lie-algebras-and-their-representations

diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -12,11 +12,11 @@ This was simple enough to do in the case of \(\mathfrak{sl}_2(K)\), but the
 reasoning behind it, as well as the mere fact equation (\ref{sym-diag}) holds,
 are harder to explain in the case of \(\mathfrak{sl}_3(K)\). The eigenspace
 decomposition associated with an operator \(V \to V\) is a very well-known
-tool, and this type of argument should be familiar to anyone familiar with
-basic concepts of linear algebra. On the other hand, the eigenspace
-decomposition of \(V\) with respect to the action of an arbitrary subalgebra
-\(\mathfrak{h} \subset \mathfrak{gl}(V)\) is neither well-known nor does it
-hold in general: as previously stated, it may very well be that
+tool, and readers familizared with basic concepts of linear algebra should be
+used to this type of argument. On the other hand, the eigenspace decomposition
+of \(V\) with respect to the action of an arbitrary subalgebra \(\mathfrak{h}
+\subset \mathfrak{gl}(V)\) is neither well-known nor does it hold in general:
+as previously stated, it may very well be that
 \[
   \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda \subsetneq V
 \]
@@ -35,13 +35,13 @@ question then is: why did we choose \(\mathfrak{h}\) with \(\dim \mathfrak{h} >
 1\) for \(\mathfrak{sl}_3(K)\)?
 
 The rational behind fixing an Abelian subalgebra is a simple one: we have seen
-in the previous chapter that representations of Abelian
-algebras are generally much simpler to understand than the general case.
-Thus it make sense to decompose a given representation \(V\) of
-\(\mathfrak{g}\) into subspaces invariant under the action of \(\mathfrak{h}\),
-and then analyze how the remaining elements of \(\mathfrak{g}\) act on this
-subspaces. The bigger \(\mathfrak{h}\) the simpler our problem gets, because
-there are fewer elements outside of \(\mathfrak{h}\) left to analyze.
+in the previous chapter that representations of Abelian algebras are generally
+much simpler to understand than the general case. Thus it make sense to
+decompose a given representation \(V\) of \(\mathfrak{g}\) into subspaces
+invariant under the action of \(\mathfrak{h}\), and then analyze how the
+remaining elements of \(\mathfrak{g}\) act on this subspaces. The bigger
+\(\mathfrak{h}\) is, the simpler our problem gets, because there are fewer
+elements outside of \(\mathfrak{h}\) left to analyze.
 
 Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h}
 \subset \mathfrak{g}\), which leads us to the following definition.
@@ -64,9 +64,9 @@ Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h}
 \begin{proof}
   Notice that \(0 \subset \mathfrak{g}\) is an Abelian subalgebra whose
   elements act as diagonal operators via the adjoint representation. Indeed,
-  \(0\) -- the only element of \(0 \subset \mathfrak{g}\) -- is such that
-  \(\operatorname{ad}(0) = 0\). Furthermore, given a chain of Abelian
-  subalgebras
+  \(0\), the only element of \(0 \subset \mathfrak{g}\), is such that
+  \(\operatorname{ad}(0) = 0\) is a diagonalizable operator. Furthermore, given
+  a chain of Abelian subalgebras
   \[
     0 \subset \mathfrak{h}_1 \subset \mathfrak{h}_2 \subset \cdots
   \]
@@ -75,7 +75,7 @@ Hence we are generally interested in maximal Abelian subalgebras \(\mathfrak{h}
   \mathfrak{g}\) is Abelian, and its elements also act diagonally in
   \(\mathfrak{g}\). It then follows from Zorn's lemma that there exists a
   subalgebra \(\mathfrak{h}\) which is maximal with respect to both these
-  properties -- i.e. a Cartan subalgebra.
+  properties, also known as a Cartan subalgebra.
 \end{proof}
 
