The Deligne-Drinfeld conjecture

Proof. For \psi(x,y)=\alpha x+\beta y, the left side minus the right side of the pentagon is -\alpha t_{12}-\beta t_{34}. The degree-one braid generators are linearly independent, so \alpha=\beta=0. The degree-two free Lie algebra is spanned by [x,y], and the three-term expression for this polynomial is 3[x,y]. Thus it too has no nonzero rational solution.□

Proof. We recall the integral free-Lie and PBW facts being used. The Hall–Lyndon construction gives a basis of the free Lie algebra over \Z: iterated use of alternation and Jacobi expresses bracket monomials in the standard Lyndon brackets, and the associative expansion of each such bracket has its Lyndon word as leading word, with coefficient one. Distinct leading words give independence over every coefficient ring. Thus base change to R identifies the abstract free Lie algebra with precisely the displayed bracket span. PBW applies to this free R-module; in particular its universal enveloping algebra is R\langle X\rangle. These are the usual integral versions of the free-Lie basis and PBW theorems [15, Theorems 0.2, 0.5, 4.9 and 5.1]; the integral Lyndon and PBW bases are also recalled in [12, Section 2].

The same Lyndon brackets form bases after base change to \Q and \F, with unchanged multidegrees. This proves the asserted injectivity and equality of dimensions. For a weighted alphabet, restriction to any bounded weight leaves finitely many letters, so the same basis argument applies.□

Proof. Any prescription on free generators extends uniquely to a derivation. Derivations D_{ij} and D_{hk} with disjoint supports commute. For three distinct indices i,j,h, put S_{ijh}=s_i+s_j+s_h. On the free subalgebra on these three letters,

Moreover D_{ij}(S_{ijh})=0, so D_{ij} commutes with this sum. On every other free generator all these derivations vanish. It follows that [D_{ij},D_{ih}+D_{jh}]=0 on the whole free Lie algebra. Hence (2.2) defines an action of the presented base algebra.

Form its semidirect product with the free Lie algebra on the s_i. The braid relations involving the last strand say precisely

They hold by (2.2). Conversely, these relations in the presented braid algebra give exactly that action. The maps between the presented algebra and the semidirect product, taking each generator to its namesake, are consequently inverse. This proves both the presentation and the assertion about the actual kernel.

Inductively the base is a free k-module. Choose Lie bases for fiber and base, and place the entire fiber basis before the base basis. PBW identifies their ordered products with a basis of the enveloping algebra. Since the enveloping algebra of a free Lie algebra is the free associative algebra, this is exactly (2.3). Induction, starting with one strand, gives the asserted word basis. Lemma 2.2 and the same basis over R, \Q, and \F give the last assertions.□

Proof. All calculations in this proof take place over \Q. We first rewrite the pentagon in a free kernel, then linearize at a+b+c=0. Equality of mixed derivatives will force the possible linearization coefficient to vanish.

The free-fiber identity. Forget strand 2 in \mathfrak t_4(\Q), and put

By Lemma 2.3, the kernel is the actual free Lie algebra on a,b,c. Define

Forgetting the fiber sends each displayed difference to zero, so F_a,F_b,F_c are elements of this free kernel. They are homogeneous of degree n.

Each of the three expressions before subtracting \psi(u,v) commutes with its corresponding letter. For example, [a,u+b]=[a,v]=0 and [b,u+a]=[b,v+c]=0; the third case is the same braid relation at c. The base action fixes S=a+b+c. Summing the resulting commutators therefore gives

Set \Delta_a=F_b-F_a and \Delta_c=F_b-F_c. We obtain the identity in the free fiber

The pentagon gives \Delta_a=\psi(a,b+c)-\psi(a,b)-\psi(b,v). The element v+b+c is central in the subalgebra on v,b,c. A Lie polynomial of degree greater than one is unchanged when a central element is added to an argument: every term in the difference contains that element in a bracket. Thus \psi(b,v)=\psi(b,-b-c). Exchanging strands 1 and 4 in the same calculation and using antisymmetry gives the second formula below:

Both differences vanish when S=0: the two remaining terms cancel by antisymmetry and \psi(z,-z)=0 in degree greater than one.

The ideal generated by S is stable under the base action, so the quotient fiber is \Lie_{\Q}\langle a,b\rangle, with c=-a-b. In this quotient the actions of u,v are \ad_c,\ad_a, respectively. For example,

and [v,a]=0, [v,b]=[b,c]=[a,b], [v,c]=[c,b]=[a,c]. It follows that \psi(u,v) acts on the quotient fiber by \ad_{\psi(c,a)}. Since the expression defining F_b+\psi(u,v) commutes with b, we have

in this free Lie quotient.

The centralizer of b in the ordinary free Lie algebra on a,b is \Q b. To see this directly, embed it in associative words. For a homogeneous polynomial P=\sum_w p_w w commuting with b, comparison in bP=Pb gives p_{av}=0 and p_{b^j av}=p_{b^{j-1}avb} for every word v and j>0. Iterating shows that every word containing a has coefficient zero. Hence P is a polynomial in b. Setting a=0 now shows that a Lie element of this form belongs to the one-dimensional ordinary Lie algebra on b. The homogeneous element at hand has degree n>1, and is therefore zero. We have proved an equality of quotient fiber elements, not just of their induced derivations:

The first variation at S=0. We have determined the quotient value of F_b. We now differentiate (2.6) transversely to this quotient and compare the result with the derivative of the expression in (2.4). All subsequent derivatives mean coefficient extraction after adjoining free direction letters and commuting scalar parameters. Thus

The degree-n Lie polynomials on a,T containing exactly one T and n-1 copies of a form the one-dimensional space spanned by \ad_a^{n-1}T: in a nonzero bracket monomial, each branch not containing T must be a single a. Consequently there is a scalar k\in\Q, depending only on \psi, such that

Differentiating the identity \psi(a,-a)=0 while varying both arguments gives

Take c=S-a-b, keep a,b fixed, and extract the coefficient linear in S=tT. In these same coordinates the two differences are

Writing c=-a-b after differentiation, their derivatives at zero are exactly

The last equality uses (2.11) and antisymmetry. In differentiating (2.6), the left side becomes [T,F_b|_{S=0}]: its other product-rule term has outside letter S=0. On the right the outside a is fixed, and the derivative of the outside c contributes [T,\Delta_c|_{S=0}]=0. Thus no outside-letter term remains, and (2.9) gives

Now define G(a,b)=[a,\psi(b,a)]+[c,\psi(b,c)], where c=-a-b. Hold b fixed and vary a in direction T, so that c varies in direction -T. The full product rule is

The three-term identity and antisymmetry give \psi(b,a)-\psi(b,c)=\psi(c,a). Comparing with (2.12), we conclude that

Vanishing of the scalar obstruction. Equation (2.13) expresses the derivative of G in terms of the single scalar k. To show that this scalar is zero, introduce two independent constant directions U,V. The coefficients of their two commuting scalar parameters in G are unchanged when the differentiations are interchanged. After forming these mixed derivatives, specialize

where a_0,T are free letters. The derivative of \ad_c^n vanishes there, since every summand contains n-1>0 copies of \ad_c. The two derivatives of the other term are

Only j=n-1 survives in the second sum. Equality of the mixed derivatives in (2.13) implies

The word a_0^nT has coefficient one in \ad_{a_0}^{n}T. Since the calculation is over \Q, it follows that k=0. This proves (2.5).

