- Commit
- 88d8fe84ac22de01e72c1abcc9ddeccc7edc23f0
- Parent
- 0e9e90d3184b316030d36c55a56c44e8a6985ee3
- Author
- Pablo <pablo-escobar@riseup.net>
- Date
Restructured the section on Verma modules
Restructured the section on Verma modules to include more general
results regarding highest weight modules, which will be needed for the
discussion of the classification of semisimple coherent families
Also added some TODO items
diff --git a/sections/fin-dim-simple.tex b/sections/fin-dim-simple.tex
@@ -666,7 +666,7 @@ uniqueness theorem analogous to Theorem~\ref{thm:sl2-exist-unique} and
Theorem~\ref{thm:sl3-existence-uniqueness}. This will be the focus of the next
section.
-\section{Verma Modules \& the Highest Weight Theorem}
+\section{Highest Weight Modules \& the Highest Weight Theorem}
It is already clear from the previous discussion that if \(\lambda\) is the
highest weight of \(M\) then \(\lambda(H_\alpha) \ge 0\) for all positive roots
@@ -675,193 +675,232 @@ highest weight of \(M\) then \(\lambda(H_\alpha) \ge 0\) for all positive roots
\(\mathfrak{g}\)-module with highest weight given by \(\lambda\). Surprisingly,
this condition is also sufficient. In other words\dots
+% TODO: Move this definition to beforehand
\begin{definition}\index{weights!dominant weight}\index{weights!integral weight}
An element \(\lambda\) of \(P\) such that \(\lambda(H_\alpha) \ge 0\) for all
\(\alpha \in \Delta^+\) is referred to as an \emph{dominant integral weight
of \(\mathfrak{g}\)}.
\end{definition}
-\begin{theorem}\label{thm:dominant-weight-theo}
+\begin{theorem}\label{thm:dominant-weight-theo}\index{weights!Highest Weight Theorem}
For each dominant integral \(\lambda \in P\) there exists precisely one
finite-dimensional simple \(\mathfrak{g}\)-module \(M\) whose highest weight
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 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 \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\). 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.
+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 \(\mathfrak{sl}_3(K)\). Indeed, as in
+Proposition~\ref{thm:sl3-mod-is-highest-weight}, one is able to show\dots
+
+\begin{proposition}\label{thm:fin-dim-simple-mod-has-singular-vector}
+ Let \(M\) be a finite-dimensional simple \(\mathfrak{g}\)-module. Then there
+ exists a nonzero weight vector \(m \in M\) which is annihilated by all
+ positive root spaces of \(\mathfrak{g}\) -- i.e. \(X \cdot m = 0\) for all
+ \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\).
+\end{proposition}
+
+\begin{proof}
+ If \(\lambda\) is the highest weight of \(M\), it suffices to take any \(m
+ \in M_\lambda\). Indeed, given \(X \in \mathfrak{g}_\alpha\) with \(\alpha
+ \in \Delta^+\), \(X \cdot m \in M_{\lambda + \alpha} = 0\) because \(\lambda
+ + \alpha \succ \lambda\).
+\end{proof}
+
+Unfortunately for us, this is where the parallels with
+Proposition~\ref{thm:sl3-mod-is-highest-weight} end. 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}
+\(\mathfrak{g}\)-module \(M\) whose highest weight is \(\lambda\)? In the
+process of answering this question we will come across a surprisingly elegant
+solution to our problem.
+
+If \(M\) is a finite-dimensional simple module with highest weight \(\lambda\)
+and \(m \in M_\lambda\), we already know that \(X \cdot m = 0\) for any \(m \in
+M_\lambda\) and \(X \in \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\). Since
+\(M = \mathcal{U}(\mathfrak{g}) \cdot m\), 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
+\emph{freely} 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 \(M(\lambda) =
- \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K m^+\), where the action of
- \(\mathfrak{b}\) on \(K m^+\) is given by \(H \cdot m^+ = \lambda(H) m^+\)
- for all \(H \in \mathfrak{h}\) and \(X \cdot m^+ = 0\) for \(X \in
- \mathfrak{g}_{\alpha}\), \(\alpha \in \Delta^+\), is called \emph{the Verma
- module of weight \(\lambda\)}.
