2 I like to call M the overlapping shue algebra in analogy with the well known shue algebra which is obtained as follows. Let U = ZhU ; U 2 ; : : :i b

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1 The Leibniz-Hopf Algebra and Lyndon Words Michiel Hazewinkel CWI P.O. Box 94079, 090 GB Amsterdam, The Netherlands Abstract Let Z denote the free associative algebra ZhZ ; Z2; : : :i over the integers. This algebra carries a Hopf algebra structure for which the comultiplication is Zn 7! i+j=nzi Zj. This the noncommutative Leibniz-Hopf algebra. It carries a natural grading for which gr(zn) = n. The Ditters-Scholtens theorem says that the graded dual, M, of Z, herein called the overlapping shue algebra (on the semigroup of natural numbers), is the free commutative polynomial algebra over Z with as polynomial generators certain words which are called elementary unreachable words (EUW). In this note unreachable words are shown to be precisely the (concatenation) powers of Lyndon words. More precisely it is shown that the block decomposition algorithm of [4] is in fact an algorithm for obtaining the Chen-Fox-Lyndon factorization of a word into decreasing Lyndon words. Further links are discussed between the shue algebra and the overlapping shue algebra. AMS Subject Classication (99): 6W30, 05A99, 7B0 Keywords & Phrases: Hopf algebra, Leibniz-Hopf algebra, Ditters-Scholtens theorem, shue algebra, Lyndon word, free Lie algebra, Chen-Fox-Lyndon theorem.. Introduction and statement of results Let Z = ZhZ ; Z 2 ; : : :i be the free associative algebra over the integers in the (noncommuting) indeterminates Z ; Z X 2 ; : : :. With "(Z i ) = 0, i = ; 2; : : :, the comultiplication (Z n ) = Z i Z j (.) i+j=n where Z 0 =, and the antipole (Z n ) = X i +:::+ik (?) k Z i Z i2 : : : Z ik (.2) where the sum is over all strings i ; i 2 ; : : : i k, i j 2 N, such that i + i 2 + : : : + i k = n, the algebra Z becomes a Hopf algebra. This is the noncommutative Leibniz-Hopf algebra. It is of course cocommutative. Giving Z n degree n turns Z into a graded Hopf algebra and in particular a cocommutative coalgebra, whose homogeneous components are nite dimensional free Z-modules. Its graded dual, M, is hence an algebra over Z. The Ditters-Scholtens theorem, proved in [4], says that as an algebra M is a free commutative polynomial algebra on certain generators which are (indexed by) certain words over the alphabeth N = f; 2; : : :g which are called elementary unreachable words (EUW). It is a main purpose of this note to point out that these are precisely the concatenation powers of elementary Lyndon words (see below for a denition of EUW and Lyndon word). At this point I would like to thank Guy Melancon, who rst suggested that EUW's should have something to do with Lyndon words when I lectured on the Ditters-Scholtens theorem in early December 995 at LABRI, Univ. de Bordeaux I. Work done while visiting LABRI, Univ. de Bordeaux I, Talence, France, in Dec. 995 with partial support from LABRI.

2 2 I like to call M the overlapping shue algebra in analogy with the well known shue algebra which is obtained as follows. Let U = ZhU ; U 2 ; : : :i be the free associative algebra over the integers in the (noncommuting) indeterminates U ; U 2 ; : : :. This time take (U i ) = 0; i = ; 2; : : : (U i ) = U i + U i (.3) (U i ) =?U i to make U into a graded Hopf algebra. The graded dual of U, as a coalgebra, is the socalled shue algebra, S, over Z. It is a well-known theorem that over Q, S becomes a free commutative polynomial algebra with generators that are (indexed by) Lyndon words, see e.g. [3], Cor. 5.5, p.. This is denitely not true over Z, that is, S is not free polynomial over Z. And thus the overlapping shue algebra M can be seen as a rather nicer version of the shue algebra S. For, indeed, over Q they become isomorphic, the isomorphism being determined by writing down the identity + Z t + Z 2 t 2 + : : : = exp(u t + U 2 r 2 + : : :) (.4) where t is a counting variable that commutes with everything. Thus the Ditters-Scholtens theorem implies the shue algebra theorem (without explicitely specifying a set of generators for the latter). The proof of the Ditters-Scholtens theorem by Astrid Scholtens in [4] rests on an algorithm called the block decomposition algorithm. A main result of the present note is that this algorithm is in fact an algorithm for the nding of the Chen-Fox-Lyndon factorization of a word into decreasing Lyndon words. It also seems to me to be a very ecient algorithm for doing so, much better than the standard algorithm. This, however, remains to be sorted out. The identication of elementary unreachable words with powers of Lyndon words is an immediate consequence of the theorem that the block decomposition algorithm gives the Chen-Fox-Lyndon factorization. 2. Block decompositions This section describes the block decomposition algorithm of [4]., which is the main tool for the proof of the Ditters-Scholtens theorem in [4]. Consider words in N and give N its natural ordering. 2. Lexicographical ordening on N Let v = a a 2 : : : a r, w = b b 2 : : : b s be two elements of N, i.e. two words in the alphabeth N. Then v < w (lexicographically ordering) i there is a k such that a = b ; : : : ; a k? = b k?, a k < b k, where nothing is less than any a 2 N. Thus for example ; ; 2 < ; ; 3 and ; < ; ;. 2.2 Block decomposition algorithm Consider a nonempty word w = a a 2 : : : a r. Let t be such that a = a 2 = : : : = a t 6= a t+. Let s t be such that a t+ > a ; : : : ; a s > a, a s+ a. Then the rst level block of w is a a 2 : : : a s. Now consider the sux a s+ : : : a r of w and repeat the procedure. This produces the level one block decomposition of w w = b () b() 2 : : : b (2) r : Now treat w as a word made up out of the symbols b () ; : : : ; b(2) r with the lexicographic ordering described above and repeat the procedure to obtain the level two block decomposition w = b (2) b(2) 2 : : : b (2) r 2 :

