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Let be the divisor function. A classical application of the Dirichlet hyperbola method gives the asymptotic

where denotes the estimate as . Much better error estimates are possible here, but we will not focus on the lower order terms in this discussion. For somewhat idiosyncratic reasons I will interpret this estimate (and the other analytic number theory estimates discussed here) through the probabilistic lens. Namely, if is a random number selected uniformly between and , then the above estimate can be written as

that is to say the random variable has mean approximately . (But, somewhat paradoxically, this is not the median or mode behaviour of this random variable, which instead concentrates near , basically thanks to the Hardy-Ramanujan theorem.)

Now we turn to the pair correlations for a fixed positive integer . There is a classical computation of Ingham that shows that

The error term in (2) has been refined by many subsequent authors, as has the uniformity of the estimates in the aspect, as these topics are related to other questions in analytic number theory, such as fourth moment estimates for the Riemann zeta function; but we will not consider these more subtle features of the estimate here. However, we will look at the next term in the asymptotic expansion for (2) below the fold.

Using our probabilistic lens, the estimate (2) can be written as

From (1) (and the asymptotic negligibility of the shift by ) we see that the random variables and both have a mean of , so the additional factor of represents some arithmetic coupling between the two random variables.

Ingham’s formula can be established in a number of ways. Firstly, one can expand out and use the hyperbola method (splitting into the cases and and removing the overlap). If one does so, one soon arrives at the task of having to estimate sums of the form

for various . For much less than this can be achieved using a further application of the hyperbola method, but for comparable to things get a bit more complicated, necessitating the use of non-trivial estimates on Kloosterman sums in order to obtain satisfactory control on error terms. A more modern approach proceeds using automorphic form methods, as discussed in this previous post. A third approach, which unfortunately is only heuristic at the current level of technology, is to apply the Hardy-Littlewood circle method (discussed in this previous post) to express (2) in terms of exponential sums for various frequencies . The contribution of “major arc” can be computed after a moderately lengthy calculation which yields the right-hand side of (2) (as well as the correct lower order terms that are currently being suppressed), but there does not appear to be an easy way to show directly that the “minor arc” contributions are of lower order, although the methods discussed previously do indirectly show that this is ultimately the case.

Each of the methods outlined above requires a fair amount of calculation, and it is not obvious while performing them that the factor will emerge at the end. One can at least explain the as a normalisation constant needed to balance the factor (at a heuristic level, at least). To see this through our probabilistic lens, introduce an independent copy of , then

using symmetry to order (discarding the diagonal case ) and making the change of variables , we see that (4) is heuristically consistent with (3) as long as the asymptotic mean of in is equal to . (This argument is not rigorous because there was an implicit interchange of limits present, but still gives a good heuristic “sanity check” of Ingham’s formula.) Indeed, if denotes the asymptotic mean in , then we have (heuristically at least)

and we obtain the desired consistency after multiplying by .

This still however does not explain the presence of the factor. Intuitively it is reasonable that if has many prime factors, and has a lot of factors, then will have slightly more factors than average, because any common factor to and will automatically be acquired by . But how to quantify this effect?

One heuristic way to proceed is through analysis of local factors. Observe from the fundamental theorem of arithmetic that we can factor

where the product is over all primes , and is the local version of at (which in this case, is just one plus the –valuation of : ). Note that all but finitely many of the terms in this product will equal , so the infinite product is well-defined. In a similar fashion, we can factor

where

(or in terms of valuations, ). Heuristically, the Chinese remainder theorem suggests that the various factors behave like independent random variables, and so the correlation between and should approximately decouple into the product of correlations between the local factors and . And indeed we do have the following local version of Ingham’s asymptotics:

Proposition 1 (Local Ingham asymptotics)For fixed and integer , we haveand

From the Euler formula

we see that

and so one can “explain” the arithmetic factor in Ingham’s asymptotic as the product of the arithmetic factors in the (much easier) local Ingham asymptotics. Unfortunately we have the usual “local-global” problem in that we do not know how to rigorously derive the global asymptotic from the local ones; this problem is essentially the same issue as the problem of controlling the minor arc contributions in the circle method, but phrased in “physical space” language rather than “frequency space”.

