\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=Saturday, September 10, 2016 00:29:31} %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}} \newenvironment{proof2}[] {\begin{proof}[#1]}{\end{proof}} \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}} \newcommand\arxiv{\href{http://www.arxiv.org/abs/#1}{\texttt{arXiv:#1}}} \let\sumnonlimits\sum \let\prodnonlimits\prod \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits} \setlength\textheight{22.5cm} \setlength\textwidth{15cm} \ihead{Errata to Partition algebras''} \ohead{\today} \begin{document} \begin{center} \textbf{Partition Algebras} \textit{Tom Halverson and Arun Ram} [\texttt{halverson ram - partition algebras - 0401314v2.pdf}] version of 11 February 2004 (arXiv preprint \href{http://www.arxiv.org/abs/math/0401314v2}{\texttt{arXiv:math/0401314v2}}) \textbf{Darij's list of errata and comments} \bigskip \end{center} \begin{itemize} \item \textbf{Page 2:} Typo: \textquotedblleft partiton\textquotedblright. \item \textbf{Page 2:} Replace \textquotedblleft the algebras $A_{k}\left( n\right)$\textquotedblright\ by \textquotedblleft the algebras $\mathbb{C}A_{k}\left( n\right)$\textquotedblright. \item \textbf{Page 3:} In the definition of $A_{k}$ and $A_{k+\dfrac{1}{2}}$, replace \textquotedblleft$\mathbb{Z}_{>0}$\textquotedblright\ by \textquotedblleft$\mathbb{Z}_{\geq0}$\textquotedblright. Similarly, in many other places throughout the article (but not everywhere), \textquotedblleft% $\mathbb{Z}_{>0}$\textquotedblright\ can and should be replaced by \textquotedblleft$\mathbb{Z}_{\geq0}$\textquotedblright. (While the first two monoids $A_{0}$ and $A_{\dfrac{1}{2}}$ are not very interesting, you do use them -- e.g., they appear in the graph on page 14.) \item \textbf{Page 4:} The set $I_{k}$ is not a submonoid of $A_{k}$, but a \textbf{nonunital} submonoid\footnote{i.e., a subsemigroup} of $A_{k}$ (unlike $S_{k}$, $P_{k}$, $B_{k}$ and $T_{k}$, all of which are unital monoids). I don't think that you want to use the word \textquotedblleft monoid\textquotedblright\ (without qualification) for nonunital monoids, because if you do, then you would have to include the element $1$ in the presentation in Theorem 1.11 (a). \item \textbf{Page 5, (1.8):} Replace \textquotedblleft$\sum\limits_{\ell \geq0}C\left( \ell-1\right) z^{\ell}$\textquotedblright\ by \textquotedblleft$\sum\limits_{\ell\geq0}C\left( \ell\right) z^{\ell}%$\textquotedblright. \item \textbf{Page 5, (1.8):} Replace \textquotedblleft$\sum\limits_{\ell \geq0}\left( 2\left( \ell-1\right) \right) !!\dfrac{z^{\ell}}{\ell!}%$\textquotedblright\ by \textquotedblleft$\sum\limits_{\ell\geq0}\left( 2\ell\right) !!\dfrac{z^{\ell}}{\left( \ell+1\right) !}$\textquotedblright. \item \textbf{Page 6:} Here you introduce the notation $d_{1}d_{2}=d_{1}\circ d_{2}$, which is perfectly fine, but it would have been better to introduce it before, since it was already used on page 5 (when you wrote \textquotedblleft% $d=\sigma_{1}t\sigma_{2}$\textquotedblright). \item \textbf{Page 6, Theorem 1.11:} I think the generator $p_{\dfrac{1}{2}}$ occuring in parts (b) and (d) of Theorem 1.11 doesn't actually exist (at least you have never defined it!) and is not needed. I have not checked the proof, but I assume it can just be removed. \item \textbf{Page 8:} On the first line of page 8, replace \textquotedblleft the the\textquotedblright\ by \textquotedblleft the\textquotedblright. \item \textbf{Page 9, \S 2:} Replace \textquotedblleft$\mathbb{C}%$span-\textquotedblright\ by \textquotedblleft$\mathbb{C}$% -span\textquotedblright. \item \textbf{Page 9, (2.2):} Please explain that whenever $k\in\dfrac{1}% {2}\mathbb{Z}_{\geq0}$, you are abbreviating $\mathbb{C}A_{k}\left( n\right)$ by $\mathbb{C}A_{k}$. \item \textbf{Page 10:} The sentence \textquotedblleft The map $\varepsilon ^{\dfrac{1}{2}}$ is the composition $\mathbb{C}A_{k-\dfrac{1}{2}% }\hookrightarrow\mathbb{C}A_{k}\overset{\varepsilon_{1}}{\longrightarrow }\mathbb{C}A_{k-1}$\textquotedblright\ should be moved to below (2.4) (because it uses the map $\varepsilon_{1}$ which is only defined in (2.4)). \item \textbf{Page 10:} The \textquotedblleft$\operatorname*{tr}%$\textquotedblright\ in (2.7) and the \textquotedblleft$tr$\textquotedblright% \ on the line above should appear in the same font. \item \textbf{Page 12, (2.19):} Replace \textquotedblleft$\lambda\vdash n$\textquotedblright\ by \textquotedblleft$\lambda\vdash k$\textquotedblright. \item \textbf{Page 12, (2.20):} This equation should end with a comma, rather than with a period. \item \textbf{Page 13:} In the picture showing the first few levels of $\widehat{S}$, the \textquotedblleft k = 2\textquotedblright\ should be in mathmode. \item \textbf{Page 13:} \textquotedblleft Young tableaux of shape $\lambda$\textquotedblright\ should be \textquotedblleft Young tableaux of shape $\mu$\textquotedblright. \item \textbf{Page 13:} \textquotedblleft the box of $\lambda$% \textquotedblright\ should be \textquotedblleft the box of $\mu$% \textquotedblright. \item \textbf{Page 14:} In the picture showing the first few levels of $\widehat{A}$, the \textquotedblleft k = 2\textquotedblright\ should be in mathmode. \item \textbf{Page 16:} Replace \textquotedblleft for some constant $p$\textquotedblright\ by \textquotedblleft for some constant $k\in\mathbb{C}%$\textquotedblright. \item \textbf{Page 16:} Replace \textquotedblleft so that there are $A$-submodules\textquotedblright\ by \textquotedblleft so that there are nonzero $A$-submodules\textquotedblright. \item \textbf{Page 16:} It would be useful to replace \textquotedblleft If $p$ is an idempotent in $A$ and $Ap$ is a simple $A$-module\textquotedblright\ by \textquotedblleft If $p$ is an idempotent in a $\mathbb{C}$-algebra $A$ and $Ap$ is a simple $A$-module\textquotedblright\ to remind the reader that $A$ is a $\mathbb{C}$-algebra (this becomes particularly important here, because the $pAp=\mathbb{C}p$ claim requires the base ring to be algebraically closed). \item I have never figured out whether you require algebras to be unital in your paper or not. Sometimes it seems that you do (for example, on page 16, you write \textquotedblleft$\mathbb{C}\left( p\cdot1\cdot p\right)$\textquotedblright, which seems to assume there exists a $1$, although you could just as well avoid this by writing \textquotedblleft$\mathbb{C}\left( p\cdot p\cdot p\right)$\textquotedblright\ instead), and sometimes you definitely do (e.g., in (4.20a) you use the $1$ of $A$), but sometimes you definitely don't (e.g., when defining the basic construction you don't assume algebras to be unital, since the basic construction for $A$ and $B$ could be non-unital even when $A$ and $B$ are unital). \item \textbf{Page 17:} On the line just above (2.38), replace \textquotedblleft define the $\mathbb{Z}\left[ x\right]$-algebra by\textquotedblright\ by \textquotedblleft define the $\mathbb{Z}\left[ x\right]$-algebra $A_{k,\mathbb{Z}}$ by\textquotedblright. \item \textbf{Page 17, (2.39):} Replace \textquotedblleft$\mathbb{Z}$-module homomorphism\textquotedblright\ by \textquotedblleft$\mathbb{Z}$-algebra homomorphism\textquotedblright. (A $\mathbb{Z}$-module homomorphism $\mathbb{Z}\left[ x\right] \rightarrow\mathbb{C}$ would not be uniquely determined by where it takes $x$.) \item \textbf{Page 18, Proposition 2.43:} Replace \textquotedblleft% $\mathbb{Z}$-module homomorphism\textquotedblright\ by \textquotedblleft% $\mathbb{Z}$-algebra