The Chowla conjecture asserts that all non-trivial correlations of the Liouville function are asymptotically negligible; for instance, it asserts that

as for any fixed natural number . This conjecture remains open, though there are a number of partial results (e.g. these two previous results of Matomaki, Radziwill, and myself).

A natural generalisation of Chowla’s conjecture was proposed by Elliott. For simplicity we will only consider Elliott’s conjecture for the pair correlations

For such correlations, the conjecture was that one had

as for any natural number , as long as was a completely multiplicative function with magnitude bounded by , and such that

for any Dirichlet character and any real number . In the language of “pretentious number theory”, as developed by Granville and Soundararajan, the hypothesis (2) asserts that the completely multiplicative function does not “pretend” to be like the completely multiplicative function for any character and real number . A condition of this form is necessary; for instance, if is precisely equal to and has period , then is equal to as and (1) clearly fails. The prime number theorem in arithmetic progressions implies that the Liouville function obeys (2), and so the Elliott conjecture contains the Chowla conjecture as a special case.

As it turns out, Elliott’s conjecture is false as stated, with the counterexample having the property that “pretends” *locally* to be the function for in various intervals , where and go to infinity in a certain prescribed sense. See this paper of Matomaki, Radziwill, and myself for details. However, we view this as a technicality, and continue to believe that certain “repaired” versions of Elliott’s conjecture still hold. For instance, our counterexample does not apply when is restricted to be real-valued rather than complex, and we believe that Elliott’s conjecture is valid in this setting. Returning to the complex-valued case, we still expect the asymptotic (1) provided that the condition (2) is replaced by the stronger condition

as for all fixed Dirichlet characters . In our paper we supported this claim by establishing a certain “averaged” version of this conjecture; see that paper for further details. (See also this recent paper of Frantzikinakis and Host which establishes a different averaged version of this conjecture.)

One can make a stronger “non-asymptotic” version of this corrected Elliott conjecture, in which the parameter does not go to infinity, or equivalently that the function is permitted to depend on :

Conjecture 1 (Non-asymptotic Elliott conjecture)Let , let be sufficiently large depending on , and let be sufficiently large depending on . Suppose that is a completely multiplicative function with magnitude bounded by , such thatfor all Dirichlet characters of period at most . Then one has

for all natural numbers .

The -dependent factor in the constraint is necessary, as can be seen by considering the completely multiplicative function (for instance). Again, the results in my previous paper with Matomaki and Radziwill can be viewed as establishing an averaged version of this conjecture.

Meanwhile, we have the following conjecture that is the focus of the Polymath5 project:

Conjecture 2 (Erdös discrepancy conjecture)For any function , the discrepancyis infinite.

It is instructive to compute some near-counterexamples to Conjecture 2 that illustrate the difficulty of the Erdös discrepancy problem. The first near-counterexample is that of a non-principal Dirichlet character that takes values in rather than . For this function, one has from the complete multiplicativity of that

If denotes the period of , then has mean zero on every interval of length , and thus

Thus has bounded discrepancy.

Of course, this is not a true counterexample to Conjecture 2 because can take the value . Let us now consider the following variant example, which is the simplest member of a family of examples studied by Borwein, Choi, and Coons. Let be the non-principal Dirichlet character of period (thus equals when , when , and when ), and define the completely multiplicative function by setting when and . This is about the simplest modification one can make to the previous near-counterexample to eliminate the zeroes. Now consider the sum

with for some large . Writing with coprime to and at most , we can write this sum as

Now observe that . The function has mean zero on every interval of length three, and is equal to mod , and thus

for every , and thus

Thus also has unbounded discrepancy, but only barely so (it grows logarithmically in ). These examples suggest that the main “enemy” to proving Conjecture 2 comes from completely multiplicative functions that somehow “pretend” to be like a Dirichlet character but do not vanish at the zeroes of that character. (Indeed, the special case of Conjecture 2 when is completely multiplicative is already open, appears to be an important subcase.)