+ Given \(\lambda \in \mathfrak{h}^*\), consider the \(\mathfrak{b}\)-module
+ \(K m^+\) where \(H \cdot m^+ = \lambda(H) m^+\) for all \(H \in
+ \mathfrak{h}\) and \(X \cdot m^+ = 0\) for \(X \in \mathfrak{g}_{\alpha}\)
+ with \(\alpha \in \Delta^+\). The \(\mathfrak{g}\)-module \(M(\lambda) =
+ \operatorname{Ind}_{\mathfrak{b}}^{\mathfrak{g}} K m^+\) is called \emph{the
+ Verma module of weight \(\lambda\)}.
+\end{definition}
+
+\begin{example}\label{ex:sl2-verma}
+ If \(\mathfrak{g} = \mathfrak{sl}_2(K)\), then we can take \(\mathfrak{h} = K
+ h\) and \(\mathfrak{b} = K e \oplus K h\). In this setting, the linear map
+ \(g : \mathfrak{h}^* \to K\) defined by \(g(h) = 1\) affords us a canonical
+ identification \(\mathfrak{h}^* = K g \cong K\), so that given \(\lambda \in
+ K\) we may denote \(M(\lambda g)\) simply by \(M(\lambda)\). Using this
+ notation \(M(\lambda) = \bigoplus_{k \ge 0} K f^k \cdot m^+\), and the action
+ of \(\mathfrak{sl}_2(K)\) on \(M(\lambda)\) is given by
+ formula (\ref{eq:sl2-verma-formulas}).
+ \begin{equation}\label{eq:sl2-verma-formulas}
+ \begin{aligned}
+ f^k \cdot m^+ & \overset{e}{\mapsto} k(\lambda+1-k) f^{k-1} \cdot m^+ &
+ f^k \cdot m^+ & \overset{f}{\mapsto} f^{k+1} \cdot m^+ &
+ f^k \cdot m^+ & \overset{h}{\mapsto} (\lambda - 2k) f^k \cdot m^+ &
+ \end{aligned}
+ \end{equation}
+\end{example}
+
+\begin{example}\label{ex:verma-is-not-irr}
+ Consider the \(\mathfrak{sl}_2(K)\)-module \(M(2)\) as described in
+ Example~\ref{ex:sl2-verma}. It follows from formula
+ (\ref{eq:sl2-verma-formulas}) that the action of \(\mathfrak{sl}_2(K)\) on
+ \(M(2)\) is given by
+ \begin{center}
+ \begin{tikzcd}
+ \cdots \rar[bend left=60]{-10}
+ & M(2)_{-6} \rar[bend left=60]{-4} \lar[bend left=60]{1}
+ & M(2)_{-4} \rar[bend left=60]{0} \lar[bend left=60]{1}
+ & M(2)_{-2} \rar[bend left=60]{2} \lar[bend left=60]{1}
+ & M(2)_0 \rar[bend left=60]{2} \lar[bend left=60]{1}
+ & M(2)_2 \lar[bend left=60]{1}
+ \end{tikzcd}
+ \end{center}
+ where \(M(2)_{2 - 2 k} = K f^k \cdot m^+\). Here the top arrows represent the
+ action of \(e\) and the bottom arrows represent the action of \(f\). The
+ scalars labeling each arrow indicate to which multiple of \(f^{k \pm 1} \cdot
+ m^+\) the elements \(e\) and \(f\) send \(f^k \cdot m^+\). The string of
+ weight spaces to the left of the diagram is infinite. Since \(e \cdot (f^3
+ \cdot m^+) = 0\), it is easy to see that subspace \(\bigoplus_{k \ge 3} K f^k
+ \cdot m^+\) is a (maximal) \(\mathfrak{sl}_2(K)\)-submodule, which is
+ isomorphic to \(M(-4)\).
+\end{example}
+
+These last examples show that, unlike most modules we have so far encountered,
+Verma modules are \emph{highly infinite-dimensional}. Indeed, it follows from
+the PBW Theorem that the regular module \(\mathcal{U}(\mathfrak{g})\) is a free
+\(\mathfrak{b}\)-module of infinite rank -- equal to the codimension of
+\(\mathcal{U}(\mathfrak{b})\) in \(\mathcal{U}(\mathfrak{g})\). Hence \(\dim
+M(\lambda)\), which is the same as the rank of \(\mathcal{U}(\mathfrak{g})\) as
+a \(\mathfrak{b}\)-module, is also infinite. Nevertheless, it turns out that
+finite-dimensional modules and Verma module may both be seen as particular
+cases of a more general pattern. This leads us to the following definitions.
+
+\begin{definition}
+ Let \(M\) be a \(\mathfrak{g}\)-module. A vector \(m \in M\) is called
+ \emph{singular} if it is annihilated by all positive weight spaces of
+ \(\mathfrak{g}\) -- i.e. \(X \cdot m = 0\) for all \(X \in
+ \mathfrak{g}_\alpha\), \(\alpha \in \Delta^+\).
\end{definition}
-It turns out that \(M(\lambda)\) enjoys many of the features we've grown used
-to in the past chapters. Explicitly\dots
+\begin{definition}\label{def:highest-weight-mod}
+ A \(\mathfrak{g}\)-module \(M\) is called \emph{a highest weight module} if
+ there exists some singular weight vector \(m^+ \in M_\lambda\) such that \(M
+ = \mathcal{U}(\mathfrak{g}) \cdot m^+\). Any such \(m^+\) is called \emph{a
+ highest weight vector}, while \(\lambda\) is called \emph{the highest weight
+ of \(M\)}.
+\end{definition}
-\begin{proposition}\label{thm:verma-is-weight-mod}
- The Verma module \(M(\lambda)\) is generated \(m^+ = 1 \otimes m^+ \in
- M(\lambda)\) as in Definition~\ref{def:verma}. The weight spaces
- decomposition
+\begin{example}
+ Proposition~\ref{thm:fin-dim-simple-mod-has-singular-vector} is equivalent to
+ the fact that every finite-dimensional simple \(\mathfrak{g}\)-module is a
+ highest weight module.
+\end{example}
+
+\begin{example}
+ It should be obvious from the definitions that \(M(\lambda)\) is a highest
+ weight module of highest weight \(\lambda\) and highest weight vector \(m^+ =
+ 1 \otimes m^+\) as in Definition~\ref{def:verma}. Indeed, \(u \otimes m^+ = u
+ \cdot m^+\) for all \(u \in \mathcal{U}(\mathfrak{g})\), which already shows
+ \(M(\lambda)\) is generated by \(m^+\). In particular,
+ \begin{align*}
+ H \cdot m^+ & = H \otimes m^+ = 1 \otimes H \cdot m^+ = \lambda(H) m^+ \\
+ X \cdot m^+ & = X \otimes m^+ = 1 \otimes X \cdot m^+ = 0
+ \end{align*}
+ for all \(H \in \mathfrak{h}\) and \(X \in \mathfrak{g}_\alpha\), \(\alpha
+ \in \Delta^+\).
+\end{example}
+
+While Verma modules show that a highest weight module needs not to be
+finite-dimensional, it turns out that highest weight modules enjoy many of the
+features we've grown used to in the past chapters. Explicitly, we may establish
+the properties described in the following proposition, whose statement should
+also explain the nomenclature of Definition~\ref{def:highest-weight-mod}.
+
+\begin{proposition}\label{thm:high-weight-mod-is-weight-mod}
+ Let \(M\) be a highest weight \(\mathfrak{g}\)-module with highest weight
+ vector \(m \in M_\lambda\). The weight spaces decomposition
\[
- M(\lambda) = \bigoplus_{\mu \in \mathfrak{h}^*} M(\lambda)_\mu
+ M = \bigoplus_{\mu \in \mathfrak{h}^*} M_\mu
\]
- 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
- \(m^+ \in M(\lambda)\).
+ holds. Furthermore, \(\dim M_\mu < \infty\) for all \(\mu \in
+ \mathfrak{h}^*\) and \(\dim M_\lambda = 1\) -- i.e. \(M_\lambda = K m\).
+ Finally, given a weight \(\mu\) of \(M\), \(\lambda \succeq \mu\) -- so that
+ the highest weight \(\lambda\) of \(M\) is unique and coincides with the
+ largest of the weights of \(M\).
