Baxter Algebras and Hopf Algebras

BAXTER ALGEBRAS AND HOPF ALGEBRAS

GEORGE E. ANDREWS, LI GUO, WILLIAM KEIGHER, AND KEN ONO

Abstract. By applying a recent construction of free Baxter algebras, we obtain a new class of Hopf algebras that generalizes the classical divided power Hopf algebra. We also study conditions under which these Hopf algebras are isomorphic.

1. Introduction Hopf algebras have their origin in Hopf’s seminal works on topological groups in the 1940s, and have become fundamental objects in many areas of mathematics and physics. For example, they are crucial to the study of algebraic groups, Lie groups, Lie algebras, and quantum groups. In turn, these areas have provided many of the most important examples of Hopf algebras. In this paper we construct new examples of Hopf algebras. Our examples arise naturally in a combinatorial study of Baxter algebras. A Baxter algebra [?] is an algebra A with a linear operator P on A that satisfies the identity P (x)P (y) = P (xP (y)) + P (yP (x)) + λP (xy) for all x and y in A, where λ, the weight, is a fixed element in the ground ring of the algebra A. Rota [?] began a systematic study of Baxter algebras from an algebraic and combinatorial perspective and suggested that they are related to hypergeometric functions, incidence algebras and symmetric functions [?, ?]. A survey of Baxter algebras with examples and applications can be found in [?, ?], as well as in [?]. Free Baxter algebras were first constructed by Rota [?] and Cartier [?] in the category of Baxter algebras with no identity (with some restrictions on the weight and the base ring). Recently, two of the authors [?, ?] have constructed free Baxter algebras in a more general context including these classical constructions. Their construction is in terms of mixable shuffle products which generalize the well-known shuffle products of path integrals as developed by Chen [?] and Ree [?]. Here we show that a special case of the construction of these new Baxter algebras provides a large supply of new Hopf algebras. The divided power Hopf algebra is one of the classical examples of a Hopf algebra, and it is not difficult to see that this algebra is the free Baxter algebra of weight zero on the empty set. The new Hopf algebras presented here generalize this classical example. In particular, we show that the free Baxter algebra of arbitrary weight on the empty set is a Hopf algebra.

1991 Mathematics Subject Classiﬁcation. 16W30, 16W99. Key words and phrases. Free Baxter algebra, Hopf algebra, divided power. The ﬁrst and fourth authors are supported by grants from the National Science Foundation, and the fourth author is supported by Alfred P. Sloan, David and Lucile Packard, and H. I. Romnes Fellowships. 1 2 GEORGE E. ANDREWS, LI GUO, WILLIAM KEIGHER, AND KEN ONO

Here we describe the construction. Let C be a commutative algebra with identity def 1, and let λ ∈ C. Deﬁne the sextuple A = Aλ = (A, µ, η, ∆, ε, S), where ∞ M (i) A = Aλ = Can, is the free C-module on the set {an}n≥0, n=0

(ii) µ = µλ : A ⊗C A → A, m X k m+n−k m am ⊗ an 7→ λ am+n−k, m k k=0

(iii) η = ηλ : C → A, 1 7→ a0, n n−k X X k (iv) ∆ = ∆λ : A → A ⊗C A, an 7→ (−λ) ai ⊗ an−k−i, k=0 i=0   1, n = 0, (v) ε = ελ : A → C, an 7→ λ1, n = 1,  0, n ≥ 2 n n X n−3 n−v (vi) S = Sλ : A → A, an 7→ (−1) λ av. v−3 v=0 x Here for any positive or negative integer x, k is deﬁned by the generating function ∞ x X x k (1 + z) = z . It was shown in [?] that (Aλ, µλ, ηλ) is a Baxter algebra of k k=0 weight λ with respect to the operator

P : Aλ → Aλ, an 7→ an+1. The main theorem in Section ?? is

Theorem 1.1. For any λ ∈ C, Aλ is a Hopf C-algebra.

When λ = 0, we have the divided power Hopf algebra. In general, Aλ will be called the λ-divided power Hopf algebra. The divided power algebra plays an important role in several areas of mathematics, including crystalline cohomology in number theory [?], umbral calculus in combinatorics [?] and Hurwitz series in differential algebra [?]. We expect that the λ-divided power algebra Aλ introduced here will play similar roles in these areas. To describe the application to umbral calculus, we recall that a sequence {pn(x) | n ∈ N} of polynomials in C[x] is called a sequence of binomial type if n X n pn(x + y) = pn−k(x)pk(y) k k=0 in C[x, y]. The classic umbral calculus studies these sequences and their general- izations, such as Sheffer sequences and cross sequences. It is well-known that the theory of classic umbral calculus can be most conceptually described in the frame- work of Hopf algebras. Furthermore, most of the main results in umbral calculus follows from properties of the divided power Hopf algebra structure on the linear dual HomC (C[x],C), usually called the umbral algebra (see [?, ?] for details). This algebra is the completion of the divided power algebra equipped with an umbral shift operator which turns out to be a Baxter operator, making the umbral algebra BAXTER ALGEBRAS AND HOPF ALGEBRAS 3 the complete free Baxter algebra of weight zero on the empty set. See [?] for details. A program carried out there gives a generalization of the umbral calculus using the λ-divided power algebra. It is natural to ask whether these Hopf algebras are isomorphic for different values of λ. We address this question in Section ?? and obtain the following result. Theorem 1.2. Let λ, ν be in C.

