From: Michael Weiss
Date: Fri, 17 Mar 95 15:56:55 -0500
To: rusin@math.niu.edu
Subject: More Ru-brik-a-brac
[deletia]
----------------------------------------------------------------------
[Note to moderator: this is a little long, so I've divided it into two
parts. ============= marks the split.]
Back when Rubik's cube was all the rage, Solomon Golomb noticed
something remarkable: the mating habits of quarks mirror the twisting
properties of the `corner cubies'. ("Rubik's Cube and a model of quark
confinement", Amer. J. Phys., 49 (11), November 1981).
If you've never played with the Cube (as rubicionados call it), you
should probably just go on to the next post--- I won't attempt a
verbal description! But even if you have, you might not have heard of
Golomb's observation. So let me explain. (You might also want have a
look-see at Douglas Hofstadter's two columns on Rubik's Cube,
reprinted in his book "Metamagical Themas".)
Rubik's cube is made up of 26 so-called `cubies': 8 corner cubies, 12
edge cubies, and 6 face-centered cubies. As you twist the Cube
around, the face-centered cubies rotate in place, and the other cubies
move from place to place while changing orientation. Instead of
`place to place', the usual lingo says 'cubicle to cubicle'--- the
cubicles are the little bits of space occupied by the cubies.
This is tailor-made for group theory, and we let Rubik's group be the
group of all possible operations you can perform on the Cube. (Well,
not *all* possible operations--- we don't include smashing it in
frustration!) Note that an element of the Group (as I'll call it)
permutes the cubies, but this permutation doesn't completely specify
the group element. In fact, some group elements don't evict *any*
cubies from their cozy home cubicles, but merely reorient them in
place, like so many couch potatoes. The set of these group elements
forms a subgroup--- I'll call it the Subgroup.
OK, let's consider an element of the Subgroup. Looking at a corner
cubie, we have three possibilities: no motion at all, or a
counterclockwise twist, or a clockwise twist. We now have a
conservation law, of a sort. If we count clockwise twists as +1,
counterclockwise as -1, and no motion as 0, then:
The sum (mod 3) of the twists of all the corner cubies
is zero for any element of the Subgroup.
(I'm not saying this is obvious!) Shall we call this the Law?
Perhaps not. I'll call it Rubik's Rubric. (Webster: "ru-bric ...an
authoritative rule...")
A similar law holds for the sum of the flips of the edge cubies, mod
2. But this smacks of parity, and parity is a deal less exotic than
triality. So I'll just forget about edge cubies from now on---
pretend we have, not a Rubik's Cube, but a Twobik Cube (credit
Hofstadter with the pun).
And we are now ready for Golomb's observation. Mesons are made with a
quark and an anti-quark, and baryons with three quarks or three
anti-quarks; well, an element of the Subgroup can create a twist plus
an anti-twist, or three twists or anti-twists, but never an isolated
twist. (Guess we should call this the Toy Model of Quark Confinement!)
You're probably muttering, "This is sheer coincidence!" (Can one
mutter an exclamation point?) Indeed, Golomb doesn't claim a deep
connection--- even though group theory plays a prominent role both in
Cube theory and QCD. Recently, though, I was reading Sternberg's
"Group Theory and Physics", and noticed to my surprise that the theory
of principal bundles (also a corner-stone of QCD) furnishes a neat
proof of Rubik's Rubric. If you already know all about bundles, then
I think you'll like this example; and if you don't, it's a fun place
to start learning!
(Sternberg's book serves up lots of other juicy examples. If you like
dodecahedra, you'll enjoy his treatment of the infrared and Raman
spectroscopy of the buckyball, C_60, by way of Frobenius reciprocity.
You'll find nice points of contact with John Baez's essay "Six" as
well.)
OK, how do we prove Rubik's Rubric? Say g is an element of the Group,
say c is a cubie, and say c stays in its own cubicle under the action
of g. I'll write twist(g,c) for the element of Z_3 that says how c is
twisted: 0 for no twist, +-1 for clockwise/counter-clockwise twists.
So the claim is that if g is an element of the Subgroup, then
Sum twist(g,c) = 0
all cubies c
An obvious line of attack: define twist(g) to be the sum on the
left-hand side of this equation; then show that twist(gh) =
twist(g)+twist(h) (i.e., twist is a homomorphism) and show that
twist(g)=0 for some set of g's that generate the Subgroup.
