Mixing
Geometrical meaning of A4 conjugacy classes of elements of order 3
$begingroup$
I know that elements of order 3 in A4 group split into two separate conjugacy classes, due to lack of odd elements to form a unique class like happens in the full S4.
Since A4 is the rotation symmetry group of the Tetrahedron, I was wondering if these two split conjugacy classes have a direct geometrical meaning. While I can see a direct meaning of transpositions, clearly distinguishing them from 3-cycles, I cannot see any apparent related split of them from a geometrical perspective.
Nevertheless, my feeling is this binding makes sense, hiding somewhere from my view.
Would you point me to the right direction, please?
Thanks in advance
group-theory geometry permutations permutation-cycles
edited Jan 1 at 13:44
Shaun
8,820113681
asked Jan 1 at 13:40
riccardoventrellariccardoventrella
266
$endgroup$
add a comment |
$begingroup$
I know that elements of order 3 in A4 group split into two separate conjugacy classes, due to lack of odd elements to form a unique class like happens in the full S4.
Since A4 is the rotation symmetry group of the Tetrahedron, I was wondering if these two split conjugacy classes have a direct geometrical meaning. While I can see a direct meaning of transpositions, clearly distinguishing them from 3-cycles, I cannot see any apparent related split of them from a geometrical perspective.
Nevertheless, my feeling is this binding makes sense, hiding somewhere from my view.
Would you point me to the right direction, please?
Thanks in advance
group-theory geometry permutations permutation-cycles
edited Jan 1 at 13:44
Shaun
8,820113681
asked Jan 1 at 13:40
riccardoventrellariccardoventrella
266
$endgroup$
$begingroup$
Non-identity solutions to $x^3=e$ come in pairs. (This is true in any group.)
$endgroup$
– Shaun
Jan 1 at 13:52
1
$begingroup$
Thank @Shaun, but this is yet another algebraic explanation of that, which makes a lot sense to me, of course. However, my doubt is more strictly related to explain that inside the context of A4 as Tetrahedron rotation symmetry group. I mean: in D12 (symmetry of hexagon) it's clear the transpositions form different classes, depending if the reflection axis is taken from vertex to vertex or from midpoint to midpoint (and this happens only on even Dihedral groups). I'm trying to find a similar geometrical on A4 as Tetrahedron rotation group, since in this case all 3-cycles seem undistinguishable
$endgroup$
– riccardoventrella
Jan 1 at 13:59
add a comment |
$begingroup$
I know that elements of order 3 in A4 group split into two separate conjugacy classes, due to lack of odd elements to form a unique class like happens in the full S4.
Since A4 is the rotation symmetry group of the Tetrahedron, I was wondering if these two split conjugacy classes have a direct geometrical meaning. While I can see a direct meaning of transpositions, clearly distinguishing them from 3-cycles, I cannot see any apparent related split of them from a geometrical perspective.
Nevertheless, my feeling is this binding makes sense, hiding somewhere from my view.
Would you point me to the right direction, please?
Thanks in advance
group-theory geometry permutations permutation-cycles
edited Jan 1 at 13:44
Shaun
8,820113681
asked Jan 1 at 13:40
riccardoventrellariccardoventrella
266
$endgroup$
I know that elements of order 3 in A4 group split into two separate conjugacy classes, due to lack of odd elements to form a unique class like happens in the full S4.
Since A4 is the rotation symmetry group of the Tetrahedron, I was wondering if these two split conjugacy classes have a direct geometrical meaning. While I can see a direct meaning of transpositions, clearly distinguishing them from 3-cycles, I cannot see any apparent related split of them from a geometrical perspective.
Nevertheless, my feeling is this binding makes sense, hiding somewhere from my view.
Would you point me to the right direction, please?
Thanks in advance
group-theory geometry permutations permutation-cycles
group-theory geometry permutations permutation-cycles
edited Jan 1 at 13:44
Shaun
8,820113681
asked Jan 1 at 13:40
riccardoventrellariccardoventrella
266
edited Jan 1 at 13:44
Shaun
8,820113681
asked Jan 1 at 13:40
riccardoventrellariccardoventrella
266
edited Jan 1 at 13:44
Shaun
8,820113681
edited Jan 1 at 13:44
Shaun
8,820113681
edited Jan 1 at 13:44
Shaun
8,820113681
8,820113681
asked Jan 1 at 13:40
riccardoventrellariccardoventrella
266
asked Jan 1 at 13:40
riccardoventrellariccardoventrella
266
asked Jan 1 at 13:40
riccardoventrellariccardoventrella
266
266
$begingroup$
Non-identity solutions to $x^3=e$ come in pairs. (This is true in any group.)
