Tying loose ends – E. Kowalski's blog

I don’t know if there is a single English word for the action of tying loose ends, but this is precisely what I’ve just done by putting together three different short notes and remarks in a single text, entitled “Two remarks on the large sieve”, now available on my home page. (The is ample precedent for a title to mention one less than the actual number of main characters). I’ve already discussed two of those notes: the first one is the new amplification-based proof of the arithmetic large sieve inequality, and the last is the amusing result that two primitive cusp forms are equal if and only if the signs of their respective Hecke eigenvalues coincide for all primes (or all primes with few exceptions). The reader will not fail to notice a certain distance between those two results, and therefore the third note, the most recent (which is not available separately, only in bundled form, like some journals) provides the glue from one to the other.

The gluing is in two parts (though there is rather less precedent for a title to forget about half of the characters of a tale…): there is a result that tries to answer statistically the question of how many primitive forms (of a given level q) may have Hecke eigenvalues with identical signs for the first few primes, which links with the second note, and there is a large-sieve inequality that gives a way to obtain this result. In turn, this inequality turns out to be much easier to prove using the amplification method than using the classical proofs of the arithmetic large sieve, thus relating it with the first result! To finally explain the title, I should say that I think those large sieve results are overall more interesting than the particular application, so I chose to emphasize them. (In fact, I wouldn’t be surprised if the statistical result could be improved by more refined or devious arguments).

The main attraction from the sieve point of view is to have a version of the large sieve which looks like a sieve, but is based directly on the “spectral” inequalities for coefficients of modular forms (originating in work of Iwaniec, and much developped later by Deshouillers and Iwaniec). Those were, until then, mostly considered as distinct beasts, with only a vague connection to sieve, coming from their resemblance with the “harmonic” large sieve inequality

$\sum_{q\leq Q}{\sum_{a\text{ mod } q, (a,q)=1}{\left|\sum_{n\leq N}{a(n)\exp(2i\pi an/q)}\right|^2}}\leq (N-1+Q^2)\sum_{n}{|a(n)|^2},$

which also looks like a purely analytic statement.

Here’s the idea for our unification: we consider as our set of objects to “sieve” the finite set S(q) of primitive cusp forms of level q (with a fixed weight, say 1728). As “reductions modulo primes”, we select the maps

$\rho_p\,:\, f\mapsto \lambda_f(p)\in \mathbf{R}$

sending each f to its p-th Hecke eigenvalue (and we omit primes dividing q for simplicity). [More intrinsically, this should be seen as mapping each f to its p-component when factored as an automorphic representation, seen as an element of the “dual” of the local group GL(2,Q_p), but for classical modular forms, this amounts to the same thing if we only consider unramified primes]. In fact, from the Ramanujan-Petersson conjecture (proved by Deligne) we know that

$\rho_p(f)\in [-2,2]$

for all primes.

Now what should a sifted set look like? Recall that for integers, we would look at sets defined by

$\{n\leq N\,\mid\, n\text{ mod } p\notin \Omega_p\text{ for } p\leq Q\}$

so we may want to do the same and look at sets of forms given by

$T=\{f\in S(q)\,\mid\, \lambda_f(p)\notin \Omega_p\text{ for } p\leq Q\}$

for some subsets Ω_p of [-2,2]. If we take

$\Omega_p=[0,2]$

we get the forms with negative Hecke eigenvalues for all primes p up to Q.

At the moment at least, I can not get a bound for such a set T in this generality, because if we try to approach the question by harmonic analysis (as seems natural), the fact that the target sets are compact intervals [-2,2] means that we need infinitely many harmonics to analyze functions defined on it, such as the characteristic function of Ω_p. Maybe one can chop down the higher frequencies efficiently enough, but for the moment I use a subterfuge: the natural basis on (square-integrable) functions on [-2,2] is given here by the Chebychev polynomials defined by

$X_n(2\cos \theta)=\frac{\sin (n+1)\theta}{\sin\theta},\text{ for } \theta\in [0,\pi],\ n\geq 0.$

Precisely, this is the natural basis if we count the forms in S(q) with a suitable weight ω(f) related to the inverse of their Petersson norm, as I will do; otherwise, one needs to use a different basis, which will vary with p: this is related to the fact that, for fixed p, the ρ_p(f) become equidistributed, when q increases, with respect to different measures depending on which weight is used; here the X_n have the feature of being an orthonormal basis of

$L^2([-2,2],\nu)$

where ν is the Sato-Tate measure:

$\nu=\frac{1}{\pi}\sqrt{1-\frac{x^2}{4}}dx.$

Then I consider sifted sets of the simpler type

$U=\{f\in S(q)\,\mid\, \lambda_f(p)\notin \Omega_p\text{ for } p\leq Q\}$

where

$\Omega_p=\{x\in [-2,2]\,\mid\, Y_p(x)\leq \beta_{p,0}-\delta_p\},\text{ with } \delta_p\geq 0,$

for some polynomials

$Y_p(x)=\sum_{i=0}^s{\beta_{p,i}X_i(x)}$

of degree at most s for all p. Note that, because of orthogonality of the Chebychev polynomials for the Sato-Tate measure, the constant coefficient is

$\beta_{p,0}=\int_{-2}^2{Y_p(x)d\nu(x)},$

and therefore the sets U above correspond to forms where, for each prime p up to Q, the polynomials take a value “far” from its average, determined according to the Sato-Tate measure. Using polynomials of bounded degree is what avoids the analytic difficulty with the infinite basis (X_n).

Now it is an easy matter to find a polynomial Y (in fact, of degree 2) and a suitable δ>0 such that

$\lambda_f(p)<0\text{ implies } Y(\lambda_f(p))\leq \beta_0-\delta.$

So controlling the sieve with sets of type U is enough to control the forms with negative Fourier coefficients. Then it turns out that if we unravel the sieve framework, the underlying inequality to prove is a consequence of

$\sum_{f\in S(q)}{\omega(f)\left|\sum_{m\leq Q^s}{\alpha(m)\lambda_f(m)}\right|^2}\ll (1+Q^sq^{-1})\sum_{m}{|\alpha(m)|^2}$

and this is a special case of the Deshouillers-Iwaniec inequalities! (Actually, a rather simple case). But going from this to the sieve inequality which bounds the (weighted) size of the sifted set U, although it is very simple with the amplification approach, does not seem to be doable with the standard proof (at least, I puzzled on it for a while without success — but of course I have a vested interest in my own argument being better…)

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