A similar finiteness result exists for Laurent series , which are the analogues of reals modulo 1. Namely, associate to the infinite matrix . Then Hankel’s theorem states that is rational if and only if the matrix is of finite rank.

We show here a quantitative version of this statement. Let

The rank of this matrix is related to Diophantine properties of .

To see this, we introduce for the fractional part and the norm , where is the largest such that . Then we have the following result.

If the rank of is at most , then there exist a decomposition and a monic polynomial of degree such that . The converse holds too.

The classical theory of Hankel matrices (see Chapter X, Paragraph 10 of Gantmacher’s book) tells us that if the rank of the matrix is , then there exist integers and such that and the first rows are linearly independent, each of the following rows is a linear combination of them,

and the minor formed of the first and last rows and columns is nonzero. Let be the rows of . Write where is the last coefficient of and the row of the first coefficients. By the above, we have a relation

. In fact, we shall show that for any , we have

This equation holds for . So we argue by induction and assume it holds for any for some and prove it for . To start with, the displayed equation implies that

Applying the induction hypothesis iteratively, we find coefficients for and such that

and

These coefficients satisfy the initial condition and the recurrence relations for and .

On the other hand, we know that there exist coefficients

such that

Comparing the last equation to the one involving , and using the linear independence of the first lines, we find that .

Thus . But by definition of the coefficients . We infer that ,

which conclude the inductive argument.

To see the connexion with Diophantine properties of , notice that is the row of the first coefficients of . Thus the validity of the identity for for all implies that

where satisfies .

That is, using the polynomial , we find that which concludes the proof of the direct statement. The converse is straightforward.

Now we consider a two-coordinates analogous problem, that is, we consider a set and define the sets

and , V and H meaning vertical and horizontal. What are the sets that satisfy both and ? Call such a set a *transverse* set. Natural exemples are cartesian products of vector spaces (which we shall call rectangles) as well as zero sets of bilinear forms. More generally, a set of the form for some subspaces and some bilinear forms is horizontally and vertically closed. Call such a set a *bilinear* set, and call its codimensions the codimensions of the subspaces and the number of bilinear forms required (the dimension of the space of bilinear forms vanishing on it). What is the relationship between transverse sets and bilinear sets? Are all transverse sets essentially bilinear sets?

For a transverse set and , let be the fiber above x. Thus and is a vector space. Actually depends only the projective class . Moreover, the stability under horizontal operations is equivalent to the property that if is on the projective line spanned by and , we have .

**The codimension 1 case**

Suppose and each subspace has codimension at most 1. Write with some vector uniquely defined up to homothety. It is easy to see that the set of such that is a vector subspace which we call . Furthermore, if , we have , so that descends to an injective map on . Thus it is enough to study the case where has codimension exactly one unless . In other words, we have a map satisfying the property that if is on the projective line spanned by and , then lies on the projective line spanned by and . Thus maps projective lines into lines. Actually it can be quickly checked that such a map is either bijective or constant on each single projective line.

If we suppose it is injective, too, then by the fundamental theorem of projective geometry, we conclude that it is a projective map. Thus it is induced by a linear map . Hence writing , we have shown that is the zero set of the bilinear form .

In general is far from being injective. It can for instance be constant on (equal to for some hyperplane ). In this case . This is neither a rectangle, nor the zero form of a quadratic form, because if it was, could be taken linear, and because , it should be injective. In fact, if it is the intersection of zero sets of a family of bilinear forms, there needs to be bilinear forms in this family, such as the forms if (wlog) is the hyperplane . However obviously contains a very large rectangle.

These are actually the only cases: such a map is either injective or constant. Indeed, if is just a line, it’s obvious, so suppose . Suppose that maps projective lines into lines (in the sense above) but is neither injective nor constant. That is, there exist two distinct points such that , and a third point satisfying . This implies that span a projective plane. Take a point on the line spanned by . Because is a bijection on both lines , you can find on such that . Now consider the intersection . Then you have , so that on the line , the map is neither constant nor injective, which is absurd.

**Arbitrary codimension**

A more general potential counterexample, communicated to me by Ben Green, is a set of the form where are sequences of increasing (resp. decreasing) subspaces. Thus for , we have where . Again you can make it a bilinear set, but you need atrociously many bilinear forms.

