# The Unapologetic Mathematician

## 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