# Heuristic reason for Polya's conjecture

Let $\lambda(n)$ be Liouville's function, so that for each positive integer $n = p_1^{m_1}\cdots p_r^{m_r}$, we have that $\lambda(n) = (-1)^{\sum^{r}_{k=1}{m_k}}$. In 1919, Polya conjectured that $L(x) = \sum_{n \leq x}{\lambda(n)} \leq 0$ for all $x \geq 2$; his reasoning was based on some limited numerical evidence (up to $x = 1500$, I believe), its connection to the Riemann Hypothesis (it implies RH and the simplicity of the zeroes of $\zeta(s)$), and Polya showed that for $p \equiv 3 \pmod{4}$ with class number $h(-p) = 1$, $L(p) = 0$. Unfortunately, Polya's conjecture is false; it is known that the first counterexample occurs at $x = 906150257$ (so one can't really blame Polya for trying), and that there exist infinitely many positive integers $n$ such that $L(n) \geq 0.061867 \ldots$.

Nevertheless, Polya's conjecture does seem to be usually true, in that $L(x) \leq 0$ "most" of the time. There are a couple of different arguments that give an indication of why one would expect $L(x)$ to often be negative. For example, standard methods show (under RH, of course) that $${\sum_{n \leq x}}'{\lambda(n)} = \frac{\sqrt{x}}{\zeta(1/2)} + \sum_{\rho}{\frac{\zeta(2\rho)}{\zeta'(\rho)}\frac{x^{\rho}}{\rho}} - 1 + O\left(\frac{1}{\sqrt{x}}\right),$$ and one expects the terms in the sum over the zeroes to generally be very small, whereas $1/\zeta(1/2) = -0.684765\ldots$, so it would be expected that $L(x)$ is "usually" negative. Another method is via Lambert series; I mentioned here that one can show that $$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \frac{1-\sqrt{2}}{2}\sqrt{x} + \frac{1}{2} + (\psi(x)-2\psi(x/2))\sqrt{x},$$ where $\psi(x) = \sum_{n=1}^{\infty}{e^{-\pi xn^2}} = O(e^{-\pi x})$; this Lambert series is in some sense a smoothed version of $L(x)$. Again, the leading term is negative, suggesting that $L(x) \leq 0$ often.

My question is: what other methods (elementary, analytic, or probabilistic) can be used to show why we would expect $L(x)$ to usually be negative?

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What happens with the sign and size of $$\sum_{m=2}^N L(m) ?$$ – Will Jagy Jul 25 '10 at 3:26
Hmmm, I'm not sure if that's been studied at all, but of course you could just put each term $L(m)$ into the explicit formula with the sum over the zeroes over $\zeta(s)$ to get an idea of the behaviour of $\sum^{N}_{m=2}{L(m)}$. As far as size goes, similarly you can just use the fact that $L(x) = \Omega_{\pm}(\sqrt{x})$ and $L(x) = O(x\exp(-c(\log x)^{1/2}))$ unconditionally, while $L(x) = O(x^{1/2 + \varepsilon})$ under RH. – Peter Humphries Jul 25 '10 at 3:43
Polya's reasoning was not connected to the Riemann hypothesis! He put it forward purely on the basis of the numerical evidence up to 1500. That it implies RH (and even simplicity of the zeros) was worked out 20+ years later by Ingham, who also showed it implied linear dependence relations among the positive imaginary parts of the nontrivial zeros of the zeta-function. That last aspect is not expected to be true, so it cast serious doubt on Polya's conjecture and about 15 years later Haselgrove showed the conjecture is false (although without finding a counterexample; they came later). – KConrad Jul 25 '10 at 4:22
The relation of Polya's conjcture to the linear independence of nontrivial Riemann's zeroes is mentioned in mathoverflow.net/questions/32324/…. – Wadim Zudilin Jul 25 '10 at 8:25
@Will: since $\sum_{n=1}^{N-1} L(n) = \sum_{n\le N} \lambda(n)(N-n)$, the analytic methods used to address L(n) can address this average as well. In the explicit formula for $\sum_{n\le N} \lambda(n)$ given by the original poster, just replace everything of the form $x^\rho$ by $x^{\rho+1}/(\rho+1)$. Further summations simply integrate this term over and over again. Therefore (under the assumptions in my answer) there will always be a negative bias. – Greg Martin Jul 27 '10 at 8:00

Suppose that the Riemann hypothesis is true and that all the imaginary parts of the zeros are linearly independent over the rational numbers. Then one can show, using the methods of Rubinstein and Sarnak ("Chebyshev's bias", Experiment. Math., 1994), that the function $L(x)/\sqrt x$ has the same limiting (logarithmic) distribution function as the random variable $$\frac1{\zeta(1/2)} + \sum_{\rho\colon \Im\rho>0} 2\bigg| \frac{\zeta(2\rho)}{\rho\zeta'(\rho)} \bigg| X_\rho,$$ where the $X_\rho$ are independent random variables each taking values in $[-1,1]$ according to the sine distribution (that is, each $X_\rho$ is the real part of a random variable uniformly distributed on the unit circle in the complex plane).
The distribution of the sum is symmetric around 0, which explains why the distribution itself is predominantly negative (since $\zeta(1/2) \lt 0$).
Thanks Greg. This is in fact why I asked my question, as I'm working on exactly what you mentioned for my undergraduate thesis. Unfortunately, I haven't yet been able to prove that the resulting measure via Rubinstein-Sarnak methods is nonvanishing on open sets (or even entire), so I can't quite show that the logarithmic density of the set $\{x \in [1,\infty) : L(x) \leq 0\}$ is strictly greater than one-half, just that it is at least one-half. Even to get this far, one has to assume conjectures about the growth of $1/\zeta'(\rho)$ as well as RH and LI. – Peter Humphries Jul 27 '10 at 8:05
Peter, can you prove what you want using results or conjectures on the average value of $1/| \zeta(\rho)|$? – David Hansen Jul 27 '10 at 18:21
@David and John: Yes, the method follows closely that of Ng, using the Gonek-Hejhal conjecture $J_{-1}(T) = \sum_{0 < \Im(\rho) < T}{|\zeta'(\rho)|^{-2}} \ll T$. But although this gives the existence of a limiting distribution, it does not quite show that the associated measure extends to an entire function (as in the Rubinstein-Sarnak case) because $J_{-1}(T) \ll T$ only seems gives subexponential bounds on the Fourier transform of this measure. Of course, I could be missing something here. – Peter Humphries Jul 30 '10 at 1:23