Don_RubielMar 8, 2025

New paper: We think of a deck of card as a linear order. Then, a (riffle)-shuffle between two decks of cards is an order in the union of deck A and deck B, where the new order preserves the order of A and the order of B.
I know I used too many words for a very simple concept. You can just google riffle shuffle and see what I mean.

Now, we can change A by a tree, and still study shuffles between A and a linear tree B. Basically, assume A has two branches, let C,D be the linear orders from the leaves to the root of A. Then a shuffle of A and a linear order B is the colimit of shuffles between the decks C and the linear order B, and shuffles of D and the linear order B.

We are basically gluing shuffles, see Figure below.

In the image, we consider the linear tree B with two black cards, and a tree A, represented by the poset {x<y, x<w<z}.
If we walk from the root to any leaf on the shuffle of A and B, we recover a shuffle of B and (a path from the root A to a leaf on A).

We can define shuffles between a poset and a linear order in a similar way.

In our recent paper, we answered the question of how many different shuffles are there between a tree and a linear tree. This question is born in dendroidal homotopy theory, since trees are closely related to operads (free tree operads). More or less the shuffle of a tree and a linear tree parametrize elements in the tensor product of a free tree operad and a tree operad on the linear tree, equivalently, we are counting maximal simplices on the geometric realization of a #dendroidal set and the geometric realization of a #simplicial set. 1/3

Show threadDon_RubielMar 8, 2025

Now, our techniques to solve this combinatorics problem:
We studied an operad (the operad of posets), and then we studied power series indexed by posets.

Since the power series are indexed by posets, endomorphisms of posets can be extended to endomorphisms of those power series. That is, if we have a series $F(P)$ for every poset $P$, and we have $Q$ a function taking $P_1,...,P_n$ posets to a poset $Q(P_1,...P_n)$, then we define $Q(F(P_1),...,F(P_n)):=F(Q(P_1,...P_n))$.

If we work with sets, then we dont have to worry too much about the properties of this induced functions. They are just set theoretical functions.

Now, the correct language here is the language of #operads. Since we work with the operad of posets, this are topological operads. It is just that the topological spaces are finite but with a topology distinct from the discrete topology!.

Returning to the problem of enumerating shuffles between a tree and a linear tree. If we consider the series $F(P)$ whose $n$ coefficient enumerates shuffles between a poset $P$ and the linear order with $n$ cards, here $P$ is the poset of vertices of the input tree,
then we have an action of the operad of posets of the form $Q(F(P_1),...,F(P_n)):=F(Q(P_1,...P_n))$.
Now, the difficult part is to compute the actions of the posets. But in the paper https://doi.org/10.1007/s10801-025-01386-7 we were able to compute the action of the posets $\{x,y\}$ and $\{x<y\}$, this is enough to compute the series of any series parallel poset, which includes trees.

Sometimes, the action of the operad preserves the combinatorial / enumerative property of the series. That is, the operad sends series solving a problem to series solving the same problem. #EnumerativeCombinatorics 2/3

Show threadDon_Rubiel

Finally, in the paper we show two algebras over the operad of posets, and each algebra is split into three isomorphic algebras.

We conjecture that this is a general result, certain algebras over the operad of posets come in triples. We are working on the formalization of this statement, but we have evidence.
I probably have to clarify that we proved that series parallel posets preserve the enumeration of shuffles, but I suposse this is true for arbitrary posets. And we believe to prove the general result we may need a version of Ehrhart theory. #Shuffles #OperadicCombinatorics 3/3

Mar 8, 2025 at 10:33amWeb