A SIMPLE PROOF OF A RESULT OF FURUSHO
ANTON ALEKSEEV AND CHARLES TOROSSIAN
DRAFT VERSION
Abstract. In this short note we recover, using material of [1], a important
result of Furusho [3], the Drinfeld pentagonal equation implies the hexagonal
and the symmetry relations.
1. Introduction
In [1] we introduced new methods to study the Kashiwara-Vergne problem and
we connected it to Drinfeld’s theory of associators. We defined a differential on the
Lie algebra of tangential derivations and calculated the second and third cohomological spaces in [1] Theorem 3.1. As a by product of this result we recover results
of [3].
2. Free Lie algebras and their derivations
2.1. Free Lie algebras and derivations. Let K be a field of characteristic zero,
and let lien = lie(x1 , . . . , xn ) be the degree completion of the graded free Lie algebra
over K with generators x1 , . . . , xn of degree one. We shall denote by dern the Lie
algebra of derivations of lien . An element u ∈ dern is completely determined by
its values on generators, u(x1 ), . . . , u(xn ) ∈ lien . The Lie algebra dern carries a
grading induced by the one of lien .
Definition 1. A derivation u ∈ dern is called tangential if there exist ai ∈ lien , i =
1, . . . , n such that u(xi ) = [xi , ai ].
Tangential derivations form a Lie subalgebra tdern ⊂ dern . Elements of tdern
are in one-to-one correspondence with n-tuples of elements of lien , (a1 , . . . , an ),
which verify the condition that ak has no linear term in xk for all k. By abuse of
notations, we shall often write u = (a1 , . . . , an ) and tdern is a graded Lie algebra.
For two elements of tdern , u = (a1 , . . . , an ) and v = (b1 , . . . , bn ), we have [u, v]tder =
(c1 , . . . , cn ) with
(1)
ck = u(bk ) − u(ak ) + [ak , bk ]lie .
Definition 2. A derivation
u = (a1 , . . . , an ) ∈ tdern is called special if u(x) =
P
Pn
[x
,
a
]
=
0
for
x
=
x
i i i
i=1 i .
We shall denote the space of special derivations by sdern . It is obvious that
sdern ⊂ tdern is a Lie subalgebra. Both tdern and sdern integrate to prouniportent
groups denoted by T Autn and SAutn , respectively. In more detail, T Autn consists
of automorphisms of lien such that xi 7→ Adgi xi = gi xi gi−1 , where gi ∈ exp(lien ).
Similarly, elements
Pn of SAutn are tangential automorphisms of lien with an extra
property x = i=1 xi 7→ x.
1
2
ANTON ALEKSEEV AND CHARLES TOROSSIAN
The family of Lie algebras tdern is equipped with simplicial Lie homomorphisms
tdern → tdern+1 . For instance, for u = (a, b) ∈ tder2 we define
u1,2
u2,3
u12,3
= (a(x, y), b(x, y), 0) ,
= (0, a(y, z), b(y, z)) ,
= (a(x + y, z), a(x + y, z), b(x + y, z)),
and similarly for other simplicial maps. These Lie homomorphisms integrate to
group homomorphisms of T Autn and SAutn .
2.2. Braid Lie algebra. Consider t = (y, x) ∈ sder2 . By composing various
simplicial maps we obtain n(n − 1)/2 elements of ti,j = tj,i ∈ tdern with nonvanishing components xi at the jth place and xj at the ith place.
Elements ti,j ∈ sdern span a Lie subalgebra isomorphic to the quotient of the
free Lie algebra with n(n − 1)/2 generators by the following relations,
(2)
[ti,j , tk,l ] = 0
for k, l 6= i, j, and
(3)
[ti,j + ti,k , tj,k ] = 0
for all triples of distinct indices i, j, k. We denote by tn the Lie algebra defined by
relations (2) P
and (3). This is the pure braid Lie algebra. Recall the following result:
the element i<j ti,j belongs to the center of sdern .
Q∞
k
2.3. Cyclic words. Let Ass+
n =
k=1 Ass (x1 , . . . , xn ) be the graded free associative algebra (without unit) with generators x1 , . P
. . , xn . Every element a ∈ Ass+
n
n
admits a unique decomposition of the form a =
i=1 (∂i a)xi , where ai ∈ Assn
(Assn is a free associative algebra with unit).
We define the graded vector space cyn as a quotient
cyn = Ass+
n /h(ab − ba); a, b ∈ Assn i.
Here h(ab−ba); a, b ∈ Assn i is the subspace of Ass+
n spanned by commutators. The
multiplication map of Ass+
does
not
descend
to
cy
n which only has a structure of
n
a graded vector space. We shall denote by tr : Ass+
n → cyn the natural projection.