Equation (2.13) now says that every directional derivative of G in a vanishes. Equivalently, setting the direction equal to a multiplies its component with j occurrences of a by j. In characteristic zero every such component with j>0 is zero. Thus G is independent of a, and evaluation at a=0,c=-b gives G=0. This is (2.4).□

Proof. An intersection of a rational subspace with a finite free R-module is saturated: if v\in K_n and 2v\in L_n, then v\in W_n, hence v\in L_n. Thus L_n\cap2K_n=2L_n, which gives the asserted injection after reduction. It is a finite free module because R is a principal ideal domain. Clearing denominators in a rational basis of W_n shows that \Q L_n=W_n, so its rank and its reduction dimension are \dim_{\Q}W_n. Exactly the same argument applies to M_n; embedding both reductions in K_n/2K_n proves their stated inclusion.

For a representative \psi\in L_n, each defining identity is an expression with coefficients in R. Its target is either a free Lie lattice or the braid Lie algebra over R. These targets embed in their rationalizations by Lemmas 2.2 and 2.3. An expression which vanishes over \Q therefore already vanishes over R, and can be reduced modulo two. The identical argument applies to the special identity and its linearization, which were established over \Q in Lemma 2.4. Adjoining free letters and commuting parameters preserves the word bases; extracting a parameter coefficient involves no division and commutes with reduction. This justifies the assertion about formal coefficient consequences.

Finally, reduce (2.4), take a=x,b=y,c=x+y, and use the reduced antisymmetry to replace \overline\psi(y,x) and \overline\psi(y,x+y) by \overline\psi(x,y) and \overline\psi(x+y,y). Expanding the outside bracket with x+y gives (2.15).□

Proof. Order words of each fixed length lexicographically with x>y. The leading words of x^2,[x,y],y are respectively xx,xy,y, each with coefficient one. These form a prefix code: none is a proper initial segment of another. Thus their concatenations have unique decodings, so distinct words in A,C,B have distinct leading words. Split a putative relation by its total x,y length and take the largest leading word in each part; its coefficient proves associative injectivity. Ordinary free Lie algebras embed in their free associative enveloping algebras, so the Lie assertion follows as well.

We use the ordinary free-Lie elimination construction [15, Theorem 0.6], whose needed instance can be seen directly. Set e_j=\ad_x^j y for j\geq0. The Lie subalgebra generated by the e_j is stable under \ad_x, because \ad_x(e_j)=e_{j+1} and \ad_x is a derivation. Its sum with the line \F x is therefore a Lie algebra containing x,y, hence the whole free Lie algebra. The subalgebra is thus an ideal with one-dimensional quotient spanned by the class of x. Every homogeneous Lie polynomial of weight greater than one consequently lies in this subalgebra. It is itself free on the e_j: their leading words x^j y form a prefix code, which proves injectivity of the corresponding free associative algebra and hence of its ordinary free Lie algebra.

In characteristic two, the associative identity \ad_x^2(P)=[x^2,P] gives

Both expressions belong to the ordinary Lie algebra on the three abstract letters A,C,B, which proves the final assertion.□

Proof. For the source, the braid relations give the following derivations on its free fiber on X,U_1,\ldots,U_{d-1}:

The derivation rule and [X,V_i]=[X^2,U_i]=R_i imply

Since \ad_{a_i}=\delta_i^2, applying \delta_i again gives (3.6). For its incident V_i entry, in particular, R_i+Q_i=[V_i,U_i] gives

Applying \epsilon_{ij} to H,V_k gives (3.7), since [X,[U_i,U_j]]=W_{ij}.

Next \ad_{v_{ij}}=[\delta_i,\epsilon_{ij}] sends X to W_{ij}, sends U_i to [U_i,V_j], sends U_j to [U_j,V_i], and kills the other U_k. Its action on H is [X,W_{ij}]=[H,[U_i,U_j]]. Its remaining actions follow uniformly from

which proves (3.8), including the nonincident entries.

For the target, the flat where z_i=z_d=0 lies in exactly the four hyperplanes with generators a_i,H,U_i^{+1},U_i^{-1}. Its relations give

Summing the second formula over t proves the V_i entry of (3.9). The signed triple flat containing u_{ij}^s,U_i^t,U_j^{st} gives

Substitute U_k^{-1}=U_k+V_k. For example,

Summing also over s gives [v_{ij},U_i]=[U_i,V_j] and [v_{ij},V_i]=[V_i,V_j]; exchanging i,j gives the other incident entries. Every remaining pair in the claimed zero entries meets in a flat contained in just those two hyperplanes, so its generators commute. This proves all target formulas.

The count assertions can now be read term by term. In the source, the terms beyond the common quadratic terms add two U’s. In the target, the only extra term is [V_i,V_j] in (3.10), whose count is zero instead of one. This also proves agreement of the parts preserving count.□

Proof of Proposition 3.2. Relabel the four braid strands 0,1,2,3. Since signs disappear over \F, the reduced pentagon says that the sum of

is zero in the source. Express these five terms using \Psi. Their exact triples of arguments (A,C,B) are, in the same order,

In particular,

Only the former square has a u_{12}^2 summand.

Expand the pentagon as a polynomial in the restricted source symbols before ordering. Its part of count exactly r is precisely the four-term polynomial P on the right of (3.5). To verify this, the higher components of \Psi already have more than r occurrences of B, each of which becomes a single u. Either extra summand u_{12}^2 or [u_{12},u_{23}] in the second triple increases count by two. The third triple, being entirely in the u alphabet, gives count n>r. All remaining replacements preserve the original B-count. Thus the full source pentagon has the form P+E=0, where every word of E has count greater than r.

The distinction between original weight and symbolic length is essential here. Assign source weight two to each a,v,H,V and weight one to each u,U. A word of length \ell and count q has original weight

All source action formulas preserve this weight. Thus a correction that raises count by two raises length by one, and every ordered word of weight n and count r has length (n+r)/2=m+r. In particular P has that length, while the higher-count expressions are allowed to have greater length.

