lie-algebras-and-their-representations

diff --git a/sections/introduction.tex b/sections/introduction.tex
@@ -1017,7 +1017,7 @@ Surprisingly, this functor has a right adjoint.
   \]
 \end{example}
 
-\begin{proposition}
+\begin{proposition}\label{thm:frobenius-reciprocity}
   Given a Lie algebra \(\mathfrak{g}\), a subalgebra \(\mathfrak{h} \subset
   \mathfrak{g}\), a \(\mathfrak{h}\)-module \(M\) and a \(\mathfrak{g}\)-module
   \(N\), the map

diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -696,9 +696,20 @@ Of course, what we are really interested in is\dots
 The ``existence'' part is more nuanced. Our first instinct is, of course, to
 try to generalize the proof used for \(\mathfrak{sl}_3(K)\). The issue is that
 our proof relied heavily on our knowledge of the roots of
-\(\mathfrak{sl}_3(K)\). Instead, we need a new strategy for the general
-setting. To that end, we introduce a special class of \(\mathfrak{g}\)-modules,
-known as \emph{Verma modules}.
+\(\mathfrak{sl}_3(K)\). This is therefore little hope of generalizing this
+argument to some arbitrary \(\mathfrak{g}\) with out current knowledge. Instead,
+we focus on a simpler question: how can we construct \emph{any} highest
+\(\mathfrak{g}\)-module \(M\) of highest weight \(\lambda\)?
+
+If \(M\) is a finite-dimensional module with highest weight vector \(m^+ \in
+M_\lambda\), we already know \(H \cdot m^+ = \lambda(H) m^+\) for all
+\(\mathfrak{h}\) and \(X \cdot m^+ = 0\) for \(X \in \mathfrak{g}_\alpha\),
+\(\alpha \in \Delta^+\). In other words, the restriction of \(M\) to the Borel
+subalgebra \(\mathfrak{b} \subset \mathfrak{g}\) has a prescribed action. On the
+other hand, we have essentially no information about the action of the rest of
+\(\mathfrak{g}\) on \(M\). Nevertheless, given a \(\mathfrak{b}\)-module we may
+obtain a \(\mathfrak{g}\)-module by formally extending the action of
+\(\mathfrak{b}\) via induction. This leads us to the following definition.
 
 \begin{definition}\label{def:verma}\index{\(\mathfrak{g}\)-module!(generalized) Verma modules}
   The \(\mathfrak{g}\)-module \(M(\lambda) =
@@ -709,14 +720,8 @@ known as \emph{Verma modules}.
   module of weight \(\lambda\)}.
 \end{definition}
 
-We should point out that, unlike most modules we have encountered so far,
-Verma modules are \emph{highly infinite-dimensional}. Indeed, the dimension of
-\(M(\lambda)\) is the same as the codimension of \(\mathcal{U}(\mathfrak{b})\)
-in \(\mathcal{U}(\mathfrak{g})\), which is always infinite. Nevertheless,
-\(M(\lambda)\) turns out to be quite well behaved. For instance, by
-construction \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot m^+\) -- where
-\(m^+ = 1 \otimes m^+ \in M(\lambda)\) is as in Definition~\ref{def:verma}.
-Moreover, we find\dots
+It turns out that \(M(\lambda)\) enjoys many of the features we've grown used to
+in the past chapters. Explicitly\dots
 
 \begin{proposition}\label{thm:verma-is-weight-mod}
   The weight spaces decomposition
@@ -828,12 +833,58 @@ Moreover, we find\dots
   \end{equation}
 \end{example}
 