The homomorphism property is pretty obvious. Not so a set of
zero-total-twist generators for the Subgroup. On the other hand,
there *is* a natural set of six generators for the Group: the 90
degree turns of the six faces. (Each generator moves four cubies on
Twobik's Cube.)
The problem is the natural generators of the Group *do* move the
cubies around. And it's not obvious how to define twist(g,c) (and
hence twist(g)) when cubies move.
Bundle theory to the rescue! (If you already know about principal
bundles, skip right to part II of this post.)
A *fiber bundle* has a base space, a total space, and a projection map
from the total space onto the base space. There's usually more to the
definition, but first let's look at the poster child of fiber bundles:
the Mobius band. The total space is the set of all points on the
band, and the base space is the set of all points on the meridian of
the band (so the base space is a circle). To project a point on the
band into the meridian, just drop a perpendicular.
The *fiber* over a base-space point is just the set of all total-space
elements that project down to that point. For the Mobius band, all the
fibers are line segments.
Usually, all the fibers "look the same". What does that mean? Well,
it depends on what sort of bundle you're talking about (or in the
usual jargon, what category you're dealing with.) More precisely, we
assume that all the fibers are objects in some category, and demand
that they all be isomorphic--- every category comes fully equipped out
of the box with a notion of isomorphism. So you can get your daily
dose of fiber from a varied menu: vector spaces, groups, topological
spaces, manifolds--- personally I recommend the category-of-the-day,
G-sets, with a dry white wine...
For a topological fiber bundle, one demands even more. For one thing,
the total space must also be a topological space, and the projection
map must be continuous. There's another condition ("local
triviality") I won't bother to spell out, but for the Mobius band it
amounts to this: if you snip off a little piece of the band, you get
something homeomorphic to a rectangle.
Now here's a key point: while we said any two fibers must be
isomorphic, we didn't say there was a "natural" or "canonical" or
"preferred" isomorphism between a pair of fibers. In a way, that's the
whole point of bundle theory. Take the Mobius band: if you "slide a
fiber a little bit along the meridian", you get a homeomorphism
between the fiber in the old position and the fiber in the new
position. But if you slide the fiber smoothly once around the band,
it *flips over*. Even without working through the details, it's clear
we have two competing homeomorphisms of the fiber to itself: the
identity map, and the flipping map.
For the Cube, discreteness reigns. So forget about topology. But we
do have some groups to consider. The category of choice is now *sets
with a group-action*. This calls for a new batch of definitions.
Let S be a set, and let G be a group of permutations of S. In modern
jargon, one says that G acts on S. If you think of g in G as a
function, you write g(s) for the result of g acting on s, but it turns
out to be more convenient to write actions like multiplication. We
have a choice of notation:
gs (G acting on the left)
sg (G acting on the right)
Say g and h are in G. If you write your action on the left, then gh
means "do h, then g"; if you write it on the right, then gh means "do
g, then h".
It turns out that when things really heat up, you want keep *both*
notations handy. For example, you might have *two* groups of
permutations, H and K, with H acting on S on the left and K on the
right.
OK, let's say G acts on S on the right. So (sg)h = s(gh). Now there
are two particularly interesting properties an action can have:
(*) Transitivity: G acts *transitively* if you can always get
from one given element of S to another given element via the
action of G. I.e., for any s and t in S, there is a g in G
such that sh=t.
(*) Freedom: G acts *freely* on S if an element of G leaving one
element of S fixed must leave all of S fixed. I.e., if sg=s
for some s, then g is the identity permutation.
These are sort of like being onto and one-to-one. Indeed, if the
action of G is both transitive and free, then we can set up a 1-1
correspondence between G and S quite easily: pick some fixed s_0 in S,
and let g correspond to (s_0)g.
A *principal G-bundle* is a fiber bundle where the fibers are sets on
which G acts both freely and transitively. G is called the *structure
group* of the bundle.
Example? The Cube--- as we will see in part II.
======================================================================
Part II
Picking up where we left off, we were about to see how the Twobik Cube
is an example of a principal bundle. The structure group is Z_3, the
integers mod 3.
