\documentclass{amsart} \usepackage{amssymb} \usepackage{amsmath} \newtheorem{lem}{Lemma} \newtheorem*{Theorem}{Theorem} \newtheorem*{Def}{Definition} \newtheorem*{cor}{Corollary} \newtheorem*{Prop}{Proposition} \newcommand*{\g}{\mathfrak{g}} \newcommand*{\gN}{\mathfrak{N}} \newcommand*{\rg}{\textnormal{Rad }\g} \newcommand*{\ad}{\textnormal{ad }} \newcommand{\F}{{\mathbb{F}}} \newcommand{\s}{{\mathfrak{s}}} \newenvironment*{Ex} {\noindent \textbf{Example.}\hspace{.1 em}} \newenvironment*{Problem}[1] {\vspace{1em} \noindent \textbf{Exercise #1:}\hspace{.5em}} \title{18.745: Lecture 23} \author{Professor: Victor Ka\v{c}\\ Scribes: David Meyer and Christopher Davis} \date{December 2nd, 2004} \begin{document} \maketitle From now on, let $\g$ be an arbitrary finite-dimensional Lie Algebra over an algebraically closed field $\F$ of characteristic 0. \begin{Theorem} (Weyl Complete Reducibility) If $\g$ is semisimple, then any finite-dimensional $\g$-module $V$ is completely reducible, ie: for any submodule $U$ there exists a complementary submodule $U'$ so that $V = U \oplus U'$. \end{Theorem} \begin{cor} Any finite-dimensional $\g$-module is isomorphic to a direct sum of irreducible $\g$-modules. \end{cor} \begin{Theorem} (Levi) Let $\g$ be any finite-dimensional Lie Algebra over $\F$, and $R(\g)$ the radical of $\g$. Then there exists a semisimple subalgebra $\s$ of $\g$ such that $\g = \s + R(\g)$. (As previously noted, $\s \cap R(\g) = \{0\}$). \end{Theorem} \begin{Theorem} (Mal'cev) In Levi's theorem, if $\s_1$ is another semisimple subalgebra of $\g$ such that $\g = \s_1 + R(\g)$, then there exists an automorphism $\phi$ of $\g$ such that $\phi(\s) = \s_1$. \end{Theorem} Recall that a ``projector'' of a vector space $V$ to a subspace $U \subset V$ is a linear operator $P$ on $V$ such that $P(V)=U$ and $P(u) = u$ for all $u \in U$. Note that $V = U \oplus Ker(P)$ for any projector $P$. Given a projector $P_0$ of $V$ to $U$, any other projector of $V$ to $U$ is of the form $P = P_0 + A$ where $A(V) \subset U$ and $A(U) = 0$. \begin{proof} (\textbf{Weyl's Theorem}) Consider $End\ V$ as a $\g$-module, with: $a*A = [a,A] = aA - Aa$, for $a \in \g$, $A \in End\ V$. Notational Remark: since $V$ is a $\g$-module, we are identifying elements of $\g$ with elements of $End\ V$. Hence, ``$aA$'' and ``$Aa$'' represent composition in $End\ V$, while $a*A$ represents the $\g$-module action defined above. Let $M$ be the subspace of $End\ V$ consisting of all $A$ such that $A(V) \subset U$ and $A(U)=0$. This is a submodule of $End\ V$: $$(a*A)V = (aA - Aa)V \subset a(A(V)) + A(a(V)) \subset a(U) + A(V) \subset U$$ $$(a*A)U = (aA - Aa)U \subset a(A(U)) + A(a(U)) \subset a(0) + A(U) = 0$$ Pick an arbitrary projector $P_0:V \longrightarrow U$, and define the following 1-cocycle: $$f: \g \longrightarrow M; f(a) = a*P_0 = a P_0 - P_0 a$$ It is clear $f$ maps into $M$ since $a$ and $P_0$ commute on U, and: $$(a P_0 - P_0 a)V \subset a(P_0(V)) + P_0(a(V)) \subset a(U) + P_0(V) \subset U$$ $f$ is also clearly a 1-cocycle, since: $$f([a,b]) = [a,b]*P_0 = a*(b*P_0) - b*(a*P_0) = a*f(b) - b*f(a).