Remark 2The relation between the local means and the global mean can also be seen heuristically through the applicationof Mertens’ theorem, where is Pólya’s magic exponent, which serves as a useful heuristic limiting threshold in situations where the product of local factors is divergent.

Let us now prove this proposition. One could brute-force the computations by observing that for any fixed , the valuation is equal to with probability , and with a little more effort one can also compute the joint distribution of and , at which point the proposition reduces to the calculation of various variants of the geometric series. I however find it cleaner to proceed in a more recursive fashion (similar to how one can prove the geometric series formula by induction); this will also make visible the vague intuition mentioned previously about how common factors of and force to have a factor also.

It is first convenient to get rid of error terms by observing that in the limit , the random variable converges vaguely to a uniform random variable on the profinite integers , or more precisely that the pair converges vaguely to . Because of this (and because of the easily verified uniform integrability properties of and their powers), it suffices to establish the exact formulae

in the profinite setting (this setting will make it easier to set up the recursion).

We begin with (5). Observe that is coprime to with probability , in which case is equal to . Conditioning to the complementary probability event that is divisible by , we can factor where is also uniformly distributed over the profinite integers, in which event we have . We arrive at the identity

As and have the same distribution, the quantities and are equal, and (5) follows by a brief amount of high-school algebra.

We use a similar method to treat (6). First treat the case when is coprime to . Then we see that with probability , and are simultaneously coprime to , in which case . Furthermore, with probability , is divisible by and is not; in which case we can write as before, with and . Finally, in the remaining event with probability , is divisible by and is not; we can then write , so that and . Putting all this together, we obtain

and the claim (6) in this case follows from (5) and a brief computation (noting that in this case).

Now suppose that is divisible by , thus for some integer . Then with probability , and are simultaneously coprime to , in which case . In the remaining event, we can write , and then and . Putting all this together we have

which by (5) (and replacing by ) leads to the recursive relation

and (6) then follows by induction on the number of powers of .

The estimate (2) of Ingham was refined by Estermann, who obtained the more accurate expansion

for certain complicated but explicit coefficients . For instance, is given by the formula

where is the Euler-Mascheroni constant,

The formula for is similar but even more complicated. The error term was improved by Heath-Brown to ; it is conjectured (for instance by Conrey and Gonek) that one in fact has square root cancellation here, but this is well out of reach of current methods.

These lower order terms are traditionally computed either from a Dirichlet series approach (using Perron’s formula) or a circle method approach. It turns out that a refinement of the above heuristics can also predict these lower order terms, thus keeping the calculation purely in physical space as opposed to the “multiplicative frequency space” of the Dirichlet series approach, or the “additive frequency space” of the circle method, although the computations are arguably as messy as the latter computations for the purposes of working out the lower order terms. We illustrate this just for the term below the fold.

Fifteen years ago, I wrote a paper entitled Global regularity of wave maps. II. Small energy in two dimensions, in which I established global regularity of wave maps from two spatial dimensions to the unit sphere, assuming that the initial data had small energy. Recently, Hao Jia (personal communication) discovered a small gap in the argument that requires a slightly non-trivial fix. The issue does not really affect the subsequent literature, because the main result has since been reproven and extended by methods that avoid the gap (see in particular this subsequent paper of Tataru), but I have decided to describe the gap and its fix on this blog.

I will assume familiarity with the notation of my paper. In Section 10, some complicated spaces are constructed for each frequency scale , and then a further space is constructed for a given frequency envelope by the formula

where is the Littlewood-Paley projection of to frequency magnitudes . Then, given a spacetime slab , we define the restrictions

where the infimum is taken over all extensions of to the Minkowski spacetime ; similarly one defines

The gap in the paper is as follows: it was implicitly assumed that one could restrict (1) to the slab to obtain the equality

(This equality is implicitly used to establish the bound (36) in the paper.) Unfortunately, (1) only gives the lower bound, not the upper bound, and it is the upper bound which is needed here. The problem is that the extensions of that are optimal for computing are not necessarily the Littlewood-Paley projections of the extensions of that are optimal for computing .

