most recent 30 from http://mathoverflow.net 2013-05-22T13:59:16Z http://mathoverflow.net/feeds/question/10846 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10846/octonionic-unitary-group Marios 2010-01-05T21:22:34Z 2010-01-06T03:03:02Z

<p>Hi all.</p> <p>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.</p> <p><a href="http://arxiv.org/abs/math/0105155" rel="nofollow">http://arxiv.org/abs/math/0105155</a></p> <p>Thanks</p>

http://mathoverflow.net/questions/10846/octonionic-unitary-group/10856#10856 Mark Hovey 2010-01-05T22:44:03Z 2010-01-05T22:44:03Z

<p>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. </p>

http://mathoverflow.net/questions/10846/octonionic-unitary-group/10858#10858 Allen Knutson 2010-01-05T22:51:28Z 2010-01-05T22:51:28Z

<p>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!).</p> <p>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.</p>

http://mathoverflow.net/questions/10846/octonionic-unitary-group/10873#10873 José Figueroa-O'Farrill 2010-01-06T03:03:02Z 2010-01-06T03:03:02Z

<p>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 <em>octonion</em> for <em>quaternion</em> 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.</p> <p>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.</p>