From 5407edfd93c4c8caaf94783dfa6df53c3db41959 Mon Sep 17 00:00:00 2001 From: Jendrik Stelzner Date: Mon, 22 Apr 2019 23:18:16 +0200 Subject: [PATCH] Added the root space decomposition. --- generalstyle.sty | 3 + sections/abstract_jordan_decomposition.tex | 34 + sections/cartan_subalgebras.tex | 382 --------- .../definition_of_semisimple_lie_algebras.tex | 1 + .../killing_form_and_cartans_criterion.tex | 22 + sections/root_space_decomposition.tex | 740 ++++++++++++++++++ sections/root_systems.tex | 3 - sections/roots_and_reflections.tex | 8 + sections/semisimple_lie_algebras.tex | 5 +- sections/sl2_theory.tex | 5 +- 10 files changed, 816 insertions(+), 387 deletions(-) delete mode 100644 sections/cartan_subalgebras.tex create mode 100644 sections/root_space_decomposition.tex delete mode 100644 sections/root_systems.tex create mode 100644 sections/roots_and_reflections.tex diff --git a/generalstyle.sty b/generalstyle.sty index a26cf97..39ddbbe 100644 --- a/generalstyle.sty +++ b/generalstyle.sty @@ -221,6 +221,7 @@ \newcommand{\hlie}{\mathfrak{h}} \newcommand{\klie}{\mathfrak{k}} \newcommand{\gllie}{\mathfrak{gl}} +\newcommand{\rlie}{\mathfrak{r}} \newcommand{\sllie}{\mathfrak{sl}} \newcommand{\tlie}{\mathfrak{t}} \newcommand{\nlie}{\mathfrak{n}} @@ -289,6 +290,8 @@ % delimiters \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} +\DeclarePairedDelimiter{\bil}{\langle}{\rangle} +\DeclarePairedDelimiter{\norm}{\lVert}{\rVert} \DeclarePairedDelimiter{\pair}{\langle}{\rangle} \NewDocumentCommand{\restrict}{smmO{}}{ \IfBooleanTF{#1} diff --git a/sections/abstract_jordan_decomposition.tex b/sections/abstract_jordan_decomposition.tex index ea24153..64ff8ad 100644 --- a/sections/abstract_jordan_decomposition.tex +++ b/sections/abstract_jordan_decomposition.tex @@ -212,6 +212,40 @@ \section{The Abstract Jordan Decomposition} \end{example} +\begin{lemma} + \label{commuting via abstract jd} + Let~$\glie$ be a finite dimensional semisimple Lie~algebra and let~$x \in \glie$. + Then any other element~$y \in \glie$ commutes with~$x$ if and only if both~$y_s$ and~$y_n$ commute with~$x$. +\end{lemma} + + +\begin{proof} + We may assume that~$\glie$ is linear by using the adjoint representation, which is faithful. + Then the assertion follows from \cref{concrete jordan decomposition} because the concrete and abstract Jordan decompositions coincide. +\end{proof} + + +\begin{lemma} + \label{abstract jordan decomposition of sum} + Let~$x$ and~$y$ be two commuting elements of a finite dimensional semisimple Lie~algebra~$\glie$. + Then the Jordan decomposition of~$x+y$ can be computed from the Jordan decompositions of~$x$ and~$y$ via + \[ + (x + y)_s + = + x_s + y_s + \quad\text{and}\quad + (x + y)_n + = + x_n + y_n \,. + \] +\end{lemma} + + +\begin{proof} + We may assume that~$\glie$ is linear and then apply \cref{concrete jordan decomposition of sum}. +\end{proof} + + % % \begin{proof}[Proof of \cref{abstract jordan decomposition}] % \leavevmode diff --git a/sections/cartan_subalgebras.tex b/sections/cartan_subalgebras.tex deleted file mode 100644 index c03e025..0000000 --- a/sections/cartan_subalgebras.tex +++ /dev/null @@ -1,382 +0,0 @@ -\section{Cartan subalgebras} -Throughout this section $\g$ denotes the finite dimensional semisimple Lie algebra. - - - - - -\subsection{Definition and root space decomposition} - - -\begin{definition} - A subalgebra $\h \subseteq \g$ is \emph{toral} if it consists of semisimple elements. -\end{definition} - - -\begin{lemma} - Let $\h \subseteq \g$ be a toral subalgebra. Then $\h$ is abelian. -\end{lemma} -\begin{proof} - Let $x \in \h$. Because $x$ is $\ad_\g$-semisimple and $\h$ is $\ad_\g(x)$-invariant it follows that $\ad_\h(x) = \ad_\g(x)|_\h$ is also semisimple. It is enough to show that all eigenvalues of $\ad_\h(x)$ are zero. - - Let $y \in \h$ be an eigenvector of $\ad_\h(x)$ with eigenvalue $\mu$ (in particular $y \neq 0$). In the same way as for $\ad_\h(x)$ it follows that $\ad_\h(y)$ is semisimple. Fer every $\lambda \in k$ let $\h_\lambda$ denote the eigenpsace of $\ad_\h(y)$ with respect to the eigenvalue $\lambda$. - - On the one hand $[y,x] \in \h_0$ because - \[ - \ad_\h(y)([y,x]) - = [y,[y,x]] - = [y, -\mu y] - = 0. - \] - On the other hand - \[ - [y,x] - = \ad_\h(y)(x) - \in \ad_\h(y)(\h) - = \ad_\h(y)\left( \bigoplus_{\lambda \in k} \h_\lambda \right) - = \bigoplus_{\lambda \neq 0} \h_\lambda. - \] - By the directness of the sum $\h = \bigoplus_{\lambda \in k} \h_\lambda$ it follows that $0 = [y,x] = -\mu y$. Because $y \neq 0$ it follows that $\mu = 0$. -\end{proof} - - -\begin{definition} - A \emph{Cartan subalgebra} of $\g$ is a maximal toral subalgebra. -\end{definition} - - -\begin{remark} - The toral subalgebra $0 \subseteq \g$ is contained in a toral subalgebra (of $\g$) of maximal dimensional, which is then a Cartan subalgebra. Therefore $\g$ contains a Cartan subalgebra. - - Also notice that if $\g \neq 0$ then any Cartan subalgebra of $\g$ is non-zero. Too see this first notice thet $\g$ then contains a non-zero semisimple element $x \in \g$, because otherwise $\g$ would be nilpotent by Engel’s Theorem. Then the one-dimensional linear subspace $kx$ is a toral subalgebra properly containing $0$. -\end{remark} - - -% TODO: Remark about usual definition. - - -\begin{definition} - Let $\h$ be a Cartan subalgebra of $\g$. For any $\alpha \in \h^*$ let - \[ - \g_\alpha \coloneqq \{y \in \g \mid \text{$[x,y] = \alpha(x)y$ for every $x \in \h$}\} - \] - be the \emph{root space} of $\g$ with respect to $\alpha$. Then $\Phi(\g, \h) \coloneqq \{\alpha \in \h^* \setminus \{0\} \mid \g_\alpha \neq 0\}$ is the set of \emph{roots} of $\g$ with respect to $\h$. -\end{definition} - - -\begin{remark} - Notice that $\g_0 = \{y \in \g \mid \text{$[x,y] = 0$ for every $x \in \h$}\} = Z_\g(\h)$. -\end{remark} - - - -\begin{lemma} - Let $\h$ be a Cartan subalgebra of $\g$ with roots $\Phi \coloneqq \Phi(\g, \h)$. Then $\g = Z_\g(\h) \oplus \bigoplus_{\alpha \in \Phi} \g_\alpha$. -\end{lemma} -\begin{proof} - Because $\h$ is abelian the same goes for $\ad_\g(\h) \subseteq \gl(V)$. Therefore $\ad_\g(\h)$ consists of semisimple pairwise commuting endomorphisms. Hence $\ad_\g(\h)$ is simultaneously diagonalizable, which is why - \[ - \g - = \bigoplus_{\lambda \in \h^*} \g_\lambda - = \g_0 \oplus \bigoplus_{\alpha \in \Phi} \g_\alpha - = Z_\g(\h) \oplus \bigoplus_{\alpha \in \Phi} \g_\alpha. - \qedhere - \] -\end{proof} - - -% TODO: Beispiele hinzufügen - - -\begin{lemma} - Let $\h$ be a Cartan subalgebra of $\g$. Then $[\g_\alpha, \g_\beta] \subseteq \g_{\alpha+\beta}$ for all $\alpha, \beta \in \h^*$. -\end{lemma} -\begin{proof} - Let $x \in \g_\alpha$ and $y \in \g_\beta$. Then for every $h \in \h$ - \[ - [h,[x,y]] - = [[h,x],y] + [x,[h,y]] - = \alpha(h)[x,y] + \beta(h)[x,y] - = (\alpha+\beta)(h) [x,y]. - \qedhere - \] -\end{proof} - - -\begin{lemma}\label{lem: root spaces orthogonal with respect to the Killing form} -Let $\h \subseteq \g$ be a Cartan subalgebra and $\alpha, \beta \in \h^*$. If $\alpha \neq -\beta$ then $\g_\alpha$ and $\g_\beta$ are orthogonal with respect to the Killing form. -\end{lemma} -\begin{proof} - Because $\alpha \neq -\beta$ it follows that there exists $h \in \h$ with $(\alpha+\beta)(h) \neq 0$. For every $x \in \g_\alpha$ and $y \in \g_\beta$ it then follows that - \[ - \alpha(h) \kappa(x,y) - = \kappa([h,x],y) - = -\kappa([x,h],y) - = -\kappa(x,[h,y]) - = -\beta(h)\kappa(x,y) - \] - and therefore $(\alpha+\beta)(h)\kappa(x,y) = 0$. Because $(\alpha+\beta)(h) \neq 0$ it follows that $\kappa(x,y) = 0$ for every $x \in \g_\alpha$ and $y \in \g_\beta$. -\end{proof} - - -\begin{corollary}\label{cor: restriction of killing form to centralizer is non-degenerate} - Let $\h \subseteq \g$ be a Cartan subalgebra. Then $\kappa_\g|_{Z_\g(\h) \times Z_\g(\h)}$ is non-degenerate. -\end{corollary} -\begin{proof} - Because $\g$ is semisimple $\kappa_\g$ is non-degenerate. Because $Z_\g(\h) = \g_0$ is orthogonal to $\g_\alpha$ for every $\alpha \in \Phi$ with $\alpha \neq 0$ the statement follows from $\g = Z_\g(\h) \oplus \bigoplus_{\alpha \in \Phi} \g_\alpha$. -\end{proof} - - -% TODO: Add warning about restriction of Killing form - - -\begin{proposition}\label{prop: CSA are self-centralizing} - Let $\h$ be a Cartan subalgebra of $\g$. Then $Z_\g(\h) = \h$, i.e.