La décomposition en poids des algèbres de Hopf

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1 La décomposition en poids des algèbres de Hopf Frédéric Patras Ann. Inst. Fourier, Grenoble 43, , pp published version Errata and remarks by Darij Grinberg The following are remarks I have made while reading the above-cited paper by Frédéric Patras. I think it is an interesting and rather readable text despite some minor typos and tersely written proofs. Some of the below remarks are just quick corrections of minor mistakes at least as far as I can tell; I can neither guarantee that these mistakes really are mistakes, nor that my corrections are correct!. Some others are detailed expositions of certain proofs which have been only vaguely sketched in Patras s paper. Finally, some others give alternative proofs for results in Patras s paper sometimes inserting additional results into Patras s paper, to be used as lemmata later on. Different remarks are separated by horizontal lines, like this: Page 1068: I think aux endomorphismes ψ k should be aux endomorphismes Ψ k here. Page 1069, Definition 1.1: There is nothing wrong here, but I think it would be helpful to notice that what Patras calls algèbre de Hopf is not the same as what modern-day algebraists call a Hopf algebra. What Patras calls algèbre de Hopf is a kind of super-version of a graded bialgebra not necessarily having an antipode!; in constrast, what modern-day algebraists call a Hopf algebra is just a bialgebra with antipode. Nevertheless, I am going to use the words Hopf algebra for what Patras calls algèbre de Hopf in the following. Page 1070, fifth line of this page: Here, Patras write: Une bigèbre graduée ou une algèbre de Hopf est connexe si H 0 K. This definition is good when K is a field, but in the general case when K is a commutative ring, it is not a reasonable definition of connected. Since Patras, in his paper, always works over a field K, this is not a problem for him, but I still prefer the following in my opinion, better definition of connected : A graded bialgebra or Hopf algebra is connected if and only if the map ɛ H0 : H 0 K is an isomorphism. Note that, when K is a field, this definition is equivalent to Patras s definition, 1

2 because we have the following equivalence of assertions: H 0 K as K-vector spaces dim H 0 1 where dim V denotes the dimension of any K-vector space V dim Ker ɛ H0 0 since we know that the map ɛ H0 : H 0 K is surjective because ɛ H0 1 ɛ 1 1 by the axioms of a bialgebra, since H is a bialgebra, and thus by the isomorphism theorem K H 0 Ker ɛ H0, so that dim K dim H 0 Ker ɛ H0 dim H 0 dim Ker ɛ H0, so that dim H 0 } dim{{ K} + dim Ker ɛ H0 1 + dim Ker ɛ H0, 1 and therefore the equation dim H 0 1 is equivalent to dim Ker ɛ H0 0 Ker ɛ H0 0 ɛ H0 is injective ɛ H0 is bijective since we know that the map ɛ H0 : H 0 K is surjective because ɛ H0 1 ɛ 1 1 by the axioms of a bialgebra, since H is a bialgebra, and thus this map ɛ H0 is injective if and only if it is bijective ɛ H0 is an isomorphism. Page 1070, two lines above Definition 1.2: Here, Patras writes: [...] l ensemble L H des endomorphismes linéaires de H [...]. I don t think that L H denotes the set of all linear endomorphisms of H throughout the text. It seems to me that L H indeed denotes the set of all linear endomorphisms of H when H is just a bialgebra not graded; but when H is a graded bialgebra or an algèbre de Hopf I would translate this by Hopf algebra, but as I said, this does not mean what people nowadays mean by a Hopf algebra, L H denotes the set of all graded 1 linear endomorphisms of H. Note that I might be wrong about this, and L H might indeed mean the set of all linear endomorphisms of H throughout the text. But in this case, the homomorphism ρ n defined on page 1074 Notons ρ n l homomorphisme de restriction de L H dans L H n. is not a simple restriction homomorphism i. e., it is not just given by ρ n f f n for every f L H, but instead requires a more subtle definition: H i i0 It must then be defined by ρ n f n p i f p i i0 for all f L H where p i : H H denotes the map which sends every element of H to its i-th graded component seen again as an element of H. Page 1071, proof of Proposition 1.4: Here, Patras writes: La deuxième partie de la proposition se ramène à établir l égalité : [k] Π [l] Π [l] k [k] l, 1 A linear map f : V W between graded vector spaces V and W is said to be graded or compatible with the grading if, for every n N, it satisfies f V n W n. 2,

3 qui est une conséquence à peu près immédiate des axiomes de structure des bigèbres commutatives. This is not totally precise. The identity [k] Π [l] Π [l] k [k] l is true in any bialgebra, not only in commutative ones it follows from the axioms of a bialgebra by a double induction over k and l. But deriving the deuxième partie de la proposition from this identity requires the bialgebra to be commutative. Here are the details of this derivation: We have [k] l [l] [lk] this holds for any coalgebra, and can be proven by induction using the coassociativity and counity axioms of a coalgebra 2 and Π [k] Π [l] k Π[lk] this holds for any commutative algebra, and is easy to see - but doesn t generally hold for noncommutative algebras!, so that Ψ k Ψ l I k Π [k] I k [k] Π [k] [k] I l Π [l] I l [l] Π [l] [l] Π [k] } [k] {{ Π [l] } [l] Π [k] Π [l] k Π [l] k [k] l Π [lk] I lk [lk] I lk Ψ lk, Π [lk] and this proves the second part of Proposition 1.4. [k] l [l] Π [lk] [lk] [lk] Page 1072: A typo: Notons Φ k the n-ième endomorphisme should be Notons Φ k the k-ième endomorphisme. Page 1073, proof of Proposition 2.3: There is nothing wrong to be corrected here, but I don t find the proof of this proposition as obvious as Patras does, so let me write down this proof here: Proof of Proposition 2.3. We can prove that any commuting x M and y M satisfy log k x + log k y log k xy this follows from the well-known fact that log 1 + X+log 1 + Y log 1 + X 1 + Y in the ring K [[X, Y ]] of formal power series, using the fact that the K-representation A is unipotent of rank k. Using this fact, we can prove by induction over n that every x M and n N satisfy n log k x log k x n. But for every x M, we have ρ Φ n x ρ Φ n x ρ x n x n 2 This proof can be found in [P3] Lemma II.8 of [P3], to be precise. 3

4 and k 1 k 1 n i ε i x n i i0 i0 k 1 i0 k 1 ε i x n i log k x i i! log k x i i0 i! n log k x i i! exp k n log k x log k x n by the definition of exp k n log k x exp k log k x n exp k log k x n ρ x n. ρ by Lemma 2.1 k 1 Hence, for every x M, we have ρ Φ n x ρ x n i ε i x. Thus, ρ Φ n k 1 i0 n i ε i. This proves Proposition 2.3. Page 1074, proof of Lemma 3.1: Let me add that the same argument which Patras used to prove Lemma 3.1 can be used to prove a more general statement: n Lemma Let ρ n : Hom K H, H Hom K H i, H be the map which takes every linear map g Hom K H, H to the restriction of g to n H i. i0 For every map f Hom K H, H satisfying f 1 1, we have ρ n f 1 n+1 0 where 1 denotes the unity of the K-algebra L H, i. e., the map η ɛ. Proof of Lemma Copy the proof of Lemma 3.1, replacing every occurence of Ψ k by f, and replacing every occurence of ρ n by ρ n. This gives a proof of Lemma Note that we replaced ρ n by ρ n in the statement of Lemma 3.11 because we didn t want to require f to be graded. If f Hom K H, H is a graded map, then ρ n f is more or less the same as ρ n f the only difference between the maps ρ n f and ρ n f is that the codomain of ρ n f is n H i, whereas the codomain of ρ n f is the i0 whole H. However, if f is not a graded map, ρ n f is either not defined or not identic with ρ n f depending on how ρ n is defined: see my remark about Page 1070, two lines above Definition 1.2 above. Here is a very useful consequence of Lemma 3.11: Lemma Let f Hom K H, H be a map satisfying f 1 1. Then, for every x H, the infinite sum 1 n+1 f 1 n x has only finitely n1 n many nonzero terms. Proof of Lemma Let x H. Since x H H i j H i, there j N i0 exists some j N such that x j H i. Consider this j. i0 4 i0 i0