To remedy the problem, one has to prove an upper bound of the form

for all Schwartz (actually we need affinely Schwartz , but one can easily normalise to the Schwartz case). Without loss of generality we may normalise the RHS to be . Thus

for each , and one has to find a single extension of such that

for each . Achieving a that obeys (4) is trivial (just extend by zero), but such extensions do not necessarily obey (5). On the other hand, from (3) we can find extensions of such that

the extension will then obey (5) (here we use Lemma 9 from my paper), but unfortunately is not guaranteed to obey (4) (the norm does control the norm, but a key point about frequency envelopes for the small energy regularity problem is that the coefficients , while bounded, are not necessarily summable).

This can be fixed as follows. For each we introduce a time cutoff supported on that equals on and obeys the usual derivative estimates in between (the time derivative of size for each ). Later we will prove the truncation estimate

Assuming this estimate, then if we set , then using Lemma 9 in my paper and (6), (7) (and the local stability of frequency envelopes) we have the required property (5). (There is a technical issue arising from the fact that is not necessarily Schwartz due to slow decay at temporal infinity, but by considering partial sums in the summation and taking limits we can check that is the strong limit of Schwartz functions, which suffices here; we omit the details for sake of exposition.) So the only issue is to establish (4), that is to say that

for all .

For this is immediate from (2). Now suppose that for some integer (the case when is treated similarly). Then we can split

where

The contribution of the term is acceptable by (6) and estimate (82) from my paper. The term sums to which is acceptable by (2). So it remains to control the norm of . By the triangle inequality and the fundamental theorem of calculus, we can bound

By hypothesis, . Using the first term in (79) of my paper and Bernstein’s inequality followed by (6) we have

and then we are done by summing the geometric series in .

It remains to prove the truncation estimate (7). This estimate is similar in spirit to the algebra estimates already in my paper, but unfortunately does not seem to follow immediately from these estimates as written, and so one has to repeat the somewhat lengthy decompositions and case checkings used to prove these estimates. We do this below the fold.

*[This blog post was written jointly by Terry Tao and Will Sawin.]*

In the previous blog post, one of us (Terry) implicitly introduced a notion of rank for tensors which is a little different from the usual notion of tensor rank, and which (following BCCGNSU) we will call “slice rank”. This notion of rank could then be used to encode the Croot-Lev-Pach-Ellenberg-Gijswijt argument that uses the polynomial method to control capsets.

Afterwards, several papers have applied the slice rank method to further problems – to control tri-colored sum-free sets in abelian groups (BCCGNSU, KSS) and from there to the triangle removal lemma in vector spaces over finite fields (FL), to control sunflowers (NS), and to bound progression-free sets in -groups (P).

In this post we investigate the notion of slice rank more systematically. In particular, we show how to give lower bounds for the slice rank. In many cases, we can show that the upper bounds on slice rank given in the aforementioned papers are sharp to within a subexponential factor. This still leaves open the possibility of getting a better bound for the original combinatorial problem using the slice rank of some other tensor, but for very long arithmetic progressions (at least eight terms), we show that the slice rank method cannot improve over the trivial bound using any tensor.

It will be convenient to work in a “basis independent” formalism, namely working in the category of abstract finite-dimensional vector spaces over a fixed field . (In the applications to the capset problem one takes to be the finite field of three elements, but most of the discussion here applies to arbitrary fields.) Given such vector spaces , we can form the tensor product , generated by the tensor products with for , subject to the constraint that the tensor product operation is multilinear. For each , we have the smaller tensor products , as well as the tensor product

defined in the obvious fashion. Elements of of the form for some and will be called *rank one functions*, and the *slice rank* (or *rank* for short) of an element of is defined to be the least nonnegative integer such that is a linear combination of rank one functions. If are finite-dimensional, then the rank is always well defined as a non-negative integer (in fact it cannot exceed . It is also clearly subadditive:

For , is when is zero, and otherwise. For , is the usual rank of the -tensor (which can for instance be identified with a linear map from to the dual space ). The usual notion of tensor rank for higher order tensors uses complete tensor products , as the rank one objects, rather than , giving a rank that is greater than or equal to the slice rank studied here.