\ $\h$ is self-centralizing. -\end{proposition} -\begin{proof} - Throughout this proof abbreviate $\cl \coloneqq Z_\g(\h)$, $\ad \coloneqq \ad_\g$ and $\kappa \coloneqq \kappa_\g$. - - \begin{claim*}\label{claim: technical and traceless} - Let $x \in \g$ be nilpotent and $y \in \g$ commuting with $x$. Then $\kappa(x,y) = 0$. - \end{claim*} - \begin{proof} - Because $x$ and $y$ commute so do $\ad(x)$ and $\ad(y)$. Because $\ad(x)$ is nilpotent it follows that $\ad(x)\ad(y)$ is also nilpotent. Therefore $\kappa(x,y) = \tr(\ad(x)\ad(y)) = 0$. - \end{proof} - - Start by noticing that $\cl$ contains the semisimple and nilpotent parts of all its elements: If $x \in \cl$ then $y \in \g$ commutes with $x$ if and only if it commutes with both $y_s$ and $y_n$. Therefore $y \in \cl$ if and only if $y_s, y_n \in \cl$. - - Let $s \in \cl$ be semisimple. Because $s$ is semisimple and commutes with $\h$ it then follows that $\h + ks$ is a Lie subalgebra of $\g$ consisting of semisimple elements. By the maximality of $\h$ it follows that $s \in \h$. - - Let $x \in \cl$ and let $x = x_s + x_n$ be the Jordan decomposition of $x$. It was already shown above that $x_s \in \h$, so $\ad_\cl x_s = 0$. Hence $\ad_\cl x = \ad_\cl x_n = \ad_\g x_n|_\cl$ is nilpotent. By Engel’s Theorem $\cl$ is nilpotent. - - Notice that $\kappa|_{\h \times \h}$ is non-degenerate. To see this let $x \in \h$ with $\kappa(x, \h) = 0$. It needs to be shown that $x = 0$, and for this it sufficies to show that $\kappa(x, \cl) = 0$ by Corollary~\ref{cor: restriction of killing form to centralizer is non-degenerate}. Because $\cl$ contains the semisimple and nilpotent parts of all its elements is sufffices to show that $\kappa(x,s) = \kappa(x,n) = 0$ for every semisimple $s \in \cl$ and nilpotent $n \in \cl$. It was already shown that $s \in \h$ so $\kappa(x,s) = 0$ by assumption. That $\kappa(x,n) = 0$ follows from the claim above. - - It follows that $\h \cap [\cl,\cl] = 0$ because $[\h,\cl] = 0$ and thus $\kappa(\h, [\cl,\cl]) = \kappa([\h, \cl], \cl) = 0$. It further follows that $\cl$ is abelian. Otherwise $\cl$ is nilpotent with $[\cl, \cl] \neq 0$. Then $Z(\cl) \cap [\cl,\cl] \neq 0$ by Corollary~\ref{cor: ideal in nilpotent has nontrivial intersection with center}. Let $x \in Z(\cl) \cap [\cl,\cl]$ be non-zero. Notice that $x$ cannot be semisimple because $\h \cap [\cl,\cl] = 0$. So $x_n$ is nonzero and it was already shown that $x_n \in \cl$. Because $x \in Z(\cl)$ it follows that $x_n \in Z(\cl)$. From the claim above it follows that $\kappa(x_n, \cl) = 0$. Because $\kappa_{\cl \times \cl}$ is non-degenerate it follows that $x_n = 0$, contradicting that $x_n$ is non-zero. - - Suppose now that $\h \subsetneq \cl$. Let $x \in \cl$ with $\h \neq 0$. Because $x_s, x_n \in \cl$ with $x_s \in \h$ it can be assumed w.l.o.g.\ that $x$ is nilpotent by replacing it with $x_n$. Then $\kappa(x_n, \cl) = 0$ by the above claim, contradicting $\kappa_{\cl \times \cl}$ being non-degenerate. -\end{proof} - - -\begin{corollary} - Let $\h \subseteq \g$ be a Cartan subalgebra. Then the restriction $\kappa_\g|_{\h \times \h}$ is non-degenerate and $\g = \h \oplus \bigoplus_{\alpha \in \Phi(\g,\h)} \g_\alpha$. -\end{corollary} - - -\begin{corollary}\label{cor: g alpha and g -alpha pair non degenerate with the Killing form} - Let $\h \subseteq \g$ be a Cartan subalgebra and $\alpha \in \Phi(\g,\h)$. Then $\kappa(\g_\alpha, \g_{-\alpha}) \neq 0$, i.e.\ there exists $x \in \g_\alpha$ and $y \in \g_{-\alpha}$ with $\kappa(x,y) \neq 0$. -\end{corollary} -\begin{proof} - Because $\alpha \in \Phi(\g,\h)$ it follows that $\g_\alpha \neq 0$. Because with respect to the Killing form $\g_\alpha$ is orthogonal to $\g_\beta$ for $\beta \neq -\alpha$ and $\h = \g_0$ by Lemma~\ref{lem: root spaces orthogonal with respect to the Killing form} and $\kappa|_{\h \times \h}$ is non-degenerate is follows that $0 \neq \kappa(\g_\alpha, \g) = \kappa(\g_\alpha, \g_{-\alpha})$. -\end{proof} - - -\begin{remark} - The proof of Proposition~\ref{prop: CSA are self-centralizing} is taken from \cite[\S 8.2]{Humphreys}. -\end{remark} - - -\begin{definition} - Let $\h \subseteq \g$ be a Cartan subalgebra. Then the decomposition - \[ - \g = \h \oplus \bigoplus_{\alpha \in \Phi} \g_\alpha - \quad\text{with}\quad - \Phi \coloneqq \Phi(\g,\h) - \] - is the \emph{root space decomposition} of $\g$ with respect to $\h$. -\end{definition} - - -\subsection{Properties of the root space decomposition} -Troughout this subsection let $\h \subseteq \g$ be a Cartan subalgebra and $\Phi \coloneqq \Phi(\g,\h)$ the associated roots. - - -\begin{definition} - Becaus $\kappa|_{\h \times \h}$ is non-degenerate the map - \[ - \h \to \h^*, \quad h \mapsto \kappa(h, \cdot) - \] - is an isomorphism of vector spaces. For every $\phi \in \h^*$ let $t_\phi \in \h$ be the unique element with $\kappa(t_\phi, \cdot) = \phi$. -\end{definition} - - -\begin{proposition} - \begin{enumerate}[leftmargin=*] - \item - $\Phi$ generates $\h^*$ as a vector space. - \item - If $\alpha \in \Phi$ then $-\alpha \in \Phi$. - \item - $[x,y] = \kappa(x,y) t_\alpha$ for every $\alpha \in \Phi$, $x \in \g_\alpha$ and $y \in \g_{-\alpha}$. - \item - Lef $\alpha \in \Phi$. Then $[\g_\alpha, \g_{-\alpha}]$ is one-dimensional and has $t_\alpha$ as basis. - \item - If $\alpha \in \Phi$ then $\alpha(t_\alpha) = \kappa(t_\alpha, t_\alpha) \neq 0$. - \end{enumerate} -\end{proposition} -\begin{proof} - \begin{enumerate}[leftmargin=*] - \item - Suppose $\Phi$ does not generates $\h^*$ as a vector space. Then there exists some $\phi \in \h^*$ with $\phi \notin \vspan_k \Phi$. Then there exists some $y \in \h^{**}$ with $y|_{\vspan_k \Phi} = 0$ but $y(\phi) \neq 0$. By the natural isomorphism $\h \xrightarrow{\sim} \h^{**}, x \mapsto (\psi \mapsto \psi(x))$ there exists some $x \in \h$ with $y(\psi) = \psi(x)$ for every $\psi \in \h^*$. In particular $\alpha(x) = 0$ for every $\alpha \in \h^*$ but $\phi(x) \neq 0$ and thus $x \neq 0$. Using the root space decomposition $\g = \h \oplus \bigoplus_{\alpha \in \Phi} \g_\alpha$ it follows that $x \in Z(\g)$. Because $Z(\g) = 0$ this contradicts $x$ being non-zero. - - \item - Let $\alpha \in \Phi$. Then $\kappa(\g_\alpha, \g_{-\alpha}) \neq 0$ by Corollary~\ref{cor: g alpha and g -alpha pair non degenerate with the Killing form}, from which it follows that $\g_{-\alpha} \neq 0$ and therefore $-\alpha \in \Phi$. - - \item - Let $h \in \h$. Then - \begin{align*} - \kappa(h, [x,y]) - &= \kappa([h,x], y) - = \alpha(h) \kappa(x,y) \\ - &= \kappa(t_\alpha, h) \kappa(x,y) - = \kappa(\kappa(x,y) t_\alpha, h) - = \kappa(h, \kappa(x,y) t_\alpha). - \end{align*} - Becasue $\kappa_{\h \times \h}$ is non-degenerate it follows that $[x,y] = \kappa(x,y) t_\alpha$. - - \item - Let $x \in \g_\alpha$ and $y \in \g_{-\alpha}$ with $\kappa(x,y) \neq 0$ (such elements exist by Corollary~\ref{cor: g alpha and g -alpha pair non degenerate with the Killing form}). Then $[x,y] = \kappa(x,y) t_\alpha \neq 0$ and therefore $[\g_\alpha, \g_{-\alpha}] = \kappa(\g_\alpha, \g_{-\alpha}) t_\alpha = k t_\alpha$. - - \item - Suppose that there exist some $\alpha \in \Phi$ with $\kappa(t_\alpha, t_\alpha) = 0$. Because $\alpha = \kappa(t_\alpha, \cdot)$ it follows that $\alpha(t_\alpha) = \kappa(t_\alpha, t_\alpha) = 0$. Let $x \in \g_\alpha$ and $y \in \g_{-\alpha}$ with $\kappa(x,y) \neq 0$. By rescaling it can be additionally assumed that $\kappa(x,y) = 1$ and thus $[x,y] = \kappa(x,y) t_\alpha = t_\alpha$. Then $L \coloneqq \vspan_k(x,t_\alpha,y)$ is a three-dimensional solvable Lie subalgebra of $\g$ because $[t_\alpha, x] = \alpha(t_\alpha)x = 0$ and $[t_\alpha, y] = -\alpha(t_\alpha)y = 0$. - - It follows that $\ad(L)$ is a three-dimensional solvable Lie subalgebra of $\gl(\g)$ and thus $[\ad(L), \ad(L)] = \ad([L,L]) = k \ad(t_\alpha)$ consists of nilpotent endomorphisms of $\g$. (To see this notice that by Lie’s theorem there exists a basis of $\g$ with respect to which $\ad(L)$ is represented by upper triangular matrices. Then with respect to this basis $[\ad(L), \ad(L)]$ is represented by strictly upper triangular matrices.) In particular $\ad(t_\alpha)$ is nilpotent. On the other hand $t_\alpha$ is semisimple (because $t_\alpha \in \h$ and $\h$ consists of semisimple elements by the definition of a toral subalgebra) and thus $\ad(t_\alpha)$ is semisimple. Because $\ad(t_\alpha)$ is both nilpotent is semisimple it follows that $\ad(t_\alpha) = 0$ and thus $t_\alpha = 0$, contradicting $\kappa(t_\alpha, \cdot) = \alpha \neq 0$. - \qedhere - \end{enumerate} -\end{proof} - - -\begin{definition} - Let $\alpha \in \Phi$ be a root. Then the associated \emph{coroot} is the unique element $\alpha^\vee \in [\g_\alpha, \g_{-\alpha}]$ with $\alpha(\alpha^\vee) = 2$. -\end{definition} - - -\begin{remark} - Let $\alpha \in \Phi$. The existence and uniqueness of $\alpha^\vee$ follows from the fact that $[\g_\alpha, \g_{-\alpha}]$ is one-dimensional with $\alpha([\g_\alpha, \g_{-\alpha}]) = k \alpha(t_\alpha) \neq 0$. Notice that $[\g_\alpha, \g_{-\alpha}] = k\alpha^\vee$ and that $(-\alpha)^\vee = -\alpha^\vee$. Also notice that $\alpha^\vee = 2 t_\alpha / \kappa(t_\alpha, t_\alpha)$ because - \[ - \alpha\left( \frac{2 t_\alpha}{\kappa(t_\alpha, t_\alpha)} \right) - = \frac{2\alpha(t_\alpha)}{\kappa(t_\alpha, t_\alpha)} - = \frac{2\kappa(t_\alpha,t_\alpha)}{\kappa(t_\alpha, t_\alpha)} - = 2. - \] -\end{remark} - - -\begin{remark}\label{rem: construction of S alpha} - Let $\alpha \in \Phi$. Let $S^\alpha \coloneqq \g_\alpha \oplus k\alpha^\vee \oplus \g_{-\alpha}$. Because $[\g_\alpha, \g_{-\alpha}] = k\alpha^\vee$ there exist $\hat{e} \in \g_\alpha$ and $\hat{f} \in \g_{-\alpha}$ with $[\hat{e},\hat{f}] = \alpha^\vee$. Because $\alpha^\vee \neq 0$ it follows that $\hat{e}, \hat{f} \neq 0$. Also notice that $[\alpha^\vee, \hat{e}] = \alpha(\alpha^\vee) \hat{e} = 2\hat{e}$ and similary $[\alpha^\vee, \hat{f}] = -2\hat{f}$. It follows that $S^\alpha$ is a Lie subalgebra of $\g$ and - \[ - \phi \colon \sll_2(k) \to S, \quad - e \mapsto \hat{e}, \quad - h \mapsto \alpha^\vee, \quad - f \mapsto \hat{f} - \] - an isomorphism of Lie algebras. Therefore $\g$ becomes a representation of $\sll_2(k)$ via $x.y = [\phi(x), y]$ for every $x \in \sll_2(k)$ and $y \in \g$. -\end{remark} - - -\begin{definition} - Let $\alpha \in \Phi$. Then $S^\alpha \subseteq \g$ is the Lie subalgebra construtcted in Remark~\ref{rem: construction of S alpha}. -\end{definition} - - -\begin{proposition} - Let $\alpha \in \Phi$ be a root. - \begin{enumerate}[leftmargin=*] - \item - The root space $\g_\alpha$ is one-dimensional. - \item - The only multiples of $\alpha$ which are roots are $\alpha$ and $-\alpha$, i.e.\ $k\alpha \cap \Phi = \{\alpha, -\alpha\}$. - \end{enumerate} -\end{proposition} -\begin{proof} - Let $S^\alpha$ acts on $\g$ as in Remark~\ref{rem: construction of S alpha}. Notice that - \[ - L \coloneqq k \alpha^\vee \oplus \bigoplus_{\substack{c \in k \\ c\alpha \in \Phi}} \g_{c\alpha} - \] - is a subrepresentation of $S^\alpha$. - - By Weyl‘s theorem $L$ is completely reducible. Notice that $h.\alpha^\vee = [\alpha^\vee, \alpha^\vee] = 0$ and - \[ - h.x = [\alpha^\vee, x] = c\alpha(\alpha^\vee)(x) = 2cx - \quad \text{for every $x \in \g_{c\alpha}$}. - \] - Because all weights of $L$ are integral it follows that $2c \in \Z$ for every $c \in k$ with $c\alpha \in \Phi$ and thus $c \in \frac{1}{2}\Z$. From the classification of the finite dimensional representations of $\sll_2(k)$ it follows that $L$ is the direct sum of irreducible submodules, of which there are two kinds: Those with even weights and those with odd weights; the number of summands of the first kind is given by $\dim L_0$ and the number of summands of the second kind is given by $\dim L_1$. - - Now $\dim L_0 = \dim k\alpha^\vee = 1$, therefore a decomposition of $L$ into irreducible subrepresentations has exactly one summand with even weights. As $S \subseteq L$ is a subrepresentation with even even weights, namely $S^\alpha_{-2} = k\hat{f}$, $S^\alpha_0 = k\alpha^\vee$ and $S^\alpha_2 = k\hat{e}$, we have found this summand. In particular $2\alpha$ is not a root, resulting in the following: - - \begin{claim*} - Let $\beta \in \Phi$ be a root. Then $2\beta$ is not it root, i.e.\ twice a root is never a root. Equivalently half a root is never a root. - \end{claim*} - - From this claim it follows that $\alpha/2$ is not root, so $L_1 = \g_{\alpha/2} = 0$. Hence $L$ contains no summand with odd weights and therefore already $L = S^\alpha$. As a direct consequence $\dim \g_\alpha = \dim L_2 = \dim S^\alpha_2 = 1$ and $k\alpha \cap \Phi = \{c \in k \setminus \{0\} \mid L_{2c} \neq 0\} = \{1,-1\}$. -\end{proof} - - -\begin{definition} - For every $\lambda \in \h^*$ and $h \in \h$ set $\pair{h, \lambda} \coloneqq \pair{\lambda, h} \coloneqq \lambda(h)$. -\end{definition} - - -\begin{proposition}\label{prop: integral and reflection properties of root pairing} - Let $\alpha, \beta \in \Phi$. Then $\pair{\alpha, \beta^\vee} \in \Z$ and $\alpha - \pair{\alpha, \beta^\vee} \beta \in \Phi$. -\end{proposition} -\begin{proof} - If $\alpha$ and $\beta$ are linearly dependent then $\beta = \pm \alpha$. Then $\beta^\vee = \pm \alpha^\vee$ and therefore $\pair{\alpha, \beta^\vee} = \pm \pair{\alpha, \alpha^\vee} = \pm 2 \in \Z$ and - \[ - \alpha - \pair{\alpha, \beta^\vee} \beta - = \alpha - 2 \alpha - = -\alpha \in \Phi. - \] - - Suppose that $\alpha$ and $\beta$ are linearly independent. Let $S^\beta$ act on $\g$ as in Remark~\ref{rem: construction of S alpha} and notice that $L \coloneqq \bigoplus_{i \in \Z} \g_{\alpha+i\beta}$ is a subrepresentation. For $x \in \g_{\alpha+i\beta}$ it follows that - \[ - h.x = [\beta^\vee, x] = (\alpha+i\beta)(\beta^\vee)x = (\pair{\alpha, \beta^\vee}+2i)x, - \] - so $\g_{\alpha+i\beta} = L_{\pair{\alpha, \beta^\vee}+2i}$ for every $i \in \Z$. Because $L_{\pair{\alpha, \beta^\vee}} = \g_\alpha \neq 0$ it