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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.

*[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 .

The twin prime conjecture, still unsolved, asserts that there are infinitely many primes such that is also prime. A more precise form of this conjecture is (a special case) of the Hardy-Littlewood prime tuples conjecture, which asserts that

as , where is the von Mangoldt function and is the twin prime constant

Because is almost entirely supported on the primes, it is not difficult to see that (1) implies the twin prime conjecture.

One can give a heuristic justification of the asymptotic (1) (and hence the twin prime conjecture) via sieve theoretic methods. Recall that the von Mangoldt function can be decomposed as a Dirichlet convolution

where is the Möbius function. Because of this, we can rewrite the left-hand side of (1) as

To compute this double sum, it is thus natural to consider sums such as

or (to simplify things by removing the logarithm)

The prime number theorem in arithmetic progressions suggests that one has an asymptotic of the form

where is the multiplicative function with for even and

for odd. Summing by parts, one then expects

and so we heuristically have

The Dirichlet series

has an Euler product factorisation

for ; comparing this with the Euler product factorisation

for the Riemann zeta function, and recalling that has a simple pole of residue at , we see that

has a simple zero at with first derivative

From this and standard multiplicative number theory manipulations, one can calculate the asymptotic

which concludes the heuristic justification of (1).

What prevents us from making the above heuristic argument rigorous, and thus proving (1) and the twin prime conjecture? Note that the variable in (2) ranges to be as large as . On the other hand, the prime number theorem in arithmetic progressions (3) is not expected to hold for anywhere that large (for instance, the left-hand side of (3) vanishes as soon as exceeds ). The best unconditional result known of the type (3) is the Siegel-Walfisz theorem, which allows to be as large as . Even the powerful generalised Riemann hypothesis (GRH) only lets one prove an estimate of the form (3) for up to about .

However, because of the averaging effect of the summation in in (2), we don’t need the asymptotic (3) to be true for *all* in a particular range; having it true for *almost all* in that range would suffice. Here the situation is much better; the celebrated Bombieri-Vinogradov theorem (sometimes known as “GRH on the average”) implies, roughly speaking, that the approximation (3) is valid for *almost all* for any fixed . While this is not enough to control (2) or (1), the Bombieri-Vinogradov theorem can at least be used to control variants of (1) such as

for various sieve weights whose associated divisor function is supposed to approximate the von Mangoldt function , although that theorem only lets one do this when the weights are supported on the range . This is still enough to obtain some partial results towards (1); for instance, by selecting weights according to the Selberg sieve, one can use the Bombieri-Vinogradov theorem to establish the upper bound

which is off from (1) by a factor of about . See for instance this blog post for details.

It has been difficult to improve upon the Bombieri-Vinogradov theorem in its full generality, although there are various improvements to certain restricted versions of the Bombieri-Vinogradov theorem, for instance in the famous work of Zhang on bounded gaps between primes. Nevertheless, it is believed that the Elliott-Halberstam conjecture (EH) holds, which roughly speaking would mean that (3) now holds for almost all for any fixed . (Unfortunately, the factor cannot be removed, as investigated in a series of papers by Friedlander, Granville, and also Hildebrand and Maier.) This comes tantalisingly close to having enough distribution to control all of (1). Unfortunately, it still falls short. Using this conjecture in place of the Bombieri-Vinogradov theorem leads to various improvements to sieve theoretic bounds; for instance, the factor of in (4) can now be improved to .

In two papers from the 1970s (which can be found online here and here respectively, the latter starting on page 255 of the pdf), Bombieri developed what is now known as the *Bombieri asymptotic sieve* to clarify the situation more precisely. First, he showed that on the Elliott-Halberstam conjecture, while one still could not establish the asymptotic (1), one could prove the generalised asymptotic

for all natural numbers , where the generalised von Mangoldt functions are defined by the formula

These functions behave like the von Mangoldt function, but are concentrated on -almost primes (numbers with at most prime factors) rather than primes. The right-hand side of (5) corresponds to what one would expect if one ran the same heuristics used to justify (1). Sadly, the case of (5), which is just (1), is just barely excluded from Bombieri’s analysis.

More generally, on the assumption of EH, the Bombieri asymptotic sieve provides the asymptotic

for any fixed and any tuple of natural numbers other than , where

is a further generalisation of the von Mangoldt function (now concentrated on -almost primes). By combining these asymptotics with some elementary identities involving the , together with the Weierstrass approximation theorem, Bombieri was able to control a wide family of sums including (1), except for one undetermined scalar . Namely, he was able to show (again on EH) that for any fixed and any continuous function on the simplex that had suitable vanishing at the boundary, the sum

when was even, where the integral on is with respect to the measure (this is Dirac measure in the case ). In particular, we have

and the twin prime conjecture would be proved if one could show that is bounded away from zero, while (1) is equivalent to the assertion that is equal to . Unfortunately, no additional bound beyond the inequalities provided by the Bombieri asymptotic sieve is known, even if one assumes all other major conjectures in number theory than the prime tuples conjecture and its variants (e.g. GRH, GEH, GUE, abc, Chowla, …).