 We have already seen some concrete examples. For instance, one can readily
@@ -100,9 +100,9 @@ self-normalizing. Hence \(\mathfrak{h}\) is a Cartan subalgebra of
 The intersection of such subalgebra with \(\mathfrak{sl}_n(K)\) -- i.e. the
 subalgebra of traceless diagonal matrices -- is a Cartan subalgebra of
 \(\mathfrak{sl}_n(K)\). In particular, if \(n = 2\) or \(n = 3\) we get to the
-subalgebras described the previous two sections. The remaining question then
-is: if \(\mathfrak{h} \subset \mathfrak{g}\) is a Cartan subalgebra and \(V\)
-is a representation of \(\mathfrak{g}\), does the eigenspace decomposition
+subalgebras described the previous chapter. The remaining question then is: if
+\(\mathfrak{h} \subset \mathfrak{g}\) is a Cartan subalgebra and \(V\) is a
+representation of \(\mathfrak{g}\), does the eigenspace decomposition
 \[
   V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda
 \]
@@ -126,9 +126,7 @@ What is simultaneous diagonalization all about then?
 
 We should point out that simultaneous diagonalization \emph{only works in the
 finite-dimensional setting}. In fact, simultaneous diagonalization is usually
-framed as an equivalent statement about diagonalizable \(n \times n\) matrices
--- where \(n\) is, of course, finite.
-
+framed as an equivalent statement about diagonalizable \(n \times n\) matrices.
 Simultaneous diagonalization implies that to show \(V = \bigoplus_\lambda
 V_\lambda\) it suffices to show that \(H\!\restriction_V : V \to V\) is a
 diagonalizable operator for each \(H \in \mathfrak{h}\). To that end, we
@@ -165,12 +163,8 @@ words\dots
 \end{proposition}
 
 This last result is known as \emph{the preservation of the Jordan form}, and a
-proof can be found in appendix C of \cite{fulton-harris}. We should point out
-this fails spectacularly in positive characteristic. Furthermore, the statement
-of proposition~\ref{thm:preservation-jordan-form} only makes sense for
-\emph{semisimple} Lie algebras -- i.e. the algebras \(\mathfrak{g}\) for which
-the abstract Jordan decomposition of \(\mathfrak{g}\) is defined. Nevertheless,
-as promised this implies\dots
+proof can be found in appendix C of \cite{fulton-harris}. As promised this
+implies\dots
 
 \begin{corollary}\label{thm:finite-dim-is-weight-mod}
   Let \(\mathfrak{g}\) be a semisimple Lie algebra, \(\mathfrak{h} \subset
@@ -214,7 +208,7 @@ Cartan subalgebra of \(\mathfrak{g}\) is \(\mathfrak{g}\) itself, and a
 \(\mathfrak{g}\)-module is simply a vector space \(V\) endowed with an operator
 \(V \to V\) -- which corresponds to the action of \(1 \in \mathfrak{g}\) on
 \(V\). In particular, if we choose an operator \(V \to V\) which is \emph{not}
-diagonalizable we find \(V \ne \bigoplus_{\lambda \in \mathfrak{h}^*}
+diagonalizable we find \(V \ne 0 = \bigoplus_{\lambda \in \mathfrak{h}^*}
 V_\lambda\).
 
 However, corollary~\ref{thm:finite-dim-is-weight-mod} does work for reductive
@@ -224,14 +218,16 @@ representations as scalar operators. The hypothesis of finite-dimensionality is
 also of huge importance. In the next chapter we will encounter
 infinite-dimensional \(\mathfrak{g}\)-modules for which the eigenspace
 decomposition \(V = \bigoplus_{\lambda \in \mathfrak{h}^*} V_\lambda\) fails.
-As a first consequence of corollary~\ref{thm:finite-dim-is-weight-mod}
+As a first consequence of corollary~\ref{thm:finite-dim-is-weight-mod} we
+show\dots
 
 \begin{corollary}
-  The restriction of \(B\) to \(\mathfrak{h}\) is non-degenerate.
+  The restriction of the Killing form \(B\) to \(\mathfrak{h}\) is
+  non-degenerate.
 \end{corollary}
 