By definition, we have tr(ab) = tr(ba) for all a, b ∈ Assn imitating the defining
property of trace. In general, graded components of cyn are spanned by words of a
given length modulo cyclic permutations.
Example 1. The space cy1 is isomorphic to the space of formal power series in
one variable without constant term, cy1 ∼
= xk[[x]]. This isomorphism is given by
the following formula,
∞
∞
X
X
fk xk 7→
fk tr(xk ).
f (x) =
k=1
k=1
2.4. Divergence. Let u = (a1 , . . . , an ) ∈ tdern . We define the divergence as
n
X
div(u) =
tr(xi (∂i ai )).
i=1
It is a 1-cocycle of tdern with values in cyn (see Proposition 3.6 in [1]). Actually
this
We define krvn ⊂ sdern ⊂ tdern as the Lie algebra of special derivation with
vanishing divergence. Hence, u = (a1 , . . . , an ) ∈ krvn is a solution of two equations:
3
Pn
Pn
= 0 and i=1 tr(xi (∂i ai )) = 0. We shall denote by KRVn = exp(krvn )
the corresponding prounipotent group.
In [1] we have defined a differential d : tdern → tdern+1 and computed some
cohomology group. Recall the formula
i=1 [xi , ai ]
(4) d u = u2,3,...,n+1 −u12,...,n,n+1 +· · ·+(−1)n u1,2,...,n−1,n(n+1) +(−1)n+1 u1,2,...,n .
Example 2. For u ∈ tder2 we get d u = u2,3 − u12,3 + u1,23 − u1,2 . For u ∈ tder3
we obtain d u = u2,3,4 − u12,3,4 + u1,23,4 − u1,2,34 + u1,2,3 .
2.5. General projection. Let krvn ⊂ sdern ⊂ tdern the Lie algebra of special
dn the extended Lie algebra of special
derivation with divergence 0. Let denote krv
derivations u, such that it exists f ∈ xk[[x]] such that
div(u) = tr f (x1 ) + f (x2 ) . . . + f (xn ) − f (x1 + . . . + xn ) .
dn .
Such f are necessarily odd by [1] Proposition 4.1 and krvn is an ideal of krv
Let’s define πn the map from sdern to sdern−1 by
πn (a1 , . . . , an )|(x1 ,...,xn−1 ) = (a1 − an , a2 − an , · · · , an−1 − an )|(x1 ,...,xn )
with xn = −x1 − . . . − xn−1 .
It is well known that derivation of degree one in sdern are linear combination of
ti,j ∈ krvn . We get for example πn (ti,j ) = ti,j for i, j 6= n and
πn (ti,n ) = −
n−1
X
tk,i ∈ krvn−1 .
k=1
dn ) ⊂ krv
\
Lemma 1. The map πn is a Lie map from sdern to sdern−1 and πn (krv
n−1 .
Proof : The Lie property is left to the reader. Actually πn is not a Lie map on
dn is valued to krv
\
tdern . Let’s verify that πn restricted to krv
n−1 .
The first equation (special derivation) is easy because
[x1 , a1 − an ] + . . . + [xn−1 , an−1 − an ] = [x1 , a1 ] + . . . + [xn−1 , an−1 ] + [xn , an ] = 0.
Let’s now consider the divergence equation and deg(u) > 1. Indeed, if deg(u) = 1
then u is a linear combinaison of ti,j and clearly πn (ti,j ) ∈ krvn−1 .
Suppose
div(a1 , . . . , an )|(x1 ,...,xn ) = tr f (x1 ) + f (x2 ) + . . . + f (xn ) − f (x1 + x2 + · · · + xn ) .
Pn
Let a ∈ lien with a = i=1 ∂i (a)xi , then at (x1 , x2 , . . . , xn−2 , −x1 − x2 . . . − xn−1 )
we get
a(x1 , x2 , . . . , xn−2 , −x1 − x2 . . . − xn−1 ) =
n−1
X
n−1
X
i=1
i=1
(∂i a − ∂n a)xi =
(∂i0 a)xi .
We deduce the corresponding relation after substitution xn 7→ −x1 − x2 . . . − xn−1 ,
∂i0 a = (∂i a − ∂n a)|(x1 ,x2 ,...,xn−2 ,−x1 −x2 ...−xn−1 ) .
