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Lehmer's conjecture for Ramanujan's tau function, $$ \Delta(q)=q\prod_{n=1}^\infty(1-q^n)^{24}=\sum_{m=1}^\infty\tau(m)q^m, $$ asserts that $\tau(m)$ never vanishes for $m=1,2,\dots$. In the recent question it was asked why it is important to have the nonvanishing.

I am wondering whether there are upper bounds, unconditional or conditional (modulo some other known conjectures), in terms of $x\in\mathbb R_+$ for the number of integers $m\le x$ satisfying $\tau(m)=0$ (maybe better, for the number of primes $p\le x$ satisfying $\tau(p)=0$)?

It looks like the series $\Delta(q)$ is very far from being "lacunary". But besides Deligne's upper bound $|\tau(m)|\le d(m)m^{11/2}$ (where $d(\ )$ counts the number of divisors) and the lower bound $$ \operatorname{card}\lbrace\tau(n):n\le x\rbrace\ge \operatorname{const}\cdot x^{1/2}e^{-4\log x/\log\log x} $$ from [M.Z. Garaev, V.C. Garcia, and S.V. Konyagin, A note on the Ramanujan $\tau$-function, Arch. Math. (Basel) 89:5 (2007) 411--418] for the distribution of tau values, I cannot find any quantitative progress towards Lehmer's original question.

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    $\begingroup$ There is a preprint today on arXiv front.math.ucdavis.edu/1406.3607 which claims to prove the conjecture (and seems serious on first look). $\endgroup$
    – BS.
    Jun 16, 2014 at 10:46
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    $\begingroup$ On second look though, Lemma 2 of the paper seems wrong; the deduction after equation (19) seems false if i<5 and indeed should be false, because all he's using is (12) which implies b_i=0 for 1<=i<=4. $\endgroup$
    – eric
    Jun 24, 2014 at 11:53

3 Answers 3

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One of the canonical references for questions like this is Serre's "Quelques applications du theoreme de densite de Chebotarev", Publ. Math. IHES 54. He proves, for example, that the number of primes $0\leq p \leq X$ with $\tau(p)=0$ is $\ll X (\log{X})^{-3/2}$ unconditionally, and is $\ll X^{\frac{3}{4}}$ under GRH.

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  • $\begingroup$ Thank you very much, David! Indeed, this seems to be the latest news towards Lehmer's question, although it doesn't even exclude the lacunarity. $\endgroup$
    – Wadim Zudilin
    Jul 20, 2010 at 22:49
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Lehmer's conjecture has an equivalent result in the theory of Harmonic Maass forms. The non-vanishing of the tau function is equivalent to the irrationality of the coefficients of Harmonic Maass forms.

Specifically there is a correspondence between the spaces $ \zeta_{2-k} : H_{2-k}(N, \chi) \rightarrow S_k(N, \chi) $. where

  1. $ \zeta_{2-k}$ is a differential operator
  2. H = Harmonic Maass forms
  3. S = cusp forms (referred to as the shadow of the Maass form)

The discriminant function $\Delta(z)$ is the shadow of the Harmonic Maass form $\frac{1}{11!} Q^+(-1, 12, 1; z) $

See Theorem 12.5 in the paper Unearthing the visions of a master: harmonic Maass forms and number theory by Ken Ono. Also see Algebraicity of Harmonic Maass forms

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  • $\begingroup$ Thanks for reminding me on this remarkable connection! Indeed, I remember Jan Bruinier mentioned this fact in one of his talks. $\endgroup$
    – Wadim Zudilin
    Jul 22, 2010 at 0:30
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There's a nice paper by Kowalski, Robert, and Wu that discusses this problem,

  • Small gaps in coefficients of L-functions and B-free numbers in small intervals, Rev. Mat. Iberoamericana 23 (1) (2007) 281–326, doi:10.4171/RMI/496, arXiv:math/0507001.
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  • $\begingroup$ Thanks, Matt! It sounds like a nice addition to Serre. The authors also take into account the Lang-Trotter conjecture and its generalisation (which they call Conjecture 1) and the strongest result they can achieve is conditional Corollary 7 that still does not exclude the lacunarity. They comment themselves: "This is of course trivial and of little practical significance towards the Lehmer conjecture." So, Lehmer's conjecture is a hard die... $\endgroup$
    – Wadim Zudilin
    Jul 20, 2010 at 22:54

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