Source code for my notes on representations of semisimple Lie algebras and Olivier Mathieu's classification of simple weight modules
Fixed the proof on the theorem on D-modules
Also clarified the definition of the ring of differential operators
1 file changed, 26 insertions, 19 deletions
| Status | File Name | N° Changes | Insertions | Deletions |
| Modified | sections/introduction.tex | 45 | 26 | 19 |
diff --git a/sections/introduction.tex b/sections/introduction.tex @@ -648,7 +648,8 @@ The construction of \(\mathcal{U}(\mathfrak{g})\) may seem like a purely algebraic affair, but the universal enveloping algebra of the Lie algebra of a Lie group \(G\) is in fact intimately related with the algebra \(\operatorname{Diff}(G)\) of differential operators \(C^\infty(G) \to -C^\infty(G)\) -- as defined in Coutinho's \citetitle{coutinho} +C^\infty(G)\) -- \(\mathbb{R}\)-linear endomorphisms \(C^\infty(G) \to +C^\infty(G)\) of finite orderd as defined in Coutinho's \citetitle{coutinho} \cite[ch.~3]{coutinho}, for example. Algebras of differential operators and their modules are the subject of the theory of \(D\)-modules, which has seen remarkable progress in the past century. Specifically, we find\dots @@ -664,32 +665,38 @@ remarkable progress in the past century. Specifically, we find\dots \end{proposition} \begin{proof} - An order \(1\) \(G\)-invariant differential operator in \(G\) is simply a - left invariant derivation \(C^\infty(G) \to C^\infty(G)\). All other - \(G\)-invariant differential operators are generated by such derivations. Now - recall that there is a canonical isomorphism of Lie algebras + An order \(0\) \(G\)-invariant differential operator in \(G\) is simply + multiplication by a constant in \(\mathbb{R}\). An order \(1\) + \(G\)-invariant differential operator in \(G\) is simply a left invariant + derivation \(C^\infty(G) \to C^\infty(G)\). All other \(G\)-invariant + differential operators are generated by invariant operators of order \(0\) + and \(1\). Hence \(\operatorname{Diff}(G)^G\) is generated by + \(\operatorname{Der}(G)^G\) -- here \(\operatorname{Der}(G)^G \subset + \operatorname{Der}(G)\) denotes the Lie subalgebra of invariant derivations. + + Now recall that there is a canonical isomorphism of Lie algebras \(\mathfrak{X}(G) \isoto \operatorname{Der}(G)\). This isomorphism takes left invariant fields to left invariant derivations, so it restricts to an isomorphism \(f : \mathfrak{g} \isoto \operatorname{Der}(G)^G \subset - \operatorname{Diff}(G)^G\) -- here \(\operatorname{Der}(G)^G \subset - \operatorname{Der}(G)\) denotes the Lie subalgebra of invariant derivations. + \operatorname{Diff}(G)^G\). Since \(f\) is a homomorphism of Lie algebras, it + can be extended to an algebra homomorphism \(\tilde f : + \mathcal{U}(\mathfrak{g}) \to \operatorname{Diff}(G)^G\). We claim \(\tilde + f\) is an isomorphism. - Since \(f\) is a homomorphism of Lie algebras, it can be extended to an - algebra homomorphism \(\bar f : \mathcal{U}(\mathfrak{g}) \to - \operatorname{Diff}(G)^G\). We claim \(\bar f\) is an isomorphism. To see - that \(\bar f\) is injective, it suffices to notice + To see that \(\tilde f\) is injective, it suffices to notice \[ - \bar f(X_1 \cdots X_n) - = \bar f(X_1) \cdots \bar f(X_n) + \tilde f(X_1 \cdots X_n) + = \tilde f(X_1) \cdots \tilde f(X_n) = f(X_1) \cdots f(X_n) \ne 0 \] - for all nonzero \(X_1, \ldots, X_n \in \mathfrak{g}\) -- - \(\operatorname{Diff}(G)^G\) is a domain. Since \(\mathcal{U}(\mathfrak{g})\) - is generated by the image of the inclusion \(\mathfrak{g} \to - \mathcal{U}(\mathfrak{g})\), this implies \(\ker \bar f = 0\). Given that - \(\operatorname{Diff}(G)^G\) is generated by \(\operatorname{Der}(G)^G\), - this also goes to show \(\bar f\) is surjective. + for all nonzero \(X_1, \ldots, X_n \in \mathfrak{g}\) -- the product of + operators of positive order has positive order and is therefore nonzero. + Since \(\mathcal{U}(\mathfrak{g})\) is generated by the image of the + inclusion \(\mathfrak{g} \to \mathcal{U}(\mathfrak{g})\), this implies \(\ker + \tilde f = 0\). Given that \(\operatorname{Diff}(G)^G\) is generated by + \(\operatorname{Der}(G)^G\), this also goes to show \(\tilde f\) is + surjective. \end{proof} As one would expect, the same holds for complex Lie groups and algebraic groups