lie-algebras-and-their-representations

diff --git a/TODO.md b/TODO.md
@@ -1,4 +1,3 @@
 # TODO
 
-* Change the notation for the Casimir element (use capital omega)
 * Make sure example 2.4 is right and find a more reliable answer

diff --git a/sections/complete-reducibility.tex b/sections/complete-reducibility.tex
@@ -272,10 +272,10 @@ to introduce some basic tools which will come in handy later on, known as\dots
 \section{Invariant Bilinear Forms}
 
 \begin{definition}\index{invariant bilinear form}
-  A symmetric bilinear \(B : \mathfrak{g} \times \mathfrak{g} \to K\) is called
-  \emph{\(\mathfrak{g}\)-invariant} if the operator \(\operatorname{ad}(X) :
-  \mathfrak{g} \to \mathfrak{g}\) is antisymmetric with respect to \(B\) for
-  all \(X \in \mathfrak{g}\).
+  A symmetric bilinear form \(B : \mathfrak{g} \times \mathfrak{g} \to K\) is
+  called \emph{\(\mathfrak{g}\)-invariant} if the operator
+  \(\operatorname{ad}(X) : \mathfrak{g} \to \mathfrak{g}\) is antisymmetric
+  with respect to \(B\) for all \(X \in \mathfrak{g}\).
   \[
     B(\operatorname{ad}(X) Y, Z) + B(Y, \operatorname{ad}(X) Z) = 0
   \]
@@ -625,18 +625,18 @@ a \(\mathfrak{g}\)-module}.
   i.e. the unique basis for \(\mathfrak{g}\) satisfying \(\kappa_M(X_i, X^j) =
   \delta_{i j}\). We call
   \[
-    C_M = X_1 X^1 + \cdots + X_r X^r \in \mathcal{U}(\mathfrak{g})
+    \Omega_M = X_1 X^1 + \cdots + X_r X^r \in \mathcal{U}(\mathfrak{g})
   \]
   the \emph{Casimir element of \(M\)}.
 \end{definition}
 
 \begin{lemma}
-  The definition of \(C_M\) is independent of the choice of basis
+  The definition of \(\Omega_M\) is independent of the choice of basis
   \(\{X_i\}_i\).
 \end{lemma}
 
 \begin{proof}
-  Whatever basis \(\{X_i\}_i\) we choose, the image of \(C_M\) under the
+  Whatever basis \(\{X_i\}_i\) we choose, the image of \(\Omega_M\) under the
   canonical isomorphism \(\mathfrak{g} \otimes \mathfrak{g} \isoto \mathfrak{g}
   \otimes \mathfrak{g}^* \isoto \operatorname{End}(\mathfrak{g})\) is the
   identity operator\footnote{Here the isomorphism $\mathfrak{g} \otimes
@@ -646,16 +646,16 @@ a \(\mathfrak{g}\)-module}.
 \end{proof}
 
 \begin{proposition}
-  The Casimir element \(C_M \in \mathcal{U}(\mathfrak{g})\) is central, so that
-  \(C_M\!\restriction_N : N \to N\) is a \(\mathfrak{g}\)-homomorphism for any
-  \(\mathfrak{g}\)-module \(N\). Furthermore, \(C_M\) acts on \(M\) as a
-  nonzero scalar operator whenever \(M\) is a non-trivial finite-dimensional
-  simple \(\mathfrak{g}\)-module.
+  The Casimir element \(\Omega_M \in \mathcal{U}(\mathfrak{g})\) is central, so
+  that \(\Omega_M\!\restriction_N : N \to N\) is a
+  \(\mathfrak{g}\)-homomorphism for any \(\mathfrak{g}\)-module \(N\).
+  Furthermore, \(\Omega_M\) acts on \(M\) as a nonzero scalar operator whenever
+  \(M\) is a non-trivial finite-dimensional simple \(\mathfrak{g}\)-module.
 \end{proposition}
 
