5. The kernel of the map QSym → A Epk 123
5.3. An M-generating set of K Epk
Another characterization of the ideal KEpk of QSym can be obtained using the monomial basis of QSym. Let us first recall how said basis is defined:
For any compositionα = (α1,α2, . . . ,α`), we let
Mα =
∑
i1<i2<···<i`
xαi1
1 xαi2
2 · · ·xαi`
`
(where the sum is over all strictly increasing `-tuples (i1,i2, . . . ,i`) of positive integers). This power seriesMαbelongs to QSym. The family(Mα)αis a composition
is a basis of theQ-vector space QSym; it is called the monomial basisof QSym.
Proposition 5.7. If J = (j1,j2, . . . ,jm) and K are two compositions, then we shall write J →
M K if there exists an ` ∈ {2, 3, . . . ,m} such that j` >2 and K = (j1,j2, . . . ,j`−1, 2,j`−2,j`+1,j`+2, . . . ,jm). (In other words, we write J →
M K ifK can be obtained from J by “splitting” some entry j` >2 into two consecutive entries 2 and j`−2, provided that this entry was not the first entry – i.e., we
had ` >1 – and that this entry was greater than 2.)
The idealKEpkof QSym is spanned (as aQ-vector space) by all sums of the form MJ +MK, where J and Kare two compositions satisfying J →
M K.
Example 5.8. We have (2, 1, 4, 4) →
M (2, 1, 2, 2, 4), since the composition (2, 1, 2, 2, 4) is obtained from (2, 1, 4, 4) by splitting the third entry (which is 4>2) into two consecutive entries 2 and 2.
Similarly,(2, 1, 4, 4) →
M (2, 1, 4, 2, 2) and (2, 1, 5, 4)→
M (2, 1, 2, 3, 4).
But we do not have (3, 1) →
M (2, 1, 1), because splitting the first entry of the composition is not allowed in the definition of the relation →
M. Two compositions J and K satisfying J →
M K must necessarily satisfy |J| =
|K|.
Here are all relations →
M between compositions of size 4:
(1, 3) →
M (1, 2, 1). Here are all relations →
M between compositions of size 5:
(1, 4) →
M (1, 2, 2), (1, 3, 1) →
M (1, 2, 1, 1), (1, 1, 3) →
M (1, 1, 2, 1), (2, 3) →
M (2, 2, 1). There are no relations→
M between compositions of size ≤3.
Before we start proving Proposition 5.7, let us recall a basic formula ([GriRei18, (5.2.2)]) that connects the monomial quasisymmetric functions with the funda-mental quasisymmetric functions:
Proposition 5.9. Let n ∈N. Letα be any composition ofn. Then,
Mα =
∑
βis a composition ofnthat refinesα
(−1)`(β)−`(α)Fβ.
Here, if γ is any composition, then `(γ) denotes the length of γ (that is, the number of entries of γ).
Proof of Proposition 5.9. This is precisely [GriRei18, (5.2.2)].
Proposition 5.10. Let nbe a positive integer. Let C be a subset of[n−1]. (a)Then,
MCompC =
∑
B⊇C
(−1)|B\C|FCompB.
(The bound variable B in this sum and any similar sums is supposed to be a subset of [n−1]; thus, the above sum ranges over all subsets B of [n−1] satisfying B⊇C.)
(b)Let k∈ [n−1] be such thatk ∈/C. Then, MCompC+MComp(C∪{k}) =
∑
B⊇C;
k∈/B
(−1)|B\C|FCompB.
(c)Let k∈ [n−1] be such thatk ∈/C and k−1 /∈ C∪ {0}. Then,
Proposition 5.9 (applied toα =CompC) yields MCompC
here, we have substituted CompBfor β, since
the map Comp :{subsets of [n−1]} → {compositions ofn}
(b)Proposition 5.10(a)(applied to C∪ {k} instead ofC) yields
Adding (62) to this equality, we obtain MCompC+MComp(C∪{k})
Proposition 5.10(b)yields (since the mapΦis a bijection))
=
∑
Proof of Proposition 5.7. We shall use the notation hfi | i ∈ Ii for the Q-linear span of a family(fi)i∈I of elements of aQ-vector space.
