lie-algebras-and-their-representations

diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -135,31 +135,40 @@ Jordan decomposition of a semisimple Lie algebra}.
 
 \begin{proposition}[Jordan]
   Given a finite-dimensional vector space \(V\) and an operator \(T : V \to
-  V\), there are unique commuting operators \(T_s, T_n : V \to V\), with
-  \(T_s\) diagonalizable and \(T_n\) nilpotent, such that \(T = T_s + T_n\).
-  The pair \((T_s, T_n)\) is known as \emph{the Jordan decomposition of \(T\)}.
+  V\), there are unique commuting operators \(T_{\operatorname{s}},
+  T_{\operatorname{n}} : V \to V\), with \(T_{\operatorname{s}}\)
+  diagonalizable and \(T_{\operatorname{n}}\) nilpotent, such that \(T =
+  T_{\operatorname{s}} + T_{\operatorname{n}}\). The pair
+  \((T_{\operatorname{s}}, T_{\operatorname{n}})\) is known as \emph{the Jordan
+  decomposition of \(T\)}.
 \end{proposition}
 
 \begin{proposition}
   Given \(\mathfrak{g}\) semisimple and \(X \in \mathfrak{g}\), there are
-  \(X_s, X_n \in \mathfrak{g}\) such that \(X = X_s + X_n\), \([X_s, X_n] =
-  0\), \(\operatorname{ad}(X_s)\) is a diagonalizable operator and
-  \(\operatorname{ad}(X_n)\) is a nilpotent operator. The pair \((X_s, X_n)\)
+  \(X_{\operatorname{s}}, X_{\operatorname{n}} \in \mathfrak{g}\) such that \(X
+  = X_{\operatorname{s}} + X_{\operatorname{n}}\), \([X_{\operatorname{s}},
+  X_{\operatorname{n}}] = 0\), \(\operatorname{ad}(X_{\operatorname{s}})\) is a
+  diagonalizable operator and \(\operatorname{ad}(X_{\operatorname{n}})\) is a
+  nilpotent operator. The pair \((X_{\operatorname{s}}, X_{\operatorname{n}})\)
   is known as \emph{the Jordan decomposition of \(X\)}.
 \end{proposition}
 
-It should be clear from the uniqueness of \(\operatorname{ad}(X)_s\) and
-\(\operatorname{ad}(X)_n\) that the Jordan decomposition of
-\(\operatorname{ad}(X)\) is \(\operatorname{ad}(X) = \operatorname{ad}(X_s) +
-\operatorname{ad}(X_n)\). What's perhaps more remarkable is the fact this holds
-for \emph{any} finite-dimensional representation of \(\mathfrak{g}\). In other
-words\dots
+It should be clear from the uniqueness of
+\(\operatorname{ad}(X)_{\operatorname{s}}\) and
+\(\operatorname{ad}(X)_{\operatorname{n}}\) that the Jordan decomposition of
+\(\operatorname{ad}(X)\) is \(\operatorname{ad}(X) =
+\operatorname{ad}(X_{\operatorname{s}}) +
+\operatorname{ad}(X_{\operatorname{n}})\). What's perhaps more remarkable is
+the fact this holds for \emph{any} finite-dimensional representation of
+\(\mathfrak{g}\). In other words\dots
 
 \begin{proposition}\label{thm:preservation-jordan-form}
   Let \(V\) be a finite-dimensional representation of \(\mathfrak{g}\) and \(X
   \in \mathfrak{g}\). Denote by \(X\!\restriction_V\) the action of \(X\) on
-  \(V\). Then \(X_s\!\restriction_V = (X\!\restriction_V)_s\) and
-  \(X_n\!\restriction_V = (X\!\restriction_V)_n\).
+  \(V\). Then \(X_{\operatorname{s}}\!\restriction_V =
+  (X\!\restriction_V)_{\operatorname{s}}\) and
+  \(X_{\operatorname{n}}\!\restriction_V =
+  (X\!\restriction_V)_{\operatorname{n}}\).
 \end{proposition}
 
 This last result is known as \emph{the preservation of the Jordan form}, and a
@@ -187,15 +196,20 @@ implies\dots
   Fix some \(H \in \mathfrak{h}\). It suffices to show that \(H\!\restriction_V
   : V \to V\) is a diagonalizable operator.
 
-  If we write \(H = H_s + H_n\) for the abstract Jordan decomposition of \(H\),
-  we know \(\operatorname{ad}(H_s) = \operatorname{ad}(H)_s\). But
-  \(\operatorname{ad}(H)\) is a diagonalizable operator, so that
-  \(\operatorname{ad}(H)_s = \operatorname{ad}(H)\). This implies
-  \(\operatorname{ad}(H_n) = \operatorname{ad}(H)_n = 0\), so that \(H_n\) is a
-  central element of \(\mathfrak{g}\). Since \(\mathfrak{g}\) is semisimple,
-  \(H_n = 0\). Proposition~\ref{thm:preservation-jordan-form} then implies
-  \((H\!\restriction_V)_n = (H_n)\!\restriction_V = 0\), so \(H\!\restriction_V
-  = (H\!\restriction_V)_s\) is a diagonalizable operator.
+  If we write \(H = H_{\operatorname{s}} + H_{\operatorname{n}}\) for the
+  abstract Jordan decomposition of \(H\), we know
+  \(\operatorname{ad}(H_{\operatorname{s}}) =
+  \operatorname{ad}(H)_{\operatorname{s}}\). But \(\operatorname{ad}(H)\) is a
+  diagonalizable operator, so that \(\operatorname{ad}(H)_{\operatorname{s}} =
+  \operatorname{ad}(H)\). This implies
+  \(\operatorname{ad}(H_{\operatorname{n}}) =
+  \operatorname{ad}(H)_{\operatorname{n}} = 0\), so that
+  \(H_{\operatorname{n}}\) is a central element of \(\mathfrak{g}\). Since
+  \(\mathfrak{g}\) is semisimple, \(H_{\operatorname{n}} = 0\).
+  Proposition~\ref{thm:preservation-jordan-form} then implies
+  \((H\!\restriction_V)_{\operatorname{n}} =
+  (H_{\operatorname{n}})\!\restriction_V = 0\), so \(H\!\restriction_V =
+  (H\!\restriction_V)_{\operatorname{s}}\) is a diagonalizable operator.
 \end{proof}
 
 We should point out that this last proof only works for semisimple Lie