5 Recall that ρ j : Hom K H, H Hom K j i0 H i, H is the map which takes every linear map g Hom K H, H to the restriction of g to j H i. Hence, for every n N, the map ρ j f 1 n is the restriction of the map f 1 n to i0 j i0 H i. Since x j H i, this yields that ρ j f 1 n x f 1 n x for every n N. But we i0 also have ρ j f 1 n ρ j f 1 n for every n N since ρ j is a K-algebra homomorphism. But Lemma 3.11 applied to j instead of n yields ρ j f 1 j+1 0. Hence, every integer n j + 1 satisfies ρ j f 1 n ρ j f 1 j+1+n j+1 ρ j f 1 j+1 ρ j f 1 n j Thus, every integer n j + 1 satisfies 1 n+1 f 1 n x 1 n+1 f 1 n x ρ 1 n+1 j f 1 n x n n n since ρ j f 1 n x f 1 n x This proves Lemma ρ 1 n+1 j f 1 n x since ρ n j f 1 n ρ j f 1 n 1 n+1 0 since ρ n j f 1 n 0 due to n j Page 1074: Two lines above Proposition 3.2, Patras writes: ε i n est donc un morphisme de E dans L H n. It would be helpful to emphasize that morphisme means a morphism of sets here, not a morphism of monoids unless I am missing something!. Page 1074: Four lines above Proposition 3.2, Patras writes: Nous noterons dans la suite ε i n, 1 i n, [...]. I think that considering the ε i n only for 1 i n but not for i 0 is a bad decision, since it leads to several minor mistakes afterwards. For example, the first identity on page 1077, ρ n n Ψ ζ ρ n ζ i e i, i1 is not completely correct, since the sum on the right hand is missing an i 0 term, but as long as e 0 is not defined, this does not make much sense. For another example, Definition 3.7 does not uniquely define Ψ ζ, because H is not the direct sum of all H n i unless we allow i to be 0. 5

6 I think the simplest way to clean up this mess is to define the maps ε i n for all 0 i n in the same as way as they are defined for all 1 i n in the text. This yields that ε 0 n x log 1 x where 1 denotes the unity of L H; this is the 0! 1 map η ɛ not the map I. Thus, in particular, ε 0 n I 1, so that e 0 n ε 0 n I Hn by the definition of e 0 { n 1, if n 0; 1 Hn 0, if n 0. Hence, for n 0, we have e 0 n 0. This is why we don t have care about e 0 n when n 0. But for n 0, we have e Now we have the following in my opinion, slightly better version of Proposition 3.2: Proposition 3.2. For every n N including the case n 0, we have Ψ k n n k i e i n. i0 Proof of Proposition 3.2. Let n N. Then, by the definition of Ψ k n, we have Ψ k n Ψ k Hn Ψ k n H i i0 ρ nψ k Hn ρ n Ψ k Hn Φ k I ρ n Φ k I Hn n ρ n Φ k I Hn k i ε i n I Hn since ρ n Φ k n k i ε i n by Proposition 2.3 applied to i0 E, L H n, ρ n, n + 1, k instead of M, A, ρ, k and n n n k i ε i n I }{{ Hn k i e i } n. e i i0 n because this is how e i n was defined i0 This proves Proposition 3.2. Proposition 3.2 is merely an obvious consequence of Proposition 3.2 : Proof of Proposition 3.2. For every n > 0, we have Ψ k n i0 n k i e i n by Proposition 3.2 i0 This proves Proposition 3.2. k 0 e 0 n + 0 since n 0 n k i e i n i1 n k i e i n. i1 6

7 Page 1075, Definition 3.3: It would not harm to add here that e i n is understood to be 0 if i > n. Otherwise, e i n would not be defined at all for i > n. Page 1075, proof of Proposition 3.4: Patras proof of Proposition 3.4 confines itself to one sentence: Les deux identités résultent respectivement de 1.4 et 3.2, et de la définition 2.2 des projecteurs de poids i. I don t think this enough, however indirectly résultent is meant. It is indeed easy i + j to conclude e i e j e i+j from 1.4, but concluding e i e j δj i e i from 3.2 is i rather difficult. Here is how I would prove Proposition 3.4: First, a rather standard combinatorial identity which we won t prove: Theorem 0.1. Let N N. Then, the equalities N N 1 k k k0 k l 0 for every l {0, 1,..., N 1} 1 and are satisfied in Z. N N 1 k k k0 k N 1 N N! 2 This Theorem 0.1 is, for example, the result of applying Theorem 1 of [DG1] to R Z. This has, as a consequence, a kind of polynomials that are zero at all nonnegative integers must be identically zero result for torsionfree abelian groups: Theorem 0.2. Let R be a torsionfree abelian group. Let n N. Let α 0, α 1,..., α n and β 0, β 1,..., β n be two n + 1-tuples of elements of R n such that every k N satisfies k m α m n k m β m. Then, α m β m for every m {0, 1,..., n}. m0 Proof of Theorem 0.2. We are going to prove that for every l {0, 1,..., n}, we have α n l β n l. 3 Proof of 3. We will prove 3 by strong induction over l. A strong induction does not need an induction base, so let us start with the induction step: Induction step: Let L {0, 1,..., n} be arbitrary. Assume that 3 is already proven for all l {0, 1,..., n} satisfying l < L. Now we must prove 3 for l L. We have m0 α n l β n l for every l {0, 1,..., n} satisfying l < L 4 since 3 is already proven for all l {0, 1,..., n} satisfying l < L. 7

8 Let k N be arbitrary. Then, n k m α m m0 and similarly n L m0 k m α m + n mn L+1 k m α m n L m0 L 1 k m α m + k n l α n l 5 l0 here, we substituted n l for m in the second sum n k m β m m0 Subtracting 6 from 5, we get n k m α m m0 Hence, n L m0 n L m0 n n L L 1 k m β m k m α m + m0 l0 m0 k m α m In other words, n L m0 n L m0 L 1 k m β m + k n l β n l. 6 k n l l0 α n l β n l by 4, since l<l L 1 k m α m + k n l β n l l0 n L n L k m α m k m β m. m0 k m β m m0 n k m α m m0 n k m β m m0 n L m0 n k m β m m0 k m α m n L m0 n L m0 n L m0 n k m β m m0 L 1 k m β m + k n l β n l l0 L 1 k m β m + k n l β n l l0 n k m β m 0. m0 k m β m. 7 Now, forget that we fixed k. We thus have shown 7 for every k N. Now, L {0, 1,..., n} yields n L, so that n L 0. Denote the nonnegative integer n L by N. Then, every k N satisfies N k m α m m0 N k m β m m0 this is just the identity 7, rewritten using N n L. Thus, N N N 1 k k m α m k k0 m0 N N N 1 k k m β m. 8 k k0 m0 8