All of these conjectures remain open. However, I would like to record in this blog post the following striking connection, illustrating the power of the Elliott conjecture (particularly in its nonasymptotic formulation):

Theorem 3 (Elliott conjecture implies unbounded discrepancy)Conjecture 1 implies Conjecture 2.

The argument relies heavily on two observations that were previously made in connection with the Polymath5 project. The first is a Fourier-analytic reduction that replaces the Erdos Discrepancy Problem with an averaged version for completely multiplicative functions . An application of Cauchy-Schwarz then shows that any counterexample to that version will violate the conclusion of Conjecture 1, so if one assumes that conjecture then must pretend to be like a function of the form . One then uses (a generalisation) of a second argument from Polymath5 to rule out this case, basically by reducing matters to a more complicated version of the Borwein-Choi-Coons analysis. Details are provided below the fold.

There is some hope that the Chowla and Elliott conjectures can be attacked, as the parity barrier which is so impervious to attack for the twin prime conjecture seems to be more permeable in this setting. (For instance, in my previous post I raised a possible approach, based on establishing expander properties of a certain random graph, which seems to get around the parity problem, in principle at least.)

(Update, Sep 25: fixed some treatment of error terms, following a suggestion of Andrew Granville.)

** — 1. Fourier reduction — **

We will prove Theorem 3 by contradiction, assuming that there is a function with bounded discrepancy and then concluding a violation of the Elliott conjecture.

The function need not have any multiplicativity properties, but by using an argument from Polymath5 we can extract a *random* completely multiplicative function which also has good discrepancy properties (albeit in an probabilistic sense only):

Proposition 4 (Fourier reduction)Suppose that is a function such thatThen there exists a random completely multiplicative function of magnitude such that

uniformly for all natural numbers (we allow implied constants to depend on ).

*Proof:* For the readers convenience, we reproduce the Polymath5 argument.

The space of completely multiplicative functions of magnitude can be identified with the infinite product since is determined by its values at the primes. In particular, this space is compact metrisable in the product topology. The functions are continuous in this topology for all . By vague compactness of probability measures on compact metrisable spaces (Prokhorov’s theorem), it thus suffices to construct, for each , a random completely multiplicative function of magnitude such that

for all , where the implied constant is uniform in and .

for all (the implied constant can depend on but is otherwise absolute). Let , and let be the primes up to . Let be a natural number that we assume to be sufficiently large depending on . Define a function by the formula

for . We also define the function by setting whenever is in (this is well defined for ). Applying (4) for and of the form with , we conclude that

for all and all but of the elements of . For the exceptional elements, we have the trivial bound

Square-summing in , we conclude (if is sufficiently large depending on ) that

By Fourier expansion, we can write

where , , and

A little Fourier-analytic calculation then allows us to write the left-hand side of (5) as

On the other hand, from the Plancherel identity we have

and so we can interpret as the probability distribution of a random frequency . The estimate (5) now takes the form

for all . If we then define the completely multiplicative function by setting for , and for all other primes, we obtain

for all , as desired.

Remark 5A similar reduction applies if the original function took values in the unit sphere of a complex Hilbert space, rather than in . Conversely, the random constructed above can be viewed as an element of a unit sphere of a suitable Hilbert space, so the conclusion of Proposition 4 is in fact logically equivalent to failure of the Hilbert space version of the Erdös discrepancy conjecture.

Remark 6From linearity of expectation, we see from Proposition 4 that for any natural number , we haveand hence for each we conclude that there exists a

deterministiccompletely multiplicative function of unit magnitude such thatThis was the original formulation of the Fourier reduction in Polymath5, however the fact that varied with made this formulation inconvenient for our argument.

** — 2. Applying the Elliott conjecture — **

Suppose for contradiction that Conjecture 1 holds but that there exists a function of bounded discrepancy in the sense of (3). By Proposition 4, we may thus find a random completely multiplicative function of magnitude such that

We now use Elliott’s conjecture as a sort of “inverse theorem” (in the spirit of the inverse sumset theorem of Frieman, and the inverse theorems for the Gowers uniformity norms) to force to pretend to behave like a modulated character quite often.

