lie-algebras-and-their-representations

 \chapter{Simple Weight Modules}\label{ch:mathieu}
 
 In this chapter we will expand our results on finite-dimensional simple modules
 of semisimple Lie algebras by considering \emph{infinite-dimensional}
 \(\mathfrak{g}\)-modules, which introduces numerous complications to our
 analysis.
 
 For instance, in the infinite-dimensional setting we can no longer take
 complete-reducibility for granted. Indeed, we have seen that even if
 \(\mathfrak{g}\) is a semisimple Lie algebra, there are infinite-dimensional
 \(\mathfrak{g}\)-modules which are not semisimple. For a counterexample look no
 further than Example~\ref{ex:regular-mod-is-not-semisimple}: the regular
 \(\mathfrak{g}\)-module \(\mathcal{U}(\mathfrak{g})\) is never semisimple.
 Nevertheless, for simplicity -- or shall we say \emph{semisimplicity} -- we
 will focus exclusively on \emph{semisimple} \(\mathfrak{g}\)-modules. Our
 strategy is, once again, that of classifying simple modules. The regular
 \(\mathfrak{g}\)-module hides further unpleasant surprises, however: recall
 from Example~\ref{ex:regular-mod-is-not-weight-mod} that
 \[
   \bigoplus_\lambda \mathcal{U}(\mathfrak{g})_\lambda
   = 0
   \subsetneq \mathcal{U}(\mathfrak{g})
 \]
 and the weight space decomposition fails for \(\mathcal{U}(\mathfrak{g})\).
 
 Indeed, our proof of the weight space decomposition in the finite-dimensional
 case relied heavily in the simultaneous diagonalization of commuting operators
 in a finite-dimensional space. Even if we restrict ourselves to simple modules,
 there is still a diverse spectrum of counterexamples to
 Corollary~\ref{thm:finite-dim-is-weight-mod} in the infinite-dimensional
 setting. For instance, any \(\mathfrak{g}\)-module \(M\) whose restriction to
 \(\mathfrak{h}\) is a free module satisfies \(M_\lambda = 0\) for all
 \(\lambda\) as in Example~\ref{ex:regular-mod-is-not-weight-mod}. These are
 called \emph{\(\mathfrak{h}\)-free \(\mathfrak{g}\)-modules}, and rank \(1\)
 simple \(\mathfrak{h}\)-free \(\mathfrak{sp}_{2 n}(K)\)-modules where first
 classified by Nilsson in \cite{nilsson}. Dimitar's construction of the so
 called \emph{exponential tensor \(\mathfrak{sl}_n(K)\)-modules} in
 \cite{dimitar-exp} is also an interesting source of counterexamples.
 
 Since the weight space decomposition was perhaps the single most instrumental
 ingredient of our previous analysis, it is only natural to restrict ourselves
 to the case it holds. This brings us to the following definition.
 
 \begin{definition}\label{def:weight-mod}\index{\(\mathfrak{g}\)-module!weight modules}\index{weights!weight modules}\index{\(\mathfrak{g}\)-module!(essential) support}
   A \(\mathfrak{g}\)-module \(M\) is called a \emph{weight
   \(\mathfrak{g}\)-module} if \(M = \bigoplus_{\lambda \in \mathfrak{h}^*}
   M_\lambda\) and \(\dim M_\lambda < \infty\) for all \(\lambda \in
   \mathfrak{h}^*\). The \emph{support of \(M\)} is the set
   \(\operatorname{supp} M = \{\lambda \in \mathfrak{h}^* : M_\lambda \ne 0\}\).
 \end{definition}
 
 \begin{example}
   Corollary~\ref{thm:finite-dim-is-weight-mod} is equivalent to the fact that
   every finite-dimensional module of a semisimple Lie algebra is a weight
   module. More generally, every finite-dimensional simple module of a reductive
   Lie algebra is a weight module.
 \end{example}
 
 \begin{example}\label{ex:reductive-alg-equivalence}
   We have seen that every finite-dimensional \(\mathfrak{g}\)-module is a
   weight module for semisimple \(\mathfrak{g}\). In particular, if
   \(\mathfrak{g}\) is finite-dimensional then the adjoint
   \(\mathfrak{g}\)-module \(\mathfrak{g}\) is a weight module. More generally,
   a finite-dimensional Lie algebra \(\mathfrak{g}\) is reductive if, and only
   if the adjoint \(\mathfrak{g}\)-module \(\mathfrak{g}\) is a weight module,
   in which case its weight spaces are given by the root spaces of
   \(\mathfrak{g}\)
 \end{example}
 
 \begin{example}
   Proposition~\ref{thm:high-weight-mod-is-weight-mod} is equivalent to the fact
   that any highest weight \(\mathfrak{g}\)-module \(M\) of highest weight
   \(\lambda\) is a weight module whose support is contained in \(\lambda +
   \mathbb{N} \Delta^- = \{\lambda - k_n \alpha_1 - \cdots - k_n \alpha_n :
   \alpha_i \in \Delta^+, k_i \in \mathbb{Z}, k_i \ge 0\}\). In particular,
   Verma modules are weight modules.
 \end{example}
 
 \begin{example}\label{ex:submod-is-weight-mod}
   Proposition~\ref{thm:max-verma-submod-is-weight} implies that the unique
   maximal submodule \(N(\lambda)\) of \(M(\lambda)\) is a weight module. In
   fact, the proof of Proposition~\ref{thm:max-verma-submod-is-weight} can be
   generalized to show that every submodule \(N \subset M\) of a weight module
   \(M\) is a weight module, and \(N_\lambda = M_\lambda \cap N\) for all
   \(\lambda \in \mathfrak{h}^*\).
 \end{example}
 
 \begin{example}\label{ex:quotient-is-weight-mod}
   Given a weight module \(M\), a submodule \(N \subset M\) and \(\lambda \in
   \mathfrak{h}^*\), it is clear that \(\mfrac{M_\lambda}{N} \subset
   \left(\mfrac{M}{N}\right)_\lambda\). In addition, \(\mfrac{M}{N} =
   \bigoplus_{\lambda \in \mathfrak{h}^*} \mfrac{M_\lambda}{N}\). Hence
   \(\mfrac{M}{N}\) is weight \(\mathfrak{g}\)-module with
   \(\left(\mfrac{M}{N}\right)_\lambda = \mfrac{M_\lambda}{N} \cong
   \mfrac{M_\lambda}{N_\lambda}\).
 \end{example}
 
 \begin{example}\label{ex:tensor-prod-of-weight-is-weight}
   Let \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\) be Lie algebras, \(M_1\) be a
   weight \(\mathfrak{g}_1\)-module and \(M_2\) a weight
   \(\mathfrak{g}_2\)-module. Recall from Example~\ref{ex:cartan-direct-sum}
   that if \(\mathfrak{h}_i \subset \mathfrak{g}_i\) are Cartan subalgebras then
   \(\mathfrak{h} = \mathfrak{h}_1 \oplus \mathfrak{h}_2\) is a Cartan
   subalgebra of \(\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2\) with
   \(\mathfrak{h}^* = \mathfrak{h}_1^* \oplus \mathfrak{h}_2^*\). In this
   setting, one can readily check that \(M_1 \otimes M_2\) is a weight
   \(\mathfrak{g}\)-module with
   \[
     (M_1 \otimes M_2)_{\lambda_1 + \lambda_2}
     = (M_1)_{\lambda_1} \otimes (M_2)_{\lambda_2}
   \]
   for all \(\lambda_i \in \mathfrak{h}_i^*\) and \(\operatorname{supp} (M_1
   \otimes M_2) = \operatorname{supp} M_1 \oplus \operatorname{supp} M_2 = \{
   \lambda_1 + \lambda_2 : \lambda_i \in \operatorname{supp} M_i \subset
   \mathfrak{h}_i^*\}\).
 \end{example}
 
 \begin{example}\label{thm:simple-weight-mod-is-tensor-prod}
   Let \(\mathfrak{g} = \mathfrak{z} \oplus \mathfrak{s}_1 \oplus \cdots \oplus
   \mathfrak{s}_r\) be a reductive Lie algebra, where \(\mathfrak{z}\) is the
   center of \(\mathfrak{g}\) and \(\mathfrak{s}_1, \ldots, \mathfrak{s}_r\) are
   its simple components. As in
   Example~\ref{ex:all-simple-reps-are-tensor-prod}, any simple weight
   \(\mathfrak{g}\)-module \(M\) can be decomposed as
   \[
     M \cong Z \otimes M_1 \otimes \cdots \otimes M_r
   \]
   where \(Z\) is a \(1\)-dimensional representation of \(\mathfrak{z}\) and
   \(M_i\) is a simple weight \(\mathfrak{s}_i\)-module. The modules \(Z\) and
   \(M_i\) are uniquely determined up to isomorphism.
 \end{example}
 
 \begin{example}\label{ex:adjoint-action-in-universal-enveloping-is-weight}
   We would like to show that the requirement of finite-dimensionality in
   Definition~\ref{def:weight-mod} is not redundant. Let \(\mathfrak{g}\) be a
   finite-dimensional reductive Lie algebra and consider the adjoint
   \(\mathfrak{g}\)-module \(\mathcal{U}(\mathfrak{g})\) -- where \(X \in
   \mathfrak{g}\) acts by taking commutators. Given \(\alpha \in Q\), a simple
   computation shows \(K \langle X_1 \cdots X_n H_1 \cdots H_m : X_i \in
   \mathfrak{g}_{\alpha_i}, H_i \in \mathfrak{h}, \alpha_i \in \Delta, \alpha =
   \alpha_1 + \cdots + \alpha_n \rangle \subset
   \mathcal{U}(\mathfrak{g})_\alpha\). The PBW Theorem and
   Example~\ref{ex:reductive-alg-equivalence} thus imply that
   \(\mathcal{U}(\mathfrak{g}) = \bigoplus_{\alpha \in Q}
   \mathcal{U}(\mathfrak{g})_\alpha\) where \(\mathcal{U}(\mathfrak{g})_\alpha =
   K \langle X_1 \cdots X_n H_1 \cdots H_m : X_i \in \mathfrak{g}_{\alpha_i},
   H_i \in \mathfrak{h}, \alpha_i \in \Delta, \alpha = \alpha_1 + \cdots +
   \alpha_n \rangle\). However, \(\dim \mathcal{U}(\mathfrak{g})_\alpha =
   \infty\). For instance, \(\mathcal{U}(\mathfrak{g})_0\) is \emph{precisely}
   the commutator of \(\mathfrak{h}\) in \(\mathcal{U}(\mathfrak{g})\), which
   contains \(\mathcal{U}(\mathfrak{h})\) and is therefore infinite-dimensional.
 \end{example}
 
 \begin{note}
   We should stress that the weight spaces \(M_\lambda \subset M\) of a given
   weight \(\mathfrak{g}\)-module \(M\) are \emph{not}
   \(\mathfrak{g}\)-submodules. Nevertheless, \(M_\lambda\) is a
   \(\mathfrak{h}\)-submodule. More generally, \(M_\lambda\) is a
   \(\mathcal{U}(\mathfrak{g})_0\)-submodule, where
   \(\mathcal{U}(\mathfrak{g})_0\) is the centralizer of \(\mathfrak{h}\) in
   \(\mathcal{U}(\mathfrak{g})\) -- which coincides with the weight space of \(0
   \in \mathfrak{h}^*\) in the adjoint \(\mathfrak{g}\)-module
   \(\mathcal{U}(\mathfrak{g})\), as seen in
   Example~\ref{ex:adjoint-action-in-universal-enveloping-is-weight}.
 \end{note}
 
 A particularly well behaved class of examples are the so called
 \emph{bounded} modules.
 
