The Unapologetic Mathematician

Mathematics for the interested outsider

Construction of E-Series Root Systems

Today we construct the last of our root systems, following our setup. These correspond to the Dynkin diagrams E_6, E_7, and E_8. But there are transformations of Dynkin diagrams that send E_6 into E_7, and E_7 on into E_8. Thus all we really have to construct is E_8, and then cut off the right simple roots in order to give E_7, and then E_6.

We start similarly to our construction of the F_4 root system; take the eight-dimensional space with the integer-coefficient lattice I, and then build up the set of half-integer coefficient vectors

\displaystyle I'=I+\frac{1}{2}(\epsilon_1+\epsilon_2+\epsilon_3+\epsilon_4+\epsilon_5+\epsilon_6+\epsilon_7+\epsilon_8)

Starting from lattice I\cup I', we can write a generic lattice vector as

\displaystyle\frac{c^0}{2}(\epsilon_1+\epsilon_2+\epsilon_3+\epsilon_4+\epsilon_5+\epsilon_6+\epsilon_7+\epsilon_8)+c^1\epsilon_1+c^2\epsilon_2+c^3\epsilon_3+c^4\epsilon_4+c^5\epsilon_5+c^6\epsilon_6+c^7\epsilon_7+c^8\epsilon_8

and we let J\subseteq I\cup I' be the collection of lattice vectors so that the sum of the coefficients c^i is even. This is well-defined even though the coefficients aren’t unique, because the only redundancy is that we can take {2} from c^0 and add {1} to each of the other eight coefficients, which preserves the total parity of all the coefficients.

Now let \Phi consist of those vectors \alpha\in J with \langle\alpha,\alpha\rangle=2. The explicit description is similar to that from the F_4 root system. From I, we get the vectors \pm(\epsilon_i\pm\epsilon_j), but not the vectors \pm\epsilon_i because these don’t make it into J. From I' we get some vectors of the form

\displaystyle\frac{1}{2}(\pm\epsilon_1\pm\epsilon_2\pm\epsilon_3\pm\epsilon_4\pm\epsilon_5\pm\epsilon_6\pm\epsilon_7\pm\epsilon_8)

Starting with the choice of all minus signs, this vector is not in J because c^0=-1 and all the other coefficients are {0}. To flip a sign, we add \epsilon_i, which flips the total parity of the coefficients. Thus the vectors of this form that make it into \Phi are exactly those with an odd number of minus signs.

We need to verify that \frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}=\langle\beta,\alpha\rangle\in\mathbb{Z} for all \alpha and \beta in \Phi (technically we should have done this yesterday for F_4, but here it is. If both \alpha and \beta come from I, this is clear since all their coefficients are integers. If \alpha=\pm\epsilon_i\pm\epsilon_j\in I and \beta\in I', then the inner product is the sum of the ith and jth coefficients of \beta, but with possibly flipped signs. No matter how we choose \alpha\in I and \beta\in I', the resulting inner product is either -1, {0}, or {1}. Finally, if both \alpha and \beta are chosen from I', then each one is c=-\frac{1}{2}(\epsilon_1+\epsilon_2+\epsilon_3+\epsilon_4+\epsilon_5+\epsilon_6+\epsilon_7+\epsilon_8) plus an odd number of the \epsilon_i, which we write as a and b, respectively. Thus the inner product is

\displaystyle\langle\alpha,\beta\rangle=\langle c+a,c+b\rangle=\langle c,c\rangle+\langle c,b\rangle+\langle a,c\rangle+\langle a,b\rangle

The first term here is 2, and the last term is also an integer because the coefficients of a and b are all integers. The middle two terms are each a sum of an odd number of \pm\frac{1}{2}, and so each of them is a half-integer. The whole inner product then is an integer, as we need.

What explicit base \Delta should we pick? We start out as we’ve did for F_4 with \epsilon_2-\epsilon_3, \epsilon_3-\epsilon_4, and so on up to \epsilon_7-\epsilon_8. These provide six of our eight vertices, and the last two of them are perfect for cutting off later to make the E_7 and E_6 root systems. We also throw in \epsilon_2+\epsilon_3, like we did for the D_n series. This provides us with the triple vertex in the E_8 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 \epsilon_2+\epsilon_3, with which it should have the inner product -1. 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 \frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4-\epsilon_5-\epsilon_6-\epsilon_7-\epsilon_8). Establishing that the reflection with respect to this vector preserves the lattice J — and thus the root system \Phi — proceeds as in the F_4 case.

The Weyl group of E_8 is again the group of symmetries of a polytope. In this case, it turns out that the vectors in \Phi are exactly the vertices of a regular eight-dimensional polytope inscribed in the sphere of radius {2}, and the Weyl group of E_8 is exactly the group of symmetries of this polyhedron! Notice that this is actually something interesting; in the A_2 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 A_2 diagram possesses a symmetry that flips it end-over-end, and we will explore this behavior further.

The Weyl groups of E_7 and E_6 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, E_8 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).

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March 10, 2010 - Posted by | Geometry, Root Systems

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