lie-algebras-and-their-representations

diff --git a/references.bib b/references.bib
@@ -243,3 +243,12 @@
    isbn =      {9780720420340},
    year =      {1970},
 }
+
+@book{coutinho,
+   title =     {A primer of algebraic D-modules},
+   author =    {S. C. Coutinho},
+   publisher = {Cambridge University Press},
+   isbn =      {0521551196},
+   year =      {1995},
+   series =    {London Mathematical Society student texts 33},
+}

diff --git a/sections/introduction.tex b/sections/introduction.tex
@@ -260,10 +260,16 @@ avoid much of delicacies of geometric objects such as real and complex Lie
 groups. As nonlinear objects, groups can be complicated beasts -- even when
 working without additional geometric considerations. In this regard, the
 linearity of Lie algebras makes them much more flexible than groups.
+
 Having thus hopefully established Lie algebras are interesting, we are now
-ready to dive deeper into them.
+ready to dive deeper into them. We begin by analysing some of their most basic
+properties.
+
+\section{Basic Structure of Lie Algebras}
 
-\section{Lie Algebras}
+However bizarre Lie algebras may seem at a first glance, they actually share a
+lot a structural features with their associative counterparts. For instance, it
+is only natural to define\dots
 
 \begin{definition}
   Given a Lie algebra \(\mathfrak{g}\), a subspace \(\mathfrak{h} \subset
@@ -275,17 +281,48 @@ ready to dive deeper into them.
   \(\mathfrak{a} \normal \mathfrak{g}\).
 \end{definition}
 
-\begin{proposition}
+\begin{note}
+  In the context of associative algebras, it's usual practice to distinguish
+  between \emph{left ideals} and \emph{right ideals}. This is not neccessary
+  when dealing with Lie algebras, however, since any ``left ideal'' of a Lie
+  algebra is also a ``right ideal'' -- \([Y, X] = - [X, Y] \in \mathfrak{a}\)
+  for all \(X \in \mathfrak{g}\) and \(Y \in \mathfrak{a}\).
+\end{note}
+
+\begin{example}
+  Let \(f : \mathfrak{g} \to \mathfrak{h}\) be a homomorphism between Lie
+  algebras \(\mathfrak{g}\) and \(\mathfrak{h}\). Then \(\ker f \subset
+  \mathfrak{g}\) and \(\operatorname{im} f \subset \mathfrak{h}\) are
+  subalgebras. Furtheremore, \(\ker f \normal \mathfrak{g}\).
+\end{example}
+
+\begin{example}
+  Let \(G\) be an affine algebraic \(K\)-group and \(H \subset G\) be a
+  connected closed subgroup. Denote by \(\mathfrak{g}\) and \(\mathfrak{h}\)
+  the Lie algebras of \(G\) and \(H\), respectively. The inclusion \(H \to G\)
+  induces an injective homomorphism \(\mathfrak{h} \to \mathfrak{g}\). We may
+  thus regard \(\mathfrak{h}\) as a subalgebra of \(\mathfrak{g}\). In
+  addition, \(\mathfrak{h} \normal \mathfrak{g}\) if, and only if \(H \normal
+  G\).
+\end{example}
+
+There is also a natural analogue of quotients.
+
+\begin{definition}
   Given a Lie algebra \(\mathfrak{g}\) and \(\mathfrak{a} \normal
   \mathfrak{g}\), the space \(\mfrac{\mathfrak{g}}{\mathfrak{a}}\) has the
   natural structure of a Lie algebra over \(K\), where
   \[
     [X + \mathfrak{a}, Y + \mathfrak{a}] = [X, Y] + \mathfrak{a}
   \]
+\end{definition}
 
