The operads of planar forests are Koszul

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The operads of planar forests are Koszul

Loïc Foissy

To cite this version:

Loïc Foissy. The operads of planar forests are Koszul. 2009. �hal-00366681�

The operads of planar forests are Koszul

L. Foissy

Laboratoire de Math´ematiques, Universit´e de Reims Moulin de la Housse - BP 1039 - 51687 REIMS Cedex 2, France

e-mail : [email protected]

ABSTRACT. We describe the Koszul dual of two quadratic operads on planar forests in- troduced to study the infinitesimal Hopf algebra of planar rooted trees and prove that these operads are Koszul.

KEYWORDS. Koszul quadratic operads, planar rooted trees.

AMS CLASSIFICATION. 05C05, 18D50.

Contents

1 Operads of planar forests 2

1.1 Presentation . . . . 2 1.2 Free algebras on these operads . . . . 3

2 The operad

is Koszul 3

2.1 Koszul dual of

. . . . 3 2.2 Free

ց

-algebras . . . . 4 2.3 Homology of a

-algebra . . . . 6 2.4 Homology of free

-algebras . . . . 7

3 The operad

is Koszul 11

3.1 Koszul dual of

. . . . 11 3.2 Free

ր

-algebras . . . . 12 3.3 Homology of a

-algebra . . . . 12 3.4 Homology of free

-algebras . . . . 14

Introduction

The Hopf algebra of planar rooted trees, described in [2, 6], is a non-commutative version of the Hopf algebra of rooted tree introduced in [1, 7, 8, 9] in the context of Quantum Field Theories and Renormalisation. An infinitesimal version of this object is introduced in [4], and is related to two operads on planar forests in [3]. These two operads, denoted by

and

, are presented in the following way:

1.

is generated by m and ց∈

(2), with relations:







m ◦ (ց, I ) = ց ◦(I, m),

m ◦ (m, I) = m ◦ (I, m),

ց ◦(m, I) = ց ◦(I, ց).

2.

is generated by m and ր∈

(2), with relations:







m ◦ (ր, I) = ր ◦(I, m), m ◦ (m, I) = m ◦ (I, m), ր ◦(ր, I ) = ր ◦(I, ր).

The algebra of planar rooted trees is both the free

- and

-algebra generated by

, with products ր and ց given by certain graftings.

The operads

and

are quadratic. Our aim in this note is to prove that they are both Koszul, in the sense of [5]. We describe their Koszul dual (it turns out that they are quotient of

and

) and the associated homology of

- or

-algebras. We compute these homologies for free objects and prove that they are concentrated in degree 0. This proves that these operads are Koszul.

1 Operads of planar forests

1.1 Presentation

We work in this text with operads, whereas we worked in [3] with non-Σ-operads. In other terms, we replace the non-Σ-operads of [3] by their symmetrization [10].

Definition 1

1.

is generated, as an operad, by m and ց, with the relations:







ց ◦(m, I) = ց ◦(I, ց), ց ◦(I, m) = m ◦ (ց, I), m ◦ (m, I) = m ◦ (I, m).

2.

is generated, as an operad, by m and ր, with the relations:







ր ◦(ր, I) = ր ◦(I, ր), ր ◦(I, m) = m ◦ (ր, I ), m ◦ (m, I) = m ◦ (I, m).

Remarks.

1. Graphically, the relations defining

can be written in the following way:

@@

ց m

1 2

3

=

@@

ց 1 ց

2 3

,

@@

m m

1 2

3

=

@@

m 1 m

2 3

,

@@

m ց

1 2

3

=

@@

ց 1 m

2 3

.

2. We denote by ˜

the sub-non-Σ-operad of

generated by m and ց. Then

is the symmetrization of ˜

.

3. Graphically, the relations of

ր

can be written in the following way:

@@

ր ր

=

@@

ր ր

,

@@

m m

=

@@

m m

,

@@

m ր

=

@@

ր m

.

4. We denote by ˜

the sub-non-Σ-operad of

generated by m and ր. Then

is the

symmetrization of ˜

.

Both of these non-Σ-operads admits a description in terms of planar forests [3]. In particular, the dimension of ˜

(n) and ˜

(n) is given by the n-th Catalan number [11, 12]. Multiplying by a factorial, for all n ≥ 1:

dim

(n) = dim

(n) = (2n)!

