A Bernoulli-harmonic-zeta identity for the free scalar on hyperbolic space

Abstract

For every natural number k, the combination of harmonic numbers, Bernoulli numbers and values of the Riemann zeta function at the non-positive integers displayed as eq. (C.12) of T. Nishioka and Y. Sato (arXiv:2101.02399, JHEP 05 (2021) 074) vanishes. The authors checked it numerically up to k = 100 and wrote that they do not know a proof; they use it to simplify the derivative of the spectral zeta function of a conformally coupled free scalar on even-dimensional hyperbolic space. That use is not formalized here: the statement below is the identity itself.

Definition 1.1 (The identity).

$$claim\iff (\forall k\in \mathbb{N},-(\frac{2^{-(2\cdot k+2)}}{k+1})\cdot \mathrm{harmonic}\left(2\cdot k+1\right)-(\sum _{m\in \mathrm{Icc}\left(1,k\right)}\frac{2^{-(2\cdot k+2)}\cdot (2^{2\cdot m}-2)}{k-m+1}\cdot (\frac{\mathrm{bernoulli}\left(2\cdot m\right)}{2\cdot m}))+(\sum _{j\in \mathrm{range}\left(2\cdot k+2\right)}\frac{(-1{)}^j}{2^{2\cdot k-j}}\cdot \mathrm{choose}\left(2\cdot k+1,j\right)\cdot \mathrm{harmonic}\left(j\right)\cdot \mathrm{riemannZeta}\left(-j\right))+(1-2^{-(2\cdot k+1)})\cdot \mathrm{harmonic}\left(2\cdot k+1\right)\cdot (\frac{\mathrm{bernoulli}\left(2\cdot k+2\right)}{k+1})=0)$$

Formalization. D5/S3/Quantum/FockSpace/HyperbolicScalarZetaIdentity.claim (✓ std3).

Citation. Tatsuma Nishioka; Yoshiki Sato (2021). Free energy and defect C-theorem in free scalar theory. DOI: 10.1007/JHEP05(2021)074. URL: https://arxiv.org/abs/2101.02399v5.

Commentary.

Eq. (C.12) of the paper, for every k. The sum over m runs over 1 <= m <= k (empty for k = 0) and the sum over j over range(2k + 2) = {0, …, 2k + 1}. H_n is the harmonic number harmonic(n) = 1 + 1/2 + … + 1/n (with H_0 = 0), B_n the Bernoulli number bernoulli(n) with B_1 = -1/2, and riemannZeta the Riemann zeta function, evaluated at -j; the rational numbers are read in the complex numbers. The exponents -(2k + 2), 2k - j and -(2k + 1) of 2 are integers, so 2^(2k - j) is 1/2^(j - 2k) for j > 2k.

Theorem 1.2 (Proof of the identity).

$$claim$$

Proof. Machine-checked in Lean as D5/S3/Quantum/FockSpace/HyperbolicScalarZetaIdentity.result (✓ std3). ∎

Resolves. Problems/nishioka-sato-2021-bernoulli-harmonic-zeta-identity (proved) by D5/S3/Quantum/FockSpace/HyperbolicScalarZetaIdentity.result.

Source. Repository-derived.

Acknowledgement. Tatsuma Nishioka; Yoshiki Sato (2021). Free energy and defect C-theorem in free scalar theory. DOI: 10.1007/JHEP05(2021)074. URL: https://arxiv.org/abs/2101.02399v5.

Commentary.

Replace riemannZeta(-j) by (-1)^j B_(j+1)/(j+1) and multiply by 2^(2k+2)(k+1); put N = 2k + 2 and b_m = sum over i of C(m, i) 2^i B_i. Since b_m = 2^m B_m(1/2), the value B_m(1/2) = (2^(1-m) - 1) B_m of the Bernoulli polynomial at 1/2 gives b_m = (2 - 2^m) B_m; in particular b_m = 0 for odd m. The middle sum becomes the sum of b_i (1/i + 1/(N - i)) over 1 <= i <= N - 1, and the zeta sum becomes the sum of C(N, i) 2^i B_i H_(i-1) over 1 <= i <= N. Two harmonic-binomial identities, C(n, i)(H_n - H_i) = sum over j = 1..n of C(n - j, i)/j (by Pascal’s rule and induction) and the transform of C(n, i) 2^i B_i / i into the sum of (b_j - 1)/j (by induction on n), rewrite the zeta sum through sums of b_i/i and b_(N-i)/i. These cancel against the middle sum, and what is left is -H_(N-1) - 1/N + H_N = 0 after b_N = (2 - 2^N) B_N is used.

References

• Truth anchor: D5/S3/Quantum/FockSpace/HyperbolicScalarZetaIdentity.claim
• Truth anchor: D5/S3/Quantum/FockSpace/HyperbolicScalarZetaIdentity.result