 \begin{definition}\index{\(\mathfrak{g}\)-module!bounded modules}\index{\(\mathfrak{g}\)-module!(essential) support}
   A weight \(\mathfrak{g}\)-module \(M\) is called \emph{bounded} if \(\dim
   M_\lambda\) is bounded. The lowest upper bound \(\deg M\) for \(\dim
   M_\lambda\) is called \emph{the degree of \(M\)}. The \emph{essential
   support} of \(M\) is the set \(\operatorname{supp}_{\operatorname{ess}} M =
   \{ \lambda \in \mathfrak{h}^* : \dim M_\lambda = \deg M \}\).
 \end{definition}
 
 \begin{example}\label{ex:supp-ess-of-tensor-is-product}
   Let \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\) be Lie algebras with Cartan
   subalgebras \(\mathfrak{h}_i \subset \mathfrak{g}_i\) and take \(\mathfrak{g}
   = \mathfrak{g}_1 \oplus \mathfrak{g}_2\). Given bounded
   \(\mathfrak{g}_i\)-modules \(M_i\), it follows from
   Example~\ref{ex:tensor-prod-of-weight-is-weight} that \(M_1 \otimes M_2\) is
   a bounded \(\mathfrak{g}\)-module with \(\deg M_1 \otimes M_2 = \deg M_1
   \cdot \deg M_2\) and
   \[
     \operatorname{supp}_{\operatorname{ess}} (M_1 \otimes M_2)
     = \operatorname{supp}_{\operatorname{ess}} M_1 \oplus
       \operatorname{supp}_{\operatorname{ess}} M_2
     = \{
         \lambda_1 + \lambda_2 : \lambda_i \in
         \operatorname{supp}_{\operatorname{ess}} M_i \subset \mathfrak{h}_i^*
       \}
   \]
 \end{example}
 
 \begin{example}\label{ex:laurent-polynomial-mod}
   There is a natural action of \(\mathfrak{sl}_2(K)\) on the space \(K[x,
   x^{-1}]\) of Laurent polynomials, given by the formulas in
   (\ref{eq:laurent-polynomials-cusp-mod}). One can quickly verify \(K[x,
   x^{-1}]_{2 k} = K x^k\) and \(K[x, x^{-1}]_\lambda = 0\) for any \(\lambda
   \notin 2 \mathbb{Z}\), so that \(K[x, x^{-1}] = \bigoplus_{k \in \mathbb{Z}}
   K x^k\) is a degree \(1\) bounded weight \(\mathfrak{sl}_2(K)\)-module. It
   follows from the remark at the end of Example~\ref{ex:submod-is-weight-mod}
   that any nonzero submodule \(N \subset K[x, x^{-1}]\) must contain a
   monomial \(x^k\). But since the operators \(-\frac{\mathrm{d}}{\mathrm{d}x} +
   \frac{x^{-1}}{2}, x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x}{2} : K[x,
   x^{-1}] \to K[x, x^{-1}]\) are both injective, this implies all other
   monomials can be found in \(N\) by successively applying \(f\) and \(e\).
   Hence \(N = K[x, x^{-1}]\) and \(K[x, x^{-1}]\) is a simple module.
   \begin{align}\label{eq:laurent-polynomials-cusp-mod}
     e \cdot p
     & = \left( x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x}{2} \right) p &
     f \cdot p
     & = \left(- \frac{\mathrm{d}}{\mathrm{d}x} + \frac{x^{-1}}{2} \right) p &
     h \cdot p
     & = 2 x \frac{\mathrm{d}}{\mathrm{d}x} p
   \end{align}
 \end{example}
 
 Notice that the support of \(K[x, x^{-1}]\) is the trivial \(2
 \mathbb{Z}\)-coset \(0 + 2 \mathbb{Z}\). This is representative of the general
 behavior in the following sense: if \(M\) is a simple weight
 \(\mathfrak{g}\)-module, since \(M[\lambda] = \bigoplus_{\alpha \in Q}
 M_{\lambda + \alpha}\) is stable under the action of \(\mathfrak{g}\) for all
 \(\lambda \in \mathfrak{h}^*\), \(\bigoplus_{\alpha \in Q} M_{\lambda +
 \alpha}\) is either \(0\) or all of \(M\). In other words, the support of a
 simple weight module is always contained in a single \(Q\)-coset.
 
 However, the behavior of \(K[x, x^{-1}]\) deviates from that of an arbitrary
 bounded \(\mathfrak{g}\)-module in the sense its essential support is
 precisely the entire \(Q\)-coset it inhabits -- i.e.
 \(\operatorname{supp}_{\operatorname{ess}} K[x, x^{-1}] = 2 \mathbb{Z}\). This
 isn't always the case. Nevertheless, in general we find\dots
 
 \begin{proposition}\label{thm:ess-supp-is-zariski-dense}
   Let \(\mathfrak{g}\) be a finite-dimensional semisimple Lie algebra and \(M\)
   be a simple infinite-dimensional bounded \(\mathfrak{g}\)-module. The
   essential support \(\operatorname{supp}_{\operatorname{ess}} M\) is
   Zariski-dense\footnote{Any choice of basis for $\mathfrak{h}^*$ induces a
   $K$-linear isomorphism $\mathfrak{h}^* \isoto K^n$. In particular, a choice
   of basis induces a unique topology in $\mathfrak{h}^*$ such that the map
   $\mathfrak{h}^* \to K^n$ is a homeomorphism onto $K^n$ with the Zariski
   topology. Any two basis induce the same topology in $\mathfrak{h}^*$, which
   we call \emph{the Zariski topology of $\mathfrak{h}^*$}.} in
   \(\mathfrak{h}^*\).
 \end{proposition}
 
 This proof was deemed too technical to be included in here, but see Proposition
 3.5 of \cite{mathieu} for the case where \(\mathfrak{g} = \mathfrak{s}\) is a
 simple Lie algebra. The general case then follows from
 Example~\ref{thm:simple-weight-mod-is-tensor-prod},
 Example~\ref{ex:supp-ess-of-tensor-is-product} and the asserting that the
 product of Zariski-dense subsets in \(K^n\) and \(K^m\) is Zariski-dense in
 \(K^{n + m} = K^n \times K^m\).
 
 We now begin a systematic investigation of the problem of classifying the
 infinite-dimensional simple weight modules of a given Lie algebra
 \(\mathfrak{g}\). As in the previous chapter, let \(\mathfrak{g}\) be a
 finite-dimensional semisimple Lie algebra. As a first approximation of a
 solution to our problem, we consider the Verma modules \(M(\lambda)\) for
 \(\lambda \in \mathfrak{h}^*\) which is not dominant integral. After all, the
 simple quotients of Verma modules form a remarkably large class of
 infinite-dimensional simple weight modules -- at least as large as
 \(\mathfrak{h}^* \setminus P^+\)! More generally, the induction functor
 \(\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} :
 \mathfrak{b}\text{-}\mathbf{Mod} \to \mathfrak{g}\text{-}\mathbf{Mod}\) has
 proven itself a powerful tool for constructing modules.
 
 We claim this is not an unmotivated guess. Specifically, there are very good
 reasons behind the choice to consider induction over the Borel subalgebra
 \(\mathfrak{b} \subset \mathfrak{g}\). First, the fact that \(\mathfrak{h}
 \subset \mathfrak{g}\) affords us great control over the weight spaces of
 \(\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} M\): by assigning a
 prescribed action of \(\mathfrak{h}\) to \(M\) we can ensure that
 \(\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} M = \bigoplus_\lambda
 (\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} M)_\lambda\). In addition, we
 have seen in the proof of Proposition~\ref{thm:high-weight-mod-is-weight-mod}
 that by requiring that the positive part of \(\mathfrak{b}\) acts on \(M\) by
 zero we can ensure that \(\dim
 (\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} M)_\lambda < \infty\). All in
 all, the nature of \(\mathfrak{b}\) affords us just enough control to guarantee
 that \(\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} M\) is a weight module
 for sufficiently well behaved \(M\).
 
 Unfortunately for us, this is still too little control: there are simple weight
 modules which are not of the form \(L(\lambda)\). More generally, we may
 consider induction over some parabolic subalgebra \(\mathfrak{p} \subset
 \mathfrak{g}\) -- i.e. some subalgebra such that \(\mathfrak{p} \supset
 \mathfrak{b}\). This leads us to the following definition.
 
 \begin{definition}\index{\(\mathfrak{g}\)-module!(generalized) Verma modules}
   Let \(\mathfrak{p} \subset \mathfrak{g}\) be a parabolic subalgebra and \(M\)
   be a simple \(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module. We
   can view \(M\) as a \(\mathfrak{p}\)-module where
   \(\mathfrak{nil}(\mathfrak{p})\) acts by zero by setting \(X \cdot m = (X +
   \mathfrak{nil}(\mathfrak{p})) \cdot m\) for all \(m \in M\) and \(X \in
   \mathfrak{p}\) -- which is the same as the \(\mathfrak{p}\)-module given by
   composing the action map \(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}
   \to \mathfrak{gl}(M)\) with the projection \(\mathfrak{p} \to
   \mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\). The module
   \(M_{\mathfrak{p}}(M) = \operatorname{Ind}_{\mathfrak{p}}^{\mathfrak{g}} M\)
   is called \emph{generalized Verma module associated with \(M\)}.
 \end{definition}
 
 \begin{example}
   It is not hard to see that
   \(\mfrac{\mathfrak{b}}{\mathfrak{nil}(\mathfrak{b})} = \mathfrak{h}\). If we
   take \(\lambda \in \mathfrak{h}^*\) and let \(K m^+\) be the
   \(1\)-dimensional \(\mathfrak{h}\)-module where \(\mathfrak{h}\) acts by
   \(\lambda\) then \(M(\lambda) = M_{\mathfrak{b}}(K m^+)\).
 \end{example}
 
 As promised, \(M_{\mathfrak{p}}(M)\) is generally well behaved for well behaved
 \(M\). In particular, if \(M\) is highest weight
 \(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module then
 \(M_{\mathfrak{p}}(M)\) is also a highest weight \(\mathfrak{g}\)-module, and
 if \(M\) is a weight
 \(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module then
 \(M_{\mathfrak{p}}(M)\) is a weight module with \(M_{\mathfrak{p}}(M)_\lambda =
 \sum_{\alpha + \mu = \lambda} \mathcal{U}(\mathfrak{g})_\alpha
 \otimes_{\mathcal{U}(\mathfrak{p})} M_\mu\), -- see Lemma 1.1 of \cite{mathieu}
 for a full proof. However, \(M_{\mathfrak{p}}(M)\) is not simple in general.
 Indeed, regular Verma modules not necessarily simple. This issue may be dealt
 with by passing to the simple quotients of \(M_{\mathfrak{p}}(M)\).
 
 Let \(M\) be a simple weight
 \(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module. As it turns out,
 the situation encountered in Proposition~\ref{thm:max-verma-submod-is-weight}
 is also verified in the general setting. Namely, since \(M_{\mathfrak{p}}(M)\)
 is generated by \(K \otimes_{\mathcal{U}(\mathfrak{p})} M = \bigoplus_{\lambda
 \in Q + \operatorname{supp} M} M_{\mathfrak{p}}(M)_\lambda\), it follows that
 any proper submodule of \(M_{\mathfrak{p}}(M)\) is contained in
 \(\bigoplus_{\lambda \notin Q + \operatorname{supp} M}
 M_{\mathfrak{p}}(M)_\lambda\). The sum \(N_{\mathfrak{p}}(M)\) of all such
 submodules is thus the unique maximal submodule of \(M_{\mathfrak{p}}(M)\) and
 \(L_{\mathfrak{p}}(M) = \mfrac{M_{\mathfrak{p}}(M)}{N_{\mathfrak{p}}(M)}\) is
 its unique simple quotient -- again, we refer the reader to \cite{mathieu} for
 a complete proof. This leads us to the following definition.
 
 \begin{definition}\index{\(\mathfrak{g}\)-module!parabolic induced modules}\index{\(\mathfrak{g}\)-module!cuspidal modules}
   A simple weight \(\mathfrak{g}\)-module is called \emph{parabolic induced} if
   it is isomorphic to \(L_{\mathfrak{p}}(M)\) for some proper parabolic
   subalgebra \(\mathfrak{p} \subsetneq \mathfrak{g}\) and some simple weight
   \(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module \(M\). A
   \emph{cuspidal \(\mathfrak{g}\)-module} is a simple weight
   \(\mathfrak{g}\)-module which is \emph{not} parabolic induced.
 \end{definition}
 
 The first breakthrough regarding our classification problem was given by
 Fernando in his now infamous paper \citetitle{fernando} \cite{fernando}, where
 he proved that every simple weight \(\mathfrak{g}\)-module is parabolic
 induced by a cuspidal module.
 
 \begin{theorem}[Fernando]
   Any simple weight \(\mathfrak{g}\)-module is isomorphic to
   \(L_{\mathfrak{p}}(M)\) for some parabolic subalgebra \(\mathfrak{p} \subset
   \mathfrak{g}\) and some cuspidal
   \(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module \(M\).
 \end{theorem}
 
