# Octonionic Unitary Group?

Hi all.

I was wondering if anyone has any references on work related to the Octonionic Unitary group. I would imagine that such a group would be generated by Octonionic skew-Hermitian matrices (at the lie-algebra level) in analogy to the Complex Unitary and the Quaternion Unitary (aka Compact Symplectic) groups. Some hints of the construction appear on page 28 of the paper "The Octonions" by Baez.

http://arxiv.org/abs/math/0105155

Thanks

Instead of asking for a group of matrices, one could try to think about the automorphism group of projective space plus its metric. For R^n, this is PO(n). For C^n, it's PU(n) x Z_2 (semidirect product), where the Z_2 is coming from complex conjugation. For H^n, it's PU(n,H) x SO(3) (again semidirect), where the SO(3) is really inner automorphisms of H. For OP^2, it's E_6, as I recall (not a semidirect product!).

Even defining OP^2 is nonobvious. Daniel Allcock has a paper in which he shows the equivalence of several peoples' definitions. There doesn't seem to be any reasonable candidate for OP^3, but some have speculated that it should be the "Monster manifold" whose associated CFT is the moonshine module.

• I believe that the automorphism group of OP^2 is F_4, as F_4/Spin(9) = OP^2 (though I'm VERY not certain about this). More on exactly what OP^2 is can be found here: mathoverflow.net/questions/1922/… Commented Jan 5, 2010 at 23:42
• The group $E_6$ is the group of collineations (aka projective transformtions) of $\mathbb{OP}^2$ while the group $F_4$ is the group of isometries since there is a naturally defined Riemannian metric on $\mathbb{OP}^2$. Commented May 15, 2012 at 11:46

As others have pointed out, the nonassociativity of the octonions prevents one from constructing a group. For example, any subgroup of the octonions lives inside of a quaternion subalgebra. Having said that, the Clifford algebra $Cl(\mathbb{R}^7)$ has two inequivalent irreducible representations which are each as real vector spaces isomorphic to the octonions (i.e., they are eight-dimensional) and provided that we identify $\mathbb{R}^7$ with the imaginary octonions, the action of $Cl(\mathbb{R}^7)$ is given by left and right octonionic multiplications. This is analogous to what happens with $Cl(\mathbb{R}^3)$ substituting octonion for quaternion in what I said above. Now the Spin group $Spin(3)$ is the one-dimensional quaternionic unitary group and lives naturally inside $Cl(\mathbb{R}^3)$, so one could think of the group $Spin(7)$ as being the analogue of the one-dimensional octonionic unitary group.

By the same token, and given the low-dimensional isomorphisms $$Spin(2,1) \cong SL(2,\mathbb{R})$$ $$Spin(3,1) \cong SL(2,\mathbb{C})$$ $$Spin(5,1) \cong SL(2,\mathbb{H})$$ one would be tempted to think of $Spin(9,1)$ as $SL(2,\mathbb{O})$, even though such a group as written does not exist.

• The reference I included in the OP (pages 28, 29) shows exactly how to construct not only $SL(2,\mathbb{O})$ but $SL(m,\mathbb{K})$ in general where $\mathbb{K}$ is $\mathbb{R, C, H, O}$. The gist of the idea is to work at the lie-algebra level and generate real-linear operators on $\mathbb{K}^m$ which are associative even for $\mathbb{K} = \mathbb{O}$. Commented Mar 15, 2017 at 23:35

Such a thing would not be a group, because the octonions are not associative. For example, the octonionic unitary group of dimension 1 would just be the unit octonions, which form S^7, the sphere in R^8. This is of course not a group. This is not to say the algebraic structure you get is not interesting, it is just not a group.

• My reading of the reference supplied by the OP was that Baez explains precisely how to get around this issue. Commented Jan 6, 2010 at 11:11
• Correct, mathoverflow.net/users/2893/john-baez explains how to get around the associativity problem and generate $SL(m, \mathbb{O})$ which is not an Octonionic Unitary "Group". Commented Mar 15, 2017 at 23:38