homomorphism\textquotedblright. \item \textbf{Page 18, (3.2):} Replace the summation index \textquotedblleft% $1\leq i_{1^{\prime}},...i_{k^{\prime}}\leq n$\textquotedblright\ by \textquotedblleft$1\leq i_{1^{\prime}},...,i_{k^{\prime}}\leq n$% \textquotedblright. \item \textbf{Page 19, proof of Theorem 3.6 (a):} Here it would be helpful to introduce the following notation you are using: The family $\left( v_{i_{1}}\otimes v_{i_{2}}\otimes\cdots\otimes v_{i_{k}% }\right) _{\left( i_{1},i_{2},\ldots,i_{k}\right) \in\left\{ 1,2,\ldots,n\right\} ^{k}}$ is a basis of the $\mathbb{C}$-vector space $V^{\otimes k}$. For every $b\in\operatorname*{End}\left( V^{\otimes k}\right)$ and every $\left( u_{1},u_{2},\ldots,u_{k}\right) \in\left\{ 1,2,\ldots,n\right\} ^{k}$ and every $\left( j_{1},j_{2},\ldots ,j_{k}\right) \in\left\{ 1,2,\ldots,n\right\} ^{k}$, we denote by $b_{j_{1},j_{2},\ldots,j_{k}}^{u_{1},u_{2},\ldots,u_{k}}$ the \newline$\left( v_{j_{1}}\otimes v_{j_{2}}\otimes\cdots\otimes v_{j_{k}}\right)$-coordinate of $b\left( v_{u_{1}}\otimes v_{u_{2}}\otimes\cdots\otimes v_{u_{k}}\right)$ (with respect to the basis $\left( v_{i_{1}}\otimes v_{i_{2}}\otimes \cdots\otimes v_{i_{k}}\right) _{\left( i_{1},i_{2},\ldots,i_{k}\right) \in\left\{ 1,2,\ldots,n\right\} ^{k}}$ of $V^{\otimes k}$). This coordinate $b_{j_{1},j_{2},\ldots,j_{k}}^{u_{1},u_{2},\ldots,u_{k}}$ is called the \textit{matrix entry} of $b$ at the \textit{matrix coordinates} \newline% $\left( \left( u_{1},u_{2},\ldots,u_{k}\right) ,\left( j_{1},j_{2}% ,\ldots,j_{k}\right) \right)$. This notation has the consequence that% $b\left( v_{i_{1}}\otimes v_{i_{2}}\otimes\cdots\otimes v_{i_{k}}\right) =\sum_{1\leq i_{1^{\prime}},i_{2^{\prime}},\ldots,i_{k^{\prime}}\leq n}b_{i_{1^{\prime}},i_{2^{\prime}},\ldots,i_{k^{\prime}}}^{i_{1},i_{2}% ,\ldots,i_{k}}v_{i_{1^{\prime}}}\otimes v_{i_{2^{\prime}}}\otimes\cdots\otimes v_{i_{k^{\prime}}}%$ for every $b\in\operatorname*{End}\left( V^{\otimes k}\right)$ and every $\left( i_{1},i_{2},\ldots,i_{k}\right) \in\left\{ 1,2,\ldots,n\right\} ^{k}$. Comparing this with (3.2), we conclude that every $d\in A_{k}$, every $\left( i_{1},i_{2},\ldots,i_{k}\right) \in\left\{ 1,2,\ldots,n\right\} ^{k}$ and every $\left( i_{1^{\prime}},i_{2^{\prime}},\ldots,i_{k^{\prime}% }\right) \in\left\{ 1,2,\ldots,n\right\} ^{k}$ satisfy% $\left( d\right) _{i_{1^{\prime}},i_{2^{\prime}},\ldots,i_{k^{\prime}}% }^{i_{1},i_{2},\ldots,i_{k}}=\left( \Phi_{k}\left( d\right) \right) _{i_{1^{\prime}},i_{2^{\prime}},\ldots,i_{k^{\prime}}}^{i_{1},i_{2}% ,\ldots,i_{k}}.$ \item \textbf{Page 20, proof of Theorem 3.6 (b):} Replace \textquotedblleft vertices $i_{k+1}$ and $i_{\left( k+1\right) ^{\prime}}$ must be in the same block of $d$\textquotedblright\ by \textquotedblleft vertices $k+1$ and $\left( k+1\right) ^{\prime}$ must be in the same block of $d$% \textquotedblright. \item \textbf{Page 25, proof of Theorem 3.27:} Replace \textquotedblleft is cannot be\textquotedblright\ by \textquotedblleft cannot be\textquotedblright. \item \textbf{Page 25, proof of Theorem 3.27:} I suppose \textquotedblleft Theorem Theorem 2.26(c)\textquotedblright\ should be \textquotedblleft Theorem 2.26(c)\textquotedblright. \item \textbf{Page 26, (3.32):} There seems to be one closing parenthesis too much here. \item \textbf{Page 31:} Replace \textquotedblleft statment\textquotedblright% \ by \textquotedblleft statement\textquotedblright. \item \textbf{Page 31:} Remove the comma at the end of (4.3). \item \textbf{Page 32:} Replace \textquotedblleft$a_{PQ}^{\mu}\leftarrow E_{PQ}^{\mu}$\textquotedblright\ by \textquotedblleft$% %TCIMACRO{\TeXButton{x}{\xymatrix{ %a^\mu_{PQ} & E^\mu_{PQ} \ar@{|->}[l] %}}}% %BeginExpansion \xymatrix{ a^\mu_{PQ} & E^\mu_{PQ} \ar@{|->}[l] }% %EndExpansion$\textquotedblright. \item \textbf{Page 32:} At the very end of (4.13), replace \textquotedblleft% $\varepsilon_{XY}^{\mu}a_{ST}^{\mu}$\textquotedblright\ by \textquotedblleft% $\delta_{\lambda\mu}\varepsilon_{XY}^{\mu}a_{ST}^{\mu}$\textquotedblright. \item \textbf{Page 33, (4.16):} Replace \textquotedblleft$\overrightarrow{a}% _{P}^{\mu}\otimes\overleftarrow{a}_{P}^{\mu}$\textquotedblright\ by \textquotedblleft$\overleftarrow{a}_{P}^{\mu}\otimes\overrightarrow{a}% _{P}^{\mu}$\textquotedblright. \item \textbf{Page 33, (4.17):} Replace \textquotedblleft$\overrightarrow{a}% _{W}^{\lambda}\otimes\overleftarrow{a}_{Z}^{\mu}$\textquotedblright\ by \textquotedblleft$\overleftarrow{a}_{W}^{\lambda}\otimes\overrightarrow{a}% _{Z}^{\mu}$\textquotedblright\ on the left-hand side of (4.17). Make similar replacements on the other sides (every time, the second tensorand should have an $\overleftarrow{a}$ and the third tensorand an $\overrightarrow{a}$). \item \textbf{Page 33:} Here you claim that \textquotedblleft$\left\{ \overline{m}_{XY}^{\mu}\ \mid\ \mu\in\widehat{A},\ X\in\widehat{R}^{\mu },\ Y\in\widehat{L}^{\mu}\right\}$ is a basis of $\overline{R}% \otimes_{\overline{A}}\overline{L}$\textquotedblright. It took me a while to understand why this holds. Here is my proof for it: Recall that $\overline {A}\cong\bigoplus\limits_{\mu\in\widehat{A}}M_{d_{\mu}}\left( \mathbb{F}% \right) =\bigoplus\limits_{\nu\in\widehat{A}}M_{d_{\nu}}\left( \mathbb{F}\right)$ as $\mathbb{F}$-algebras. Use this isomorphism to identify $\overline{A}$ with $\bigoplus\limits_{\nu\in\widehat{A}}M_{d_{\nu}% }\left( \mathbb{F}\right)$. Fix $\mu\in\widehat{A}$. Then, $\overleftarrow{A}_{\mu}$ is isomorphic to the right $\overline{A}$-module of length-$d_{\mu}$ row vectors over $\mathbb{F}$ on which the $M_{d_{\mu}% }\left( \mathbb{F}\right)$ addend of the direct sum $\bigoplus \limits_{\nu\in\widehat{A}}M_{d_{\nu}}\left( \mathbb{F}\right)$ acts by vector-matrix multiplication, whereas all other addends act as $0$. Similarly, $\overrightarrow{A}_{\mu}$ is isomorphic to the left $\overline{A}$-module of length-$d_{\mu}$ column vectors over $\mathbb{F}$ on which the $M_{d_{\mu}% }\left( \mathbb{F}\right)$ addend of the direct sum $\bigoplus \limits_{\nu\in\widehat{A}}M_{d_{\nu}}\left( \mathbb{F}\right)$ acts by matrix-vector multiplication, whereas all other addends act as $0$. From these descriptions of $\overleftarrow{A}_{\mu}$ and $\overrightarrow{A}_{\mu}$, it is easy to see that $\overleftarrow{A}_{\mu}\otimes_{\overline{A}% }\overrightarrow{A}_{\mu}\cong\mathbb{F}$ (as $\mathbb{F}$-vector spaces), and more precisely, that the one-element family $\left( \overleftarrow{a}% _{P}^{\mu}\otimes\overrightarrow{a}_{P}^{\mu}\right)$ is an $\mathbb{F}%$-vector space basis of $\overleftarrow{A}_{\mu}\otimes_{\overline{A}% }\overrightarrow{A}_{\mu}$ for every $P\in\widehat{A}^{\mu}$. Now, if we fix some $P\in\widehat{A}^{\mu}$, then the $\mathbb{F}$-vector space% $\underbrace{R^{\mu}}_{\substack{\text{this }\mathbb{F}\text{-vector space}\\\text{has basis }\left( r_{Y}^{\mu}\right) _{Y\in\widehat{R}^{\mu}}% }}\otimes\underbrace{\overleftarrow{A}^{\mu}\otimes_{\overline{A}% }\overrightarrow{A}^{\mu}}_{\substack{\text{this }\mathbb{F}\text{-vector space}\\\text{has basis }\left( \overleftarrow{a}_{P}^{\mu}\otimes \overrightarrow{a}_{P}^{\mu}\right) }}\otimes\underbrace{L^{\mu}% }_{\substack{\text{this }\mathbb{F}\text{-vector space}\\\text{has basis }\left( \ell_{X}^{\mu}\right) _{X\in\widehat{L}^{\mu}}}}$ clearly has basis% \begin{align*} & \left( r_{Y}^{\mu}\otimes\overleftarrow{a}_{P}^{\mu}\otimes \overrightarrow{a}_{P}^{\mu}\otimes\ell_{X}^{\mu}\right) _{Y\in \widehat{R}^{\mu},\ X\in\widehat{L}^{\mu}}\\ & =\left( \underbrace{r_{X}^{\mu}\otimes\overleftarrow{a}_{P}^{\mu}% \otimes\overrightarrow{a}_{P}^{\mu}\otimes\ell_{Y}^{\mu}}_{=\overline{m}% _{XY}^{\mu}}\right) _{X\in\widehat{R}^{\mu},\ Y\in\widehat{L}^{\mu}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we have renamed the indices }Y\text{ and }X\text{ as }X\text{ and }Y\right) \\ & =\left( \overline{m}_{XY}^{\mu}\right) _{X\in\widehat{R}^{\mu}% ,\ Y\in\widehat{L}^{\mu}}. \end{align*} Now, let us forget that we fixed $\mu$. We thus see that for every $\mu \in\widehat{A}$, the $\mathbb{F}$-vector space $R^{\mu}\otimes \overleftarrow{A}^{\mu}\otimes_{\overline{A}}\overrightarrow{A}^{\mu}\otimes L^{\mu}$ has basis $\left( \overline{m}_{XY}^{\mu}\right) _{X\in \widehat{R}^{\mu},\ Y\in\widehat{L}^{\mu}}$. Now, \begin{align*} & \underbrace{\overline{R}}_{=\bigoplus_{\mu\in\widehat{A}}R^{\mu}% \otimes\overleftarrow{A}^{\mu}}\otimes_{\overline{A}}\underbrace{\overline{L}% }_{=\bigoplus_{\mu\in\widehat{A}}\overrightarrow{A}^{\mu}\otimes L^{\mu}}\\ & =\left( \bigoplus_{\mu\in\widehat{A}}R^{\mu}\otimes\overleftarrow{A}^{\mu }\right) \otimes_{\overline{A}}\left( \bigoplus_{\mu\in\widehat{A}% }\overrightarrow{A}^{\mu}\otimes L^{\mu}\right) \cong\bigoplus_{\mu \in\widehat{A},\ \nu\in\widehat{A}}R^{\mu}\otimes\overleftarrow{A}^{\mu }\otimes_{\overline{A}}\overrightarrow{A}^{\nu}\otimes L^{\nu}\\ & =\bigoplus_{\mu\in\widehat{A}}\underbrace{R^{\mu}\otimes\overleftarrow{A}% ^{\mu}\otimes_{\overline{A}}\overrightarrow{A}^{\mu}\otimes L^{\mu}% }_{\substack{\text{this }\mathbb{F}\text{-vector space has basis}\\\left( \overline{m}_{XY}^{\mu}\right) _{X\in\widehat{R}^{\mu},\ Y\in\widehat{L}% ^{\mu}}}}\ \ \ \ \ \ \ \ \ \ \left( \text{since }\overleftarrow{A}^{\mu }\otimes_{\overline{A}}\overrightarrow{A}^{\nu}=0\text{ whenever }\mu\neq \nu\right) . \end{align*} If we regard the isomorphisms in this equality as identities, we thus conclude that the $\mathbb{F}$-vector space $\overline{R}\otimes_{\overline{A}% }\overline{L}$ has basis $\left( \overline{m}_{XY}^{\mu}\right) _{\mu \in\widehat{A},\ X\in\widehat{R}^{\mu},\ Y\in\widehat{L}^{\mu}}$, qed. \item \textbf{Page 34:} In the first displayed equation on this page, replace \textquotedblleft$\overline{n}_{XY}$\textquotedblright\ by \textquotedblleft% $\overline{n}_{XY}^{\mu}$\textquotedblright, and replace \textquotedblleft% $\overline{m}_{Q_{1}Q_{2}}$\textquotedblright\ by \textquotedblleft% $\overline{m}_{Q_{1}Q_{2}}^{\mu}$\textquotedblright. \item \textbf{Page 34:} Replace \textquotedblleft using (4.10) and (4.12)\textquotedblright\ by \textquotedblleft using (4.10) and (4.13)\textquotedblright. \item \textbf{Page 34:} Replace \textquotedblleft$\overrightarrow{a}% _{W}^{\lambda}\otimes\overleftarrow{a}_{W}^{\lambda}$\textquotedblright\ by \textquotedblleft$\overleftarrow{a}_{W}^{\lambda}\otimes\overrightarrow{a}% _{W}^{\lambda}$\textquotedblright\ in the chain of equalities below the words \textquotedblleft By direct computations\textquotedblright. Make similar replacements throughout this chain of equalities. \item \textbf{Page 34:} Replace \textquotedblleft$\overline{a}_{WZ}^{\lambda}%$\textquotedblright\ by \textquotedblleft$a_{WZ}^{\lambda}$\textquotedblright. \item \textbf{Page 34:} In \textquotedblleft$\dfrac{1}{\varepsilon _{T}^{\lambda}}\dfrac{1}{\varepsilon_{V}^{\mu}}n_{YT}^{\lambda}n_{UV}^{\mu }=\delta_{\lambda\mu}\delta_{TU}\dfrac{1}{\varepsilon_{T}^{\lambda}% \varepsilon_{V}^{\lambda}}\varepsilon_{T}^{\lambda}n_{YV}^{\lambda}%$\textquotedblright, replace the \textquotedblleft$=$\textquotedblright\ sign by an \textquotedblleft$\equiv$\textquotedblright\ sign. \item \textbf{Page 34:} You claim that \textquotedblleft the images of the elements $e_{YT}^{\lambda}$ in (4.7) form a set of matrix units in the algebra $\left( R\otimes_{A}L\right) /I$\textquotedblright. First, I think you should remove the words \textquotedblleft in (4.7)\textquotedblright\ here, because they are confusing (they sounds as if you mean the images under $\pi$, but instead you actually mean the images under the projection $R\otimes _{A}L\rightarrow\left( R\otimes_{A}L\right) /I$). Second, this might need some further explanation. You have proven that the images of the elements $e_{YT}^{\lambda}$ under the projection $R\otimes_{A}L\rightarrow\left( R\otimes_{A}L\right) /I$ multiply like matrix units, but it remains to show that these images form a basis of the $\mathbb{F}$-vector space $\left( R\otimes_{A}L\right) /I$ (in fact, a family of $0$'s also multiplies like matrix units, but does not constitute matrix units unless it is empty). However, this is not hard to show: We already know that $\left\{ \overline {m}_{XY}^{\mu}\ \mid\ \mu\in\widehat{A},\ X\in\widehat{R}^{\mu},\ Y\in \widehat{L}^{\mu}\right\}$ is a basis of $\overline{R}\otimes_{\overline{A}% }\overline{L}$. Consequently, $\left\{ \overline{n}_{XY}^{\mu}\ \mid\ \mu \in\widehat{A},\ X\in\widehat{R}^{\mu},\ Y\in\widehat{L}^{\mu}\right\}$ is a basis of $\overline{R}\otimes_{\overline{A}}\overline{L}$ as well (because the definition of $\overline{n}_{XY}^{\mu}$ shows that for every $\mu \in\widehat{A}$, we have the matrix equality \begin{align*} \left( \overline{n}_{XY}^{\mu}\right) _{X\in\widehat{R}^{\mu},\ Y\in \widehat{L}^{\mu}} & =\underbrace{\left( C_{ZW}^{\mu}\right) _{W\in\widehat{R}^{\mu},\ Z\in\widehat{R}^{\mu}}}_{\substack{\text{this is an invertible matrix}\\\text{(being the transpose of the invertible matrix }C^{\mu}\text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ \cdot\left( \overline{m}_{XY}^{\mu}\right) _{X\in\widehat{R}^{\mu},\ Y\in\widehat{L}^{\mu}}\cdot\underbrace{\left( D_{ST}^{\mu}\right) _{T\in\widehat{L}^{\mu},\ S\in\widehat{L}^{\mu}}% }_{\substack{\text{this is an invertible matrix}\\\text{(being the transpose of the invertible matrix }D^{\mu}\text{)}}} \end{align*} ). In other words, $\left\{ \pi\left( n_{XY}^{\mu}\right) \ \mid\ \mu \in\widehat{A},\ X\in\widehat{R}^{\mu},\ Y\in\widehat{L}^{\mu}\right\}$ is a basis of $\pi\left( R\otimes_{A}L\right)$ (since $\overline{n}_{XY}^{\mu }=\pi\left( n_{XY}^{\mu}\right)$ and $\overline{R}\otimes_{\overline{A}% }\overline{L}=\pi\left( R\otimes_{A}L\right)$). In other words, \newline$\left\{ \pi\left( n_{YT}^{\mu}\right) \ \mid\ \mu\in \widehat{A},\ Y\in\widehat{R}^{\mu},\ T\in\widehat{L}^{\mu}\right\}$ is a basis of $\pi\left( R\otimes_{A}L\right)$ (here, we renamed the indices $X$ and $Y$ as $Y$ and $T$). Therefore, the family% $\mathfrak{F}:=\left\{ k_{i},\ n_{YT}^{\mu}\ \mid\ \mu\in\widehat{A}% ,\ Y\in\widehat{R}^{\mu},\ T\in\widehat{L}^{\mu}\right\}$ is a basis of $R\otimes_{A}L$ (because $\left\{ k_{i}\right\}$ is a basis of $\ker\pi$). But the subfamily% $\mathfrak{G}:=\left\{ k_{i},\ n_{YT}^{\mu}\ \mid\ \mu\in\widehat{A}% ,\ Y\in\widehat{R}^{\mu},\ T\in\widehat{L}^{\mu},\ \left( \varepsilon _{Y}^{\mu}=0\text{ or }\varepsilon_{T}^{\mu}=0\right) \right\}$ of this latter family is a basis of $I$ (because $I$ was defined as the $\mathbb{F}$-span of $\mathfrak{G}$). Hence, the images of the elements of $\mathfrak{F}\setminus\mathfrak{G}$ under the projection $R\otimes _{A}L\rightarrow\left( R\otimes_{A}L\right) /I$ form a basis of $\left( R\otimes_{A}L\right) /I$. Since% \begin{align*} & \mathfrak{F}\setminus\mathfrak{G}\\ & =\left\{ k_{i},\ n_{YT}^{\mu}\ \mid\ \mu\in\widehat{A},\ Y\in \widehat{R}^{\mu},\ T\in\widehat{L}^{\mu}\right\} \\ & \ \ \ \ \ \ \ \ \ \ \setminus\left\{ k_{i},\ n_{YT}^{\mu}\ \mid\ \mu \in\widehat{A},\ Y\in\widehat{R}^{\mu},\ T\in\widehat{L}^{\mu},\ \left( \varepsilon_{Y}^{\mu}=0\text{ or }\varepsilon_{T}^{\mu}=0\right) \right\} \\ & =\left\{ n_{YT}^{\mu}\ \mid\ \mu\in\widehat{A},\ Y\in\widehat{R}^{\mu },\ T\in\widehat{L}^{\mu},\ \left( \text{neither }\varepsilon_{Y}^{\mu }=0\text{ nor }\varepsilon_{T}^{\mu}=0\right) \right\} , \end{align*} this rewrites as follows: The images of the elements $n_{YT}^{\mu}\text{ for }\mu\in\widehat{A},\ Y\in\widehat{R}^{\mu}% ,\ T\in\widehat{L}^{\mu}\text{ satisfying}\ \left( \text{neither }% \varepsilon_{Y}^{\mu}=0\text{ nor }\varepsilon_{T}^{\mu}=0\right)$ under the projection $R\otimes_{A}L\rightarrow\left( R\otimes_{A}L\right) /I$ form a basis of $\left( R\otimes_{A}L\right) /I$. But recall that we need to prove that the images of the elements% $e_{YT}^{\mu}\text{ for }\mu\in\widehat{A},\ Y\in\widehat{R}^{\mu}% ,\ T\in\widehat{L}^{\mu}\text{ satisfying}\ \left( \text{neither }% \varepsilon_{Y}^{\mu}=0\text{ nor }\varepsilon_{T}^{\mu}=0\right)$ under the projection $R\otimes_{A}L\rightarrow\left( R\otimes_{A}L\right) /I$ form a basis of $\left( R\otimes_{A}L\right) /I$. This immediately follows from the fact that the images of the elements $n_{YT}^{\mu}\text{ for }\mu\in\widehat{A},\ Y\in\widehat{R}^{\mu}% ,\ T\in\widehat{L}^{\mu}\text{ satisfying}\ \left( \text{neither }% \varepsilon_{Y}^{\mu}=0\text{ nor }\varepsilon_{T}^{\mu}=0\right)$ under the projection $R\otimes_{A}L\rightarrow\left( R\otimes_{A}L\right) /I$ form a basis of $\left( R\otimes_{A}L\right) /I$ (because $e_{YT}^{\mu }=\dfrac{1}{\varepsilon_{T}^{\mu}}n_{YT}^{\mu}$ differs from $n_{YT}^{\mu}$ only in a nonzero multiplicative factor). This completes the proof of your claim that \textquotedblleft the images of the elements $e_{YT}^{\lambda}$ in (4.7) form a set of matrix units in the algebra $\left( R\otimes_{A}L\right) /I$\textquotedblright. \item \textbf{Page 35:} You write: \textquotedblleft Let $A\subseteq B$ be an inclusion of algebras\textquotedblright. I think this is one of the places where you want $A$ and $B$ (or $B$ at least) to be unital, or else (4.20a) and (4.20c) don't make sense. \item \textbf{Page 35, (4.20c):} After \textquotedblleft$pAp=\mathbb{F}% p$\textquotedblright, add \textquotedblleft and $p$ is an idempotent\textquotedblright. \item \textbf{Page 35, (4.22):} It would help to explain that your notation $P\rightarrow\mu\rightarrow\lambda$ is shorthand for a pair $\left( P\rightarrow\mu,\mu\rightarrow\lambda\right)$ of an element $P\rightarrow \mu$ of $\widehat{A}^{\mu}$ and an edge $\mu\rightarrow\lambda$ of $\Gamma$. (Anyway, I am wondering why you don't define an extended graph $\widehat{\Gamma}$ which consists of $\Gamma$ and an additional vertex $\mathbb{F}$, and which has the same edges as $\Gamma$ and, additionally, $\left\vert \widehat{A}^{\mu}\right\vert$ edges from $\mathbb{F}$ to $\mu$ for every $\mu\in\widehat{A}$. Then, you could identify $\widehat{B}^{\lambda }$ with the set of edges from $\mathbb{F}$ to $\lambda$ in this graph $\widehat{\Gamma}$ for every $\lambda\in\widehat{B}$.) \item \textbf{Page 36, (4.24):} Replace \textquotedblleft$\delta _{\lambda\sigma}\delta_{QS}\delta_{\gamma\tau}b_{\substack{PT\\\mu\nu \\\lambda}}$\textquotedblright\ by \textquotedblleft$\delta_{\lambda\sigma }\delta_{Q\rightarrow\gamma,S\rightarrow\tau}\delta_{\gamma\tau}\delta _{\gamma\rightarrow\lambda,\ \tau\rightarrow\sigma}b_{\substack{PT\\\mu \nu\\\lambda}}$\textquotedblright. (The $\delta_{\gamma\rightarrow \lambda,\ \tau\rightarrow\sigma}$ factor is important; there might be several edges from $\gamma$ to $\lambda$, and they give rise to different matrix elements.) \item \textbf{Page 38:} Replace \textquotedblleft The rese\textquotedblright% \ by \textquotedblleft The rest\textquotedblright. \item \textbf{Page 39, \S 5:} In the definition of \textquotedblleft trace\textquotedblright, replace \textquotedblleft linear\textquotedblright% \ by \textquotedblleft$\overline{\mathbb{F}}$-linear\textquotedblright. \item \textbf{Page 39, \S 5:} In the definition of \textquotedblleft nondegenerate\textquotedblright, replace \textquotedblleft for each $b\in A$\textquotedblright\ by \textquotedblleft for each nonzero $b\in A$\textquotedblright. \item \textbf{Page 39, Lemma 5.1:} The notations here conflict with the notations introduced just a few moments earlier. For example, you want the trace $\overrightarrow{t}$ in Lemma 5.1 to be an $\mathbb{F}$-linear map $A\rightarrow\mathbb{F}$, whereas you previously defined a trace as an $\overline{\mathbb{F}}$-linear map $\overline{A}\rightarrow\overline {\mathbb{F}}$. It would probably best to define the notions of \textquotedblleft trace\textquotedblright\ and \textquotedblleft nondegenerate\textquotedblright\ over arbitrary fields first, and only then apply them to the case of $\overline{\mathbb{F}}$. \item \textbf{Page 39, proof of Lemma 5.1:} Replace \textquotedblleft% $\overline{\mathbb{F}}$\textquotedblright\ by \textquotedblleft$\mathbb{F}%$\textquotedblright. \item \textbf{Page 39, proof of Lemma 5.1:} Replace \textquotedblleft the columns of $G$ are linearly dependent\textquotedblright\ by \textquotedblleft the rows of $G$ are linearly dependent\textquotedblright. \item \textbf{Page 40, Proposition 5.2:} In part (a), replace \textquotedblleft$\operatorname*{Hom}\nolimits_{\overline{\mathbb{F}}}%$\textquotedblright\ by \textquotedblleft$\operatorname*{Hom}% \nolimits_{\mathbb{F}}$\textquotedblright. \item \textbf{Page 43, proof of Theorem 5.8:} I would replace \textquotedblleft vacuously true\textquotedblright\ by \textquotedblleft obviously true\textquotedblright. (\textquotedblleft Vacuously true\textquotedblright\ means that the conditions can never be satisfied; this is probably not what you meant.) \item \textbf{Page 43, proof of Theorem 5.8:} Replace \textquotedblleft a proper submodule $N$\textquotedblright\ by \textquotedblleft a proper nonzero submodule $N$\textquotedblright. \item \textbf{Page 44, proof of Theorem 5.8:} Replace \textquotedblleft complementary to $M$\textquotedblright\ by \textquotedblleft complementary to $N$ in $M$\textquotedblright. \item \textbf{Page 44, Theorem 5.10:} Remove the comma in \textquotedblleft% $\mathbb{F}$, the field of fractions\textquotedblright. \item \textbf{Page 44, Theorem 5.10:} Remove the comma in \textquotedblleft and $\overline{R}$, the integral closure\textquotedblright. \item \textbf{Page 44, Theorem 5.10:} Replace \textquotedblleft$t_{1}% \overrightarrow{A}\left( b_{1}\right) +\cdots t_{d}\overrightarrow{A}\left( b_{d}\right)$\textquotedblright\ by \textquotedblleft$t_{1}% \overrightarrow{A}\left( b_{1}\right) +\cdots+t_{d}\overrightarrow{A}\left( b_{d}\right)$\textquotedblright. \item \textbf{Page 45, Theorem 5.10 (a):} Replace the \textquotedblleft% $\longmapsto$\textquotedblright\ arrow by a \textquotedblleft$\longrightarrow$\textquotedblright\ arrow in \textquotedblleft$A_{\overline{\mathbb{F}}% }\longmapsto\overline{\mathbb{F}}$\textquotedblright. \item \textbf{Page 45, Theorem 5.10 (a):} Replace \textquotedblleft be the extension\textquotedblright\ by \textquotedblleft be an extension\textquotedblright. \item \textbf{Page 45, Theorem 5.10 (b):} Replace the \textquotedblleft% $\longmapsto$\textquotedblright\ arrow by a \textquotedblleft$\longrightarrow$\textquotedblright\ arrow in \textquotedblleft$A_{\overline{\mathbb{K}}% }\longmapsto\overline{\mathbb{K}}$\textquotedblright. \end{itemize} \end{document}