lie-algebras-and-their-representations

diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -862,13 +862,62 @@ To finish the proof, we now show\dots
   the union of Zariski-open subsets and is therefore open. We are done.
 \end{proof}
 
+The major remaining question for us to tackle is that of existence of coherent
+extensions, which will be the focus of our next section.
+
 \section{Localizations \& the Existance of Coherent Extensions}
 
-% TODO: Comment on the intuition behind the proof: we can get vectors in a
-% given eigenspace by translating by the F's and E's, but neither of those are
-% injective in general, so the translation could take nonzero vectors to zero.
-% If the F's were invertible this problem wouldn't exist, so we might as well
-% invert them by force!
+Let \(V\) be an irreducible admissible \(\mathfrak{g}\)-module of degree \(d\).
+Our goal is to prove that \(V\) has a (unique) simple completely reducible
+coherent extension \(\mathcal{M}\). Since \(V\) is irreducible, we know \(V
+\subset \mathcal{M}[\lambda]\) for any \(\lambda \in \operatorname{supp} V\).
+Our first task is constructing \(\mathcal{M}[\lambda]\). The issue here is that
+\(\operatorname{supp}_{\operatorname{ess}} V\) may not be all of \(\lambda + Q
+= \operatorname{supp}_{\operatorname{ess}} \mathcal{M}[\lambda]\), so we may
+find \(V \subsetneq \mathcal{M}[\lambda]\). In fact, we may find
+\(\operatorname{supp} V \subsetneq \lambda + Q\).
+
+This wasn't an issue an example~\ref{ex:laurent-polynomial-mod} because we
+verified that the action of \(f \in \mathfrak{sl}_2(K)\) in \(K[x, x^{-1}]\) is
+injective. Since all weight spaces of \(K[x, x^{-1}]\) are \(1\)-dimensional,
+this implies the action of \(f\) is actually bijective, so we can obtain a
+non-zero vector in \(K[x, x^{-1}]_{2 k} = K x^k\) for any \(k \in \mathbb{Z}\)
+by translating between weight spaced using \(f\) and \(f^{-1}\) -- here
+\(f^{-1}\) denote the differential operator \((-
+\sfrac{\mathrm{d}}{\mathrm{d}x} + \sfrac{x^{-1}}{2})^{-1}\), which is the
+inverse of the action of \(f\) in \(K[x, x^{-1}]\).
+\begin{center}
+  \begin{tikzcd}
+    \cdots     \arrow[bend left=60]{r}{f^{-1}}
+    & K x^{-2} \arrow[bend left=60]{r}{f^{-1}} \arrow[bend left=60]{l}{f}
+    & K x^{-1} \arrow[bend left=60]{r}{f^{-1}} \arrow[bend left=60]{l}{f}
+    & K x^0    \arrow[bend left=60]{r}{f^{-1}} \arrow[bend left=60]{l}{f}
+    & K x^1    \arrow[bend left=60]{r}{f^{-1}} \arrow[bend left=60]{l}{f}
+    & K x^2    \arrow[bend left=60]{r}{f^{-1}} \arrow[bend left=60]{l}{f}
+    & \cdots   \arrow[bend left=60]{l}{f}
+  \end{tikzcd}
+\end{center}
+
+In our case, the action of some \(F_\alpha \in \mathfrak{g}\) with \(\alpha \in
+\Delta\) in \(V\) may not be injective. In fact, we have seen that the action
+of \(F_\alpha\) is injective for all \(\alpha \in \Delta\) if, and only if
+\(V\) is cuspidal. Nevertheless, we could intuitively \emph{make it injective}
+by formally inverting the elements \(F_\alpha \in \mathcal{U}(\mathfrak{g})\).
+This would allow us to obtain nonzero vectors in \(V_\mu\) for all \(\mu \in
+\lambda + Q\) by succecively applying elements of \(\{F_\alpha^{\pm
+1}\}_{\alpha \in \Delta}\) to a nonzero weight vector \(v \in V_\lambda\).
+Moreover, if the actions of the \(F_\alpha\) were to be invertible, we would
+find that all \(V_\mu\) are \(d\)-dimensional for \(\mu \in \lambda + Q\).
+
+In a commutative domain, this can be achieved by tensoring our module by the
+field of fractions. However, \(\mathcal{U}(\mathfrak{g})\) is hardly ever
+commutative -- \(\mathcal{U}(\mathfrak{g})\) is commutative if, and only if
+\(\mathfrak{g}\) is Abelian -- and the situation is more delicate in the
+noncommutative case. For starters, a noncommutative ring \(R\) may not even
+have a ``field of fractions'' -- i.e. an over-ring where all elements of \(R\)
+have inverses. Nevertheless, it is possible to formally invert elements of
+certain subsets of \(R\) via a proccess known as \emph{localization}, which we
+now describe.
 
 \begin{definition}
   Let \(R\) be a ring. A subset \(S \subset R\) is called \emph{multiplicative}
@@ -894,19 +943,41 @@ To finish the proof, we now show\dots
   \end{center}
 \end{theorem}
 
-% TODO: Cite the discussion of goodearl-warfield, chap 6, on how to derive the
-% localization condition
-% TODO: In general checking that a set satisfies Ore's condition can be tricky,
-% but there is an easyer condition given by the lemma
+If we identify an element with its image under the localization map, it follows
+directly from Ore's construction that every element of \(S^{-1} R\) has the
+form \(s^{-1} r_1\) for some \(s \in S\) and \(r_1 \in R\). Likewise, any
+element of \(S^{-1} R\) can also be written as \(r_2 t^{-1}\) for some \(t \in
+S\), \(r_2 \in R\).
+
+Ore's localization condition may seem a bit arbitrary at first, but a more
+thorough investigation reveals the intuition behind it. The issue in question
+here is that in the noncommtative case we can no longer take the existance of
+common denominators for granted. However, the existance of common denominators
+is fundamental to the proof of the fact the field of fractions is a ring -- it
+is used, for example, to define the sum of two elements in the field of
+fractions. We thus need to impose their existance for us to have any hope of
+defining consistant arithmetics in the localization of a ring, and Ore's
+condition is actually equivalent to the existance of common denominators --
+see the discussion in the introduction of \cite[ch.~6]{goodearl-warfield} for
+further details.
+
+We should also point out that there are numerous other conditions -- which may
+be easyer to check than Ore's -- known to imply Ore's condition. For
+instance\dots
 
 \begin{lemma}
   Let \(S \subset R\) be a multiplicative subset generated by locally
   \(\operatorname{ad}\)-nilpotent elements -- i.e. elements \(s \in S\) such
-  that for each \(r \in R\) there exists \(n > 0\) such that \(\operatorname{ad}(s)^n r = [s, [s, \cdots
-  [s, r]]\cdots] = 0\). Then \(S\) satisfies Ore's
-  localization condition.
+  that for each \(r \in R\) there exists \(n > 0\) such that
+  \(\operatorname{ad}(s)^n r = [s, [s, \cdots [s, r]]\cdots] = 0\). Then \(S\)
+  satisfies Ore's localization condition.
 \end{lemma}
 
+In our case, we are more interested in formally inverting the action of the
+action of \(F_\alpha\) in \(V\) than in inverting \(F_\alpha\) itself. To that
+end, we introduce one further construction, kwon as \emph{the localization of a
+module}.
+
 \begin{definition}
   Let \(S \subset R\) be a multiplicative subset satisfying Ore's localization
   condition and \(M\) be a \(R\)-module. The \(S^{-1} R\)-module \(S^{-1} M =
@@ -919,9 +990,21 @@ To finish the proof, we now show\dots
   is called \emph{the localization map of \(M\)}.
 \end{definition}
 
-% TODO: Point out that the localization of modules is functorial
+Notice that the \(S^{-1} R\)-module \(S^{-1} M\) has the natural structure of
+an \(R\)-module, where the action of \(R\) is given by the localization map \(R
+\to S^{-1} R\).
+
+It is interesting to observe that, unlike in the case of the field of fractions
+of a commutative domain, in general the localization map \(R \to S^{-1} R\) --
+i.e. the map \(r \mapsto \frac{r}{1}\) -- may not be injective. For instance,
+if \(S\) contains a divisor of zero \(s\), its image under the localization map
+cannot be a divisor of zero in \(S^{-1} R\) -- since it is invertible. In
+particular, if \(r \in R\) is nonzero and such that \(s r = 0\) then its image
+under the localization map has to be \(0\), given that the image of \(s r = 0\)
+is \(0\). However, the existance of divisors of zero in \(S\) turns out to be
+the only obstruction to the injectivity of the localization map, as shown
+in\dots
 