Define aQ-vector subspaceM of QSym by M =
(because if J and K are two compositions satisfying J →
M K, then J and K have the same size).
Consider the binary relation→ defined in Proposition 5.5. Then, Proposition 5.5 yields
KEpk=FJ −FK | J andK are compositions satisfying J →K
=
∑
n∈N
FJ−FK | J and K are compositions ofnsatisfying J →K (because if J and K are two compositions satisfying J → K, then J and K have the same size).
Now, fix n ∈ N. Let Ω be the set of all pairs (C,k) in whichC is a subset of [n−1] and k is an element of [n−1] satisfying k ∈/ C, k−1 ∈ C and k+1 /∈ C∪ {n}.
For every(C,k) ∈Ω, we define two elementsmC,k and fC,k of QSym by mC,k = MCompC+MComp(C∪{k+1}) and (63)
fC,k =FCompC−FComp(C∪{k}). (64)
60
We have the following:
Claim 1: We have
MJ +MK | J andK are compositions ofn satisfying J →
M K
=hmC,k | (C,k) ∈Ωi.
[Proof of Claim 1: It is easy to see that two subsets C and D of [n−1] satisfy CompC →
M CompD if and only if there exists some k ∈ [n−1] satisfying D =
60These two elements are well-defined, because both C∪ {k} and C∪ {k+1} are subsets of [n−1](sincek+1 /∈C∪ {n}shows thatk+16=n).
C∪ {k+1}, k∈/C,k−1∈ Cand k+1 /∈ C∪ {n}. 61. Thus, MCompC+MCompD | C and D are subsets of [n−1]
satisfying CompC→
M CompD
=MCompC+MCompD | C and D are subsets of [n−1] such that there exists some k∈ [n−1]
satisfying D =C∪ {k+1}, k∈/ C, k−1 ∈C and k+1 /∈ C∪ {n}i
=DMCompC+MComp(C∪{k+1}) | C⊆[n−1] and k∈ [n−1] are such that k∈/ C, k−1 ∈C and k+1 /∈ C∪ {n}i
=
*
MCompC+MComp(C∪{k+1})
| {z }
=mC,k (by (63))
| (C,k)∈ Ω +
(by the definition ofΩ)
=hmC,k | (C,k)∈ Ωi.
Now, recall that Comp is a bijection between the subsets of[n−1]and the com-positions ofn. Hence,
MJ +MK | J andK are compositions ofn satisfying J →
M K
=MCompC+MCompD | Cand D are subsets of [n−1] satisfying CompC →
M CompD
=hmC,k | (C,k) ∈Ωi. This proves Claim 1.]
Claim 2: We have
FJ −FK | J and Kare compositions ofn satisfying J →K
=hfC,k | (C,k)∈ Ωi.
[Proof of Claim 2: It is easy to see that two subsets C and D of [n−1] satisfy CompC → CompD if and only if there exists some k ∈ [n−1] satisfying D =
61To prove this, recall that “splitting” an entry of a composition J into two consecutive entries (summing up to the original entry) is always tantamount to adding a new element to DesJ. It suffices to show that the conditions under which an entry of a compositionJcan be split in the definition of the relation→
M are precisely the conditionsk∈/C,k−1∈Candk+1 /∈C∪ {n} on C = DesJ, where k+1 denotes the new element that we are adding to DesJ. This is straightforward.
C∪ {k}, k∈/C, k−1∈ Cand k+1 /∈C∪ {n}. 62. Thus, FCompC−FCompD | C and Dare subsets of [n−1]
satisfying CompC →CompDi
=FCompC−FCompD | C and Dare subsets of [n−1] such that there exists somek ∈ [n−1]
satisfyingD =C∪ {k}, k ∈/C, k−1∈ Cand k+1 /∈C∪ {n}i
=DFCompC−FComp(C∪{k}) | C⊆[n−1] and k∈ [n−1]
are such thatk ∈/C, k−1∈ C andk+1 /∈ C∪ {n}i
=
*
FCompC−FComp(C∪{k})
| {z }
=fC,k (by (64))
| (C,k) ∈ Ω +
(by the definition ofΩ)
=hfC,k | (C,k) ∈ Ωi.