Proposition 7Let the hypotheses and notation be as above. Let , and suppose that is sufficiently large depending on . Then with probability , one can find a Dirichlet character of period and a real number such that

*Proof:* We use the van der Corput trick. Let be a moderately large natural number depending on to be chosen later, and suppose that is sufficiently large depending on . From (6) and the triangle inequality we have

so from Markov’s inequality we see with probability that

Let us condition to this event. Shifting by we conclude (for large enough) that

and hence by the triangle inequality

which we rewrite as

We can square the left-hand side out as

The diagonal term contributes to this expression. Thus, for sufficiently large depending on , we can apply the triangle inequality and pigeonhole principle to find *distinct* such that

By symmetry we can take . Setting , we conclude (for large enough) that

Applying Conjecture 1 in the contrapositive, we obtain the claim.

The conclusion (8) asserts that in some sense, “pretends” to be like the function ; as it has magnitude one, it should resemble the function discussed in the introduction. The remaining task is to find some generalisation of the argument that shows that had (logarithmically) large discrepancy to show that likewise fails to obey (6).

** — 3. Ruling out correlation with modulated characters — **

We now use (a generalisation of) this Polymath5 argument. Let be the random completely multiplicative function provided by Proposition 4. We will need the following parameters:

- A sufficiently small quantity .
- A natural number that is sufficiently large depending on .
- A quantity that is sufficiently small depending on .
- A natural number that is sufficiently large depending on .
- A real number that is sufficiently large depending on .

By Proposition 7, we see with probability that there exists a Dirichlet character of period and a real number such that

By reducing if necessary we may assume that is primitive.

It will be convenient to cut down the size of .

Lemma 8With probability , one has

*Proof:* By Proposition 7 with replaced by , we see that with probability , one can find a Dirichlet character of period and a real number such that

We may assume that , since we are done otherwise. Applying the pretentious triangle inequality (see Lemma 3.1 of this paper of Granville and Soundararajan), we conclude that

However, from the Vinogradov-Korobov zero-free region for (see this previous blog post) it is not difficult to show that

if is sufficiently large depending on , a contradiction. The claim follows.

Let us now condition to the probability event that , exist obeying (8) and the bound (9).

The bound (8) asserts that “pretends” to be like the completely multiplicative function . We can formalise this by making the factorisation

where is the completely multiplicative function of magnitude defined by setting for and for , and is the completely multiplicative function of magnitude defined by setting for , and for . The function should be compared with the function of the same name studied in the introduction.

The bound (8) then becomes

We now perform some manipulations to remove the and factors from and isolate the factor, which is more tractable to compute with; then we will perform more computations to arrive at an expression just involving which we will be able to evaluate fairly easily.

From (6) and the triangle inequality we have

for all (even after conditioning to the event). The averaging will not be used until much later in the argument, and the reader may wish to ignore it for now.

By (10), the above estimate can be written as

For we can use (9) to conclude that . The contribution of the error term is negligible, thus

for all . We can factor out the to obtain

For we can crudely bound the left-hand side by . If is sufficiently small, we can then sum weighted by and conclude that

(The zeta function type weight of will be convenient later in the argument when one has to perform some multiplicative number theory, as the relevant sums can be computed quite directly and easily using Euler products.) Thus, with probability , one has

We condition to this event. We have successfully eliminated the role of ; we now work to eliminate . Call a residue class *bad* if is divisible by for some and . and *good* otherwise. We restrict to good residue classes, thus

