What is the rotation group of a cube?

[lugita15]

I'm trying to find out what the rotation group of a cube is. It seems natural to view it as a subgroup of $S_{6}$, because any rotation is just a permutation of the faces of the cube. The sources I've seen say that the rotation group of a cube is isomorphic to $S_{4}$, because rotations can be seen as permutations of the diagonals of a cube. It's hard for me to simultaneously visualize all four diagonals of a cube, much less how they are affected by rotations.

But the one thing I do know is that $S_{4}$ has 24 elements, and that seems like too many. The rotation group of a cube is generated by the following three elements: a 90 degree rotation about x, a 90 degree rotation about y, and a 90 degree rotation about z. But any rotation about z can be written as a composition of rotations about x and y, so it seems that the rotation group is just generated by the x-rotation and the y-rotation. If that's the case, then any rotation of a cube can be written as $a^{m}b^{n}$, where a is the rotation about x, b is the rotation about y, and m and n range from 1 to 4. Thus the rotation group should only have 16 elements. Where am I going wrong?

Any help would be greatly appreciated.

Thank You in Advance.

 

[micromass]

Hi lugita15!

To be honest, I can't quite follow your argument. I'll try to explain why the answer is indeed 24, and I hope you'll find you're mistake yourself.

When following a course on advanced group theory, the professor usually worked this out with the orbit-stabilizer theorem. However, he also told us a very handy trick:

Take a cube, it is made out of 6 faces. Every face is a square. Now draw the diagonals of each square, this will divide the square in 4 parts. So each face of the cube is divided in 4 parts. In total, the entire cube is divided in 24 parts (=4*6).

Now, what is the trick? Well each of the 24 parts can be sent to each of the other 24 parts by a rotation. Furthermore, the rotation is uniquely determined by how it acts on one of the 24 parts.

Lets number the 24 parts. There is a unique rotation that sends part 1 to part 1 (=the identity). There is unique rotation that sends part 1 to part 2. ... There is a unique rotation that sends part 1 to part 24.

This determines all the rotations, thus there are 24 rotations!

Note that the number of symmetries of the cube is larger, it is 48. The point is that we also need to add reflections to the symmetries. To see why the number of symmetries is 48, we apply the trick as above. Except that we now divide each face of the cube in 8 parts (draw 2 diagonals, and draw 2 lines that connect the midpoints of opposite sides). Then apply thesame reasoning.

I sincerely hope that this was a bit clear. If not, feel free to ask questions!

 

[chogg]

Here's an educated guess: rotations about x (i.e. in the y-z plane), and rotations about y (i.e. in the z-x plane) don't commute. So you can't only write them as $a^mb^n$, because (for instance) $ab \ne ba$.

 

[micromass]

I read your post again, and I think I understand now

then any rotation of a cube can be written as $a^{m}b^{n}$, where a is the rotation about x, b is the rotation about y, and m and n range from 1 to 4.

This is not true. The group of rotations is not commutative, thus I see no a priori reason why ba or abbaab could be written in the form $a^mb^n$. Maybe these two rotations can be written in that form, but cannot all be written in that form!

 

[lugita15]

I read your post again, and I think I understand now



This is not true. The group of rotations is not commutative, thus I see no a priori reason why ba or abbaab could be written in the form $a^mb^n$. Maybe these two rotations can be written in that form, but cannot all be written in that form!

Haha, the whole reason I'm interested in the rotation group of a cube is that I'm trying to develop some intuition about why three-dimensional rotations don't commute. Anyway, thanks micromass and chogg for your help.

 

[lugita15]

One more question: a group presentation of $S_{4}$ is given http://www.weddslist.com/groups/misc/serpres.html" , but I'm not sure what rotations of a cube k and r correspond to. What is the group presentation of the rotation group of a cube in terms of the rotations a and b I specified before, where a is a 90 degree rotation about x and b is a 90 degree rotation about y?

 

[micromass]

The group in terms of your rotations is

$$\{1,a,a^2,a^3,b,ba,ba^2,ba^3,b^2,b^2a,b^2a^2,b^2a^3,b^3,b^3a,b^3a^2,b^3a^3,ab,aba,aba^2,aba^3,a^3b,a^3ba,a^3ba^2,a^3ba^3\}$$

Where a means rotating wrt the x-axis and b is rotating wrt the y-axis. So, for example, ab means first rotate wrt the y-axis and then wrt to the x-axis.