\end{proposition}
\begin{proof}
- The PBW Theorem implies that \(M(\lambda)\) is spanned by the vectors
- \(F_{\alpha_{i_1}} F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+\) for
- \(\Delta^+ = \{\alpha_1, \ldots, \alpha_r\}\) and \(F_{\alpha_i} \in
- \mathfrak{g}_{- \alpha_i}\) as in the proof of
- Proposition~\ref{thm:distinguished-subalgebra}. But
+ Since \(M = \mathcal{U}(\mathfrak{g}) \cdot m\), the PBW Theorem implies
+ that \(M\) is spanned by the vectors \(F_{\alpha_{i_1}} F_{\alpha_{i_2}}
+ \cdots F_{\alpha_{i_s}} \cdot m\) for \(\Delta^+ = \{\alpha_1, \ldots,
+ \alpha_r\}\) and \(F_{\alpha_i} \in \mathfrak{g}_{- \alpha_i}\) as in the
+ proof of Proposition~\ref{thm:distinguished-subalgebra}. But
\[
\begin{split}
- H \cdot (F_{\alpha_{i_1}} F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+)
+ H \cdot (F_{\alpha_{i_1}} F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}}
+ \cdot m)
& = ([H, F_{\alpha_{i_1}}] + F_{\alpha_{i_1}} H)
- F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\
- & = - \alpha_{i_1}(H) F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+
+ F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m \\
+ & = - \alpha_{i_1}(H) F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m
+ F_{\alpha_{i_1}} ([H, F_{\alpha_{i_2}}] + F_{\alpha_{i_2}} H)
- F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\
+ F_{\alpha_{i_2}} \cdots F_{\alpha_{i_s}} \cdot m \\
& \;\; \vdots \\
& = (- \alpha_{i_1} - \cdots - \alpha_{i_s})(H)
- F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+
- + F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} H \cdot m^+ \\
+ F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m
+ + F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} H \cdot m \\
& = (\lambda - \alpha_{i_1} - \cdots - \alpha_{i_s})(H)
- F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+ \\
- & \therefore F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m^+
- \in M(\lambda)_{\lambda - \alpha_{i_1} - \cdots - \alpha_{i_s}}
+ F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m \\
+ & \therefore F_{\alpha_{i_1}} \cdots F_{\alpha_{i_s}} \cdot m
+ \in M_{\lambda - \alpha_{i_1} - \cdots - \alpha_{i_s}}
\end{split}
\]
- Hence \(M(\lambda) \subset \bigoplus_{\mu \in \mathfrak{h}^*}
- M(\lambda)_\mu\), as desired. In fact we have established
+ Hence \(M \subset \bigoplus_{\mu \in \mathfrak{h}^*} M_\mu\), as desired. In
+ fact we have established
\[
- M(\lambda)
+ M
\subset
\bigoplus_{k_i \in \mathbb{N}}
- M(\lambda)_{\lambda - k_1 \cdot \alpha_1 - \cdots - k_r \cdot \alpha_r}
+ M_{\lambda - k_1 \cdot \alpha_1 - \cdots - k_r \cdot \alpha_r}
\]
where \(\{\alpha_1, \ldots, \alpha_r\} = \Delta^+\), so that all weights of
- \(M(\lambda)\) have the form \(\mu = \lambda - k_1 \cdot \alpha_1 - \cdots -
- k_r \cdot \alpha_r\).
-
- 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
- \(m^+\) is nonzero weight vector. Clearly \(m^+ \in M_\lambda\). The
- Poincaré-Birkhoff-Witt Theorem implies \(\mathcal{U}(\mathfrak{g})\) is a
- free \(\mathfrak{b}\)-module, so that
- \[
- M(\lambda)
- \cong \left(\bigoplus_i \mathcal{U}(\mathfrak{b}) \right)
- \otimes_{\mathcal{U}(\mathfrak{b})} K m^+
- \cong \bigoplus_i \mathcal{U}(\mathfrak{b})
- \otimes_{\mathcal{U}(\mathfrak{b})} K m^+
- \cong \bigoplus_i K m^+
- \ne 0
- \]
- 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
- weight vector \(m^+\).
-
- 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
+ \(M\) have the form \(\mu = \lambda - k_1 \cdot \alpha_1 - \cdots - k_r \cdot
+ \alpha_r\). This already gives us that the weights of \(M\) are bounded by
+ \(\lambda\).
+
+ To see that \(\dim M_\mu < \infty\), simply note that there are only finitely
+ many monomials \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots
F_{\alpha_s}^{k_s}\) such that \(\mu = \lambda + k_1 \cdot \alpha_1 + \cdots
- + k_s \cdot \alpha_s\). Since \(M(\lambda)_\mu\) is spanned by the images of
- \(m^+\) under such monomials, we conclude \(\dim M(\lambda) < \infty\). In
- particular, there is a single monomials \(F_{\alpha_1}^{k_1}
- F_{\alpha_2}^{k_2} \cdots F_{\alpha_s}^{k_s}\) such that \(\lambda = \lambda
- + k_1 \cdot \alpha_1 + \cdots + k_s \cdot \alpha_s\) -- which is, of course,
- the monomial where \(k_1 = \cdots = k_n = 0\). Hence \(\dim M_\lambda = 1\).