Apply the deterministic source ordering to P+E=0. Since count never decreases, E contributes nothing at count r. Independence of the ordered restricted source products, proved above from the free fibers, implies that every ordered coefficient of count r in P is zero separately. This conclusion uses independence across all the restricted words, so expressions of other lengths or counts cannot cancel a coefficient under consideration.

Now interpret only the polynomial P, of length m+r, in the target. Apply the same deterministic ordering, discarding a branch as soon as its count falls below r. Such a branch can never return to count r. The retained branches use the identical quadratic formulas that retained count r in the source. Their ordered coefficient vector is therefore the zero vector just obtained. Every target replacement preserves length, so every discarded branch has length m+r and count less than r. Their sum belongs to \mathcal U_{m+r,<r}, proving (3.5).

Only source independence enters this argument. In the target the ordering procedure is a sequence of valid equalities yielding a spanning expression; no independence or free-fiber assertion for \mathfrak h is required.□

Proof. For distinct letters put \omega_{PQ}=d\log(P-Q), and put \omega_{PP}=0. Signs of differences have no effect on these forms. The nonconstant differences, up to sign, factor as

The remaining nonzero difference 1-0 is constant. Consequently every \omega_{PQ} is an integer linear combination of d\log\ell_H, where \ell_H is a linear defining equation of H.

Consider the matrix-valued rational one-form

The coefficient of a deletion at P_j is exactly \omega_{P_jP_{j+1}}-\omega_{P_{j-1}P_j}. We claim that \Omega\wedge\Omega=0. For two nonadjacent deleted positions, neither deletion changes the neighbors of the other, so the contributions of the two orders cancel by antisymmetry of the wedge product. For two adjacent positions, denote the four successive letters by L,P,Q,R. The sum of the contributions, with the rightmost matrix acting first, is

Expanding this expression cancels the cross terms involving \omega_{LP} and \omega_{QR}. The remaining terms are the two logarithmic triangle identities for (L,P,Q) and (P,Q,R). The triangle identity is

For three distinct letters it follows by differentiating the relation (I-J)+(J-K)+(K-I)=0 and putting the terms over a common denominator. If exactly two letters agree, one form is zero and the other two are equal; if all three agree, all three forms are zero. Thus it also holds in every repeated-letter case. This proves that (4.5) vanishes, including when other pairs among L,P,Q,R agree. It proves the claimed matrix identity.

Take its residue first along a fixed H_0. Terms involving no d\log\ell_{H_0} have no pole at a generic point of H_0, and the result is the rational one-form identity on H_0

Now take the residue on H_0 along a codimension-one subspace X\subset H_0 which is a flat of the arrangement. For H\ne H_0, the restricted nonzero linear form \ell_H|_{H_0} has order one along X precisely when H\supset X, and has order zero otherwise. This gives (4.3); including H_0 in the sum adds a zero commutator.

This residue calculation is performed over \Q. Distinct hyperplanes may restrict to multiples of the same equation on H_0; each matrix then contributes separately with coefficient one. In particular, restrictions proportional to z, -z, or 2z all have residue one. There is no division by the number of coincident restrictions, nor by the constant 2. Finally every matrix in (4.3) has integer entries. Its vanishing over \Q is therefore its vanishing over \Z, which permits reduction modulo two without changing the characteristic-zero incidence relations.□

Proof. If a difference has orders \lambda_- and \lambda_+ along z_i-z_j=0 and z_i+z_j=0, its evaluations on u_{ij} and v_{ij} are respectively \lambda_- and \lambda_-+\lambda_+. Applying this to (4.4) gives every entry of (4.6). For example p has orders 1,-1, so its v_{23} weight is zero already integrally; the factor 2 in p-w contributes no hyperplane valuation.

Every a- or v-weight on an edge touching 0 is zero. Thus an a- or v-operation cannot delete a zero, regardless of preceding deletions, because all remaining letters and endpoints still belong to \{0,e,p,w\}. Each u-operation removes at most one zero. Emptying r zeros consequently requires at least r such operations. If there are exactly r, using even one to remove a nonzero letter makes it impossible to remove the remaining zeros. The same conclusion holds for linear combinations within the indicated slots.□

Proof. The associative multihomogeneous components of an ordinary Lie polynomial are again ordinary Lie polynomials. Since the component with no A vanishes, suppose that h>0 is the least A-count of a nonzero component \chi_h. We will rule out h<m.

Fix any internal word s with h letters e, m-h letters p, and r zeros. In each summand of (3.5), an A-, C-, or B-slot means the corresponding first, second, or third argument of \chi. Every B-slot must delete a zero by Lemma 4.2. The other relevant actions follow directly from the edge table; the four rows are in the order of that equation:

Here “terminal” means the last remaining internal letter, adjacent to w. Each indicated boundary contributes weight one, so two such boundaries contribute zero. For instance, the third A-slot is a_1+a_2+v_{12}: its weights on ep,ew,pw are respectively 0,1+1,1+1, and its weights on edges touching zero vanish. Its deletion operator is identically zero. In the second row the C-slot is v_{23} and the B-slot is u_{23}.

A nonzero contribution from an A-count-t component must use each A-slot to delete an e, so t\leq h. Minimality gives t\geq h. Hence only \chi_h can contribute, and all h letters e must be deleted by its A-slots. In particular, the possible C-deletions of e in the table cannot occur in a surviving contribution. This also excludes contributions from every larger A-count.

In the fourth summand every deletion is therefore at the right endpoint, and a slot word has at most one complete deletion sequence. It succeeds precisely when it is the word s' obtained by e\mapsto A, p\mapsto C, 0\mapsto B, in which case its weight is one. Thus the fourth summand evaluates to the coefficient [s']\chi_h. The first summand cannot delete any p, so it vanishes because m-h>0. The third summand vanishes because h>0 and its A-operator is zero.

Consider the second summand. Suppose an e is followed by a nonempty maximal block consisting of p’s and zeros, ending at the next e or at w. Its left e cannot disappear while any of this block remains: all e’s must be removed by A-slots, which remove only a terminal e. If the right boundary is an e and that e disappears before the block does, it must already be terminal, so the new right boundary is w. Thus a surviving nonempty part of the block always has boundaries e,e or e,w. The final surviving letter of this block cannot be removed. If that letter is p, its C-weight is 1+1=0; if it is 0, its B-weight is 0+0=0. Consequently the second summand has no complete deletion sequence whenever some e occurs before a non-e. Figure 1 records the two possible boundaries and the last-letter obstruction.

