Bounded discrepancy multiplicative functions do not correlate with characters

From Polymath Wiki
Revision as of 20:09, 30 January 2010 by Teorth (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search
Theorem Let [math]g: {\Bbb N} \to S^1[/math] be a completely multiplicative function of bounded discrepancy. Let [math]\chi[/math] be a non-principal Dirichlet character of some conductor q. Then [math]\sum_{p} \frac{|1-g(p)\overline{\chi(p)}|}{p} = \infty[/math].

Remark By a theorem of Wintner on mean values of completely multiplicative functions, this implies that [math] \sum_{n \leq N} g(n) \overline{\chi(n)} = o(N)[/math], thus g asymptotically does not correlate with any Dirichlet character.

Proof By deleting the exceptional prime 2 from the character, we may assume that the conductor q is even. Suppose for contradiction that the sum wqas finite. We let the implied constants in the O() notation depend on q, the discrepancy of g, and on the sum, thus q = O(1),

[math] g(1)+\ldots+g(n) = O(1)[/math], (0)


[math] \sum_{p} \frac{|1-g(p)\overline{\chi(p)}|}{p} = O(1)[/math]. (1)

the sum was finite. Let [math]\tilde \chi[/math] be the completely multiplicative function which equals [math]\chi[/math] for numbers coprime to q, and such that [math]\tilde \chi(p) = g(p)[/math] for p dividing q. Let h be the completely multiplicative function with [math]h(p) = g(p) \overline{\chi(p)}[/math] for p coprime to q, and h(p)=1 otherwise. Then we have the factorisation

[math]g(n) = h(n) \tilde \chi(n)[/math]. (2)

By hypothesis, we have

[math] \sum_p \frac{|1-h(p)|}{p} = O(1)[/math]. (3)

As for [math]\tilde \chi[/math], we observe the following limited amount of periodicity:

Lemma Let k be a positive integer, and let Q be a sufficiently large power of q (depending on k). Then for every [math]1 \leq i \leq q^k[/math], one has [math]\tilde \chi(Qn+i) = \tilde \chi(i)[/math] for all n.

Proof Let r be the lcm of Q and i, then it suffices by multiplicativity to show that [math]\tilde \chi(Qn/r + i/r) = \tilde \chi(i/r)[/math]. But by construction of i and Q, i/r is coprime to q, and Q/r is a multiple of q. We then have

[math] \tilde \chi(Qn/r + i/r) = \chi(Qn/r + i/r) = \chi(i/r) = \tilde \chi(i/r)[/math]

as desired. [math]\Box[/math]

We're going to use the Maier matrix method. Let k be a large integer, and let Q be as in the lemma. Let M be a large number (depending on k, Q) to be chosen later, and let N be an enormous number (depending on k, Q, M). It may be helpful to keep the hierarchy

[math] k \ll Q \ll M \ll N [/math]

in mind.

For any n,

[math] \frac{1}{q^k} \sum_{a=1}^{q^k} |\sum_{i=1}^{a} g( Q n + i )|^2 = O(1) [/math]

by bounded discrepancy (0). We average this over n in an interval [n,n+M] to get

[math] \frac{1}{q^k} \sum_{a=1}^{q^k} |\sum_{i=1}^{a} \frac{1}{M} \sum_{m=1}^M g( Q(n+m) + i )|^2 = O(1).[/math]

Pick n at random from 1 to N. By the lemma, we have [math]\tilde \chi(Q(n+m)+i) = \tilde \chi(i)[/math]. Thus by (2) we have

[math] \frac{1}{q^k} \sum_{a=1}^{q^k} |\sum_{i=1}^{a} \tilde \chi(i) \frac{1}{M} \sum_{m=1}^M h( Q(n+m) + i )|^2 = O(1).[/math]

Now we perform the m summation. By construction, Q is coprime to all the primes for which h(p) is different from 1. An application of the second moment method shows that

[math]\frac{1}{M} \sum_{m=1}^M h(Q(n+m)+i) = c + o(1)[/math]

with probability 1-o(1) for each i, where c is the singular series

[math]c = \prod_{p \in {\mathcal P}} \frac{1-1/p}{1-h(p)/p}[/math]

and o(1) denotes a quantity that goes to zero if M is large enough (and N is huge enough depending on M), for fixed k and Q. Thus, with probability 1-o(1), one has

[math] c \frac{1}{q^k} \sum_{a=1}^{q^k} |\sum_{i=1}^{a} \tilde \chi(i)|^2 = O(1).[/math]

However, by (3), c is non-zero (this is why we needed to eliminate the exceptional prime 2), thus

[math] \frac{1}{q^k} \sum_{a=1}^{q^k} |\sum_{i=1}^{a} \tilde \chi(i)|^2 = O(1).[/math]

In other words, if a is chosen at random from 1 to q^k, then [math]\sum_{i=1}^{a} \tilde \chi(i)[/math] has bounded variance. But in fact this variance grows linearly in k, which gives a contradiction by setting k large enough.

It's easiest to prove this linear growth in the case when q is prime. Then we can split this random sum as the sum of k terms

[math] \tilde \chi(q)^l \sum_{1 \leq i \leq a/q^l: (a,q)=1} \chi(i)[/math]

for [math]l=0,\ldots,k-1[/math] (plus a O(1) error), but each such term has a variance comparable to 1 and depends only on the l^th digit in the base q expansion of a, so these terms have negligible correlation with respect to each other and their sum has a variance growing linearly in k. I'm pretty sure one can adapt this argument to composite k.