+ + k_s \cdot \alpha_s\). Since \(M_\mu\) is spanned by the images of \(m\)
+ under such monomials, we conclude \(\dim M_\mu < \infty\). In particular,
+ there is a single monomial \(F_{\alpha_1}^{k_1} F_{\alpha_2}^{k_2} \cdots
+ F_{\alpha_s}^{k_s}\) such that \(\lambda = \lambda + k_1 \cdot \alpha_1 +
+ \cdots + k_s \cdot \alpha_s\) -- which is, of course, the monomial where
+ \(k_1 = \cdots = k_n = 0\). Hence \(\dim M_\lambda = 1\).
\end{proof}
-\begin{example}\label{ex:sl2-verma}
- If \(\mathfrak{g} = \mathfrak{sl}_2(K)\), then we can take \(\mathfrak{h} = K
- h\) and \(\mathfrak{b} = K e \oplus K h\). If \(\lambda \in
- \mathfrak{h}^*\) is the map \(h \mapsto 2\) then \(M(\lambda) =
- \bigoplus_{k \ge 0} K f^k \cdot m^+\), and the action of
- \(\mathfrak{sl}_2(K)\) on \(M(\lambda)\) is given by the formulas in
- (\ref{eq:sl2-verma-formulas}). Visually,
- \begin{center}
- \begin{tikzcd}
- \cdots \rar[bend left=60]{-10}
- & M(\lambda)_{-6} \rar[bend left=60]{-4} \lar[bend left=60]{1}
- & M(\lambda)_{-4} \rar[bend left=60]{0} \lar[bend left=60]{1}
- & M(\lambda)_{-2} \rar[bend left=60]{2} \lar[bend left=60]{1}
- & M(\lambda)_0 \rar[bend left=60]{2} \lar[bend left=60]{1}
- & M(\lambda)_2 \lar[bend left=60]{1}
- \end{tikzcd}
- \end{center}
- where \(M(\lambda)_{2 - 2 k} = K f^k \cdot m^+\). Here the top arrows
- represent the action of \(e\) and the bottom arrows represent the action of
- \(f\). The scalars labeling each arrow indicate to which multiple of \(f^{k
- \pm 1} \cdot m^+\) the elements \(e\) and \(f\) send \(f^k \cdot m^+\). The
- string of weight spaces to the left of the diagram is infinite.
- \begin{equation}\label{eq:sl2-verma-formulas}
- \begin{aligned}
- f^k \cdot m^+ & \overset{e}{\mapsto} (2 - k(k + 1)) f^{k - 1} \cdot m^+ &
- f^k \cdot m^+ & \overset{f}{\mapsto} f^{k + 1} \cdot m^+ &
- f^k \cdot m^+ & \overset{h}{\mapsto} (2 - 2k) f^k \cdot m^+ &
- \end{aligned}
- \end{equation}
-\end{example}
-
-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}
+At this point it is important to note that, far from a ``misbehaved'' class of
+examples, Verma modules hold a very special place in the theory of highest
+weight modules. Intuitively speaking, the Verma module \(M(\lambda)\) should
+really be though-of as ``the freest highest weight \(\mathfrak{g}\)-module of
+highest weight \(\lambda\)''. In practice, this translates to the following
+universal property.
\begin{proposition}
Let \(M\) be a \(\mathfrak{g}\)-module and \(m \in M_\lambda\) be a singular
@@ -881,35 +920,51 @@ universal property of \(M(\lambda)\).
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.
+ such that \(f(u \otimes m^+) = u \cdot g(m^+)\) for all \(u \in
+ \mathcal{U}(\mathfrak{g})\), 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.
+ Any \(K\)-linear map \(g : K m^+ \to M\) is determined by \(m = g(m^+)\).
+ Finally, notice that \(g\) is a \(\mathfrak{b}\)-homomorphism if, and only if
+ \(m\) is a singular vector lying in \(M_\lambda\).
\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.
+Why is any of this interesting to us, however? After all, Verma modules are not
+specially well suited candidates for a proof of the Highest Weight Theorem.