To put it another way, the Bombieri asymptotic sieve is able (on EH) to compute asymptotics for sums

without needing to know the unknown scalar , when is a function supported on almost primes of the form

for and some fixed , with vanishing elsewhere and for some continuous (symmetric) functions obeying some vanishing at the boundary, so long as the parity condition

is obeyed (informally: gives the same weight to products of an odd number of primes as to products of an even number of primes, or to put it another way, is asymptotically orthogonal to the Möbius function ). But when violates the parity condition, the asymptotic involves the unknown . This scalar thus embodies the “parity problem” for the twin prime conjecture (discussed in these previous blog posts).

Because the obstruction to the parity problem is only one-dimensional (on EH), one can replace any parity-violating weight (such as ) with any other parity-violating weight and obtain a logically equivalent estimate. For instance, to prove the twin prime conjecture on EH, it would suffice to show that

for some fixed , or equivalently that there are solutions to the equation in primes with and . (In some cases, this sort of reduction can also be made using other sieves than the Bombieri asymptotic sieve, as was observed by Ng.) As another example, the Bombieri asymptotic sieve can be used to show that the asymptotic (1) is equivalent to the asymptotic

where is the set of numbers that are *rough* in the sense that they have no prime factors less than for some fixed (the function clearly correlates with and so must violate the parity condition). One can replace with similar sieve weights (e.g. a Selberg sieve) that concentrate on almost primes if desired.

As it turns out, if one is willing to strengthen the assumption of the Elliott-Halberstam (EH) conjecture to the assumption of the *generalised Elliott-Halberstam (GEH) conjecture* (as formulated for instance in Claim 2.6 of the Polymath8b paper), one can also swap the factor in the above asymptotics with other parity-violating weights and obtain a logically equivalent estimate, as the Bombieri asymptotic sieve also applies to weights such as under the assumption of GEH. For instance, on GEH one can use two such applications of the Bombieri asymptotic sieve to show that the twin prime conjecture would follow if one could show that there are solutions to the equation

in primes with and , for some . Similarly, on GEH the asymptotic (1) is equivalent to the asymptotic

for some fixed , and similarly with replaced by other sieves. This form of the quantitative twin primes conjecture is appealingly similar to the (special case)

of the Chowla conjecture, for which there has been some recent progress (discussed for instance in these recent posts). Informally, the Bombieri asymptotic sieve lets us (on GEH) view the twin prime conjecture as a sort of Chowla conjecture restricted to almost primes. Unfortunately, the recent progress on the Chowla conjecture relies heavily on the multiplicativity of at small primes, which is completely destroyed by inserting a weight such as , so this does not yet yield a viable path towards the twin prime conjecture even assuming GEH. Still, the similarity is striking, and one can hope that further ways to attack the Chowla conjecture may emerge that could impact the twin prime conjecture. (Alternatively, if one assumes a sufficiently optimistic version of the GEH, one could perhaps relax the notion of “almost prime” to the extent that one could start usefully using multiplicativity at smallish primes, though this seems rather wishful at present, particularly since the most optimistic versions of GEH are known to be false.)

The Bombieri asymptotic sieve is already well explained in the original two papers of Bombieri; there is also a slightly different treatment of the sieve by Friedlander and Iwaniec, as well as a simplified version in the book of Friedlander and Iwaniec (in which the distribution hypothesis is strengthened in order to shorten the arguments. I’ve decided though to write up my own notes on the sieve below the fold; this is primarily for my own benefit, but may be useful to some readers also. I largely follow the treatment of Bombieri, with the one idiosyncratic twist of replacing the usual “elementary” Selberg sieve with the “analytic” Selberg sieve used in particular in many of the breakthrough works in small gaps between primes; I prefer working with the latter due to its Fourier-analytic flavour.

** — 1. Controlling generalised von Mangoldt sums — **

To prove (5), we shall first generalise it, by replacing the sequence by a more general sequence obeying the following axioms:

- (i) (Non-negativity) One has for all .
- (ii) (Crude size bound) One has for all , where is the divisor function.
- (iii) (Size) We have for some constant .
- (iv) (Elliott-Halberstam type conjecture) For any , one has
where is a multiplicative function with for all primes and .

These axioms are a little bit stronger than what is actually needed to make the Bombieri asymptotic sieve work, but we will not attempt to work with the weakest possible axioms here.

We introduce the function

which is analytic for ; in particular it can be evaluated at to yield

There are two model examples of data to keep in mind. The first, discussed in the introduction, is when , then and is as in the introduction; one of course needs EH to justify axiom (iv) in this case. The other is when , in which case and for all . We will later take advantage of the second example to avoid doing some (routine, but messy) main term computations.

The main result of this section is then

Theorem 1Let be as above. Let be a tuple of natural numbers (independent of ) that is not equal to . Then one has the asymptoticas , where .

Note that this recovers (5) (on EH) as a special case.

We now begin the proof of this theorem. Henceforth we allow implied constants in the or notation to depend on and .

