Double Shuffle and Kashiwara-Vergne Lie algebras

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by  Leila Schneps

Double shuffle and Kashiwara-Vergne Lie algebras
Leila Schneps
Abstract. We prove that the double shuffle Lie algebra ds, dual to the space of new
formal multiple zeta values, injects into the Kashiwara-Vergne Lie algebra krv defined
2
andstudiedbyAlekseev-Torossian. Theproofisbasedonareformulationofthedefinition
of krv , and uses a theorem of Ecalle on a property of elements of ds.
2
2
1
0
§1. Definitions and main results
2

n
a Let Qhx,yi denote the ring of polynomials in non-commutative variables x and y, and Lie[x,y] the Lie
J algebra of Lie polynomials inside it. For each n ≥ 1, let Q hx,yi (resp. Lie [x,y]) denote the subspace of
n n

5 homogeneouspolynomials(resp. Lie polynomials)ofdegreen. Fork ≥1,letQ hx,yi(resp. Lie [x,y])
n≥k n≥k
2 denote the space of polynomials (resp. Lie polynomials) all of whose monomials are of degree ≥ k, i.e. the
] direct sum of the Qnhx,yi (resp. Lien[x,y]) for n≥k.
G ThemaintheoremofthispapergivesaninjectivemapbetweentwoLiealgebrasstudiedintheliterature
concerning formal multiple zeta values: the double shuffle Lie algebra ds, investigated in papers by Racinet
A
and Ecalle amongst others (the associated graded of ds is also studied in papers by Zagier, Kaneko and
.
h others), andthe Kashiwara-VergneLie algebraintroduced in workofAlekseev and Torossian(cf. [AT]). We
at begin by recalling the definitions of these two Lie algebras. As vector spaces, both are subspaces of the free
m Lie algebra Lie[x,y].
Foranynon-trivialmonomialwandpolynomialf ∈Qhx,yi,weusethenotation(f|w)forthecoefficient
[
of the monomial w in the polynomial f, and extend it by linearity to polynomials w without constantterm.
1 Set y = xi−1y for all i ≥ 1; then all words ending in y can be written as words in the variables y . The
v i i
stuffle product st(u,v)∈Qhx,yi of two such words u and v is defined recursively by
6
1
3 st(1,u)=st(u,1)=u and st(yiu,yjv)=yist(u,yjv)+yjst(yiu,v)+yi+jst(u,v).
5
.
1
Definition 1.1. The double shuffle Lie algebra ds∗ is the vector space of elements f ∈ Lie [x,y] such
0 n≥3
2 that
1 f st(u,v) =0
v: for all words u,v∈Qhx,yi ending in y but no(cid:0)t b(cid:12)(cid:12)oth simu(cid:1)ltaneously powers of y.
i
X
IthasbeenshownbyRacinet[R](see alsoasimplifiedversionofRacinet’sproofinthe appendixof[F])
r
a and Ecalle [E]that ds is actually closed, i.e. a Lie algebra,under the Poissonbracketdefined onLie[x,y] by
{f,g}=[f,g]+D (g)−D (f), (1.1)
f g
where for any f ∈Lie[x,y], the associatedderivationD ofLie[x,y] is defined by D (x)=0, D (y)=[y,f].
f f f
This Lie bracket corresponds to identifying f with D and taking the natural Lie bracket on derivations:
f
[D ,D ]=D . (1.2)
f g {f,g}
∗ The equivalence of the present definition with the usual definition introduced in [R] is proven in [CS],
Theorem 2, which proves that if a polynomial f ∈ Lie [x,y] has the property of the present definition,
n
then f + (−1)n−1(f|xn−1y)yn satisfies the stuffle relations for all pairs of words u,v ending in y. Since
n
the words ending in x are not involved in this condition, this is equivalent to the assertion that π (f)+
y
(−1)n−1(f|xn−1y)yn satisfiesstuffle, where π (f)denotes the projectionoff onto just its wordsending iny.
n y
This is the standard form of the defining property of elements of ds.
1
Let us now recall the definition of krv . Following [AT], let TR denote the vector space quotient of
2
Qhx,yi by relations ab = ba. The image in TR of a monomial w is the equivalence class of monomials
obtained by cyclically permutating the letters of w. The trace map Qhx,yi→TR is denoted by tr.
Definition 1.2. For any pair of elements F, G ∈ Lie [x,y] with n ≥ 1, let D denote the derivation of
n F,G
Lie[x,y] defined by x 7→ [x,G] and y 7→ [y,F]. Such a derivation is said to be special if x+y 7→ 0, i.e. if
[x,G]+[y,F]=0. The underlying vector space of the Kashiwara-VergneLie algebrais spanned by those of
thesespecialderivationsD thatalsosatisfythepropertythatwritingF =F x+F y andG=G x+G y,
F,G x y x y
there exists a constant A such that
tr(F y+G x)≡Atr (x+y)n−xn−yn ∈TR. (1.3)
y x
(cid:0) (cid:1)
It is shown in [AT] that krv is a Lie algebra under the natural bracket on derivations. The degree provides
2
a grading on krv , for which (krv ) is spanned by the D with F,G∈Lie [x,y].
