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
36b566c9870c636181c2723f27521cc9e48135ac
Parent
e2ec70dd6089a6518c1c393efb9b33b1bf4843ec
Author
Pablo <pablo-escobar@riseup.net>
Date

Minor improvement in the notation of a proof

Diffstat

1 file changed, 16 insertions, 16 deletions

Status File Name N° Changes Insertions Deletions
Modified sections/introduction.tex 32 16 16
diff --git a/sections/introduction.tex b/sections/introduction.tex
@@ -554,37 +554,37 @@ containing \(\mathfrak{g}\) as a Lie subalgebra. In practice this means\dots
 
 \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
+  universal property of free algebras, there is a homomorphism of algebras
+  \(\tilde f : T \mathfrak{g} \to A\) such that
   \begin{center}
     \begin{tikzcd}
-      T \mathfrak{g} \arrow[dotted]{dr}{g} & \\
+      T \mathfrak{g} \arrow[dotted]{dr}{\tilde f} & \\
       \mathfrak{g} \uar \rar[swap]{f}      & A
     \end{tikzcd}
   \end{center}
 
   Since \(f\) is a homomorphism of Lie algebras,
   \[
-    g([X, Y])
+    \tilde f([X, Y])
     = f([X, Y])
     = [f(X), f(Y)]
-    = [g(X), g(Y)]
-    = g(X \otimes Y - Y \otimes X)
+    = [\tilde f(X), \tilde f(Y)]
+    = \tilde f(X \otimes Y - Y \otimes X)
   \]
   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 \(\mathcal{U}(\mathfrak{g}) = \mfrac{T
-  \mathfrak{g}}{I}\).
+  \otimes X) : X, Y \in \mathfrak{g}) \subset \ker \tilde f\) and therefore
+  \(\tilde f\) factors through the quotient \(\mathcal{U}(\mathfrak{g}) =
+  \mfrac{T \mathfrak{g}}{I}\).
   \begin{center}
     \begin{tikzcd}
-      T \mathfrak{g} \rar{g} \dar                                 & A \\
-      \mathcal{U}(\mathfrak{g}) \arrow[swap, dotted]{ur}{\bar{g}} &
+      T \mathfrak{g} \rar{\tilde f} \dar                                 & A \\
+      \mathcal{U}(\mathfrak{g}) \arrow[swap, dotted]{ur}{\bar{\tilde f}} &
     \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 \(\bar{g}\).
+  Combining the two previous diagrams, we can see that \(\bar{\tilde f}\) is
+  indeed an extension of \(f\). The uniqueness of the extension then follows
+  from the uniqueness of \(\tilde f\) and \(\bar{\tilde f}\).
 \end{proof}
 
 We should point out this construction is functorial. Indeed, if
@@ -603,8 +603,8 @@ algebras \(\mathcal{U}(f) : \mathcal{U}(\mathfrak{g}) \to
 \end{center}
 
 It is important to note, however, that \(\mathcal{U} : K\text{-}\mathbf{LieAlg}
-\to K\text{-}\mathbf{Alg}\) is not the ``inverse'' of \(\operatorname{Lie} :
-K\text{-}\mathbf{Alg} \to K\text{-}\mathbf{LieAlg}\). For instance, if
+\to K\text{-}\mathbf{Alg}\) is not the ``inverse'' of our functor
+\(K\text{-}\mathbf{Alg} \to K\text{-}\mathbf{LieAlg}\). For instance, if
 \(\mathfrak{g} = K\) is the \(1\)-dimensional Abelian Lie algebra then
 \(\mathcal{U}(\mathfrak{g}) \cong K[x]\), which is infinite-dimensional.
 Nevertheless, proposition~\ref{thm:universal-env-uni-prop} may be restated