lie-algebras-and-their-representations

diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -2137,31 +2137,32 @@ theorem~\ref{thm:sl3-existence-uniqueness}. Lo and behold\dots
 \end{theorem}
 
 Fix some dominant integral \(\lambda \in P\). The ``uniqueness'' part of the
-theorem follows at once from the argument used for \(\mathfrak{sl}_3(K)\).
-
-\begin{proposition}
-  Let \(V\) and \(W\) be irreducible finite-dimensional
-  \(\mathfrak{g}\)-modules whose highest weight is \(\lambda\). Then \(V \cong
-  W\).
-\end{proposition}
-
-The ``existance'' 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.
-
+theorem follows at once from the argument used for \(\mathfrak{sl}_3(K)\). The
+``existance'' 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}.
+
+% TODO: Define the induced representation beforehand
 \begin{definition}\label{def:verma}
   The \(\mathfrak{g}\)-module \(M(\lambda) =
-  \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K v\), where the action of
-  \(\mathfrak{b}\) in \(K v\) is given by \(H v = \lambda(H) \cdot v\) for all
-  \(H \in \mathfrak{h}\) and \(X v = 0\) for \(X \in \mathfrak{g}_{\alpha}\),
-  \(\alpha \in \Delta^+\), is called \emph{the Verma module of weight
-  \(\lambda\)}
+  \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K v^+\), where the action of
+  \(\mathfrak{b}\) in \(K v^+\) is given by \(H v^+ = \lambda(H) \cdot v^+\)
+  for all \(H \in \mathfrak{h}\) and \(X v^+ = 0\) for \(X \in
+  \mathfrak{g}_{\alpha}\), \(\alpha \in \Delta^+\), is called \emph{the Verma
+  module of weight \(\lambda\)}
 \end{definition}
 
-% TODO: Point out that the Verma module is usually highly infinite-dimensional,
-% yet very well behaved
+We should point out that, unlike most representations we've 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 \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+\) -- where
+\(v^+ = 1 \otimes v^+ \in M(\lambda)\) is as in definition~\ref{def:verma}.
+Moreover, we find\dots
 
 % TODO: State the PBW theorem in the introduction
 \begin{proposition}\label{thm:verma-is-weight-mod}
@@ -2171,29 +2172,29 @@ setting.
   \]
   holds. Furthermore, \(\dim M(\lambda)_\mu < \infty\) for all \(\mu \in
   \mathfrak{h}^*\) and \(\dim M(\lambda) = 1\). Finally, \(\lambda\) is the
-  highest weight of \(M(\lambda)\), with highest weight vetor given by \(v = 1
-  \otimes v \in M(\lambda)\) as in definition~\ref{def:verma}.
+  highest weight of \(M(\lambda)\), with highest weight vetor given by \(v^+ =
+  1 \otimes v^+ \in M(\lambda)\) as in definition~\ref{def:verma}.
 \end{proposition}
 
 \begin{proof}
   The Poincaré-Birkhoff-Witt theorem implies that \(M(\lambda)\) is spanned by
-  the vectors \(F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v\) for
+  the vectors \(F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v^+\) for
   \(\alpha_i \in \Delta^-\) and \(F_{\alpha_i} \in \mathfrak{g}_{\alpha_i}\) as
   in the proof of proposition~\ref{thm:distinguished-subalgebra}. But
   \[
     \begin{split}
-      H F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v
+      H F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v^+
       & = ([H, F_{\alpha_1}] + F_{\alpha_1} H)
-          F_{\alpha_2} \cdots F_{\alpha_n} v \\
-      & = \alpha_1(H) \cdot F_{\alpha_1} \cdots F_{\alpha_n} v
+          F_{\alpha_2} \cdots F_{\alpha_n} v^+ \\
+      & = \alpha_1(H) \cdot F_{\alpha_1} \cdots F_{\alpha_n} v^+
         + F_{\alpha_1} ([H, F_{\alpha_2}] + F_{\alpha_2} H)
-          F_{\alpha_2} \cdots F_{\alpha_n} v \\
+          F_{\alpha_2} \cdots F_{\alpha_n} v^+ \\
       & \;\; \vdots \\
       & = (\alpha_1 + \cdots + \alpha_n)(H) \cdot
-          F_{\alpha_1} \cdots F_{\alpha_n} H v \\
+          F_{\alpha_1} \cdots F_{\alpha_n} H v^+ \\
       & = (\lambda + \alpha_1 + \cdots + \alpha_n)(H) \cdot
-          F_{\alpha_1} \cdots F_{\alpha_n} v \\
-      & \therefore F_{\alpha_1} \cdots F_{\alpha_n} v
+          F_{\alpha_1} \cdots F_{\alpha_n} v^+ \\
+      & \therefore F_{\alpha_1} \cdots F_{\alpha_n} v^+
         \in M(\lambda)_{\lambda + \alpha_1 + \cdots + \alpha_n}
     \end{split}
   \]
@@ -2211,28 +2212,28 @@ setting.
   k_n \cdot \alpha_n\).
 