2 2 n F,G n
Thefirstgradedpiece,(krv ) ,is1-dimensional,generatedbyD . Thesecondgradedpiece(krv ) =0.
2 1 y,x 2 2
Now let n ≥ 3. Note that for any F ∈ Lie [x,y], if there exists G ∈ Lie [x,y] such that [y,F]+[x,G] =0,
n n
then G is unique. Indeed, G is defined up to a centralizer of x, but that can only be x, which is of degree 1.
One of the most useful results of this paper is the precise determination of the elements F admitting such a
G, together with an explicit formula for G (theorem 2.1, see also (1.5)).
Let ∂ denote the derivation of Qhx,yi defined by ∂ (x)=1, ∂ (y)=0. Following Racinet [R], for any
x x x
polynomial h in x and y, set
(−1)i
s(h)= ∂i(h)yxi. (1.4)
i! x
Xi≥0
Racinetshowsthatiff =f x+f y isanelementofLie[x,y],orindeedanypolynomialsuchthat∂ (f)=0,
x y x
then
f =s(f ). (1.5)
y
The main result of this paper is the following.
Theorem 1.1. Let f˜(x,y) ∈ ds, and set f(x,y) = f˜(x,−y) and F(x,y) = f(z,y) with z = −x−y. Write
F =F x+F y =xFx+yFy in Qhx,yi. Set G=s(Fx). Then the map f˜7→D yields an injective map
x y F,G
of Lie algebras
ds֒→krv .
2
Remark. The map defined in theorem 1.1. from ds to the space of derivations D mapping x 7→ [x,G]
F,G
and y 7→ [y,F] is injective. Indeed, because D (y) = [y,F] is a Lie element in which no word starts and
F,G
ends with x, we can recover F from D (y) by applying proposition 2.2 (with x and y exchanged in the
F,G
statement), and then we recover f˜by F(x,y)=f˜(z,−y).
Furthermore, this injection of vector spaces is in fact an injection of Lie algebras, since ds is equipped
with the Poisson bracket, which is compatible with the natural bracket on derivations (cf. (1.2)).
Thus, to prove theorem 1.1, it remains only to prove that the derivations D arising from elements
F,G
f˜∈ds actually lie in krv , i.e. are special and satisfy the trace formula (1.3).
2
One ofthe mainingredients inour proofof theorem1.1 is a combinatorialreformulationofthe defining
properties of krv , given in theorem 1.2 below. First we need some definitions.
2
Definition 1.3. Let w = xa0y···yxar be a monomial in Qhx,yi of depth r (i.e. containing r y’s), with
a ≥0 for 0≤i≤r. Let anti denote the palindrome or backwards-writing operator on monomials, and let
i
push denote the cyclic permutation of x-powers operator on monomials, defined respectively by
anti(xa0y···yxar−1yxar)=xaryxar−1y···yxa0 (1.6)
2
push(xa0y···yxar−1yxar)=xaryxa0···yxar−1. (1.7)
Foranywordw,wedefinethelistPush(w)tobethelistof(r+1)wordsobtainedfromwbyiteratingthepush
operator. Note thatPush(w) is alist, not aset; it may containrepeatedwords. Forexample,if w =x2yxy,
then Push(w)=[x2yxy,yx2yx,xy2x2], and if w =xyxyx, then Push(w)=[xyxyx,xyxyx,xyxyx].
Definition 1.4. We extend the anti and push operatorsto operatorson polynomials by linearity; it makes
sensetoapplytheseoperatorstoapolynomialevenifthemonomialsinthepolynomialhavedifferentdegrees
and depths. If f is a polynomial in x and y of homogeneous degree n≥3, we say that f is
• palindromic if f =(−1)n−1anti(f),
• antipalindromic if f =(−1)nanti(f),
• push-invariant if push(f)=f,
• push-constant if there exists a constant A such that (f|v)=A for all w6=yn, and (f|yn)=0.
v∈Push(w)
P
The following statement contains our reformulation of the definition of krv that appears in [AT].