$endgroup$
– Shaun
Jan 1 at 13:52
1
$begingroup$
Thank @Shaun, but this is yet another algebraic explanation of that, which makes a lot sense to me, of course. However, my doubt is more strictly related to explain that inside the context of A4 as Tetrahedron rotation symmetry group. I mean: in D12 (symmetry of hexagon) it's clear the transpositions form different classes, depending if the reflection axis is taken from vertex to vertex or from midpoint to midpoint (and this happens only on even Dihedral groups). I'm trying to find a similar geometrical on A4 as Tetrahedron rotation group, since in this case all 3-cycles seem undistinguishable
$endgroup$
– riccardoventrella
Jan 1 at 13:59
add a comment |
$begingroup$
Non-identity solutions to $x^3=e$ come in pairs. (This is true in any group.)
$endgroup$
– Shaun
Jan 1 at 13:52
1
$begingroup$
Thank @Shaun, but this is yet another algebraic explanation of that, which makes a lot sense to me, of course. However, my doubt is more strictly related to explain that inside the context of A4 as Tetrahedron rotation symmetry group. I mean: in D12 (symmetry of hexagon) it's clear the transpositions form different classes, depending if the reflection axis is taken from vertex to vertex or from midpoint to midpoint (and this happens only on even Dihedral groups). I'm trying to find a similar geometrical on A4 as Tetrahedron rotation group, since in this case all 3-cycles seem undistinguishable
$endgroup$
– riccardoventrella
Jan 1 at 13:59
$begingroup$
Non-identity solutions to $x^3=e$ come in pairs. (This is true in any group.)
$endgroup$
– Shaun
Jan 1 at 13:52
$begingroup$
Non-identity solutions to $x^3=e$ come in pairs. (This is true in any group.)
$endgroup$
– Shaun
Jan 1 at 13:52
1
1
$begingroup$
Thank @Shaun, but this is yet another algebraic explanation of that, which makes a lot sense to me, of course. However, my doubt is more strictly related to explain that inside the context of A4 as Tetrahedron rotation symmetry group. I mean: in D12 (symmetry of hexagon) it's clear the transpositions form different classes, depending if the reflection axis is taken from vertex to vertex or from midpoint to midpoint (and this happens only on even Dihedral groups). I'm trying to find a similar geometrical on A4 as Tetrahedron rotation group, since in this case all 3-cycles seem undistinguishable
$endgroup$
– riccardoventrella
Jan 1 at 13:59
$begingroup$
Thank @Shaun, but this is yet another algebraic explanation of that, which makes a lot sense to me, of course. However, my doubt is more strictly related to explain that inside the context of A4 as Tetrahedron rotation symmetry group. I mean: in D12 (symmetry of hexagon) it's clear the transpositions form different classes, depending if the reflection axis is taken from vertex to vertex or from midpoint to midpoint (and this happens only on even Dihedral groups). I'm trying to find a similar geometrical on A4 as Tetrahedron rotation group, since in this case all 3-cycles seem undistinguishable
$endgroup$
– riccardoventrella
Jan 1 at 13:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For me, it is helpful to see the conjugacy classes of $A_4$ listed out:
Now, let there be a tetrahedron sitting on a desk. Label the three vertices on the desk $1,2,3$, going clockwise, and label the vertex at the top as $4$. Now, if you rotate the tetrahedron $120^circ$ degrees clockwise, that would be like the cycle $(123)$. Rotate it counterclockwise, however, and that would be like the cycle $(132)$.
Now, turn the tetrahedron so that the vertex $1$ is now on top. You will now notice that, in clockwise order, the vertices on the desk are $2,4,3$. Thus, rotate the tetrahedron $120^circ$ degrees clockwise, and that would be like the cycle $(243)$. Rotate it counterclockwise, and that would be like the cycle $(234)$.