 We should point out that the relationship between simple weight
 \(\mathfrak{g}\)-modules and pairs \((\mathfrak{p}, M)\) is not one-to-one.
 Nevertheless, this relationship is well understood. Namely, Fernando himself
 established\dots
 
 \begin{proposition}[Fernando]
   Given a parabolic subalgebra \(\mathfrak{p} \subset \mathfrak{g}\), there
   exists a basis \(\Sigma\) for \(\Delta\) such that \(\Sigma \subset
   \Delta_{\mathfrak{p}} \subset \Delta\), where \(\Delta_{\mathfrak{p}}\)
   denotes the set of roots of \(\mathfrak{p}\). Furthermore, if \(\mathfrak{p}'
   \subset \mathfrak{g}\) is another parabolic subalgebra, \(M\) is a cuspidal
   \(\mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}\)-module and \(N\) is a
   cuspidal \(\mfrac{\mathfrak{p}'}{\mathfrak{nil}(\mathfrak{p}')}\)-module then
   \(L_{\mathfrak{p}}(M) \cong L_{\mathfrak{p}'}(N)\) if, and only if
   \(\mathfrak{p}' = \twisted{\mathfrak{p}}{\sigma}\) and \(M \cong
   \twisted{N}{\sigma}\) as \(\mathfrak{p}\)-modules for some\footnote{Here
   $\twisted{\mathfrak{p}}{\sigma}$ denotes the image of $\mathfrak{p}$ under
   the automorphism of $\sigma : \mathfrak{g} \to \mathfrak{g}$ given by the
   canonical action of $W$ on $\mathfrak{g}$ and $\twisted{N}{\sigma}$ is the
   $\mathfrak{p}$-module given by composing the map $\mathfrak{p}' \to
   \mathfrak{gl}(N)$ with the restriction $\sigma\!\restriction_{\mathfrak{p}} :
   \mathfrak{p} \to \mathfrak{p}'$.} \(\sigma \in W_M\), where
   \[
     W_M
     = \langle
       \sigma_\beta : \beta \in \Sigma, H_\beta + \mathfrak{nil}(\mathfrak{p})
       \ \text{is central in}\ \mfrac{\mathfrak{p}}{\mathfrak{nil}(\mathfrak{p})}
       \ \text{and}\ H_\beta\ \text{acts on \(M\) as a positive integer}
       \rangle
     \subset W
   \]
 \end{proposition}
 
 \begin{note}
   The definition of the subgroup \(W_M \subset W\) is independent of the choice
   of basis \(\Sigma\).
 \end{note}
 
 As a first consequence of Fernando's Theorem, we provide two alternative
 characterizations of cuspidal modules.
 
 \begin{corollary}[Fernando]\label{thm:cuspidal-mod-equivs}
   Let \(M\) be a simple weight \(\mathfrak{g}\)-module. The following
   conditions are equivalent.
   \begin{enumerate}
     \item \(M\) is cuspidal.
     \item \(F_\alpha\) acts injectively on \(M\) for all
       \(\alpha \in \Delta\).
     \item The support of \(M\) is precisely one \(Q\)-coset.
   \end{enumerate}
 \end{corollary}
 
 \begin{example}
   As noted in Example~\ref{ex:laurent-polynomial-mod}, the element \(f \in
   \mathfrak{sl}_2(K)\) acts injectively on the space of Laurent polynomials.
   Hence \(K[x, x^{-1}]\) is a cuspidal \(\mathfrak{sl}_2(K)\)-module.
 \end{example}
 
 Having reduced our classification problem to that of classifying cuspidal
 modules, we are now faced the daunting task of actually classifying them.
 Historically, this was first achieved by Olivier Mathieu in the early 2000's in
 his paper \citetitle{mathieu} \cite{mathieu}. To do so, Mathieu introduced new
 tools which have since proved themselves remarkably useful throughout the
 field, known as\dots
 
 \section{Coherent Families}
 
 We begin our analysis with a simple question: how to do we go about
 constructing cuspidal modules? Specifically, given a cuspidal
 \(\mathfrak{g}\)-module, how can we use it to produce new cuspidal modules? To
 answer this question, we look back at the single example of a cuspidal module
 we have encountered so far: the \(\mathfrak{sl}_2(K)\)-module \(K[x, x^{-1}]\)
 of Laurent polynomials -- i.e. Example~\ref{ex:laurent-polynomial-mod}.
 
 Our first observation is that \(\mathfrak{sl}_2(K)\) acts on \(K[x, x^{-1}]\)
 via differential operators. In other words, the action map
 \(\mathcal{U}(\mathfrak{sl}_2(K)) \to \operatorname{End}(K[x, x^{-1}])\)
 factors through the inclusion of the algebra \(\operatorname{Diff}(K[x,
 x^{-1}]) = K\left[x, x^{-1}, \frac{\mathrm{d}}{\mathrm{d}x}\right]\) of
 differential operators in \(K[x, x^{-1}]\).
 \begin{center}
   \begin{tikzcd}
     \mathcal{U}(\mathfrak{sl}_2(K))   \rar &
     \operatorname{Diff}(K[x, x^{-1}]) \rar &
     \operatorname{End}(K[x, x^{-1}])
   \end{tikzcd}
 \end{center}
 
 The space \(K[x, x^{-1}]\) can be regarded as a \(\operatorname{Diff}(K[x,
 x^{-1}])\)-module in the natural way, and we can produce new
 \(\operatorname{Diff}(K[x, x^{-1}])\)-modules by twisting \(K[x, x^{-1}]\) by
 automorphisms of \(\operatorname{Diff}(K[x, x^{-1}])\). For example, given
 \(\lambda \in K\) we may take the automorphism
 \begin{align*}
   \varphi_\lambda : \operatorname{Diff}(K[x, x^{-1}]) &
   \to \operatorname{Diff}(K[x, x^{-1}]) \\
   x & \mapsto x \\
   x^{-1} & \mapsto x^{-1} \\
   \frac{\mathrm{d}}{\mathrm{d} x} & \mapsto \frac{\mathrm{d}}{\mathrm{d} x} +
   \frac{\lambda}{2} x^{-1}
 \end{align*}
 and consider the twisted module \(\twisted{K[x, x^{-1}]}{\varphi_\lambda} =
 K[x, x^{-1}]\), where some operator \(P \in \operatorname{Diff}(K[x, x^{-1}])\)
 acts as \(\varphi_\lambda(P)\).
 
 By composing the action map \(\operatorname{Diff}(K[x, x^{-1}]) \to
 \operatorname{End}(\twisted{K[x, x^{-1}]}{\varphi_\lambda})\) with the
 homomorphism of algebras \(\mathcal{U}(\mathfrak{sl}_2(K)) \to
 \operatorname{Diff}(K[x, x^{-1}])\) we can give \(\twisted{K[x,
 x^{-1}]}{\varphi_\lambda}\) the structure of an \(\mathfrak{sl}_2(K)\)-module.
 Diagrammatically, we have
 \begin{center}
   \begin{tikzcd}
     \mathcal{U}(\mathfrak{sl}_2(K))   \rar                  &
     \operatorname{Diff}(K[x, x^{-1}]) \rar{\varphi_\lambda} &
     \operatorname{Diff}(K[x, x^{-1}]) \rar                  &
     \operatorname{End}(K[x, x^{-1}])
   \end{tikzcd},
 \end{center}
 where the maps \(\mathcal{U}(\mathfrak{sl}_2(K)) \to \operatorname{Diff}(K[x,
 x^{-1}])\) and \(\operatorname{Diff}(K[x, x^{1}]) \to \operatorname{End}(K[x,
 x^{-1}])\) are the ones from the previous diagram.
 
 Explicitly, we find that the action of \(\mathfrak{sl}_2(K)\) on
 \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is given by
 \begin{align*}
   p & \overset{e}{\mapsto}
   \left(
   x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \frac{1 + \lambda}{2} x
   \right) p &
   p & \overset{f}{\mapsto}
   \left(
   - \frac{\mathrm{d}}{\mathrm{d}x} + \frac{1 - \lambda}{2} x^{-1}
   \right) p &
   p & \overset{h}{\mapsto}
   \left( 2 x \frac{\mathrm{d}}{\mathrm{d}x} + \lambda \right) p,
 \end{align*}
 so we can see \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}_{2 k +
 \frac{\lambda}{2}} = K x^k\) for all \(k \in \mathbb{Z}\) and \(\twisted{K[x,
 x^{-1}]}{\varphi_\lambda}_\mu = 0\) for all other \(\mu \in \mathfrak{h}^*\).
 
 Hence \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is a degree \(1\) bounded
 \(\mathfrak{sl}_2(K)\)-module with \(\operatorname{supp} \twisted{K[x,
 x^{-1}]}{\varphi_\lambda} = \frac{\lambda}{2} + 2 \mathbb{Z}\). One can also
 quickly check that if \(\lambda \notin 1 + 2 \mathbb{Z}\) then \(e\) and \(f\)
 act injectively in \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\), so that
 \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is simple. In particular, if
 \(\lambda, \mu \notin 1 + 2 \mathbb{Z}\) with \(\lambda \notin \mu + 2
 \mathbb{Z}\) then \(\twisted{K[x, x^{-1}]}{\varphi_\lambda}\) and
 \(\twisted{K[x, x^{-1}]}{\varphi_\mu}\) are non-isomorphic cuspidal
 \(\mathfrak{sl}_2(K)\)-modules, since their supports differ. These cuspidal
 modules can be ``glued together'' in a \emph{monstrous concoction} by summing
 over \(\lambda \in K\), as in
 \[
   \mathcal{M}
   = \bigoplus_{\lambda + 2 \mathbb{Z} \in \mfrac{K}{2 \mathbb{Z}}}
     \twisted{K[x, x^{-1}]}{\varphi_\lambda},
 \]
 
 To a distracted spectator, \(\mathcal{M}\) may look like just another,
 innocent, \(\mathfrak{sl}_2(K)\)-module. However, the attentive reader may have
 already noticed some of the its bizarre features, most noticeable of which is
 the fact that \(\mathcal{M}\) is very big. In fact, \(\mathcal{M}\) is as big a
 degree \(1\) bounded module gets: \(\operatorname{supp} \mathcal{M}
 = \operatorname{supp}_{\operatorname{ess}} \mathcal{M}\) is the entirety of
 \(\mathfrak{h}^*\). This may look very alien the reader familiarized with the
 finite-dimensional setting, where the configuration of weights is very rigid.
 For this reason, \(\mathcal{M}\) deserves to be called ``a monstrous
 concoction''.
 
 On a perhaps less derogatory note, \(\mathcal{M}\) also deserves to be called
 \emph{a family}. This is because \(\mathcal{M}\) consists of lots of smaller
 cuspidal modules which fit together inside of it in a \emph{coherent} fashion.
 Mathieu's ingenious breakthrough was the realization that \(\mathcal{M}\) is a
 particular example of a more general pattern, which he named \emph{coherent
 families}.
 
 \begin{definition}\index{coherent family}
   A \emph{coherent family \(\mathcal{M}\) of degree \(d\)} is a weight
   \(\mathfrak{g}\)-module \(\mathcal{M}\) such that
   \begin{enumerate}
     \item \(\dim \mathcal{M}_\lambda = d\) for \emph{all} \(\lambda \in
       \mathfrak{h}^*\) -- i.e. \(\operatorname{supp}_{\operatorname{ess}}
       \mathcal{M} = \mathfrak{h}^*\).
     \item For any \(u \in \mathcal{U}(\mathfrak{g})\) in the centralizer
       \(\mathcal{U}(\mathfrak{g})_0\) of \(\mathfrak{h}\) in
       \(\mathcal{U}(\mathfrak{g})\), the map
       \begin{align*}
         \mathfrak{h}^* & \to K \\
                \lambda & \mapsto
                \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\lambda})
       \end{align*}
       is polynomial in \(\lambda\).
   \end{enumerate}
 \end{definition}
 
 \begin{example}\label{ex:sl-laurent-family}
   The module \(\mathcal{M} = \bigoplus_{\lambda + 2 \mathbb{Z} \in
   \mfrac{K}{2 \mathbb{Z}}} \twisted{K[x, x^{-1}]}{\varphi_\lambda}\) is a
   degree \(1\) coherent \(\mathfrak{sl}_2(K)\)-family.
 \end{example}
 
 \begin{example}
   Given \(\lambda \in K\), \(\mathcal{M}(\lambda) = \bigoplus_{\mu \in K} K
   x^\mu\) with
   \begin{align*}
     p & \overset{e}{\mapsto}
         \left(x^2 \frac{\mathrm{d}}{\mathrm{d}x} + \lambda x\right) p &
     p & \overset{f}{\mapsto}
         \left(-\frac{\mathrm{d}}{\mathrm{d}x} + \lambda x^{-1}\right) p &
     p & \overset{h}{\mapsto} 2 x \frac{\mathrm{d}}{\mathrm{d}x} p,
   \end{align*}
   is a degree \(1\) coherent \(\mathfrak{sl}_2(K)\)-family -- where \(x^{\pm
   1}, \sfrac{\mathrm{d}}{\mathrm{d}x} : \mathcal{M}(\lambda) \to
   \mathcal{M}(\lambda)\) are given by \(x^{\pm 1} x^\mu = x^{\mu \pm 1}\) and
   \(\sfrac{\mathrm{d}}{\mathrm{d}x} x^\mu = \mu x^{\mu - 1}\). It is easy to
   check \(\mathcal{M}\) from Example~\ref{ex:sl-laurent-family} is isomorphic
   to \(\mathcal{M}(\sfrac{1}{2})\) and \((\mathcal{M}(\sfrac{1}{2}))[0] \cong
   K[x, x^{-1}]\).
 \end{example}
 
 \begin{note}
   We would like to stress that coherent families have proven themselves useful
   for problems other than the classification of cuspidal
   \(\mathfrak{g}\)-modules. For instance, Nilsson's classification of rank 1
   \(\mathfrak{h}\)-free \(\mathfrak{sp}_{2 n}(K)\)-modules is based on the
   notion of coherent families and the so called \emph{weighting functor}.
 \end{note}
 