-  Furtheremore, every homomorphism of Lie algebras \(f : \mathfrak{g} \to
-  \mathfrak{h}\) such that \(a \subset \ker f\) uniquely factors trought the
-  projection \(\mathfrak{g} \to \mfrac{\mathfrak{g}}{\mathfrak{a}}\).
+\begin{proposition}
+  Given a Lie algebra \(\mathfrak{g}\) and \(\mathfrak{a} \normal
+  \mathfrak{g}\), every homomorphism of Lie algebras \(f : \mathfrak{g} \to
+  \mathfrak{h}\) such that \(\mathfrak{a} \subset \ker f\) uniquely factors
+  trought the projection \(\mathfrak{g} \to
+  \mfrac{\mathfrak{g}}{\mathfrak{a}}\).
   \begin{center}
     \begin{tikzcd}
       \mathfrak{g} \rar{f} \dar                             & \mathfrak{h} \\
@@ -294,6 +331,20 @@ ready to dive deeper into them.
   \end{center}
 \end{proposition}
 
+Due to their relationship with Lie groups and algebraic groups, Lie algebras
+also share structural features with groups. For example\dots
+
+\begin{definition}
+  A Lie algebra \(\mathfrak{g}\) is called \emph{Abelian}  if \([X, Y] = 0\)
+  for all \(X, Y \in \mathfrak{g}\).
+\end{definition}
+
+\begin{example}
+  Let \(G\) be a connected algebraic \(K\)-group and \(\mathfrak{g}\) be its
+  Lie algebra. Then \(G\) is Abelian if, and only if \(\mathfrak{g}\) is
+  Abelian.
+\end{example}
+
 \begin{definition}
   A Lie algebra \(\mathfrak{g}\) is called \emph{solvable} if its derived
   series
@@ -311,6 +362,11 @@ ready to dive deeper into them.
   converges to \(0\) in finite time.
 \end{definition}
 
+\begin{example}
+  Let \(G\) be a connected affine algebraic \(K\)-group and \(\mathfrak{g}\) be
+  its Lie algebra. Then \(G\) is solvable if, and only if \(\mathfrak{g}\) is.
+\end{example}
+
 \begin{definition}
   A Lie algebra \(\mathfrak{g}\) is called \emph{nilpotent} if its derived
   series
@@ -324,36 +380,17 @@ ready to dive deeper into them.
   converges to \(0\) in finite time.
 \end{definition}
 
-\begin{definition}
-  Let \(\mathfrak{g}\) be a Lie algebra. The sum \(\mathfrak{a} +
-  \mathfrak{b}\) of solvable ideals \(\mathfrak{a}, \mathfrak{b} \normal
-  \mathfrak{g}\) is again a solvable ideal. Hence the sum of all solvable
-  ideals of \(\mathfrak{g}\) is a maximal solvable ideal, known as \emph{the
-  radical \(\mathfrak{rad}(\mathfrak{g})\) of \(\mathfrak{g}\)}.
-  \[
-    \mathfrak{rad}(\mathfrak{g})
-    = \sum_{\substack{\mathfrak{a} \normal \mathfrak{g}\\\text{solvable}}}
-      \mathfrak{a}
-  \]
-\end{definition}
+\begin{example}
+  Every nilpotent Lie algebra if solvable.
+\end{example}
 
-\begin{definition}
-  Let \(\mathfrak{g}\) be a Lie algebra. The sum of nilpotent ideals is a
-  nilpotent ideal. Hence the sum of all nilpotent ideals of \(\mathfrak{g}\) is
-  a maximal nilpotent ideal, known as \emph{the nilradical
-  \(\mathfrak{nil}(\mathfrak{g})\) of \(\mathfrak{g}\)}.
-  \[
-    \mathfrak{nil}(\mathfrak{g})
-    = \sum_{\substack{\mathfrak{a} \normal \mathfrak{g}\\\text{nilpotent}}}
-      \mathfrak{a}
-  \]
-\end{definition}
+\begin{example}
+  Let \(G\) be a connected affine algebraic \(K\)-group and \(\mathfrak{g}\) be
+  its Lie algebra. Then \(G\) is nilpotent if, and only if \(\mathfrak{g}\) is.
+\end{example}
 
-\begin{proposition}\label{thm:quotients-by-rads}
-  Let \(\mathfrak{g}\) be a Lie algebra. Then
-  \(\mfrac{\mathfrak{g}}{\mathfrak{rad}(\mathfrak{g})}\) is semisimple and
-  \(\mfrac{\mathfrak{g}}{\mathfrak{nil}(\mathfrak{g})}\) is reductive.
-\end{proposition}
+Other interesting classes of Lie algebras are the so called \emph{simple} and
+\emph{semisimple} Lie algebras.
 
