# From the octahedron to $E_8$¶

Try to work out the problem in John Carlos Baez's post https://johncarlosbaez.wordpress.com/2020/03/20/from-the-octahedron-to-e8/.

$a+b\sqrt{2}+(c+d\sqrt{2})i+(e+f\sqrt{2})j+(g+h\sqrt{2})k$

In [1]:
from IPython.display import display, HTML
import sympy as sy

In [2]:
sy.init_printing()

In [3]:
h = sy.S(1)/2
h

Out[3]:
$\displaystyle \frac{1}{2}$

Express the vectors in the orthonormal basis.

In [4]:
v1 = sy.Matrix([h, 0, h, 0, h, 0, h, 0])
v2 = sy.Matrix([0, h, 0, h, 0, 0, 0, 0])
v3 = sy.Matrix([0, 0, 1, 0, 0, 0, 0, 0])
v4 = sy.Matrix([0, 0, 0, h, 0, h, 0, 0])
v5 = sy.Matrix([0, 0, 0, 0, 1, 0, 0, 0])
v6 = sy.Matrix([0, 0, 0, 0, 0, h, 0, h])
v7 = sy.Matrix([0, 0, 0, 0, 0, 0, 1, 0])
v8 = sy.Matrix([0, 0, 0, 0, 0, 0, 0, 1])

In [5]:
vs = {1: v1, 2: v2, 3: v3, 4: v4, 5: v5, 6: v6, 7: v7, 8: v8}


Verify that they look like we mean them to.

In [6]:
def v_string(v):
i, j, k = sy.symbols('i j k')
s = sy.sqrt(2)
vec = sy.Matrix([1, s, i, i*s, j, j*s, k, k*s])
return sy.latex(sy.simplify((v.T @ vec)[0]))

def display_v(v):
display(HTML('${}$'.format(v_string(v))))

In [7]:
display(HTML(r'\begin{align}'
+ r'\\'.join(['v_{{{}}}&={}'.format(n, v_string(vs[n])) for n in range(1, 8+1)])
+ r'\end{align}'))

\begin{align}v_{1}&=\frac{i}{2} + \frac{j}{2} + \frac{k}{2} + \frac{1}{2}\\v_{2}&=\frac{\sqrt{2} \left(i + 1\right)}{2}\\v_{3}&=i\\v_{4}&=\frac{\sqrt{2} \left(i + j\right)}{2}\\v_{5}&=j\\v_{6}&=\frac{\sqrt{2} \left(j + k\right)}{2}\\v_{7}&=k\\v_{8}&=\sqrt{2} k\end{align}

Calculate the Gram matrix.

In [8]:
bad_gram_matrix = sy.Matrix([[sy.simplify((vs[m].T @ vs[n])[0])
for n in range(1, 8+1)]
for m in range(1, 8+1)])

Out[8]:
$\displaystyle \left[\begin{matrix}1 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0\\0 & \frac{1}{2} & 0 & \frac{1}{4} & 0 & 0 & 0 & 0\\\frac{1}{2} & 0 & 1 & 0 & 0 & 0 & 0 & 0\\0 & \frac{1}{4} & 0 & \frac{1}{2} & 0 & \frac{1}{4} & 0 & 0\\\frac{1}{2} & 0 & 0 & 0 & 1 & 0 & 0 & 0\\0 & 0 & 0 & \frac{1}{4} & 0 & \frac{1}{2} & 0 & \frac{1}{2}\\\frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 1 & 0\\0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 1\end{matrix}\right]$

I think the inner product JCB gives in his post is not the same inner product as David Harden is using in his MathOverflow question https://mathoverflow.net/q/353486.

\begin{align} \mathrm{QN}\left(a+b\sqrt{2}+(c+d\sqrt{2})i+(e+f\sqrt{2})j+(g+h\sqrt{2})k\right) &= a^2+2b^2+2ab\sqrt{2}+c^2+2d^2+2cd\sqrt{2} \\ & +e^2+2f^2+2ef\sqrt{2}+g^2+2h^2+2gh\sqrt{2} \end{align}
In [9]:
def QN(v):
return sy.Matrix([sum([v[n]**2 for n in [0, 2, 4, 6]])
+ 2*sum([v[n]**2 for n in [1, 3, 5, 7]]),
sum([2*v[n]*v[n+1] for n in [0, 2, 4, 6]])])

def EN(u):
return sum(u)

In [10]:
QN(vs[1])

Out[10]:
$\displaystyle \left[\begin{matrix}1\\0\end{matrix}\right]$
In [11]:
EN(QN(vs[1] + vs[1])) - EN(QN(vs[1])) - EN(QN(vs[1]))

Out[11]:
$\displaystyle 2$
In [12]:
def inner_prod(x, y):
return EN(QN(x + y)) - EN(QN(x)) - EN(QN(y))

In [13]:
good_gram_matrix = sy.Matrix([[sy.simplify(inner_prod(vs[m], vs[n]))
for n in range(1, 8+1)]
for m in range(1, 8+1)])
good_gram_matrix

Out[13]:
$\displaystyle \left[\begin{matrix}2 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\1 & 2 & 1 & 1 & 0 & 0 & 0 & 0\\1 & 1 & 2 & 1 & 0 & 0 & 0 & 0\\1 & 1 & 1 & 2 & 1 & 1 & 0 & 0\\1 & 0 & 0 & 1 & 2 & 1 & 0 & 0\\1 & 0 & 0 & 1 & 1 & 2 & 1 & 2\\1 & 0 & 0 & 0 & 0 & 1 & 2 & 2\\1 & 0 & 0 & 0 & 0 & 2 & 2 & 4\end{matrix}\right]$
In [14]:
gen_matrix = sy.Matrix([vs[n].T for n in range(1, 8+1)])
gen_matrix

Out[14]:
$\displaystyle \left[\begin{matrix}\frac{1}{2} & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0\\0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0\\0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2}\\0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\end{matrix}\right]$
In [15]:
gen_matrix.det()

Out[15]:
$\displaystyle \frac{1}{16}$

David Harden indicates that it's the Gram matrix that needs to have determinant 1:

In [16]:
good_gram_matrix.det()

Out[16]:
$\displaystyle 1$
In [ ]: