\chapter{Introduction}

\pagenumbering{arabic}
\setcounter{page}{1}

Associative algebras have proven themselves remarkably useful throughout
mathematics. There is no lack of natural and interesting examples coming from a
diverse spectrum of different fields: topology, number theory, analysis, you
name it. Associative algebras have thus been studied at length, specially the
commutative ones. On the other hand, non-associative algebras have never
sustained the same degree of scrutiny. To this day, non-associative algebras
remain remarkably mysterious. Many have given up on attempting a systematic
investigation and focus instead on understanding particular classes of
non-associative algebras -- i.e. algebras satisfying
\emph{pseudo-associativity} conditions.

Perhaps the most fascinating class of non-associative algebras are the so
called \emph{Lie algebras}, and these will be the focus of these notes.

\begin{definition}\index{Lie algebra}
  Given a field \(K\), a Lie algebra over \(K\) is a \(K\)-vector space
  \(\mathfrak{g}\) endowed with an antisymmetric bilinear map \([\, ,] :
  \mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}\) -- which we call its
  \emph{Lie bracket} -- satisfying the Jacobi identity
  \[
    [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0
  \]
\end{definition}

\begin{definition}\index{Lie algebra!homomorphism}
  Given two Lie algebras \(\mathfrak{g}\) and \(\mathfrak{h}\) over \(K\), a
  homomorphism of Lie algebras \(\mathfrak{g} \to \mathfrak{h}\) is a
  \(K\)-linear map \(f : \mathfrak{g} \to \mathfrak{h}\) which \emph{preserves
  bracket} in the sense that
  \[
    f([X, Y]) = [f(X), f(Y)]
  \]
  for all \(X, Y \in \mathfrak{g}\). The dimension \(\dim \mathfrak{g}\) of
  \(\mathfrak{g}\) is its dimension as a \(K\)-vector space.
\end{definition}

The collection of Lie algebras over a fixed field \(K\) thus form a category,
which we call \(K\text{-}\mathbf{LieAlg}\). We are primarily interested in
finite-dimensional Lie algebras over algebraically closed fields of
characteristic \(0\). Hence from now on we assume \(K\) is algebraically closed
and \(\operatorname{char} K = 0\) unless explicitly stated otherwise.
Ironically, perhaps the most basic examples of Lie algebras are derived from
associative algebras.

\begin{example}\label{ex:inclusion-alg-in-lie-alg}\index{Lie algebra!Lie algebra of an associative algebra}
  Given an associative \(K\)-algebra \(A\), we can view \(A\) as a Lie algebra
  over \(K\) with the Lie bracket given by the commutator \([a, b] = ab -
  ba\). In particular, given a \(K\)-vector space \(V\) we may view the
  \(K\)-algebra \(\operatorname{End}(V)\) as a Lie algebra, which we call
  \(\mathfrak{gl}(V)\). We may also regard the Lie algebra \(\mathfrak{gl}_n(K)
  = \mathfrak{gl}(K^n)\) as the space of \(n \times n\) matrices with
  coefficients in \(K\).
\end{example}

\begin{example}\label{ex:gln-inclusions}
  Let \(n \le m\). Then the map
  \begin{align*}
    \mathfrak{gl}_n(K) & \to     \mathfrak{gl}_m(K)                          \\
                     X & \mapsto \begin{pmatrix} X & 0 \\ 0 & 0 \end{pmatrix}
  \end{align*}
  is a homomorphism of Lie algebras.
\end{example}

While straightforward enough, I always found the definition of a Lie algebra
unconvincing on its own. Specifically, the Jacobi identity can look very alien
to someone who has never ventured outside of the realms of associativity.
Traditional abstract algebra courses offer little in the way of a motivation
for studying non-associative algebras in general. Why should we drop the
assumption of associativity if every example of an algebraic structure we have
ever seen is an associative one? Instead, the most natural examples of Lie
algebras often come from an entirely different field: geometry.

Here the meaning of \emph{geometry} is somewhat vague. Topics such as
differential and algebraic geometry are prominently featured, but examples
from fields such as the theory of differential operators and \(D\)-modules also
show up a lot in the theory of representations -- which we will soon discuss.
Perhaps one of the most fundamental themes of the study of Lie algebras is
their relationship with groups, specially in geometric contexts. We will now
provide a brief description of this relationship through a series of examples.

\begin{example}\index{Lie algebra!Lie algebra of derivations}
  Let \(A\) be an associative \(K\)-algebra and \(\operatorname{Der}(A)\) be
  the space of all derivations on \(A\) -- i.e. all linear maps \(D : A \to A\)
  satisfying the Leibniz rule \(D(a \cdot b) = a \cdot D b + (D a) \cdot b\).
  The commutator \([D, D']\) of two derivations \(D, D' \in
  \operatorname{Der}(A)\) in the ring \(\operatorname{End}(A)\) of \(K\)-linear
  endomorphisms of \(A\) is a derivation. Hence \(\operatorname{Der}(A)\) is a
  Lie algebra.
\end{example}

One specific instance of this last example is\dots

\begin{example}\index{Lie algebra!Lie algebra of vector fields}
  Given a smooth manifold \(M\), the space \(\mathfrak{X}(M)\) of all smooth
  vector fields is canonically identified with \(\operatorname{Der}(M) =
  \operatorname{Der}(C^\infty(M))\) -- where a field \(X \in \mathfrak{X}(M)\)
  is identified with the map \(C^\infty(M) \to C^\infty(M)\) which takes a
  function \(f \in C^\infty(M)\) to its derivative in the direction of \(X\).
  This gives \(\mathfrak{X}(M)\) the structure of a Lie algebra over
  \(\mathbb{R}\).
\end{example}

\begin{example}\label{ex:lie-alg-of-lie-grp}\index{Lie algebra!Lie algebra of a Lie group}
  Given a Lie group \(G\) -- i.e. a smooth manifold endowed with smooth group
  operations -- we call \(X \in \mathfrak{X}(G)\) left invariant if \((d
  \ell_g)_1 X_1 = X_g\) for all \(g \in G\), where \(\ell_g : G \to G\) denotes
  the left translation by \(g\). The commutator of invariant fields is
  invariant, so the space \(\mathfrak{g} = \operatorname{Lie}(G)\) of all
  invariant vector fields has the structure of a Lie algebra over
  \(\mathbb{R}\) with bracket given by the usual commutator of fields. Notice
  that an invariant field \(X\) is completely determined by \(X_1 \in T_1 G\).
  Hence there is a linear isomorphism \(\mathfrak{g} \isoto T_1 G\). In
  particular, \(\mathfrak{g}\) is finite-dimensional.
\end{example}

We should point out that the Lie algebra \(\mathfrak{g}\) of a complex Lie
group \(G\) -- i.e. a complex manifold endowed with holomorphic group
operations -- has the natural structure of a complex Lie algebra. Indeed, every
left invariant field \(X \in \mathfrak{X}(G)\) is holomorphic, so
\(\mathfrak{g}\) is a (complex) subspace of the complex vector space of
holomorphic vector fields over \(G\). There is also an algebraic analogue of
this last construction.