 \begin{definition}
   A non-Abelian Lie algebra \(\mathfrak{s}\) over \(K\) is called \emph{simple}
@@ -377,11 +414,19 @@ ready to dive deeper into them.
 \end{example}
 
 \begin{definition}\label{thm:sesimple-algebra}
-  A Lie algebra \(\mathfrak{g}\) is called \emph{semisimple} if it has no
-  non-zero solvable ideals. Equivalently, a Lie algebra \(\mathfrak{g}\) is
-  called \emph{semisimple} if it is the direct sum of simple Lie algebras.
+  A Lie algebra \(\mathfrak{g}\) is called \emph{semisimple} if it is the
+  direct sum of simple Lie algebras. Equivalently, a Lie algebra
+  \(\mathfrak{g}\) is called \emph{semisimple} if it has no non-zero solvable
+  ideals.
 \end{definition}
 
+\begin{example}
+  Let \(G\) be a connected affinite algebraic \(K\)-group. Then \(G\) is
+  semisimple if, and only if \(\mathfrak{g}\) semisimple.
+\end{example}
+
+A slight generalization is\dots
+
 \begin{definition}
   A Lie algebra \(\mathfrak{g}\) is called \emph{reductive} if \(\mathfrak{g}\)
   is the direct sum of a semisimple Lie algebra and an Abelian Lie algebra.
@@ -409,7 +454,51 @@ ready to dive deeper into them.
   \mathfrak{sl}_n(K) \oplus K\).
 \end{example}
 
-\section{The Universal Enveloping Algebra}
+As suggested by their names, simple and semisimple algebras are quite well
+behaved when compared with the general case. To a lesser degree, reductive
+algebras are also unusualy well behaved. In the next chapter we will explore
+the question of why this is the case, but for now we note that we can get
+semisimple and reductive algebras by modding out by certain ideals, known as
+\emph{radicals}.
+
+\begin{definition}
+  Let \(\mathfrak{g}\) be a Lie algebra. The sum \(\mathfrak{a} +
+  \mathfrak{b}\) of solvable ideals \(\mathfrak{a}, \mathfrak{b} \normal
+  \mathfrak{g}\) is again a solvable ideal. Hence the sum of all solvable
+  ideals of \(\mathfrak{g}\) is a maximal solvable ideal, known as \emph{the
+  radical \(\mathfrak{rad}(\mathfrak{g})\) of \(\mathfrak{g}\)}.
+  \[
+    \mathfrak{rad}(\mathfrak{g})
+    = \sum_{\substack{\mathfrak{a} \normal \mathfrak{g}\\\text{solvable}}}
+      \mathfrak{a}
+  \]
+\end{definition}
+
+\begin{definition}
+  Let \(\mathfrak{g}\) be a Lie algebra. The sum of nilpotent ideals is a
+  nilpotent ideal. Hence the sum of all nilpotent ideals of \(\mathfrak{g}\) is
+  a maximal nilpotent ideal, known as \emph{the nilradical
+  \(\mathfrak{nil}(\mathfrak{g})\) of \(\mathfrak{g}\)}.
+  \[
+    \mathfrak{nil}(\mathfrak{g})
+    = \sum_{\substack{\mathfrak{a} \normal \mathfrak{g}\\\text{nilpotent}}}
+      \mathfrak{a}
+  \]
+\end{definition}
+
+As promised, we finds\dots
+
+\begin{proposition}\label{thm:quotients-by-rads}
+  Let \(\mathfrak{g}\) be a Lie algebra. Then
+  \(\mfrac{\mathfrak{g}}{\mathfrak{rad}(\mathfrak{g})}\) is semisimple and
+  \(\mfrac{\mathfrak{g}}{\mathfrak{nil}(\mathfrak{g})}\) is reductive.
+\end{proposition}
+
+We've seen in example~\ref{ex:inclusion-alg-in-lie-alg} that we can pass from
+associative algebras to Lie algebras using the functor \(\operatorname{Lie} :
+K\text{-}\mathbf{Alg} \to K\text{-}\mathbf{LieAlg}\). We can also go the other
+direction by embedding a Lie algebra \(\mathfrak{g}\) in an associative
+algebra, kwown as \emph{the universal enveloping algebra of \(\mathfrak{g}\)}.
 