$$ (the second equality is just the defining property of a $\g$-module) Thus, by the main theorem of Lecture 22, $f$ is trivial: there exists $A \in M$ such that $f(a) = a*A$ for all $a \in \g$. Equivalently: $a*P_0 = a*A$, or $a*(P_0 - A) = 0$. Let $P = P_0 - A$. Observe that $P$ is just a projector of $V$ to $U$, and we have $a*P = 0$, ie, $aP = Pa$, for all $a \in \g$. Hence, $Ker(P)$ is $a$-invariant for all $a \in \g$, and thus $Ker(P)$ is a $\g$-submodule of V. Since $V = U \oplus Ker(P)$, $Ker(P)$ is a $\g$-submodule of $V$ complementary to $U$, as desired. \end{proof} \begin{proof} (\textbf{Levi's Theorem}, using induction on dim $\g$) If $Rad(\g)$ is not abelian, consider $\bar{\g} = \g/[Rad(\g), Rad(\g)]$. Since $dim(\bar{\g}) < dim(\g)$, by the inductive assumption there exists semisimple $\bar{\s} \subset \bar{\g}$ such that $\bar{\g} = \bar{\s} + Rad(\bar{\g})$. Hence $\g = \g_1 + Rad(\g)$, where $\g_1$ is the preimage of $\bar{\s}$ in $\g$, and $dim(\g_1) < dim(\g)$. Now apply the inductive assumption to $\g_1$, so that we can write $\g_1 = \s + Rad(\g_1)$, with $\s$ semisimple. Thus $\g = \s + (Rad(\g_1) + Rad(\g))$. Note that $Rad(\g_1) + Rad(\g)$ is an ideal of $\g$: Write an arbitrary element of $Rad(\g_1) + Rad(\g)$ as $a+b$, where $a \in Rad(\g_1)$ and $b \in Rad(\g)$. Write an arbitrary element of $\g$ as $c+d$, where $c \in \g_1$ and $d \in Rad(\g)$. Then \begin{align*} [a+b,c+d] &= [a, c+d] + [b, c+d] \\ &= [a, c] + [a,d] + [b, c+d], \end{align*} \noindent We have $[a,c] \in Rad(\g_1)$ because $Rad(\g_1)$ is an ideal of $\g_1$, $[a,d],[b,c+d] \in Rad(\g)$ because $Rad(\g)$ is an ideal. Furthermore, $Rad(\g_1) + Rad(\g)$ is solvable, because $(Rad(\g_1) + Rad(\g))^{(n)} \subseteq Rad(\g_1)^{(n)} + Rad(\g)$, which follows from the fact that $Rad(\g_1)$ is a sub-algebra and $Rad(\g)$ is an ideal of $\g$. Hence, if $Rad(\g_1)^{(n)} = 0$, and $Rad(\g)^{(m)} = 0$, then $(Rad(\g_1) + Rad(\g))^{(m+n)} = 0$. Hence, we have a solvable ideal of $\g$ which contains $Rad(\g)$, so we conclude that our ideal is in fact $Rad(\g)$, yielding the equality $$\g = s + Rad(\g).