follows that $\pair{\alpha, \beta^\vee}$ is a weight of $L$ and therefore integral by $\sll_2$-theory. From $\sll_2$-theory it also follows that $-\pair{\alpha, \beta^\vee}$ is also a weight of $L$. Therefore there exists some $i \in \Z$ with $\pair{\alpha, \beta^\vee} + 2i = -\pair{\alpha, \beta^\vee}$, namely $i = -\pair{\alpha, \beta^\vee}$. Then - \[ - \g_{\alpha - \pair{\alpha, \beta^\vee} \beta} - = \g_{\alpha + i\beta} - = L_{\pair{\alpha, \beta^\vee}+2i} - = L_{\pair{\alpha, \beta^\vee} -2\pair{\alpha, \beta^\vee}} - = L_{-\pair{\alpha, \beta^\vee}} - \neq 0, - \] - which shows that $\alpha - \pair{\alpha, \beta^\vee} \beta \in \Phi$. -\end{proof} - - -\begin{corollary} - Let $\alpha, \beta \in \Phi$. Then - \[ - [\g_\alpha, \g_\beta] = - \begin{cases} - \g_{\alpha+\beta} & \text{if $\alpha+\beta \in \Phi$}, \\ - 0 & \text{otherwise}. - \end{cases} - \] -\end{corollary} -\begin{proof} - It is already known that $[\g_\alpha, \g_\beta] \subseteq \g_{\alpha + \beta}$. If $\alpha+\beta \notin \Phi$ then $\g_{\alpha + \beta} = 0$ and thus $[\g_\alpha, \g_\beta] = 0$. Suppose now that $\alpha + \beta \in \Phi$. Then $\alpha$ and $\beta$ are linearly independent (because otherwise $\beta = \pm \alpha$ and therefore $\alpha+\beta \in \{-2\alpha, 0, 2\alpha\}$, none of which is a root). Let $\g^\alpha$ act on $\g$ as in Remark~\ref{rem: construction of S alpha} and consider the subrepresentation $L \coloneqq \bigoplus_{i \in \Z} \g_{\beta + i\alpha}$. - - As seen in the proof of Proposition~\ref{prop: integral and reflection properties of root pairing} it follows that $\g_{\beta + i\alpha} = L_{\pair{\beta, \alpha^\vee}+2i}$ for every $i \in \Z$. In particular every nonzero weight space of $L$ is one-dimensional and all weights have the same parity. Hence $L$ is an irreducible representation of $\g^\alpha$. - - Let $\hat{e} \in \g_\alpha \subseteq S^\alpha$ be the element corresponding to $e \in \sll_2(k)$ under the isomorphism $\sll_2(k) \cong S^\alpha$ which was used to construct the action of $\sll_2(k)$ on $\g$. Because $\beta, \alpha + \beta \in \Phi$ are both roots it follows that $L_{\pair{\beta, \alpha^\vee}} = \g_\beta$ and $L_{\pair{\beta, \alpha^\vee}+2} = \g_{\alpha+\beta}$ are both nonzero. Therefore it follows from $\sll_2$-theory that $e.L_{\pair{\beta, \alpha^\vee}} = L_{\pair{\beta, \alpha^\vee}+2}$. But this is the same as saying that $[\hat{e}, \g_\beta] = \g_{\alpha+\beta}$ and thus $[\g_\alpha, \g_\beta] = \g_{\alpha+\beta}$. -\end{proof} - - - - - - - - - - - - - - - - - - - diff --git a/sections/definition_of_semisimple_lie_algebras.tex b/sections/definition_of_semisimple_lie_algebras.tex index 5f1e1b3..75e40a0 100644 --- a/sections/definition_of_semisimple_lie_algebras.tex +++ b/sections/definition_of_semisimple_lie_algebras.tex @@ -314,6 +314,7 @@ \subsection{More on Bilinear Forms} \begin{recall}[Orthogonals and non-degeneracy] + \label{reviewing orthogonals} If~$V$ is a finite dimensional (real or complex) inner product space then for every linear subspace~$U$ of~$V$ the restriction of the inner product to~$U$ is again an inner product,~$V = U \oplus U^\perp$ and~$(U^\perp)^\perp = U$. That the restriction is again an inner product holds because positive definiteness is a pointwise property. That~$U \cap U^\perp = 0$ also follows from the positive definiteness, and that~$V = U \oplus U^\perp$ can then be concluded from the dimension formula~$\dim V = \dim U + \dim U^\perp$. diff --git a/sections/killing_form_and_cartans_criterion.tex b/sections/killing_form_and_cartans_criterion.tex index 46ef9ad..00a98e4 100644 --- a/sections/killing_form_and_cartans_criterion.tex +++ b/sections/killing_form_and_cartans_criterion.tex @@ -696,6 +696,28 @@ \subsection{Concrete Jordan Decomposition} \end{warning} +\begin{lemma} + \label{concrete jordan decomposition of sum} + Let~$x$ and~$y$ be two commuting endomorphisms of a finite dimensional~{\vectorspace{$\kf$}}. + Then the Jordan decomposition of~$x+y$ can be computed from that of~$x$ and~$y$ via + \[ + (x + y)_s + = + x_s + y_s + \quad\text{and}\quad + (x + y)_n + = + x_n + y_n \,. + \] +\end{lemma} + + +\begin{proof} + It follows from the commutativity of~$x$ and~$y$ that each part~$x_s$ and~$x_n$ of~$x$ commutes with each part~$y_s$ and~$y_n$ of~$y$. + It follows that~$x_s + y_s$ is again semisimple and that~$x_n + y_n$ is again nilpotent. +\end{proof} + + \begin{definition} An element~$x \in \glie$ of a finite dimensional Lie~algebra~$\glie$ is~\defemph{\adsemisimple} if the endomorphism~$\ad(x)$ of~$\glie$ is semisimple. \end{definition} diff --git a/sections/root_space_decomposition.tex b/sections/root_space_decomposition.tex new file mode 100644 index 0000000..da79a6c --- /dev/null +++ b/sections/root_space_decomposition.tex @@ -0,0 +1,740 @@ +\section{Root Space Decomposition} + + +\begin{convention} + Throughout this section~$\glie$ denotes a finite dimensional semisimple Lie~algebra. +\end{convention} + + + + + +\subsection{Construction of the Root Space Decomposition} + + +\begin{definition} + A Lie~subalgebra~$\hlie$ of~$\glie$ is \defemph{toral}\index{toral subalgebra} if it consists of semisimple elements. +\end{definition} + + +\begin{lemma} + Every toral Lie~subalgebra of~$\glie$ is abelian. +\end{lemma} + + +\begin{proof} + Let~$x \in \hlie$. + Then~$\ad_{\hlie}(x) = \restrict{\ad_{\glie}(x)}{\hlie}$ is semisimple because it is the restriction of the semisimple endomorphism~$\ad_{\glie}(x)$. + We show that all eigenvalues of~$\ad_{\hlie}(x)$ are zero. + + Let~$y \in \hlie$ be an eigenvector of~$\ad_{\hlie}(x)$ with corresponding eigenvalue~$\mu \in \kf$. + Then~$\ad_{\hlie}(y)$ is semisimple by the the above argumentation. + For every~$\lambda \in \kf$ let~$\hlie_\lambda$ denote the eigenspace of~$\ad_{\hlie}(y)$ with respect to the eigenvalue~$\lambda$. + On the one hand $[y,x] \in \hlie_0$ because + \[ + \ad_{\hlie}(y)([y,x]) + = + [y,[y,x]] + = + [y, -\mu y] + = + 0 \,. + \] + On the other hand + \[ + [y,x] + = + \ad_{\hlie}(y)(x) + \in + \ad_{\hlie}(y)(\hlie) + = + \ad_{\hlie}(y)\left( \bigoplus_{\lambda \in \kf} \hlie_\lambda \right) + = + \bigoplus_{\lambda \neq 0} \hlie_\lambda \,. + \] + It follows from the directness of the sum~$\hlie = \bigoplus_{\lambda \in \kf} \hlie_\lambda$ that $0 = [y,x] = -\mu y$. + It follows that~$\mu = 0$ because~$y \neq 0$. +\end{proof} + + +\begin{definition} + A \defemph{Cartan~subalgebra}\index{Cartan!subalgebra} of~$\glie$ is an inclusion maximal toral subalgebra. +\end{definition} + + +\begin{remark} + \leavevmode + \begin{enumerate} + \item + The toral Lie~subalgebra~$0$ is contained in a toral Lie~subalgebra of~$\glie$ of maximal dimension, which is then a Cartan~subalgebra of~$\glie$. + Therefore~$\glie$ contains a Cartan~subalgebra. + \item + If~$\glie \neq 0$ then any Cartan~subalgebra of~$\glie$ is non-zero. + Indeed,~$\glie$ contains some nonzero semisimple element~$x$ because otherwise~$\glie$ would need to be nilpotent by Engel’s~theorem. + The linear subspace~$\kf x$ is then a toral Lie~subalgebra of~$\glie$. + \item + It can be shown (see \cite[\S 16.2]{humphreys}) that Cartan subalgebras are unique up to automorphism: + If~$\hlie$ and~$\hlie'$ are two Cartan subalgebra of~$\glie$ then there exists an automorphism~$\sigma$ of~$\glie$ with~$\sigma(\hlie) = \hlie'$. + It follows in particular that any two Cartan subalgebras are of the same dimension, and hence that Cartan subalgebras can equivalently be characterized as those toral Lie~subalgebras of maximal dimension. + \end{enumerate} +\end{remark} + + +\begin{definition} + Let~$\hlie$ be a Cartan~subalgebra of~$\glie$. + For every~$\alpha \in \hlie^*$ the linear subspace + \[ + \glie_\alpha + \defined + \{ + y \in \glie + \suchthat + \text{$[x,y] = \alpha(x)y$ for every~$x \in \hlie$} + \} + \] + is the \defemph{root space}\index{root space} of~$\glie$ with respect to~$\alpha$. + If~$\alpha \neq 0$ then~$\alpha$ is a \defemph{root}\index{root} and the set of roots is denoted by~$\Phi(\glie, \hlie) \defined \{\alpha \in \hlie^* \smallsetminus \{0\} \suchthat \glie_\alpha \neq 0\}$. +\end{definition} + + +\begin{remark} + The root space~$\glie_0$ is the centralizer of~$\hlie$ in~$\glie$, i.e.