("Lazy" because monkey wonders, monkey posts without thinking about question.) As you may know, the division algebras 1D reals R, 2D complexes C, 4D quaternions H, and 8D octonions O, each have their projective spaces: RP^n, CP^n, HP^n, and OP^n, respectively. (The standard construction is to let A* be the nonzero elements of an associative division algebra A, and to use the equivalence relation (a_1,...,a_(n+1)) ~ lambda*(a_1,...,a_(n+1)), where lambda and all a_j are in A, with at least one a_j being non-zero.) This is an equivalence relation when A is associative, and the set of equivalence classes is defined as RA^n.) It is often stated, however, that OP^n exists solely for n <= 2, since the relation (o_1,...,,o_n,o_(n+1)) ~ lambda*((o_1,...,,o_n,o_(n+1)) is said to be an equivalence relation only when multiplication is associative -- which is not true for arbitrary octonions. There is, however, a careful construction that yields what is generally agreed upon to be OP^2. QUESTION: What exactly goes wrong when one tries to construct OP^n, n >= 3 ??? In particular, suppose one defines a relation among all (n+1)-tuples (o_1,...,o+(n+1)) of octonions by the usual means, as above: (*) (o_1,...,,o_n,o_(n+1)) ~ lambda*((o_1,...,,o_n,o_(n+1)) where lambda, and the o_j's are octonions ... and lambda -- and at least one o_j -- is nonzero. How does this relation fail to be an equivalence relation? Clearly it satisfies reflexivity and symmetry, so the problem is presumably with transitivity. One can always take a relation and discuss the least equivalence relation it defines. SO: What is the least equivalence relation defined by (*) above? Does it make all elements of O^(n+1)* (all (n+1)-tuples of octonions that are not all 0) equivalent? Or what? Be merry, --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele