lie-algebras-and-their-representations

diff --git a/sections/complete-reducibility.tex b/sections/complete-reducibility.tex
@@ -3,67 +3,257 @@
 % TODO: Remove this?
 \epigraph{Nobody has ever bet enough on a winning horse.}{Some gambler}
 
-% TODOOO: Point out we are now working with finite-dimensional Lie algebras
-% over an algebraicly closed field of characteristic zero
-
-% TODO: Update the 40 pages thing when we're done
-% TODO: Have we seen the fact representations are useful?
-Having hopefully established in the previous chapter that Lie algebras are
-indeed useful, we are now faced with the Herculean task of trying to
-understand them. We have seen that representations are a remarkably effective
-way to derive information about groups -- and therefore algebras -- but the
-question remains: how to we go about classifying the representations of a given
-Lie algebra? This is a question that have sparked an entire field of research,
-and we cannot hope to provide a comprehensive answer the 40 pages we have left.
-Nevertheless, we can work on particular cases.
-
-Like any sane mathematician would do, we begin by studying a simpler case,
-which is that of \emph{semisimple} Lie algebras algebras. The first question we
-have is thus: why are semisimple algebras simpler -- or perhaps
-\emph{semisimpler} -- to understand than any old Lie algebra? Well, the special
-thing about semisimple algebras is that the relationship between their
-indecomposable representations and their irreducible representations is much
-clearer -- at least in finite dimension. Namely\dots
+% TODO: Update the 40 pages thing when we're done TODO: Have we seen the fact
+% representations are useful?
+Having hopefully established in the previous chapter that Lie algebras and
+their representations are indeed useful, we are now faced with the Herculean
+task of trying to understand them. We have seen that representations are a
+remarkably effective way to derive information about groups -- and therefore
+algebras -- but the question remains: how to we go about classifying the
+representations of a given Lie algebra? This is a question that have sparked an
+entire field of research, and we cannot hope to provide a comprehensive answer
+the 40 pages we have left. Nevertheless, we can work on particular cases.
+
+For instance, one can redily check that a representation \(V\) of the
+\(n\)-dimensional Abelian Lie algebra \(K^n\) is nothing more than a choice of
+\(n\) commuting operators \(V \to V\) -- correspoding to the action of the
+canonical basis elements \(e_1, \ldots, e_n \in K^n\). In other words,
+classifying the representations of Abelian algebras is a trivial affair.
+Instead, we focus on the the finite-dimensional representations of a
+finite-dimensional Lie algebra \(\mathfrak{g}\) over an algebraicly closed
+field of characteristic \(0\). But why are the representations semisimple
+algebras simpler -- or perhaps \emph{semisimpler} -- to understand than those
+of any old Lie algebra?
+
+We will come back to this question in a moment, but for now we simply note
+that, when solving a classification problem, it is often profitable to break
+down our structure is smaller peaces. This leads us to the following
+definitions.
+
+\begin{definition}
+  A representation of \(\mathfrak{g}\) is called \emph{indecomposable} if it is
+  not isomorphic to the direct sum of two non-zero representations.
+\end{definition}
+
+\begin{definition}
+  A representation of \(\mathfrak{g}\) is called \emph{irreducible} if it has
+  no non-zero subrepresentations.
+\end{definition}
+
+\begin{example}
+  The trivial representation \(K\) is an example of an irreducible
+  representations. In fact, every \(1\)-dimensional representation \(V\) of a
+  Lie algebra \(\mathfrak{g}\) is irreducible: \(V\) has no non-zero proper
+  subspaces, let alone \(\mathfrak{g}\)-invariant subspaces.
+\end{example}
+
+The general strategy for classifying finite-dimensional representations of an
+algebra is to classify the indecomposable representations. This is because\dots
+
+\begin{theorem}[Krull-Schmidt]\label{thm:krull-schmidt}
+  Every finite-dimensional representation of a Lie algebra can be uniquely --
+  up the isomorphisms and reordering of the summands -- decomposed into a
+  direct sum of indecomposable representations.
+\end{theorem}
+
+Hence finding the indecomposable representations suffices to find \emph{all}
+finite-dimensional representations: they are the direct sum of indecomposable
+representations. The existence of the decomposition should be clear from the
+definitions. Indeed, if \(V\) is representation of \(\mathfrak{g}\) a simple
+argument via induction in \(\dim V\) suffices to prove the existence: if \(V\)
+is indecomposable then there is nothing to prove, and if \(V\) is not
+indecomposable then \(V = W \oplus U\) for some \(W, U \subsetneq V\) non-zero
+subrepresentations, so that their dimensions are both strictly smaller than
+\(\dim V\) and the existence follows from the induction hypothesis. For a proof
+of uniqueness please refer to \cite{etingof}.
+
+Finding the indecomposable representations of an arbitrary Lie algebra,
+however, turns out to be a bit of a circular problem: the indecomposable
+representations are the ones that cannot be decomposed, which is to say, those
+that are \emph{not} decomposable. Ideally, we would like to find some other
+condition, equivalent to indecomposability, but which is easier to work with.
+It is clear from the definitions that every irreducible representation is
+indecomposable, but there is no reason to believe the converse is true. Indeed,
+this is not always the case. For instance\dots
+
+\begin{example}\label{ex:indecomposable-not-irr}
+  The space \(V = K^2\) endowed with the homorphism of Lie algebras
+  \begin{align*}
+    \rho : K[x] & \to     \mathfrak{gl}(V) \\
+              x & \mapsto \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}
+  \end{align*}
+  is a representation of the Lie algebra \(K[x]\). Notice \(V\) has a single
+  non-zero proper subrepresentation, which is spanned by the vector \((1, 0)\).
+  This is because if \((a + b, b) = \rho(x) \ (a, b) = \lambda (a, b)\) for
+  some \(\lambda \in \CC\) then \(\lambda = 1\) and \(b = 0\). Hence \(V\) is
+  indecomposable -- it cannot be broken into a direct sum of \(1\)-dimensional
+  subrepresentations -- but it is evidently not irreducible.
+\end{example}
+
+This counterexample poses an interesting question: are there conditions one can
+impose on an algebra \(\mathfrak{g}\) under which every indecomposable
+representation of \(\mathfrak{g}\) is irreducible? This is what is known in
+representation theory as \emph{complete reducibility}.
+
+\begin{definition}
+  A \(\mathfrak{g}\)-module \(V\) is called \emph{completely reducible} if it
+  the direct sum of irreducible representations.
+\end{definition}
+
+In case the relationship between complete reducibility and the irreducibility
+of indecomposable representations is unclear, the following results should
+clear things up.
 