(n + 1)! .

In particular, dim

(2) = dim

(2) = 4 and dim

(3) = dim

(3) = 30.

1.2 Free algebras on these operads

We described in [3] the free

- and

-algebras on one generators, using planar rooted trees.

We here generalise (without proof) these results. Let D be any set. We denote by T

the set of planar trees decorated by D and by F

the set of non-empty planar forests decorated by D.

1. The free

-algebra generated by D has the set F

as a basis. The product m is given by concatenation of forests. For all F, G ∈ F

, the product F ց G is obtained by grafting F on the root of G, on the left.

2. The free

-algebra generated by D has the set F

as a basis. The product m is given by concatenation of forests. For all F, G ∈ F

, the product F ր G is obtained by grafting F on the left leaf of G.

In both cases, we identified d ∈ D with

∈ F

. Moreover, for all F ∈ F

, F ց

= F ր

is the tree obtained by grafting the trees of F on a common root decorated by d: this tree will be denoted by B

(F).

2 The operad P

is Koszul

2.1 Koszul dual of P

(See [5, 10] for the notion of Koszul duality for quadratic operads). We denote by

ց

the Koszul dual of

.

Theorem 2 The operad

ց

is generated by m and ց∈

ց

(2), with the relations:



 

 



 

 



ց ◦(m, I) = ց ◦(I, ց), m ◦ (m, I) = m ◦ (I, m), m ◦ (ց, I) = ց ◦(I, m), ց ◦(ց, I) = 0,

m ◦ (I, ց) = 0.

Proof. Let

(E) be the operad freely generated by the S

-module freely generated by m and ց. Then

can be written

=

(E)/(R), where R is a sub-S

-module of

(E)(3).

As dim(

(E)) = 48 and dim(

(3)) = 30, dim(R) = 18. So dim(R

) = 48 − 18 = 30. We then verify that the given relations for

ց

are indeed in R

, that each of them generates a free S

-module, which are in direct sum. So these relations generate entirely

(E)(3).

Remarks.

1. So

ց

is a quotient of

.

2. Moreover,

ց

is the symmetrisation of the non-Σ-operad ˜

ց

generated by m and ց and

the relations: 

 

 



 

 



ց ◦(m, I) = ց ◦(I, ց), m ◦ (m, I) = m ◦ (I, m), m ◦ (ց, I) = ց ◦(I, m), ց ◦(ց, I ) = 0,

m ◦ (I, ց) = 0.

This is a general fact: the Koszul dual of the symmetrisation of a quadratic non-Σ operad is itself the symmetrisation of a certain quadratic non-Σ-operad.

3. Graphically, the relations defining

ց

can be written in the following way:

@@

ց m

1 2

3

=

@@

ց 1 ց

2 3

,

@@

m m

1 2

3

=

@@

m 1 m

2 3

,

@@

m ց

1 2

3

=

@@

ց 1 m

2 3

,

@@

ց ց

1 2

3

= 0,

@@

m 1 ց

2 3

= 0.

2.2 Free P

ց

-algebras

Let V be finite-dimensional vector space. We put:



 

 



 

 



T

(V )(n) = M

V

for all n ≥ 1, T

(V ) =

M

T

(V )(n).

In order to distinguish the different copies of V

, we put:

T (V )(n) = M

A ⊗ . . . ⊗ A ⊗ A ˙ ⊗ A ⊗ . . . ⊗ A

| {z }

the k-th copy of A is pointed.

.

The elements of A⊗. . .⊗A⊗ A⊗A⊗. . .⊗A ˙ will be denoted by v

⊗. . .⊗v

⊗ v ˙

⊗v

⊗. . .⊗v

. We define m and ց over T

(V ) in the following way: for v = v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

and w = w

⊗ . . . ⊗ w ˙

⊗ . . . ⊗ w

,

vw =

0 if l 6= 1,

v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

⊗ w

⊗ . . . ⊗ w

if l = 1;

v ց w =

0 if k 6= 1,

v

⊗ . . . ⊗ v

⊗ w

⊗ . . . ⊗ w ˙

⊗ . . . ⊗ w

if k = 1.

Lemma 3 T

(V ) is a

ց

-algebra generated by V . Proof. Let us first show that the relations of the

ց

-algebras are satisfied. Let u = u

⊗ . . . ⊗ u ˙

⊗ . . . ⊗ u

, v = v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

and w = w

⊗ . . . ⊗ w ˙

⊗ . . . ⊗ w

.