The scalar leading pentagon now gives [s']\chi_h=0 for every such s. The only remaining words have the form v e^h, with v a word of p’s and zeros. Because m-h>0, the word v is nonempty. Let \operatorname{rev} denote reversal of associative words. For every ordinary homogeneous Lie polynomial q of associative length d,

Indeed reversal sends [q_1,q_2] to -[\operatorname{rev}(q_1),\operatorname{rev}(q_2)], proving the formula by induction on Lie monomials. In characteristic two it fixes \chi_h. The coefficient of v e^h therefore equals the coefficient of e^h\operatorname{rev}(v), which has already been shown to vanish. Every coefficient of \chi_h is zero, a contradiction.

Thus a least nonzero A-count can only be m. All A-counts are at most m, so only the component using A,B remains. It is an ordinary Lie polynomial as asserted.□

Proof. The map is well defined because changing a representative by an element of F^{r+1} does not change its count-r component. The parity assertion and the multidegree of its image follow from \wt(A)=\wt(C)=2 and \wt(B)=1. If m=0, the leading part is an ordinary Lie polynomial in B alone and is zero in the weights at issue. We may therefore assume m>0.

Suppose that \chi(0,C,B)=0. By Lemma 4.3, \chi=P(A,B). If r=0, this is an ordinary Lie polynomial in A alone, and hence zero except possibly in weight two. Thus a nonzero kernel would require r\geq1.

Apply the reduction of Lemma 2.4 and antisymmetry with z=x+y. They give

Indeed the special identity with a=x, b=y, c=z becomes [x,\bar\psi(y,x)]+[z,\bar\psi(y,z)]=0, and antisymmetry in characteristic two interchanges each pair of arguments.

Under x\mapsto x+y, the three associative elements become

Moreover, on their free associative algebra,

The part of \ad_x that lowers B-count by one is consequently the derivation

its other part raises B-count by one. The substitution A\mapsto A+C+B^2 never lowers that count. Therefore components of \bar\psi with count greater than r cannot contribute to count r-1 in (4.8), and its outside bracket with y has count at least r+1. Within the leading part, using any B^2 in the substitution also raises the count by two. The count-r-1 component of (4.8) is exactly

Write c(v) for the coefficient of a word v with m letters A and r letters B in P. Fix any target word with m-1 letters A, r-1 letters B, and two letters C. Its coefficient on the left of (4.9) is the sum of exactly two coefficients of P: fill the two marked C-positions with (A,B) or with (B,A). In each case the original A is replaced by C in the substitution and the original B by C under D; these are the only possibilities. The term D(P(A,B)) has only one C and contributes nothing here. Thus the coefficients of P are invariant under exchanging any chosen A-position and B-position. These exchanges connect all words with the specified multiplicities, so

for one \lambda\in\F.

If m\geq2, fix a target word with m-2 letters A, r-1 letters B, and three letters C. Exactly one of its three C-positions must have been the B differentiated by D; the other two must have been A’s replaced in the substitution. There are exactly three choices. Its coefficient is therefore 3\lambda=\lambda, which (4.9) forces to vanish. Hence P=0.

It remains to consider m=1. The polynomial P has original (x,y)-bidegree (2,r). An A,C,B word with counts (a,c,b) has original bidegree (2a+c,c+b). Solving for bidegree (2,r) gives only

The latter possibility is excluded by the assumption \bar\psi\in F^r, and the former is precisely P. Thus if P\ne0, the entire bidegree-(2,r) component of \bar\psi is nonzero. Symmetry under x\leftrightarrow y forces a nonzero component of bidegree (r,2). Its B-count is at most its y-count, namely two, so F^r forces r\leq2.

For r=2, the ordinary free Lie component with one A and two B’s is spanned by

To see the spanning assertion directly, every three-letter bracketing with two identical B’s either has the inner bracket [B,B]=0, or is [B,[A,B]] up to sign and skew-symmetry. The coefficients on ABB,BAB,BBA are therefore (\mu,0,\mu), which are all equal as required by (4.10) only when \mu=0.

For r=1, the forced swapped bidegree is (1,2). The count equations 2a+c=1, c+b=2 have the unique nonnegative solution (a,c,b)=(0,1,1). Thus this nonzero component is a multiple of [C,B], lies in count one, and survives the projection A=0. No higher-filtration component has this bidegree, so it cannot cancel this projection. This contradicts \chi(0,C,B)=0. Both exceptional cases are excluded, proving injectivity.□

Proof. By the elimination argument in Lemma 3.1, every homogeneous Lie polynomial in x,y of weight greater than one is a Lie polynomial in U_j=\ad_x^j y, j\geq0. In characteristic two these elimination letters are

Associative words in the E_i,O_i are linearly independent. Indeed, order the letters with A largest and use degree followed by lexicographic order. The leading words of these letters are A^iB,A^iC, respectively, with coefficient one. They form a prefix code: one reads a string of A’s up to its next B or C, thereby determining each successive codeword. Distinct products consequently have distinct leading words.

Each E_i is homogeneous of B-count one, whereas each O_i has B-count zero. Thus the expansion of \psi in independent associative words in this alphabet is graded by the number of E-factors. The hypothesis \psi\in F^r forces the coefficients of all words with fewer than r such factors to vanish separately. This establishes a termwise restriction before any substitution or quotient is taken.

Select the component of original bidegree (m+r,m). Every U_j has exactly one y, so every product in this component has exactly m factors. The reduced antisymmetry equation gives \psi(x,y)=\psi(y,x). Swapping x and y therefore gives the desired expression (5.1) for bidegree (m,m+r), with the same lower bound on the number of even indices. If r>m, no such product exists.

For completeness, an A,C,B-word with respective counts (a,c,b) has original bidegree (2a+c,c+b). In bidegree (m,m+r) this means

The terms with a>0 have higher B-count and vanish under A=0; the terms with a=0 have c=m,b=r and are precisely f. Thus the even-index bound applies to the whole bidegree component, including every term subsequently removed by the projection.□

Proof. Let R be any commutative \F-algebra, and work in S=R\langle C,B\rangle. Define an R-linear derivation D by

In characteristic two the square of a derivation is again a derivation: the two middle terms in the twice-applied product rule cancel. Since D^2 vanishes on B,C, it vanishes on all of S. Consequently the Ore extension, followed by the indicated quotient,

has underlying left S-module S\oplus SX. Indeed X^2q=qX^2+D^2q=qX^2, so X^2 is central in the Ore extension and its quotient introduces no relation in S. In particular, S embeds in \mathcal E. The substitutions x\mapsto X, y\mapsto B give

Equivalently, \mathcal E acts on S with X acting by D and elements of S acting by left multiplication. For an element written q+q'X, its value on 1 is its constant term q.

