\documentclass[numbers=enddot,12pt,final,onecolumn,notitlepage]{scrartcl}% \usepackage[headsepline,footsepline,manualmark]{scrlayer-scrpage} \usepackage[all,cmtip]{xy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{framed} \usepackage{amsmath} \usepackage{comment} \usepackage{color} \usepackage{hyperref} \usepackage[sc]{mathpazo} \usepackage[T1]{fontenc} \usepackage{amsthm} %TCIDATA{OutputFilter=latex2.dll} %TCIDATA{Version=5.50.0.2960} %TCIDATA{LastRevised=Wednesday, July 17, 2019 10:56:52} %TCIDATA{SuppressPackageManagement} %TCIDATA{} %TCIDATA{} %TCIDATA{BibliographyScheme=Manual} %BeginMSIPreambleData \providecommand{\U}{\protect\rule{.1in}{.1in}} %EndMSIPreambleData \theoremstyle{definition} \newtheorem{theo}{Theorem}[section] \newenvironment{theorem}[] {\begin{theo}[#1]\begin{leftbar}} {\end{leftbar}\end{theo}} \newtheorem{lem}[theo]{Lemma} \newenvironment{lemma}[] {\begin{lem}[#1]\begin{leftbar}} {\end{leftbar}\end{lem}} \newtheorem{prop}[theo]{Proposition} \newenvironment{proposition}[] {\begin{prop}[#1]\begin{leftbar}} {\end{leftbar}\end{prop}} \newtheorem{defi}[theo]{Definition} \newenvironment{definition}[] {\begin{defi}[#1]\begin{leftbar}} {\end{leftbar}\end{defi}} \newtheorem{remk}[theo]{Remark} \newenvironment{remark}[] {\begin{remk}[#1]\begin{leftbar}} {\end{leftbar}\end{remk}} \newtheorem{coro}[theo]{Corollary} \newenvironment{corollary}[] {\begin{coro}[#1]\begin{leftbar}} {\end{leftbar}\end{coro}} \newtheorem{conv}[theo]{Convention} \newenvironment{condition}[] {\begin{conv}[#1]\begin{leftbar}} {\end{leftbar}\end{conv}} \newtheorem{quest}[theo]{Question} \newenvironment{algorithm}[] {\begin{quest}[#1]\begin{leftbar}} {\end{leftbar}\end{quest}} \newtheorem{warn}[theo]{Warning} \newenvironment{conclusion}[] {\begin{warn}[#1]\begin{leftbar}} {\end{leftbar}\end{warn}} \newtheorem{conj}[theo]{Conjecture} \newenvironment{conjecture}[] {\begin{conj}[#1]\begin{leftbar}} {\end{leftbar}\end{conj}} \newtheorem{exmp}[theo]{Example} \newenvironment{example}[] {\begin{exmp}[#1]\begin{leftbar}} {\end{leftbar}\end{exmp}} \iffalse \newenvironment{proof}[Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \fi \newenvironment{verlong}{}{} \newenvironment{vershort}{}{} \newenvironment{noncompile}{}{} \excludecomment{verlong} \includecomment{vershort} \excludecomment{noncompile} \newcommand{\kk}{\mathbf{k}} \newcommand{\id}{\operatorname{id}} \newcommand{\ev}{\operatorname{ev}} \newcommand{\Comp}{\operatorname{Comp}} \newcommand{\bk}{\mathbf{k}} \newcommand{\Nplus}{\mathbb{N}_{+}} \newcommand{\NN}{\mathbb{N}} \let\sumnonlimits\sum \let\prodnonlimits\prod \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits} \DeclareSymbolFont{bbold}{U}{bbold}{m}{n} \DeclareSymbolFontAlphabet{\mathbbold}{bbold} \setlength\textheight{22.5cm} \setlength\textwidth{15cm} \ihead{Errata to Introduction to Kac-Moody Lie algebras''} \ohead{\today} \begin{document} \begin{center} \textbf{Introduction to Kac-Moody Lie algebras} \textit{Nicolas Perrin} \url{http://www.hcm.uni-bonn.de/?id=961} version of 2011 \textbf{Errata and addenda by Darij Grinberg} \bigskip \end{center} %\setcounter{section}{} \subsection{Errata} I am not an expert in Lie theory; hence, please approach the corrections below with a critical eye. \begin{itemize} \item \textbf{Definition 2.2.1:} Replace $x\otimes y-z\otimes x-\left[ x,y\right]$ by $x\otimes y-y\otimes x-\left[ x,y\right]$. (Could this be due to the switched \textquotedblleft$y$\textquotedblright\ and \textquotedblleft$z$\textquotedblright\ keys on the German keyboard layout?) \item \textbf{Theorem 2.2.4:} The product $x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}$ should be $e_{1}^{a_{1}}\cdots e_{n}^{a_{n}}$ here. \item \textbf{Proposition 3.1.2:} Replace $\ell$ by $n$ (in size $\ell$''). \item \textbf{Proof of Proposition 3.1.2:} These is easy'' should be This is easy''. \item \textbf{Proof of Proposition 3.1.2:} I don't understand the part of this proof that begins with For this, we may assume that $A_{3}=0$ and $A_{4}% =0$'' and ends with and the matrix $C$ is non degenerate''. Why can we assume that $A_{3}=0$ and $A_{4}=0$ without changing things, and why do we have the $\left( \operatorname*{Vect}\left( ...\right) \right) ^{\perp }=\operatorname*{Vect}\left( ...\right)$ relations (particularly the second one)? (Here is how I would show that the matrix $C$ is nondegenerate: Since $\alpha_{1},\alpha_{2},...,\alpha_{n}$ are linearly independent and $\left\langle \cdot,\cdot\right\rangle$ is a nondegenerate bilinear form, the block matrix $\left( \begin{array} [c]{cc}% A_{1} & A_{2}\\ A_{3} & A_{4}\\ X_{1} & X_{2}% \end{array} \right)$ has rank $n$. But each row of the \textquotedblleft middle part\textquotedblright\ (by this I mean the $\left( \begin{array} [c]{cc}% A_{3} & A_{4}% \end{array} \right)$ part) of this matrix is a linear combination of the rows of the \textquotedblleft upper part\textquotedblright\ (the $\left( \begin{array} [c]{cc}% A_{1} & A_{2}% \end{array} \right)$ part) (because $\operatorname*{rank}\left( \begin{array} [c]{cc}% A_{1} & A_{2}\\ A_{3} & A_{4}% \end{array} \right) =\operatorname*{rank}A=\ell=\operatorname*{rank}A_{1}\leq \operatorname*{rank}\left( \begin{array} [c]{cc}% A_{1} & A_{2}% \end{array} \right)$). Hence, by performing row operations to the matrix $\left( \begin{array} [c]{cc}% A_{1} & A_{2}\\ A_{3} & A_{4}\\ X_{1} & X_{2}% \end{array} \right)$, we can replace the $\left( \begin{array} [c]{cc}% A_{3} & A_{4}% \end{array} \right)$ part by zeroes.\footnote{Is this what you mean by \textquotedblleft assume that $A_{3}=0$ and $A_{4}=0$ \textquotedblright?} Since row operations don't change the rank, this yields that $\operatorname*{rank}\left( \begin{array} [c]{cc}% A_{1} & A_{2}\\ 0 & 0\\ X_{1} & X_{2}% \end{array} \right) =\operatorname*{rank}\left( \begin{array} [c]{cc}% A_{1} & A_{2}\\ A_{3} & A_{4}\\ X_{1} & X_{2}% \end{array} \right)$. Thus,% $n=\operatorname*{rank}\left( \begin{array} [c]{cc}% A_{1} & A_{2}\\ A_{3} & A_{4}\\ X_{1} & X_{2}% \end{array} \right) =\operatorname*{rank}\left( \begin{array} [c]{cc}% A_{1} & A_{2}\\ 0 & 0\\ X_{1} & X_{2}% \end{array} \right) =\operatorname*{rank}\left( \begin{array} [c]{cc}% A_{1} & A_{2}\\ X_{1} & X_{2}% \end{array} \right) .