   This already gives us that the weights of \(M(\lambda)\) are bounded by
-  \(\lambda\). To see that \(\lambda\) is indeed a weight, we show that \(v\)
-  is nonzero weight vector. Clearly \(v \in V_\lambda\). The
+  \(\lambda\). To see that \(\lambda\) is indeed a weight, we show that \(v^+\)
+  is nonzero weight vector. Clearly \(v^+ \in V_\lambda\). The
   Poincaré-Birkhoff-Witt theorem implies
   \[
     M(\lambda)
     \cong \left(\bigoplus_i \mathcal{U}(\mathfrak{b}) \right)
-    \otimes_{\mathcal{U}(\mathfrak{b})} K v
+    \otimes_{\mathcal{U}(\mathfrak{b})} K v^+
     \cong \bigoplus_i \mathcal{U}(\mathfrak{b})
-    \otimes_{\mathcal{U}(\mathfrak{b})} K v
-    \cong \bigoplus_i K v
+    \otimes_{\mathcal{U}(\mathfrak{b})} K v^+
+    \cong \bigoplus_i K v^+
     \ne 0
   \]
-  as \(\mathcal{U}(\mathfrak{b})\)-modules, so \(v \ne 0\) -- for if this was
-  not the case we would find \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v =
-  0\). Hence \(V_\lambda \ne 0\) and therefore \(\lambda\) is the highest
-  weight of \(M(\lambda)\), with highest weight vector \(v\).
+  as \(\mathcal{U}(\mathfrak{b})\)-modules, so \(v^+ \ne 0\) -- for if this was
+  not the case we would find \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+
+  = 0\). Hence \(V_\lambda \ne 0\) and therefore \(\lambda\) is the highest
+  weight of \(M(\lambda)\), with highest weight vector \(v^+\).
 
   To see that \(\dim M(\lambda)_\mu < \infty\), simply note that there are only
   finitely many monomials \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots
   F_{\alpha_n}^{k_n}\) such that \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots
   + k_n \cdot \alpha_n\). Since \(M(\lambda)_\mu\) is spanned by the images of
-  \(v\) under such monimials, we conclude \(\dim M(\lambda) < \infty\). In
+  \(v^+\) under such monimials, we conclude \(\dim M(\lambda) < \infty\). In
   particular, there is a single monimials \(F_{\alpha_1}^{k_1}
   F_{\alpha_2}^{k_2} \cdots F_{\alpha_n}^{k_n}\) such that \(\lambda = \lambda
   + k_1 \cdot \alpha_1 + \cdots + k_n \cdot \alpha_n\) -- which is, of course,
@@ -2241,20 +2242,27 @@ setting.
 
 % TODO: Give an example for sl2?
 