 \begin{proof}
-  Consider the eigenspace decomposition \(\mathfrak{g} = \mathfrak{g}_0 \oplus
+  Consider the root space decomposition \(\mathfrak{g} = \mathfrak{g}_0 \oplus
   \bigoplus_\alpha \mathfrak{g}_\alpha\) of the adjoint representation, where
   \(\alpha\) ranges over all nonzero eigenvalues of the adjoint action of
   \(\mathfrak{h}\). We claim \(\mathfrak{g}_0 = \mathfrak{h}\).
@@ -270,7 +266,7 @@ 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
-  \(\mathfrak{h}^*\). We denote this form by \(B\).
+  \(\mathfrak{h}^*\). We denote this form by \(B\) as well.
 \end{note}
 
 We now have most of the necessary tools to reproduce the results of the
@@ -278,8 +274,8 @@ previous chapter in a general setting. Let \(\mathfrak{g}\) be a
 finite-dimensional semisimple algebra with a Cartan subalgebra \(\mathfrak{h}\)
 and let \(V\) be a finite-dimensional irreducible representation of
 \(\mathfrak{g}\). We will proceed, as we did before, by generalizing the
-results about of the previous two sections in order. By now the pattern should
-be starting become clear, so we will mostly omit technical details and proofs
+results of the previous two sections in order. By now the pattern should be
+starting to become clear, so we will mostly omit technical details and proofs
 analogous to the ones on the previous sections. Further details can be found in
 appendix D of \cite{fulton-harris} and in \cite{humphreys}.
 
@@ -290,24 +286,23 @@ We begin our analysis, as we did for \(\mathfrak{sl}_2(K)\) and
 \(\mathfrak{g}\). Throughout chapter~\ref{ch:sl3} we've seen that the weights
 of any given finite-dimensional representation of \(\mathfrak{sl}_2(K)\) or
 \(\mathfrak{sl}_3(K)\) can only assume very rigid configurations. For instance,
-we've seen that the weights of any given representation are symmetric with
-respect to the origin. In this chapter we will generalize most results from
-chapter~\ref{ch:sl3} the rigidity of the geometry of the set of weights of a
-given representations.
+we've seen that the roots of \(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\)
+are symmetric with respect to the origin. In this chapter we will generalize
+most results from chapter~\ref{ch:sl3} regarding the rigidity of the geometry
+of the set of weights of a given representations.
 
 As for the afford mentioned result on the symmetry of roots, this turns out to
 be a general fact, which is a consequence of the non-degeneracy of the
 restriction of the Killing form to the Cartan subalgebra.
 
 \begin{proposition}\label{thm:weights-symmetric-span}
-  The eigenvalues \(\alpha\) of the adjoint action of \(\mathfrak{h}\) on
-  \(\mathfrak{g}\) are symmetrical about the origin -- i.e. \(- \alpha\) is
-  also an eigenvalue -- and they span all of \(\mathfrak{h}^*\).
+  The roots \(\alpha\) of \(\mathfrak{g}\) are symmetrical about the origin --
+  i.e. \(- \alpha\) is also a root -- and they span all of \(\mathfrak{h}^*\).
 \end{proposition}
 
 \begin{proof}
   We'll start with the first claim. Let \(\alpha\) and \(\beta\) be two
-  eigenvalues of the adjoint action of \(\mathfrak{h}\). Notice
+  roots. Notice
   \([\mathfrak{g}_\alpha, \mathfrak{g}_\beta] \subset \mathfrak{g}_{\alpha +
   \beta}\). Indeed, if \(X \in \mathfrak{g}_\alpha\) and \(Y \in
   \mathfrak{g}_\beta\) then
@@ -329,17 +324,16 @@ restriction of the Killing form to the Cartan subalgebra.
   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 \(B\).
+  for all roots \(\beta\), 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 \(H \in \mathfrak{h}\)
-  nonzero such that \(\alpha(H) = 0\) for all eigenvalues \(\alpha\), which is
+  For the second statement, note that if the roots of \(\mathfrak{g}\) do not
+  span all of \(\mathfrak{h}^*\) then there is some nonzero \(H \in
+  \mathfrak{h}\) such that \(\alpha(H) = 0\) for all roots \(\alpha\), 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.
+  the center \(\mathfrak{z} = 0\) of \(\mathfrak{g}\), a contradiction.
 \end{proof}
 