Again, turn the tetrahedron so that the vertex $2$ is now on top. In clockwise order, the vertices on the desk are now $1,3,4$. Thus, rotate the tetrahedron $120^circ$ degrees clockwise, and that would be like the cycle $(134)$. Rotate it counterclockwise, and that would be like the cycle $(143)$.
Hopefully, you can guess what will happen if you turn the tetrahedron so that the vertex $3$ is on top. Anyway, I hope you can now see how the conjugacy class with $(123)$ corresponds to clockwise rotations when looking from above while the conjugacy class with $(132)$ corresponds to counter-clockwise rotations when looking from above.
answered Jan 1 at 13:54
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
$begingroup$
thanks @Noble Mushtak, this was exactly the meaning I was looking for. Cool.
$endgroup$
– riccardoventrella
Jan 1 at 14:00
$begingroup$
However, of course (132) could be seen still as CW rotations. I'm still wondering why this CW/CCW trick does not apply to rotations in dihedral groups. I mean, taking D12 (symmetry of the Tetrahedron), the rotations a split in 3 conjugacy classes (naming the Hexagon vertices as the usual 1---6): {(135)(246)}, {(123456)(165432)} and {(14)(25)(36)} which belongs to the Z(D12) as forms a class by itself. In this case, the CW and CCW rotations belong to the same pertaining classes. Is this because rotations commute in plain 2D space, while rotations in 3D are not? Thanks
$endgroup$
– riccardoventrella
Jan 1 at 14:08
$begingroup$
@riccardoventrella I think the reason the CW/CCW trick doesn't work for conjugacy classes of $D_{12}$ is because $D_{12}$ has a reflection permutation which allows you to turn clockwise into counterclockwise roations. Meanwhile, in the above scenario, there's no way to reflect the whole tetrahedron over some plane. Yes, you can rotate the tetrahedron so that, from some perspective, $(132)$ looks clockwise and $(123)$ looks counterclockwise. However, by such rotation, you are changing the location of vertex $4$, which changes things.
$endgroup$
– Noble Mushtak
Jan 1 at 14:17
1
$begingroup$
However in full CW reasoning, we can still apply yours replacing CW by PI/3 rotations and CCW by 2/3PI ones, having all them as plain CW. So yes, we have still a split where now conjugacy classes mean "PI/3 CW rotations" Vs "2/3PI CW rotations". It's a mere renaming of that.
$endgroup$
– riccardoventrella
Jan 1 at 14:30
1
$begingroup$
@riccardoventrella Yes, you can also label the conjugacy classes as both clockwise rotations, but with different angles. In any case, there is still a clear distinction between the different geometric rotations.
$endgroup$
– Noble Mushtak
Jan 1 at 14:32
|
show 3 more comments
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$begingroup$
For me, it is helpful to see the conjugacy classes of $A_4$ listed out:
Now, let there be a tetrahedron sitting on a desk. Label the three vertices on the desk $1,2,3$, going clockwise, and label the vertex at the top as $4$. Now, if you rotate the tetrahedron $120^circ$ degrees clockwise, that would be like the cycle $(123)$. Rotate it counterclockwise, however, and that would be like the cycle $(132)$.
Now, turn the tetrahedron so that the vertex $1$ is now on top. You will now notice that, in clockwise order, the vertices on the desk are $2,4,3$. Thus, rotate the tetrahedron $120^circ$ degrees clockwise, and that would be like the cycle $(243)$. Rotate it counterclockwise, and that would be like the cycle $(234)$.
Again, turn the tetrahedron so that the vertex $2$ is now on top. In clockwise order, the vertices on the desk are now $1,3,4$. Thus, rotate the tetrahedron $120^circ$ degrees clockwise, and that would be like the cycle $(134)$. Rotate it counterclockwise, and that would be like the cycle $(143)$.
Hopefully, you can guess what will happen if you turn the tetrahedron so that the vertex $3$ is on top. Anyway, I hope you can now see how the conjugacy class with $(123)$ corresponds to clockwise rotations when looking from above while the conjugacy class with $(132)$ corresponds to counter-clockwise rotations when looking from above.
answered Jan 1 at 13:54
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
$begingroup$
thanks @Noble Mushtak, this was exactly the meaning I was looking for. Cool.