 \begin{proposition}\label{thm:complete-reducibility-equiv}
-  Given a finite-dimensional Lie algebra \(\mathfrak{g}\) over \(K\), the
-  following conditions are equivalent.
+  The following conditions are equivalent.
   \begin{enumerate}
-    \item \(\mathfrak{g}\) is semisimple.
-
-    \item Given a finite-dimensional representation \(V\) of \(\mathfrak{g}\)
-      and a subrepresentation \(W \subset V\), \(W\) has a
-      \(\mathfrak{g}\)-invariant complement in \(V\).
+    \item Every subrepresentation of a finite-dimensional representation of
+      \(\mathfrak{g}\) has a \(\mathfrak{g}\)-invariant complement -- i.e.
+      given \(W \subset V\) there is a subrepresentation \(U \subset V\) such
+      that \(V = W \oplus U\).
 
     \item Every exact sequence of finite-dimensional representations of
       \(\mathfrak{g}\) splits.
 
-    \item Every finite-dimensional indecomposable representation of
+    \item Every indecomposable finite-dimensional representation of
       \(\mathfrak{g}\) is irreducible.
 
-    \item Every finite-dimensional representation of \(\mathfrak{g}\) can be
-      uniquely decomposed as a direct sum of irreducible representations.
+    \item Every finite-dimensional representation of \(\mathfrak{g}\) is
+      completely reducible.
   \end{enumerate}
 \end{proposition}
 
-Condition \textbf{(ii)} is known as \emph{complete reducibility}. The
-equivalence between conditions \textbf{(ii)} to \textbf{(iv)} follows at once
-from simple arguments. Furthermore, the equivalence between \textbf{(ii)} and
-\textbf{(v)} is a direct consequence of the Krull-Schmidt theorem. On the other
-hand, the equivalence between \textbf{(i)} and the other items is more subtle.
-We are particularly interested in the proof that \textbf{(i)} implies
-\textbf{(ii)}. In other words, we are interested in the fact that every
-finite-dimensional representation of a semisimple Lie algebra is
-\emph{completely reducible}.
-
-This is because if every finite-dimensional representation of \(\mathfrak{g}\)
-is completely reducible, the equivalence between \textbf{(ii)} and \textbf{(v)}
-implies a classification of the finite-dimensional irreducible representations
-of \(\mathfrak{g}\) leads to a classification of \emph{all} finite-dimensional
-representation of \(\mathfrak{g}\) -- it suffices to take direct sums of the
-already classified irreducible modules. This leads us to the third restriction
-we will impose: for now, we will focus our attention exclusively on
-finite-dimensional representations.
+\begin{proof}
+  We begin by \(\textbf{(i)} \Rightarrow \textbf{(ii)}\). Let
+  \begin{center}
+    \begin{tikzcd}
+      0 \arrow{r} &
+      W \arrow{r}{f} &
+      V \arrow{r}{g} &
+      U \arrow{r} &
+      0
+    \end{tikzcd}
+  \end{center}
+  be an exact sequence of representations of \(\mathfrak{g}\). We can suppose
+  without any loss of generality that \(W \subset V\) is a subrepresentation
+  and \(f\) is the usual inclusion, for if this is not the case there is an
+  isomorphism of sequences
+  \begin{center}
+    \begin{tikzcd}
+         0 \arrow{r} &
+         W \arrow{r}{f} \arrow[swap]{d}{f} &
+         V \arrow{r}{g} \arrow[Rightarrow, no head]{d} &
+         U \arrow{r}    \arrow[Rightarrow, no head]{d} &
+         0 \\
+         0 \arrow{r} &
+      f(W) \arrow{r} &
+         V \arrow[swap]{r}{g} &
+         U \arrow{r} &
+         0
+    \end{tikzcd}
+  \end{center}
+
+  It then follows from \textbf{(i)} that there exists a subrepresentation \(U'
+  \subset V\) such that \(V = W \oplus U'\). Finally, the projection \(\pi : V
+  \to W\) is an intertwiner satisfying
+  \begin{center}
+    \begin{tikzcd}
+      0 \arrow{r} &
+      W \arrow{r}{f} &
+      V \arrow{r}{g} \arrow[bend left=30]{l}{\pi} &
+      U \arrow{r} &
+      0
+    \end{tikzcd}
+  \end{center}
+
+  Next is \(\textbf{(ii)} \Rightarrow \textbf{(iii)}\). If \(V\) is an
+  indecomposable \(\mathfrak{g}\)-module and \(W \subset V\) is a
+  subrepresentation, we have an exact sequence
+  \begin{center}
+    \begin{tikzcd}
+                 0 \arrow{r}      &
+                 W \arrow{r}{i}   &
+                 V \arrow{r}{\pi} &
+      \mfrac{V}{W} \arrow{r}      &
+                 0
+    \end{tikzcd}
+  \end{center}
+  of representations of \(\mathfrak{g}\).
+
+  Since our sequence splits, we must have \(V \cong W \oplus \mfrac{V}{W}\).
+  But \(V\) is indecomposable, so that either \(W = V\) or \(W = 0\). Since
