The Unapologetic Mathematician

Mathematics for the interested outsider

Construction of the F4 Root System

Today we construct the F_4 root system starting from our setup.

As we might see, this root system lives in four-dimensional space, and so we start with this space and its integer-component lattice I. However, we now take another copy of I and push it off by the vector \frac{1}{2}(\epsilon_1+\epsilon_2+\epsilon_3+\epsilon_4). This set I' consists of all vectors each of whose components is half an odd integer (a “half-integer” for short). Together with I, we get a new lattice J=I\cup I' consisting of vectors whose components are either all integers or all half-integers. Within this lattice J, we let \Phi consist of those vectors of squared-length 2 or 1: \langle\alpha,\alpha\rangle=2 or \langle\alpha,\alpha\rangle=1; we want to describe these vectors explicitly.

When we constructed the B_n and C_n series, we saw that the vectors of squared-length 1 and 2 in I are those of the form \pm\epsilon_i (squared-length 1) and of the form \pm(\epsilon_i\pm\epsilon_j) for i\neq j (squared-length 2). But what about the vectors in I'? We definitely have \left(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2}\right) — with squared-length 1 — but can we have any others? The next longest vector in I' will have one component \pm\frac{3}{2} and the rest \pm\frac{1}{2}, but this has squared-length 3 and won’t fit into \Phi! We thus have twenty-four long roots of squared-length 2 and twenty-four short roots of squared-length 1.

Now, of course we need an explicit base \Delta, and we can guess from the diagram F_4 that two must be long and two must be short. In fact, in a similar way to the B_3 root system, we start by picking \epsilon_2-\epsilon_3 and \epsilon_3-\epsilon_4 as two long roots, along with \epsilon_4 as one short root. Indeed, we can see a transformation of Dynkin diagrams sending B_3 into F_4, and sending the specified base of B_3 to these three vectors.

But we need another short root which will both give a component in the direction of \epsilon_1 and will give us access to I'. Further, it should be orthogonal to both \epsilon_2-\epsilon_3 and \epsilon_3-\epsilon_4, and should have a Cartan integer of -1 with \epsilon_4 in either order. For this purpose, we pick \frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4), which then gives us the last vertex of the F_4 Dynkin diagram.

Does the reflection with respect to this last vector preserve the root system, though? What is its effect on vectors in J? We calculate

\displaystyle\begin{aligned}\sigma_{\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)}(v)&=v-\frac{2\left\langle v,\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\right\rangle}{\left\langle\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4),\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\right\rangle}\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\\&=v-\left\langle v,\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\right\rangle(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\\&=v-\frac{v^1-v^2-v^3-v^4}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\end{aligned}

Now the sum v^1-v^2-v^3-v^4 is always an integer, whether the components of v are integers or half-integers. If the sum is even, then we are changing each component of v by an integer, which sends I and I' back to themselves. If the sum is off, then we are changing each component of v by a half-integer, which swaps I and I'. In either case, the lattice J is sent back to itself, and so this reflection fixes \Phi.

Like we say for G_2 it’s difficult to understand the Weyl group of F_4 in terms of its action on the components of v. However, also like G_2, we can understand it geometrically. But instead of a hexagon, now the long and short roots each make up a four-dimensional polytope called the “24-cell”. It’s a shape with 24 vertices, 96 edges, 96 equilateral triangular faces, and 24 three-dimensional “cells”, each of which is a regular octahedron; the Weyl group of F_4 is its group of symmetries, just like the Weyl group of G_2 was the group of symmetries of the hexagon.

Also like the G_2 case, the F_4 root system is isomorphic to its own dual. The long roots stay the same length when dualized, while the short roots double in length and become the long roots of the dual root system. Again, a scaling and rotation sends the dual system back to the one we constructed.

March 9, 2010 Posted by | Geometry, Root Systems | 2 Comments

   

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