Also, we might, in order to give ourselves a bit more flexibility, the opportunity for smooth cutoffs, etc., like to replace the sum over by a sum with weights. That is, we might like to consider an expression of the form

Again, if we are searching for coefficients that are independent of we might like to place restrictions on Note that we are already hoping that will be concentrated in the set of such that , which is the set of such that and are small multiples of So it feels natural to make a function of and A nice aspect of that is that the integral is also a function of and

If we do this, then I think the entire expression depends on only. Let me just check that. Yes, it’s clear, since if we multiply and by we can just substitute for and end up with the original expression.

I now need to do a little bit of paper calculation in order to work out a nice way of expressing the matrix as a function of Then one can think about how to make that function 1 at 1 and 0 everywhere else.

]]>I see. It didn’t occur to me that you were allowing your HAPs to have length greater than so that they could actually repeat themselves. It seems to me that asking exactly the same question but restricting the length of the HAP to be at most leads to a more natural question.

Actually, I take that back. In we can express any AP with common difference coprime to as a difference of two HAPs, so if e.g. is prime then the question becomes equivalent to Roth’s theorem.

]]>Here’s another way to think about it which may be clearer. Consider functions that are periodic with period . Obviously the discrepancy of such a function with respect to HAPs of common difference is unbounded if is a multiple of . But we can ask whether it is possible for the discrepancy to be bounded with respect to all other .

This is only possible if is a power of 2. For if where is odd, then has unbounded discrepancy with respect to HAPs with common difference (since , which tends to as because the factored-out sum is the sum of an odd number of terms and so non-zero).

The second observation is that it *is* possible when is a power of 2 — and that we can get the discrepancy as low as if , by simply repeating the first terms of the function (if is even) or of the function (if is odd).

It may be more interesting to consider functions . Then of course the first observation is no longer valid.

]]>Let’s suppose we want to write the identity on as a linear combination of products of HAPs. What I want to think about here is whether there is some “beautiful formula” that “magically works”. I’ll ignore, for the moment, any concerns about the fact that is infinite and that certain sums might not converge.

First of all, here is a very quick argument that we cannot get away with HAPs that are all of the form for some fixed To see this, note that we could find a multiplicative function that happens to have a small partial sum at and it will not correlate with any such HAP. But a representation-of-diagonals proof shows that every function correlates with an HAP that is used in the decomposition.

The next simplest thing one could hope for is to use HAPs that are of length bounded above by Since one can subtract one HAP from another, it seems nice to allow HAPs of the form where both and are at most However, choosing coefficients that depend on and seems nasty.

It can be made a bit nicer if we consider functions of the form which I define to take to if and to 0 otherwise. That is, is the pointwise product of the HAP with a trigonometric function. If we do this, it is important to restrict in such a way that this function can be efficiently decomposed into HAPs. A sufficient condition for this is that should be bounded above by a constant that tends to zero (so that is approximately constant on intervals of unbounded length).

So let’s ask the following question. Can we find a decomposition of the form

The value of the right-hand side at then works out to be

where is the set of all such that both and are of the form for some So we would like this to be 1 if and 0 otherwise.

I’ll say more about this later, but have to go now.

]]>Alec, I find this thought interesting (despite taking ages to respond to it). But at the moment I’m being slow to understand your point (1). Would it be possible to make both the statement and its proof more precise?

A slight variant of what you suggest could also fit in rather well with what I was trying to do over while avoiding some of the technical problems. That would be to work in for some very large prime but to restrict the lengths of the HAPs to at most for some much smaller, but unbounded, Basically, would be thought of as a fixed, but large, constant, while tended to infinity. I don’t instantly see how a lower bound for this problem would give a lower bound for EDP, but the converse clearly holds, and it’s hard to believe that a non-trivial lower bound for this problem wouldn’t shed light on EDP.

]]>A quick reminder of what I was trying to do in some of the comments above. I wanted to find a representation-of-diagonal approach, and had noted an idea that potentially made the task much easier. That idea was that working over the rationals could eliminate one of the main difficulties of the approach — understanding what the diagonal matrix should be. The point is that if one works with the integers up to then there is a hugely important distinction between smooth integers and non-smooth ones. But in all numbers are equal, so it becomes reasonable to search for a representation of the \textit{identity}.