However it contains a large rectangle. In fact, while so that the density of is at most . Thus . Finally by the pigeonhole principle, one of the cartesian products have density at least .

In general, we have a vector space . Suppose for instance each has codimension 2. If we manage to find maps such that for each , we are in good shape. Though it is not clear how to show that such a consistent choice of bases can be made…

**The conjecture**

If a has cardinality at least , it contains a bilinear set of (linear and bilinear) codimensions .

]]>

Let be system of linear forms, i.e and

where for any and . For a measurable function , one may define the -count of as

Here by a slight abuse of notation, the measure on denotes in fact the product measure of the Haar measure on .

Let

When a system is fixed, we may omit it from the notation . Given a sequence of finite groups of cardinality tending to infinity, it is natural to consider the sequence . When the infimum is taken only over -valued measurable functions of density at least , this is the minimum number of APs a set of density at least can have. But these infimum are actually the same (basically by convexity).

The case was introduced by Croot. He proved that the minimal density of three-term arithmetic progressions converged as through the primes. Sisask, in his PhD thesis, was able to express this limit as an integral over the corresponding limit object, namely the circle , thus

As customary, the corresponding problem in finite field models is cleaner. Thus we set . Whereas Croot noticed that the sequence was in some sense approximately decreasing, it is easy to see that for any system , the sequence is genuinely non-increasing. This is because embeds naturally in ; thus if is a map from to , it can be extended into by putting .

Now , and . This proves that , which implies that this sequence has a limit.

Let be now fixed. We want to find an appealing expression for the limit of the sequence . It is thus tempting to look for a natural topological compact group that would play the role of limit object of the sequence , just as played the role of limit object for the sequence .

The analogy leads one to introducing and . The group equipped with the product topology is a compact group by Tychonoff’s theorem, and equipped with the discrete topology is locally compact; it is the Pontryagin dual of . It is now reasonable to ask whether

One easily notices that for any by extending, as above, a function into a function on with the same density and the same -count. Indeed, one can define

and then is a measurable function which has same density and same -count.

We prove the other inequality. For this we prove two intermediate propositions. We write from henceforth .

**Proposition 1**

If is a sequence of -valued functions on that converges in to a -valued function , then .

**Proof**

Let . We show that converges in to 0, which will conclude. In fact we prove a stronger convergence, a convergence in . We show that . For that we simply remark that

Now, any non-trivial linear map , i.e. any map of the form

with , preserves the Haar measure, which implies that

Using this, the triangle equality and the fact that the functions are 1-bounded, we conclude.

**Proposition 2**

Let be the set of functions for of density at least Then is dense in

**Proof**

Let . We construct a sequence of functions that converge to We use duality and Fourier transform. We have and

For , write

Then write

This defines a function , which obviously satisfies

.

The same holds for its extension . Notice that is also the conditional expectation with respect to the -algebra . More precisely,

as both equal .

This implies in particular that , and thus , has its values in .

To check that , we use the conditional expectation interpretation

Now we check the convergence. It follows from basic harmonic analysis: being in , its Fourier series converges to in . This concludes the proof of Proposition 2.

Thus even for 4-APs, this shows that

exists and equals

This is in contrast to a paper of Candela-Szegedy, for the same pattern (4-APs) but with where they obtain as a limit a much more complicated infimum, an infimum over functions on of the -count, where is another pattern. The part that holds in the setting of and not in the setting of is Proposition 2. The truncated Fourier series of , which is the natural approximation, does not give a function on cyclic groups with the same density and AP-count as .

We remark that the paper of Hatami-Hatami-Hirst implies that the sequence of functions admits a *limit object* which is defined on a finite cartesian power of .

In fact, if is the -APs, pattern of complexity , and , the limit object is defined on . In particular, if , this means that the infimum above is attained, i.e.

for some measurable. When , it is still not known whether such a result (i.e. the existence of a limit object ) holds. However, Szegedy proves the existence of a limit object on a non-explicit group .

]]>A recent breakthrough due to Croot, Lev and Pach, recycled by Ellenberg-Gijswijt and streamlined by Tao on this blog post, revolutionised the landscape of Roth’s type problems in vector spaces over finite fields. Indeed, Ellenberg-Gijswijt obtained that a set free of triple satisfying can have at most elements out of the out there. Here we show the same method can deal with sets free of solutions to various other equations in any (given that if , we have as a vector space, there’s no point in considering general ) and we make explicit the dependence on of the resulting bound. We refer in the sequel mostly to Tao’s symmetric proof instead of the original articles. One great merit of this exposition is to reveal, thanks to the extension of the notion of rank from matrices (seen as functions of two variables) to functions of more variables, how the scope of the method can be extended.

