Sixth Roots of Unity.
Let's go back for a bit to the X-product. Martin Cederwall introduced a couple of
very useful identities:
(AX)(X*B) = X((X*A)B) = (A(BX))X*.
Now let X =U3, where U is a unit octonion, and let's apply these identities
a few times (in what follows it may seem I am occasionally ignoring the nonassociativity
of the octonions, but it only occurs where it is allowed):
The automorphism group of the octonions is G2, a 14-dimensional group, and one
of the five exceptional Lie groups associated with the octonions. If g is in G2,
and g[A] is the action of g on A an octonion, then
g[AB] = g[A]g[B].
This property defines g as an automorphism.
Ok, look at the last U-identity at the bottom on the left. Suppose U6 = U-6 = 1,
ie., U is an octonion 6th root of unity. In that case, U3 = U-3 = ±1, implying
(AU3)(U-3B) = AB,
which implies that if U6 = U-6 = 1, then
(U-1AU)(U-1BU) = U-1(AB)U. That is, the action A --> U-1AU
is in G2. In fact, it is not too difficult to prove that one can nest actions of this sort
and still have automorphisms. That is, if U, V, and W are all octonion 6th roots of unity, then
A --> W(V(UAU-1)V-1)W-1
is an element of G2 (clearly it doesn't matter which side the inverses go on; I've switched
since the calculator below uses the inverses on the right). An open question, at least to me,
is how many nested actions does it take to account for all of G2? I've made a
perfunctory effort to derive the index cycling automorphism using 3 nested actions as above.
I haven't succeeded yet. If anyone out there does, please let me know (email@example.com).
Note, this calculator does not require 6th roots of unity U, V and W as inputs, but if you
decide to go that root, one very easy collection of 6th roots are elements of the form
(±1 ±ea ±eb ±ec) / 2
(I haven't always explicitly said so, but indices represented by distinct letters are assumed
to be distinct indices). These elements of X2 are therefore 6th roots of unity.