# nLab shuffle

Contents

### Context

#### Combinatorics

combinatorics

enumerative combinatorics

graph theory

rewriting

Basic structures

Generating functions

Proof techniques

Combinatorial identities

Polytopes

category: combinatorics

# Contents

## Idea

In mathematics, by “shuffles” or “un-shuffles” one means certain permutations , namely those which (when thought of as permuting the natural numbers $\{1, 2, \cdots, n \}$) are linear order-preserving along the first $p$ steps and then again along the remaining $q = n - p$ steps.

The term ‘shuffle’ conjures up the idea of shuffling a pack of cards. In fact the mathematical idea is nearer to shuffling two packs of cards one through the other. Suppose we have a pack of $p$ cards and a pack of $q$ cards and we build a pack of $p+q$ cards, whilst retaining the order on the two ‘sub-packs’. The result is a $(p,q)$-shuffle.

Or rather, the permutation in question may be seen as describing the un-doing of such a shuffling, whence some authors say “un-shuffle” for the same concept.

In this vein, the following graphics shows an illustration of an example of a $(4,3)$-(un-)shuffle of an ordered list of 7 elements:

## Definitions

###### Definition

For $p,q \in \mathbb{N}$ two natural numbers, a $(p,q)$-shuffle is a permutation

$(\mu_1, \cdots, \mu_p, \nu_1, \cdots, \nu_q)$

of $(1,2, \cdots, p+q)$ subject to the condition that

$\mu_1 \lt \mu_2 \lt \cdots \lt \mu_p$

and

$\nu_1 \lt \nu_2 \lt \cdots \lt \nu_q \,.$

In more formal terms, a $(p,q)$-shuffle is a permutation $\sigma \,\in\, \Sigma_{p + q}$ (an element of the symmetric group acting on the finite set $\{1, \cdots, p + q\}$ of $p + q$ elements), such that

$\sigma(1) \lt \sigma(2) \lt \cdots \lt \sigma(p) \;\;\;\;\;\;\;\; \text{and} \;\;\;\;\;\;\;\; \sigma(p+1) \lt \sigma(p+2) \lt \cdots \lt \sigma(p+q) \,;$

hence, equivalently, such that

$\underset{1 \leq i \lt p+q}{\forall} \;\;\;\;\;\;\;\;\;\; i \neq p \;\;\;\; \Rightarrow \;\;\;\; \sigma(i) \lt \sigma(i + 1) \,.$

The signature of a $(p,q)$-shuffle is the signature of the corresponding permutation.

Two other equivalent (and dual) ways of defining the notion of $(p,q)$-shuffle are as follows (e.g. Hoffbeck-Moerdijk 17, section 1.1), naturally understood as characterizing the non-degenerate simplices in a product of simplices (e.g. Kerodon 2.5.7.2: 00RH):

• Consider $p$ and $q$ as the linear orders $(p) = \{ 1 \lt \dots \lt p \}$ and $(q) = \{ 1 \lt \dots \lt q \}$. Then a $(p,q)$-shuffle is a way of extending the partial order on the coproduct $(p) + (q)$ to a linear order, or equivalently, a surjective monotone function

$(p) + (q) \to (p+q)$
• Consider $p$ and $q$ as non-empty linear orders $[p] = \{ 0 \lt \dots \lt p \}$ and $[q] = \{ 0 \lt \dots \lt q \}$. Then a $(p,q)$-shuffle is a maximal chain within the product partial order $[p] \times [q]$, or equivalently, an injective monotone function:

$[p + q] \to [p] \times [q] \,.$

## Applications

### Products of simplices

Shuffles control the combinatorics of products of simplices. See products of simplices for details.

### Eilenberg-Zilber map

Related to the product of simplices: shuffles control the Eilenberg-Zilber map. See there for details.

Shuffles are used in defining the pre-cgc structure on $\bigwedge V$ in the theory of differential graded coalgebras

Shuffles are also used for defining the shuffle product on $T(V)$, see differential graded Hopf algebra.

### In $L_\infty$-algebras

In the definition of L-∞ algebras the unshuffle side of shuffles is used.

## Properties

### The poset of $(p,q)$-shuffles

Any pair $(p,q)$ yields a poset relating the various $(p,q)$-shuffles.

