Editing Lie algebra representation (section)

== Induced representation ==
Let <math>\mathfrak{g}</math> be a finite-dimensional Lie algebra over a field of characteristic zero and <math>\mathfrak{h} \subset \mathfrak{g}</math> a subalgebra. <math>U(\mathfrak{h})</math> acts on <math>U(\mathfrak{g})</math> from the right and thus, for any <math>\mathfrak{h}</math>-module ''W'', one can form the left <math>U(\mathfrak{g})</math>-module <math>U(\mathfrak{g}) \otimes_{U(\mathfrak{h})} W</math>. It is a <math>\mathfrak{g}</math>-module denoted by <math>\operatorname{Ind}_\mathfrak{h}^\mathfrak{g} W</math> and called the <math>\mathfrak{g}</math>-module induced by ''W''. It satisfies (and is in fact characterized by) the universal property: for any <math>\mathfrak{g}</math>-module ''E''
:<math>\operatorname{Hom}_\mathfrak{g}(\operatorname{Ind}_\mathfrak{h}^\mathfrak{g} W, E) \simeq \operatorname{Hom}_\mathfrak{h}(W, \operatorname{Res}^\mathfrak{g}_\mathfrak{h} E)</math>.
Furthermore, <math>\operatorname{Ind}_\mathfrak{h}^\mathfrak{g}</math> is an exact functor from the category of <math>\mathfrak{h}</math>-modules to the category of <math>\mathfrak{g}</math>-modules. These uses the fact that <math>U(\mathfrak{g})</math> is a free right module over <math>U(\mathfrak{h})</math>. In particular, if <math>\operatorname{Ind}_\mathfrak{h}^\mathfrak{g} W</math> is simple (resp. absolutely simple), then ''W'' is simple (resp. absolutely simple). Here, a <math>\mathfrak{g}</math>-module ''V'' is absolutely simple if <math>V \otimes_k F</math> is simple for any field extension <math>F/k</math>.

The induction is transitive: <math>\operatorname{Ind}_\mathfrak{h}^\mathfrak{g} \simeq \operatorname{Ind}_\mathfrak{h'}^\mathfrak{g} \circ \operatorname{Ind}_\mathfrak{h}^\mathfrak{h'}</math>
for any Lie subalgebra <math>\mathfrak{h'} \subset \mathfrak{g}</math> and any Lie subalgebra <math>\mathfrak{h} \subset \mathfrak{h}'</math>. The induction commutes with restriction: let <math>\mathfrak{h} \subset \mathfrak{g}</math> be subalgebra and <math>\mathfrak{n}</math> an ideal of <math>\mathfrak{g}</math> that is contained in <math>\mathfrak{h}</math>. Set <math>\mathfrak{g}_1 = \mathfrak{g}/\mathfrak{n}</math> and <math>\mathfrak{h}_1 = \mathfrak{h}/\mathfrak{n}</math>. Then <math>\operatorname{Ind}^\mathfrak{g}_\mathfrak{h} \circ \operatorname{Res}_\mathfrak{h} \simeq \operatorname{Res}_\mathfrak{g} \circ \operatorname{Ind}^\mathfrak{g_1}_\mathfrak{h_1}</math>.