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Preliminary Notes (Click Me) %$\mathbb{L,M,N,Q,R,S,T,Z}$ \section{Representation theory} Fix a field $k$ and an element $\delta \in k$. This fixes a partition category $\caP$ and, for each $n \in \mathbb{N}$, the partition algebra $P_n$ as the $n$-th End-set $\caP(n,n)$ in $\caP$.
Write $\circc$ for the monoidal composition in $\caP$, given by side-by-side (as opposed to vertical) composition of partition diagrams.
Write $1_n$ for the identity element in $\caP(n,n)$.
\subsection{Notations, pictures and decompositions for set partitions} %Write $P^{\circ}(n,m)$ denotes the basis of hom-set $\caP(n,m)$ consisting of set partitions (of the set consisting of a row of $n$ and a row of $m$ `vertices', labelled $1,2,...,n$ and $1',2',...,m'$ respectively).
%Write ${\pbase}^l (n,m)$ denotes the subset of ${\pbase} (n,m)$ consisting of partitions with $l$ propagating parts.
Note that in $p \in {\pbase}^l (n,l)$ each primed vertex lies in a distinct propagating part.
Remark: Fix a partition $p$ in ${\pbase} (n,m)$. Each propagating part has a lowest numbered vertext from $1,2,...,n$ and a lowest numbered vertex from $1',2',...,m'$.
One could draw a picture for $p$ (i.e. a relative graph $g$ such that $\pp(g)=p$) in which the only lines drawn between the unprimed and the primed side correspond to propagating parts, and are drawn between lowest-numbered unprimed vertex and lowest-numbered primed vertex in each case.
Indeed one could draw a picture of $p$ in which the only lines drawn between unprimed (resp. primed) vertices are between pairs adjacent in the natural order in the same part. See Fig.\ref{fig0x} for example. We will call pictures drawn in this way low-neat . \begin{proposition} Every $p \in {\pbase}^l (n,m)$ can be expressed in the form $p = p_L * p_R$ where $p_L \in {\pbase}^l (n,l)$ and $p_R \in {\pbase}^l (l,m)$. \end{proposition} Proof: use the remarks about pictures above.

Remark: the decomposition $p = p_L * p_R$ above is not unique in general. In order to have a useful unique decomposition we need to organise partitions in a slightly non-canonical way, for example as follows. The `low-shell' of $p \in {\pbase} (n,m)$ is the partition obtained by replacing each propagating part by a part containing just the two lowest numbered elements (one primed, one unprimed), with all other elements being singletons. We say two propagating parts are low-non-crossing if their image in the low-shell is non-crossing in the obvious sense. We say $p$ is low-non-crossing if each pair of propagating parts is low-non-crossing. For example $1_n$ is (equal to its low-shell and) low-non-crossing. Low-shell example: \begin{figure} \caption{\label{fig0x} A partition drawn low-neat, and its low-shell} \includegraphics[width=3in]{xfig/partit-low1.gif} \end{figure} \begin{figure} \caption{\label{fig1x} A partition in $\pbase^{3}(7,3)$ and its low-shell} \includegraphics[width=2in]{xfig/partit-low2.gif} \end{figure} Note that the middle (dashed) horizontal line drawn in Fig.\ref{fig1x} separates the partition into a low-non-crossing partition and a permutation (in $S_3$, or in $\pbase^{3}(3,3)$, as it were). Note that $p$ can be low-non-crossing without being a plane partition.

