lie-algebras-and-their-representations

diff --git a/TODO.md b/TODO.md
@@ -1,6 +1,5 @@
 # TODO
 
-* Change the notation for the Killing form (use kappa)
 * Comment on the Chevalieu-Eilenberg complex
 * 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
@@ -298,7 +298,7 @@ An interesting example of an invariant bilinear form is the so called
   Given a finite-dimensional Lie algebra \(\mathfrak{g}\), the symmetric
   bilinear form
   \begin{align*}
-    B : \mathfrak{g} \times \mathfrak{g} & \to K \\
+    \kappa : \mathfrak{g} \times \mathfrak{g} & \to K \\
     (X, Y) &
     \mapsto \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y))
   \end{align*}
@@ -313,7 +313,7 @@ Z \in \mathfrak{gl}_n(K)\). In fact this same identity show\dots
   Given a finite-dimensional \(\mathfrak{g}\)-module \(M\), the symmetric
   bilinear form
   \begin{align*}
-    B_M : \mathfrak{g} \times \mathfrak{g} & \to K \\
+    \kappa_M : \mathfrak{g} \times \mathfrak{g} & \to K \\
     (X, Y) & \mapsto \operatorname{Tr}(X\!\restriction_M \, Y\!\restriction_M)
   \end{align*}
   is \(\mathfrak{g}\)-invariant.
@@ -331,7 +331,7 @@ characterization of finite-dimensional semisimple Lie algebras, known as
     \item For each non-trivial finite-dimensional \(\mathfrak{g}\)-module
       \(M\), the \(\mathfrak{g}\)-invariant bilinear form
       \begin{align*}
-        B_M : \mathfrak{g} \times \mathfrak{g}                               &
+        \kappa_M : \mathfrak{g} \times \mathfrak{g}                          &
         \to K                                                                \\
         (X, Y)                                                               &
         \mapsto \operatorname{Tr}(X\!\restriction_M \circ Y\!\restriction_M)
@@ -339,24 +339,24 @@ characterization of finite-dimensional semisimple Lie algebras, known as
       is non-degenerate\footnote{A symmetric bilinear form $B : \mathfrak{g}
       \times \mathfrak{g} \to K$ is called non-degenerate if $B(X, Y) = 0$ for
       all $Y \in \mathfrak{g}$ implies $X = 0$.}.
-    \item The Killing form \(B\) is non-degenerate.
+    \item The Killing form \(\kappa\) is non-degenerate.
   \end{enumerate}
 \end{proposition}
 
 This proof is somewhat technical, but the idea behind it is simple. First, for
 \strong{(i)} \(\implies\) \strong{(ii)} we show that \(\mathfrak{a} = \{ X \in
-\mathfrak{g} : B_M(X, Y) = 0 \, \forall Y \in \mathfrak{g}\}\) is a solvable
-ideal of \(\mathfrak{g}\). Hence \(\mathfrak{a} = 0\). For \strong{(ii)}
-\(\implies\) \strong{(iii)} it suffices to take \(M = \mathfrak{g}\) the
-adjoint \(\mathfrak{g}\)-module. Finally, for \strong{(iii)} \(\implies\)
-\strong{(i)} we note that the orthogonal complement of any \(\mathfrak{a}
-\normal \mathfrak{g}\) with respect to the Killing form \(B\) is an ideal
-\(\mathfrak{b}\) of \(\mathfrak{g}\) with \(\mathfrak{g} = \mathfrak{a} \oplus
-\mathfrak{b}\). Furthermore, the Killing form of \(\mathfrak{a}\) is the
-restriction \(B\!\restriction_{\mathfrak{a}}\) of the Killing form of
-\(\mathfrak{g}\) to \(\mathfrak{a} \times \mathfrak{a}\), which is
-non-degenerate. It then follows from induction in \(\dim \mathfrak{a}\) that
-\(\mathfrak{g}\) is the sum of simple ideals.
+\mathfrak{g} : \kappa_M(X, Y) = 0 \, \forall Y \in \mathfrak{g}\}\) is a
+solvable ideal of \(\mathfrak{g}\). Hence \(\mathfrak{a} = 0\). For
+\strong{(ii)} \(\implies\) \strong{(iii)} it suffices to take \(M =
+\mathfrak{g}\) the adjoint \(\mathfrak{g}\)-module. Finally, for \strong{(iii)}
+\(\implies\) \strong{(i)} we note that the orthogonal complement of any
+\(\mathfrak{a} \normal \mathfrak{g}\) with respect to the Killing form
+\(\kappa\) is an ideal \(\mathfrak{b}\) of \(\mathfrak{g}\) with \(\mathfrak{g}
+= \mathfrak{a} \oplus \mathfrak{b}\). Furthermore, the Killing form of
+\(\mathfrak{a}\) is the restriction \(\kappa\!\restriction_{\mathfrak{a}}\) of
+the Killing form of \(\mathfrak{g}\) to \(\mathfrak{a} \times \mathfrak{a}\),
+which is non-degenerate. It then follows from induction in \(\dim
+\mathfrak{a}\) that \(\mathfrak{g}\) is the sum of simple ideals.
 
