Source code for my notes on representations of semisimple Lie algebras and Olivier Mathieu's classification of simple weight modules
Added the statement of a lemma
Added the statement of the relevant section of Lemma 6.1 to the chapter on the classification of coherent families
2 files changed, 23 insertions, 15 deletions
| Status | File Name | N° Changes | Insertions | Deletions |
| Modified | sections/coherent-families.tex | 15 | 11 | 4 |
| Modified | sections/simple-weight.tex | 23 | 12 | 11 |
diff --git a/sections/coherent-families.tex b/sections/coherent-families.tex @@ -91,7 +91,14 @@ combinatorial counterpart. Then \(\chi_\lambda = \chi_\mu\). \end{proposition} -% TODOO: State Lemma 6.1 of Mathieu +% TODO: Note that if σ_β ∙ λ is not dominant integral then L(σ_β ∙ λ) is +% infinite-dimensional and 𝓔𝔁𝓽(L(σ_β ∙ λ)) ≅ 𝓔𝔁𝓽(L(λ)) +\begin{proposition} + Let \(\beta \in \Sigma\) and \(\lambda \notin P^+\) be such that + \(L(\lambda)\) is bounded and \(\lambda(H_\beta) \notin \mathbb{N}\). Then + \(L(\sigma_\beta \bullet \lambda) \subset \mExt(L(\lambda))\). +\end{proposition} + % TODOO: Treat the case of sl(2) here? \section{Coherent \(\mathfrak{sp}_{2n}(K)\)-families} @@ -250,9 +257,9 @@ Example~\ref{ex:sl-canonical-basis}. Also fix \(\rho = \sfrac{1}{2} \beta_1 + % implies L(μ) is contained in 𝓔𝔁𝓽(L(λ)) - so that L(μ) is also bounded and % 𝓔𝔁𝓽(L(λ)) ≅ 𝓔𝔁𝓽(L(μ)) \begin{definition} - Given \(m, m' \in \mathscr{B}\), say there is an arrow \(m \to m'\) if - the unique \(i\) such that \(m_i - m_{i + 1}\) is not a positive integer is - such that \(m' = \sigma_i \cdot m\). + Given \(m, m' \in \mathscr{B}\), say there is an arrow \(m \to m'\) if the + unique \(i\) such that \(m_i - m_{i + 1}\) is not a positive integer is such + that \(m' = \sigma_i \cdot m\). \end{definition} It should then be obvious that\dots
diff --git a/sections/simple-weight.tex b/sections/simple-weight.tex @@ -1572,27 +1572,28 @@ A sort of ``reciprocal'' of Theorem~\ref{thm:mathieu-ext-exists-unique} also holds. Namely\dots \begin{proposition} - Let \(\mathcal{M}\) be a semisimple irreducible coherent extension. Then - there exists some simple bounded \(\mathfrak{g}\)-module \(M\) such that - \(\mathcal{M} \cong \mExt(M)\). + Let \(\mathcal{M}\) be a semisimple irreducible coherent family and \(M + \subset \mathcal{M}\) be an infinite-dimensional simple submodule. Then + \(\mathcal{M} \cong \mExt(M)\). In particular, all semisimple coherent + families have the form \(\mathcal{M} \cong \mExt(M)\) for some simple bounded + \(\mathfrak{g}\)-module \(M\). \end{proposition} \begin{proof} - Let \(M \subset \mathcal{M}\) be any simple submodule and \(d = \deg - \mathcal{M}\). Since \(M\) is bounded, + Since \(M \subset \mathcal{M}\), \(M\) is bounded and \(\operatorname{supp}_{\operatorname{ess}} M\) is Zariski-dense. In addition, it follows from Lemma~\ref{thm:set-of-simple-u0-mods-is-open} that \(U = \{\lambda \in \mathfrak{h}^* : \mathcal{M}_\lambda \ \text{is a simple - $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is a Zariski-open subset - -- which is non-empty since \(\mathcal{M}\) is irreducible. + $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is a Zariski-open subset -- which + is non-empty since \(\mathcal{M}\) is irreducible. Hence there is some \(\lambda \in \operatorname{supp}_{\operatorname{ess}} M \cap U\). In particular, there is some \(\lambda \in \operatorname{supp}_{\operatorname{ess}} M\) such that \(M_\lambda = - \mathcal{M}_\lambda\) and thus \(\deg M = \dim \mathcal{M}_\lambda = d\). - This implies that \(\mathcal{M}\) is a coherent extension of \(M\), so that - by the uniqueness of semisimple irreducible coherent extensions we get - \(\mathcal{M} \cong \mExt(M)\). + \mathcal{M}_\lambda\) and thus \(\deg M = \dim \mathcal{M}_\lambda = \deg + \mathcal{M}\). This implies that \(\mathcal{M}\) is a coherent extension of + \(M\), so that by the uniqueness of semisimple irreducible coherent + extensions we get \(\mathcal{M} \cong \mExt(M)\). \end{proof} Having thus reduced the problem of classifying the cuspidal