2
Theorem 1.2. Let V be the vector space spanned by all polynomials F ∈ Lie [x,y] for n ≥ 3 such that,
kv n
writing F =F x+F y, we have
x y
i) F is antipalindromic, or equivalently, F is push-invariant;
y
ii) F −F is push-constant.
y x
For each such F, set G=s(Fx). Then the map F 7→D extends to a vector space isomorphism
F,G
V →∼ krv . (1.8)
kv 2
Themainresultof§2,theorem2.1,isanenumerationofseveralconditionsequivalenttothe specialness
property. Using this result,theorems 1.1 and1.2are provedin§3. The proofoftheorem1.1 is basedontwo
previously known results for ds, each implying one of the two properties of theorem 1.2. The first of these
theorems,theorem3.3,isatranslationintothestandardtermsofx,y variablesofatheoremduetoJ.Ecalle
[E]. Because this result is couched in Ecalle’s own original language, we give not only the reference to the
precise statement, but also an appendix giving the complete calculation-translation which brings it to the
form of theorem 3.3. The second, theorem 3.4, appeared as Theorem 1 of [CS], with a complete elementary
proof which was also based on an idea of Ecalle.
§2. Characterizing special derivations
The main theorem of this section characterizes special derivations D of Lie[x,y]. From now on, if
F,G
f is an element of Lie[x,y], we say that f is special if setting F = f(z,y) with z = −x−y, there exists
a G ∈ Lie[x,y] such that D is special. By additivity, we may restrict ourselves to homogeneous Lie
F,G
elements.
Notation. For any f ∈Qhx,yi, we will use the notation
f =f x+f y =xfx+yfy.
x y
Observe that since every Lie element is palindromic, if f ∈Lie [x,y], we have
n
f =(−1)n−1anti(f)=f x+f y =(−1)n−1xanti(f )+(−1)n−1yanti(f )=xfx+yfy,
x y x y
so in fact
fx =(−1)n−1anti(f ), fy =(−1)n−1anti(f ). (2.1)
x y
3
Recall also the definition of the map s:Qhx,yi→Qhx,yi from (1.4). We will also use the similar map
(−1)i
s′(h)= xiy∂i(h). (2.2)
i! x
Xi≥0
When f ∈ Lie [x,y] (n ≥ 2), it follows by symmetry from Racinet’s result f = s(f ) that if we write
n y
f =xfx+yfy, then f =s′(fy).
Theorem 2.1. Let n ≥ 3, and let f ∈ Lie [x,y]. Set F = f(−x−y,y), and write f = f x+f y and
n x y
F =F x+F y. Then the following are equivalent:
x y
i) f is special, i.e. there exists a unique G∈Lie [x,y] such that [y,F]+[x,G]=0.
n
ii) Setting G=s′(F ), the derivation D is special.
x F,G
iii) F is antipalindromic.
y
iv) F is push-invariant.
v) f −f is antipalindromic.
y x
The equivalence of i), ii) and iii) is given in proposition 2.3. the equivalence of iii) and iv) is proven in
proposition 2.4, and the equivalence of iii) and v) is given following proposition 2.6. Some of these results,
in particular propositions 2.2 and 2.6, will also be used in the proofs of the main theorems in §3.
Proposition2.2. Letn≥3,andletf ∈Lie [x,y]havethepropertythatexpandedasapolynomial, f hasno
n
terms that start and end in y, so that writing f =f x+f y, we have f y =xPy. Then s(P)∈Lie [x,y]
x y y n−1
and f =[x,s(P)].
Proof. By hypothesis, f has no terms starting and ending in y, so we can write f y = xPy. By Racinet’s
y
result, we have g = s(g ) for all g ∈ Lie [x,y] with n ≥ 2, so in particular we have f = s(xP). Now, since
y n
the partial derivative satisfies ∂i(xP) = i∂i−1(P)+x∂i(P), and ∂n(P) = 0 since P is of degree n−1, we
compute
n (−1)i
f =s(xP)= ∂i(xP)yxi
i!
Xi=0
n (−1)i
= (i∂i−1(P)+x∂i(P))yxi
i!
Xi=0
n (−1)i n (−1)i
= ∂i−1(P)yxi+ (x∂i(P))yxi+1
(i−1)! i!
Xi=0 Xi=0
n−1(−1)i−1 n−1(−1)i
= ∂i(P)yxi+ (x∂i(P))yxi
i! i!
Xi=0 Xi=0
=−s(P)x+xs(P).
Thus, f =[x,s(P)].
It remains only to show that s(P) is a Lie element. Let Φ : Q hx,yi → Lie[x,y] be the linear map
n≥1
sending a non-trivial word w = x x x ···x to [x ,[x ,[x ,···]]], where x ∈ {x,y}, and let θ : Qhx,yi →
1 2 3 m 1 2 3 i
End Lie[x,y] be the algebra homomorphism mapping x to ad(x) and y to ad(y). By [B, Ch 2, §3, no. 2]
Q
the following properties hold:
• a polynomial h∈Q hx,yi is Lie if and only if Φ(h)=nh;
n
• Φ(uv)=θ(u)Φ(v) for u∈Qhx,yi and v ∈Q hx,yi.
n≥1
• θ(u)(v)=[u,v] if u is Lie.
