diff --git a/sections/complete-reducibility.tex b/sections/complete-reducibility.tex
@@ -11,116 +11,116 @@ is a question that have 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 representation \(V\) of the
-\(n\)-dimensional Abelian Lie algebra \(K^n\) is nothing more than a choice of
-\(n\) commuting operators \(V \to V\) -- corresponding to the action of the
-canonical basis elements \(e_1, \ldots, e_n \in K^n\). In particular, a
-\(1\)-dimensional representation of \(K^n\) is just a choice of \(n\) scalars
-\(\lambda_1, \ldots, \lambda_n\). Different choices of scalars yield
-non-isomorphic representations, so that the \(1\)-dimensional representations of
-\(K^n\) are parameterized by points in \(K^n\).
+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 representations of a
+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
-representations of semisimple algebras simpler -- or perhaps \emph{semisimpler}
+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}
- A representation of \(\mathfrak{g}\) is called \emph{indecomposable} if it is
- not isomorphic to the direct sum of two nonzero representations.
+ 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}
- A representation of \(\mathfrak{g}\) is called \emph{irreducible} if it has
- no nonzero proper subrepresentations.
+ A \(\mathfrak{g}\)-module is called \emph{simple} if it has no nonzero proper
+ \(\mathfrak{g}\)-modules.
\end{definition}
\begin{example}
- The trivial representation \(K\) is an example of an irreducible
- representations. In fact, every \(1\)-dimensional representation \(V\) of a
- Lie algebra \(\mathfrak{g}\) is irreducible: \(V\) has no nonzero proper
- subspaces, let alone \(\mathfrak{g}\)-invariant subspaces.
+ 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}
-The general strategy for classifying finite-dimensional representations of an
-algebra is to classify the indecomposable representations. This is because\dots
+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}
- Every finite-dimensional representation of a Lie algebra can be uniquely --
- up to isomorphism and reordering of the summands -- decomposed into a direct
- sum of indecomposable representations.
+ 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 representations suffices to find \emph{all}
-finite-dimensional representations: they are the direct sum of indecomposable
-representations. The existence of the decomposition should be clear from the
-definitions. Indeed, if \(V\) is a finite-dimensional representation of
-\(\mathfrak{g}\) a simple argument via induction in \(\dim V\) suffices to
-prove the existence: if \(V\) is indecomposable then there is nothing to prove,
-and if \(V\) is not indecomposable then \(V = W \oplus U\) for some \(W, U
-\subsetneq V\) nonzero subrepresentations, so that their dimensions are both
-strictly smaller than \(\dim V\) and the existence follows from the induction
+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 representations of an arbitrary Lie algebra,
-however, turns out to be a bit of a circular problem: the indecomposable
-representations are the ones that cannot be decomposed, which is to say, those
-that are \emph{not} decomposable. Ideally, we would like to find some other
-condition, equivalent to indecomposability, but which is easier to work with.
-It is clear from the definitions that every irreducible representation is
-indecomposable, but there is no reason to believe the converse is true. Indeed,
-this is not always the case. For instance\dots
+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 \(V = K^2\) endowed with the homomorphism of Lie algebras
+ 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*}
- is a representation of the Lie algebra \(K[x]\). Notice \(V\) has a single
- nonzero proper subrepresentation, 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 \(V\) is indecomposable -- it cannot be broken into a direct
- sum of \(1\)-dimensional subrepresentations -- but it is evidently not
- irreducible.
+ 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
-representation of \(\mathfrak{g}\) is irreducible? This is what is known in
-representation theory as \emph{complete reducibility}.
+\(\mathfrak{g}\)-module is simple? This is what is known in representation
+theory as \emph{complete reducibility}.
\begin{definition}
- A \(\mathfrak{g}\)-module \(V\) is called \emph{completely reducible} if it
- the direct sum of irreducible representations.
+ A \(\mathfrak{g}\)-module \(M\) is called \emph{completely reducible}, or
+ \emph{semisimple}, if it is the direct sum of simple \(\mathfrak{g}\)-modules.
\end{definition}
-In case the relationship between complete reducibility and the irreducibility
-of indecomposable representations is unclear, the following results should
-clear things up.
+In case the relationship between complete reducibility and the simplicity of
+indecomposable \(\mathfrak{g}\)-modules is unclear, the following results
+should clear things up.
\begin{proposition}\label{thm:complete-reducibility-equiv}
The following conditions are equivalent.
\begin{enumerate}
- \item Every subrepresentation of a finite-dimensional representation of
- \(\mathfrak{g}\) has a \(\mathfrak{g}\)-invariant complement -- i.e.
- given \(W \subset V\) there is a subrepresentation \(U \subset V\) such
- that \(V = W \oplus U\).
+ \item Every submodule of a finite-dimensional \(\mathfrak{g}\)-module has a
+ \(\mathfrak{g}\)-invariant complement -- i.e. given \(N \subset M\) there
+ is a \(\mathfrak{g}\)-submodule \(L \subset M\) such that \(M = N \oplus
+ L\).
- \item Every exact sequence of finite-dimensional representations of
- \(\mathfrak{g}\) splits.
+ \item Every exact sequence of finite-dimensional \(\mathfrak{g}\)-modules
+ splits.
- \item Every indecomposable finite-dimensional representation of
- \(\mathfrak{g}\) is irreducible.
+ \item Every indecomposable finite-dimensional \(\mathfrak{g}\)-module is
+ simple.
- \item Every finite-dimensional representation of \(\mathfrak{g}\) is
- completely reducible.
+ \item Every finite-dimensional \(\mathfrak{g}\)-module is completely
+ reducible.
