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\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}