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

Commit
51e9c3007ff31a835c1a6bfe39bdb8aac2251db1
Parent
e439b14d73a79d1c7006a8da078c5aa988d10213
Author
Pablo <pablo-escobar@riseup.net>
Date

Finished hydrating the section on the basics of coherent families and coherent extensions

Diffstat

1 file changed, 137 insertions, 95 deletions

Status File Name N° Changes Insertions Deletions
Modified sections/mathieu.tex 232 137 95
diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -475,7 +475,7 @@ On a perhaps less derogatory note, \(\mathcal{M}\) also deserves to be called
 \emph{a family}. This is because \(\mathcal{M}\) consists of lots of smaller
 cuspidal representations which fit together inside of it in a \emph{coherent}
 fashion. Mathieu's engineous breaktrough was the realization that
-\(\mathcal{M}\) is a particular example of a more general pattern, which ha
+\(\mathcal{M}\) is a particular example of a more general pattern, which he
 named \emph{coherent families}.
 
 \begin{definition}
@@ -517,7 +517,32 @@ named \emph{coherent families}.
   \(x^k\) we can see \((\mathcal{M}(\sfrac{1}{2}))[0] \cong K[x, x^{-1}]\).
 \end{example}
 
-% TODO: Move this somewhere else
+Our hope is that given an irreducible cuspidal representation \(V\), we can
+somehow find \(V\) inside of a coherent \(\mathfrak{g}\)-family, such as in the
+case of \(K[x, x^{-1}]\) and \(\mathcal{M}\) from
+example~\ref{ex:sl-laurent-family}. This leads us to the following definition.
+
+\begin{definition}
+  Given an admissible representation \(V\) of \(\mathfrak{g}\) of degree \(d\),
+  a coherent extension \(\mathcal{M}\) of \(V\) is a coherent family
+  \(\mathcal{M}\) of degree \(d\) that contains \(V\) as a subquotient.
+\end{definition}
+
+Our goal is now showing that every admissible representation has a coherent
+extension. The idea then is to classify coherent extensions, and classify which
+submodules of a given coherent extension are actually irreducible cuspidal
+representations. If every admissible \(\mathfrak{g}\)-module fits inside a
+coherent extension, this would lead to classification of all irreducible
+cuspidal representations, which we now know is the key for the solution of out
+classification problem. However, there are some complications to this scheme.
+
+Leaving aside the question of existence for a second, we should point out that
+coherent families turn out to be rather complicated on their own. In fact they
+are too complicated to classify in general. Idealy, we would like to find
+\emph{nice} coherent extensions -- ones we can actually classify. For instance,
+we may search for \emph{simple} coherent extensions, which are defined as
+follows.
+
 \begin{definition}
   A coherent family \(\mathcal{M}\) is called \emph{simple} if it contains no
   proper coherent subfamilies -- i.e. \(\mathcal{M}\) is a simple object in the
@@ -534,86 +559,86 @@ named \emph{coherent families}.
   \mathfrak{h}^*\).
 \end{definition}
 
-\begin{definition}
-  Given an admissible representation \(V\) of \(\mathfrak{g}\) of degree \(d\),
-  a coherent extension \(\mathcal{M}\) of \(V\) is a coherent family
-  \(\mathcal{M}\) of degree \(d\) that contains \(V\) as a subquotient.
-\end{definition}
+Another natural cadidate for the role of ``nice extensions'' are the completely
+reducible coherent families -- i.e. families wich are completely reducible as
+\(\mathfrak{g}\)-modules. These turn out to be very easy to produce. Namely,
+there is a construction, known as \emph{the semisimplification\footnote{Recall
+that a ``semisimple'' is a synonim for ``completely reducible'' in the context
+of modules.} of a coherent family}, which takes a coherent extension of \(V\)
+to a completely reducible coherent extension of \(V\).
 
+% TODO: Note somewhere that M[mu] is a submodule
 % Mathieu's proof of this is somewhat profane, I don't think it's worth
 % including it in here
-% TODO: Define the notation for M[mu] somewhere else
-% TODO: Note somewhere that M[mu] is a submodule
 \begin{lemma}
   Given a coherent family \(\mathcal{M}\) and \(\lambda \in \mathfrak{h}^*\),
   \(\mathcal{M}[\lambda]\) has finite length as a \(\mathfrak{g}\)-module.
 \end{lemma}
 
-% TODO: From this we may conclude that any admissible submodule is a submodule
-% of the semisimplification of any of its coherent extensions
+% TODO: Point out this construction is NOT functorial, since it depends on the
+% choice of composition series
 \begin{corollary}
   Let \(\mathcal{M}\) be a coherent family of degree \(d\). There exists a
   unique completely reducible coherent family
   \(\mathcal{M}^{\operatorname{ss}}\) of degree \(d\) such that the composition
   series of \(\mathcal{M}^{\operatorname{ss}}[\lambda]\) is the same as that of
   \(\mathcal{M}[\lambda]\) for all \(\lambda \in \mathfrak{h}^*\), called
-  \emph{the semisimplification\footnote{Recall that a ``semisimple'' is a
-  synonim for ``completely reducible'' in the context of modules.} of
-  \(\mathcal{M}\)}.
-
-  Namely, if \(\{\lambda_i\}_i\) is a set of representatives of the
-  \(Q\)-cosets of \(\mathfrak{h}^*\) and \(0 = \mathcal{M}_{i 0} \subset
-  \mathcal{M}_{i 1} \subset \cdots \subset \mathcal{M}_{i n_i} =
-  \mathcal{M}[\lambda_i]\) is a composition series,
+  \emph{the semisimplification of \(\mathcal{M}\)}.
+
+  Namely, if \(\lambda \in \mathfrak{h}^*\) and \(0 = \mathcal{M}_{\lambda 0}
+  \subset \mathcal{M}_{\lambda 1} \subset \cdots \subset \mathcal{M}_{\lambda
+  n_\lambda} = \mathcal{M}[\lambda]\) is a composition series,
   \[
     \mathcal{M}^{\operatorname{ss}}
-    \cong \bigoplus_{i j} \mfrac{\mathcal{M}_{i j + 1}}{\mathcal{M}_{i j}}
+    \cong \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}}
+          \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}
   \]
 \end{corollary}
 
 \begin{proof}
   The uniqueness of \(\mathcal{M}^{\operatorname{ss}}\) should be clear:
   since \(\mathcal{M}^{\operatorname{ss}}\) is completely reducible, so is
-  \(\mathcal{M}^{\operatorname{ss}}[\lambda_i]\). Hence
+  \(\mathcal{M}^{\operatorname{ss}}[\lambda]\). Hence
   \[
-    \mathcal{M}^{\operatorname{ss}}[\lambda_i]
-    \cong \bigoplus_j \mfrac{\mathcal{M}_{i j + 1}}{\mathcal{M}_{i j}}
+    \mathcal{M}^{\operatorname{ss}}[\lambda]
+    \cong
+    \bigoplus_i \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}
   \]
 
   As for the existence of the semisimplification, it suffices to show
   \[
     \mathcal{M}^{\operatorname{ss}}
-    = \bigoplus_{i j} \mfrac{\mathcal{M}_{i j + 1}}{\mathcal{M}_{i j}}
+    = \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}}
+    \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}
   \]
   is indeed a completely reducible coherent family of degree \(d\).
 