-What is interesting to us about all this is that we have just constructed a
-\(\mathfrak{g}\)-module whose highest weight is \(\lambda\). This is not a
-proof of Theorem~\ref{thm:dominant-weight-theo}, however, since \(M(\lambda)\)
-is neither simple nor finite-dimensional. Nevertheless, we can use
-\(M(\lambda)\) to construct a simple \(\mathfrak{g}\)-module whose highest
-weight is \(\lambda\).
+The Verma module \(M(\lambda)\) should really be though-of as ``the freest
+highest weight \(\mathfrak{g}\)-module of weight \(\lambda\)''. Unfortunately
+for us, this is not a proof of Theorem~\ref{thm:dominant-weight-theo}, since in
+general \(M(\lambda)\) is neither simple nor finite-dimensional. Indeed, the
+dimension of \(M(\lambda)\) is the same as the codimension of
+\(\mathcal{U}(\mathfrak{b})\) in \(\mathcal{U}(\mathfrak{g})\), which is always
+infinite. Nevertheless, we may use \(M(\lambda)\) to prove
+Theorem~\ref{thm:dominant-weight-theo} as follows.
+
+Given a \(\mathfrak{g}\)-module \(M\), any \(\mathfrak{g}\)-homomorphism \(f :
+M(\lambda) \to M\) is determined by the image of \(m^+\). Indeed, \(f(u \cdot
+m^+) = u \cdot f(m^+)\) for all \(u \in \mathcal{U}(\mathfrak{g})\). In
+addition, it is clear that
+\[
+  H \cdot f(m^+) = f(H \cdot m^+) = f(\lambda(H) m^+) = \lambda(H) f(m^+)
+\]
+for all \(H \in \mathfrak{h}\) and, similarly, \(X \cdot f(m^+) = 0\) for all
+\(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\). This leads us to the
+universal property of \(M(\lambda)\).
+
+\begin{definition}
+  Let \(M\) be a \(\mathfrak{g}\)-module and \(m \in M\). If \(X \cdot m = 0\)
+  for all \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\), then \(m\) is
+  called \emph{a singular vector of \(M\)}.
+\end{definition}
+
+\begin{proposition}
+  Let \(M\) be a \(\mathfrak{g}\)-module and \(m \in M_\lambda\) be a singular
+  vector. Then there exists a unique \(\mathfrak{g}\)-homomorphism \(f :
+  M(\lambda) \to M\) such that \(f(m^+) = m\). Furthermore, all homomorphisms
+  \(M(\lambda) \to M\) are given in this fashion.
+\end{proposition}
+
+\begin{proof}
+  The result follows directly from Proposition~\ref{thm:frobenius-reciprocity}.
+  Indeed, by Frobenius reciprocity, a \(\mathfrak{g}\)-homomorphism \(f :
+  M(\lambda) \to M\) is the same as a \(\mathfrak{b}\)-homomorphism \(g : K m^+
+  \to M = \operatorname{Res}_{\mathfrak{b}}^{\mathfrak{g}} M\). Any
+  \(K\)-linear map \(g : K m^+ \to M\) is determined by the image \(g(m^+)\) of
+  \(m^+\) and such an image is a singular vector if, and only if \(g\) is a
+  \(\mathfrak{b}\)-homomorphism.
+\end{proof}
+
+Notice that any highest weight vector is a singular vector. Now suppose \(M\)
+is a simple finite-dimensional \(\mathfrak{g}\)-module of highest weight vector
+\(m \in M_\lambda\). By the last proposition, there is a
+\(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\) such that \(f(m^+) = m\).
+Since \(M\) is simple, \(M = \mathcal{U}(\mathfrak{g}) \cdot m\) and therefore
+\(M \cong \mfrac{M(\lambda)}{\ker f}\). It then follows from the simplicity of
+\(M\) that \(\ker f \subset M(\lambda)\) is a maximal
+\(\mathfrak{g}\)-submodule. Maximal submodules of Verma modules are thus of
+primary interest to us. As it turns out, these can be easily classified.
 
 \begin{proposition}\label{thm:max-verma-submod-is-weight}
   Every submodule \(N \subset M(\lambda)\) is the direct sum of its weight
@@ -901,11 +952,9 @@ weight is \(\lambda\).
   with highest weight \(\lambda\) constructed in chapter~\ref{ch:sl3}.
 \end{example}
 
-This last example is particularly interesting to us, since it show we can
-realize the finite-dimensional simple \(\mathfrak{sl}_2(K)\)-module as
-quotients of Verma modules. This is because the quotient \(L(\lambda)\) in
-Example~\ref{ex:sl2-verma-quotient} is finite-dimensional. As it turns out,
-this is not a coincidence.
+All its left to prove the Highest Weight Theorem is verifying that the
+situation encountered in Example~\ref{ex:sl2-verma-quotient} holds for any
+\(\lambda \in P\). In other words, we need to show\dots
 
 \begin{proposition}\label{thm:verma-is-finite-dim}
   If \(\mathfrak{g}\) is semisimple and \(\lambda\) is dominant integral then
@@ -924,8 +973,8 @@ weights of the unique simple quotient of \(M(\lambda)\) is finite. But
 each weight space is finite-dimensional. Hence so is the simple quotient
 \(L(\lambda)\).
 
-We refer the reader to \cite[ch. 21]{humphreys} for further details. What we
-are really interested in is\dots
+We refer the reader to \cite[ch. 21]{humphreys} for further details. As
+promised, it follows\dots
 
 \begin{corollary}
   Let \(\lambda\) be a dominant integral weight of \(\mathfrak{g}\). Then there