User rlo - MathOverflow

Is this extension of the Selberg class trivial?

2012-11-30T20:27:36Z

I came across the following modification of the Selberg class in some of my work (see below), and while I've moved on in some sense -- I submitted the paper in question -- I can't get it off of my mind.

The question: Consider the following modification of the axioms of the Selberg class. Let $L(s)$ be a function such that:

1) $L(s)$ possesses a Dirichlet series and Euler product as in the standard axioms for the Selberg class (thus, the Dirichlet coefficients are bounded by $n^\epsilon$, e.g.).

2) There are rational $a_j$ such that the completion $\Lambda(s):=L(s) Q^s \prod_{j=1}^k \Gamma\left(\lambda_j s + \mu_j\right)^{a_j}$ satisfies the standard sort of functional equation, namely $\Lambda(1-s) = \epsilon \cdot \overline{\Lambda}(s)$ valid when $s,1-s$ are both in the region of analyticity (see 3). This is the first place where the definition differs from the Selberg class. There, the $a_j$'s are taken to be positive integers, whereas here they could be both negative and non-integral, provided that they are still rational.

3) Via the functional equation in 2, suppose that $L(s)$ can be analytically continued to all of $\mathbb{C}$, with the possible exception of a simple pole at $s=1$, and allowing for the (finitely many) branch cuts $(-\infty+iy_j,1/2+iy_j)$ where each $y_j$ is the imaginary part of $\mu_j$.

An example of something which is almost in this class, but fails 3, is $L(s,f)^{1/2}$ where $L(s,f)$ is a primitive element of the Selberg class. Conjecturally, $L(s,f)$ will have infinitely many simple zeros, so there would need to be infinitely many branches to continue $L(s,f)^{1/2}$ beyond the half-line.

My question is what you'd probably expect: if $L(s)$ satisfies the above properties, must it actually be in the Selberg class? Roughly speaking, you might expect this for a couple of reasons. First, the analytic properties of the class of functions satisfying 1-3 are similar enough to those of the Selberg class that, believing that elements of the Selberg class come from arithmetic, probably the same is true for this class. Second, believing in a proof of the classification of the Selberg class -- nevermind that such a thing is a long way off -- the proof must only rely on the analytic properties of the functions (obviously! what else could it depend on?), and probably any particular proof can be modified to work for this extension. In particular, I did this for the $d=1$ case for my work, and I'd imagine that $0\leq d < 2$ should also work, but I haven't looked at this.

I expect that this question is hard (unless it's false), so I'd be happy with an answer that assumes big conjectures.

Also, maybe this is more of a meta-question, but is this interesting for a reason beyond my application? That is, are there interesting consequences if this turns out to be true? One can deduce some simple consequences about divisibility, e.g., that if the zero-set of one element (with multiplicity) is contained in the zero-set of another, then the first function must divide the second. But I'm not sure that's so interesting.

My application: If we look at multiplicative functions $|f(n)|\leq 1$, complex-valued, what must be true if $\sum_{n\leq x} f(n) \ll x^{1/2-\delta}$ for some fixed $\delta>0$? Certainly we want to assume that $f(n)$ is not too small (i.e., we want to rule out things like $f(n)=1/n$), so let's impose the condition that $\sum_{n\leq x} |f(n)| \gg x$. If such a function exhibits more than square root cancellation, must it come from a Dirichlet character? I don't know of a counterexample, and I suspect that this is true (but I will also readily admit that I'm not sure I'm thinking about the right pathologies). Believing that it's true, it's probably very hard to prove.

Based on this, we want to look at a class of multiplicative functions for which we can see Dirichlet characters naturally appearing, and for which we can provide an answer. The class of functions I looked at depended on the arithmetic of a finite, Galois extension $K/\mathbb{Q}$. In particular, look at functions satisfying the following properties:

1) $f(n)$ is completely multiplicative, 2) $|f(p)|=1$ for all primes $p$ that split completely in $K$, and 3) if $\mathrm{Frob}_p = \mathrm{Frob}_q$ (up to conjugacy -- crucially, we'll want $K$ to be non-abelian), then $f(p)=f(q)$.

We can see Dirichlet characters arising by taking $K$ to be cyclotomic. It turns out that for functions defined as above, more than square root cancellation implies that $f(n)$ coincides with a Dirichlet character. The argument essentially shows that some product $$\prod_{\rho} L(s,\rho)^{a_\rho}$$ of Artin $L$-functions associated to the irreducible representations of $\mathrm{Gal}(K/\mathbb{Q})$, each $a_\rho\in\mathbb{Q}$, is in the above defined class. The restriction that $|f(p)|=1$ for a prime that splits completely actually forces the degree to be 1, and following the proof of the classification of degree 1 elements of the Selberg class, one obtains the result.

