DISTRIBUTION RELATION FOR BERNOULLI POLYNOMIALS ATTACHED TO FORMAL GROUP

JUNYA SATOH

Abstract. We extend a well-known distribution relation for ordinary Bernoulli polynomials to that of Bernoulli polynomials attached to formal group.

Let q be an indeterminate and let 𝔬 be the formal power series ring in q − 1 over some ℚ-algebra. Furthermore let 𝔉(X, Y) be a 1-dimensional commutative formal group defined over 𝔬 and let 𝔣(X) be an isomorphism from the additive formal group X + Y to 𝔉(X, Y). We note that there exists a unique isomorphism 𝔣_{𝔉}(X) from X + Y to 𝔉(X, Y) defined over 𝔬 such that 𝔣′_{𝔉}(0) = 1. And 𝔣(X) is equal to 𝔣_{𝔉}(cX) for some invertible element c ∈ 𝔬^{×}. Conversely for any c ∈ 𝔬^{×}, 𝔣_{𝔉}(cX) is an isomorphism from X + Y to 𝔉(X, Y). We note that we also denote 𝔣_{𝔉}(cX) by [X]_{𝔉,c}. Throughout this paper we assume that

Assumption 1.

ord_{q−1}𝔣_{𝔉}^{(n)}(0) ≥ n − 1 for all n ≥ 1,

where 𝔣_{𝔉}^{(n)} means the n-th derivative of 𝔣_{𝔉}.

We note that by this assumption 𝔉_{n}(A, B) (see Definition 1 below) and 𝔣(a) are convergent in 𝔬 for any A, B ∈ 𝔬[[X]] and a ∈ 𝔬 as formal power series (see [6, Remark 3]).

Definition 1. For each non-negative integer n we denote the expansion of 𝔉(X, Y)^{n} by

𝔉(X, Y)^{n} = ∑_{i,j≥0} \binom{n}{i, j}_{𝔉}X^{i}Y^{j} ,

and we set

𝔉_{n}(A, B) = ∑_{i,j≥0} \binom{n}{i, j}_{𝔉}a_{i}b_{j}

for any power series A = ∑_{n≥0} a_{n} \frac{X^{n}}{n!} and B = ∑_{n≥0} b_{n} \frac{X^{n}}{n!} in 𝔬[[X]]. And we define

the ∗_{𝔉}-product by

A ∗_{𝔉} B = ∑_{n≥0} 𝔉_{n}(A, B)\frac{X^{n}}{n!} .

We note that we also denote 𝔉_{n}(A, B) by (a +_{𝔉} b)^{n}.

We can prove (𝔬[[X]], +, ∗_{𝔉}) is an 𝔬-algebra (see [6, Proposition 1]). Next we extend the following map:

X^{n} ↦ c^{n} \underbrace{X ∗_{𝔉} ⋯ ∗_{𝔉} X}_{n times}

𝔬-linearly. Hence we can get a natural homomorphism from (𝔬[[X]], +, ·) to (𝔬[[X]], +, ∗_{𝔉}). We denote the image of A ∈ 𝔬[[X]] under this homomorphism by A_{𝔉,c}. And we define an analogue of power series A attached to 𝔉 and c by A_{𝔉,c}. Next we define Bernoulli numbers β_{n}(𝔉, c) and Bernoulli polynomials β_{n}(t; 𝔉, c) attached to 𝔉 and c as follows:

Definition 2. For each non-negative integer n we define the n-th Bernoulli number β_{n}(𝔉, c) and Bernoulli polynomial β_{n}(t; 𝔉, c) attached to 𝔉(X, Y) ∈ 𝔬[[X, Y]] and c ∈ 𝔬^{×} by the coefficient of \frac{X^{n}}{n!} in (\frac{X}{e^{X}−1})_{𝔉,c} and that of \frac{X^{n}}{n!} in (\frac{Xe^{tX}}{e^{X}−1})_{𝔉,c} respectively.

We note that β_{n}(0; 𝔉, c) = β_{n}(𝔉, c). And because of

(\frac{Xe^{tX}}{e^{X} − 1})_{𝔉,c} = (\frac{X}{e^{X} − 1})_{𝔉,c} ∗_{𝔉} (e^{tX})_{𝔉,c}

= (\frac{X}{e^{X} − 1})_{𝔉,c} ∗_{𝔉} e^{[t]_{𝔉,c}X}

= ∑_{n≥0} β_{n}(𝔉, c)\frac{X^{n}}{n!} ∗_{𝔉} ∑_{n≥0}[t]_{𝔉,c}^{n} \frac{X^{n}}{n!} ,

we can denote β_{n}(t; 𝔉, c) by (β(𝔉, c) +_{𝔉} [t]_{𝔉})^{n}.

