A RECURRENCE FORMULA FOR THE q-BERNOULLI NUMBERS ATTACHED TO FORMAL GROUP

JUNYA SATOH

Abstract. Kaneko [2] proved a new recurrence formula for the Bernoulli numbers and gave two proofs. One of them was due to Don Zagier. We shall apply Zagier’s idea to the q-Bernoulli numbers attached to formal group.

1. Generalization of Kaneko’s recurrence formula

Let 𝔅 = 𝔅(X) be the generating function of the Bernoulli numbers, i.e. ,

𝔅 = \frac{X}{e^{X} − 1} ,

then it is anti-invariant under a map: 𝔅 ↦ 𝔅e^{X}, i.e. ,

𝔅(X)e^{X} = 𝔅(−X) .

Zagier gave another proof of Kaneko’s recurrence formula for the Bernoulli numbers by using this property [2]. On the other hand because of 𝔅(−X) = 𝔅(X)+X, we can see that 𝔅 is transformed to the sum of a polynomial and itself under the above map. We use the second property in order to generalize Kaneko’s recurrence formula and prove a formula for the q-Bernoulli numbers attached to formal group.

First we suppose a power series B in X which satisfies the following:

Assumption 1.

Be^{X} = B + C ,

where C is a polynomial.

If C = X, then B is equal to 𝔅, and if C = X^{2}, then B is essentially equal to the generating function of B̃_{n} which was defined in [2], i.e. ,

∑_{n≥0} B̃_{n} \frac{X^{n}}{n!} = (\frac{X^{2}}{e^{X} − 1})′ .

The starting point of our argument is the following trivial lemma:

Lemma 1. For any power series A and non-negative integer n, we have

A^{(n)}e^{X} = ∑_{i=0}^{n} \binom{n}{i}(−1)^{n−i}(Ae^{X})^{(i)} ,

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

Proof. Because of A = (Ae^{X})e^{−X}, we can get what we want.

Set A = B and compare the coefficient of \frac{X^{m}}{m!} for any non-negative integer m, then we have a generalization of Kaneko’s recurrence formula as follows:

Proposition 1. If B satisfies Assumption1, then we have

∑_{i=0}^{m} \binom{m}{i}b_{n+i} = ∑_{i=0}^{n} \binom{n}{i}(−1)^{n−i}(b_{m+i} + c_{m+i}) ,

where B = ∑_{n≥0} b_{n} \frac{X^{n}}{n!} and C = ∑_{n≥0} c_{n} \frac{X^{n}}{n!} .

If m > deg C, then we have

Corollary 1.

∑_{i=0}^{m} \binom{m}{i}b_{n+i} = ∑_{i=0}^{n} \binom{n}{i}(−1)^{n−i}b_{m+i} .

Furthermore if m = n, then we have

Corollary 2.

∑_{\begin{matrix}i=0 \\ i≢n mod 2 \end{matrix}}^{n} \binom{n}{i}b_{n+i} = 0 .

If C = X, then there is no new information about the Bernoulli numbers. But if C = X^{2}, then this is equivalent to Kaneko’s recurrence formula.

2. q-recurrence formula

In this section we shall extend results in the previous section for the q-Bernoulli numbers 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). Throughout this paper we assume that Assumption 2.

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

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]]. Then we define the ∗_{𝔉}-product by

A ∗_{𝔉} B = ∑_{n≥0} 𝔉_{n}(A, B)\frac{X^{n}}{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 call this map q-operator and denote the image of A ∈ 𝔬[[X]] under the q-operator by A_{𝔉,c}. We define a q-analogue of power series A attached to 𝔉 and c by A_{𝔉,c}. The following proposition is essential for our theory of q-analogue (see [6, Theorem 1 and Proposition 2]).

Proposition 2. For any a, b ∈ 𝔬, we have (i) (e^{aX})_{𝔉,c} = e^{𝔣(a)X} ,

(ii) e^{𝔣(a)X} ∗_{𝔉} e^{𝔣(b)X} = e^{𝔣(a+b)X} .

