Are H$_n$(O) (n>3) Jordan algebras?

Question

As we know, H$_3$(O) is a 27-dimensional exceptional Jordan algebra, here O is Cayley octonion algebra.But how about n>3? I guess that when n>3, H$_n$(O) are not Jordan algebras. But I only have a virtual reason: if they are also Jordan algebras, why do our books not refer to it? Am I right? And I want to know the real reasons.

When $n=2$, $H_n(O)$ is water. – Gerry Myerson Jun 11 '13 at 10:13 — Gerry Myerson, Jun 11 '13 at 10:13

score 1 · Answer 1 · answered Jun 11 '13 at 04:59

To answer your question: No. They are not Jordan algebras.

Theorem 2.7.6 (on page 54) of the text "Jordan operator algebras" by Harald Hanche-Olsen and Erling Størmer addresses this. By the way, this text is available (free) online at: http://www.math.ntnu.no/~hanche/joa/

The Theorem states that if $H_3(R)$ is a Jordan algebra then $R$ is an alternative algebra (which is true for the octonions) and if $H_n(R)$ is a Jordan algebra for some $n>3$, then $R$ is associative algebra (which is not true for the octonions). So $H_n(\mathbb{O})$ is not a Jordan algebra when $n>3$.

Are H$_n$(O) (n>3) Jordan algebras?

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