From basic linear algebra we have the following equivalences:

Lemma 1Let be finite-dimensional vector spaces over a field , let be an element of , and let be a non-negative integer. Then the following are equivalent:

- (i) One has .
- (ii) One has a representation of the form
where are finite sets of total cardinality at most , and for each and , and .

- (iii) One has
where for each , is a subspace of of total dimension at most , and we view as a subspace of in the obvious fashion.

- (iv) (Dual formulation) There exist subspaces of the dual space for , of total dimension at least , such that is orthogonal to , in the sense that one has the vanishing
for all , where is the obvious pairing.

*Proof:* The equivalence of (i) and (ii) is clear from definition. To get from (ii) to (iii) one simply takes to be the span of the , and conversely to get from (iii) to (ii) one takes the to be a basis of the and computes by using a basis for the tensor product consisting entirely of functions of the form for various . To pass from (iii) to (iv) one takes to be the annihilator of , and conversely to pass from (iv) to (iii).

One corollary of the formulation (iv), is that the set of tensors of slice rank at most is Zariski closed (if the field is algebraically closed), and so the slice rank itself is a lower semi-continuous function. This is in contrast to the usual tensor rank, which is not necessarily semicontinuous.

Corollary 2Let be finite-dimensional vector spaces over an algebraically closed field . Let be a nonnegative integer. The set of elements of of slice rank at most is closed in the Zariski topology.

*Proof:* In view of Lemma 1(i and iv), this set is the union over tuples of integers with of the projection from of the set of tuples with orthogonal to , where is the Grassmanian parameterizing -dimensional subspaces of .

One can check directly that the set of tuples with orthogonal to is Zariski closed in using a set of equations of the form locally on . Hence because the Grassmanian is a complete variety, the projection of this set to is also Zariski closed. So the finite union over tuples of these projections is also Zariski closed.

We also have good behaviour with respect to linear transformations:

Lemma 3Let be finite-dimensional vector spaces over a field , let be an element of , and for each , let be a linear transformation, with the tensor product of these maps. Then

Furthermore, if the are all injective, then one has equality in (2).

Thus, for instance, the rank of a tensor is intrinsic in the sense that it is unaffected by any enlargements of the spaces .

*Proof:* The bound (2) is clear from the formulation (ii) of rank in Lemma 1. For equality, apply (2) to the injective , as well as to some arbitrarily chosen left inverses of the .

Computing the rank of a tensor is difficult in general; however, the problem becomes a combinatorial one if one has a suitably sparse representation of that tensor in some basis, where we will measure sparsity by the property of being an antichain.

Proposition 4Let be finite-dimensional vector spaces over a field . For each , let be a linearly independent set in indexed by some finite set . Let be a subset of .

where for each , is a coefficient in . Then one has

where the minimum ranges over all coverings of by sets , and for are the projection maps.

Now suppose that the coefficients are all non-zero, that each of the are equipped with a total ordering , and is the set of maximal elements of , thus there do not exist distinct , such that for all . Then one has

In particular, if is an antichain (i.e. every element is maximal), then equality holds in (4).

*Proof:* By Lemma 3 (or by enlarging the bases ), we may assume without loss of generality that each of the is spanned by the . By relabeling, we can also assume that each is of the form

with the usual ordering, and by Lemma 3 we may take each to be , with the standard basis.

Let denote the rank of . To show (4), it suffices to show the inequality

for any covering of by . By removing repeated elements we may assume that the are disjoint. For each , the tensor

can (after collecting terms) be written as

for some . Summing and using (1), we conclude the inequality (6).