We define the q-Bernoulli numbers β_{n}(𝔉, c) attached to 𝔉 and c as follows:

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

We note that if 𝔉 = X + Y + (q − 1)XY and c = \frac{log q}{q−1} , then 𝔣(X) = \frac{q^{X}−1}{q−1} and β_{n}(𝔉, c) satisfies the following recurrence formula:

β_{0}(𝔉, c) = 1, (qβ(𝔉, c) + 1)^{n} − β_{n}(𝔉, c) = \begin{cases}\frac{log q}{q−1} & for n = 1 , \\ 0 & for n > 1 , \end{cases}

where we use the usual convention about replacing β(𝔉, c)^{i} by β_{i}(𝔉, c) for each non-negative integer i.

Proof. Apply the q-operator to 𝔅e^{X} −𝔅 = X, then we have 𝔅_{𝔉,c} ∗_{𝔉} e^{X} −𝔅_{𝔉,c} = cX and 𝔉_{n}(𝔅_{𝔉,c}, e^{X}) = (qβ(𝔉, c) + 1)^{n}. Hence the above recurrence formula holds.

Now we may get a q-analogue of Proposition 1 by applying the q-operator to Lemma 1, but it is unknown the commutativity of the q-operator and the derivative on 𝔬[[X]]. So we need to take another approach to get a q-analogue of Lemma 1.

Lemma 2. For any power series A, B and non-negative integer n, we have

(A ∗_{𝔉} B)^{(n)} = ∑_{i,j≥0} \binom{n}{i, j}_{𝔉}A^{(i)} ∗_{𝔉} B^{(j)} .  (1)

Proof. For any non-negative integer m, the coefficient of \frac{X^{m}}{m!} in the left hand side of (1) is equal to

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

On the other hand that in the right hand of (1) is equal to

∑_{i,j≥0} \binom{n}{i, j}_{𝔉}𝔉(A^{(i)}, B^{(j)}) = ∑_{i,j≥0} \binom{n}{i, j}_{𝔉} ∑_{k,l≥0} \binom{m}{k, l}_{𝔉}a_{i+k}b_{j+l} .

Hence it is sufficient to show that

\binom{m + n}{i, j}_{𝔉} = ∑_{\begin{matrix}0≤k≤i \\ 0≤l≤j \end{matrix}} \binom{m}{k, l}_{𝔉}\binom{n}{i − k, j − l}_{𝔉}

for all i ≥ 0 and j ≥ 0. Because of 𝔉(X, Y)^{m+n} = 𝔉(X, Y)^{m}𝔉(X, Y)^{n}, we can get what we want.

Apply this lemma to A ∗_{𝔉} e^{𝔣(1)X} and e^{𝔣(−1)X}, then we have

Lemma 3.

A^{(n)} ∗_{𝔉} e^{𝔣(1)X} = ∑_{i,j≥0} \binom{n}{i, j}_{𝔉}𝔣(−1)^{j}(A ∗_{𝔉} e^{𝔣(1)X})^{(i)} .

This is a q-analogue of Lemma 1. If B satisfies Assumption 1, by applying the q-operator, we have

B_{𝔉,c} ∗_{𝔉} e^{𝔣(1)X} = B_{𝔉,c} + C_{𝔉,c} .

If C is a polynomial, then C_{𝔉,c} is also a polynomial and deg C = deg C_{𝔉,c} (see [6, Lemma 2]). Hence we have the following:

Proposition 3. If B satisfies Assumption 1， then we have

∑_{i,j≥0} \binom{m}{i, j}_{𝔉}𝔣(1)^{j}β_{n+i} = ∑_{i,j≥0} \binom{n}{i, j}_{𝔉}𝔣(−1)^{j}(β_{m+i} + γ_{m+i})

for any non-negative integer m and n, where B_{𝔉,c} = ∑_{n≥0} β_{n} \frac{X^{n}}{n!} and C_{𝔉,c} = ∑_{n≥0} γ_{n} \frac{X^{n}}{n!} .

If m > deg C, then we have

Corollary 3.

∑_{i,j≥0} \binom{m}{i, j}_{𝔉}𝔣(1)^{j}β_{n+i} = ∑_{i,j≥0} \binom{n}{i, j}_{𝔉}𝔣(−1)^{j}β_{m+i} .

Furthermore if m = n, then we have

Corollary 4.

∑_{i,j≥0} \binom{n}{i, j}_{𝔉}\{𝔣(1)^{j} − 𝔣(−1)^{j}\}β_{n+i} = 0 .

