Here are some beginner questions on partition algebras...

I am trying to understand the monoid called $P_k$ in Tom Halverson, Arun Ram, *Partition Algebras*. For the sake of simplicity, let $k$ be a nonnegative integer (not just a half-integer). Then, $P_k$ is defined as the monoid of all "planar" set partitions of $\left\lbrace 1,2,...,k,1^{\prime},2^{\prime},...,k^{\prime} \right\rbrace$. Here are my questions:

`UPDATE:` Question 1 and Question 2 are answered.

**Question 1:**

What exactly does "planar" mean combinatorially? (I don't really trust pictures, particularly if they are not unique.) Am I seeing it right that a set partition $\pi$ is planar if and only if it satisfies the following five conditions, where $\equiv$ denotes the equivalence relation "lie in the same part of $\pi$":

**(A)** If $a$, $b$, $c$, $d$ belong to $\left\lbrace 1,2,...,k\right\rbrace$ and satisfy $a < b < c$, $a \equiv c$ and $b\equiv d$ but not $a < d < c$, then $a \equiv b \equiv c \equiv d$. [This is a noncrossing property for edges between vertices on the upper edge of the diagram.]

**(B)** If $a$, $b$, $c$, $d$ belong to $\left\lbrace 1,2,...,k\right\rbrace$ and satisfy $a < b < c$, $a^{\prime} \equiv c^{\prime}$ and $b^{\prime}\equiv d^{\prime}$ but not $a < d < c$, then $a^{\prime} \equiv b^{\prime} \equiv c^{\prime} \equiv d^{\prime}$. [This is a noncrossing property for edges between vertices on the lower edge of the diagram.]

**(C)** If $a$, $b$, $c$, $d$ belong to $\left\lbrace 1,2,...,k\right\rbrace$ and satisfy $a < b < c$, $a \equiv c$ and $b\equiv d^{\prime}$, then $a \equiv b \equiv c \equiv d^{\prime}$. [This is a noncrossing property for edges between vertices on the upper edge of the diagram and up-down edges.]

**(D)** If $a$, $b$, $c$, $d$ belong to $\left\lbrace 1,2,...,k\right\rbrace$ and satisfy $a < b < c$, $a^{\prime} \equiv c^{\prime}$ and $b^{\prime}\equiv d$, then $a^{\prime} \equiv b^{\prime} \equiv c^{\prime} \equiv d$. [This is a noncrossing property for edges between vertices on the upper edge of the diagram and up-down edges.]

**(E)** If $a$, $b$, $c$, $d$ belong to $\left\lbrace 1,2,...,k\right\rbrace$ and satisfy $a < c$, $d < b$, $a \equiv b^{\prime}$ and $c\equiv d^{\prime}$, then $a \equiv b^{\prime} \equiv c \equiv d^{\prime}$. [This is a noncrossing property for up-down edges.]

Sorry for not being able to draw what I mean...

`UPDATE:` Here is another way to restate my question: Am I seeing it right that a set partition $\pi$ of $\left\lbrace 1,2,...,k,1^{\prime},2^{\prime},...,k^{\prime} \right\rbrace$ is planar if and only if labelling the vertices of a regular $2k$-gon by the labels $1$, $2$, ..., $k$, $k^{\prime}$, $\left(k-1\right)^{\prime}$, ..., $1^{\prime}$ in this order and connecting every pair of equivalent vertices (equivalent with respect to the equivalence relation that is $\pi$) gives us a bunch of segments with the following property: If two of the segments intersect at an interior point, then all four of their endpoints are equivalent with respect to $\pi$.

`UPDATE:` **Answer to Question 1:** The answer to Question 1 is "Yes". This follows from the fact that if $A$, $B$, $C$ and $D$ are four distinct points lying on a circle **in this order**, then any path from $A$ to $C$ which is contained in the (closed) disk encompassed by the circle must intersect any path from $B$ to $D$ which is contained in the (closed) disk encompassed by the circle. I don't really know how this fact is proven, but I don't think it has any chances to be wrong, and I suspect it somehow follows from the Jordan curve theorem.

For an alternative argument, one could notice that if $C\left(\ell\right)$ denotes the $\ell$-th Catalan number for every nonnegative integer $\ell$, then there are precisely $C\left(2k\right)$ partitions $\pi$ of $\left\lbrace 1,2,...,k,1^{\prime},2^{\prime},...,k^{\prime} \right\rbrace$ satisfying the conditions **(A)**, **(B)**, **(C)**, **(D)** and **(E)** (this can be proven combinatorially by more or less the argument by which Halverson and Ram prove (1.5) in their paper), whereas we know (from formula (1.9) in Halverson-Ram) that there are precisely $C\left(2k\right)$ planar partitions of $\left\lbrace 1,2,...,k,1^{\prime},2^{\prime},...,k^{\prime} \right\rbrace$. Since every partition $\pi$ of $\left\lbrace 1,2,...,k,1^{\prime},2^{\prime},...,k^{\prime} \right\rbrace$ satisfying the conditions **(A)**, **(B)**, **(C)**, **(D)** and **(E)** is planar, this yields that every planar partition $\pi$ of $\left\lbrace 1,2,...,k,1^{\prime},2^{\prime},...,k^{\prime} \right\rbrace$ satisfies the conditions **(A)**, **(B)**, **(C)**, **(D)** and **(E)**. In other words, these conditions determine exactly the planar partitions. Of course, this argument is roundabout because the proof of Halverson-Ram's (1.9) probably goes through something like my conditions, but when it comes to arguments like this I'm more inclined to trust Halverson-Ram than my lying eyes.

**Question 2:**

Are there any subsets of these five properties which already determine a monoid under composition? (I used to believe that the partitions satisfying **(E)** form a monoid, but this has proven illusory.) Can we enumerate the partitions satisfying some of these subsets of conditions? (Admittedly this is quite a fishing expedition, with $2^5$ possible subsets of five conditions to choose from.)

`UPDATE:` **Answer to Question 2:** Question **2.** was stupid. For any nonempty proper subset of {**(A)**, **(B)**, **(C)**, **(D)**, **(E)**}, the partitions $\pi$ satisfying the conditions in that subset don't form a monoid. This holds even if the partitions are required to be perfect matchings, i. e., belong to the Brauer monoid.

**Question 3:**

Do these "planar" versions of partition algebras have applications to the study of the "non-geometric" ones (by "non-geometric" I mean those whose definitions are easier to cast in combinatorial than in geometric terms)? Are there Schur-Weyl dualities for planar partition algebras? What is a non-physicist's (I assume physicists care for the Temperley-Lieb algebra) motivation to study them, apart from combinatorial curiosity (which is my motivation right now)?

3.yesterday; once I've understood it well enough, I'll post it here as CW. $\endgroup$ – darij grinberg Apr 4 '13 at 16:38