-% TODO: Point out the the localization map is in not injective in general
 \begin{lemma}
   Let \(S \subset R\) be a multiplicative subset satisfying Ore's localization
   condition and \(M\) be a \(R\)-module. If \(S\) acts injectively in \(M\)
@@ -929,18 +1012,30 @@ To finish the proof, we now show\dots
   \(S\) has no zero divisors then \(R\) is a subring of \(S^{-1} R\).
 \end{lemma}
 
-% TODO: Point out that S^-1 M can be seen as a R-module, where R acts via the
-% localization map
-% TODO: Point out that each element of the localization has the form s^-1 r
+Again, in our case we are interested in inverting the actions of the
+\(F_\alpha\) in \(V\). However, for us to be able to translate between all
+weight spaces associated with elements of \(\lambda + Q\), \(\lambda \in
+\operatorname{supp} V\), we only need to invert the \(F_\alpha\)'s for
+\(\alpha\) in some subset of \(\Delta\) which spans all of \(Q = \mathbb{Z}
+\Delta\). In other words, it suffices to invert \(F_\beta\) for all \(\beta\)
+in some basis \(\Sigma\) for \(\Delta\). We can choose such a basis to be
+well-behaved. For example, we can show\dots
 
-% TODO: Point out that Sigma depends on V!
 \begin{lemma}\label{thm:nice-basis-for-inversion}
   Let \(V\) be an irreducible infinite-dimensional admissible
   \(\mathfrak{g}\)-module. There is a basis \(\Sigma = \{\beta_1, \ldots,
-  \beta_n\}\) of \(\Delta\) such that the elements \(F_{\beta_i}\) all act
+  \beta_n\}\) for \(\Delta\) such that the elements \(F_{\beta_i}\) all act
   injectively on \(V\) and satisfy \([F_{\beta_i}, F_{\beta_j}] = 0\).
 \end{lemma}
 
+\begin{note}
+  The basis \(\Sigma\) may very well depend on the representation \(V\)! This
+  is another obstacle to showing the functoriality of our constructions.
+\end{note}
+
+Since \(F_\alpha\) is locally \(\operatorname{ad}\)-nilpotent for all \(\alpha
+\in \Delta\), we can see\dots
+
 \begin{corollary}
   Let \(\Sigma\) be as in lemma~\ref{thm:nice-basis-for-inversion} and
   \((F_\beta)_{\beta \in \Sigma} \subset \mathcal{U}(\mathfrak{g})\) be the
@@ -951,7 +1046,11 @@ To finish the proof, we now show\dots
   \Sigma}\), the localization map \(V \to \Sigma^{-1} V\) is injective.
 \end{corollary}
 
-% TODO: Fix V and Sigma beforehand
+From now on let \(\Sigma\) be some fixed basis for \(\Delta\) satisfying the
+hypothesis of lemma~\ref{thm:nice-basis-for-inversion}. As promise, we now show
+that \(\Sigma^{-1} V\) contains \(V\) and that its support is an entire
+\(Q\)-coset.
+
 \begin{proposition}\label{thm:irr-admissible-is-contained-in-nice-mod}
   The the restriction of the localization \(\Sigma^{-1} V\) is an admissible
   \(\mathfrak{g}\)-module of degree \(d\) with \(\operatorname{supp}
@@ -995,7 +1094,7 @@ To finish the proof, we now show\dots
   s^{-1} \otimes v\) for \(s \in (F_\beta)_{\beta \in \Sigma}\) and \(v \in
   V\), we can see that \(\Sigma^{-1} V = \bigoplus_\lambda \Sigma^{-1}
   V_\lambda\). Furtheremore, since the action of each \(F_\beta\) in
-  \(\Sigma^{-1} V\) is bijective and \(\Sigma\) is a basis of \(Q\) we obtain
+  \(\Sigma^{-1} V\) is bijective and \(\Sigma\) is a basis for \(Q\) we obtain
   \(\operatorname{supp} \Sigma^{-1} V = Q + \operatorname{supp} V\).
 