~$\glie_0 = \centerlie_{\glie}(\hlie)$. +\end{remark} + + +\begin{lemma} + \label{pre root space decomposition} + If~$\hlie$ is a Cartan subalgebra of~$\glie$ with roots~$\Phi \defined \Phi(\glie, \hlie)$ then~$\glie = \centerlie_{\glie}(\hlie) \oplus \bigoplus_{\alpha \in \Phi} \glie_\alpha$. +\end{lemma} + + +\begin{proof} + The linear subspace~$\ad_{\glie}(\hlie)$ of~$\gllie(\glie)$ consists of pairwise commuting semisimple endomorphisms. + Hence~$\ad_{\glie}(\hlie)$ is simultaneously diagonalizable, so that + \[ + \glie + = + \bigoplus_{\lambda \in \hlie^*} \glie_\lambda + = + \glie_0 + \oplus + \bigoplus_{\alpha \in \Phi} \glie_\alpha + = + \centerlie_{\glie}(\hlie) + \oplus + \bigoplus_{\alpha \in \Phi} \glie_\alpha + \] + as claimed. +\end{proof} + + +% TODO: Add examples. + + +\begin{lemma} + Let~$\hlie$ be a Cartan subalgebra of~$\glie$. + Then~$[\glie_\alpha, \glie_\beta] \subseteq \glie_{\alpha+\beta}$ for all~$\alpha, \beta \in \hlie^*$. +\end{lemma} + + +\begin{proof} + We have for all~$h \in \hlie$ and~$x \in \glie_\alpha$,~$y \in \glie_\beta$ that + \[ + [h,[x,y]] + = + [[h,x],y] + [x,[h,y]] + = + \alpha(h)[x,y] + \beta(h)[x,y] + = + (\alpha+\beta)(h) [x,y] + \] + and hence~$[x,y] \in \glie_{\alpha+\beta}$. +\end{proof} + + +\begin{lemma} + \label{root spaces orthogonal with respect to killing form} + Let~$\hlie$ be a Cartan subalgebra of~$\glie$. + If~$\alpha, \beta \in \hlie^*$ with~$\alpha \neq -\beta$ then the root spaces~$\glie_\alpha$ and~$\glie_\beta$ are orthogonal with respect to the Killing form~$\kappa$ of~$\glie$. +\end{lemma} + + +\begin{proof} + There exists some~$h \in \hlie$ with~$\alpha(h) \neq -\beta(h)$. + We have for all~$x \in \glie_\alpha$ and~$y \in \glie_\beta$ that + \[ + \alpha(h) \kappa(x,y) + = + \kappa([h,x],y) + = + -\kappa([x,h],y) + = + -\kappa(x,[h,y]) + = + -\beta(h) \kappa(x,y) + \] + and hence~$\kappa(x,y) = 0$. +\end{proof} + + +\begin{corollary} + \label{restriction of killing form to centralizer is non-degenerate} + Let~$\hlie$ be a Cartan subalgebra of~$\glie$. + Then the restriction of the Killing form~$\kappa$ of~$\glie$ to the centralizer~$\centerlie_{\glie}(\hlie)$ is non-degenerate. +\end{corollary} + + +\begin{proof} + The Killing form~$\kappa$ is non-degenerate because~$\glie$ is semisimple. + It follows from \cref{root spaces orthogonal with respect to killing form} that~$\centerlie_{\glie}(\hlie) = \glie_0$ is orthogonal to every root space~$\glie_\alpha$ with~$\alpha \neq 0$ whence~$\glie = \centerlie_{\glie}(\hlie) \oplus \bigoplus_{\alpha \in \Phi} \glie_\alpha$ is the decompsition~$\glie = \centerlie_{\glie}(\hlie) \oplus \centerlie_{\glie}(\hlie)^\perp$. + It follows from \cref{pre root space decomposition} that the restriction of~$\kappa$ to~$\centerlie_{\glie}(\hlie)$ is non-degenerate. +\end{proof} + + +\begin{proposition} + \label{csa are self-centralizing} + Let~$\hlie$ be a Cartan subalgebra of~$\glie$. + Then~$\hlie$ is self-centralizing, i.e.~$\centerlie_{\glie}(\hlie) = \hlie$. +\end{proposition} + + +\begin{proof} + Throughout this proof abbreviate $\clie \defined \centerlie_{\glie}(\hlie)$,~$\ad \defined \ad_{\glie}$ and~$\kappa \defined \kappa_{\glie}$. + + \begin{claim*} + \label{technical and traceless} + If~$x \in \glie$ is nilpotent and~$y \in \glie$ commutes with~$x$ then~$\kappa(x,y) = 0$. + \end{claim*} + + \begin{proof} + The endomorphisms~$\ad(x)$ and~$\ad(y)$ commute and~$\ad(x)$ is nilpotent. + Therefore~$\ad(x)\ad(y)$ is again nilpotent and hence~$0 = \tr(\ad(x)\ad(y)) = \kappa(x,y)$. + \end{proof} + + We know from \cref{commuting via abstract jd} that the centralizer~$\clie$ contains the semisimple and nilpotent parts of all its elements. + + Every semisimple element of~$\clie$ is already contained in~$\hlie$: + If~$s \in \clie$ is semisimple then~$\hlie + \kf s$ is again a toral Lie~subalgebra of~$\glie$ by \cref{abstract jordan decomposition of sum}. + Then~$\hlie = \hlie + \kf s$ by the maximality of~$\hlie$ and hence~$s \in \hlie$. + + The Lie~algebra~$\clie$ is nilpotent: + Let~$x \in \clie$ with Jordan decomposition~$x = x_s + x_n$. + We have shown that~$x_s \in \hlie$, so~$\ad_{\clie}(x_s) = 0$ because~$\clie$ centralizes~$\hlie$ and thus~$\ad_{\clie}(x) = \ad_{\clie}(x_n) = \restrict{\ad_{\glie}(x_n)}{\clie}$. + This shows that~$\ad_{\clie}(x)$ is nilpotent for every~$x \in \clie$ whence~$\clie$ is nilpotent by Engel’s theorem. + + The restriction of~$\kappa$ to~$\hlie$ is non-degenerate: + Let~$x \in \hlie$ with $\kappa(x, \hlie) = 0$. + We have seen that every semisimple~$s \in \clie$ is already contained in~$\hlie$ thus~$\kappa(x,s) = 0$ for every such~$s$. + We have also seen in the above claim that~$\kappa(x,n) = 0$ for every nilpotent~$n \in \clie$. + It follows that~$\kappa(x,y) = 0$ for every~$y \in \clie$ because~$\clie$ contains the semisimple and nilpotent parts of all its elements. + We can now conclude from \cref{restriction of killing form to centralizer is non-degenerate} that~$x = 0$. + + It follows that~$\hlie \cap [\clie,\clie] = 0$ because~$[\hlie,\clie] = 0$ and thus~$\kappa(\hlie, [\clie,\clie]) = \kappa([\hlie, \clie], \clie) = 0$. + + It further follows that $\clie$ is abelian: + Otherwise~$[\clie, \clie] \neq 0$. + Then $\centerlie(\clie) \cap [\clie,\clie] \neq 0$ because~$\clie$ is nilpotent. + Let~$x \in \centerlie(\clie) \cap [\clie,\clie]$ be non-zero. + We observe that~$x$ cannot be semisimple because otherwise~$x \in \hlie$ as seen above, but $\hlie \cap [\clie,\clie] = 0$. + The nilpotent part~$x_n$ must therefore be nonzero, and we have seen that~$x_n \in \clie$. + It follows from~$x \in \centerlie(\clie)$ that also~$x_n \in \centerlie(\clie)$ by \cref{commuting via abstract jd}. + Thus~$\kappa(x_n, \clie) = 0$ by the above claim. + It follows that~$x_n = 0$ because the restriction of~$\kappa$ to~$\clie$ is non-degenerate, contradicting that~$x_n$ is non-zero. + + Suppose now that~$\hlie$ is a proper Lie~subalgebra of~$\clie$ and let~$x \in \clie$ with~$x \notin \hlie$. + Both~$x_s$ and~$x_n$ are again contained in~$\clie$, and~$x_s$ is even contained in~$\hlie$. + We may replace~$x$ by~$x_n$ to assume that~$x$ is nilpotent. + Then~$\kappa(x, \clie) = 0$ by the above claim, which contradict~$\kappa$ being non-degenerate on~$\clie$. +\end{proof} + + +\begin{remark} + The proof of \cref{csa are self-centralizing} is taken from \cite[\S 8.2]{humphreys}. +\end{remark} + + +\begin{corollary}[Existence of root space decomposition] + Let~$\hlie$ be a Cartan subalgebra of~$\glie$. + Then + \[ + \glie + = + \hlie + \oplus + \bigoplus_{\alpha \in \Phi(\glie,\hlie)} + \glie_\alpha + \] + and the restriction of the Killing form of~$\glie$ to~$\hlie$ is non-degenerate. +\end{corollary} + + +\begin{proof} + This follows from \cref{pre root space decomposition} and \cref{restriction of killing form to centralizer is non-degenerate} because~$\hlie$ is self centralizing. +\end{proof} + + +\begin{corollary} + \label{g alpha and g -alpha pair non degenerate with the killing form} + Let~$\hlie$ be a Cartan subalgebra of~$\glie$ and let~$\alpha \in \Phi(\glie,\hlie)$. + Then~$\kappa_{\glie}(\glie_\alpha, \glie_{-\alpha}) \neq 0$, i.e.