lie-algebras-and-their-representations

diff --git a/preamble.tex b/preamble.tex
@@ -167,5 +167,8 @@
 % the appropriate spacing
 \renewcommand{\chi}{\ensuremath \raisebox{\depth}{$\mathchar"11F$}}
 
+% Isomorphism arrow
+\newcommand{\isoto}{\xlongrightarrow{\sim}}
+
 % Quotient object
 \newcommand{\mfrac}[2]{\mathlarger{\sfrac{#1}{#2}}}

diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -770,7 +770,7 @@ irreducible representation of \(\mathfrak{sl}_2(K)\) in terms of this vector. We
 reproduce these steps in the context of \(\mathfrak{sl}_3(K)\) by fixing a direction in
 the place an considering the weight lying the furthest in that direction.
 
-% TODO: This doesn't make any sence in field other than C
+% TODO: This doesn't make any sence in a field other than C
 In practice this means we'll choose a linear functional \(f : \mathfrak{h}^*
 \to \RR\) and pick the weight that maximizes \(f\). To avoid any ambiguity we
 should choose the direction of a line irrational with respect to the root
@@ -918,7 +918,7 @@ a contradiction.
 
 There are a number of important consequences to this, of the first being that
 the weights of \(V\) appearing on \(W\) must be symmetric with respect to the
-the line \(\langle \alpha_1 - \alpha_2, \alpha \rangle = 0\). The picture is
+the line \(B(\alpha_1 - \alpha_2, \alpha) =  0\). The picture is
 thus
 \begin{center}
   \begin{tikzpicture}
@@ -932,7 +932,7 @@ thus
       \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\(\langle \alpha_1 - \alpha_2, \alpha \rangle=0\)};
+      node[above]{\small\(B(\alpha_1 - \alpha_2, \alpha) = 0\)};
     \end{rootSystem}
   \end{tikzpicture}
 \end{center}
@@ -940,9 +940,9 @@ thus
 Notice we could apply this same argument to the subspace \(\bigoplus_k
 V_{\lambda + k (\alpha_3 - \alpha_2)}\): this subspace is invariant under the
 action of the subalgebra spanned by \(E_{2 3}\), \(E_{3 2}\) and \([E_{2 3},
-E_{3 2}]\), which is again isomorphic to \(\mathfrak{sl}_2(K)\), so that the weights in
-this subspace must be symmetric with respect to the line \(\langle \alpha_3 -
-\alpha_2, \alpha \rangle = 0\). The picture is now
+E_{3 2}]\), which is again isomorphic to \(\mathfrak{sl}_2(K)\), so that the
+weights in this subspace must be symmetric with respect to the line
+\(B(\alpha_3 - \alpha_2, \alpha) = 0\). The picture is now
 \begin{center}
   \begin{tikzpicture}
     \AutoSizeWeightLatticefalse
@@ -957,9 +957,9 @@ this subspace must be symmetric with respect to the line \(\langle \alpha_3 -
       \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\(\langle \alpha_1 - \alpha_2, \alpha \rangle=0\)};
+      node[above]{\small\(B(\alpha_1 - \alpha_2, \alpha) = 0\)};
       \draw[very thick] \weight{-4}{0} -- \weight{4}{0}
-      node[right]{\small\(\langle \alpha_3 - \alpha_2, \alpha \rangle=0\)};
+      node[right]{\small\(B(\alpha_3 - \alpha_2, \alpha) = 0\)};
     \end{rootSystem}
   \end{tikzpicture}
 \end{center}
@@ -1125,8 +1125,7 @@ Finally\dots
   The weights of \(V\) 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 \(\langle \alpha_i - \alpha_j, \alpha \rangle =
-  0\).
+  reflections across the lines \(B(\alpha_i - \alpha_j, \alpha) = 0\).
 \end{theorem}
 
 Once more there's a clear parallel between the case of \(\mathfrak{sl}_3(K)\) and that
@@ -1456,16 +1455,20 @@ We begin our analysis by remarking that in both \(\mathfrak{sl}_2(K)\) and
 of the following theorem.
 
 % TODO: Add a refenrence to a proof (probably Humphreys)
-% TODO: Changed the notation for the Killing form so that there is no conflict
-% with the notation for the base field
 % TODO: Clarify the meaning of "non-degenerate"
 % TODO: Move this to before the analysis of sl3
 \begin{theorem}
-  If \(\mathfrak g\) is semisimple then its Killing form \(K\) is
-  non-degenerate. Furthermore, the restriction of \(K\) to \(\mathfrak{h}\) is
+  If \(\mathfrak g\) is semisimple then its Killing form \(B\) is
+  non-degenerate. Furthermore, the restriction of \(B\) to \(\mathfrak{h}\) is
   non-degenerate.
 \end{theorem}
 
+\begin{note}
+  Since \(B\) is induces an isomorphism \(\mathfrak{h} \isoto \mathfrak{h}^*\),
+  it induces a bilinear form \((B(X, \cdot), B(Y, \cdot)) \mapsto B(X, Y)\) in
+  \(\mathfrak{h}^*\). We denote this form by \(B\).
+\end{note}
+
 \begin{proposition}\label{thm:weights-symmetric-span}
   The eigenvalues \(\alpha\) of the adjoint action of \(\mathfrak{h}\) in
   \(\mathfrak{g}\) are symmetrical about the origin -- i.e. \(- \alpha\) is
@@ -1485,29 +1488,32 @@ of the following theorem.
   \]
   for all \(H \in \mathfrak{h}\).
 