(1) If (λ) = (ν), then Aλ and Aν are isomorphic Hopf algebras. (2) Suppose C is a Q-algebra. If Aλ and Aν are isomorphic Hopf algebras, then (λ) = (ν). p p (3) If Aλ and Aν are isomorphic Hopf algebras, then (λ) = (ν).

We also prove a stronger but more technical version of the third statement in Proposition ??. According to Theorem ??, for any C, there are at least two non-isomorphic λ- divided power Hopf algebras, namely A0 and A1. Furthermore, if C is a Q-algebra, then there is a one-one correspondence between isomorphic classes of λ-divided power Hopf algebras and principal ideals of C. For other examples, see Example ??.

Acknowledgements We thank Jacob Sturm for helpful discussions.

2. On λ-divided power Hopf algebras Since the constant λ ∈ C will be fixed throughout this section, we will drop the dependence on the subscript λ. In section ?? we recall the defining properties of a Hopf algebra. The remainder of the section establishes that Aλ satisfies these properties. Since two of the authors proved that (A, µ, η) is a C-algebra [?], the proof of Theorem 1.1 reduces to a step by step verification of the defining properties of a Hopf algebra.

2.1. Preliminaries. Here we recall some basic deﬁnitions and facts for later reference. All tensor products in this paper are taken over the ﬁxed commutative ring C. Recall that a cocommutative C-coalgebra is a triple (A, ∆, ε) where A is a C-module, ∆ : A → A ⊗ A and ε : A → C are C-linear maps that make the following diagrams commute.

A −→∆ A ⊗ A (2.1) ↓ ∆ ↓ id⊗∆ A ⊗ A ∆−→⊗id A ⊗ A ⊗ A

C ⊗ A ε←−⊗id A ⊗ A id−→⊗ε A ⊗ C (2.2) ∼=- ↑ ∆ %∼= A

A (2.3) ∆.&∆ τ A ⊗ A −→A,A A ⊗ A 4 GEORGE E. ANDREWS, LI GUO, WILLIAM KEIGHER, AND KEN ONO where τA,A : A ⊗ A → A ⊗ A is deﬁned by τA,A(x ⊗ y) = y ⊗ x. The C-algebra C has a natural structure of a C-coalgebra with

∆C : C → C ⊗ C, c 7→ c ⊗ 1, c ∈ C and

εC = idC : C → C.

We also denote the multiplication in C by µC . Recall that a C-bialgebra is a quintuple (A, µ, η, ∆, ε) where (A, µ, η) is a C- algebra and (A, ∆, ε) is a C-coalgebra such that µ and η are morphisms of C- coalgebras. In other words, we have the commutativity of the following diagrams.

µ A ⊗ A −→ A (2.4) (id⊗τ⊗id)(∆⊗∆) ↓ ↓ ∆ µ⊗µ A ⊗ A ⊗ A ⊗ A −→ A ⊗ A

A ⊗ A −→ε⊗ε C ⊗ C

(2.5) ↓ µ ↓ µC A −→ε C

η C −→ A

(2.6) ↓ ∆C ↓ ∆ η⊗η C ⊗ C −→ A ⊗ A

η C −→ A (2.7) id &.ε C Let (A, µ, η, ∆, ε) be a C-bialgebra. For C-linear maps f, g : A → A, the convo- lution f ? g of f and g is the composition of the maps

f⊗g µ A −→∆ A ⊗ A −→ A ⊗ A −→ A. A C-linear endomorphism S of A is called an antipode for A if

(2.8) S? idA = idA ?S = η ◦ ε. A Hopf algebra is a bialgebra A with an antipode S.

2.2. Coalgebra Properties. We verify that (A, ∆, ε) satisﬁes the axioms of a coalgebra characterized by the diagrams (??), (??) and (??). To prove (??), we only need to verify that, for each n ≥ 0,

(2.9) (∆ ⊗ id)(∆(an)) = (id ⊗ ∆)(∆(an)). Unwinding deﬁnitions on the left hand side we get

n n−k i i−` X X X X k+` (∆ ⊗ id)(∆(an)) = (−λ) aj ⊗ ai−`−j ⊗ an−k−i. k=0 i=0 `=0 j=0 BAXTER ALGEBRAS AND HOPF ALGEBRAS 5

