lie-algebras-and-their-representations

Source code for my notes on representations of semisimple Lie algebras and Olivier Mathieu's classification of simple weight modules

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sections/simple-weight.tex	81392B	-rw-r--r--

\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.