   Again, because of the bijectivity of the \(F_\beta\)'s, to see that \(\dim
@@ -1013,6 +1112,36 @@ To finish the proof, we now show\dots
   \Sigma^{-1} V_\lambda\) and therefore \(\dim \Sigma^{-1} V_\lambda = d\).
 \end{proof}
 
+We now have a good canditate for a coherent extension of \(V\), but
+\(\Sigma^{-1} V\) is still not a coherent extension since its support is
+contained in a single coset. In particular, \(\operatorname{supp} \Sigma^{-1} V
+\ne \mathfrak{h}^*\) and \(\Sigma^{-1} V\) is not a coherent family. To obtain
+a coherent family we thus need somehow extend \(\Sigma^{-1} V\). To that end,
+we'll attempt to replicate the construction of the coherent extension of the
+\(\mathfrak{sl}_2(K)\)-module \(K[x, x^{-1}]\). Specifically, the idea is that
+if twist \(\Sigma^{-1} V\) by an automorphism which shifts its support by some
+\(\lambda \in \mathfrak{h}^*\), we can construct a coherent family by summing
+this modules over \(\lambda\) as in example~\ref{ex:sl-laurent-family}.
+
+For \(\lambda = \beta \in \Sigma\) the map
+\begin{align*}
+  \theta_\beta : \Sigma^{-1} \mathcal{U}(\mathfrak{g}) & \to
+                 \Sigma^{-1} \mathcal{U}(\mathfrak{g}) \\
+                 r & \mapsto F_\beta r F_\beta^{-1}
+\end{align*}
+is a natural candidate for such a twisting automorphism. Indeed, we will soon
+see that \((\theta_\beta \Sigma^{-1} V)_\lambda = \Sigma^{-1} V_{\lambda +
+\beta}\). However, this is hardly useful to us, since \(\beta \in Q\) and
+therefore \(\beta + \operatorname{supp} \Sigma^{-1} V = \operatorname{supp}
+\Sigma^{-1} V\). If we want to expand the support of \(\Sigma^{-1} V\) we will
+have to twist by automorphisms that shift its support by \(\lambda \in
+\mathfrak{h}^*\) lying \emph{outside} of \(Q\).
+
+The situation is much less obvious in this case. Nevertheless, it turns out we
+can extend the family \(\{\theta_\beta\}_{\beta \in \Sigma}\) to a family of
+automorphisms \(\{\theta_\lambda\}_{\lambda \in \mathfrak{h}^*}\).
+Explicitly\dots
+
 \begin{proposition}\label{thm:nice-automorphisms-exist}
   There is a family of automorphisms \(\{\theta_\lambda : \Sigma^{-1}
   \mathcal{U}(\mathfrak{g}) \to \Sigma^{-1}
@@ -1132,25 +1261,28 @@ To finish the proof, we now show\dots
   so that \((\theta_\lambda M)_{\mu + \lambda} = M_\mu\).
 \end{proof}
 
+It should now be obvious\dots
+
 \begin{proposition}[Mathieu]
   There exists a coherent extension \(\mathcal{M}\) of \(V\).
 \end{proposition}
 
+% TODO: Point out that here we have to fix representatives, so this
+% construction is, once again, not functorial
 \begin{proof}
-  Let \(\Lambda\) be a set of representatives of the \(Q\)-cosets in
-  \(\mathfrak{h}^*\) with \(0 \in \Lambda\) and take
+  Take
   \[
     \mathcal{M}
-    = \bigoplus_{\lambda \in \Lambda} \theta_\lambda \Sigma^{-1} V
+    = \bigoplus_{\lambda + Q \in \mfrac{\mathfrak{h}^*}{Q}}
+      \theta_\lambda \Sigma^{-1} V
   \]
 