However, there is a natural objection to the above setup. We need to talk about non-abelian number fields for things that are legitimately different from Dirichlet characters to arise, but once we're in this setting, the condition that $|f(p)|=1$ for primes that split completely is totally unnatural. If the above class is actually equal to the Selberg class, then one could remove this condition and conclude that the Dirichlet series $L(s,f)$ is in the Selberg class. Maybe that's not so interesting, but I think it would be more satisfying than what I have now.

Answer by rlo for The behavior of a certain greedy algorithm for Erdős Discrepancy Problem

2012-08-24T20:41:47Z

Update 2: Original answer below. I've put together graphs showing more than just champions, using every $N\leq 10^4$ and also every $N\equiv 0\pmod{100}$ up to $10^5$. This is for the original version, not the variant, but I'd expect that to be essentially the same. I am starting to be somewhat skeptical of my $1/3$ estimate. I'll collect some data further out, but it'll be less complete since it's more costly to collect.

Here's the graphs of $D(N)$ versus $N$. The added curves are $N^{1/3}$ and $\log N$.

Here's the graph of $\log D/\log N$ versus $N$. The horizontal line is at $1/3$. (Note the different scales.)

Original answer: I have some basic numerical observations. I hacked together some code in c++ to work on this, and would be happy to collect more focused data or to share the code. Also, whenever there was a choice of whether to assign the value $+1$ or $-1$ to $f(p)$, I chose $-1$ for consistency of output. This yields a well-defined function $f_N$ for each $N$. Let $D(N)$ denote the discrepancy of $f_N$ up to $N$.

(I've also looked at choosing $f(p)=+1$ if it's undetermined, and at $f(p)=\pm 1$ according to whether $p\equiv 1,3\pmod{4}$. In these cases, the data is essentially the same as below.)

$D(N)$ is roughly increasing, but is not monotonic. Its champion values for $N\leq10000$ are, in the form $(N,D(N))$, $(1,1)$,$(10,2)$,$(24,3)$,$(70,5)$,$(91,6)$,$(391,7)$,$(553,8)$,$(668,9)$,$(961,10)$,$(1235,11)$,$(1265,13)$,$(2561,14)$,$(2604,17)$,$(6275,18)$,$(6276,19),\dots$. This growth is more than logarithmic, and is probably polynomial -- $\log D/\log N$ hovers pretty close to $1/3$ for each of these points, so that may be the answer for the $\Omega$ result.
It appears that the functions $f_N$ may converge as $N\to\infty$, but I'm not sure of this and need more data. Let $l(N)$ denote the least prime $p$ for which the value of $f_N(p)$ is undetermined. Certainly $l(N)\geq 5$ once $N\geq 4$, and while there is a great deal of fluctuation, it appears that maybe $l(N)\geq 7$ once $N\geq40500$; I will be seeing if this is (numerically) true.

I hope to update this answer once I have more data.

Update: For the variation where we only look at the values of $f_N$ on squarefree integers, the behavior appears to be the same.

$(N,D(N),\log D/\log N)$:

$(1,1,\text{NaN})$
$(30,2,0.2037950471)$
$(42,3,0.2939297479)$
$(77,4,0.3191428313)$
$(190,5,0.3067334722)$
$(238,6,0.3274252273)$
$(319,8,0.3606890916)$
$(939,9,0.3210056698)$
$(1358,10,0.3191931033)$
$(1461,11,0.3290703914)$
$(2185,13,0.3335707591)$
$(2769,14,0.3329519195)$
$(3354,15,0.3335896252)$
$(3689,17,0.3449622741)$

Answer by rlo for Duality of eta product identities: a new idea?

2012-08-14T16:12:18Z

This looks like the Fricke involution to me. Given any positive integer $M$, define $W_M$ to be the operator given by slashing with the matrix $$\left(\begin{array}{cc} 0 & -1 \newline M & 0 \end{array}\right),$$ or, equivalently, if $f(z)$ is weight $k$ modular, by $f(z)\mid_k W_M = (-iz\sqrt{M})^{-k}f\left(\frac{-1}{Mz}\right)$. This is the weight $k$, level $M$ Fricke involution. It preserves $M_k(\Gamma_0(M))$, albeit changing the character.

