In the previous video, we used Burnside’s lemma to tackle the cube painting problem,
and in this video, we are going to use the lemma to tackle the non-mathematical applications,
i.e. the chemistry and music problems briefly mentioned in the previous video. Burnside’s
lemma gives us this formula, and to use this properly, we simply need to have appropriate
choices of the group G and the set X. Let’s look at the chemistry application first.
In this video, we will be considering a molecule called naphthalene, which has this chemical
structure. More precisely, we are considering naphthalene with 4 chlorine substituents,
or tetrachloronaphthalene. If you don’t know organic chemistry, here is a brief explanation
of what this diagram actually means. All the vertices here represent a carbon atom, and
under stable circumstances, each carbon atom is like this one in the middle, having four
bonds with something else. However, if we consider this carbon atom instead, it looks
like it only has 3 bonds, so where does the hidden bond go? In a simplified chemical formula,
any implicit bonds must be with hydrogen. We can also stick a chlorine atom here to
substitute the hydrogen atom, which is why chlorine is usually called a substituent.
So the question we are interested in is how many naphthalene molecules there are with
4 chlorine substituents. There are altogether 8 places for the chlorine substituents, and
we choose 4 of them.
You are right to think that the 3D structure might affect how we consider symmetries, but
don’t worry because luckily all the atoms are on the same plane for this molecule. With
this in mind, there are 4 symmetries that we need to consider, including the identity,
a reflection along the vertical axis, a reflection along the horizontal axis and a combination
of the two, which is a rotation by 180 degrees. We now use Burnside’s lemma to actually
compute the number of naphthalene with four chlorine substituents. The first thing we
need is the size of the group. Well, we have already listed all of them at the bottom,
so we can substitute size of G as 4 at the top here. The next thing is the size of the
fixes.
Firstly, we will consider the fix of the identity. By definition it fixes everything, so its
fix size would be the whole set. There are altogether 8 places that we can place our
chlorine substituents, and we choose four out of it, so its fix size is 8 choose 4.
The next symmetry is the horizontal reflection. In this symmetry, 1 is swapped with 4, so
in order for a molecule to be fixed by the reflection, both 1 and 4 need to attach to
the same thing, i.e. both must have the chlorine, or both not having the chlorine, and default
to hydrogen. Similarly, we have other pairs of carbons swapping with each other under
the reflection, and again in each pair, both carbons have to attach to hydrogen together,
or chlorine together. So out of the 4 pairs of places for the chlorine substituents, we
choose 2 pairs, making the fix size 4 choose 2.
Next up, we consider the vertical reflection. But similar to what we saw before, pairs of
carbons are swapped under the reflection. So the argument for the fix size being 4 choose
2 still applies. Now what about the 180 degree rotation? It might be a bit more difficult
to see, but again pairs of carbon atoms swap under the rotation, so this 4 choose 2 argument
still stands in this case. So finally, to add up all the fix sizes, we first add the
8 choose 4 for the identity, then add three times 4 choose 2 because for all the three
other symmetries, their fix sizes are 4 choose 2, which totals to 88. By Burnside’s lemma,
the number of orbits will be 22 after dividing by 4, the size of the group. So there are
altogether 22 tetrachloronaphthalene molecules.
Another application of Burnside’s lemma is in music theory, which starts off with
this circle. In Western music, there are 12 tones, and the pitch goes up when we go clockwise
around this circle, so if you start with middle C [play note] for example, going around the
ring clockwise, you will end up at another C an octave higher [play note]. In this case,
we are only concerned with the tones, not specific octaves, so there are 12 tones and
that’s it. In this video, we want to know how many trichords there are, or in other
words, we choose 3 tones out of the 12. But there is a catch. If we can shift a chord
[play chord], to another chord [play chord] just by shifting each note an equal amount,
we would count the two chords the same. So in counting how many trichords, we need to
take rotation into account, because if we can rotate one chord to another, we count
them the same.
Similar to what we did before, we would need Burnside’s lemma, and the first thing we
need to do is to list all the symmetries we will consider, and compute the fix size of
each of the symmetries. We first consider the identity, which always fixes everything
in the set. In this case, out of the 12 tones available, we choose 3 of them to form the
trichord, so there are altogether 12 choose 3 things in the set, which is 220, which we
record as the fix size. What about the fix size of the 30 degree rotation, clockwise
or counterclockwise? C is transformed to C sharp, so if C is chosen, C# must also be
chosen, but C# goes to D, so then D must also be chosen, and so on and so forth. So either
every tone is chosen, or nothing is chosen. Since we are aiming for trichords only, the
fix size is 0 because no trichords get fixed in this symmetry.
For 60 degree rotations, C will go to D, so if C is chosen, D must be chosen as well,
and so are E, F sharp, G sharp and A sharp. But we only need to choose 3 of them so we
cannot choose the C note. If C sharp is chosen, D sharp must also be chosen, and in a similar
way, the other *four* notes have to be chosen as well. So actually again, the 60 degree
rotation does not fix any trichords. 90 degree rotations? Well, if C is chosen, then all
these four have to be chosen as well, but the same thing happens if we choose any other
note as part of the trichord, so the fix size of the 90 degree rotation is still 0. But
things start to change when it comes to 120 degree rotation, because if C is chosen, only
three notes need to be chosen because of it. There are altogether 4 trichords fixed by
the 120 degree rotation, so the fix size is 4. However, since both clockwise and anticlockwise
120 degree rotations would have the same fix size, we multiply this by 2.
Next up, we have the 150 degree rotations. If C is chosen, then it turns out that through
this logic, we would require every single tone to be chosen. So similar to what we did
with the 30 degree rotation, the fix size is 0 as well. The final symmetry is the 180
degree rotation. In this case, pairs of notes are swapped, so pairs of notes get chosen
or not, but since we are considering three-note chords, and three is odd, the fix size is
0 again. So we can finally use Burnside’s lemma to tackle this problem. There are altogether
12 rotations that we have considered, and the total fix size is simply adding everything
in this column, which is 228. So there are altogether 19 unique trichords.
While both non-mathematical applications do prove the use of Burnside’s lemma, they
are often cited as examples of an even more general result known as Polya enumeration
theorem, but since it is a general result, it is bound to be notationally heavy, so I
would direct those who are interested in finding more to the description, where you will find
resources to learn about this theorem.
Before you click away, here is something that I haven’t talked about in my videos for
a while now, which is the Google form linked in every video’s description. It allows
me to know your math levels and make the planning of future videos better. So please do fill
in the form to help me create better videos.
In any case, like, subscribe, hit the bell, and leave a comment if you enjoyed this video.
See you next time! Bye!