(uv) ց w = 0 if j 6= 1 or k 6= 1,

= u

⊗ . . . ⊗ u

⊗ v

⊗ . . . ⊗ v

⊗ w

⊗ . . . ⊗ w ˙

⊗ . . . ⊗ w

if j = k = 1,

u ց (v ց w) = 0 if j 6= 1 or k 6= 1,

= u

⊗ . . . ⊗ u

⊗ v

⊗ . . . ⊗ v

⊗ w

⊗ . . . ⊗ w ˙

⊗ . . . ⊗ w

if j = k = 1, (uv)w = 0 if k 6= 1 or l 6= 1,

= u

⊗ . . . ⊗ u ˙

. . . ⊗ u

⊗ v

⊗ . . . ⊗ v

⊗ w

⊗ . . . ⊗ w

if k = l = 1, u(vw) = 0 if k 6= 1 or l 6= 1,

= u

⊗ . . . ⊗ u ˙

. . . ⊗ u

⊗ v

⊗ . . . ⊗ v

⊗ w

⊗ . . . ⊗ w

if k = l = 1, (u ց v)w = 0 if j 6= 1 or l 6= 1,

= u

⊗ . . . ⊗ u

⊗ v

⊗ . . . ⊗ v ˙

. . . ⊗ v

⊗ w

⊗ . . . ⊗ w

if j = l = 1, u ց (vw) = 0 if j 6= 1 or l 6= 1,

= u

⊗ . . . ⊗ u

⊗ v

⊗ . . . ⊗ v ˙

. . . ⊗ v

⊗ w

⊗ . . . ⊗ w

if j = l = 1, (u ց v) ց w = 0,

u(v ց w) = 0.

So (T

(V ), m, ց) is a

ց

-algebra. Moreover, for all v

, . . . , v

∈ V : v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

= ( ˙ v

. . . v ˙

) ց ( ˙ v

. . . v ˙

).

Hence, T

(V ) is generated by V .

The

ց

-algebra T

(V ) is also graded by putting V in degree 1. It is then a quotient of the free

ց

-algebra generated by V , which is:

M

P

˜

ց

(n) ⊗ V

. So, for all n ∈

, dim(˜

ց

(n) ⊗ V

) ≥ dim(T

(V )(n)), so dim(˜

ց

(n)) ≥ n and dim(

ց

(n)) ≥ nn!.

Lemma 4 For all n ∈

, dim(

ց

(n)) ≤ nn!.

Proof.

ց

(n) is linearly generated by the binary trees with n indexed leaves, whose internal vertices are decorated by m and ց. By the four first relations of

ց

, we obtain that

ց

(n) is generated by the trees of the following form:

@@ ^@

...

@

σ(1) σ(2)

σ(n−2) σ(n−1)σ(n)

a1 a2

a_n

−2 an−1

,

with σ ∈ S

, a

, . . . a

∈ {m, ց}. With the last relation, we deduce that

ց

(n) is generated by the trees of the following form:

@@ ^@

...

@

σ(1) σ(2)

σ(i−1) σ(i)

ց ց

ց m

@

...

σ(n−2) σ(n−1)σ(n)

m m

,

where σ ∈ S

, 1 ≤ i ≤ n. Hence, dim(

ց

(n)) ≤ nn!.

As a consequence:

Theorem 5 Let n ≥ 1.

1. dim(

ց

(n)) = nn!.

2.

ց

(n) is freely generated, as a S

-module, by the following trees:

@@ ^@

...

@

1 2

i−1 i

ց ց

ց m

@

...

n−2 n−1 n

m m

, where 1 ≤ i ≤ n.

3. T

(V ) is the free

ց

-algebra generated by V . 2.3 Homology of a P

-algebra

Let us now describe the cofree

-algebra cogenerated by V . By duality, it is equal to T

(V ) as a vector space, with coproducts given in the following way: for v = v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

,

∆(v) =

m−1

X

i=k

(v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

) ⊗ ( ˙ v

⊗ . . . ⊗ v

),

∆

(v) = X

i=1

( ˙ v

⊗ . . . ⊗ v

) ⊗ (v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

).