+What's interesting to us about all this is that we've 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 irreducible nor finite-dimensional. Nevertheless, we can use
+\(M(\lambda)\) to construct an irreducible representation of \(\mathfrak{g}\)
+whose highest weight is \(\lambda\).
+
 % TODO: Adjust the notation for the maximal submodule
 \begin{proposition}
-  Every subrepresentation \(W \subset M(\lambda)\) is the direct sum of its
+  Every subrepresentation \(V \subset M(\lambda)\) is the direct sum of its
   weight spaces. In particular, \(M(\lambda)\) has a unique maximal
   subrepresentation \(N(\lambda)\) and a unique irreducible quotient
   \(\sfrac{M(\lambda)}{N(\lambda)}\).
 \end{proposition}
 
 \begin{proof}
-  Let \(W \subset M(\lambda)\) be a subrepresentation and take any nonzero \(w
-  \in W\). Because of proposition~\ref{thm:verma-is-weight-mod}, we know there
+  Let \(V \subset M(\lambda)\) be a subrepresentation and take any nonzero \(v
+  \in V\). Because of proposition~\ref{thm:verma-is-weight-mod}, we know there
   are \(\mu_1, \ldots, \mu_n \in \mathfrak{h}^*\) and nonzero \(v_i \in
-  M(\lambda)_{\mu_i}\) such that \(w = v_1 + \cdots + v_n\). We want to show
-  \(v_i \in W\) for all \(i\).
+  M(\lambda)_{\mu_i}\) such that \(v = v_1 + \cdots + v_n\). We want to show
+  \(v_i \in V\) for all \(i\).
 
   Fix some \(H_2 \in \mathfrak{h}\) such that \(\mu_1(H_2) \ne \mu_2(H_2)\).
   Then
@@ -2263,8 +2271,8 @@ setting.
     - \frac{(\mu_3 - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_3
     - \cdots
     - \frac{(\mu_n - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_n
-    = \left( 1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) w
-    \in W
+    = \left( 1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) v
+    \in V
   \]
 
   Now take \(H_3 \in \mathfrak{h}\) such that \(\mu_1(H_3) \ne \mu_3(H_3)\). By
@@ -2272,62 +2280,80 @@ setting.
   \begin{multline*}
     v_1
     -
-    \frac{(\mu_4 - \mu_1)(H_2) \cdot (\mu_4 - \mu_3)(H_3)}
+    \frac{(\mu_4 - \mu_3)(H_3) \cdot (\mu_4 - \mu_1)(H_2)}
          {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} v_4
     - \cdots -
-    \frac{(\mu_n - \mu_1)(H_2) \cdot (\mu_n - \mu_3)(H_3)}
+    \frac{(\mu_n - \mu_3)(H_3) \cdot (\mu_n - \mu_1)(H_2)}
          {(\mu_3 - \mu_1)(H_3) \cdot (\mu_2 - \mu_1)(H_2)} v_n \\
     =
     \left(1 - \frac{H_3 - \mu_1(H_3)}{(\mu_3 - \mu_1)(H_3)} \right)
-    \left(1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) w
-    \in W
+    \left(1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) v
+    \in V
   \end{multline*}
 
   By applying the same procedure over and over again we can see that \(v_1 = X
-  w \in W\) for some \(X \in \mathcal{U}(\mathfrak{g})\). Furtheremore, if we
-  reproduce all this for \(v_2 + \cdots + v_n = w - v_1 \in W\) we get that
-  \(v_2 \in W\). Now by applying the same procesude over and over we find
-  \(v_1, \ldots, v_n \in W\). Hence
+  v \in V\) for some \(X \in \mathcal{U}(\mathfrak{g})\). Furtheremore, if we
+  reproduce all this for \(v_2 + \cdots + v_n = v - v_1 \in V\) we get that
+  \(v_2 \in V\). Now by applying the same procesude over and over we find
+  \(v_1, \ldots, v_n \in V\). Hence
   \[
-    W = \bigoplus_\mu W_\mu = \bigoplus_\mu M(\lambda)_\mu \cap W
+    V = \bigoplus_\mu V_\mu = \bigoplus_\mu M(\lambda)_\mu \cap V
   \]
 
-  Since \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v\), if \(v\) is the
-  highest weight vector of \(M(\lambda)\) and \(W\) is a proper
-  subrepresentation then \(v \notin W\), for if this is not case \(M(\lambda) =
-  \mathcal{U}(\mathfrak{g}) \cdot v \subset W\). Hence any proper submodule
-  lies in the sum of weight spaces other than \(V_\lambda\), so the sum
+  Since \(M(\lambda) = \mathcal{U}(\mathfrak{g}) \cdot v^+\), \(V\) is a proper
+  subrepresentation then \(v^+ \notin V\). Hence any proper submodule lies in
+  the sum of weight spaces other than \(M(\lambda)_\lambda\), so the sum
   \(N(\lambda)\) of all such submodules is still proper. In fact, this implies
   \(N(\lambda)\) is the unique maximal subrepresentation of \(M(\lambda)\) and
   \(\sfrac{M(\lambda)}{N(\lambda)}\) is its unique irreducible quotient.
 \end{proof}
 
