lie-algebras-and-their-representations

 \chapter{Finite-Dimensional Simple Modules}
 
 In this chapter we classify the finite-dimensional simple
 \(\mathfrak{g}\)-modules for a finite-dimensional semisimple Lie algebra
 \(\mathfrak{g}\) over \(K\). At the heart of our analysis of
 \(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\) was the decision to consider
 the eigenspace decomposition
 \begin{equation}\label{sym-diag}
   M = \bigoplus_\lambda M_\lambda
 \end{equation}
 
 This was simple enough to do in the case of \(\mathfrak{sl}_2(K)\), but the
 rational behind it and the reason why equation (\ref{sym-diag}) holds are
 harder to explain in the case of \(\mathfrak{sl}_3(K)\). The eigenspace
 decomposition associated with an operator \(M \to M\) is a very well-known
 tool, and readers familiarized with basic concepts of linear algebra should be
 used to this type of argument. On the other hand, the eigenspace decomposition
 of \(M\) with respect to the action of an arbitrary subalgebra \(\mathfrak{h}
 \subset \mathfrak{gl}(M)\) is neither well-known nor does it hold in general:
 as indicated in the previous chapter, it may very well be that
 \[
   \bigoplus_{\lambda \in \mathfrak{h}^*} M_\lambda \subsetneq M
 \]
 
 We should note, however, that these two cases are not as different as they may
 sound at first glance. Specifically, we can regard the eigenspace decomposition
 of a \(\mathfrak{sl}_2(K)\)-module \(M\) with respect to the eigenvalues of the
 action of \(h\) as the eigenvalue decomposition of \(M\) with respect to the
 action of the subalgebra \(\mathfrak{h} = K h \subset \mathfrak{sl}_2(K)\).
 Furthermore, in both cases \(\mathfrak{h} \subset \mathfrak{sl}_n(K)\) is the
 subalgebra of diagonal matrices, which is Abelian. The fundamental difference
 between these two cases is thus the fact that \(\dim \mathfrak{h} = 1\) for
 \(\mathfrak{h} \subset \mathfrak{sl}_2(K)\) while \(\dim \mathfrak{h} > 1\) for
 \(\mathfrak{h} \subset \mathfrak{sl}_3(K)\). The 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 \(\mathfrak{h}\) 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 \(\mathfrak{g}\)-module \(M\) of into
 subspaces invariant under the action of \(\mathfrak{h}\), and then analyze how
 the remaining elements of \(\mathfrak{g}\) act on these 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.
 
 \begin{definition}\index{Lie subalgebra!Cartan subalgebra}
   A subalgebra \(\mathfrak{h} \subset \mathfrak{g}\) is called \emph{a Cartan
   subalgebra of \(\mathfrak{g}\)} if is self-normalizing -- i.e. \([X, H] \in
   \mathfrak{h}\) for all \(H \in \mathfrak{h}\) if, and only if \(X \in
   \mathfrak{h}\) -- and nilpotent. Equivalently for reductive \(\mathfrak{g}\),
   \(\mathfrak{h}\) is called \emph{a Cartan subalgebra of \(\mathfrak{g}\)} if
   it is Abelian, \(\operatorname{ad}(H)\) is diagonalizable for each \(H \in
   \mathfrak{h}\) and if \(\mathfrak{h}\) is maximal with respect to the former
   two properties.
 \end{definition}
 
 \begin{proposition}
   There exists a Cartan subalgebra \(\mathfrak{h} \subset \mathfrak{g}\).
 \end{proposition}
 
 \begin{proof}
   Notice that \(0 \subset \mathfrak{g}\) is an Abelian subalgebra whose
   elements act as diagonal operators via the adjoint \(\mathfrak{g}\)-module.
   Indeed, \(0\), 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
   \[
     0 \subset \mathfrak{h}_1 \subset \mathfrak{h}_2 \subset \cdots
   \]
   such that \(\operatorname{ad}(H)\) is a diagonal operator for each \(H \in
   \mathfrak{h}_i\), the subalgebra \(\bigcup_i \mathfrak{h}_i \subset
   \mathfrak{g}\) 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, also known as a Cartan subalgebra.
 \end{proof}
 
 We have already seen some concrete examples. Namely\dots
 
 \begin{example}\label{ex:cartan-of-gl}
   The Lie subalgebra
   \[
     \mathfrak{h} =
     \begin{pmatrix}
            K &      0 & \cdots &      0 \\
            0 &      K & \cdots &      0 \\
       \vdots & \vdots & \ddots & \vdots \\
            0 &      0 & \cdots &      K
     \end{pmatrix}
     \subset \mathfrak{gl}_n(K)
   \]
   of diagonal matrices is a Cartan subalgebra.
   Indeed, every pair of diagonal matrices commutes, so that \(\mathfrak{h}\)
   is an Abelian -- and hence nilpotent -- subalgebra. A
   simple calculation also shows that if \(i \ne j\) then the coefficient of
   \(E_{i j}\) in \([E_{i i}, X]\) is the same as the coefficient of \(E_{i j}\)
   in \(X\), for all \(X \in \mathfrak{gl}_n(K)\). In particular, if \([E_{i i},
   X]\) is diagonal for all \(i\), then so is \(X\) -- i.e. \(\mathfrak{h}\) is
   self-normalizing.
 \end{example}
 
 \begin{example}\label{ex:cartan-of-sl}
   Let \(\mathfrak{h}\) be as in Example~\ref{ex:cartan-of-gl}. Then the
   subalgebra \(\mathfrak{h} \cap \mathfrak{sl}_n(K)\) of traceless diagonal
   matrices is a Cartan subalgebra of \(\mathfrak{sl}_n(K)\).
 \end{example}
 
 \begin{example}\label{ex:cartan-of-sp}
   It is easy to see from Example~\ref{ex:sp2n} that \(\mathfrak{h} = \{X \in
   \mathfrak{sp}_{2n}(K) : X\ \text{is diagonal} \}\) is a Cartan subalgebra.
 \end{example}
 
 \begin{example}\label{ex:cartan-direct-sum}
   Let \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\) be Lie algebras and
   \(\mathfrak{h}_i \subset \mathfrak{g}_i\) be Cartan subalgebras. Then
   \(\mathfrak{h}_1 \oplus \mathfrak{h}_2\) is a Cartan subalgebra of
   \(\mathfrak{g}_1 \oplus \mathfrak{g}_2\).
 \end{example}
 
 \index{Cartan subalgebra!simultaneous diagonalization}
 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 chapter. The remaining question then is: if
 \(\mathfrak{h} \subset \mathfrak{g}\) is a Cartan subalgebra and \(M\) is a
 \(\mathfrak{g}\)-module, does the eigenspace decomposition
 \[
   M = \bigoplus_\lambda M_\lambda
 \]
 of \(M\) hold? The answer to this question turns out to be yes. This is a
 consequence of something known as \emph{simultaneous diagonalization}, which is
 the primary tool we will use to generalize the results of the previous section.
 What is simultaneous diagonalization all about then?
 
 \begin{definition}\label{def:sim-diag}
   Given a \(K\)-vector space \(V\), a set of operators \(\{T_j : V \to V\}_j\)
   is called \emph{simultaneously diagonalizable} if there is a basis \(\{v_1,
   \ldots, v_n\}\) for \(V\) such that \(T_j v_i\) is a scalar multiple of
   \(v_i\), for all \(i, j\).
 \end{definition}
 
 \begin{proposition}
   Given a \emph{finite-dimensional} vector space \(V\), a set of diagonalizable
   operators \(V \to V\) is simultaneously diagonalizable if, and only if all of
   its elements commute with one another.
 \end{proposition}
 
 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.
 Simultaneous diagonalization implies that to show \(M = \bigoplus_\lambda
 M_\lambda\) it suffices to show that \(H\!\restriction_M : M \to M\) is a
 diagonalizable operator for each \(H \in \mathfrak{h}\). To that end, we
 introduce \emph{the Jordan decomposition of an operator} and \emph{the abstract
 Jordan decomposition of a semisimple Lie algebra}.
 