By Cauchy-Schwarz, we conclude that

Now we claim that for a in a good residue class , the quantity does not depend on . Indeed, by hypothesis, is not divisible by for any and is thus a factor of , and is coprime to . We then factor

where in the last line we use the periodicity of . Thus we have , and so

Shifting by we see that

Now, we perform some multiplicative number theory to understand the innermost sum. From taking Euler products we have

for ; from (11) and Mertens’ theorem one can easily verify that

More generally, for any Dirichlet character we have

Since

we have

which after using (11), Cauchy-Schwarz (using and Mertens theorem gives

for any non-principal character of period dividing ; for a principal character of period dividing we have

since and hence for all , where we have also used (13). By expansion into Dirichlet characters we conclude that

for all and primitive residue classes . For non-primitive residue classes , we write and . The previous arguments then give

which since gives (again using (13))

for all (not necessarily primitive). Inserting this back into (12) we see that

and thus by (13) we conclude (for large enough) that

We have now eliminated both and . The remaining task is to establish some lower bound on the discrepancy of that will contradict (14). As mentioned above, this will be a more complicated variant of the Borwein-Choi-Coons analysis in the introduction. The square in (14) will be helpful in dealing with the fact that we don’t have good control on the for (as we shall see, the squaring introduces two terms of this type that end up cancelling each other).

We expand (14) to obtain

Write and , thus and for we have

We thus have

We reinstate the bad . The number of such is at most , so their total contribution here is which is negligible, thus

For or , the inner sum is , which by the divisor bound gives a negligible contribution. Thus we may restrict to . Note that as is already restricted to numbers coprime to , and divide , we may replace the constraints with for .

We consider the contribution of an off-diagonal term for a fixed choice of . To handle these terms we expand the non-principal character as a linear combination of for with Fourier coefficients . Thus we can expand out

as a linear combination of expressions of the form

with and coefficients of size .

The constraints are either inconsistent, or constrain to a single residue class . Writing , we have

for some phase that can depend on but is independent of . If , then at least one of the two quantities and is divisible by a prime that does not divide the other quantity. Therefore cannot be divisible by , and thus by . We can then sum the geometric series in (or ) and conclude that

and so by the divisor bound the off-diagonal terms contribute at most to (15). For large, this is negligible, and thus we only need to consider the diagonal contribution . Here the terms helpfully cancel, and we obtain

We have now eliminated , leaving only the Dirichlet character which is much easier to work with. We gather terms and write the left-hand side as

The summand in is now non-negative. We can thus throw away all the except of the form with , to conclude that

It is now that we finally take advantage of the averaging to simplify the summation. Observe from the triangle inequality that for any and one has

summing over we conclude that

In particular, by the pigeonhole principle there exists such that

Shifting by and discarding some terms, we conclude that

Observe that for a fixed there is exactly one in the inner sum, and . Thus we have

Making the change of variables , we thus have

But is periodic of period with mean , thus

and thus

which leads to a contradiction for large enough (note the logarithmic growth in here, matching the logarithmic growth in the Borwein-Choi-Coons analysis). The claim follows.

## 46 comments

Comments feed for this article

11 September, 2015 at 7:19 am

AnonymousSince the LHS of (2) is dependent on , it is not sufficiently clear which values of should satisfy (2).

[Oops, there was a typo: (2) was meant to hold for all values of . I’ve edited the text to reflect this. -T.]12 September, 2015 at 3:39 am

Uwe StroinskiIn the proof of Proposition 4 near the end, the definition of should contain instead of .

[Corrected, thanks – T.]12 September, 2015 at 8:26 am

Gil KalaiDear Terry,

This is very cool. Two questions: first, there is some probabilistic heuristics that for Erdos discrepancy problem the actual Sup behaves like sqrt (log n). Does this makes sense/fits other heuristics, when you transfer it to Elliot conjecture? Second, Do Elliott or Chowla conjectures (perhaps in some strong forms or for some special cases) follow from GRH or similar conjectures?

12 September, 2015 at 11:34 am

Terence TaoLet’s restrict to completely multiplicative functions for the sake of discussion (the Fourier reduction already more or less does this, at the cost of making the completely multiplicative function stochastic instead of deterministic). One can make a distinction between the “pretentious” case when the multiplicative function behaves like a modulated character , and the “non-pretentious” case when it doesn’t. For instance, the Borwein-Choi-Coons example (and the randomised version of that example that is giving the discrepancy) are in the “pretentious” category.

