The Unapologetic Mathematician

Mathematics for the interested outsider

Construction of Root Systems (setup)

Now that we’ve proven the classification theorem, we know all about root systems, right? No! All we know is which Dynkin diagrams could possibly arise from root systems. We don’t know whether there actually exists a root system for any given one of them. The situation is sort of like what we found way back when we solved Rubik’s magic cube: first we established some restrictions on allowable moves, and then we showed that everything else actually happened.

And so we must construct some actual root systems. For this task, we let E stand for a finite-dimensional real vector space \mathbb{R}^n for various n, equipped with its usual inner product. We pick an orthonormal basis \{\epsilon_1,\dots,\epsilon_n\} and let the integral linear combinations of these basis vectors form the lattice I. Here, I do not mean “lattice” in the order-theory sense. I mean that this is a discrete collection of points in the vector space that is closed under addition.

In every case we’re going to take either the lattice I, or a slightly modified lattice J. We’ll define our root system \Phi to be the collection of vectors in the lattice of either one or two specified lengths (since there can be at most two root lengths). That is, we’re considering the intersection of a discrete collection of points with one or two spheres. These spheres are closed and bounded, and thus compact. The collection \Phi must be finite or else it would have an accumulation point by Bolzano-Weierstrass, and thus wouldn’t be discrete!

Any one of our constructed collections will span E, and in fact an explicit basis will be shown in each case, in case it’s not clear. It should also be clear that none of them can contain the vector {0}, and so the first condition of being a root system will hold. Our choice of lengths will make it clear that there are no possible scalar multiples of a root besides itself and its negative. On the other hand, it should be clear that if \alpha is in a lattice I and on a sphere \lVert\alpha\rVert^2=r^2, then -\alpha is also in both, and thus the second condition holds.

The reflection \sigma_\alpha preserves lengths, and so it sends the spheres back to themselves. We’ll have to check in each case that \sigma_\alpha sends every vector in our collection back into the lattice, which will establish the third condition.

As to the fourth condition, the inner product \langle\alpha,\beta\rangle is automatically going to be in \mathbb{Z} when we pick \alpha and \beta from a lattice, and so picking the squared radii of our spheres to divide 2 should be enough to guarantee that \frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}\in\mathbb{Z}.

Tomorrow we start in constructing our root systems, towards the theorem: For each Dynkin diagam allowed by the classification theorem, there exists an irreducible root system having that diagram.


March 1, 2010 Posted by | Geometry, Root Systems | 7 Comments