Fix s\in R[C], and put F=B+s(C). The substitution B\mapsto F, C\mapsto C commutes with D, since D(s)=0. It therefore extends to an automorphism of \mathcal E fixing X. Let h(x,y) be the whole bidegree-(m,m+r) component in Lemma 5.1. Evaluation at x=X,y=B kills the terms containing A in its A,C,B expression, so that h(X,B)=f(C,B) in the embedded subalgebra S of \mathcal E. Applying the shift automorphism just constructed gives h(X,F)=f(C,F). We may therefore evaluate the even-index product expansion of h term by term in \mathcal E. Its sum belongs to S, so its value on 1 is exactly f(C,F). The factors of positive index become elements of S:

whereas the index-zero factor is X. We shall bound the B-degree of the constant term of each product separately.

Write

Here L raises B-degree by one and Q preserves it. The relations needed below are

For the two smallest positive indices one has

For j\geq2, every nonzero operator word in (L+Q)^{j-1}C has rightmost operator L, because Q(C)=0. Thus P_j has no B-degree-zero component, and its degree-one component is

This is killed by D, since DLC=KC=[C,C]=0. As D lowers B-degree by one, DP_j has no degree-zero component for any j\geq1.

For j\geq3, the degree-two component is explicitly

Upon applying D, differentiating the rightmost L gives zero, and differentiating the other L gives Q^aKQ^bLC. The commutation of Q and K therefore gives

For even j this is zero in R. Together with (5.4), these calculations prove the following lower bounds; the zero polynomial satisfies every listed bound:

The bound for odd P_j is deliberately weak, so that it includes P_1=C. All these calculations allow constant terms in s and arbitrary scalar coefficients in R.

Now consider one product with q index-zero factors and e positive even-index factors. Move each X to the right using XP=PX+DP, and apply the resulting expression to 1. Every surviving term must use each of the original q copies of X as a differentiation, since a terminal X kills 1. Repeated use of the product rule assigns these differentiations to positive-index factors to their right. An allocation assigning two differentiations to the same factor vanishes, because D^2=0. Hence every surviving term is a product in which exactly q distinct positive-index factors have been differentiated once. An undifferentiated positive even-index factor contributes a lower bound of one, and an odd-index factor contributes a lower bound of zero. Differentiating either type increases the respective lower bound by one. The resulting B-degree is therefore at least

which is precisely the number of all even indices, including zero. If no such allocation is possible the constant term is zero, which also satisfies the bound. Cancellations among allocations cannot create lower-degree terms.

Lemma 5.1 now shows that f(C,B+s(C)) has no B-degree below r. On the other hand, f is homogeneous of B-degree r, and replacing B by B+s(C) cannot increase that degree. Its degree-r component is exactly f(C,B), obtained by choosing B at every occurrence. This proves (5.3) over R, and in particular over any polynomial ring in scalar indeterminates.□

Proof. First suppose r\geq1. Encode the associative words of B-degree r by the vector-space isomorphism

Put P=\mathcal P_r(f). For 1\leq i\leq r, let R_iP be its restriction obtained by identifying s_{i-1} and s_i, written in the same polynomial ring \F[t_0,\ldots,t_{r-1}] for every i:

Empty initial or final strings in this formula are omitted. In the coefficient of \lambda in f(C,B+\lambda C^j), replacing the ith B merges its two neighboring powers of C and adds j to their exponent. Its encoding with r-1 remaining B’s is consequently t_{i-1}^{j}R_iP. Lemma 5.2, applied with the independent scalar \lambda, gives

Taking j=0,\ldots,r-1 produces a Vandermonde matrix whose determinant is

It is a nonzero polynomial over \F, because the t_i are independent indeterminates. Thus the matrix is invertible over \F(t_0,\ldots,t_{r-1}), and all R_iP vanish. This argument also applies to even r; it never divides by r or specializes the t_i to elements of the two-element field. For r=1 the matrix is simply (1).

The kernel of the ith identification is the principal ideal generated by d_i=s_{i-1}-s_i. These d_i are distinct nonassociate prime linear polynomials. Their individual divisibility therefore implies

It remains to translate this associative statement into an ordinary Lie statement.

Temporarily include g_0=B. The ordinary Lie ideal generated by B in \Lie_{\F}\langle C,B\rangle is the subalgebra generated by g_k=\ad_C^kB, k\geq0. Indeed, this subalgebra contains B and is stable under \ad_C by the derivation rule and [C,g_k]=g_{k+1}; the reverse inclusion is immediate. Since f has positive B-degree, it belongs to this ordinary Lie subalgebra.

Expanding the iterated adjoints gives the useful product formula

The variables d_1,\ldots,d_r,s_r are an invertible linear change of coordinates from s_0,\ldots,s_r. In particular, the monomials in the right side of (5.8) are linearly independent. For different product lengths the B-degrees differ, so all associative words in the g_k, k\geq0, are independent. Their ordinary Lie algebra is therefore the ordinary free Lie algebra on these letters, by its embedding in the free associative algebra.

Write the associative expansion of f uniquely in these independent words. Its encoding lies in \F[d_1,\ldots,d_r]. By (5.7), every monomial appearing with nonzero coefficient has positive exponent of every d_i. Equivalently, the expansion of f contains no word with any factor g_0. The ordinary free Lie algebra on g_0,g_1,\ldots has a retraction

Its extension to the free associative algebra fixes the expansion of f. Hence \rho(f)=f, proving that f belongs to the ordinary free Lie algebra on g_k, k\geq1. This retraction preserves ordinary Lie membership throughout; no identification with all primitive elements in characteristic two is being made.

Each g_k contains one B and k copies of C. The original weight is therefore 2k+1, and the B-count of a Lie word is its length in these generators. This proves the asserted weight and length statement.