Hence if C = X, then we get a recurrence formula for β_{n} = β_{n}(𝔉, c). On the other hand if C = X^{2}, then β_{n} is the coefficient of \frac{X^{n}}{n!} in

(\frac{X^{2}}{e^{X} − 1})_{𝔉,c} = cX ∗_{𝔉} 𝔅_{𝔉,c} = cXd_{𝔉}(𝔅_{𝔉,c}) ,

where d_{𝔉} = ∑_{i≥0} \binom{1}{i,1}_{𝔉} \frac{d^{i}}{dX^{i}} (see [6, Lemma 1]). Furthermore if 𝔉(X, Y) = X + Y + (q − 1)XY and

c = \frac{log q}{q−1} , then \frac{1}{c}d_{𝔉}(𝔅_{𝔉,c}) is equal to the generating function of Carlitz’s q-Bernoulli numbers (see the next section). This means that we get a Kaneko’s type of recurrence formula for Carlitz’s q-Bernoulli numbers.

3. X + Y + (q − 1)XY

If 𝔉(X, Y) = X + Y + (q − 1)XY and c = \frac{log q}{q−1} , then we can state results in the previous section as follows:

Corollary 5. If B satisfies Assumption 1, then we have

(i) ∑_{i=0}^{m} \binom{m}{i}q^{n+i}β_{n+i} = ∑_{i=0}^{n} \binom{n}{i}(−1)^{n−i}(β_{m+i} + γ_{m+i}) ,

(ii) ∑_{i=0}^{m} \binom{m}{i}q^{n+i}β_{n+i} = ∑_{i=0}^{n} \binom{n}{i}(−1)^{n−i}β_{m+i} if m > deg C ,

(iii) ∑_{i=0}^{n} \binom{n}{i}\{q^{n+i} − (−1)^{n−i}\}β_{n+i} = 0 if m = n > deg C ,

where B_{𝔉,c} = ∑_{n≥0} β_{n} \frac{X^{n}}{n!} and C_{𝔉,c} = ∑_{n≥0} γ_{n} \frac{X^{n}}{n!} .

Proof. If 𝔉(X, Y) = X + Y + (q − 1)XY, then, by the definition of \binom{m}{i,j}_{𝔉}, we have

\binom{m}{i, j}_{𝔉} = \binom{m}{i}\binom{i}{i + j − m}(q − 1)^{i+j−m} .  (2)

6 JUNYA SATOH

Hence for any a ∈ 𝔬, we have

∑_{i,j≥0} \binom{m}{i, j}_{𝔉}𝔣(a)^{j}β_{n+i} = ∑_{i,j≥0} \binom{m}{i}\binom{i}{i + j − m}(q − 1)^{i+j−m}𝔣(a)^{j}β_{n+i}
= ∑_{i=0}^{m} \binom{m}{i}𝔣(a)^{n−i}β_{n+i} ∑_{j=m−i}^{m} \binom{i}{i + j − m}\{(q − 1)𝔣(a)\}^{i+j−m}
= ∑_{i=0}^{m} \binom{m}{i}𝔣(a)^{n−i}β_{n+i}q^{ai} .

Hence we can get what we want.

Let β̄_{n}(𝔉, c) be the coefficient of \frac{X^{n}}{n!} in \frac{1}{c}d_{𝔉}(𝔅_{𝔉,c}), i.e. , that of c^{2} \frac{X^{n+1}}{n!} in (\frac{X^{2}}{e^{X}−1})_{𝔉,c}, then β̄_{n}(𝔉, c) satisfies the following Carlitz’s recurrence formula ([1]):

β̄_{0}(𝔉, c) = 1, q(qβ̄(𝔉, c) + 1)^{n} − β̄_{n}(𝔉, c) = \begin{cases}1 & for n = 1 , \\ 0 & for n > 1 . \end{cases}