 Our hope is that given a cuspidal module \(M\), we can somehow fit \(M\) inside
 of a coherent \(\mathfrak{g}\)-family, such as in the case of \(K[x, x^{-1}]\)
 and \(\mathcal{M}\) from Example~\ref{ex:sl-laurent-family}. In addition, we
 hope that such coherent families are somehow \emph{uniquely determined} by
 \(M\). This leads us to the following definition.
 
 \begin{definition}\index{coherent family!coherent extension}
   Given a bounded \(\mathfrak{g}\)-module \(M\) of degree \(d\), a
   \emph{coherent extension \(\mathcal{M}\) of \(M\)} is a coherent family
   \(\mathcal{M}\) of degree \(d\) that contains \(M\) as a subquotient.
 \end{definition}
 
 Our goal is now showing that every simple bounded module has a coherent
 extension. The idea then is to classify coherent families, and classify which
 submodules of a given coherent family are actually cuspidal modules. If every
 simple bounded \(\mathfrak{g}\)-module fits inside a coherent extension, this
 would lead to classification of all cuspidal \(\mathfrak{g}\)-modules, which we
 now know is the key for the solution of our classification problem. However,
 there are some complications to this scheme.
 
 Leaving aside the question of existence for a second, we should point out that
 coherent families turn out to be rather complicated on their own. In fact they
 are too complicated to classify in general. Ideally, we would like to find
 \emph{nice} coherent extensions -- ones we can actually classify. For instance,
 we may search for \emph{irreducible} coherent extensions, which are defined as
 follows.
 
 \begin{definition}\index{coherent family!irreducible coherent family}
   A coherent family \(\mathcal{M}\) is called \emph{irreducible} if it contains
   no proper coherent subfamilies -- i.e. \(\mathcal{M}\) is a simple object in
   the full subcategory of \(\mathfrak{g}\text{-}\mathbf{Mod}\) consisting of
   coherent families. Equivalently, we call \(\mathcal{M}\) irreducible if
   \(\mathcal{M}_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module
   for some \(\lambda \in \mathfrak{h}^*\).
 \end{definition}
 
 Another natural candidate for the role of ``nice extensions'' are the
 semisimple coherent families -- i.e. families which are semisimple as
 \(\mathfrak{g}\)-modules. These turn out to be very easy to produce. Namely,
 there is a construction, known as \emph{the semisimplification of a coherent
 family}, which takes a coherent extension of \(M\) to a semisimple coherent
 extension of \(M\).
 
 % Mathieu's proof of this is somewhat profane, I don't think it's worth
 % including it in here
 % TODO: Move this somewhere else? This holds in general for weight modules
 % whose suppert is contained in a single Q-coset
 \begin{lemma}\label{thm:component-coh-family-has-finite-length}
   Given a coherent family \(\mathcal{M}\) and \(\lambda \in \mathfrak{h}^*\),
   \(\mathcal{M}[\lambda]\) has finite length as a \(\mathfrak{g}\)-module.
 \end{lemma}
 
 \begin{proposition}\index{coherent family!semisimplification}
   Let \(\mathcal{M}\) be a coherent family of degree \(d\). There exists a
   unique semisimple coherent family \(\mathcal{M}^{\operatorname{ss}}\) of
   degree \(d\) such that the composition series of
   \(\mathcal{M}^{\operatorname{ss}}[\lambda]\) is the same as that of
   \(\mathcal{M}[\lambda]\) for all \(\lambda \in \mathfrak{h}^*\), called
   \emph{the semisimplification of \(\mathcal{M}\)}.
 
   Namely, if \(\lambda \in \mathfrak{h}^*\) and \(0 = \mathcal{M}_{\lambda 0}
   \subset \mathcal{M}_{\lambda 1} \subset \cdots \subset \mathcal{M}_{\lambda
   r_\lambda} = \mathcal{M}[\lambda]\) is a composition series\footnote{Notice
   that $\mathcal{M}[\lambda] = \mathcal{M}[\mu]$ for any $\mu \in \lambda + Q$.
   Hence the sum $\bigoplus_{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q}}
   \bigoplus_i \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}$ is
   independent of the choice of representative for $\lambda + Q$ -- provided we
   choose $\mathcal{M}_{\mu i} = \mathcal{M}_{\lambda i}$ for all $\mu \in
   \lambda + Q$ and $i$.},
   \[
     \mathcal{M}^{\operatorname{ss}}
     \cong \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}}
           \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}
   \]
 \end{proposition}
 
 \begin{proof}
   The uniqueness of \(\mathcal{M}^{\operatorname{ss}}\) should be clear:
   since \(\mathcal{M}^{\operatorname{ss}}\) is semisimple, so is
   \(\mathcal{M}^{\operatorname{ss}}[\lambda]\). Hence by the Jordan-Hölder
   Theorem
   \[
     \mathcal{M}^{\operatorname{ss}}[\lambda]
     \cong
     \bigoplus_i \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}
   \]
 
   As for the existence of the semisimplification, it suffices to show
   \[
     \mathcal{M}^{\operatorname{ss}}
     = \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}}
     \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}
   \]
   is indeed a semisimple coherent family of degree \(d\).
 
   We know from Examples~\ref{ex:submod-is-weight-mod} and
   \ref{ex:quotient-is-weight-mod} that each quotient
   \(\mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}\) is a weight
   module. Hence \(\mathcal{M}^{\operatorname{ss}}\) is a weight module.
   Furthermore, given \(\mu \in \mathfrak{h}^*\)
   \[
     \mathcal{M}_\mu^{\operatorname{ss}}
     = \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}}
       \left(
       \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}
       \right)_\mu
     = \bigoplus_i
       \left(
       \mfrac{\mathcal{M}_{\mu i + 1}}{\mathcal{M}_{\mu i}}
       \right)_\mu
     \cong \bigoplus_i
       \mfrac{(\mathcal{M}_{\mu i + 1})_\mu}
             {(\mathcal{M}_{\mu i})_\mu}
   \]
 
   In particular,
   \[
     \dim \mathcal{M}_\mu^{\operatorname{ss}}
     = \sum_i
       \dim (\mathcal{M}_{\mu i + 1})_\mu - \dim (\mathcal{M}_{\mu i})_\mu
     = \dim \mathcal{M}[\mu]_\mu
     = \dim \mathcal{M}_\mu
     = d
   \]
 
   Likewise, given \(u \in \mathcal{U}(\mathfrak{g})_0\) the value
   \[
     \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu^{\operatorname{ss}}})
     = \sum_i
       \operatorname{Tr}(u\!\restriction_{(\mathcal{M}_{\mu i + 1})_\mu})
     - \operatorname{Tr}(u\!\restriction_{(\mathcal{M}_{\mu i})_\mu})
     = \operatorname{Tr}(u\!\restriction_{\mathcal{M}[\mu]_\mu})
     = \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu})
   \]
   is polynomial in \(\mu \in \mathfrak{h}^*\).
 \end{proof}
 
 \begin{note}
   Although we have provided an explicit construction of
   \(\mathcal{M}^{\operatorname{ss}}\) in terms of \(\mathcal{M}\), we should
   point out this construction is not functorial. First, given a
   \(\mathfrak{g}\)-homomorphism \(f : \mathcal{M} \to \mathcal{N}\) between
   coherent families, it is unclear what \(f^{\operatorname{ss}} :
   \mathcal{M}^{\operatorname{ss}} \to \mathcal{N}^{\operatorname{ss}}\) is
   supposed to be. Secondly, and this is more relevant, our construction depends
   on the choice of composition series \(0 = \mathcal{M}_{\lambda 0} \subset
   \cdots \subset \mathcal{M}_{\lambda r_\lambda} = \mathcal{M}[\lambda]\).
   While different choices of composition series yield isomorphic results, there
   is no canonical isomorphism. In addition, there is no canonical choice of
   composition series.
 \end{note}
 
 The proof of Lemma~\ref{thm:component-coh-family-has-finite-length} is
 extremely technical and will not be included in here. It suffices to note that,
 as in Proposition~\ref{thm:ess-supp-is-zariski-dense}, the general case follows
 from the case where \(\mathfrak{g}\) is simple, which may be found in
 \cite{mathieu} -- see Lemma 3.3. As promised, if \(\mathcal{M}\) is a coherent
 extension of \(M\) then so is \(\mathcal{M}^{\operatorname{ss}}\).
 
 \begin{proposition}
   Let \(M\) be a simple bounded \(\mathfrak{g}\)-module and \(\mathcal{M}\)
   be a coherent extension of \(M\). Then \(\mathcal{M}^{\operatorname{ss}}\) is
   a coherent extension of \(M\), and \(M\) is in fact a submodule of
   \(\mathcal{M}^{\operatorname{ss}}\).
 \end{proposition}
 
 \begin{proof}
   Since \(M\) is simple, its support is contained in a single \(Q\)-coset.
   This implies that \(M\) is a subquotient of \(\mathcal{M}[\lambda]\) for any
   \(\lambda \in \operatorname{supp} M\). If we fix some composition series \(0
   = \mathcal{M}_0 \subset \mathcal{M}_1 \subset \cdots \subset \mathcal{M}_r =
   \mathcal{M}[\lambda]\) of \(\mathcal{M}[\lambda]\) with \(M \cong
   \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i}\), there is a natural inclusion
   \[
     M
     \isoto \mfrac{\mathcal{M}_{i + 1}}{\mathcal{M}_i}
     \to \bigoplus_j \mfrac{\mathcal{M}_{j + 1}}{\mathcal{M}_j}
     \cong \mathcal{M}^{\operatorname{ss}}[\lambda]
   \]
 \end{proof}
 
 Given the uniqueness of the semisimplification, the semisimplification of any
 semisimple coherent extension \(\mathcal{M}\) is \(\mathcal{M}\)
 itself and therefore\dots
 
 \begin{corollary}\label{thm:bounded-is-submod-of-extension}
   Let \(M\) be a simple bounded \(\mathfrak{g}\)-module and \(\mathcal{M}\)
   be a semisimple coherent extension of \(M\). Then \(M\) is
   contained in \(\mathcal{M}\).
 \end{corollary}
 
 These last results provide a partial answer to the question of existence of
 well behaved coherent extensions. As for the uniqueness \(\mathcal{M}\) in
 Corollary~\ref{thm:bounded-is-submod-of-extension}, it suffices to show that
 the multiplicities of the simple weight \(\mathfrak{g}\)-modules in
 \(\mathcal{M}\) are uniquely determined by \(M\). These multiplicities may be
 computed via the following lemma.
 
 \begin{lemma}\label{thm:centralizer-multiplicity}
   Let \(M\) be a semisimple weight \(\mathfrak{g}\)-module. Then \(M_\lambda\)
   is a semisimple \(\mathcal{U}(\mathfrak{g})_0\)-module for any \(\lambda \in
   \operatorname{supp} M\). Moreover, if \(L\) is a simple weight
   \(\mathfrak{g}\)-module such that \(\lambda \in \operatorname{supp} L\) then
   \(L_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module and the
   multiplicity \(L\) in \(M\) coincides with the multiplicity of \(L_\lambda\)
   in \(M_\lambda\) as a \(\mathcal{U}(\mathfrak{g})_0\)-module.
 \end{lemma}
 
 \begin{proof}
   We begin by showing that \(L_\lambda\) is simple. Let \(N \subset L_\lambda\)
   be a nontrivial \(\mathcal{U}(\mathfrak{g})_0\)-submodule. We want to
   establish that \(N = L_\lambda\).
 