Now, using the transformation properties of $\eta(z)$, in particular that $\eta(-1/z)=(-iz)^{1/2}\eta(z)$, it's possible to see that hitting any $\eta$-product with the appropriate Fricke involution takes it to another $\eta$-product, perhaps after scaling. Thus, hitting an $\eta$-product identity with the Fricke involution produces another identity, which is moreover dual to the original. At least the first example you give has this property, i.e., it follows from the Fricke involution.

As to the question of unique duals, you've already seen that things get wonky if you allow for self-dual identities. But even if you require the identities to not be self-dual, things are still bad (although perhaps unsatisfyingly so). Let $I_1$ and $I_2$ be two linearly independent, not dual to each other, and not self-dual identities. Let $I_1^\prime$ and $I_2^\prime$ denote their duals, respectively (if you want, just take them to be images under the Fricke involution). Consider now the identity $I=I_1+I_2$. By your definition of dual, any of the identities $I^\prime=aI_1^\prime+bI_2^\prime$ will be dual to $I$, and so the dual-space is at least two-dimensional. Modifying this in the obvious way suggests that the dimension of the dual-space can be arbitrarily large, with the requisite $I_1, \dots, I_n$ being of the form $I_1^aI_2^b$ varying over $a+b=n-1$, say.

I think the right way to phrase the uniqueness question is by asking whether there is a set of primitive $\eta$-product identities from which all other identities can be obtained, with the property that every dual of an identity is generated by Fricke involutions of the constituent primitive elements of the original identity. In particular, if an identity is primitive, is its dual-space one-dimensional? I don't know the answer to this question.

Answer by rlo for Axioms for Riemann $\zeta$ function

2012-07-18T13:35:04Z

I don't know if this is in the spirit you're looking for, but there is the Selberg class -- an attempt at axiomatizing $L$-functions, requiring a Dirichlet series, functional equation of a certain type, analyticity, and an Euler product (typically) -- and it would be possible to impose extra conditions to isolate the zeta function. In particular, all degree 1 elements are known to come from Dirichlet L-functions (this was proved by Kaczorowski and Perelli, and then reproved by Soundararajan). Thus, requiring the degree and the conductor to both be 1 should isolate the zeta function.

Distinct simple zeros of Dirichlet L-functions

2011-09-20T15:21:56Z

Given a finite set of distinct primitive Dirichlet characters, $\chi_1, \dots, \chi_r$, is it known that the product of the L-functions, $$L(s):=\prod_{i=1}^r L(s,\chi_i),$$ has a simple zero? It's conjectured that all of the zeros of the $L(s,\chi_i)$ are distinct and simple, but I don't know what is known unconditionally except in the case that $r\leq 2$.

It's known that each $L(s,\chi)$ has infinitely many simple zeros (in fact, a positive proportion of its zeros are simple and lie on the half-line), which immediately answers the question in the case that $r=1$. If $r=2$, the answer to my question seems to be provided by work of Conrey, Ghosh, and Gonek (Simple zeros of the zeta function of a quadratic number field, I. Invent. Math., MR0860683). They prove that the Dedekind zeta function associated to a quadratic field has $\gg T^{6/11}$ simple zeros with imaginary part up to $T$, all arising from $\zeta(s)$. It appears that their method can be adapted to consider the product of any two Dirichlet L-functions, and this is confirmed by a statement of Bombieri and Perelli (Distinct zeros of L-functions. Acta Arith., MR1611193), who additionally write that $r=2$ is the limit of the Conrey-Ghosh-Gonek method.

I have not been able to find any work which applies to my question in the case that $r\geq 3$. The paper of Bombieri and Perelli referenced above discusses counting distinct zeros of more general L-functions, but it is not obvious to me how to isolate the simple zeros in their argument.

I also don't know if the fact that I'm only looking for a single simple zero of $L(s)$ saves me anything. That is, I don't know of techniques that detect the existence of such a zero without proving that there are an infinite number. Nevertheless, this could prove to be useful, since it seems entirely possible that showing that $L(s)$ has infinitely many simple zeros when $r\geq 3$ could be quite difficult.

Multiplicative functions whose Dirichlet series have essential singularities

2011-07-24T21:31:14Z

What can be said about the partial sums of a complex-valued completely multiplicative function, let's say bounded by 1 in absolute value, if its Dirichlet series has an essential singularity?

