6. Dendriform structures 149
6.4. Application to K Epk
We now claim the following:
Theorem 6.15. The idealKEpkof QSym is a ź-ideal, a Á-ideal, a ≺-ideal and a -ideal of QSym.
Proof of Theorem 6.15. Let A =QSym. Corollary 5.4 shows that KEpk is an ideal of QSym.
Let us recall the binary relation→on the set of compositions defined in Propo-sition 5.5.
Claim 1: Let J and K be two compositions satisfying J →K. LetG be a further composition. Then,[G,J] →[G,K].
[Proof of Claim 1: Write the composition J in the form J = (j1,j2, . . . ,jm). Write the composition Gin the form G= g1,g2, . . . ,gp
.
We have J → K. In other words, there exists an ` ∈ {2, 3, . . . ,m} such that j` >2 and K= (j1,j2, . . . ,j`−1, 1,j`−1,j`+1,j`+2, . . . ,jm) (by the definition of the relation →). Consider this `. Clearly, ` > 1 (since ` ∈ {2, 3, . . . ,m}), so that
p+` > p
|{z}≥0
+1≥1.
FromG= g1,g2, . . . ,gp
and J = (j1,j2, . . . ,jm), we obtain [G,J] = g1,g2, . . . ,gp,j1,j2, . . . ,jm
. (84)
FromG = g1,g2, . . . ,gp
and K = (j1,j2, . . . ,j`−1, 1,j`−1,j`+1,j`+2, . . . ,jm), we obtain
[G,K] = g1,g2, . . . ,gp,j1,j2, . . . ,j`−1, 1,j`−1,j`+1,j`+2, . . . ,jm
. (85) From looking at (84) and (85), we conclude immediately that the composition [G,K] is obtained from[G,J] by “splitting” the entryj` >2 into two consecutive entries 1 and j`−1, and that this entry j` was not the first entry (indeed, this entry is the (p+`)-th entry, but p+` > 1). Hence, [G,J] → [G,K] (by the definition of the relation→). This proves Claim 1.]
Claim 2: We have AźKEpk ⊆ KEpk.
[Proof of Claim 2: We must show that aźm ∈ KEpk for every a ∈ A and m∈ KEpk. So let us fix a ∈ A andm ∈ KEpk.
Proposition 5.5 shows that the Q-vector space KEpk is spanned by all differ-ences of the formFJ −FK, where Jand Kare two compositions satisfying J →K.
Hence, we can WLOG assume that m is such a difference (because the relation aźm ∈ KEpk, which we must prove, is Q-linear in m). Assume this. Thus, m = FJ −FK for some two compositions J and K satisfying J → K. Consider these J and K.
From J → K, we easily conclude that the composition J is nonempty. Thus,
|J| 6=0. But from J → K, we also obtain|J| =|K|. Hence, |K| = |J| 6=0. Thus, the compositionK is nonempty.
Recall that the family(FL)Lis a compositionis a basis of theQ-vector space QSym= A. Hence, we can WLOG assume that a belongs to this family (since the rela-tion aźm ∈ KEpk, which we must prove, is Q-linear in a). Assume this. Thus, a= FG for some composition G. Consider thisG.
If G is the empty composition, then a = FG = 1, and therefore a
|{z}
=1
źm = 1źm=m ∈ KEpkholds. Thus, for the rest of this proof, we WLOG assume that G is not the empty composition. Thus,G is nonempty.
Recall that for any two compositions α and β, we have Fαź Fβ = F[α,β]. Applying this to α = G and β = J, we obtain FG źFJ = F[G,J]. Similarly, FG źFK =F[G,K].
But Claim 1 yields [G,J] → [G,K]. Hence, the difference F[G,J]−F[G,K] is one of the differences which span the idealKEpk according to Proposition 5.5. Thus, in particular, this difference lies inKEpk. In other words, F[G,J]−F[G,K] ∈ KEpk.
Now,
a
|{z}=FG
ź m
=|{z}FJ−FK
=FG ź(FJ−FK) = FGźFJ
| {z }
=F[G,J]
−FG źFK
| {z }
=F[G,K]
=F[G,J]−F[G,K] ∈ KEpk. This proves Claim 2.]
Claim 3: Let J and K be two compositions satisfying J →K. LetG be a further composition. Then,[J,G] →[K,G].
[Proof of Claim 3: This is proven in the same way as we proved Claim 1, with the only difference that j` is now the `-th entry of [J,G] and not the (p+`)-th entry (but this is still sufficient, since` >1).]
Claim 4: We have KEpkź A⊆ KEpk.
[Proof of Claim 4: This is proven in the same way as we proved Claim 2, with the only difference that now we need to use Claim 3 instead of Claim 1.]
Combining Claim 2 and Claim 4, we conclude thatKEpk is a ź-ideal of A = QSym.
Claim 5: Let JandKbe two nonempty compositions satisfyingJ → K.
Let Gbe a further nonempty composition. Then, GJ → GK.
[Proof of Claim 5: Write the composition J in the form J = (j1,j2, . . . ,jm). Write the composition G in the form G = g1,g2, . . . ,gp
. Thus, p > 0 (since the composition Gis nonempty).
We have J → K. In other words, there exists an ` ∈ {2, 3, . . . ,m} such that j` >2 and K= (j1,j2, . . . ,j`−1, 1,j`−1,j`+1,j`+2, . . . ,jm) (by the definition of the
relation →). Consider this `. Clearly, ` ≥ 2 (since ` ∈ {2, 3, . . . ,m}), so that p
|{z}>0
+ `
|{z}≥2
−1>0+2−1 =1.