   We know from examples~\ref{ex:submod-is-weight-mod} and
-  \ref{ex:quotient-is-weight-mod} that each quotient \(\mfrac{\mathcal{M}_{i j
-  + 1}}{\mathcal{M}_{i j}}\) is a weight module. Hence
-  \(\mathcal{M}^{\operatorname{ss}}\) is a weight module. Furthermore, given
-  \(\mu \in \lambda_k + Q\)
+  \ref{ex:quotient-is-weight-mod} that each quotient
+  \(\mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}\) is a weight
+  module. Hence \(\mathcal{M}^{\operatorname{ss}}\) is a weight module.
+  Furthermore, given \(\mu \in \mathfrak{h}^*\)
   \[
     \mathcal{M}_\mu^{\operatorname{ss}}
-    = \bigoplus_{i j}
+    = \bigoplus_{\substack{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q} \\ i}}
       \left(
-      \mfrac{\mathcal{M}_{i j + 1}}{\mathcal{M}_{i j}}
+      \mfrac{\mathcal{M}_{\lambda i + 1}}{\mathcal{M}_{\lambda i}}
       \right)_\mu
-    = \bigoplus_j
+    = \bigoplus_i
       \left(
-      \mfrac{\mathcal{M}_{k j + 1}}{\mathcal{M}_{k j}}
+      \mfrac{\mathcal{M}_{\mu i + 1}}{\mathcal{M}_{\mu i}}
       \right)_\mu
-    \cong \bigoplus_j
-      \mfrac{(\mathcal{M}_{k j + 1})_\mu}
-            {(\mathcal{M}_{k j})_\mu}
+    \cong \bigoplus_i
+      \mfrac{(\mathcal{M}_{\mu i + 1})_\mu}
+            {(\mathcal{M}_{\mu i})_\mu}
   \]
 
   In particular,
   \[
     \dim \mathcal{M}_\mu^{\operatorname{ss}}
-    = \sum_j
-      \dim (\mathcal{M}_{k j + 1})_\mu
-    - \dim (\mathcal{M}_{k j})_\mu
-    = \dim \mathcal{M}[\lambda_k]_\mu
+    = \sum_i
+      \dim (\mathcal{M}_{\mu i + 1})_\mu - \dim (\mathcal{M}_{\mu i})_\mu
+    = \dim \mathcal{M}[\mu]_\mu
     = \dim \mathcal{M}_\mu
     = d
   \]
@@ -621,20 +646,24 @@ named \emph{coherent families}.
   Likewise, given \(u \in \mathcal{U}(\mathfrak{g})_0\) the value
   \[
     \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu^{\operatorname{ss}}})
-    = \sum_j
-      \operatorname{Tr}(u\!\restriction_{(\mathcal{M}_{k j + 1})_\mu})
-    - \operatorname{Tr}(u\!\restriction_{(\mathcal{M}_{k j})_\mu})
-    = \operatorname{Tr}(u\!\restriction_{\mathcal{M}[\lambda_k]_\mu})
+    = \sum_i
+      \operatorname{Tr}(u\!\restriction_{(\mathcal{M}_{\mu i + 1})_\mu})
+    - \operatorname{Tr}(u\!\restriction_{(\mathcal{M}_{\mu i})_\mu})
+    = \operatorname{Tr}(u\!\restriction_{\mathcal{M}[\mu]_\mu})
     = \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu})
   \]
   is polynomial in \(\mu \in \mathfrak{h}^*\).
 \end{proof}
 
-\begin{corollary}\label{thm:admissible-is-submod-of-extension}
+As promised, if \(\mathcal{M}\) is a coherent extension of \(V\) then so is
+\(\mathcal{M}^{\operatorname{ss}}\).
+
+\begin{proposition}
   Let \(V\) be an irreducible admissible \(\mathfrak{g}\)-module and
-  \(\mathcal{M}\) be a completely reducible coherent extension of \(V\). Then
-  \(V\) is contained in \(\mathcal{M}\).
-\end{corollary}
+  \(\mathcal{M}\) be a coherent extension of \(V\). Then
+  \(\mathcal{M}^{\operatorname{ss}}\) is a coherent extension of \(V\) and
+  \(V\) is in fact a subrepresentation of \(\mathcal{M}^{\operatorname{ss}}\).
+\end{proposition}
 