As a concrete example, consider the completely multiplicative function defined by $f(p)=i$ for all primes $p$. The Dirichlet series of this function has the Euler product $$L(s,f)=\prod_p \left(1-\frac{i}{p^s}\right)^{-1},$$ and by taking logs, we see that $$\log L(s,f) = -\sum_p \log\left(1-\frac{i}{p^s}\right) = i \sum_p \frac{1}{p^s} + O(1) = i\log\left(\frac{1}{s-1}\right)+\theta(s),$$ where $\theta(s)$ is analytic if $\Re(s)\geq 1$. By taking the limit as $s\to 1$ in this right half-plane, we see that the argument of $L(s,f)$ goes to infinity while its absolute value converges to something non-zero (at least along the real axis), whence $L(s,f)$ has an essential singularity at $s=1$. Standard Dirichlet series techniques (Perron's formula, for example) let us say that the partial sums of $f$, $$S_f(x):=\sum_{n\leq x} f(n),$$ are not $O(x^{1-\epsilon})$ for any $\epsilon>0$, but to my knowledge, this is the best these techniques can achieve. By using a quantitative version of Halasz's theorem, I imagine it should be possible to show that $$S_f(x) \ll x \frac{\log\log^A x}{\log x}$$ for some $A\geq 1$. This bound is pretty good, in that it achieves quantitative savings over the trivial bound $O(x)$, but I have no idea if it is the truth in some sense, and it's also easy to imagine situations in which using Halasz's theorem is not feasible (essentially, $f(p)=i$ is a nice, consistent choice).

My question is this: What can be said about $S_f(x)$? Are there good lower bounds for it? Is there a way, even heuristically, to determine its order of magnitude, or even just to get better upper bounds? More generally, are there techniques other than the standard Dirichlet series approach and Halasz's theorem to get good upper bounds on partial sums of such functions?

The number of pairings between multisets

2011-06-09T17:17:13Z

Given two multisets $A$ and $B$ of the same finite cardinality $n$, how many ways are there of pairing the two sets together?

If both sets consist of distinct elements, the answer is $n!$: there are $n$ ways to pair the first element of $A$ with something from $B$, $n-1$ for the second element, etc. If one of the sets has distinct elements and the other is allowed to have repeated elements, again the answer is well-understood. If $A$ has distinct elements and the elements of $B$ have multiplicities $b_1,\dots,b_s$ with $b_1+\dots+b_s=n$, then the number of pairings is $n!/b_1!\dots b_s!$. What's not obvious to me is what happens when both sets are allowed to have repeated elements.

As a simple example, suppose $A={1,2,3}$ and $B={a,a,b}$. Either per the above formula or by simple counting, one sees that there are 3 pairings - $[1a,2a,3b],[1a,2b,3a]$, and $[1b,2a,3a]$. However, if $A={1,1,2}$ and $B={a,a,b}$, then there are only 2 pairings - $[1a,1a,2b]$ and $[1a,1b,2a]$. This example is noteworthy in that it shows that the number of pairings doesn't have to divide $n!/b_1!\dots b_s!$. In particular, if $a_1,\dots,a_r$ are the multiplicities of the elements of $A$, the number of pairings is not $n!/a_1!\dots a_r! b_1! \dots b_s!$, a quantity which does not even have to be an integer.

For my purposes, I'd like to have a way to write this in terms of fairly simple combinatorial objects (multinomial coefficients, Bell or Stirling numbers, etc.), but I'm not convinced this is possible, at least without resorting to a heinous sum. In fact, I only care about the parity of this count, so even a characterization of the $a_i$ and $b_i$ which make this even or odd would be of use to me. The only restriction I have on $A$ and $B$ is that at least one $b_i$, say, must be 1, but I'm not sure how to take advantage of that here.

Answer by rlo for Exact formulas for the partition function?

2010-11-29T21:33:49Z

This doesn't really answer the question, so perhaps it would be better as a comment, but alas, I don't have the necessary reputation.

Following up on Thomas Bloom's reference to the work of Bringmann and Ono, there is a paper of Folsom and Masri (Mathematische Annalen, available here: http://www.math.yale.edu/~alf8/Folsom-Masri-MathAnn07-10.pdf) which considers the main term one would get in an asymptotic formula arising from BO's Poincare series formula. In particular, they also consider the problem of the error arising from truncating the infinite sum at $O(n^{1/2})$, obtaining power savings over the best known results of $O(n^{-1/2+\epsilon})$ if one truncates at $\lfloor \sqrt{n/6} \rfloor$.