Let A be a

-algebra. The homology complex of A is given by the shifted cofree coalgebra T

(V )[−1], with differential d : T

(V )(n) −→ T

(V )(n − 1), uniquely determined by the following conditions:

1. for all a, b ∈ A, d( ˙ a ⊗ b) = ab.

2. for all a, b ∈ A, d(a ⊗ b) = ˙ a ց b.

3. Let θ : T

(A) −→ T

(A) be the following application:

θ :

T

(A) −→ T

(A)

x −→ (−1)

x for all homogeneous x.

Then d is a θ-coderivation: for all x ∈ T

(A),

∆(d(x)) = (d ⊗ Id + θ ⊗ Id) ◦ ∆(x),

∆

(d(x)) = (d ⊗ Id + θ ⊗ Id) ◦ ∆

(x).

So, d is the application which sends the element v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

to:

k−2

X

i=1

(−1)

v

⊗ . . . ⊗ v

v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

+(−1)

v

⊗ . . . ⊗

.

z }| {

v

ց v

⊗ . . . ⊗ v

+(−1)

v

⊗ . . . ⊗

.

z }| {

v

v

⊗ . . . ⊗ v

+

n−1

X

i=k+1

(−1)

v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

v

⊗ . . . ⊗ v

.

The homology of this complex will be denoted by H

(A). More clearly, for all n ∈

:

H

(A) = Ker

d

Im

d

. Examples. Let v

, v

, v

∈ A.



 

 

 

 

 



 

 

 

 

 



d(v

) = 0, d( ˙ v

⊗ v

) = v

v

, d(v

⊗ v ˙

) = v

ց v

, d( ˙ v

⊗ v

⊗ v

) =

.

z}|{ v

v

⊗v

− v ˙

⊗ v

v

, d(v

⊗ v ˙

⊗ v

) =

.

z }| {

v

ց v

⊗v

− v

⊗

.

z}|{ v

v

, d(v

⊗ v

⊗ v ˙

) = v

v

⊗ v ˙

− v

⊗

.

z }| { v

ց v

.

So: 





d

( ˙ v

⊗ v

⊗ v

) = (v

v

)v

− v

(v

v

),

d

(v

⊗ v ˙

⊗ v

) = (v

ց v

)v

− v

ց (v

v

), d

(v

⊗ v

⊗ v ˙

) = (v

v

) ց v

− v

ց (v

ց v

).

So the nullity of d

on T

(A)(3) is equivalent to the three relations defining

-algebras (this is a general fact [5]). In particular:

H

(A) = A

A.A + A ց A . 2.4 Homology of free P

-algebras

The aim of this paragraph is to prove the following result:

Theorem 6 let N ≥ 1 and let A be the free

-algebra generated by D elements. Then H

(A) is D-dimensional; if n ≥ 1, H

(A) = (0).

Proof. Preliminaries. We put, for all k, n ∈

:



 

 



 

 



C

= T

(A)(n),

C

= A ⊗ . . . ⊗ A ˙ ⊗ . . . ⊗ A

| {z } A in position k

⊆ C

if k ≤ n, C

= L

i≤k,n

C

⊆ C

.

For all k ∈

, C

is a subcomplex of C

. In particular, C

is isomorphic to the complex defined by C

= A

, with a differential defined by:

d

:



 

 

A

−→ A

v

⊗ . . . ⊗ v

−→

n−1

X

i=1

(−1)

v

⊗ . . . ⊗ v

⊗ v

v

⊗ v

⊗ . . . ⊗ v

.

The homology of C

is then the shifted Hochschild homology of A. As A is a free (non unitary) associative algebra, this homology is concentrated in degree 1. So, Ker

d

⊆ Im(d) if n ≥ 2.

First step. Let us fix n ≥ 2 and let us show that Ker d

n

⊆ Im(d) for all 1 ≤ k ≤ n − 1 by induction on k. For k = 1, this is already done. Let us assume that 2 ≤ k < n and Ker

d

⊆ Im(d). Let x = X

i=1

x

∈ Ker d

n

, with x

∈ C

. If x

= 0, then x ∈ Ker

d

by the induction hypothesis. Otherwise, we put:

x

= X

v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

. We project d(x) over C

. we obtain:

0 =

k−1

X

i=1

π

(d(x

)) + X

X

i=1

(−1)

π

(v

⊗ . . . ⊗ v

⊗ v

v

⊗ v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

)