$ Now,% \begin{align*} \operatorname*{rank}C & =\operatorname*{rank}\left( \begin{array} [c]{ccc}% A_{1} & A_{2} & 0\\ A_{3} & A_{4} & I_{n-\ell}\\ X_{1} & X_{2} & 0 \end{array} \right) =\operatorname*{rank}\left( \begin{array} [c]{ccc}% A_{1} & A_{2} & 0\\ X_{1} & X_{2} & 0\\ A_{3} & A_{4} & I_{n-\ell}% \end{array} \right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since permutations of rows don't change the rank of a matrix}\right) \\ & =\underbrace{\operatorname*{rank}\left( \begin{array} [c]{cc}% A_{1} & A_{2}\\ X_{1} & X_{2}% \end{array} \right) }_{=n}+\underbrace{\operatorname*{rank}\left( I_{n-\ell}\right) }_{=n-\ell}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since any block matrix of the form }\left( \begin{array} [c]{cc}% U & 0\\ V & I_{m}% \end{array} \right) \\ \text{satisfies }\operatorname*{rank}\left( \begin{array} [c]{cc}% U & 0\\ V & I_{m}% \end{array} \right) =\operatorname*{rank}U+m \end{array} \right) \\ & =n+n-\ell=2n-\ell, \end{align*} so that $C$ is nondegenerate, qed.) \item \textbf{Definition 3.1.4:} You say: \textquotedblleft Any matrix can be decomposed as a direct sum of indecomposable matrix\textquotedblright. Maybe you should add \textquotedblleft(up to simultaenous permutation of rows and columns)\textquotedblright\ here. \item \textbf{Definition 3.1.5:} Maybe add \textquotedblleft for all $h\in\mathfrak{h}$ and $h^{\prime}\in\mathfrak{h}$\textquotedblright\ to the defining relations. \item \textbf{Proof of Theorem 3.1.6:} It would be better to explicitly distinguish between the vector space $\mathfrak{h}$ which belongs to the realization of $A$, and the subspace $\mathfrak{h}$ of the Lie algebra $\widetilde{\mathfrak{g}}\left( A\right)$. It is clear that there is a canonical surjection from the former space to the latter space, but it is not a priori clear that this surjection is a bijection (i. e., that the relations given in Definition 3.1.5 don't force some elements of $\mathfrak{h}$ to become zero). This does not become clear until the following argument in your proof: \textquotedblleft Assume there is a relation $n_{-}+h+n_{-}=0$ with $n_{-}% \in\widetilde{\mathfrak{n}}_{-}$, $h\in\mathfrak{h}$ and $n_{+}\in \widetilde{\mathfrak{n}}_{+}$ [...] so that $h=0$\textquotedblright The $h$ in the beginning of this argument means an element of $\widetilde{\mathfrak{g}}\left( A\right)$, whereas the $h$ in the end of this argument means a corresponding element of the original vector space $\mathfrak{h}$. Hence, this argument actually shows that if some element $h$ of our original vector space $\mathfrak{h}$ becomes $0$ in $\widetilde{\mathfrak{g}}\left( A\right)$, then it must have been $0$ to begin with. This justifies the identification of $\mathfrak{h}$ with the image of $\mathfrak{h}\rightarrow\widetilde{\mathfrak{g}}\left( A\right)$. It would help a lot if you make this explicit. (Maybe even in the theorem itself, not just in the proof...) \item \textbf{Proof of Theorem 3.1.6:} In the formula% \begin{align*} \left( he_{i}-e_{i}h\right) \left( a\otimes v_{j}\right) & =\text{[...]