Since f ∈Lie[x,y], we have
[f,x]=θ(f)(x)=θ([x,s(P)])(x) = ad(x),θ s(P) (x)= x,θ s(P) (x) =− θ s(P) (x),x .
(cid:2) (cid:0) (cid:1)(cid:3) (cid:2) (cid:0) (cid:1) (cid:3) (cid:2) (cid:0) (cid:1) (cid:3)
4
Thus, f +θ s(P) (x),x = 0, so since both f and θ s(P) (x) are Lie elements of degree > 1, we have
f =−θ(cid:2) s(P)(cid:0)(x). T(cid:1)hus, (cid:3) (cid:0) (cid:1)
(cid:0) (cid:1)
nf =Φ(f)=Φ([x,s(P)])=θ(x)Φ s(P) −θ s(P) Φ(x)= x,Φ s(P) −θ s(P) (x)= x,Φ s(P) +f.
(cid:0) (cid:1) (cid:0) (cid:1) (cid:2) (cid:0) (cid:1)(cid:3) (cid:0) (cid:1) (cid:2) (cid:0) (cid:1)(cid:3)
Thus x,Φ s(P) =(n−1)f =(n−1)[x,s(P)], so x,Φ s(P) −(n−1)s(P) =0. Since s(P) is of degree
n−1(cid:2)>1,(cid:0)we mu(cid:1)s(cid:3)t have Φ s(P) =(n−1)s(P), bu(cid:2)t this(cid:0)mean(cid:1)s that s(P)∈L(cid:3)ie [x,y]. ♦
n−1
(cid:0) (cid:1)
Proposition 2.3. Let f ∈ Lie [x,y], and set F = f(z,y) = F x+F y and G = s′(F ). Then D is
n≥3 x y x F,G
special if and only if f is special, and this is the case if and only if F is antipalindromic.
y
Proof. If setting G=s′(F ), the derivationD is special, then f is specialby definition. Conversely,if f
x F,G
isspecial,thereexistsaunique G∈Lie [x,y]suchthat[y,F]+[x,G]=0. SettingH =yF−Fy =Gx−xG
n
and writing F =F x+F y =xFx+yFy and G=G x+G y =xGx+yGy, this means that
x y x y
H =yF y+yF x−yFyy−xFxy =xGxx+yGyx−xG x−xG y, (2.3)
y x x y
so comparing the terms starting with x and ending with y, we find that −xFxy =−xG y, so Fx =G . By
y y
a result of Racinet [R], since G is a Lie element, we must have G = s(G ) = s(Fx) = s′(F ). This proves
y x
the first equivalence.
Let us now assume that F is antipalindromic, i.e. by (2.1), Fy =F . Set
y y
H =yF −Fy =y(F y+F x)−(yFy+xFx)y =yF y−yFyy+yF x−xFxy. (2.4)
y x y x
ThisshowsthatH hasnowordsstartingandendinginy,sobyproposition2.2,thereexistsG∈Lie [x,y]
n−1
such that H =Gx−xG. But then the derivation D is special, so f is special.
F,G
Finally, assume that f is special, and set H = yF −Fy, so that there exists G with H = yF −Fy =
Gx−xG. Then (2.3) holds. The expression H = Gx−xG shows that H can have no terms starting and
ending in y, and the left-hand expression for H in (2.3) then shows that we must have F = Fy, i.e. by
y
(2.1), F is antipalindromic. ♦
y
Proposition 2.4. Let F ∈Lie hx,yi. Then F is antipalindromic if and only if F is push-invariant.
n y
Proof. As usual, we write F =F x+F y =xFx+yFy. Assume first that F is antipalindromic, i.e. that
x y y
F =Fy. Since F is a Lie polynomial, we have F =s(F )=s′(Fy)=s′(F ), i.e.
y y y
(−1)i (−1)i (−1)i
F = ∂i(F )yxi = xiy∂i(Fy)= xiy∂i(F ). (2.5)
i! x y i! x i! x y
Xi≥0 Xi≥0 Xi≥0
Using the second and fourth terms of (2.5), we compute the coefficient of a word in F as
(−1)ar
(F|xa0y···xar−1yxar)= (∂ar(F )yxar|xa0y···yxar−1yxar)
(a )! x y
r
(−1)ar
= (∂ar(F )|xa0y···yxar−1)
(a )! x y
r
(−1)ar
= (xary∂ar(F )|xaryxa0y···yxar−1)
(a )! x y
r
=(F|xaryxa0y···yxar−1),
so F is push-invariant.
In the other direction, suppose that F is push-invariant, and let’s show that F =Fy. By assumption,
y
we have
(F|xa0y···yxar)=(F|xaryxa0y···xar−1).
5
In particular, for all words with ar =0, we have (F|xa0y···yxar−1y)=(F|yxa0y···yxar−1), i.e.
