diff --git a/sections/introduction.tex b/sections/introduction.tex
@@ -45,12 +45,12 @@
\begin{example}
Given a smooth manifold \(M\), the space \(\mathfrak{X}(M)\) of all smooth
- vector fields is canonically identifyed with
- \(\operatorname{Der}(C^\infty(M))\) -- where a field \(X \in
- \mathfrak{X}(M)\) is identified with the map \(C^\infty(M) \to C^\infty(M)\)
- which takes a function \(f \in C^\infty(M)\) to its derivative in the
- direction of \(X\). This gives \(\mathfrak{X}(M)\) the structure of a Lie
- algebra over \(\mathbb{R}\).
+ vector fields is canonically identifyed with \(\operatorname{Der}(M) =
+ \operatorname{Der}(C^\infty(M))\) -- where a field \(X \in \mathfrak{X}(M)\)
+ is identified with the map \(C^\infty(M) \to C^\infty(M)\) which takes a
+ function \(f \in C^\infty(M)\) to its derivative in the direction of \(X\).
+ This gives \(\mathfrak{X}(M)\) the structure of a Lie algebra over
+ \(\mathbb{R}\).
\end{example}
\begin{example}
@@ -373,13 +373,23 @@
\end{definition}
\begin{lemma}[Schur]
- Let \(V\) and \(W\) be two irreducible representations of \(\mathfrak{g}\).
- and \(T : V \to W\) be an intertwiner. If \(V \not\cong W\) then \(T = 0\) --
- i.e. \(\operatorname{Hom}_{\mathfrak{g}}(V, W) = 0\). If \(V = W\) then \(T\)
- is a scalar operator -- i.e. \(\operatorname{End}_{\mathfrak{g}}(V) = K
- \operatorname{Id}\).
+ Let \(\mathfrak{g}\) be a Lie algebra over a field \(K\). If \(V\) and \(W\)
+ irreducible representations of \(\mathfrak{g}\). and \(T : V \to W\) be an
+ intertwiner then \(T\) is either \(0\) or an isomorphism. Furtheremore, if
+ \(K\) is algebraicly closed and \(V = W\) then \(T\) is a scalar operator.
\end{lemma}
+\begin{proof}
+ For the first statement, it suffices to notice that \(\ker T\) and
+ \(\operatorname{im} T\) are both subrepresentations. In particular, either
+ \(\ker T = 0\) and \(\operatorname{im} T = W\) or \(\ker T = V\) and
+ \(\operatorname{im} T = 0\). Now suppose \(K\) is algebraicly closed and \(V
+ = W\). Let \(\lambda \in K\) be an eigenvalue of \(T\) and \(V_\lambda\) be
+ its corresponding eigenspace. Given \(v \in V_\lambda\), \(T X v = X T v =
+ \lambda \cdot X v\). In other words, \(V_\lambda\) is a subrepresentation.
+ It then follows \(V_\lambda = V\), given that \(V_\lambda \ne 0\).
+\end{proof}
+
\section{The Universal Enveloping Algebra}
\begin{definition}
@@ -407,6 +417,41 @@
\end{center}
\end{proposition}
+\begin{proof}
+ Let \(f : \mathfrak{g} \to A\) be a homomorphism of Lie algebras. By the
+ universal property of free algebras, there is a homomorphism of algebras \(g
+ : T \mathfrak{g} \to A\) such that
+ \begin{center}
+ \begin{tikzcd}
+ T \mathfrak{g} \arrow{dr}{g} & \\
+ \mathfrak{g} \uar \rar[swap]{f} & A
+ \end{tikzcd}
+ \end{center}
+
+ Since \(f\) is a homomorphism of Lie algebras,
+ \[
+ g([X, Y])
+ = f([X, Y])
+ = [f(X), f(Y)]
+ = [g(X), g(Y)]
+ = g(X \otimes Y - Y \otimes X)
+ \]
+ for all \(X, Y \in \mathfrak{g}\). Hence \(I = ([X, Y] - (X \otimes Y - Y
+ \otimes X) : X, Y \in \mathfrak{g}) \subset \ker g\) and therefore \(g\)
+ factors through the quotient \(\mathcal{U}(\mathfrak{g}) = \mfrac{T
+ \mathfrak{g}}{I}\).
+ \begin{center}
+ \begin{tikzcd}
+ T \mathfrak{g} \arrow{dr}{\bar{g}} & \\
+ \mathcal{U}(\mathfrak{g}) \uar \rar[swap]{g} & A
+ \end{tikzcd}
+ \end{center}
+
+ Combining the two previous diagrams, we can see that \(\bar{g}\) is indeed an
+ extension of \(f\). The uniqueness of the extension then follows from the
+ uniqueness of \(g\) and \(\bar{g}\).
+\end{proof}
+
% TODO: Remark this construction is functorial
\begin{center}
@@ -485,7 +530,7 @@
\operatorname{Ind}_{\mathfrak{h}}^{\mathfrak{g}}\).
\end{proposition}
-\begin{theorem}
+\begin{proposition}
Let \(G\) be a Lie group and \(\mathfrak{g}\) be its Lie algebra. Denote by
\(\operatorname{Diff}(G)^G\) the algebra of \(G\)-invariant differential
operators in \(G\) -- i.e. the algebra of all differential operators \(L :
@@ -493,7 +538,36 @@
\ell_{g^{-1}} = L f\) for all \(f \in C^\infty(G)\) and \(g \in G\). There is
a canonical isomorphism of algebras \(\mathcal{U}(\mathfrak{g}) \isoto
\operatorname{Diff}(G)^G\).
-\end{theorem}
+\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 differetial operators are generated by such 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.
+
+ Since \(f\) is a homomorphism of Lie algebras, it extends to an algebra
+ homomorphism \(g : \mathcal{U}(\mathfrak{g}) \to \operatorname{Diff}(G)^G\).
+ We claim \(g\) is an isomorphim. To see that \(g\) is injective, it suffices
+ to notice
+ \[
+ g(X_1 \cdots X_n)
+ = g(X_1) \cdots g(X_n)
+ = f(X_1) \cdots f(X_n)
+ \ne 0
+ \]
+ for all nonzero \(X_1, \cdots, X_n \in \mathfrak{g}\) --
+ \(\operatorname{Diff}(G)^G\) 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 \(\ker g = 0\). Given that
+ \(\operatorname{Diff}(G)^G\) is generated by \(\operatorname{Der}(G)^G\),
+ this also goes to show \(g\) is surjective.
+\end{proof}
% TODO: Comment on the fact this holds for algebraic groups too
@@ -506,5 +580,3 @@
\(\{X_{i_1} \cdot X_{i_2} \cdots X_{i_n} : n \ge 0, i_1 \le i_2 \le \cdots
\le i_n\}\) is a basis for \(\mathcal{U}(\mathfrak{g})\).
\end{theorem}
-
-% TODO: The analytic proof of PBW only works for finite-dimensional algebras