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\textbf{On blocks of Deligne's category }$\underline{\operatorname*{Rep}%
}\left( S_{t}\right) $\textbf{.}
\textit{Jonathan Comes and Victor Ostrik}
arXiv:0910.5695v2, version of 15 Oct 2010
\texttt{0910.5695v2.pdf}
\textbf{Errata I}
\bigskip
\end{center}
This list refers to the version of the paper on arXiv, not the (currently)
older version on Jonny Comes's website.
\begin{itemize}
\item \textbf{Page 2, \S 1.1:} Terminological question: is ``$F$-linear
bifunctor $\otimes:\mathcal{T}\times\mathcal{T}\rightarrow\mathcal{T}$'' really
the right terminology, or should ``$F$-linear'' be ``$F$-bilinear'' here? The
latter seems to give more google hits, but this isn't exactly an argument...
\item \textbf{Page 3:} You write: ``By a \emph{partition} $\pi$ of a finite set
$S$ we mean a collection $\pi_{1},\ldots,\pi_{n}$ of disjoint subsets of $S$
with $S=\bigcup_{i}\pi_{i}$.'' I think you should add ``nonempty'' before
``subsets'' here.
\item \textbf{Page 5:} When you write
\[
g(v_{\mbox{\boldmath${i}$}}):=\sum_{{\mbox{\boldmath${i}$}}^{\prime}\in\lbrack
m,d]}g_{{\mbox{\boldmath${i}$}}^{\prime}}^{\mbox{\boldmath${i}$}}%
v_{{\mbox{\boldmath${i}$}}^{\prime}}\qquad({\mbox{\boldmath${i}$}}\in\lbrack
n,d]),
\]
I think the ``$:=$'' should be an ``$=:$'', since it is the
$g_{{\mbox{\boldmath${i}$}}^{\prime}}^{\mbox{\boldmath${i}$}}$'s that you are
defining herein.
\item \textbf{Pages 6, 7 and 8:} The notation ``not among'' appears once on each
of these three pages, and I fear it is slightly ambiguous. When you say that
some objects are not among some other objects, you mean that the set of the
former objects is disjoint from the set of the latter objects. However, one
could misunderstand this notation as meaning that the set of the former
objects is not contained in the set of the latter objects...
\item \textbf{Page 11, Proposition 2.20:} Replace ``are two decomposition'' by
``are two decompositions''.
\item \textbf{Page 11, 2.3:} I am wondering why you write ``(not equal to
$\operatorname*{id}\nolimits_{0}$)'' here;\ I don't see how the $\pi
=\operatorname*{id}\nolimits_{0}$ case is any different from the rest...
\item \textbf{Page 12, Example 2.22:} In (ii), you can replace ``positive'' by
``nonnegative'' and nothing goes wrong (but I guess you don't care).
\item \textbf{Page 12, proof of Lemma 3.1:} I don't think the claim that
``$FS_{n}\cap\left( \zeta\right) =0$'' is completely obvious. The simplest
proof of $FS_{n}\cap\left( \zeta\right) =0$ that I know of uses a nontrivial
idea which does not appear in your paper (that of the propagating number of a
partition)\footnote{The proof goes as follows:
\par
For any partition $\pi\in P_{m,\ell}$, define the \textit{propagating number}
of $\pi$ as the number of all parts $S$ of $\pi$ satisfying both
$S\cap\left\{ 1,2,...,m\right\} \neq\varnothing$ and $S\cap\left\{
1^{\prime},2^{\prime},...,\ell^{\prime}\right\} \neq\varnothing$. Denote this
number by $\operatorname*{prop}\pi$. Then, it is easy to see that any two
partitions $\pi\in P_{m,\ell}$ and $\chi\in P_{\ell,k}$ satisfy
$\operatorname*{prop}\left( \chi\star\pi\right) \leq\operatorname*{prop}%
\chi$ and $\operatorname*{prop}\left( \chi\star\pi\right) \leq
\operatorname*{prop}\pi$. As a consequence, if we denote by $L_{n}$ the
$F$-vector subspace of $FP_{n}\left( t\right) $ spanned by all $\pi$ with
$\operatorname*{prop}\pi1$ have several
properties that $1$-cycles don't have. For example, given an $r$-cycle
$\left( k_{1},k_{2},...,k_{r}\right) $, we can uniquely reconstruct the
elements $k_{1},k_{2},...,k_{r}$ as being the elements not fixed under the
$r$-cycle.
This breaks down for $r=1$, and the proof breaks here.
So the formula (4.5) needs to be corrected to $S\left( \pi,r,d\right) =1$ in
the case when $r=1$. The interesting thing is that this seems to give two
different versions of $\omega_{n}^{1}\left( t\right) $ depending on whether
you modify the definition of $q_{\pi,r,t}$ accordingly (namely, by setting
$q_{\pi,1,t}=1$ whenever $S\left( \pi,1,d\right) \neq0$) or leave it as it
is. The one obtained by modifying the definition $q_{\pi,r,t}$ is simply
$\operatorname*{id}\nolimits_{n}\in FP_{n}\left( t\right) $. The one
obtained by leaving the definition of $q_{\pi,r,t}$ as it is, unfortunately,
does not satisfy Proposition 4.6 (not even its part (ii), so it does not
induce an endomorphism of the identity functor).