}\\ & =\text{[...]}\\ & =\text{[...]}=\left\langle \alpha_{i},h\right\rangle e_{i}\left( a\otimes v_{j}\right) \end{align*} (the parts that I have omitted are correct), replace both occurrences of $a\otimes v_{j}$ by $v_{j}\otimes a$. \item \textbf{Proof of Theorem 3.1.6:} When you say there is a surjective map $U\left( \mathfrak{n}_{-}\right) \rightarrow T\left( V\right)$'', you might want to add and this map is an algebra homomorphism'' (since otherwise the next sentence is not clear). \item \textbf{Theorem 3.1.6 and its Proof:} It seems that you write $\mathfrak{n}_{+}$ for $\widetilde{\mathfrak{n}}_{+}$ (and, similarly, $\mathfrak{n}_{-}$ for $\widetilde{\mathfrak{n}}_{-}$) several times here (e. g., in part (iv) of the theorem). \item \textbf{Proof of Theorem 3.1.6:} When you write $n_{-}\left( 1\right) =\left\langle \alpha,h\right\rangle 1$, I think you mean $n_{-}\left( 1\right) =-\left\langle \alpha,h\right\rangle 1$. \item \textbf{Proof of Theorem 3.1.6:} In the formula% $\mathfrak{n}_{\pm}=\bigoplus\limits_{\alpha\in Q_{+},\ \alpha\neq 0}\widetilde{\mathfrak{g}}_{\pm},$ the $\widetilde{\mathfrak{g}}_{\pm}$ should be $\widetilde{\mathfrak{g}}% _{\pm\alpha}$. \item \textbf{Proof of Theorem 3.1.6:} When you say we have the inequality $\dim\widetilde{\mathfrak{g}}_{\alpha}\leq n^{\left\vert \operatorname*{ht}% \alpha\right\vert }$\ \ \ \ '', you might want to add for $\alpha\neq0$''. \item \textbf{Definition 3.1.8:} Add a whitespace after (i)''. \item \textbf{Definition 3.1.8:} Replace is it also contained'' by it is also contained''. \item \textbf{Between Definition 3.1.8 and Definition 3.1.9:} When you say We have the estimate $\dim\mathfrak{g}_{\alpha}0$ and lie in $\mathfrak{i}$ whenever $t=0$. \item \textbf{Proof of Lemma 3.2.7:} Replace $a=0$'' by $x=0$''. \item \textbf{Proof of Proposition 3.2.6 (continued after proof of Lemma 3.2.7):} Replace Lemme 3.2.5'' by Proposition 3.2.5 (ii)''. In the formula directly below this, replace $f_{i}^{a_{i,j}}$ by $f_{i}^{-a_{i,j}}$. \item \textbf{Proof of Proposition 3.2.6 (continued after proof of Lemma 3.2.7):} I think that $-a_{i,j}\left( \operatorname*{ad}f_{i}\right) ^{-a_{i,j}}\left( f_{i}\right)$ should be $a_{j,i}\left( \operatorname*{ad}f_{i}\right) ^{-a_{i,j}}\left( f_{i}\right)$ here. Now you use the $a_{i,j}=0$ $\Longrightarrow$ $a_{j,i}=0$'' condition from Definition 3.2.1 to see that this vanishes. (But I may very well be mistaken.) \item \textbf{Proof of Lemma 3.2.8:} By block matrix'' you mean block-diagonal matrix''. \item \textbf{Proposition 3.2.9:} In part (ii), replace $g^{\prime}\left( A\right)$ by $\mathfrak{g}^{\prime}\left( A\right)$. \item \textbf{Proposition 3.2.9:} The conclusion of part (iii) should not be $\mathfrak{g}^{\prime}\left( A\right) /\mathfrak{c}=0$'' but it should be $\mathfrak{g}^{\prime}\left( A\right) /\mathfrak{c}$ is simple''. \item \textbf{Proof