- Commit
- dba0653482d86747f1446e39262fb9b0506aba4c
- Parent
- ee93c98538ba2196d20fe88f3daef859eeab2ec6
- Author
- Pablo <pablo-escobar@riseup.net>
- Date
Added proofs to some of the theorems of the introduction
Source code for my notes on representations of semisimple Lie algebras and Olivier Mathieu's classification of simple weight modules
Added proofs to some of the theorems of the introduction
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