 We refer the reader to \cite[ch. 5]{humphreys} for a complete proof. Without
 further ado, we may proceed to our\dots
@@ -621,8 +621,8 @@ a \(\mathfrak{g}\)-module}.
 \begin{definition}\label{def:casimir-element}\index{Casimir element}
   Let \(M\) be a finite-dimensional \(\mathfrak{g}\)-module. Let
   \(\{X_i\}_i\) be a basis for \(\mathfrak{g}\) and denote by \(\{X^i\}_i
-  \subset \mathfrak{g}\) its dual basis with respect to the form \(B_M\) --
-  i.e. the unique basis for \(\mathfrak{g}\) satisfying \(B_M(X_i, X^j) =
+  \subset \mathfrak{g}\) its dual basis with respect to the form \(\kappa_M\) --
+  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})
@@ -642,7 +642,7 @@ a \(\mathfrak{g}\)-module}.
   identity operator\footnote{Here the isomorphism $\mathfrak{g} \otimes
   \mathfrak{g} \isoto \mathfrak{g} \otimes \mathfrak{g}^*$ is given by
   tensoring the identity $\mathfrak{g} \to \mathfrak{g}$ with the isomorphism
-  $\mathfrak{g} \isoto \mathfrak{g}^*$ induced by the form $B_M$.}.
+  $\mathfrak{g} \isoto \mathfrak{g}^*$ induced by the form $\kappa_M$.}.
 \end{proof}
 
 \begin{proposition}
@@ -660,12 +660,12 @@ a \(\mathfrak{g}\)-module}.
   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.
 
-  The invariance of \(B_M\) implies
+  The invariance of \(\kappa_M\) implies
   \[
     \lambda_{i k}
-    = B_M([X, X_i], X^k)
-    = B_M(-[X_i, X], X^k)
-    = B_M(X_i, -[X, X^k])
+    = \kappa_M([X, X_i], X^k)
+    = \kappa_M(-[X_i, X], X^k)
+    = \kappa_M(X_i, -[X, X^k])
     = - \mu_{k i}
   \]
 

diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -186,7 +186,7 @@ implies\dots
   a diagonal matrix in this basis. In other words, there are linear functionals
   \(\lambda_i \in \mathfrak{h}^*\) so that
   \(
-    H m_i = \lambda_i(H) m_i
+    H \cdot m_i = \lambda_i(H) m_i
   \)
   for all \(H \in \mathfrak{h}\). In particular,
   \[
@@ -238,7 +238,7 @@ M_\lambda\) fails. As a first consequence of
 Corollary~\ref{thm:finite-dim-is-weight-mod} we show\dots
 
 \begin{corollary}
-  The restriction of the Killing form \(B\) to \(\mathfrak{h}\) is
+  The restriction of the Killing form \(\kappa\) to \(\mathfrak{h}\) is
   non-degenerate.
 \end{corollary}
 
@@ -260,31 +260,31 @@ Corollary~\ref{thm:finite-dim-is-weight-mod} we show\dots
   \]
 