Comment by rlo

2013-04-17T03:08:53Z

Have you done any computations yourself? While I'm dubious that this should be true for almost any form, it's worth noting that $L(s,\theta_\chi)=L(2s-1/2,\chi)$ for a non-trivial Dirichlet character $\chi$, so RH presumably holds in this case. In general, though, the multiplicative structure of half-integral weight eigenforms is more complex, and I'd be very surprised if it were to hold if the form is orthogonal to the space of unary theta functions.

Comment by rlo

2012-11-30T23:37:11Z

You're absolutely right that there are issues with $L(s,f)^{1/2}$, which is why it's not actually something I want to consider. I brought it up mostly to clarify points 1 and 2. Maybe a prototype question would be this: Can $L(s,f)^{1/2}L(s,g)^{1/2}$ ever be sensibly continued to an entire function, where $L(s,f)$ and $L(s,g)$ are primitive elements of the Selberg class? It's known that each has zeros disjoint from the other, but maybe all the simple zeros coincide, or are there are no simple zeros, or... Conjecturally, this can't happen, but that's the sort of thing I'm imagining.

Comment by rlo

2012-11-30T23:29:26Z

Right, this absolutely falls under the purview of pretentiousness; in fact, my question about more than square root cancellation can be viewed as a counterpart to Halasz's theorem which says that anything with large sums must come from (pretend to be) one of a natural set of examples. There, though, the natural examples are not Dirichlet characters. Instead, they are the additive characters, $n^{it}$. Indeed, non-principal Dirichlet characters don't have large partial sums - the partial sums are bounded! Thus, they are the natural examples for exceptional (more than square root) cancellation.

Comment by rlo

2012-08-25T16:15:21Z

I will add data for the squarefree problem to my answer in a second, but let me just quickly give you the gist -- it's the same basic behavior, with $1/3$ appearing just as clearly.

Comment by rlo

2012-08-25T00:30:15Z

His. And the ratio of $\log D/\log N$ I alluded to, not in graph form: 0.3010299957, 0.345687124, 0.3788254007, 0.3972102138, 0.3260186782, 0.3302216042, 0.3378116896, 0.3352640758, 0.3368385727, 0.3590944053, 0.336264762, 0.3602395359, 0.3305424391, 0.3367194215. Close enough to 1/3 that I suspect something is happening.

Comment by rlo

2012-08-24T15:03:04Z

Are we requiring $f(n)$ to be $\pm 1$ only?

Comment by rlo

2012-07-19T02:29:43Z

As I stated it, yes, that is necessary. However, Kaczorowski and Perelli define what they call the extended Selberg class, where there is no Euler product, and they provide a classification of the elements of degree up to 1. Degree 0 elements are Dirichlet polynomials satisfying a certain symmetry condition, and degree 1 elements are linear combinations of the product of a degree 0 element with a (potentially shifted) Dirichlet L-function. Sound's proof also works for this class, and shows that the Dirichlet coefficients are periodic, so that multiplicativity implies Dirichlet character.

Comment by rlo

2012-07-18T18:18:09Z

Fair enough. It's also worth noting that Hamburger's theorem, as mentioned by Micah and Stopple, is the better way of stating my answer, since the conditions that the degree and conductor are 1 restrict the functional equation to being exactly the one satisfied by zeta, and is thus subject to Hamburger's result.

Comment by rlo

2012-03-04T17:00:17Z

That is the conclusion I reached, yeah, especially after some conversations with people and a couple of failed attempts to prove something. In the application I had in mind (which actually required Artin L-functions is addition to Dirichlet), I ended up just using that it is simple to check in each case.

Comment by rlo

2011-07-25T15:02:44Z

This seems great. I was interested in that specific example as well as the general case, so this was definitely helpful. Suppose I choose f(p) such that the sum of f(p)/p doesn't converge but also doesn't diverge to infinity in any direction. It seems like this shouldn't be amenable to study using the Selberg-Delange method, at least as presented in Tenenbaum. Are you aware of any techniques which might apply?

Comment by rlo

2011-06-09T21:12:16Z

Thanks, I was sure this had to be well-studied. It's unfortunate that there's no particularly nice formula, but I'll make do.

Comment by rlo

2011-06-09T17:56:21Z

Sorry, that is perhaps unclear. I included the examples to try to clear things up, but I guess I was unsuccessful. I mean a pairing to be a way to associate to each element of A a unique element of B. Perhaps one-to-one correspondence would be better?