of Proposition 3.2.9:} You write: If for any $\alpha$ we have $\mathfrak{i}\cap\mathfrak{g}_{\alpha}=0$ then $\mathfrak{i}% \subset\mathfrak{h}$''. Here, it would be better to replace any'' by all'', since any'' could also mean some''. Also, again you should say that you are only considering $\alpha\neq0$. \item \textbf{Proof of Proposition 3.2.9:} You write: We can therefore take $\alpha$ a root minimal''. By minimal'' you mean minimal among the roots in $Q_{+}$'' (not all of $Q$). \item \textbf{Proof of Proposition 3.2.9:} Replace colinear'' by collinear''. \item \textbf{Proof of Proposition 3.2.9:} It would be better not to speak of the center'' here, but just say $\mathfrak{c}$, because the center'' might also mean the center of $\mathfrak{g}^{\prime}\left( A\right)$ (and I am not sure whether this is the same center). \item \textbf{Proof of Proposition 3.2.9:} Replace where $n\in \mathfrak{n}_{-}\oplus\mathfrak{n}_{+}$ and $h\in\mathfrak{h}$'' by where $n\in\mathfrak{n}_{-}\oplus\mathfrak{n}_{+}$ and $h\in\mathfrak{h}^{\prime}$''. \item \textbf{Proof of Proposition 3.2.9:} At the moment when you write By minimality, this implies that $\gamma=\alpha_{i}$'', I am losing track of what you are doing. However, it is not hard to complete the proof from here: Since $\left[ f_{i},x\right] \in\mathfrak{i}$ and $\left[ f_{i},x\right] _{\gamma-\alpha_{i}}\neq0$, we get a contradiction to the minimality of $\gamma$ unless either $\left[ f_{i},x\right] _{0}\neq0$ or $\left[ f_{i},x\right] \in\mathfrak{c}$. So we conclude that either $\left[ f_{i},x\right] _{0}\neq0$ or $\left[ f_{i},x\right] \in\mathfrak{c}$. In the former case, we must have $x_{\alpha_{i}}\neq0$ (since $\left[ f_{i},x_{\alpha_{i}}\right] =\left[ f_{i},x\right] _{0}\neq0$). In the latter case, we must have $x_{\alpha_{i}}\neq0$ as well (since $\left[ f_{i},x\right] \in\mathfrak{c}\subseteq\mathfrak{h}$ and thus $\left[ f_{i},x\right] =\left[ f_{i},x\right] _{0}$, so that $\left[ f_{i},x_{\alpha_{i}}\right] =\left[ f_{i},x\right] _{0}\neq0$). Hence, in both cases, we have $x_{\alpha_{i}}\neq0$. Thus, $x_{\alpha_{i}}$ is a nonzero scalar multiple of $e_{i}$ (since $x_{\alpha_{i}}\in\mathfrak{g}_{\alpha_{i}}%$). Hence, $\left[ f_{i},x_{\alpha_{i}}\right]$ is a nonzero scalar multiple of $\left[ f_{i},e_{i}\right] =-\alpha_{i}^{\vee}$, therefore a nonzero scalar multiple of $\alpha_{i}^{\vee}$. Since $\left[ f_{i}% ,x_{\alpha_{i}}\right] =\left[ f_{i},x\right] _{0}$, this shows that $\left[ f_{i},x\right] _{0}$ is a nonzero scalar multiple of $\alpha _{i}^{\vee}$. Since $\left[ f_{i},x\right] _{0}\in\mathfrak{i}$ (because $\left[ f_{i},x\right] \in\mathfrak{i}$ and by Lemma 3.1.7), this yields $\alpha_{i}^{\vee}\in\mathfrak{i}$. Since $\alpha_{i}^{\vee}\notin% \mathfrak{c}$ (this is easy to prove using Proposition 3.1.12 and the fact that $A$ is an indecomposable Cartan matrix), this yields that there exists an element $h\in\mathfrak{i}\cap\mathfrak{h}$ not in $\mathfrak{c}$ (namely, $h=\alpha_{i}^{\vee}$). As you already have shown above, this