 \begin{proof}
-  To see that \(C_M\) is central fix a basis \(\{X_i\}_i\) for \(\mathfrak{g}\)
-  and denote by \(\{X^i\}_i\) its dual basis as in
+  To see that \(\Omega_M\) is central fix a basis \(\{X_i\}_i\) for
+  \(\mathfrak{g}\) and denote by \(\{X^i\}_i\) its dual basis as in
   Definition~\ref{def:casimir-element}. Let \(X \in \mathfrak{g}\) and denote
   by \(\lambda_{i j}, \mu_{i j} \in K\) the coefficients of \(X_j\) and \(X^j\)
   in \([X, X_i]\) and \([X, X^i]\), respectively.
@@ -672,29 +672,30 @@ a \(\mathfrak{g}\)-module}.
   Hence
   \[
     \begin{split}
-      [X, C_M]
+      [X, \Omega_M]
       & = \sum_i [X, X_i X^i] \\
       & = \sum_i [X, X_i] X^i + \sum_i X_i [X, X^i] \\
       & = \sum_{i j} \lambda_{i j} X_j X^i + \sum_{i j} \mu_{i j} X_i X^j \\
       & = 0
     \end{split},
   \]
-  and \(C_M\) is central. This implies that \(C_M\!\restriction_N : N \to N\)
-  is a \(\mathfrak{g}\)-homomorphism for all \(\mathfrak{g}\)-modules \(N\):
-  its action commutes with the action of any other element of \(\mathfrak{g}\).
+  and \(\Omega_M\) is central. This implies that \(\Omega_M\!\restriction_N : N
+  \to N\) is a \(\mathfrak{g}\)-homomorphism for all \(\mathfrak{g}\)-modules
+  \(N\): its action commutes with the action of any other element of
+  \(\mathfrak{g}\).
 
   In particular, it follows from Schur's Lemma that if \(M\) is
-  finite-dimensional and simple then \(C_M\) acts on \(M\) as a scalar
+  finite-dimensional and simple then \(\Omega_M\) acts on \(M\) as a scalar
   operator. To see that this scalar is nonzero we compute
   \[
-    \operatorname{Tr}(C_M\!\restriction_M)
+    \operatorname{Tr}(\Omega_M\!\restriction_M)
     = \operatorname{Tr}(X_1\!\restriction_M X^1\!\restriction_M)
     + \cdots
     + \operatorname{Tr}(X_r\!\restriction_M X^r\!\restriction_M)
     = \dim \mathfrak{g},
   \]
-  so that \(C_M\!\restriction_M = \lambda \operatorname{Id}\) for \(\lambda =
-  \frac{\dim \mathfrak{g}}{\dim M} \ne 0\).
+  so that \(\Omega_M\!\restriction_M = \lambda \operatorname{Id}\) for
+  \(\lambda = \frac{\dim \mathfrak{g}}{\dim M} \ne 0\).
 \end{proof}
 
 As promised, the Casimir element of a \(\mathfrak{g}\)-module can be used to
@@ -732,60 +733,59 @@ establish\dots
   % because the action of every element of g is strictly upper triangular -- and
   % semisimple -- because it is a quotient of g, which is semisimple. We thus
   % have rho(g) = 0, so that W is trivial
-  Since \(\dim N = 2\), the simple component \(\mathcal{U}(\mathfrak{g})
-  \cdot n\) of \(n\) in \(N\) is either \(K n\) or \(N\) itself. But this
-  component cannot be \(N\), since the image of \(f\) is a
-  \(1\)-dimensional \(\mathfrak{g}\)-module -- i.e. a proper nonzero submodule.
-  Hence \(K n\) is invariant under the action of \(\mathfrak{g}\). In
-  particular, \(X \cdot n = 0\) for all \(X \in \mathfrak{g}\). Since \(n\) lies
-  outside the image of \(f\), \(g(w) \ne 0\) -- which is
-  to say, \(n \notin \ker g = \operatorname{im} f\). This implies the map \(K
-  \to N\) that takes \(1\) to \(\sfrac{n}{g(n)}\) is a splitting of
-  (\ref{eq:trivial-extrems-exact-seq}).
-
-  Now suppose that \(M\) is non-trivial, so that \(C_M\) acts on \(M\) as
-  \(\lambda\) for some \(\lambda \ne 0\). Denote by \(N^\mu\)
-  the generalized eigenspace of \(C_M\!\restriction_N : N \to N\) associated
-  with \(\mu \in K\). If we identify \(M\) with \(f(M)\),
-  it is clear that \(M \subset N^\lambda\). The exactness of
-  (\ref{eq:exact-seq-h1-vanishes}) implies \(\dim N = \dim M + 1\), so
-  that either \(N^\lambda = M\) or \(N^\lambda = N\). But if \(N^\lambda = N\)
-  then there is some nonzero \(n \in N^\lambda\) with \(n \notin M = \ker g\)
-  such that
+  Since \(\dim N = 2\), the simple component \(\mathcal{U}(\mathfrak{g}) \cdot
+  n\) of \(n\) in \(N\) is either \(K n\) or \(N\) itself. But this component
+  cannot be \(N\), since the image of \(f\) is a \(1\)-dimensional
+  \(\mathfrak{g}\)-module -- i.e. a proper nonzero submodule. Hence \(K n\) is
+  invariant under the action of \(\mathfrak{g}\). In particular, \(X \cdot n =
+  0\) for all \(X \in \mathfrak{g}\). Since \(n\) lies outside the image of
+  \(f\), \(g(n) \ne 0\) -- which is to say, \(n \notin \ker g =
+  \operatorname{im} f\). This implies the map \(K \to N\) that takes \(1\) to
+  \(\sfrac{n}{g(n)}\) is a splitting of (\ref{eq:trivial-extrems-exact-seq}).
+
+  Now suppose that \(M\) is non-trivial, so that \(\Omega_M\) acts on \(M\) as
+  \(\lambda\) for some \(\lambda \ne 0\). Denote by \(N^\mu\) the generalized
+  eigenspace of \(\Omega_M\!\restriction_N : N \to N\) associated with \(\mu
+  \in K\). If we identify \(M\) with \(f(M)\), it is clear that \(M \subset
+  N^\lambda\). The exactness of (\ref{eq:exact-seq-h1-vanishes}) implies \(\dim
+  N = \dim M + 1\), so that either \(N^\lambda = M\) or \(N^\lambda = N\). But
+  if \(N^\lambda = N\) then there is some nonzero \(n \in N^\lambda\) with \(n
+  \notin M = \ker g\) such that
   \[
     0
-    = (C_M - \lambda)^r \cdot n
-    = \sum_{k = 0}^r (-1)^k \binom{r}{k} \lambda^k C_M^{r - k} \cdot n
+    = (\Omega_M - \lambda)^r \cdot n
+    = \sum_{k = 0}^r (-1)^k \binom{r}{k} \lambda^k \Omega_M^{r - k} \cdot n
   \]
   for some \(r \ge 1\).
 
