3. LR-shuffle-compatibility 85
3.6. Left- and right-shuffle-compatibility
In this section, we shall study two notions closely related to LR-shuffle-compatibility:
Definition 3.16. Let st be a permutation statistic.
(a) We say that st is left-shuffle-compatible if for any two disjoint nonempty permutations π and σ having the property that π1 > σ1, the multiset {stτ | τ ∈ S≺(π,σ)}multi depends only on stπ, stσ, |π|and |σ|.
(b)We say that st isright-shuffle-compatibleif for any two disjoint nonempty permutations π and σ having the property that π1 > σ1, the multiset {stτ | τ ∈ S(π,σ)}multi depends only on stπ, stσ, |π|and |σ|.
For a shuffle-compatible permutation statistic, these two notions are equiva-lent to the notions of LR-shuffle-compatibility and head-graft-compatibility, as the following proposition reveals:
Proposition 3.17. Let st be a shuffle-compatible permutation statistic. Then, the following assertions are equivalent:
• AssertionA1: The statistic st is LR-shuffle-compatible.
• AssertionA2: The statistic st is left-shuffle-compatible.
• AssertionA3: The statistic st is right-shuffle-compatible.
• AssertionA4: The statistic st is head-graft-compatible.
Proof of Proposition 3.17 (sketched). We shall prove the implications A1 =⇒ A2, A2 =⇒ A3, A3 =⇒ A2, A3 =⇒ A4 and A4 =⇒ A1.
The implicationA4 =⇒ A1 follows from Theorem 3.9.
Proof of the implicationA1 =⇒ A2: Assume that Assertion A1 holds. In other words, the statistic st is LR-shuffle-compatible. We shall show that AssertionA2 holds.
The statistic st is LR-shuffle-compatible. In other words, for any two disjoint nonempty permutationsπ and σ, the multisets
{stτ | τ ∈ S≺(π,σ)}multi and {stτ | τ ∈ S(π,σ)}multi depend only on stπ, stσ, |π|, |σ| and [π1 >σ1]. Hence, for any two disjoint nonempty permutations π and σ, the multiset {stτ | τ ∈ S≺(π,σ)}multi de-pends only on stπ, stσ, |π|, |σ| and [π1 >σ1]. Hence, for any two disjoint nonempty permutationsπ and σ having the property thatπ1 >σ1, the multiset {stτ | τ ∈ S≺(π,σ)}multi depends only on stπ, stσ, |π| and |σ| (indeed, it no longer depends on [π1 >σ1], because our condition π1 > σ1 ensures that [π1 >σ1] = 1). In other words, the statistic st is left-shuffle-compatible (by the definition of “left-shuffle-compatible”). In other words, AssertionA2holds. This proves the implicationA1=⇒ A2.
Proof of the implicationA2 =⇒ A3: Assume that Assertion A2 holds. In other words, the statistic st is left-shuffle-compatible. We shall show that AssertionA3 holds.
Ifπ and σare two disjoint nonempty permutations, then {stτ | τ ∈ S(π,σ)}multi
={stτ | τ ∈ S(π,σ)}multi− {stτ | τ ∈ S≺(π,σ)}multi
(by Lemma 3.11 (b)). Hence, if π and σ are two disjoint nonempty permuta-tions, then the multiset {stτ | τ ∈ S(π,σ)}multi is uniquely determined by {stτ | τ ∈ S(π,σ)}multi and {stτ | τ ∈ S≺(π,σ)}multi.
Thus, altogether, we know that for any two disjoint nonempty permutations π and σhaving the property that π1 >σ1, the following holds:
• The multiset{stτ | τ ∈S≺(π,σ)}multi depends only on stπ, stσ, |π|and
|σ| (since the statistic st is left-shuffle-compatible);
• The multiset {stτ | τ ∈ S(π,σ)}multi depends only on stπ, stσ, |π| and
|σ| (since the statistic st is shuffle-compatible);
• The multiset{stτ | τ ∈ S(π,σ)}multi is uniquely determined by {stτ | τ ∈S(π,σ)}multi and {stτ | τ ∈ S≺(π,σ)}multi.
Hence, the multiset{stτ | τ ∈ S(π,σ)}multi also depends only on stπ, stσ,
|π| and |σ|. In other words, the statistic st is right-shuffle-compatible (by the
definition of “right-shuffle-compatible”). In other words, Assertion A3 holds.
This proves the implicationA2 =⇒ A3.
Proof of the implication A3 =⇒ A2: The proof of the implication A3 =⇒ A2 is entirely analogous to the above proof of the implicationA2 =⇒ A3 (except that we need to interchange “left” and “right” and likewise interchange “S≺” and
“S”).
Proof of the implicationA3 =⇒ A4: Assume that Assertion A3 holds. We shall show that AssertionA4 holds.
We have already proven the implicationA3 =⇒ A2. Thus, AssertionA2holds (since A3 holds). In other words, the statistic st is left-shuffle-compatible. In other words, the following claim holds:
Claim 1: Let π and σ be two disjoint nonempty permutations having the property that π1 > σ1. Let π0 and σ0 be two disjoint nonempty permutations having the property thatπ01>σ10. Assume that
stπ =st π0
, stσ=st σ0 ,
|π| =π0
and |σ|=σ0 . Then,
{stτ | τ ∈ S≺(π,σ)}multi =stτ | τ ∈ S≺ π0,σ0 multi.