3 3 Write a i = b (0) i, r = r 0. Note that r i+ r i and that if r i+ = r i then b (i+) j = b (i) j, j = ; : : : ; r i. Thus after some time the procedure stops. The smallest i for which r i+ = r i is the complexity of w, and the nal stabilized decomposition of w w = b (c) b(c) 2 : : : b (c) r c ; c = complexity (w) is the block decomposition of w. For example 3,2, has complexity zero and block decomposition b (c) = 3; b (c) 2 = 2; b (c) 3 =. Here is another example level 0 w = ; ; 3; ; ; 3; ; ; 4; ; ; 2; ; ; 2; ; ; ; 2; 2; ; ; 4; ; 2; 2; 3; ; ; ; level w = (; ; 3)(; ; 3)(; ; 4)(; ; 2)(; ; 2)(; ; ; 2; 2)(; ; 4)(; 2; 2; 3)(; ; ; ) level 2 w = (; ; 3; ; ; 3; ; ; 4)(; ; 2; ; ; 2)(; ; ; 2; 2; ; ; 4; ; 2; 2; 3)(; ; ; ; ) and this is the block decomposition of w, which hence has complexity Elementary unreachable words A nonempty word w = a a 2 : : : a r is elementary if the greatest common divisor of a ; : : : ; a r is. The nonempty word w is unreachable if its (nal) block decomposition consists of one block. 2.4 Lyndon words A strict sux of a nonempty word w = a : : : a r is a word a i : : : a r, < i r. A Lyndon word is a word w such that w < v for each strict sux v of w. A central theorem about Lyndon words is the Chen-Fox-Lyndon theorem, [], which says that every word w has a unique factorization in decreasing Lyndon words u ; : : : ; u s. w = u u 2 : : : u s ; u i u i+ ; i = ; : : : ; s? : of the block decomposition of w are (concate- = v s i i = v i v i : : : v i (s i factors) and v i > v i+, i = ; : : : ; r?. As will be shown in section 4 the blocks b (c) ; : : : ; b(c) r nation) powers of Lyndon words, b (c) i Thus the block decomposition algorithm of [4] is a (left to right) algorithm for obtaining the Chen- Fox-Lyndon factorization of a word. (The standard algorithm is right to left). It also looks like the block algorithm is a much more ecient algorithm but that still needs to be sorted out in detail. 3. Overlapping shuffle algebra As in the introduction, let M be the graded dual of Z = ZhZ ; Z 2 ; : : :i. As an Abelian group M has as basis all words over N where hw; Z i Z i2 : : : Z ir i =( if w = i i 2 : : : i r 0 otherwise and the empty word is on Z 0 = 2 Z and zero on all nontrivial monomials in the Z i. The comultiplication on Z induces a multiplication on M, the overlapping shue product. Explicitely the overlapping shue products looks as follows. Let v = a : : : a r, w = b : : : b s be two elements of N, i.e. two basis elements of M. Then the overlapping shue product of v and w is equal to (3.) v osh w = X f;g c c 2 : : : c t (3.2) where the sum is over all t and pairs of maps f : f; : : : ; rg! f; : : : ; tg g : f; : : : ; sg! f; : : : ; tg such that f and g are order preserving and injective and Im(f) [ Im(g) = f; : : : ; tg, and where c i = a f? (i) + b g? (i); i = ; : : : ; t (3.3)