\end{enumerate}
\end{proposition}
@@ -129,130 +129,129 @@ clear things up.
\begin{center}
\begin{tikzcd}
0 \arrow{r} &
- W \arrow{r}{i} &
- V \arrow{r}{\pi} &
- U \arrow{r} &
+ N \arrow{r}{f} &
+ M \arrow{r}{g} &
+ L \arrow{r} &
0
\end{tikzcd}
\end{center}
- be an exact sequence of representations of \(\mathfrak{g}\). We can suppose
- without loss of generality that \(W \subset V\) is a subrepresentation and
- \(i\) is its inclusion in \(V\), for if this is not the case there is an
- isomorphism of sequences
+ 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 \arrow{r} &
- W \arrow{r}{i} \arrow[swap]{d}{i} &
- V \arrow{r}{\pi} \arrow[Rightarrow, no head]{d} &
- U \arrow{r} \arrow[Rightarrow, no head]{d} &
- 0 \\
- 0 \arrow{r} &
- i(W) \arrow{r} &
- V \arrow[swap]{r}{\pi} &
- U \arrow{r} &
+ 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 subrepresentation \(U'
- \subset V\) such that \(V = W \oplus U'\). Finally, the projection \(T : V
- \to W\) is an intertwiner satisfying
+ 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 \arrow{r} &
- W \arrow{r}{i} &
- V \arrow{r}{\pi} \arrow[bend left=30]{l}{T} &
- U \arrow{r} &
+ 0 \arrow{r} &
+ N \arrow{r}{f} &
+ M \arrow{r}{g} \arrow[bend left=30]{l}{s} &
+ L \arrow{r} &
0
\end{tikzcd}
\end{center}
- Next is \(\textbf{(ii)} \implies \textbf{(iii)}\). If \(V\) is an
- indecomposable \(\mathfrak{g}\)-module and \(W \subset V\) is a
- subrepresentation, we have an exact sequence
+ 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 \arrow{r} &
- W \arrow{r}{i} &
- V \arrow{r}{\pi} &
- \mfrac{V}{W} \arrow{r} &
+ 0 \arrow{r} &
+ N \arrow{r} &
+ M \arrow{r} &
+ \mfrac{M}{N} \arrow{r} &
0
\end{tikzcd}
\end{center}
- of representations of \(\mathfrak{g}\).
+ of \(\mathfrak{g}\)-modules.
- Since our sequence splits, we must have \(V \cong W \oplus \mfrac{V}{W}\).
- But \(V\) is indecomposable, so that either \(W = V\) or \(W = 0\). Since
- this holds for all \(W \subset V\), \(V\) is irreducible. For
- \(\textbf{(iii)} \implies \textbf{(iv)}\) it suffices to apply
- Theorem~\ref{thm:krull-schmidt}.
+ 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 \(V\) be a representation of \(\mathfrak{g}\) with
- decomposition into irreducible subrepresentations
+ \(\textbf{(iv)}\) and let \(M\) be a \(\mathfrak{g}\)-module with
+ decomposition into simple submodules
\[
- V = \bigoplus_i V_i
+ M = \bigoplus_i M_i
\]
- and \(W \subset V\) is a subrepresentation. Take some maximal set of indexes
- \(\{i_1, \ldots, i_n\}\) so that \(\left( \bigoplus_k V_{i_k}
- \right) \cap W = 0\) and let \(U = \bigoplus_k V_{i_k}\). We want to
- establish \(V = W \oplus U\).
-
- Suppose without any loss in generality that \(i_k = k\) for all \(k\) and
- let \(j > n\). By the maximality of our set of indexes, there is some
- nonzero \(w \in (V_j \oplus U) \cap W\). Say \(w = v_j + v_1 + \cdots +
- v_n\) with each \(v_i \in V_i\). Then \(v_j = w - v_1 - \cdots - v_n \in V_j
- \cap (W \oplus U)\) is nonzero. Indeed, if this is not the case we find \(0
- \ne w = v_1 + \cdots + v_n \in \left( \bigoplus_{i = 1}^n V_i \right) \cap
- W\), a contradiction. This implies \(V_j \cap (W \oplus U)\) is a nonzero
- subrepresentation of \(V_j\). Since \(V_j\) is irreducible, \(V_j = V_j \cap
- (W \oplus U)\) and therefore \(V_j \subset W \oplus U\). Given the arbitrary
- choice of \(j\), it then follows \(V = W \oplus U\).
+ 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 are primarily interested in indecomposable representations -- which is
-usually a strictly larger class of representations than that of irreducible
-representations -- it is important to note that irreducible representations are
-generally much easier to find. The relationship between irreducible
-representations is also well understood. This is because of the following
-result, known as \emph{Schur's Lemma}.
+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 \(V\) and \(W\) be irreducible representations of \(\mathfrak{g}\) and
- \(T : V \to W\) be an intertwiner. Then \(T\) is either \(0\) or an
- isomorphism. Furthermore, if \(V = W\) then \(T\) is a scalar operator.
+ 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 \(\ker T\) and
- \(\operatorname{im} T\) are both subrepresentations. In particular, either
- \(\ker T = 0\) and \(\operatorname{im} T = W\) or \(\ker T = V\) and
- \(\operatorname{im} T = 0\). Now suppose \(V = W\). Let \(\lambda \in K\) be
- an eigenvalue of \(T\) -- which exists because \(K\) is algebraically closed --
- and \(V_\lambda\) be its corresponding eigenspace. Given \(v \in V_\lambda\),
- \(T X v = X T v = \lambda \cdot X v\). In other words, \(V_\lambda\) is a
- subrepresentation. It then follows \(V_\lambda = V\), given that \(V_\lambda
- \ne 0\).
+ 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
-representations and their irreducible representations is much clearer.
-Namely\dots
+modules and their simple modules is much clearer. Namely\dots
\begin{proposition}\label{thm:complete-reducibility-equiv}
Given a finite-dimensional Lie algebra \(\mathfrak{g}\) over \(K\),
\(\mathfrak{g}\) is semisimple if, and only if every finite-dimensional
- representation of \(\mathfrak{g}\) is completely reducible.
+ \(\mathfrak{g}\)-module is completely reducible.
\end{proposition}
The proof of the fact that a finite-dimensional Lie algebra \(\mathfrak{g}\)
-whose finite-dimensional representations are completely reducible is semisimple
-is actually pretty simple. Namely, it suffices to note that the adjoint
-representation \(\mathfrak{g}\) is the direct sum of irreducible
-subrepresentations, which are all simple ideals of \(\mathfrak{g}\) -- so
-\(\mathfrak{g}\) is the direct sum of simple Lie algebras. The proof of the
-converse is more nuanced, and this will be our next milestone.