Now, recall that Comp is a bijection between the subsets of[n−1]and the com-positions ofn. Hence,
FJ −FK | J and Kare compositions ofn satisfying J →K
=FCompC−FCompD | Cand D are subsets of [n−1] satisfying CompC→CompDi
=hfC,k | (C,k)∈ Ωi. This proves Claim 2.]
We define a partial order on the setΩ by setting
(B,k) ≥(C,`) if and only if (k =`and B⊇C). Thus,Ω is a finite poset.
Claim 3: For every(C,`) ∈ Ω, we have mC,` =
∑
(B,k)∈Ω;
(B,k)≥(C,`)
(−1)|B\C|fB,k.
62To prove this, recall that “splitting” an entry of a composition J into two consecutive entries (summing up to the original entry) is always tantamount to adding a new element to DesJ.
It suffices to show that the conditions under which an entry of a composition Jcan be split in the definition of the relation→are precisely the conditionsk∈/ C,k−1 ∈ Candk+1 /∈ C∪ {n}onC=DesJ, wherekdenotes the new element that we are adding to DesJ. This is straightforward.
[Proof of Claim 3: Let (C,`) ∈ Ω. Thus, C is a subset of [n−1] and ` is an C⊆ B). Hence, we can manipulate summation signs as follows:
B
∑
⊇C;63Here and in the following, the bound variableBin a sum always is understood to be a subset of[n−1].
Now, the definition ofmC,` yields
Now, Claim 3 shows that the family (mC,k)(C,k)∈Ω expands triangularly with respect to the family(fC,k)(C,k)∈Ω with respect to the poset structure onΩ. More-over, the expansion is unitriangular (because if (B,k) = (C,`), then B = C and thus (−1)|B\C| = (−1)|C\C| = (−1)0 = 1) and thus invertibly triangular (this means that the diagonal entries are invertible). Therefore, by a standard fact from linear algebra (see, e.g., [GriRei18, Corollary 11.1.19(b)]), we conclude that the span of the family (mC,k)(C,k)∈Ω equals the span of the family (fC,k)(C,k)∈Ω.
(by Claim 2).
Now, forget that we fixedn. We thus have proven that
MJ +MK | J andK are compositions ofn satisfying J →
M K
=FJ −FK | J and Kare compositions of nsatisfying J →K for eachn∈ N. Thus,
n
∑
∈N
MJ +MK | J andK are compositions ofn satisfying J →
M K
=
∑
n∈N
FJ −FK | J and Kare compositions ofn satisfying J →K . In light of
KEpk=
∑
n∈N
FJ−FK | J and K are compositions ofnsatisfying J →K and
M =
∑
n∈N
MJ+MK | J and K are compositions ofnsatisfying J →
M K
, this rewrites asM =KEpk. In other words, KEpk =M. This proves Proposition 5.7.
Question 5.11. It is worth analyzing the kernels of other known descent statis-tics (shuffle-compatible or not). Let us say that a descent statistic st is M-binomialif its kernel Kst can be spanned by elements of the formλMJ +µMK
with λ,µ ∈ Qand compositions J,K. Then, Proposition 5.7 yields that Epk is M-binomial. It is easy to see that the statistics Des and des are M-binomial as well. Computations using SageMath suggest that the statistics Lpk, Rpk, Pk, Val, pk, lpk, rpk and val (see [GesZhu17] for some of their definitions) are M-binomial, too (at least for compositions of size ≤ 9); this would be nice to prove. On the other hand, the statistics maj, (des, maj) and (val, des) (again, see [GesZhu17] for definitions) are not M-binomial.