$$ What remains is to consider the case when $Rad(\g)$ is abelian. Consider the $\g$-module structure $End\ \g$ defined by: $$a*m = (ad\ a)m - m(ad\ a)$$ for $a \in \g$, $m \in End\ \g$. Further, consider the following submodule of $End\ \g$: $$\tilde{M} = \{ m \in End\ \g | m(\g) \subset Rad(\g) \textrm{ and } m(Rad(\g)) = 0 \}$$ Let $M = \tilde{M}/\tilde{R}$ where $\tilde{R} = \{ ad(a) | a \in Rad(\g) \}$. Note that since $Rad(\g)$ is abelian, it acts trivially on $M$. Hence, $M$ is actually an $\s$-module. (where $\s = \g/Rad(\g)$) \begin{Problem}{23.1} Show that $\tilde{M}$ is a submodule of $End(\g)$, and $\tilde{R}$ is a submodule of $\tilde{M}$. \end{Problem} \begin{proof} $\tilde{M}$ is clearly a subspace of $End(\g)$. Given $a \in \g$ and $m \in \tilde{M}$: $$(a*m)(\g) \subset (ad\ a)m(\g) + m((ad\ a) \g) \subset (ad\ a)Rad(\g) + m(\g) \subset Rad(\g)$$ $$(a*m)(Rad(\g)) \subset (ad\ a)m(Rad(\g)) + m(ad\ a)(Rad(\g)) \subset (ad\ a)0 + m(Rad(\g)) = 0$$ Hence $a*m \in \tilde{M}$. So $\tilde{M}$ is a submodule of $End(\g)$.\\ $\tilde{R}$ is also clearly a subspace of $End(\g)$. Given $ad\ a \in \tilde{R}$, with $a \in Rad(\g)$: $$(ad\ a)(\g) = [a,\g] \subset Rad(\g)$$ $$(ad\ a)(Rad(\g)) = [a, Rad(\g)] = 0$$ Hence $\tilde{R}$ is a subspace of $\tilde{M}$. To see that it is a submodule, note that for $ad\ a \in \tilde{R}$ and $b \in \g$: $$b*(ad\ a) = (ad\ b)(ad\ a) - (ad\ a)(ad\ b) = [ad\ b,ad\ a] = ad([b,a]) \in \tilde{R}$$ \end{proof} Let $P_{0}$ be a projector of $\g$ to $\rg$, and consider the following 1-cocycle: \begin{align*} f:s &\rightarrow M \\ a &\rightarrow (\ad \tilde{a})P_{0} - P_{0}(\ad \tilde{a}), \end{align*} \noindent where $\tilde{a}$ is any preimage of $a$ under the map $\g \rightarrow \g/\rg$. \begin{Problem}{23.2} The map $f$ above is a well-defined 1-cocycle. \end{Problem} \begin{proof} Let $b \in \rg$. \begin{align*} (\ad \tilde{a})P_{0} - P_{0}(\ad \tilde{a}) -((\ad \tilde{a} + \tilde{b})P_{0} -P_{0}(\ad \tilde{a} + \tilde{b})) &=\\ (\ad \tilde{b})P_{0} - P_{0}(\ad \tilde{b}) &= \ad -\tilde{b}, \end{align*} \noindent since we are assuming $\rg$ is abelian, and since $P_{0}$ acts as the identity on $\rg$. Hence, the map $f$ is well-defined, since the target of our map is $\tilde{M}/\tilde{R}$. It remains to make sure that our map is indeed a 1-cocycle. In the notation above, it's clear that $[\tilde{a},\tilde{b}]$ maps to $[a,b]$. So, on one hand, \begin{align*} f([a,b]) &= (\ad [\tilde{a}, \tilde{b}])P_{0} - P_{0}(\ad [\tilde{a}, \tilde{b}]) \\ \intertext{and on the other hand} a f(b) - b f(a) &= (\ad a)f(b) -f(b)(\ad a) - (\ad b)f(a) + f(a)(\ad b)\\ &= (\ad a)((\ad \tilde{b})P_{0} - P_{0} (\ad \tilde{b})) - ((\ad \tilde{b})P_{0} -P_{0}(\ad \tilde{b}))(\ad a)\\ & \hspace{2 em} - (\ad b)((\ad \tilde{a})P_{0} - P_{0}(\ad \tilde{a})) + ((\ad \tilde{a})P_{0} - P_{0}(\ad \tilde{a}))(\ad b)\\ &= (\ad [\tilde{a}, \tilde{b}])P_{0} - P_{0}(\ad [\tilde{a}, \tilde{b}]). \end{align*} \end{proof} By the Fundamental Theorem on Cohomology, which may be applied because $s$ is semi-simple, there exists $m \in M$ such that $a_{m} = (\ad \widetilde{a})P_{0} - P_{0}(\ad \widetilde{a})$. But, simply writing out $a_{m}$ and subtracting, this means that \begin{equation} (\ad \widetilde{a})(P_{0} - \widetilde{m}) -(P_{0} - \widetilde{m})(\ad \widetilde{a}) = \ad r_{a}, \label{eqn1} \end{equation} where $\widetilde{m}$ is a preimage of $m$ in $\widetilde{M}$, and $r_{a} \in \rg$. Consider the projector $P = P_{0} - \widetilde{m}$ of $\g$ to $\rg$. \vspace{1 em} \noindent Case 1: If all $r_{a} = 0$, then equation \ref{eqn1} above implies that $\ad \tilde{a}$ commutes with $P$ for all $a \in \g$. In this case, let $s = \textnormal{Ker } P$. Because $P$ is a projector, $s \cap \rg = 0$ and $\g = s + \rg$. So we get that $\g = s \oplus \rg$ as vector spaces. This is in fact a direct sum of ideals: that $\rg$ is an ideal is clear, and $s$ is an ideal because, by the commutativity mentioned above, for every $a \in \g, s_{1} \in s$, $$0 = [a,P(s_{1})] = P([a,s_{1}]).$$ \vspace{1 em} \noindent Case 2: Now assume that not all $r_{a} = 0$. Let $\g_{1} = \{a\in \g \mid P(\ad a) = (\ad a)P.\}$ This is a subalgebra by the first properties of the adjoint representation. It is a \textit{proper} subalgebra by our assumption, and equation \ref{eqn1} together with the facts that $(\ad r_{a})P = 0$ since $\rg$ is abelian and $P(\ad r_{a}) = \ad r_{a}$ since $P$ is a projector, implies $$P(\ad a - r_{a}) = (\ad a - r_{a})P.$$ Hence, $\g = \g_{1} + \rg$. Applying the inductive assumption, we can find a subalgebra $s$ of $\g_{1}$ such that $\g_{1} = s + \textnormal{Rad } \g_{1}$, a direct sum of vector spaces. Hence, $\g = s + \rg + \textnormal{Rad } \g_{1}$. As in the proof of Weyl's Theorem, this is equivalent to $\g = s + \rg$. \end{proof} \begin{proof} (\textbf{Mal'cev's Theorem}) Suppose the radical is abelian, and, from Levi's Theorem, we have $\g = s \oplus \rg$. Let $P_{s}$ and $P_{r}$ be the canonical projectors of $\g$ onto $s$ and $\rg$, respectively. Let $a_{r}, b_{r} \in \rg$, and $a_{s}, b_{s} \in s$. Say $a = a_{r} + a_{s}$, $b = b_{r} + b_{s}$. Such decompositions exist and are unique, because of our decomposition of $\g$. Because $\rg$ is an ideal, $P_{s}([a,b]) = [s_{1},s_{2}]$, and thus $P_{s}$ is a Lie algebra homomorphism. Furthermore, by again considering the decomposition $a = a_{r} + a_{s}$, it's clear that $P_{r} + P_{s} = 1$. Let $s_{1} \oplus \rg = \g$ be another Levi decomposition. By the adjoint representation, since it is an ideal, $\rg$ is an $s_{1}$ module. Define the 1-cocycle $f: s_{1} \rightarrow \rg$ by the formula $f(a) = P_{r}(a)$. \begin{Problem}{23.3} Check that $f$ is indeed a 1-cocycle. \end{Problem} \begin{proof} Write decompositions of two elements $a, b \in s_{1}$ as above (in particular, $a_{s} \in s$, not $s_{1})$. Then \begin{align*} P_{r}([a_{r}+a_{s},b_{r}+b_{s}]) &= [a_{r},b_{r}]+[a_{s},b_{r}] + [a_{r},b_{s}] \\ &= [a_{s},b_{r}] + [a_{r},b_{s}] \end{align*} \noindent since we are assuming $\rg$ to be abelian. This clearly