If r=0, then f is an ordinary Lie polynomial in C alone. The ordinary Lie algebra on one generator is its one-dimensional linear span, of original weight two; thus f=0 for n>2. Proposition 4.4 then gives \operatorname{gr}^0_F\overline L_n=0. The possible boundary m=0 also contributes nothing: its leading component would be an ordinary Lie polynomial in B alone, whose only possible positive weight is one.□

Proof. For n>2, combine Proposition 4.4 with Proposition 5.3. If n-r is odd or negative the source piece is zero by the weights of A,C,B, and the target piece is likewise zero since the weight of a length-r word in odd-weight letters has parity r. The cases r=0 and m=0 were disposed of in Proposition 5.3. Weights one and two have zero solution space by Lemma 2.1, and have no generators or Lie words in \mathfrak f_{\mathrm{odd}}. Finally, F^{n+1}\overline L_n=0, so this is a finite filtration in weight n and the dimensions of its associated graded pieces sum to \dim_{\F}\overline L_n. The equality of this dimension with \dim_{\Q}W_n is Proposition 2.5.□

Proof. We first prove the filtration assertion before applying A=0. The derivation \delta=\ad_x acts on the free A,C,B algebra by

Write \delta=\delta_-+\delta_+, where

These two derivations respectively lower and raise the B-count by one. In particular, \delta(F^r)\subset F^{r-1}. Directly from the definition of the Ihara derivation,

The last formula is the derivation identity applied to [x,B]. Thus D_\psi B\in F^{r+1} and D_\psi C\in F^r: the outside B in [B,\delta\psi] restores the one count that \delta_- can remove. Replacing a B in a word of count s adds at least r to its count, and replacing a C also adds at least r; the derivative of an A vanishes. It follows that

Interchanging \psi,\phi and including their ordinary bracket proves the asserted containment of the Ihara bracket.

Let \chi,\eta be the components of \psi,\phi of counts r,s. In (5.12), only \chi can contribute to the part of the derivation that raises count by exactly r. Its action is

Indeed, a higher-count component of \psi still has higher count after applying [B,\delta_-(-)], while [B,\delta_+\chi] raises the count by two more. This derivation preserves the ideal generated by A, so it descends under A=0. The projection commutes with \delta_-, whose induced action on \F\langle C,B\rangle is the derivation D from Lemma 5.2. The induced leading action is consequently

By Proposition 5.3, f is an ordinary Lie polynomial in g_k, k\geq1; each of these is killed by D, because

Thus the extra term [B,Df] vanishes, and the induced derivation on \F\langle C,B\rangle is q\mapsto[q,f]. One can also read Df=0 directly as the coefficient of \lambda in (5.3) with s(C)=\lambda C.

The two Ihara derivations therefore project to [g,f] and [f,g], respectively, and the ordinary bracket projects to [f,g]. Their specified combination is

over \F. All higher components of \psi or \phi have already been excluded by the filtration calculation, so this proves (5.11).□

Proof. Write \psi=\delta(\alpha). The coderivation rule at the group-like arrow \alpha says

Primitives in a characteristic-zero enveloping algebra are its Lie algebra: by the PBW filtration, a primitive of filtration degree d>1 would give a primitive homogeneous polynomial of degree d in a symmetric algebra, whereas its coproduct has a nonzero component of bidegree (1,d-1). Induction on filtration degree proves the assertion. Now

with T central: replacing t_{13} by T-x-y transforms exactly the three defining relations into the centrality of T. Thus a primitive of weight n\geq2 is a Lie polynomial \psi(x,y).

Reversing the two children at any fork is insertion, followed if necessary by an inner insertion, of a unit arrow in arity two. Its \delta-value is zero. The values of unit coefficient arrows add under composition, and inverse unit arrows have opposite values. Relabel \alpha by 1\leftrightarrow3. Its source ((32)1) is carried to b by fork reversals, and its target (3(21)) is carried to a in the same way. Its value is therefore both \psi(y,x) and -\psi(x,y), giving

For the cyclic relation, use the successive trees ((12)3),((23)1),((31)2). Reassociation on the ordered triples (1,2,3),(2,3,1),(3,1,2), each followed by a root-fork reversal, forms a cycle of unit arrows. Consequently

An occurrence of the central T in a Lie monomial of degree greater than one contributes zero. Substitution of t_{31}=T-x-y gives

Figure 2 shows the five four-leaf trees and the two paths of reassociation arrows:

The two edge values on the first path are \psi(t_{13}+t_{23},t_{34}) and \psi(t_{12},t_{23}+t_{24}). Those on the second are \psi(t_{12},t_{23}), \psi(t_{12}+t_{13},t_{24}+t_{34}), and \psi(t_{23},t_{34}). Both composites are 1_{AE}, hence

This proves membership in W_n using only the stated operations.

It remains to check the bracket, including its sign. Let d_a,d_b be the loop derivations at a,b. Inner insertion gives d_a(x)=0 and d_b(y)=0. Differentiating the equality of arrows \alpha y_a=y_b\alpha gives

Outer arity-two insertion also kills t_{13}+t_{23} at a, so d_a(T)=0. In the decomposition (6.2) its restriction is exactly D_\psi. An arrow from a to b with coefficient h equals \alpha h_a; therefore

If another derivation \varepsilon has value \phi, their commutator has value

Evaluation is thus a Lie homomorphism onto its image. No injectivity of evaluation is required.□

Proof. At each arity choose a base object o. For a composition derivation put h_p=\delta(1_{op}), with h_o=0, and let d be its derivation on loops at o. Since every arrow has the factorization 1_{oq}\,u_o\,1_{po}, the product and inverse rules force

Conversely, an algebra derivation d and elements h_p define a composition derivation by this formula. In weight n the data are

There are finitely many objects and generators, and every displayed space is finite dimensional over \Q.

The conditions that d preserve the defining braid relations are rational linear equations in these data. The coderivation conditions are equivalent to primitivity of all h_p and d(t_{ij}): necessity follows on the reference unit arrows and generator loops, and sufficiency follows by the product rule and (6.7). These are again finitely many rational linear equations, since coproducts in the indicated weights take values in finite sums of finite-dimensional spaces.

Finally, there are only finitely many label permutations, object graftings, and inner placements through arity four. For each associated functor \rho, the condition \delta\rho=\rho\delta need only be checked on the reference unit arrows and generator loops: all other arrows follow by composition, the product rule, and continuity. These checks, and the requirement of zero arity-two action, are rational linear equations. Thus \mathcal D_n is the kernel of a linear map between finite-dimensional rational spaces. Kernels and images of such maps commute with scalar extension. Evaluation and coefficient extraction are rational linear maps, proving the last assertion as well.□

Proof. Convergence at each endpoint. If d is the lowest common ancestor of i,j, formula (7.2) gives

where r_{ij} is a finite signed sum of powers \epsilon^{a_e-a_d} with positive exponents. Choose a positive lower bound \eta for all these exponent gaps. Then

where every coefficient of R is O(\epsilon^{\eta-1}). The sign \sigma_{ij} contributes no logarithmic differential.

Here and below estimates are at a fixed weight cutoff N, using any norm on the finite-dimensional quotient by weights greater than N. Changing from g to E_p^{-1}g removes the singular term exactly, because the K_d commute. The remaining form is E_p^{-1}R E_p\,d\epsilon. In weight at most N, conjugation adds only polynomial powers of \log\epsilon, so its norm is bounded by

Its integral between 0 and \epsilon is O_N(\epsilon^\eta(1+|\log\epsilon|)^N). The finite iterated-integral formula, or the differential equation in the truncated algebra, therefore proves convergence and gives the same bound for the difference between collar transport in these frames and the identity.