To state our main result, we need to define a new formal group.

Definition 3. For any a ∈ 𝔬^{×} we define new 1-dimensional commutative formal group over 𝔬 by

𝔉_{(a)}(X, Y) = \frac{𝔉([a]_{𝔉,c}X, [a]_{𝔉,c}Y)}{[a]_{𝔉,c}}.

Let [X]_{𝔉_{(a)},d} be an isomorphism from the additive formal group X + Y to 𝔉_{(a)}(X, Y) for d ∈ 𝔬^{×}.

Lemma 1. If d = \frac{a}{[a]_{𝔉,c}} c, then we have

[X]_{𝔉_{(a)},d} = \frac{[aX]_{𝔉,c}}{[a]_{𝔉,c}} .

Proof. If we put 𝔤 = \frac{[aX]_{𝔉,c}}{[a]_{𝔉,c}}, then 𝔤 is a power series in 𝔬[[X]] and 𝔤(0) = 0, 𝔤′(0) = d. Hence it is sufficient to prove that

𝔤(𝔤^{−1}(X) + 𝔤^{−1}(Y)) = 𝔉_{(a)}(X, Y).

Now we again note [X]_{𝔉,c} = 𝔣_{𝔉}(cX). Because of

𝔤^{−1}(X) = \frac{𝔣_{𝔉}^{−1}(𝔣_{𝔉}(ac)X)}{ac} and 𝔉(X, Y) = 𝔣_{𝔉}(𝔣_{𝔉}^{−1}(X) + 𝔣_{𝔉}^{−1}(Y)),

we have

𝔤(𝔤^{−1}(X) + 𝔤^{−1}(Y)) = \frac{𝔣_{𝔉}(ac(\frac{𝔣_{𝔉}^{−1}(𝔣_{𝔉}(ac)X)}{ac} + \frac{𝔣_{𝔉}^{−1}(𝔣_{𝔉}(ac)Y)}{ac}))}{𝔣_{𝔉}(ac)}

= \frac{𝔣_{𝔉}(𝔣_{𝔉}^{−1}(𝔣_{𝔉}(ac)X) + 𝔣_{𝔉}^{−1}(𝔣_{𝔉}(ac)Y)}{𝔣_{𝔉}(ac)}

= \frac{𝔉([a]_{𝔉,c}X, [a]_{𝔉,c}Y)}{[a]_{𝔉,c}}

= 𝔉_{(a)}(X, Y).

This completes the proof of Lemma 1

Our main result of this paper is as follows:

Theorem 1. For each non-negative integer n and positive integer m we have

β_{n}(t; 𝔉, c) = \frac{[m]_{𝔉,c}^{n}}{m} ∑_{l=0}^{m−1} β_{n}(\frac{t + l}{m}; 𝔉_{(m)}, d),

where d = \frac{m}{[m]_{𝔉,c}} c

The distribution relation for the ordinary Bernoulli polynomials B_{n}(t) is

B_{n}(t) = m^{n−1} ∑_{l=0}^{m−1} B_{n}(\frac{t + l}{m} ).

By considering the generating function of B_{n}(t), the distribution relation is equivalent to

\frac{Xe^{tX}}{e^{X} − 1} = \frac{1}{m} ∑_{l=0}^{m−1} \frac{mXe^{\frac{t+l}{m} mX}}{e^{mX} − 1}

and this is a trivial equality. Hence if we apply our homomorphism to the above equality, we can get

(\frac{Xe^{tX}}{e^{X} − 1})_{𝔉,c} = \frac{1}{m} ∑_{l=0}^{m−1} (\frac{mXe^{\frac{t+l}{m} mX}}{e^{mX} − 1})_{𝔉,c}.  (1)

However in general it is not clear that the relationship of A(X)_{𝔉,c} and A(aX)_{𝔉,c} for A ∈ 𝔬[[X]] and a ∈ 𝔬. Hence we provide several lemmas.