-  On the one hand, \(V\) lies in \(\Sigma^{-1} V = \theta_0 \Sigma^{-1} V\) --
-  notice that \(\theta_0\) is just the identity operator -- and therefore \(V
-  \subset \mathcal{M}\). On the other hand, \(\dim \mathcal{M}_\mu = \dim
-  \theta_\lambda \Sigma^{-1} V_\mu = \dim \Sigma^{-1} V_{\mu - \lambda} = d\)
-  for all \(\mu \in \lambda + Q\), \(\lambda \in \Lambda\). Furtheremore, given
-  \(u \in \mathcal{U}(\mathfrak{g})_0\) and \(\mu \in \lambda +
-  Q\),
+  It is clear \(V\) lies in \(\Sigma^{-1} V = \theta_0 \Sigma^{-1} V\) and
+  therefore \(V \subset \mathcal{M}\). On the other hand, \(\dim
+  \mathcal{M}_\mu = \dim \theta_\lambda \Sigma^{-1} V_\mu = \dim \Sigma^{-1}
+  V_{\mu - \lambda} = d\) for all \(\mu \in \lambda + Q\), \(\lambda \in
+  \Lambda\). Furtheremore, given \(u \in \mathcal{U}(\mathfrak{g})_0\) and
+  \(\mu \in \lambda + Q\),
   \[
     \operatorname{Tr}(u\!\restriction_{\mathcal{M}_\mu})
     = \operatorname{Tr}
@@ -1160,6 +1292,8 @@ To finish the proof, we now show\dots
   proposition~\ref{thm:nice-automorphisms-exist}.
 \end{proof}
 
+Lo and behold\dots
+
 \begin{theorem}[Mathieu]
   There exists a unique completely reducible coherent extension
   \(\operatorname{Ext}(V)\) of \(V\). More precisely, if \(\mathcal{M}\) is any
@@ -1194,21 +1328,28 @@ To finish the proof, we now show\dots
     \mapsto \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})
   \end{align*}
 
-  % TODO: Point out that this multiplicity is determined by the characters
-  % beforehand
-  Indeed, given \(\lambda \in \operatorname{supp} V\) the multiplicity of \(W\)
-  in \(\mathcal{N}\) is the same as the multiplicity of \(W_\lambda\) in
-  \(\mathcal{N}_\lambda\), which is determined by the character
-  \(\chi_{\mathcal{N}_\lambda} : \mathcal{U}(\mathfrak{g})_0
-  \to K\) -- see proposition~\ref{thm:centralizer-multiplicity}. We now claim
-  that the trace function of \(\mathcal{N}\) is the same as that of
-  \(\operatorname{Ext}(V)\). Clearly,
-  \(\operatorname{Tr}(u\!\restriction_{\operatorname{Ext}(V)_\lambda})
-  = \operatorname{Tr}(u\!\restriction_{V_\lambda})
-  = \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\) for all
-  \(\lambda \in \operatorname{supp}_{\operatorname{ess}} V\), \(u \in
-  \mathcal{U}(\mathfrak{g})_0\). Since the essential support of
-  \(V\) is Zariski-dense and the maps \(\lambda \mapsto
+  It is a well known fact of the theory of modules that, given a \(K\)-algebra
+  \(A\), a finite-dimensional completely reducible \(A\)-module \(M\) is
+  determined, up to isomorphism, by its \emph{character}
+  \begin{align*}
+    \chi_M : A & \to     K                                    \\
+             a & \mapsto \operatorname{Tr}(a\!\restriction_M)
+  \end{align*}
+
+  In particular, the multiplicity of \(W\) in \(\mathcal{N}\), which is the
+  same as the multiplicity of \(W_\lambda\) in \(\mathcal{N}_\lambda\), is
+  determined by the character \(\chi_{\mathcal{N}_\lambda} :
+  \mathcal{U}(\mathfrak{g})_0 \to K\). Since this holds for all irreducible
+  weight \(\mathfrak{g}\)-modules, it follows that \(\mathcal{N}\) is
+  determined by its trace function. Of course, the same holds for
+  \(\operatorname{Ext}(V)\). We now claim that the trace function of
+  \(\mathcal{N}\) is the same as that of \(\operatorname{Ext}(V)\). Clearly,
+  \(\operatorname{Tr}(u\!\restriction_{\operatorname{Ext}(V)_\lambda}) =
+  \operatorname{Tr}(u\!\restriction_{V_\lambda}) =
+  \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\) for all \(\lambda
+  \in \operatorname{supp}_{\operatorname{ess}} V\), \(u \in
+  \mathcal{U}(\mathfrak{g})_0\). Since the essential support of \(V\) is
+  Zariski-dense and the maps \(\lambda \mapsto
   \operatorname{Tr}(u\!\restriction_{\operatorname{Ext}(V)_\lambda})\) and
   \(\lambda \mapsto \operatorname{Tr}(u\!\restriction_{\mathcal{N}_\lambda})\)
   are polynomial in \(\lambda \in \mathfrak{h}^*\), it follows that this maps