+ X

(−1)

π

(v

⊗ . . . ⊗ v

⊗

.

z }| {

v

ց v

⊗v

⊗ . . . ⊗ v

)

+ X

(−1)

π

(v

⊗ . . . ⊗ v

⊗

.

z }| {

v

v

⊗v

⊗ . . . ⊗ v

) + X

X

i=k+1

(−1)

π

(v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

⊗ v

v

⊗ v

⊗ . . . ⊗ v

)

= 0 + 0 + 0 + X

(−1)

v

⊗ . . . ⊗ ⊗v

⊗

.

z }| {

v

v

⊗v

⊗ . . . ⊗ v

+ X

X

i=k+1

(−1)

v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

⊗ v

v

⊗ v

⊗ . . . ⊗ v

= (−1)

X

v

⊗ . . . ⊗ v

⊗ d

(v

⊗ . . . ⊗ v

).

Hence, we can suppose that d

(v

⊗ . . . ⊗ v

) = 0. As n − k + 1 ≥ 2 and the complex C

is exact in degree n − k + 1 ≥ 2, there exists P

w

⊗ . . . ⊗ w

∈ A

, such that:

d

X

w

⊗ . . . ⊗ w

= v

⊗ . . . ⊗ v

. We put w = P

v

⊗. . .⊗v

⊗( P

˙

w

⊗ . . . ⊗ w

). Then d(w) = x

+C

, so x−d(w) ∈ C

. As Im(d) ⊆ Ker(d), x − d(w) ∈ Ker

d

⊆ Im(d) by the induction hypothesis. So, x ∈ Im(d).

Second step. Let us show that Ker d

n

⊆ Im(d) if n ≥ 3. Let x ∈ Ker d

n

. As before, we put x =

X

x

, with x

∈ C

and:

x

= X

i

v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

.

We can assume that the v

’s are homogeneous. Let us fix an integer N , greater than the degree of x

and an integer M , smaller than max

i

{weight(v

)}. Let us show by decreasing induction the following property: For allx ∈ Ker

d

n

of weight ≤ N and such that max

i

{weight(v

)} ≥ M , then x ∈ Im(d). If M > N, such an x is zero and the result is trivial. Let us assume the property at rank M + 1 and let us prove it at rank M . Let A

be the homogeneous component (for the weight of forests) of degree M of A. We project d(x) over A ⊗ . . . ⊗ A ˙

. Then:

0 = ̟

(d(x)) = X

i, weight(vnⁱ)=M

d

(v

⊗ . . . ⊗ v

) ⊗ v ˙

.

Hence, we can suppose that, for all i such that weight(v

) = M , d

(v

⊗ . . . ⊗ v

) = 0. As n ≥ 3 and C

is exact at n − 1 ≥ 2, there exists X

j

w

⊗ . . . ⊗ w

∈ A

such that:

d



 X

j

w

⊗ . . . ⊗ w



 = v

⊗ . . . ⊗ v

.

As d

is homogeneous for the weight, the weight of this element can be supposed smallest than the weight of v

⊗ . . . ⊗ v

. We then put w = X

i, poids(vⁱn)=M



 X

j

w

⊗ . . . ⊗ w



 ⊗ v ˙

. So x − d(w) is in x ∈ Ker

d

, with weight ≤ N , and satisfies the property on the v

’s for M + 1. By induction hypothesis, x − d(w) ∈ Im(d), so x ∈ Im(d).

Hence, if n ≥ 3, Ker d

n

⊆ Im(d). As C

= C

, for all n ≥ 3, d(C

) ⊆ Ker d

⊆ d(C

). Consequently, if n ≥ 2, H

(A) = (0).

Third step. We now compute H

(A). We take an element x ∈ C

and show that it belongs to Im(d). This x can be written under the form:

x = X

F,G∈F^D−{1}

a

F ⊗ G ˙ − X

F,G∈F^D−{1}

b

F ˙ ⊗ G.

So:

d(x) = X

F,G∈F^D−{1}

a

F ց G − X

F,G∈F−{1}

b

F G.

Hence, the following assertions are equivalent:

1. d(x) = 0.

2. For all H ∈ F

− {1}, X

FցG=H

a

= X

F G=H

b

.