Lemma 2. For any A, B ∈ 𝔬[[X]] we have

A([a]_{𝔉,c}X) ∗_{𝔉} B([a]_{𝔉,c}X) = (A ∗_{𝔉_{(a)}} B)([a]_{𝔉,c}X).  (2)

Proof. We put A = ∑_{n≥0} a_{n} \frac{X^{n}}{n!} and B = ∑_{n≥0} b_{n} \frac{X^{n}}{n!} . By the definition of ∗_{𝔉}-product,

the left hand side of (2) is equal to

A([a]_{𝔉,c}X) ∗_{𝔉} B([a]_{𝔉,c}X) = ∑_{n≥0} 𝔉_{n}(A([a]_{𝔉,c}X), B([a]_{𝔉,c}X))\frac{X^{n}}{n!}

= ∑_{n≥0} ∑_{i,j≥0} \binom{n}{i, j}_{𝔉}[a]_{𝔉,c}^{i+j}a_{i}b_{j} \frac{X^{n}}{n!} .

On the other hand the right hand side of (2) is equal to

(A ∗_{𝔉_{(a)}} B)([a]_{𝔉,c}X) = ∑_{n≥0} 𝔉_{(a)n}(A, B)[a]_{𝔉,c}^{n} \frac{X^{n}}{n!}

= ∑_{n≥0} ∑_{i,j≥0} \binom{n}{i, j}_{𝔉_{(a)}}a_{i}b_{j}[a]_{𝔉,c}^{n} \frac{X^{n}}{n!} .

Now by the definition of 𝔉_{(a)} and \binom{n}{i,j}_{𝔉_{(a)}}, we have

∑_{i,j≥0} \binom{n}{i, j}_{𝔉_{(a)}}X^{i}Y^{j} = 𝔉_{(a)}(X, Y)^{n}

= \frac{𝔉([a]_{𝔉,c}X, [a]_{𝔉,c}Y)^{n}}{[a]_{𝔉,c}^{n}}

= \frac{1}{[a]_{𝔉,c}^{n}} ∑_{i,j≥0} \binom{n}{i, j}_{𝔉}[a]_{𝔉,c}^{i+j}X^{i}Y^{j}.

Hence we have

\binom{n}{i, j}_{𝔉}[a]_{𝔉,c}^{i+j} = \binom{n}{i, j}_{𝔉_{(a)}}[a]_{𝔉,c}^{n}.

This completes the proof of Lemma 2.

Lemma 3. Let d = \frac{a}{[a]_{𝔉,c}} c. For any a ∈ 𝔬 and A ∈ 𝔬[[X]] we have

A(aX)_{𝔉,c} = A_{𝔉_{(a)},d}([a]_{𝔉,c}X).

Proof. It is sufficient to prove for A = X^{n}. By Lemma 2 and the mathematical induction, we have

\underbrace{[a]_{𝔉,c}X ∗_{𝔉} ⋯ ∗_{𝔉} [a]_{𝔉,c}X}_{n times} = \underbrace{X ∗_{𝔉_{(a)}} ⋯ ∗_{𝔉_{(a)}} X}_{n times}([a]_{𝔉,c}X).

This means

(aX)_{𝔉,c}^{n} = (X^{n})_{𝔉_{(a)},d}([a]_{𝔉,c}(X))

and completes the proof of Lemma 3.

Now we return to the proof of theorem. Let B(t; X) be the generating function of Bernoulli polynomials, i.e. ,

B(t; X) = \frac{Xe^{tX}}{e^{X} − 1}.

By (1) and Lemma 3 we have

B_{𝔉,c}(t; X) = \frac{1}{m} ∑_{l=0}^{m−1} (\frac{mXe^{\frac{t+l}{m} mX}}{e^{mX} − 1})_{𝔉,c}

= \frac{1}{m} ∑_{l=0}^{m−1} B(\frac{t + l}{m}; mX)_{𝔉,c}

= \frac{1}{m} ∑_{l=0}^{m−1} B_{𝔉_{(m)},d}(\frac{t + l}{m}; [m]_{𝔉,c}X).

By comparing the coefficient of \frac{X^{n}}{n!} , we cat get what we want.

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Graduate School of Human Informatics, Nagoya University. Furo-cho Chikusa-ku, Nagoya 464-8601, Japan

E-mail address: jsatoh@math.nagoya-u.ac.jp