\begin{example}\index{Lie algebra!Lie algebra of an algebraic group}
  Let \(G\) be an affine algebraic \(K\)-group -- i.e. an affine variety over
  \(K\) with rational group operations -- and \(K[G]\) denote the ring of
  regular functions \(G \to K\). We call a derivation \(D : K[G] \to K[G]\)
  left invariant if \(D(g \cdot f) = g \cdot D f\) for all \(g \in G\) and \(f
  \in K[G]\) -- where the action of \(G\) on \(K[G]\) is given by \((g \cdot
  f)(h) = f(g^{-1} h)\). The commutator of left invariant derivations is
  invariant too, so the space \(\operatorname{Lie}(G) =
  \operatorname{Der}(G)^G\) of invariant derivations in \(K[G]\) has the
  structure of a Lie algebra over \(K\) with bracket given by the commutator
  of derivations. Again, \(\operatorname{Lie}(G)\) is isomorphic to the Zariski
  tangent space \(T_1 G\), which is finite-dimensional.
\end{example}

\begin{example}
  The Lie algebra \(\operatorname{Lie}(\operatorname{GL}_n(K))\) is canonically
  isomorphic to the Lie algebra \(\mathfrak{gl}_n(K)\). Likewise, the Lie
  algebra \(\operatorname{Lie}(\operatorname{SL}_n(K))\) is canonically
  isomorphic to the Lie algebra \(\mathfrak{sl}_n(K)\) of traceless \(n \times
  n\) matrices.
  \[
    \mathfrak{sl}_n(K)
    = \{ X \in \mathfrak{gl}_n(K) : \operatorname{Tr} X = 0 \}
  \]
\end{example}

\begin{example}\label{ex:sl2-basis}
  The elements
  \begin{align*}
    e & = \begin{pmatrix} 0 & 1 \\ 0 &  0 \end{pmatrix} &
    f & = \begin{pmatrix} 0 & 0 \\ 1 &  0 \end{pmatrix} &
    h & = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
  \end{align*}
  form a basis for \(\mathfrak{sl}_2(K)\) and are subject to the following
  relations.
  \begin{align*}
    [e, f] & = h & [h, f] & = -2 f & [h, e] = 2 e
  \end{align*}
\end{example}

\begin{example}\label{ex:sp2n}
  The Lie algebra of the affine algebraic group
  \[
    \operatorname{Sp}_{2 n}(K)
    = \{
        g \in \operatorname{GL}_{2 n}(K) :
        \omega(g \cdot v, g \cdot w) = \omega(v, w) \, \forall v, w \in K^{2n}
      \}
  \]
  is canonically isomorphic to the Lie algebra
  \[
    \mathfrak{sp}_{2 n}(K) =
    \left\{
      \begin{pmatrix}
        X &  Y      \\
        Z & -X^\top
      \end{pmatrix}
      : X, Y, Z \in \mathfrak{gl}_n(K), Y = Y^\top, Z = Z^\top
    \right\},
  \]
  with bracket given by the usual commutator of matrices -- where
  \[
    \omega(
      (v_1, \ldots, v_n, \dot v_1, \ldots, \dot v_n),
      (w_1, \ldots, w_n, \dot w_1, \ldots, \dot w_n)
    )
    = v_1 \dot w_1 + \cdots + v_n \dot w_n
    - \dot v_1 w_1 - \cdots - \dot v_n w_n
  \]
  is, of course, the standard symplectic form of \(K^{2n}\).
\end{example}

It is important to point out that the construction of the Lie algebra
\(\mathfrak{g}\) of a Lie group \(G\) in Example~\ref{ex:lie-alg-of-lie-grp} is
functorial. Specifically, one can show the derivative \(d f_1 : \mathfrak{g}
\cong T_1 G \to T_1 H \cong \mathfrak{h}\) of a smooth group homomorphism \(f :
G \to H\) is a homomorphism of Lie algebras, and the chain rule implies \(d (f
\circ g)_1 = d f_1 \circ d g_1\). This is known as the \emph{the Lie functor}
\(\operatorname{Lie} : \mathbf{LieGrp} \to \mathbb{R}\text{-}\mathbf{LieAlg}\)
between the category of Lie groups and smooth group homomorphisms and the
category of Lie algebras.

This goes to show Lie algebras are invariants of Lie groups. What is perhaps
more surprising is the fact that, in certain contexts, Lie algebras are perfect
invariants. Even more so\dots

\begin{theorem}[Lie]\label{thm:lie-theorems}
  The restriction \(\operatorname{Lie} : \mathbf{LieGrp}_{\operatorname{simpl}}
  \to \mathbb{R}\text{-}\mathbf{LieAlg}\) of the Lie functor to the full
  subcategory of simply connected Lie groups is an equivalence of categories
  onto the full subcategory of finite-dimensional real Lie algebras.
\end{theorem}

This last theorem is a direct corollary of the so called \emph{first and third
fundamental Lie Theorems}. Lie's first Theorem establishes that if \(G\) is a
simply connected Lie group and \(H\) is a connected Lie group then the induced
map \(\operatorname{Hom}(G, H) \to \operatorname{Hom}(\mathfrak{g},
\mathfrak{h})\) is bijective, which implies the Lie functor is fully faithful.
On the other hand, Lie's third Theorem states that every finite-dimensional
real Lie algebra is the Lie algebra of a simply connected Lie group -- i.e. the
Lie functor is essentially surjective.

This goes to show that the relationship between Lie groups and Lie algebras is
deeper than the fact they share a name: in a very strong sense, studying simply
connected Lie groups is \emph{precisely} the same as studying
finite-dimensional Lie algebras. Such a vital connection between apparently
distant subjects is bound to produce interesting results. Indeed, the passage
from the geometric setting to its algebraic counterpart and vice-versa has
proven itself a fruitful one.

This correspondence can be extended to the complex case too. In other words,
the Lie functor \(\mathbf{CLieGrp}_{\operatorname{simpl}} \to
\mathbb{C}\text{-}\mathbf{LieAlg}\) is also an equivalence of categories
between the category of simply connected complex Lie groups and the full
subcategory of finite-dimensional complex Lie algebras. The situation is more
delicate in the algebraic case. For instance, consider the complex Lie algebra
homomorphism
\begin{align*}
  f : \mathbb{C} & \to     \mathfrak{sl}_2(\mathbb{C}) \\
         \lambda & \mapsto \lambda h =
                   \begin{pmatrix} \lambda & 0 \\ 0 & - \lambda \end{pmatrix}
\end{align*}

Since \(\mathfrak{sl}_2(\mathbb{C}) =
\operatorname{Lie}(\operatorname{SL}_2(\mathbb{C}))\) and
\(\operatorname{SL}_2(\mathbb{C})\) is simply connected, we know there exists a
unique holomorphic group homomorphism \(g : \mathbb{C} \to
\operatorname{SL}_2(\mathbb{C})\) between the affine line \(\mathbb{C}\) and
the complex \emph{algebraic} group \(\operatorname{SL}_2(\mathbb{C})\) such
that \(f = d g_1\). Indeed, this homomorphism is
\begin{align*}
  g : \mathbb{C} & \to     \operatorname{SL}_2(\mathbb{C}) \\
         \lambda & \mapsto \operatorname{exp}(\lambda h) =
                   \begin{pmatrix} e^\lambda & 0 \\ 0 & e^{-\lambda} \end{pmatrix},
\end{align*}
which is not a rational map. It then follows from the uniqueness of \(g\) that
there is no rational group homomorphism \(\mathbb{C} \to
\operatorname{SL}_2(\mathbb{C})\) whose derivative at the identity is \(f\).