$endgroup$
– riccardoventrella
Jan 1 at 14:00
$begingroup$
However, of course (132) could be seen still as CW rotations. I'm still wondering why this CW/CCW trick does not apply to rotations in dihedral groups. I mean, taking D12 (symmetry of the Tetrahedron), the rotations a split in 3 conjugacy classes (naming the Hexagon vertices as the usual 1---6): {(135)(246)}, {(123456)(165432)} and {(14)(25)(36)} which belongs to the Z(D12) as forms a class by itself. In this case, the CW and CCW rotations belong to the same pertaining classes. Is this because rotations commute in plain 2D space, while rotations in 3D are not? Thanks
$endgroup$
– riccardoventrella
Jan 1 at 14:08
$begingroup$
@riccardoventrella I think the reason the CW/CCW trick doesn't work for conjugacy classes of $D_{12}$ is because $D_{12}$ has a reflection permutation which allows you to turn clockwise into counterclockwise roations. Meanwhile, in the above scenario, there's no way to reflect the whole tetrahedron over some plane. Yes, you can rotate the tetrahedron so that, from some perspective, $(132)$ looks clockwise and $(123)$ looks counterclockwise. However, by such rotation, you are changing the location of vertex $4$, which changes things.
$endgroup$
– Noble Mushtak
Jan 1 at 14:17
1
$begingroup$
However in full CW reasoning, we can still apply yours replacing CW by PI/3 rotations and CCW by 2/3PI ones, having all them as plain CW. So yes, we have still a split where now conjugacy classes mean "PI/3 CW rotations" Vs "2/3PI CW rotations". It's a mere renaming of that.
$endgroup$
– riccardoventrella
Jan 1 at 14:30
1
$begingroup$
@riccardoventrella Yes, you can also label the conjugacy classes as both clockwise rotations, but with different angles. In any case, there is still a clear distinction between the different geometric rotations.
$endgroup$
– Noble Mushtak
Jan 1 at 14:32
|
show 3 more comments
$begingroup$
For me, it is helpful to see the conjugacy classes of $A_4$ listed out:
Now, let there be a tetrahedron sitting on a desk. Label the three vertices on the desk $1,2,3$, going clockwise, and label the vertex at the top as $4$. Now, if you rotate the tetrahedron $120^circ$ degrees clockwise, that would be like the cycle $(123)$. Rotate it counterclockwise, however, and that would be like the cycle $(132)$.
Now, turn the tetrahedron so that the vertex $1$ is now on top. You will now notice that, in clockwise order, the vertices on the desk are $2,4,3$. Thus, rotate the tetrahedron $120^circ$ degrees clockwise, and that would be like the cycle $(243)$. Rotate it counterclockwise, and that would be like the cycle $(234)$.
Again, turn the tetrahedron so that the vertex $2$ is now on top. In clockwise order, the vertices on the desk are now $1,3,4$. Thus, rotate the tetrahedron $120^circ$ degrees clockwise, and that would be like the cycle $(134)$. Rotate it counterclockwise, and that would be like the cycle $(143)$.
Hopefully, you can guess what will happen if you turn the tetrahedron so that the vertex $3$ is on top. Anyway, I hope you can now see how the conjugacy class with $(123)$ corresponds to clockwise rotations when looking from above while the conjugacy class with $(132)$ corresponds to counter-clockwise rotations when looking from above.
answered Jan 1 at 13:54
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
$begingroup$
thanks @Noble Mushtak, this was exactly the meaning I was looking for. Cool.
$endgroup$
– riccardoventrella
Jan 1 at 14:00
$begingroup$
However, of course (132) could be seen still as CW rotations. I'm still wondering why this CW/CCW trick does not apply to rotations in dihedral groups. I mean, taking D12 (symmetry of the Tetrahedron), the rotations a split in 3 conjugacy classes (naming the Hexagon vertices as the usual 1---6): {(135)(246)}, {(123456)(165432)} and {(14)(25)(36)} which belongs to the Z(D12) as forms a class by itself. In this case, the CW and CCW rotations belong to the same pertaining classes. Is this because rotations commute in plain 2D space, while rotations in 3D are not? Thanks
$endgroup$
– riccardoventrella
Jan 1 at 14:08
$begingroup$
@riccardoventrella I think the reason the CW/CCW trick doesn't work for conjugacy classes of $D_{12}$ is because $D_{12}$ has a reflection permutation which allows you to turn clockwise into counterclockwise roations. Meanwhile, in the above scenario, there's no way to reflect the whole tetrahedron over some plane. Yes, you can rotate the tetrahedron so that, from some perspective, $(132)$ looks clockwise and $(123)$ looks counterclockwise. However, by such rotation, you are changing the location of vertex $4$, which changes things.