+Indeed, we have seen in Example~\ref{ex:verma-is-not-irr} that in general
+\(M(\lambda)\) is not simple, nor is it ever finite-dimensional. Nevertheless,
+we may use \(M(\lambda)\) to establish Theorem~\ref{thm:dominant-weight-theo}
+as follows.
+
+Suppose \(M\) is a highest weight \(\mathfrak{g}\)-module of highest weight
+\(\lambda\) with highest weight vector \(m\). By the last proposition, there is
+a \(\mathfrak{g}\)-homomorphism \(f : M(\lambda) \to M\) such that \(f(m^+) =
+m\). Since \(M = \mathcal{U}(\mathfrak{g}) \cdot m\), \(f\) is surjective and
+therefore \(M \cong \mfrac{M(\lambda)}{\ker f}\). Hence\dots
+
+\begin{proposition}
+ Let \(M\) be a highest weight \(\mathfrak{g}\)-module of highest weight
+ \(\lambda\). Then \(M\) is quotient of \(M(\lambda)\). If \(M\) is simple
+ then \(M\) is the quotient of \(M(\lambda)\) by a maximal
+ \(\mathfrak{g}\)-submodule.
+\end{proposition}
+
+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
spaces. In particular, \(M(\lambda)\) has a unique maximal submodule
\(N(\lambda)\) and a unique simple quotient \(L(\lambda) =
- \sfrac{M(\lambda)}{N(\lambda)}\).
+ \sfrac{M(\lambda)}{N(\lambda)}\). Any simple highest weight
+ \(\mathfrak{g}\)-module has the form \(L(\lambda)\) for some unique \(\lambda
+ \in \mathfrak{h}^*\).
\end{proposition}
\begin{proof}
Let \(N \subset M(\lambda)\) be a submodule and take any nonzero \(n \in N\).
- Because of Proposition~\ref{thm:verma-is-weight-mod}, we know there are
- \(\mu_1, \ldots, \mu_r \in \mathfrak{h}^*\) and nonzero \(m_i \in
+ Because of Proposition~\ref{thm:high-weight-mod-is-weight-mod}, we know there
+ are \(\mu_1, \ldots, \mu_r \in \mathfrak{h}^*\) and nonzero \(m_i \in
M(\lambda)_{\mu_i}\) such that \(n = m_1 + \cdots + m_r\). We want to show
\(m_i \in N\) for all \(i\).
@@ -957,17 +1012,29 @@ primary interest to us. As it turns out, these can be easily classified.
quotient.
\end{proof}
+\begin{corollary}\label{thm:classification-of-simple-high-weight-mods}
+ Let \(M\) be a simple weight \(\mathfrak{g}\)-module of weight \(\lambda\).
+ Then \(M \cong L(\lambda)\).
+\end{corollary}
+
+We thus know that \(L(\lambda)\) is the only possible candidate for the
+\(\mathfrak{g}\)-module \(M\) in the statement of
+Theorem~\ref{thm:dominant-weight-theo}. We should also note that our past
+examples indicate that \(L(\lambda)\) does fulfill its required role.
+Indeed\dots
+
\begin{example}\label{ex:sl2-verma-quotient}
- If \(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto 2\), we
- can see from Example~\ref{ex:sl2-verma} that \(N(\lambda) = \bigoplus_{k \ge
- 3} K f^k \cdot m^+\), so that \(L(\lambda)\) is the \(3\)-dimensional simple
- \(\mathfrak{sl}_2(K)\)-module -- i.e. the finite-dimensional simple module
- with highest weight \(\lambda\) constructed in chapter~\ref{ch:sl3}.
+ Consider the \(\mathfrak{sl}_2(K)\) module \(M(2)\) as described in
+ Example~\ref{ex:sl2-verma}. We can see from Example~\ref{ex:verma-is-not-irr}
+ that \(N(2) = \bigoplus_{k \ge 3} K f^k \cdot m^+\), so that \(L(2)\) is the
+ \(3\)-dimensional simple \(\mathfrak{sl}_2(K)\)-module -- i.e. the
+ finite-dimensional simple module with highest weight \(2\) constructed in
+ chapter~\ref{ch:sl3}.
\end{example}
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
+dominant \(\lambda \in P\). 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
@@ -990,55 +1057,50 @@ We refer the reader to \cite[ch. 21]{humphreys} for further details. We are now
ready to prove the Highest Weight Theorem.
\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 \(m^+ + N(\lambda) \in L(\lambda)_\lambda\). 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
- \(L(\lambda)\), 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.
