Added the root space decomposition.

generalstyle.sty

@@ -221,6 +221,7 @@
 \newcommand{\hlie}{\mathfrak{h}}
 \newcommand{\klie}{\mathfrak{k}}
 \newcommand{\gllie}{\mathfrak{gl}}
+\newcommand{\rlie}{\mathfrak{r}}
 \newcommand{\sllie}{\mathfrak{sl}}
 \newcommand{\tlie}{\mathfrak{t}}
 \newcommand{\nlie}{\mathfrak{n}}
@@ -289,6 +290,8 @@
 
 % delimiters
 \DeclarePairedDelimiter{\abs}{\lvert}{\rvert}
+\DeclarePairedDelimiter{\bil}{\langle}{\rangle}
+\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
 \DeclarePairedDelimiter{\pair}{\langle}{\rangle}
 \NewDocumentCommand{\restrict}{smmO{}}{
   \IfBooleanTF{#1}

sections/abstract_jordan_decomposition.tex

@@ -212,6 +212,40 @@ \section{The Abstract Jordan Decomposition}
 \end{example}
 
 
+\begin{lemma}
+  \label{commuting via abstract jd}
+  Let~$\glie$ be a finite dimensional semisimple Lie~algebra and let~$x \in \glie$.
+  Then any other element~$y \in \glie$ commutes with~$x$ if and only if both~$y_s$ and~$y_n$ commute with~$x$.
+\end{lemma}
+
+
+\begin{proof}
+  We may assume that~$\glie$ is linear by using the adjoint representation, which is faithful.
+  Then the assertion follows from \cref{concrete jordan decomposition} because the concrete and abstract Jordan decompositions coincide.
+\end{proof}
+
+
+\begin{lemma}
+  \label{abstract jordan decomposition of sum}
+    Let~$x$ and~$y$ be two commuting elements of a finite dimensional semisimple Lie~algebra~$\glie$.
+    Then the Jordan decomposition of~$x+y$ can be computed from the Jordan decompositions of~$x$ and~$y$ via
+    \[
+      (x + y)_s
+      =
+      x_s + y_s
+      \quad\text{and}\quad
+      (x + y)_n
+      =
+      x_n + y_n \,.
+    \]
+\end{lemma}
+
+
+\begin{proof}
+  We may assume that~$\glie$ is linear and then apply \cref{concrete jordan decomposition of sum}.
+\end{proof}
+
+
 % 
 % \begin{proof}[Proof of \cref{abstract jordan decomposition}]
 %   \leavevmode

sections/definition_of_semisimple_lie_algebras.tex

@@ -314,6 +314,7 @@ \subsection{More on Bilinear Forms}
 
 
 \begin{recall}[Orthogonals and non-degeneracy]
+  \label{reviewing orthogonals}
   If~$V$ is a finite dimensional (real or complex) inner product space then for every linear subspace~$U$ of~$V$ the restriction of the inner product to~$U$ is again an inner product,~$V = U \oplus U^\perp$ and~$(U^\perp)^\perp = U$.
   That the restriction is again an inner product holds because positive definiteness is a pointwise property.
   That~$U \cap U^\perp = 0$ also follows from the positive definiteness, and that~$V = U \oplus U^\perp$ can then be concluded from the dimension formula~$\dim V = \dim U + \dim U^\perp$.

sections/killing_form_and_cartans_criterion.tex

@@ -696,6 +696,28 @@ \subsection{Concrete Jordan Decomposition}
 \end{warning}
 
 
+\begin{lemma}
+  \label{concrete jordan decomposition of sum}
+  Let~$x$ and~$y$ be two commuting endomorphisms of a finite dimensional~{\vectorspace{$\kf$}}.
+  Then the Jordan decomposition of~$x+y$ can be computed from that of~$x$ and~$y$ via
+  \[
+    (x + y)_s
+    =
+    x_s + y_s
+    \quad\text{and}\quad
+    (x + y)_n
+    =
+    x_n + y_n \,.
+  \]
+\end{lemma}
+
+
+\begin{proof}
+  It follows from the commutativity of~$x$ and~$y$ that each part~$x_s$ and~$x_n$ of~$x$ commutes with each part~$y_s$ and~$y_n$ of~$y$.
+  It follows that~$x_s + y_s$ is again semisimple and that~$x_n + y_n$ is again nilpotent.
+\end{proof}
+
+
 \begin{definition}
   An element~$x \in \glie$ of a finite dimensional Lie~algebra~$\glie$ is~\defemph{\adsemisimple} if the endomorphism~$\ad(x)$ of~$\glie$ is semisimple.
 \end{definition}