   If \(\mathcal{U}(\mathfrak{g})_\alpha\) denotes the root space of \(\alpha\)
   in \(\mathcal{U}(\mathfrak{g})\) under the adjoint action of \(\mathfrak{g}\)
   as in Example~\ref{ex:adjoint-action-in-universal-enveloping-is-weight},
   \(\alpha \in Q\), a simple calculation shows
   \(\mathcal{U}(\mathfrak{g})_\alpha \cdot N \subset L_{\lambda + \alpha}\).
   Since \(L\) is simple and \(N\) is nonzero, it follows from
   Example~\ref{ex:adjoint-action-in-universal-enveloping-is-weight} that
   \[
     L
     = \mathcal{U}(\mathfrak{g}) \cdot N
     = \bigoplus_{\alpha \in Q} \mathcal{U}(\mathfrak{g})_\alpha \cdot N
   \]
   and thus \(L_{\lambda + \alpha} = \mathcal{U}(\mathfrak{g})_\alpha \cdot N\).
   In particular, \(L_\lambda = \mathcal{U}(\mathfrak{g})_0 \cdot N \subset N\)
   and \(N = L_\lambda\).
 
   Now given a semisimple weight \(\mathfrak{g}\)-module \(M = \bigoplus_i M_i\)
   with \(M_i\) simple, it is clear \(M_\lambda = \bigoplus_i (M_i)_\lambda\).
   Each \((M_i)_\lambda\) is either \(0\) or a simple
   \(\mathcal{U}(\mathfrak{g})_0\)-module, so that \(M_\lambda\) is a semisimple
   \(\mathcal{U}(\mathfrak{g})_0\)-module. In addition, to see that the
   multiplicity of \(L\) in \(M\) coincides with the multiplicity of
   \(L_\lambda\) in \(M_\lambda\) it suffices to show that if \((M_i)_\lambda
   \cong (M_j)_\lambda\) are both nonzero then \(M_i \cong M_j\).
 
   If \(I(M_i) = \mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{g})_0}
   (M_i)_\lambda\), the inclusion of \(\mathcal{U}(\mathfrak{g})_0\)-modules
   \((M_i)_\lambda \to M_i\) induces a \(\mathfrak{g}\)-homomorphism
   \begin{align*}
             I(M_i) & \to     M_i       \\
     u \otimes m & \mapsto u \cdot m
   \end{align*}
 
   Since \(M_i\) is simple and \(\lambda \in \operatorname{supp} M_i\), \(M_i =
   \mathcal{U}(\mathfrak{g}) \cdot (M_i)_\lambda\). The homomorphism \(I(M_i)
   \to M_i\) is thus surjective. Similarly, if \(I(M_j) =
   \mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{g})_0}
   (M_j)_\lambda\) then there is a natural surjective
   \(\mathfrak{g}\)-homomorphism \(I(M_j) \to M_j\). Now suppose there is an
   isomorphism of \(\mathcal{U}(\mathfrak{g})_0\)-modules \(f: (M_i)_\lambda
   \isoto (M_j)_\lambda\). Such an isomorphism induces an isomorphism of
   \(\mathfrak{g}\)-modules
   \begin{align*}
     \tilde f : I(M_i) & \isoto  I(M_j)            \\
        u \otimes m & \mapsto u \otimes f(m)
   \end{align*}
 
   By composing \(\tilde f\) with the projection \(I(M_j) \to M_j\) we get a
   surjective homomorphism \(I(M_i) \to M_j\). We claim \(\ker (I(M_i) \to M_i)
   = \ker (I(M_i) \to M_j)\).  To see this, notice that \(\ker(I(M_i) \to M_i)\)
   coincides with the largest submodule \(Z(M_i) \subset I(M_i)\) contained in
   \(\bigoplus_{\alpha \ne 0} \mathcal{U}(\mathfrak{g})_\alpha
   \otimes_{\mathcal{U}(\mathfrak{g})_0} (M_i)_\lambda\). Indeed, a simple
   computation shows \(\ker (I(M_i) \to M_i) \cap (\mathcal{U}(\mathfrak{g})_0
   \otimes_{\mathcal{U}(\mathfrak{g})_0} (M_i)_\lambda) = 0\), which implies
   \(\ker(I(M_i) \to M_i) \subset Z(M_i)\). Since \(M_i\) is simple, \(\ker
   (I(M_i) \to M_i)\) is maximal and thus \(\ker(I(M_i) \to M_i) = Z(M_i)\). By
   the same token, \(\ker (I(M_j) \to M_j)\) is the largest submodule of
   \(I(M_j)\) contained in \(\bigoplus_{\alpha \ne 0}
   \mathcal{U}(\mathfrak{g})_\alpha \otimes_{\mathcal{U}(\mathfrak{g})_0}
   (M_j)_\lambda\) and therefore \(\ker(I(M_i) \to M_i) =
   \tilde{f}^{-1}(\ker(I(M_j) \to M_j)) = \ker(I(M_i) \to M_j)\).
 
   Hence there is an isomorphism \(\mfrac{I(M_i)}{\ker(I(M_i) \to M_i)} \isoto
   M_j\) satisfying
   \begin{center}
     \begin{tikzcd}
       I(M_i) \rar{\tilde f} \dar & I(M_j) \dar \\
       \mfrac{I(M_i)}{\ker(I(M_i) \to M_i)} \rar{\sim} & M_j
     \end{tikzcd}
   \end{center}
  and finally \(M_i \cong \mfrac{I(M_i)}{\ker(I(M_i) \to M_i)} \cong M_j\).
 \end{proof}
 
 A complementary question now is: which submodules of a \emph{nice} coherent
 family are cuspidal?
 
 \begin{proposition}[Mathieu]
   Let \(\mathcal{M}\) be an irreducible coherent family of degree \(d\) and
   \(\lambda \in \mathfrak{h}^*\). The following conditions are equivalent.
   \begin{enumerate}
     \item \(\mathcal{M}[\lambda]\) is simple.
     \item \(F_\alpha\!\restriction_{\mathcal{M}[\lambda]}\) is injective for
       all \(\alpha \in \Delta\).
     \item \(\mathcal{M}[\lambda]\) is cuspidal.
   \end{enumerate}
 \end{proposition}
 
 \begin{proof}
   The fact that \strong{(i)} and \strong{(iii)} are equivalent follows directly
   from Corollary~\ref{thm:cuspidal-mod-equivs}. Likewise, it is clear from the
   corollary that \strong{(iii)} implies \strong{(ii)}. All it is left is to
   show \strong{(ii)} implies \strong{(iii)}. This isn't already clear from
   Corollary~\ref{thm:cuspidal-mod-equivs} because, at first glance,
   $\mathcal{M}[\lambda]$ may not be simple for some $\lambda$ satisfying
   \strong{(ii)}. We will show this is never the case.
 
   Suppose \(F_\alpha\) acts injectively on the submodule
   \(\mathcal{M}[\lambda]\), for all \(\alpha \in \Delta\). Since
   \(\mathcal{M}[\lambda]\) has finite length, \(\mathcal{M}[\lambda]\) contains
   an infinite-dimensional simple \(\mathfrak{g}\)-submodule \(M\). Moreover,
   again by Corollary~\ref{thm:cuspidal-mod-equivs} we conclude \(M\) is a
   cuspidal module, and its degree is bounded by \(d\). We want to show
   \(\mathcal{M}[\lambda] = M\).
 
   We claim the set \(U = \{\mu \in \mathfrak{h}^* : \mathcal{M}_\mu \ \text{is
   a simple $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is Zariski-open. If we
   suppose this is the case for a moment or two, it follows from the fact that
   \(M\) is simple and \(\operatorname{supp}_{\operatorname{ess}} M\) is
   Zariski-dense that \(U \cap \operatorname{supp}_{\operatorname{ess}} M\) is
   non-empty. In other words, there is some \(\mu \in \mathfrak{h}^*\) such that
   \(\mathcal{M}_\mu\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module and
   \(\dim M_\mu = \deg M\).
 
   In particular, \(M_\mu \ne 0\), so \(M_\mu = \mathcal{M}_\mu\). Now given any
   simple \(\mathfrak{g}\)-module \(L\), it follows from
   Lemma~\ref{thm:centralizer-multiplicity} that the multiplicity of \(L\)
   in \(\mathcal{M}[\lambda]\) is the same as the multiplicity \(L_\mu\) in
   \(\mathcal{M}_\mu\) as a \(\mathcal{U}(\mathfrak{g})_0\)-module -- which is,
   of course, \(1\) if \(L \cong M\) and \(0\) otherwise. Hence
   \(\mathcal{M}[\lambda] = M\) and \(\mathcal{M}[\lambda]\) is cuspidal.
 \end{proof}
 
 To finish the proof, we now show\dots
 
 \begin{lemma}\label{thm:set-of-simple-u0-mods-is-open}
   Let \(\mathcal{M}\) be a coherent family. The set \(U = \{\lambda \in
   \mathfrak{h}^* : \mathcal{M}_\lambda \ \text{is a simple
   $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is Zariski-open.
 \end{lemma}
 
 \begin{proof}
   For each \(\lambda \in \mathfrak{h}^*\) we introduce the bilinear form
   \begin{align*}
     B_\lambda : \mathcal{U}(\mathfrak{g})_0 \times \mathcal{U}(\mathfrak{g})_0
     & \to K \\
     (u, v)
     & \mapsto \operatorname{Tr}(u v \!\restriction_{\mathcal{M}_\lambda})
   \end{align*}
   and consider its rank -- i.e. the dimension of the image of the induced
   operator
   \begin{align*}
     \mathcal{U}(\mathfrak{g})_0 & \to     \mathcal{U}(\mathfrak{g})_0^* \\
                               u & \mapsto B_\lambda(u, \cdot)
   \end{align*}
 
   Our first observation is that \(\operatorname{rank} B_\lambda \le d^2\). This
   follows from the commutativity of
   \begin{center}
     \begin{tikzcd}
       \mathcal{U}(\mathfrak{g})_0               \rar \dar  &
       \mathcal{U}(\mathfrak{g})_0^*                        \\
       \operatorname{End}(\mathcal{M}_\lambda)   \rar{\sim} &
       \operatorname{End}(\mathcal{M}_\lambda)^* \uar
     \end{tikzcd},
   \end{center}
   where the map \(\mathcal{U}(\mathfrak{g})_0 \to
   \operatorname{End}(\mathcal{M}_\lambda)\) is given by the action of
   \(\mathcal{U}(\mathfrak{g})_0\), the map
   \(\operatorname{End}(\mathcal{M}_\lambda)^* \to
   \mathcal{U}(\mathfrak{g})_0^*\) is its dual, and the isomorphism
   \(\operatorname{End}(\mathcal{M}_\lambda) \isoto
   \operatorname{End}(\mathcal{M}_\lambda)^*\) is induced by the trace form
   \begin{align*}
     \operatorname{End}(\mathcal{M}_\lambda) \times
     \operatorname{End}(\mathcal{M}_\lambda) & \to K \\
     (T, S) & \mapsto \operatorname{Tr}(T S)
   \end{align*}
 
   Indeed, \(\operatorname{rank} B_\lambda \le
   \operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to
   \operatorname{End}(\mathcal{M}_\lambda)) \le \dim
   \operatorname{End}(\mathcal{M}_\lambda) = d^2\). Furthermore, if
   \(\operatorname{rank} B_\lambda = d^2\) then we must have
   \(\operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to
   \operatorname{End}(\mathcal{M}_\lambda)) = d^2\) -- i.e. the map
   \(\mathcal{U}(\mathfrak{g})_0 \to \operatorname{End}(\mathcal{M}_\lambda)\)
   is surjective. In particular, if \(\operatorname{rank} B_\lambda = d^2\) then
   \(\mathcal{M}_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module,
   for if \(M \subset \mathcal{M}_\lambda\) is invariant under the action of
   \(\mathcal{U}(\mathfrak{g})_0\) then \(M\) is invariant under any
   \(K\)-linear operator \(\mathcal{M}_\lambda \to \mathcal{M}_\lambda\), so
   that \(M = 0\) or \(M = \mathcal{M}_\lambda\).
 
