(Third and final part.) REPRESENTATIONS, WEIGHTS, AND ROOTS Suppose we have a Lie group and its corresponding Lie algebra which, remember, is kinda like a blo

(Third and final part.) REPRESENTATIONS, WEIGHTS, AND ROOTS Suppose we have a Lie group and its corresponding Lie algebra which, remember, is kinda like a blown-up copy of an infinitesimally tiny bit of the original group. Then there's a natural way for the group to "act on" the algebra, which roughly speaking corresponds to the ordinary "do one symmetry and then another" operation in the group. But the Lie algebra is, among other things, a vector space (what I've described above as a euclidean space), and it turns out that each symmetry in the group gives us a linear transformation of that space. Another way to say this is that we're taking our abstract symmetry group, and exhibiting its elements as symmetries of a particular object, namely the Lie algebra. This situation, where you have a group and you find a way to turn its elements into linear transformations of a vector space, is called a representation of the group. We can also have the Lie algebra act on itself in a basically-equivalent way, and then what we get is a representation of the Lie algebra. (In quantum physics, the state of the universe is described by a vector in an enormous vector space, symmetries of the laws of physics correspond to linear transformations on that vector space, and so representations arise quite naturally. If instead of asking about the whole universe we ask about some much smaller configuration of particles, we get a much smaller vector space; if the symmetries we're interested in leave that configuration of particles alone, we get a representation on the smaller vector space.) The next bit is kinda unmotivated. I'm sorry. So, suppose you have a representation of a Lie algebra -- a manifestation of its elements as linear transformations on a vector space. Consider what's called a "Cartan subalgebra", which is to say as large as possible a subspace of the Lie algebra with the property that within it all the brackets are zero. (Meaning that the corresponding things in the Lie group commute, meaning that it doesn't matter whether you do A then B or B then A.) Let's give names to some key objects here. G consists of the linear transformations corresponding (under our representation) to the original Lie algebra. H consists of the ones coming from the Cartan subalgebra; so it's just a portion of G, usually a lot smaller than G itself. Now, suppose we have g in G and h in H. We can apply our bracket operation to them -- in fact, since we're doing all this via the representation, the bracket is always just gh-hg in terms of matrix (= linear transformation) multiplication. If we're lucky, the following amazing thing might happen: this might (1) always be a scalar multiple of g, where (2) the actual multiple only depends on h. That is, we might have [g,h] = f(h)g for some function f, which in fact will always have to be linear. That would be a hell of a coincidence, but perhaps a smaller coincidence might happen: the same might hold but only for some subspace of G. The linear functions on H for which this does happen for a nontrivial subspace of G are called weights for the representation. (It's a pretty terrible name. I'm sorry.) There's a particularly important representation called the "adjoint representation", which is the one I described above immediately under the heading "REPRESENTATIONS, WEIGHTS, AND ROOTS". The weights of the adjoint representation are called "roots", which is also a terrible name. What happens for E8? Well, the Cartan subalgebra is pretty small; it's 8-dimensional (versus 248-dimensional, as it happens, for E8 itself). To describe a linear function that maps an 8-dimensional space onto numbers, you need 8 numbers -- e.g., the first says what number (1,0,0,0,0,0,0,0) maps to. So you can think of its roots as also living in an 8-dimensional space. It turns out that there are exactly 240 roots for E8; they are as follows. First, there are all the 8-dimensional integer vectors of length sqrt(2); that is, the vectors of 8 integers consisting of six 0s and two +-1s. There are 112 of these. Second, there are the 8-dimensional vectors whose entries are all +1/2 or -1/2, where we have an even number of +1/2 and an even number of -1/2. There are 128 of these. THE ROOT LATTICE Finally: the root lattice for a Lie algebra consists of all integral combinations of its roots. So, e.g., for E8 we have the 240 roots listed above; call them r1,r2,...,r240. The root lattice consists of everything you can get by picking integers a1,a2,...,a240 and computing a1.r1 + ... + a240.r240. And lo, the root lattice is exactly the E8 lattice I described way up above.