For example, consider this $(2,1)$-shuffle:

$\begin{matrix} (0\lt 1) &-& (0\lt 2) &-& (1\lt 2) \end{matrix}$

(This Hasse diagram has been laid out horizontally to save space. The bottom is to the left. The vertex $(0\lt 1)$ corresponds to the shuffle with $\mu_1 = 0, \mu_1 = 1$, and so on.)

We need here to explain the partial order. We take the ‘$\mu$-signature’ of the shuffle, that is, the ordered set $\mu_1\lt\ldots \lt \mu_p$. (Of course, this determines the $\nu$-signature as that is the complement of $\mu$.)

###### Definition

Given two $(p,q)$-shuffles, represented by $(\mu, \nu)$ and $(\mu\prime,\nu\prime)$, we set

$(\mu, \nu) \leq (\mu\prime,\nu\prime)$

if, for each $i$ in the range $1\leq i\leq p$, $\mu_i \leq \mu_i\prime.$ We refer to this poset as $(Shuff(p,q),\leq)$.

Going to $(Shuff(3,1),\leq)$, the poset for $(3,1)$-shuffles, the fact that $q = 1$ will mean that this poset is again linear:

$\begin{matrix} (0 \lt 1 \lt 2) &-& (0 \lt 1 \lt 3)& - &(0 \lt 2 \lt 3)& - &(1 \lt 2 \lt 3) \end{matrix}$

This is true in general:

###### Lemma

If $p = 1$ or $q = 1$, then $(Shuff(p,q),\leq)$ is a linear poset.

###### Proof

If $p = 1$, $\mu = (\mu_1)$ is a singleton, and the poset will be:

$\begin{matrix} (0) & - & (1) & - & \ldots & - & (q). \end{matrix}$

For $q = 1$, $\nu = (\nu_1)$, and the poset is

$\begin{matrix} (0 \lt 1 \lt \ldots \lt p-1) & - & (0 \lt 1 \lt \ldots \lt p-2 \lt p) & - & \ldots & - & (1 \lt \ldots \lt p), \end{matrix}$

where at each stage one misses out the single $\nu$-value.

In all cases, each position is obtained from the one immediately to its ‘left’ by increasing one value, yet remaining with a shuffle. This is more clearly seen in the (2,2) example, which is no longer linear. First we display the grid in which things are happening.

$\begin{matrix} (0,2) & - & (1,2) & - & (2,2) \\ | & & | & & | \\ (0,1) & - & (1,1) & - & (2,1) \\ | & & | & & | \\ (0,0) & - & (1,0) & - & (2,0) \end{matrix}$

We can then look at the shuffle poset, noting again that it is not linear:

The left hand shuffle, labelled $(0\lt1)$, corresponds to $\left(\begin{array}{ccccc}0&1&2&2&2\\0&0&0&1&2\end{array}\right)$, so gives the path along the bottom of the square and then up the right hand side.

The first change goes to $(0\lt 2)$ and gives a path with 2 steps,

and corresponds to the shuffle, $\left(\begin{array}{ccccc}0&1&1&2&2\\0&0&1&1&2\end{array}\right)$, so at this position, $(0\lt2)$, there is a choice, either increase 0 by 1 to get $(1\lt 2)$ or increase 2 by 1 to get $(0\lt3)$. In the first case, we get the path

going horizontally across the square, and the shuffle, $\left(\begin{array}{ccccc}0&0&1&2&2\\0&1&1&1&2\end{array}\right)$, and in the second case, we get the shuffle: $\left(\begin{array}{ccccc}0&1&1&1&2\\0&0&1&2&2\end{array}\right)$, and the obvious path up the middle of the square:

From $(1 \lt 2)$ or $(0\lt3)$, there is only one way to go, namely to $(1 \lt 3)$ and a 2-step path (left to you), and, finally it is just one step from $(1 \lt 3)$ to $(2 \lt 3)$ and the other extremal path.

In summary, each path corresponds to a simplex of maximal dimension in the product. The poset encodes the simplest relationships between those paths with the links in the Hasse diagram corresponding to simple changes in the path, and adjacency of the two simplices in the product, but note that the poset is usually not linear.

### The Anti-Lex order

There is a total order on $Shuff(p,q)$ related to the above partial order and which is useful when checking cancellation of terms in non-commutative contexts, as occurs in the Whitehead product formula given by Curtis and attributed to Kan. This is the anti-lexicographic order.