 \begin{proof}
   Since \(V\) is irreducible, its support is contained in a single \(Q\)-coset.
@@ -649,12 +678,70 @@ named \emph{coherent families}.
     \to \bigoplus_j \mfrac{\mathcal{M}_{j + 1}}{\mathcal{M}_j}
     \cong \mathcal{M}^{\operatorname{ss}}[\lambda]
   \]
+\end{proof}
+
+Given the uniqueness of the semisimplification, the semisimplification of any
+completely reducible coherent extension \(\mathcal{M}\) is \(\mathcal{M}\)
+itself and therefore\dots
+
+\begin{corollary}\label{thm:admissible-is-submod-of-extension}
+  Let \(V\) be an irreducible admissible \(\mathfrak{g}\)-module and
+  \(\mathcal{M}\) be a completely reducible coherent extension of \(V\). Then
+  \(V\) is contained in \(\mathcal{M}\).
+\end{corollary}
+
+This last results provide a partial answer to the question of existence of nice
+coherent extensions. A complementary question now is: wich submodules of a nice
+coherent family are cuspidal representations?
+
+\begin{theorem}[Mathieu]
+  Let \(\mathcal{M}\) be a simple coherent family of degree \(d\) and
+  \(\lambda \in \mathfrak{h}^*\). The following conditions are equivalent.
+  \begin{enumerate}
+    \item \(\mathcal{M}[\lambda]\) is irreducible.
+    \item \(F_\alpha\!\restriction_{\mathcal{M}[\lambda]}\) is injective for
+      all \(\alpha \in \Delta\).
+    \item \(\mathcal{M}[\lambda]\) is cuspidal.
+  \end{enumerate}
+\end{theorem}
+
+\begin{proof}
+  The fact that \strong{(i)} and \strong{(iii)} are equivalent follows directly
+  from corollary~\ref{thm:cuspidal-mod-equivs}. Likewise, it is clear from the
+  corollary that \strong{(iii)} implies \strong{(ii)}. All it's left is to show
+  \strong{(ii)} implies \strong{(iii)}\footnote{This isn't already clear from
+  corollary~\ref{thm:cuspidal-mod-equivs} because, at first glance,
+  $\mathcal{M}[\lambda]$ may not be irreducible for some $\lambda$ satisfying
+  \strong{(ii)}. We will show this is never the case.}.
 
-  It then follows from the uniqueness of the semisimplification of
-  \(\mathcal{M}\) that \(\mathcal{M} \cong \mathcal{M}^{\operatorname{ss}}\),
-  so we have an inclusion \(V \to \mathcal{M}\).
+  Suppose \(F_\alpha\) acts injectively in the subrepresentation
+  \(\mathcal{M}[\lambda]\), for all \(\alpha \in \Delta\). Since
+  \(\mathcal{M}[\lambda]\) has finite length, \(\mathcal{M}[\lambda]\) contains
+  an infinite-dimensiona irreducible \(\mathfrak{g}\)-submodule \(V\).
+  Moreover, again by corollary~\ref{thm:cuspidal-mod-equivs} we conclude \(V\)
+  is a cuspidal representation, and its degree is bounded by \(d\). We want to
+  show \(\mathcal{M}[\lambda] = V\).
+
+  We claim the set \(U = \{\mu \in \mathfrak{h}^* : \mathcal{M}_\mu \ \text{is
+  a simple $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is Zariski-open. If we
+  suppose this is the case for a moment or two, it follows from the fact that
+  \(\mathcal{M}\) is simple and \(\operatorname{supp}_{\operatorname{ess}} V\)
+  is Zariski-dense that \(U \cap \operatorname{supp}_{\operatorname{ess}} V\)
+  is non-empty. In other words, there is some \(\mu \in \mathfrak{h}^*\) such
+  that \(\mathcal{M}_\mu\) is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module
+  and \(\dim V_\mu = \deg V\).
+
+  % Here we use thm:centralizer-multiplicity
+  In particular, \(V_\mu \ne 0\), so \(V_\mu = \mathcal{M}_\mu\). Now given any
+  irreducible \(\mathfrak{g}\)-module \(W\), the multiplicity of \(W\) in
+  \(\mathcal{M}[\lambda]\) is the same as the multiplicity \(W_\mu\) in
+  \(\mathcal{M}_\mu\) as a \(\mathcal{U}(\mathfrak{g})_0\)-module, which is, of
+  course, \(1\) if \(W \cong V\) and \(0\) otherwise. Hence
+  \(\mathcal{M}[\lambda] = V\) and \(\mathcal{M}[\lambda]\) is cuspidal.
 \end{proof}
 