\ there exist~$x \in \glie_\alpha$ and~$y \in \glie_{-\alpha}$ with $\kappa_{\glie}(x,y) \neq 0$. +\end{corollary} + + +\begin{proof} + The root space~$\glie_\alpha$ is nonzero, hence there exists some nonzero~$x \in \glie_\alpha$. + It follows from the non-degeneracy of~$\kappa_{\glie}$ that there exists some~$y \in \glie$ with~$\kappa(x,y) = 0$. + The root space~$\glie_\alpha$ is orthogonal to every root space~$\glie_\beta$ with~$\beta \neq -\alpha$. + Hence we may assume that~$y \in \glie_{-\alpha}$. +\end{proof} + + +\begin{definition} + If~$\hlie$ is a Cartan subalgebra of~$\glie$ with roots~$\Phi \defined \Phi(\glie, \hlie)$ then the decomposition + \[ + \glie + = + \hlie \oplus \bigoplus_{\alpha \in \Phi} \glie_\alpha + \] + is the \defemph{root space decomposition} of~$\glie$ with respect to $\hlie$. +\end{definition} + + + + + +\subsection{Properties of the Root Space Decomposition} + + +\begin{convention} + Troughout this subsection let~$\hlie$ be a Cartan subalgebra of~$\glie$ with associated set of roots~$\Phi \defined \Phi(\glie,\hlie)$. + Let~$\kappa$ be the Killing form of~$\glie$. +\end{convention} + + +\begin{definition} + \label{def of t_phi} + For every~$\phi \in \hlie^*$ let~$t_\phi \in \hlie$ be the unique element with~$\kappa(t_{\phi}, -) = \phi$. +\end{definition} + + +\begin{remark} + \Cref{def of t_phi} makes sense because because~$\kappa$ is non-degenerate on~$\hlie$ and thus induced an isomorphism of vector spaces~$\hlie \to \hlie^*$ given by~$x \mapsto \kappa(x,-)$. +\end{remark} + + +\begin{proposition} + \leavevmode + \begin{enumerate} + \item + The set of roots~$\Phi$ spans the dual space~$\hlie^*$. + \item + If~$\alpha \in \Phi$ then also~$-\alpha \in \Phi$, i.e.\ the negative of any root is again a root. + \item + \label{bracket via kappa} + $[x,y] = \kappa(x,y) t_\alpha$ for every root~$\alpha \in \Phi$ and all~$x \in \glie_\alpha$ and~$y \in \glie_{-\alpha}$. + \item + For every root~$\alpha \in \Phi$ then the linear subspace~$[\glie_\alpha, \glie_{-\alpha}]$ of~$\hlie$ is {\onedimensional} with basis~$t_\alpha$. + \item + If~$\alpha \in \Phi$ is any root then~$\alpha(t_\alpha) = \kappa(t_\alpha, t_\alpha) \neq 0$. + \end{enumerate} +\end{proposition} + + +\begin{proof} + \leavevmode + \begin{enumerate} + \item + Suppose~$\Phi$ does not span~$\hlie^*$. + Then there exists some~$\phi \in \hlie^*$ with~$\phi \notin \gen{\Phi}_{\kf}$. + Then there exists some~$x' \in \hlie^{**}$ with~$x'(\alpha) = 0$ for every~$\alpha \in \Phi$ but~$x'(\phi) \neq 0$. + Using the natural isomorphism~$\hlie \to \hlie^{**}$ there exists some nonzero~$x \in \hlie$ such that~$x'$ is given by evaluation at~$x$. + Then~$\alpha(x) = 0$ for every~$\alpha \in \hlie^*$ but~$\phi(x) \neq 0$; + in particular~$x \neq 0$. + In the root space decomposition~$\glie = \hlie \oplus \bigoplus_{\alpha \in \Phi} \glie_\alpha$ we see that~$x$ commutes with every root space~$\glie_\alpha$,~$\alpha \in \Phi$. + And~$x$ also commutes with~$\hlie$ because~$\hlie$ is abelian. + This means that~$x \in \centerlie(\glie)$. + But~$\centerlie(\glie) = 0$ because~$\glie$ is semisimple while~$x$ is supposed to be nonzero. + \item + Let $\alpha \in \Phi$. + Then~$\kappa(\glie_\alpha, \glie_{-\alpha}) \neq 0$ by \cref{g alpha and g -alpha pair non degenerate with the killing form} whence~$\glie_{-\alpha} \neq 0$, which means that~$-\alpha \in \Phi$. + \item + We find for~$h \in \hlie$ that + \begin{align*} + \kappa(h, [x,y]) + &= + \kappa([h,x], y) + \\ + &= + \alpha(h) \kappa(x,y) + \\ + &= + \kappa(t_\alpha, h) \kappa(x,y) + \\ + &= + \kappa(\kappa(x,y) t_\alpha, h) + \\ + &= + \kappa(h, \kappa(x,y) t_\alpha) \,. + \end{align*} + It follows that~$[x,y] = \kappa(x,y) t_\alpha$ because the restriction of~$\kappa$ to~$\hlie$ is non-degenerate. + \item + We have seen in part~\ref*{bracket via kappa} that~$[\glie_\alpha, \glie_{-\alpha}]$ is contained in the span of~$t_\alpha$. + It follows from \cref{g alpha and g -alpha pair non degenerate with the killing form} that~$[\glie_\alpha, \glie_{-\alpha}]$ already contains some nonzero multiple of~$t_\alpha$, and hence the span of~$t_\alpha$. + \item + Let us suppose that there exist some~$\alpha \in \Phi$ with~$\alpha(\alpha, t_\alpha) = \kappa(t_\alpha, t_\alpha) = 0$. + We know from \cref{g alpha and g -alpha pair non degenerate with the killing form} that there exists~$x \in \glie_\alpha$ and~$y \in \glie_{-\alpha}$ with~$\kappa(x,y) \neq 0$. + We can choose~$x$ and~$y$ so that~$\kappa(x,y) = 1$ and thus~$[x,y] = \kappa(x,y) t_\alpha = t_\alpha$. + Then~$\rlie \defined \gen{x, t_\alpha, y}_{\kf}$ is a {\threedimensional} solvable Lie~subalgebra of~$\glie$ because~$[t_\alpha, x] = \alpha(t_\alpha) x = 0$ and~$[t_\alpha, y] = -\alpha(t_\alpha) y = 0$. + + It follows that~$\ad(\rlie)$ is a {\threedimensional} solvable Lie~subalgebra of~$\gllie(\glie)$, and is hence given by upper triangular matrices with respect to some suitable basis of~$\glie$ by \cref{triangularization of solvable linear lie algebras}. + Then~$[\ad(\rlie), \ad(\rlie)] = \ad([\rlie, \rlie]) = \kf \ad(t_\alpha)$ consists of nilpotent endomorphisms whence~$\ad(t_\alpha)$ is nilpotent. + But the element~$t_\alpha$ is semisimple because it is contained in the Cartan subalgebra~$\hlie$ whence the endomorphism~$\ad(t_\alpha)$ is semisimple. + This shows that the endomorphism~$\ad(t_\alpha)$ is both nilpotend and semisimple whence~$\ad(t_\alpha) = 0$. + It follows that~$t_\alpha = 0$ because the adjoint representation is faithful, but this contradicts~$\kappa(t_\alpha, -) = \alpha$ being nonzero. + \qedhere + \end{enumerate} +\end{proof} + + +% \begin{recall}[Reflections] +% \leavevmode +% \begin{enumerate} +% \item +% Let~$V$ be a finite dimensional euclidian vector space, i.e.\ a real vector space endowed with an inner product~$\bil{-,-}$. +% +% We can consider for any linear subspace~$U$ of~$V$ the orthogonal reflection at~$U$. +% This reflection~$s$ is uniquely determined by the property that~$s(u) = u$ for every~$u \in U$ and~$s(u') = -u'$ for every~$u' \in U^\perp$. +% So with respect to the decomposition~$V = U \oplus U^\perp$ the reflection~$s$ is given by~$s = (\id_U) \oplus (-\id_{U^\perp})$. +% If~$p \colon V \to V$ is the orthogonal projection onto~$U^\perp$ then the reflection~$s$ can also be expressed as +% \[ +% s(x) +% = +% x - 2 p(x) +% \] +% for every~$x \in V$. +% +% If~$H$ is a hyperplane in~$V$, i.e.\ a linear subspace of codimension~$1$, then the orthogonal complement~$H^\perp$ is {\onedimensional}. +% For every nonzero~$v \in H^\perp$ the orthogonal projection~$p$ onto~$H^\perp$ is given by +% \[ +% p_v(x) +% = +% \frac{\bil{x,v}}{\bil{v,v}} v +% \] +% for every~$x \in V$. +% The reflection at~$H$ is then given by +% \[ +% s_v(x) +% = +% x - 2 p_v(x) +% = +% x - 2 \frac{\bil{x,v}}{\bil{v,v}} v \,. +% \] +% There are two special cases of these formulae, depending on choices of~$v$: +% If~$v$ is normalized so that~$\bil{v,v} = 1$ (i.e.