 \begin{definition}
   Let \(\mathfrak{g}\) be a Lie algebra and \(T \mathfrak{g} = \bigoplus_n
@@ -422,7 +511,28 @@ ready to dive deeper into them.
   \otimes X)\).
 \end{definition}
 
-\begin{proposition}
+Notice there is a canonical homomorphism \(\mathfrak{g} \to
+\mathcal{U}(\mathfrak{g})\) given by the composition
+\begin{center}
+  \begin{tikzcd}
+    \mathfrak{g} \rar                                     &
+    T \mathfrak{g} \rar                                   &
+    \mfrac{T \mathfrak{g}}{I} = \mathcal{U}(\mathfrak{g})
+  \end{tikzcd}
+\end{center}
+
+We denote the image of some \(X \in \mathfrak{g}\) under the inclusion
+\(\mathfrak{g} \to T \mathfrak{g}\) simply by \(X \in
+\mathcal{U}(\mathfrak{g})\), and we write \(u \cdot v\) for \((u + I) \otimes
+(v + I)\). Since the projection \(T \mathfrak{g} \to
+\mathcal{U}(\mathfrak{g})\) is not injective, it is not at all clear that the
+homomorphism \(\mathfrak{g} \to \mathcal{U}(\mathfrak{g})\) is injective -- as
+suggested by the notation we've just highlighted. However, we will soon see
+this is the case. Intuitively, \(\mathcal{U}(\mathfrak{g})\) is the smallest
+associative \(K\)-algebra containing \(\mathfrak{g}\) as a Lie subalgebra. In
+practice this means\dots
+
+\begin{proposition}\label{thm:universal-env-uni-prop}
   Let \(\mathfrak{g}\) be a Lie algebra and \(A\) be an associative
   \(K\)-algebra. Then every homomorphism of Lie algebras \(f : \mathfrak{g} \to
   A\) -- where \(A\) is endowed with the structure of a Lie algebra as in
@@ -471,8 +581,11 @@ ready to dive deeper into them.
   uniqueness of \(g\) and \(\bar{g}\).
 \end{proof}
 
-% TODO: Remark this construction is functorial
-
+We should point out this construction is functorial. Indeed, if
+\(f : \mathfrak{g} \to \mathfrak{h}\) is a homomorphism of Lie algebras then
+proposition~\ref{thm:universal-env-uni-prop} implies there is a homomorphism of
+algebras \(\mathcal{U}(f) : \mathcal{U}(\mathfrak{g}) \to
+\mathcal(\mathfrak{h})\) satisfying
 \begin{center}
   \begin{tikzcd}
     \mathcal{U}(\mathfrak{g}) \arrow[dotted]{rr}{\mathcal{U}(f)} & &
@@ -483,8 +596,13 @@ ready to dive deeper into them.
   \end{tikzcd}
 \end{center}
 
-% TODO: Point out U is not the "inverse" of K-Alg -> K-LieAlg, but there is an
-% adjunction
+It is important to note, however, that \(\mathcal{U} : K\text{-}\mathbf{LieAlg}
+\to K\text{-}\mathbf{Alg}\) is not the ``inverse'' of \(\operatorname{Lie} :
+K\text{-}\mathbf{Alg} \to K\text{-}\mathbf{LieAlg}\). For instance, if
+\(\mathfrak{g} = K\) is the \(1\)-dimensional Abelian Lie algebra then
+\(\mathcal{U}(\mathfrak{g}) \cong K[x]\), which is infinite-dimensional.
+Nevertheless, proposition~\ref{thm:universal-env-uni-prop} may be restated
+as\dots
 
 \begin{corollary}
   If \(\operatorname{Lie} : K\text{-}\mathbf{Alg} \to
@@ -493,8 +611,16 @@ ready to dive deeper into them.
   \(\operatorname{Lie} \vdash \mathcal{U}\).
 \end{corollary}
 