+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
@@ -269,6 +268,8 @@ to introduce some basic tools which will come in handy later on, known as\dots
\]
\end{definition}
+% TODO: Note that antisymmetric matrices are the infinitesimal version of
+% ortogonal matrices
\begin{note}
The etymology of the term \emph{invariant form} comes from group
representation theory. If \(G \subset \operatorname{GL}(V)\) is a group of
@@ -296,11 +297,11 @@ 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}
- Given a finite-dimensional representation \(V\) of \(\mathfrak{g}\), the
- symmetric bilinear form
+ Given a finite-dimensional \(\mathfrak{g}\)-module \(M\), the symmetric
+ bilinear form
\begin{align*}
- B_V : \mathfrak{g} \times \mathfrak{g} & \to K \\
- (X, Y) & \mapsto \operatorname{Tr}(X\!\restriction_V \, Y\!\restriction_V)
+ B_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}
@@ -314,12 +315,12 @@ characterization of finite-dimensional semisimple Lie algebras, known as
equivalent.
\begin{enumerate}
\item \(\mathfrak{g}\) is semisimple.
- \item For each non-trivial finite-dimensional representation \(V\) of
- \(\mathfrak{g}\), the \(\mathfrak{g}\)-invariant bilinear form
+ \item For each non-trivial finite-dimensional \(\mathfrak{g}\)-module
+ \(M\), the \(\mathfrak{g}\)-invariant bilinear form
\begin{align*}
- B_V : \mathfrak{g} \times \mathfrak{g} & \to K \\
+ B_M : \mathfrak{g} \times \mathfrak{g} & \to K \\
(X, Y) &
- \mapsto \operatorname{Tr}(X\!\restriction_V \circ Y\!\restriction_V)
+ \mapsto \operatorname{Tr}(X\!\restriction_M \circ 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
@@ -330,12 +331,12 @@ characterization of finite-dimensional semisimple Lie algebras, known as
This proof is somewhat technical, but the idea behind it is simple. First, for
\strong{(i)} \(\implies\) \strong{(ii)} we show that \(\mathfrak{a} = \{ X \in
-\mathfrak{g} : B_V(X, Y) = 0 \, \forall Y \in \mathfrak{g}\}\) is a solvable
+\mathfrak{g} : B_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 \(V = \mathfrak{g}\) the
-adjoint representation. 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 \(B\) is an ideal
+\(\implies\) \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 \(B\) 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 \(B\!\restriction_{\mathfrak{a}}\) of the Killing form of
@@ -349,17 +350,17 @@ further ado, we may proceed to our\dots
\section{Proof of Complete Reducibility}
Let \(\mathfrak{g}\) be a finite-dimensional semisimple Lie algebra over \(K\).
-We want to establish that all finite-dimensional representations of
-\(\mathfrak{g}\) are completely reducible. 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 representations 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.
+We want to establish that all finite-dimensional \(\mathfrak{g}\)-modules are
+completely reducible. 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
@@ -382,38 +383,38 @@ basic}. In fact, all we need to know is\dots
There is a sequence of bifunctors \(\operatorname{Ext}^i :
\mathfrak{g}\text{-}\mathbf{Mod} \times \mathfrak{g}\text{-}\mathbf{Mod} \to
K\text{-}\mathbf{Vect}\), \(i \ge 0\) such that, given a
- \(\mathfrak{g}\)-module \(S\), every exact sequence of
+ \(\mathfrak{g}\)-module \(L'\), every exact sequence of
\(\mathfrak{g}\)-modules
\begin{center}
\begin{tikzcd}
- 0 \arrow{r} & W \arrow{r}{i} & V \arrow{r}{\pi} & U \arrow{r} & 0
+ 0 \arrow{r} & N \arrow{r}{f} & M \arrow{r}{g} & L \arrow{r} & 0
\end{tikzcd}
\end{center}
induces long exact sequences
\begin{center}
\begin{tikzcd}
0 \arrow[r] &
- \operatorname{Hom}_{\mathfrak{g}}(S, W)
- \arrow[r, "i \circ -"', swap]\ar[draw=none]{d}[name=X, anchor=center]{} &
- \operatorname{Hom}_{\mathfrak{g}}(S, V) \arrow[r, "\pi \circ -"', swap] &
- \operatorname{Hom}_{\mathfrak{g}}(S, U)
+ \operatorname{Hom}_{\mathfrak{g}}(L', N)
+ \arrow[r, "f \circ -"', swap]\ar[draw=none]{d}[name=X, anchor=center]{} &
+ \operatorname{Hom}_{\mathfrak{g}}(L', M) \arrow[r, "g \circ -"', swap] &
+ \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(S, W)
+ \operatorname{Ext}^1(L', N)
\arrow[r]\ar[draw=none]{d}[name=Y, anchor=center]{} &
- \operatorname{Ext}^1(S, V) \arrow[r] &
- \operatorname{Ext}^1(S, U)
+ \operatorname{Ext}^1(L', M) \arrow[r] &
+ \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(S, W) \arrow[r] &
- \operatorname{Ext}^2(S, V) \arrow[r] &
- \operatorname{Ext}^2(S, U) \arrow[r, dashed] &
+ \operatorname{Ext}^2(L', N) \arrow[r] &
+ \operatorname{Ext}^2(L', M) \arrow[r] &
+ \operatorname{Ext}^2(L', L) \arrow[r, dashed] &
\cdots
\end{tikzcd}
\end{center}