$endgroup$
– Noble Mushtak
Jan 1 at 14:17
1
$begingroup$
However in full CW reasoning, we can still apply yours replacing CW by PI/3 rotations and CCW by 2/3PI ones, having all them as plain CW. So yes, we have still a split where now conjugacy classes mean "PI/3 CW rotations" Vs "2/3PI CW rotations". It's a mere renaming of that.
$endgroup$
– riccardoventrella
Jan 1 at 14:30
1
$begingroup$
@riccardoventrella Yes, you can also label the conjugacy classes as both clockwise rotations, but with different angles. In any case, there is still a clear distinction between the different geometric rotations.
$endgroup$
– Noble Mushtak
Jan 1 at 14:32
|
show 3 more comments
$begingroup$
For me, it is helpful to see the conjugacy classes of $A_4$ listed out:
Now, let there be a tetrahedron sitting on a desk. Label the three vertices on the desk $1,2,3$, going clockwise, and label the vertex at the top as $4$. Now, if you rotate the tetrahedron $120^circ$ degrees clockwise, that would be like the cycle $(123)$. Rotate it counterclockwise, however, and that would be like the cycle $(132)$.
Now, turn the tetrahedron so that the vertex $1$ is now on top. You will now notice that, in clockwise order, the vertices on the desk are $2,4,3$. Thus, rotate the tetrahedron $120^circ$ degrees clockwise, and that would be like the cycle $(243)$. Rotate it counterclockwise, and that would be like the cycle $(234)$.
Again, turn the tetrahedron so that the vertex $2$ is now on top. In clockwise order, the vertices on the desk are now $1,3,4$. Thus, rotate the tetrahedron $120^circ$ degrees clockwise, and that would be like the cycle $(134)$. Rotate it counterclockwise, and that would be like the cycle $(143)$.
Hopefully, you can guess what will happen if you turn the tetrahedron so that the vertex $3$ is on top. Anyway, I hope you can now see how the conjugacy class with $(123)$ corresponds to clockwise rotations when looking from above while the conjugacy class with $(132)$ corresponds to counter-clockwise rotations when looking from above.
answered Jan 1 at 13:54
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
For me, it is helpful to see the conjugacy classes of $A_4$ listed out:
Now, let there be a tetrahedron sitting on a desk. Label the three vertices on the desk $1,2,3$, going clockwise, and label the vertex at the top as $4$. Now, if you rotate the tetrahedron $120^circ$ degrees clockwise, that would be like the cycle $(123)$. Rotate it counterclockwise, however, and that would be like the cycle $(132)$.
Now, turn the tetrahedron so that the vertex $1$ is now on top. You will now notice that, in clockwise order, the vertices on the desk are $2,4,3$. Thus, rotate the tetrahedron $120^circ$ degrees clockwise, and that would be like the cycle $(243)$. Rotate it counterclockwise, and that would be like the cycle $(234)$.
Again, turn the tetrahedron so that the vertex $2$ is now on top. In clockwise order, the vertices on the desk are now $1,3,4$. Thus, rotate the tetrahedron $120^circ$ degrees clockwise, and that would be like the cycle $(134)$. Rotate it counterclockwise, and that would be like the cycle $(143)$.
Hopefully, you can guess what will happen if you turn the tetrahedron so that the vertex $3$ is on top. Anyway, I hope you can now see how the conjugacy class with $(123)$ corresponds to clockwise rotations when looking from above while the conjugacy class with $(132)$ corresponds to counter-clockwise rotations when looking from above.
answered Jan 1 at 13:54
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 1 at 13:54
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 1 at 13:54
Noble MushtakNoble Mushtak
15.2k1735
answered Jan 1 at 13:54
Noble MushtakNoble Mushtak
15.2k1735
15.2k1735
$begingroup$
thanks @Noble Mushtak, this was exactly the meaning I was looking for. Cool.