 \begin{proposition}[Jordan]
   Given a finite-dimensional vector space \(V\) and an operator \(T : V \to
   V\), there are unique commuting operators \(T_{\operatorname{ss}},
   T_{\operatorname{nil}} : V \to V\), with \(T_{\operatorname{ss}}\)
   diagonalizable and \(T_{\operatorname{nil}}\) nilpotent, such that \(T =
   T_{\operatorname{ss}} + T_{\operatorname{nil}}\). The pair
   \((T_{\operatorname{ss}}, T_{\operatorname{nil}})\) is known as \emph{the Jordan
   decomposition of \(T\)}.
 \end{proposition}
 
 \begin{proposition}\index{abstract Jordan decomposition}
   Given \(\mathfrak{g}\) semisimple and \(X \in \mathfrak{g}\), there are
   \(X_{\operatorname{ss}}, X_{\operatorname{nil}} \in \mathfrak{g}\) such that \(X
   = X_{\operatorname{ss}} + X_{\operatorname{nil}}\), \([X_{\operatorname{ss}},
   X_{\operatorname{nil}}] = 0\), \(\operatorname{ad}(X_{\operatorname{ss}})\) is a
   diagonalizable operator and \(\operatorname{ad}(X_{\operatorname{nil}})\) is a
   nilpotent operator. The pair \((X_{\operatorname{ss}}, X_{\operatorname{nil}})\)
   is known as \emph{the Jordan decomposition of \(X\)}.
 \end{proposition}
 
 It should be clear from the uniqueness of
 \(\operatorname{ad}(X)_{\operatorname{ss}}\) and
 \(\operatorname{ad}(X)_{\operatorname{nil}}\) that the Jordan decomposition of
 \(\operatorname{ad}(X)\) is \(\operatorname{ad}(X) =
 \operatorname{ad}(X_{\operatorname{ss}}) +
 \operatorname{ad}(X_{\operatorname{nil}})\). What is perhaps more remarkable is
 the fact this holds for \emph{any} finite-dimensional \(\mathfrak{g}\)-module.
 In other words\dots
 
 \begin{proposition}\label{thm:preservation-jordan-form}
   Let \(M\) be a finite-dimensional \(\mathfrak{g}\)-module and \(X
   \in \mathfrak{g}\). Denote by \(X\!\restriction_M\) the action of \(X\) on
   \(M\). Then \(X_{\operatorname{ss}}\!\restriction_M =
   (X\!\restriction_M)_{\operatorname{ss}}\) and
   \(X_{\operatorname{nil}}\!\restriction_M =
   (X\!\restriction_M)_{\operatorname{nil}}\).
 \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}. As promised this
 implies\dots
 
 \begin{corollary}\label{thm:finite-dim-is-weight-mod}
   Let \(\mathfrak{g}\) be a semisimple Lie algebra, \(\mathfrak{h} \subset
   \mathfrak{g}\) be a Cartan subalgebra and \(M\) be any finite-dimensional
   \(\mathfrak{g}\)-module. Then there is a basis \(\{m_1, \ldots,
   m_r\}\) of \(M\) so that each \(m_i\) is simultaneously an eigenvector of all
   elements of \(\mathfrak{h}\) -- i.e. each element of \(\mathfrak{h}\) acts as
   a diagonal matrix in this basis. In other words, there are linear functionals
   \(\lambda_i \in \mathfrak{h}^*\) so that
   \(
     H \cdot m_i = \lambda_i(H) m_i
   \)
   for all \(H \in \mathfrak{h}\). In particular,
   \[
     M = \bigoplus_{\lambda \in \mathfrak{h}^*} M_\lambda
   \]
 \end{corollary}
 
 \begin{proof}
   Fix some \(H \in \mathfrak{h}\). It suffices to show that \(H\!\restriction_M
   : M \to M\) is a diagonalizable operator.
 
   If we write \(H = H_{\operatorname{ss}} + H_{\operatorname{nil}}\) for the
   abstract Jordan decomposition of \(H\), we know
   \(\operatorname{ad}(H_{\operatorname{ss}}) =
   \operatorname{ad}(H)_{\operatorname{ss}}\). But \(\operatorname{ad}(H)\) is a
   diagonalizable operator, so that \(\operatorname{ad}(H)_{\operatorname{ss}} =
   \operatorname{ad}(H)\). This implies
   \(\operatorname{ad}(H_{\operatorname{nil}}) =
   \operatorname{ad}(H)_{\operatorname{nil}} = 0\), so that
   \(H_{\operatorname{nil}}\) is a central element of \(\mathfrak{g}\). Since
   \(\mathfrak{g}\) is semisimple, \(H_{\operatorname{nil}} = 0\).
   Proposition~\ref{thm:preservation-jordan-form} then implies
   \((H\!\restriction_M)_{\operatorname{nil}} =
   H_{\operatorname{nil}}\!\restriction_M = 0\), so \(H\!\restriction_M =
   (H\!\restriction_M)_{\operatorname{ss}}\) is a diagonalizable operator.
 \end{proof}
 
 We should point out that this last proof only works for semisimple Lie
 algebras. This is because we rely heavily on
 Proposition~\ref{thm:preservation-jordan-form}, as well in the fact that
 semisimple Lie algebras are centerless. In fact,
 Corollary~\ref{thm:finite-dim-is-weight-mod} fails even for reductive Lie
 algebras. For a counterexample, consider the algebra \(\mathfrak{g} = K\): the
 Cartan subalgebra of \(\mathfrak{g}\) is \(\mathfrak{g}\) itself, and a
 \(\mathfrak{g}\)-module is simply a vector space \(M\) endowed with an operator
 \(M \to M\) -- which corresponds to the action of \(1 \in \mathfrak{g}\) on
 \(M\). In particular, if we choose an operator \(M \to M\) which is \emph{not}
 diagonalizable we find \(M \ne \bigoplus_{\lambda \in \mathfrak{h}^*}
 M_\lambda\).
 
 However, Corollary~\ref{thm:finite-dim-is-weight-mod} does work for reductive
 \(\mathfrak{g}\) if we assume that the \(\mathfrak{g}\)-module \(M\) in
 question is simple, since central elements of \(\mathfrak{g}\) act on simple
 \(\mathfrak{g}\)-modules as scalar operators. The hypothesis of
 finite-dimensionality is also of huge importance. For instance, consider\dots
 
 \begin{example}\label{ex:regular-mod-is-not-weight-mod}
   Let \(\mathcal{U}(\mathfrak{g})\) denote the regular \(\mathfrak{g}\)-module.
   Notice that \(\mathcal{U}(\mathfrak{g})_\lambda = 0\) for all \(\lambda \in
   \mathfrak{h}^*\). Indeed, since \(\mathcal{U}(\mathfrak{g})\) is a domain, if
   \((H - \lambda(H)) u = 0\) for some nonzero \(H \in \mathfrak{h}\) then \(u =
   0\). In particular,
   \[
     \bigoplus_{\lambda \in \mathfrak{h}^*} \mathcal{U}(\mathfrak{g})_\lambda
     = 0 \neq \mathcal{U}(\mathfrak{g})
   \]
 \end{example}
 
 As a first consequence of Corollary~\ref{thm:finite-dim-is-weight-mod} we
 show\dots
 
 \begin{corollary}
   The restriction of the Killing form \(\kappa\) to \(\mathfrak{h}\) is
   non-degenerate.
 \end{corollary}
 
 \begin{proof}
   Consider the root space decomposition \(\mathfrak{g} = \mathfrak{g}_0 \oplus
   \bigoplus_\alpha \mathfrak{g}_\alpha\) of the adjoint
   \(\mathfrak{g}\)-module, where \(\alpha\) ranges over all nonzero eigenvalues
   of the adjoint action of \(\mathfrak{h}\). We claim \(\mathfrak{g}_0 =
   \mathfrak{h}\).
 