4 4 with a f? (i) = 0 if f? (i) = ; and similarly for b g? (i). The Ditters-Scholtens theorem now says that the overlapping shue algebra OSH(N) = M = w Zw with this multiplication is the free commutative polynomial algebra over Z with as polynomial generators the elementary unreachable words: M = Z[w : w 2 EUW ]. The shue algebra Sh(N) = S also has the words over N as basis, but the multiplication diers v sh w = X f;g c c 2 : : : c r+s (3.4) where f; g and c i are as before. The dierence is that t is required to be equal to r + s so that each c i is equal to an a j or b k. Obviously a shue algebra Sh(A) can be dened for any alphabeth A instead of N. Similarly an overlapping shue algebra OSh(S) is dened for any semigroup S. A basis of OSh(S) as a free abelian group over Z is formed by S, the words over S. The multiplication is given by (3.2) and (3.3) with in (3.3) the \+" replaced by the multiplication in S. I have done some preliminary exploratory calculations on OSh(S), expecially in the case that S is a free semigroup. It looks like the OSh(S) will well repay further study. 4. Chen-Fox-Lyndon factorization In this section is shown that the block decomposition of a word in fact yields the Chen-Fox-Lyndon factorization. 4.. Lemma. Let u; v 2 N be two Lyndon words and u < v (lexicographically) then uv is a Lyndon words. This is a known lemma, cf. e.g. [3], p. 06 or [2], p. 6. A slight extension is: 4.2. Lemma. Let u : : : u n be Lyndon words and let u < u n. Then u u 2 : : : u n is Lyndon. Proof. With induction on n, the case n = 2 being taken care of by lemma 4.. Now let n > 2. If u 2 < u n, then u 2 : : : u n is Lyndon by induction and u u 2 < u 2 : : : u n so that u : : : u n is Lyndon by lemma 4.. If u 2 = : : : = u n, then u u 2 is Lyndon by lemma 4. and u u 2 < u 2 = u 3 u 3 : : : u n and so u u 2 : : : u n is Lyndon by lemma Lemma. Let u; v be two Lyndon words and suppose that u r v s for some r; s 2 N. Then u v. Proof. There are three cases to distinguish. Case. length(u) = length(v). Let u = a : : : a m, v = b : : : b m. Now suppose that u r v s. There are two possibilities : (i) a = b ; : : : ; a m = b m (the rst inequality of u r v s shows up after m); (ii) a = b ; : : : ; a k?, a k < b k, k m, the rst inequality shows up at or before m. In the rst case we have u = v (and r s). In the second case u < v. Case 2. length(u) < length(v), u = a : : : a m, v = b : : : b n, m < n. Again it could happen that a = b ; : : : ; a m = b m (the rst inequality of u r < v s shows up after m) and then v = ub m+ : : : v n > u. Or it could happen that a = b ; : : : ; a k? = b k?, a k < v k for k m and then also u < v. Case 3. length(u) > length(v), u = a : : : a m, v = b : : : b n, m > n. Write m = kn + t, with 0 t n?. Then u = a : : : a n a (2) n : : : a (k) : : : a n (k) a (k+) : : : a (k+) t v = b : : : b n Now if b : : : b n < a : : : a n, then denitely not u r v s. Therefore a : : : a n b : : : b n. If a n : : : a n <

5 5 b : : : b n also u < v and we are through. There remains the case that a : : : a n = b : : : b n. Then necessarily s 2 and a (2) n : : : a (k) : : : a (k) n a (k+) : : : a (k+) t : : : b : : : b n : : : This implies that a (2) n b : : : b n. If a (2) n < b : : : b n = a () : : : a () n then the sux a (2) n : : : a (k+) t < a () : : : a () n : : : a (k+) t = u which would contradict that u is Lyndon. Hence a (2) n = b : : : b n. Continuing in this way we see that and a () a() 2 : : : a () n = a (2) n = : : : = a (k) : : : a n (k) = b : : : b n a (k+) : : : a (k+) t a () : : : a () n?t : : : b : : : b t b t+ : : : b n : : : ; It follows that a (k+) : : : a (k+) t b : : : b t = a () Lyndon. Hence a (k+) : : : a (k+) t t < n : : : a () t. If a (k+) : : : a (k+) t < b : : : b t, u would not be = b : : : b t. Thus u is of the form u = v k where is a strict prex of v, or empty. If is not empty < v < v k = u contradicting that u is Lyndon (because is a strict sux of u). Hence is empty and u = v k. But u is Lyndon and no power 2 of a word can be Lyndon. Hence in this case k = and u = v Remark. The lemma still holds if v is not necessarily Lyndon. Indeed the proof did not use that v is Lyndon Theorem. The block decomposition algorithm of section 2 yields the Chen-Fox-Lyndon factorization. More precisely, for a nonempty word w 2 N (i) All blocks formed during the block decomposition algorithm are (concatenation) power of Lyndon words (ii) If w = b (c) : : : b (c) r, b (c) i = u r i i, is the block decomposition of w, then u > u 2 > : : : > u r. Proof. With induction. At level zero all blocks are single letters and hence Lyndon. Now if b is a block at level i +, then b is of one of the forms b = c : : : c r ; c = c 2 = : : : = c r b = c : : : c k : : : c r ; c = : : : = c k ; c k+ > c ; : : : ; c r > c ; r > k where in both cases all the c i are Lyndon words. In the rst case b is a power of a Lyndon word. In the second case b is a Lyndon word by lemma 4.2. This proves (i). Now let w = b (c) : : : b(c) r ; b (c) i = u r i i (4.5) be the block decomposition of w. By (i), b (c) i (nal) block decomposition is the power of a Lyndon word u i. Because this is the b (c) > b (c) 2 > : : : > b (c) r and it now follows from lemma 4.3 that u > u 2 > : : : u r proving (ii) and that w = u n : : : u n r r is the Chen-Fox-Lyndon factorization of w (which is know to be unique) Corollary. A word is unreachable if and only if it is a power of a Lyndon word. A word is elementary unreachable if and only if it is a power of an elementary Lyndon word.