   We furthermore claim that \(\mathfrak{h} = \mathfrak{g}_0\) is orthogonal to
-  \(\mathfrak{g}_\alpha\) with respect to \(B\) for any \(\alpha \ne 0\).
+  \(\mathfrak{g}_\alpha\) with respect to \(\kappa\) for any \(\alpha \ne 0\).
   Indeed, given \(X \in \mathfrak{g}_\alpha\) and \(H_1, H_2 \in \mathfrak{h}\)
   with \(\alpha(H_1) \ne 0\) we have
   \[
-    \alpha(H_1) \cdot B(X, H_2)
-    = B([H_1, X], H_2)
-    = - B([X, H_1], H_2)
-    = - B(X, [H_1, H_2])
+    \alpha(H_1) \cdot \kappa(X, H_2)
+    = \kappa([H_1, X], H_2)
+    = - \kappa([X, H_1], H_2)
+    = - \kappa(X, [H_1, H_2])
     = 0
   \]
 
-  Hence the non-degeneracy of \(B\) implies the non-degeneracy of its
+  Hence the non-degeneracy of \(\kappa\) implies the non-degeneracy of its
   restriction.
 \end{proof}
 
-We should point out that the restriction of \(B\) to \(\mathfrak{h}\) is
+We should point out that the restriction of \(\kappa\) to \(\mathfrak{h}\) is
 \emph{not} the Killing form of \(\mathfrak{h}\). In fact, since
 \(\mathfrak{h}\) is Abelian, its Killing form is identically zero -- which is
 hardly ever a non-degenerate form.
 
 \begin{note}
-  Since \(B\) induces an isomorphism \(\mathfrak{h} \isoto \mathfrak{h}^*\), it
-  induces a bilinear form \((B(X, \cdot), B(Y, \cdot)) \mapsto B(X, Y)\) in
-  \(\mathfrak{h}^*\). As in section~\ref{sec:sl3-reps}, we denote this form by
-  \(B\) as well.
+  Since \(\kappa\) induces an isomorphism \(\mathfrak{h} \isoto
+  \mathfrak{h}^*\), it induces a bilinear form \((\kappa(X, \cdot), \kappa(Y,
+  \cdot)) \mapsto \kappa(X, Y)\) in \(\mathfrak{h}^*\). As in
+  section~\ref{sec:sl3-reps}, we denote this form by \(\kappa\) as well.
 \end{note}
 
 We now have most of the necessary tools to reproduce the results of the
@@ -338,11 +338,11 @@ of the Killing form to the Cartan subalgebra.
     \in \mathfrak{g}_{r \alpha + r \beta + \gamma}
     = 0
   \]
-  for \(r\) large enough. In particular, \(B(X, Y) =
+  for \(r\) large enough. In particular, \(\kappa(X, Y) =
   \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\). Now if
-  \(- \alpha\) is not an eigenvalue we find \(B(X, \mathfrak{g}_\beta) = 0\)
-  for all roots \(\beta\), which contradicts the non-degeneracy of \(B\).
-  Hence \(- \alpha\) must be an eigenvalue of the adjoint action of
+  \(- \alpha\) is not an eigenvalue we find \(\kappa(X, \mathfrak{g}_\beta) =
+  0\) for all roots \(\beta\), which contradicts the non-degeneracy of
+  \(\kappa\). Hence \(- \alpha\) must be an eigenvalue of the adjoint action of
   \(\mathfrak{h}\).
 
   For the second statement, note that if the roots of \(\mathfrak{g}\) do not
@@ -505,17 +505,17 @@ for \(\Delta\)?
 