The collar bound proves the limit in (7.3); concatenation, homotopy invariance and group-likeness pass to the limit. Different choices of middle representatives of a continued homotopy class give the same answer by flatness.

Independence of endpoint scales. To check exponent independence, interpolate two systems of exponents linearly with a parameter u\in[0,1]. The inequalities between parent and child exponents persist, and their finitely many gaps have a uniform positive lower bound \eta. On this interpolation at fixed \epsilon,

Again the singular part is precisely E^{-1}dE before changing frame. The remaining form on the interpolation has integral norm at most C_N\epsilon^\eta(1+|\log\epsilon|)^{N+1} after changing frame. Its transport tends to the identity. Thus identifying endpoints by these real interpolation paths does not change (7.3).

The two insertion identities. We have constructed transport on \mathcal P(I), independently of the endpoint exponents. It remains to verify the insertion operations required of a functor to the chord categories in Section 6. We first compare the connection forms for each kind of insertion, then check the operations on path classes. On a truncated outer path of the above form, there are constants c,K>0 such that all distinct outer coordinates are separated by at least c\epsilon^K. Their total variations are bounded independently of \epsilon, and the integral norms of their logarithmic differences are O(1+|\log\epsilon|). These statements follow from (7.4) on collars and compactness on the middle path.

Insert fixed configurations at offsets of size O(\epsilon^M), with their own smaller nested scales, and choose M>2K, also larger than all outer scale exponents. Write D=z_i-z_j for an outer difference and e for the constant difference of the two inserted offsets. Then

Internal differences are constant along the outer path. Replacing all cross-cluster forms by the outer forms gives precisely the connection obtained from (7.1) by t_{ij}\mapsto t_{B_iB_j}. A difference of iterated integrals of length at most N is expanded by changing one factor at a time. The error bound (7.6) and the logarithmic bounds for the other factors give an error bounded by a positive power of \epsilon times a power of 1+|\log\epsilon|. Multiplication by the two endpoint frames preserves convergence to zero: their coefficients of weight at most N are polynomials in \log\epsilon. A single bound sufficient for all these errors is

The full endpoint frame factors exactly as the outer frame under the block substitution, times the internal frames of the fixed inserted trees. The latter frames are the same at both endpoints, and every inner chord commutes with every outer block sum, as checked in Section 6. They therefore cancel from normalized transport. This proves compatibility with outer insertion.

For inner insertion, keep the outer configuration fixed and insert a varying inner path \xi scaled uniformly by \epsilon^M. Choose M>K, where c\epsilon^K bounds the distance between its anchor and outside points from below. Inner coordinate variations are bounded as above. Each cross-cluster differential now has integral norm at most C\epsilon^{M-K}; the inner logarithmic differentials are unchanged by uniform scaling. The same iterated-integral argument gives (7.7). There is one additional frame factor. If B is the inner label set, it is

Indeed scaling multiplies every inner scale by \epsilon^M. The total chord T_B is central in \mathfrak t_B, by summing its three-label relations. All outer frame factors commute with inner chords as well. These common factors at the two endpoints cancel, leaving the unscaled inner transport embedded with its original labels. The argument includes insertion into a single leaf. For sufficiently small \epsilon all the cabled paths remain in the configuration space and vary continuously with \epsilon; after collar identification they consequently define a single continued homotopy class. A homotopy of representatives has a compact middle part, so the inserted scales can be chosen sufficiently small uniformly on it; applying the same insertion throughout gives a homotopy of the cabled paths. Increasing the scale exponent also gives a homotopy through collision-free paths. Thus the path operations are well defined on the continued classes and compatible with concatenation, and the limiting identities are identities for these operations.

Coherence of the path operations. For truncated outer and inner paths, with parameters s,t\in[0,1] and the uniform bounds above, use the homotopy square

If |z_i(s)-z_j(s)|\ge c\epsilon^K and |w_{ia}(t)|\le C, then distinct clusters remain separated by c\epsilon^K-2C\epsilon^M>c\epsilon^K/2 for sufficiently small \epsilon and M>2K; within a cluster, differences are \epsilon^M(w_{ia}(t)-w_{ib}(t))\ne0. Thus the two boundary routes give the same path class. Nested insertions agree after flattening, since z+\epsilon^M(w+\epsilon^Nv)=z+\epsilon^Mw+\epsilon^{M+N}v. The exponent-interpolation estimate identifies other admissible scale choices with these, with normalized transport tending to the identity. The commuting frame factors above therefore identify these operations exactly. For a one-leaf outer tree, restoring root scale one uses a positive real dilation by \rho; its transport is \exp(\lambda\log\rho\,T_B) and cancels from normalized transport at the two endpoints by centrality of T_B.

The second transport rule and arity two. For H_-, apply the same construction to \bar\gamma with t_{ij} replaced by -t_{ij} and endpoint frame E_p^{-1}. All estimates and cancellation identities just proved remain valid; relabeling is immediate for both rules. In arity two the root scale is 1, and the difference of the two endpoint coordinates is 1 or -1. Every path has logarithmic-difference integral k\pi i for some integer k, of the parity determined by the endpoint orders. Its first holonomy is \exp(kt_{12}/2). Conjugating the path negates k, and negating the chord negates it once again, so the second holonomy is the same. This checks both full turns and half-turns.□

Proof. Fix an object, let P=\pi_1(X_I) and let I_P be the augmentation ideal in \CC[P]. Pair-collision meridians normally generate P. To see this directly, fill a loop by a disk in \CC^I and perturb its interior to meet the collision hyperplanes transversely in finitely many smooth points, avoiding their complex-codimension-two intersections. Removing small disks about these points expresses the loop as a product of conjugated positive or negative meridians. Meridians around the same hyperplane are conjugate, since its smooth part outside the other hyperplanes is path connected. It follows that their classes generate I_P/I_P^2. They are also independent: the winding numbers of the functions z_i-z_j take value 1 on the corresponding positive meridian and 0 on all the others.