   On the other hand, if \(\mathcal{M}_\lambda\) is simple then by Burnside's
   Theorem on matrix algebras the map \(\mathcal{U}(\mathfrak{g})_0 \to
   \operatorname{End}(\mathcal{M}_\lambda)\) is surjective. Hence the
   commutativity of the previously drawn diagram, as well as the fact that
   \(\operatorname{rank}(\mathcal{U}(\mathfrak{g})_0 \to
   \operatorname{End}(\mathcal{M}_\lambda)) =
   \operatorname{rank}(\operatorname{End}(\mathcal{M}_\lambda)^* \to
   \mathcal{U}(\mathfrak{g})_0^*)\), imply that \(\operatorname{rank} B_\lambda
   = d^2\). This goes to show that \(U\) is precisely the set of all \(\lambda\)
   such that \(B_\lambda\) has maximal rank \(d^2\). We now show that \(U\) is
   Zariski-open. First, notice that
   \[
     U =
     \bigcup_{\substack{V \subset \mathcal{U}(\mathfrak{g})_0 \\ \dim V = d}}
     U_V,
   \]
   where \(U_V = \{\lambda \in \mathfrak{h}^* : \operatorname{rank}
   B_\lambda\!\restriction_V = d^2 \}\). Here \(V\) ranges over all
   \(d\)-dimensional subspaces of \(\mathcal{U}(\mathfrak{g})_0\) -- \(V\) is
   not necessarily a \(\mathcal{U}(\mathfrak{g})_0\)-submodule.
 
   Indeed, if \(\operatorname{rank} B_\lambda = d^2\) it follows from the
   subjectivity of the map \(\mathcal{U}(\mathfrak{g})_0 \to
   \operatorname{End}(\mathcal{M}_\lambda)\) that there is some \(V \subset
   \mathcal{U}(\mathfrak{g})_0\) with \(\dim V = d\) such that the restriction
   \(V \to \operatorname{End}(\mathcal{M}_\lambda)\) is surjective. The
   commutativity of
   \begin{center}
     \begin{tikzcd}
       V                                         \rar \dar  & V^* \\
       \operatorname{End}(\mathcal{M}_\lambda)   \rar{\sim} &
       \operatorname{End}(\mathcal{M}_\lambda)^* \uar
     \end{tikzcd}
   \end{center}
   then implies \(\operatorname{rank} B_\lambda\!\restriction_V = d^2\). In
   other words, \(U \subset \bigcup_V U_V\).
 
   Likewise, if \(\operatorname{rank} B_\lambda\!\restriction_V = d^2\) for some
   \(V\), then the commutativity of
   \begin{center}
     \begin{tikzcd}
       V                             \rar \dar & V^* \\
       \mathcal{U}(\mathfrak{g})_0   \rar      &
       \mathcal{U}(\mathfrak{g})_0^* \uar
     \end{tikzcd}
   \end{center}
   implies \(\operatorname{rank} B_\lambda \ge d^2\), which goes to show
   \(\bigcup_V U_V \subset U\).
 
   Given \(\lambda \in U_V\), the surjectivity of \(V \to
   \operatorname{End}(\mathcal{M}_\lambda)\) and the fact that \(\dim V <
   \infty\) imply \(V \to V^*\) is invertible. Since \(\mathcal{M}\) is a
   coherent family, \(B_\lambda\) depends polynomially in \(\lambda\). Hence so
   does the induced maps \(V \to V^*\). In particular, there is some Zariski
   neighborhood \(U'\) of \(\lambda\) such that the map \(V \to V^*\) induced by
   \(B_\mu\!\restriction_V\) is invertible for all \(\mu \in U'\).
 
   But the surjectivity of the map induced by \(B_\mu\!\restriction_V\) implies
   \(\operatorname{rank} B_\mu = d^2\), so \(\mu \in U_V\) and therefore \(U'
   \subset U_V\). This implies \(U_V\) is open for all \(V\). Finally, \(U\) is
   the union of Zariski-open subsets and is therefore open. We are done.
 \end{proof}
 
 The major remaining question for us to tackle is that of the existence of
 coherent extensions, which will be the focus of our next section.
 
 \section{Localizations \& the Existence of Coherent Extensions}
 
 Let \(M\) be a simple bounded \(\mathfrak{g}\)-module of degree \(d\). Our
 goal is to prove that \(M\) has a (unique) irreducible semisimple coherent
 extension \(\mathcal{M}\). Since \(M\) is simple, we know \(M \subset
 \mathcal{M}[\lambda]\) for any \(\lambda \in \operatorname{supp} M\). Our first
 task is constructing \(\mathcal{M}[\lambda]\). The issue here is that
 \(\operatorname{supp}_{\operatorname{ess}} M\) may not be all of \(\lambda + Q
 = \operatorname{supp}_{\operatorname{ess}} \mathcal{M}[\lambda]\), so we may
 find \(M \subsetneq \mathcal{M}[\lambda]\). In fact, we may find
 \(\operatorname{supp} M \subsetneq \lambda + Q\).
 
 This wasn't an issue an Example~\ref{ex:laurent-polynomial-mod} because we
 verified that the action of \(f \in \mathfrak{sl}_2(K)\) on \(K[x, x^{-1}]\) is
 injective. Since all weight spaces of \(K[x, x^{-1}]\) are \(1\)-dimensional,
 this implies the action of \(f\) is actually bijective, so we can obtain a
 nonzero vector in \(K[x, x^{-1}]_{2 k} = K x^k\) for any \(k \in \mathbb{Z}\)
 by translating between weight spaced using \(f\) and \(f^{-1}\) -- here
 \(f^{-1}\) denotes the \(K\)-linear operator \((-
 \sfrac{\mathrm{d}}{\mathrm{d}x} + \sfrac{x^{-1}}{2})^{-1}\), which is the
 inverse of the action of \(f\) on \(K[x, x^{-1}]\).
 \begin{center}
   \begin{tikzcd}
     \cdots     \rar[bend left=60]{f^{-1}}
     & K x^{-2} \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f}
     & K x^{-1} \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f}
     & K        \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f}
     & K x      \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f}
     & K x^2    \rar[bend left=60]{f^{-1}} \lar[bend left=60]{f}
     & \cdots   \lar[bend left=60]{f}
   \end{tikzcd}
 \end{center}
 
 In the general case, the action of some \(F_\alpha \in \mathfrak{g}\) with
 \(\alpha \in \Delta\) in \(M\) may not be injective. In fact, we have seen that
 the action of \(F_\alpha\) is injective for all \(\alpha \in \Delta^+\) if, and
 only if \(M\) is cuspidal. Nevertheless, we could intuitively \emph{make it
 injective} by formally inverting the elements \(F_\alpha \in
 \mathcal{U}(\mathfrak{g})\). This would allow us to obtain nonzero vectors in
 \(M_\mu\) for all \(\mu \in \lambda + Q\) by successively applying elements of
 \(\{F_\alpha^{\pm 1}\}_{\alpha \in \Delta}\) to a nonzero weight vector \(m \in
 M_\lambda\). Moreover, if the actions of the \(F_\alpha\) were to be
 invertible, we would find that all \(M_\mu\) are \(d\)-dimensional for \(\mu
 \in \lambda + Q\).
 
 In a commutative domain, this can be achieved by tensoring our module by the
 field of fractions. However, \(\mathcal{U}(\mathfrak{g})\) is hardly ever
 commutative -- \(\mathcal{U}(\mathfrak{g})\) is commutative if, and only if
 \(\mathfrak{g}\) is Abelian -- and the situation is more delicate in the
 non-commutative case. For starters, a non-commutative \(K\)-algebra \(A\) may
 not even have a ``field of fractions'' -- i.e. an over-ring where all elements
 of \(A\) have inverses. Nevertheless, it is possible to formally invert
 elements of certain subsets of \(A\) via a process known as
 \emph{localization}, which we now describe.
 
 \begin{definition}\index{localization!multiplicative subsets}\index{localization!Ore's condition}
   Let \(A\) be a \(K\)-algebra. A subset \(S \subset A\) is called
   \emph{multiplicative} if \(s \cdot t \in S\) for all \(s, t \in S\) and \(0
   \notin S\). A multiplicative subset \(S\) is said to satisfy \emph{Ore's
   localization condition} if for each \(a \in A\) and \(s \in S\) there exists
   \(b, c \in A\) and \(t, t' \in S\) such that \(s a = b t\) and \(a s = t'
   c\).
 \end{definition}
 
 \begin{theorem}[Ore-Asano]\index{localization!Ore-Asano Theorem}
   Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization
   condition. Then there exists a (unique) \(K\)-algebra \(S^{-1} A\), with a
   canonical algebra homomorphism \(A \to S^{-1} A\), enjoying the universal
   property that each algebra homomorphism \(f : A \to B\) such that \(f(s)\) is
   invertible for all \(s \in S\) can be uniquely extended to an algebra
   homomorphism \(S^{-1} A \to B\). \(S^{-1} A\) is called \emph{the
   localization of \(A\) by \(S\)}, and the map \(A \to S^{-1} A\) is called
   \emph{the localization map}.
   \begin{center}
     \begin{tikzcd}
       A        \dar \rar{f}        & B \\
       S^{-1} A \urar[swap, dotted] &
     \end{tikzcd}
   \end{center}
 \end{theorem}
 
 If we identify an element with its image under the localization map, it follows
 directly from Ore's construction that every element of \(S^{-1} A\) has the
 form \(s^{-1} a\) for some \(s \in S\) and \(a \in A\). Likewise, any element
 of \(S^{-1} A\) can also be written as \(b t^{-1}\) for some \(t \in S\), \(b
 \in A\).
 
 Ore's localization condition may seem a bit arbitrary at first, but a more
 thorough investigation reveals the intuition behind it. The issue in question
 here is that in the non-commutative case we can no longer take the existence of
 common denominators for granted. However, the existence of common denominators
 is fundamental to the proof of the fact the field of fractions is a ring -- it
 is used, for example, to define the sum of two elements in the field of
 fractions. We thus need to impose their existence for us to have any hope of
 defining consistent arithmetics in the localization of an algebra, and Ore's
 condition is actually equivalent to the existence of common denominators --
 see the discussion in the introduction of \cite[ch.~6]{goodearl-warfield} for
 further details.
 
 We should also point out that there are numerous other conditions -- which may
 be easier to check than Ore's -- known to imply Ore's condition. For
 instance\dots
 
 \begin{lemma}
   Let \(S \subset A\) be a multiplicative subset generated by finitely many
   locally \(\operatorname{ad}\)-nilpotent elements -- i.e. elements \(s \in S\)
   such that for each \(a \in A\) there exists \(r > 0\) such that
   \(\operatorname{ad}(s)^r a = [s, [s, \cdots [s, a]]\cdots] = 0\). Then \(S\)
   satisfies Ore's localization condition.
 \end{lemma}
 
 In our case, we are more interested in formally inverting the action of
 \(F_\alpha\) on \(M\) than in inverting \(F_\alpha\) itself. To that end, we
 introduce one further construction, known as \emph{the localization of a
 module}.
 
 \begin{definition}\index{localization!localization of modules}
   Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization
   condition and \(M\) be an \(A\)-module. The \(S^{-1} A\)-module \(S^{-1} M =
   S^{-1} A \otimes_A M\) is called \emph{the localization of \(M\) by \(S\)},
   and the homomorphism of \(A\)-modules
   \begin{align*}
     M & \to     S^{-1} M    \\
     m & \mapsto 1 \otimes m
   \end{align*}
   is called \emph{the localization map of \(M\)}.
 \end{definition}
 
 Notice that the \(S^{-1} A\)-module \(S^{-1} M\) has the natural structure of
 an \(A\)-module, where the action of \(A\) is given by the localization map \(A
 \to S^{-1} A\).
 