%Let ${\pbase}^{=l}(n,m)$ denotes the subset of ${\pbase}^l (n,m)$ consisting of low-non-crossing elements. \begin{proposition} Every $p \in {\pbase}^{l} (n,l)$ can be expressed in the form $p_L * p_R$ where $p_L \in {\pbase}^{=l} (n,l)$ and $p_R \in {\pbase}^l (l,l)$. \end{proposition} Proof: Elements $p_R$ of the given form can be used to `uncross any crossing'. And they are invertible. \begin{proposition} The category composition defines bijections \begin{eqnarray} * : {\pbase}^{l} (n,l) \times {\pbase}^{=l} (l,m) & \rightarrow & {\pbase}^{l} (n,m) \label{eq01} \\ (p_L , p_R ) & \mapsto & p_L * p_R \end{eqnarray} \begin{eqnarray} ** : {\pbase}^{=l} (n,l) \times {\pbase}^l (l,l) \times {\pbase}^{=l} (l,m) & \rightarrow & {\pbase}^{l} (n,m) \\ (p_L , p_C , p_R) & \mapsto & p_L * p_C * p_R \end{eqnarray} \end{proposition} \subsection{Useful elements of $P_n$} Fix a natural inclusion of algebras \[ P_{n-1} \subset P_{n} \] by $ p \mapsto p \cup \{ \{ n , n' \}\} \; = p \circc 1_1 $. The identity element in $P_0$ is the empty partition. Note that this and the (iterated) inclusion map above determines the identity in $P_n$. Define an element \[ %a_1 = \{ \{ 1 \}, \{ 1' \}, ... \} a_1 = \{ \{ 1 \}, \{ 1' \} \} \in P_1 \] Define $a_1$ in $P_n$ by inclusion (so it is otherwise the same as the identity element). That is $a_1 = \{ \{ 1 \} , \{ 1' \} \} \otimes 1_{n-1}$ in $P_n$. Define $a_i$ similarly. Assuming (for now) that $\delta$ is a unit in $k$, define \[ e_i = \frac{1}{\delta} a_i \] Note that this is idempotent. Define $E_n = e_1 e_2 ... e_n$ and $E_n^l = e_1 e_2 ... e_{n-l}$. (The point of this notation is that, in $P_n$, $E_n^l$ is a scalar multiple of a partition with $l$ propagating components.) Note the useful isomorphism \begin{equation} E_n^l P_n E_n^l \cong P_l . \end{equation} In particular we have \[ %e_1 e_2 ... e_n E_n P_n E_n = k E_n \] Now for any ring $R$ and idempotent $e \in R$, $eRe$ a local ring implies that $e$ is primitive. Thus, $E_n$ is a primitive idempotent. \subsection{Ideals} Consider the chain of ideals \[ P_n \supset P_n e_1 P_n \supset P_n e_1 e_2 P_n \supset ... \supset P_n e_1 e_2 ... e_n P_n \] Define $P_n^l = P_n e_1 e_2 ... e_{n-l} P_n$. (Explicitly, $P_n^n = P_n$, $P_n^{n-1} = P_n e_1 P_n$, ... , $P_n^{0} = P_n e_1 e_2 ... e_n P_n$, and we take also $P_n^{-1} = 0$.)
Proposition. The ideal $P_n^l$ is spanned by partitions with at most $l$ propagating components.
Proof: Exercise.
Define quotient algebra $$ P_n^{/l} = P_n / P_n^l . $$ This is like $P_n$, except that any partition basis element with $l$ or fewer propagating components is congruent to 0.
In particular the idempotent $E_n^m = 0$ unless $m > l$ in $P_n^{/l}$. Let $S_n$ denote the symmetric group. \begin{proposition} We have isomorphisms of $k$-algebras \[ E_n^{l+1} P_n^{/l} E_n^{l+1} \cong P_{l+1}^{/l} \cong k S_{l+1} \] \end{proposition} Proof: We can choose a subset of the partition basis of $P_n$ to form a basis for the algebra on the left. No such element on the left can have more than $l+1$ propagating components by virtue of the factor $E_n^{l+1}$; nor less than $l+1$ by virtue of the quotient. Consider a complete orthogonal idempotent decomposition of $1 \in kS_{l+1}$: $$ 1 = \sum_\lambda f_\lambda $$ % in $kS_{l+1}$ Recall that such a decomposition exists for any artinian ring; and that each $f_\lambda kS_{l+1} f_\lambda$ is a local ring - see e.g. [AndersonFuller]. In our case, if $k = \C$ say, the idempotents may be indexed by $\lambda \in \Lambda_{l+1}^{st}$, the set of standard Young tableau associated to all the integer partitions of $l+1$.