In particular, the Lie functor
\(\mathbb{C}\text{-}\mathbf{Grp}_{\operatorname{simpl}} \to
\mathbb{C}\text{-}\mathbf{LieAlg}\) -- between the category
\(\mathbb{C}\text{-}\mathbf{Grp}_{\operatorname{simpl}}\) of simply connected
complex algebraic groups and the category of complex Lie algebras -- fails to
be full. Similarly, the functor
\(\mathbb{C}\text{-}\mathbf{Grp}_{\operatorname{simpl}} \to
\mathbb{C}\text{-}\mathbf{LieAlg}\) is \emph{not} essentially surjective onto
the subcategory of finite-dimensional algebras: every finite-dimensional
complex Lie algebra is isomorphic to the Lie algebra of a unique simply
connected complex Lie group, but there are simply connected complex Lie groups
which are not algebraic groups. Nevertheless, Lie algebras are still powerful
invariants of algebraic groups. An interesting discussion of some of these
delicacies can be found in sixth section of \cite[ch.~II]{demazure-gabriel}.

All in all, there is a profound connection between groups and
finite-dimensional Lie algebras throughout multiple fields. While perhaps
unintuitive at first, the advantages of working with Lie algebras over their
group-theoretic counterparts are numerous. First, Lie algebras allow us to
avoid much of the delicacies of geometric objects such as real and complex Lie
groups. Even when working without additional geometric considerations, groups
can be complicated beasts themselves. They are, after all, nonlinear objects.
On the other hand, Lie algebras are linear by nature, which makes them much
more flexible than groups.

Having thus hopefully established that Lie algebras are interesting, we are now
ready to dive deeper into them. We begin by analyzing some of their most basic
properties.

\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}\index{Lie subalgebra}\index{Lie subalgebra!ideals}
  Given a Lie algebra \(\mathfrak{g}\), a subspace \(\mathfrak{h} \subset
  \mathfrak{g}\) is called \emph{a subalgebra of \(\mathfrak{g}\)} if \([X, Y]
  \in \mathfrak{h}\) for all \(X, Y \in \mathfrak{h}\). A subalgebra
  \(\mathfrak{a} \subset \mathfrak{g}\) is called \emph{an ideal of
  \(\mathfrak{g}\)} if \([X, Y] \in \mathfrak{a}\) for all \(X \in
  \mathfrak{g}\) and \(Y \in \mathfrak{a}\), in which case we write
  \(\mathfrak{a} \normal \mathfrak{g}\).
\end{definition}

\begin{note}
  In the context of associative algebras, it is usual practice to distinguish
  between \emph{left ideals} and \emph{right ideals}. This is not necessary
  when dealing with Lie algebras, however, since any ``left ideal'' of a Lie
  algebra is also a ``right ideal'': given \(\mathfrak{a} \normal
  \mathfrak{g}\), \([Y, X] = - [X, Y] \in \mathfrak{a}\) 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. Furthermore, \(\ker f \normal \mathfrak{g}\).
\end{example}

\begin{example}
  Let \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\) be a Lie algebras over \(K\).
  Then the space \(\mathfrak{g}_1 \oplus \mathfrak{g}_2\) is a Lie algebra with
  bracket
  \[
    [X_1 + X_2, Y_1 + Y_2] = [X_1, Y_1] + [X_2, Y_2],
  \]
  and \(\mathfrak{g}_1, \mathfrak{g}_2 \normal \mathfrak{g}_1 \oplus
  \mathfrak{g}_2\).
\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}

\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
  through the projection \(\mathfrak{g} \to
  \mfrac{\mathfrak{g}}{\mathfrak{a}}\).
  \begin{center}
    \begin{tikzcd}
      \mathfrak{g}                       \rar{f} \dar       & \mathfrak{h} \\
      \mfrac{\mathfrak{g}}{\mathfrak{a}} \arrow[dotted]{ur} &
    \end{tikzcd}
  \end{center}
\end{proposition}

\begin{definition}\index{Lie algebra!Abelian Lie algebra}
  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{note}
  Notice that an Abelian Lie algebra is determined by its dimension. Indeed,
  any linear map \(\mathfrak{g} \to \mathfrak{h}\) between Abelian Lie algebras
  \(\mathfrak{g}\) and \(\mathfrak{h}\) is a homomorphism of Lie algebras. In
  particular, any linear isomorphism \(\mathfrak{g} \isoto K^n\) -- where
  \(K^n\) is endowed with the trivial bracket \([v, w] = 0\), \(v, w \in K^n\)
  -- is an isomorphism of Lie algebras for Abelian \(\mathfrak{g}\).
\end{note}

\begin{example}\index{Lie algebra!center}
  Let \(\mathfrak{g}\) be a Lie algebra and \(\mathfrak{z} = \{ X \in
  \mathfrak{g} : [X, Y] = 0, Y \in \mathfrak{g}\}\). Then \(\mathfrak{z}\) is
  an Abelian ideal of \(\mathfrak{g}\), known as \emph{the center of
  \(\mathfrak{z}\)}.
\end{example}

Due to their relationship with Lie groups and algebraic groups, Lie algebras
also share structural features with groups. For example\dots

\begin{definition}\index{Lie algebra!solvable Lie algebra}
  A Lie algebra \(\mathfrak{g}\) is called \emph{solvable} if its derived
  series
  \[
    \mathfrak{g}
    \supseteq [\mathfrak{g}, \mathfrak{g}]
    \supseteq [[\mathfrak{g}, \mathfrak{g}], [\mathfrak{g}, \mathfrak{g}]]
    \supseteq
    [
      [[\mathfrak{g}, \mathfrak{g}], [\mathfrak{g}, \mathfrak{g}]],
      [[\mathfrak{g}, \mathfrak{g}], [\mathfrak{g}, \mathfrak{g}]]
    ]
    \supseteq \cdots
  \]
  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}\index{Lie algebra!nilpotent Lie algebra}
  A Lie algebra \(\mathfrak{g}\) is called \emph{nilpotent} if its lower
  central series
  \[
    \mathfrak{g}
    \supseteq [\mathfrak{g}, \mathfrak{g}]
    \supseteq [\mathfrak{g}, [\mathfrak{g}, \mathfrak{g}]]
    \supseteq [\mathfrak{g}, [\mathfrak{g}, [\mathfrak{g}, \mathfrak{g}]]]
    \supseteq \cdots
  \]
  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 nilpotent if, and only if \(\mathfrak{g}\) is.
\end{example}

Other interesting classes of Lie algebras are the so called \emph{simple} and
\emph{semisimple} Lie algebras.