$endgroup$
– riccardoventrella
Jan 1 at 14:00
$begingroup$
However, of course (132) could be seen still as CW rotations. I'm still wondering why this CW/CCW trick does not apply to rotations in dihedral groups. I mean, taking D12 (symmetry of the Tetrahedron), the rotations a split in 3 conjugacy classes (naming the Hexagon vertices as the usual 1---6): {(135)(246)}, {(123456)(165432)} and {(14)(25)(36)} which belongs to the Z(D12) as forms a class by itself. In this case, the CW and CCW rotations belong to the same pertaining classes. Is this because rotations commute in plain 2D space, while rotations in 3D are not? Thanks
$endgroup$
– riccardoventrella
Jan 1 at 14:08
$begingroup$
@riccardoventrella I think the reason the CW/CCW trick doesn't work for conjugacy classes of $D_{12}$ is because $D_{12}$ has a reflection permutation which allows you to turn clockwise into counterclockwise roations. Meanwhile, in the above scenario, there's no way to reflect the whole tetrahedron over some plane. Yes, you can rotate the tetrahedron so that, from some perspective, $(132)$ looks clockwise and $(123)$ looks counterclockwise. However, by such rotation, you are changing the location of vertex $4$, which changes things.
$endgroup$
– Noble Mushtak
Jan 1 at 14:17
1
$begingroup$
However in full CW reasoning, we can still apply yours replacing CW by PI/3 rotations and CCW by 2/3PI ones, having all them as plain CW. So yes, we have still a split where now conjugacy classes mean "PI/3 CW rotations" Vs "2/3PI CW rotations". It's a mere renaming of that.
$endgroup$
– riccardoventrella
Jan 1 at 14:30
1
$begingroup$
@riccardoventrella Yes, you can also label the conjugacy classes as both clockwise rotations, but with different angles. In any case, there is still a clear distinction between the different geometric rotations.
$endgroup$
– Noble Mushtak
Jan 1 at 14:32
|
show 3 more comments
$begingroup$
thanks @Noble Mushtak, this was exactly the meaning I was looking for. Cool.
$endgroup$
– riccardoventrella
Jan 1 at 14:00
$begingroup$
However, of course (132) could be seen still as CW rotations. I'm still wondering why this CW/CCW trick does not apply to rotations in dihedral groups. I mean, taking D12 (symmetry of the Tetrahedron), the rotations a split in 3 conjugacy classes (naming the Hexagon vertices as the usual 1---6): {(135)(246)}, {(123456)(165432)} and {(14)(25)(36)} which belongs to the Z(D12) as forms a class by itself. In this case, the CW and CCW rotations belong to the same pertaining classes. Is this because rotations commute in plain 2D space, while rotations in 3D are not? Thanks
$endgroup$
– riccardoventrella
Jan 1 at 14:08
$begingroup$
@riccardoventrella I think the reason the CW/CCW trick doesn't work for conjugacy classes of $D_{12}$ is because $D_{12}$ has a reflection permutation which allows you to turn clockwise into counterclockwise roations. Meanwhile, in the above scenario, there's no way to reflect the whole tetrahedron over some plane. Yes, you can rotate the tetrahedron so that, from some perspective, $(132)$ looks clockwise and $(123)$ looks counterclockwise. However, by such rotation, you are changing the location of vertex $4$, which changes things.
$endgroup$
– Noble Mushtak
Jan 1 at 14:17
1
$begingroup$
However in full CW reasoning, we can still apply yours replacing CW by PI/3 rotations and CCW by 2/3PI ones, having all them as plain CW. So yes, we have still a split where now conjugacy classes mean "PI/3 CW rotations" Vs "2/3PI CW rotations". It's a mere renaming of that.
$endgroup$
– riccardoventrella
Jan 1 at 14:30
1
$begingroup$
@riccardoventrella Yes, you can also label the conjugacy classes as both clockwise rotations, but with different angles. In any case, there is still a clear distinction between the different geometric rotations.
$endgroup$
– Noble Mushtak
Jan 1 at 14:32
$begingroup$
thanks @Noble Mushtak, this was exactly the meaning I was looking for. Cool.
$endgroup$
– riccardoventrella
Jan 1 at 14:00
$begingroup$
thanks @Noble Mushtak, this was exactly the meaning I was looking for. Cool.
