\chapter{Semisimplicity \& Complete Reducibility}

\label{start-47}

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 can be
used to derive geometric information about groups, but the question remains:
how do we go about classifying the representations of a given Lie algebra? This
question has sparked an entire field of research, and we cannot hope to provide
a comprehensive answer in the \pagedifference{start-47}{end-47} pages we have
left. Nevertheless, we can work on particular cases.

For instance, one can readily check that a \(K^n\)-module \(M\) -- here \(K^n\)
denotes the \(n\)-dimensional Abelian Lie algebra -- is nothing more than a
choice of \(n\) commuting operators \(M \to M\) -- corresponding to the action
of the canonical basis elements \(e_1, \ldots, e_n \in K^n\). In particular, a
\(1\)-dimensional \(K^n\)-module is just a choice of \(n\) scalars \(\lambda_1,
\ldots, \lambda_n\). Different choices of scalars yield non-isomorphic modules,
so that the \(1\)-dimensional \(K^n\)-modules are parameterized by points in
\(K^n\).

This goes to show that classifying the representations of Abelian algebras is
not that interesting of a problem. Instead, we focus on a less trivial, yet
reasonably well behaved case: the finite-dimensional modules of a
finite-dimensional semisimple Lie algebra \(\mathfrak{g}\) over an
algebraically closed field \(K\) of characteristic \(0\). But why are the
modules of a semisimple Lie algebras simpler -- or perhaps \emph{semisimpler}
-- to understand than those of any old Lie algebra? We will get 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 pieces. This leads us to the following definitions.

\begin{definition}\index{\(\mathfrak{g}\)-module!indecomposable module}
  A \(\mathfrak{g}\)-module is called \emph{indecomposable} if it is
  not isomorphic to the direct sum of two nonzero \(\mathfrak{g}\)-modules.
\end{definition}

\begin{definition}\index{\(\mathfrak{g}\)-module!simple module}\index{simple!\(\mathfrak{g}\)-module}
  A \(\mathfrak{g}\)-module is called \emph{simple} if it has no nonzero proper
  \(\mathfrak{g}\)-modules.
\end{definition}

\begin{example}
  The trivial \(\mathfrak{g}\)-module \(K\) is an example of a simple
  \(\mathfrak{g}\)-module. In fact, every \(1\)-dimensional
  \(\mathfrak{g}\)-module \(M\) is simple: \(M\) has no nonzero proper
  \(K\)-subspaces, let alone \(\mathfrak{g}\)-submodules.
\end{example}

\begin{example}\label{ex:all-simple-reps-are-tensor-prod}
  Given a finite-dimensional simple \(\mathfrak{g}_1\)-module \(M_1\) and a
  finite-dimensional simple \(\mathfrak{g}_2\)-module \(M_2\), the tensor
  product \(M_1 \otimes M_2\) is a simple \(\mathfrak{g}_1 \oplus
  \mathfrak{g}_2\)-module. All finite-dimensional simple \(\mathfrak{g}_1
  \oplus \mathfrak{g}_2\)-modules have the form \(M_1 \otimes M_2\) for unique
  (up to isomorphism) \(M_1\) and \(M_2\). In light of
  Example~\ref{ex:univ-enveloping-of-sum-is-tensor}, this is a particular case
  of the fact that, given \(K\)-algebras \(A\) and \(B\), all
  finite-dimensional simple \(A \otimes_K B\)-modules are given tensor products
  of simple \(A\)-modules with simple \(B\)-modules -- see
  \cite[ch.~3]{etingof}.
\end{example}

The general strategy for classifying finite-dimensional modules over an algebra
is to classify the indecomposable modules. This is because\dots

\begin{theorem}[Krull-Schmidt]\label{thm:krull-schmidt}
  Let \(\mathfrak{g}\) be a Lie algebra.
  Then every finite-dimensional \(\mathfrak{g}\)-module can be uniquely --
  up to isomorphisms and reordering of the summands -- decomposed into a direct
  sum of indecomposable \(\mathfrak{g}\)-modules.
\end{theorem}

Hence finding the indecomposable \(\mathfrak{g}\)-modules suffices to find
\emph{all} finite-dimensional \(\mathfrak{g}\)-modules: they are the direct sum
of indecomposable \(\mathfrak{g}\)-modules. The existence of the decomposition
should be clear from the definitions. Indeed, if \(M\) is a finite-dimensional
\(\mathfrak{g}\)-modules a simple argument via induction in \(\dim M\) suffices
to prove the existence: if \(M\) is indecomposable then there is nothing to
prove, and if \(M\) is not indecomposable then \(M = N \oplus L\) for some
nonzero submodules \(N, L \subsetneq M\), so that their dimensions are both
strictly smaller than \(\dim M\) and the existence follows from the induction
hypothesis. For a proof of uniqueness please refer to \cite{etingof}.

Finding the indecomposable modules of an arbitrary Lie algebra, however, turns
out to be a bit of a circular problem: the indecomposable
\(\mathfrak{g}\)-modules 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 simple
\(\mathfrak{g}\)-module 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 \(M = K^2\) endowed with the action
  \begin{align*}
    x \cdot e_1 & = e_1 & x \cdot e_2 = e_1 + e_2
  \end{align*}
  of the Lie algebra \(K[x]\) is a \(K[x]\)-module. Notice \(M\) has a single
  nonzero proper submodule, which is spanned by the vector \(e_1\). This is
  because if \((a + b) e_1 + b e_2 = x \cdot (a e_1 + b e_2) = \lambda \cdot (a
  e_1 + b e_2)\) for some \(\lambda \in K\) then \(\lambda = 1\) and \(b = 0\).
  Hence \(M\) is indecomposable -- it cannot be broken into a direct sum of
  \(1\)-dimensional submodules -- but it is evidently not simple.
\end{example}

This counterexample poses an interesting question: are there conditions one can
impose on an algebra \(\mathfrak{g}\) under which every indecomposable
\(\mathfrak{g}\)-module is simple? This is what is known in representation
theory as \emph{complete reducibility}.

\begin{definition}\index{\(\mathfrak{g}\)-module!completely reducible module}
  A \(\mathfrak{g}\)-module \(M\) is called \emph{completely reducible} if
  every \(\mathfrak{g}\)-submodule of \(M\) has a \(\mathfrak{g}\)-invariant
  complement -- i.e. given \(N \subset M\), there is a submodule \(L \subset
  M\) such that \(M = N \oplus L\).
\end{definition}

\begin{definition}\index{\(\mathfrak{g}\)-module!semisimple module}\index{semisimple!\(\mathfrak{g}\)-module}
  A \(\mathfrak{g}\)-module \(M\) is called \emph{semisimple} if it is the
  direct sum of simple \(\mathfrak{g}\)-modules.
\end{definition}

In case the relationship between complete reducibility, semisimplicity of
\(\mathfrak{g}\)-modules and the simplicity of indecomposable modules is
unclear, the following results should clear things up.

\begin{proposition}
  The following conditions are equivalent.
  \begin{enumerate}
    \item Every submodule of a finite-dimensional \(\mathfrak{g}\)-module is
      completely reducible.