concludes the proof. \item \textbf{Lemma 4.1.2:} In part (i), replace $x$, $y$ and $z$'' by $x$ and $y$''. \item \textbf{Proof of Lemma 4.1.2:} You write: Applying it to the adjoint representation gives the result.'' Why? If you apply the formula% $\left( \operatorname*{ad}x\right) ^{k}\left[ y,z\right] =\sum \limits_{i=0}^{k}\dbinom{k}{i}\left[ \left( \operatorname*{ad}x\right) ^{i}y,\left( \operatorname*{ad}x\right) ^{k-i}z\right] \ \ \ \ \ \ \ \ \ \ \text{in }U\left( \mathfrak{g}\right)$ to the adjoint representation, you get% $\operatorname*{ad}\left( \left( \operatorname*{ad}x\right) ^{k}\left[ y,z\right] \right) =\operatorname*{ad}\left( \sum\limits_{i=0}^{k}% \dbinom{k}{i}\left[ \left( \operatorname*{ad}x\right) ^{i}y,\left( \operatorname*{ad}x\right) ^{k-i}z\right] \right) \ \ \ \ \ \ \ \ \ \ \text{in }\mathfrak{g},$ which does not immediately yield $\left( \operatorname*{ad}x\right) ^{k}\left[ y,z\right] =\sum\limits_{i=0}^{k}\dbinom{k}{i}\left[ \left( \operatorname*{ad}x\right) ^{i}y,\left( \operatorname*{ad}x\right) ^{k-i}z\right]$ in $\mathfrak{g}$ unless we know that $\mathfrak{g}$ has trivial center. Maybe you wanted to use Corollary 2.2.5 (i), but then you wouldn't need the adjoint representation. Am I understanding something wrong? \item \textbf{Proof of Corollary 4.1.3:} You write: In particular both parts of the equality are well defined.'' Why is the left hand side well-defined? \item \textbf{Lemma 4.1.4:} In part (ii), replace (resp. locally nilpotent element)'' by (resp. locally nilpotent) element''. \item \textbf{Proof of Lemma 4.1.5:} Replace $t\in C$'' by $t\in \mathbb{C}$''. \item \textbf{Proof of Lemma 4.1.5:} There are some opening brackets missing and/or some closing brackets too much in certain equations in this proof. For example: $\exp\left( \operatorname*{ad}y\right) )\left( x\right) )$. \item \textbf{Corollary 4.1.7:} Replace and locally nilpotent'' by are locally nilpotent''. \item \textbf{Proposition 4.2.2:} Replace integral'' by integrable''. \item \textbf{Proposition 4.2.2:} Replace $g_{\left( i\right) }$'' by $\mathfrak{g}_{\left( i\right) }$''. \item \textbf{Proposition 6.1.2:} In part (i), replace symmetrisable generalised Cartan matrix'' by a symmetrisable generalised Cartan matrix''. \item \textbf{Proposition 6.1.2:} In part (ii), replace symmetric'' by symmetrisable''. \item \textbf{Proposition 6.1.2:} In part (iii), replace symmetric indecomposable'' by symmetrisable indecomposable''. \item \textbf{Proof of Proposition 6.1.2:} Replace These solutions'' by These equations''. \item \textbf{Proof of Proposition 6.1.2:} Replace Furthermore because all the $a_{i_{j},i_{j+1}}$ are non negative'' by Furthermore because all the $a_{i_{j},i_{j+1}}$ and $a_{i_{j+1},i_{j}}$ are negative''. \item \textbf{Proof of Proposition 6.1.2:} Replace me may assume'' by we may assume''. \item \textbf{Proposition 6.1.3:} Replace for all sequence $i_{1}\cdots i_{k}$'' by for all sequences $\left( i_{1},\cdots,i_{k}\right)$''. \item Your use of American English vs. British English (realization'' vs. realisation'') is inconsistent. \item \textbf{Proposition 6.2.1:} The sentence Let $A$ be symmetrizable and indecomposable.'' could be better placed at the very beginning of this proposition, not inside part (i), because it concerns all three parts (i), (ii) and (iii). \item \textbf{Proposition 6.2.1:} In part (ii), replace resctriction'' by restriction''. \item \textbf{Proof of Proposition 6.2.1:} Replace Let $D=\operatorname*{Diag}\left( \epsilon_{i}\right)$ be a diagonal matrix'' by Let $D=\operatorname*{Diag}\left( \epsilon_{i}\right)$ be a nondegenerate diagonal matrix''. \item \textbf{Proof of Proposition 6.2.1:} When you write $\left( \alpha_{i}^{\vee},\alpha_{j}^{\vee}\right) =\left\langle \alpha_{i}% ,\alpha_{j}^{\vee}\right\rangle \epsilon_{i}=\left\langle \alpha_{j}^{\vee },\alpha_{i}\right\rangle \epsilon_{j}=\left( \alpha_{j}^{\vee},\alpha _{i}^{\vee}\right)$'', you should replace $\left\langle \alpha_{j}^{\vee },\alpha_{i}\right\rangle$ by $\left\langle \alpha_{j},\alpha_{i}^{\vee }\right\rangle$. \item \textbf{Proof of Proposition 6.2.1:} Replace $\left\langle \sum _{i}c_{i}\epsilon_{i}\alpha_{i}^{\vee},h^{\prime}\right\rangle =0$'' by $\left\langle \sum_{i}c_{i}\epsilon_{i}\alpha_{i},h^{\prime}\right\rangle =0$''. \item \textbf{Proof of Proposition 6.2.1:} In your proof of $\left( s_{i}\left( h\right) ,s_{i}\left( h^{\prime}\right) \right) =\left( h,h^{\prime}\right)$, you should replace $\left\langle \alpha_{i},h^{\prime }\right\rangle \left\langle \alpha_{i},h^{\prime}\right\rangle$ by $\left\langle \alpha_{i},h\right\rangle \left\langle \alpha_{i},h^{\prime }\right\rangle$. (This typo appears twice.) Also, replace $\left\langle \alpha_{i},h^{\prime}\right\rangle \left\langle h^{\prime},\alpha _{i}\right\rangle$ by $\left\langle \alpha_{i},h^{\prime}\right\rangle \left\langle h,\alpha_{i}\right\rangle$. \item \textbf{Proof of Proposition 6.2.1:} Replace Let us set $\epsilon _{i}=\left( \left( \alpha_{i},\alpha_{i}^{\vee}\right) \right) /2$'' by Let us set $\epsilon_{i}=\left( \left( \alpha_{i}^{\vee},\alpha_{i}^{\vee }\right) \right) /2$''. \item \textbf{Remark 6.2.3:} I do not see why $\left( \alpha_{i},\alpha _{i}\right) >0$ should hold unless we choose the $\epsilon_{i}$ positive in the construction of the form $\left( \cdot,\cdot\right)$. \item \textbf{Proof of Theorem 6.2.5:} Replace For $\alpha=\sum_{i}% \alpha_{i}$'' by For $\alpha=\sum_{i}k_{i}\alpha_{i}$''. \item \textbf{Proof of Theorem 6.2.5:} I think what you call $\left\vert \alpha\right\vert$ here is what you called $\operatorname*{ht}\alpha$ in Chapter 3. \item \textbf{Proof of Theorem 6.2.5:} You say: this proves the invariance since the other conditions all vanish''. This is not exactly the case (for example, the condition $\left( \left[ e_{i},h\right] ,f_{j}\right) =\left( e_{i},\left[ h,f_{j}\right] \right)$ does not vanish, nor does the condition $\left( \left[ f_{j},e_{i}\right] ,h\right) =\left( f_{j},\left[ e_{i},h\right] \right)$). Still it is probably fair to say that the other conditions