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