 Furthermore, as in the case of \(\mathfrak{sl}_2(K)\) and
@@ -351,26 +345,24 @@ Furthermore, as in the case of \(\mathfrak{sl}_2(K)\) and
 
 The proof of the first statement of
 proposition~\ref{thm:weights-symmetric-span} highlights something interesting:
-if we fix some some eigenvalue \(\alpha\) of the adjoint action of
-\(\mathfrak{h}\) on \(\mathfrak{g}\) and a eigenvector \(X \in
-\mathfrak{g}_\alpha\), then for each \(H \in \mathfrak{h}\) and \(v \in
-V_\lambda\) we find
+if we fix some eigenvalue \(\alpha\) of the adjoint action of \(\mathfrak{h}\)
+on \(\mathfrak{g}\) and a eigenvector \(X \in \mathfrak{g}_\alpha\), then for
+each \(H \in \mathfrak{h}\) and \(v \in V_\lambda\) we find
 \[
   H (X v)
   = X (H v) + [H, X] v
   = (\lambda + \alpha)(H) \cdot X v
 \]
 so that \(X\) carries \(v\) to \(V_{\lambda + \alpha}\). We have encountered
-this formula twice in this chapter: again, we find \(\mathfrak{g}_\alpha\)
-\emph{acts on \(V\) by translating vectors between eigenspaces}. In other
-words, if we denote by \(\Delta\) the set of all roots of \(\mathfrak{g}\)
-then\dots
+this formula twice in these notes: again, we find \(\mathfrak{g}_\alpha\)
+\emph{acts on \(V\) by translating vectors between eigenspaces}. In particular,
+if we denote by \(\Delta\) the set of all roots of \(\mathfrak{g}\) then\dots
 
 \begin{theorem}\label{thm:weights-congruent-mod-root}
   The weights of an irreducible representation \(V\) of \(\mathfrak{g}\) are
   all congruent module the root lattice \(Q = \ZZ \Delta\) of \(\mathfrak{g}\).
-  In other words, all weights of \(V\) lie in the same \(Q\)-coset in
-  \(\mfrac{\mathfrak{h}^*}{Q}\).
+  In other words, all weights of \(V\) lie in the same \(Q\)-coset
+  \(t \in \mfrac{\mathfrak{h}^*}{Q}\).
 \end{theorem}
 
 Again, we may leverage our knowledge of \(\mathfrak{sl}_2(K)\) to obtain further
@@ -418,7 +410,8 @@ restrictions on the weights of \(V\). Namely, if \(\lambda\) is a weight,
 \begin{definition}\label{def:weight-lattice}
   The lattice \(P = \{ \lambda \in \mathfrak{h}^* : \lambda(H_\alpha) \in
   \mathbb{Z} \, \forall \alpha \in \Delta \} \subset \mathfrak{h}^*\) is called
-  \emph{the weight lattice of \(\mathfrak{g}\)}.
+  \emph{the weight lattice of \(\mathfrak{g}\)}. We call the elements of \(P\)
+  \emph{integral}.
 \end{definition}
 
 \begin{proposition}\label{thm:weights-fit-in-weight-lattice}
@@ -426,20 +419,17 @@ restrictions on the weights of \(V\). Namely, if \(\lambda\) is a weight,
   in the weight lattice \(P\).
 \end{proposition}
 