-  This implies that if \(\alpha + \beta \ne 0\) then \(\operatorname{ad}(X) \operatorname{ad}(Y)\) is
-  nilpotent: if \(Z \in \mathfrak{g}_\gamma\) then
+  This implies that if \(\alpha + \beta \ne 0\) then \(\operatorname{ad}(X)
+  \operatorname{ad}(Y)\) is nilpotent: if \(Z \in \mathfrak{g}_\gamma\) then
   \[
     (\operatorname{ad}(X) \operatorname{ad}(Y))^n Z
     = [X, [Y, [ \ldots, [X, [Y, Z]]] \ldots ]
     \in \mathfrak{g}_{n \alpha + n \beta + \gamma}
     = 0
   \]
-  for \(n\) large enough. In particular, \(K(X, Y) = \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\).
-  Now if \(- \alpha\) is not an eigenvalue we find \(K(X, \mathfrak{g}_\beta) =
-  0\) for all eigenvalues \(\beta\), which contradicts the non-degeneracy of
-  \(K\). Hence \(- \alpha\) must be an eigenvalue of the adjoint action of
+  for \(n\) large enough. In particular, \(B(X, Y) =
+  \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y)) = 0\). Now if
+  \(- \alpha\) is not an eigenvalue we find \(B(X, \mathfrak{g}_\beta) = 0\)
+  for all eigenvalues \(\beta\), which contradicts the non-degeneracy of \(B\).
+  Hence \(- \alpha\) must be an eigenvalue of the adjoint action of
   \(\mathfrak{h}\).
 
   For the second statement, note that if the eigenvalues of \(\mathfrak{h}\) do
   not span all of \(\mathfrak{h}^*\) then there is some \(H \in \mathfrak{h}\)
   non-zero such that \(\alpha(H) = 0\) for all eigenvalues \(\alpha\), which is
-  to say, \(\operatorname{ad}(H) X = [H, X] = 0\) for all \(X \in \mathfrak{g}\). Another way
-  of putting it is to say \(H\) is an element of the center \(\mathfrak{z}\) of
-  \(\mathfrak{g}\), which is zero by the semisimplicity -- a contradiction.
+  to say, \(\operatorname{ad}(H) X = [H, X] = 0\) for all \(X \in
+  \mathfrak{g}\). Another way of putting it is to say \(H\) is an element of
+  the center \(\mathfrak{z}\) of \(\mathfrak{g}\), which is zero by the
+  semisimplicity -- a contradiction.
 \end{proof}
 
-Furthermore, as in the case of \(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\) one can show\dots
+Furthermore, as in the case of \(\mathfrak{sl}_2(K)\) and
+\(\mathfrak{sl}_3(K)\) one can show\dots
 
 \begin{proposition}\label{thm:root-space-dim-1}
   The eigenspaces \(\mathfrak{g}_\alpha\) are all 1-dimensional.
@@ -1536,11 +1542,11 @@ then\dots
 \end{theorem}
 
 % TODO: Rewrite this: the concept of direct has no sence in the general setting
-To proceed further, as in the case of \(\mathfrak{sl}_3(K)\) we have to fix a direction
-in \(\mathfrak{h}^*\) -- i.e. we fix a linear function \(\mathfrak{h}^* \to
-\RR\) such that \(Q\) lies outside of its kernel. This choice induces a
-partition \(\Delta = \Delta^+ \cup \Delta^-\) of the set of roots of
-\(\mathfrak{g}\) and once more we find\dots
+To proceed further, as in the case of \(\mathfrak{sl}_3(K)\) we have to fix a
+direction in \(\mathfrak{h}^*\) -- i.e. we fix a linear function
+\(\mathfrak{h}^* \to \RR\) such that \(Q\) lies outside of its kernel. This
+choice induces a partition \(\Delta = \Delta^+ \cup \Delta^-\) of the set of
+roots of \(\mathfrak{g}\) and once more we find\dots
 
 \begin{theorem}
   There is a weight vector \(v \in V\) that is killed by all positive root
@@ -1630,12 +1636,10 @@ is\dots
   \(\mathfrak{g}\)}.
 \end{definition}
 
-% TODO: Note that this is the line orthogonal to alpha_i - alpha_j with respect
-% to the Killing form
-This is entirely analogous to the situation of \(\mathfrak{sl}_3(K)\), where we found
-that the weights of the irreducible representations were symmetric with respect
-to the lines \(\langle \alpha_i - \alpha_j, \alpha \rangle = 0\). Indeed, the
-same argument leads us to the conclusion\dots
+This is entirely analogous to the situation of \(\mathfrak{sl}_3(K)\), where we
+found that the weights of the irreducible representations were symmetric with
+respect to the lines \(K \alpha\) with \(B(\alpha_i - \alpha_j, \alpha) = 0\).
+Indeed, the same argument leads us to the conclusion\dots
 
 \begin{theorem}\label{thm:irr-weight-class}
   The weights of an irreducible representation \(V\) of \(\mathfrak{g}\) with