First case. For all F, G ∈ F

− {1}, a

= 0, that is to say x ∈ A ˙ ⊗ A. So d(x) = d

(x

). As C

is exact in degree 2, there exists v

⊗ v

⊗ v

∈ A

such that d

(v

⊗v

⊗ v

) = X

F,G

b

F ⊗ G.

Consequently, d( ˙ v

⊗ v

⊗ v

) = X

F,G

b

F ˙ ⊗ G = x.

Second case. x = F

⊗ F ˙

− G ˙

⊗ G

, F

, F

, G

, G

∈ F

, such that F

ց F

= G

G

=

H. We put H = t

. . . t

and t

= B

(s

. . . s

), t

, . . . , t

, s

, . . . , s

∈ T

. There exists

i ∈ {1, . . . , n − 1} such that G

= t

. . . t

and G

= t

. . . t

; there exists j ∈ {1, . . . , m − 1}

such that F

= s

. . . s

and F

= B

(s

. . . s

)t

. . . t

. Then:

d(s

. . . s

⊗

.

z }| {

B

(s

. . . s

)t

. . . t

⊗t

. . . t

)

=

.

z }| {

(s

. . . s

) ց B

(s

. . . s

)t

. . . t

⊗t

. . . t

−s

. . . s

⊗

.

z }| {

B

(s

. . . s

)t

. . . t

t

. . . t

= G ˙

⊗ G

− F

⊗ F ˙

. So, x ∈ Im(d).

Third case. We suppose now the following condition:

(a

6= 0) = ⇒ (G / ∈ T

).

So, x can be written:

x = X

F,G∈F^D, t∈T^D

a

F ⊗

.

z}|{ tG − X

F,G∈F^D

b

F ˙ ⊗ G.

By the second case, F ⊗

.

z}|{ tG −

.

z }| {

F ց t ⊗G ∈ Im(d) ⊆ Ker(d). So the following element belongs to Ker(d):

x − X

F,G∈F^D, t∈T^D

a

(F ⊗

.

z}|{ tG −

.

z }| { F ց t ⊗G)

= − X

F,G∈F^D

b

F ˙ ⊗ G + X

F,G∈F^D, t∈T^D

a

.

z }| { F ց t ⊗G.

By the first case, this element belongs to Im(d), so x ∈ Im(d).

Fourth case. We suppose now the following condition:

(a

6= 0) = ⇒ (G / ∈ T

ou G =

, d ∈ D).

Let H = B

(t

. . . t

) ∈ T

, different from a single root. Then:

0 = X

FցG=H

a

− X

F G=H

b

= X

i=1

a

− 0 = a

+ 0 = a

. Consequently, for all F ∈ F

, d ∈ D, a

= 0. By the third case, x ∈ Im(d).

General case. The following element belongs to Ker(d):

x

= x + X

F,G∈F^D, d∈D

a

d(F ⊗ G ⊗

˙

)

= x + X

F,G∈F^D, d∈D

a

F G ⊗

˙

− X

F,G∈F^D, d∈D

a

F ⊗

.

z }| { G ց

= x + X

F,G∈F^D, d∈D

a

F G ⊗

˙

− X

F,G∈F^D, d∈D

a

F ⊗ B

(G) ˙

= X

F∈F^D, G∈F^D−T^D

a

F ⊗ G ˙ + X

F∈F^D, d∈D

a

F ⊗

˙

− X

F,G∈F^D

b

F ˙ ⊗ G + X

F,G∈F^D, d∈D

a

F G ⊗

˙

.

So x

satisfies the condition of the fourth case, so x

∈ Im(d). Hence, x ∈ Im(d). This proves finally that Ker(d

) = d(C

), so H

(A) = (0)

It remains to compute H

(A). This is equal to A/(A.A + A ց A), so a basis of H

(A) is

given by the trees of weight 1, so dim(H

(A)) = D.

As an immediate corollary:

Corollary 7 The operad

is Koszul.

3 The operad P

is Koszul

3.1 Koszul dual of P

We denote by

ր

the Koszul dual of

. Theorem 8 The operad

ր

is generated by m and ր∈

ր

(2), with the relations:



 

 



 

 



ր ◦(ր, I) = ր ◦(I, ր), m ◦ (m, I) = m ◦ (I, m), m ◦ (ր, I) = ր ◦(I, m),

ր ◦(m, I) = 0, m ◦ (I, ր) = 0.

Proof. Similar as the proof of theorem 2.

Remarks.