9 But since and similarly we have N N N 1 k k m α m k k0 m0 N N N N α m 1 k k m k 1 N N!α N so that m0 N 1 l0 N 1 l0 k0 l0 α l N k0 1 k N k k l here, we renamed the index m as l in the first sum N N N N 1 k k l +α N 1 k k N k k k0 } {{ } k0 } {{ } 0 by 1 α l 1 N N! by 2 α l 0 +α N 1 N N! α N 1 N N! 1 N N!α N 0 N N N 1 k k m β m 1 N N!β N, k k0 m0 N N N 1 k k m α m k k0 1 N N!β N, m0 N N N 1 k k m β m by 8 k k0 m0 0 1 N N!α N 1 N N!β N 1 N N! α N β N. Since 1 N N! is a nonzero integer, this yields 0 α N β N since R is torsionfree, so that α N β N. Since N n L, this rewrites as α n L β n L. In other words, 3 is proven for l L. This completes the induction step. Thus, the induction proof of 3 is complete. Now, for every m {0, 1,..., n}, we have α m α n n m β n n m by 3, applied to l n m β m. This proves Theorem 0.2. Proof of Proposition 3.4. We prove Proposition 3.4 in several steps. a For every n N, every k N and every l N we have n n n k i e i n l j e j n kl i e i n. 9 i0 j0 Proof of 9. Let n N, k N and l N. Then, Ψ k Ψ l Ψ kl by Proposition 1.4. This yields Ψ k n Ψ l n Ψ kl n since Ψ k n, Ψ l n and Ψ kl n are just the restrictions of Ψ k, Ψ l and Ψ kl to H n. 9 i0

10 Proposition 3.2 yields Ψ k n k yields Ψ l n n l i e i n i0 n j0 n i0 k i e i n. Proposition 3.2 applied to l instead of l j e j n here we renamed the index i as j in the sum. Proposition 3.2 applied to kl instead of k yields Ψ kl n n i0 n n k i e i n l j e j n Ψ k Ψ l Ψ kl i0 } {{ } Ψ k n j0 } {{ } Ψ l n kl i e i n. Thus, n kl i e i n, so that 9 is proven. b For every n N, every l N and every i {0, 1,..., n}, we have n e i n l j e j n l i e i n. 10 Proof of 10. Let n N and l N. For every k N, we have n n n n k e m m n l j e j n k m e m n l j e j n m0 j0 m0 j0 j0 since composition of K-linear maps is K-bilinear n n i0 k i e i n j0 l j e j n i0 here, we renamed the index m as i in the first sum n kl i e i n by 9 i0 n kl m e m n here, we renamed the index i as m in the sum k m l m n k m l m e m n. m0 m0 n Thus, Theorem 0.2 applied to R End K H n, α m e m n l j e j n and β m j0 n l m e m n yields that e m n l j e j n l m e m n for every m {0, 1,..., n}. Renaming the j0 n index m as i in this result, we obtain: e i n l j e j n l i ei n for every i {0, 1,..., n}. Thus, 10 is proven. c For every n N, every i {0, 1,..., n} and every j {0, 1,..., n}, we have j0 e i n e j n δ i j e i n

11 Proof of 11. Let n N and i {0, 1,..., n}. For every k N, we have n k m δm i m0 m {0,1,...,n} k i δi i + 1 k m δ i m m {0,1,...,n}; m i } {{ } 0 Now, for every k N, we have n n k m e i n e m n e i n k m e m n m0 m0 m {0,1,...,n}; mi k m δm i + } {{ } k i δi i k m 0 k i. m {0,1,...,n}; m i k m δ i m 0 since m i since composition of K-linear maps is K-bilinear n e i n k j e j n here, we renamed the index m as j in the sum k i j0 n k m δm i m0 n k m δm i e i n m0 e i n by 10, applied to k instead of l n k m δm i e i n. m0 Thus, Theorem 0.2 applied to R End K H n, α m e i n e m n and β m δm i e i n yields that e i n e m n δm i e i n for every m {0, 1,..., n}. Renaming the index m as j in this result, we obtain: e i n e j n δj i e i n for every j {0, 1,..., n}. Thus, 11 is proven. d We have e i n e j n δ i j e i n for every i N, j N and n N. 12 Proof of 12. Let i N, j N and n N. We must have one of the following three cases: Case 1: We have i {0, 1,..., n} and j {0, 1,..., n}. Case 2: We have i / {0, 1,..., n}. Case 3: We have i {0, 1,..., n} and j / {0, 1,..., n}. In Case 1, we notice that 12 follows directly from 11. In Case 2, we find that e i n is 0, and thus 12 is trivially true. In Case 3, both e j n and δj i are 0 in fact, δj i 0 because i j, and thus 12 is trivially true. Hence, we have seen that 12 in each of the three Cases 1, 2 and 3. This proves 12. e For every i N and j N, we have e i e j δj i e i. Proof. Let i N and j N. The maps e i and e j are defined as graded endomorphisms of L H, whose n-th graded components are e i n and e j n respectively for each n N. Hence, in order to prove that e i e j δj i e i, it is enough to show that 11

12 e i n e j n δj i e i n for every n N. But this has already been shown in 12. Thus, we are done proving that e i e j δj i e i. f For every i N, we have log I i e i, 13 i! where log I is to be understood as the result of applying the formal power series of the logarithm to I. This result is well-defined, since for every x H, only finitely many terms of the formal power series log I x are nonzero. Proof. Let i N. Also, let n N be arbitrary. By the definition of ρ n, every f E such that f 1 1 satisfies ρ n log f ρ n 1 m+1 f 1 m m m 1 since log f 1 m+1 f 1 m by the definition of the logarithm m m 1 1 m+1 ρ n f 1 m since ρ n is a K-algebra homomorphism, and can m easily be seen to commute with reasonable infinite series m 1 1 m+1 ρ n f 1 m + 1 m+1 ρ n f 1 m m m 1; m 1; }{{ m } m<n+1 m n+1 0 since Lemma 3.1 yields n ρ nf 1 n+1 0, so that m1 ρ nf 1 m 0 since m n+1 n 1 m+1 ρ n f 1 m + n 1 m m+1 ρ n f 1 m m m m1 m 1; m1 m n+1 0 n log n+1 f since log n+1 f 1 m+1 ρ n f 1 m by the definition of log m n+1 f. m1 Applied to f I, this yields ρ n log I log n+1 I. g n But every graded map g : H H satisfies g Hn H i i0 g n H i i0 Hn 3 and ρ n g because this is how ρ n was defined. Hence, every graded map g : H H 3 In fact, if we restrict the map g to H n, we get the same result as if we restrict the map g to n first and then restrict this restriction to H n. H i i0 12

13 satisfies g Hn g n H i i0 ρ ng Hn ρ n g Hn. Applied to g log I i, this yields i! log I i Hn i! log I i ρ n Hn ρ n log I i Hn i! i! log I since i ρ n is a K-algebra homomorphism, so that ρ n logn+1 I i i! Combined with Hn since ρn log I log n+1 I. e i Hn e i n ε i n I logn+1 I i i! by the definition of ε i n by the definition of e i Hn logn+1 I i i! i! Hn, ρ n log I i i! logn+1 I i this yields e i log I i Hn Hn Hn. i! i! Now forget that we fixed n N. We thus have proven that every n N satisfies e i log I i Hn Hn. Therefore, e i log I i. We have thus proven 13. i! i! i + j g For every i N and j N, we have e i e j e i+j. Proof. Let i N and j N. By 13, we have e i j instead of i, we have e j e i+j log I i+j. Now, i + j! i log I i. By 13 applied to i! log I j. By 13 applied to i + j instead of i, we have j! log I i e i i! e j log I j j! log I i i! i + j! i!j! } {{} i + j i log I j j! log I i+j i + j! } {{ } e i+j i + j e i+j, i 13

