A Complete Idiot's Guide to E8
No one seems to have stepped up to the plate here, so let's have an amateur take a swing at the ball.
I'm going to talk about E8 as compared with the more familiar symmetry group SO(3) or SO(3,R). Wikipedia entries:
http://en.wikipedia.org/wiki/Rotation_group
http://en.wikipedia.org/wiki/E8_(mathematics)
The Symmetry Manifold and its Size
SO(3) is the group of rotations around the origin in 3 dimensions. When you rotate something, you get to choose an axis of rotation and how much to rotate around that axis. The axis of rotations choice is like picking a direction away from the origin. Let's count them.
You can move two perpendicular directions which gives you 2 dimensions. Or you can rotate around a spot and that gives you 1 more. Thus the SO(3) rotations are a 3 dimensional "manifold". Another way of saying the "3" is that if you begin with no rotations, there are basically three small movements you can make.
Get a globe and find Albuquerque, my home town. Think about what you can do to the globe, symmetry wise, relative to Albuquerque. You can move Albuquerque to the north/south, or to the east/west, or you can spin the globe on an axis around Albuquerque in a clockwise/counterclockwise direction. That is 3 dimensions and so SO(3) is a 3-manifold.
Our purpose is to talk about quantum numbers, but first let's talk about the dimensionality of the quantum numbers. That means "how many" quantum numbers each particle gets.
The Dimensionality of the Quantum Numbers
You get to have one quantum number for every motion you can do with your symmetry that is "independent" sort of. Two small motions are independent if it doesn't matter what order you do them in (i.e. they "commute" as in obey the commutation law of multiplication AB = BA so order doesn't matter). Independent motions are great. They're easier to analyze because you can fiddle with one without screwing up the other.
For the example of SO(3), the three small rotations do not commute. It might be obvious that rotation around Albuquerque doesn't commute with moving Albuquerque North/South. To see that moving Albuquerque North/South doesn't commute with moving Albuquerque East/West we can discuss the puzzle:
Suppose two people have good GPS systems and start hiking from the same point. Person X goes 1 mile East, and then 1 mile North. Person Y goes 1 mile North and then 1 mile East. Do they end up at the exact same point?
The answer is that, in general, they do not. To see why, get a globe, and see what happens if you increase the 1 mile to 1000 miles. Assuming that the starting point is Albuquerque (which is in the Northern hemisphere), you will find that the person who starts going North first, will end up farther to the east. The reason is that when you travel East at a higher latitude (i.e. more northerly) you cross more lines of longitude.
The same effect occurs for very small rotations. And the result of careful calculations is that none of the small rotations in SO(3) commute and so you can't break things up. By contrast, with E8 you can pick out 8 small rotations that commute. Therefore the quantum numbers of an E8 state requires 8 quantum numbers to specify.
Operators In Quantum Mechanics
The subject we are applying this theory to is quantum mechanics and so should discuss it a little. In quantum mechanics, the quantum states are "eigenvectors of operators". What that means is that if you write the operator as a matrix A, the quantum states are vectors \psi that satisfy the equation:
A \psi = \lambda_A\psi where \lambda_A is the quantum number. This will be familiar to people who've studied even the most elementary quantum mechanics book.
[note]Author is a proponent of the density operator formalism. In that formalism, the above is much sexier, but to discuss it here would unduly confuse most readers. Accordingly, with effort, he will suppress the urge to preach to you sinners.[/note]
To fully characterize a quantum state, you first choose as many operators A, B, C, ... which commute, and then define the quantum states as being eigenvectors of ALL these commuting operators. The reason for doing this is that it is always possible (due to some math theorems), and it gives you a nice clean way to describe the quantum states, namely their eigenvectors for A, B, C, which we can write as a vector (\lambda_A,\lambda_B,\lambda_C,...).
With SO(3), only one operator can be chosen, so there is only one quantum number. SO(3) isn't used as much in QM as the very similar symmetry group SU(2). In SU(2), that quantum number is called "spin". With E8, you can pick out 8 commuting operators, so to define a quantum state, you have to define 8 quantum numbers. With SU(2), you only have one commuting operator (which is usually chosen to be "spin in the z direction") so an SU(2) state gets only one quantum number.
Now if you've been paying attention in your elementary particles classes, you know that to distinguish an electron from a neutrino requires more than just 1 quantum number. To get all the particles into one group requires a more complicated symmetry group than SU(2). What Garrett did was to fit the known elementary particles into E8 by carefully assigning their quantum numbers. And he did it in a way that somehow respects gravity in a way that I do not understand yet but certainly got the Perimeter Institute to applaud.
Those who have studied beginning quantum mechanics learned that spin comes in various "representations". In spin 1/2, the quantum number (spin) is either -1/2 or +1/2. The difference between these two quantum numbers is 1.
In spin 1, the spin is either -1, 0, or +1. The difference between consecutive spins is 1. In spin 3/2, the spin is either -3/2, -1/2, +1/2, or +3/2. The difference between each is again 1. This difference between quantum numbers is consistent, and this is true in general. Note that with spin 3/2, you could talk about a difference of 2 or 3 between spin values instead of 1, but that would be a waste of time because 2 and 3 are multiples of 1.
For E8, there are 8 quantum numbers, so in any representation of E8, the difference between two different quantum numbers has to be given by an 8-vector. Similar to the differences between the quantum numbers of 3/2, you can choose a set of differences between quantum numbers of E8 (which are therefore 8-vectors), that are sufficient to get you anywhere you want to go, and are minimal in that you couldn't get rid of one. These weights are the origin of those diagrams with the little circles connected by lines.
Getting back to the pretty applet, the 8-vectors each correspond to the quantum numbers of a particle (in a specific representation of E8).
Did that help? By the way, note that the "Complete Idiot" in the title of this post is me.