Algebra out of a set of modules of a Lie algebra? Fusion - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:34:29Z http://mathoverflow.net/feeds/question/104760 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104760/algebra-out-of-a-set-of-modules-of-a-lie-algebra-fusion Algebra out of a set of modules of a Lie algebra? Fusion Eugene Starling 2012-08-15T12:29:26Z 2012-08-15T12:29:26Z <p>The problem I faced is how to organize a set of finite-dimensional irreducible representations $U_\alpha$ of some simple Lie algebra $g$ into an associative algebra $A$ that contains $g$ as a Lie subalgebra under commutators and such that $A$ as a $g$-module decomposes into $\bigoplus_\alpha U_\alpha$. The set of irreps $U_\alpha$ of $g$ is the input data (the adjoint of $g$ can be thought of as one of $U_\alpha$). </p> <p>A simple positive answer is given by the universal enveloping algebra evaluated at some other irreducible representation, say $V$, then $A=U(g)|_{V}\sim V\otimes V^*$ is an algebra and $A$ as a $g$-module is given simply by decomposing $V\otimes V^*=\bigoplus_\alpha U_\alpha$ into irreps $U_\alpha$. So given $U_\alpha$ that come out of $V\otimes V^*$ there exists an algebra $A$, whose 'generators' are given by $U_\alpha$. This gives a lot of algebras of this type. I wonder if this is the complete answer. This is reminiscent of the fusion problem in conformal field theory.</p>