slug: nishioka-sato-2021-bernoulli-harmonic-zeta-identity bibkey: nishioka2021freescalar doi: 10.1007/JHEP05(2021)074 url: https://arxiv.org/abs/2101.02399v5 triage: theorem motivation_gids:
- D5/S3/Quantum/FockSpace/HyperbolicScalarZetaIdentity.result
A Bernoulli–harmonic–zeta identity for the free scalar on hyperbolic space
Problem
T. Nishioka and Y. Sato (arXiv:2101.02399, hep-th; JHEP 05 (2021) 074)
compute the free energy of a conformally coupled free scalar on hyperbolic
space. In the derivative of the spectral zeta function for even dimension they
need, for every k ≥ 0,
`−2^{−2k−2}/(k+1) H_{2k+1} − Σ_{m=1}^{k} 2^{−2k−2}(2^{2m}−2)/(k−m+1) · B_{2m}/(2m)
- Σ_{j=0}^{2k+1} (−1)^j/2^{2k−j} · C(2k+1, j) H_j ζ(−j) + (1 − 2^{−2k−1}) H_{2k+1} B_{2k+2}/(k+1) = 0`
(eq. (C.12) of the published text), and state: “We confirmed (C.12) up to
k = 100 numerically. However, we do not know a proof of (C.12).” Issue #11322
fixes the reading: H_j the harmonic numbers, B_n the Bernoulli numbers
with B_1 = −1/2, ζ the Riemann zeta function.
Motivation
The identity removes all terms other than ζ'(−j) from ∂_s ζ_{H^d}(0, 1/2),
which the paper uses for the free energy on H^d and its defect C-theorem.
The follow-up on the free fermion (arXiv:2102.11468) records that the scalar
case was checked only numerically. The frozen declaration
D5/S3/Quantum/FockSpace/HyperbolicScalarZetaIdentity.result proves the
identity for every k; the physical consequence is not formalized.
Gap
Issue #11322 preregisters the reading, the route and the literature check.
arXiv v5 and the published JHEP text state the identity without proof.
INSPIRE lists 38 citing records; the 34 arXiv full texts were searched and
none proves or restates it. Boyadzhiev (arXiv:1701.00544) evaluates
Σ C(n, k) H_k B_k, a different sum.
These readings are not-found-in-searched-scope; they do not establish an
exhaustive worldwide literature search, priority, or the absence of an
independent answer.
Route
Put N = 2k + 2 and b_m = Σ_i C(m, i) 2^i B_i.
b_m = 2^m B_m(1/2) = (2 − 2^m) B_mby the value of the Bernoulli polynomial at1/2(Mathlib’sbernoulliFun_eval_half); in particularb_m = 0for oddm.- With
ζ(−j) = (−1)^j B_{j+1}/(j+1)and after multiplying by2^N (k+1), the middle sum isΣ_{1≤i≤N−1} b_i (1/i + 1/(N−i))and the zeta sum isΣ_{1≤i≤N} C(N, i) 2^i B_i H_{i−1}. C(n, i)(H_n − H_i) = Σ_{j=1}^{n} C(n−j, i)/jandΣ_i C(n, i) 2^i B_i / i = Σ_{j=1}^{n} (b_j − 1)/jturn the zeta sum into sums ofb_i/iandb_{N−i}/i, which cancel against the middle sum; what is left is−H_{N−1} − 1/N + H_N = 0.
Falsifier
The answer would change if B_1 were taken as +1/2 or ζ(0) differed from
−1/2; or if H_j were shifted: replacing H_j by H_{j+1} gives
−5/12 at k = 0.
Evidence
Exact rational arithmetic (issue #11322): the left side is 0 for
k = 0, …, 60. Dropping 2^{−2k−1} from the last coefficient gives
1/12, −11/2880, …, and no zero for k ≤ 20.
The canonical source is
D5/S3/Quantum/FockSpace/HyperbolicScalarZetaIdentity.lean. Its public
declarations are claim and result. The frozen module state has statement
identity
sha256:13a21460fdd4da0d876280b4e0f84827b2d98c97b0be77795f0d5d5a9008ec0b.
The result declaration has statement identity
sha256:fc7f2aa577e096bb0f6203e18e5592dcdc8f994af0f4e8cdca30688a856ea95c.
The Freeze event is
sha256:1cb4a9a2fe4e397bd01733f9ff42f200f41e199960e0b6d0743f11386d008f81
and has no project-level frozen prerequisites. The proof uses only the standard
axioms propext, Classical.choice and Quot.sound; no sorry,
native_decide, or new axiom.
Triage
Tier 1 external named conjecture, preregistered in issue #11322 before any
Lean. theorem; resolution proved. The public theorem has
proof_shape: content: the harmonic–binomial identities and the telescoping
are proved in the module, and the value of the Bernoulli polynomials at 1/2
is taken from Mathlib. Its escape witness is form (1), and its admission basis is open-problem-resolution. Utility
none: the identity is proved for every k.
ASSUMED-UNVERIFIED
The bounded literature check does not establish exhaustive worldwide novelty, priority, or the absence of an independent answer. The Lean kernel does not authenticate the external source or its version history.