Exchanging the third and the fourth summations, and then exchanging the second and the third summations, we obtain n n−k n−k i−j X X X X k+` (∆ ⊗ id)(∆(an)) = (−λ) aj ⊗ ai−`−j ⊗ an−k−i. k=0 j=0 i=j `=0 Replacing i by i + j gives us n n−k n−k−j i X X X X k+` (∆ ⊗ id)(∆(an)) = (−λ) aj ⊗ ai−` ⊗ an−k−i−j. k=0 j=0 i=0 `=0 Exchanging the third and the fourth summations, followed by a substitution of i by i + `, gives us n n−k n−k−j n−k−j−` X X X X k+` (∆ ⊗ id)(∆(an)) = (−λ) aj ⊗ ai ⊗ an−k−i−`−j. k=0 j=0 `=0 i=0 We get the same expression after unwinding deﬁnitions on the right hand side. This proves equation (??). Before proving the commutativity of diagram (??) we display a lemma. Lemma 2.1. For any integers n and ` with n ≥ ` ≥ 0, we have

n−` X 1, ` = n, (2.10) (−λ)kε(a ) = n−`−k 0, ` < n. k=0

Proof. When ` = n, we have n−` X k (−λ) ε(an−`−k) = ε(a0) = 1. k=0 When ` < n, we have n−` X k n−`−1 n−` (−λ) ε(an−`−k) = (−λ) ε(a1) + (−λ) ε(a0). k=0

By the deﬁnition of ε(an), the right hand side is (−λ)n−`−1λ1 + (−λ)n−`1 = 0. We can now prove (??). Consider the left triangle in (??). For each n ≥ 0, we have

n n−k X X k (ε ⊗ id)(∆(an)) = (−λ) ε(ai) ⊗ an−k−i. k=0 i=0 A substitution i = n − k − ` and then an exchange of the order of summations give us n n−` X X k (ε ⊗ id)(∆(an)) = ( (−λ) ε(an−k−`)) ⊗ a` `=0 k=0 6 GEORGE E. ANDREWS, LI GUO, WILLIAM KEIGHER, AND KEN ONO which is 1 ⊗ an by Lemma ??. This proves the commutativity of the left triangle in (??). The proof of the right triangle in (??) is similar. The cocommutativity of diagram (??) is easy to verify:

n n−k X X k τA,A(∆(an)) = (−λ) an−k−i ⊗ ai. k=0 i=0 After replacing i by n − k − j, we see that it is the same as ∆(an). Hence we have shown that (A, ∆, ε) is a cocommutative C-coalgebra. 2.3. Compatibility. We now prove that the algebra and coalgebra structures on A are compatible so that they give a bialgebra structure on A. Since ε(η(1)) = ε(1) = ε(a0) = 1 = id(1), we have veriﬁed the commutativity of diagram (??). We also have

(η ⊗ η)(∆C (1)) = (η ⊗ η)(1 ⊗ 1) = a0 ⊗ a0 and 0 0−k X X k ∆(η(1)) = ∆(a0) = (−λ) ai ⊗ a0−k−i = a0 ⊗ a0. k=0 i=0 This proves the commutativity of diagram (??). We next prove the commutativity of diagram (??). We have   1 ⊗ 1, (m, n) = (0, 0),   λ1 ⊗ 1, (m, n) = (1, 0), (ε ⊗ ε)(am ⊗ an) = 1 ⊗ λ1, (m, n) = (0, 1),   λ1 ⊗ λ1, (m, n) = (1, 1),  0, m ≥ 2 or n ≥ 2. So   1, (m, n) = (0, 0),  λ1, (m, n) = (1, 0) or (0, 1), µC ((ε ⊗ ε)(am ⊗ an)) = 2  λ 1, (m, n) = (1, 1),  0, m ≥ 2 or n ≥ 2. On the other hand, we have m ! X i m+n−i m ε(µ(am ⊗ an)) = ε λ am+n−i m i i=0 m X i m+n−i m = λ ε(am+n−i). m i i=0 So when (m, n) = (0, 0), we have

ε(µ(a0 ⊗ a0)) = ε(a0) = 1. When (m, n) = (0, 1), we have 0 X i 1−i 0 ε(µ(a0 ⊗ a1)) = λ ε(a1−i) = ε(a1) = λ1. 0 i i=0 BAXTER ALGEBRAS AND HOPF ALGEBRAS 7

By the commutativity of the multiplication µ in A, we also have

ε(µ(a1 ⊗ a0)) = λ1. When (m, n) = (1, 1), we have 1 X i 2−i 1 ε(µ(a1 ⊗ a1)) = λ ε(a2−i) 1 i i=0 0 2 1 1 1 = λ ε(a2) + λ ε(a1) 1 0 1 1 = λ21.