FromG= g1,g2, . . . ,gp
and J = (j1,j2, . . . ,jm), we obtain GJ = g1,g2, . . . ,gp−1,gp+j1,j2,j3, . . . ,jm
. (86)
FromG = g1,g2, . . . ,gp
and K = (j1,j2, . . . ,j`−1, 1,j`−1,j`+1,j`+2, . . . ,jm), we obtain
GK = g1,g2, . . . ,gp−1,gp+j1,j2,j3, . . . ,j`−1, 1,j`−1,j`+1,j`+2, . . . ,jm
(87) (notice that the gp+j1term isnot agp+1 term, because`≥2).
From looking at (86) and (87), we conclude immediately that the composition GKis obtained fromGJby “splitting” the entry j` >2 into two consecutive entries 1 and j`−1, and that this entry j` was not the first entry (indeed, this entry is the (p+`−1)-th entry, but p+`−1 > 1). Hence, GJ → GK (by the definition of the relation→). This proves Claim 5.]
Claim 6: We have AÁ KEpk ⊆ KEpk.
[Proof of Claim 6: This is proven in the same way as we proved Claim 2, with the only difference that now we need to use Claim 5 instead of Claim 1 and that we need to use the formulaFα Á Fβ =Fαβ instead of Fα źFβ =F[α,β].]
Claim 7: Let JandKbe two nonempty compositions satisfyingJ → K.
Let Gbe a further nonempty composition. Then, JG→ KG.
[Proof of Claim 7: Write the composition J in the form J = (j1,j2, . . . ,jm). Write the composition G in the form G = g1,g2, . . . ,gp
. Thus, p > 0 (since the composition Gis nonempty).
We have J → K. In other words, there exists an ` ∈ {2, 3, . . . ,m} such that j` >2 and K= (j1,j2, . . . ,j`−1, 1,j`−1,j`+1,j`+2, . . . ,jm) (by the definition of the relation →). Consider this `. Clearly, ` ≥ 2 (since ` ∈ {2, 3, . . . ,m}), so that
` >1.
FromG= g1,g2, . . . ,gp
and J = (j1,j2, . . . ,jm), we obtain JG= j1,j2, . . . ,jm−1,jm+g1,g2,g3, . . . ,gp
. (88)
Now, we distinguish between the following two cases:
Case 1: We have `=m.
Case 2: We have `6=m.
Let us first consider Case 1. In this case, we have`=m. Thus, m=` ≥2 >1 and jm+g1 = j`
|{z}
>2
+ g1
|{z}≥0
>2.
FromG= g1,g2, . . . ,gp and
K = (j1,j2, . . . ,j`−1, 1,j`−1,j`+1,j`+2, . . . ,jm)
= (j1,j2, . . . ,jm−1, 1,jm −1) (since `=m), we obtain
KG = j1,j2, . . . ,jm−1, 1,(jm−1) +g1,g2,g3, . . . ,gp
= j1,j2, . . . ,jm−1, 1,jm+g1−1,g2,g3, . . . ,gp
. (89)
From looking at (88) and (89), we conclude immediately that the composition KG is obtained from J G by “splitting” the entry jm +g1 > 2 into two consecutive entries 1 andjm+g1−1, and that this entry jm+g1was not the first entry (indeed, this entry is the m-th entry, but m > 1). Hence, JG → KG (by the definition of the relation→). This proves Claim 7 in Case 1.
Let us next consider Case 2. In this case, we have ` 6= m. Hence, ` ∈ {2, 3, . . . ,m−1} (since`∈ {2, 3, . . . ,m}).
From G = g1,g2, . . . ,gp
and K = (j1,j2, . . . ,j`−1, 1,j`−1,j`+1,j`+2, . . . ,jm), we obtain
KG
= j1,j2, . . . ,j`−1, 1,j`−1,j`+1,j`+2, . . . ,jm−1,jm+g1,g2,g3, . . . ,gp
(90) (notice that the jm+g1term isnot a(j`−1) +g1 term, because`6=m).
From looking at (88) and (90), we conclude immediately that the composition KGis obtained from JGby “splitting” the entry j` >2 into two consecutive entries 1 and j`−1, and that this entry j` was not the first entry (indeed, this entry is the `-th entry, but ` > 1). Hence, JG → KG (by the definition of the relation→). This proves Claim 7 in Case 2.
We have now proven Claim 7 in both Cases 1 and 2. Thus, Claim 7 always holds.]
Claim 8: We have KEpkÁ A ⊆ KEpk.
[Proof of Claim 8: This is proven in the same way as we proved Claim 6, with the only difference that now we need to use Claim 7 instead of Claim 5.]
Combining Claim 6 and Claim 8, we conclude that KEpk is a Á-ideal of A = QSym.
Finally, Theorem 6.12(c)(applied to M=KEpk) shows thatKEpk is a ≺-ideal and a -ideal of QSym.
Thus, altogether, we have proven thatKEpk is a ź-ideal, a Á-ideal, a ≺-ideal and a -ideal of QSym. This proves Theorem 6.15.
Question 6.16. What other descent statistics st have the property that Kst is an ź-ideal, Á-ideal, ≺-ideal and/or -ideal? We will see some answers in Subsection 6.7, but a more systematic study would be interesting.