\begin{definition}\index{simple!Lie algebra}\index{Lie algebra!simple Lie algebra}
  A non-Abelian Lie algebra \(\mathfrak{s}\) over \(K\) is called \emph{simple}
  if its only ideals are \(0\) and \(\mathfrak{s}\).
\end{definition}

\begin{example}
  The Lie algebra \(\mathfrak{sl}_2(K)\) is simple. To see this, notice that
  any ideal \(\mathfrak{a} \normal \mathfrak{sl}_2(K)\) must be stable under
  the operator \(\operatorname{ad}(h) : \mathfrak{sl}_2(K) \to
  \mathfrak{sl}_2(K)\) given by \(\operatorname{ad}(h) X = [h, X]\). But
  Example~\ref{ex:sl2-basis} implies \(\operatorname{ad}(h)\) is
  diagonalizable, with eigenvalues \(0\) and \(\pm 2\). Hence \(\mathfrak{a}\)
  must be spanned by some of the eigenvectors \(e, f, h\) of
  \(\operatorname{ad}(h)\). If \(h \in \mathfrak{a}\), then \([e, h] = - 2 e
  \in \mathfrak{a}\) and \([f, h] = 2 f \in \mathfrak{a}\), so \(\mathfrak{a} =
  \mathfrak{sl}_2(K)\). If \(e \in \mathfrak{a}\) then \([f, e] = - h \in
  \mathfrak{a}\), so again \(\mathfrak{a} = \mathfrak{sl}_2(K)\). Similarly, if
  \(f \in \mathfrak{a}\) then \([e, f] = h \in \mathfrak{a}\) and
  \(\mathfrak{a} = \mathfrak{sl}_2(K)\). More generally, the Lie algebra
  \(\mathfrak{sl}_n(K)\) is simple for each \(n > 1\) -- see the section of
  \cite[ch. 6]{kirillov} on invariant bilinear forms and the semisimplicity of
  classical Lie algebras.
\end{example}

\begin{example}
  The Lie algebras \(\mathfrak{sp}_{2n}(K)\) are simple for all \(n \ge 1\) --
  agina, see \cite[ch. 6]{kirillov}.
\end{example}

\begin{definition}\label{thm:sesimple-algebra}\index{semisimple!Lie algebra}\index{Lie algebra!semisimple Lie algebra}
  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 nonzero solvable
  ideals.
\end{definition}

\begin{example}
  Let \(G\) be a connected affine algebraic \(K\)-group. Then \(G\) is
  semisimple if, and only if \(\mathfrak{g}\) semisimple.
\end{example}

A slight generalization is\dots

\begin{definition}\index{Lie algebra!reductive Lie algebra}
  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.
\end{definition}

\begin{example}
  The Lie algebra \(\mathfrak{gl}_n(K)\) is reductive. Indeed,
  \[
    X
    =
    \begin{pmatrix}
      a_{1 1} - \frac{\operatorname{Tr}(X)}{n} & \cdots & a_{1 n} \\
       \vdots & \ddots &  \vdots \\
      a_{n 1} & \cdots & a_{n n} - \frac{\operatorname{Tr}(X)}{n}
    \end{pmatrix}
    +
    \begin{pmatrix}
      \frac{\operatorname{Tr}(X)}{n} & \cdots & 0 \\
      \vdots & \ddots & \vdots \\
      0 & \cdots & \frac{\operatorname{Tr}(X)}{n}
    \end{pmatrix}
  \]
  for each matrix \(X = (a_{i j})_{i j}\). In other words,
  \(\mathfrak{gl}_n(K) = \mathfrak{sl}_n(K) \oplus K \operatorname{Id} \cong
  \mathfrak{sl}_n(K) \oplus K\).
\end{example}

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 unusually 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}\index{Lie algebra!radical}
  Let \(\mathfrak{g}\) be a finite-dimensional 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}\index{Lie algebra!nilradical}
  Let \(\mathfrak{g}\) be a finite-dimensional 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{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 have seen in Example~\ref{ex:inclusion-alg-in-lie-alg} that we can pass from
an associative algebra \(A\) to a Lie algebra by taking its bracket as the
commutator \([a, b] = ab - ba\). We should also not that any homomorphism of
\(K\)-algebras \(f : A \to B\) preserves commutators, so that \(f\) is also a
homomorphism of Lie algebras. Hence we have a 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, known as \emph{the universal enveloping algebra of \(\mathfrak{g}\)}.

\begin{definition}\index{universal enveloping algebra}
  Let \(\mathfrak{g}\) be a Lie algebra and \(T \mathfrak{g} = \bigoplus_n
  \mathfrak{g}^{\otimes n}\) be its tensor algebra -- i.e. the free
  \(K\)-algebra generated by the elements of \(\mathfrak{g}\). We call the
  \(K\)-algebra \(\mathcal{U}(\mathfrak{g}) = \mfrac{T \mathfrak{g}}{I}\)
  \emph{the universal enveloping algebra of \(\mathfrak{g}\)}, where \(I\) is
  the left ideal of \(T \mathfrak{g}\) generated by the elements \([X, Y] - (X
  \otimes Y - Y \otimes X)\).
\end{definition}

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}

Given \(X_1, \ldots, X_n \in \mathfrak{g}\), we identify \(X_i\) with its image
under the inclusion \(\mathfrak{g} \to T \mathfrak{g}\) and we write \(X_1
\cdots X_n\) for \((X_1 \otimes \cdots \otimes X_n) + I\). This notation
suggests the map \(\mathfrak{g} \to \mathcal{U}(\mathfrak{g})\) is injective,
but at this point this is not at all clear -- given that the projection \(T
\mathfrak{g} \to \mathcal{U}(\mathfrak{g})\) is not injective. 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
  Example~\ref{ex:inclusion-alg-in-lie-alg} -- can be uniquely extended to a
  homomorphism of algebras \(\mathcal{U}(\mathfrak{g}) \to A\).
  \begin{center}
    \begin{tikzcd}
      \mathfrak{g}              \rar{f} \dar  & A \\
      \mathcal{U}(\mathfrak{g}) \urar[dotted] &
    \end{tikzcd}
  \end{center}
\end{proposition}

\begin{proof}
  Let \(f : \mathfrak{g} \to A\) be a homomorphism of Lie algebras. By the
  universal property of free algebras, there is a homomorphism of algebras
  \(\tilde f : T \mathfrak{g} \to A\) such that
  \begin{center}
    \begin{tikzcd}
      \mathfrak{g}   \dar \rar{f}                  & A \\
      T \mathfrak{g} \urar[swap, dotted]{\tilde f} &
    \end{tikzcd}
  \end{center}

  Since \(f\) is a homomorphism of Lie algebras,
  \[
    \tilde f([X, Y])
    = f([X, Y])
    = [f(X), f(Y)]
    = [\tilde f(X), \tilde f(Y)]
    = \tilde f(X \otimes Y - Y \otimes X)
  \]
  for all \(X, Y \in \mathfrak{g}\). Hence \(I = ([X, Y] - (X \otimes Y - Y
  \otimes X) : X, Y \in \mathfrak{g}) \subset \ker \tilde f\) and therefore
  \(\tilde f\) factors through the quotient \(\mathcal{U}(\mathfrak{g}) =
  \mfrac{T \mathfrak{g}}{I}\).
  \begin{center}
    \begin{tikzcd}
      T \mathfrak{g} \rar{\tilde f} \dar                                 & A \\
      \mathcal{U}(\mathfrak{g}) \arrow[swap, dotted]{ur}{\bar{\tilde f}} &
    \end{tikzcd}
  \end{center}