The argument I give in Section 3 to handle the pretentious case does indeed seem to give a type lower bound (there is a squaring at some point, and then the final contradiction is by a logarithmic factor). To get such a bound overall, though, one needs a quantitative version of the Elliott conjecture (i.e. some info on how depends on ). The averaged version of Elliott’s conjecture in my paper with Matomaki and Radziwill has such a quantitative version but there are some error terms that involve small powers of which would prevent one from getting the discrepancy lower bound all the way up to . But these are probably artefacts of our method and if one assumed a sufficiently strong quantitative version of Elliott’s conjecture one could imagine that one could get close to the conjectural .

Conjectures like GRH are very helpful in multiplicative number theory, for instance in controlling sums such as or for multiplicative f. However they are not nearly as helpful for non-multiplicative situations such as controlling . Of course, they certainly don’t hurt, but I think the improvements to existing Chowla type results from assuming GRH are actually relatively modest.

13 September, 2015 at 7:39 am

AnonymousIn conjecture 1, the “sufficiently large” and are independent of the function . Is this independence really important?

13 September, 2015 at 8:12 am

Terence TaoYes, for instance for the application to the discrepancy problem it is essential that the parameters A and X do not depend on g. If they are allowed to depend on g then the conjecture more or less reverts back to the qualitative version of the Elliott conjecture discussed in the introduction to the blog post.

13 September, 2015 at 9:42 am

John MangualPresumably, Polymath 5 is active??

I am surprised by Conjecture 2. No idea why bounded self-correlation (say for the Mobius function would lead to infinite discrepancy.

17 September, 2015 at 11:16 pm

mixedmathI’m afraid not. Polymath5 has been quite dormant for a few years now. At least, online it has. Sometimes questions follow you around, wake you up at night, and the like.

13 September, 2015 at 6:36 pm

The Erdos discrepancy problem | Exploratoria[…] via the Elliot Conjecture ~ via Terry Tao. […]

14 September, 2015 at 11:00 pm

AnonymousConcerning conjecture 2, what is known about best lower/upper bounds of

?

15 September, 2015 at 7:09 am

Terence TaoThe best upper bound is (possibly with an iterated logarithm correction) for some absolute constant , using a randomised version of the Borwein-Choi-Coons example (which already gives a bound of the form ); I don’t have a reference for this though (perhaps some of the Polymath5 participants would know). The best lower bound, remarkably, is just (for ); see this paper of Konev and Lisitsa.

EDIT: See also Tim Gowers’ survey on these problems, recently made available on the arXiv at http://arxiv.org/abs/1509.03421

20 September, 2015 at 12:24 pm

JosephI am wondering how one can randomise the Borwein-Choi-Coons example in order to get (I don’t know how to do it and I see in your paper (p. 5-6): “although it is unclear to the author whether such a slowly diverging example can also be attained for Corollary 1.2”). What is more precisely known about this question?

20 September, 2015 at 2:10 pm

Terence TaoYes, the situation here appears to be a bit subtle. One can construct a random multiplicative function such that for all , which gives growth on the average, by randomising Borwein-Choi-Coons, but this isn’t the same as locating a deterministic such that for all n. As I said, it may be that some other Polymath5 participants would know the precise status of this question.

17 September, 2015 at 11:14 pm

mixedmathDear Terry,

This is very interesting. I’ve kept thinking about the Erdos Discrepancy Problem on and off since the Polymath Project. I notice your new preprint on the arXiv, which is very exciting.

I wonder: can you extend to a complexified Erdos Discrepancy Problem, where the underlying function perhaps is of form where are the th roots of unity? Or perhaps all the way to ?