Playing a bit with GAP yields the following presentation for the rotation group:

$$<a,b~\vert~a^4, b^4, (a^2b)^2, (ba)^3>$$

Playing a bit further gives me that you can take k=ba, r=a², and this will be the presentation you want...

 

[lugita15]

The group in terms of your rotations is

$$\{1,a,a^2,a^3,b,ba,ba^2,ba^3,b^2,b^2a,b^2a^2,b^2a^3,b^3,b^3a,b^3a^2,b^3a^3,ab,aba,aba^2,aba^3,a^3b,a^3ba,a^3ba^2,a^3ba^3\}$$

Where a means rotating wrt the x-axis and b is rotating wrt the y-axis. So, for example, ab means first rotate wrt the y-axis and then wrt to the x-axis.

Playing a bit with GAP yields the following presentation for the rotation group:

$$<a,b~\vert~a^4, b^4, (a^2b)^2, (ba)^3>$$

Playing a bit further gives me that you can take k=ba, r=a², and this will be the presentation you want...

What's GAP? Also, why aren't a and b on equal footing in the presentation?

 

[micromass]

What's GAP? Also, why aren't a and b on equal footing in the presentation?

GAP is a computer algebra system that allows you to calculate things in groups very easily.
See http://www.gap-system.org/

Hmm, you can exchange the roles of a and b if you want, that won't change anything. Or you can make the definition symmetric by

$$<a,b~\vert~a^4,b^4,(a^2b)^2,(b^2a)^2,(ba)^3>$$

Maybe there are nicer presentations possible...

 

[lpetrich]

The group of all rotations and reflections of a cube and its n-D versions can be constructed as follows:

• Construct the n! permutations of the rows (or columns) of the n-D identity matrix. This is permutations of the cube faces' directions.
• Construct 2ⁿ diagonal matrices whose elements are +1 or -1. This is inversions of the cube faces' directions.
• Multiply them.

This group has these subgroups:

• Even permutations only
• Signs multiply to +1 only
• Determinant = +1 only: (even,+1) and (odd,-1)
• Even permutations and signs multiplying to +1

For 3D, these groups are

• Octahedral with reflections: Oh -- S4 * Z2
• Pyritohedral group: Th -- A4 * Z2
• Tetrahedral group with reflections: Td -- S4
• Octahedral group with rotations only: O -- S4
• Tetrahedral group with rotations only: T -- A4

For 2D, these groups are

• Rotations and reflections of a square: diihedral group D4
• Horizontal and vertical flips, 180d rotation: dihedral group D2 -- Z2 * Z2
• Horizontal-vertical interchange, 180d rotation: dihedral group D2 -- Z2 * Z2
• Rotations of a square (90d, 180d, 270d): cyclic group C4 -- Z4
• 180d rotation of a square: cyclic group C2 -- Z2

If one has some computer-algebra software, one can explicitly demonstrate these results by construction.

 

[D H]

Haha, the whole reason I'm interested in the rotation group of a cube is that I'm trying to develop some intuition about why three-dimensional rotations don't commute. Anyway, thanks micromass and chogg for your help.

To develop some intuition about this:

Get a book and place it on a table, closed, and with the front cover of the book facing up and in an orientation so you can read the print naturally (not upside down, not sideways). Now let's label some axes: The book x-axis is parallel to the print on the cover page. The z axis points upward out of the book, and the y-axis completes a right handed system.

Now pick the book up, holding it at about chest level and in the same orientation it was on the table. Next give it a 90 degree rotation about the book x-axis and then a 90 degree rotation about the (rotated) book y axis. You should be looking down on the top edge of the book, with the spine of the book facing to your chest and the front cover pointing to your right. Now restart the procedure, but this time do the rotation about the book y-axis first rather than last. The book will be in a very different orientation.

Compare that to vector addition. Take two steps forward and one to your right and mark your new location. Now repeat, but switching the order: Take one step to your right and two steps forward. You end up at the same spot. Vector addition commutes, rotation does not.

 

[lugita15]

To develop some intuition about this:

Get a book and place it on a table, closed, and with the front cover of the book facing up and in an orientation so you can read the print naturally (not upside down, not sideways). Now let's label some axes: The book x-axis is parallel to the print on the cover page. The z axis points upward out of the book, and the y-axis completes a right handed system.

Now pick the book up, holding it at about chest level and in the same orientation it was on the table. Next give it a 90 degree rotation about the book x-axis and then a 90 degree rotation about the (rotated) book y axis. You should be looking down on the top edge of the book, with the spine of the book facing to your chest and the front cover pointing to your right. Now restart the procedure, but this time do the rotation about the book y-axis first rather than last. The book will be in a very different orientation.

