Re: Group theory riddle

Hello:

If this is a good solution to the riddle, all the steps, no matter how
odd, should have a compelling reason behind them. Let's hope this
works out.

Here is the riddle again:

An observer is sitting in a Universe at the point in spacetime
t=x=y=z=0, also refered to as here and now. The observer sees an
event. No one knows what caused this event, nor does one care.
Describe, as completely as possible, the event, using group theory.

As my hint implied, it is best to approach this problem with the math
hat on. Therefore we need to state the basic ideas behind group
theory. Let's quote http://en.wikipedia.org/wiki/Group_theory
directly:

A group (G, *) is a set G closed under a binary operation *
satisfying the following 3 axioms:

* Associativity: For all a, b and c in G, (a * b) * c = a * (b * c).
* Identity element: There exists an e∈G such that for all a in G,
e * a = a * e = a.
* Inverse element: For each a in G, there is an element b in G
such that a * b = b * a = e, where e is the identity element.


We are given one specific point in the riddle, (0, 0, 0, 0). This
looks like an additive identity for 4-vectors if our binary operation
is addition, good old '+'. Let's call the other event 'A' for now.
So:

1. + Associativity: (0 + A) + A = 0 + (A + A)
2. 0 Identity element: 0 + A = A + 0 = A
3. + Inverse element: A + (-A) = (-A) + A = 0

That was good and boring, but shows the machinery of group theory.

Let's try and do something more with the other point, A. The other
point could be in the future, it could be in the past, it could be
near, it could be 10 billion light years away. The riddle says to
deal with all these situations. That's a large number of
possibilities, more than I can count using both my fingers and toes.
Let's simplify the problem. Let's work with the event normalized to
itself, norm(A/|A|) = (1, 0, 0, 0). If we could work with a sphere
that had a norm of one, then blow up or shrink the sphere to the right
scale, then A will live on that sphere. Spheres are a central actor
in the world of group theory, so this is promising.

Let's try and find a group for (1, 0, 0, 0). With the real numbers, 1
and multiplication form a group so long as the additive identity -
zero - is removed. Can the same be done with a 4-vector? The answer
is no, because vectors can only be added, subtracted or multiplied by
a scalar. By the definition of group theory, we need inverses. There
is only one choice here, known as the quaternions. If you are
unfamiliar with them, do your wiki reading here, http://en.wikipedia.org/wiki/Quaternions.
The super short answer is that quaternions behave exactly like the
numbers you know - they can be added, subtracted, multiplied and
divided - but do so with 4 numbers. The gang of four creates two
problems. Like the complex numbers, the quaternions are not a
completely ordered set as the real numbers are, where one real number
is greater than, less than or equal to another real number. A
quaternion could be to the left, up, or back of another quaternion,
and one cannot say that one quaternion is any "greater" than the
other. The second difference is that quaternions usually do not
commute, AB != BA. The "usually" is important, because quaternions
commute with themselves, and with polynomials made up of themselves,
because they all point in the same direction. It is the cross product
that makes quaternions products non-communitive, so if the cross
product is zero, AB=BA.

Now that we have been forced to use quaternions, let's use "the most
famous quaternion of all" in physics, the unit quaternion SU(2) that
sits in the middle of the stage of the standard model. This is one of
those things one looks up in a book, but a way to write the unit
quaternions is to use the 3-vector part of a quaternion, A - A*, and
take the exponential of the 3-vector: exp(A - A*) is an element of
SU(2). This is pretty good, but not good enough because it uses only
three of the four numbers sitting inside the quaternion. Let's
multiply this by A again, because A has the forth number in it. We
need to have this still live on the sphere of radius 1, so normalize
it:

(1, 0, 0, 0) = (A/|A| exp(A - A*))* A/|A| exp(A - A*)

We have a representative of the group SU(2) with the exponential, but
what about the other part, A/|A|? Notice:

A/|A| exp(A - A*) = exp(A - A*) A/|A|

Because there is only one quaternion in this expression, it always
points in the same way, the cross product is always zero, so these
commute. This additional factor has only one degree of freedom
sitting as it does next to this particular element of SU(2). This
translates into a description of the group U(1). The canonical
example for the group U(1) is the complex numbers with a norm of 1
which form a circle in the complex plane. For the quaternions, we
again have a circle with a norm of 1 in a plane that is determined by
the 3-vector A-A*. Notice that q/|q| is by itself not a
representative of U(1) because it has four degrees of freedom.

A/|A| exp(A - A*) is an element of U(1)xSU(2)

Now we are talking physics, specifically electroweak symmetry. This
is odd, since I have not been talking about photons or beta decay. I
shouldn't claim a connection, just a fun accident.

A pile up happens when an accident involves more than two cars. A
third vehicle to look for would be SU(3), the third musketeer of the
standard model. The group SU(3) has the Lie algebra su(3) which has 8
generators. We have 8 numbers in this universe, four 0's and four
numbers in A, but the zeroes have to be excluded from a group with the
multiplication operator. Bummer.

If the patient reader will indulge me to add one more event somewhere
in this Universe, we may proceed. Let's call this new event B. Let
us think about the following product:

(1, 0, 0, 0) = (B/|B| exp(B - B*))* A/|A| exp(A - A*)

This has eight degrees of freedom needed for the Lie algebra su(3).
This has a norm of 1. Technically, that conjugate in there is
important. Without it, the multiplication table would be the same,
and we would just get another element of U(1)xSU(2), because
U(1)xSU(2) is a group after all! Quaternion multiplication with a
conjugate is not associative, (A B)* C != A* (B C). That trips the
alarms, because a group is associative. Let's check the three rules:

1. * Associativity: ((A* B) (C* D)) (E* F) = (A* B) ((C* D) (E*
F))
2. 1 Identity element: 1 (A* B) = (A* B) 1 = A* B
3. * Inverse element: (A* B) (A* B)* = (A* B)* (A* B) = 1

where A = A/|A| exp(A - A*), likewise for B, C, D, E, and F
1 = (1, 0, 0, 0)

The three rules for a group are satisfied, the norm is one, there are
eight degrees of freedom, so this is a way to represent the group
SU(3). Cool!