I don’t quite have the intuition to know when these should be true, but I do feel that something more general should be true. If I recall correctly, the Fourier transition from general to multiplicative functions relied on Plancherel’s identity, and that . I suppose this would be the first question to consider.

18 September, 2015 at 9:10 am

mixedmathI realize I commented too soon. I’ll use the excuse that it was late, and I was tired.

Could you say a few more words about how to extend proposition 4 to the unit sphere?

18 September, 2015 at 4:11 am

Updates and plans III. | Combinatorics and more[…] by Terry Tao (1), (2) (reporting also on subsequent works by Matomaki, Radzwill, and Tao) NEW (3) (4)). Update (Sept 18, 2015): Terry Tao have just uploaded a paper to the arxive where he solves the […]

18 September, 2015 at 4:46 pm

The logarithmically averaged Chowla and Elliott conjectures for two-point correlations; the Erdos discrepancy problem | What's new[…] pair of papers is an outgrowth of these two recent blog posts and the ensuing discussion. In the first paper, we establish the following logarithmically averaged […]

19 September, 2015 at 10:37 am

Gergely HarcosThis is truly beautiful. In the Erdős discrepancy paper, in the last display of page 10, I think that should be in order to apply Theorem 1.8. (Similarly, in the next display, should be , but this display is not even needed as one can apply the last display of page 10 directly).

[This will be corrected in the next revision of the MS – T.]20 September, 2015 at 2:49 am

EDP28 — problem solved by Terence Tao! | Gowers's Weblog[…] Tao has solved the Erdős discrepancy problem. He has blogged about the solution in two posts, a first that shows how to reduce the problem to the Elliott conjecture in number theory, and a second that […]

20 September, 2015 at 10:03 pm

Uwe StroinskiCongratulation! A quick skim through your xarchive papers shows that the Fourier-reduction part is essentially a copy of your post here. Since it is published now I would like to ask you whether it is possible to spell out the routine Fourier argument. As far as I can see in (2.2) (or (5) here) is not always in the domain of (take eg and j = 2). Is that a problem?

21 September, 2015 at 7:35 am

Terence TaoThe domain of is the finite abelian group , not the discrete cube (which is a fundamental domain for that group). In particular, the addition takes place in this group and so does not leave the domain of . (The corresponding addition operation on the discrete cube can leave that cube, which is why there are the exceptions to the inequality two displays before (5).)

21 September, 2015 at 9:12 am

Uwe StroinskiGot it. The result then follows from linearity, translation property (applied to $\pi(j)$ and Plancherel. Thanks a lot.

20 September, 2015 at 11:27 pm

Royce PengOn reddit, the following interesting question was asked concerning the Erdos discrepancy problem: Can one always choose a single infinite sequence whose partial sums are unbounded? i.e. does there exist a d such that ?

Sorry if this questions is obvious or already answered.

21 September, 2015 at 7:46 am

Terence TaoThis is a nice question. Looks like there is a counterexample as follows. Set , and then perform the following recursive construction for :

Informally, the block is alternating copies of the block .

Thus the sequence is

.

If is a natural number and is any odd number larger than , then we see that the block is D+1 alternating copies of and thus sums to zero; in fact if we divide into consecutive blocks of length then all such blocks sum to zero, and so is finite for all d.

24 September, 2015 at 8:46 am

Frogs and Lily Pads and Discrepancy | Gödel's Lost Letter and P=NP[…] Conjecture” (DC) of Paul Erdős. This emerged from discussion of his two earlier posts this month on his blog. They and his 9/18 announcement post re-create much of the content of the […]

26 September, 2015 at 1:16 pm

Математик Теренс Тао обогнал компьютер в решении проблемы несоответствия Эрдеша | news.siras.ru[…] литературы). Аргументы математика использовали специального вида гипотезу Эллиота-Халберстама (о распределении […]