-We call the elements of \(P\) \emph{integral}.
 Proposition~\ref{thm:weights-fit-in-weight-lattice} is clearly analogous to
 corollary~\ref{thm:sl3-weights-fit-in-weight-lattice}. In fact, the weight
 lattice of \(\mathfrak{sl}_3(K)\) -- as in definition~\ref{def:weight-lattice}
 -- is precisely \(\mathbb{Z} \alpha_1 \oplus \mathbb{Z} \alpha_2 \oplus
-\mathbb{Z} \alpha_3\).
-
-To proceed further, we would like to take \emph{the highest weight of \(V\)} as
-in section~\ref{sec:sl3-reps}, but the meaning of \emph{highest} is again
-unclear in this situation. We could simply fix a linear function \(\mathbb{Q} P
-\to \mathbb{Q}\) -- as we did in section~\ref{sec:sl3-reps} -- and choose a
-weight \(\lambda\) of \(V\) that maximizes this functional, but at this point
-it is convenient to introduce some additional tools to our arsenal. This tools
-are called \emph{basis}.
+\mathbb{Z} \alpha_3\). To proceed further, we would like to take \emph{the
+highest weight of \(V\)} as in section~\ref{sec:sl3-reps}, but the meaning of
+\emph{highest} is again unclear in this situation. We could simply fix a linear
+function \(\mathbb{Q} P \to \mathbb{Q}\) -- as we did in
+section~\ref{sec:sl3-reps} -- and choose a weight \(\lambda\) of \(V\) that
+maximizes this functional, but at this point it is convenient to introduce some
+additional tools to our arsenal. This tools are called \emph{basis}.
 
 \begin{definition}\label{def:basis-of-root}
   A subset \(\Sigma = \{\beta_1, \ldots, \beta_k\} \subset \Delta\) of linearly
@@ -450,9 +440,9 @@ are called \emph{basis}.
 
 The interesting thing about basis for \(\Delta\) is that they allow us to
 compare weights of a given representation. At this point the reader should be
-asking himself: how? Definition~\ref{def:basis-of-root} doesn't exactly scream
-``comparison''. Well, the thing is that any choice of basis induces a partial
-order in \(Q\), where elements are ordered by their \emph{heights}.
+asking himself: how? Definition~\ref{def:basis-of-root} isn't exactly all that
+intuitive. Well, the thing is that any choice of basis induces a partial order
+in \(Q\), where elements are ordered by their \emph{heights}.
 
 \begin{definition}
   Let \(\Sigma = \{\beta_1, \ldots, \beta_k\}\) be a basis for \(\Delta\).
@@ -484,8 +474,8 @@ order in \(Q\), where elements are ordered by their \emph{heights}.
 It should be obvious that the binary relation \(\preceq\) in \(Q\) is a partial
 order. In addition, we may compare the elements of a given \(Q\)-coset
 \(\lambda + Q\) by comparing their difference with \(0 \in Q\). In other words,
-we say \(\lambda \preceq \mu\) if \(\lambda - \mu \preceq 0\) for \(\lambda \in
-\mu + Q\). In particular, since the weights of \(V\) all lie in a single
+given \(\lambda \in \mu + Q\), we say \(\lambda \preceq \mu\) if \(\lambda -
+\mu \preceq 0\). In particular, since the weights of \(V\) all lie in a single
 \(Q\)-coset, we may compare them in this fashion. Given a basis \(\Sigma\) for
 \(\Delta\) we may take ``the highest weight of \(V\)'' as a maximal weight
 \(\lambda\) of \(V\). The obvious question then is: can we always find a basis
@@ -499,14 +489,15 @@ 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 \(B(\alpha, \beta) \in
 \mathbb{Z}\) for all \(\alpha, \beta \in \Delta\), so that the Killing form
-\(B\) restricts to a nondegenerate 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.
+\(B\) restricts to a nondegenerate \(\mathbb{Q}\)-linear 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
@@ -558,8 +549,8 @@ This has a number of important consequences. For instance\dots
 \end{proof}
 
 This is entirely analogous to the situation of \(\mathfrak{sl}_3(K)\), where we
-found that the weights of the irreducible representations formed a continuous
-string symmetric with respect to the lines \(K \alpha\) with \(B(\alpha_i -
+found that the weights of the irreducible representations formed continuous
+strings symmetric with respect to the lines \(K \alpha\) with \(B(\alpha_i -
 \alpha_j, \alpha) = 0\). As in the case of \(\mathfrak{sl}_3(K)\), the same
 class of arguments leads us to the conclusion\dots
 