It will be convenient to replace the range by a shorter range by the following standard localisation trick. Let be a large quantity depending on to be chosen later, and let denote the interval . We will show the estimate

from which the original claim follows by a routine summation argument. Observe from axiom (iv) and the triangle inequality that

for any .

Write for the logarithm function , thus for any . Without loss of generality we may assume that ; we then factor , where

This function is just when . When the function is more complicated, but we at least have the following crude bound:

*Proof:* We induct on . The case is obvious, so suppose and the claim has already been proven for . Since , we see from induction hypothesis and the triangle inequality that

Since by Möbius inversion, the claim follows.

We can write

In the region , we have . Thus

for . The contribution of the error term to to (10) is easily seen to be negligible if is large enough, so we may freely replace with with little difficulty.

If we insert this replacement directly into the left-hand side of (10) and rearrange, we get

We can’t quite control this using axiom (iv) because the range of is a bit too big, as explained in the introduction. So let us introduce a truncated function

where is a small quantity to be chosen later, and is a smooth function that equals on and equals on . Suppose one could establish the following two estimates for any fixed :

where is a quantity that depends on but not on . Then on combining the two estimates we would have

One could in principle compute explicitly from the proof of (13), but one can avoid doing so by the following comparison trick. In the special case , standard multiplicative number theory (noting that the Dirichlet series has a pole of order at , with top Laurent coefficient ) gives the asymptotic

which when compared with (14) for (recalling that in this case) gives the formula

Inserting this back into (14) and recalling that can be made arbitrarily small, we obtain (10).

As it turns out, the estimate (13) is easy to establish, but the estimate (12) is not, roughly speaking because the typical number in has too many divisors in the range , each of which gives a contribution to the error term. (In the book of Friedlander and Iwaniec, the estimate (13) is established anyway, but only after assuming a stronger version of (iv), roughly speaking in which is allowed to be as large as .) To resolve this issue, we will insert a preliminary sieve that will remove most of the potential divisors i the range (leaving only about such divisors on the average for typical ), making the analogue of (12) easier to prove (at the cost of making the analogue of (13) more difficult). Namely, if one can find a function for which one has the estimates

for some quantity that depends on but not on , then by repeating the previous arguments we will again be able to establish (10).

The key estimate is (16). As we shall see, when comparing with , the weight will cost us a factor of , but the term in the definitions of and will recover a factor of , which will give the desired bound since we are assuming .

One has some flexibility in how to select the weight : basically any standard sieve that uses divisors of size at most to localise (at least approximately) to numbers that are rough in the sense that they have no (or at least very few) factors less than , will do. We will use the analytic Selberg sieve choice

where is a smooth function supported on that equals on .

It remains to establish the bounds (15), (16), (17). To warm up and introduce the various methods needed, we begin with the standard bound

where denotes the derivative of . Note the loss of that had previously been pointed out. In the arguments that follows I will be a little brief with the details, as they are standard (see e.g. this previous post).

We now prove (19). The left-hand side can be expanded as

where denotes the least common multiple of and . From the support of we see that the summand is only non-vanishing when . We now use axiom (iv) and split the left-hand side into a main term

and an error term that is at most

From axiom (ii) and elementary multiplicative number theory, we have the bound

so from axiom (iv) and Cauchy-Schwarz we see that the error term (20) is acceptable. Thus it will suffice to establish the bound

The summand here is almost, but not quite, multiplicative in . To make it genuinely multiplicative, we perform a (shifted) Fourier expansion

for some rapidly decreasing function (essentially the Fourier transform of ). Thus

and so the left-hand side of (21) can be rearranged using Fubini’s theorem as

We can factorise as an Euler product:

Taking absolute values and using Mertens’ theorem leads to the crude bound

which when combined with the rapid decrease of , allows us to restrict the region of integration in (23) to the square (say) with negligible error. Next, we use the Euler product

for to factorise

where

For with nonnegative real part, one has

and so by the Weierstrass -test, is continuous at . Since

we thus have

Also, since has a pole of order at with residue , we have

and thus

The quantity (23) can thus be written, up to errors of , as

Using the rapid decrease of , we may remove the restriction on , and it will now suffice to prove the identity

But on differentiating and then squaring (22) we have

and the claim follows by integrating in from zero to infinity (noting that vanishes for ).

We have the following variant of (19):

for any . We also have the variant

If in addition has no prime factors less than for some fixed , one has

Roughly speaking, the above estimates assert that is concentrated on those numbers with no prime factors much less than , but factors without such small prime divisors occur with about the same relative density as they do in the integers.

*Proof:* The left-hand side of (24) can be expanded as

If we define

then the previous expression can be written as

while one has

which gives (25) from Axiom (iv). To prove (24), it now suffices to show that

Arguing as before, the left-hand side is

where

From Mertens’ theorem we have

when , so the contribution of the terms where can be absorbed into the error (after increasing that error slightly). For the remaining contributions, we see that

where if does not divide , and

if divides times for some . In the latter case, Taylor expansion gives the bounds

and the claim (28) follows. When and we have

and (27) follows by repeating the previous calculations. Finally, (26) is proven similarly to (24) (using in place of ).