Compare that to vector addition. Take two steps forward and one to your right and mark your new location. Now repeat, but switching the order: Take one step to your right and two steps forward. You end up at the same spot. Vector addition commutes, rotation does not.

Yes, since I first learned about the non-commutativity of rotations several years ago, I've done that simple demonstration countless times. So I thoroughly accept that rotations don't commute, but I don't understand why. Currently, it just seems to me like a mysterious phenomenon that rotations performed in different orders yield different results. But suppose humans had never discovered this property. Without actually composing two rotations and determining whether they commute, what a priori reason would you have for supposing that they don't?

In fact, I think it would be more natural to assume that rotations do commute. After all, angular velocity is a vector, so it's integral is also a vector. And wouldn't you assume that the integral of angular velocity from t1 to t2 is a rotation? Of course, we know that this integral is not actually a rotation, but that's only because integrals are additive and we do know about the non-commutativity of rotations.

So in order to better understand why rotations don't commute, I thought that instead of struggling with SO(3), the rotation group of a sphere, it would be simpler to deal with the rotation group of a cube.

 

[D H]

Strictly speaking, angular velocity is *not* a vector. It is a pseudovector. Even better, it is a skew-symmetric, rank 2 contravariant tensor, or a bivector. A 2D skew-symmetric matrix is fully specified by one parameter. 2D rotation is characterized by a pseudoscalar. In takes three parameters to fully specify a 3D skew-symmetric matrix. Those three parameters look like a vector -- if you don't poke too hard. Go up to four-space and now you need six parameters to describe something that is rotating.

It is only in three space that you can pretend that angular velocity is a vector.

 

[lugita15]

Strictly speaking, angular velocity is *not* a vector. It is a pseudovector. Even better, it is a skew-symmetric, rank 2 contravariant tensor, or a bivector. A 2D skew-symmetric matrix is fully specified by one parameter. 2D rotation is characterized by a pseudoscalar. In takes three parameters to fully specify a 3D skew-symmetric matrix. Those three parameters look like a vector -- if you don't poke too hard. Go up to four-space and now you need six parameters to describe something that is rotating.

It is only in three space that you can pretend that angular velocity is a vector.

OK, some of this stuff rings a bell from Goldstein. I know about how rotations can be represented as orthogonal matrices, and in particular infinitesimal rotations can be represented by skew-symmetric matrices. So it makes sense that you can divide an infinitesimal rotation matrix by dt and get a skew-symmetric matrix for angular velocity.

But instead of using matrices, why can't we instead deal with pseudovectors as you mentioned? What kind of beast is the integral of a pseudovector like angular velocity? Would it be a polar or axial vector, or something else entirely? And if the integral of angular velocity is indeed some kind of vector, why does it not make sense to call this integral a rotation (other than the fact that we know rotations don't commute)?

 

[lugita15]

The group of all rotations and reflections of a cube and its n-D versions can be constructed as follows:

• Construct the n! permutations of the rows (or columns) of the n-D identity matrix. This is permutations of the cube faces' directions.
• Construct 2ⁿ diagonal matrices whose elements are +1 or -1. This is inversions of the cube faces' directions.
• Multiply them.

This group has these subgroups:

• Even permutations only
• Signs multiply to +1 only
• Determinant = +1 only: (even,+1) and (odd,-1)
• Even permutations and signs multiplying to +1

For 3D, these groups are

• Octahedral with reflections: Oh -- S4 * Z2
• Pyritohedral group: Th -- A4 * Z2
• Tetrahedral group with reflections: Td -- S4
• Octahedral group with rotations only: O -- S4
• Tetrahedral group with rotations only: T -- A4

For 2D, these groups are

• Rotations and reflections of a square: diihedral group D4
• Horizontal and vertical flips, 180d rotation: dihedral group D2 -- Z2 * Z2
• Horizontal-vertical interchange, 180d rotation: dihedral group D2 -- Z2 * Z2
• Rotations of a square (90d, 180d, 270d): cyclic group C4 -- Z4
• 180d rotation of a square: cyclic group C2 -- Z2

If one has some computer-algebra software, one can explicitly demonstrate these results by construction.

What would be the smallest or simplest rotation group which exhibits the non-commutativity of three-dimensional rotations? Is there anything simpler than the rotation group of a cube?

 

[lpetrich]

There are some infinite families of what may be called the axial rotation groups, those built around some axis of symmetry. Here are their elements.