   Indeed, since \(\mathfrak{h}\) is Abelian, \(\operatorname{ad}(\mathfrak{h})
   \mathfrak{h} = 0\) -- i.e. \(\mathfrak{h} \subset \mathfrak{g}_0\). On the
   other hand, since \(\mathfrak{h}\) is self-normalizing, if \([X, H] = 0 \in
   \mathfrak{h}\) for all \(H \in \mathfrak{h}\) then \(X \in \mathfrak{h}\) --
   i.e. \(\mathfrak{g}_0 \subset \mathfrak{h}\). So the eigenspace decomposition
   becomes
   \[
     \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_\alpha \mathfrak{g}_\alpha
   \]
 
   We furthermore claim that \(\mathfrak{h} = \mathfrak{g}_0\) is orthogonal to
   \(\mathfrak{g}_\alpha\) 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
   \[
     \alpha(H_1) \cdot \kappa(X, H_2)
     = \kappa([H_1, X], H_2)
     = - \kappa([X, H_1], H_2)
     = - \kappa(X, [H_1, H_2])
     = 0
   \]
 
   Hence the non-degeneracy of \(\kappa\) implies the non-degeneracy of its
   restriction.
 \end{proof}
 
 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 \(\kappa\) 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
 previous chapter in a general setting. Let \(\mathfrak{g}\) be a
 finite-dimensional semisimple algebra with a Cartan subalgebra \(\mathfrak{h}\)
 and let \(M\) be a finite-dimensional simple \(\mathfrak{g}\)-module. We will
 proceed, as we did before, by generalizing the 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}.
 
 \section{The Geometry of Roots and Weights}
 
 We begin our analysis, as we did for \(\mathfrak{sl}_2(K)\) and
 \(\mathfrak{sl}_3(K)\), by investigating the locus of roots of and weights of
 \(\mathfrak{g}\). Throughout chapter~\ref{ch:sl3} we have seen that the weights
 of any given finite-dimensional module of \(\mathfrak{sl}_2(K)\) or
 \(\mathfrak{sl}_3(K)\) can only assume very rigid configurations. For instance,
 we have 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 module.
 
 As for the aforementioned 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 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 will start with the first claim. Let \(\alpha\) and \(\beta\) be two
   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
   \[
     [H, [X, Y]]
     = [X, [H, Y]] - [Y, [H, X]]
     = (\alpha + \beta)(H) \cdot [X, Y]
   \]
   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
   \[
     (\operatorname{ad}(X) \operatorname{ad}(Y))^r Z
     = [X, [Y, [ \ldots, [X, [Y, Z]]] \ldots ]
     \in \mathfrak{g}_{r \alpha + r \beta + \gamma}
     = 0
   \]
   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 \(\kappa(X, \mathfrak{g}_\beta) =
   0\) 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
   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 \(\mathfrak{z} = 0\) of \(\mathfrak{g}\), a contradiction.
 \end{proof}
 
 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 root spaces \(\mathfrak{g}_\alpha\) are all \(1\)-dimensional.
 \end{proposition}
 
 The proof of the first statement of
 Proposition~\ref{thm:weights-symmetric-span} highlights something interesting:
 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 \(m \in M_\lambda\) we find
 \[
   H \cdot (X \cdot m)
   = X H \cdot m + [H, X] \cdot m
   = (\lambda + \alpha)(H) X \cdot m
 \]
 
 Thus \(X\) sends \(m\) to \(M_{\lambda + \alpha}\). We have encountered this
 formula twice in these notes: again, we find \(\mathfrak{g}_\alpha\) \emph{acts
 on \(M\) 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}\index{weights!root lattice}
   The weights of a finite-dimensional simple \(\mathfrak{g}\)-module \(M\) are
   all congruent modulo the root lattice \(Q = \mathbb{Z} \Delta\) of
   \(\mathfrak{g}\). In other words, all weights of \(M\) lie in the same
   \(Q\)-coset \(\xi \in \mfrac{\mathfrak{h}^*}{Q}\).
 \end{theorem}
 
 Again, we may leverage our knowledge of \(\mathfrak{sl}_2(K)\) to obtain
 further restrictions on the geometry of the locus of weights of \(M\). Namely,
 as in the case of \(\mathfrak{sl}_3(K)\) we show\dots
 
 \begin{proposition}\label{thm:distinguished-subalgebra}
   Given a root \(\alpha\) of \(\mathfrak{g}\) the subspace
   \(\mathfrak{s}_\alpha = \mathfrak{g}_\alpha \oplus \mathfrak{g}_{- \alpha}
   \oplus [\mathfrak{g}_\alpha, \mathfrak{g}_{- \alpha}]\) is a subalgebra
   isomorphic to \(\mathfrak{sl}_2(K)\).
 \end{proposition}
 
 \begin{corollary}\label{thm:distinguished-subalg-rep}
   For all weights \(\mu\), the subspace
   \[
     \bigoplus_k M_{\mu - k \alpha}
   \]
   is invariant under the action of the subalgebra \(\mathfrak{s}_\alpha\)
   and the weight spaces in this string match the eigenspaces of \(h\).
 \end{corollary}
 
 The proof of Proposition~\ref{thm:distinguished-subalgebra} is very technical
 in nature and we won't include it here, but the idea behind it is simple:
 recall that \(\mathfrak{g}_\alpha\) and \(\mathfrak{g}_{- \alpha}\) are both
 \(1\)-dimensional, so that \(\dim [\mathfrak{g}_\alpha, \mathfrak{g}_{-
 \alpha}]\) is at most 1. We check that \([\mathfrak{g}_\alpha, \mathfrak{g}_{-
 \alpha}] \ne 0\) and that no generator of \([\mathfrak{g}_\alpha,
 \mathfrak{g}_{- \alpha}]\) is annihilated by \(\alpha\), so that by adjusting
 scalars we can find \(E_\alpha \in \mathfrak{g}_\alpha\) and \(F_\alpha \in
 \mathfrak{g}_{- \alpha}\) such that \(H_\alpha = [E_\alpha, F_\alpha]\)
 satisfies
 \begin{align*}
   [H_\alpha, F_\alpha] & = -2 F_\alpha &
   [H_\alpha, E_\alpha] & =  2 E_\alpha
 \end{align*}
 
 The elements \(E_\alpha, F_\alpha \in \mathfrak{g}\) are not uniquely
 determined by this condition, but \(H_\alpha\) is. As promised, the second
 statement of Corollary~\ref{thm:distinguished-subalg-rep} imposes strong
 restrictions on the weights of \(M\). Namely, if \(\lambda\) is a weight,
 \(\lambda(H_\alpha)\) is an eigenvalue of \(h\) on some
 \(\mathfrak{sl}_2(K)\)-module, so it must be an integer. In other words\dots
 
 \begin{definition}\label{def:weight-lattice}\index{weights!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 \(\mathfrak{g}\)}. We call the elements of \(P\)
   \emph{integral}.
 \end{definition}
 
 \begin{proposition}\label{thm:weights-fit-in-weight-lattice}
   The weights of a finite-dimensional simple \(\mathfrak{g}\)-module \(M\) of
   all lie in the weight lattice \(P\).
 \end{proposition}
 
 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 \(\mathbb{Z} \langle \epsilon_1, \epsilon_2, \epsilon_3 \rangle\). To
 proceed further, we would like to take \emph{the highest weight of \(M\)} 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 \(M\) that maximizes this functional, but at this point it is
 convenient to introduce some additional tools to our arsenal. These tools are
 called \emph{basis}.
 
 \begin{definition}\label{def:basis-of-root}\index{weights!basis}
   A subset \(\Sigma = \{\beta_1, \ldots, \beta_r\} \subset \Delta\) of linearly
   independent roots is called \emph{a basis for \(\Delta\)} if, given \(\alpha
   \in \Delta\), there are unique \(k_1, \ldots, k_r \in \mathbb{N}\) such that
   \(\alpha = \pm(k_1 \beta_1 + \cdots + k_r \beta_r)\).
 \end{definition}
 
 \begin{example}\label{ex:sl-canonical-basis}
   Suppose \(\mathfrak{g} = \mathfrak{sl}_n(K)\) and \(\mathfrak{h} \subset
   \mathfrak{g}\) is the subalgebra of diagonal matrices, as in
   Example~\ref{ex:cartan-of-sl}. Consider the linear functionals \(\epsilon_1,
   \ldots, \epsilon_n \in \mathfrak{h}^*\) such that \(\epsilon_i(H)\) is the
   \(i\)-th entry of the diagonal of \(H\). As observed in
   section~\ref{sec:sl3-reps} for \(n = 3\), the roots of \(\mathfrak{sl}_n(K)\)
   are \(\epsilon_i - \epsilon_j\) for \(i \ne j\) -- with root vectors given by
   \(E_{i j}\) -- and we may take the basis \(\Sigma = \{\beta_1, \ldots,
   \beta_{n-1}\}\) with \(\beta_i = \epsilon_i - \epsilon_{i+1}\).
 \end{example}
 
 \begin{example}\label{ex:sp-canonical-basis}
   Suppose \(\mathfrak{g} = \mathfrak{sp}_{2n}(K)\) and \(\mathfrak{h} \subset
   \mathfrak{g}\) is the subalgebra of diagonal matrices, as in
   Example~\ref{ex:cartan-of-sp}. Consider the linear functionals \(\epsilon_1,
   \ldots, \epsilon_n \in \mathfrak{h}^*\) such that \(\epsilon_i(H)\) is the
   \(i\)-th entry of the diagonal of \(H\). Then the roots of
   \(\mathfrak{sp}_{2n}(K)\) are \(\pm \epsilon_i \pm \epsilon_j\) for \(i \ne
   j\) and \(\pm 2 \epsilon_i\) -- see \cite[ch.~16]{fulton-harris}. In this
   case, we may take the basis \(\Sigma = \{\beta_1, \ldots, \beta_n\}\) with
   \(\beta_i = \epsilon_i - \epsilon_{i+1}\) for \(i < n\) and \(\beta_n = 2
   \epsilon_n\).
 \end{example}
 