Now we can prove (15), (16), (17). We begin with (15). Using the Leibniz rule applied to the identity and using and Möbius inversion (and the associativity and commutativity of Dirichlet convolution) we see that

Next, by applying the Leibniz rule to for some and using (29) we see that

and hence we have the recursive identity

In particular, from induction we see that is supported on numbers with at most distinct prime factors, and hence is supported on numbers with at most distinct prime factors. In particular, from (18) we see that on the support of . Thus it will suffice to show that

If and , then has at most distinct prime factors , with . If we factor , where is the contribution of those with , and is the contribution of those with , then at least one of the following two statements hold:

- (a) (and hence ) is divisible by a square number of size at least .
- (b) .

The contribution of case (a) is easily seen to be acceptable by axiom (ii). For case (b), we observe from (30) and induction that

and so it will suffice to show that

where ranges over numbers bounded by with at most distinct prime factors, the smallest of which is at most , and consists of those numbers with no prime factor less than or equal to . Applying (26) (with replaced by ) gives the bound

so by (25) it suffices to show that

subject to the same constraints on as before. The contribution of those with distinct prime factors can be bounded by

applying Mertens’ theorem and summing over , one obtains the claim.

Now we show (16). As discussed previously in this section, we can replace by with negligible error. Comparing this with (16) and (11), we see that it suffices to show that

From the support of , the summand on the left-hand side is only non-zero when , which makes , where we use the crucial hypothesis to gain enough powers of to make the argument here work. Applying Lemma 2, we reduce to showing that

We can make the change of variables to flip the sum

and then swap the sums to reduce to showing that

By Lemma 3, it suffices to show that

To prove this, we use the Rankin trick, bounding the implied weight by . We can then bound the left-hand side by the Euler product

which can be bounded by

and the claim follows from Mertens’ theorem.

Finally, we show (17). By (11), the left-hand side expands as

We let be a small constant to be chosen later. We divide the outer sum into two ranges, depending on whether only has prime factors greater than or not. In the former case, we can apply (27) to write this contribution as

plus a negligible error, where the is implicitly restricted to numbers with all prime factors greater than . The main term is messy, but it is of the required form up to an acceptable error, so there is no need to compute it any further. It remains to consider those that have at least one prime factor less than . Here we use (24) instead of (27) as well as Lemma 3 to dominate this contribution by

up to negligible errors, where is now restricted to have at least one prime factor less than . This makes at least one of the factors to be at most . A routine application of Rankin’s trick shows that

and so the total contribution of this case is . Since can be made arbitrarily small, (17) follows.

** — 2. Weierstrass approximation — **

Having proved Theorem 1, we now take linear combinations of this theorem, combined with the Weierstrass approximation theorem, to give the asymptotics (7), (8) described in the introduction.

Let , , , be as in that theorem. It will be convenient to normalise the weights by to make their mean value comparable to . From Theorem 1 and summation by parts we have

whenever does not consist entirely of ones.

We now take a closer look at what happens when does consist entirely of ones. Let denote the -tuple . Convolving the case of (30) with copies of for some and using the Leibniz rule, we see that

and hence

Multiplying by and summing over , and using (31) to control the term, one has

If we define (up to an error of ) by the formula

then an induction then shows that

for odd , and

for even . In particular, after adjusting by if necessary, we have since the left-hand sides are non-negative.

If we now define the comparison sequence , standard multiplicative number theory shows that the above estimates also hold when is replaced by ; thus

for both odd and even . The bound (31) also holds for when does not consist entirely of ones, and hence

for any fixed (which may or may not consist entirely of ones).

Next, from induction (on ), the Leibniz rule, and (30), we see that for any and , , the function

is a finite linear combination of functions of the form for tuples that may possibly consist entirely of ones. We thus have

whenever is one of these functions (32). Specialising to the case , we thus have

where . The contribution of those that are powers of primes can be easily seen to be negligible, leading to

where now . The contribution of the case where two of the primes agree can also be seen to be negligible, as can the error when replacing with , and then by symmetry

By linearity, this implies that

for any polynomial that vanishes on the coordinate hyperplanes . The right-hand side can also be evaluated by Mertens’ theorem as

when is odd and

when is even. Using the Weierstrass approximation theorem, we then have

for any continuous function that is compactly supported in the interior of . Computing the right-hand side using Mertens’ theorem as before, we obtain the claimed asymptotics (7), (8).

Remark 4The Bombieri asymptotic sieve has to use the full power of EH (or GEH); there are constructions due to Ford that show that if one only has a distributional hypothesis up to for some fixed constant , then the asymptotics of sums such as (5), or more generally (9), are not determined by a single scalar parameter , but can also vary in other ways as well. Thus the Bombieri asymptotic sieve really is asymptotic; in order to get type error terms one needs the level of distribution to be asymptotically equal to as . Related to this, the quantitative decay of the error terms in the Bombieri asymptotic sieve are extremely poor; in particular, they depend on the dependence of implied constant in axiom (iv) on the parameters , for which there is no consensus on what one should conjecturally expect.