 It is interesting to observe that, unlike in the case of the field of fractions
 of a commutative domain, in general the localization map \(A \to S^{-1} A\) --
 i.e. the map \(a \mapsto \frac{a}{1}\) -- may not be injective. For instance,
 if \(S\) contains a divisor of zero \(s\), its image under the localization map
 is invertible and therefore cannot be a divisor of zero in \(S^{-1} A\). In
 particular, if \(a \in A\) is nonzero and such that \(s a = 0\) or \(a s = 0\)
 then its image under the localization map has to be \(0\). However, the
 existence of divisors of zero in \(S\) turns out to be the only obstruction to
 the injectivity of the localization map, as shown in\dots
 
 \begin{lemma}
   Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization
   condition and \(M\) be an \(A\)-module. If \(S\) acts injectively on \(M\)
   then the localization map \(M \to S^{-1} M\) is injective. In particular, if
   \(S\) has no zero divisors then \(A\) is a subalgebra of \(S^{-1} A\).
 \end{lemma}
 
 Again, in our case we are interested in inverting the actions of the
 \(F_\alpha\) on \(M\). However, for us to be able to translate between all
 weight spaces associated with elements of \(\lambda + Q\), \(\lambda \in
 \operatorname{supp} M\), we only need to invert the \(F_\alpha\)'s for
 \(\alpha\) in some subset of \(\Delta\) which spans all of \(Q = \mathbb{Z}
 \Delta\). In other words, it suffices to invert \(F_\beta\) for all \(\beta\)
 in some basis \(\Sigma\) for \(\Delta\). We can choose such a basis to be
 well-behaved. For example, we can show\dots
 
 \begin{lemma}\label{thm:nice-basis-for-inversion}
   Let \(M\) be a simple infinite-dimensional bounded \(\mathfrak{g}\)-module.
   There is a basis \(\Sigma = \{\beta_1, \ldots, \beta_r\}\) for \(\Delta\)
   such that the elements \(F_{\beta_i}\) all act injectively on \(M\) and
   satisfy \([F_{\beta_i}, F_{\beta_j}] = 0\).
 \end{lemma}
 
 \begin{note}
   The basis \(\Sigma\) in Lemma~\ref{thm:nice-basis-for-inversion} may very
   well depend on the representation \(M\)! This is another obstruction to the
   functoriality of our constructions.
 \end{note}
 
 The proof of the previous Lemma is quite technical and was deemed too tedious
 to be included in here. See Lemma 4.4 of \cite{mathieu} for a full proof. Since
 \(F_\alpha\) is locally \(\operatorname{ad}\)-nilpotent for all \(\alpha \in
 \Delta\), we can see\dots
 
 \begin{corollary}
   Let \(\Sigma\) be as in Lemma~\ref{thm:nice-basis-for-inversion} and
   \((F_\beta)_{\beta \in \Sigma} \subset \mathcal{U}(\mathfrak{g})\) be the
   multiplicative subset generated by the \(F_\beta\)'s. The \(K\)-algebra
   \(\Sigma^{-1} \mathcal{U}(\mathfrak{g}) = (F_\beta)_{\beta \in \Sigma}^{-1}
   \mathcal{U}(\mathfrak{g})\) is well defined. Moreover, if we denote by
   \(\Sigma^{-1} M\) the localization of \(M\) by \((F_\beta)_{\beta \in
   \Sigma}\), the localization map \(M \to \Sigma^{-1} M\) is injective.
 \end{corollary}
 
 From now on let \(\Sigma\) be some fixed basis for \(\Delta\) satisfying the
 hypothesis of Lemma~\ref{thm:nice-basis-for-inversion}. We now show that
 \(\Sigma^{-1} M\) is a weight \(\mathfrak{g}\)-module whose support is an
 entire \(Q\)-coset.
 
 \begin{proposition}\label{thm:irr-bounded-is-contained-in-nice-mod}
   The restriction of the localization \(\Sigma^{-1} M\) is a bounded
   \(\mathfrak{g}\)-module of degree \(d\) with \(\operatorname{supp}
   \Sigma^{-1} M = Q + \operatorname{supp} M\) and \(\dim \Sigma^{-1} M_\lambda
   = d\) for all \(\lambda \in \operatorname{supp} \Sigma^{-1} M\).
 \end{proposition}
 
 \begin{proof}
   Fix some \(\beta \in \Sigma\). We begin by showing that \(F_\beta\) and
   \(F_\beta^{-1}\) map the weight space \(\Sigma^{-1} M_\lambda\) to
   \(\Sigma^{-1} M_{\lambda - \beta}\) and \(\Sigma^{-1} M_{\lambda + \beta}\),
   respectively. Indeed, given \(m \in M_\lambda\) and \(H \in \mathfrak{h}\) we
   have
   \[
     H \cdot (F_\beta \cdot m)
     = ([H, F_\beta] + F_\beta H) \cdot m
     = F_\beta (-\beta(H) + H) \cdot m
     = (\lambda - \beta)(H) F_\beta \cdot m
   \]
 
   On the other hand,
   \[
     0
     = [H, 1]
     = [H, F_\beta F_\beta^{-1}]
     = F_\beta [H, F_\beta^{-1}] + [H, F_\beta] F_\beta^{-1}
     = F_\beta [H, F_\beta^{-1}] - \beta(H) F_\beta F_\beta^{-1},
   \]
   so that \([H, F_\beta^{-1}] = \beta(H) \cdot F_\beta^{-1}\) and therefore
   \[
     H \cdot (F_\beta^{-1} \cdot m)
     = ([H, F_\beta^{-1}] + F_\beta^{-1} H) \cdot m
     = F_\beta^{-1} (\beta(H) + H) \cdot m
     = (\lambda + \beta)(H) F_\beta^{-1} \cdot m
   \]
 
   From the fact that \(F_\beta^{\pm 1}\) maps \(M_\lambda\) to \(\Sigma^{-1}
   M_{\lambda \pm \beta}\) follows our first conclusion: since \(M\) is a weight
   module and every element of \(\Sigma^{-1} M\) has the form \(s^{-1} \cdot m =
   s^{-1} \otimes m\) for \(s \in (F_\beta)_{\beta \in \Sigma}\) and \(m \in
   M\), we can see that \(\Sigma^{-1} M = \bigoplus_\lambda \Sigma^{-1}
   M_\lambda\). Furthermore, since the action of each \(F_\beta\) on
   \(\Sigma^{-1} M\) is bijective and \(\Sigma\) is a basis for \(Q\) we obtain
   \(\operatorname{supp} \Sigma^{-1} M = Q + \operatorname{supp} M\).
 
   Again, because of the bijectivity of the \(F_\beta\)'s, to see that \(\dim
   \Sigma^{-1} M_\lambda = d\) for all \(\lambda \in \operatorname{supp}
   \Sigma^{-1} M\) it suffices to show that \(\dim \Sigma^{-1} M_\lambda = d\)
   for some \(\lambda \in \operatorname{supp} \Sigma^{-1} M\). We may take
   \(\lambda \in \operatorname{supp} M\) with \(\dim M_\lambda = d\). For any
   finite-dimensional subspace \(V \subset \Sigma^{-1} M_\lambda\) we can find
   \(s \in (F_\beta)_{\beta \in \Sigma}\) such that \(s \cdot V \subset M\). If
   \(s = F_{\beta_{i_1}} \cdots F_{\beta_{i_r}}\), it is clear \(s \cdot V
   \subset M_{\lambda - \beta_{i_1} - \cdots - \beta_{i_r}}\), so \(\dim V =
   \dim s \cdot V \le d\). This holds for all finite-dimensional \(V \subset
   \Sigma^{-1} M_\lambda\), so \(\dim \Sigma^{-1} M_\lambda \le d\). It then
   follows from the fact that \(M_\lambda \subset \Sigma^{-1} M_\lambda\) that
   \(M_\lambda = \Sigma^{-1} M_\lambda\) and therefore \(\dim \Sigma^{-1}
   M_\lambda = d\).
 \end{proof}
 
 We now have a good candidate for a coherent extension of \(M\), but
 \(\Sigma^{-1} M\) is still not a coherent extension since its support is
 contained in a single \(Q\)-coset. In particular, \(\operatorname{supp}
 \Sigma^{-1} M \ne \mathfrak{h}^*\) and \(\Sigma^{-1} M\) is not a coherent
 family. To obtain a coherent family we thus need somehow extend \(\Sigma^{-1}
 M\). To that end, we will attempt to replicate the construction of the coherent
 extension of the \(\mathfrak{sl}_2(K)\)-module \(K[x, x^{-1}]\). Specifically,
 the idea is that if twist \(\Sigma^{-1} M\) by an automorphism which shifts its
 support by some \(\lambda \in \mathfrak{h}^*\), we can construct a coherent
 family by summing these modules over \(\lambda\) as in
 Example~\ref{ex:sl-laurent-family}.
 
 For \(K[x, x^{-1}]\) this was achieved by twisting the
 \(\operatorname{Diff}(K[x, x^{-1}])\)-module \(K[x, x^{-1}]\) by the
 automorphisms \(\varphi_\lambda : \operatorname{Diff}(K[x, x^{-1}]) \to
 \operatorname{Diff}(K[x, x^{-1}])\) and restricting the results to
 \(\mathcal{U}(\mathfrak{sl}_2(K))\) via the map
 \(\mathcal{U}(\mathfrak{sl}_2(K)) \to \operatorname{Diff}(K[x, x^{-1}])\), but
 this approach is inflexible since not every \(\mathfrak{sl}_2(K)\)-module
 factors through \(\operatorname{Diff}(K[x, x^{-1}])\). Nevertheless, we could
 just as well twist \(K[x, x^{-1}]\) by automorphisms of
 \(\mathcal{U}(\mathfrak{sl}_2(K))_f\) directly -- where
 \(\mathcal{U}(\mathfrak{sl}_2(K))_f = (f)^{-1} \mathcal{U}(\mathfrak{g})\) is
 the localization of \(\mathcal{U}(\mathfrak{sl}_2(K))\) by the multiplicative
 subset generated by \(f\).
 
 In general, we may twist the \(\Sigma^{-1} \mathcal{U}(\mathfrak{g})\)-module
 \(\Sigma^{-1} M\) by automorphisms of \(\Sigma^{-1}
 \mathcal{U}(\mathfrak{g})\). For \(\lambda = \beta \in \Sigma\) the map
 \begin{align*}
   \theta_\beta : \Sigma^{-1} \mathcal{U}(\mathfrak{g}) & \to
                  \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\
                  u & \mapsto F_\beta u F_\beta^{-1}
 \end{align*}
 is a natural candidate for such a twisting automorphism. Indeed, we will soon
 see that \(\twisted{(\Sigma^{-1} M)}{\theta_\beta}_\lambda = \Sigma^{-1}
 M_{\lambda + \beta}\). However, this is hardly useful to us, since \(\beta \in
 Q\) and therefore \(\beta + \operatorname{supp} \Sigma^{-1} M =
 \operatorname{supp} \Sigma^{-1} M\). If we want to expand the support of
 \(\Sigma^{-1} M\) we will have to twist by automorphisms that shift its support
 by \(\lambda \in \mathfrak{h}^*\) lying \emph{outside} of \(Q\).
 
 The situation is much less obvious in this case. Nevertheless, it turns out we
 can extend the family \(\{\theta_\beta\}_{\beta \in \Sigma}\) to a family of
 automorphisms \(\{\theta_\lambda\}_{\lambda \in \mathfrak{h}^*}\).
 Explicitly\dots
 
 \begin{proposition}\label{thm:nice-automorphisms-exist}
   There is a family of automorphisms \(\{\theta_\lambda : \Sigma^{-1}
   \mathcal{U}(\mathfrak{g}) \to \Sigma^{-1}
   \mathcal{U}(\mathfrak{g})\}_{\lambda \in \mathfrak{h}^*}\) such that
   \begin{enumerate}
     \item \(\theta_{k_1 \beta_1 + \cdots + k_r \beta_r}(u) = F_{\beta_1}^{k_1}
       \cdots F_{\beta_r}^{k_r} u F_{\beta_r}^{- k_r} \cdots F_{\beta_1}^{-
       k_1}\) for all \(u \in \Sigma^{-1} \mathcal{U}(\mathfrak{g})\) and \(k_1,
       \ldots, k_r \in \mathbb{Z}\).
 
     \item For each \(u \in \Sigma^{-1} \mathcal{U}(\mathfrak{g})\) the map
       \begin{align*}
         \mathfrak{h}^* & \to     \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\
                \lambda & \mapsto \theta_\lambda(u)
       \end{align*}
       is polynomial.
 
     \item If \(\lambda, \mu \in \mathfrak{h}^*\), \(N\) is a \(\Sigma^{-1}
       \mathcal{U}(\mathfrak{g})\)-module whose restriction to
       \(\mathcal{U}(\mathfrak{g})\) is a weight \(\mathfrak{g}\)-module and
       \(\twisted{N}{\theta_\lambda}\) is the \(\Sigma^{-1}
       \mathcal{U}(\mathfrak{g})\)-module \(N\) twisted by the automorphism
       \(\theta_\lambda\) then \(N_\mu = \twisted{N}{\theta_\lambda}_{\mu +
       \lambda}\). In particular, \(\operatorname{supp}
       \twisted{N}{\theta_\lambda} = \lambda + \operatorname{supp} N\).
   \end{enumerate}
 \end{proposition}
 
 \begin{proof}
   Since the elements \(F_\beta\), \(\beta \in \Sigma\) commute with one
   another, the endomorphisms
   \begin{align*}
     \theta_{k_1 \beta_1 + \cdots + k_r \beta_r}
     : \Sigma^{-1} \mathcal{U}(\mathfrak{g}) &
     \to \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\
     u & \mapsto
     F_{\beta_1}^{k_1} \cdots F_{\beta_r}^{k_r}
     u
     F_{\beta_1}^{- k_r} \cdots F_{\beta_r}^{- k_1}
   \end{align*}
   are well defined for all \(k_1, \ldots, k_r \in \mathbb{Z}\).
 
