composition algebra

Noun

 * 1)  A non-associative (not necessarily associative), A, over some field, together with a nondegenerate quadratic form, N, such that N(xy) = N(x)N(y) for all x, y ∈ A.
 * 2) * 1993, F. L. Zak (translator and original author), Simeon Ivanov (editor), Tangents and Secants of Algebraic Varieties,, page 11,
 * More precisely, $$X^n\subset\mathbb{P}^N$$ is a Severi variety if and only if $$\mathbb{P}^N = \mathbb{P}(\mathfrak{J})$$, where $$\mathfrak J$$ is the Jordan algebra of Hermitian (3 × 3)-matrices over a composition algebra $$\mathfrak A$$, and $$X$$ corresponds to the cone of Hermitian matrices of rank $$\le 1$$ (in that case $$SX$$ corresponds to the cone of Hermitian matrices with vanishing determinant; cf. Theorem 4.8). In other words, $$X$$ is a Severi variety if and only if $$X$$ is the “Veronese surface” over one of the composition algebras over the field $$K$$ (Theorem 4.9).
 * 1) * 2006, Alberto Elduque, Chapter 12: A new look at Freudenthal's Magic Square, Lev Sabinin, Larissa Sbitneva, Ivan Shestakov (editors, Non-Associative Algebra and Its Applications, Taylor & Francis Group (Chapman & Hall/CRC), page 150,
 * At least in the split cases, this is a construction that depends on two unital composition algebras, since the Jordan algebra involved consists of the 3 x 3-hermitian matrices over a unital composition algebra.
 * At least in the split cases, this is a construction that depends on two unital composition algebras, since the Jordan algebra involved consists of the 3 x 3-hermitian matrices over a unital composition algebra.

Usage notes

 * Formally, a, $$(A,\ ^*, N)$$, where $$A$$ is a nonassociative algebra, the mapping $$x\to x^*$$ is an involution, called a , and $$N$$ is the quadratic form $$N(x) = x x^*\!\!$$, called the of the algebra.
 * A composition algebra may be:
 * A if there exists some $$\nu\in A:\nu\ne 0\and N(\nu)=0$$ (called a ). In this case, $$N$$ is called an  and the algebra is said to split.
 * A otherwise; so named because division, except by 0, is possible: the multiplicative inverse of $$x$$ is $$\textstyle x^*/N(x)$$. In this case, $$N$$ is an.

Translations

 * French: algèbre de composition