But first the 1D and 2D rotation/reflection group elements.

1D: {{1}}, {{-1}}

2D: R(a,s) = {{cos(a), -s*sin(a)}, {sin(a), s*cos(a)}}

Pure rotation:
R(a,1) = {{cos(a), -sin(a)}, {sin(a), cos(a)}}
Reflection:
R(a,-1) = {{cos(a), sin(a)}, {sin(a), -cos(a)}}

Discrete groups: a = 2*pi*k/n, for cycle number n.
Continuous groups: a is continuous

Pure rotation: R(a,1): abstract group Z(n) / SO(2) ~ U(1)
Rotation and reflection (rotoreflection): R(a,1), R(a,-1): abstract group D(n) / O(2)

3D: R(a,sp,sa) = {{cos(a), -sp*sin(a), 0}, {sin(a), sp*cos(a), 0}, {0,0,sa}}

Pure rotation:
R(a,1,1) = {{cos(a), -sin(a), 0}, {sin(a), cos(a), 0}, {0,0,1}}
Reflection along axis:
R(a,1,-1) = {{cos(a), -sin(a), 0}, {sin(a), cos(a), 0}, {0,0,-1}}
Reflection in rotation plane:
R(a,-1,1) = {{cos(a), sin(a), 0}, {sin(a), -cos(a), 0}, {0,0,1}}
Both reflections (another pure rotation):
R(a,-1,-1) = {{cos(a), sin(a), 0}, {sin(a), -cos(a), 0}, {0,0,-1}}

There are seven families of discrete axial groups, which may be associated with the seven frieze groups:

p1: R(a,1,1), abstract group Z(n)
p1m: R(a,1,1), R(a,1,-1), abstract group Z(n) * Z(2)
p1g: R(a,1,1), R(a',1,-1), a' = 2*pi*(k+1/2)/n, abstract group Z(2n)
pm1: R(a,1,1), R(a,-1,1), abstract group D(n)
p2: R(a,1,1), R(a,-1,-1), abstract group Z(n) * Z(2)
p2m: R(a,1,1), R(a,1,-1), R(a,-1,1), R(a,-1,-1), abstract group D(n) * Z(2)
p2g: R(a,1,1), R(a',1,-1), R(a,-1,1), R(a',-1,-1), abstract group D(2n)

The continuous axial ones are
p1: R(a,1,1), abstract group SO(2)
p1m: R(a,1,1), R(a,1,-1), abstract group SO(2) * Z(2)
pm1: R(a,1,1), R(a,-1,1), abstract group O(2)
p2: R(a,1,1), R(a,-1,-1), abstract group SO(2) * Z(2)
p2m: R(a,1,1), R(a,1,-1), R(a,-1,1), R(a,-1,-1), abstract group O(2) * Z(2)

There are also quasi-spherical groups. These include the five tetrahedral, pyritohedral, and octahedral groups I'd mentioned earlier, and also the two icosahedral groups (pure rotation and rotoreflection). I'm not going to give how to construct the icosahedral-group elements unless someone really wants me to.

Their continuous versions are:
Pure rotation: SO(3)
Rotoreflection: O(3) = SO(3) * Z(2)
All the O(2n+1)'s break down like that; the O(2n)'s don't.

So I've given how to construct all the 1D, 2D, and 3D point groups, with the exception of the icosahedral ones. I hope that my descriptions have not been too difficult to follow.

 

[lugita15]

There are some infinite families of what may be called the axial rotation groups, those built around some axis of symmetry. Here are their elements.

But first the 1D and 2D rotation/reflection group elements.

1D: {{1}}, {{-1}}

2D: R(a,s) = {{cos(a), -s*sin(a)}, {sin(a), s*cos(a)}}

Pure rotation:
R(a,1) = {{cos(a), -sin(a)}, {sin(a), cos(a)}}
Reflection:
R(a,-1) = {{cos(a), sin(a)}, {sin(a), -cos(a)}}

Discrete groups: a = 2*pi*k/n, for cycle number n.
Continuous groups: a is continuous

Pure rotation: R(a,1): abstract group Z(n) / SO(2) ~ U(1)
Rotation and reflection (rotoreflection): R(a,1), R(a,-1): abstract group D(n) / O(2)