14 qed. From the results of steps e and g, we conclude that Proposition 3.4 holds. Page 1076, Corollary 3.6: Here is an alternative proof of Corollary 3.6 without using eigenspaces. Before we even begin proving this corollary, let us record some lemmata that could just as well have been stated and proven in Section 1: a First, an elementary lemma: Lemma 1.5. For any two K-coalgebras C and D, any two K-algebras A and B, and any four K-linear maps p : C A, q : C A, r : D B and s : D B, we have p r q s p q r s. Here, p r q s means the convolution of the two K-linear maps p r : C D A B and q s : C D A B. This lemma is easy to prove particularly if you are using Sweedler s notation, but even without it. The following two lemmata are easy consequences of Lemma 1.5: Lemma 1.6. For any two K-coalgebras C and D, any two K-algebras A and B, and any two K-linear maps f : C A and g : D B, we have f 1HomK D,B 1HomK C,A g f g 1 HomK C,A g f 1 HomK D,B. 4 Here, both Hom K D, B and Hom K C, A are made into K-algebras by the convolution. Lemma 1.7. For any two K-coalgebras C and D, any two K-algebras A and B, and any two K-linear maps f : C A and g : C A, we have f 1HomK D,B g 1HomK D,B f g 1HomK D,B. Here, Hom K D, B is made into a K-algebra by the convolution. By repeated application of Lemma 1.7, we get: Lemma 1.8. a For any two K-coalgebras C and D, any two K-algebras A and B, any n N and any K-linear map f : C A, we have n f 1HomK D,B f n 1 HomK D,B. Here, Hom K D, B is made into a K-algebra by the convolution. b For any two K-coalgebras C and D, any two K-algebras A and B, any n N and any K-linear map g : D B, we have 1HomK C,A g n 1HomK C,A g n. Here, Hom K C, A is made into a K-algebra by the convolution. 4 Here and in the following, for any K-algebra U, we denote by 1 U the unity of U. 14

15 Proof of Lemma 1.8. Lemma 1.8 a is proven by induction over n and use of Lemma 1.7 in the induction step. Lemma 1.8 b is completely analogous to Lemma 1.8 a the only difference is the order of the tensorands, so the proof is analogous as well. The details of these proofs are left to the reader. b On the other hand, we recall that for every connected graded bialgebra A and any K-linear map f : A A satisfying f 1 1, the K-linear map log f : A A is welldefined. Namely, this map log f is defined by applying the formal power series of the logarithm to f; in other words, log f is defined as the infinite sum 1 n+1 f 1 n where 1 denotes the unity of the K-algebra L A, i. e., the map η ɛ. 5 The following lemma is easy to check: Lemma 1.9. For every connected graded bialgebra A and any two K- linear maps f : A A and g : A A satisfying f 1 g 1 1 and f g g f, we have log f g log f + log g. In fact, Lemma 1.9 follows from the identity log 1 + X 1 + Y log 1 + X + log 1 + Y in the ring K [[X, Y ]] of formal power series. c Next, we have: Lemma For any two connected graded K-bialgebras A and B and any K-linear map f : A A satisfying f 1 1, we have f 1 LB 1 1 and log f 1 LB log f 1LB. Proof of Lemma Checking f 1 LB 1 1 is very easy and left to the reader. Now let us prove that log f 1 LB log f 1LB : For every n N, we have f 1 LA 1HomK B,B n f 1LA n 1HomK B,B by Lemma 1.8 a, applied to A, B and f 1 LA instead of C, D and f. Since 1 HomK B,B 1 LB, this rewrites as f 1 LA 1LB n f 1LA n 1LB for every n N. 5 This sum is infinite, but it still gives us a well-defined map A A, because for every x A, the infinite sum 1 n+1 f 1 n x has only finitely many nonzero terms by Lemma 3.12, applied n1 n to A instead of H and thus has a well-defined value in A. n1 n 15

16 By the definition of log f 1 LB, we have log f 1 LB n f 1 n+1 1LB 1 LA B n f 1 n+1 1LA 1LB n n n1 n1 since f 1 LB 1 LA B f 1 LB 1 LA 1 LB f 1LA 1LB 1 LA 1 LB n1 1 n+1 f 1LA n 1LB n since f 1LA 1LB n f 1LA n 1LB for every n N n f 1 n+1 1LA 1 LB. n n1 n On the other hand, by the definition of log f, we have log f f 1 n+1 1LA, n1 n so that n f log f 1 LB 1 n+1 1LA n f 1 LB 1 n+1 1LA 1LB n n n1 here, we are using the fact that the tensor product commutes with convergent infinite sums; this can be easily checked. Thus, n1 log f 1 LB n1 1 n+1 f 1LA n 1LB n log f 1 LB. This proves Lemma We now have: Lemma For any two connected graded K-bialgebras A and B, any K-linear map f : A A satisfying f 1 1, and any K-linear map g : B B satisfying g 1 1, we have f g 1 1 and log f g log f 1 LB + 1 LA log g. Proof of Lemma Again, checking that f g 1 1 is very easy. Let us now prove that log f g log f 1 LB + 1 LA log g. By Lemma 1.6 applied to C A and D B, we get f 1HomK B,B 1HomK A,A g f g 1 HomK A,A g f 1 HomK B,B. Since 1 HomK A,A 1 LA and 1 HomK B,B 1 LB, this rewrites as f 1LB 1LA g f g 1 LA g f 1 LB. 16

17 By Lemma 1.10, we have f 1 LB 1 1 and log f 1LB log f 1LB. Similarly, 1 LA g 1 1 and log 1 LA g 1 LA log g. Since f 1 LB 1LA g 1 LA g f 1 LB, f 1LB 1 1 and 1 LA g 1 1, we can apply Lemma 1.9 to f 1 LB, 1 LA g and A B instead of f, g and A. We obtain log f 1 LB 1LA g log f 1 LB + log 1 LA g. Hence, log f g f 1 LB 1 LA g log f 1 LB 1LA g log f 1 LB + log 1LA g log f 1 LB 1 LA log g log f 1 LB + 1 LA log g. This proves Lemma d Now, finally, let us come to the alternative proof of Corollary 3.6: Assume that H is a graded bialgebra. The case of H being a Hopf algebra is similar, in that the same argument works but we have to interpret H H as a tensor product of graded superalgebras rather than as a plain tensor product of graded algebras. For every i N, let e i H H denote the i-th projecteur de poids i of the graded bialgebra or Hopf algebra H H. Note that e i still denotes the i-th projecteur de poids i of the graded bialgebra or Hopf algebra H. By 13 applied to i 1, we have e 1 log I 1 1! Note also that every u N satisfies log I 1 log I. e u log I u u! e1 u u! by 13, applied to i u since log I e 1, so that e 1 u u! e u. 14 log Denote by I H H the identity map H H H H. Then, I H H I I log I I log I 1 LH + 1 LH log I e 1 e 1 by Lemma 1.11, applied to A H, B H, f I and g I e 1 1 LH + 1 LH e 1. 17