Now assume that the are all non-zero and that is the set of maximal elements of . To conclude the proposition, it suffices to show that the reverse inequality

holds for some covering . By Lemma 1(iv), there exist subspaces of whose dimension sums to

Let . Using Gaussian elimination, one can find a basis of whose representation in the standard dual basis of is in row-echelon form. That is to say, there exist natural numbers

such that for all , is a linear combination of the dual vectors , with the coefficient equal to one.

We now claim that is disjoint from . Suppose for contradiction that this were not the case, thus there exists for each such that

As is the set of maximal elements of , this implies that

for any tuple other than . On the other hand, we know that is a linear combination of , with the coefficient one. We conclude that the tensor product is equal to

plus a linear combination of other tensor products with not in . Taking inner products with (3), we conclude that , contradicting the fact that is orthogonal to . Thus we have disjoint from .

For each , let denote the set of tuples in with not of the form . From the previous discussion we see that the cover , and we clearly have , and hence from (8) we have (7) as claimed.

As an instance of this proposition, we recover the computation of diagonal rank from the previous blog post:

Example 5Let be finite-dimensional vector spaces over a field for some . Let be a natural number, and for , let be a linearly independent set in . Let be non-zero coefficients in . Thenhas rank . Indeed, one applies the proposition with all equal to , with the diagonal in ; this is an antichain if we give one of the the standard ordering, and another of the the opposite ordering (and ordering the remaining arbitrarily). In this case, the are all bijective, and so it is clear that the minimum in (4) is simply .

The combinatorial minimisation problem in the above proposition can be solved asymptotically when working with tensor powers, using the notion of the Shannon entropy of a discrete random variable .

Proposition 6Let be finite-dimensional vector spaces over a field . For each , let be a linearly independent set in indexed by some finite set . Let be a non-empty subset of .Let be a tensor of the form (3) for some coefficients . For each natural number , let be the tensor power of copies of , viewed as an element of . Then

and range over the random variables taking values in .

Now suppose that the coefficients are all non-zero and that each of the are equipped with a total ordering . Let be the set of maximal elements of in the product ordering, and let where range over random variables taking values in . Then

as . In particular, if the maximizer in (10) is supported on the maximal elements of (which always holds if is an antichain in the product ordering), then equality holds in (9).

*Proof:*

as , where is the projection map. Then the same thing will apply to and . Then applying Proposition 4, using the lexicographical ordering on and noting that, if are the maximal elements of , then are the maximal elements of , we obtain both (9) and (11).

We first prove the lower bound. By compactness (and the continuity properties of entropy), we can find a random variable taking values in such that

Let be a small positive quantity that goes to zero sufficiently slowly with . Let denote the set of all tuples in that are within of being distributed according to the law of , in the sense that for all , one has

By the asymptotic equipartition property, the cardinality of can be computed to be

if goes to zero slowly enough. Similarly one has

Now let be an arbitrary covering of . By the pigeonhole principle, there exists such that

which by (13) implies that

noting that the factor can be absorbed into the error). This gives the lower bound in (12).

Now we prove the upper bound. We can cover by sets of the form for various choices of random variables taking values in . For each such random variable , we can find such that ; we then place all of in . It is then clear that the cover and that

for all , giving the required upper bound.

It is of interest to compute the quantity in (10). We have the following criterion for when a maximiser occurs:

Proposition 7Let be finite sets, and be non-empty. Let be the quantity in (10). Let be a random variable taking values in , and let denote the essential range of , that is to say the set of tuples such that is non-zero. Then the following are equivalent:

- (i) attains the maximum in (10).
- (ii) There exist weights and a finite quantity , such that whenever , and such that
for all , with equality if . (In particular, must vanish if there exists a with .)

Furthermore, when (i) and (ii) holds, one has

*Proof:* We first show that (i) implies (ii). The function is concave on . As a consequence, if we define to be the set of tuples such that there exists a random variable taking values in with , then is convex. On the other hand, by (10), is disjoint from the orthant . Thus, by the hyperplane separation theorem, we conclude that there exists a half-space

where are reals that are not all zero, and is another real, which contains on its boundary and in its interior, such that avoids the interior of the half-space. Since is also on the boundary of , we see that the are non-negative, and that whenever .