A *capset* in the vector space over the finite field of three elements is a subset of that does not contain any lines , where and . A basic problem in additive combinatorics (discussed in one of the very first posts on this blog) is to obtain good upper and lower bounds for the maximal size of a capset in .

Trivially, one has . Using Fourier methods (and the density increment argument of Roth), the bound of was obtained by Meshulam, and improved only as late as 2012 to for some absolute constant by Bateman and Katz. But in a very recent breakthrough, Ellenberg (and independently Gijswijt) obtained the exponentially superior bound , using a version of the polynomial method recently introduced by Croot, Lev, and Pach. (In the converse direction, a construction of Edel gives capsets as large as .) Given the success of the polynomial method in superficially similar problems such as the finite field Kakeya problem (discussed in this previous post), it was natural to wonder that this method could be applicable to the cap set problem (see for instance this MathOverflow comment of mine on this from 2010), but it took a surprisingly long time before Croot, Lev, and Pach were able to identify the precise variant of the polynomial method that would actually work here.

The proof of the capset bound is very short (Ellenberg’s and Gijswijt’s preprints are both 3 pages long, and Croot-Lev-Pach is 6 pages), but I thought I would present a slight reformulation of the argument which treats the three points on a line in symmetrically (as opposed to treating the third point differently from the first two, as is done in the Ellenberg and Gijswijt papers; Croot-Lev-Pach also treat the middle point of a three-term arithmetic progression differently from the two endpoints, although this is a very natural thing to do in their context of ). The basic starting point is this: if is a capset, then one has the identity

for all , where is the Kronecker delta function, which we view as taking values in . Indeed, (1) reflects the fact that the equation has solutions precisely when are either all equal, or form a line, and the latter is ruled out precisely when is a capset.

To exploit (1), we will show that the left-hand side of (1) is “low rank” in some sense, while the right-hand side is “high rank”. Recall that a function taking values in a field is of *rank one* if it is non-zero and of the form for some , and that the rank of a general function is the least number of rank one functions needed to express as a linear combination. More generally, if , we define the *rank* of a function to be the least number of “rank one” functions of the form

for some and some functions , , that are needed to generate as a linear combination. For instance, when , the rank one functions take the form , , , and linear combinations of such rank one functions will give a function of rank at most .

It is a standard fact in linear algebra that the rank of a diagonal matrix is equal to the number of non-zero entries. This phenomenon extends to higher dimensions:

Lemma 1 (Rank of diagonal hypermatrices)Let , let be a finite set, let be a field, and for each , let be a coefficient. Then the rank of the function

*Proof:* We induct on . As mentioned above, the case follows from standard linear algebra, so suppose now that and the claim has already been proven for .

It is clear that the function (2) has rank at most equal to the number of non-zero (since the summands on the right-hand side are rank one functions), so it suffices to establish the lower bound. By deleting from those elements with (which cannot increase the rank), we may assume without loss of generality that all the are non-zero. Now suppose for contradiction that (2) has rank at most , then we obtain a representation

for some sets of cardinalities adding up to at most , and some functions and .

Consider the space of functions that are orthogonal to all the , in the sense that

for all . This space is a vector space whose dimension is at least . A basis of this space generates a coordinate matrix of full rank, which implies that there is at least one non-singular minor. This implies that there exists a function in this space which is nowhere vanishing on some subset of of cardinality at least .

If we multiply (3) by and sum in , we conclude that

where

The right-hand side has rank at most , since the summands are rank one functions. On the other hand, from induction hypothesis the left-hand side has rank at least , giving the required contradiction.

On the other hand, we have the following (symmetrised version of a) beautifully simple observation of Croot, Lev, and Pach:

*Proof:* Using the identity for , we have

The right-hand side is clearly a polynomial of degree in , which is then a linear combination of monomials

with with

In particular, from the pigeonhole principle, at least one of is at most .

Consider the contribution of the monomials for which . We can regroup this contribution as

where ranges over those with , is the monomial

and is some explicitly computable function whose exact form will not be of relevance to our argument. The number of such is equal to , so this contribution has rank at most . The remaining contributions arising from the cases and similarly have rank at most (grouping the monomials so that each monomial is only counted once), so the claim follows.

Upon restricting from to , the rank of is still at most . The two lemmas then combine to give the Ellenberg-Gijswijt bound

All that remains is to compute the asymptotic behaviour of . This can be done using the general tool of Cramer’s theorem, but can also be derived from Stirling’s formula (discussed in this previous post). Indeed, if , , for some summing to , Stirling’s formula gives

where is the entropy function

We then have

where is the maximum entropy subject to the constraints

A routine Lagrange multiplier computation shows that the maximum occurs when

and is approximately , giving rise to the claimed bound of .

Remark 3As noted in the Ellenberg and Gijswijt papers, the above argument extends readily to other fields than to control the maximal size of subset of that has no non-trivial solutions to the equation , where are non-zero constants that sum to zero. Of course one replaces the function in Lemma 2 by in this case.

Remark 4This symmetrised formulation suggests that one possible way to improve slightly on the numerical quantity by finding a more efficient way to decompose into rank one functions, however I was not able to do so (though such improvements are reminiscent of the Strassen type algorithms for fast matrix multiplication).