    \item Every exact sequence of finite-dimensional \(\mathfrak{g}\)-modules
      splits.

    \item Every indecomposable finite-dimensional \(\mathfrak{g}\)-module is
      simple.

    \item Every finite-dimensional \(\mathfrak{g}\)-module is semisimple.
  \end{enumerate}
\end{proposition}

\begin{proof}
  We begin by \(\textbf{(i)} \implies \textbf{(ii)}\). Let
  \begin{center}
    \begin{tikzcd}
      0 \rar    &
      N \rar{f} &
      M \rar{g} &
      L \rar    &
      0
    \end{tikzcd}
  \end{center}
  be an exact sequence of \(\mathfrak{g}\)-modules. We can suppose without loss
  of generality that \(N \subset M\) is a submodule and \(f\) is its inclusion
  in \(M\), for if this is not the case there is an isomorphism of sequences
  \begin{center}
    \begin{tikzcd}
         0 \rar                                    &
         N \rar{f}       \dar[swap]{f}             &
         M \rar{g}       \dar[Rightarrow, no head] &
         L \rar          \dar[Rightarrow, no head] &
         0                                         \\
         0 \rar                                    &
      f(N) \rar                                    &
         M \rar[swap]{g}                           &
         L \rar                                    &
         0
    \end{tikzcd}
  \end{center}

  It then follows from \textbf{(i)} that there exists a
  \(\mathfrak{g}\)-submodule \(L' \subset M\) such that \(M = N \oplus L'\).
  Finally, the projection \(s : M \to N\) is \(\mathfrak{g}\)-homomorphism
  satisfying
  \begin{center}
    \begin{tikzcd}
      0 \rar                          &
      N \rar{f}                       &
      M \rar{g} \lar[bend left=30]{s} &
      L \rar                          &
      0
    \end{tikzcd}
  \end{center}

  Next is \(\textbf{(ii)} \implies \textbf{(iii)}\). If \(M\) is an
  indecomposable \(\mathfrak{g}\)-module and \(N \subset M\) is a submodule, we
  have an exact sequence
  \begin{center}
    \begin{tikzcd}
                 0 \rar &
                 N \rar &
                 M \rar &
      \mfrac{M}{N} \rar &
                 0
    \end{tikzcd}
  \end{center}
  of \(\mathfrak{g}\)-modules.

  Since our sequence splits, we must have \(M \cong N \oplus \mfrac{M}{N}\).
  But \(M\) is indecomposable, so that either \(M = N\) or \(M \cong
  \mfrac{M}{N}\), in which case \(N = 0\). Since this holds for all \(N \subset
  M\), \(M\) is simple. For \(\textbf{(iii)} \implies \textbf{(iv)}\) it
  suffices to apply Theorem~\ref{thm:krull-schmidt}.

  Finally, for \(\textbf{(iv)} \implies \textbf{(i)}\), if we assume
  \(\textbf{(iv)}\) and let \(M\) be a \(\mathfrak{g}\)-module with
  decomposition into simple submodules
  \[
    M = \bigoplus_i M_i
  \]
  and \(N \subset M\) is a submodule. Take some maximal set of indexes \(\{i_1,
  \ldots, i_r\}\) so that \(\left( \bigoplus_k M_{i_k} \right) \cap M = 0\) and
  let \(L = \bigoplus_k M_{i_k}\). We want to establish \(M = N \oplus L\).

  Suppose without any loss in generality that \(i_k = k\) for all \(k\) and let
  \(j > r\). By the maximality of our set of indexes, there is some nonzero \(n
  \in (M_j \oplus L) \cap N\). Say \(n = m_j + m_1 + \cdots + m_r\) with each
  \(m_i \in M_i\). Then \(m_j = n - m_1 - \cdots - m_r \in M_j \cap (N \oplus
  L)\) is nonzero. Indeed, if this is not the case we find \(0 \ne n = m_1 +
  \cdots + m_r \in \left( \bigoplus_{i = 1}^r M_i \right) \cap N\), a
  contradiction. This implies \(M_j \cap (N \oplus L)\) is a nonzero submodule
  of \(M_j\). Since \(M_j\) is simple, \(M_j = M_j \cap (N \oplus L)\) and
  therefore \(M_j \subset N \oplus L\). Given the arbitrary choice of \(j\), it
  then follows \(M = N \oplus L\).
\end{proof}

While we are primarily interested in indecomposable \(\mathfrak{g}\)-modules --
which is usually a strictly larger class of representations than that of simple
\(\mathfrak{g}\)-modules -- it is important to note that simple
\(\mathfrak{g}\)-modules are generally much easier to find. The relationship
between simple \(\mathfrak{g}\)-modules is also well understood. This is
because of the following result, known as \emph{Schur's Lemma}.

\begin{lemma}[Schur]
  Let \(M\) and \(N\) be simple \(\mathfrak{g}\)-modules and \(f : M \to N\) be
  a \(\mathfrak{g}\)-homomorphism. Then \(f\) is either \(0\) or an
  isomorphism. Furthermore, if \(M = N\) is finite-dimensional then \(f\) is a
  scalar operator.
\end{lemma}

\begin{proof}
  For the first statement, it suffices to notice that \(\ker f\) and
  \(\operatorname{im} f\) are both submodules. In particular, either \(\ker f =
  0\) and \(\operatorname{im} f = N\) or \(\ker f = M\) and \(\operatorname{im}
  f = 0\). Now suppose \(M = N\) is finite-dimensional. Let \(\lambda \in K\)
  be an eigenvalue of \(f\) -- which exists because \(K\) is algebraically
  closed -- and \(M_\lambda\) be its corresponding eigenspace. Given \(m \in
  M_\lambda\), \(f(X \cdot m) = X \cdot f(m) = \lambda X \cdot m\). In other
  words, \(M_\lambda\) is a \(\mathfrak{g}\)-submodule. It then follows
  \(M_\lambda = M\), given that \(M_\lambda \ne 0\).
\end{proof}

We are now ready to answer our first question: the special thing about
semisimple algebras is that the relationship between their indecomposable
modules and their simple modules is much clearer. Namely\dots

\begin{proposition}
  Given a finite-dimensional Lie algebra \(\mathfrak{g}\) over \(K\),
  \(\mathfrak{g}\) is semisimple if, and only if every finite-dimensional
  \(\mathfrak{g}\)-module is completely reducible.
\end{proposition}

