The Unapologetic Mathematician

Mathematics for the interested outsider

Irreducible Modules

Sorry for the delay; it’s getting crowded around here again.

Anyway, an irreducible module for a Lie algebra L is a pretty straightforward concept: it’s a module M such that its only submodules are 0 and M. As usual, Schur’s lemma tells us that any morphism between two irreducible modules is either 0 or an isomorphism. And, as we’ve seen in other examples involving linear transformations, all automorphisms of an irreducible module are scalars times the identity transformation. This, of course, doesn’t depend on any choice of basis.

A one-dimensional module will always be irreducible, if it exists. And a unique — up to isomorphism, of course — one-dimensional module will always exist for simple Lie algebras. Indeed, if L is simple then we know that [L,L]=L. Any one-dimensional representation \phi:L\to\mathfrak{gl}(1,\mathbb{F}) must have its image in [\mathfrak{gl}(1,\mathbb{F}),\mathfrak{gl}(1,\mathbb{F})]=\mathfrak{sl}(1,\mathbb{F}). But the only traceless 1\times1 matrix is the zero matrix. Setting \phi(x)=0 for all x\in L does indeed give a valid representation of L.

September 15, 2012 - Posted by | Algebra, Lie Algebras, Representation Theory

1 Comment »

  1. […] might be surmised from irreducible modules, a reducible module for a Lie algebra is one that contains a nontrivial proper submodule — […]

    Pingback by Reducible Modules « The Unapologetic Mathematician | September 16, 2012 | Reply

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