@@ -577,19 +568,19 @@ class of arguments leads us to the conclusion\dots
   \(\mathcal{W}\).
 \end{theorem}
 
-Aside from showing up the previous theorem, the Weyl group will also play an
+Aside from showing up in the previous theorem, the Weyl group will also play an
 important role in chapter~\ref{ch:mathieu} by virtue of the existence of a
-canonical action of \(\mathcal{W}\) on \(\mathfrak{h}\). By its very nature
+canonical action of \(\mathcal{W}\) on \(\mathfrak{h}\). By its very nature,
 \(\mathcal{W}\) acts in \(\mathfrak{h}^*\). If we conjugate the action
 \(\sigma\!\restriction_{\mathfrak{h}^*} : \mathfrak{h}^* \isoto
 \mathfrak{h}^*\) of some \(\sigma \in \mathcal{W}\) by the isomorphism
 \(\mathfrak{h}^* \isoto \mathfrak{h}\) afforded by the restriction of the
-Killing for to \(\mathfrak{h}\) we get a linear automorphism \(\mathfrak{h}
-\isoto \mathfrak{h}\). As it turns out, the automorphism
-\(\sigma\!\restriction_{\mathfrak{h}} : \mathfrak{h} \isoto \mathfrak{h}\) can
-be extended to an automorphism of Lie algebras \(\mathfrak{g} \isoto
-\mathfrak{g}\). This translates into the following results, which we do not
-prove -- but see \cite[sec.~14.3]{humphreys}.
+Killing for to \(\mathfrak{h}\) we get a linear automorphism
+\(\sigma\!\restriction_{\mathfrak{h}} : \mathfrak{h} \isoto \mathfrak{h}\). As
+it turns out, the \(\sigma\!\restriction_{\mathfrak{h}}\) can be extended to an
+automorphism of Lie algebras \(\mathfrak{g} \isoto \mathfrak{g}\). This
+translates into the following results, which we do not prove -- but see
+\cite[sec.~14.3]{humphreys}.
 
 \begin{proposition}\label{thm:weyl-group-action}
   Given \(\alpha \in \Delta^+\), let\footnote{Notice that since $\mathfrak{g}$
@@ -741,7 +732,7 @@ Moreover, we find\dots
   \[
     M(\lambda)
     \subset
-    \bigoplus_{\substack{k_i \in \ZZ \\ k_i \ge 0}}
+    \bigoplus_{k_i \in \mathbb{N}}
     M(\lambda)_{\lambda + k_1 \cdot \alpha_1 + \cdots + k_n \cdot \alpha_n}
   \]
   where \(\{\alpha_1, \ldots, \alpha_m\} = \Delta^-\), so that all weights of
@@ -786,9 +777,9 @@ Moreover, we find\dots
   \bigoplus_{k \ge 0} K f^k v^+\), and the action of \(\mathfrak{sl}_2(K)\) on
   \(M(\lambda)\) is given by
   \begin{align*}
-    f^{k + 1} v^+ & \overset{e}{\mapsto} (2 - k (k - 1)) f^k v^+ &
-    f^{k + 1} v^+ & \overset{f}{\mapsto} f^{k + 2} v^+ &
-    f^{k + 1} v^+ & \overset{h}{\mapsto} - 2 k f^{k + 1} v^+ &
+    f^k v^+ & \overset{e}{\mapsto} (2 - k (k + 1)) f^{k - 1} v^+ &
+    f^k v^+ & \overset{f}{\mapsto} f^{k + 1} v^+                 &
+    f^k v^+ & \overset{h}{\mapsto} - 2 (k - 1) f^k v^+           &
   \end{align*}
 