 The interesting thing about basis for \(\Delta\) is that they allow us to
 compare weights of a given \(\mathfrak{g}\)-module. At this point the reader
 should be asking himself: how? Definition~\ref{def:basis-of-root} isn't exactly
 all that intuitive. Well, the thing is that any choice of basis \(\Sigma\)
 induces an order in \(Q\), where elements are ordered by their
 \emph{\(\Sigma\)-coordinates}.
 
 \begin{definition}\index{weights!orderings of roots}
   Let \(\Sigma = \{\beta_1, \ldots, \beta_r\}\) be a basis for \(\Delta\).
   Given \(\alpha = k_1 \beta_1 + \cdots + k_r \beta_r \in Q\) with \(k_1,
   \ldots, k_r \in \mathbb{Z}\), we call the vector \(\alpha_\Sigma = (k_1,
   \ldots, k_r) \in \mathbb{Z}^r\) \emph{the \(\Sigma\)-coordinate of
   \(\alpha\)}. We say that \(\alpha \preceq \beta\) if \(\alpha_\Sigma \le
   \beta_\Sigma\) in the lexicographical order.
 \end{definition}
 
 \begin{definition}
   Given a basis \(\Sigma\) for \(\Delta\), there is a canonical
   partition\footnote{Notice that $\operatorname{ht}(\alpha) = 0$ if, and only
   if $\alpha = 0$. Since $0$ is, by definition, not a root, the sets $\Delta^+$
   and $\Delta^-$ account for all roots.} \(\Delta^+ \cup \Delta^- = \Delta\),
   where \(\Delta^+ = \{ \alpha \in \Delta : \alpha \succ 0 \}\) and \(\Delta^-
   = \{ \alpha \in \Delta : \alpha \prec 0 \}\). The elements of \(\Delta^+\)
   and \(\Delta^-\) are called \emph{positive} and \emph{negative roots},
   respectively.
 \end{definition}
 
 \begin{example}
   If \(\mathfrak{g} = \mathfrak{sl}_3(K)\) and \(\Sigma\) is as in
   Example~\ref{ex:sl-canonical-basis} then the partition \(\Delta^+ \cup
   \Delta^-\) induced by \(\Sigma\) is the same as the one described in
   section~\ref{sec:sl3-reps}.
 \end{example}
 
 \begin{definition}\index{Lie subalgebra!Borel subalgebra}\index{Lie subalgebra!parabolic subalgebra}
   Let \(\Sigma\) be a basis for \(\Delta\). The subalgebra \(\mathfrak{b} =
   \mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+} \mathfrak{g}_\alpha\) is
   called \emph{the Borel subalgebra associated with \(\mathfrak{h}\) and
   \(\Sigma\)}. A subalgebra \(\mathfrak{p} \subset \mathfrak{g}\) is called
   \emph{parabolic} if \(\mathfrak{p} \supset \mathfrak{b}\).
 \end{definition}
 
 It should be obvious that the binary relation \(\preceq\) in \(Q\) is a total
 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,
 given \(\lambda \in \mu + Q\), we say \(\lambda \preceq \mu\) if \(\lambda -
 \mu \preceq 0\). In particular, since the weights of \(M\) all lie in a single
 \(Q\)-coset, we may compare them in this way. Given a basis \(\Sigma\) for
 \(\Delta\) we may take ``the highest weight of \(M\)'' as a maximal weight
 \(\lambda\) of \(M\). The obvious question then is: can we always find a basis
 for \(\Delta\)?
 
 \begin{proposition}
   There is a basis \(\Sigma\) for \(\Delta\).
 \end{proposition}
 
 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 \(\kappa(\alpha, \beta) \in
 \mathbb{Z}\) for all \(\alpha, \beta \in \Delta\), so that the Killing form
 \(\kappa\) 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
 \(M\). We call \(\lambda\) \emph{the highest weight of \(M\)}, and we call any
 nonzero \(m \in M_\lambda\) \emph{a highest weight vector}. The strategy then
 is to describe all weight spaces of \(M\) in terms of \(\lambda\) and \(m\), as
 in Theorem~\ref{thm:sl3-irr-weights-class}. Unsurprisingly we do so by
 reproducing the proof of the case of \(\mathfrak{sl}_3(K)\).
 
 First, we note that any highest weight vector \(m \in M_\lambda\) is
 annihilated by all positive root spaces, for if \(\alpha \in \Delta^+\) then
 \(E_\alpha \cdot m \in M_{\lambda + \alpha}\) must be zero -- or otherwise we
 would have that \(\lambda + \alpha\) is a weight with \(\lambda \prec \lambda +
 \alpha\). In particular,
 \[
   \bigoplus_{k \in \mathbb{Z}}   M_{\lambda - k \alpha}
   = \bigoplus_{k \in \mathbb{N}} M_{\lambda - k \alpha}
 \]
 and \(\lambda(H_\alpha)\) is the right-most eigenvalue of the action of \(h\)
 on the \(\mathfrak{sl}_2(K)\)-module \(\bigoplus_k M_{\lambda - k \alpha}\).
 
 This has a number of important consequences. For instance\dots
 
 \begin{corollary}
   If \(\alpha \in \Delta^+\) and \(\sigma_\alpha : \mathfrak{h}^* \to
   \mathfrak{h}^*\) is the reflection in the hyperplane perpendicular to
   \(\alpha\) with respect to the Killing form, the weights of \(M\) occurring
   in the line joining \(\lambda\) and \(\sigma_\alpha\) are precisely the \(\mu
   \in P\) lying between \(\lambda\) and \(\sigma_\alpha(\lambda)\).
 \end{corollary}
 
 \begin{proof}
   Notice that any \(\mu \in P\) in the line joining \(\lambda\) and
   \(\sigma_\alpha(\lambda)\) has the form \(\mu = \lambda - k \alpha\) for some
   \(k\), so that \(M_\mu\) corresponds the eigenspace associated with the
   eigenvalue \(\lambda(H_\alpha) - 2k\) of the action of \(h\) on \(\bigoplus_k
   M_{\lambda - k \alpha}\). If \(\mu\) lies between \(\lambda\) and
   \(\sigma_\alpha(\lambda)\) then \(k\) lies between \(0\) and
   \(\lambda(H_\alpha)\), in which case \(M_\mu \neq 0\) and therefore \(\mu\)
   is a weight.
 