1. So

ր

is a quotient of

. 2. The operad

ր

is the symmetrization of the non-Σ-operad ˜

ր

, generated by m and ր, with relations:



 

 



 

 



ր ◦(ր, I ) = ր ◦(I, ր), m ◦ (m, I) = m ◦ (I, m), m ◦ (ր, I ) = ր ◦(I, m),

ր ◦(m, I) = 0, m ◦ (I, ր) = 0.

3. Graphically, the relations of

ր

can be written in the following way:

@@

ր ր

=

@@

ր ր

,

@@

m m

=

@@

m m

,

@@

m ր

=

@@

ր m

,

@@

ր m

= 0,

@@

m ր

= 0.

3.2 Free P

ր

-algebras

Let V be finite-dimensional vector space. We put:



 

 



 

 



T

(V )(n) = M

V

for all n ≥ 1, T

(V ) =

M

T

(V )(n).

In order to distinguish the different copies of V

, we put:

T (V )(n) = M



 

 A ր ⊗ . . . ր ⊗ A ր ⊗ A ⊗ A ⊗ . . . ⊗ A

| {z }

(k − 1) signs ր ⊗



 

 .

The elements of A ր ⊗ . . . ր ⊗ A ⊗ . . . ⊗ A will be denoted by v

ր ⊗ . . . ր ⊗ v

⊗ . . . ⊗ v

. We define m and ր over T

(V ) in the following way: for v = v

ր ⊗ . . . ր ⊗ v

⊗ . . . ⊗ v

and w = w

ր ⊗ . . . ր ⊗ w

⊗ . . . ⊗ w

,

vw =

0 if l 6= 1,

v

ր ⊗ . . . ր ⊗ v

⊗ . . . ⊗ v

⊗ w

⊗ . . . ⊗ w

if l = 1;

v ր w =

0 if k 6= m − 1,

v

ր ⊗ . . . ր ⊗ v

ր ⊗ w

ր ⊗ . . . ր ⊗ w

⊗ . . . ⊗ w

if k = 1.

As for

, we can prove the following result:

Theorem 9 Let n ≥ 1.

1. dim(

ր

(n)) = nn!.

2.

ր

(n) is freely generated, as a S

-module, by the following trees:

@@ ^@

...

@

σ(1) σ(2)

σ(i−1) σ(i)

ր ր

ր m

@

...

σ(n−2) σ(n−1)σ(n)

m m

, where 1 ≤ i ≤ n.

3. T

(V ) is the free

ր

-algebra generated by V . 3.3 Homology of a P

-algebra

Let us now describe the cofree

-algebra cogenerated by V . By duality, it is equal to T

(V ) as a vector space, with coproducts given in the following way: for v = v

ր ⊗ . . . ր ⊗ v

⊗ . . . ⊗ v

,

∆(v) =

m−1

X

i=k

(v

ր ⊗ . . . ր ⊗ v

⊗ . . . ⊗ v

) ⊗ (v

⊗ . . . ⊗ v

),

∆

(v) =

k−1

X

i=1

(v

ր ⊗ . . . ր ⊗ v

) ⊗ (v

ր ⊗ . . . ր ⊗ v

⊗ . . . ⊗ v

).

Let A be a

-algebra. The homology complex of A is given by the shifted cofree coalgebra T

(V )[−1], with differential d : T

(V )(n) −→ T

(V )(n − 1), uniquely determined by the following conditions:

1. for all a, b ∈ A, d(a ⊗ b) = ab.

2. for all a, b ∈ A, d(a ր ⊗ b) = a ր b.

3. Let θ : T

(A) −→ T

(A) be the following application:

θ :

T

(A) −→ T

(A)

x −→ (−1)

x for all homogeneous x.

Then d is a θ-coderivation: for all x ∈ T

(A),

∆(d(x)) = (d ⊗ Id + θ ⊗ Id) ◦ ∆(x),

∆

(d(x)) = (d ⊗ Id + θ ⊗ Id) ◦ ∆

(x).

So, d is the application which sends the element v

⊗ . . . ⊗ v ˙

⊗ . . . ⊗ v

to:

d(v

ր ⊗ . . . ր ⊗ v

⊗ . . . ⊗ v

)

= X

i=1

(−1)

v

ր ⊗ . . . ր ⊗ v

ր ⊗ v

ր v

ր ⊗ v

ր ⊗ . . . ր ⊗ v

⊗ . . . ⊗ v

+

n−1

X

i=k

(−1)

v

ր ⊗ . . . ր ⊗ v

⊗ . . . ⊗ v

⊗ v

v

⊗ v

⊗ . . . ⊗ v

.