Every augmentation associated graded algebra is generated by its degree-one part, since I_P^r/I_P^{r+1} is spanned by products of r classes from I_P/I_P^2. These meridian classes satisfy the infinitesimal braid relations. Near a disjoint pair of collisions the two local meridians commute. Near a triple collision, consider the orbit loop rotating the small three-point cluster simultaneously through 2\pi. It commutes with every local meridian: a circle action carries any such loop through a homotopy whose two boundary composites are the two orders of these loops. Its degree-one class is the sum of the three pair-meridian classes, because each of the three differences winds once and all other differences wind zero times. The degree-two commutator therefore says [t_{ij},t_{ik}+t_{jk}]=0 (the commutator with t_{ij} itself is zero). Transporting these local loops to the basepoint only changes them by conjugation, which changes their classes by terms in I_P^2 and does not affect their degree-two commutators. Hence there is a surjective graded algebra map

Residues show that H_+ of a positive pair meridian is 1+t_{ij}+\text{terms of weight at least two}: the integral of the corresponding logarithmic form is 2\pi i. For H_- both the turn and the chord change sign, giving the same result. Endpoint normalization conjugates loop transport by a series with constant term 1 and hence does not change its degree-one term. Thus both holonomies are filtered and each induces a graded map

Since the chords generate the enveloping algebra, gf=\id. Surjectivity of f now implies that f and g are inverse isomorphisms. In particular both associated graded holonomies are the same inverse of (7.9).

An isomorphism on associated graded spaces lifts inductively to an isomorphism on every finite filtration quotient: lift the constant term, then correct the error successively in each weight; injectivity follows by the first nonzero weight. Passing to inverse limits proves the completed assertion on loop algebras. A reference path identifies each other arrow space with a loop-algebra torsor, so the same assertion holds on the whole category. Coproduct preservation follows from group-likeness, and preservation of insertions and relabeling follows from Proposition 7.1, first on paths and then on their linear completions by continuity.□

Proof. Use the objects a,b and unit arrow \alpha from Section 6. Let \beta be the real order-preserving path of reassociation from a to b, and put

We identify \Phi,U with arrow coefficients. Since an arrow with coefficient h is \alpha h_a, one has exactly

The path \beta is real, so its conjugate is itself and H_-(\beta)=\Phi(-t_{ij}). Hence

Source x and target y are fixed by the arity-two insertion operations. Applying S to \alpha y_a=y_b\alpha therefore yields

The source total chord T is fixed as well.

The linear term of \Phi is zero. Along a real order-preserving path the sign of each difference is constant, so its integral is the logarithm of the ratio of endpoint absolute differences. Normalizing by the frames subtracts the logarithms of their lowest-common-ancestor scales. By (7.4), each remaining ratio tends to 1. This proves the claim for every chord separately. As \theta preserves filtration, equation (7.12) implies that U too has zero linear term. Both series are group-like. Their logarithms are primitive and have weight at least two, so (6.2) places these logarithms in the completed free Lie algebra on x,y. Every such Lie monomial uses both letters, since the free Lie algebra on one letter has only weight one. Consequently \Phi,U depend only on x,y and

where J is the closed two-sided ideal of series having at least one y in \CC\langle\!\langle x,y\rangle\!\rangle.

Conjugation in (7.13) changes y only by terms in J^2, because U-1\in J. Substitution therefore gives

on the full series algebra (it fixes the pure-x terms). Reducing (7.14) modulo J^2 gives

This is a statement about the image of the unit arrow. To pass to its infinitesimal value, put N=S_{ab}-\id on this arrow space. By (7.11), for h\in J,

More generally N(J^r)\subseteq J^{r+1} for r\geq1. Since N(1)=U-1\in J, all N^j(1) with j\geq2 belong to J^2. The operator logarithm consequently gives

It remains to compute a single convergent coefficient of \Phi. Set the central T to zero and use s=(z_2-z_1)/(z_3-z_1). Writing the three differences as a common factor times s,s-1,1 shows that the common logarithmic differential has coefficient T, while the remaining connection is

The reassociation path runs from s=0 to s=1 with positive real tangents. With the convention dg=\Omega g, the coefficient of x^{n-1}y has y at the earliest integration time. Thus for n\geq2 it is

Here [w] denotes the coefficient of the associative word w. To justify convergence and the last equality, expand (1-t)^{-1}=\sum_{q\geq0}t^q and apply monotone convergence to the nonnegative integrand; substitution t=e^{-u} gives the individual integrals (n-1)!/(q+1)^n. The series converges for n\geq2. Regularization does not change this coefficient. Its frame factors are pure-y series on the left and pure-x series on the right; a nonconstant factor would make a word begin with y or end with x, whereas x^{n-1}y does neither.

Negating both arguments multiplies a weight-n word by (-1)^n. Equations (7.16) and (7.18) therefore give, for odd n\geq3,

This only uses positivity of the real series, with no arithmetic independence assertion. The functional [x^{n-1}y] is rational on the finite data of Lemma 6.2. Its nonvanishing on \delta_n\in\mathcal D_n(\CC) implies nonvanishing on some \mathcal D_n(\Q), and evaluation gives the required \psi_n\in V_n. Finally, Lie elimination, or direct induction on bracketed words with one y, shows that the weight-n, depth-one free Lie space is the line spanned by \ad_x^{n-1}y; its x^{n-1}y coefficient is 1.□

Proof. An application of D_\psi replaces one occurrence of y by [y,\psi], while it kills x. Each nonzero resulting word therefore has at least p more occurrences of y than the original word. Thus D_\psi(\phi)\in\mathcal D^{p+q}L. The same reasoning applies to D_\phi(\psi), and the ordinary bracket also adds depth. Equation (1.6) proves the assertion. Its consequence follows by induction on the length of the Lie word.□

Proof. In a nonzero rational relation, multiply by a rational scalar so that every coefficient belongs to \Z_{(2)} and at least one is a unit. Reduction modulo two gives a contradiction.

Now let v=\sum_i c_iw_i\in K, with c_i\in\Q. If some c_i\notin\Z_{(2)}, let e\ge1 be the smallest integer for which every 2^ec_i belongs to \Z_{(2)}. At least one of these coefficients is a unit, whereas

Reduction gives the same contradiction. Thus all the c_i belong to \Z_{(2)}, proving the equality.□

Proof of Theorem 1.1. Send e_{2k+1} to the elements \sigma_{2k+1} just constructed. Since the target V is a Lie algebra, the universal property gives a graded Lie homomorphism

In each weight, its Hall-basis images have independent reductions by the preceding leading-term argument. They are therefore rationally independent, so the map is injective. Their number in weight n is d_n, and (8.3) shows that they span both V_n and W_n. Thus V=W and the map is surjective. This also proves closure of W under the stated Ihara bracket without having to assume closure during the upper-bound argument.

Each homogeneous piece is finite-dimensional. The isomorphism therefore extends componentwise to the products of the weight pieces. The bracket in a fixed weight involves only finitely many pairs of positive weights, so it extends to these products and the two mutually inverse maps are continuous for the weight filtration. This proves the completed assertion.□