   On the other hand, if \(\mu\) does not lie between \(\lambda\) and
   \(\sigma_\alpha(\lambda)\) then either \(k < 0\) or \(k >
   \lambda(H_\alpha)\). Suppose \(\mu\) is a weight. In the first case \(\mu
   \succ \lambda\), a contradiction. On the second case the fact that \(M_\mu
   \ne 0\) implies \(M_{\lambda  + (k - \lambda(H_\alpha)) \alpha} =
   M_{\sigma_\alpha(\mu)} \ne 0\), which contradicts the fact that \(M_{\lambda
   + \ell \alpha} = 0\) for all \(\ell \ge 0\).
 \end{proof}
 
 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
 \(\kappa(\epsilon_i - \epsilon_j, \alpha) = 0\). As in the case of
 \(\mathfrak{sl}_3(K)\), the same sort of arguments leads us to the
 conclusion\dots
 
 \begin{definition}\index{Weyl group}
   We refer to the (finite) group \(W = \langle \sigma_\alpha : \alpha \in
   \Delta \rangle = \langle \sigma_\beta : \beta \in \Sigma \rangle \subset
   \operatorname{O}(\mathfrak{h}^*)\) as \emph{the Weyl group of
   \(\mathfrak{g}\)}.
 \end{definition}
 
 \begin{theorem}\label{thm:irr-weight-class}
   The weights of a simple \(\mathfrak{g}\)-module \(M\) with highest weight
   \(\lambda\) are precisely the elements of the weight lattice \(P\) congruent
   to \(\lambda\) modulo the root lattice \(Q\) lying inside the convex hull of
   the orbit of \(\lambda\) under the action of the Weyl group \(W\).
 \end{theorem}
 
 At this point we are basically done with results regarding the geometry of the
 weights of \(M\), but it is convenient to introduce some further notation.
 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 \(W\) on \(\mathfrak{h}\).
 
 \begin{definition}\index{Weyl group!natural action}\index{Weyl group!dot action}
   The canonical action of \(W\) on \(\mathfrak{h}^*\) given by \(\sigma \cdot
   \lambda = \sigma(\lambda)\) is called \emph{the natural action of \(W\)}. We
   also consider the equivalent ``shifted'' action \(\sigma \bullet \lambda =
   \sigma(\lambda + \rho) - \rho\) of \(W\) on \(\mathfrak{h}^*\), known as
   \emph{the dot action of \(W\)} -- here \(\rho =  \sfrac{1}{2} \beta_1 +
   \cdots \sfrac{1}{2} \beta_r\).
 \end{definition}
 
 This already allow us to compute some examples of Weyl groups.
 
 \begin{example}\label{ex:sl-weyl-group}
   Suppose \(\mathfrak{g} = \mathfrak{sl}_n(K)\) and \(\mathfrak{h} \subset
   \mathfrak{g}\) is as in Example~\ref{ex:cartan-of-sl}. Let \(\epsilon_1,
   \ldots, \epsilon_n \in \mathfrak{h}^*\) be as in
   Example~\ref{ex:sl-canonical-basis} and take the associated basis \(\Sigma =
   \{\beta_1, \ldots, \beta_{n-1}\}\) for \(\Delta\), \(\beta_i = \epsilon_i -
   \epsilon_{i + 1}\). Then a simple calculation shows that \(\sigma_{\beta_i}\)
   permutes \(\epsilon_i\) and \(\epsilon_{i+1}\) and fixes the other
   \(\epsilon_j\). This translates to a canonical isomorphism
   \begin{align*}
                    W & \isoto  S_n                       \\
     \sigma_{\beta_i} & \mapsto \sigma_i = (i \; i\!+\!1)
   \end{align*}
 \end{example}
 
 \begin{example}\label{ex:sp-weyl-group}
   Suppose \(\mathfrak{g} = \mathfrak{sp}_{2n}(K)\) and \(\mathfrak{h} \subset
   \mathfrak{g}\) is as in Example~\ref{ex:cartan-of-sp}. Let \(\epsilon_1,
   \ldots, \epsilon_n \in \mathfrak{h}^*\) be as in
   Example~\ref{ex:sp-canonical-basis} and take the associated basis \(\Sigma =
   \{\beta_1, \ldots, \beta_n\}\) for \(\Delta\). Then a simple calculation
   shows that \(\sigma_{\beta_i}\) permutes \(\epsilon_i\) and
   \(\epsilon_{i+1}\) for \(i < n\) and \(\sigma_{\beta_n}\) switches the sign
   of \(\epsilon_n\). This translates to a canonical isomorphism
   \begin{align*}
                    W & \isoto  S_n \ltimes (\mathbb{Z}/2\mathbb{Z})^n \\
     \sigma_{\beta_i} & \mapsto (\sigma_i, (\bar 0, \ldots, \bar 0))   \\
     \sigma_{\beta_n} & \mapsto (1, (\bar 0, \ldots, \bar 0, \bar 1)),
   \end{align*}
   where \(\sigma_i = (i \ i\!+\!1)\) are the canonical transpositions.
 \end{example}
 
 If we conjugate some \(\sigma \in W\) by the isomorphism \(\mathfrak{h}^*
 \isoto \mathfrak{h}\) afforded by the restriction of the Killing for to
 \(\mathfrak{h}\) we get a linear action of \(W\) on \(\mathfrak{h}\), which is
 given by \(\kappa(\sigma \cdot H, \cdot) = \sigma \cdot \kappa(H, \cdot)\). As
 it turns out, this action can be extended to an action of \(W\) on
 \(\mathfrak{g}\) by automorphisms of Lie algebras. 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 \(\alpha \in \Delta^+\), there is an automorphism of Lie algebras
   \(f_\alpha : \mathfrak{g} \isoto \mathfrak{g}\) such that
   \(f_\alpha(H) = \sigma_\alpha \cdot H\) for all \(H \in \mathfrak{h}\). In
   addition, these automorphisms can be chosen in such a way that the family
   \(\{f_\alpha\}_{\alpha \in \Delta^+}\) defines an action of \(W\) on
   \(\mathfrak{g}\) -- which is obviously compatible with the natural action of
   \(W\) on \(\mathfrak{h}\).
 \end{proposition}
 
 \begin{note}
   We should notice the action of \(W\) on \(\mathfrak{g}\) from
   Proposition~\ref{thm:weyl-group-action} is not canonical, since it depends on
   the choice of \(E_\alpha\) and \(F_\alpha\). Nevertheless, different choices
   of \(E_\alpha\) and \(F_\alpha\) yield isomorphic actions and the restriction
   of these actions to \(\mathfrak{h}\) is independent of any choices.
 \end{note}
 
 We should point out that the results in this section regarding the geometry
 roots and weights are only the beginning of a well develop axiomatic theory of
 the so called \emph{root systems}, which was used by Cartan in the early 20th
 century to classify all finite-dimensional simple complex Lie algebras in terms
 of Dynking diagrams. This and much more can be found in \cite[III]{humphreys}
 and \cite[ch.~21]{fulton-harris}. Having found all of the weights of \(M\), the
 only thing we are missing for a complete classification is an existence and
 uniqueness theorem analogous to Theorem~\ref{thm:sl2-exist-unique} and
 Theorem~\ref{thm:sl3-existence-uniqueness}. This will be the focus of the next
 section.
 
 \section{Highest Weight Modules \& the Highest Weight Theorem}
 
 It is already clear from the previous discussion that if \(\lambda\) is the
 highest weight of \(M\) then \(\lambda(H_\alpha) \ge 0\) for all positive roots
 \(\alpha\). Indeed, as in the \(\mathfrak{sl}_3(K)\), for each \(\alpha \in
 \Delta^+\) we know \(\lambda(H_\alpha)\) is the highest eigenvalue of the
 action of \(h\) in the \(\mathfrak{sl}_2(K)\)-module \(\bigoplus_k M_{\lambda -
 k \alpha}\) -- which must be a non-negative integer. This fact may be
 summarized in the following proposition.
 
 \begin{definition}\index{weights!dominant weight}\index{weights!integral weight}
   An element \(\lambda\) of \(P\) such that \(\lambda(H_\alpha) \ge 0\) for all
   \(\alpha \in \Delta^+\) is referred to as an \emph{dominant integral weight
   of \(\mathfrak{g}\)}. The set of all dominant integral weights is denotes by
   \(P^+\).
 \end{definition}
 
 \begin{proposition}\label{thm:highes-weight-of-fin-dim-is-dominant}
   Suppose \(M\) is a finite-dimensional simple \(\mathfrak{g}\)-module and
   \(\lambda\) is its highest weight. Then \(\lambda\) is a dominant integral
   weight of \(\mathfrak{g}\).
 \end{proposition}
 
 The condition that \(\lambda \in P^+\) is thus necessary for the existence of a
 simple \(\mathfrak{g}\)-module with highest weight given by \(\lambda\). Given
 our previous experience with \(\mathfrak{sl}_2(K)\) and \(\mathfrak{sl}_3(K)\),
 it is perhaps unsurprising that this condition is also sufficient.
 
 \begin{theorem}\label{thm:dominant-weight-theo}\index{weights!Highest Weight Theorem}
   For each dominant integral \(\lambda \in P^+\) there exists precisely one
   finite-dimensional simple \(\mathfrak{g}\)-module \(M\) whose highest weight
   is \(\lambda\).
 \end{theorem}
 