The proof of the fact that a finite-dimensional Lie algebra \(\mathfrak{g}\)
whose finite-dimensional modules are completely reducible is semisimple is
actually pretty simple. Namely, it suffices to note that the adjoint
\(\mathfrak{g}\)-module is the direct sum of simple submodules, 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 proceeding to the proof of complete reducibility, however, we would like
to introduce some basic tools which will come in handy later on, known as\dots

\section{Invariant Bilinear Forms}

\begin{definition}\index{invariant bilinear form}
  A symmetric bilinear form \(B : \mathfrak{g} \times \mathfrak{g} \to K\) is
  called \emph{\(\mathfrak{g}\)-invariant} if the operator
  \(\operatorname{ad}(X) : \mathfrak{g} \to \mathfrak{g}\) is antisymmetric
  with respect to \(B\) for all \(X \in \mathfrak{g}\).
  \[
    B(\operatorname{ad}(X) Y, Z) + B(Y, \operatorname{ad}(X) Z) = 0
  \]
\end{definition}

\begin{note}
  The etymology of the term \emph{invariant form} comes from group
  representation theory. Namely, given a linear action of a group \(G\) on a
  vector space \(V\) equipped with a bilinear form \(B\), \(B\) is called
  \(G\)-invariant if all \(g \in G\) act via \(B\)-orthogonal operators. The
  condition of \(\mathfrak{g}\)-invariance can thus be though-of as an
  \emph{infinitesimal approximation} of the notion of a \(G\)-invariant form.
  Indeed \(\operatorname{Lie}(\operatorname{O}(B))\) is precisely the Lie
  subalgebra of \(\mathfrak{gl}(V)\) consisting of antisymmetric operators \(V
  \to V\).
\end{note}

An interesting example of an invariant bilinear form is the so called
\emph{Killing form}.

\begin{definition}\index{invariant bilinear form!Killing form}\index{Killing form}
  Given a finite-dimensional Lie algebra \(\mathfrak{g}\), the symmetric
  bilinear form
  \begin{align*}
    \kappa : \mathfrak{g} \times \mathfrak{g} & \to K \\
    (X, Y) &
    \mapsto \operatorname{Tr}(\operatorname{ad}(X) \operatorname{ad}(Y))
  \end{align*}
  is called \emph{the Killing form of \(\mathfrak{g}\)}.
\end{definition}

The fact that the Killing form is an invariant form follows directly from the
identity \(\operatorname{Tr}([X, Y] Z) = \operatorname{Tr}(X [Y, Z])\), \(X, Y,
Z \in \mathfrak{gl}_n(K)\). In fact this same identity show\dots

\begin{lemma}\index{invariant bilinear form!bilinear form of a \(\mathfrak{g}\)-module}
  Given a finite-dimensional \(\mathfrak{g}\)-module \(M\), the symmetric
  bilinear form
  \begin{align*}
    \kappa_M : \mathfrak{g} \times \mathfrak{g} & \to K \\
    (X, Y) & \mapsto \operatorname{Tr}(X\!\restriction_M \, Y\!\restriction_M)
  \end{align*}
  is \(\mathfrak{g}\)-invariant.
\end{lemma}

The reason why we are discussing invariant bilinear forms is the following
characterization of finite-dimensional semisimple Lie algebras, known as
\emph{Cartan's criterion for semisimplicity}.

\begin{proposition}
  Let \(\mathfrak{g}\) be a Lie algebra. The following conditions are
  equivalent.
  \begin{enumerate}
    \item \(\mathfrak{g}\) is semisimple.
    \item For each non-trivial finite-dimensional \(\mathfrak{g}\)-module
      \(M\), the \(\mathfrak{g}\)-invariant bilinear form
      \begin{align*}
        \kappa_M : \mathfrak{g} \times \mathfrak{g}                          &
        \to K                                                                \\
        (X, Y)                                                               &
        \mapsto \operatorname{Tr}(X\!\restriction_M \, Y\!\restriction_M)
      \end{align*}
      is non-degenerate\footnote{A symmetric bilinear form $B : \mathfrak{g}
      \times \mathfrak{g} \to K$ is called non-degenerate if $B(X, Y) = 0$ for
      all $Y \in \mathfrak{g}$ implies $X = 0$.}.
    \item The Killing form \(\kappa\) is non-degenerate.
  \end{enumerate}
\end{proposition}

This proof is somewhat technical, but the idea behind it is simple. First, for
\strong{(i)} \(\implies\) \strong{(ii)} we show that \(\mathfrak{a} = \{ X \in
\mathfrak{g} : \kappa_M(X, Y) = 0 \, \forall Y \in \mathfrak{g}\}\) is a
solvable ideal of \(\mathfrak{g}\). Hence \(\mathfrak{a} = 0\). For
\strong{(ii)} \(\implies\) \strong{(iii)} it suffices to take \(M =
\mathfrak{g}\) the adjoint \(\mathfrak{g}\)-module. Finally, for \strong{(iii)}
\(\implies\) \strong{(i)} we note that the orthogonal complement of any
\(\mathfrak{a} \normal \mathfrak{g}\) with respect to the Killing form
\(\kappa\) is an ideal \(\mathfrak{b}\) of \(\mathfrak{g}\) with \(\mathfrak{g}
= \mathfrak{a} \oplus \mathfrak{b}\). Furthermore, the Killing form of
\(\mathfrak{a}\) is the restriction \(\kappa\!\restriction_{\mathfrak{a}}\) of
the Killing form of \(\mathfrak{g}\) to \(\mathfrak{a} \times \mathfrak{a}\),
which is non-degenerate. It then follows from induction in \(\dim
\mathfrak{a}\) that \(\mathfrak{g}\) is the sum of simple ideals.

We refer the reader to \cite[ch. 5]{humphreys} for a complete proof. Without
further ado, we may proceed to our\dots

\section{Proof of Complete Reducibility}

Let \(\mathfrak{g}\) be a finite-dimensional Lie algebra over \(K\). We want to
establish that if \(\mathfrak{g}\) is semisimple then all finite-dimensional
\(\mathfrak{g}\)-modules are semisimple. Historically, this was first proved by
Herman Weyl for \(K = \mathbb{C}\), using his knowledge of smooth
representations of compact Lie groups. Namely, Weyl showed that any
finite-dimensional semisimple complex Lie algebra is (isomorphic to) the
complexification of the Lie algebra of a unique simply connected compact Lie
group, known as its \emph{compact form}. Hence the category of the
finite-dimensional modules of a given complex semisimple algebra is equivalent
to that of the finite-dimensional smooth representations of its compact form,
whose representations are known to be completely reducible because of Maschke's
Theorem -- see \cite[ch. 3]{serganova} for instance.

