diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -658,60 +658,31 @@ this condition is also sufficient. In other words\dots
is \(\lambda\).
\end{theorem}
-\index{weights!Highest Weight Theorem}
-This is known as \emph{the Highest Weight Theorem}, and its proof is the focus
-of this section. The ``uniqueness'' part of the theorem follows at once from
-the argument used for \(\mathfrak{sl}_3(K)\). Namely\dots
-
-\begin{proposition}\label{thm:irr-subrep-generated-by-vec}
- Let \(M\) be a finite-dimensional \(\mathfrak{g}\)-module and \(m \in M\) be
- a highest weight vector. Then the \(\mathfrak{g}\)-submodule
- \(\mathcal{U}(\mathfrak{g}) \cdot m \subset M\) generated by \(m\) is simple.
-\end{proposition}
-
-The proof of Proposition~\ref{thm:irr-subrep-generated-by-vec} is very similar
-in spirit to that of Proposition~\ref{thm:sl3-positive-roots-span-all-irr-rep}:
-we use the commutator relations of \(\mathfrak{g}\) to inductively show that
-the subspace spanned by the images of a highest weight vector under successive
-applications of negative root vectors is invariant under the action of
-\(\mathfrak{g}\) -- please refer to \cite{fulton-harris} for further details.
-Of course, what we are really interested in is\dots
-
-\begin{corollary}
- Let \(M\) and \(N\) be finite-dimensional simple \(\mathfrak{g}\)-modules
- with highest weight given by some common \(\lambda \in P\). Then \(M \cong
- N\).
-\end{corollary}
-
-\begin{proof}
- Let \(m \in M\) and \(n \in N\) be highest weight vectors and \(L =
- \mathcal{U}(\mathfrak{g}) \cdot m + n \subset M \oplus N\). It is clear that
- \(m + n\) is a highest weight vector of \(M \oplus N\). Hence by
- Proposition~\ref{thm:irr-subrep-generated-by-vec} \(L\) is simple. The
- projections \(\pi_1 : L \to M\) and \(\pi_2 : L \to N\) are thus nonzero
- \(\mathfrak{g}\)-homomorphisms between simple \(\mathfrak{g}\)-modules and
- are therefore isomorphisms. Hence \(M \cong L \cong N\).
-\end{proof}
-
-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
-\(\mathfrak{sl}_3(K)\). 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}
-(potentially infinite-dimensional) \(\mathfrak{g}\)-module \(M\) of highest
-weight \(\lambda\)?
+\index{weights!Highest Weight Theorem} This is known as \emph{the Highest
+Weight Theorem}, and its proof is the focus of this section. The ``uniqueness''
+part of the theorem follows at once from the arguments used for
+\(\mathfrak{sl}_3(K)\). However, the ``existence'' part of the theorem is more
+nuanced.
+
+Our first instinct is, of course, to try to generalize the proof used for
+\(\mathfrak{sl}_3(K)\). The issue is that our proof relied heavily on our
+knowledge of the roots of \(\mathfrak{sl}_3(K)\). It is thus clear that we need
+a more systematic approach for the general setting. We begin by asking a
+simpler question: how can we construct \emph{any} -- potentially
+infinite-dimensional -- \(\mathfrak{g}\)-module \(M\) of highest weight
+\(\lambda\)? In the process of answering this question we will come across a
+surprisingly elegant solution to our problem.
If \(M\) is a 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 \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\).
-Since \(M = \mathcal{U}(\mathfrak{g}) \cdot m^+\), this implies 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.
+\cdot m^+ = 0\) for \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\). If
+\(M\) is simple we find \(M = \mathcal{U}(\mathfrak{g}) \cdot m^+\), which
+implies 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) =
@@ -726,14 +697,16 @@ 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
+ The Verma module \(M(\lambda)\) is generated \(m^+ = 1 \otimes m^+ \in
+ M(\lambda)\) as in Definition~\ref{def:verma}. The weight spaces
+ decomposition
\[
M(\lambda) = \bigoplus_{\mu \in \mathfrak{h}^*} M(\lambda)_\mu
\]
holds. Furthermore, \(\dim M(\lambda)_\mu < \infty\) for all \(\mu \in
\mathfrak{h}^*\) and \(\dim M(\lambda)_\lambda = 1\). Finally, \(\lambda\) is
the highest weight of \(M(\lambda)\), with highest weight vector given by
- \(m^+ = 1 \otimes m^+ \in M(\lambda)\) as in Definition~\ref{def:verma}.
+ \(m^+ \in M(\lambda)\).
\end{proposition}
\begin{proof}
@@ -787,7 +760,7 @@ to in the past chapters. Explicitly\dots
\cong \bigoplus_i K m^+
\ne 0
\]
- as \(\mathcal{U}(\mathfrak{b})\)-modules. We then conclude \(m^+ \ne 0\) in
+ as \(\mathfrak{b}\)-modules. We then conclude \(m^+ \ne 0\) in
\(M(\lambda)\), for if this was not the case we would find \(M(\lambda) =
\mathcal{U}(\mathfrak{g}) \cdot m^+ = 0\). Hence \(M_\lambda \ne 0\) and
therefore \(\lambda\) is the highest weight of \(M(\lambda)\), with highest
@@ -866,16 +839,25 @@ universal property of \(M(\lambda)\).
vector. Then there exists a unique \(\mathfrak{g}\)-homomorphism \(f :
M(\lambda) \to M\) such that \(f(m^+) = m\). Furthermore, all homomorphisms
\(M(\lambda) \to M\) are given in this fashion.