 This is known as \emph{the Highest Weight Theorem}, and its proof is the focus
 of this section. The ``uniqueness'' part of the theorem follows at once from
 the arguments used for \(\mathfrak{sl}_3(K)\). However, the ``existence'' part
 of the theorem is more nuanced. Our first instinct is, of course, to try to
 generalize the proof used for \(\mathfrak{sl}_3(K)\). Indeed, as in
 Proposition~\ref{thm:sl3-mod-is-highest-weight}, one is able to show\dots
 
 \begin{proposition}\label{thm:fin-dim-simple-mod-has-singular-vector}
   Let \(M\) be a finite-dimensional simple \(\mathfrak{g}\)-module. Then there
   exists a nonzero weight vector \(m \in M\) which is annihilated by all
   positive root spaces of \(\mathfrak{g}\) -- i.e. \(X \cdot m = 0\) for all
   \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\).
 \end{proposition}
 
 \begin{proof}
   If \(\lambda\) is the highest weight of \(M\), it suffices to take any \(m
   \in M_\lambda\). Indeed, given \(X \in \mathfrak{g}_\alpha\) with \(\alpha
   \in \Delta^+\), \(X \cdot m \in M_{\lambda + \alpha} = 0\) because \(\lambda
   + \alpha \succ \lambda\).
 \end{proof}
 
 Unfortunately for us, this is where the parallels with
 Proposition~\ref{thm:sl3-mod-is-highest-weight} end. The issue is that our
 proof relied heavily on our knowledge of the roots of \(\mathfrak{sl}_3(K)\).
 It is thus clear that we need a more systematic approach for the general
 setting. We begin by asking a simpler question: how can we construct \emph{any}
 \(\mathfrak{g}\)-module \(M\) whose highest weight is \(\lambda\)? In the
 process of answering this question we will come across a surprisingly elegant
 solution to our problem.
 
 If \(M\) is a finite-dimensional simple module with highest weight \(\lambda\)
 and \(m \in M_\lambda\), we already know that \(X \cdot m = 0\) for any \(m \in
 M_\lambda\) and \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\). Since
 \(M = \mathcal{U}(\mathfrak{g}) \cdot m\), the restriction of \(M\) to the
 Borel subalgebra \(\mathfrak{b} \subset \mathfrak{g}\) has a prescribed action.
 On the other hand, we have essentially no information about the action of the
 rest of \(\mathfrak{g}\) on \(M\). Nevertheless, given a
 \(\mathfrak{b}\)-module we may obtain a \(\mathfrak{g}\)-module by
 \emph{freely} extending the action of \(\mathfrak{b}\) via induction. This
 leads us to the following definition.
 
 \begin{definition}\label{def:verma}\index{\(\mathfrak{g}\)-module!(generalized) Verma modules}
   Given \(\lambda \in \mathfrak{h}^*\), consider the \(\mathfrak{b}\)-module
   \(K m^+\) where \(H \cdot m^+ = \lambda(H) m^+\) for all \(H \in
   \mathfrak{h}\) and \(X \cdot m^+ = 0\) for \(X \in \mathfrak{g}_{\alpha}\)
   with \(\alpha \in \Delta^+\). The \(\mathfrak{g}\)-module \(M(\lambda) =
   \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K m^+\) is called \emph{the
   Verma module of weight \(\lambda\)}.
 \end{definition}
 
 \begin{example}\label{ex:sl2-verma}
   If \(\mathfrak{g} = \mathfrak{sl}_2(K)\), then we can take \(\mathfrak{h} = K
   h\) and \(\mathfrak{b} = K e \oplus K h\). In this setting, the linear map
   \(g : \mathfrak{h}^* \to K\) defined by \(g(h) = 1\) affords us a canonical
   identification \(\mathfrak{h}^* = K g \cong K\), so that given \(\lambda \in
   K\) we may denote \(M(\lambda g)\) simply by \(M(\lambda)\). Using this
   notation \(M(\lambda) = \bigoplus_{k \ge 0} K f^k \cdot m^+\), and the action
   of \(\mathfrak{sl}_2(K)\) on \(M(\lambda)\) is given by
   formula (\ref{eq:sl2-verma-formulas}).
   \begin{equation}\label{eq:sl2-verma-formulas}
     \begin{aligned}
       f^k \cdot m^+ & \overset{e}{\mapsto} k(\lambda+1-k) f^{k-1} \cdot m^+ &
       f^k \cdot m^+ & \overset{f}{\mapsto} f^{k+1} \cdot m^+                &
       f^k \cdot m^+ & \overset{h}{\mapsto} (\lambda - 2k) f^k \cdot m^+     &
     \end{aligned}
   \end{equation}
 \end{example}
 
 \begin{example}\label{ex:verma-is-not-irr}
   Consider the \(\mathfrak{sl}_2(K)\)-module \(M(2)\) as described in
   Example~\ref{ex:sl2-verma}. It follows from formula
   (\ref{eq:sl2-verma-formulas}) that the action of \(\mathfrak{sl}_2(K)\) on
   \(M(2)\) is given by
   \begin{center}
     \begin{tikzcd}
       \cdots      \rar[bend left=60]{-10}
       & M(2)_{-6} \rar[bend left=60]{-4}  \lar[bend left=60]{1}
       & M(2)_{-4} \rar[bend left=60]{0}   \lar[bend left=60]{1}
       & M(2)_{-2} \rar[bend left=60]{2}   \lar[bend left=60]{1}
       & M(2)_0    \rar[bend left=60]{2}   \lar[bend left=60]{1}
       & M(2)_2                            \lar[bend left=60]{1}
     \end{tikzcd}
   \end{center}
   where \(M(2)_{2 - 2 k} = K f^k \cdot m^+\). Here the top arrows represent the
   action of \(e\) and the bottom arrows represent the action of \(f\). The
   scalars labeling each arrow indicate to which multiple of \(f^{k \pm 1} \cdot
   m^+\) the elements \(e\) and \(f\) send \(f^k \cdot m^+\). The string of
   weight spaces to the left of the diagram is infinite. Since \(e \cdot (f^3
   \cdot m^+) = 0\), it is easy to see that subspace \(\bigoplus_{k \ge 3} K f^k
   \cdot m^+\) is a (maximal) \(\mathfrak{sl}_2(K)\)-submodule, which is
   isomorphic to \(M(-4)\).
 \end{example}
 
 These last examples show that, unlike most modules we have so far encountered,
 Verma modules are \emph{highly infinite-dimensional}. Indeed, it follows from
 the PBW Theorem that the regular module \(\mathcal{U}(\mathfrak{g})\) is a free
 \(\mathfrak{b}\)-module of infinite rank -- equal to the codimension of
 \(\mathcal{U}(\mathfrak{b})\) in \(\mathcal{U}(\mathfrak{g})\). Hence \(\dim
 M(\lambda)\), which is the same as the rank of \(\mathcal{U}(\mathfrak{g})\) as
 a \(\mathfrak{b}\)-module, is also infinite. Nevertheless, it turns out that
 finite-dimensional modules and Verma module may both be seen as particular
 cases of a more general pattern. This leads us to the following definitions.
 
 \begin{definition}
   Let \(M\) be a \(\mathfrak{g}\)-module. A vector \(m \in M\) is called
   \emph{singular} if it is annihilated by all positive weight spaces of
   \(\mathfrak{g}\) -- i.e. \(X \cdot m = 0\) for all \(X \in
   \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\).
 \end{definition}
 
 \begin{definition}\label{def:highest-weight-mod}
   A \(\mathfrak{g}\)-module \(M\) is called \emph{a highest weight module} if
   there exists some singular weight vector \(m^+ \in M_\lambda\) such that \(M
   = \mathcal{U}(\mathfrak{g}) \cdot m^+\). Any such \(m^+\) is called \emph{a
   highest weight vector}, while \(\lambda\) is called \emph{the highest weight
   of \(M\)}.
 \end{definition}
 
 \begin{example}
   Proposition~\ref{thm:fin-dim-simple-mod-has-singular-vector} is equivalent to
   the fact that every finite-dimensional simple \(\mathfrak{g}\)-module is a
   highest weight module.
 \end{example}
 
 \begin{example}
   It should be obvious from the definitions that \(M(\lambda)\) is a highest
   weight module of highest weight \(\lambda\) and highest weight vector \(m^+ =
   1 \otimes m^+\) as in Definition~\ref{def:verma}. Indeed, \(u \otimes m^+ = u
   \cdot m^+\) for all \(u \in \mathcal{U}(\mathfrak{g})\), which already shows
   \(M(\lambda)\) is generated by \(m^+\). In particular,
   \begin{align*}
     H \cdot m^+ & = H \otimes m^+ = 1 \otimes H \cdot m^+ = \lambda(H) m^+ \\
     X \cdot m^+ & = X \otimes m^+ = 1 \otimes X \cdot m^+ = 0
   \end{align*}
   for all \(H \in \mathfrak{h}\) and \(X \in \mathfrak{g}_\alpha\), \(\alpha
   \in \Delta^+\).
 \end{example}
 
