Given three colours, red, green and blue, if you are asked to paint the faces of a cube,
how many ways are there to do it? Well, for each face, we have 3 choices of colours, and
there are altogether 6 faces of a cube, so the total number of ways should be 3 to the
6, or 729 colouring patterns. However, what if we count any two colourings the same if
one can be transformed into another by a rotation, so that even though these two look different,
we still count them the same. With that consideration in mind, the question of how many ways of
coloring becomes more difficult.
This kind of counting problems where we consider symmetries is not only applicable in this
mathematical investigation, but also notably in counting the number of isomers of an organic
molecule, because apart from the letters “Cl” upside down, the molecule after rotation is
the same molecule. A perhaps less expected application is in music theory, where we count
the number of three-note chords, or trichords, where chords obtained via transpositions and
inversions are considered the same. But we will leave these non-mathematical applications
for another video.
Before we tackle all these questions, we need to first revisit orbit-stabiliser theorem
in Chapter 2 of the video series on group theory on this channel. Let’s say we have
a group G acting on a set X, and here we illustrate the set X as these dots. Let’s say we pick
this dot as little x, then we will define its orbit to be where little x can go under
the action of symmetries, so all it contains has to come from the set X. On the other hand,
symmetries that fix x form the stabiliser of X. The orbit-stabiliser theorem simply
states that the product of the sizes of the orbit and stabiliser is the size of the whole
group. The proof is basically this flow chart, which is fleshed out with much more detail
in this video.
Here is one simple fact about orbits that we will need. Let’s say these dim spots
represent elements of the set X, and pick an element to be our little x. Let’s say
these bright spots represent the orbit of x, or where x can go, and pick one of these
spots to be called x star. What is the orbit of x star? Can it, say, go to some other bright
spots as well? Of course. We can first go back to the little x, then go to the bright
spot that we want. This is because, by definition of being in the orbit of little x, there is
at least a symmetry to get x to each of these two spots. For the first step, we just revert
the symmetry, and keep this symmetry for the second step. Basically that means the orbit
of x star at least includes every bright spot. What about some dim spots? Can x star go there?
No, because if x star can go there, then little x can also go there, which is not possible
because the dim spot is not in the orbit of x, so x cannot go there. So the orbit of x
star is exactly the bright spots, no more no less.
If we pick x somewhere else instead, then its orbit would be different. Eventually,
everything in the set would belong to exactly one orbit. With a bit more knowledge about
orbits, let’s see what we actually want. We want to count the number of ways to color
a cube with symmetries in mind. The natural action to take is to let the group of all
rotational symmetries of a cube act on the set of all the possible colorings. Representing
the set of all colorings as these little dots, what does being in the same orbit mean? Or
what does being able to transform from x to x star mean? Because G is the group of rotational
symmetries of a cube, that means we can go from x to x star via a rotation. But we say
that we would count them the same if one can be transformed to another by a rotation, so
the entire orbit represents the same coloring. The same thing applies in other orbits, so
counting the number of colorings means counting the number of orbits in group theoretic terms.
So we will now focus on counting the number of orbits.
We will use a trick to find the number of orbits, which starts off by lining up the
elements of X, as well as the symmetries in the group G. Then we start to draw some lines
between G and X, but with the criterion that we only draw lines between little g and little
x if the little symmetry g does not move little x under the action. So for example, if this
is the identity symmetry, then by definition it fixes everything, so we will draw lines
from this identity symmetry to every element of X. After we draw all the lines, we want
to count how many lines we have drawn. There are actually two ways to count.
The first method is to count the number of lines from each symmetry in group G, then
add them all up. But we need to first understand what each number we are adding actually means.
Based on the criterion on how we draw the lines, these numbers are the numbers of the
x’s each symmetry fixes. In this case, it fixes this green dot, and we call the collection
of things that is fixed by g the fix of g. So each number here represents the size of
the fix of g, and we are summing up all these sizes of the fixes. But there is another way
to count the total number of lines, which is counting the number of lines to each element
in X then adding them up.