@@ -421,44 +422,44 @@ basic}. In fact, all we need to know is\dots
\begin{center}
\begin{tikzcd}
0 \arrow[r] &
- \operatorname{Hom}_{\mathfrak{g}}(U, S)
- \arrow[r, "- \circ \pi"', swap]\ar[draw=none]{d}[name=X, anchor=center]{} &
- \operatorname{Hom}_{\mathfrak{g}}(V, S) \arrow[r, "- \circ i"', swap] &
- \operatorname{Hom}_{\mathfrak{g}}(W, S)
+ \operatorname{Hom}_{\mathfrak{g}}(L, L')
+ \arrow[r, "- \circ g"', swap]\ar[draw=none]{d}[name=X, anchor=center]{} &
+ \operatorname{Hom}_{\mathfrak{g}}(M, L') \arrow[r, "- \circ f"', swap] &
+ \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(U, S)
+ \operatorname{Ext}^1(L, L')
\arrow[r]\ar[draw=none]{d}[name=Y, anchor=center]{} &
- \operatorname{Ext}^1(V, S) \arrow[r] &
- \operatorname{Ext}^1(W, S)
+ \operatorname{Ext}^1(M, L') \arrow[r] &
+ \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(U, S) \arrow[r] &
- \operatorname{Ext}^2(V, S) \arrow[r] &
- \operatorname{Ext}^2(W, S) \arrow[r, dashed] &
+ \operatorname{Ext}^2(L, L') \arrow[r] &
+ \operatorname{Ext}^2(M, L') \arrow[r] &
+ \operatorname{Ext}^2(N, L') \arrow[r, dashed] &
\cdots
\end{tikzcd}
\end{center}
\end{theorem}
\begin{theorem}\label{thm:ext-1-classify-short-seqs}
- Given \(\mathfrak{g}\)-modules \(W\) and \(U\), there is a one-to-one
- correspondence between elements of \(\operatorname{Ext}^1(U, W)\) and
+ Given \(\mathfrak{g}\)-modules \(N\) and \(N\), 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 \arrow{r} & W \arrow{r} & V \arrow{r} & U \arrow{r} & 0
+ 0 \arrow{r} & N \arrow{r} & M \arrow{r} & L \arrow{r} & 0
\end{tikzcd}
\end{center}
- In particular, \(\operatorname{Ext}^1(U, W) = 0\) if, and only if every short
- exact sequence of \(\mathfrak{g}\)-modules with \(W\) and \(U\) in the
+ 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}
@@ -469,57 +470,70 @@ basic}. In fact, all we need to know is\dots
a more modern account using derived categories.
\end{note}
-We are particularly interested in the case where \(S = K\) is the trivial
-representation of \(\mathfrak{g}\). Namely, we may define\dots
+We are particularly interested in the case where \(L' = K\) is the trivial
+\(\mathfrak{g}\)-module. Namely, we may define\dots
+
+\begin{definition}
+ Given 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}
- Given a \(\mathfrak{g}\)-module \(V\), we refer to the Abelian group
- \(H^i(\mathfrak{g}, V) = \operatorname{Ext}^i(K, V)\) as \emph{the \(i\)-th
- Lie algebra cohomology group of \(\mathfrak{g}\) with coefficients in \(V\)}.
+ 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}
-Given a \(\mathfrak{g}\)-module \(V\), we call the vector space
-\(V^{\mathfrak{g}} = \{v \in V : X v = 0 \; \forall X \in \mathfrak{g}\}\)
-\emph{the space of invariants of \(V\)}. The Lie algebra cohomology groups are
-very much related to invariants of representations. Namely, the canonical
-isomorphism of functors
+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 \(f(1)\). 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, V) & \isoto V^{\mathfrak{g}} \\
- T & \mapsto T(1)
+ \operatorname{Hom}_{\mathfrak{g}}(K, M) & \isoto M^{\mathfrak{g}} \\
+ f & \mapsto f(1)
\end{align*}
-implies\dots
+
+This implies\dots
\begin{corollary}
Every short exact sequence of \(\mathfrak{g}\)-modules
\begin{center}
\begin{tikzcd}
- 0 \arrow{r} & W \arrow{r}{i} & V \arrow{r}{\pi} & U \arrow{r} & 0
+ 0 \arrow{r} & N \arrow{r}{f} & M \arrow{r}{g} & L \arrow{r} & 0
\end{tikzcd}
\end{center}
induces a long exact sequence
\begin{center}
\begin{tikzcd}
0 \arrow[r] &
- W^{\mathfrak{g}} \arrow[r, "i"', swap]\ar[draw=none]{d}[name=X, anchor=center]{} &
- V^{\mathfrak{g}} \arrow[r, "\pi"', swap] &
- U^{\mathfrak{g}}
+ N^{\mathfrak{g}} \arrow[r, "f"', swap]\ar[draw=none]{d}[name=X, anchor=center]{} &
+ M^{\mathfrak{g}} \arrow[r, "g"', swap] &
+ 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}, W) \arrow[r]\ar[draw=none]{d}[name=Y, anchor=center]{} &
- H^1(\mathfrak{g}, V) \arrow[r] &
- H^1(\mathfrak{g}, U)
+ H^1(\mathfrak{g}, N) \arrow[r]\ar[draw=none]{d}[name=Y, anchor=center]{} &
+ H^1(\mathfrak{g}, M) \arrow[r] &
+ 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}, W) \arrow[r] &
- H^2(\mathfrak{g}, V) \arrow[r] &
- H^2(\mathfrak{g}, U) \arrow[r, dashed] &
+ H^2(\mathfrak{g}, N) \arrow[r] &
+ H^2(\mathfrak{g}, M) \arrow[r] &
+ H^2(\mathfrak{g}, L) \arrow[r, dashed] &
\cdots
\end{tikzcd}
\end{center}
@@ -530,18 +544,18 @@ implies\dots
\begin{center}
\begin{tikzcd}
0 \arrow{r} &
- \operatorname{Hom}_{\mathfrak{g}}(K, W)
- \arrow{r}{i \circ -} \arrow{d} &
- \operatorname{Hom}_{\mathfrak{g}}(K, V)
- \arrow{r}{\pi \circ -} \arrow{d} &