When n ≥ 2, we have m + n − i ≥ 2 for 0 ≤ i ≤ m. Thus ε(am+n−i) = 0 and ε(µ(am ⊗ an)) = 0. The same is true when m ≥ 2 by the commutativity of µ. Thus we have veriﬁed the commutativity of diagram (??). The rest of this section is devoted to the veriﬁcation the commutativity of diagram (??). In other words, we want to prove the identity

(2.11) ∆(µ(am ⊗ an)) = (µ ⊗ µ)(id ⊗ τA,A ⊗ id)(∆ ⊗ ∆)(am ⊗ an), for any m, n ≥ 0. The left hand side can be simpliﬁed as follows.

m m+n−i m+n−i−k X X X k i+k m+n−i m (−1) λ aj ⊗ am+n−i−k−j m i i=0 k=0 j=0 m+n m+n−i m+n−i−k X X X k i+k m+n−i m = (−1) λ aj ⊗ am+n−i−k−j m i i=0 k=0 j=0 m = 0 for i > m i m+n m+n−j m+n−j−k X X X k i+k m+n−i m = (−1) λ aj ⊗ am+n−i−k−j m i j=0 k=0 i=0 (exchanging the ﬁrst and third summations ) m+n m+n−i m+n−j−k X X X k m+n−j−` j+k+` m = (−1) λ aj ⊗ a` m m+n−j−k−` j=0 k=0 `=0 (letting i = m + n − j − k − ` in the third sum) m+n m+n−j m+n−j−` ! X X m+n−j−` X k j+k+` m = λ (−1) aj ⊗ a` m m+n−j−k−` j=0 `=0 k=0 (exchanging the second and third summations ). We next simplify the right hand side of equation (??). Unwinding deﬁnitions we see that the right hand side is

m m−k n n−` i m−k−i X X X X X X (−1)k+`λk+`+u+v k=0 i=0 `=0 j=0 u=0 v=0 i+j−u i m+n−k−i−`−j−v m−k−i × ai+j−u ⊗ am+n−k−i−`−j−v. i u m−k−i v 8 GEORGE E. ANDREWS, LI GUO, WILLIAM KEIGHER, AND KEN ONO

By substitutions b = i + j − u, e = m + n − k − i − j − ` − v where we treat u and v as the variables, we obtain

m m−k n n−` i+j n+m−i−j−k−` X X X X X X (−1)k+`λm+n−b−e k=0 i=0 `=0 j=0 b=j e=n−`−j b i e m−k−i × ab ⊗ ae. i i+j−b m−k−i m+n−i−j−k−`−e Because of the nature of the summation limits, we cannot yet exchange the order of the summations as we did for the left hand side of equation (??). But we have Lemma 2.2. m m−k n n−` i+j n+m−i−j−k−` X X X X X X (−1)k+`λm+n−b−e k=0 i=0 `=0 j=0 b=j e=n−`−j b i e m−k−i × ab ⊗ ae i i+j−b m−k−i m+n−i−j−k−`−e

m+n m+n m+n m+n m+n m+n X X X X X X = (−1)k+`λm+n−b−e k=0 i=0 `=0 j=0 b=0 e=0 b i e m−k−i × ab ⊗ ae. i i+j−b m−k−i m+n−i−j−k−`−e Proof. Note that we have m, n, b, e ≥ 0 by assumption. Also for any integers x, y x with x ≥ 0 and y < 0 or x ≥ 0 and y > x, we have y = 0. So

e k > m ⇒ m − k − i < 0 ⇒ = 0. m−k−i This shows that we can replace the ﬁrst sum on the left hand side of the equation in the lemma by the ﬁrst sum on the right hand side. Similarly, we have

e i > m − k ⇒ m − k − i < 0 ⇒ m−k−i = 0, e ` > n ⇒ n − ` − j < 0 ⇒ n−`−j = 0, e j > n − ` ⇒ n − ` − j < 0 ⇒ n−`−j = 0, b b < j ⇒ j = 0, i b > i + j ⇒ i + j − b < 0 ⇒ i+j−b = 0, e e < n − j − ` ⇒ n−j−` = 0, m−k−i e > m + n − i − j − k − ` ⇒ m+n−i−j−k−`−e = 0. Considering in addition

b i b j = i i+j−b j i+j−b BAXTER ALGEBRAS AND HOPF ALGEBRAS 9 and e m−k−i e n−`−j = , m−k−i m+n−i−j−k−`−e n−`−j m+n−i−j−k−`−e we see that each of the other sums on the left hand side of the equation can be replaced by the corresponding sum on the right hand side of the equation. This proves the lemma.

Continuing with the proof of the commutativity of the diagram (??), we see that the limits of the sums on the right hand side of the equation in Lemma ?? are given by the same constants. Thus we can exchange the order of the summations and get

m+n m+n m+n m+n m+n m+n X X X X X X (−1)k+`λm+n−b−e k=0 i=0 `=0 j=0 b=0 e=0 b i e m−k−i × ab ⊗ ae i i+j−b m−k−i m+n−i−j−k−`−e

m+n m+n m+n m+n m+n m+n X X X X X X = (−1)k+`λm+n−b−e b=0 e=0 i=0 j=0 k=0 `=0 b i e m−k−i × ab ⊗ ae. i i+j−b m−k−i m+n−i−j−k−`−e Now by the same argument as in the proof of Lemma ??, we obtain

m+n m+n m+n m+n m+n m+n X X X X X X (−1)k+`λm+n−b−e b=0 e=0 k=0 i=0 `=0 j=0 b i e m−k−i × ab ⊗ ae i i+j−b m−k−i m+n−i−j−k−`−e

m+n m+n−b  b b m−i m+n−i−j−k−e X X X X X X k+` m+n−b−e =  (−1) λ b=0 e=0 i=0 j=b−i k=0 `=0

b i e m−k−i × ab ⊗ ae. i i+j−b m−k−i m+n−i−j−k−`−e Comparing the right hand side of the above equation with the simpliﬁed form of the left hand side of equation (??), we see that to prove equation (??), we only need to prove