+ We begin by the ``existence'' part of the theorem. Let \(\lambda\) be a
+ dominant integral weight of \(\mathfrak{g}\). Since \(\dim L(\lambda) <
+ \infty\), all its left is to show that \(M = L(\lambda)\) is indeed a highest
+ weight module of highest weight \(\lambda\). It is clear from the definitions
+ that \(m^+ + N(\lambda) \in L(\lambda)_\lambda\) is singular and generates
+ all of \(L(\lambda)\). Hence it suffices to show that \(m^+ + N(\lambda)\) is
+ nonzero. But this is the same as checking that \(m^+ \notin N(\lambda)\),
+ which is also clear from the previous definitions. As for the uniqueness of
+ \(M\), it suffices to apply
+ Corollary~\ref{thm:classification-of-simple-high-weight-mods}.
\end{proof}
-We should point out that Proposition~\ref{thm:verma-is-finite-dim} fails for
-non-dominant \(\lambda \in P\). While \(\lambda\) is always a maximal weight of
-\(M(\lambda)\), one can show that if \(\lambda \in P\) is not dominant then
-\(N(\lambda) = 0\) and \(M(\lambda)\) is simple. For instance, if
-\(\mathfrak{g} = \mathfrak{sl}_2(K)\) and \(\lambda : h \mapsto -2\) then the
-action of \(\mathfrak{g}\) on \(M(\lambda)\) is given by
-\begin{center}
- \begin{tikzcd}
- \cdots \rar[bend left=60]{-20}
- & M(\lambda)_{-8} \rar[bend left=60]{-12} \lar[bend left=60]{1}
- & M(\lambda)_{-6} \rar[bend left=60]{-6} \lar[bend left=60]{1}
- & M(\lambda)_{-4} \rar[bend left=60]{-2} \lar[bend left=60]{1}
- & M(\lambda)_{-2} \lar[bend left=60]{1}
- \end{tikzcd},
-\end{center}
-so we can see that \(M(\lambda)\) has no proper submodules. Verma modules can
-thus serve as examples of infinite-dimensional simple modules. Our next
-question is: what are \emph{all} the infinite-dimensional simple
+We would now like to conclude this chapter by describing the situation where
+\(\lambda \in \mathfrak{h}^*\) is not dominant integral. We begin by pointing
+out that Proposition~\ref{thm:verma-is-finite-dim} fails in the general
+setting. For instance, consider\dots
+
+\begin{example}
+ The action of \(\mathfrak{sl}_2(K)\) on \(M(-4)\) is given by the following
+ diagram. In general, it is possible to check using formula
+ (\ref{eq:sl2-verma-formulas}) that \(e\) always maps \(f^{k + 1} \cdot m^+\)
+ to a nonzero multiple of \(f^k \cdot m^+\), so we can see that \(M(-4)\) has
+ no proper submodules and \(M(-4) \cong L(-4)\).
+ \begin{center}
+ \begin{tikzcd}
+ \cdots \rar[bend left=60]{-28}
+ & M(-4)_{-10} \rar[bend left=60]{-18} \lar[bend left=60]{1}
+ & M(-4)_{-8} \rar[bend left=60]{-10} \lar[bend left=60]{1}
+ & M(-4)_{-6} \rar[bend left=60]{-4} \lar[bend left=60]{1}
+ & M(-4)_{-4} \lar[bend left=60]{1}
+ \end{tikzcd},
+ \end{center}
+\end{example}
+
+While \(L(\lambda)\) is always a highest weight module of highest weight
+\(\lambda\), one can show that if \(\lambda\) is not dominant integral then
+\(L(\lambda) \cong M(\lambda)\) is infinite-dimensional. Indeed, since the
+highest weight of a finite-dimensional simple \(\mathfrak{g}\)-module is always
+dominant integral, \(L(\lambda)\) is infinite-dimensional for any \(\lambda\)
+which is not dominant integral. Since the only \(\mathfrak{g}\)-submodules of
+\(M(\lambda)\) of infinite codimension is \(0\), it follows that \(N(\lambda) =
+0\) and \(L(\lambda) \cong M(\lambda)\).
+
+Verma modules can thus serve as examples of infinite-dimensional simple
+modules. In the next chapter we expand our previous results by exploring the
+question: what are \emph{all} the infinite-dimensional simple
\(\mathfrak{g}\)-modules?