Source code for my notes on representations of semisimple Lie algebras and Olivier Mathieu's classification of simple weight modules
Switched the notation for modules/representations in the second chapter
Also added some clarifications on g-module invariants
Also added a TODO note
Fixed some typos along the way
1 file changed, 397 insertions, 382 deletions
| Status | File Name | N° Changes | Insertions | Deletions |
| Modified | sections/complete-reducibility.tex | 779 | 397 | 382 |
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.