By construction, the quantity

is maximised when . At this point we could use the method of Lagrange multipliers to obtain the required constraints, but because we have some boundary conditions on the (namely, that the probability that they attain a given element of has to be non-negative) we will work things out by hand. Let be an element of , and an element of . For small enough, we can form a random variable taking values in , whose probability distribution is the same as that for except that the probability of attaining is increased by , and the probability of attaining is decreased by . If there is any for which and , then one can check that

for sufficiently small , contradicting the maximality of ; thus we have whenever . Taylor expansion then gives

for small , where

and similarly for . We conclude that for all and , thus there exists a quantity such that for all , and for all . By construction must be nonnegative. Sampling using the distribution of , one has

almost surely; taking expectations we conclude that

The inner sum is , which equals when is non-zero, giving (17).

Now we show conversely that (ii) implies (i). As noted previously, the function is concave on , with derivative . This gives the inequality

for any (note the right-hand side may be infinite when and ). Let be any random variable taking values in , then on applying the above inequality with and , multiplying by , and summing over and gives

By construction, one has

and

so to prove that (which would give (i)), it suffices to show that

or equivalently that the quantity

is maximised when . Since

it suffices to show this claim for the quantity

One can view this quantity as

By (ii), this quantity is bounded by , with equality if is equal to (and is in particular ranging in ), giving the claim.

The second half of the proof of Proposition 7 only uses the marginal distributions and the equation(16), not the actual distribution of , so it can also be used to prove an upper bound on when the exact maximizing distribution is not known, given suitable probability distributions in each variable. The logarithm of the probability distribution here plays the role that the weight functions do in BCCGNSU.

Remark 8Suppose one is in the situation of (i) and (ii) above; assume the nondegeneracy condition that is positive (or equivalently that is positive). We can assign a “degree” to each element by the formula

then every tuple in has total degree at most , and those tuples in have degree exactly . In particular, every tuple in has degree at most , and hence by (17), each such tuple has a -component of degree less than or equal to for some with . On the other hand, we can compute from (19) and the fact that for that . Thus, by asymptotic equipartition, and assuming , the number of “monomials” in of total degree at most is at most ; one can in fact use (19) and (18) to show that this is in fact an equality. This gives a direct way to cover by sets with , which is in the spirit of the Croot-Lev-Pach-Ellenberg-Gijswijt arguments from the previous post.

We can now show that the rank computation for the capset problem is sharp:

Proposition 9Let denote the space of functions from to . Then the function from to , viewed as an element of , has rank as , where is given by the formula

*Proof:* In , we have

Thus, if we let be the space of functions from to (with domain variable denoted respectively), and define the basis functions

of indexed by (with the usual ordering), respectively, and set to be the set

then is a linear combination of the with , and all coefficients non-zero. Then we have . We will show that the quantity of (10) agrees with the quantity of (20), and that the optimizing distribution is supported on , so that by Proposition 6 the rank of is .

To compute the quantity at (10), we use the criterion in Proposition 7. We take to be the random variable taking values in that attains each of the values with a probability of , and each of with a probability of ; then each of the attains the values of with probabilities respectively, so in particular is equal to the quantity in (20). If we now set and

we can verify the condition (16) with equality for all , which from (17) gives as desired.

This statement already follows from the result of Kleinberg-Sawin-Speyer, which gives a “tri-colored sum-free set” in of size , as the slice rank of this tensor is an upper bound for the size of a tri-colored sum-free set. If one were to go over the proofs more carefully to evaluate the subexponential factors, this argument would give a stronger lower bound than KSS, as it does not deal with the substantial loss that comes from Behrend’s construction. However, because it actually constructs a set, the KSS result rules out more possible approaches to give an exponential improvement of the upper bound for capsets. The lower bound on slice rank shows that the bound cannot be improved using only the slice rank of this particular tensor, whereas KSS shows that the bound cannot be improved using any method that does not take advantage of the “single-colored” nature of the problem.