The base space is the set of cubicles. The fiber over a cubicle is
the set of three ways a cubie can sit in that cubicle. To picture
these, it's convenient to label the three visible faces in some
arbitrary way, say x, y, and z. (But this labelling is not itself
part of the fiber!) So the three elements of the fiber might look
something like this:
x z y
. . .
y z x y z x
That dot in the middle stands for the vertex of the cubie; I haven't
tried to draw the edges. So Z_3 cyclically permutes these three
elements, twisting the cubie from one orientation to another. There
are eight fibers (remember we have only corner cubicles), and so the
total space of the bundle consists of the 8*3=24 possible positions
for a cubie.
You can see how any two fibers are isomorphic. Pick a random element
of the first fiber; call it p_1. Pick a random element of the
second fiber; call it q_1. Now give p_1 and q_1 each a clockwise
twist, and call the results p_2 and q_2. One more twist gives us p_3
and q_3, and the isomorphism is
p_1 <--> q_1
p_2 <--> q_2
p_3 <--> q_3
You can construct three different isomorphisms this way. Nothing
singles out any one of these isomorphisms as the preferred one.
So what does this have to do with Rubik's Rubric? Remember the
problem we faced: if g is an element of the Group (*not* of Z_3), and
c is a cubie, we wanted to define twist(g,c), the amount that cubie c
is twisted by g.
Our cubie has become an element of the fiber over its cubicle.
How about g? Elements of the Group move cubies from one position to
another. So if g is an element of the Group, then g defines a map
from the total space to itself.
The Group is, in fact, a subgroup of the *automorphism group* of the
bundle. This just means that if you take a cubie, and twist it and
then apply a Group element, you get the same result as when you apply
the Group element first and the twist second. A little notation
helps: let t be in Z_3, let g be in the Group, and let c be a cubie
position (an element of the total space). We let Z_3 act on the right
and the Group on the left, so ct stands for c with a twist, and gc
stands for c after being moved by g. So the automorphism property
takes the pretty form:
(gc)t = g(ct)
Now how about defining twist(g,c)? If gc happens to belong to same
fiber as c, then no problem: g just twists c by some amount. In other
words, gc = ct for a unique t in Z_3, and we can define twist(g,c)=t.
But the hard part remains: what if gc is in a different fiber? In
this case, we need some reference cubie in the new fiber to compare gc
to. It seems we need a "global coordinate system" for the bundle.
Bundle theory interrupts: "Don't use a coordinate system, use a
*section* of the bundle!" (Such has been one of the lessons of bundle
theory over the years: coordinate systems are more-or-less equivalent
to sections. Which you use is often a matter of taste, but modern
mathematical fashion much prefers to talk of sections.)
A *section* of a bundle simply picks an element from each fiber. Some
more notation: if s is a section and x is a point in the base space,
then s_x is the chosen element of the fiber over x. So a section
in our example assigns a cubie position to each cubicle.
The Group is pretty active: it acts on cubicles, on cubie positions,
and on sections. I won't write out descriptions, just a little
notation. Let g be a Group element, x a cubicle, s a section, and s_x
(as we just said) a cubie position in the fiber over x. Then:
gx is a cubicle
g(s_x) is in the fiber over gx
gs is a section, defined by the
equation (gs)_(gx) = g(s_x)
for all x
That last equation may look like a jumble of symbols the first time
you see it, but after working through it, I'll bet you say "Of
course!"
Still sticking with the same meanings for g, s, and x, define:
twist(g,s,x) = the twist needed to change the position of x's
old occupant to the position of the new
occupant
or more formally:
twist(g,s,x) = the unique t in Z_3 such that (s_x)t = (gs)_x
Hmm, those parameters to the twist function keep piling up... Well,
let's check out the dependence on the choice of section s. Let's say
r and s are sections. You can turn r into s by twisting each cubie in
place. In other words, we have yet another group to worry about: the
direct product of eight copies of Z_3, one for each cubicle. Letting
this new group (call it T) act on the right, we have rt = s, for some
t in T. Borrowing some notation, we let t_x stand for the twist at
cubicle x. (Notice that the Subgroup is a subgroup of T. Note also
that twist(g,s,x), as x varies over the eight cubicles, defines an
element of T.)
OK, let's compare twist(g,r,x) with twist(g,s,x). We need to keep
track of old occupants and new occupants. Let's say the new occupant
at cubicle x originally resided at cubicle y. (Really you should work
this out for yourself. But I'm feeling generous today, so here goes.)