This proof, however, is heavily reliant on the geometric structure of
\(\mathbb{C}\). In other words, there is no hope for generalizing this for some
arbitrary \(K\). Fortunately for us, there is a much simpler, completely
algebraic proof of complete reducibility, which works for algebras over any
algebraically closed field of characteristic zero. The algebraic proof included
in here is mainly based on that of \cite[ch. 6]{kirillov}, and uses some basic
homological algebra. Admittedly, much of the homological algebra used in here
could be concealed from the reader, which would make the exposition more
accessible -- see \cite{humphreys} for instance.

However, this does not change the fact the arguments used in this proof are
essentially homological in nature. Hence we consider it more productive to use
the full force of the language of homological algebra, instead of burring the
reader in a pile of unmotivated, yet entirely elementary arguments.
Furthermore, the homological algebra used in here is actually \emph{very
basic}. In fact, all we need to know is\dots

\begin{theorem}\label{thm:ext-exacts-seqs}\index{\(\operatorname{Ext}\) functors}
  There is a sequence of bifunctors \(\operatorname{Ext}^i :
  \mathfrak{g}\text{-}\mathbf{Mod}^{\operatorname{op}} \times
  \mathfrak{g}\text{-}\mathbf{Mod} \to K\text{-}\mathbf{Vect}\), \(i \ge 0\)
  such that, given a \(\mathfrak{g}\)-module \(L'\), every exact sequence of
  \(\mathfrak{g}\)-modules
  \begin{center}
    \begin{tikzcd}
      0 \rar & N \rar{f} & M \rar{g} & L \rar & 0
    \end{tikzcd}
  \end{center}
  induces long exact sequences
  \begin{center}
    \begin{tikzcd}
      0 \rar                                                    &
      \operatorname{Hom}_{\mathfrak{g}}(L', N)
      \rar{f \circ -}\ar[draw=none]{d}[name=X, anchor=center]{} &
      \operatorname{Hom}_{\mathfrak{g}}(L', M) \rar{g \circ -}  &
      \operatorname{Hom}_{\mathfrak{g}}(L', L)
      \ar[rounded corners,
          to path={ -- ([xshift=2ex]\tikztostart.east)
                    |- (X.center) \tikztonodes
                    -| ([xshift=-2ex]\tikztotarget.west)
                    -- (\tikztotarget)}]{dll}[at end]{} \\      &
      \operatorname{Ext}^1(L', N)
      \rar\ar[draw=none]{d}[name=Y, anchor=center]{}            &
      \operatorname{Ext}^1(L', M) \rar                          &
      \operatorname{Ext}^1(L', L)
      \ar[rounded corners,
          to path={ -- ([xshift=2ex]\tikztostart.east)
                    |- (Y.center) \tikztonodes
                    -| ([xshift=-2ex]\tikztotarget.west)
                    -- (\tikztotarget)}]{dll}[at end]{} \\      &
      \operatorname{Ext}^2(L', N) \rar                          &
      \operatorname{Ext}^2(L', M) \rar                          &
      \operatorname{Ext}^2(L', L) \rar[dashed]                  &
      \cdots
    \end{tikzcd}
  \end{center}
  and
  \begin{center}
    \begin{tikzcd}
      0 \rar                                                    &
      \operatorname{Hom}_{\mathfrak{g}}(L, L')
      \rar{- \circ g}\ar[draw=none]{d}[name=X, anchor=center]{} &
      \operatorname{Hom}_{\mathfrak{g}}(M, L') \rar{- \circ f}  &
      \operatorname{Hom}_{\mathfrak{g}}(N, L')
      \ar[rounded corners,
          to path={ -- ([xshift=2ex]\tikztostart.east)
                    |- (X.center) \tikztonodes
                    -| ([xshift=-2ex]\tikztotarget.west)
                    -- (\tikztotarget)}]{dll}[at end]{} \\      &
      \operatorname{Ext}^1(L, L')
      \rar\ar[draw=none]{d}[name=Y, anchor=center]{}            &
      \operatorname{Ext}^1(M, L') \rar                          &
      \operatorname{Ext}^1(N, L')
      \ar[rounded corners,
          to path={ -- ([xshift=2ex]\tikztostart.east)
                    |- (Y.center) \tikztonodes
                    -| ([xshift=-2ex]\tikztotarget.west)
                    -- (\tikztotarget)}]{dll}[at end]{} \\      &
      \operatorname{Ext}^2(L, L') \rar                          &
      \operatorname{Ext}^2(M, L') \rar                          &
      \operatorname{Ext}^2(N, L') \rar[dashed]                  &
      \cdots
    \end{tikzcd}
  \end{center}
\end{theorem}

\begin{theorem}\label{thm:ext-1-classify-short-seqs}
  Given \(\mathfrak{g}\)-modules \(N\) and \(L\), there is a one-to-one
  correspondence between elements of \(\operatorname{Ext}^1(L, N)\) and
  isomorphism classes of short exact sequences
  \begin{center}
    \begin{tikzcd}
      0 \rar & N \rar & M \rar & L \rar & 0
    \end{tikzcd}
  \end{center}

  In particular, \(\operatorname{Ext}^1(L, N) = 0\) if, and only if every short
  exact sequence of \(\mathfrak{g}\)-modules with \(N\) and \(L\) in the
  extremes splits.
\end{theorem}

We should point out that, although we have not provided an explicit definition
of the bifunctors \(\operatorname{Ext}^i\), they are uniquely determined by
the conditions of Theorem~\ref{thm:ext-exacts-seqs} and some additional
minimality constraints. This is, of course, \emph{far} from a comprehensive
account of homological algebra. Nevertheless, this is all we need. We refer the
reader to \cite{harder} for a complete exposition, or to part II of
\cite{ribeiro} for a more modern account using derived categories.

We are particularly interested in the case where \(L' = K\) is the trivial
\(\mathfrak{g}\)-module. Namely, we may define\dots

\begin{definition}\index{Lie algebra!cohomology}\index{cohomology of Lie algebras}
  Given a Lie algebra \(\mathfrak{g}\) and a \(\mathfrak{g}\)-module \(M\), we
  refer to the Abelian group \(H^i(\mathfrak{g}, M) = \operatorname{Ext}^i(K,
  M)\) as \emph{the \(i\)-th Lie algebra cohomology group of \(\mathfrak{g}\)
  with coefficients in \(M\)}.
\end{definition}

\begin{definition}\index{cohomology of Lie algebras!invariants}
  Given a \(\mathfrak{g}\)-module \(M\), we call the vector space
  \(M^{\mathfrak{g}} = \{m \in M : X \cdot m = 0 \; \forall X \in
  \mathfrak{g}\}\) \emph{the space of invariants of \(M\)}. A simple
  calculations shows that a \(\mathfrak{g}\)-homomorphism \(f : M \to N\) takes
  invariants to invariants, so that \(f\) restricts to a map \(M^{\mathfrak{g}}
  \to N^{\mathfrak{g}}\). This construction thus yields a functor
  \(-^{\mathfrak{g}} : \mathfrak{g}\text{-}\mathbf{Mod} \to
  K\text{-}\mathbf{Vect}\).
\end{definition}

