Dec 9 at 21:43
|
show 1 more comment
1
Credit to @Amnotwhy, he essentially posted my answer but more in depth.
– thegrouptheorist
Dec 9 at 21:31
Yes trolling does seem to have become an issue it's pretty much the one guy... he always has bob in his name I wrote a post on the meta about it. some people have nothing better to do i suppose. As for your answer could you tell me then if the following is correct say we have a 20 sided die , if we rotate about a vertex we have 5 rotations by $72^0$ and then that means we have 12 elements of order 5 because there are 12 vertices right ?
– can'tcauchy
Dec 9 at 21:36
1
You are close: there are 24 elements of order 5.
– thegrouptheorist
Dec 9 at 21:38
1
Don't forget to upvote the answers :)
– thegrouptheorist
Dec 9 at 21:38
1
No problem, glad I could help! Also, that bob phenomenon does seem pretty strange.
– thegrouptheorist
Dec 9 at 21:43
1
1
Credit to @Amnotwhy, he essentially posted my answer but more in depth.
– thegrouptheorist
Dec 9 at 21:31
Credit to @Amnotwhy, he essentially posted my answer but more in depth.
– thegrouptheorist
Dec 9 at 21:31
Yes trolling does seem to have become an issue it's pretty much the one guy... he always has bob in his name I wrote a post on the meta about it. some people have nothing better to do i suppose. As for your answer could you tell me then if the following is correct say we have a 20 sided die , if we rotate about a vertex we have 5 rotations by $72^0$ and then that means we have 12 elements of order 5 because there are 12 vertices right ?
– can'tcauchy
Dec 9 at 21:36
Yes trolling does seem to have become an issue it's pretty much the one guy... he always has bob in his name I wrote a post on the meta about it. some people have nothing better to do i suppose. As for your answer could you tell me then if the following is correct say we have a 20 sided die , if we rotate about a vertex we have 5 rotations by $72^0$ and then that means we have 12 elements of order 5 because there are 12 vertices right ?
– can'tcauchy
Dec 9 at 21:36
1
1
You are close: there are 24 elements of order 5.
– thegrouptheorist
Dec 9 at 21:38
You are close: there are 24 elements of order 5.
– thegrouptheorist
Dec 9 at 21:38
1
1
Don't forget to upvote the answers :)
– thegrouptheorist
Dec 9 at 21:38
Don't forget to upvote the answers :)
– thegrouptheorist
Dec 9 at 21:38
1
1
No problem, glad I could help! Also, that bob phenomenon does seem pretty strange.
– thegrouptheorist
Dec 9 at 21:43
No problem, glad I could help! Also, that bob phenomenon does seem pretty strange.
– thegrouptheorist
Dec 9 at 21:43
|
show 1 more comment
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