   In the language of the diagrams used in chapter~\ref{ch:sl3}, we write
@@ -819,7 +810,7 @@ whose highest weight is \(\lambda\).
   Every subrepresentation \(V \subset M(\lambda)\) is the direct sum of its
   weight spaces. In particular, \(M(\lambda)\) has a unique maximal
   subrepresentation \(N(\lambda)\) and a unique irreducible quotient
-  \(\sfrac{M(\lambda)}{N(\lambda)}\).
+  \(L(\lambda) = \sfrac{M(\lambda)}{N(\lambda)}\).
 \end{proposition}
 
 \begin{proof}
@@ -856,11 +847,11 @@ whose highest weight is \(\lambda\).
     \in V
   \end{multline*}
 
-  By applying the same procedure over and over again we can see that \(v_1 = X
-  v \in V\) for some \(X \in \mathcal{U}(\mathfrak{g})\). Furthermore, if we
+  By applying the same procedure over and over again we can see that \(v_1 = u
+  v \in V\) for some \(u \in \mathcal{U}(\mathfrak{g})\). Furthermore, if we
   reproduce all this for \(v_2 + \cdots + v_n = v - v_1 \in V\) we get that
-  \(v_2 \in V\). Now by applying the same procedure over and over we find
-  \(v_1, \ldots, v_n \in V\). Hence
+  \(v_2 \in V\). Now by interating this procedure we find \(v_1, \ldots, v_n
+  \in V\). Hence
   \[
     V = \bigoplus_\mu V_\mu = \bigoplus_\mu M(\lambda)_\mu \cap V
   \]
@@ -868,57 +859,58 @@ whose highest weight is \(\lambda\).
   Since \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+\), \(V\) is a proper
   subrepresentation then \(v^+ \notin V\). Hence any proper submodule lies in
   the sum of weight spaces other than \(M(\lambda)_\lambda\), so the sum
-  \(N(\lambda)\) of all such submodules is still proper. In fact, this implies
+  \(N(\lambda)\) of all such submodules is still proper. This implies
   \(N(\lambda)\) is the unique maximal subrepresentation of \(M(\lambda)\) and
-  \(\sfrac{M(\lambda)}{N(\lambda)}\) is its unique irreducible quotient.
+  \(L(\lambda) = \sfrac{M(\lambda)}{N(\lambda)}\) is its unique irreducible
+  quotient.
 \end{proof}
 
 \begin{example}\label{ex:sl2-verma-quotient}
   If \(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto 2\), we
   can see from example~\ref{ex:sl2-verma} that \(N(\lambda) = \bigoplus_{k \ge
-  3} K f^k v^+\), so that \(\sfrac{M(\lambda)}{N(\lambda)}\) is the
-  \(3\)-dimensional irreducible representation of \(\mathfrak{sl}_2(K)\) --
-  i.e. the finite-dimensional irreducible representation with highest weight
-  \(\lambda\) constructed in chapter~\ref{ch:sl3}.
+  3} K f^k v^+\), so that \(L(\lambda)\) is the \(3\)-dimensional irreducible
+  representation of \(\mathfrak{sl}_2(K)\) -- i.e. the finite-dimensional
+  irreducible representation with highest weight \(\lambda\) constructed in
+  chapter~\ref{ch:sl3}.
 \end{example}
 
 This last example is particularly interesting to us, since it indicates that
 the finite-dimensional irreducible representations of \(\mathfrak{sl}_2(K)\) as
 quotients of Verma modules. This is because the quotient
 \(\sfrac{M(\lambda)}{N(\lambda)}\) in example~\ref{ex:sl2-verma-quotient}
-happened to be finite-dimensional. As it turns out, this is always the case for
-semisimple \(\mathfrak{g}\). Namely\dots
+is finite-dimensional. As it turns out, this is not a coincidence.
 
 \begin{proposition}\label{thm:verma-is-finite-dim}
-  If \(\lambda\) is dominant integral then the unique irreducible quotient of
-  \(M(\lambda)\) is finite-dimensional.
+  If \(\mathfrak{g}\) is semisimple and \(\lambda\) is dominant integral then
+  the unique irreducible quotient of \(M(\lambda)\) is finite-dimensional.
 \end{proposition}
 