(F y|xa0y···yxar−1y)=(yFy|yxa0y···yxar−1),
y
so
(F |xa0y···yxar−1)=(Fy|xa0y···yxar−1).
y
Thus F =Fy. ♦
y
Lemma 2.5. Let g ∈ Q hx,yi, let φ(x,y) and ψ(x,y) be linear expressions of the form ax+by, a,b ∈ Q,
n
and let h(x,y)=g φ(x,y),ψ(x,y) . If g is antipalindromic, then h is antipalindromic.
(cid:0) (cid:1)
Proof. The operator anti is an anti-automorphismof the ring Qhx,yi, so
anti(h)=anti g anti(φ),anti(ψ) .
(cid:16) (cid:0) (cid:1)(cid:17)
But anti fixes linear expressions ax+by, so since g is antipalindromic, we have
anti(h)=anti g φ,ψ =anti(g)(φ,ψ)=(−1)n−1g(φ,ψ)=(−1)n−1h.
(cid:16) (cid:0) (cid:1)(cid:17)
Thus h is antipalindromic. ♦
Proposition 2.6. For any g ∈ Lie [x,y], set z = −x−y and G = g(z,y). Write g = g x+g y and
n x y
G=G x+G y. Then
x y
G −G =g (z,y).
y x y
In particular, g is antipalindromic if and only if G −G is antipalindromic.
y y x
Proof. We have g(x,y)=g (x,y)x+g (x,y)y, so
x y
G=g(z,y)=g (z,y)z+g (z,y)y =−g (z,y)x−g (z,y)y+g (z,y)y.
x y x x y
Thus G =−g (z,y)+g (z,y) and G =−g (z,y), so G −G =g (z,y). Then by Lemma 2.5, since g is
y x y x x y x y y
antipalindromic, so is G −G , and the converse holds as well since (G −G )(z,y)=g . ♦
y x y x y
We cannow conclude the proofoftheorem2.1 by showing the equivalence of iii) andv). To do this, we
simply apply proposition2.6 with f =G and g =F, to see that F is antipalindromic if and only if f −f
y y x
is antipalindromic. This completes the proof.
§3. Proofs of theorems 1.2 and 1.1
Proof of theorem 1.2. Let F ∈ V . We may assume that F is homogeneous of degree n ≥ 3, i.e.
kv
F ∈Lie [x,y] with n≥ 3. Set G= s′(F )=s(Fx). By theorem 2.1, F is antipalindromic if and only if F
n x y
is push-invariant, and these conditions are equivalent to the fact that G∈Lie [x,y] and D is special.
n F,G
NowconsiderthemapF 7→D fromV tothevectorspaceofspecialderivations,andletusshowthat
F,G kv
itisinjective. SupposethatF,F′ ∈Vkv andDF,G =DF′,G′. ThenDF,G(y)=DF′,G′(y),i.e. [y,F]=[y,F′],
so F −F′ commutes with y. Since F −F′ is of degree >1, this means that F −F′ =0.
LetusnowshowthatD satisfiesthetraceformula(1.3). Notethatby(2.1),F −F =(−1)n−1anti(Fy−
F,G y x
Fx), so by symmetry, F −F is push-constant if and only if Fy−Fx is push-constant. It is convenient to
y x
use the latter condition.
Since any Lie polynomial of degree >1 is a sum of terms of the form fg−gf, Lie polynomials map to
zero in TR. Thus, we have tr(G x)=−tr(G y), so
x y
tr(F y+G x)=tr(F y−G y)
y x y y
=tr(F y−Fxy) since G =Fx
y y
(3.1)
=tr(Fyy−Fxy) since Fy =F
y
=tr (Fy−Fx)y .
(cid:0) (cid:1)
6
Rephrasing the trace formula (1.3) via (3.1) as
tr((Fy −Fx)y)=Atr (x+y)n−xn−yn , (3.2)
(cid:0) (cid:1)
we can now show that a special derivation D satisfies the trace formula in TR if and only if Fy −Fx is
F,G
push-constant. In fact, these are just two ways of making the identical statement. To see this, let C denote
the list of words in the cyclic permutation class of w, so that C contains exactly n words; then C consists
of n/|C| copies of C. For any word v =uy ending in y, let C denote the associated cyclic permutation list,
and C the list obtained from C by removing all the words ending in x. Write C = [u y,...,u y]. Then
y y 1 r
by definition, we have the equality of lists
[u ,...,u ]=Push(u). (3.3)
1 r
Now, the trace condition tr((Fy −Fx)y)=Atr (x+y)n−xn−yn means firstly that (Fy −Fx)y|yn =
0, which is equivalent to Fy −F |yn−1) = 0,(cid:0)and secondly that(cid:1)for each equivalenc(cid:0)e class C of cy(cid:1)clic
x
permutations of a given wo(cid:0)rd w 6=yn, we have
tr (Fy −Fx)y C := (Fy−Fx)y v =|C|A,
(cid:16) (cid:0) (cid:1)(cid:12) (cid:17) vX∈C(cid:16) (cid:12) (cid:17)
(cid:12) (cid:12)
where the first equality is just the definition of the coefficient of an equivalence class in a trace polynomial.
