lie-algebras-and-their-representations

Source code for my notes on representations of semisimple Lie algebras and Olivier Mathieu's classification of simple weight modules

Commit
dba0653482d86747f1446e39262fb9b0506aba4c
Parent
ee93c98538ba2196d20fe88f3daef859eeab2ec6
Author
Pablo <pablo-escobar@riseup.net>
Date

Added proofs to some of the theorems of the introduction

Diffstat

1 file changed, 87 insertions, 15 deletions

Status File Name N° Changes Insertions Deletions
Modified sections/introduction.tex 102 87 15
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