- \operatorname{Hom}_{\mathfrak{g}}(K, U) \arrow{r} \arrow{d} &
- H^1(\mathfrak{g}, W) \arrow[dashed]{r} \arrow[Rightarrow, no head]{d} &
+ \operatorname{Hom}_{\mathfrak{g}}(K, N)
+ \arrow{r}{f \circ -} \arrow{d} &
+ \operatorname{Hom}_{\mathfrak{g}}(K, M)
+ \arrow{r}{g \circ -} \arrow{d} &
+ \operatorname{Hom}_{\mathfrak{g}}(K, L) \arrow{r} \arrow{d} &
+ H^1(\mathfrak{g}, N) \arrow[dashed]{r} \arrow[Rightarrow, no head]{d} &
\cdots \\
0 \arrow{r} &
- W^{\mathfrak{g}} \arrow[swap]{r}{i} &
- V^{\mathfrak{g}} \arrow[swap]{r}{\pi} &
- U^{\mathfrak{g}} \arrow{r} &
- H^1(\mathfrak{g}, W) \arrow[dashed]{r} &
+ N^{\mathfrak{g}} \arrow[swap]{r}{f} &
+ M^{\mathfrak{g}} \arrow[swap]{r}{g} &
+ L^{\mathfrak{g}} \arrow{r} &
+ H^1(\mathfrak{g}, N) \arrow[dashed]{r} &
\cdots
\end{tikzcd}
\end{center}
@@ -559,12 +573,12 @@ trying to control obstructions of some kind. In our case, the bifunctor
Explicitly\dots
\begin{theorem}
- Given \(\mathfrak{g}\)-modules \(W\) and \(U\), there is a one-to-one
- correspondence between elements of \(H^1(\mathfrak{g}, \operatorname{Hom}(W,
- U))\) and isomorphism classes of short exact sequences
+ Given \(\mathfrak{g}\)-modules \(N\) and \(L\), there is a one-to-one
+ correspondence between elements of \(H^1(\mathfrak{g}, \operatorname{Hom}(N,
+ L))\) and isomorphism classes of short exact sequences
\begin{center}
\begin{tikzcd}
- 0 \arrow{r} & W \arrow{r} & V \arrow{r} & U \arrow{r} & 0
+ 0 \arrow{r} & N \arrow{r} & M \arrow{r} & L \arrow{r} & 0
\end{tikzcd}
\end{center}
\end{theorem}
@@ -581,123 +595,124 @@ can be computed very concretely by considering a canonical acyclic resolution
0
\end{tikzcd}
\end{center}
-of the trivial representation \(K\), which provides an explicit construction of
-the cohomology groups -- see \cite[sec.~9]{lie-groups-serganova-student} or
+of the trivial \(\mathfrak{g}\)-module \(K\), which provides an explicit
+construction of the cohomology groups -- see
+\cite[sec.~9]{lie-groups-serganova-student} 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 \(H^1(\mathfrak{g}, V) = 0\) for all
-finite-dimensional \(V\), and that the fact that \(H^1(\mathfrak{g},
-\operatorname{Hom}(W, U)) = 0\) for all finite-dimensional \(W\) and \(U\)
+force. Namely, we will show that \(H^1(\mathfrak{g}, M) = 0\) for all
+finite-dimensional \(M\), and that the fact that \(H^1(\mathfrak{g},
+\operatorname{Hom}(N, L)) = 0\) 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 representation}.
+a \(\mathfrak{g}\)-module}.
\begin{definition}\label{def:casimir-element}
- Let \(V\) be a finite-dimensional representation of \(\mathfrak{g}\). Let
+ Let \(M\) 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 \(B_V\) --
- i.e. the unique basis for \(\mathfrak{g}\) satisfying \(B_V(X_i, X^j) =
+ \subset \mathfrak{g}\) its dual basis with respect to the form \(B_M\) --
+ i.e. the unique basis for \(\mathfrak{g}\) satisfying \(B_M(X_i, X^j) =
\delta_{i j}\). We call
\[
- C_V = X_1 X^1 + \cdots + X_n X^n \in \mathcal{U}(\mathfrak{g})
+ C_M = X_1 X^1 + \cdots + X_r X^r \in \mathcal{U}(\mathfrak{g})
\]
- the \emph{Casimir element of \(V\)}.
+ the \emph{Casimir element of \(M\)}.
\end{definition}
\begin{lemma}
- The definition of \(C_V\) is independent of the choice of basis
+ The definition of \(C_M\) is independent of the choice of basis
\(\{X_i\}_i\).
\end{lemma}
\begin{proof}
- Whatever basis \(\{X_i\}_i\) we choose, the image of \(C_V\) under the
+ Whatever basis \(\{X_i\}_i\) we choose, the image of \(C_M\) 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 $B_V$.}.
+ $\mathfrak{g} \isoto \mathfrak{g}^*$ induced by the form $B_M$.}.
\end{proof}
\begin{proposition}
- The Casimir element \(C_V \in \mathcal{U}(\mathfrak{g})\) is central, so that
- \(C_V : W \to W\) is an intertwining operator for any \(\mathfrak{g}\)-module
- \(W\). Furthermore, \(C_V\) acts on \(V\) as a nonzero scalar operator
- whenever \(V\) is a non-trivial finite-dimensional irreducible representation
- of \(\mathfrak{g}\).
+ The Casimir element \(C_M \in \mathcal{U}(\mathfrak{g})\) is central, so that
+ \(C_M\!\restriction_N : N \to N\) is a \(\mathfrak{g}\)-homomorphism for any
+ \(\mathfrak{g}\)-module \(N\). Furthermore, \(C_M\) 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 \(C_V\) is central fix a basis \(\{X_i\}_i\) for \(\mathfrak{g}\)
+ To see that \(C_M\) is central fix a basis \(\{X_i\}_i\) for \(\mathfrak{g}\)
and denote by \(\{X^i\}_i\) its dual basis as in
Definition~\ref{def:casimir-element}. Let \(X \in \mathfrak{g}\) and denote
by \(\lambda_{i j}, \mu_{i j} \in K\) the coefficients of \(X_j\) and \(X^j\)
in \([X, X_i]\) and \([X, X^i]\), respectively.