are similarly proven. \item \textbf{Proof of Theorem 6.2.5:} You write: where all the elements $a$, $b$, $c$ and $d$ as well as the brackets $\left[ \left[ a,b\right] ,c\right]$, $\left[ b,\left[ c,d\right] \right]$, $\left[ \left[ a,c\right] ,b\right]$, $\left[ a,\left[ b,c\right] \right]$, $\left[ a,c\right]$, $\left[ b,d\right]$, $\left[ \left[ b,c\right] ,d\right]$ and $\left[ c,\left[ b,d\right] \right]$ are in $\mathfrak{g}\left( N-1\right)$''. This condition is not enough (for the proof at least); you also need $\left[ b,c\right]$ to lie in $\mathfrak{g}\left( N-1\right)$. \item \textbf{Proof of Theorem 6.2.5:} Replace $\left( \left[ \left[ s_{j},t_{j}\right] ,u_{i}\right] ,v_{j}\right)$ by $\left( \left[ \left[ s_{j},t_{j}\right] ,u_{i}\right] ,v_{i}\right)$. \item \textbf{Proof of Theorem 6.2.5:} You write: Then we have to define $\left( x,y\right)$ and $\left( y,x\right)$ for $x\in\mathfrak{g}_{N}$ and $y\in\mathfrak{g}_{-N}$''. But you define only $\left( x,y\right)$. This, of course, is easy to fix: just define $\left( y,x\right)$ to mean $\left( x,y\right)$. As a consequence of this definition, we see by induction that the form $\left( \cdot,\cdot\right)$ on $\mathfrak{g}\left( N\right) \times\mathfrak{g}\left( N\right)$ is symmetric. \item \textbf{Proof of Theorem 6.2.5:} You write: For the invariance, we still need to prove that for $x\in\mathfrak{g}_{N}$, for $y\in\mathfrak{g}_{-N}$ and for all $h$ we have the relations% $\left( x,\left[ h,y\right] \right) =\left( \left[ x,h\right] ,y\right) \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \left( \left[ x,y\right] ,h\right) =\left( x,\left[ y,h\right] \right) .$ '' This is not enough. First of all, I think you need also to prove the relation $\left( h,\left[ x,y\right] \right) =\left( \left[ h,x\right] ,y\right)$ (but that's easy: it follows from $\left( \left[ x,y\right] ,h\right) =\left( x,\left[ y,h\right] \right)$ using the symmetry of $\left( \cdot,\cdot\right)$ and the antisymmetry of $\left[ \cdot ,\cdot\right]$). Secondly, you also need to show that $\left( \left[ x,y\right] ,z\right) =\left( x,\left[ y,z\right] \right)$ holds whenever one of the vectors $x,y,z$ lies in either $\mathfrak{g}_{N}$ or $\mathfrak{g}_{-N}$ and the other two lie in $\mathfrak{g}\left( N-1\right)$. It seems to me that the latter part is easy, but I am not sure whether it immediately follows from the definition of $\left( x,y\right)$ as $\sum\limits_{i}\left( \left[ x,u_{i}\right] ,v_{i}\right) =\sum \limits_{j}\left( s_{j},\left[ t_{j},y\right] \right)$ (at least it does not follow without some rewriting using the symmetry of $\left( \cdot ,\cdot\right)$ and the antisymmetry of $\left[ \cdot,\cdot\right]$; and even then there are a lot of cases to consider). \item \textbf{Proof of Theorem 6.2.5:} In your proof of $\left( x,\left[ h,y\right] \right) =\left( \left[ x,h\right] ,y\right)$ (for $x\in\mathfrak{g}_{N}$, for $y\in\mathfrak{g}_{-N}$ and for all $h$), you should replace all $\sum_{i}$ signs by $\sum_{j}$ signs. \end{itemize} \end{document}