$endgroup$
– riccardoventrella
Jan 1 at 14:00
$begingroup$
However, of course (132) could be seen still as CW rotations. I'm still wondering why this CW/CCW trick does not apply to rotations in dihedral groups. I mean, taking D12 (symmetry of the Tetrahedron), the rotations a split in 3 conjugacy classes (naming the Hexagon vertices as the usual 1---6): {(135)(246)}, {(123456)(165432)} and {(14)(25)(36)} which belongs to the Z(D12) as forms a class by itself. In this case, the CW and CCW rotations belong to the same pertaining classes. Is this because rotations commute in plain 2D space, while rotations in 3D are not? Thanks
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– riccardoventrella
Jan 1 at 14:08
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However, of course (132) could be seen still as CW rotations. I'm still wondering why this CW/CCW trick does not apply to rotations in dihedral groups. I mean, taking D12 (symmetry of the Tetrahedron), the rotations a split in 3 conjugacy classes (naming the Hexagon vertices as the usual 1---6): {(135)(246)}, {(123456)(165432)} and {(14)(25)(36)} which belongs to the Z(D12) as forms a class by itself. In this case, the CW and CCW rotations belong to the same pertaining classes. Is this because rotations commute in plain 2D space, while rotations in 3D are not? Thanks
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– riccardoventrella
Jan 1 at 14:08
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@riccardoventrella I think the reason the CW/CCW trick doesn't work for conjugacy classes of $D_{12}$ is because $D_{12}$ has a reflection permutation which allows you to turn clockwise into counterclockwise roations. Meanwhile, in the above scenario, there's no way to reflect the whole tetrahedron over some plane. Yes, you can rotate the tetrahedron so that, from some perspective, $(132)$ looks clockwise and $(123)$ looks counterclockwise. However, by such rotation, you are changing the location of vertex $4$, which changes things.
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– Noble Mushtak
Jan 1 at 14:17
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@riccardoventrella I think the reason the CW/CCW trick doesn't work for conjugacy classes of $D_{12}$ is because $D_{12}$ has a reflection permutation which allows you to turn clockwise into counterclockwise roations. Meanwhile, in the above scenario, there's no way to reflect the whole tetrahedron over some plane. Yes, you can rotate the tetrahedron so that, from some perspective, $(132)$ looks clockwise and $(123)$ looks counterclockwise. However, by such rotation, you are changing the location of vertex $4$, which changes things.
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– Noble Mushtak
Jan 1 at 14:17
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However in full CW reasoning, we can still apply yours replacing CW by PI/3 rotations and CCW by 2/3PI ones, having all them as plain CW. So yes, we have still a split where now conjugacy classes mean "PI/3 CW rotations" Vs "2/3PI CW rotations". It's a mere renaming of that.
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– riccardoventrella
Jan 1 at 14:30
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However in full CW reasoning, we can still apply yours replacing CW by PI/3 rotations and CCW by 2/3PI ones, having all them as plain CW. So yes, we have still a split where now conjugacy classes mean "PI/3 CW rotations" Vs "2/3PI CW rotations". It's a mere renaming of that.
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– riccardoventrella
Jan 1 at 14:30
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@riccardoventrella Yes, you can also label the conjugacy classes as both clockwise rotations, but with different angles. In any case, there is still a clear distinction between the different geometric rotations.
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– Noble Mushtak
Jan 1 at 14:32
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@riccardoventrella Yes, you can also label the conjugacy classes as both clockwise rotations, but with different angles. In any case, there is still a clear distinction between the different geometric rotations.
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– Noble Mushtak
Jan 1 at 14:32
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Non-identity solutions to $x^3=e$ come in pairs. (This is true in any group.)
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– Shaun
Jan 1 at 13:52
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Thank @Shaun, but this is yet another algebraic explanation of that, which makes a lot sense to me, of course. However, my doubt is more strictly related to explain that inside the context of A4 as Tetrahedron rotation symmetry group. I mean: in D12 (symmetry of hexagon) it's clear the transpositions form different classes, depending if the reflection axis is taken from vertex to vertex or from midpoint to midpoint (and this happens only on even Dihedral groups). I'm trying to find a similar geometrical on A4 as Tetrahedron rotation group, since in this case all 3-cycles seem undistinguishable
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– riccardoventrella
Jan 1 at 13:59