\begin{example}
  Let \(M\) be a \(\mathfrak{g}\)-module. Then \(M\) is a direct sum of copies
  of the trivial \(\mathfrak{g}\)-module if, and only if \(M =
  M^{\mathfrak{g}}\).
\end{example}

\begin{example}\label{ex:hom-invariants-are-g-homs}
  Let \(M\) and \(N\) be \(\mathfrak{g}\)-modules. Then \(\operatorname{Hom}(M,
  N)^{\mathfrak{g}} = \operatorname{Hom}_{\mathfrak{g}}(M, N)\). Indeed, given
  a \(K\)-linear map \(f : M \to N\) we find
  \[
    \begin{split}
      f \in \operatorname{Hom}(M, N)^{\mathfrak{g}}
      & \iff X \cdot f(m) - f(X \cdot m) = (X \cdot f)(m) = 0
      \; \forall X \in \mathfrak{g}, m \in M \\
      & \iff X \cdot f(m) = f(X \cdot m)
      \; \forall X \in \mathfrak{g}, m \in M \\
      & \iff f \in \operatorname{Hom}_{\mathfrak{g}}(M, N)
    \end{split}
  \]
\end{example}

The Lie algebra cohomology groups are very much related to invariants of
\(\mathfrak{g}\)-modules. Namely, constructing a \(\mathfrak{g}\)-homomorphism
\(f : K \to M\) is precisely the same as fixing an invariant of \(M\) --
corresponding to \(f(1)\), which must be an invariant for \(f\) to be a
\(\mathfrak{g}\)-homomorphism. Formally, this translates to the existence of a
canonical isomorphism of functors
\(\operatorname{Hom}_{\mathfrak{g}}(K, -) \isoto {-}^{\mathfrak{g}}\) given by
\begin{align*}
  \operatorname{Hom}_{\mathfrak{g}}(K, M) & \isoto  M^{\mathfrak{g}} \\
                                        f & \mapsto f(1)
\end{align*}

This implies\dots

\begin{corollary}
  Every short exact sequence of \(\mathfrak{g}\)-modules
  \begin{center}
    \begin{tikzcd}
      0 \rar & N \rar{f} & M \rar{g} & L \rar & 0
    \end{tikzcd}
  \end{center}
  induces a long exact sequence
  \begin{center}
    \begin{tikzcd}
      0 \rar                                                              &
      N^{\mathfrak{g}} \rar{f}
      \ar[draw=none]{d}[name=X, anchor=center]{}                          &
      M^{\mathfrak{g}} \rar{g}                                            &
      L^{\mathfrak{g}}
      \ar[rounded corners,
          to path={ -- ([xshift=2ex]\tikztostart.east)
                    |- (X.center) \tikztonodes
                    -| ([xshift=-2ex]\tikztotarget.west)
                    -- (\tikztotarget)}]{dll}[at end]{} \\                &
      H^1(\mathfrak{g}, N) \rar\ar[draw=none]{d}[name=Y, anchor=center]{} &
      H^1(\mathfrak{g}, M) \rar                                           &
      H^1(\mathfrak{g}, L)
      \ar[rounded corners,
          to path={ -- ([xshift=2ex]\tikztostart.east)
                    |- (Y.center) \tikztonodes
                    -| ([xshift=-2ex]\tikztotarget.west)
                    -- (\tikztotarget)}]{dll}[at end]{} \\                &
      H^2(\mathfrak{g}, N) \rar                                           &
      H^2(\mathfrak{g}, M) \rar                                           &
      H^2(\mathfrak{g}, L) \rar[dashed]                                   &
      \cdots
    \end{tikzcd}
  \end{center}
\end{corollary}

\begin{proof}
  We have an isomorphism of sequences
  \begin{center}
    \begin{tikzcd}
      0 \rar &
      \operatorname{Hom}_{\mathfrak{g}}(K, N) \rar{f \circ -} \dar &
      \operatorname{Hom}_{\mathfrak{g}}(K, M) \rar{g \circ -} \dar &
      \operatorname{Hom}_{\mathfrak{g}}(K, L) \rar \dar            &
      H^1(\mathfrak{g}, N) \rar[dashed]\dar[Rightarrow, no head]   & \cdots \\
      0 \rar                                                       &
      N^{\mathfrak{g}} \rar[swap]{f}                               &
      M^{\mathfrak{g}} \rar[swap]{g}                               &
      L^{\mathfrak{g}} \rar                                        &
      H^1(\mathfrak{g}, N) \rar[dashed]                            &
      \cdots
    \end{tikzcd}
  \end{center}

  By Theorem~\ref{thm:ext-exacts-seqs} the sequence on the top is exact. Hence
  so is the sequence on the bottom.
\end{proof}

This is all well and good, but what does any of this have to do with complete
reducibility? Well, in general cohomology theories really shine when one is
trying to control obstructions of some kind. In our case, the bifunctor
\(H^1(\mathfrak{g}, \operatorname{Hom}(-, -)) :
\mathfrak{g}\text{-}\mathbf{Mod}^{\operatorname{op}} \times
\mathfrak{g}\text{-}\mathbf{Mod} \to K\text{-}\mathbf{Vect}\) classifies
obstructions to complete reducibility. Explicitly\dots

\begin{theorem}
  There is a natural isomorphism \(\operatorname{Ext}^1 \isoto
  H^1(\mathfrak{g}, \operatorname{Hom}(-, -))\). In particular, given
  \(\mathfrak{g}\)-modules \(N\) and \(L\), there is a one-to-one
  correspondence between elements of \(H^1(\mathfrak{g}, \operatorname{Hom}(L,
  N))\) and isomorphism classes of short exact sequences
  \begin{center}
    \begin{tikzcd}
      0 \rar & N \rar & M \rar & L \rar & 0
    \end{tikzcd}
  \end{center}
\end{theorem}

This is essentially a consequence of Example~\ref{ex:hom-invariants-are-g-homs}
and Theorem~\ref{thm:ext-1-classify-short-seqs}, as well as the minimality
conditions that characterize \(\operatorname{Ext}^1\). For the readers already
familiar with homological algebra: the correspondence between
\(H^1(\mathfrak{g}, \operatorname{Hom}(L, N))\) and short exact sequences of
\(\mathfrak{g}\)-modules can be described in very concrete terms by considering
a canonical free resolution
\begin{center}
  \begin{tikzcd}
    \cdots                                                    \rar[dashed] &
    \mathcal{U}(\mathfrak{g}) \otimes (\wedge^2 \mathfrak{g}) \rar         &
    \mathcal{U}(\mathfrak{g}) \otimes \mathfrak{g}            \rar         &
    \mathcal{U}(\mathfrak{g})                                 \rar         &
    K                                                         \rar         &
    0
  \end{tikzcd}
\end{center}
of the trivial \(\mathfrak{g}\)-module \(K\), known as \emph{the
Chevalley-Eilenberg resolution}, which provides an explicit construction of the
cohomology groups -- see \cite[sec.~1.3C]{cohomologies-lie} or
\cite[sec.~24]{symplectic-physics} for further details.