   Fix some \(u \in \Sigma^{-1} \mathcal{U}(\mathfrak{g})\). For any \(s \in
   (F_\beta)_{\beta \in \Sigma}\) and \(k > 0\) we have \(s^k u = \binom{k}{0}
   \operatorname{ad}(s)^0 u s^{k - 0} + \cdots + \binom{k}{k}
   \operatorname{ad}(s)^k u s^{k - k}\). Now if we take \(\ell\) such
   \(\operatorname{ad}(F_\beta)^{\ell + 1} u = 0\) for all \(\beta \in \Sigma\)
   we find
   \[
     \theta_{k_1 \beta_1 + \cdots + k_r \beta_r}(u)
     = \sum_{i_1, \ldots, i_r = 1, \ldots, \ell}
     \binom{k_1}{i_1} \cdots \binom{k_r}{i_r}
     \operatorname{ad}(F_{\beta_1})^{i_1} \cdots
     \operatorname{ad}(F_{\beta_r})^{i_r}
     u
     F_{\beta_1}^{- i_1} \cdots F_{\beta_r}^{- i_r}
   \]
   for all \(k_1, \ldots, k_r \in \mathbb{N}\).
 
   Since the binomial coefficients \(\binom{x}{k} = \frac{x (x-1) \cdots (x - k
   + 1)}{k!}\) can be uniquely extended to polynomial functions in \(x \in K\),
   we may in general define
   \[
     \theta_\lambda(u)
     = \sum_{i_1, \ldots, i_r \ge 0}
     \binom{\lambda_1}{i_1} \cdots \binom{\lambda_r}{i_r}
     \operatorname{ad}(F_{\beta_1})^{i_1} \cdots
     \operatorname{ad}(F_{\beta_r})^{i_r}
     r
     F_{\beta_1}^{- i_1} \cdots F_{\beta_r}^{- i_r}
   \]
   for \(\lambda_1, \ldots, \lambda_r \in K\), \(\lambda = \lambda_1 \beta_1 +
   \cdots + \lambda_r \beta_r \in \mathfrak{h}^*\).
 
   It is clear that the \(\theta_\lambda\) are endomorphisms. To see that the
   \(\theta_\lambda\) are indeed automorphisms, notice \(\theta_{- k_1 \beta_1 -
   \cdots - k_r \beta_r} = \theta_{k_1 \beta_1 + \cdots + k_r \beta_r}^{-1}\).
   The uniqueness of the polynomial extensions then implies \(\theta_{- \lambda}
   = \theta_\lambda^{-1}\) in general: given \(u \in \Sigma^{-1}
   \mathcal{U}(\mathfrak{g})\), the map
   \begin{align*}
     \mathfrak{h}^* & \to \Sigma^{-1} \mathcal{U}(\mathfrak{g})        \\
            \lambda & \mapsto \theta_\lambda(\theta_{-\lambda}(u)) - u
   \end{align*}
   is a polynomial extension of the zero map \(\mathbb{Z} \beta_1 \oplus \cdots
   \oplus \mathbb{Z} \beta_r \to \Sigma^{-1} \mathcal{U}(\mathfrak{g})\) and is
   therefore identically zero.
 
   Finally, let \(N\) be a \(\Sigma^{-1} \mathcal{U}(\mathfrak{g})\)-module
   whose restriction is a weight module. If \(n \in N\) then
   \[
     n \in \twisted{N}{\theta_\lambda}_{\mu + \lambda}
     \iff \theta_\lambda(H) \cdot n = (\mu + \lambda)(H) n
     \, \forall H \in \mathfrak{h}
   \]
 
   But
   \[
     \theta_\beta(H)
     = F_\beta H F_\beta^{-1}
     = ([F_\beta, H] + H F_\beta) F_\beta^{-1}
     = (\beta(H) + H) F_\beta F_\beta^{-1}
     = \beta(H) + H
   \]
   for all \(H \in \mathfrak{h}\) and \(\beta \in \Sigma\). In general,
   \(\theta_\lambda(H) = \lambda(H) + H\) for all \(\lambda \in \mathfrak{h}^*\)
   and hence
   \[
     \begin{split}
       n \in \twisted{N}{\theta_\lambda}_{\mu + \lambda}
       & \iff (\lambda(H) + H) \cdot n = (\mu + \lambda)(H) n
         \; \forall H \in \mathfrak{h} \\
       & \iff H \cdot n = \mu(H) n \; \forall H \in \mathfrak{h} \\
       & \iff n \in N_\mu
     \end{split},
   \]
   so that \(\twisted{N}{\theta_\lambda}_{\mu + \lambda} = N_\mu\).
 \end{proof}
 
 It should now be obvious\dots
 
 \begin{proposition}[Mathieu]\label{thm:coh-ext-exists}
   There exists a coherent extension \(\mathcal{M}\) of \(M\).
 \end{proposition}
 
 \begin{proof}
   Take\footnote{Here we fix some $\lambda_\xi \in \xi$ for each $Q$-coset $\xi
   \in \mfrac{\mathfrak{h}^*}{Q}$. While there is a natural isomorphism
   $\twisted{(\Sigma^{-1} M)}{\theta_\lambda} \isoto \twisted{(\Sigma^{-1}
   M)}{\theta_\mu}$ for each $\mu \in \lambda + Q$, they are not the same
   \(\mathfrak{g}\)-modules strictly speaking. This is yet another obstruction
   to the functoriality of our constructions.}
   \[
     \mathcal{M}
     = \bigoplus_{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q}}
       \twisted{(\Sigma^{-1} M)}{\theta_\lambda}
   \]
 
   It is clear \(M\) lies in \(\Sigma^{-1} M = \twisted{(\Sigma^{-1}
   M)}{\theta_0}\) and therefore \(M \subset \mathcal{M}\). On the other hand,
   \(\dim \mathcal{M}_\mu = \dim \twisted{(\Sigma^{-1} M)}{\theta_\lambda}_\mu =
   \dim \Sigma^{-1} M_{\mu - \lambda} = d\) for all \(\mu \in \lambda + Q\) --
   \(\lambda\) standing for some fixed representative of its \(Q\)-coset.
   Furthermore, given \(u \in \mathcal{U}(\mathfrak{g})_0\) and \(\mu \in
   \lambda + Q\),
   \[
     \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu})
     = \operatorname{Tr}
       (\theta_\lambda(u)\!\restriction_{\Sigma^{-1} M_{\mu - \lambda}})
   \]
   is polynomial in \(\mu\) because of the second item of
   Proposition~\ref{thm:nice-automorphisms-exist}.
 \end{proof}
 
 Lo and behold\dots
 
 \begin{theorem}[Mathieu]\label{thm:mathieu-ext-exists-unique}\index{coherent family!Mathieu's \(\mExt\) coherent extension}
   There exists a unique semisimple coherent extension \(\mExt(M)\) of \(M\).
   More precisely, if \(\mathcal{M}\) is any coherent extension of \(M\), then
   \(\mathcal{M}^{\operatorname{ss}} \cong \mExt(M)\). Furthermore, \(\mExt(M)\)
   is an irreducible coherent family.
 \end{theorem}
 
 \begin{proof}
   The existence part should be clear from the previous discussion: it suffices
   to fix some coherent extension \(\mathcal{M}\) of \(M\) and take
   \(\mExt(M) = \mathcal{M}^{\operatorname{ss}}\).
 
   To see that \(\mExt(M)\) is irreducible, recall from
   Corollary~\ref{thm:bounded-is-submod-of-extension} that \(M\) is a
   \(\mathfrak{g}\)-submodule of \(\mExt(M)\). Since the degree of \(M\) is the
   same as the degree of \(\mExt(M)\), some of its weight spaces have maximal
   dimension inside of \(\mExt(M)\). In particular, it follows from
   Lemma~\ref{thm:centralizer-multiplicity} that \(\mExt(M)_\lambda =
   M_\lambda\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module for some
   \(\lambda \in \operatorname{supp} M\).
 
   As for the uniqueness of \(\mExt(M)\), fix some other semisimple coherent
   extension \(\mathcal{N}\) of \(M\). We claim that the multiplicity of a given
   simple \(\mathfrak{g}\)-module \(L\) in \(\mathcal{N}\) is determined by its
   \emph{trace function}
   \begin{align*}
     \mathfrak{h}^* \times \mathcal{U}(\mathfrak{g})_0 &
     \to K \\
     (\lambda, u) &
     \mapsto \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})
   \end{align*}
 
   It is a well known fact of the theory of modules that, given an associative
   \(K\)-algebra \(A\), a finite-dimensional semisimple \(A\)-module \(L\) is
   determined, up to isomorphism, by its \emph{character}
   \begin{align*}
     \chi_L : A & \to     K                                    \\
              a & \mapsto \operatorname{Tr}(a\!\restriction_L)
   \end{align*}
 
   In particular, the multiplicity of \(L\) in \(\mathcal{N}\), which is the
   same as the multiplicity of \(L_\lambda\) in \(\mathcal{N}_\lambda\), is
   determined by the character \(\chi_{\mathcal{N}_\lambda} :
   \mathcal{U}(\mathfrak{g})_0 \to K\). Since this holds for all simple weight
   \(\mathfrak{g}\)-modules, it follows that \(\mathcal{N}\) is determined by
   its trace function. Of course, the same holds for \(\mExt(M)\). We now claim
   that the trace function of \(\mathcal{N}\) is the same as that of
   \(\mExt(M)\). Clearly,
   \(\operatorname{Tr}(u\!\restriction_{\mExt(M)_\lambda}) =
   \operatorname{Tr}(u\!\restriction_{M_\lambda}) =
   \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\) for all \(\lambda
   \in \operatorname{supp}_{\operatorname{ess}} M\), \(u \in
   \mathcal{U}(\mathfrak{g})_0\). Since the essential support of \(M\) is
   Zariski-dense and the maps \(\lambda \mapsto
   \operatorname{Tr}(u\!\restriction_{\mExt(M)_\lambda})\) and \(\lambda \mapsto
   \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\) are polynomial in
   \(\lambda \in \mathfrak{h}^*\), it follows that these maps coincide for all
   \(u\).
 
   In conclusion, \(\mathcal{N} \cong \mExt(M)\) and \(\mExt(M)\) is unique.
 \end{proof}
 
 A sort of ``reciprocal'' of Theorem~\ref{thm:mathieu-ext-exists-unique} also
 holds. Namely\dots
 
 \begin{proposition}\label{thm:coherent-families-are-all-ext}
   Let \(\mathcal{M}\) be a semisimple irreducible coherent family and \(M
   \subset \mathcal{M}\) be an infinite-dimensional simple submodule. Then
   \(\mathcal{M} \cong \mExt(M)\). In particular, all semisimple coherent
   families have the form \(\mathcal{M} \cong \mExt(M)\) for some simple bounded
   \(\mathfrak{g}\)-module \(M\).
 \end{proposition}
 
 \begin{proof}
   Since \(M \subset \mathcal{M}\), \(M\) is bounded and
   \(\operatorname{supp}_{\operatorname{ess}} M\) is Zariski-dense. In addition,
   it follows from Lemma~\ref{thm:set-of-simple-u0-mods-is-open} that \(U =
   \{\lambda \in \mathfrak{h}^* : \mathcal{M}_\lambda \ \text{is a simple
   $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is a Zariski-open subset -- which
   is non-empty since \(\mathcal{M}\) is irreducible.
 
   Hence there is some \(\lambda \in \operatorname{supp}_{\operatorname{ess}} M
   \cap U\). In particular, there is some \(\lambda \in
   \operatorname{supp}_{\operatorname{ess}} M\) such that \(M_\lambda =
   \mathcal{M}_\lambda\) and thus \(\deg M = \dim \mathcal{M}_\lambda = \deg
   \mathcal{M}\). This implies that \(\mathcal{M}\) is a coherent extension of
   \(M\), so that by the uniqueness of semisimple irreducible coherent
   extensions we get \(\mathcal{M} \cong \mExt(M)\).
 \end{proof}
 
 Having thus reduced the problem of classifying the cuspidal
 \(\mathfrak{g}\)-modules to that of understanding semisimple irreducible
 coherent families, the only remaining question for us to tackle is: what are
 the coherent \(\mathfrak{g}\)-families? This turns out to be a decently
 complicated question on its own, and we will require a full chapter to answer
 it. This will be the focus of our final chapter.