18 Now, for every l N, we have e l H H log I H H l 1 l! 1 l! 1 l! 1 l! l! log I H H e 1 1 LH +1 LH e 1 l k0 k0 k0 l k by 13, applied to l and H H instead of i and H l 1 l! e 1 1 LH k l k0 e 1 k 1 LH by Lemma 1.8 a, applied to CH, DH, AH, BH, fe 1 and nk e 1 1 LH + 1 LH e 1 l } {{} l k e 1 1 LH k 1 LH e 1 l k by the binomial formula 1LH e 1 l k 1 LH e 1 l k by Lemma 1.8 a, applied to CH, DH, AH, BH, ge 1 and nl k l l e 1 k 1LH 1 LH e 1 l k k l l e 1 k 1HomKH,H 1 HomK H,H e 1 l k k } {{ } e 1 k e 1 l k because Lemma 1.6 applied to CH, DH, AH, BH, fe 1 k and ge 1 l k e 1 k e 1 l k since 1LH 1 HomK H,H yields e1 k 1 HomK H,H 1 HomK H,H e 1 l k 1 HomK H,H e 1 l k e1 k 1 HomK H,H 1 l! 1 l! l k0 l k0 l k l! k! l k! e 1 k k! e k by 14, applied to uk l! k! l k! k! e k l k! e l k e 1 l k l k! e l k by 14, applied to ul k l e k e l k. 15 k0 Now, let us consider two cases: Case 1: The bialgebra H is commutative. Case 2: The bialgebra H is cocommutative. Note that this is not really a case distinction, because in each of the two cases we have to prove a different claim: In Case 1, we have to prove that the decomposition of Theorem 3.5 endows H with a structure of a bigraded algebra, whereas in Case 2, we have to prove that the decomposition of Theorem 3.5 endows H with a structure of a bigraded coalgebra. First let us consider Case 1. In this case, H is commutative. Thus, the map Π : H H H is an algebra homomorphism. Since Π also is a coalgebra homomorphism by the axioms of a bialgebra and a graded map since H is a graded bialgebra, this 18

19 yields that Π is a graded bialgebra homomorphism. Thus, since the definition of e i was functorial with respect to H, the diagram H H Π H e u H H e u H H H Π is commutative for every u N. In other words, Π e u H H e u Π for every u N. 16 Now, let i N and j N be arbitrary. By 15 applied to l i + j, we have e i+j H H i+j e k e i+j k, so that k0 e i+j H H e i e j i+j e k e i+j k e i e j i+j e k e i+j k e i e j k0 k0 e k e i e i+j k e j i+j k0 k {0,1,...,i+j} k {0,1,...,i+j} k {0,1,...,i+j}; ki e k e i δ k i ek by Proposition 3.4, applied to k and i instead of i and j δi k e k δ i+j k j e i+j k δi k e k δ i+j k j e i+j k δ i i 1 δi iei δ i+j i j e i+j i since i {0,1,...,i+j} e i δ i+j i j + i+j i e + δ j j 1 e j k {0,1,...,i+j}; k i } {{ } 0 e i+j k e j δ i+j k j e i+j k by Proposition 3.4, applied to i+j k and j instead of i and j k {0,1,...,i+j}; k i δi k e k δ i+j k j e i+j k 0 since k i, so that δ k i 0 0 e i e j + 0 e i e j. 17 Now, fix n N and m N. By the definitions of H i n and H j m, we have H i n 19

20 e i H n and H m j e j H m. Thus, H i n e i H n Hence, Π H i n H m j e i H n e j H m e i e j e j H m H m j e i+j H H ei e j by 17 e i+j H H e i e j H n H m e i+j H H e i+j H H e i H n H n since e i is a graded map e i+j H H H n H m. Π e i+j H n H m e j H m H m since e j is a graded map Π e i+j H H H n H m H H H n H m e i+j Π by 16, applied to ui+j e i e j H n H m e i H n e j H m e i+j Π H n H m e i+j Π H n H m e i+j H n H m H n H m since Π is the multiplication map H n+m e i+j H n+m H i+j n+m H i+j n+m since H i+j n+m was defined as e i+j H n+m Let us now forget that we fixed i, j, n and m. We have thus proved that Π H n i since H is a graded algebra. H m j for every i N, j N, n N and m N. In other words, we have proved that the multiplication map Π is bigraded. Due to this and to the trivial fact that 1 H H 0 0, the decomposition of Theorem 3.5 endows H with a structure of a bigraded algebra. Thus, we have proven our claim in Case 1. Next, let us consider Case 2. In this case, H is cocommutative. Thus, the map : H H H is a coalgebra homomorphism. Since also is an algebra homomorphism by the axioms of a bialgebra and a graded map since H is a graded bialgebra, this yields that is a graded bialgebra homomorphism. Thus, since the definition of e i was functorial with respect to H, the diagram H H H e u e u H H H H H is commutative for every u N. In other words, e u e u H H for every u N

21 Now, fix n N and i N. Since H is a graded coalgebra, we have H n H m H n m H m H n m since direct sums are sums. m n m n Let us notice that the direct sum H m j H i j n m is well-defined. 6 Moreover, it satisfies m n, j i H j m H i j n m 6 Proof. We know that H m,j N 2 m n, j i m,j N 2 H j m H j m H H H m j H m n, j i m n, j i H m j H i j n m e j H m e i j H n m by the definition of H m j by the definition of H i j n m since direct sums are sums e j H m e i j H n m. m n, j i m,j N 2 H j m, so that H H because tensor products commute with direct sums. H m j H m,j N 2 Thus, the direct sum is well-defined because it is a partial sum of the well-defined direct sum. Hence, whenever x m,j m n, j i is a family of elements of H H such that x m,j 0 and such that every m n and j i satisfy m n, j i x m,j H j m H, then x m,j m n, j i 0 m n, j i As a consequence of this, whenever x m,j m n, j i is a family of elements of H H such that x m,j 0 and such that every m n and j i satisfy m n, j i x m,j H j m H i j n m, then x m,j m n, j i 0 m n, j i because if x m,j m n, j i is a family of elements of H H such that m n, j i. 19 x m,j 0 and such that every m n and j i satisfy x m,j H m j H i j n m, then every m n and j i satisfy x m,j H m j H i j n m H m j H, and thus we can apply 19 to obtain x m,j m n, j i 0 m n, j i. H In other words, the direct sum m n, j i H j m H i j n m is well-defined, qed. 21

22 But by the definition of H n i, we have H n i e i H n e i H i n e i H H by 18, applied to ui e i H n, so that H n e i H H H n e i H H H n e i H H H m H n m m n H m H n m m n i e k e i k H m H n m m n k0 i k0 m n m n, k i m n, j i since 15 applied to l i yields e i H H i e k e i k k0 e k e i k H m H n m e k H m e i k H n m e k H m e i k m n, k i H n m e j H m e i j H n m here, we renamed the index k as j in the sum. Hence, H n i e j H m e i j H n m m n, j i m n, j i H j m H i j n m. Now forget that we fixed n and i. We have shown that every n N and i N satisfy H n i H m j H i j n m. In other words, the comultiplication map m n, j i is bigraded. Due to this and to the easily checked fact that the map ɛ is 0 on H i n for all n, i 0, 0, the decomposition of Theorem 3.5 endows H with a structure of a bigraded coalgebra. Thus, we have proven our claim in Case 2. The proof of Corollary 3.6 is now complete since the claim of Corollary 3.6 has been proven in each of the Cases 1 and 2. Page 1076, Definition 3.7: As explained above, this definition is only correct if i is allowed to be 0. Page 1077, Proof of Proposition 3.8: The formulas in this proof are slightly wrong: Replace the n sign by n n. Also, replace each of the signs by n. i1 i0 Page 1077, Proposition 3.9: Let me give some hints for the proof of this proposition. First, in order to prove that Les opérations caractéristiques, les opérations caractéristiques généralisées et les projecteurs de poids i sur H et H gr sont alors deux à deux adjoints, the main step to make is to show that f g f g for any two 22 i,j1 i,j0