We can also show that the slice rank upper bound in a result of Naslund-Sawin is similarly sharp:

Proposition 10Let denote the space of functions from to . Then the function from , viewed as an element of , has slice rank

*Proof:* Let and be a basis for the space of functions on , itself indexed by . Choose similar bases for and , with and .

Set . Then is a linear combination of the with , and all coefficients non-zero. Order the usual way so that is an antichain. We will show that the quantity of (10) is , so that applying the last statement of Proposition 6, we conclude that the rank of is ,

Let be the random variable taking values in that attains each of the values with a probability of . Then each of the attains the value with probability and with probability , so

Setting and , we can verify the condition (16) with equality for all , which from (17) gives as desired.

We used a slightly different method in each of the last two results. In the first one, we use the most natural bases for all three vector spaces, and distinguish from its set of maximal elements . In the second one we modify one basis element slightly, with instead of the more obvious choice , which allows us to work with instead of . Because is an antichain, we do not need to distinguish and . Both methods in fact work with either problem, and they are both about equally difficult, but we include both as either might turn out to be substantially more convenient in future work.

Proposition 11Let be a natural number and let be a finite abelian group. Let be any field. Let denote the space of functions from to .Let be any -valued function on that is nonzero only when the elements of form a -term arithmetic progression, and is nonzero on every -term constant progression.

Then the slice rank of is .

*Proof:* We apply Proposition 4, using the standard bases of . Let be the support of . Suppose that we have orderings on such that the constant progressions are maximal elements of and thus all constant progressions lie in . Then for any partition of , can contain at most constant progressions, and as all constant progressions must lie in one of the , we must have . By Proposition 4, this implies that the slice rank of is at least . Since is a tensor, the slice rank is at most , hence exactly .

So it is sufficient to find orderings on such that the constant progressions are maximal element of . We make several simplifying reductions: We may as well assume that consists of all the -term arithmetic progressions, because if the constant progressions are maximal among the set of all progressions then they are maximal among its subset . So we are looking for an ordering in which the constant progressions are maximal among all -term arithmetic progressions. We may as well assume that is cyclic, because if for each cyclic group we have an ordering where constant progressions are maximal, on an arbitrary finite abelian group the lexicographic product of these orderings is an ordering for which the constant progressions are maximal. We may assume , as if we have an -tuple of orderings where constant progressions are maximal, we may add arbitrary orderings and the constant progressions will remain maximal.

So it is sufficient to find orderings on the cyclic group such that the constant progressions are maximal elements of the set of -term progressions in in the -fold product ordering. To do that, let the first, second, third, and fifth orderings be the usual order on and let the fourth, sixth, seventh, and eighth orderings be the reverse of the usual order on .

Then let be a constant progression and for contradiction assume that is a progression greater than in this ordering. We may assume that , because otherwise we may reverse the order of the progression, which has the effect of reversing all eight orderings, and then apply the transformation , which again reverses the eight orderings, bringing us back to the original problem but with .

Take a representative of the residue class in the interval . We will abuse notation and call this . Observe that , and are all contained in the interval modulo . Take a representative of the residue class in the interval . Then is in the interval for some . The distance between any distinct pair of intervals of this type is greater than , but the distance between and is at most , so is in the interval . By the same reasoning, is in the interval . Therefore . But then the distance between and is at most , so by the same reasoning is in the interval . Because is between and , it also lies in the interval . Because is in the interval , and by assumption it is congruent mod to a number in the set greater than or equal to , it must be exactly . Then, remembering that and lie in , we have and , so , hence , thus , which contradicts the assumption that .

In fact, given a -term progressions mod and a constant, we can form a -term binary sequence with a for each step of the progression that is greater than the constant and a for each step that is less. Because a rotation map, viewed as a dynamical system, has zero topological entropy, the number of -term binary sequences that appear grows subexponentially in . Hence there must be, for large enough , at least one sequence that does not appear. In this proof we exploit a sequence that does not appear for .

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