\item \textbf{Page 24, proof of Proposition 4.3:} Remove the ``in'' from ``Now,
an $r$-cycle in $\sigma\in S_{d}$''.
\item \textbf{Page 26, proof of Proposition 4.6:} Here you write: ``On the
other hand, (2.2) shows that $f(\omega_{n}^{r}(t))$ maps
$v_{\mbox{\boldmath${i}$}}\mapsto\sum_{{\mbox{\boldmath${i}$}}^{\prime}%
\in\lbrack n,d]}q_{\pi({\mbox{\boldmath${i}$}},{\mbox{\boldmath${i}$}}%
^{\prime}),r,d}v_{{\mbox{\boldmath${i}$}}^{\prime}}$.'' The $f(\omega_{n}%
^{r}(t))$ should be an $f(\omega_{n}^{r}(d))$ here.
\item \textbf{Page 26, one line above Proposition 4.9:} Replace comma by
semicolon in ``$\underline{\operatorname*{Rep}}\left( S_{t},F\right) $''.
\item \textbf{Page 27, Theorem 4.8:} Numerous things are slightly wrong here.
First, in (4.7), the product $(\mu_{i}+k-1)(\mu_{i}+k-2)\cdots(\mu_{i}+k-r)$
should be $(\mu_{i}+k)(\mu_{i}+k-1)\cdots(\mu_{i}+k-r+1)$. Check on $\lambda$
being a $1$-row partition (and $r\neq1$, see below).
Second, the formula, even corrected this way, does not hold for $r=1$;
instead, we have $\xi_{1,k}^{\lambda}=1$ (for obvious reasons).
Finally, this is not a real error, but I am pretty sure that ``$k$ is any
positive integer'' should be ``$k$ is any nonnegative integer''.
\item \textbf{Page 27, after Theorem 4.8:} I think that a good reference for
Theorem 4.8 (not in the form it appears in your text, but in a form very close
to the one in Fulton-Harris) is: Tullio Ceccherini-Silberstein, Fabio
Scarabotti, Filippo Tolli, \textit{Representation Theory of the Symmetric
Groups -- The Okounkov-Vershik Approach, Character Formulas, and Partition
Algebras}, CUP 2010, Proposition 4.2.11.
\item \textbf{Page 27, Proposition 4.9:} Replace comma by semicolon in
``$\underline{\operatorname*{Rep}}\left( S_{t},F\right) $''.
\item \textbf{Page 28, proof of Lemma 5.4:} There is a minor hole in the
proof. The formula
\[
\xi_{r,k}^{\lambda(t)}=\frac{1}{r}\sum_{i=0}^{k}\mu_{i}^{r}+\left(
\text{terms of total degree less than }r\right)
\]
holds only for $r>1$, so you need an extra argument to get the first power sum
of the $\mu_{0},...,\mu_{k}$ equal to the first power sum of the $\mu
_{0}^{\prime},...,\mu_{k}^{\prime}$. Fortunately, this argument is very
simple: Since%
\[
\sum_{i=0}^{k}\mu_{i}^{1}=\sum_{i=0}^{k}\mu_{i}=\underbrace{\sum_{i=0}%
^{k}\lambda_{i}}_{=t-\left\vert \lambda\right\vert +\left\vert \lambda
\right\vert =t}-\dfrac{k\left( k+1\right) }{2}=t-\dfrac{k\left( k+1\right)
}{2}%
\]
and (similarly) $\sum\limits_{i=0}^{k}\mu_{i}^{\prime1}=t-\dfrac{k\left(
k+1\right) }{2}$, we have $\sum\limits_{i=0}^{k}\mu_{i}^{1}=\sum
\limits_{i=0}^{k}\mu_{i}^{\prime1}$.
\item \textbf{Page 29, Proposition 5.5:} Replace ``$\lambda_{\mu}\left(
t\right) $'' by ``$\mu_{\lambda}\left( t\right) $''.
\item \textbf{Page 29, proof of Proposition 5.5:} Replace ``suppose $\mu
_{i}^{\prime}>\mu_{i+1}^{\prime}$ for all $i>0$'' by ``suppose $\mu_{i}^{\prime
}\in\mathbb{Z}$ and $\mu_{i}^{\prime}>\mu_{i+1}^{\prime}$ for all $i>0$''.
\item \textbf{Page 29, proof of Proposition 5.5:} Replace ``Set $\lambda
_{i}^{\prime}=\mu_{i}+i$'' by ``Set $\lambda_{i}^{\prime}=\mu_{i}^{\prime}+i$''.
\item \textbf{Page 29, proof of Proposition 5.5:} Replace ``Moreover, $\mu
_{i}>\mu_{i+1}$ for $i>0$'' by ``Moreover, $\mu_{i}^{\prime}>\mu_{i+1}^{\prime}$
for $i>0$''.