- The invariance of \(B_V\) implies
+ The invariance of \(B_M\) implies
\[
\lambda_{i k}
- = B_V([X, X_i], X^k)
- = B_V(-[X_i, X], X^k)
- = B_V(X_i, -[X, X^k])
+ = B_M([X, X_i], X^k)
+ = B_M(-[X_i, X], X^k)
+ = B_M(X_i, -[X, X^k])
= - \mu_{k i}
\]
Hence
\[
\begin{split}
- [X, C_V]
+ [X, C_M]
& = \sum_i [X, X_i X^i] \\
& = \sum_i [X, X_i] X^i + \sum_i X_i [X, X^i] \\
& = \sum_{i j} \lambda_{i j} X_j X^i + \sum_{i j} \mu_{i j} X_i X^j \\
& = 0
\end{split},
\]
- and \(C_V\) is central. This implies that \(C_V : W \to W\) is an intertwiner
- for all representations \(W\) of \(\mathfrak{g}\): its action commutes with
- the action of any other element of \(\mathfrak{g}\).
+ and \(C_M\) is central. This implies that \(C_M\!\restriction_N : N \to N\)
+ 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 \(V\) is
- finite-dimensional and irreducible then \(C_V\) acts on \(V\) as a scalar
+ In particular, it follows from Schur's Lemma that if \(M\) is
+ finite-dimensional and simple then \(C_M\) acts on \(M\) as a scalar
operator. To see that this scalar is nonzero we compute
\[
- \operatorname{Tr}(C_V\!\restriction_V)
- = \operatorname{Tr}(X_1\!\restriction_V X^1\!\restriction_V)
+ \operatorname{Tr}(C_M\!\restriction_M)
+ = \operatorname{Tr}(X_1\!\restriction_M X^1\!\restriction_M)
+ \cdots
- + \operatorname{Tr}(X_n\!\restriction_V X^n\!\restriction_V)
+ + \operatorname{Tr}(X_r\!\restriction_M X^r\!\restriction_M)
= \dim \mathfrak{g},
\]
- so that \(C_V\!\restriction_V = \lambda \operatorname{Id}\) for \(\lambda =
- \frac{\dim \mathfrak{g}}{\dim V} \ne 0\).
+ so that \(C_M\!\restriction_M = \lambda \operatorname{Id}\) for \(\lambda =
+ \frac{\dim \mathfrak{g}}{\dim M} \ne 0\).
\end{proof}
-As promised, the Casimir element of a representation can be used to
+As promised, the Casimir element of a \(\mathfrak{g}\)-module can be used to
establish\dots
\begin{proposition}\label{thm:first-cohomology-vanishes}
- Let \(V\) be a finite-dimensional representation of \(\mathfrak{g}\). Then
- \(H^1(\mathfrak{g}, V) = 0\).
+ 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 \(V\) is irreducible. Due to
+ 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 \arrow{r} & V \arrow{r} & W \arrow{r}{\pi} & K \arrow{r} & 0
+ 0 \rar & M \rar{f} & N \rar{g} & K \rar & 0
\end{tikzcd}
\end{equation}
splits.
- If \(V = K\) is the trivial representation then the exactness of
+ If \(M = K\) is the trivial \(\mathfrak{g}\)-module then the exactness of
\begin{equation}\label{eq:trivial-extrems-exact-seq}
\begin{tikzcd}
- 0 \arrow{r} & K \arrow{r} & W \arrow{r}{\pi} & K \arrow{r} & 0
+ 0 \rar & K \rar{f} & N \rar{g} & K \rar & 0
\end{tikzcd}
\end{equation}
- implies \(W\) is 2-dimensional. Take any nonzero \(w \in W\) outside of the
- image of the inclusion \(K \to W\).
+ 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
@@ -706,169 +721,170 @@ establish\dots
% 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 \(\dim W = 2\), the irreducible component \(\mathcal{U}(\mathfrak{g})
- \cdot w\) of \(w\) in \(W\) is either \(K w\) or \(W\) itself. But this
- component cannot be \(W\), since the image the inclusion \(K \to W\) is a
- \(1\)-dimensional representation -- i.e. a proper nonzero subrepresentation.
- Hence \(K w\) is invariant under the action of \(\mathfrak{g}\). In
- particular, \(X w = 0\) for all \(X \in \mathfrak{g}\). Since \(w\) lies
- outside the image of the inclusion \(K \to W\), \(\pi(w) \ne 0\) -- which is
- to say, \(w \notin \ker \pi\). This implies the map \(K \to W\) that takes
- \(1\) to \(\sfrac{w}{\pi(w)}\) is a splitting of
+ Since \(\dim W = 2\), 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 \(f\), \(g(w) \ne 0\) -- 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 \(V\) is non-trivial, so that \(C_V\) acts on \(V\) as
- \(\lambda \operatorname{Id}\) for some \(\lambda \ne 0\). Denote by \(W^\mu\)
- the generalized eigenspace of \(C_V\!\restriction_W : W \to W\) associated
- with \(\mu \in K\). If we identify \(V\) with its image under the inclusion
- \(V \to W\), it is clear that \(V \subset W^\lambda\). The exactness of
- (\ref{eq:exact-seq-h1-vanishes}) then implies \(\dim W = \dim V + 1\), so
- that either \(W^\lambda = V\) or \(W^\lambda = W\). But if \(W^\lambda = W\)
- then there is some nonzero \(w \in W^\lambda\) with \(w \notin V = \ker
- \pi\) such that
+ Now suppose that \(M\) is non-trivial, so that \(C_M\) acts on \(M\) as
+ \(\lambda\) for some \(\lambda \ne 0\). Denote by \(N^\mu\)
+ the generalized eigenspace of \(C_M\!\restriction_N : N \to N\) 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
\[
0
- = (C_V - \lambda)^n w
- = \sum_{k = 0}^n (-1)^k \binom{n}{k} \lambda^k C_V^{n - k} w
+ = (C_V - \lambda)^r \cdot n
+ = \sum_{k = 0}^r (-1)^k \binom{r}{k} \lambda^k C_M^{r - k} \cdot n
\]
- for some \(n \ge 1\) -- given that \(C_V \in \mathcal{U}(\mathfrak{g})\) is
- central.
+ for some \(r \ge 1\).