  Combining the two previous diagrams, we can see that \(\bar{\tilde f}\) is
  indeed an extension of \(f\). The uniqueness of the extension then follows
  from the uniqueness of \(\tilde f\) and \(\bar{\tilde f}\).
\end{proof}

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 \(\mathcal{U}(f) : \mathcal{U}(\mathfrak{g}) \to
\mathcal{U}(\mathfrak{h})\) satisfying
\begin{center}
  \begin{tikzcd}
    \mathfrak{g}              \rar{f} \dar                              &
    \mathfrak{h}              \rar                                      &
    \mathcal{U}(\mathfrak{h})                                           \\
    \mathcal{U}(\mathfrak{g}) \arrow[swap, dotted]{urr}{\mathcal{U}(f)} & &
  \end{tikzcd}
\end{center}

It is important to note, however, that \(\mathcal{U} : K\text{-}\mathbf{LieAlg}
\to K\text{-}\mathbf{Alg}\) is not the ``inverse'' of our functor
\(K\text{-}\mathbf{Alg} \to K\text{-}\mathbf{LieAlg}\). 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
using the language of adjoint functors -- as described in \cite{maclane} for
instance.

\begin{corollary}
  If \(\operatorname{Lie} : K\text{-}\mathbf{Alg} \to
  K\text{-}\mathbf{LieAlg}\) is the functor described in
  Example~\ref{ex:inclusion-alg-in-lie-alg}, there is an adjunction
  \(\operatorname{Lie} \vdash \mathcal{U}\).
\end{corollary}

The structure of \(\mathcal{U}(\mathfrak{g})\) can often be described in terms
of the structure of \(\mathfrak{g}\). For instance, \(\mathfrak{g}\) is Abelian
if, and only if \(\mathcal{U}(\mathfrak{g})\) is commutative, in which case any
basis \(\{X_i\}_i\) for \(\mathfrak{g}\) induces an isomorphism
\(\mathcal{U}(\mathfrak{g}) \cong K[x_1, x_2, \ldots, x_i, \ldots]\). More
generally, we find\dots

\begin{theorem}[Poincaré-Birkoff-Witt]\index{PBW Theorem}
  Let \(\mathfrak{g}\) be a Lie algebra over \(K\) and \(\{X_i\}_i \subset
  \mathfrak{g}\) be an ordered basis for \(\mathfrak{g}\) -- i.e. a basis
  indexed by an ordered set. 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}

This last result is known as \emph{the PBW Theorem}. It is hugely important and
will come up again and again throughout these notes. Among other things, it
implies\dots

\begin{corollary}
  Let \(\mathfrak{g}\) be a Lie algebra over \(K\). Then
  \(\mathcal{U}(\mathfrak{g})\) is a domain and the inclusion \(\mathfrak{g}
  \to \mathcal{U}(\mathfrak{g})\) is injective.
\end{corollary}

The PBW Theorem can also be used to compute a series of
examples.

\begin{example}
  Consider the Lie algebra \(\mathfrak{gl}_n(K)\) and its canonical basis
  \(\{E_{i j}\}_{i j}\). Even though \(E_{i j} E_{j k} = E_{i k}\) in the
  associative algebra \(\operatorname{End}(K^n)\), the PBW
  Theorem implies \(E_{i j} E_{j k} \ne E_{i k}\) in
  \(\mathcal{U}(\mathfrak{gl}_n(K))\). In general, if \(A\) is an associative
  \(K\)-algebra then the elements in the image of the inclusion \(A \to
  \mathcal{U}(A)\) do not satisfy the same relations as the elements of \(A\).
\end{example}

\begin{example}
  Let \(\mathfrak{g}\) be an Abelian Lie algebra. As previously stated, any
  choice of basis \(\{X_i\}_i \subset \mathfrak{g}\) induces an isomorphism of
  algebras \(\mathcal{U}(\mathfrak{g}) \isoto K[x_1, x_2, \ldots, x_i,
  \ldots]\) which takes \(X_i \in \mathfrak{g}\) to the variable \(x_i \in
  K[x_1, x_2, \ldots, x_i, \ldots]\).
\end{example}

\begin{example}\label{ex:univ-enveloping-of-sum-is-tensor}
  Let \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\) be Lie algebras over \(K\). We
  claim that the natural map
  \begin{align*}
    f: \mathcal{U}(\mathfrak{g}_1) \otimes_K \mathcal{U}(\mathfrak{g}_2) &
    \to \mathcal{U}(\mathfrak{g}_1 \oplus \mathfrak{g}_2) \\
    u \otimes v & \mapsto u \cdot v
  \end{align*}
  is an isomorphism of algebras. Since the elements of \(\mathfrak{g}_1\)
  commute with the elements of \(\mathfrak{g}_2\) in \(\mathfrak{g}_1 \oplus
  \mathfrak{g}_2\), a simple calculation shows that \(f\) is indeed a
  homomorphism of algebras. In addition, the PBW Theorem implies that \(f\) is
  a linear isomorphism.
\end{example}

The construction of \(\mathcal{U}(\mathfrak{g})\) may seem like a purely
algebraic affair, but the universal enveloping algebra of the Lie algebra of a
Lie group \(G\) is in fact intimately related with the algebra
\(\operatorname{Diff}(G)\) of differential operators \(C^\infty(G) \to
C^\infty(G)\) -- i.e. \(\mathbb{R}\)-linear endomorphisms \(C^\infty(G) \to
C^\infty(G)\) of finite order, as defined in \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 \(P :
  C^\infty(G) \to C^\infty(G)\) such that \(g \cdot P f = P (g \cdot f)\) for
  all \(f \in C^\infty(G)\) and \(g \in G\). There is a canonical isomorphism
  of algebras \(\mathcal{U}(\mathfrak{g}) \isoto \operatorname{Diff}(G)^G\).
\end{proposition}

\begin{proof}
  An order \(0\) \(G\)-invariant differential operator in \(G\) is simply
  multiplication by a constant in \(\mathbb{R}\). A homogeneous order \(1\)
  \(G\)-invariant differential operator in \(G\) is simply a left invariant
  derivation \(C^\infty(G) \to C^\infty(G)\). All other \(G\)-invariant
  differential operators are generated by invariant operators of order \(0\)
  and \(1\). Hence \(\operatorname{Diff}(G)^G\) is generated by
  \(\operatorname{Der}(G)^G + \mathbb{R}\) -- here \(\operatorname{Der}(G)^G
  \subset \operatorname{Der}(G)\) denotes the Lie subalgebra of invariant
  derivations.

  Now recall that there is a canonical isomorphism of Lie algebras
  \(\mathfrak{X}(G) \isoto \operatorname{Der}(G)\). This isomorphism takes left
  invariant fields to left invariant derivations, so it restricts to an
  isomorphism \(f : \mathfrak{g} \isoto \operatorname{Der}(G)^G \subset
  \operatorname{Diff}(G)^G\). Since \(f\) is a homomorphism of Lie algebras, it
  can be extended to an algebra homomorphism \(\tilde f :
  \mathcal{U}(\mathfrak{g}) \to \operatorname{Diff}(G)^G\). We claim \(\tilde
  f\) is an isomorphism.