26 September, 2015 at 2:23 pm

MichaelI am a big fan of Paul Erdos and his legacy to mathematics. I am Physics major and was recently reading and discussing this paper with some friends (Pysicists and Mathematicians) but could not get grasp of it all. I asked the people at Fermat’s Library (platform to annotate and discuss scientific papers) to upload the paper and was wondering if anyone wanted to help me annotate it: http://fermatslibrary.com/s/the-erdos-discrepancy-problem

26 September, 2015 at 5:58 pm

Richard StanleyThe Erdős discrepancy problem is vaguely similar to another long-standing problem, the circulant Hadamard matrix conjecture. Namely, if then there does not exist a sequence of ‘s that is orthogonal to every non-identity cyclic shift of itself. Can any of these discrepancy techniques be applied to Hadamard circulants?

27 September, 2015 at 2:09 pm

Terence TaoThis is a nice question! I spent a bit of time looking at the literature on this problem (though it seems one has to be careful of the claimed proofs of the conjecture that didn’t work out). It’s unlikely that the methods here can be applied directly to the circulant problem, since multiplicative sequences seem to play no special role in the circulant problem while being central to the discrepancy problem. But one could imagine the following strategy, analogous to the EDP strategy of generalising +1/-1 sequences to Hilbert-space valued sequences, then reducing to (stochastic) multiplicative sequences, and thence to multiplicative sequences that pretend to be modulated Dirichlet characters.

Let’s first generalise the situation by considering sequences taking values in rather than . Now there are a family of counterexamples for every odd coming from the quadratic phases for integers with coprime to . Naively, one could conjecture that these are essentially the only counterexamples, in that any -valued sequence orthogonal to its cyclic shifts must “pretend” to be a quadratic phase in some sense. This would reduce the problem to figuring out which sequences pretend to be quadratic, and checking them individually for orthogonality, which looks like a relatively tractable problem. But this could be too naive, and there could be some more exotic sequences than quadratic which are also orthogonal to their cyclic shifts. Perhaps the recent advances in design theory may help clarify whether such exotic sequences would be expected to exist? (I haven’t followed the design theory literature too closely though.)

27 September, 2015 at 4:18 pm

Mark LewkoIt is interesting to note that a Dirichlet character mod a prime is nearly a counterexample. By this I mean that the shifts of such a Dirichlet character are nearly orthogonal in the sense that for . This is a bit unintuitive, at least to me, as one might expect this sum to be . On the other hand, Dirichlet characters do not correlate with quadratic phases (by Weil’s bound). Thus a reduction will need a mechanism for distinguishing between orthogonal and nearly orthogonal sequences (or, perhaps, one should consider twisted quadratic phases).

28 September, 2015 at 6:32 am

Richard StanleyWe can border the “Dirichlet character matrix” with a row and column of 1’s to obtain a Hadamard matrix. This is a construction of Paley. See

https://en.wikipedia.org/wiki/Paley_construction.

28 September, 2015 at 6:48 am

Mark LewkoYes but of course that destroys the circulant property.

From Freddie Manners’ mathoverflow post (linked below) one sees that there are in fact “exotic” complex circulant Hadamard matrices. On the other hand the example he gives there correlates with a quadratic Dirichlet character. Perhaps one can naively move the goal to something like: a complex circulant matrix either “pretends” to be a quadratic phase or a Dirichlet character.

27 September, 2015 at 11:20 pm

Mark LewkoOn further thought, here is some support for your proposed strategy:

Let be a unimodular circulant sequence. Define the shift of by as . By hypothesis . By Plancherel one then has . Using the fact that the Fourier transform takes translations to modulations this implies for . In turn, this implies that =1. One is thus left with the question of when a function and its Fourier transform have constant modulus. It seems there is some literature on this question, for instance see this mathoverflow post (http://mathoverflow.net/questions/181102/functions-f-on-mathbbz-n-mathbbz-with-f-and-widehatf-constan). In particular, it seems to be known that if for then must be a quadratic polynomial. That seems very close to what you are looking for.

(I see Yemon Choi pointed the mathoverflow discussion referenced to a paper on “circulant complex Hadamard matrices” so it’s likely everything I just stated is well know in the context of this problem.)