Remark 5It is tempting to see if this method can get non-trivial upper bounds for sets with no length progressions, in (say) . One can run the above arguments, replacing the functionwith

this leads to the bound where

Unfortunately, is asymptotic to and so this bound is in fact slightly worse than the trivial bound ! However, there is a slim chance that there is a more efficient way to decompose into rank one functions that would give a non-trivial bound on . I experimented with a few possible such decompositions but unfortunately without success.

Remark 6Return now to the capset problem. Since Lemma 1 is valid for any field , one could perhaps hope to get better bounds by viewing the Kronecker delta function as taking values in another field than , such as the complex numbers . However, as soon as one works in a field of characteristic other than , one can adjoin a cube root of unity, and one now has the Fourier decompositionMoving to the Fourier basis, we conclude from Lemma 1 that the function on now has rank exactly , and so one cannot improve upon the trivial bound of by this method using fields of characteristic other than three as the range field. So it seems one has to stick with (or the algebraic completion thereof).

Thanks to Jordan Ellenberg and Ben Green for helpful discussions.

When teaching mathematics, the traditional method of lecturing in front of a blackboard is still hard to improve upon, despite all the advances in modern technology. However, there are some nice things one can do in an electronic medium, such as this blog. Here, I would like to experiment with the ability to animate images, which I think can convey some mathematical concepts in ways that cannot be easily replicated by traditional static text and images. Given that many readers may find these animations annoying, I am placing the rest of the post below the fold.

In functional analysis, it is common to endow various (infinite-dimensional) vector spaces with a variety of topologies. For instance, a normed vector space can be given the strong topology as well as the weak topology; if the vector space has a predual, it also has a weak-* topology. Similarly, spaces of operators have a number of useful topologies on them, including the operator norm topology, strong operator topology, and the weak operator topology. For function spaces, one can use topologies associated to various modes of convergence, such as uniform convergence, pointwise convergence, locally uniform convergence, or convergence in the sense of distributions. (A small minority of such modes are not topologisable, though, the most common of which is pointwise almost everywhere convergence; see Exercise 8 of this previous post).

Some of these topologies are much stronger than others (in that they contain many more open sets, or equivalently that they have many fewer convergent sequences and nets). However, even the weakest topologies used in analysis (e.g. convergence in distributions) tend to be Hausdorff, since this at least ensures the uniqueness of limits of sequences and nets, which is a fundamentally useful feature for analysis. On the other hand, some Hausdorff topologies used are “better” than others in that many more analysis tools are available for those topologies. In particular, topologies that come from Banach space norms are particularly valued, as such topologies (and their attendant norm and metric structures) grant access to many convenient additional results such as the Baire category theorem, the uniform boundedness principle, the open mapping theorem, and the closed graph theorem.

Of course, most topologies placed on a vector space will not come from Banach space norms. For instance, if one takes the space of continuous functions on that converge to zero at infinity, the topology of uniform convergence comes from a Banach space norm on this space (namely, the uniform norm ), but the topology of pointwise convergence does not; and indeed all the other usual modes of convergence one could use here (e.g. convergence, locally uniform convergence, convergence in measure, etc.) do not arise from Banach space norms.

I recently realised (while teaching a graduate class in real analysis) that the closed graph theorem provides a quick explanation for why Banach space topologies are so rare:

Proposition 1Let be a Hausdorff topological vector space. Then, up to equivalence of norms, there is at most one norm one can place on so that is a Banach space whose topology is at least as strong as . In particular, there is at most one topology stronger than that comes from a Banach space norm.

*Proof:* Suppose one had two norms on such that and were both Banach spaces with topologies stronger than . Now consider the graph of the identity function from the Banach space to the Banach space . This graph is closed; indeed, if is a sequence in this graph that converged in the product topology to , then converges to in norm and hence in , and similarly converges to in norm and hence in . But limits are unique in the Hausdorff topology , so . Applying the closed graph theorem (see also previous discussions on this theorem), we see that the identity map is continuous from to ; similarly for the inverse. Thus the norms are equivalent as claimed.

By using various generalisations of the closed graph theorem, one can generalise the above proposition to Fréchet spaces, or even to F-spaces. The proposition can fail if one drops the requirement that the norms be stronger than a specified Hausdorff topology; indeed, if is infinite dimensional, one can use a Hamel basis of to construct a linear bijection on that is unbounded with respect to a given Banach space norm , and which can then be used to give an inequivalent Banach space structure on .

One can interpret Proposition 1 as follows: once one equips a vector space with some “weak” (but still Hausdorff) topology, there is a *canonical* choice of “strong” topology one can place on that space that is stronger than the “weak” topology but arises from a Banach space structure (or at least a Fréchet or F-space structure), provided that at least one such structure exists. In the case of function spaces, one can usually use the topology of convergence in distribution as the “weak” Hausdorff topology for this purpose, since this topology is weaker than almost all of the other topologies used in analysis. This helps justify the common practice of describing a Banach or Fréchet function space just by giving the set of functions that belong to that space (e.g. is the space of Schwartz functions on ) without bothering to specify the precise topology to serve as the “strong” topology, since it is usually understood that one is using the canonical such topology (e.g. the Fréchet space structure on given by the usual Schwartz space seminorms).