23 graded K-linear maps f : H H and g : H H. This is easy to show using the definition of the convolution: f g Π f g. Once this is shown, it yields that f i f i for every graded K-linear map f : H H and every i N, and that log f log f whenever both sides of this equation are well-defined, etc. - and ultimately the adjointness part of Proposition 3.9. Now to the proof that H i H grj : j i For every n N, let e n denote the projecteur de poids i associé à H gr. We know that e n is adjoint to e n for every n N. For every f H gr, we have the following equivalence of assertions: f H grj e i f 0 f e i 0 j i since e i is adjoint to e i, so that e i f f e i f e i H 0 f H i 0 Hence, j i H grj H i, thus completing our proof of Proposition 3.9. f H i. Section 3: Let me add a lemma into this section which I am going to use further below: Lemma Let n N and i {0, 1,..., n}. There exists some N N and some elements α 0, α 1,..., α N of K such that every graded bialgebra or Hopf algebra H satisfies N e i n α k Ψ k n. 20 k0 Note that we are using the notations of Section 3 here, i. e., the e i n and Ψ k n 23

24 in Lemma 3.10 are the endomorphisms of H n defined on page Page 1078: One line above Lemma 4.1, Patras writes: éléments de H. This should be éléments x de H. Page 1078, proof of Lemma 4.1: I wouldn t agree that L inclusion Prim H H 1 est immédiate. Here is how I would prove that Prim H H 1 : 7 Proof of Lemma By the definition of e i n, we have logn+1 I i e i n ε i n I logn+1 I i Hn i! Hn i! by the definition of ε i n 1 i! n m1 ρ n 1 i! log n+1 I ρ 1 m+1 n I 1 m m by the definition of log n+1 n 1 m+1 I 1 m i Hn m m1 i Hn 1 n 1 m+1 ρ n I 1 m i Hn i! m m1 because ρ n is a K-algebra homomorphism 1 n 1 m+1 I 1 m i Hn i! m m1 because any f L H satisfies ρ n f Hn f Hn since ρ n f is the restriction of f to n H i, and restricting this restriction i0 further to H n gives the same result as just restricting f itself to H n in every graded bialgebra or Hopf algebra H. But 1 n 1 m+1 I 1 m i can be seen as a i! m1 m polynomial whose coefficients lie in K and don t depend on H applied to I. In other words, there exists some N N and some appropriate elements α 0, α 1,..., α N of K which don t depend on H such that 1 n 1 m+1 I 1 m i N α k I k. Consider this N and these elements α 0, α 1, i! m1 m k0..., α N of K. Then, in every graded bialgebra or Hopf algebra H, we have e i n 1 n 1 m+1 I 1 m i N Hn α k I k i! m m1 k0 thus proving Lemma N k0 α k N α k I k k0 I k Hn Ψ k N k0 24 α k Ψ k Hn Ψ k n N α k Ψ k n, k0 Hn

25 a First, a lemma: For any K-algebra A and any two K-linear maps f : H A and g : H A satisfying f 1 g 1 0, we have f g 1 0 and f g Prim H 0 21 Proof of 21. Let A be a K-algebra, and let f : H A and g : H A be two K-linear maps satisfying f 1 g 1 0. Let x Prim H. Then, x is primitive by the definition of Prim H, so that x x x. Now, by the definition of convolution, f g Π f g, so that f g x Π f g x Π f g x x 1+1 x. Π f g x x fx g1+f1 gx Π f x g 1 + f 1 g x Π f x g x Now forget that we fixed x. We thus have shown that f g x 0 for every x Prim H. Thus, f g Prim H 0. Besides, f g 1 Π f g 1 Π f g 1 Π f g 1 1 Π f g 1 1 f1 g1 Π f 1 g 1 Π Thus, 21 is proven. b As a consequence of the previous lemma, we can prove the next lemma: For any K-algebra A, any integer i 2 and any K-linear map f : H A satisfying f 1 0, we have f i 1 0 and f i Prim H 0 22 Proof of 22. We will prove 22 by induction over i where the induction base is the case i 2: Induction base: For any K-algebra A and any K-linear map f : H A satisfying f 1 0, we have and f 2 f f f 2 f f 1 f f 1 0 by 21, applied to g f Prim H f f Prim H 0 by 21, applied to g f. In other words, 22 holds for i 2. This completes the induction base.. 25

26 Induction step: Let n be an integer 2. Assume that 22 holds for i n. We must prove that 22 also holds for i n + 1. Let A be a K-algebra, and f : H A be a K-linear map satisfying f 1 0. We assumed that 22 holds for i n. Hence, by 22 applied to i n, we get f n 1 0 and f n Prim H 0. Since f 1 f n 1 0, we can apply 21 to g f n, and conclude that f f n 1 0 and f f n Prim H 0. Since f f n f n+1, this rewrites as follows: We have f n and f n+1 Prim H 0. Forget that we fixed A and f. We thus have showed that for any K-algebra A and any K-linear map f : H A satisfying f 1 0, we have f n and f n+1 Prim H 0. In other words, we have proven that 22 holds for i n + 1. This completes the induction step. Thus, the induction proof of 22 is complete. c Now, we will prove: For any K-algebra A and any K-linear map f : H A satisfying 23 f 1 1, we have log f Prim H f Prim H. Note that, in this assertion, log f is defined by applying the formal power series of the logarithm to f; in other words, log f is defined as the infinite sum 1 n+1 f 1 n n1 n where 1 denotes the unity of the K-algebra Hom K H, A 8, i. e., the map η ɛ. This sum is infinite, but it still gives us a well-defined map H A, because for every x H, the infinite sum 1 n+1 f 1 n x has only finitely many nonzero terms 9 n1 n and thus has a well-defined value in A. Proof of 23. Let A be a K-algebra, and let f : H A be a K-linear map satisfying f 1 1. Then, f 1 1 f Hence, 1 1 for any integer i 2, we have f 1 i 1 0 and f 1 i Prim H 0 24 by 22, applied to f 1 instead of f. Now let x Prim H. Then, for any integer i 2, we have f 1 i x 0 25 since for any integer i 2, we have f 1 i Prim H 0 by 24, but f 1 i x f 1 i Prim H since x Prim H, so that f 1 i x f 1 i Prim H 0 and thus f 1 i x 0. On the other hand, x Prim H, so that x is primitive, and thus ɛ x 0 indeed, there is a well-known fact that every primitive element ξ of any bialgebra satisfies ɛ ξ 0. Recall that by the definition of log f we have log f 1 n+1 f 1 n 1 i+1 f 1 i n i n1 i1 here, we renamed the index n as i in the sum, 8 This K-algebra Hom K H, A is the K-vector space of all K-linear maps H A, with convolution as multiplication. 9 This follows from Lemma 3.12 if A H, and is proven exactly in the same way in the general case. 26