In particular,
\[
- (- \lambda)^n \pi(w)
- = \sum_{k = 0}^{n - 1} (-1)^k \binom{n}{k} \lambda^k \pi(C_V^{n - k} w)
- = \sum_{k = 0}^{n - 1} (-1)^k \binom{n}{k} \lambda^k
- \underbrace{C_V^{n - k} \pi(w)}_{= \; 0}
+ (- \lambda)^{r - 1} g(n)
+ = \sum_{k = 0}^{r - 1} (-1)^k \binom{r}{k} \lambda^k g(C_M^{r - k} \cdot n)
+ = \sum_{k = 0}^{r - 1} (-1)^k \binom{r}{k} \lambda^k
+ \underbrace{C_M^{r - k} \cdot g(n)}_{= \; 0}
= 0,
\]
- which is a contradiction in light of the fact that neither \((-\lambda)^n\)
- nor \(\pi(w)\) are nil. Hence \(V = W^\lambda\) and there must be some other
- eigenvalue \(\mu\) of \(C_V\!\restriction_W\). For any such \(\mu\) and any
- \(w \in W^\mu\),
+ 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 \(C_M\!\restriction_N\). For any such \(\mu\) and any
+ eigenvector \(n \in N_\mu\),
\[
- \mu \pi(w)
- = \pi(\mu w)
- = \pi(C_V w)
- = C_V \pi(w)
+ \mu g(n)
+ = g(\mu n)
+ = g(C_M \cdot n)
+ = C_M \cdot g(n)
= 0
\]
- implies \(\mu = 0\), so that the eigenvalues of the action of \(C_V\) in
- \(W\) are precisely \(\lambda\) and \(0\).
+ implies \(\mu = 0\), so that the eigenvalues of the action of \(C_M\) on
+ \(N\) are precisely \(\lambda\) and \(0\).
- Now notice that \(W^0\) is in fact a subrepresentation of \(W\). Indeed,
- given \(w \in W^0\) and \(X \in \mathfrak{g}\), it follows from the fact that
- \(C_V\) is central that
+ 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
+ \(C_M\) is central that
\[
- C_V^n X w = X C_V^n w = X \cdot 0 = 0
+ C_M^r \cdot (X \cdot n) = X \cdot (C_V^r \cdot n) = X \cdot 0 = 0
\]
- for some \(n\). Hence \(W = V \oplus W^0\) as representations. The
- homomorphism \(\pi\) thus induces an isomorphism \(W^0 \cong \mfrac{W}{V}
+ 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 \(V\) is not irreducible. Suppose
- \(H^1(\mathfrak{g}, W) = 0\) for all \(\mathfrak{g}\)-modules with \(\dim W <
- \dim V\) and let \(W \subset V\) be a proper nonzero subrepresentation. Then
- the exact sequence
+ 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 \arrow{r} & W \arrow{r} & V \arrow{r} & \sfrac{V}{W} \arrow{r} & 0
+ 0 \arrow{r} & N \arrow{r} & M \arrow{r} & \sfrac{M}{N} \arrow{r} & 0
\end{tikzcd}
\end{center}
induces a long exact sequence of the form
\begin{center}
\begin{tikzcd}
\cdots \arrow[dashed]{r} &
- H^1(\mathfrak{g}, W) \arrow{r} &
- H^1(\mathfrak{g}, V) \arrow{r} &
- H^1(\mathfrak{g}, \sfrac{V}{W}) \arrow[dashed]{r} &
+ H^1(\mathfrak{g}, N) \arrow{r} &
+ H^1(\mathfrak{g}, M) \arrow{r} &
+ H^1(\mathfrak{g}, \sfrac{M}{N}) \arrow[dashed]{r} &
\cdots
\end{tikzcd}
\end{center}
- Since \(0 < \dim W, \dim \sfrac{V}{W} < \dim V\) it follows
- \(H^1(\mathfrak{g}, W) = H^1(\mathfrak{g}, \sfrac{V}{W}) = 0\). The exactness
+ Since \(0 < \dim N, \dim \sfrac{M}{N} < \dim M\) it follows
+ \(H^1(\mathfrak{g}, N) = H^1(\mathfrak{g}, \sfrac{M}{N}) = 0\). The exactness
of
\begin{center}
\begin{tikzcd}
0 \arrow{r} &
- H^1(\mathfrak{g}, V) \arrow{r} &
+ H^1(\mathfrak{g}, M) \arrow{r} &
0
\end{tikzcd}
\end{center}
- then implies \(H^1(\mathfrak{g}, V) = 0\). Hence by induction in \(\dim V\)
- we find \(H^1(\mathfrak{g}, V) = 0\) for all finite-dimensional \(V\). We are
+ then implies \(H^1(\mathfrak{g}, M) = 0\). Hence by induction in \(\dim V\)
+ 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}
- Every representation of a semisimple Lie algebra is completely reducible.
+ Given a semisimple Lie algebra \(\mathfrak{g}\), every finite-dimensional
+ \(\mathfrak{g}\)-module is completely reducible.
\end{theorem}
\begin{proof}
Let
\begin{equation}\label{eq:generict-exact-sequence}
\begin{tikzcd}
- 0 \arrow{r} & W \arrow{r} & V \arrow{r}{\pi} & U \arrow{r} & 0
+ 0 \arrow{r} & N \arrow{r}{f} & M \arrow{r}{g} & L \arrow{r} & 0
\end{tikzcd}
\end{equation}
- be a short exact sequence of finite-dimensional representations of
- \(\mathfrak{g}\). We want to establish that
- (\ref{eq:generict-exact-sequence}) splits.