We will use the previous result implicitly in our proof, but we will not prove
it in its full force. Namely, we will show that if \(\mathfrak{g}\) is
semisimple then \(H^1(\mathfrak{g}, M) = 0\) for all finite-dimensional \(M\),
and that the fact that \(H^1(\mathfrak{g}, \operatorname{Hom}(L, N)) = 0\) for
all finite-dimensional \(N\) and \(L\) implies complete reducibility. To that
end, we introduce a distinguished element of \(\mathcal{U}(\mathfrak{g})\),
known as \emph{the Casimir element of a \(\mathfrak{g}\)-module}.

\begin{definition}\label{def:casimir-element}\index{Casimir element}
  Let \(\mathfrak{g}\) be a finite-dimensional semisimple Lie algebra and \(M\)
  be a finite-dimensional \(\mathfrak{g}\)-module. Let \(\{X_i\}_i\) be a basis
  for \(\mathfrak{g}\) and denote by \(\{X^i\}_i \subset \mathfrak{g}\) its
  dual basis with respect to the form \(\kappa_M\) -- i.e. the unique basis for
  \(\mathfrak{g}\) satisfying \(\kappa_M(X_i, X^j) = \delta_{i j}\), whose
  existence is a consequence of the non-degeneracy of \(\kappa_M\). We call
  \[
    \Omega_M = X_1 X^1 + \cdots + X_r X^r \in \mathcal{U}(\mathfrak{g})
  \]
  the \emph{Casimir element of \(M\)}.
\end{definition}

\begin{lemma}
  The definition of \(\Omega_M\) is independent of the choice of basis
  \(\{X_i\}_i\).
\end{lemma}

\begin{proof}
  Whatever basis \(\{X_i\}_i\) we choose, the image of \(\Omega_M\) under the
  canonical isomorphism \(\mathfrak{g} \otimes \mathfrak{g} \isoto \mathfrak{g}
  \otimes \mathfrak{g}^* \isoto \operatorname{End}(\mathfrak{g})\) is the
  identity operator\footnote{Here the isomorphism $\mathfrak{g} \otimes
  \mathfrak{g} \isoto \mathfrak{g} \otimes \mathfrak{g}^*$ is given by
  tensoring the identity $\mathfrak{g} \to \mathfrak{g}$ with the isomorphism
  $\mathfrak{g} \isoto \mathfrak{g}^*$ induced by the form $\kappa_M$.}.
\end{proof}

\begin{proposition}
  The Casimir element \(\Omega_M \in \mathcal{U}(\mathfrak{g})\) is central, so
  that \(\Omega_M\!\restriction_N : N \to N\) is a
  \(\mathfrak{g}\)-homomorphism for any \(\mathfrak{g}\)-module \(N\).
  Furthermore, \(\Omega_M\) acts on \(M\) as a nonzero scalar operator whenever
  \(M\) is a non-trivial finite-dimensional simple \(\mathfrak{g}\)-module.
\end{proposition}

\begin{proof}
  To see that \(\Omega_M\) is central fix a basis \(\{X_i\}_i\) for
  \(\mathfrak{g}\) and denote by \(\{X^i\}_i\) its dual basis with respect to
  \(\kappa_M\), as in
  Definition~\ref{def:casimir-element}. Given any \(X \in \mathfrak{g}\), it
  follows from definition of the \(X^i\) that \(X = \kappa_M(X, X^1) X_1 +
  \cdots + \kappa_M(X, X^r) X_r = \kappa_M(X, X_1) X^1 + \cdots + \kappa_M(X,
  X_r) X^r\).

  In particular, it follows from the invariance of \(\kappa_M\) that
  \[
    \begin{split}
      [X, \Omega_M]
      & = \sum_i [X, X_i X^i] \\
      & = \sum_i [X, X_i] X^i + \sum_i X_i [X, X^i] \\
      & = \sum_{i j} \kappa_M([X, X_i], X^j) X_j X^i
        + \sum_{i j} \kappa_M([X, X^i], X_j) X_i X^j \\
      & = \sum_{i j} (\kappa_M([X, X_j], X^i) + \kappa_M(X_j, [X, X^i]))
          X_i X^j \\
      & = 0
    \end{split},
  \]
  and \(\Omega_M\) is central. This implies that \(\Omega_M\!\restriction_N : N
  \to N\) is a \(\mathfrak{g}\)-homomorphism for all \(\mathfrak{g}\)-modules
  \(N\): its action commutes with the action of any other element of
  \(\mathfrak{g}\).

  In particular, it follows from Schur's Lemma that if \(M\) is
  finite-dimensional and simple then \(\Omega_M\) acts on \(M\) as a scalar
  operator. To see that this scalar is nonzero we compute
  \[
    \operatorname{Tr}(\Omega_M\!\restriction_M)
    = \operatorname{Tr}(X_1\!\restriction_M X^1\!\restriction_M)
    + \cdots
    + \operatorname{Tr}(X_r\!\restriction_M X^r\!\restriction_M)
    = \dim \mathfrak{g},
  \]
  so that \(\Omega_M\!\restriction_M = \lambda \operatorname{Id}\) for
  \(\lambda = \frac{\dim \mathfrak{g}}{\dim M} \ne 0\).
\end{proof}

As promised, the Casimir element of a \(\mathfrak{g}\)-module can be used to
establish\dots

\begin{proposition}\label{thm:first-cohomology-vanishes}
  Suppose \(\mathfrak{g}\) is semisimple and let \(M\) be a finite-dimensional
  \(\mathfrak{g}\)-module. Then \(H^1(\mathfrak{g}, M) = 0\).
\end{proposition}

\begin{proof}
  We begin by the case where \(M\) is simple. Due to
  Theorem~\ref{thm:ext-1-classify-short-seqs}, it suffices to show that any
  exact sequence of the form
  \begin{equation}\label{eq:exact-seq-h1-vanishes}
    \begin{tikzcd}
      0 \rar & M \rar{f} & N \rar{g} & K \rar & 0
    \end{tikzcd}
  \end{equation}
  splits.

   If \(M = K\) is the trivial \(\mathfrak{g}\)-module then the exactness of
  \begin{equation}\label{eq:trivial-extrems-exact-seq}
    \begin{tikzcd}
      0 \rar & K \rar{f} & N \rar{g} & K \rar & 0
    \end{tikzcd}
  \end{equation}
  implies \(N\) is 2-dimensional. Take any nonzero \(n \in N\) outside of the
  image of \(f\).