Using the notationC andC asaboveand(3.3),this means thatfor everywordv 6=yn ending iny,writing
y
v =uy, we have
|C| |C| |C|
|C|A= (Fy −Fx)y v = (Fy −Fx)y v = (Fy −Fx) u′ . (3.4)
n n n
vX∈C(cid:16) (cid:12)(cid:12) (cid:17) vX∈Cy(cid:0) (cid:12)(cid:12) (cid:1) u′∈PXush(u)(cid:0) (cid:12)(cid:12) (cid:1)
But this is equivalent to
(Fy−Fx) u′ =nA (3.5)
u′∈PXush(u)(cid:0) (cid:12) (cid:1)
(cid:12)
for all u′ 6= yn−1, which, together with the fact that (Fy −Fx|yn−1) = 0, is precisely equivalent to the
statement that Fy −Fx is push-constant (for the constant nA).
So far we have proven that F 7→D for homogeneous F extends to an injective map V ֒→krv .
F,s(Fx) kv 2
Let us show that it is an isomorphism, i.e. also surjective. It is enough to consider derivations D ∈krv
F,G 2
with F, G homogeneous of degree n. Then D is special, so F is antipalindromic by theorem 2.1, and
F,G y
D satisfies the trace formula (3.2), which as we just saw is equivalent to the property that Fy −Fx is
F,G
push-constant. Finally, since F −F =(−1)n−1anti(Fy−Fx), we see that F −F is also push-constant,
y x y x
so F ∈V , completing the proof. ♦
kv
Let us now provetheorem 1.1. The proof is based onthe fact that two previously knowncombinatorial
results about double shuffle elements f˜∈ds make it possible to deduce that F =f˜(x,−y) satisfies the two
defining propertiesof V givenin theorem1.2. Thus f˜7→F yields aninjection ds֒→V , and the injection
kv kv
V ֒→krv of theorem 1.2 completes the argument.
kv 2
The two known results are given in theorems 3.3 and 3.4. As the original statement of theorem 3.3 is
extremely different in appearance(theorem A.1 below), the translationfrom the originalterminology to the
statementgivenhereis providedinthe appendix, whichalsoservesasaninitiationtoEcalle’slanguage. We
writeds forthehomogeneousweightnpartofds,consistingofpolynomialsindswhichareofhomogeneous
n
degree n.
Theorem 3.3. [E, cf. Appendix] Let f˜∈ds , and write f˜=f˜x+f˜y. Then f˜ +f˜ is antipalindromic.
n x y x y
Theorem 3.4. [CS] Let f˜= f˜x+f˜y ∈ ds , and set A = (f˜|xn−1y). Then f˜ satisfies the property that
x y n y
(f˜|yn−1)=0 and for each degree n monomial w 6=yn−1 containing r y’s, we have
y
f˜|v =(−1)rA.
y
v∈PXush(w)(cid:0) (cid:1)
7
Proof of theorem 1.1. Let n ≥ 3 and assume that f˜ ∈ ds , i.e. f˜ is a homogeneous Lie polynomial
n
of degree n. Set f(x,y) = f˜(x,−y). It follows directly from theorem 3.4 that f is push-constant. Let
y
us deduce from theorem 3.3 that f −f is antipalindromic. Indeed, f(x,y) = f (x,y)x+f (x,y)y and
y x x y
f˜(x,y)=f(x,−y), so f˜(x,y)=f (x,−y)x−f (x,−y)y, i.e. f˜ =f (x,−y), f˜ =−f (x,−y). Thus
x y x x y y
f˜ +f˜ =f (x,−y)−f (x,−y)=(f −f )(x,−y). (3.6)
x y x y x y
The left-hand side is antipalindromic by theorem 3.3,so the right-handside is antipalindromic,and then by
Lemma 2.5 f −f and thus also f −f are antipalindromic.
x y y x
Set F = f(z,y). We will use the two properties on f to show that f 7→ F is an injection from ds into
V . By proposition 2.6 with g = F and G = f, we see that f −f antipalindromic implies that F is
kv y x y
antipalindromic. Itremainsonlytoshowthatf push-constantimpliesthatF −F ispush-constant,which
y y x
is a little more delicate. We prove it in the following lemma.
Lemma 3.5. For any f ∈Lie [x,y], set F =f(z,y) and write f =f x+f y and F =F x+F y. Suppose
n x y x y
that f is push-constant for a constant A, and that A=0 if n is even. Then F −F is also push-constant
y y x
for A.
