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

git clone: git://git.pablopie.xyz/lie-algebras-and-their-representations
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Commit: 17698441329687ba14bf0502955e2a2425e8fafc
Parent: 930a094e555decdb730a170be36307b75291ad54
Author: Pablo <pablo-escobar@riseup.net>
Date: Mon, 26 Jun 2023 22:05:44 +0000

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

Diffstat

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