Also, Assertion A3 holds. In other words, the statistic st is right-shuffle-compatible. In other words, the following claim holds:
Claim 2: Let π and σ be two disjoint nonempty permutations having the property that π1 > σ1. Let π0 and σ0 be two disjoint nonempty permutations having the property thatπ01>σ10. Assume that
stπ =st π0
, stσ=st σ0 ,
|π| =π0
and |σ|=σ0 . Then,
{stτ | τ ∈ S(π,σ)}multi =stτ | τ ∈ S π0,σ0 multi. We are now going to prove the following claim:
Claim 3: Let π be a nonempty permutation, and let a be a letter that does not appear in π. Let π0 be a nonempty permutation, and let a0 be a letter that does not appear inπ0. Assume that
stπ =st π0
, |π| =π0
and [a>π1] = a0 >π01 . Then, st(a : π) =st(a0 : π0).
[Proof of Claim 3: The 1-permutations(a)and (a0) are order-isomorphic. Thus, st((a)) =st((a0))(since st is a permutation statistic). Also, |(a)| =1=|(a0)|.
We are in one of the following two cases:
Case 1: We have a>π1. Case 2: We have a≤π1.
Let us first consider Case 1. In this case, we havea > π1. Thus, [a >π1] =1.
Comparing this with [a>π1] = [a0 >π10], we obtain [a0 >π10] = 1, so that a0 >π01.
Consider the 1-permutations (a) and (a0). Then, the first entry of (a) is (a)1 = a > π1. Also, (a0)1 = a0 > π10. Also, the permutations (a) and π are disjoint (since a does not appear in π). Similarly, the permutations (a0) and π0 are disjoint. Furthermore, st((a)) = st((a0)), stπ = st(π0), |(a)| = |(a0)| and
|π|=|π0|. Hence, Claim 1 (applied to (a),π, (a0) and π0 instead ofπ, σ, π0 and σ0) yields
{stτ | τ ∈ S≺((a),π)}multi =stτ | τ ∈ S≺ a0
,π0 multi. (49) ButS≺((a),π) = {a: π}. Hence,
{stτ | τ ∈ S≺((a),π)}multi ={stτ | τ ∈ {a : π}}multi ={st(a : π)}multi. Similarly,
stτ | τ ∈ S≺ a0
,π0 multi =st a0 : π0 multi.
In light of the last two equalities, the equality (49) rewrites as {st(a: π)}multi = {st(a0 : π0)}multi. In other words, st(a: π) = st(a0 : π0). Thus, we have proven st(a : π) =st(a0 : π0) in Case 1.
Let us now consider Case 2. In this case, we have a ≤ π1. But a does not appear in π; therefore, a 6= π1. Combined with a ≤ π1, this yields a < π1. In other words,π1 >a. Also, from a≤π1, we obtain[a >π1] =0. Comparing this with [a>π1] = [a0 >π10], we obtain [a0 >π10] = 0, so that a0 ≤ π01. But a0 does not appear in π0; thus, a0 6=π01. Combined with a0 ≤ π10, this yieldsa0 < π10. In other words,π10 >a0.
Consider the 1-permutations(a)and (a0). Then, the first entry of(a)is(a)1 = a; similarly, (a0)1 = a0. Now, π1 > a = (a)1 and π01 > a0 = (a0)1. Also, the permutationsπand (a)are disjoint (since adoes not appear inπ). Similarly, the permutations π0 and (a0) are disjoint. Furthermore, st((a)) = st((a0)), stπ = st(π0), |(a)| = |(a0)| and |π| = |π0|. Hence, Claim 2 (applied to π, (a), π0 and (a0) instead ofπ, σ, π0 and σ0) yields
{stτ | τ ∈ S(π,(a))}multi =stτ | τ ∈ S π0, a0 multi. (50) ButS(π,(a)) = {a: π}. Hence,
{stτ | τ ∈ S(π,(a))}multi ={stτ | τ ∈ {a: π}}multi ={st(a: π)}multi. Similarly,
stτ | τ ∈ S π0, a0 multi =st a0 : π0 multi.
In light of the last two equalities, the equality (50) rewrites as {st(a: π)}multi = {st(a0 : π0)}multi. In other words, st(a: π) = st(a0 : π0). Thus, we have proven st(a : π) =st(a0 : π0) in Case 2.
We have now proven st(a : π) =st(a0 : π0) in both Cases 1 and 2. Therefore, st(a : π) =st(a0 : π0) always holds. This proves Claim 3.]
Claim 3 shows that the statistic st is head-graft-compatible. In other words, AssertionA4 holds. This proves the implicationA3 =⇒ A4.
We have now proven the implicationsA1 =⇒ A2, A2 =⇒ A3, A3 =⇒ A4 and A4 =⇒ A1. Thus, all four statements A1, A2, A3 and A4 are equivalent. This proves Proposition 3.17.
Note that on their own, the properties of left-shuffle-compatibility and right-shuffle-compatibility are not equivalent. For example, the permutation statistic that sends each nonempty permutationπto the truth value[π1>πi for alli >1] (and the 0-permutation () to 0) is right-shuffle-compatible (because in the def-inition of right-shuffle-compatibility, all the stτ will be 0), but not left-shuffle-compatible.