Hilbert–Pólya conjecture

Let ζ\zeta denote the Riemann zeta function, that is, the meromorphic continuation to C\mathbb{C} of the function ζ(s)=n=1ns\zeta(s)=\sum_{n=1}^{\infty}n^{-s}, Res>1\operatorname{Re} s>1. Call ρC\rho\in\mathbb{C} a non-trivial zero of ζ\zeta if ζ(ρ)=0\zeta(\rho)=0 and 0<Reρ<10<\operatorname{Re}\rho<1, and let ZZ be the multiset of non-trivial zeros, each zero counted with its multiplicity as a zero of ζ\zeta. For ρZ\rho\in Z define

tρ=i(ρ12)C,so thatρ=12+itρ.t_{\rho}=-i\left(\rho-\tfrac{1}{2}\right)\in\mathbb{C},\qquad\text{so that}\qquad \rho=\tfrac{1}{2}+it_{\rho}.

There exist a separable complex Hilbert space H\mathcal{H}, a dense linear subspace DHD\subseteq\mathcal{H}, and a self-adjoint (possibly unbounded) operator H ⁣:DHH\colon D\to\mathcal{H} such that

spec(H)={tρ  :  ρZ},\operatorname{spec}(H)=\left\{\,t_{\rho}\;:\;\rho\in Z\,\right\},

the spectrum consisting purely of eigenvalues, and such that for every tspec(H)t\in\operatorname{spec}(H) the dimension of the eigenspace ker(Htid)\ker(H-t\,\mathrm{id}) equals the multiplicity of 12+it\tfrac{1}{2}+it as a zero of ζ\zeta.

Equivalently, in terms of the operator L=12+iHL=\tfrac{1}{2}+iH defined on DD: spec(L)=Z\operatorname{spec}(L)=Z with matching multiplicities, where H=i(L12)H=-i\left(L-\tfrac{1}{2}\right) is self-adjoint on H\mathcal{H}.

Equivalent formulations 1

Other statements of this same problem, merged from separate entries. Each is equivalent to the statement above — proving any one settles them all.

  1. Hilbert–Pólya conjecture

    In mathematics, the Hilbert–Pólya conjecture states that the non-trivial zeros of the Riemann zeta function correspond to eigenvalues of a self-adjoint operator. It is a possible approach to the Riemann hypothesis, by means of spectral theory.

    source: Wikipedia

Sources & referencesView supporting material

Primary source

Wikipedia

Additional references

  1. Wikipedia, Hilbert–Pólya conjecture, the article this problem comes from.

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.