 The intuition behind the proof of this proposition is similar to our original
 idea of fixing a direction in \(\mathfrak{h}^*\) in the case of
-\(\mathfrak{sl}_3(K)\). Namely, one can show that \(B(\alpha, \beta) \in
+\(\mathfrak{sl}_3(K)\). Namely, one can show that \(\kappa(\alpha, \beta) \in
 \mathbb{Z}\) for all \(\alpha, \beta \in \Delta\), so that the Killing form
-\(B\) restricts to a nondegenerate \(\mathbb{Q}\)-bilinear form \(\mathbb{Q}
-\Delta \times \mathbb{Q} \Delta \to \mathbb{Q}\). We can then fix a nonzero
-vector \(\gamma \in \mathbb{Q} \Delta\) and consider the orthogonal projection
-\(f : \mathbb{Q} \Delta \to \mathbb{Q} \gamma \cong \mathbb{Q}\). We say a root
-\(\alpha \in \Delta\) is \emph{positive} if \(f(\alpha) > 0\), and we call a
-positive root \(\alpha\) \emph{simple} if it cannot be written as the sum two
-other positive roots. The subset \(\Sigma \subset \Delta\) of all simple roots
-is a basis for \(\Delta\), and all other basis can be shown to arise in this
-way.
+\(\kappa\) restricts to a nondegenerate \(\mathbb{Q}\)-bilinear form
+\(\mathbb{Q} \Delta \times \mathbb{Q} \Delta \to \mathbb{Q}\). We can then fix
+a nonzero vector \(\gamma \in \mathbb{Q} \Delta\) and consider the orthogonal
+projection \(f : \mathbb{Q} \Delta \to \mathbb{Q} \gamma \cong \mathbb{Q}\). We
+say a root \(\alpha \in \Delta\) is \emph{positive} if \(f(\alpha) > 0\), and
+we call a positive root \(\alpha\) \emph{simple} if it cannot be written as the
+sum two other positive roots. The subset \(\Sigma \subset \Delta\) of all
+simple roots is a basis for \(\Delta\), and all other basis can be shown to
+arise in this way.
 
 Fix some basis \(\Sigma\) for \(\Delta\), with corresponding decomposition
 \(\Delta^+ \cup \Delta^- = \Delta\). Let \(\lambda\) be a maximal weight of
@@ -569,7 +569,7 @@ This has a number of important consequences. For instance\dots
 This is entirely analogous to the situation of \(\mathfrak{sl}_3(K)\), where we
 found that the weights of the simple \(\mathfrak{sl}_3(K)\)-modules formed
 continuous strings symmetric with respect to the lines \(K \alpha\) with
-\(B(\alpha_i - \alpha_j, \alpha) = 0\). As in the case of
+\(\kappa(\alpha_i - \alpha_j, \alpha) = 0\). As in the case of
 \(\mathfrak{sl}_3(K)\), the same class of arguments leads us to the
 conclusion\dots
 

diff --git a/sections/sl2-sl3.tex b/sections/sl2-sl3.tex
@@ -672,16 +672,16 @@ Among other things, this last result goes to show that the diagrams we have
 been drawing are in fact consistent with the theory we have developed. Namely,
 since all weights lie in the rational span of \(\{\alpha_1, \alpha_2,
 \alpha_3\}\), we may as well draw them in the Cartesian plane. In fact, the
-attentive reader may notice that \(B(E_{1 2}, E_{2 3}) = - \sfrac{1}{2}\), so
-that the angle -- with respect to the Killing form \(B\) -- between the root
-vectors \(E_{1 2}\) and \(E_{2 3}\) is precisely the same as the angle between
-the points representing their roots \(\alpha_1 - \alpha_2\) and \(\alpha_2 -
-\alpha_3\) in the Cartesian plane. Since \(\alpha_1 - \alpha_2\) and \(\alpha_2
-- \alpha_3\) span \(\mathfrak{h}^*\), this implies the diagrams we've been
-drawing are given by an isometry \(\mathbb{Q} P \isoto \mathbb{Q}^2\), where
-\(\mathbb{Q} P\) is endowed with the bilinear form defined by \((\alpha_i -
-\alpha_j, \alpha_k - \alpha_\ell) \mapsto B(E_{i j}, E_{k \ell})\) -- which we
-denote by \(B\) as well.
+attentive reader may notice that \(\kappa(E_{1 2}, E_{2 3}) = - \sfrac{1}{2}\),
+so that the angle -- with respect to the Killing form \(\kappa\) -- between the
+root vectors \(E_{1 2}\) and \(E_{2 3}\) is precisely the same as the angle
+between the points representing their roots \(\alpha_1 - \alpha_2\) and
+\(\alpha_2 - \alpha_3\) in the Cartesian plane. Since \(\alpha_1 - \alpha_2\)
+and \(\alpha_2 - \alpha_3\) span \(\mathfrak{h}^*\), this implies the diagrams
+we've been drawing are given by an isometry \(\mathbb{Q} P \isoto
+\mathbb{Q}^2\), where \(\mathbb{Q} P\) is endowed with the bilinear form
+defined by \((\alpha_i - \alpha_j, \alpha_k - \alpha_\ell) \mapsto \kappa(E_{i
+j}, E_{k \ell})\) -- which we denote by \(\kappa\) as well.
 