4
ANTON ALEKSEEV AND CHARLES TOROSSIAN
Then
(5)
0
x1 ∂10 (a1 −an )+. . .+xn−1 ∂n−1
(an−1 −an ) = x1 (∂1 a1 −∂n a1 )+. . .+xn−1 (∂n−1 an−1 −∂n an−1 )
+ x1 (−∂1 an + ∂n an ) + · · · + xn−1 (−∂n−1 an + ∂n an )
= x1 ∂1 a1 + . . . + xn−1 ∂n−1 an−1 − x1 ∂n a1 − · · · − xn−1 ∂n an−1 − xn ∂n an
− x1 ∂1 an − . . . − xn−1 ∂n−1 an
Use the relation [x1 , a1 ] + . . . + [xn , an ] = 0, we get
0 = ∂n ([x1 , a1 ] + . . . + [xn , an ]) = x1 ∂n (a1 ) + . . . xn ∂n (an ) − an .
By (5) and tr(x1 ∂1 an + . . . + xn−1 ∂n−1 an ) = tr(an − xn ∂n (an )) and after evaluating at (x1 , x2 , . . . , xn−2 , −x1 − x2 . . . − xn−1 ) we get
(6) div(πn (u)) = tr(x1 ∂1 a1 + . . . + xn−1 ∂n an ) − 2tr(an ) =
tr f (x1 )+. . .+f (xn−1 )+f (xn ) = tr f (x1 )+. . .+f (xn−1 )−f (x1 +. . .+xn−1 )
because f is odd and deg(an ) > 1.
Our maps extend to formal groups KRVn . In particular the kernel of π3 is a
sub-group of KRV3 . Set KRV3∗ = ker(π3 ). Any Drinfeld associators lives in KRV3∗
because π3 (Φ(t1,2 , t2,3 )) = Φ(t1,2 , −t1,2 ) = 1.
2.6. Symmetries of order n + 1 in krvn . Let u ∈ krvn and define
Sn+1 (u) = πn+1 (u2,3,...,n+1 ).
dn the previous computation ensures that Sn+1 (u) ∈ krv
dn with the same
For u ∈ krv
f in the divergence’s formula.
Lemma 2. Suppose u = (a, b) ∈ tder2 , then d u ∈ krv3 , if and only if u = cr + v,
d2 .
with r = (y, 0) and v ∈ krv
Proof : Indeed f (x, y) = u(x + y) = [x, a] + [y, b] verifies δf = (d u)(x + y + z) = 0,
so by [1] Theorem 3.1, f = 0 if deg(u) > 1, else u = c(y, 0) + αt with t = (y, x).
Now put g(x, y) = div(u), then δ(g) = div(d u) = 0 and g = δ(h) by [1], Theorem
2.1.
Lemma 3. The map Sn+1 is a Lie map of order n + 1.
Proof : It’s an easy verification. Our map is a transformation corresponding to the
companion matrix


0 0 0 . . . −1
 1 0 0 . . . −1 


 0 1 0 . . . −1 




..


.
0 0 . . . 1 −1
5
3. Furusho’s result
We now recover Furusho’s result [3] as consequence of [1] Theorem 3.1.
Theorem 1. Let ϕ ∈ lie2 a solution of the Drinfeld’s pentagonal equation. Then
ϕ is in grt1 .
Proof : Recall that ϕ solves the tangential pentagone equation if
ϕ(t2,3 , t3,4 )+ϕ(t1,2 +t1,3 , t4,2 +t4,3 )+ϕ(t1,2 , t2,3 ) = ϕ(t1,3 +t2,3 , t3,4 )+ϕ(t1,2 , t2,3 +t2,4 ).
Let’s consider φ = ϕ(t1,2 , t2,3 ) ∈ krv3 . Then
(7)
dφ = ϕ(t2,3 , t3,4 ) + ϕ(t1,2 + t1,3 , t4,2 + t4,3 ) + ϕ(t1,2 , t2,3 )
− ϕ(t1,3 + t2,3 , t3,4 ) − ϕ(t1,2 , t2,3 + t2,4 ) = 0
by the pentagonal equation with d the differential define on tdern [1]. Then by
Lemma1 π4 is a Lie map, we get
π4 (dφ) = ϕ(t1,2 , t2,3 ) + ϕ(t2,3 , t1,2 ),
because as element of t3 ⊂ sder3 (modulo c = t1,2 + t1,3 + t2,3 a central element
in sder3 ) π4 (t3,4 ) = −t1,3 − t2,3 = t1,2 , π4 (t4,2 + t4,3 ) = −t3,2 = t1,3 + t1,2 ,
π4 (t3,4 ) = −t1,3 − t2,3 and π4 (t2,3 + t2,4 ) = t2,3 − t2,1 − t2,3 = −t1,2 . So ϕ is
antisymmetric and φ1,2,3 + φ3,2,1 = 0.