27 so that log f x 1 i+1 f 1 i x i i1 i i+1 1 f 1 x + f 1 1 x 1 f 1 1 xf 1x i2 1 i+1 f 1 i x i f 1 i x } {{ i } 0 since f 1 i x0 by 25, since i 2 1 i+1 0 f 1 x f x 1 i2 } {{ } 0 f x η ɛ x f x η 0 f x. 0 0 η ɛ x f x η ɛ x ηɛx Now forget that we fixed x. We thus have shown that every x Prim H satisfies log f x f x. In other words, log f Prim H f Prim H. This proves 23. d Now to the proof of Prim H H 1 : First, we notice that every i N satisfies e i H H i. This is because, for every i N, we have e i H n N n i n i e i n H n e i n H n H i n H i n H i. because the n-th graded component of e i is e i n by Definition 3.3 here, we removed the addends with n < i from our direct sum; this did not change the sum since e i n 0 when n < i Applied to i 1, this yields e 1 H H 1. On the other hand, by 13 applied to i 1, we have e 1 log I 1 log I 1! 1 log I. Thus, e 1 Prim H log I Prim H I Prim H by 23, applied to f I and A H. Hence, for every x Prim H, we have e 1 x I x. Thus, for every x Prim H, we have x I x e 1 x e 1 H H 1. In other words, Prim H H 1, qed. Page 1079, line 7: Typo in this line: sructure should be structure. Page 1079, the equation after line 7: This equation is This should instead be x A n, y A m, [x, y] x y 1 n m y x. x A n, y A m, [x, y] xy 1 n m yx. 27

28 Page 1079, Proposition 4.3: A proof of Proposition 4.3 is sketched in [C] more precisely, in the proof of Theorem in [C]. Page 1080, Section 5: Here, Patras claims that x i1... x ik sgn β x iβ1... x iβp p+qk β x iβp+1... x iβp+q, 26 where the sum is over all opérateurs d interclassement d indices p et q 10. β He defines an opérateur d interclassement d indices p et q as an element σ of S p+q satisfying σ 1 1 <... < σ 1 p and σ 1 p + 1 <... < σ 1 p + q. 27 This is wrong. formula 26 by x i1... x ik There are two ways to correct this mistake: Either replace the p+qk β sgn β x iβ x iβ 1p x iβ 1p+1... x iβ 1p+q, 28 where the sum is still over all opérateurs d interclassement d indices p et q. Or β replace the formula 27 in the definition of an opérateur d interclassement d indices p et q by σ 1 <... < σ p and σ p + 1 <... < σ p + q. In the following, I am going to assume that the mistake has been corrected in the first way i. e., that the formula 26 has been replaced by 28. Page 1080, Section 5: Just a remark on T X. Patras gave the formula 26 for the comultiplication of T h X, and we corrected it to 28. There is an analogous formula for the comultiplication of T X, namely: x i1... x ik x iβ x iβ 1p x iβ 1p+1... x iβ 1p+q. Here, the β p+qk β sum is still over all opérateurs d interclassement d indices p et q. Page 1081, Proof of Proposition 5.1: Here, Patras writes: En explicitant les formules pour le produit et le coproduit dans T X resp. T h X en termes d opérateurs d interclassement, on vérifie facilement que : Ψ k n K [S n ] End T n X 10 I think that the proper English translation of the notion opérateur d interclassement d indices p et q is p, q-shuffle ; but I am not sure about this. Anyway there seem to exist at least three non-equivalent definitions of p, q-shuffle in literature, so one should be careful when using this notion

29 resp. : Ψ k n K [S n ] End T h n X. Let me detail this argument: First, let us work in T X. By repeated application of 29, we see that any l N, any k N and any i 1, i 2,..., i l {1, 2,..., n} l satisfy [k] x i1... x il p 1 +p p k l σ S l ; σ1<σ2<...<σp 1 ; σp 1 +1<σp 1 +2<...<σp 1 +p 2 ; σp 1 +p 2 +1<σp 1 +p 2 +2<...<σp 1 +p 2 +p 3 ;...; σp 1 +p p k 1 +1<σp 1 +p p k 1 +2<...<σp 1 +p p k 1 +p k x iσ1... x iσp1... x iσp x i σp1 +p 2 x iσp1 +p p k x i σp1 +p p k 1 +p k Hence, any k N and any i 1, i 2,..., i n {1, 2,..., n} n satisfy Ψ k n x i1... x in Ψ k x i1... x in Π [k] [k] x i1... x in I k Π [k] [k] Π [k] [k] x i1... x in p 1 +p p k n σ S n; σ1<σ2<...<σp 1 ; σp 1 +1<σp 1 +2<...<σp 1 +p 2 ; σp 1 +p 2 +1<σp 1 +p 2 +2<...<σp 1 +p 2 +p 3 ;...; σp 1 +p p k 1 +1<σp 1 +p p k 1 +2<...<σp 1 +p p k 1 +p k x iσ1... x iσp1... x iσp x i σp1 +p 2 x iσp1 +p p k x i σp1 +p p k 1 +p k by 30 and since Π is the multiplication map p 1 +p p k n σ S n; σ1<σ2<...<σp 1 ; σp 1 +1<σp 1 +2<...<σp 1 +p 2 ; σp 1 +p 2 +1<σp 1 +p 2 +2<...<σp 1 +p 2 +p 3 ;...; σp 1 +p p k 1 +1<σp 1 +p p k 1 +2<...<σp 1 +p p k 1 +p k p 1 +p p k n σ S n; σ1<σ2<...<σp 1 ; σp 1 +1<σp 1 +2<...<σp 1 +p 2 ; σp 1 +p 2 +1<σp 1 +p 2 +2<...<σp 1 +p 2 +p 3 ;...; σp 1 +p p k 1 +1<σp 1 +p p k 1 +2<...<σp 1 +p p k 1 +p k x... x iσp1 +p 2 +1 i σp1 +p 2 +p x... x iσp1 +p 2 +1 i σp1 +p 2 +p 3 x iσ1 x iσ2... x iσn σ 1 x i1... x in σ 1 x i1... x in. 29

30 Thus, for every k N, the map Ψ k n End T n X is the image of the element σ 1 K [S n ] p 1 +p p k n σ S n; σ1<σ2<...<σp 1 ; σp 1 +1<σp 1 +2<...<σp 1 +p 2 ; σp 1 +p 2 +1<σp 1 +p 2 +2<...<σp 1 +p 2 +p 3 ;...; σp 1 +p p k 1 +1<σp 1 +p p k 1 +2<...<σp 1 +p p k 1 +p k under the map K [S n ] End T n X. Similarly, we can show the analogous result for T h n X instead of T n X: Namely, for every k N, the map Ψ k n End T h n X is the image of the element sgn σ σ 1 K [S n ] p 1 +p p k n σ S n; σ1<σ2<...<σp 1 ; σp 1 +1<σp 1 +2<...<σp 1 +p 2 ; σp 1 +p 2 +1<σp 1 +p 2 +2<...<σp 1 +p 2 +p 3 ;...; σp 1 +p p k 1 +1<σp 1 +p p k 1 +2<...<σp 1 +p p k 1 +p k under the map K [S n ] End T h n X. Page 1082, Proof of Proposition 5.1: Here, Patras writes: Le reste de la proposition ne présente pas de difficultés. Let me try to make this part of the proof a bit more precise. Namely, let me show that the idempotents e i n and f i n of the algebra K [S n ] are obtained from each other by the involution K [S n ] K [S n ] σ sgn σ σ of the K-algebra K [S n ]. In fact, let inv denote the K-vector space homomorphism K [S n ] K [S n ] σ sgn σ σ for every σ S n. It is easy to see that inv is a K-algebra homomorphism this is more or less because sgn : S n { 1, 1} is a group homomorphism and an involution since sgn σ 2 1 for every σ S n. We must now prove that inv fn i e i n. We will do this in two steps: a For every k N, let Ψ k T,n denote the element of K [S n] whose image under the map K [S n ] End T n X is the Ψ k n of T X where by Ψ k n of T X, we mean the map Ψ k n defined with respect to the graded bialgebra T X. Then, for every k N, we have Ψ k T,n σ 1. p 1 +p p k n σ S n; σ1<σ2<...<σp 1 ; σp 1 +1<σp 1 +2<...<σp 1 +p 2 ; σp 1 +p 2 +1<σp 1 +p 2 +2<...<σp 1 +p 2 +p 3 ;...; σp 1 +p p k 1 +1<σp 1 +p p k 1 +2<...<σp 1 +p p k 1 +p k 30 31