(2.12) m+n−b−e X b+e+k m (−1)k m m+n−b−e−k k=0 b b m−i m+n−i−j−k−e X X X X b i e m−k−i = (−1)k+` i i+j−b m−k−i m+n−i−j−k−`−e i=0 j=b−i k=0 `=0 for all m, n, b, e ≥ 0 with b + e ≤ m + n. 10 GEORGE E. ANDREWS, LI GUO, WILLIAM KEIGHER, AND KEN ONO

Using the substitutions   j = b − i + c, k = m − i − a,  ` = n − b − c − e + i + a − d on the right hand side of equation (??) gives us

b b m−i m+n−i−j−k−e X X X X b i e m−k−i (−1)k+` i i+j−b m−k−i m+n−i−j−k−`−e i=0 j=b−i k=0 `=0 b i m−i n−b−c−e+i+a X X X X b i e a = (−1)m+n−e−b−c−d . i c a d i=0 c=0 a=0 d=0 Thus to prove equation (??), and hence equation (??), we only need to prove the following theorem. Theorem 2.3. If m, n, b and e are nonnegative integers satisfying m + n ≥ b + e, then m+n−b−e X b + e + k m (−1)m+n−e−b (−1)k m m + n − b − e − k k=0

b i m−i n−e−b+i−c+a X X X X b i e a = (−1)c+d . i c a d i=0 c=0 a=0 d=0 Proof. By replacing k by m + n − b − e − k in the sum on the left hand side of the above equation and using the fact that [?]

j X a a − 1 (−1)d = (−1)j d j d=0 −1 (Notice that this means that = (−1)j), the problem is reduced to showing j that b i m−i X X X b i e a−1 (−1)n−e−b+i+a i c a n−e−b+i−c+a i=0 c=0 a=0 (2.13) m+n−b−e X m+n−k m = (−1)k m k k=0 Using the classical summation of Vandermonde [?] m X m n m + n = k i − k i k=0 in the summation on c on the left hand side of (??), we ﬁnd that (??) reduces to BAXTER ALGEBRAS AND HOPF ALGEBRAS 11

b m−i X X b e a + i − 1 (−1)n−e−b+i+a i a n − e − b + i + a i=0 a=0 (2.14) m+n−b−e X m + n − k m = (−1)k . m k k=0

Now set T = a + i. By the Vandermonde again, we ﬁnd that the left hand side of (??) becomes

b X X b e T − 1 (−1)n−e−b+T i T − i n − e − b + T T ≤m i=0 X b + e T − 1 = (−1)n−e−b+T . T n − e − b + T T ≤m

Therefore, it suﬃces to prove, by letting H = b + e in (??), the following identity

m+n−H X m + n − k m (−1)k m k k=0 (2.15) X H T − 1 = (−1)n−H+T . T n − H + T T ≤m

For brevity, write (??) as

(2.16) L(m, n) = R(m, n).

Clearly we have the following

(2.17) L(m, 0) = L(0, n) = R(0, n) = R(m, 0) = 1, H m − 1 (2.18) R(m, n) − R(m − 1, n) = (−1)n+m−H . m n + m − H

Therefore, we have

R(m, n) − R(m − 1, n) − R(m, n − 1) + R(m − 1, n − 1) H m − 1 m − 1 = (−1)n+m−H + m n + m − H n − 1 + m − H (2.19) H m = (−1)n+m−H . m n + m − H

Using the fact that

m + n − k m n m + n − k = , m k k n 12 GEORGE E. ANDREWS, LI GUO, WILLIAM KEIGHER, AND KEN ONO we ﬁnd that

L(m, n) − L(m − 1, n)

H m = (−1)m+n−H m m + n − H

m+n−H−1 X n m + n − k m + n − k − 1 + (−1)k − k n n k=0 H m = (−1)m+n−H m m + n − H

m+n−H−1 X n m + n − k − 1 + (−1)k k m − k k=0 H m = (−1)m+n−H m m + n − H

m+n−H−1 X n − 1 n − 1 m + n − k − 1 + (−1)k + k k − 1 m − k k=0 H m = (−1)m+n−H + L(m, n − 1) m m + n − H

m+n−H−2 X n − 1 m + n − k − 2 + (−1)k+1 k m − 1 − k k=0 H m = (−1)m+n−H + L(m, n − 1) − L(m − 1, n − 1). m m + n − H

Therefore,

L(m, n) − L(m − 1, n) − L(m, n − 1) + L(m − 1, n − 1) (2.20) H m = (−1)n+m−H . m n + m − H

Thus we see that (??) and (??) show that L(m, n) and R(m, n) satisfy the same bilinear recurrence. Since they have the same initial values, we have that L(m, n) = R(m, n) for all nonnegative n and m.