\item \textbf{Page 29, Example 5.7:} In part (2), replace ``$\lambda$'' by
``$\varnothing$''.
\item \textbf{Page 30, proof of Corollary 5.9:} After ``$\lambda_{1}^{(i)}%
=\mu_{0}+1=d-|\lambda^{(0)}|+1$'', add ``(for $i>0$)''.
\item \textbf{Page 32, Corollary 5.17:} Remove a superfluous period from ``a
nonnegative integer..''.
\item \textbf{Page 35, proof of Lemma 5.20:} Replace ``as in as in'' by ``as in''.
\item \textbf{Page 37, proof of Proposition 6.1:} Replace ``$\tau\mu=\pi$'' by
``$\tau\mu=\mu$''. Similarly, replace ``$-s_{n}\pi s_{n}$'' by ``$-s_{n}\mu s_{n}$''.
\item \textbf{Page 38, proof of Lemma 6.2:} You write: ``notice that for
$\sigma\in S_{n+1}$ there are exactly $(n-1)!$ pairs $(\tau_{1},\tau_{2})$
with $\tau_{1},\tau_{2}\in S_{n}$ such that $\tau_{1}x_{n}^{n-1}x_{n-1}%
^{n}\tau_{2}=x_{n}^{n+1}\sigma x_{n+1}^{n}$''. This is not precisely true. What
you mean is: ``notice that for $\sigma\in S_{n+1}$ satisfying $\sigma\left(
n+1\right) \neq n+1$ there are exactly $(n-1)!$ pairs $(\tau_{1},\tau_{2})$
with $\tau_{1},\tau_{2}\in S_{n}$ such that $\tau_{1}x_{n}^{n-1}x_{n-1}%
^{n}\tau_{2}=x_{n}^{n+1}\sigma x_{n+1}^{n}$ (whereas those $\sigma\in S_{n+1}$
which satisfy $\sigma\left( n+1\right) =n+1$ satisfy $x_{n}^{n+1}\sigma
x_{n+1}^{n}=0$ in $\underline{\operatorname*{Rep}}\left( S_{0};F\right) $)''.
\item \textbf{Page 38, proof of Lemma 6.2:} You write: ``Conversely, given any
$\tau_{1},\tau_{2}\in S_{n}$, either $\tau_{1}x_{n}^{n-1}x_{n-1}^{n}\tau
_{2}=0$ or there exists a unique $\sigma\in S_{n+1}$ with $\tau_{1}x_{n}%
^{n-1}x_{n-1}^{n}\tau_{2}=x_{n}^{n+1}\sigma x_{n+1}^{n}$.'' The first of these
two alternatives cannot happen. I think you want to say ``Conversely, given any
$\tau_{1},\tau_{2}\in S_{n}$, there exists a unique $\sigma\in S_{n+1}$ with
$\sigma\left( n+1\right) \neq n+1$ and $\tau_{1}x_{n}^{n-1}x_{n-1}^{n}%
\tau_{2}=x_{n}^{n+1}\sigma x_{n+1}^{n}$.''
\item \textbf{Page 38, Lemma 6.3:} I think the ``$\gamma_{n}$'' in (6) and in
(7) should be a ``$-\gamma_{n}$'', and the proof should be modified accordingly.
\item \textbf{Page 38, proof of Lemma 6.3:} In the proof of (3), you write:
``$\gamma_{n}=\sum_{\pi\in P_{n+1,n}}c_{\pi}\pi$''. The sum should run over
$P_{n,n}$, not $P_{n+1,n}$.
\item \textbf{Page 39, Proof of Proposition 6.6:} You write: ``First, notice
$d\not =0$ as we are assuming a nontrivial block exists in
$\underline{\operatorname*{Rep}}(S_{d};F)$.'' Don't you mean ``a non-minimal
nontrivial block'' rather than ``a nontrivial block'' here? After all,
$\underline{\operatorname*{Rep}}(S_{0};F)$ has a non-trivial block, too.
\item \textbf{Page 41, Appendix A:} You write: ``$A$ is a free $F[[u]]$-algebra
of finite rank''. What exactly is a free $F[[u]]$-algebra? Do you mean an
$F[[u]]$-algebra which is free as an $F[[u]]$-module? (I think so...)
\item \textbf{Page 41, proof of Lemma A.1:} Replace ``$ab\left( 1-\sum
_{i=1}^{\infty}x^{i}u^{i}\right) =1$'' by ``$ab\left( 1+\sum_{i=1}^{\infty
}\left( -x\right) ^{i}u^{i}\right) =1$''.
\item \textbf{Page 42, proof of Theorem A.2:} Replace ``$a_{i}^{2}%
-2(2a_{i}-1)b_{i}$'' by ``$a_{i}^{2}-2(2a_{i}-1)a_{i}b_{i}$'' in the long equation.
\item \textbf{References, [Fro68]:} The word ``B\"{a}nde'' is German for
``volumes''. I think you want the singular form ``Band'' (``volume'').
\end{itemize}
\end{document}