Is this a way to represent the standard model since we can see all
three groups, U(1), SU(2), and SU(3)? The answer is unequivocally
no. The standard model has Lie algebras with 1+3+8=12 degrees of
freedom. These two quaternions have only 8. Many folks have
researched larger groups for the standard model, from SU(5) and
SU(10), to the recent work by Garrett Lisi on the exceptional Lie
group E8. To go the other way, to a group that is smaller, hardly
makes any sense, unless you worry about the issue of color
confinement. One is never going to see an isolated quark. As a
tensor product, there is no trivial reason why the SU(3) of
U(1)xSU(2)xSU(3) should be different from the other two. I have no
doubt quark confinement is part of the description of Nature given the
data from atom smashers, but feel like there is no reason for this
starkly different behavior. Where there is symmetry, their are
conservation laws, but there is not necessarily a new particle that
can live in isolation. For something like A/|A| exp(A - A*), there
are particles one can isolate for the U(1)xSU(2) symmetry, namely the
photon, the +/-W and the Z. The same applies for B, no change at
all. I cannot think about SU(3) without having an A and B. Lose one
or the other and the representation of SU(3) symmetry goes away
completely. That has some of the flavor of confinement.

In the world of programming, it is considered bad form to write code
with a hardwired constant. In the preceding analysis, there was a
specific number written in, the 1 of (1, 0, 0, 0). The 1 arose from
the normalization quite naturally. Now put on the physics hat, and
see if we can find a situation where things might be viewed
differently. Let's put another observer in this Universe who also
happens to be closer to a gravitational source mass that is also added
to said Universe. Both observers are looking at the same event, and
both observers calculate:

first observer:
(1, 0, 0, 0) = (A/|A| exp(A - A*))* A/|A| exp(A - A*)
second observer:
(1, 0, 0, 0) = (A'/|A'| exp(A' - A'*))* A'/|A'| exp(A' - A'*)

That is the way the math machine works. Yet we know that the metrics
of spacetime are different for the two observers. The norm is not an
invariant under a Lorentz transformation, but it will depend on the
g_00, g_11, g_22, and g_33 components of the metric, because it is
these terms that are getting mashed up. If the metric has a signature
+---, the norm would be:

norm(A) = g_00 A_0^2 - g_11 A_1^2 - g_22 A_2^2 - g_33 A_3^2

= A_0^2 + A_1^2 + A_2^2 + A_3^2 in flat spacetime

Without the gravitational mass, the metric is constant everywhere in
spacetime, and we get the expected norm, A_0^2+A_1^2+A_2^2+A_3^2,
which does not equal the norm of A' because the measurements start
from different places. With a gravitational mass, the second observer
works in a more curved spacetime. The second observer will naturally
think this is just the way the Universe happens to be. If we want to
use only one coordinate system for the Universe, the one for the first
observer, then by the time the coordinate system is transported from
the first observer to the second, the length of the norm of A' should
change. The only way that can happen is if the metric changes in a
smooth and continuous way. This is a quality of the group Diff(M) of
continuous diffeomorphisms of the manifold: lengths change because the
metric changes in a smooth way due to spacetime curvature.

All these groups can be pictured. The group U(1) is a unit circle in
the complex plane. You could draw it out on a piece of paper, with
time on one axis, and space on the other. If you were adventurous,
you would write software that would scan this graph, like a pencil
traveling from left to right. That would appear as a pair of dots
that separated from each other quickly at first, slowed down at the
maximum separation, then smashed into each other and disappeared. I
have written software to do just that.

For the group SU(2), I wrote above that one representation is exp(A' -
A'*). Take a thousand quaternions at random, toss them into this bit
of algebra, sort by time, collect those falling within particular time
zones, and use a 3D graphics program like POVRay to make a frame of an
animation. I've done that too. It is an odd animation, a series of
eight points that grow into each other until they form a sphere that
shrinks to nothing. I bought an ipod to carry around a visual
representation of SU(2), probably the only person on the planet to buy
an ipod for that reason.

It is just a different equation in the same software to calculate
U(1)xSU(2) and SU(3). The animations are all up on YouTube (listed at
end).


Unfortunately, no one was able to provide the complete answer I was
looking for, so the button will stay in my possession. The buttons
are for sale on line. Over a 5 year period of time, two people have
purchased the button. The sale last week to someone in California
motivated me to change the original design of the "Justify Your Love
for the Standard Model" button to be consistent with the algebra
discussed here. The graphic design work led to the idea for this
riddle.

Thanks everyone for giving it go :-)

doug <sweetser@xxxxxxxxxxxx>


Videos of group representations
The Group U(1): http://www.youtube.com/watch?v=KZeULLKHE7w
The Group SU(2): http://www.youtube.com/watch?v=OMnNyyZruuE
The Group U(1)xSU(2): http://www.youtube.com/watch?v=Jbdj3Xd_nmI
The Group SU(3): http://www.youtube.com/watch?v=8T_aNL8LvCs
The Group Diff(M)xSU(3): http://www.youtube.com/watch?v=pYiEV8yEZYA
All 5 in in: http://www.youtube.com/watch?v=ExNPiMcVXww

.