2.4. Existence of an Antipode. We now show that the linear map S deﬁned in Section ?? is an antipode on the bialgebra Aλ, thus making the bialgebra into a Hopf algebra. Since A is commutative, we only need to prove

µ ◦ (S ⊗ id) ◦ ∆ = η ◦ ε. BAXTER ALGEBRAS AND HOPF ALGEBRAS 13

¿From the definitions of ε and η, we have   a0, n = 0, η ◦ ε(an) = λa0, n = 1,  0, n > 1. Thus to prove that S is an antipode on A, we only need to show   a0, n = 0, (2.21) µ(S ⊗ id(∆(an))) = λa0, n = 1,  0, n > 1. Recall that we have defined the C-linear map S : A → A by n n X n−3 n−v S(an) = (−1) λ av. v−3 v=0 Using this and the definitions of µ and ∆, we have

µ(S ⊗ id(∆(an))) n n−k i v X X X X k+i k+i−v+` i−3 n−k−i+v−` v = (−1) λ an−k−i+v−`. v−3 v ` k=0 i=0 v=0 `=0 With a change of variable ` = n − k − i + v − w (w = n − k − i + v − `)), we get n n−k i n−k−i+v X X X X k+i n−w i−3 w v µ(S⊗id(∆(an))) = (−1) λ aw. v−3 v n−k−i+v−w k=0 i=0 v=0 w=n−k−i As in the proof of diagram (??), we want constant limits for the summations. x Since y = 0 for integers x, y with x ≥ 0 and either y < 0 or y > x, we have i > n − k ⇒ n − k − i < 0

( n − k − i + v − w < 0 ⇒ v = 0, if v ≤ w, ⇒ n−k−i+v−w w v = 0, if v > w. w < n − k − i ⇒ n − k − i − w > 0

v ⇒ n − k − i + v − w > v ⇒ n−k−i+v−w = 0, w > n − k − i + v ⇒ n − k − i + v − w < 0

v ⇒ n−k−i+v−w = 0. Therefore the last nested sum can be replaced by n n i n X X X X k+i n−w i−3 w v (−1) λ aw v−3 v n−k−i+v−w k=0 i=0 v=0 w=0 n " n n i # X n−w X X X k+i i−3 w v = λ (−1) aw. v−3 v n−k−i+v−w w=0 k=0 i=0 v=0 Thus to prove equation (??) we only need to prove 14 GEORGE E. ANDREWS, LI GUO, WILLIAM KEIGHER, AND KEN ONO

Theorem 2.4. For any integers n ≥ w ≥ 0, we have n n i X X X i−3 w v (−1)k+i v−3 v n−k−i+v−w k=0 i=0 v=0 (2.22) 1, if (n, w) = (0, 0) or (1, 0), = 0, otherwise . Proof. The proof of this identity requires only three facts [?]: −A A + i − 1 (2.23) = (−1)i , i i X A B A + B (2.24) = , j C − j C j≥0 N X N 1, if N = 0, (2.25) (−1)j = j 0, otherwise. j=0 Hence we have that n n i X X X i − 3 w v (−1)k+i v − 3 v n − k − i + v − w k=0 i=0 v=0 n n n X X X i − 3 w v = (−1)k+i v − 3 v n − k − i + v − w k=0 v=0 i=v n n n−v X X X i + v − 3 w v = (−1)k+i+v . v − 3 v n − k − i − w k=0 v=0 i=0 Now by (??) and (??) we ﬁnd that this sum equals

n n n−v X X w X 2 − v v (−1)k+v v i n − k − i − w k=0 v=0 i=0 n n X X 2 w = (−1)k+v . n − k − w v k=0 v=0 By (??) we ﬁnd that this sum equals  0, if w > 0,  n X 2 (−1)k , if w = 0.  n − k  k=0   0, if w > 0,  1, if w = 0 and n = 0, =  1, if w = 0 and n = 1,  0, if n > 1.

This completes the proof of Theorem 1.1. BAXTER ALGEBRAS AND HOPF ALGEBRAS 15

3. Isomorphisms between λ-divided power Hopf algebras We will prove the three statements in Theorem ?? by proving Proposition ??, ?? and ??. To distinguish elements in λ-divided power Hopf algebras with various values of λ, we write M Aλ = Caλ,n n≥0 where {aλ,n} is the standard basis.

Proposition 3.1. Let λ, ν be in C. If (λ) = (ν), then Aλ and Aν are isomorphic Hopf algebras. Proof. Suppose (λ) = (ν). Then λ = ων for a unit ω in C. Deﬁne n ϕ : Aλ → Aν , aλ,n 7→ ω aν,n. Then it is straightforward to verify that ϕ is a homomorphism of Hopf algebras. More precisely, we have

ϕ ◦ µλ = µν ◦ (ϕ ⊗ ϕ),

(ϕ ⊗ ϕ) ◦ ∆λ = ∆ν ◦ ϕ,

ϕ ◦ Sλ = Sν ◦ ϕ. −n It is also easily seen that ϕ has inverse given by sending aν,n to ω aλ,n.