31 11 Similarly, for every k N, let Ψ k T h,n denote the element of K [S n] whose image under the map K [S n ] End T h n X is the Ψ k n of T h X where by Ψ k n of T h X, we mean the map Ψ k n defined with respect to the Hopf algebra T h X. Then, for every k N, we have Ψ k T h,n sgn σ σ p 1 +p p k n σ S n; σ1<σ2<...<σp 1 ; σp 1 +1<σp 1 +2<...<σp 1 +p 2 ; σp 1 +p 2 +1<σp 1 +p 2 +2<...<σp 1 +p 2 +p 3 ;...; σp 1 +p p k 1 +1<σp 1 +p p k 1 +2<...<σp 1 +p p k 1 +p k Comparing 31 with 32, we immediately see that inv ΨT,n k Ψ k T h,n. b Fix some n N and some i {0, 1,..., n}. Consider the N and the elements α 0, α 1,..., α N whose existence is guaranteed by Lemma Note that the equality 20 is a concretization of Patras claim that Ce morphisme peut, d après 1.3 et 3.1, se réécrire comme une combinaison linéaire finie d endomorphismes caractéristiques on page Patras assertion doesn t make it clear that the coefficients of this combinaison linéaire don t depend on H, but our Lemma 3.10 does, and we are going to use this now. Applying 20 to H T X, we get fn i N α k Ψ k T,n. Applying 20 to H 11 This is because we showed above that for every k N, the map Ψ k n End T n X is the image of the element p 1+p p k n σ S n; σ1<σ2<...<σp 1; σp 1+1<σp 1+2<...<σp 1+p 2; σp 1+p 2+1<σp 1+p 2+2<...<σp 1+p 2+p 3;...; σp 1+p p k 1 +1<σp 1+p p k 1 +2<...<σp 1+p p k 1 +p k k0 σ 1 K [S n ] under the map K [S n ] End T n X. 12 This is because we showed above that for every k N, the map Ψ k n End T h n X is the image of the element p 1+p p k n σ S n; σ1<σ2<...<σp 1; σp 1+1<σp 1+2<...<σp 1+p 2; σp 1+p 2+1<σp 1+p 2+2<...<σp 1+p 2+p 3;...; σp 1+p p k 1 +1<σp 1+p p k 1 +2<...<σp 1+p p k 1 +p k under the map K [S n ] End T h n X. 32 sgn σ σ 1 K [S n ] 31

32 T h X, we get e i n N α k Ψ k T h,n. Thus, qed. k0 inv N fn i inv α k Ψ k T,n since fn i k0 k0 N α k inv Ψ k T,n Ψ k T h,n N α k Ψ k T h,n e i n, k0 N α k Ψ k T,n k0 since inv is K-linear Page 1082, three lines above Lemma 5.2: Patras writes: Par définition, l algèbre de Lie libre resp. graduée est la plus petite sous-algèbre de Lie resp. graduée de T X resp. de T h X contenant X. I don t think this really follows from the definition but it follows from the Poincaré-Birkhoff-Witt theorem 13. Page 1082: On this page, it should be said somewhere that T h X is just an abbreviation for T h gr X and not the dual of T h X as an ungraded K-vector space. Page 1082: On the penultimate line of page 1082, Patras writes: l algèbre graduée T. This is inaccurate, since T is not a graded algebra but the completion of a graded algebra with respect to the canonical topology induced by the grading. Fortunately this does not prevent the conclusion that the logarithm of S is well-defined from being valid it actually would not be valid if T was just a graded algebra!. Page 1083, proof of Proposition 5.3: On the second line of this page, there is a typo: T g X should be T h X. Page 1083, proof of Proposition 5.3: Here, I don t understand why Patras claims that Q m i 1,...,i k Π [m] I η ɛ m [m] x i1... x ik. But I think there is a alternative proof of Proposition 5.3 anyway. It is rather simple, but it uses a lot of notation and commonplace facts from linear algebra. So let us begin with some definitions that could just as well stand in a linear algebra text: Definition 0.3. Whenever V and W are two K-vector spaces, then we denote by ρ V,W the K-linear map V W Hom K V, W, f ξ the map V W which sends every x V to f x ξ 13 The paper [BF] gives a different proof of the fact that the free Lie algebra over X is the smallest Lie subalgebra of T X containing X. This proof doesn t use the Poincaré-Birkhoff-Witt theorem but still is far from being trivial. 32

33 where V denotes the dual of V. This map ρ V,W is injective this can be proven by standard linear algebra, i. e., working with bases. In general, it is not surjective but it is surjective if dim V <. Next, a similarly elementary definition related to graded vector spaces: Definition 0.4. a Whenever V and W are two graded K-vector spaces, we let dirsum V,W denote the canonical injection Hom K V i, W i Hom K V, W which takes every family f i Hom K V i, W i of maps to the direct sum f i : V i W i this direct sum is, of course, a map V W, since V i V and W i W. b Whenever V is a graded K-vector space and W is a K-vector space, let us define a topology on the K-vector space Hom K V, W this is the space of all K-linear maps not only the graded ones from V to W as follows: Let N denote the set { 1, 0, 1, 2,...}. For every i N, let V i denote the subspace V 0 + V V i of V. For any i N and any g Hom K V i, W, let Hom K,i,g V, W denote the subset {f Hom K V, W f V i g} of Hom K V, W. Then, we define the topology on the K-vector space Hom K V, W to be the topology generated by the sets Hom K,i,g V, W with i N and g Hom K V i, W as open sets. Note that Hom K, 1,0 V, W Hom K V, W. This topology will be called the right degree topology on Hom K V, W. I am pretty sure that this topology has a different, more standard name, but I don t know it. Let us summarize a few easy-to-prove properties of this right degree topology: Let V be a graded K-vector space and W a K-vector space. First of all, the right degree topology makes Hom K V, W into a Hausdorff topological space, so it makes sense to speak of the limit of a convergent sequence. Second, the right degree topology makes Hom K V, W into a complete set, as can be easily seen. Furthermore, it is easy to see that a sequence f i of K-linear maps f i : V W converges to a K-linear map f : V W if and only if for every i N, there exists some N N such that every n N satisfies f n V i f V i. Thus, an infinite sum g i of K-linear maps g i : V W converges to a K-linear map g : V W if and only if for every i N, there exists some N N such that g N V i g N+1 V i g N+2 V i... 0 and ; i N 1 g i V i g V. i 33