**Other linear equations**

In fact the bound found for the sets without 3-term arithmetic progressions is also valid

for a set having no non-trivial solutions to

where are three linear automorphisms of summing to 0. Indeed,

we can then still write

Even more generally, for any integer , if are automorphisms of which sum to 0, and if does not have any non-trivial solution to the equation , we can still write

The right-hand side is still a function on of rank . Meanwhile, the left-hand side may still be expanded using the formula valid for all , which implies that the LHS equals

where means the -th coordinate of ; it is

thus a linear polynomial. This implies that the whole expression is a polynomial of degree . The treatment shown by Tao on his blog to bound the rank of this expression can thus be applied here too.

Roth’s theorem is simply the case where and for , or more generally and .

The more variables, the smaller the bound becomes. For instance, a set without any solution to the equation , in particular without any 3-APs, has size at most , less than the bound for sets without 3-APS which is .

**Even non-linear equations**

Upon scrutinizing the polynomial method at work in Ellenberg-Gijswijt, one can be struck by the fact that the translation-invariance of the equation is not even used. In Meshulam’s theorem, as in all form of density increment strategy, we crucially use the fact that if is a solution and if is any constant, then is a solution too. What is used in the polynomial strategy is that any constant tuple is a solution, which is vastly weaker (though it is synonymous for linear patterns). Thus if are any bounded-degree polynomial maps (i.e. maps whose coordinates are polynomials of degree at most ) summing to 0, then the method should, in principle, spit out a bound for the maximal size of a set free of solutions to , depending on the degree. The only thing that changes in Tao’s write-up is that the polynomial

whose rank we try to bound is now of degree up to . Thus the parameter of relevance and which determines whether the method will succeed or not is . It has to be significantly less than . Otherwise, the method requiring a good bound (exponential if possible) on

where are independent, identically distributed variable on , we are hopeless.

Thus, considering , we need variables.

**Equations in a polynomial ring**

As advocated by Tom Bloom (for instance in this paper), it might be of interest to ask for analogues of theorems in polynomial ring. However, a relation of the form among three polynomials over of degree less than is simply a relation among three vectors, the vectors of their coefficients. But we can form much more general equations, such as

for some . We require again . Now the map on the polynomials of degree less than translates, in term of the coefficients of a polynomial, into a linear map from to where .

Let be the maximum of the degrees of the and let us introduce the maps corresponding to asking for the size of without any solution to the equation above amounts to asking for the size of a set without any solution to the equation , a situation we have already analysed.

Even better, translating the map into a a polynomial map , namely

we’re back to the previous paragraph. Here , so that to ensure , we need . We then get a bound . This number of variables is terrible. In fact, in the integers, Browning and Prendiville showed that 5 variables suffice for such an equation, that is, they got a bound for the size of a set without solution to the equation . It is not satisfactory to end up needing more variables in the function field setting than in the integer case. The polynomial method does not seem good at tackling polynomial equations.

**Dependence on **

Let’s return to the case of Roth’s theorem.

Following Tao, one gets a bound of the form

where is the maximum of

under the constraints

It could be interesting to know how evolves with .

Using the help of a computer to compute for upto , I got the feeling that . This would mean that a subset free of 3-term arithmetic progressions necessarily satisfies .

We now proceed to show that the empirically conjectured behaviour of is indeed the true one.

The constrained extrema theorem implies that the maximum of is attained on a point where

belongs to . This condition amounts to the existence of so that for all indices . So we have only two degrees of freedom for the tuple . And because we have two equations

there should be a unique such tuple.

Putting , we transform the first constraint in

thus the second one amounts to

Here the LHS equals

Thus the equation to solve is

in other words

It clearly imposes that as , we must have , for if it remains bounded away from 1, the left-hand side is asymptotic to , which is not asymptotic to , the right-hand side.

We now rearrange the equation as

that is

Let us introduce and determine the behaviour of , or equivalently of . One can prove by contradiction that cannot tend to 0. Indeed, supposing it did, one could write

Thus

and comparing the term in we see an absurdity.