equals $$[a_{r} + a_{s}, P_{r}(b_{r} + b_{s})] - [b_{r}+b_{s}, P_{r}(a_{r}+a_{s})],$$ and so $f$ is a 1-cocycle. \end{proof} By the Fundamental Theorem, this cocycle is trivial. So, there exists $r \in \rg$ such that $P_{r}(a) = [a,r]$, for any $a \in s_{1}$. So we have that $P_{r} = -\ad r$ on $s_{1}$. Since $\rg$ is abelian, $(\ad r)^{2} = 0$. Hence $\exp{(\ad r)} = 1 + (\ad r)$, and, from Exercise 9.2, that must be an automorphism of $\g$. Call it $\sigma$. Let $a \in s_{1}$. We have $$\sigma(a) = (1 + \ad r)a = (1 - P_{r})a = P_{s}(a) \in s.$$ Hence, we have an automorphism $\sigma$ of $\g$ such that $\sigma(s_{1}) \subseteq s$. An automorphism is, of course, injective, and by the vector space decompositions $\g = s \oplus \rg$ and $\g = s_{1} \oplus \rg$, we have that the dimensions of $s$ and $s_{1}$ are equal. Hence, $\sigma(s_{1}) = s$. \begin{Problem}{23.4} As in the proof of Levi's Theorem, reduce Mal'cev's Theorem in the case of non-abelian radical $\rg$ to the case where $\rg$ is abelian. \end{Problem} \begin{proof} We take a slightly different route, following Jacobson, \cite{Ja62}. Let $\mathfrak{N}$ denote the sum of all the nilpotent ideals of $\g$. That $\gN$ is a nilpotent ideal is obvious. We will prove inductively that for every $i \geq 1$, there is an automorphism $A_{i}$ of $\g$ such that $A_{i}(s_{1}) \subseteq s + \gN^{(i)}$. Considering dimensions and the fact that $\gN$ is solvable, we will be able to conclude Mal'cev's Theorem. Considering the equation $$P_{r}([a_{r}+a_{s},b_{r}+b_{s}]) = [a_{r},b_{r}]+[a_{s},b_{r}] + [a_{r},b_{s}],$$ we see that $P_{r}([a,b]) \in [\g, \rg]$ for each $a,b \in s_{1}$. \begin{lem} With notation as above, we have that $[\g, \rg] \subseteq \gN$. \end{lem} \begin{proof} We know from lecture five that $[\rg, \rg] \subseteq \gN$. By Engel's Theorem, this is equivalent to $\ad [a_{r}, b_{r}]$ being nilpotent for $a_{r},b_{r} \in \rg$. Using the Jacobi identity, and the fact that $\rg$ is an ideal, we get that $\ad [a, b_{r}]$ is nilpotent for any $a \in \g$. \end{proof} \begin{lem} As $s_{1}$ is semi-simple, we have $[s_{1},s_{1}] = s_{1}$. \end{lem} \begin{proof} This follows from the structure theorem: the result holds for a simple Lie algebra $t$ because it contain no non-trivial proper ideals, and thus $[t,t]$ of such an ideal is either $0$ or the whole simple algebra, and as $t$ is not abelian, it must be the latter. The result continues to hold under passage to direct sums. \end{proof} Now, consider any $a \in s_{1}$. We have just seen that we may write $a = [a_{1},a_{2}]$, for some $a_{1}, a_{2} \in s_{1}$. Hence, from what we proved above, $P_{r}(a) \in \gN$. We know that $P_{r} + P_{s} = 1$, so we have that $s_{1} \subseteq s + \gN^{(1)}$. This provides the first step in our induction, with the identity automorphism