This will now give us a certain collection of primitive idempotents in $P_n^{/l}$: for any $\lambda \in \Lambda_{l+1}^{st}$ setting $$ F_\lambda = E_{n-({l+1})} \circc f_{\lambda} = E_n ^{l+1} (1_{n-(l+1)} \circc f_\lambda ) $$ (in the side-by-side notation) we have \[ F_\lambda P_n^{/l} F_\lambda = ( E_{n-({l+1})} \circc f_\lambda ) P_n^{/l} ( E_{n-({l+1})} \circc f_\lambda ) = E_{n -( {l+1})} \circc ( f_\lambda P_{l+1}^{/l} f_\lambda ) \cong f_\lambda (k S_{l+1} ) f_\lambda %= kf_\lambda \] a local ring. Thus the left ideal \[ \Pi_\lambda = P_n^{/l} F_\lambda \] is indecomposable projective in the quotient $P_n^{/l}$. Hence for each $l\in \{ -1,0,...,n-1 \}$ and each $\lambda \in \Lambda_{l+1}$ the module $\Pi_\lambda$ is also an indecomposable $P_n$-module with simple head. \subsection{Modules} Define $P_n,P_n$-bimodule \[ \bar{P}_n^l = P_n^l / P_n^{l-2} \] Note that this has a basis $\pbase^l(n,n)$ (of the $(n,n)$-partitions with $l$ propagating parts). %`lines' (components). We have the following useful decomposition as a left-module \[ \bar{P}_n^l = \oplus_{p_R \in \pbase^{=l}(l,n)} k \pbase^{l}(n,l) * p_R \] Cf. Equation(\ref{eq01}), noting that each $k \pbase^{l}(n,l) * p_R$ is a $P_n$-module by virtue of the quotient (action of $P_n$ from the left on some $p_L * p_R$ does not change $p_R$ unless it reduces the number of propagating lines, in which case the action gives 0). It will be clear that each of these summands is isomorphic, so we can study $\bar{P}_n^l$ as a left-module by studying $k \pbase^{l}(n,l) * p_R$, or indeed $k \pbase^{l}(n,l) $. Note that the latter is also a right $kS_l$-module in an obvious way, indeed it is a free module. Thus we have a functor \[ S : kS_l - mod \rightarrow P_n-mod \] given by $M \mapsto k \pbase^{l}(n,l) \otimes_{kS_l} M$. The idempotent decomposition of 1 in $kS_l$ as above induces a decomposition \[ k \pbase^{l}(n,l) \cong \oplus_\lambda \Pi_\lambda \] Altogether then, we see that the left regular module may be filtered by the indecomposable modules of form $\Pi_\lambda$ (all possible $\lambda$). In the complex case (for example) it will be clear that $kS_l f_\lambda \cong kS_l f_\mu$ if and only if the underlying Young diagrams of $\lambda$ and $\mu$ are the same. Thus a complete set of $\Pi_\lambda$ modules is obtained by taking one of each shape. It will be clear that $e_1 P_n e_1 \cong P_{n-1}$. Indeed $e_i P_n e_i \cong P_{n-1}$ for any suitable $i$. It follows that we have a functor \[ G : P_{n-1}-mod \rightarrow P_n-mod \] given by $M \mapsto P_n e_n \otimes M$. Next we will show that $G(\Pi_\lambda ) = \Pi_\lambda$ (the reader should take care to consider the meaning of this identity). We will be able to use this to determine the structure of $P_n$ by induction on $n$ (using the fact that $P_0$ and $P_1$ are simple base cases which may be analysed by brute force). ...