A diagram may help:
g
y ============> x
g new occupant: old occupant:
r_y ============> g(r_y) r_x
g
s_y ============> g(s_y) s_x
E.g., with the r section, r_x was the old occupant at x, and g(r_y)
is the new occupant.
We want to compute the twist between the old and new occupants. The
following diagram says it all:
twist(g,r,x) g
r_x ---------------> g(r_y) <============ r_y
| | |
t_x| |t_y |t_y
| | |
v v v
s_x ---------------> g(s_y) <============ s_y
twist(g,s,x) g
Single-weight arrows (--->) stay in the same fiber, double-weight
arrows (===>) move from fiber y to fiber x. The two vertical arrows
on the right are both labelled t_y because g[(r_y)(t_y)] =
[g(r_y)](t_y) --- g is a bundle automorphism, remember! Look now at
two ways to get from s_x to g(s_y): take the twist(g,s,x) route, or
take t_x in reverse, then twist(g,r,x), then t_y down. So we have the
following equation in Z_3:
twist(g,s,x) = twist(g,r,x) + t_y - t_x
So twist(g,s,x) is *not* independent of s. But here's the punchline:
summing over x removes the dependence.
sum twist(g,s,x) = sum twist(g,r,x)
x x
For in the sum, we add t_x and subtract t_x once for each x.
And that's all we need! We *don't* ever define twist(g,x)--- we go
directly to twist(g), our goal all along.
The rest is a smooth coast down the bunny slope. If you pick the
right section, a generator g of the Group (a 90 degree twist of one
face) has twist(g,s,x) = 0 for all x. As for the homomorphism
property:
twist(gh) = twist(g) + twist(h)
the trick here is to pick one section s to compute twist(h) and
twist(gh), and another section to compute twist(g). Work it out if
you're skeptical. QED (or `\omicron\epsilon\delta, as Euclid wrote).
Three groups played a role in our little drama: Z_3, the Group, and a
group I called T earlier--- the direct product of eight Z_3's.
Z_3 is called the structure group of the bundle. We saw that the
Group was a subgroup of the automorphism group of the bundle, Aut(E).
The group T, which twists cubies in place, also has a special name:
it's called the *gauge group* of the bundle. So the Subgroup is a
subgroup of the gauge group.
Now, why did things work out so nicely? A mathematician's question.
A story has it that a grad student was walking on the campus, when
he met a jubilant physics prof. Said the prof, "I finally finished my
calculations, and everything magically clicked into place! Now I
understand it all!" The student congratulated him, and walked on.
Next he met a gloomy math prof, who said, "I finally finished my
calculations, and everything magically clicked into place! I don't
understand it at all!"
In this case, the answer is: because the structure group is abelian.
That's really all we need. If you trace through the argument
carefully, you'll find it depends on three things:
(*) The structure group is abelian.
(*) The automorphism group acts transitively on the base space.
(*) The base space is finite.
If the base space is *not* finite, I imagine you can push the argument
through with an integral instead of a sum--- provided you pick the
right categories, of course!
Now the structure group of QED is U(1), an abelian group. Perhaps
Rubik's Rubric corresponds to some well-known fact of E&M, maybe
conservation of charge?
But quarks--- well, the big thing about quarks is that the structure
group, SU(3), is *not* abelian. (If there's anything big about
quarks!)
So my feeling is that Golomb was quarking up the wrong tree. And the
skeptical reader, who doubted that Rubik's Cube could have anything to
do with quarks, seems to have the last laugh.
Except we have this tantalizing remark by John Baez:
One other little observation... Z_3 is deeply related to
SU(3), because the center of SU(3), consisting of the diagonal
matrices exp(2 pi i n/3), is isomorphic to Z_3. More generally,
the center of SU(n) is Z_n in the same way. For SU(2) it's
Z_2, and when we mod out by this Z_2 we get good old SO(3).
From this one would suspect that Z_3 representation theory is
important in SU(3) representation theory. My guess: Each irrep of
SU(3) is obviously (or should I say Schurly) an irrep of Z_3,
of which there are 3, and to tell which one, just count
the number of boxes in your Young diagram mod 3. I'm finally
really learning about Young diagrams and I love 'em!
The corresponding Z_2 invariant of SU(2) irreps is just
the "integer/half-integer spin" distinction!