   In particular,
   \[
     (- \lambda)^{r - 1} g(n)
-    = \sum_{k = 0}^{r - 1} (-1)^k \binom{r}{k} \lambda^k g(C_M^{r - k} \cdot n)
     = \sum_{k = 0}^{r - 1} (-1)^k \binom{r}{k} \lambda^k
-      \underbrace{C_M^{r - k} \cdot g(n)}_{= \; 0}
+                           g(\Omega_M^{r - k} \cdot n)
+    = \sum_{k = 0}^{r - 1} (-1)^k \binom{r}{k} \lambda^k
+      \underbrace{\Omega_M^{r - k} \cdot g(n)}_{= \; 0}
     = 0,
   \]
-  which is a contradiction -- given that neither \((-\lambda)^{r - 1}\)
-  nor \(g(n)\) are nil. Hence \(M = N^\lambda\) and there must be some other
-  eigenvalue \(\mu\) of \(C_M\!\restriction_N\). For any such \(\mu\) and any
-  eigenvector \(n \in N_\mu\),
+  which is a contradiction -- given that neither \((-\lambda)^{r - 1}\) nor
+  \(g(n)\) are nil. Hence \(M = N^\lambda\) and there must be some other
+  eigenvalue \(\mu\) of \(\Omega_M\!\restriction_N\). For any such \(\mu\) and
+  any eigenvector \(n \in N_\mu\),
   \[
     \mu g(n)
     = g(\mu n)
-    = g(C_M \cdot n)
-    = C_M \cdot g(n)
+    = g(\Omega_M \cdot n)
+    = \Omega_M \cdot g(n)
     = 0
   \]
-  implies \(\mu = 0\), so that the eigenvalues of the action of \(C_M\) on
+  implies \(\mu = 0\), so that the eigenvalues of the action of \(\Omega_M\) on
   \(N\) are precisely \(\lambda\) and \(0\).
 
   Now notice that \(N^0\) is in fact a submodule of \(N\). Indeed,
   given \(n \in N^0\) and \(X \in \mathfrak{g}\), it follows from the fact that
-  \(C_M\) is central that
+  \(\Omega_M\) is central that
   \[
-    C_M^r \cdot (X \cdot n) = X \cdot (C_V^r \cdot n) = X \cdot 0 = 0
+    \Omega_M^r \cdot (X \cdot n) = X \cdot (\Omega_M^r \cdot n) = X \cdot 0 = 0
   \]
   for some \(r\). Hence \(N = M \oplus N^0\) as \(\mathfrak{g}\)-modules. The
   homomorphism \(g\) thus induces an isomorphism \(N^0 \cong \mfrac{N}{M}
@@ -888,8 +888,7 @@ We are now finally ready to prove\dots
 
   Now notice \(\operatorname{Hom}(L, L')^{\mathfrak{g}} =
   \operatorname{Hom}_{\mathfrak{g}}(L, L')\) for all \(\mathfrak{g}\)-modules
-  \(L'\). Indeed, given 
-  a \(K\)-linear map \(f : L \to L'\)
+  \(L'\). Indeed, given a \(K\)-linear map \(f : L \to L'\)
   \[
     \begin{split}
       f \in \operatorname{Hom}(L, L')^{\mathfrak{g}}