Proof. To show that F −F is push-constant for A, let us first show that (F −F |yn−1) = 0. As we
y x y x
saw in the proof of theorem 1.2, the condition that f is push-constant is equivalent to the condition that
y
tr(f y)=Atr (x+y)n−xn−yn . By proposition 2.6 with g =f and G=F, we have F −F =f (z,y),
y y x y
so (Fy −Fx)(z(cid:0),y)=fy. Multiplyin(cid:1)g by y on the right of both sides and taking the trace yields
tr (F −F )(z,y)y =tr(f y)=Atr (x+y)n−xn−yn .
y x y
(cid:16) (cid:17) (cid:16) (cid:17)
Making the variable change x7→z =−x−y on both sides, this gives
tr (F −F )y =Atr (−1)nxn−(−1)n(x+y)n−yn .
y x
(cid:16) (cid:17) (cid:16) (cid:17)
Whenn isodd, the right-handside doesnotcontainthe equivalenceclassofyn,sothe left-handside cannot
contain it either, which means that (F −F |yn−1) = 0. When n is even, A = 0 by assumption, so the
y x
equivalence class of yn−1 cannot appear in the left-hand side, which again means that (F −F |yn−1)=0.
y x
Now let us prove that F −F is push-constant. Write
y x
fy = caxa0y···yxar = cvv,
Xa Xv
where a runs over the tuples a = (a ,...,a ) with r ≥ 1 and a +···+a = n−r−1, and v runs over
0 r 0 r
degree n−1 words. If v =xa0y···yxar, we write cv =ca. For a given tuple a=(a0,...,ar), let
Push(a)=[(a ,...,a ),(a ,a ,...,a ),...,(a ,...,a ,a )]
0 r r 0 r−1 1 r 0
be the list of its r+1 cyclic permutations. The fact that f is push-constant means that for all w 6= yn−1,
y
we have
(fy|v)= cv = ca′ =A. (3.7)
v∈PXush(w) v∈PXush(w) a′∈PXush(a)
Let us now compute the coefficient in Fy − Fx of a given word w = xb0y···yxbd, w 6= yn−1. By
proposition 2.6, we have
Fy −Fx =fy(z,y)= caza0y···yzar = (−1)n−r−1ca(x+y)a0y···y(x+y)ar, (3.8)
Xa Xa
8
so
Fy −Fx w =(−1)n−1 (−1)rca(x+y)a0y···y(x+y)ar w . (3.9)
(cid:0) (cid:12)(cid:12) (cid:1) (cid:18)Xa (cid:12)(cid:12) (cid:19)
Clearly if r >d then the expansion of (x+y)a0y···y(x+y)ar cannot contain the(cid:12)word w, so (3.9) is equal
to
Fy −Fx w =(−1)n−1 (−1)rca (x+y)a0y···y(x+y)ar xb0y···yxbd . (3.10)
(cid:0) (cid:12) (cid:1) a s.tX. 0≤r≤d (cid:18) (cid:12)(cid:12) (cid:19)
(cid:12) (cid:12)
The only terms (x+y)a0y···(x+y)ar in which w will appear with a positive coefficient (necessarily equal
to 1) are the 2d terms (x+y)a0y···(x+y)ar constructed as follows: choose any of the 2d subsets of the y’s
in w, and change the y’s in that subset to x’s; then substitute x7→(x+y) in the resulting word.
Let w = xb0y···yxbd be a monomial, and set b = (b0,...,bd). Write Xb for the set of 2d sequences
(a0,...,ar), 0≤r ≤d, such that the corresponding word xa0y···yxar is obtained from w by changing any
subset of y’s into x’s. Then the coefficient (3.10) is equal to
Fy −Fx w =(−1)n−1 (−1)rca. (3.11)
(cid:0) (cid:12) (cid:1) aX∈Xb
(cid:12)
By (3.11), we have
Fy−Fx|v =(−1)n−1 (−1)rca. (3.12)
v∈PXush(w)(cid:0) (cid:1) c∈PXush(b)aX∈Xc
Let us write
Xb = Xc
c∈Paush(b)
for the disjoint union, i.e. the list-union of the words in the lists Xc, where c runs through the cyclic
permutations of b. There are (d+1)2d words in Xb. Let us count the words in Xb of each given depth
0≤r ≤d.
For each tuple c ∈ Push(b), let wc be the word associated to c. The list Xb is exactly the list of all
words obtained by changing k of the d y’s in wc to x’s, for all 0 ≤ k ≤ d and all c ∈ Push(b). Thus, Xb
contains (d+1) words of depth d, which are the words wc for c ∈ Push(b), and for each smaller depth
r =d−k for 1≤k ≤d, Xb contains the words obtained by changing k y’s to x’s in each of the d+1 words
(all of depth d) of Push(b). Thus, there are exactly (d+1) d words of depth r =d−k in Xb, and these
k
words fall into exactly (cid:0) (cid:1)
d+1 d d+1
=
d−k+1 k k
(cid:16) (cid:17) (cid:16) (cid:17)
cycles of length r+1=d−k+1, of words of depth r =d−k.