  To see that \(\tilde f\) is injective, it suffices to notice
  \[
    \tilde f(X_1 \cdots X_n)
    = \tilde f(X_1) \cdots \tilde f(X_n)
    = f(X_1) \cdots f(X_n)
    \ne 0
  \]
  for all nonzero \(X_1, \ldots, X_n \in \mathfrak{g}\) -- the product of
  operators of positive order has positive order and is therefore nonzero.
  Since \(\mathcal{U}(\mathfrak{g})\) is generated by the image of the
  inclusion \(\mathfrak{g} \to \mathcal{U}(\mathfrak{g})\), this implies \(\ker
  \tilde f = 0\). Given that \(\operatorname{Diff}(G)^G\) is generated by
  \(\operatorname{Der}(G)^G + \mathbb{R}\), this also goes to show \(\tilde f\)
  is surjective.
\end{proof}

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 proposition has profound implications. For example, it
affords us an analytic proof of certain particular cases of the
PBW Theorem. Most surprising of all,
Proposition~\ref{thm:geometric-realization-of-uni-env} implies
\(\mathcal{U}(\mathfrak{g})\)-modules are \emph{precisely} the same as modules
over the ring of \(G\)-invariant differential operators -- i.e.
\(\operatorname{Diff}(G)^G\)-modules. We can thus use
\(\mathcal{U}(\mathfrak{g})\) and its modules to understand the geometry of
\(G\).

Proposition~\ref{thm:geometric-realization-of-uni-env} is in fact only the
beginning of a profound connection between the theory of \(D\)-modules and
\emph{representation theory}, the latter of which we now explore in the
following section.

\section{Representation Theory}

First introduced in 1896 by Georg Frobenius in his paper \citetitle{frobenius}
\cite{frobenius} in the context of group theory, representation theory is now
one of the cornerstones of modern mathematics. In this section we provide a
brief overview of basic concepts of the representation theory of Lie algebras.
We should stress, however, that the representation theory of Lie algebras is
only a small fragment of what is today known as ``representation theory'',
which is in general concerned with a diverse spectrum of algebraic and
combinatorial structures -- such as groups, quivers and associative algebras.
An introductory exploration of some of these structures can be found in
\cite{etingof}.

We begin by noting that any \(\mathcal{U}(\mathfrak{g})\)-module \(M\) may be
regarded as a \(K\)-vector space endowed with a ``linear action'' of
\(\mathfrak{g}\). Indeed, by restricting the action map
\(\mathcal{U}(\mathfrak{g}) \to \operatorname{End}(M)\) to \(\mathfrak{g}
\subset \mathcal{U}(\mathfrak{g})\) yields a homomorphism of Lie algebras
\(\mathfrak{g} \to \mathfrak{gl}(M) = \operatorname{End}(M)\). In fact
Proposition~\ref{thm:universal-env-uni-prop} implies that given a vector space
\(M\) there is a one-to-one correspondence between
\(\mathcal{U}(\mathfrak{g})\)-module structures for \(M\) and homomorphisms
\(\mathfrak{g} \to \mathfrak{gl}(M)\). This leads us to the following
definition.

\begin{definition}
  Given a Lie algebra \(\mathfrak{g}\) over \(K\), \emph{a representation \(V\)
  of \(\mathfrak{g}\)} is a \(K\)-vector space endowed with a homomorphism of
  Lie algebras \(\rho : \mathfrak{g} \to \mathfrak{gl}(V)\).
\end{definition}

Hence there is a one-to-one correspondence between representations of
\(\mathfrak{g}\) and \(\mathcal{U}(\mathfrak{g})\)-modules.

\begin{example}\index{\(\mathfrak{g}\)-module!trivial module}
  Given a Lie algebra \(\mathfrak{g}\), the zero map \(0 : \mathfrak{g} \to K\)
  gives \(K\) the structure of a representation of \(\mathfrak{g}\), known as
  \emph{the trivial representation}.
\end{example}

\begin{example}\index{\(\mathfrak{g}\)-module!adjoint module}
  Given a Lie algebra \(\mathfrak{g}\), consider the homomorphism
  \(\operatorname{ad} : \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})\) given by
  \(\operatorname{ad}(X) Y = [X, Y]\). This gives \(\mathfrak{g}\) the
  structure of a representation of \(\mathfrak{g}\), known as \emph{the adjoint
  representation}.
\end{example}

\begin{example}\index{\(\mathfrak{g}\)-module!regular module}
  Given a Lie algebra \(\mathfrak{g}\), the map \(\rho : \mathfrak{g} \to
  \mathfrak{gl}(\mathcal{U}(\mathfrak{g}))\) given by left multiplication
  endows \(\mathcal{U}(\mathfrak{g})\) with the structure of a representation
  of \(\mathfrak{g}\), known as \emph{the regular representation of
  \(\mathfrak{g}\)}.
  \[
    \arraycolsep=1.4pt
    \begin{array}[t]{rl}
      \rho : \mathfrak{g} & \to \mathfrak{gl}(\mathcal{U}(\mathfrak{g})) \\
      X & \mapsto
      \begin{array}[t]{rl}
        \rho(X) : \mathcal{U}(\mathfrak{g}) & \to \mathcal{U}(\mathfrak{g}) \\
        u & \mapsto X \cdot u
      \end{array}
    \end{array}
  \]
\end{example}

\begin{example}\index{\(\mathfrak{g}\)-module!natural module}
  Given a subalgebra \(\mathfrak{g} \subset \mathfrak{gl}_n(K)\), the inclusion
  \(\mathfrak{g} \to \mathfrak{gl}_n(K)\) endows \(K^n\) with the structure of
  a representation of \(\mathfrak{g}\), known as \emph{the natural
  representation of \(\mathfrak{g}\)}.
\end{example}

When the map \(\rho : \mathfrak{g} \to \mathfrak{gl}(V)\) is clear from context
it is usual practice to denote the \(K\)-endomorphism \(\rho(X) : V \to V\),
\(X \in \mathfrak{g}\), simply by \(X\!\restriction_V\). This leads us to the
natural notion of \emph{transformations} between representations.

\begin{definition}
  Given a Lie algebra \(\mathfrak{g}\) and two representations \(V\) and \(W\)
  of \(\mathfrak{g}\), we call a \(K\)-linear map \(f : V \to W\) \emph{an
  intertwining operator}, or \emph{an intertwiner}, if it commutes with the
  action of \(\mathfrak{g}\) on \(V\) and \(W\), in the sense that the diagram
  \begin{center}
    \begin{tikzcd}
      V \rar{f} \dar[swap]{X\!\restriction_V} & W \dar{X\!\restriction_W} \\
      V \rar[swap]{f}                         & W
    \end{tikzcd}
  \end{center}
  commutes for all \(X \in \mathfrak{g}\). We denote the space of all
  intertwiners \(V \to W\) by \(\operatorname{Hom}_{\mathfrak{g}}(V, W)\) -- as
  opposed the space \(\operatorname{Hom}(V, W)\) of all \(K\)-linear maps
  \(V \to W\).
\end{definition}