playing the role of $A_{1}$. Simplify notation by assuming that $s_{1} \subseteq s + \gN^{(k)}$ (i.e., if necessary, replace $s_{1}$ with the isomorphic Lie Algebra $A_{k}(s_{1})$). Because $s$ is semi-simple, $s \cap \rg = \emptyset$, and hence $P_{r}(s_{1}) \subseteq \gN^{(k)}$. We can make $\gN^{(k)}$ into a $s_{1}$-module by defining $a\cdot z= [P_{s}(a),z]$. We check that this is indeed a module: \begin{align*} [a,b] \cdot z & = [P_{s}([a,b]),z] \\ & = [[P_{s}(a),P_{s}(b)],z] \\ & = -[[P_{s}(b),z],P_{s}(a)] - [[z,P_{s}(a)],P_{s}(b)] \\ & = [P_{s}(a),[P_{s}(b),z]] + [P_{s}(b),[z,P_{s}(a)]] \\ & = [P_{s}(a),[P_{s}(b),z]] - [P_{s}(b),[P_{s}(a),z]] \\ & = a \cdot (b \cdot z) - b \cdot (a \cdot z), \end{align*} \noindent as required. As $\gN^{(k+1)}$ is an ideal, we can consider also $\gN^{(k)}/\gN^{(k+1)}$ as an $s_{1}$-module, with action $a \cdot \overline{z} = \overline{[P_{s}(a),z]}$. From what we saw two paragraphs ago, $[P_{r}(a),P_{r}(b)] \in \gN^{(k+1)}$, for $a,b \in s_{1}$. So, $$\overline{P_{r}([a,b])} = \overline{[a_{s},b_{r}]} - \overline{[b_{s},a_{r}]} = a \cdot \overline{b_{r}} - b \cdot \overline{a_{r}}.$$ Define the map $f: a \rightarrow \overline{P_{r}(a)}$. The map is linear because the projector is linear, and the above equation says precisely that $f$ is a one-cocycle. Hence, by the fundamental theorem on cohomology, there is some element $\overline{z} \in \gN^{(k)}/\gN^{(k+1)}$ such that, for any $a \in s_{1}$, we have $\overline{P_{r}(a)} = a \cdot \overline{z} = \overline{[P_{r}(a),z]}$, where $z$ is any lift of $\overline{z}$. Let $A = \exp(\ad z)$. Again, from exercise 9.2, this is an automorphism. We have \begin{align*} A(a) & = a + [z,a] + \frac{1}{2!}[z,[z,a]] + \cdots \\ & \equiv a + [z,a] \bmod \gN^{(k+1)} \\ & \equiv a_{r} + a_{s} + [z,a_{r}] + [z,a_{s}] \bmod \gN^{(k+1)} \\ \intertext{and we know $a_{s} \in \gN^{(k)}$ and $\overline{a_{r}} = \overline{[a_{r},z]}$, so} & \equiv a_{s} \bmod \gN^{(k+1)}. \end{align*} \noindent Thus, we have that $A(s_{1}) \subseteq s + \gN^{(k+1)}$, and we proceed by induction. \end{proof} \noindent As already mentioned, the proof is now complete, when we consider that the dimensions of $s$ and $s_{1}$ must be the same. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{thebibliography}{99} \newcommand{\au}[1]{{#1},} \newcommand{\rawti}[1]{\textit{#1}} \newcommand{\ti}[1]{\rawti{#1},} \newcommand{\jo}[1]{{#1}} \newcommand{\vo}[1]{\textbf{#1}} \newcommand{\yr}[1]{(#1),} \newcommand{\pp}[1]{#1.} \newcommand{\ppc}[1]{#1,} \newcommand{\pps}[1]{#1;} \newcommand{\bk}[1]{{#1},} \newcommand{\inbk}[1]{in{#1}} \newcommand{\plain}[1]{#1} \newcommand{\xxx}[1]{{arXiv:#1}} \bibitem[Ja62]{Ja62} \au{N. Jacobson} \ti{Lie Algebras} \pp{Interscience Publishers, 1962} \end{thebibliography} %%%%%%%%%%%%%%%%%%%%%% \end{document}