6 Remark. There is a very natural bijection between the set of Lyndon words and the set of elementary unreachable words (i.e. the powers of elementary Lyndon words.) It is given by a : : : a r 7! ((a =d) : : : (a r =d)) d where d is the greatest common divisor of a ; : : : ; a r. 5. Comparing U and Z over Q Over the rational numbers the Hopf algebras U and Z become isomorphic. To see this consider the identity + Z t + Z 2 t 2 + : : : = exp(u t + U 2 t 2 + : : :) (5.) The explicit formula for Z n in terms of the U i is Z n = X i + : : : + i k = n i j 2 N U i U i2 : : : U ik k! 5.3. Theorem. Dene : Z Q! U Q by (5.3), i.e. (Z n) is equal to the right hand side of (5.2). Then is an isomorphism of Hopf algebras. Proof. Because Z is free, certainly denes a unique homomorphism of algebras. It is also degree preserving and (Z n ) U n mod(u ; : : : ; U n? ) so that is an isomorphism. Further respects the augmentation ". It remains to show that respects the comultiplication (it is then automatic in this case that respects the antipode). Thus it remains to show that if (U i ) = U i + U i is applied to (the right hand side) of (5.2) then the result is i+j=n (Z i ) (Z j ). Now (5.2) X (U i : : : U ik ) = U a : : : U ar U b : : : U bs where r +s = k and the a ; : : : ; a r ; b ; : : : ; b s are all pairs of complementary subsequences of i ; : : : ; i k. In other words i : : : i k is one of the terms of the shue product (or merge) of a : : : a r and b : : : b s. It remains to gure out how many i : : : i k there are that yield the term U a : : : U ar U b : : : U bs. This amounts precisely to choosing r places of k (where the a ; : : : ; a r are inserted in that order, and the b ; : : : ; b s are inserted in the remaining k? r = s places in that order). This number is k!. Hence r!s! ((Z n )) = X k! U a : : : U ar U b : : : U bs X Ua : : : U ar r! U b : : : U bs s! k! r!s! = ( )( X r+s=n Z r Z s ) which proves what is desired. Another way, more conceptual, but less immediately convincing to me, is as follows. Note that U i and U j are commuting variables in QhUi QhUi. Hence A = ( U )t + ( U 2 )t 2 + : : : and B = (U )t + (U 2 )t 2 + : : : are commuting elements in QhUiQhUi[t]. Hence exp(a + B) = exp(a) exp(b) and the desired results follows by applying to (5.) Corollary. The overlapping shue algebra M = OSh(N) and the shue algebra S = Sh(N) become isomorphic over Q. If now follows as a corollary of the Ditters-Scholtens theorem that S Z Q = S Q is a free commutative polynomial algebra over Q. This does not yet specify a set of polynomial generators for S Q. It is

7 7 somewhat natural at this stage to suspect that the isomorphism given by theorem 5.3 recovers a correspondence like that of Remark 4.8 by taking suitable highest order terms. References. K.T. Chen, R.H. Fox, R.C. Lyndon, Free dierential calculus IV, Ann. Math, 68 (958), 8{ M. Lothaire (ed.), Combinatorics on words, Chapter 5, Addison-Wesley, Chr. Reutenauer, Free Lie algebras, Oxford UP, A.C.J. Scholtens, S-typical curves in non-commutative Hopf algebras, thesis, Free Univ. of A'dam, March 996.