The collection of representations of a fixed Lie algebra \(\mathfrak{g}\) thus
forms a category, which we call \(\mathbf{Rep}(\mathfrak{g})\). This allow us
formulate the correspondence between representations of \(\mathfrak{g}\) and
\(\mathcal{U}(\mathfrak{g})\)-modules in more precise terms.

\begin{proposition}
  There is a natural isomorphism of categories \(\mathbf{Rep}(\mathfrak{g})
  \isoto \mathcal{U}(\mathfrak{g})\text{-}\mathbf{Mod}\).
\end{proposition}

\begin{proof}
  We have seen that given a \(K\)-vector space \(M\) there is a one-to-one
  correspondence between \(\mathfrak{g}\)-representation structures for \(M\)
  -- i.e. homomorphisms \(\mathfrak{g} \to \mathfrak{gl}(M)\) -- and
  \(\mathcal{U}(\mathfrak{g})\)-module structures for \(M\) -- i.e.
  homomorphisms \(\mathcal{U}(\mathfrak{g}) \to \operatorname{End}(M)\). This
  gives us a surjective map that takes objects in
  \(\mathbf{Rep}(\mathfrak{g})\) to objects in
  \(\mathcal{U}(\mathfrak{g})\text{-}\mathbf{Mod}\).

  As for the corresponding maps \(\operatorname{Hom}_{\mathfrak{g}}(M, N) \to
  \operatorname{Hom}_{\mathcal{U}(\mathfrak{g})}(M, N)\), it suffices to note
  that a \(K\)-linear map between representations \(M\) and \(N\) is an
  intertwiner if, and only if it is a homomorphism of
  \(\mathcal{U}(\mathfrak{g})\)-modules. Our functor thus takes an intertwiner
  \(M \to N\) to itself. It should then be clear that our functor
  \(\mathbf{Rep}(\mathfrak{g}) \to \mathfrak{g}\text{-}\mathbf{Mod}\) is
  invertible.
\end{proof}

The language of representation is thus equivalent to that of
\(\mathcal{U}(\mathfrak{g})\)-modules, which we call
\emph{\(\mathfrak{g}\)-modules}. Correspondingly, we refer to the category
\(\mathcal{U}(\mathfrak{g})\text{-}\mathbf{Mod}\) as
\(\mathfrak{g}\text{-}\mathbf{Mod}\). The terms
\emph{\(\mathfrak{g}\)-submodule} and \emph{\(\mathfrak{g}\)-homomorphism}
should also be self-explanatory. To avoid any confusion, we will, for the most
part, exclusively use the language of \(\mathfrak{g}\)-modules. It should be
noted, however, that both points of view are profitable.

For starters, the notation for \(\mathfrak{g}\)-modules is much cleaner than
that of representations: it is much easier to write ``\(X \cdot m\)'' than
``\((\rho(X))(m)\)'' or even ``\(X\!\restriction_M(m)\)''. By using the
language of \(\mathfrak{g}\)-modules we can also rely on the general theory of
modules over associative algebras -- which we assume the reader is already
familiarized with. On the other hand, it is usually easier to express geometric
considerations in terms of the language representations, particularly in group
representation theory.

Often times it is easier to define a \(\mathfrak{g}\)-module \(M\) in terms of
the corresponding map \(\mathfrak{g} \to \mathfrak{gl}(M)\) -- this is
technique we will use throughout the text. In general, the equivalence between
both languages makes it clear that to understand the action of
\(\mathcal{U}(\mathfrak{g})\) on \(M\) it suffices to understand the action of
\(\mathfrak{g} \subset \mathcal{U}(\mathfrak{g})\). For instance, for defining
a \(\mathfrak{g}\)-module \(M\) it suffices to define the action of each \(X
\in \mathfrak{g}\) and verify this action respects the commutator relations of
\(\mathfrak{g}\) -- indeed, \(\mathfrak{g}\) generates
\(\mathcal{U}(\mathfrak{g})\) as an algebra, and the only relations between
elements of \(\mathfrak{g}\) are the ones derived from the commutator
relations.

\begin{example}\label{ex:sl2-polynomial-rep}
  The space \(K[x, y]\) is a \(\mathfrak{sl}_2(K)\)-module with
  \begin{align*}
    e \cdot p & = x \frac{\mathrm{d}}{\mathrm{d}y} p &
    f \cdot p & = y \frac{\mathrm{d}}{\mathrm{d}x} p &
    h \cdot p & =
    \left(
      x \frac{\mathrm{d}}{\mathrm{d}x} - y \frac{\mathrm{d}}{\mathrm{d}y}
    \right) p
  \end{align*}
\end{example}

\begin{example}
  Given a Lie algebra \(\mathfrak{g}\) and \(\mathfrak{g}\)-modules \(M\) and
  \(N\), the space \(\operatorname{Hom}(M, N)\) of \(K\)-linear maps \(M \to
  N\) is a \(\mathfrak{g}\)-module where \((X \cdot f)(m) = X \cdot f(m) - f(X
  \cdot m)\) for all \(X \in \mathfrak{g}\) and \(f \in \operatorname{Hom}(M,
  N)\). In particular, if we take \(N = K\) the trivial
  \(\mathfrak{g}\)-module, we can view \(M^*\) -- the dual of \(M\) in the
  category of \(K\)-vector spaces -- as a \(\mathfrak{g}\)-module where \((X
  \cdot f)(m) = - f(X \cdot m)\) for all \(f : M \to K\).
\end{example}

The fundamental problem of representation theory is a simple one: classifying
all representations of a given Lie algebra up to isomorphism. However,
understanding the relationship between representations is also of huge
importance. In other words, to understand the whole of
\(\mathfrak{g}\text{-}\mathbf{Mod}\) we need to study the collective behavior
of representations -- as opposed to individual examples. For instance, we may
consider \(\mathfrak{g}\)-submodules, quotients and tensor products.

\begin{example}\label{ex:sl2-polynomial-subrep}
  Let \(K[x, y]\) be the \(\mathfrak{sl}_2(K)\)-module as in
  Example~\ref{ex:sl2-polynomial-rep}. Since \(e\), \(f\) and \(h\) all
  preserve the degree of monomials, the space \(K[x, y]^{(d)} = \bigoplus_{k +
  \ell = d} K x^k y^\ell\) of homogeneous polynomials of degree \(d\) is a
  finite-dimensional \(\mathfrak{sl}_2(K)\)-submodule of \(K[x, y]\).
\end{example}

\begin{example}
  Given a Lie algebra \(\mathfrak{g}\), a \(\mathfrak{g}\)-module \(M\) and \(m
  \in M\), the subspace \(\mathcal{U}(\mathfrak{g}) \cdot m = \{ u \cdot m : u
  \in \mathcal{U}(\mathfrak{g}) \}\) is a \(\mathfrak{g}\)-submodule of \(M\),
  which we call \emph{the submodule generated by \(m\)}.
\end{example}

\begin{example}\index{\(\mathfrak{g}\)-module!tensor product}
  Given a Lie algebra \(\mathfrak{g}\) and \(\mathfrak{g}\)-modules \(M\) and
  \(N\), the space \(M \otimes N = M \otimes_K N\) is a \(\mathfrak{g}\)-module
  where \(X \cdot (m \otimes n) = X \cdot m \otimes n + m \otimes X \cdot n\).
  The exterior and symmetric products \(M \wedge N\) and \(M \odot N\) are both
  quotients of \(M \otimes N\) by \(\mathfrak{g}\)-submodules. In particular,
  the exterior and symmetric powers \(\wedge^r M\) and \(\operatorname{Sym}^r
  M\) are \(\mathfrak{g}\)-modules.
\end{example}

\begin{note}
  We would like to stress that the monoidal structure of
  \(\mathfrak{g}\text{-}\mathbf{Mod}\) we've just described is \emph{not} given
  by the usual tensor product of modules. In other words, \(M \otimes N\) is
  not the same as \(M \otimes_{\mathcal{U}(\mathfrak{g})} N\).
\end{note}