 While Verma modules show that a highest weight module needs not to be
 finite-dimensional, it turns out that highest weight modules enjoy many of the
 features we've grown used to in the past chapters. Explicitly, we may establish
 the properties described in the following proposition, whose statement should
 also explain the nomenclature of Definition~\ref{def:highest-weight-mod}.
 
 \begin{proposition}\label{thm:high-weight-mod-is-weight-mod}
   Let \(M\) be a highest weight \(\mathfrak{g}\)-module with highest weight
   vector \(m \in M_\lambda\). The weight spaces decomposition
   \[
     M = \bigoplus_{\mu \in \mathfrak{h}^*} M_\mu
   \]
   holds. Furthermore, \(\dim M_\mu < \infty\) for all \(\mu \in
   \mathfrak{h}^*\) and \(\dim M_\lambda = 1\) -- i.e. \(M_\lambda = K m\).
   Finally, given a weight \(\mu\) of \(M\), \(\lambda \succeq \mu\) -- so that
   the highest weight \(\lambda\) of \(M\) is unique and coincides with the
   largest of the weights of \(M\).
 \end{proposition}
 
 \begin{proof}
   Since \(M = \mathcal{U}(\mathfrak{g}) \cdot m\), the PBW Theorem implies
   that \(M\) is spanned by the vectors \(F_{\alpha_{i_1}} F_{\alpha_{i_2}}
   \cdots F_{\alpha_{i_s}} \cdot m\) for \(\Delta^+ = \{\alpha_1, \ldots,
   \alpha_r\}\) and \(F_{\alpha_i} \in \mathfrak{g}_{- \alpha_i}\) as in the
   proof of Proposition~\ref{thm:distinguished-subalgebra}. But
   \[
     \begin{split}
       H \cdot (F_{\alpha_{i_1}} F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}}
                \cdot m)
       & = ([H, F_{\alpha_{i_1}}] + F_{\alpha_{i_1}} H)
           F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m \\
       & = - \alpha_{i_1}(H) F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m
         + F_{\alpha_{i_1}} ([H, F_{\alpha_{i_2}}] + F_{\alpha_{i_2}} H)
           F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m \\
       & \;\; \vdots \\
       & = (- \alpha_{i_1} - \cdots - \alpha_{i_s})(H)
           F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m
         + F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} H \cdot m \\
       & = (\lambda - \alpha_{i_1} - \cdots - \alpha_{i_s})(H)
           F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m \\
       & \therefore F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m
         \in M_{\lambda - \alpha_{i_1} - \cdots - \alpha_{i_s}}
     \end{split}
   \]
 
   Hence \(M \subset \bigoplus_{\mu \in \mathfrak{h}^*} M_\mu\), as desired. In
   fact we have established
   \[
     M
     \subset
     \bigoplus_{k_i \in \mathbb{N}}
     M_{\lambda - k_1 \cdot \alpha_1 - \cdots - k_r \cdot \alpha_r}
   \]
   where \(\{\alpha_1, \ldots, \alpha_r\} = \Delta^+\), so that all weights of
   \(M\) have the form \(\mu = \lambda - k_1 \cdot \alpha_1 - \cdots - k_r \cdot
   \alpha_r\). This already gives us that the weights of \(M\) are bounded by
   \(\lambda\).
 
   To see that \(\dim M_\mu < \infty\), simply note that there are only finitely
   many monomials \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots
   F_{\alpha_s}^{k_s}\) such that \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots
   + k_s \cdot \alpha_s\). Since \(M_\mu\) is spanned by the images of \(m\)
   under such monomials, we conclude \(\dim M_\mu < \infty\). In particular,
   there is a single monomial \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots
   F_{\alpha_s}^{k_s}\) such that \(\lambda = \lambda + k_1 \cdot \alpha_1 +
   \cdots + k_s \cdot \alpha_s\) -- which is, of course, the monomial where
   \(k_1 = \cdots = k_n = 0\). Hence \(\dim M_\lambda = 1\).
 \end{proof}
 
 At this point it is important to note that, far from a ``misbehaved'' class of
 examples, Verma modules hold a very special place in the theory of highest
 weight modules. Intuitively speaking, the Verma module \(M(\lambda)\) should
 really be though-of as ``the freest highest weight \(\mathfrak{g}\)-module of
 highest weight \(\lambda\)''. In practice, this translates to the following
 universal property.
 
 \begin{proposition}
   Let \(M\) be a \(\mathfrak{g}\)-module and \(m \in M_\lambda\) be a singular
   vector. Then there exists a unique \(\mathfrak{g}\)-homomorphism \(f :
   M(\lambda) \to M\) such that \(f(m^+) = m\). Furthermore, all homomorphisms
   \(M(\lambda) \to M\) are given in this fashion.
   \[
     \operatorname{Hom}_{\mathfrak{g}}(M(\lambda), M)
     \cong \{ m \in M_\lambda : m \ \text{is singular}\}
   \]
 \end{proposition}
 
 \begin{proof}
   The result follows directly from Proposition~\ref{thm:frobenius-reciprocity}.
   Indeed, by the Frobenius Reciprocity Theorem, a \(\mathfrak{g}\)-homomorphism
   \(f : M(\lambda) \to M\) is the same as a \(\mathfrak{b}\)-homomorphism \(g :
   K m^+ \to M = \operatorname{Res}_{\mathfrak{b}}^{\mathfrak{g}} M\). More
   specifically, given a \(\mathfrak{b}\)-homomorphism \(g : K m^+ \to M\),
   there exists a unique \(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\)
   such that \(f(u \otimes m^+) = u \cdot g(m^+)\) for all \(u \in
   \mathcal{U}(\mathfrak{g})\), and all \(\mathfrak{g}\)-homomorphism
   \(M(\lambda) \to M\) arise in this fashion.
 
   Any \(K\)-linear map \(g : K m^+ \to M\) is determined by \(m = g(m^+)\).
   Finally, notice that \(g\) is a \(\mathfrak{b}\)-homomorphism if, and only if
   \(m\) is a singular vector lying in \(M_\lambda\).
 \end{proof}
 
 Why is any of this interesting to us, however? After all, Verma modules are not
 specially well suited candidates for a proof of the Highest Weight Theorem.
 Indeed, we have seen in Example~\ref{ex:verma-is-not-irr} that in general
 \(M(\lambda)\) is not simple, nor is it ever finite-dimensional. Nevertheless,
 we may use \(M(\lambda)\) to establish Theorem~\ref{thm:dominant-weight-theo}
 as follows.
 
