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
git clone: git://git.pablopie.xyz/lie-algebras-and-their-representations
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Commit: b4e78e5336c3b58a91c59fc138460dce95e86dcf
Parent: ee6b5f89855b8c3b0683455cd8015911a9165bf7
Author: Pablo <pablo-escobar@riseup.net>
Date: Thu, 8 Sep 2022 21:38:17 +0000
Comecei a trabalhar na prova do teorema do peso integral dominante
Diffstat

1 file changed, 182 insertions, 19 deletions
Status	File Name	N° Changes	Insertions	Deletions
Modified	sections/semisimple-algebras.tex	201	182	19
diff --git a/sections/semisimple-algebras.tex b/sections/semisimple-algebras.tex
@@ -1859,6 +1859,9 @@ As promised, this implies\dots
   \]
 \end{corollary}
 
+% TODOO: Point out that simultaneous diagonalization only works in the
+% finite-dimensional setting: not all modules are weight modules!
+
 \begin{corollary}
   The restriction of \(B\) to \(\mathfrak{h}\) is non-degenerate.
 \end{corollary}
@@ -2148,23 +2151,183 @@ 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.
 
-% TODOO: Add further details. Turn this into a proper proof?
-Alternatively, one could construct  a potentially infinite-dimensional
-representation of \(\mathfrak{g}\) whose highest weight is some fixed dominant
-integral weight \(\lambda\) by taking the induced representation
-\(\operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} V_\lambda =
-\mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{b})} V_\lambda\),
-where \(\mathfrak{b} = \mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta^+}
-\mathfrak{g}_\alpha \subset \mathfrak{g}\) is the so called \emph{Borel
-subalgebra of \(\mathfrak{g}\)}, \(\mathcal{U}(\mathfrak{g})\) denotes the
-\emph{universal enveloping algebra of \(\mathfrak{g}\)} and \(\mathfrak{b}\)
-acts on \(V_\lambda = K v\) via \(H v = \lambda(H) \cdot v\) and \(X v = 0\)
-for \(X \in \mathfrak{g}_\alpha\), as does \cite{humphreys} in his proof. The
-fact that \(v\) is annihilated by all positive root spaces guarantees that the
-maximal weight of \(V\) is at most \(\lambda\), while the
-Poincare-Birkhoff-Witt \cite{humphreys} theorem guarantees that \(v = 1 \otimes
-v \in V\) is a non-zero weight vector of \(\lambda\) -- so that \(\lambda\) is
-the highest weight of \(V\). The challenge then is to show that the irreducible
-component of \(v\) in \(V\) is finite-dimensional -- see chapter 20 of
-\cite{humphreys} for a proof.
+\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
+  \(\lambda\)}
+\end{definition}
+
+% TODO: Point out that the Verma module is usually highly infinite-dimensional,
+% yet very well behaved
+
+% TODO: State the PBW theorem in the introduction
+\begin{proposition}\label{thm:verma-is-weight-mod}
+  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 \(\dim M(\lambda) = 1\). 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}.
+\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
+  \(\alpha_i \in \Delta^-\) and \(F_{\alpha_i} \in \mathfrak{g}_{\alpha_i}\) as
+  in the proof of proposition~\ref{thm:distinguished-subalgebra}. But
+  \[
+    \begin{split}
+      H F_{\alpha_1} F_{\alpha_2} \cdots F_{\alpha_n} v
+      & = ([H, F_{\alpha_1}] + F_{\alpha_1} H)
+          F_{\alpha_2} \cdots F_{\alpha_n} v \\
+      & = \alpha_1(H) \cdot F_{\alpha_1} \cdots F_{\alpha_n} v
+        + F_{\alpha_1} ([H, F_{\alpha_2}] + F_{\alpha_2} H)
+          F_{\alpha_2} \cdots F_{\alpha_n} v \\
+      & \;\; \vdots \\
+      & = (\alpha_1 + \cdots + \alpha_n)(H) \cdot
+          F_{\alpha_1} \cdots F_{\alpha_n} H v \\
+      & = (\lambda + \alpha_1 + \cdots + \alpha_n)(H) \cdot
+          F_{\alpha_1} \cdots F_{\alpha_n} v \\
+      & \therefore F_{\alpha_1} \cdots F_{\alpha_n} v
+        \in M(\lambda)_{\lambda + \alpha_1 + \cdots + \alpha_n}
+    \end{split}
+  \]
+
+  Hence \(M(\lambda) \subset \bigoplus_{\mu \in \mathfrak{h}^*}
+  M(\lambda)_\mu\), as desired. In fact we have established
+  \[
+    M(\lambda)
+    \subset
+    \bigoplus_{\substack{k_i \in \ZZ \\ k_i \ge 0}}
+    M(\lambda)_{\lambda + k_1 \cdot \alpha_1 + \cdots + k_n \cdot \alpha_n}
+  \]
+  where \(\{\alpha_1, \ldots, \alpha_m\} = \Delta^-\), so that all weights of
+  \(M(\lambda)\) have the form \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots +
+  k_n \cdot \alpha_n\).
+
+  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
+  Poincaré-Birkhoff-Witt theorem implies
+  \[
+    M(\lambda)
+    \cong \left(\bigoplus_i \mathcal{U}(\mathfrak{b}) \right)
+    \otimes_{\mathcal{U}(\mathfrak{b})} K v
+    \cong \bigoplus_i \mathcal{U}(\mathfrak{b})
+    \otimes_{\mathcal{U}(\mathfrak{b})} K v
+    \cong \bigoplus_i K v
+    \ne 0
+  \]
+  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 \(v\).
+
+  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
+  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,
+  the monomial where \(k_1 = \cdots = k_n = 0\). Hence \(\dim V_\lambda = 1\).
+\end{proof}
+
+% TODO: Give an example for sl2?
+
+% TODO: Adjust the notation for the maximal submodule
+\begin{proposition}
+  Every subrepresentation \(W \subset M(\lambda)\) 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
+  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 \(i\).
+
+  Fix some \(H_2 \in \mathfrak{h}\) such that \(\mu_1(H_2) \ne \mu_2(H_2)\).
+  Then
+  \[
+    v_1
+    - \frac{(\mu_3 - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_3
+    - \cdots
+    - \frac{(\mu_n - \mu_1)(H_2)}{(\mu_2 - \mu_1)(H_2)} v_n
+    = \left( 1 - \frac{H_2 - \mu_1(H_2)}{(\mu_2 - \mu_1)(H_2)} \right) w
+    \in W
+  \]
+
+  Now take \(H_3 \in \mathfrak{h}\) such that \(\mu_1(H_3) \ne \mu_3(H_3)\). By
+  applying the same procedure again we get
+  \begin{multline*}
+    v_1
+    -
+    \frac{(\mu_4 - \mu_1)(H_2) \cdot (\mu_4 - \mu_3)(H_3)}
+         {(\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)}
+         {(\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
+  \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
+  \[
+    W = \bigoplus_\mu W_\mu = \bigoplus_\mu M(\lambda)_\mu \cap W
+  \]
+
+  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
+  \(N(\lambda)\) 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 unique irreducible quotient of \(M(\lambda)\) is finite-dimensional.
+\end{proposition}
+
+\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,
+  \[
+    H (v + N(\lambda))
+    = H v + N(\lambda)
+    = \lambda(H) \cdot (v + N(\lambda))
+  \]
+  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 \(\lambda\).
+\end{proof}