very very basic question on semi-simple Lie algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T01:05:57Z http://mathoverflow.net/feeds/question/92105 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92105/very-very-basic-question-on-semi-simple-lie-algebras very very basic question on semi-simple Lie algebras unknown (google) 2012-03-24T20:43:21Z 2012-03-25T20:30:19Z <p>I have a very basic question on Lie algebras. I'm doing particle physics, and a lot of emphasis seems to be placed on the weight diagrams of simple Lie algebras. But these simple Lie algebras are all that ever seems to be discussed. Can anyone tell me in intuitive terms how the weight diagrams of semi-simple Lie algebras relate to those of the simple Lie algebras that sum to make the semi-simple algebras? Does one somehow 'add' the weight spaces together? (Sorry if this sounds really stoopid!)</p> http://mathoverflow.net/questions/92105/very-very-basic-question-on-semi-simple-lie-algebras/92116#92116 Answer by Allen Knutson for very very basic question on semi-simple Lie algebras Allen Knutson 2012-03-24T23:25:31Z 2012-03-25T08:04:01Z <p>I think this is more stackexchange-worthy, but here goes.</p> <p>I'm a little afraid you're mixing up general weight diagrams with the root system, which is the weight diagram of the adjoint representation. The root system of $G\times H$ lives in a space that's just the Cartesian product of the two individual spaces, and is the disjoint union: $$\Delta_{G\times H} = (\Delta_G \times 0) \cup (0 \times \Delta_H).$$</p> <p>If one's talking about weight diagrams of <em>irreducible</em> representations of $G\times H$ (which the adjoint representation is not), it helps to know that they're all of the form $V\otimes W$. So then, one takes the convolution of the two weight multiplicity diagrams, again inside that Cartesian product:</p> <p><code>$$mult(V\otimes W)_{\lambda,\mu} = mult(V)_\lambda \ mult(W)_\mu.$$</code></p>