Of course, there are still some topological vector spaces which have no “strong topology” arising from a Banach space at all. Consider for instance the space of finitely supported sequences. A weak, but still Hausdorff, topology to place on this space is the topology of pointwise convergence. But there is no norm stronger than this topology that makes this space a Banach space. For, if there were, then letting be the standard basis of , the series would have to converge in , and hence pointwise, to an element of , but the only available pointwise limit for this series lies outside of . But I do not know if there is an easily checkable criterion to test whether a given vector space (equipped with a Hausdorff “weak” toplogy) can be equipped with a stronger Banach space (or Fréchet space or -space) topology.

There is a very nice recent paper by Lemke Oliver and Soundararajan (complete with a popular science article about it by the consistently excellent Erica Klarreich for Quanta) about a surprising (but now satisfactorily explained) bias in the distribution of pairs of consecutive primes when reduced to a small modulus .

This phenomenon is superficially similar to the more well known Chebyshev bias concerning the reduction of a single prime to a small modulus , but is in fact a rather different (and much stronger) bias than the Chebyshev bias, and seems to arise from a completely different source. The Chebyshev bias asserts, roughly speaking, that a randomly selected prime of a large magnitude will typically (though not always) be slightly more likely to be a quadratic non-residue modulo than a quadratic residue, but the bias is small (the difference in probabilities is only about for typical choices of ), and certainly consistent with known or conjectured positive results such as Dirichlet’s theorem or the generalised Riemann hypothesis. The reason for the Chebyshev bias can be traced back to the von Mangoldt explicit formula which relates the distribution of the von Mangoldt function modulo with the zeroes of the -functions with period . This formula predicts (assuming some standard conjectures like GRH) that the von Mangoldt function is quite unbiased modulo . The von Mangoldt function is *mostly* concentrated in the primes, but it also has a medium-sized contribution coming from *squares* of primes, which are of course all located in the quadratic residues modulo . (Cubes and higher powers of primes also make a small contribution, but these are quite negligible asymptotically.) To balance everything out, the contribution of the primes must then exhibit a small preference towards quadratic non-residues, and this is the Chebyshev bias. (See this article of Rubinstein and Sarnak for a more technical discussion of the Chebyshev bias, and this survey of Granville and Martin for an accessible introduction. The story of the Chebyshev bias is also related to Skewes’ number, once considered the largest explicit constant to naturally appear in a mathematical argument.)

The paper of Lemke Oliver and Soundararajan considers instead the distribution of the pairs for small and for large consecutive primes , say drawn at random from the primes comparable to some large . For sake of discussion let us just take . Then all primes larger than are either or ; Chebyshev’s bias gives a very slight preference to the latter (of order , as discussed above), but apart from this, we expect the primes to be more or less equally distributed in both classes. For instance, assuming GRH, the probability that lands in would be , and similarly for .

In view of this, one would expect that up to errors of or so, the pair should be equally distributed amongst the four options , , , , thus for instance the probability that this pair is would naively be expected to be , and similarly for the other three tuples. These assertions are not yet proven (although some non-trivial upper and lower bounds for such probabilities can be obtained from recent work of Maynard).

However, Lemke Oliver and Soundararajan argue (backed by both plausible heuristic arguments (based ultimately on the Hardy-Littlewood prime tuples conjecture), as well as substantial numerical evidence) that there is a significant bias away from the tuples and – informally, adjacent primes don’t like being in the same residue class! For instance, they predict that the probability of attaining is in fact

with similar predictions for the other three pairs (in fact they give a somewhat more precise prediction than this). The magnitude of this bias, being comparable to , is significantly stronger than the Chebyshev bias of .

One consequence of this prediction is that the prime gaps are slightly less likely to be divisible by than naive random models of the primes would predict. Indeed, if the four options , , , all occurred with equal probability , then should equal with probability , and and with probability each (as would be the case when taking the difference of two random numbers drawn from those integers not divisible by ); but the Lemke Oliver-Soundararajan bias predicts that the probability of being divisible by three should be slightly lower, being approximately .

Below the fold we will give a somewhat informal justification of (a simplified version of) this phenomenon, based on the Lemke Oliver-Soundararajan calculation using the prime tuples conjecture.

I’ve been meaning to return to fluids for some time now, in order to build upon my construction two years ago of a solution to an averaged Navier-Stokes equation that exhibited finite time blowup. (I recently spoke on this work in the recent conference in Princeton in honour of Sergiu Klainerman; my slides for that talk are here.)

One of the biggest deficiencies with my previous result is the fact that the averaged Navier-Stokes equation does not enjoy any good equation for the vorticity , in contrast to the true Navier-Stokes equations which, when written in vorticity-stream formulation, become

(Throughout this post we will be working in three spatial dimensions .) So one of my main near-term goals in this area is to exhibit an equation resembling Navier-Stokes as much as possible which enjoys a vorticity equation, and for which there is finite time blowup.