-% TODO: Define the algebra of differential operators of a given algebra
-\begin{proposition}
+This construction may seem like a purely algebraic affair, but the universal
+enveloping algebra of the Lie algebra of a Lie group \(G\) is in fact
+intemately related with the algebra \(\operatorname{Diff}(G)\) of differential
+operators \(C^\infty(G) \to C^\infty(G)\) -- as defined in Coutinho's
+\citetitle{coutinho} \cite[ch.~3]{coutinho}, for example. Algebras of
+differential operators and their modules are the subject of the theory of
+\(D\)-modules, which has seen remarkable progress in the past century.
+Specifically, we find\dots
+
+\begin{proposition}\label{thm:geometric-realization-of-uni-env}
   Let \(G\) be a Lie group and \(\mathfrak{g}\) be its Lie algebra. Denote by
   \(\operatorname{Diff}(G)^G\) the algebra of \(G\)-invariant differential
   operators in \(G\) -- i.e. the algebra of all differential operators \(L :
@@ -533,15 +659,38 @@ ready to dive deeper into them.
   this also goes to show \(g\) is surjective.
 \end{proof}
 
-% TODO: Comment on the fact this holds for algebraic groups too
+As one would expect, the same holds for complex Lie groups and algebraic groups
+too -- if we replace \(C^\infty(G)\) by \(\mathcal{O}(G)\) and \(K[G]\),
+respectively. This last theorem has profound implications regarding the
+structure of \(\mathcal{U}(\mathfrak{g})\). For one, since
+\(\operatorname{Diff}(G)\) is a domain, so is \(\mathcal{U}(\mathfrak{g}) \cong
+\operatorname{Diff}(G)^G\). In adition, also
+proposition~\ref{thm:geometric-realization-of-uni-env} implies the inclusion
+\(\mathfrak{g} \to \mathcal{U}(\mathfrak{g})\) is in fact injective.
+
+Of course, this are results concerning arbitrary Lie algebras and
+proposition~\ref{thm:geometric-realization-of-uni-env} only applies for
+algebras which come from Lie groups -- as well as complex Lie groups and
+algebraic groups. Nevertheless, these are still lots of Lie algebras. For
+instance, we've seen every finite-dimensional complex Lie algebra is the Lie
+algebra of some simply connected complex Lie group.
+Proposition~\ref{thm:geometric-realization-of-uni-env} thus affords us an
+analytic proof of a particular case -- the case where \(\mathfrak{g}\) is a
+finite-dimensional complex Lie algebra -- of the following result.
 
 \begin{theorem}[Poincaré-Birkoff-Witt]
-  Let \(\mathfrak{g}\) be a finite-dimensional Lie algebra over \(K\) and
-  \(\{X_i\}_i \subset \mathfrak{g}\) be a basis for \(\mathfrak{g}\). Then
-  \(\{X_{i_1} \cdot X_{i_2} \cdots X_{i_n} : n \ge 0, i_1 \le i_2 \le \cdots
-  \le i_n\}\) is a basis for \(\mathcal{U}(\mathfrak{g})\).
+  Let \(\mathfrak{g}\) be a Lie algebra over \(K\) and \(\{X_i\}_i \subset
+  \mathfrak{g}\) be a basis for \(\mathfrak{g}\). Then \(\{X_{i_1} \cdot
+  X_{i_2} \cdots X_{i_n} : n \ge 0, i_1 \le i_2 \le \cdots \le i_n\}\) is a
+  basis for \(\mathcal{U}(\mathfrak{g})\).
 \end{theorem}
 
+We would like to stress that the Poincaré-Birkoff-Witt applies for arbitrary
+Lie algebras and that its analytic proof only works in a very particular case.
+We should also note that the fact the inclusion \(\mathfrak{g} \to
+\mathcal{U}(\mathfrak{g})\) is injective and \(\mathcal{U}(\mathfrak{g})\) is a
+domain are immediate consequences of the Poincaré-Birkoff-Witt theorem.
+
 % TODO: Comment on the fact that modules of invariant differential operators
 % over G are precisely the same as representations of g