By [1] Theorem 3.1, there exists a unique f ∈ tder2 such that d f = φ, because
φ has no terms of degree less than 2. We have (d f )3,2,1 = −d(f 21 ), because
d f = f 2,3 − f 12,3 + f 1,23 − f 1,2 . So by uniqueness we get f = f 2,1 . The hexagonal
equation is then perfectly automatic :
(8) (d f )1,2,3 + (d f )2,3,1 + (d f )3,1,2 = ψ(t1,2 , t2,3 ) + ψ(t2,3 , t3,1 ) + ψ(t3,1 , t1,2 ) = 0.
Indeed suppose f = (a, b) then
d f = −(a1,2 , b1,2 , 0) − (a12,3 , a12,3 , b12,3 ) + (0, a2,3 , b2,3 ) + (a1,23 , b1,23 , b1,23 )
(d f )3,1,2 = −(b3,1 , 0, a3,1 ) − (a31,2 , b31,2 , a31,2 ) + (a1,2 , b1,2 , 0) + (b3,12 , b3,12 , a3,12 )
(d f )2,3,1 = −(0, a2,3 , b2,3 ) − (b23,1 , a23,1 , a23,1 ) + (b3,1 , 0, a3,1 ) + (b2,31 , a2,31 , b2,31 ),
and it’s easy to verify the hexagon equation.
Lemma 4. Reciprocally suppose d f solve the hexagone equation (8) then f = f 2,1
if f is of degree at least 3.
Proof : Take f = (a, b) and g = a − b21 . Then you get by hexagonal equation
g 12,3 + g 13,2 − g 1,23 = 0 id.
g(x + y, z) − g(x, y + z) + g(x + z, y) = 0.
Put x = 0 you get g antisymmetric. Put x = y, you get g(2x, z) = 2g(x, x + z),
then g(2n x, z) = 2n g(x, (2n − 1)x + z) and by x 7→ 2−n x you get g(x, z) =
2n g(2−n x, (1 − 2−n )x + z). Put n 7→ ∞, you get g is degree one in x and by antisymmetry you get g(x, y) = [x, y]. If f is of degree at least 3, g = 0 and f 21 = f .
6
ANTON ALEKSEEV AND CHARLES TOROSSIAN
4. Symmetries of order 3
In this section we investigate some properties of our symmetries.
4.1. Symmetries and differential.
d2 and φ ∈ krv3 we have
Proposition 1. For f ∈ krv
(9)
S4 (d f ) + d (S3 (f )) = (S3 f − f )2,3
(10)
S5 (dφ) + d (S4 (φ)) = (S4 φ + φ)2,3,4
Proof : We first prove (9). Take f = (a, b), then d f = f 2,3 − f 12,3 + f 1,23 − f 1,2 ,
i.e.
d f = (0, a, b)y,z − (a, a, b)x+y,z + (a, b, b)x,y+z − (a, b, 0)x,y
and S3 (f ) = (−b, a − b)y,−x−y . For x + y + z + t = 0 you get then
d S3 (f ) = (0, −b, a−b)z,x+t −(−b, −b, a−b)z,t +(−b, a−b, a−b)y+z,t −(−b, a−b, 0)y,z+t .
By definition S4 (d f ) = π4 ((d f )2,3,4 ) so
S4 (d f ) = (−b, −b, a − b)z,t − (−b, a − b, a − b)y+z,t + (−b, a − b, 0)y,z+t − (0, a, b)y,z
you deduce
S4 (d f ) + d (S3 (f )) = (0, −b, a − b)z,x+t − (0, a, b)y,z = (S3 f − f )2,3 .
We prove now (10). Take φ = (a, b, c)x,y,z . Then S4 (φ) = (−c, a−c, b−c)y,z,−x−y−z ,
and
d φ = (0, a, b, c)y,z,t −(a, a, b, c)x+y,z,t +(a, b, b, c)x,y+z,t −(a, b, c, c)x,y,z+t +(a, b, c, 0)x,y,z .
For x + y + z + t + w = 0 you get
d S4 (φ) = (0, −c, a−c, b−c)z,t,x+w −(−c, −c, a−c, b−c)z,t,w +(−c, a−c, a−c, b−c)y+z,t,w
− (−c, a − c, b − c, b − c)y,z+t,w + (−c, a − c, b − c)y,z,w+t
and
S5 (d φ) = (−c, −c, a−c, b−c)z,t,w −(−c, a−c, a−c, b−c)y+z,t,w +(−c, a−c, b−c, b−c)y,z+t,w
− (−c, a − c, b − c, 0)y,z,t+w + (0, a, b, c)y,z,t .