 Suppose \(M\) is a highest weight \(\mathfrak{g}\)-module of highest weight
 \(\lambda\) with highest weight vector \(m\). By the last proposition, there is
 a \(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\) such that \(f(m^+) =
 m\). Since \(M = \mathcal{U}(\mathfrak{g}) \cdot m\), \(f\) is surjective and
 therefore \(M \cong \mfrac{M(\lambda)}{\ker f}\). Hence\dots
 
 \begin{proposition}
   Let \(M\) be a highest weight \(\mathfrak{g}\)-module of highest weight
   \(\lambda\). Then \(M\) is quotient of \(M(\lambda)\). If \(M\) is simple
   then \(M\) is the quotient of \(M(\lambda)\) by a maximal
   \(\mathfrak{g}\)-submodule.
 \end{proposition}
 
 Maximal submodules of Verma modules are thus of primary interest to us. As it
 turns out, these can be easily classified.
 
 \begin{proposition}\label{thm:max-verma-submod-is-weight}
   Every submodule \(N \subset M(\lambda)\) is the direct sum of its weight
   spaces. In particular, \(M(\lambda)\) has a unique maximal submodule
   \(N(\lambda)\) and a unique simple quotient \(L(\lambda) =
   \sfrac{M(\lambda)}{N(\lambda)}\). Any simple highest weight
   \(\mathfrak{g}\)-module has the form \(L(\lambda)\) for some unique \(\lambda
   \in \mathfrak{h}^*\).
 \end{proposition}
 
 \begin{proof}
   Let \(N \subset M(\lambda)\) be a submodule and take any nonzero \(n \in N\).
   Because of Proposition~\ref{thm:high-weight-mod-is-weight-mod}, we know there
   are \(\mu_1, \ldots, \mu_r \in \mathfrak{h}^*\) and nonzero \(m_i \in
   M(\lambda)_{\mu_i}\) such that \(n = m_1 + \cdots + m_r\). We want to show
   \(m_i \in N\) for all \(i\).
 
   Fix some \(H_2 \in \mathfrak{h}\) such that \(\mu_1(H_2) \ne \mu_2(H_2)\).
   Then
   \[
     m_1
     - \frac{(\mu_3 - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} \cdot m_3
     - \cdots
     - \frac{(\mu_r - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} \cdot m_r
     = \left( 1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) \cdot n
     \in N
   \]
 
   Now take \(H_3 \in \mathfrak{h}\) such that \(\mu_1(H_3) \ne \mu_3(H_3)\). By
   applying the same procedure again we get
   \begin{multline*}
     m_1
     -
     \frac{(\mu_4 - \mu_3)(H_3) \cdot (\mu_4 - \mu_1)(H_2)}
          {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} \cdot m_4
     - \cdots -
     \frac{(\mu_r - \mu_3)(H_3) \cdot (\mu_r - \mu_1)(H_2)}
          {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} \cdot m_r \\
     =
     \left(1 - \frac{H_3 - \mu_1(H_3)}{(\mu_3 - \mu_1)(H_3)} \right)
     \left(1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) \cdot n
     \in N
   \end{multline*}
 
   By applying the same procedure over and over again we can see that \(m_1 = u
   \cdot n \in N\) for some \(u \in \mathcal{U}(\mathfrak{g})\). Furthermore, if
   we reproduce all this for \(m_2 + \cdots + m_r = n - m_1 \in N\) we get that
   \(m_2 \in N\). All in all we find \(m_1, \ldots, m_r \in N\). Hence
   \[
     N = \bigoplus_\mu N_\mu = \bigoplus_\mu M(\lambda)_\mu \cap N
   \]
 
   Since \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot m^+\), if \(N\) is a
   proper submodule then \(m^+ \notin N\). 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. This implies
   \(N(\lambda)\) is the unique maximal submodule of \(M(\lambda)\) and
   \(L(\lambda) = \sfrac{M(\lambda)}{N(\lambda)}\) is its unique simple
   quotient.
 \end{proof}
 
 \begin{corollary}\label{thm:classification-of-simple-high-weight-mods}
   Let \(M\) be a simple weight \(\mathfrak{g}\)-module of weight \(\lambda\).
   Then \(M \cong L(\lambda)\).
 \end{corollary}
 
 We thus know that \(L(\lambda)\) is the only possible candidate for the
 \(\mathfrak{g}\)-module \(M\) in the statement of
 Theorem~\ref{thm:dominant-weight-theo}. We should also note that our past
 examples indicate that \(L(\lambda)\) does fulfill its required role.
 Indeed\dots
 
 \begin{example}\label{ex:sl2-verma-quotient}
   Consider the \(\mathfrak{sl}_2(K)\) module \(M(2)\) as described in
   Example~\ref{ex:sl2-verma}. We can see from Example~\ref{ex:verma-is-not-irr}
   that \(N(2) = \bigoplus_{k \ge 3} K f^k \cdot m^+\), so that \(L(2)\) is the
   \(3\)-dimensional simple \(\mathfrak{sl}_2(K)\)-module -- i.e. the
   finite-dimensional simple module with highest weight \(2\) constructed in
   chapter~\ref{ch:sl3}.
 \end{example}
 
 All its left to prove the Highest Weight Theorem is verifying that the
 situation encountered in Example~\ref{ex:sl2-verma-quotient} holds for any
 dominant integral \(\lambda \in P^+\). In other words, we need to show\dots
 
 \begin{proposition}\label{thm:verma-is-finite-dim}
   If \(\mathfrak{g}\) is semisimple and \(\lambda\) is dominant integral then
   the unique simple quotient \(L(\lambda)\) 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 \(L(\lambda)\) is stable under the natural action of the Weyl group
 \(W\) on \(\mathfrak{h}^*\). One can then show that the every weight
 of \(L(\lambda)\) is conjugate to a single dominant integral weight of
 \(L(\lambda)\), and that the set of dominant integral weights of \(L(\lambda)\)
 is finite. Since \(W\) is finitely generated, this implies the set of
 weights of the unique simple quotient of \(M(\lambda)\) is finite. But
 each weight space is finite-dimensional. Hence so is the simple quotient
 \(L(\lambda)\).
 
 We refer the reader to \cite[ch. 21]{humphreys} for further details. We are now
 ready to prove the Highest Weight Theorem.
 
 \begin{proof}[Proof of Theorem~\ref{thm:dominant-weight-theo}]
   We begin by the ``existence'' part of the theorem. Let \(\lambda\) be a
   dominant integral weight of \(\mathfrak{g}\). Since \(\dim L(\lambda) <
   \infty\), all its left is to show that \(M = L(\lambda)\) is indeed a highest
   weight module of highest weight \(\lambda\). It is clear from the definitions
   that \(m^+ + N(\lambda) \in L(\lambda)_\lambda\) is singular and generates
   all of \(L(\lambda)\). Hence it suffices to show that \(m^+ + N(\lambda)\) is
   nonzero. But this is the same as checking that \(m^+ \notin N(\lambda)\),
   which is also clear from the previous definitions. As for the uniqueness of
   \(M\), it suffices to apply
   Corollary~\ref{thm:classification-of-simple-high-weight-mods}.
 \end{proof}
 
 We would now like to conclude this chapter by describing the situation where
 \(\lambda \notin P^+\). We begin by pointing out that
 Proposition~\ref{thm:verma-is-finite-dim} fails in the general setting. For
 instance, consider\dots
 
 \begin{example}\label{ex:antidominant-verma}
   The action of \(\mathfrak{sl}_2(K)\) on \(M(-4)\) is given by the following
   diagram. In general, it is possible to check using formula
   (\ref{eq:sl2-verma-formulas}) that \(e\) always maps \(f^{k + 1} \cdot m^+\)
   to a nonzero multiple of \(f^k \cdot m^+\), so we can see that \(M(-4)\) has
   no proper submodules, \(N(-4) = 0\) and thus \(L(-4) \cong M(-4)\).
   \begin{center}
     \begin{tikzcd}
       \cdots         \rar[bend left=60]{-28}
       & M(-4)_{-10}  \rar[bend left=60]{-18} \lar[bend left=60]{1}
       & M(-4)_{-8}   \rar[bend left=60]{-10} \lar[bend left=60]{1}
       & M(-4)_{-6}   \rar[bend left=60]{-4}  \lar[bend left=60]{1}
       & M(-4)_{-4}                           \lar[bend left=60]{1}
     \end{tikzcd},
   \end{center}
 \end{example}
 
 While \(L(\lambda)\) is always a highest weight module of highest weight
 \(\lambda\), we can easily see that if \(\lambda \notin P^+\) then
 \(L(\lambda)\) is infinite-dimensional. Indeed, this is precisely the
 counterpositive of Proposition~\ref{thm:highes-weight-of-fin-dim-is-dominant}!
 If \(\lambda = k_1 \beta_1 + \cdots + k_r \beta_r \in P\) is integral and \(k_i
 < 0\) for all \(i\), then one is additionally able to show that \(M(\lambda)
 \cong L(\lambda)\) as in Example~\ref{ex:antidominant-verma}. Verma modules can
 thus serve as examples of infinite-dimensional simple modules.
 
 In the next chapter we expand our previous results by exploring the question:
 what are \emph{all} the infinite-dimensional simple \(\mathfrak{g}\)-modules?