28 September, 2015 at 1:26 pm

Terence TaoInteresting that these results are ultimately powered by Segre’s theorem in combinatorial finite field incidence geometry! (Though it seems they are currently restricted to cases when N has very few prime factors.) There was already some interest in trying to find more robust versions of Segre type theorems (perhaps somehow using the modern polynomial method) but as far as I know there isn’t a breakthrough in that area yet. One may have to wait for (or to encourage) a further advance in this area (which would be more or less an exact analogue of the situation with Polymath5 and the Erdos discrepancy problem).

28 September, 2015 at 10:35 pm

Daniel TungHi,

I wonder if the discrepancy problem would have any impact to the meaning of probability (especially its frequency interpretation).

29 September, 2015 at 9:54 pm

275A, Notes 0: Foundations of probability theory | What's new[…] and Elliott conjectures, and random multiplicative functions similarly played a central role in the paper on the Erdos discrepancy problem), this will actually be the first time I will be teaching a course on probability itself (although […]

2 October, 2015 at 10:10 pm

TECNOLOGÍA » 275A, Notes 0: Foundations of probability theory[…] and Elliott conjectures, and random multiplicative functions similarly played a central role in the paper on the Erdos discrepancy problem), this will actually be the first time I will be teaching a course on probability itself (although […]

6 October, 2015 at 9:06 am

Terence Tao magic breakthrough again with EDP, Erdos Discrepancy Problem | Turing Machine[…] 7. The Erdos discrepancy problem via the Elliott conjecture / Tao blog […]

7 October, 2015 at 10:25 am

Anonymousif you represent a decision problem with 1 and -1 as outputs instead of 0 and 1 does this problem contribute to analyse decision problems?

19 October, 2015 at 1:01 am

Il problema della discrepanza: provata una congettura di Erdős? | OggiScienza[…] dibatte, Tao ha fatto una cronologia dei suggerimenti più utili sia sul blog di Tim Gowers che sul proprio (ha scritto anche una seconda parte) dove la peer-review ha preceduto quella della rivista, […]

21 October, 2015 at 10:12 pm

EDP Reflections and Celebrations | Combinatorics and more[…] argument proving EDC based on a number-theoretic conjecture (The Eliott Conjecture) can be found in this very readable blog post by Tao. In 2010 Tim Gowers ran a polymath project devoted to the Erdős discrepancy problem (EDP). A […]

2 November, 2015 at 7:21 am

William BourisTao Number, t(n) = (18n)^n – (4(n+1))^n +3^[(1/2)(n-1)(n+2)]; t(0)=0, t(1)=11, t(2)=1161, t(3)=153,611>130,000, etc.

11 December, 2015 at 6:29 am

AnonymousConsider a generalization of Erdos discrepancy problem where takes values in a set of positive numbers. The problem is to determine all possible such sets for which the generalized discrepancy problem has an affirmative answer. Is it possible for such to have more than one element ?

8 November, 2016 at 3:02 am

Eulogio GarciaCounterexample to the Erdös discrepency problem.

We know that:

So:

(i)

Being the greated integer that exists.

hence the sum of a finite set of terms always converges; that is to say.

To have the expression (i) if we have that

this is obtained as follows.

10 November, 2016 at 4:50 am

Eulogio GarciaA more concrete and precise explanation of the counterexample.

(i)

This is always that the value of in the last summing is ; in .

Now if we apply the Erdos conjecture ( sum of terms equidistant in (i) we will have.

Because: it does not allow the consecutive sum of and therefore we will never a value tan .

example: .

for the sum we have.

13 March, 2018 at 1:06 pm

Terence Tao dimostra una congettura di Erdős (with a little help from his friends) - Maddmaths![…] giorno fa Terence Tao ha annunciato (qui e qui) di aver dimostrato una congettura di Pál (Paul) Erdős, nota come il problema della discrepanza. […]