I was trying to do this by using what we had learned from the representation-of-identity proof of Roth’s theorem about AP-discrepancy. The basic idea there is to start by decomposing the identity using characters and then decomposing the characters. This seemed reasonably promising, but eventually I got a bit stuck. But it now occurs to me that we could attack the problem from the other end. We could ask ourselves a question of the following kind: if the approach works, what properties will the resulting decomposition have, and does there actually exist a decomposition with those properties?

The real question there is the second one. We could use the previous ideas as a guide to tell us, non-rigorously, what sort of decomposition we might expect, and then we could search directly for a decomposition that is much less general than an arbitrary decomposition, or else try to prove that the much more specific decomposition cannot exist (which would then inform all subsequent efforts to find a decomposition).

In order to keep my comments bite-sized, I’ll stop this one here, and make a more technical comment on this theme separately.

]]>For , define the (zero-based) discrepancy

,

and let

.

A couple of observations:

(1) If , then is a power of 2.

(Suppose is not a power of 2. Write where is odd. Then has an odd number of terms, so is non-zero, so the discrepancy with respect to the difference is unbounded.)

(2) .

To show this, first define the function by , , , and (for ). Then define by if is even, and if is odd. Then . This follows from the observations: that is completely multiplicative (on the positive integers); that is equal to except when (when it equals ) or (when it equals ); and that .

It would be interesting to know whether this bound is best possible. In other words, is it true that for every function , ? If it is true, this seems like a simple enough problem to be tractable, and I wonder if the proof could shed some light on EDP for when we let .

In particular, I wonder if a ‘representation of the diagonal’ proof might work over , which is isomorphic to its own character group. I haven’t yet thought about this in any detail.

]]>One idea I have is to assume the character function is completely multiplicative use that to get a result that implies that any completely multiplicative function is unbounded and then prove the result that any bounded EDP implies the existence of a bounded completely completely multiplicative function and thus get the desired result. I don’t know if assuming the character function is completely multiplicative improves the situation.

]]>But if we were dealing with completely multiplicative functions, then the functions themselves would have long-range correlations all over the place. So couldn’t we get rid of our difficulties by looking at the multiplicative case? And since that is morally equivalent to the whole problem (in the sense that any proof is likely to give not just a single HAP with big discrepancy, but unbounded discrepancy on average, and that would imply EDP), isn’t it worth thinking rather hard about that case, and in particular about whether it is possible to get the ROI (on the rationals) approach to work for it?

The question is, how would one exploit the fact that we only cared about completely multiplicative functions? I think the answer would be something like this. When we express the identity as we do so because we need a matrix for which we can argue that is large for *every* function But what if we just wanted a statement like that for completely multiplicative functions ? It seems that we could then afford to throw away a lot of (additive) characters because we would know that they couldn’t correlate with a completely multiplicative function.

Let me be a bit speculative. Suppose we identify a class of characters that have “multiplicative structure” and that we can show that any that does not belong to has only a tiny correlation with any completely multiplicative function. Then the matrix would appear to be almost indistinguishable from the identity as long as one applied it only to completely multiplicative functions.

Now we do have some examples of characters that correlate with completely multiplicative functions. For instance, the …

… actually, I’m not sure I like what I was about to write there. The example I had in mind ( and the function $x\mapsto e(x/3)$) is too much of an integers example as opposed to a rationals example. Also, I’m a bit worried that when we express a function as an integral combination of characters, we may have to use lots of characters that don’t correlate well with the function (just as the point has norm 1 despite having only small correlation with any vector from the standard basis). It seems that we would have to show that the correlation between a non-special additive character and a completely multiplicative function was not just small but *very* small.

Suppose we could somehow do that. The hope would be that special additive characters had long-range correlations for HAPs — not as strong as the perfect correlations you get with completely multiplicative functions, but good enough to be exploited.

Here’s a very brief thought about what would count as “very small” correlation. Suppose we have a function that takes values of modulus 1 everywhere on Then its norm (on that set) is So if we are representing as then we can afford to disregard all for which is substantially smaller than

So a preliminary question is this. Suppose you have an additive character on the rationals and you know that its correlation with some completely multiplicative function is at least on the set . What can you say about ? I have an unpleasant feeling that the answer may be nothing at all: if you look at the mean square over all completely multiplicative functions, then I think you will get (by representing the identity in terms of multiplicative characters).

