most recent 30 from http://mathoverflow.net 2013-05-18T20:31:10Z http://mathoverflow.net/feeds/question/71017 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71017/irreps-of-this-algebra Arne Alex 2011-07-22T20:05:42Z 2011-07-23T00:36:37Z

<p>Hello!</p> <p>I stumbled upon the following commutative, non-associative, three-dimensional algebra (with basis $\{A, B, C\}$):</p> <p>$A\times A = 0$<br> $A\times B = A$<br> $A\times C = 2B$<br> $B\times A = A$<br> $B\times B = B$<br> $B\times C = C$<br> $C\times A = 2B$<br> $C\times B = C$<br> $C\times C = 0$</p> <p>Is anything known about its irreducible representations? In particular, how many nonequivalent irreps do exist? What are their dimensions? Can we construct explicit representation matrices for a given irrep?</p>

http://mathoverflow.net/questions/71017/irreps-of-this-algebra/71033#71033 Mariano Suárez-Alvarez 2011-07-23T00:36:37Z 2011-07-23T00:36:37Z

<p>Your algebra is a Jordan algebra.</p> <p>Armed with Jacobson's <em>Structure and representations of Jordan algebras</em>, for example, you will be able to find the sensible notion of representation of your algebra (there are, in fact, a couple of sensible notions...) Using that, and since the algebra is of dimension $3$, which is hopefully small, one can possibly describe the irreducible ones.</p> <p>Of course, this is more or less arbitrary: you have to choose in what sense you want to represent your algebra—this is what the comments above are hinting at—but, well, looking at it as a Jordan algebra makes this choice for you.</p>