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}