-\begin{proposition}
+The only issue standing between us and a proof of
+theorem~\ref{thm:dominant-weight-theo} is, of course, that of
+finite-dimensionality. In other words, the question now is: is the unique
+irreducible quotient of \(M(\lambda)\) finite-dimensional? The answer to this
+question turns out to be yes whenever \(\lambda\) is dominant-integral.
+
+\begin{proposition}\label{thm:verma-is-finite-dim}
   The unique irreducible quotient of \(M(\lambda)\) is finite-dimensional.
 \end{proposition}
 
+The proof of proposition~\ref{thm:verma-is-finite-dim} is very technical and we
+won't include it here, but the is simple: we show that the set of weights of
+\(\sfrac{M(\lambda)}{N(\lambda)}\) is stable under the natural action of the
+Weyl group \(W\) in \(\mathfrak{h}^*\). One can then show that the every weight
+of \(V\) is conjugate to a single dominant integral weight of
+\(\sfrac{M(\lambda)}{N(\lambda)}\), and that the set of dominant integral
+weights of such irreducible quotient is finite. Since \(W\) is finitely
+generated, this implies the set of weights of the unique irreducible quotient
+of \(M(\lambda)\) is finite. But each weight space is finite-dimensional. Hence
+so is the irreducible quotient.
+
+We refer the reader to \cite[ch. 21]{humphreys} for further details. What we
+are really interested in is\dots
+
 \begin{corollary}
   There is a finite-dimensional irreducible \(\mathfrak{g}\)-module \(V\) whose
   highest weight is \(\lambda\).
 \end{corollary}
 
 \begin{proof}
-  Let \(V\) the unique irreducible quotient of \(M(\lambda)\). It suffices to
-  show that its highest weight is \(\lambda\).
-  We have already seen that \(v \in M(\lambda)_\lambda\) is a highest weight
-  vector. In particular, \(v \ne 0\). Furthermore, since \(v\) lies outside of
-  the maximal subrepresentation of \(M(\lambda)\), the projection \(v +
-  N(\lambda) \in V\) is also nonzero.
-
-  We now claim that \(v + N(\lambda) \in V_\lambda\). Indeed,
+  Let \(V = \sfrac{M(\lambda)}{N(\lambda)}\). It suffices to show that its
+  highest weight is \(\lambda\). We have already seen that \(v^+ \in
+  M(\lambda)_\lambda\) is a highest weight vector. Now since \(v\) lies outside
+  of the maximal subrepresentation of \(M(\lambda)\), the projection \(v^+ +
+  N(\lambda) \in V\) is nonzero.
+
+  % TODO: Why is V_mu = M(lambda)_mu + N(lambda)? Turn this into a proposition?
+  We now claim that \(v^+ + N(\lambda) \in V_\lambda\). Indeed,
   \[
-    H (v + N(\lambda))
-    = H v + N(\lambda)
-    = \lambda(H) \cdot (v + N(\lambda))
+    H (v^+ + N(\lambda))
+    = H v^+ + N(\lambda)
+    = \lambda(H) \cdot (v^+ + N(\lambda))
   \]
   for all \(H \in \mathfrak{h}\). Hence \(\lambda\) is a weight of \(V\), with
-  weight vector \(v + N(\lambda)\). Finally, we remark that \(\lambda\) is the
-  highest weight of \(V\), for if this was not the case we could find a weight
-  \(\mu\) of \(M(\lambda)\) which is higher than \(\lambda\).
+  weight vector \(v^+ + N(\lambda)\). Finally, we remark that \(\lambda\) is
+  the highest weight of \(V\), for if this was not the case we could find a
+  weight \(\mu\) of \(M(\lambda)\) which is higher than \(\lambda\).
 \end{proof}