# The Unapologetic Mathematician

## 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 $i$th and $j$th 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).