However, instead of adding them individually, we add them in batches. We are going to have
the subtotal by counting everything from the same orbit, then adding the subtotals up to
get the grand total. Because of the criterion to determine how we draw the lines, the lines
are drawn from the stabilizers. That means each element contributes exactly the size
of the stabilizer to the subtotal. The reason why we consider subtotals first can be seen
from the orbit-stabilizer theorem. Everything inside the orbit has the same orbit, so their
orbit sizes must be the same. This in turn means that their stabiliser sizes must also
be the same, so each element in the same orbit contributes equally to the subtotal. Since
the number of elements in the orbit is precisely the size of the orbit, we multiply by the
size of the orbit to get the subtotal. This is precisely the size of the group G by the
orbit-stabilizer theorem. Using exactly the same argument, every orbit will contribute
the size of G as the subtotal. So the grand total is the number of orbits times the size
of the group.
Because these two expressions come from counting the same thing, they should be equal. And
by simple rearrangement, we can get the expression for the number of orbits. This result is known
as Burnside’s lemma, the topic of this video.
Let’s go back to the original problem. We want to find the number of orbits under the
action of rotations on the set of all colorings of a cube. Let’s look again at the formula
from Burnside’s lemma. The first thing we need to evaluate is the size of the group.
Of course you can list them all out, but in practice, it is very likely we miss some symmetries,
so we can use orbit-stabilizer theorem, but on a much simpler action. Instead of acting
on the set of colorings, we simply act on the faces. In this cube, we pick this top
face as the little x. It is not difficult to see that x can be rotated to any other
face, so the orbit size is 6. Symmetries that fix this face would contain only the rotations
around this axis, which are precisely identity symmetry or rotations of a multiple of 90
degrees, which means the stabilizer has size 4. So by orbit-stabilizer theorem, the size
of the group of rotations is 24.
Now, we need to find the fixes of symmetries. Let’s first consider the fix of the identity
symmetry, that is, no rotation at all, then it fixes every coloring, so its fix size would
be the size of the whole set of colorings, which is 3 to the 6, where 3 is the number
of choices of colors for each face, and 6 represents the number of faces of a cube.
Anyway, let’s move on to another symmetry that actually involves rotation. The simplest
is a rotation about this axis. First we think of the rotation by 90 degrees, clockwise or
counterclockwise. What does this rotation fix? If this face is red, and we are considering
such an anticlockwise rotation, the face on the left would be rotated to the red face,
which means the face on the left has to be red in the first place.
Using the same argument, all four lateral faces of the cube must also be of the same
color. There are no restrictions on the top and bottom faces, so the fix size is 3 cubed,
where the base denotes the number of colors available, and the exponent denotes the number
of faces that we can actually choose colors, that is the top and bottom faces, and one
of the lateral faces. But there are three such axes, and for each axis, we can rotate
90 degrees either clockwise or anticlockwise, so we need to count this 6 times. The rest
of the argument works in a similar way, so I will walk it through with a little bit less
detail. Using the same three axes, we can consider the 180 degree rotations. It turns
out the fix size is 3 to the 4, because we can freely choose the colors for the top,
bottom, and 2 of the four lateral faces.
Next up, we have rotations about the long diagonals by 120 degrees. This refers to rotation
along this axis. Since there are 4 such axes, and we can turn 120 degrees clockwise and
counterclockwise, there are 8 such rotations. It turns out the fix size is 3 squared, because
faces around a point need to be of the same color in order to be fixed. The final category
of rotation axis passes through the midpoint of opposite edges, which refers to rotations
like this. There are altogether 6 such axes, passing through each pair of opposite edges.
So there are 6 such rotations because one axis only corresponds to one rotation. It
turns out the fix size is 3 cubed, because we can choose the colors for 3 of the faces,
then the rest has to follow.
The final step is to add up these fix sizes, considering facts like there are 6 90 degree
rotations, so we need to multiply the corresponding fix size by 6, and so on. The total now turns
out to be thirteen sixty eight. Putting this directly into the formula, we see that we
also need jn the size of the group G. So we put that 24 in as well, and we get 57. So
the answer to our original question - the number of rotationally distinct ways of coloring
a cube when we are given three colors is 57. However, this is just a very classic example
of using Burnside’s lemma, it does have applications in a non-mathematical context,
like organic chemistry, or even music theory. But we will explore those in another video.
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