Hence β̄_{n}(𝔉, c) is equal to the n-th Carlitz’s q-Bernoulli number. To prove this we need the following: Lemma 4. If 𝔉(X, Y) = X + Y + (q − 1)XY, then d_{𝔉} is a homomorphism on (𝔬[[X]], +, ∗_{𝔉}). Proof. In this case we can write

d_{𝔉}(A) = A + (q − 1)A′

for any power series A ∈ 𝔬[[X]]. Hence by Lemma 2 we have

d_{𝔉}(A ∗_{𝔉} B) = A ∗_{𝔉} B + (q − 1)(A ∗_{𝔉} B)′
= A ∗_{𝔉} B + (q − 1){A′ ∗_{𝔉} B + A ∗_{𝔉} B′ + (q − 1)A′ ∗_{𝔉} B′}
= (A + (q − 1)A′) ∗_{𝔉} (B + (q − 1)B′)
= d_{𝔉}(A) ∗_{𝔉} d_{𝔉}(B)

for any A and B in 𝔬[[X]]. Hence d_{𝔉} is a homomorphism on (𝔬[[X]], +, ∗_{𝔉}). Proof (Carlitz’s recurrence formula). Apply d_{𝔉} to 𝔅_{𝔉,c} ∗_{𝔉} e^{X} − 𝔅_{𝔉,c} = cX, then we have

d_{𝔉}(𝔅_{𝔉,c}) ∗_{𝔉} qe^{X} − d_{𝔉}(𝔅_{𝔉,c}) = cX − log q .

Hence β̄_{n}(𝔉, c) satisfies Carlitz’s recurrence formula.

Finally we give another proof of Corollary 5. Lemma 5. If 𝔉(X, Y) = X + Y + (q − 1)XY, then the ∗_{𝔉}-product is written by

A ∗_{𝔉} X^{n} = d_{𝔉}^{n}(A)X^{n}

for any non-negative integer n.

Proof. It is sufficient to prove for \frac{X^{i}}{i!} ∗_{𝔉} \frac{X^{j}}{j!} (i ≥ 0 and j ≥ 0). By the definition of ∗_{𝔉} and (2), we have

\frac{X^{i}}{i!} ∗_{𝔉} \frac{X^{j}}{j!} = ∑_{m≥0} \binom{m}{i, j}_{𝔉} \frac{X^{m}}{m!}

= ∑_{m=j}^{i+j} \binom{m}{i}\binom{i}{i + j − m}(q − 1)^{i+j−m} \frac{X^{m}}{m!} .

On the other hand

d_{𝔉}^{i}(\frac{X^{j}}{j!})\frac{X^{i}}{i!} = ∑_{m=0}^{i} \binom{i}{m}(q − 1)^{m} \frac{d^{m}}{dX^{m}} (\frac{X^{j}}{j!})\frac{X^{i}}{i!}

= ∑_{m=0}^{i} \binom{i}{m}\binom{i + j − m}{i}(q − 1)^{m} \frac{X^{i+j−m}}{(i + j − m)!} .

Hence we have what we want.

Remark 1. If 𝔉 = X + Y + (q − 1)XY and c = \frac{log q}{q−1} , then, by Lemma 5, we can get

A(X) ∗_{𝔉} e^{𝔣(a)} = A(q^{a}X)e^{𝔣(a)}

for any power series A ∈ 𝔬[[X]] and a ∈ 𝔬. By this we can get Corollary 5 from Lemma 1 not using Lemma 2.

Remark 2. Lemma 4 and Lemma 5 hold only for 𝔉(X, Y) = X + Y + (q − 1)XY (see [6, Lemma 1 and Proposition 4]).

References

1. L. Carlitz, q-Bernoulli numbers and polynomials, J. Duke Math. J. Vol. 15 (1948), 987-1000.

2. M. Kaneko, A recurrence formula for the Bernoulli numbers, Proc. Japan Acad. Vol. 71, Ser. A, No. 8 (1995), 192-193.

3. J. Satoh, q-analogue of Riemann’s ζ-function and q-Euler numbers, J. Number Theory 31 (1989), 346-362.

4. J. Satoh, A construction of q-analogue of Dedekind sums, Nagoya Math. J. Vol. 127 (1992), 129-143.

5. J. Satoh, Construction of q-analogue by using Stirling numbers, Japan. J. Math. Vol. 20, No. 1 (1994), 73-91.

6. J. Satoh, Another look at the q-analogue from the viewpoint of Formal groups (preprint).

Graduate School of Human Informatics, Nagoya University. Furo-cho Chikusa-ku, Nagoya 464-8601, Japan E-mail address: jsatoh@math.human.nagoya-u.ac.jp