+ \[
+ \operatorname{Hom}_{\mathfrak{g}}(M(\lambda), M)
+ \cong \{ m \in M_\lambda : m \ \text{is singular}\}
+ \]
\end{proposition}
\begin{proof}
The result follows directly from Proposition~\ref{thm:frobenius-reciprocity}.
- Indeed, by 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.
+ Indeed, by the Frobenius Reciprocity Theorem, a \(\mathfrak{g}\)-homomorphism
+ \(f : M(\lambda) \to M\) is the same as a \(\mathfrak{b}\)-homomorphism \(g :
+ K m^+ \to M = \operatorname{Res}_{\mathfrak{b}}^{\mathfrak{g}} M\). More
+ specifically, given a \(\mathfrak{b}\)-homomorphism \(g : K m^+ \to M\),
+ there exists a unique \(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\)
+ such that \(f(u \otimes m) = u \cdot g(m)\) for all \(m \in K m^+\), and all
+ \(\mathfrak{g}\)-homomorphism \(M(\lambda) \to M\) arise in this fashion.
+
+ Any \(K\)-linear map \(g : K m^+ \to M\) is determined by 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 \(M\)
@@ -975,32 +957,41 @@ 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. As
-promised, it follows\dots
+We refer the reader to \cite[ch. 21]{humphreys} for further details. We are now
+ready to prove the Highest Weight Theorem.
-\begin{corollary}
- Let \(\lambda\) be a dominant integral weight of \(\mathfrak{g}\). Then there
- is a finite-dimensional simple \(\mathfrak{g}\)-module \(M\) whose highest
- weight is \(\lambda\).
-\end{corollary}
-
-\begin{proof}
- Let \(M = L(\lambda)\). It suffices to show that its highest weight is
- \(\lambda\). We have already seen that \(m^+ \in M(\lambda)_\lambda\) is a
- highest weight vector. Now since \(m^+\) lies outside of the maximal
- submodule of \(M(\lambda)\), the projection \(m^+ + N(\lambda) \in
- M\) is nonzero.
+\begin{proof}[Proof of Theorem~\ref{thm:dominant-weight-theo}]
+ We begin with the ``existence'' part of the theorem by showing that
+ \(L(\lambda)\) is indeed a finite-dimensional simple module whose
+ highest-weight is \(\lambda\). It suffices to show that the highest weight of
+ \(L(\lambda)\) is \(\lambda\). We have already seen that \(m^+ \in
+ M(\lambda)_\lambda\) is a highest weight vector. Now since \(m^+\) lies
+ outside of the maximal submodule of \(M(\lambda)\), the projection \(m^+ +
+ N(\lambda) \in L(\lambda)\) is nonzero.
- We now claim that \(m^+ + N(\lambda) \in M_\lambda\). Indeed,
+ We now claim that \(m^+ + N(\lambda) \in L(\lambda)_\lambda\). Indeed,
\[
H \cdot (m^+ + N(\lambda))
= H \cdot m^+ + N(\lambda)
= \lambda(H) (m^+ + N(\lambda))
\]
- for all \(H \in \mathfrak{h}\). Hence \(\lambda\) is a weight of \(M\), with
- weight vector \(m^+ + N(\lambda)\). Finally, we remark that \(\lambda\) is
- the highest weight of \(M\), for if this was not the case we could find a
- weight \(\mu\) of \(M(\lambda)\) with \(\mu \succ \lambda\).
+ for all \(H \in \mathfrak{h}\). Hence \(\lambda\) is a weight of
+ \(L(\lambda)\), with weight vector \(m^+ + N(\lambda)\). Finally, we remark
+ that \(\lambda\) is the highest weight of \(L(\lambda)\), for if this was not
+ the case we could find a weight \(\mu\) of \(M(\lambda)\) with \(\mu \succ
+ \lambda\).
+
+ Now suppose \(M\) is some other finite-dimensional simple
+ \(\mathfrak{g}\)-module with highest weight vector \(m \in M_\lambda\). By
+ the universal property of the Verma module, there is a
+ \(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\) such that \(f(m^+) =
+ m\). As indicated before, since \(M\) is simple, \(M =
+ \mathcal{U}(\mathfrak{g}) \cdot m\) and therefore \(f\) is surjective. It
+ then follows \(M \cong \mfrac{M(\lambda)}{\ker f}\).
+
+ Since \(M\) is simple, \(\ker f \subset M(\lambda)\) is maximal and therefore
+ \(\ker f = N(\lambda)\). In other words, \(M \cong \mfrac{M(\lambda)}{\ker f}
+ = L(\lambda)\). We are done.
\end{proof}
We should point out that Proposition~\ref{thm:verma-is-finite-dim} fails for