This homology will be denoted by H

(A). More clearly, for all n ∈

:

H

(A) = Ker

d

Im

d

.

Examples. Let v

, v

, v

∈ A.



 

 

 



 

 

 



d(v

) = 0, d(v

⊗ v

) = v

v

, d(v

ր ⊗ v

) = v

ր v

,

d(v

⊗ v

⊗ v

) = v

v

⊗ v

− v

⊗ v

v

, d(v

ր ⊗ v

⊗ v

) = v

ր v

⊗ v

− v

ր ⊗ v

v

, d(v

ր ⊗ v

ր ⊗ v

) = v

ր v

ր ⊗ v

− v

ր ⊗ v

ր v

.

So: 





d

(v

⊗ v

⊗ v

) = (v

v

)v

− v

(v

v

),

d

(v

ր ⊗ v

⊗ v

) = (v

ր v

)v

− v

ր (v

v

),

d

(v

ր ⊗ v

ր ⊗ v

) = (v

ր v

) ր v

− v

ր (v

ր v

).

So the nullity of d

on T

(A)(3) is equivalent to the three relations defining

-algebras, as for

. In particular:

H

(A) = A

A.A + A ր A .

3.4 Homology of free P

-algebras

The aim of this paragraph is to prove the following result:

Theorem 10 let N ≥ 1 and let A be the free

-algebra generated by D elements. Then H

(A) is D-dimensional; if n ≥ 1, H

(A) = (0).

Proof. Preliminaries. We put, for k, n ∈

:



 

 

 

 

 

 

C

= T

(A)(n),

C

= A ր ⊗ . . . ր ⊗ A ⊗ . . . ⊗ A

| {z } k − 1 signs ր ⊗

⊆ C

if k ≤ n,

C

= L

i≤k,n

C

⊆ C

.

For all k ∈

, C

is a subcomplex of C

. In particular, C

is isomorphic to the complex defined by C

= A

, with differential given by:

d

:



 

 

A

−→ A

a

⊗ . . . ⊗ a

−→

n−1

X

i=1

(−1)

a

⊗ . . . ⊗ a

⊗ a

a

⊗ a

⊗ . . . ⊗ a

.

Hence, the homology of C

is the (shifted) Hochschild homology of A. As A is a free (non unitary) associative algebra, this homology is concentrated in degree 1. So:

Ker d

n

⊆ Im(d) if n ≥ 2. (1)

Moreover, C

admits a subcomplex defined by C

(n) = A ր ⊗ . . . ր ⊗ A, with differential given by:

d :



 

 

C

(n) −→ C

(n − 1) v

ր ⊗ . . . ր ⊗ v

−→

n−1

X

i=1

(−1)

v

ր ⊗ . . . ր ⊗ v

⊗ v

ր v

ր ⊗ v

ր ⊗ . . . ր ⊗ v

. Hence, the homology of this subcomplex is the shifted Hochschild homology of the associative algebra (A, ր).

Lemma 11 Every forest F ∈ F

− {1} can be uniquely written as F

ր . . . ր F

, where the F

’s are elements of F

of the form F

=

G

.

Proof. Existence. By induction on the weight of F . If weight(F ) = 1, F =

and the result is obvious. If weight(F ) ≥ 2, we put F = B

(H

)H

, with weight(H

) < weight(F). If H

= 1, the result is obvious. If H

6= 1, we apply the induction hypothesis on H

, so it can be written as H

= F

ր . . . ր F

, with F

=

G

. We put F

=

H

, so F = F

ր . . . ր F

.

Unicity. By induction on the weight of F . If weight(F ) = 1, then F =

and this is obvious. If weight(F ) ≥ 2, we put F = B

(H

)H

, with weight(H

) < weight(F ). If F = F

ր . . . ր F

, then F

=

H

and F

ր . . . ր F

= H

. Hence, F

is unique. We

conclude with the induction hypothesis.

This lemma implies that (A, ր) is freely generated by forests of the form

G. So:

Ker d

⊆ Im(d) if n ≥ 2. (2)