~$\norm{v} = 1$) then +% \[ +% p_v(x) +% = +% \frac{x,v} v +% \quad\text{and}\quad +% s_v(x) +% = +% x - 2 \bil{x,v} v \,. +% \] +% If~$v$ is normalized so that~$\bil{v,v} = 2$ then +% \[ +% p_v(x) +% = +% \frac{\bil{x,v}}{2} v +% \quad\text{and}\quad +% s_v(x) +% = +% x - \bil{x,v} v \,. +% \] +% \item +% Suppose now that~$V$ is any vector space endowed wih a non-degenerate symmetric bilinear form~$\bil{-,-}$. +% For an arbitrary linear subspace~$U$ of~$V$ it does not have to hold that~$V = U \oplus U^\perp$. +% But if it does then we can again define the reflection at~$U$ as in the euclidian case. +% Recall from \cref{reviewing orthogonals} that~$V = U \oplus U^\perp$ if and only if the restriction of~$\beta$ to~$U$ is again non-degenerate, if and only if the restriction of~$\beta$ to~$U^\perp$ is again non-degenerate. +% +% Suppose that~$H$ is a hyperplane in~$V$. +% Then~$V = H \oplus H^\perp$ if and only if the restriction of~$H^\perp$ is non-degenerate. +% If~$v \in H^\perp$ then this means that~$\beta(v,v) \neq 0$. +% If this condition holds then define the reflection~$s_v \colon V \to V$ along~$v$ as +% \[ +% s_v(x) +% \defined +% x - 2 \frac{\bil{x,v}}{\bil{v,v}} v \,. +% \] +% Then~$\bil{s_v(x), s_v(y)} = \bil{x,y}$ for all~$x, y \in V$~$\det s = -1$ and~$s_v^2 = 0$. +% \item +% If~$\bil{-,-}$ is a non-degenerate bilinear form on a vector space~$V$ then we +% \end{enumerate} +% \end{recall} + + +% +% +% \begin{remark} +% Let $\alpha \in \Phi$. The existence and uniqueness of $h_\alpha$ follows from the fact that $[\glie_\alpha, \glie_{-\alpha}]$ is one-dimensional with $\alpha([\glie_\alpha, \glie_{-\alpha}]) = k \alpha(t_\alpha) \neq 0$. Notice that $[\glie_\alpha, \glie_{-\alpha}] = kh_\alpha$ and that $(-\alpha)^\vee = -h_\alpha$. Also notice that $h_\alpha = 2 t_\alpha / \kappa(t_\alpha, t_\alpha)$ because +% \[ +% \alpha\left( \frac{2 t_\alpha}{\kappa(t_\alpha, t_\alpha)} \right) +% = \frac{2\alpha(t_\alpha)}{\kappa(t_\alpha, t_\alpha)} +% = \frac{2\kappa(t_\alpha,t_\alpha)}{\kappa(t_\alpha, t_\alpha)} +% = 2. +% \] +% \end{remark} +% +% + + +\begin{construction} + \label{construction of S alpha} + Let~$\alpha \in \Phi$ be any root. + + We have for every~$h' \in [\glie_\alpha, \glie_\beta]$ and all~$e' \in \glie_\alpha$ and~$f' \in \glie_{-\alpha}$ that + \[ + [h', e'] + = + \alpha(h') e' + \quad\text{and}\quad + [h', f'] + = + -\alpha(h') \spacing f' \,. + \] + We have seen that the linear subspace~$[\glie_\alpha, \glie_\beta]$ of~$\hlie$ is {\onedimensional} with basis vector~$t_\alpha$, for which~$\alpha(t_\alpha) = \kappa(t_\alpha, t_\alpha) \neq 0$. + Thus there exists a unique element~$h_\alpha \in [\glie_\alpha, \glie_\beta]$ with~$\alpha(h_\alpha) = 2$, which can be gained by normalizing~$t_\alpha$ as~$h_\alpha = 2 t_\alpha / \kappa(t_\alpha, t_\alpha)$. + We then have + \[ + [h_\alpha, e'] + = + 2 e' + \quad\text{and}\quad + [h_\alpha, f'] + = + -2 \spacing f' + \] + for all~$e' \in \glie_\alpha$ and~$f' \in \glie_{-\alpha}$. + We may choose~$e_\alpha \in \glie_\alpha$ and~$f_\alpha \in \glie_\beta$ such that~$[e_\alpha, f_\alpha] = h_\alpha$ because~$[\glie_\alpha, \glie_\beta]$ is {\onedimensional} and spanned by~$t_\alpha$. + + We note that~$h_\alpha \neq 0$ because~$\alpha(h_\alpha) = 2$ and hence also~$e_\alpha, f_\alpha \neq 0$. + The elements~$e_\alpha$,~$h_\alpha$ and~$f_\alpha$ are hence linearly independent as they live in distinct root spaces. + We see altogether that~$\sllie_2^\alpha \defined \gen{e_\alpha, h_\alpha, f_\alpha}$ is a {\threedimensional} Lie~subalgebra of~$\glie$ that is isomophiic to~$\sllie_2(\kf)$ via + \[ + \phi + \colon + \sllie_2(\kf) + \to + \sllie_2^\alpha \,, + \quad + e + \mapsto + e_\alpha \,, + \quad + h + \mapsto + h_\alpha \,, + \quad + f + \mapsto + f_\alpha \,. + \] + We see in particular that~$\glie$ becomes a representation of~$\sllie_2(\kf)$ via~$x.y = [\phi(x), y]$ for all~$x \in \sllie_2(\kf)$ and~$y \in \glie$. + We can now use our understanding of finite dimensional~{\representations{$\sllie_2(\kf)$}} to better understand the root space decomposition of~$\glie$: +\end{construction} + + +\begin{proposition} + \label{roots spaces are onedimensional and reduced} + Let~$\alpha \in \Phi$ be any root. + \begin{enumerate} + \item + The root space~$\glie_\alpha$ is one-dimensional. + \item + The only multiples of~$\alpha$ which are themselves roots are~$\alpha$ and~$-\alpha$, i.e.~$\kf \alpha \cap \Phi = \{\alpha, -\alpha\}$. + \end{enumerate} +\end{proposition} + + +\begin{proof} + We consider the copy~$\sllie_2^\alpha$ of~$\sllie_2(\kf)$ in~$\glie$ and observe that + \[ + V + \defined + \kf h_\alpha + \oplus + \bigoplus_{\substack{c \neq 0}} \glie_{c\alpha} + \] + is an~{\subrepresentation{$\sllie_2^\alpha$}} of~$\glie$: + The know that~$V$ is closed under the action of~$h_\alpha$ because~$h_\alpha$ acts on every direct summand by some scalar (namely~$2c$ for~$c \neq 0$ and~$0$ for~$\kf h_\alpha$). + It is closed under the action of~$e_\alpha$ because~$[e_\alpha, \glie_{c\alpha}] \subseteq \glie_{(c+1)\alpha}$ for~$c \neq -1$ and~$[e_\alpha, \glie_{-\alpha}] \subseteq [\glie_\alpha, \glie_{-\alpha}] \subseteq \kf h_\alpha$. + That~$V$ is closed under the action of~$f_\alpha$ can be seen similarly. + + It follows from Weyl’s theorem that~$V$ is completely reducible, and it follows from the classification of finite dimensional~$\sllie_2(\kf)$ representations that~$V = \bigoplus_{n \in \Integer} V_n$ with~$h_\alpha$ acting on~$V_n$ by the scalar~$n$. + The element~$h_\alpha$ acts on the direct summand~$\glie_{c\alpha}$ by the scalar~$2c$ and on~$\kf h_\alpha$ by the scalar$~0$. + The weight space~$V_n$ does therefore coincide with the root space~$\glie_{n \alpha / 2}$ for all nonzero~$n \in \Integer$, and the weight space~$V_0$ coincides with the span~$\kf h_\alpha$. + + This already shows that the only multiples of~$\alpha$ that can again be roots are the nonzero half-integer multiples of~$\alpha$, i.e.\ those of the form~$n \alpha / 2$ with nonzero~$n \in \Integer$. + + We know from the classification of finite dimensional~{\representations{$\sllie_2(\kf)$}} that the dimension of~$V_0$ counts the total multiplicities of the irreducible representaions~$\irr(n)$ where~$n \geq 0$ is even. + We have that~$\dim V_0 = \dim \kf h_\alpha = 1$. + Hence only one of the irreducible representation~$\irr(n)$ where~$n \geq 0$ is even appears in~$V$, and only with multiplicity~$1$. + We see that~$\sllie_2^\alpha \cong \irr(2)$ itself is such an irreducible subrepresentation of~$V$. + Hence + \[ + \kf h_\alpha + \oplus + \bigoplus_{\substack{n \in \Integer \\ \text{$n \neq 0$ even}}} + \glie_{n \alpha/2} + = + \bigoplus_{\substack{n \in \Integer \\ \text{$n$ even}}} + V_n + = + \sllie_2^\alpha + \] + and therefore + \[ + V + = + \sllie_2^\alpha + \oplus + \bigoplus_{\substack{n \in \Integer \\ \text{$n$ odd}}} + V_n + = + \sllie_2^\alpha + \oplus + \bigoplus_{\substack{n \in \Integer \\ \text{$n$ odd}}} + \glie_{n \alpha/2} \,. + \] + We observe that in particular~$2\alpha$ is not again a root: + + \begin{claim*} + If~$\beta \in \Phi$ is any root then~$2 \beta$ is not it root, i.e.\ twice a root is never a root. + \end{claim*} + + The dimension~$\dim V_1$ counts the total multiplicities of the irreducible representations~$\irr(n)$ where~$n \geq 0$ is odd. + It follows from the above claim that~$\alpha/2$ cannot be a root (because otherwise~$\alpha$ could not be a root), so~$\dim V_1 = \dim \glie_{\alpha/2} = 0$. + Hence no irreducible representation~$\irr(n)$ where~$n \geq 0$ is odd appears in~$V$. + This shows that~$V_n = 0$ for every odd~$n \in \Integer$, i.e.