+  this holds for all \(W \subset V\), \(V\) is irreducible. For
+  \(\textbf{(iii)} \Rightarrow \textbf{(iv)}\) it suffices to apply
+  theorem~\ref{thm:krull-schmidt}.
+
+  Finally, for \(\textbf{(iv)} \Rightarrow \textbf{(i)}\), if we assume
+  \(\textbf{(iii)}\) and let \(V\) be a representation of \(\mathfrak{g}\) with
+  decomposition into irreducible subrepresentations
+  \[
+    V = \bigoplus_i V_i
+  \]
+  and \(W \subset V\) is a subrepresentation. Take some maximal set of indexes
+  \(\{i_1, \ldots, i_n\}\) so that \(\left( \bigoplus_k V_{i_k}
+  \right) \cap W = 0\) and let \(U = \bigoplus_k V_{i_k}\). We want to
+  establish \(V = W \oplus U\).
+
+  Suppose without any loss in generality that \(i_k = k\) for all \(k\) and
+  let \(j > n\). By the maximality of our set of indexes, there is some
+  non-zero \(w \in (V_j \oplus U) \cap W\). Say \(w = v_j + v_1 + \cdots +
+  v_n\) with each \(v_i \in V_i\). Then \(v_j = w - v_1 - \cdots - v_n \in V_j
+  \cap (W \oplus U)\) is non-zero. Indeed, if this is not the case we find \(0
+  \ne w = v_1 + \cdots + v_n \in \left( \bigoplus_{i = 1}^n V_i \right) \cap
+  W\), a contradiction. This implies \(V_j \cap (W \oplus U)\) is a non-zero
+  subrepresentation of \(V_j\). Since \(V_j\) is irreducible, \(V_j = V_j \cap
+  (W \oplus U)\) and therefore \(V_j \subset W \oplus U\). Given the arbitrary
+  choice of \(j\), it then follows \(V = W \oplus U\).
+\end{proof}
+
+The advantage of working with irreducible representations as opposed to
+indecomposable ones is that they are generally much easier to find. The
+relationship between irreducible representations is also well understood. This
+is because of the following result, known as \emph{Schur's lemma}.
+
+\begin{lemma}[Schur]
+  Let \(V\) and \(W\) be irreducible representations of \(\mathfrak{g}\) and
+  \(T : V \to W\) be an intertwiner. Then \(T\) is either \(0\) or an
+  isomorphism. Furtheremore, if \(V = W\) then \(T\) is a scalar operator.
+\end{lemma}
+
+\begin{proof}
+  For the first statement, it suffices to notice that \(\ker T\) and
+  \(\operatorname{im} T\) are both subrepresentations. In particular, either
+  \(\ker T = 0\) and \(\operatorname{im} T = W\) or \(\ker T = V\) and
+  \(\operatorname{im} T = 0\). Now suppose \(V = W\). Let \(\lambda \in K\) be
+  an eigenvalue of \(T\) -- which exists because \(K\) is algebraicly closed --
+  and \(V_\lambda\) be its corresponding eigenspace. Given \(v \in V_\lambda\),
+  \(T X v = X T v = \lambda \cdot X v\). In other words, \(V_\lambda\) is a
+  subrepresentation. It then follows \(V_\lambda = V\), given that \(V_\lambda
+  \ne 0\).
+\end{proof}
+
+We are now ready to answer our first question: the special thing about
+semisimple algebras is that the relationship between their indecomposable
+representations and their irreducible representations is much clearer.
+Namely\dots
+
+\begin{proposition}\label{thm:complete-reducibility-equiv}
+  Given a finite-dimensional Lie algebra \(\mathfrak{g}\) over \(K\),
+  \(\mathfrak{g}\) is semisimple if, and only if every finite-dimensional
+  representation of \(\mathfrak{g}\) is completely reducible.
+\end{proposition}
+
+The proof of the fact that a finite-dimensional Lie algebra \(\mathfrak{g}\)
+whose finite-dimensional representations are completely reducible is semisimple
+is actually pretty simple. Namely, it suffices to note that the adjoint
+representation \(\mathfrak{g}\) is the direct sum of irreducible
+subrepresentations, which are all simple ideals of \(\mathfrak{g}\) -- so
+\(\mathfrak{g}\) is the direct sum of simple Lie algebras. The proof of the
+converse is more nuanced, and this will be our next milestone. Before proceding
+to the proof of complete reducibility, however, we would like to review some
+basic tools which will come in handy later on, known as\dots
+
+\section{Invariant Bilinear Forms}
 
 Another interesting characterization of semisimple Lie algebras, which will
 come in handy later on, is the following.
@@ -92,7 +282,7 @@ come in handy later on, is the following.
 We refer the reader for \cite[ch. 5]{humphreys} for a proof of this last
 result. Without further ado, we may proceed to a proof of\dots
 
-\section{Complete Reducibility}
+\section{Proof of Complete Reducibility}
 
 Historically, complete reducibility was first proved by Herman Weyl for \(K =
 \mathbb{C}\), using his knowledge of smooth representations of compact Lie