 The proof of proposition~\ref{thm:verma-is-finite-dim} is very technical and we
 won't include it here, but the idea behind it is to show that the set of
-weights of \(\sfrac{M(\lambda)}{N(\lambda)}\) is stable under the natural
-action of the Weyl group \(\mathcal{W}\) on \(\mathfrak{h}^*\). One can then
-show that the every weight of \(V\) is conjugate to a single dominant integral
-weight of \(\sfrac{M(\lambda)}{N(\lambda)}\), and that the set of dominant
-integral weights of such irreducible quotient is finite. Since \(W\) is
-finitely generated, this implies the set of weights of the unique irreducible
-quotient of \(M(\lambda)\) is finite. But each weight space is
-finite-dimensional. Hence so is the irreducible quotient.
+weights of \(L(\lambda)\) is stable under the natural action of the Weyl group
+\(\mathcal{W}\) on \(\mathfrak{h}^*\). One can then show that the every weight
+of \(V\) is conjugate to a single dominant integral weight of
+\(\sfrac{M(\lambda)}{N(\lambda)}\), and that the set of dominant integral
+weights of such irreducible quotient is finite. Since \(W\) is finitely
+generated, this implies the set of weights of the unique irreducible quotient
+of \(M(\lambda)\) is finite. But each weight space is finite-dimensional. Hence
+so is the irreducible quotient.
 
 We refer the reader to \cite[ch. 21]{humphreys} for further details. What we
 are really interested in is\dots
 
 \begin{corollary}
-  There is a finite-dimensional irreducible \(\mathfrak{g}\)-module \(V\) whose
+  Let \(\lambda\) be a dominant integral weight of \(\mathfrak{g}\). Then there
+  is a finite-dimensional irreducible \(\mathfrak{g}\)-module \(V\) whose
   highest weight is \(\lambda\).
 \end{corollary}
 
 \begin{proof}
-  Let \(V = \mfrac{M(\lambda)}{N(\lambda)}\). It suffices to show that its
-  highest weight is \(\lambda\). We have already seen that \(v^+ \in
-  M(\lambda)_\lambda\) is a highest weight vector. Now since \(v\) lies outside
-  of the maximal subrepresentation of \(M(\lambda)\), the projection \(v^+ +
-  N(\lambda) \in V\) is nonzero.
+  Let \(V = L(\lambda)\). It suffices to show that its highest weight is
+  \(\lambda\). We have already seen that \(v^+ \in M(\lambda)_\lambda\) is a
+  highest weight vector. Now since \(v\) lies outside of the maximal
+  subrepresentation of \(M(\lambda)\), the projection \(v^+ + N(\lambda) \in
+  V\) is nonzero.
 
   We now claim that \(v^+ + N(\lambda) \in V_\lambda\). Indeed,
   \[
@@ -929,15 +921,15 @@ are really interested in is\dots
   for all \(H \in \mathfrak{h}\). Hence \(\lambda\) is a weight of \(V\), with
   weight vector \(v^+ + N(\lambda)\). Finally, we remark that \(\lambda\) is
   the highest weight of \(V\), for if this was not the case we could find a
-  weight \(\mu\) of \(M(\lambda)\) which is higher than \(\lambda\).
+  weight \(\mu\) of \(M(\lambda)\) with \(\mu \succ \lambda\).
 \end{proof}
 
 We should point out that proposition~\ref{thm:verma-is-finite-dim} fails for
 non-dominant \(\lambda \in P\). While \(\lambda\) is always a maximal weight of
 \(M(\lambda)\), one can show show that if \(\lambda \in P\) is not dominant
 then \(N(\lambda) = 0\) and \(M(\lambda)\) is irreducible. For instance, if
-\(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda = -2\) then the action of
-\(\mathfrak{g}\) on \(M(\lambda)\) is given by
+\(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto -2\) then the
+action of \(\mathfrak{g}\) on \(M(\lambda)\) is given by
 \begin{center}
   \begin{tikzcd}
     \cdots \arrow[bend left=60]{r}{-20}