 To proceed we once more refer to the previously established framework: next we
 saw that the eigenvalues of \(h\) form an unbroken string of integers symmetric
@@ -809,7 +809,7 @@ eigenvalue \(\lambda([E_{1 2}, E_{2 1}]) - 2k\) of the action of \(h\) on
 \(M_{\lambda - k (\alpha_1 - \alpha_2)}\), the weights of \(M\) appearing the
 string \(\lambda, \lambda + (\alpha_1 - \alpha_2), \ldots, \lambda + k
 (\alpha_1 - \alpha_2), \ldots\) must be symmetric with respect to the line
-\(B(\alpha_1 - \alpha_2, \alpha) =  0\). The picture is thus
+\(\kappa(\alpha_1 - \alpha_2, \alpha) =  0\). The picture is thus
 \begin{center}
   \begin{tikzpicture}
     \AutoSizeWeightLatticefalse
@@ -822,15 +822,15 @@ string \(\lambda, \lambda + (\alpha_1 - \alpha_2), \ldots, \lambda + k
       \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}}
       \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)};
       \draw[very thick] \weight{0}{-4} -- \weight{0}{4}
-      node[above]{\small\(B(\alpha_1 - \alpha_2, \alpha) = 0\)};
+      node[above]{\small\(\kappa(\alpha_1 - \alpha_2, \alpha) = 0\)};
     \end{rootSystem}
   \end{tikzpicture}
 \end{center}
 
 We could apply this same argument to the subspace \(\bigoplus_k M_{\lambda - k
 (\alpha_2 - \alpha_3)}\), so that the weights in this subspace must be
-symmetric with respect to the line \(B(\alpha_2 - \alpha_3, \alpha) = 0\). The
-picture is now
+symmetric with respect to the line \(\kappa(\alpha_2 - \alpha_3, \alpha) = 0\).
+The picture is now
 \begin{center}
   \begin{tikzpicture}
     \AutoSizeWeightLatticefalse
@@ -845,9 +845,9 @@ picture is now
       \foreach \i in {1,...,4}{\wt[black]{5-2*\i}{\i}}
       \node[above right=-2pt] at (hex cs:x=3,y=1){\small\(\lambda\)};
       \draw[very thick] \weight{0}{-4} -- \weight{0}{4}
-      node[above]{\small\(B(\alpha_1 - \alpha_2, \alpha) = 0\)};
+      node[above]{\small\(\kappa(\alpha_1 - \alpha_2, \alpha) = 0\)};
       \draw[very thick] \weight{-4}{0} -- \weight{4}{0}
-      node[right]{\small\(B(\alpha_2 - \alpha_3, \alpha) = 0\)};
+      node[right]{\small\(\kappa(\alpha_2 - \alpha_3, \alpha) = 0\)};
     \end{rootSystem}
   \end{tikzpicture}
 \end{center}
@@ -970,7 +970,7 @@ This final picture is known as \emph{the weight diagram of \(M\)}. Finally\dots
   The weights of \(M\) are precisely the elements of the weight lattice \(P\)
   congruent to \(\lambda\) module the sublattice \(Q\) and lying inside hexagon
   with vertices the images of \(\lambda\) under the group generated by
-  reflections across the lines \(B(\alpha_i - \alpha_j, \alpha) = 0\).
+  reflections across the lines \(\kappa(\alpha_i - \alpha_j, \alpha) = 0\).
 \end{theorem}
 
 Having found all of the weights of \(M\), the only thing we are missing is an