Heuristically, this task should be easier for the Euler equations (i.e. the zero viscosity case of Navier-Stokes) than the viscous Navier-Stokes equation, as one expects the viscosity to only make it easier for the solution to stay regular. Indeed, morally speaking, the assertion that finite time blowup solutions of Navier-Stokes exist should be roughly equivalent to the assertion that finite time blowup solutions of Euler exist which are “Type I” in the sense that all Navier-Stokes-critical and Navier-Stokes-subcritical norms of this solution go to infinity (which, as explained in the above slides, heuristically means that the effects of viscosity are negligible when compared against the nonlinear components of the equation). In vorticity-stream formulation, the Euler equations can be written as

As discussed in this previous blog post, a natural generalisation of this system of equations is the system

where is a linear operator on divergence-free vector fields that is “zeroth order” in some sense; ideally it should also be invertible, self-adjoint, and positive definite (in order to have a Hamiltonian that is comparable to the kinetic energy ). (In the previous blog post, it was observed that the surface quasi-geostrophic (SQG) equation could be embedded in a system of the form (1).) The system (1) has many features in common with the Euler equations; for instance vortex lines are transported by the velocity field , and Kelvin’s circulation theorem is still valid.

So far, I have not been able to fully achieve this goal. However, I have the following partial result, stated somewhat informally:

Theorem 1There is a “zeroth order” linear operator (which, unfortunately, is not invertible, self-adjoint, or positive definite) for which the system (1) exhibits smooth solutions that blowup in finite time.

The operator constructed is not quite a zeroth-order pseudodifferential operator; it is instead merely in the “forbidden” symbol class , and more precisely it takes the form

for some compactly supported divergence-free of mean zero with

being rescalings of . This operator is still bounded on all spaces , and so is arguably still a zeroth order operator, though not as convincingly as I would like. Another, less significant, issue with the result is that the solution constructed does not have good spatial decay properties, but this is mostly for convenience and it is likely that the construction can be localised to give solutions that have reasonable decay in space. But the biggest drawback of this theorem is the fact that is not invertible, self-adjoint, or positive definite, so in particular there is no non-negative Hamiltonian for this equation. It may be that some modification of the arguments below can fix these issues, but I have so far been unable to do so. Still, the construction does show that the circulation theorem is insufficient by itself to prevent blowup.

We sketch the proof of the above theorem as follows. We use the barrier method, introducing the time-varying hyperboloid domains

for (expressed in cylindrical coordinates ). We will select initial data to be for some non-negative even bump function supported on , normalised so that

in particular is divergence-free supported in , with vortex lines connecting to . Suppose for contradiction that we have a smooth solution to (1) with this initial data; to simplify the discussion we assume that the solution behaves well at spatial infinity (this can be justified with the choice (2) of vorticity-stream operator, but we will not do so here). Since the domains disconnect from at time , there must exist a time which is the first time where the support of touches the boundary of , with supported in .

From (1) we see that the support of is transported by the velocity field . Thus, at the point of contact of the support of with the boundary of , the inward component of the velocity field cannot exceed the inward velocity of . We will construct the functions so that this is not the case, leading to the desired contradiction. (Geometrically, what is going on here is that the operator is pinching the flow to pass through the narrow cylinder , leading to a singularity by time at the latest.)

First we observe from conservation of circulation, and from the fact that is supported in , that the integrals

are constant in both space and time for . From the choice of initial data we thus have

for all and all . On the other hand, if is of the form (2) with for some bump function that only has -components, then is divergence-free with mean zero, and

where . We choose to be supported in the slab for some large constant , and to equal a function depending only on on the cylinder , normalised so that . If , then passes through this cylinder, and we conclude that

Inserting ths into (2), (1) we conclude that

for some coefficients . We will not be able to control these coefficients , but fortunately we only need to understand on the boundary , for which . So, if happens to be supported on an annulus , then vanishes on if is large enough. We then have

on the boundary of .

Let be a function of the form

where is a bump function supported on that equals on . We can perform a dyadic decomposition where

where is a bump function supported on with . If we then set

then one can check that for a function that is divergence-free and mean zero, and supported on the annulus , and

so on (where ) we have

One can manually check that the inward velocity of this vector on exceeds the inward velocity of if is large enough, and the claim follows.

Remark 2The type of blowup suggested by this construction, where a unit amount of circulation is squeezed into a narrow cylinder, is of “Type II” with respect to the Navier-Stokes scaling, because Navier-Stokes-critical norms such (or at least ) look like they stay bounded during this squeezing procedure (the velocity field is of size about in cylinders of radius and length about ). So even if the various issues with are repaired, it does not seem likely that this construction can be directly adapted to obtain a corresponding blowup for a Navier-Stokes type equation. To get a “Type I” blowup that is consistent with Kelvin’s circulation theorem, it seems that one needs to coil the vortex lines around a loop multiple times in order to get increased circulation in a small space. This seems possible to pull off to me – there don’t appear to be any unavoidable obstructions coming from topology, scaling, or conservation laws – but would require a more complicated construction than the one given above.

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