  % TODOOOOOOOOO: Fix this
  % TODO: U(g) w doesn't need to be irreducible a priori. In fact we will show
  % U(g) w = 0, so this whole argument is inconsistant
  % TODO: The way to fix this is to prove that rho(g) is both nilpotent --
  % because the action of every element of g is strictly upper triangular -- and
  % semisimple -- because it is a quotient of g, which is semisimple. We thus
  % have rho(g) = 0, so that W is trivial
  Since \(\dim N = 2\), the simple component \(\mathcal{U}(\mathfrak{g}) \cdot
  n\) of \(n\) in \(N\) is either \(K n\) or \(N\) itself. But this component
  cannot be \(N\), since the image of \(f\) is a \(1\)-dimensional
  \(\mathfrak{g}\)-module -- i.e. a proper nonzero submodule. Hence \(K n\) is
  invariant under the action of \(\mathfrak{g}\). In particular, \(X \cdot n =
  0\) for all \(X \in \mathfrak{g}\). Since \(n\) lies outside the image of
  \(f\), \(g(n) \ne 0\) -- which is to say, \(n \notin \ker g =
  \operatorname{im} f\). This implies the map \(K \to N\) that takes \(1\) to
  \(\sfrac{n}{g(n)}\) is a splitting of (\ref{eq:trivial-extrems-exact-seq}).

  Now suppose that \(M\) is non-trivial, so that \(\Omega_M\) acts on \(M\) as
  \(\lambda\) for some \(\lambda \ne 0\). Denote by \(N^\mu\) the generalized
  eigenspace of \(\Omega_M\!\restriction_N : N \to N\) associated with \(\mu
  \in K\). If we identify \(M\) with \(f(M)\), it is clear that \(M \subset
  N^\lambda\). The exactness of (\ref{eq:exact-seq-h1-vanishes}) implies \(\dim
  N = \dim M + 1\), so that either \(N^\lambda = M\) or \(N^\lambda = N\). But
  if \(N^\lambda = N\) then there is some nonzero \(n \in N^\lambda\) with \(n
  \notin M = \ker g\) such that
  \[
    0
    = (\Omega_M - \lambda)^r \cdot n
    = \sum_{k = 0}^r (-1)^k \binom{r}{k} \lambda^k \Omega_M^{r - k} \cdot n
  \]
  for some \(r \ge 1\).

  In particular,
  \[
    (- \lambda)^{r - 1} g(n)
    = \sum_{k = 0}^{r - 1} (-1)^k \binom{r}{k} \lambda^k
                           g(\Omega_M^{r - k} \cdot n)
    = \sum_{k = 0}^{r - 1} (-1)^k \binom{r}{k} \lambda^k
      \underbrace{\Omega_M^{r - k} \cdot g(n)}_{= \; 0}
    = 0,
  \]
  which is a contradiction -- given that neither \((-\lambda)^{r - 1}\) nor
  \(g(n)\) are nil. Hence \(M = N^\lambda\) and there must be some other
  eigenvalue \(\mu\) of \(\Omega_M\!\restriction_N\). For any such \(\mu\) and
  any eigenvector \(n \in N_\mu\),
  \[
    \mu g(n)
    = g(\mu n)
    = g(\Omega_M \cdot n)
    = \Omega_M \cdot g(n)
    = 0
  \]
  implies \(\mu = 0\), so that the eigenvalues of the action of \(\Omega_M\) on
  \(N\) are precisely \(\lambda\) and \(0\).

  Now notice that \(N^0\) is in fact a submodule of \(N\). Indeed,
  given \(n \in N^0\) and \(X \in \mathfrak{g}\), it follows from the fact that
  \(\Omega_M\) is central that
  \[
    \Omega_M^r \cdot (X \cdot n) = X \cdot (\Omega_M^r \cdot n) = X \cdot 0 = 0
  \]
  for some \(r\). Hence \(N = M \oplus N^0\) as \(\mathfrak{g}\)-modules. The
  homomorphism \(g\) thus induces an isomorphism \(N^0 \cong \mfrac{N}{M}
  \isoto K\), which translates to a splitting of
  (\ref{eq:exact-seq-h1-vanishes}).

  Finally, we consider the case where \(M\) is not simple. Suppose
  \(H^1(\mathfrak{g}, N) = 0\) for all \(\mathfrak{g}\)-modules with \(\dim N <
  \dim M\) and let \(N \subset M\) be a proper nonzero submodule. Then the
  exact sequence
  \begin{center}
    \begin{tikzcd}
      0 \rar & N \rar & M \rar & \sfrac{M}{N} \rar & 0
    \end{tikzcd}
  \end{center}
  induces a long exact sequence of the form
  \begin{equation}\label{eq:standard-h1-ext-seq}
    \begin{tikzcd}
      \cdots                          \rar[dashed] &
      H^1(\mathfrak{g}, N)            \rar         &
      H^1(\mathfrak{g}, M)            \rar         &
      H^1(\mathfrak{g}, \sfrac{M}{N}) \rar[dashed] &
      \cdots
    \end{tikzcd}
  \end{equation}

  Since \(\dim N < \dim M\), it follows \(H^1(\mathfrak{g}, N) = 0\). In
  addition, since \(\dim N > 0\), we find \(\dim \mfrac{M}{N} < \dim M\) and
  thus \(H^1(\mathfrak{g}, \sfrac{M}{N}) = 0\). The exactness of
  (\ref{eq:standard-h1-ext-seq}) then implies \(H^1(\mathfrak{g}, M) = 0\).
  Hence by induction in \(\dim M\) we find \(H^1(\mathfrak{g}, M) = 0\) for all
  finite-dimensional \(M\). We are done.
\end{proof}

We are now finally ready to prove\dots

\begin{theorem}[Weyl]\label{thm:weyl-theorem}
  Given a semisimple Lie algebra \(\mathfrak{g}\), every finite-dimensional
  \(\mathfrak{g}\)-module is semisimple.
\end{theorem}

\begin{proof}
  Let
  \begin{equation}\label{eq:generict-exact-sequence}
    \begin{tikzcd}
      0 \rar & N \rar{f} & M \rar{g} & L \rar & 0
    \end{tikzcd}
  \end{equation}
  be a short exact sequence of finite-dimensional \(\mathfrak{g}\)-modules. We
  want to establish that (\ref{eq:generict-exact-sequence}) splits.