Proposition 3.2. Let C be a Q-algebra, and let λ, ν be in C. Then Aλ and Aν are isomorphic Hopf algebras if and only if (λ) = (ν).

Proof. By Proposition ??, we only need to prove that if Aλ and Aν are isomorphic Hopf algebras, then (λ) = (ν). ∼ Suppose C is a Q-algebra and Aλ = Aν . Let ϕ : Aλ → Aν be a Hopf algebra isomorphism. By [?, Proposition 3.2], both Aλ and Aν are isomorphic to C[x] as C- P algebras with aλ,1 and aν,1 as the generators. It follows that ϕ(aλ,1) = i≥1 ciaν,i ∗ −1 P ∗ with c1 ∈ C . Likewise, ϕ (aν,1) = i≥1 diaλ,i with d1 ∈ C . Then we have

((ϕ ⊗ ϕ) ◦ ∆λ)(aλ,1) = (ϕ ⊗ ϕ)(1 ⊗ aλ,1 + aλ,1 ⊗ 1 − λ ⊗ 1)     X X = 1 ⊗  ciaν,i +  ciaν,i ⊗ 1 − λ ⊗ 1 i≥1 i≥1 = −λ ⊗ 1 + higher degree terms and

(∆ν ◦ ϕ)(aλ,1)   X = ∆ν  ciaν,i i≥1

= c1(1 ⊗ aν,1 + aν,1 ⊗ 1 − ν ⊗ 1) 2 + c2(1 ⊗ aν,2 + aν,1 ⊗ aν,1 + aν,2 ⊗ 1 − ν(1 ⊗ aν,1 + aν,1 ⊗ 1) + ν ⊗ 1) + ...   X i−1 = −ν  (−ν) ci (1 ⊗ 1) + higher degree terms. i≥1 16 GEORGE E. ANDREWS, LI GUO, WILLIAM KEIGHER, AND KEN ONO

Here the degree is the natural one given by the standard basis aν,i. That is, the degree of aν,i ⊗ aν,j is (i, j) with the lexicographic order. Since ϕ is a Hopf algebra isomorphism, we have   X i−1 λ = ν  (−ν) ci . i≥1 So Cλ ⊆ Cν. Considering ϕ−1, we similarly get   X i−1 ν = λ  (−λ) di . i≥1 So Cν ⊆ Cλ. Then Cν = Cλ. For future reference, we record the following easy corollary of Proposition ??. ∼ Corollary 3.3. Let C be a Q-algebra and let ν be a unit in C. Then Aν =6 A0. Proposition 3.4. Let λ and ν be in C. If either (1) ν is not contained in p(λ), or (2) there is a prime number p such that νp−1 is not contained in (λ, p), then Aλ is not isomorphic to Aν as C-algebras The third statement of Theorem ?? follows from the ﬁrst case of the proposition. To display the utility of the second case, we provide the following examples. r ∼ Example 3.5. (1) Let C = Fp[x]/(x ) with r ≥ p. Then Ax¯ =6 A0. Herex ¯ is r the image of x in Fp[x]/(x ). r ∼ (2) Let C = Z[x]/(x ) with r ≥ 2. Then Ax¯ =6 A0 (taking p = 2). Proof of Proposition ??. To consider the case when ν is not contained in p(λ), we begin with a special situation. Lemma 3.6. Let C be reduced with characteristic p for a prime number p. Let ν ∼ be a unit in C. Then A0 =6 Aν . L Proof. By [?, Theorem 4.8], the nilradical N(A0) of A0 is n≥1 Ca0,n. By [?, p Lemma 4.9], we have N(A0) = 0. So we only need to show that Aν has no non-zero nilpotent element x with xp = 0. P∞ Let there be such an element x ∈ Aν . Write x = i=k ciaν,i with ck 6= 0. Then by the product formula in Aν (see (ii) in the introduction), ∞ p p p M x = ckaν,k + a term in Caν,i i=k+1 and ∞ p (p−1)k M aν,k = ν aν,k + a term in Caν,i. i=k+1 Therefore, ∞ p p p−1 M x = ckν aν,k + a term in Caν,i. i=k+1 p p−1 p So we must have ckν = 0. Since ν is a unit, we have ck = 0. Since C is reduced, we have ck = 0. This is a contradiction. BAXTER ALGEBRAS AND HOPF ALGEBRAS 17

Continuing with the proof of Proposition ??, we now suppose ν 6∈ p(λ), and ∼ n Aλ = Aν . Then ν is not nilpotent and (λ) ∩ {ν } is empty. By [?, Example 4], the image λ/1 of λ in the localization {νn}−1C is not a unit. Let P be a maximal n −1 ideal of {ν } C containing λ/1. Then the image λe of λ/1 (resp. of νe of ν/1) in n −1 ∼ the ﬁeld {ν } C/P is zero (resp. a unit). From Aλ = Aν , we have n −1 ∼ n −1 ({ν } C/P ) ⊗C Aλ = ({ν } C/P ) ⊗C Aν . That is A ∼ A . This is a contradiction by Corollary ?? (when the characteristic λe = νe of the ﬁeld {νn}−1C/P is 0) and Lemma ?? (when the characteristic is a prime number p). We next consider the case when there is a prime number p such that νp−1 is not contained in the ideal (λ, p) of C. We again start with a special situation. p−1 Lemma 3.7. Let C be of characteristic p. If ν is not zero, then Aν is not isomorphic to A0.