It is also interesting to consider the relationship between representations of
separate algebras. In particular, we may define\dots

\begin{example}\index{\(\mathfrak{g}\)-module!restriction}
  Let \(\mathfrak{g}\) be a Lie algebra and \(\mathfrak{h}\) be a subalgebra.
  Given a \(\mathfrak{g}\)-module \(M\), denote by
  \(\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}} M = M\) the
  \(\mathfrak{h}\)-module where the action of \(\mathfrak{h}\) is given by
  restricting the map \(\mathfrak{g} \to \mathfrak{gl}(M)\) to
  \(\mathfrak{h}\). Any homomorphism of \(\mathfrak{g}\)-modules \(M \to N\) is
  also a homomorphism of \(\mathfrak{h}\)-modules and this construction is
  clearly functorial.
  \[
    \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}} :
    \mathfrak{g}\text{-}\mathbf{Mod} \to
    \mathfrak{h}\text{-}\mathbf{Mod}
  \]
\end{example}

\begin{example}
  Given a Lie algebra \(\mathfrak{g}\), the adjoint \(\mathfrak{g}\)-module is
  a submodule of the restriction of the adjoint
  \(\mathcal{U}(\mathfrak{g})\)-module -- where we consider
  \(\mathcal{U}(\mathfrak{g})\) a Lie algebra as in
  Example~\ref{ex:inclusion-alg-in-lie-alg}, not as an associative algebra --
  to \(\mathfrak{g}\).
\end{example}

Surprisingly, this functor has a right adjoint.

\begin{example}\index{\(\mathfrak{g}\)-module!induction}
  Let \(\mathfrak{g}\) be a Lie algebra and \(\mathfrak{h}\) be a subalgebra.
  Given a \(\mathfrak{h}\)-module \(M\), let
  \(\operatorname{Ind}_{\mathfrak{h}}^{\mathfrak{g}} M =
  \mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{h})} M\) -- where
  the right \(\mathfrak{h}\)-module structure of \(\mathcal{U}(\mathfrak{g})\)
  is given by right multiplication. Any \(\mathfrak{h}\)-homomorphism \(f : M
  \to N\) induces a \(\mathfrak{g}\)-homomorphism
  \(\operatorname{Ind}_{\mathfrak{h}}^{\mathfrak{g}} f = \operatorname{id}
  \otimes f : \operatorname{Ind}_{\mathfrak{h}}^{\mathfrak{g}} M \to
  \operatorname{Ind}_{\mathfrak{h}}^{\mathfrak{g}} N\) and this construction is
  clearly functorial.
  \[
    \operatorname{Ind}_{\mathfrak{h}}^{\mathfrak{g}} :
    \mathfrak{h}\text{-}\mathbf{Mod} \to \mathfrak{g}\text{-}\mathbf{Mod}
  \]
\end{example}

\begin{proposition}\label{thm:frobenius-reciprocity}
  Given a Lie algebra \(\mathfrak{g}\), a subalgebra \(\mathfrak{h} \subset
  \mathfrak{g}\), a \(\mathfrak{h}\)-module \(M\) and a \(\mathfrak{g}\)-module
  \(N\), the map
  \[
    \arraycolsep=1.4pt
    \begin{array}[t]{rl}
    \alpha :
    \operatorname{Hom}_{\mathfrak{g}}(
      \operatorname{Ind}_{\mathfrak{h}}^{\mathfrak{g}} M,
      N
    ) & \to
    \operatorname{Hom}_{\mathfrak{h}}(
      M,
      \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}} N
    ) \\
    f & \mapsto
    \begin{array}[t]{rl}
    \alpha(f) : M & \to \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}} N \\
                m & \mapsto f(1 \otimes m)
    \end{array}
    \end{array}
  \]
  is a \(K\)-linear isomorphism. In other words, there is an adjunction
  \(\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}} \vdash
  \operatorname{Ind}_{\mathfrak{h}}^{\mathfrak{g}}\).
\end{proposition}

\begin{proof}
  It suffices to note that the map
  \[
    \arraycolsep=1.4pt
    \begin{array}[t]{rl}
    \beta :
    \operatorname{Hom}_{\mathfrak{h}}(
      M,
      \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}} N
    ) & \to
    \operatorname{Hom}_{\mathfrak{g}}(
      \operatorname{Ind}_{\mathfrak{h}}^{\mathfrak{g}} M,
      N
    ) \\
    f & \mapsto
    \begin{array}[t]{rl}
    \beta(f) : \operatorname{Ind}_{\mathfrak{h}}^{\mathfrak{g}} M & \to N \\
               u \otimes m & \mapsto u \cdot f(m)
    \end{array}
    \end{array}
  \]
  is the inverse of \(\alpha\).
\end{proof}

This last proposition is known as \emph{Frobenius reciprocity}, and was first
proved by Frobenius himself in the context of finite groups. Another
interesting construction is\dots

\begin{example}\label{ex:tensor-prod-separate-algs}\index{\(\mathfrak{g}\)-module!tensor product}
  Given two \(K\)-algebras \(A\) and \(B\), an \(A\)-module \(M\) and a
  \(B\)-module \(N\), \(M \otimes N = M \otimes_K N\) has the natural structure
  of an \(A \otimes_K B\)-module. In light of
  Example~\ref{ex:univ-enveloping-of-sum-is-tensor}, this implies that given
  Lie algebras \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\), a
  \(\mathfrak{g}_1\)-module \(M_1\) and a \(\mathfrak{g}_2\)-module \(M_2\),
  the space \(M_1 \otimes M_2\) has the natural structure of a \(\mathfrak{g}_1
  \oplus \mathfrak{g}_2\)-module, where the action of \(\mathfrak{g}_1 \oplus
  \mathfrak{g}_2\) is given by
  \[
    (X_1 + X_2) \cdot (m \otimes n)
    = X_1 \cdot m \otimes n + m \otimes X_2 \cdot n
  \]
\end{example}

Example~\ref{ex:tensor-prod-separate-algs} thus provides a way to describe
representations of \(\mathfrak{g}_1 \oplus \mathfrak{g}_2\) in terms of the
representations of \(\mathfrak{g}_1\) and \(\mathfrak{g}_2\). We will soon see
that in many cases \emph{all} (simple) \(\mathfrak{g}_1 \oplus
\mathfrak{g}_2\)-modules can be constructed in such a manner. This concludes
our initial remarks on \(\mathfrak{g}\)-modules. In the following chapters we
will explore the fundamental problem of representation theory: that of
classifying all representations of a given algebra up to isomorphism.
a given algebra up to isomorphism.
