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: 273cce63a0b48114113abb5f983feb3287b2b4a2
Parent: 4487ff4715d5035a825266620ccfc88ef4a4bb4e
Author: Pablo <pablo-escobar@riseup.net>
Date: Mon, 21 Nov 2022 20:46:27 +0000

Removed unnecessary comment

Diffstat

1 file changed, 0 insertions, 30 deletions

Status	File Name	N° Changes	Insertions	Deletions
Modified	sections/introduction.tex	30	0	30

diff --git a/sections/introduction.tex b/sections/introduction.tex
@@ -969,33 +969,3 @@ Surprisingly, this functor has right adjoint.
 \end{proof}
 
 % TODO: Add a conclusion
-
-% TODO: Move this to the next chapter
-%\begin{definition}
-%  A representation of \(\mathfrak{g}\) is called \emph{indecomposable} if it is
-%  not isomorphic to the direct sum of two non-zero representations.
-%\end{definition}
-%
-%\begin{definition}
-%  A representation of \(\mathfrak{g}\) is called \emph{irreducible} if it has
-%  no non-zero subrepresentations.
-%\end{definition}
-%
-%\begin{lemma}[Schur]
-%  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}
-