Proof. Suppose Aν is isomorphic to A0. Let ϕ : A0 → Aν be an isomorphism of C- L L algebras. By [?, Theorem 4.8], the nilradical N(A0) of A0 is N(C) n≥1 Ca0,n . On the other hand, by the product formula for aν,n ∈ Aν and the fact that L L n≥1 Caν,n is an ideal of Aν , we have N(Aν ) = N(C) L for an ideal L of Aν L contained in n≥1 Caν,n. Then ϕ induces an isomorphism between the C/N(C)- ∼ ∼ L algebras A0/N(A0) = C/N(C) and Aν /N(Aν ) = C/N(C) (⊕n≥1Caν,n) /L. It L p follows that L = n≥1 Caν,n. Thus aν,1 is nilpotent. By [?, Lemma 4.9], N(A0) p p p p is contained in N(C) . So we must have aν,1 ∈ N(C) . In particular, aν,1 ∈ C. But this cannot be true since ∞ p p−1 M aν,1 = ν aν,1 + a term in Caν,i i=k+1 p−1 and ν 6= 0. In the general case, consider C = C/(λ, p) and let λ¯ (resp.ν ¯) be the image of λ ∼ ∼ (resp. ν) in C. Then by Lemma ??, Aν¯ =6 Aλ¯. So Aν =6 Aλ. References [1] G. Baxter, An analytic problem whose solution follows from a simple algebraic identity, Pacific J. Math. 10 (1960), 731-742. [2] P. Berthelot and A. Ogus, Notes on Crystalline Cohomology, Princeton University Press, 1978. [3] P. Cartier, On the structure of free Baxter algebras, Adv. in Math. 9 (1972), 253-265. [4] K.T. Chen, Integration of paths, geometric invariants and a generalized Baker- Hausdorff formula, Ann. of Math. 65 (1957), 163-178. [5] L. Guo, Properties of Baxter algebras, Adv. in Math. 151 (2000), 346-374. [6] L. Guo, Baxter algebras and the umbral calculus, Adv. in Appl. Math. 27 (2001), 405-426. [7] L. Guo, Baxter algebras and differential algebras, In: Differential Algebra and Related Topics, World Scientific (2002), 281-305. [8] L. Guo and W. Keigher, Baxter algebras and shuffle products, Adv. in Math. 150 (2000), 117-149. [9] L. Guo and W. Keigher, On free Baxter algebras: completions and the internal construction, Adv. in Math. 151 (2000), 101-127. [10] W. Keigher, On the ring of Hurwitz series, Comm. Algebra 25 (1997), 1845-1859. [11] H. Matsumura, Commutative Ring Theory, Cambridge University Press, 1994. 18 GEORGE E. ANDREWS, LI GUO, WILLIAM KEIGHER, AND KEN ONO

[12] W. Nichols and M. Sweedler, Hopf algebras and combinatorics, In: Umbral Calculus and Hopf Algebras, Contemporary Mathematics 6, Amer. Math. Soc. (1982), 49-84. [13] R. Ree, Lie elements and an algebra associated with shuﬄes, Ann. Math. 68 (1958), 210-220. [14] J. Riodan, Combinatorial Identities, Wiley, 1968. [15] S. Roman, The Umbral Calculus, Academic Press, Orlando, FL, 1984. [16] S. Roman and G.-C. Rota, The umbral calculus, Adv. Math. 27(1978), 95–188. [17] G.-C. Rota, Baxter algebras and combinatorial identities I, II, Bull. Amer. Math. Soc. 75 (1969), 325–329, 330–334. [18] G.-C. Rota, Baxter operators, an introduction, In: Gian-Carlo Rota on Combina- torics, Introductory Papers and Commentaries, Joseph P.S. Kung, Editor, Birkh¨auser, Boston, 1995. [19] G.-C. Rota, Ten mathematics problems I will never solve, Invited address at the joint meeting of the American Mathematical Society and the Mexican Mathematical Society, Oaxaca, Mexico, December 6, 1997. DMV Mittellungen Heft 2, 1998, 45–52. [20] D. Stanton and D. White, Constructive Combinatorics, Springer-Verlag, New York, 1986.

Department of Mathematics, Pennsylvania State University, University Park, PA 16802 E-mail address: [email protected]

Department of Mathematics and Computer Science, Rutgers University at Newark, Newark, NJ 07102 E-mail address: [email protected]

Department of Mathematics, University of Wisconsin, Madison, WI 53706 E-mail address: [email protected]