We represent a $(p,q)$-shuffle $(\mu,\nu)$ by an increasing $\mu$-sequence $\mu_1\lt \ldots \lt \mu_p$, (so with complementary $\nu$-sequence $\nu_1\lt \ldots \lt \nu_q$).

We order the $(\mu,\nu)$ using just the $\mu$-sequence as, of course, that determines the $\nu$-sequence completely. Inductively in $p$, we set

$\mu = (\mu_1,\ldots, \mu_p)\prec \mu\prime = (\mu\prime_1,\ldots, \mu\prime_p),$

(i) if $\mu_p\lt \mu\prime_p$,

or

(ii) if $\mu_p = \mu\prime_p$ and $\mu_p$ is odd,

$(\mu_1,\ldots, \mu_{p-1})\prec (\mu\prime_1,\ldots, \mu\prime_{p-1}),$

whilst if $\mu_p = \mu\prime_p$ and $\mu_p$ is even,

$(\mu_1,\ldots, \mu_{p-1})\succ (\mu\prime_1,\ldots, \mu\prime_{p-1}),$

where we adopt the notation $(\mu_1,\ldots, \mu_p)$ instead of $(\mu_1\lt\ldots \lt \mu_p)$.

For example, on our (2,2)-shuffles, the total order is

$(0,1) - (1,2) - (0,2) - (0,3) - (1,3) - (2,3).$

We note the lexicographic order on the sector with $\mu_2 = 2$ is reversed.

For illustrative purposes, we will look at two other examples.

First a generic $(p,1)$ case: the shuffle poset for $(p,1)$ is, more or less,

$(0,\ldots , p-2) - (0,\ldots, p-4,p-3,p-1) - (0, \ldots, p-4,p-2,p-1) - \ldots - (1, \ldots , p-1)$

as, at each position, there is only one transposition that can be applied. The anti-lex total order will correspond exactly to this if $p-1$ is odd, but, if $p-1$ is even, it will reverse the order on the last $p-1$ positions giving

$(0,\ldots , p-2) - (1, \ldots , p-1) - (0,2, \ldots, p-1) - \ldots - (0,\ldots, p-3,p-1).$

There are various points to note. Firstly that if $p$ is even, the permutation corresponding to $(1, \ldots , p-1)$ is odd. Secondly, the geometric picture is simply a prism, $\Delta[p]\times \Delta[1]$, and we can easily interpret the above order as a filling scheme for the simplices of $(\Delta[p]\times \Delta[1])\times \Delta[1]$, starting with the empty shell of

$(\Delta[p]\times \Delta[1])\times \{0\} \cup \partial(\Delta[p]\times \Delta[1])\times \Delta[1],$

together with part of the top of the ‘canister’. (In general the total order seems to correspond to some sort of filling scheme although this is not always as clear as here.)

Our second case will be (3,2). Again we will write $(i,j,k)$ instead of $i\lt j\ltk$ in order to save confusion over the various different orders being considered. The shuffle poset, $\mathrm{Shuff}(3,2)$, has Hasse diagram given below:

The subgraph of those with $\mu_3 = 4$, of course, is a copy of our earlier $(2,2)$-example, whilst that with $\mu_3 = 3$, is a copy of $Shuff(3,1)$. Within that we have, on deleting the 3s from the final positions, a copy of $Shuff(2,1)$ and so on. Taking $Shuff(2,1) \times \{3\}$, we can match it with a $Shuff(2,1) \times \{4\}$ within the $\mu_3 = 4$ block, the matching being given by, for instance, $(0,1,3)\leftrightarrow(0,1,4)$, etc. The anti-lex order gives

$(0,1,2)-(0,1,3) - (1,2,3)-(0,2,3)-(2,3,4)-(1,3,4)-(0,3,4) - (0,2,4) - (1,2,4) - (0,1,4),$

which is

$Shuff(3,1)\oplus Shuff(2,2)^{op}\times \{4\},$

the ordinal sum, or join, of the anti-lex ordered shuffle sets. This sort of decomposition is quite general.

## References

Original articles:

Review:

The two dual equivalent characterizations of $(p,q)$-shuffles (called shuffles of linear trees or shuffles of linear orders) are discussed in section 1.1 of

Last revised on December 4, 2022 at 14:56:00. See the history of this page for a list of all contributions to it.