~$\glie_{n \alpha/2} = 0$ for every odd~$n \in \Integer$. + + We find altogether that + \[ + V + = + \sllie_2^\alpha + = + \kf h_\alpha \oplus \kf e_\alpha \oplus \kf \spacing f_\alpha + = + \kf h_\alpha \oplus \glie_\alpha \oplus \glie_{-\alpha} \,. + \] + This shows that the only nonzero multiples~$\beta$ of~$\alpha$ with~$\glie_\beta \neq 0$ are~$\alpha$ and~$-\alpha$, and that both~$\glie_\alpha$ and~$\glie_{-\alpha}$ are {\onedimensional}. +\end{proof} + + +\begin{definition} + For all~$\lambda \in \hlie^*$ and~$h \in \hlie$ the evaluation of~$\lambda$ at~$h$ is denoted by + \[ + \pair{h, \lambda} + \defined + \pair{\lambda, h} + \defined + \lambda(h) \,. + \] +\end{definition} + + +% \begin{definition} +% For any root~$\alpha \in \Phi$ the element~$h_\alpha \defined h_\alpha$ is the associated \defemph{coroot}\index{coroot} to~$\alpha$. +% \end{definition} + + +\begin{proposition} +\label{integral and reflection properties of root pairing} + Let~$\alpha, \beta \in \Phi$. + Then~$\pair{\alpha, h_\beta} \in \Integer$ and~$\alpha - \pair{\alpha, h_\beta} \beta \in \Phi$. +\end{proposition} + + +\begin{proof} + If the roots~$\alpha$ and~$\beta$ are linearly dependent then~$\beta = \pm \alpha$ by \cref{roots spaces are onedimensional and reduced}. + Then~$h_\beta = \pm h_\alpha$ and therefore~$\pair{\alpha, h_\beta} = \pm \pair{\alpha, h_\alpha} = \pm 2 \in \Integer$ and + \[ + \alpha - \pair{\alpha, h_\beta} \beta + = + \alpha - \pair{\alpha, \pm h_\alpha} (\pm \alpha) + = + \alpha - \pair{\alpha, h_\alpha} \alpha + = + \alpha - 2 \alpha + = + -\alpha \in \Phi \,. + \] + + Suppose now that the roots~$\alpha$ and~$\beta$ are linearly independent. + The copy~$\sllie_2^\beta$ of~$\sllie_2(\kf)$ in~$\glie$ acts on $\glie$ and~$V \defined \bigoplus_{i \in \Integer} \glie_{\alpha+i\beta}$ is a~{\subrepresentation{$\sllie_2^\beta$}}. + We find that for~$x \in \glie_{\alpha + i \beta}$, + \[ + h.x + = + [h_\beta, x] + = + (\alpha + i\beta)(h_\beta)x + = + (\pair{\alpha, h_\beta} + 2i) x \,, + \] + so~$\glie_{\alpha+i\beta} = V_{\pair{\alpha, h_\beta}+2i}$ for every~$i \in \Integer$. + All weight of~$V$ are integral and we find for~$i = 0$ that~$V_{\pair{\alpha, h_\beta}} = \glie_\alpha \neq 0$. + Thus~$\pair{\alpha, \beta} \in \Integer$. + + We know from the structure of~{\representations{$\sllie_2(\kf)$}} that the negative of~$\pair{\alpha, \beta}$ is again weight of~$V$. + There hence exists some~$i \in \Integer$ with~$\pair{\alpha, h_\beta} + 2i = -\pair{\alpha, h_\beta}$; + necessarily~$i = -\pair{\alpha, h_\beta}$. + Then + \[ + \glie_{\alpha - \pair{\alpha, h_\beta} \beta} + = + \glie_{\alpha + i \beta} + = + V_{\pair{\alpha, h_\beta}+2i} + = + V_{\pair{\alpha, h_\beta} - 2\pair{\alpha, h_\beta}} + = + V_{-\pair{\alpha, h_\beta}} + \neq + 0, + \] + which shows that~$\alpha - \pair{\alpha, h_\beta} \beta \in \Phi$ is again a root. +\end{proof} + + +\begin{corollary} + For any two roots~$\alpha, \beta \in \Phi$, + \[ + [\glie_\alpha, \glie_\beta] + = + \begin{cases} + \glie_{\alpha+\beta} & \text{if $\alpha+\beta \in \Phi$} \,, \\ + 0 & \text{otherwise} \,. + \end{cases} + \] +\end{corollary} + + +\begin{proof} + We have already seen that~$[\glie_\alpha, \glie_\beta] \subseteq \glie_{\alpha + \beta}$. + If~$\alpha+\beta \notin \Phi$ then~$\glie_{\alpha + \beta} = 0$ and thus~$[\glie_\alpha, \glie_\beta] = 0$. + We consider in the following the case~$\alpha + \beta \in \Phi$. + Then~$\alpha$ and~$\beta$ are linearly independent because otherwise~$\beta = \pm \alpha$ and therefore~$\alpha + \beta \in \{-2 \alpha, 0, 2 \alpha\}$ none of which is again a root. + + We consider again the~{\subrepresentation{$\sllie_2^\alpha$}}~$V \defined \bigoplus_{i \in \Integer} \glie_{\beta + i\alpha}$ of~$\glie$. + Then again~$\glie_{\beta + i\alpha} = V_{\pair{\beta, h_\alpha}+2i}$ for every~$i \in \Integer$. + Every nonzero weight space of~$V$ is {\onedimensional} by \cref{roots spaces are onedimensional and reduced} and all weights have the same parity as they are of the form~$\pair{\beta, h_\alpha} + 2i$ with~$i \in \Integer$. + Hence $V$ is irreducible as an~{\representation{$\sllie_2^\alpha$}} (as can be seen from the weight diagram we just described). + + Both~$V_{\pair{\beta, h_\alpha}} = \glie_\beta$ and~$V_{\pair{\beta, h_\alpha}+2} = \glie_{\alpha+\beta}$ are nonzero because both~$\alpha$ and~$\alpha+\beta$ are roots. + We find with our understanding of (irreducible)~{\representations{$\sllie_2(\kf)$}} that~$e_\alpha.V_{\pair{\beta, h_\alpha}} = V_{\pair{\beta, h_\alpha}+2}$. + This means that~$[e_\alpha, \glie_\beta] = \glie_{\alpha+\beta}$ and thus~$[\glie_\alpha, \glie_\beta] = \glie_{\alpha+\beta}$. +\end{proof} + + + + diff --git a/sections/root_systems.tex b/sections/root_systems.tex deleted file mode 100644 index 2ef3e66..0000000 --- a/sections/root_systems.tex +++ /dev/null @@ -1,3 +0,0 @@ -\chapter{Root Systems} - -\input{sections/abstract_root_systems} diff --git a/sections/roots_and_reflections.tex b/sections/roots_and_reflections.tex new file mode 100644 index 0000000..559f60e --- /dev/null +++ b/sections/roots_and_reflections.tex @@ -0,0 +1,8 @@ +\chapter{Roots and Reflections} + +\input{sections/root_space_decomposition} +% \input{sections/abstract_root_systems} + + + + diff --git a/sections/semisimple_lie_algebras.tex b/sections/semisimple_lie_algebras.tex index d9cee78..963db6b 100644 --- a/sections/semisimple_lie_algebras.tex +++ b/sections/semisimple_lie_algebras.tex @@ -4,4 +4,7 @@ \chapter{Semisimple Lie Algebras} \input{sections/weyls_theorem} \input{sections/sl2_theory} \input{sections/abstract_jordan_decomposition} -% \input{sections/cartan_subalgebras} + + + + diff --git a/sections/sl2_theory.tex b/sections/sl2_theory.tex index 602d6a2..7ab3cad 100644 --- a/sections/sl2_theory.tex +++ b/sections/sl2_theory.tex @@ -1067,6 +1067,8 @@ \subsubsection{Classification of Finite Dimensional Representations} \item \label{calculation of multiplicities} The multiplicities~$n_\lambda$ for~$\lambda \geq 0$ are given by~$n_\lambda = \dim V_\lambda - \dim V_{\lambda+2}$. + \item + The total multiplicity~$\sum_{\lambda \geq 0} n_\lambda$ is given by~$\dim V_0 + \dim V_1$. \end{enumerate} \end{theorem} @@ -1078,7 +1080,8 @@ \subsubsection{Classification of Finite Dimensional Representations} It follows that~$\dim V_\mu = \dim V_{-\mu}$. For~$\mu \geq 0$ the dimension of the corresponding weight space~$V_\mu$ counts how many direct summands~$W^{\lambda,i}$ with~$\lambda \geq \mu$ and~$\mu \equiv \lambda \pmod{2}$ occur in the given decomposition. - The difference~$\dim V_\mu - \dim V_{\mu + 2}$ therefore counts the how many direct summands of the form~$W^{\lambda,i}$ with exactly~$\lambda = \mu$ appear in the given decomposition. + It follows that the difference~$\dim V_\mu - \dim V_{\mu + 2}$ counts how many direct summands of the form~$W^{\lambda,i}$ with exactly~$\lambda = \mu$ appear in the given decomposition. + It also follows that~$\dim V_0 + \dim V_1$ counts the total number of irreducible direct summands, where~$\dim V_0$ counts those summands~$W^{\lambda,i}$ with~$\lambda$ even and~$\dim V_1$ counts those summands~$W^{\lambda,i}$ with~$\lambda$ odd. \end{proof}