So you get
S5 (dφ) + d (S4 (φ)) = (0, −c, a − c, b − c)z,t,−y−z−t + (0, a, b, c)y,z,t = (φ + S4 φ)2,3,4 .
7
d2 , a symme4.2. Σ3 -invariant action on associators. We have described on krv
2,1
try of order 2 , f 7→ f
and a symmetry of order 3, f 7→ S3 (f ). The group
generated by those symmetries is of order 6, isomorphic to the permutations group
Σ3 . Similar symmetries appear in Drinfeld’s seminal work [2].
Lemma 5. The Lie algebra grt1 has the two symmetries.
d2 by
Proof : Recall that we constructed a Lie map from grt1 into krv
ψ ∈ grt1 7→ Ψ = (ψ(−x − y, x), ψ(−x − y, y)
and we proved that dΨ = ψ(t1,2 , t2,3 ) ∈ krv3 . We have Ψ = Ψ2,1 and
π3 (dΨ) = Ψ − S3 (Ψ) = π3 (ψ(t1,2 , t2,3 )) = 0
because π3 (t2,3 ) = −t1,2 . So Ψ is S3 -invariant.
\2
We investigate the action of KRV
ker(π3 ).
on associators of KRV3 . Let’s put KRV3∗ =
S3
acts almost freely and transitively on PentaΣ
\2 3 acts almost freely and transitively on
group KRV
Drinfeld’s associator lives in KRV3∗ .
\2
Proposition 2. The group KRV
gon solution in KRV3∗ . The
associators of KRV3∗ . Every
Σ3
Proof : Let’s recall the Pentagonal equation for Φ ∈ KRV3 :
(11)
Φ2,3,4 Φ1,23,4 Φ1,2,3 = Φ12,3,4 Φ1,2,34 .
\2 acts almost freely and transitively on
We know by Theorem 7.1 [1] that KRV
S
\2 2 acts almost
Pentagon solution in KRV3 and by Proposition 8.8 [1] that KRV
freely and transitively on associators of KRV3 .
For π3 (Φ) = 1 then π4 (∆1,2 Φ) = π4 (Φ12,3,4 ) = ∆1,2 π3 (Φ) = 1, because π4 ◦
∆
= ∆1,2 ◦ π3 . Apply π4 to equation (11), you get S4 (Φ)Φ = 1.1 For Φ an
associator Φ3,2,1 Φ = 1, S4 (Φ) = Φ3,2,1 and S4 acts on associators as a symmetry of
order 2.
1,2
\2 and Φ ∈ KRV3∗ . Then g · Φ = g 2,3 g 1,23 Φ(g 1,2 g 12,3 )−1 and
Take g ∈ KRV
S
\2 3 acts
π3 (g · Φ) = g −1 S3 (g). So g · Φ ∈ KRV3∗ iff S3 (g) = g. We conclude KRV
almost freely and transitively on Pentagon solution in KRV3∗ .
Remark 1. Recall that any associator in KRV3 is solved by a Kashiwara-Vergne
twist, that is ΦF = F 1,23 F 2,3 (F 12,3 F 1,2 )−1 . Probably there is a matter to write
down directly an Σ3 action on F as we did for the Σ2 action in [1] §8.
1For the tangential version, you get S (ϕ) + ϕ = 0
4
8
ANTON ALEKSEEV AND CHARLES TOROSSIAN
References
[1] Alekseev, A.; Torossian, C.,– The Kashiwara-Vergne conjecture and Drinfeld’s associators. arxiv: 0802.4300
[2] Drinfeld, V. G.,– On quasitriangular quasi-Hopf algebras and on a group that is closely
connected with Gal(Q/Q). (Russian) Algebra i Analiz 2 (1990), no. 4, 149–181; translation in Leningrad Math. J. 2 (1991), no. 4, 829–860
[3] H. Furusho, Pentagon and hexagon equations, preprint arXiv:math/0702128.
[4] Kashiwara, M. ; Vergne, M.,– The Campbell-Hausdorff formula and invariant hyperfunctions. Inventiones Math. 47 (1978), 249–272.
Section de mathématiques, Université de Genève, 2-4 rue du Lièvre, c.p. 64, 1211
Genève 4, Switzerland
E-mail address: [email protected]
Institut Math?matiques de Jussieu, Université Paris 7, CNRS; Case 7012, 2 place
Jussieu, 75005 Paris, France
E-mail address: [email protected]