+ 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 \arrow{r} &
- \operatorname{Hom}(U, W) \arrow{r} &
- \operatorname{Hom}(U, V) \arrow{r}{\pi \circ -} &
- \operatorname{Hom}(U, U) \arrow{r} & 0
+ \operatorname{Hom}(L, N) \arrow{r}{f \circ -} &
+ \operatorname{Hom}(L, M) \arrow{r}{g \circ -} &
+ \operatorname{Hom}(L, L) \arrow{r} & 0
\end{tikzcd}
\end{center}
- of vector spaces. Since all maps involved are intertwiners, this is an exact
- sequence of \(\mathfrak{g}\)-modules. This then induces a long exact sequence
+ 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 \arrow[r] &
- \operatorname{Hom}(U, W)^{\mathfrak{g}} \arrow[r]\ar[draw=none]{d}[name=X, anchor=center]{} &
- \operatorname{Hom}(U, V)^{\mathfrak{g}} \arrow[r, "\pi \circ -"', swap] &
- \operatorname{Hom}(U, U)^{\mathfrak{g}}
+ \operatorname{Hom}(L, N)^{\mathfrak{g}} \arrow[r, "f \circ -"', swap]\ar[draw=none]{d}[name=X, anchor=center]{} &
+ \operatorname{Hom}(L, M)^{\mathfrak{g}} \arrow[r, "g \circ -"', swap] &
+ \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}(U, W)) \arrow[r] &
- H^1(\mathfrak{g}, \operatorname{Hom}(U, V)) \arrow[r] &
- H^1(\mathfrak{g}, \operatorname{Hom}(U, U)) \arrow[r, dashed] &
+ H^1(\mathfrak{g}, \operatorname{Hom}(L, N)) \arrow[r] &
+ H^1(\mathfrak{g}, \operatorname{Hom}(L, M)) \arrow[r] &
+ H^1(\mathfrak{g}, \operatorname{Hom}(L, L)) \arrow[r, dashed] &
\cdots
\end{tikzcd}
\end{center}
- of vector spaces. But \(H^1(\mathfrak{g}, \operatorname{Hom}(U, W))\)
+ of vector spaces. But \(H^1(\mathfrak{g}, \operatorname{Hom}(L, N))\)
vanishes because of Proposition~\ref{thm:first-cohomology-vanishes}. Hence we
have an exact sequence
\begin{center}
\begin{tikzcd}
0 \arrow{r} &
- \operatorname{Hom}(U, W)^{\mathfrak{g}} \arrow{r} &
- \operatorname{Hom}(U, V)^{\mathfrak{g}} \arrow{r}{\pi \circ -} &
- \operatorname{Hom}(U, U)^{\mathfrak{g}} \arrow{r} &
+ \operatorname{Hom}(L, N)^{\mathfrak{g}} \arrow{r}{f \circ -} &
+ \operatorname{Hom}(L, M)^{\mathfrak{g}} \arrow{r}{g \circ -} &
+ \operatorname{Hom}(L, L)^{\mathfrak{g}} \arrow{r} &
0
\end{tikzcd}
\end{center}
- Now notice \(\operatorname{Hom}(U, -)^{\mathfrak{g}} =
- \operatorname{Hom}_{\mathfrak{g}}(U, -)\). Indeed, given a
- \(\mathfrak{g}\)-module \(S\) and a \(K\)-linear map \(T : U \to S\)
+ Now notice \(\operatorname{Hom}(L, L')^{\mathfrak{g}} =
+ \operatorname{Hom}_{\mathfrak{g}}(L, L')\) for all \(\mathfrak{g}\)-modules
+ \(L'\). Indeed, given
+ a \(K\)-linear map \(f : L \to L'\)
\[
\begin{split}
- T \in \operatorname{Hom}(U, S)^{\mathfrak{g}}
- & \iff X T - T X = 0 \quad \forall X \in \mathfrak{g} \\
- & \iff X T = T X \quad \forall X \in \mathfrak{g} \\
- & \iff T \in \operatorname{Hom}_{\mathfrak{g}}(U, S)
+ f \in \operatorname{Hom}(L, L')^{\mathfrak{g}}
+ & \iff X f - f X = X \cdot f = 0 \quad \forall X \in \mathfrak{g} \\
+ & \iff X f = f X \quad \forall X \in \mathfrak{g} \\
+ & \iff f \in \operatorname{Hom}_{\mathfrak{g}}(L, L')
\end{split}
\]
@@ -876,21 +892,21 @@ We are now finally ready to prove\dots
\begin{center}
\begin{tikzcd}
0 \arrow{r} &
- \operatorname{Hom}_{\mathfrak{g}}(U, W) \arrow{r} &
- \operatorname{Hom}_{\mathfrak{g}}(U, V) \arrow{r}{\pi \circ -} &
- \operatorname{Hom}_{\mathfrak{g}}(U, U) \arrow{r} &
+ \operatorname{Hom}_{\mathfrak{g}}(L, N) \arrow{r}{f \circ -} &
+ \operatorname{Hom}_{\mathfrak{g}}(L, M) \arrow{r}{g \circ -} &
+ \operatorname{Hom}_{\mathfrak{g}}(L, L) \arrow{r} &
0
\end{tikzcd}
\end{center}
- In particular, there is some intertwiner \(T : U \to V\) such that \(\pi
- \circ T : U \to U\) is the identity operator. In other words
+ 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 \arrow{r} &
- W \arrow{r} &
- V \arrow{r}{\pi} &
- U \arrow{r} \arrow[bend left]{l}{T} &
+ N \arrow{r}{f} &
+ M \arrow{r}{g} &
+ L \arrow{r} \arrow[bend left]{l}{s} &
0
\end{tikzcd}
\end{center}
@@ -899,9 +915,9 @@ We are now finally ready to prove\dots
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 \(H^i(\mathfrak{g}, V) = 0\) for all
-non-trivial finite-dimensional irreducible \(V\), \(i > 0\). For \(K =
-\mathbb{C}\), the Lie algebra cohomology groups of an algebra \(\mathfrak{g} =
+Casimir elements can be used to show that \(H^i(\mathfrak{g}, M) = 0\) for 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
@@ -921,21 +937,20 @@ sequence
\end{tikzcd}
\end{center}
-This sequence always splits, which implies we can deduce information about the
-representations of \(\mathfrak{g}\) by studying those of its ``semisimple
+This sequence always splits, which 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{theorem}\label{thm:semi-simple-part-decomposition}
- Every irreducible representation of \(\mathfrak{g}\) is the tensor product of
- an irreducible representation of its semisimple part
- \(\mfrac{\mathfrak{g}}{\mathfrak{rad}(\mathfrak{g})}\) and a
- \(1\)-dimensional representation of \(\mathfrak{g}\).
+ Every 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{theorem}
Having finally reduced our initial classification problem to that of
-classifying the finite-dimensional irreducible representations of
-\(\mathfrak{g}\), we can now focus exclusively in irreducible
-\(\mathfrak{g}\)-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.
+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.