Today we construct the last of our root systems, following our setup. These correspond to the Dynkin diagrams , , and . But there are transformations of Dynkin diagrams that send into , and on into . Thus all we really have to construct is , and then cut off the right simple roots in order to give , and then .
We start similarly to our construction of the root system; take the eight-dimensional space with the integer-coefficient lattice , and then build up the set of half-integer coefficient vectors
Starting from lattice , we can write a generic lattice vector as
and we let be the collection of lattice vectors so that the sum of the coefficients is even. This is well-defined even though the coefficients aren’t unique, because the only redundancy is that we can take from and add to each of the other eight coefficients, which preserves the total parity of all the coefficients.
Now let consist of those vectors with . The explicit description is similar to that from the root system. From , we get the vectors , but not the vectors because these don’t make it into . From we get some vectors of the form
Starting with the choice of all minus signs, this vector is not in because and all the other coefficients are . To flip a sign, we add , which flips the total parity of the coefficients. Thus the vectors of this form that make it into are exactly those with an odd number of minus signs.
We need to verify that for all and in (technically we should have done this yesterday for , but here it is. If both and come from , this is clear since all their coefficients are integers. If and , then the inner product is the sum of the th and th coefficients of , but with possibly flipped signs. No matter how we choose and , the resulting inner product is either , , or . Finally, if both and are chosen from , then each one is plus an odd number of the , which we write as and , respectively. Thus the inner product is
The first term here is , and the last term is also an integer because the coefficients of and are all integers. The middle two terms are each a sum of an odd number of , and so each of them is a half-integer. The whole inner product then is an integer, as we need.
What explicit base should we pick? We start out as we’ve did for with , , and so on up to . These provide six of our eight vertices, and the last two of them are perfect for cutting off later to make the and root systems. We also throw in , like we did for the series. This provides us with the triple vertex in the Dynkin diagram.
We need one more vertex off to the left. It should be orthogonal to every one of the simple roots we’ve chosen so far except for , with which it should have the inner product . It should also be a half-integer root, so that we can get access to the rest of them. For this purpose, we choose the root . Establishing that the reflection with respect to this vector preserves the lattice — and thus the root system — proceeds as in the case.
The Weyl group of is again the group of symmetries of a polytope. In this case, it turns out that the vectors in are exactly the vertices of a regular eight-dimensional polytope inscribed in the sphere of radius , and the Weyl group of is exactly the group of symmetries of this polyhedron! Notice that this is actually something interesting; in the case the roots formed the vertices of a hexagon, but the Weyl group wasn’t the whole group of symmetries of the hexagon. This is related to the fact that the diagram possesses a symmetry that flips it end-over-end, and we will explore this behavior further.
The Weyl groups of and are also the symmetries of seven- and six-dimensional polytopes, respectively, but these aren’t quite so nicely apparent from their root systems.
As the most intricate (in a sense) of these root systems, has inspired quite a lot of study and effort to visualize its structure. I’ll leave you with an animation I found on Garrett Lisi’s notewiki, Deferential Geometry (with the help of Sarah Kavassalis).