I’m not sure this line of thought is going anywhere.

]]>The little observation is this. Suppose you have a rational and a character and you know, for some integer that is close to 1. What can you say about ? You can’t say that it is close to 1, but you can say that it is close to 1 with probability 1/2 (if is chosen randomly), since it has to be close to a square root of 1. More generally, is close to 1 with probability So the expected number of such that is close to 1 is about

The problem we were having earlier was that whereas we had “long-range correlations” in the Roth proof (if is roughly constant on then it is roughly constant on all translates of ) there seemed to be nothing comparable here. But perhaps the observation above is saying that there are some tiny correlations after all, and perhaps they can be exploited to get a correspondingly (and necessarily) tiny improvement on the trivial bound.

]]>I’ve thought about Kristall’s suggestion a bit more and I now think that it cannot work. Or perhaps it is more accurate to say that it can work if and only if we manage to solve some problems that we are already stuck on.

Let me try to explain. The characters on the multiplicative group of are just the completely multiplicative functions from to (each of which is determined by its behaviour on ). To represent a completely multiplicative function as a linear combination of HAPs, we would need at least some kind of correlation between and some HAPs. But that requires one to solve EDP (positively) for completely multiplicative functions.

That does, however, suggest that if we knew a sufficiently strong statement about completely multiplicative functions, then we could use the Roth method to deduce EDP for general functions. It seems that for this to work we would need the partial sums of a completely multiplicative complex-valued function to be large on average. Does this ring any bells anyone? It seems that we are being inexorably drawn to Terry’s observation that EDP follows from a suitable statement about completely multiplicative functions taking values in My guess is that if we pushed this idea, we would end up with a proof that is essentially the same as the argument Terry gave.

]]>The thought is that whereas Roth’s theorem gives you an AP of length on which the discrepancy is we know that we cannot achieve that for HAPs and functions on the rationals. The reason is that we have our old friends, the completely multiplicative functions and which extend easily to the rationals. As a reminder of how this is done, let me define the simpler of the two (I can’t remember whether this was or ). Given a rational $p/q,$ we can write it in the form where and neither nor is divisible by 3. We then define to be 1 if mod 3 and -1 if mod 3.

This function is completely multiplicative, and as we have worked out many times, the sum along any HAP of length is at most or thereabouts.

It follows that the best improvement we can hope for over the trivial bound is logarithmic in the lengths of the HAPs that are used in the decomposition.

]]>Are there any theorems saying that a -valued function on the positive integers must have large discrepancy either along HAPs or geometric progressions? That is certainly true for multiplicative functions.

Given that the multiplicative and additive groups are connected by the log-mapping we know what the characters of the multiplicative group are. But using that right seems to require quite a bit of untangling.

]]>That’s certainly an interesting thought — to use characters on the *multiplicative* group of non-negative rationals instead. It wouldn’t instantly solve the problem, because writing a multiplicative character in terms of APs (let alone HAPs) is likely to be much more challenging than decomposing an additive character. But it may be that there is some way of doing it, and then the fact that the set of HAPs of length naturally gives rise to a multiplicative structure would fit in well with the decomposition.

In a way, this illustrates the fundamental difficulty of EDP: that it mixes multiplication and addition in a strange way. Somehow we have to get from one to the other, and what you are suggesting could be thought of as trying to deal with the difficulty in a different place. Definitely worth thinking about.

]]>Could \chi be chosen so it satisfies \chi (x)\chi (y)=\chi (xy)? Then the above problem looks like it could be solved by multiplying by y/x.

]]>In the Roth argument, there are two things one might think of doing in order to approximate the rank-1 matrix where is a character on At our disposal is the fact that for each the function is a non-negative real multiple of and it is a linear combination of HAPs. Moreover, for some values of it is quite a large multiple. So either we could take the function (which will be a non-negative multiple of ) and then tensor it with itself, or we could look at the function It turns out to be very important to do the latter. The reason it helps is that in the decomposition we only ever have products of APs with the same common difference.

I haven’t checked, but I’m pretty sure that if we went down the first route, so that we would have to consider arbitrary pairs of common differences up to then we would lose a factor in the sum of the absolute values of the coefficients. This factor of is precisely what gives us a non-trivial bound at the end of the proof.

