diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -155,40 +155,40 @@ 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 \(T_{\operatorname{s}},
- T_{\operatorname{n}} : V \to V\), 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
+ V\), there are unique commuting operators \(T_{\operatorname{ss}},
+ T_{\operatorname{nil}} : V \to V\), with \(T_{\operatorname{ss}}\)
+ diagonalizable and \(T_{\operatorname{nil}}\) nilpotent, such that \(T =
+ T_{\operatorname{ss}} + T_{\operatorname{nil}}\). The pair
+ \((T_{\operatorname{ss}}, T_{\operatorname{nil}})\) is known as \emph{the Jordan
decomposition of \(T\)}.
\end{proposition}
\begin{proposition}\index{abstract Jordan decomposition}
Given \(\mathfrak{g}\) semisimple and \(X \in \mathfrak{g}\), there are
- \(X_{\operatorname{s}}, X_{\operatorname{n}} \in \mathfrak{g}\) 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 \((X_{\operatorname{s}}, X_{\operatorname{n}})\)
+ \(X_{\operatorname{ss}}, X_{\operatorname{nil}} \in \mathfrak{g}\) such that \(X
+ = X_{\operatorname{ss}} + X_{\operatorname{nil}}\), \([X_{\operatorname{ss}},
+ X_{\operatorname{nil}}] = 0\), \(\operatorname{ad}(X_{\operatorname{ss}})\) is a
+ diagonalizable operator and \(\operatorname{ad}(X_{\operatorname{nil}})\) is a
+ nilpotent operator. The pair \((X_{\operatorname{ss}}, X_{\operatorname{nil}})\)
is known as \emph{the Jordan decomposition of \(X\)}.
\end{proposition}
It should be clear from the uniqueness of
-\(\operatorname{ad}(X)_{\operatorname{s}}\) and
-\(\operatorname{ad}(X)_{\operatorname{n}}\) that the Jordan decomposition of
+\(\operatorname{ad}(X)_{\operatorname{ss}}\) and
+\(\operatorname{ad}(X)_{\operatorname{nil}}\) that the Jordan decomposition of
\(\operatorname{ad}(X)\) is \(\operatorname{ad}(X) =
-\operatorname{ad}(X_{\operatorname{s}}) +
-\operatorname{ad}(X_{\operatorname{n}})\). What is perhaps more remarkable is
+\operatorname{ad}(X_{\operatorname{ss}}) +
+\operatorname{ad}(X_{\operatorname{nil}})\). What is perhaps more remarkable is
the fact this holds for \emph{any} finite-dimensional \(\mathfrak{g}\)-module.
In other words\dots
\begin{proposition}\label{thm:preservation-jordan-form}
Let \(M\) be a finite-dimensional \(\mathfrak{g}\)-module and \(X
\in \mathfrak{g}\). Denote by \(X\!\restriction_M\) the action of \(X\) on
- \(M\). Then \(X_{\operatorname{s}}\!\restriction_M =
- (X\!\restriction_M)_{\operatorname{s}}\) and
- \(X_{\operatorname{n}}\!\restriction_M =
- (X\!\restriction_M)_{\operatorname{n}}\).
+ \(M\). Then \(X_{\operatorname{ss}}\!\restriction_M =
+ (X\!\restriction_M)_{\operatorname{ss}}\) and
+ \(X_{\operatorname{nil}}\!\restriction_M =
+ (X\!\restriction_M)_{\operatorname{nil}}\).
\end{proposition}
This last result is known as \emph{the preservation of the Jordan form}, and a
@@ -216,20 +216,20 @@ implies\dots
Fix some \(H \in \mathfrak{h}\). It suffices to show that \(H\!\restriction_M
: M \to M\) is a diagonalizable operator.
- If we write \(H = H_{\operatorname{s}} + H_{\operatorname{n}}\) for the
+ If we write \(H = H_{\operatorname{ss}} + H_{\operatorname{nil}}\) 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}} =
+ \(\operatorname{ad}(H_{\operatorname{ss}}) =
+ \operatorname{ad}(H)_{\operatorname{ss}}\). But \(\operatorname{ad}(H)\) is a
+ diagonalizable operator, so that \(\operatorname{ad}(H)_{\operatorname{ss}} =
\operatorname{ad}(H)\). 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, \(H_{\operatorname{n}} = 0\).
+ \(\operatorname{ad}(H_{\operatorname{nil}}) =
+ \operatorname{ad}(H)_{\operatorname{nil}} = 0\), so that
+ \(H_{\operatorname{nil}}\) is a central element of \(\mathfrak{g}\). Since
+ \(\mathfrak{g}\) is semisimple, \(H_{\operatorname{nil}} = 0\).
Proposition~\ref{thm:preservation-jordan-form} then implies
- \((H\!\restriction_M)_{\operatorname{n}} =
- H_{\operatorname{n}}\!\restriction_M = 0\), so \(H\!\restriction_M =
- (H\!\restriction_M)_{\operatorname{s}}\) is a diagonalizable operator.
+ \((H\!\restriction_M)_{\operatorname{nil}} =
+ H_{\operatorname{nil}}\!\restriction_M = 0\), so \(H\!\restriction_M =
+ (H\!\restriction_M)_{\operatorname{ss}}\) is a diagonalizable operator.
\end{proof}
We should point out that this last proof only works for semisimple Lie