Is there a name for this algebraic structure? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T03:06:06Z http://mathoverflow.net/feeds/question/16048 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16048/is-there-a-name-for-this-algebraic-structure Is there a name for this algebraic structure? Andrea Altomani 2010-02-22T11:57:16Z 2010-02-22T15:20:13Z <p>I found myself "naturally" dealing with an object of this form:</p> <p>X is a complex vector space, with a "product" (a,b) &rarr; {aba} which is quadratic in the first variable, linear in the second, and satisfies some associativity conditions. These conditions are actually complicated, but more or less say that {aba} looks like the product (aba) in an alternative algebra Y containing X as a subspace.</p> <p>For example, the main "associativity condition" I am interested in is: {a{b{aca}b}a}={{aba}c{aba}}</p> <p>Examples</p> <ol> <li>Symmetric matrices</li> <li>Octonions, or indeed any alternative algebra</li> <li>Let J belong to GL(n,&#x2102;), with <sup>t</sup>J=-J and J²=-Id, and W={w&isin;M(n&times;n,&#x2102;)|J<sup>t</sup>wJ=-w}</li> </ol> <p>all with the standard product {aba}=aba.</p> <p>All of these examples are Jordan algebras, with respect to the symmetrized product a&#x2218;b=&frac12;(ab+ba), but I cannot see any direct link between the Jordan product and my product.</p> http://mathoverflow.net/questions/16048/is-there-a-name-for-this-algebraic-structure/16053#16053 Answer by Kurt Luoto for Is there a name for this algebraic structure? Kurt Luoto 2010-02-22T12:29:41Z 2010-02-22T12:29:41Z <p>How hard/lengthy would it be to give a concrete example of your product?</p> http://mathoverflow.net/questions/16048/is-there-a-name-for-this-algebraic-structure/16057#16057 Answer by Mariano Suárez-Alvarez for Is there a name for this algebraic structure? Mariano Suárez-Alvarez 2010-02-22T14:06:11Z 2010-02-22T15:20:13Z <p>In a Jordan algebra with product $\cdot$, a triple product is defined by $$\{abc\}=(a\cdot b)\cdot c+(b\cdot c)\cdot a-(a\cdot c)\cdot b.$$ In a special Jordan algebra (constructed by symmetrising an associative product) one has $\{aba\}=aba$, and it is easy to show that in such algebras one always has the identity $$\{\{aba\}c\{aba\}\}=\{a\{b\{aca\}b\}a\}.$$ Now, there is an amazing general theorem of Macdonald's that states that any identity in three variables which is of degree at most one in one of them and which is valid in special Jordan algebras actually holds in all Jordan algebras. This is proved in Jacobson's breath-taking <em>Structure and representations of Jordan algebras</em>.</p> <p>So your identity holds in all Jordan algebras. As a consequence, from the information you give it is more or less impossible to distinguish your structure from Jordan algebras, as far as I can see.</p> <p>By the way, in his book, Jacobson notes that McCrimmon has developed the theory of Jordan algebras based exclusively on the composition $(a,b)\mapsto aba$, and gives [McCrimmon, Kevin. A general theory of Jordan rings. Proc. Nat. Acad. Sci. U.S.A. 56 1966 1072--1079. MR0202783 (34 #2643)] as reference. <strike>I do not have access to the paper, though.</strike> The paper can be gotten from this <a href="http://www.pnas.org/content/56/4/1072.full.pdf" rel="nofollow">link</a> Andrea provided in a comment below.</p>