Now let’s switch to HAPs and the rationals. Here it is essential to consider products of HAPs with different common differences, or at least it is essential if we want a “local” argument. The reason is that if and are far apart in the graph, then there just isn’t a common difference such that both and are contained in local HAPs of common difference The result of this is that in the expression near the beginning of the previous comment, we ended up having to sum over all pairs

But what if we could somehow get away with looking only at pairs? This is still a slightly vague thought, so let me say what I can and then stop. In the Roth case, we exploit the fact that if is roughly constant on a progression of common difference then it is roughly constant on any translate of that progression. That allows us to restrict attention to pairs of progressions with the same common difference.

What I would like to have is something similar for HAPs and characters. It seems to be far too much to hope that if is roughly constant on a HAP that contains then we can deduce that it is roughly constant on some other HAP that contains However, perhaps we can at least find a class of HAPs of size and say that on at least one HAP in that class must be roughly constant, or perhaps something weaker such as that that is true for almost all

The trouble is that if and are far apart in the graph, then it feels as though we don’t have any correlation at all. So I’m not altogether sure what it is that I am trying to say here. But the general point remains that we seem to have a non-trivial decomposition of something like the identity into HAPs, and if we can make it just the teeniest bit more efficient then we should be in extremely good shape.

]]>In this argument, was playing the role of So let’s set it to 1. Then we get

But so this gives us (That is because for each there is a unique such that at least if we ignore edge effects, as we are allowed to do.)

This looks a bit disastrous, given that we were hoping for a sum of coefficients that was small compared with However, I am not quite sure that it is disastrous. The first piece of potentially good news is that the decomposition we create could well be of a decent-sized multiple of the identity. Let me try to explain why.

We are in some sense approximating each character by the function

Setting to be identically 1, we get a function whose value at is … actually, for technical reasons this doesn’t work — we need to do a double convolution as in the Roth proof. So let me try to reason informally. We are interested in the value of this function at The only for which belongs to some are If is not within about of then the sum of over all such that is small. Otherwise, it is about

We would ordinarily expect to be within of 1 about of the time. So we might expect to get a contribution of for about values of And this is them multiplied by so the total contribution at ought to be around $m\chi(x)/s,$ up to a constant. But that means that we ought to be approximating times the identity rather than the identity itself. Therefore, we want our sum of coefficients to be small compared with

Tantalizingly, what we actually have is *equal* to this, up to a constant.

What I hope this means is that any observation that gives even a tiny improvement to what is going on will solve EDP.

But what possible room is there for improvement? The one place I can see is this: I’ve been talking about representing the identity — indeed, the fact that one could talk about the identity was the whole reason for getting excited about the rationals — but we still have considerable flexibility in our choice of matrix. Indeed, it seems that we have not in fact represented the identity, and my impression is that what we have actually done is represent something slightly “larger” than the identity (because for some and some we have “too many” such that is close to 1, but I think we never have “too few”). Can we somehow squeeze out just the tiniest bit more from the argument?

I have another idea for getting an improvement, but it is a bit harder to explain. I’ll save it for another comment.

]]>In the Roth proof, a (small) technical complication came from the need to choose non-negative coefficients The three criteria that lay behind the choice of these coefficients were as follows.

(i) For each the function should be “nice”, meaning that it has non-negative real Fourier coefficients that sum to 1.

(ii) should be non-zero only if is small.

(iii) For each there should exist such that

I realized last night (and checked this morning) that we can dispense with (ii). All we actually use in the proof is (i) and (iii). Intuitively, (ii) feels necessary, because every time is large, it is contributing to the sum of the coefficients that we are trying to make small. However, that intuitive argument is wrong: it is (iii) that controls the cancellation in the coefficients, and as long as they cancel, we don’t care how big the sum is that cancels.

The upshot of all that is that we can make the sort of simplification that usually sets alarm bells ringing: it is enough to take for every and

Let me say what happens to the Roth write-up if you do that. Up to the end of page 8, the only change needed is to cross out wherever it appears. Also, Lemma 3.1 is no longer needed. Then at the top of page 9, the calculations now look like this. (Recall that is being summed from 1 to )

In my next comment, I’ll say what the picture now looks like for rationals and HAPs.

]]>