Since fy is push-constant, the coefficients ca of each of the d+k1 cycles of depth r = d−k in fy add
up to A. Thus, for all b6=(1,...,1), (3.12) is given by (cid:0) (cid:1)
d d
d+1 d+1
(−1)n−1 (−1)rca =(−1)n−1 (−1)d−kA=(−1)n (−1)d+1−kA=(−1)n−1A.
k k
aX∈Xb Xk=0(cid:16) (cid:17) kX=0(cid:16) (cid:17)
This proves that F −F is push-constant for the value (−1)n−1A. ♦
y x
We cannowconclude the proofoftheorem1.1. Using the well-knownresultonds (cf. [E],[R], [IKZ]...)
that the coefficient of xn−1y is zero for all even-degree elements of ds, we see that when n is even, A=0 in
theorem 3.4, so if f˜∈ds, then f =f˜(x,−y) satisfies the hypotheses of Lemma 3.5.
Thus, we have shown so far that if f˜∈ ds, setting f(x,y)= f˜(x,−y) and F =f(z,y), F is antipalin-
y
dromic by the argument of of the first paragraphof the proof of theorem 1.1, and F −F is push-constant
y x
byLemma3.5. Thus,themapf˜7→F isaninjectivemapfromds→V . By(1.8),wethenhaveaninjective
kv
composition of maps
ds֒→V →∼ krv .
kv 2
This concludes the proof of theorem 1.1. ♦
9
§4. The prounipotent version
LetV be a gradedvectorspace,andletun(V)denotethe Lie algebraof(pro)unipotentendomorphisms
of V, i.e. linear endomorphisms D such that D(V )⊂V . The usual exponentiation
≥n ≥n+1
1
exp(D)= Dn (4.1)
n!
nX≥0
maps un(V) bijectively to the group UN(V) of (pro)unipotent linear automorphisms of V.
ρ
Suppose we now have a Lie algebra g equipped with an injective Lie algebra map g ֒→ un(V). The
universal enveloping algebra Ug is a ring whose multiplication we denote by ⊙. The exponential associated
to g is given by the formula
1
exp⊙(f)= f⊙n; (4.2)
n!
nX≥0
it maps g bijectively to the associated group G⊂Ug, and the following diagram commutes:
c
G //UN(V) (4.3)
OO OO
exp⊙ exp
ρ
g //un(V).
NowletV denotetheunderlyingvectorspaceofLie[x,y]. Followingthenotationof[AT],lettder denote
2
theLiealgebraoftangentialderivations ofLie[x,y],i.e. derivationsD havingthepropertythatD(x)=[x,a]
and D(y) = [y,b] for elements a,b ∈ Lie[x,y]. There is an injective map of Lie algebras tder ֒→ un(V).
2
Indeed, if V = Lie[x,y] is equipped with the grading given by the degree, then any derivation D ∈ tder
2
increases the degree, i.e. D(V ) ⊂ V . Let TAut denote the group of automorphisms of V obtained
≥n ≥n+1 2
by exponentiating tder :
2
exp:tder →TAut ⊂UN(V)
2 2
D 7→exp(D)= 1 Dn. (4.4)
n!
nX≥0
Let sder denote the subalgebraof tder consisting of derivations D such that D(x+y)=0, andSAut
2 2 2
the corresponding subgroup of TAut consisting of automorphisms such that A(x+y) = x+y, so that
2
exp(sder ) = SAut . According to [AT], the exponential map (4.1) not only restricts to (4.4), but also to
2 2
maps from the following subspaces to subgroups:
KRV // SAut // TAut //UN(V) (4.5)
2 2 2
OO OO OO OO
exp exp exp exp
krv //sder //tder // un(V)
2 2 2
wheretheupperleft-handgroup,KRV ,istheprounipotentgroupactuallydefinedastheimageinSAut of
2 2
krv ⊂sder undertheexponentialmap,althoughtheauthorsthengoontoalsoprovideadirectdescription
2 2
of KRV [AT, §5.1].
2
Let us now recall the definition of the prounipotent group version DS of the double shuffle Lie algebra
ds originally given by Racinet in [R, Chap. 4, §1]; this is the group that Racinet denotes DM (k), but we
0
take the base field k=Q; note that he also writes dm (k) for ds.
0
For any monomials u,v ∈Qhx,yi, let the shuffle product sh(u,v)∈Qhx,yi be defined recursively by
sh(1,u)=sh(u,1)=u, sh(Xu,Yv)=xsh(u,Yv)+ysh(Xu,v). (4.6)
10