+To finish the proof, we now show\dots
+
 \begin{lemma}
   Let \(\mathcal{M}\) be a coherent family. The set \(U = \{\lambda \in
   \mathfrak{h}^* : \mathcal{M}_\lambda \ \text{is a simple
@@ -775,51 +862,6 @@ named \emph{coherent families}.
   the union of Zariski-open subsets and is therefore open. We are done.
 \end{proof}
 
-\begin{theorem}[Mathieu]
-  Let \(\mathcal{M}\) be a simple coherent family of degree \(d\) and
-  \(\lambda \in \mathfrak{h}^*\). The following conditions are equivalent.
-  \begin{enumerate}
-    \item \(\mathcal{M}[\lambda]\) is irreducible.
-    \item \(F_\alpha\!\restriction_{\mathcal{M}[\lambda]}\) is injective for
-      all \(\alpha \in \Delta\).
-    \item \(\mathcal{M}[\lambda]\) is cuspidal.
-  \end{enumerate}
-\end{theorem}
-
-\begin{proof}
-  The fact that \strong{(i)} and \strong{(iii)} are equivalent follows directly
-  from corollary~\ref{thm:cuspidal-mod-equivs}. Likewise, it is clear from the
-  corollary that \strong{(iii)} implies \strong{(ii)}. All it's left is to show
-  \strong{(ii)} implies \strong{(iii)}\footnote{This isn't already clear from
-  corollary~\ref{thm:cuspidal-mod-equivs} because, at first glance,
-  $\mathcal{M}[\lambda]$ may not be irreducible for some $\lambda$ satisfying
-  \strong{(ii)}. We will show this is never the case.}.
-
-  Suppose \(F_\alpha\) acts injectively in the subrepresentation
-  \(\mathcal{M}[\lambda]\), for all \(\alpha \in \Delta\). Since
-  \(\mathcal{M}[\lambda]\) has finite length, \(\mathcal{M}[\lambda]\) contains
-  an infinite-dimensiona irreducible \(\mathfrak{g}\)-submodule \(V\).
-  Moreover, again by corollary~\ref{thm:cuspidal-mod-equivs} we conclude \(V\)
-  is a cuspidal representation, and its degree is bounded by \(d\). We claim
-  \(\mathcal{M}[\lambda] = V\).
-
-  Since \(\mathcal{M}\) is simple and
-  \(\operatorname{supp}_{\operatorname{ess}} V\) is Zariski-dense, \(U = \{\mu
-  \in \mathfrak{h}^* : \mathcal{M}_\mu \ \text{is a simple
-  $\mathcal{U}(\mathfrak{g})_0$-module}\}\) is a non-empty open set, and \(U
-  \cap \operatorname{supp}_{\operatorname{ess}} V\) is non-empty. In other
-  words, there is some \(\mu \in \mathfrak{h}^*\) such that \(\mathcal{M}_\mu\)
-  is a simple \(\mathcal{U}(\mathfrak{g})_0\)-module and \(\dim V_\mu = \deg
-  V\).
-
-  In particular, \(V_\mu \ne 0\), so \(V_\mu = \mathcal{M}_\mu\). Now given any
-  irreducible \(\mathfrak{g}\)-module \(W\), the multiplicity of \(W\) in
-  \(\mathcal{M}[\lambda]\) is the same as the multiplicity \(W_\mu\) in
-  \(\mathcal{M}_\mu\) as a \(\mathcal{U}(\mathfrak{g})_0\)-module, which is, of
-  course, \(1\) if \(W \cong V\) and \(0\) otherwise. Hence
-  \(\mathcal{M}[\lambda] = V\) and \(\mathcal{M}[\lambda]\) is cuspidal.
-\end{proof}
-
 \section{Localizations \& the Existance of Coherent Extensions}
 
 % TODO: Comment on the intuition behind the proof: we can get vectors in a