  We have an exact sequence
  \begin{center}
    \begin{tikzcd}
      0                        \rar            &
      \operatorname{Hom}(L, N) \rar{f \circ -} &
      \operatorname{Hom}(L, M) \rar{g \circ -} &
      \operatorname{Hom}(L, L) \rar            &
      0
    \end{tikzcd}
  \end{center}
  of vector spaces. Since all maps involved are \(\mathfrak{g}\)-homomorphisms,
  this is an exact sequence of \(\mathfrak{g}\)-modules. This then induces a
  long exact sequence
  \begin{center}
    \begin{tikzcd}
      0 \rar                                                   &
      \operatorname{Hom}(L, N)^{\mathfrak{g}} \rar{f \circ -}
      \ar[draw=none]{d}[name=X, anchor=center]{}               &
      \operatorname{Hom}(L, M)^{\mathfrak{g}} \rar{g \circ -}  &
      \operatorname{Hom}(L, L)^{\mathfrak{g}}
      \ar[rounded corners,
          to path={ -- ([xshift=2ex]\tikztostart.east)
                    |- (X.center) \tikztonodes
                    -| ([xshift=-2ex]\tikztotarget.west)
                    -- (\tikztotarget)}]{dll}[at end]{} \\     &
      H^1(\mathfrak{g}, \operatorname{Hom}(L, N)) \rar         &
      H^1(\mathfrak{g}, \operatorname{Hom}(L, M)) \rar         &
      H^1(\mathfrak{g}, \operatorname{Hom}(L, L)) \rar[dashed] &
      \cdots
    \end{tikzcd}
  \end{center}
  of vector spaces.

  But \(H^1(\mathfrak{g}, \operatorname{Hom}(L, N))\) vanishes because of
  Proposition~\ref{thm:first-cohomology-vanishes}. In addition, recall from
  Example~\ref{ex:hom-invariants-are-g-homs} that \(\operatorname{Hom}(L,
  L')^{\mathfrak{g}} = \operatorname{Hom}_{\mathfrak{g}}(L, L')\). We thus have
  a short exact sequence
  \begin{center}
    \begin{tikzcd}
      0 \rar &
      \operatorname{Hom}_{\mathfrak{g}}(L, N) \rar{f \circ -} &
      \operatorname{Hom}_{\mathfrak{g}}(L, M) \rar{g \circ -} &
      \operatorname{Hom}_{\mathfrak{g}}(L, L) \rar &
      0
    \end{tikzcd}
  \end{center}

  In particular, there is some \(\mathfrak{g}\)-homomorphism \(s : L \to M\)
  such that \(g \circ s : L \to L\) is the identity operator. In other words
  \begin{center}
    \begin{tikzcd}
      0 \rar & N \rar{f} & M \rar{g} & L \rar \lar[bend left]{s} & 0
    \end{tikzcd}
  \end{center}
  is a splitting of (\ref{eq:generict-exact-sequence}).
\end{proof}

Theorem~\ref{thm:weyl-theorem} typically fails in the infinite-dimensional
setting. For instance, consider\dots

\begin{example}\label{ex:regular-mod-is-not-semisimple}
  The regular \(\mathfrak{g}\)-module \(\mathcal{U}(\mathfrak{g})\) is an
  indecomposable module which is not simple. In particular,
  \(\mathcal{U}(\mathfrak{g})\) is not semisimple. To see this, notice that the
  submodules of \(\mathcal{U}(\mathfrak{g})\) are precisely its left ideals. If
  we suppose that \(I, J \normal \mathcal{U}(\mathfrak{g})\) are such that
  \(\mathcal{U}(\mathfrak{g}) = I \oplus J\) as \(\mathfrak{g}\)-modules, we
  can find \(u \in I\) and \(v \in J\) such that \(1 = u + v\). The PBW Theorem
  then implies that \(u\) and \(v\) commute, so that \(uv = vu \in I \cap J =
  0\). Since \(\mathcal{U}(\mathfrak{g})\) is a domain, either \(u = 0\) or \(v
  = 0\). Given that \(1 = u + v\), \(u = 1\) or \(v = 1\). Hence either \(I =
  \mathcal{U}(\mathfrak{g})\) and \(J = 0\) or \(I = 0\) and \(J =
  \mathcal{U}(\mathfrak{g})\), as required.
\end{example}

We should point out that these last results are just the beginning of a well
developed cohomology theory. For example, a similar argument involving the
Casimir elements can be used to show that \(H^i(\mathfrak{g}, M) = 0\) for all
semisimple \(\mathfrak{g}\) and all non-trivial finite-dimensional simple
\(M\), \(i > 0\). For \(K = \mathbb{C}\), the Lie algebra cohomology groups of
the algebra \(\mathfrak{g} = \mathbb{C} \otimes \operatorname{Lie}(G)\) are
intimately related with the topological cohomologies -- i.e. singular
cohomology, de Rham cohomology, etc. -- of \(G\) with coefficients in
\(\mathbb{C}\). We refer the reader to \cite{cohomologies-lie} and
\cite[sec.~24]{symplectic-physics} for further details.

Complete reducibility can be generalized for arbitrary -- not necessarily
semisimple -- \(\mathfrak{g}\), to a certain extent, by considering the exact
sequence
\begin{center}
  \begin{tikzcd}
    0                                                  \rar &
    \mathfrak{rad}(\mathfrak{g})                       \rar &
    \mathfrak{g}                                       \rar &
    \mfrac{\mathfrak{g}}{\mathfrak{rad}(\mathfrak{g})} \rar &
    0
  \end{tikzcd}
\end{center}

This sequence always splits for finite-dimensional \(\mathfrak{g}\), which in
light of Example~\ref{ex:all-simple-reps-are-tensor-prod} implies we can deduce
information about \(\mathfrak{g}\)-modules by studying the modules of its
``semisimple part'' \(\mfrac{\mathfrak{g}}{\mathfrak{rad}(\mathfrak{g})}\) --
see Proposition~\ref{thm:quotients-by-rads}. In practice this translates
to\dots

\begin{proposition}[Lie]\label{thm:lie-thm-solvable-reps}
  Let \(\mathfrak{g}\) be a solvable Lie algebra. Every finite-dimensional
  simple \(\mathfrak{g}\)-module is \(1\)-dimensional.
\end{proposition}

\begin{corollary}
  Let \(\mathfrak{g}\) be a Lie algebra. Every finite-dimensional simple
  \(\mathfrak{g}\)-module is the tensor product of a simple
  \(\mfrac{\mathfrak{g}}{\mathfrak{rad}(\mathfrak{g})}\)-module and a
  \(1\)-dimensional \(\mathfrak{rad}(\mathfrak{g})\)-module.
\end{corollary}

\begin{proof}
  This follows at once from Proposition~\ref{thm:lie-thm-solvable-reps} and
  Example~\ref{ex:all-simple-reps-are-tensor-prod}.
\end{proof}

Having finally reduced our initial classification problem to that of
classifying the finite-dimensional simple \(\mathfrak{g}\)-modules, we can now
focus exclusively in this particular class of \(\mathfrak{g}\)-modules.
However, there is so far no indication on how we could go about understanding
them. In the next chapter we will explore some concrete examples in the hopes
of finding a solution to our general problem.
s
of finding a solution to our general problem.
