# Semi-planar partition monoid/algebra

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)?

-
Ben Elias has given me an answer to 3. yesterday; once I've understood it well enough, I'll post it here as CW. – darij grinberg Apr 4 '13 at 16:38

I hope I am answering the question:

1. Why not trust the diagrams?

2. It seems to me that the source of your confusion is that planar refers to the whole diagram and not just the top or bottom row. So planar means that all your conditions are required. You should identify $P_k$ with set partitions of {$1,\cdots ,2k$}.

3. The planar partition algebra is isomorphic to a Temperley-Lieb algebra. The planar partition algebra arises in the Potts model and the Temperley-Lieb algebras in the ice model. [ You can see this using the diagrams! Start with a Temperley-Lieb diagram with $2k$ points on the top and bottom of a rectangle. This is a non-crossing matching of the $4k$ boundary points. Now colour the regions black and white alternately with the two sides of the rectangle coloured white. Shrink the black intervals on the top and bottom edges to be very short. Then this is an element of $P_k$. ]

4. The partition algebras map surjectively to the centraliser algebras of the tensor powers of the defining permutation representations of the symmetric groups.

-
Hmm. Brief answers: 1. because they tend to deceive me. 2. The source of my question isn't confusion but curiosity; I am wondering whether (E) alone determines a monoid and what can be said about it -- I am aware of it not being $P_k$. 3. That's interesting -- I always thought the Temperley-Lieb algebra had only matchings instead of arbitrary planar partitions? 4. I know they do, but not the planar ones -- or do they? – darij grinberg Mar 27 '13 at 21:44
Ok, my claim on (E) being enough was actually wrong. One example of why I don't trust the diagrams... – darij grinberg Mar 27 '13 at 21:49