3D: R(a,sp,sa) = {{cos(a), -sp*sin(a), 0}, {sin(a), sp*cos(a), 0}, {0,0,sa}}

Pure rotation:
R(a,1,1) = {{cos(a), -sin(a), 0}, {sin(a), cos(a), 0}, {0,0,1}}
Reflection along axis:
R(a,1,-1) = {{cos(a), -sin(a), 0}, {sin(a), cos(a), 0}, {0,0,-1}}
Reflection in rotation plane:
R(a,-1,1) = {{cos(a), sin(a), 0}, {sin(a), -cos(a), 0}, {0,0,1}}
Both reflections (another pure rotation):
R(a,-1,-1) = {{cos(a), sin(a), 0}, {sin(a), -cos(a), 0}, {0,0,-1}}

There are seven families of discrete axial groups, which may be associated with the seven frieze groups:

p1: R(a,1,1), abstract group Z(n)
p1m: R(a,1,1), R(a,1,-1), abstract group Z(n) * Z(2)
p1g: R(a,1,1), R(a',1,-1), a' = 2*pi*(k+1/2)/n, abstract group Z(2n)
pm1: R(a,1,1), R(a,-1,1), abstract group D(n)
p2: R(a,1,1), R(a,-1,-1), abstract group Z(n) * Z(2)
p2m: R(a,1,1), R(a,1,-1), R(a,-1,1), R(a,-1,-1), abstract group D(n) * Z(2)
p2g: R(a,1,1), R(a',1,-1), R(a,-1,1), R(a',-1,-1), abstract group D(2n)

The continuous axial ones are
p1: R(a,1,1), abstract group SO(2)
p1m: R(a,1,1), R(a,1,-1), abstract group SO(2) * Z(2)
pm1: R(a,1,1), R(a,-1,1), abstract group O(2)
p2: R(a,1,1), R(a,-1,-1), abstract group SO(2) * Z(2)
p2m: R(a,1,1), R(a,1,-1), R(a,-1,1), R(a,-1,-1), abstract group O(2) * Z(2)

There are also quasi-spherical groups. These include the five tetrahedral, pyritohedral, and octahedral groups I'd mentioned earlier, and also the two icosahedral groups (pure rotation and rotoreflection). I'm not going to give how to construct the icosahedral-group elements unless someone really wants me to.

Their continuous versions are:
Pure rotation: SO(3)
Rotoreflection: O(3) = SO(3) * Z(2)
All the O(2n+1)'s break down like that; the O(2n)'s don't.

So I've given how to construct all the 1D, 2D, and 3D point groups, with the exception of the icosahedral ones. I hope that my descriptions have not been too difficult to follow.

Thank you, you've been quite thorough. But let me reiterate my question: what is the simplest discrete group of proper three-dimensional rotations which exhibits the non-commutativity property?

 

[lpetrich]

What would be the smallest or simplest rotation group which exhibits the non-commutativity of three-dimensional rotations? Is there anything simpler than the rotation group of a cube?

The smallest non-commutative group is D(3), of order 6.

It can be manifested as the rotoreflection symmetry group of the equilateral triangle. Its elements:
Don't do anything to it: identity
Rotate it 120d either clockwise or counterclockwise
Reflect it across any of three lines, each one going vertex - triangle center - line center

It's easy to extend this to 3D for an equlateral-triangular prism or extruded equilateral triangle (cycle 3, pm1). That object also has some additional symmetry groups:

(cycle 3, p2) -- D(3)
Don't do anything to it: identity
Rotate it 120d around the prism axis, either clockwise or counterclockwise
Rotate it 180d around any of three lines, each one going vertex - triangle center - line center

(cycle 3, p1m) -- Z(3) * Z(2)
Don't do anything to it: identity
Rotate it 120d around the prism axis, either clockwise or counterclockwise
The first three, with flipping along the axis added

(cycle 3, p2m) -- D(3) * Z(2)
The (cycle 3, pm1) ones
Those ones, with flipping along the axis added

 

[lpetrich]

First, p2m ought to be pmm and p2g ought to be pmg.

Some Wikipedia articles:
Point group
Point groups in two dimensions
Point groups in three dimensions

All the axial groups:

Schoenflies notation matched up with frieze group
C(n) -- p1
C(n,h) -- p1m
C(n,v) -- pm1
D(n) -- p2
D(n,h) -- pmm
D(n,d) -- pmg
S(n) -- p1g

All the quasi-spherical groups:

T -- Tetrahedral, reflection
Td -- Tetrahedral, rotoreflection
Th -- Pyritohedral
O -- Octahedral, reflection
Oh -- Octahedral, rotoreflection
I -- Icosahedral, reflection
Ih -- Icosahedral, rotoreflection