It can’t either tend to . If it has a limit point , it has to be a solution to

The computer declares that the solution to this equations equals 2.149125799907062 (which is very close, fortunately enough, to values observed for for large ). Thus tends to .

Injecting the expression for the in terms of , the optimum is

which we can equate, remembering earlier formulae, to

.

So let us derive an asymptotic expression for and . We quickly obtain .

So the maximum we are after equals

This implies that

Injecting the value we found for , the computer spits out 0.84, as guessed.

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So instead one could ask if when coloring the primes in finitely many colours, one is guaranteed to find a colour class containing infinitely many pairs of primes whose difference is bounded by a common constant. This does not follow immediately from pigeonhole principle, which only says that one colour class contains infinitely many primes.

However, this follows easily if we recall that the work of Maynard does not only give infinitely many bounded gaps, that is , but also for any that . Let be this liminf.

In other words there are infinitely many (disjoint) intervals of size which contains at least primes. So if you colour the primes in colours, and you consider these intervals, you find that each one must contains at least two primes of the same colour. By pigeonhole principle again, there must be a colour for which this occurs infinitely often. Whence the existence in this colour class of infinitely gaps of size at most .

]]>**Hardy-Littlewood squarefree tuples**

The Hardy-Littlewood conjecture consists, in its strongest form, in an asymptotic, for every , of the form

where is a constant possibly equal to 0. It is quite far from being proven, although interesting steps towards it have been made in the last ten years, by Goldston-Pintz-Yildirim and Maynard.

In the squarefree numbers, the analogous problem is rather easy and was settled by Mirsky in 1947. He proved that

where

and is the cardinality of the size of the projection of the set in , that is the number of forbidden classes modulo for .

I show a quick way of deriving it with a much worse error term . Fix , a large number; ultimately, should be ok. Fix . Then the number of such that none of have a square factor with is easy to compute: in or any interval of length there are by chinese remainder theorem this many of them

Thus in , there are this many of them

There are also people who are not squarefree, but escape our sieving by primes less than . These are the guys such that at least one of the have a divisor with prime. Now there are are at most multiples of in . So at most

should be removed. Let’s try to balance the error terms and ; take .

This way both seem to be .

**Green-Tao type theorem**

We can also easily get asymptotics of the form

where is a convex body and a system of affine-linear forms of which no two are affinely related. Moreover is again the proportion of non-forbidden vectors of residues modulo . One way to do it would be to simply use the Hardy-Littlewood type asymptotics proved above coordinate by coordinate. We can also prove it quite easily, by first observing that the set of n where at least one of the vanishes has size because the 0-set of an affine-linear form is an affine hyperplane. Then as , the divisors will have to satisfy . So now we restrict the set of to the ones where the forms don’t vanish. Then we partition the sum into one where is restricted to be at most (we fix some small ), and the remaining one. Using the well-known fact that , we remark that the remaining one equals

Now the number of in the sum is . Given that , we can bound this sum by . We proceed similarly for , so that the sum to estimate is

Quite gross volume packing arguments indicate that the number of integral points (lattice points) to estimate above is , where

is the density of zeroes modulo the . Moreover we can extend the sum beyond at the cost of a mere . Hence finally for any we can write that

Indeed by multiplicativity it is easy to transform into a product over primes.

]]>

Here pattern stands for images of a system of polynomial forms. Given and a convex set of volume increasing to infinity, we are thus interested in evaluating

for N large, where is the von Mangoldt function. The case where and is the original question of Hardy and Littlewood, who proposed a tantalizing asymptotic behaviour but is still completely out of reach (even the question whether there are infinitely many such that are both primes is not settled). But the case where the system is affine-linear (thus the polynomials are all of degree 1) and no two forms are affinely dependent was solved by Green and Tao in the celebrated article Linear equations in primes.

Similar results for more general polynomial forms are rare. We have to mention the famous work of Friedlander and Iwaniec yielding an asymptotic for the number of primes of the form , where it appears that

for some constant .

I have uploaded yesterday an article on the ArXiv which provides asymptotics of the same shape as the ones in the Hardy-Littlewood for a few exceptional polynomial patterns. Thus for instance, I can tell how many arithmetic progressions of three primes smaller than N exist whose common difference is a sum of two squares – well not quite, because I have to weigh these arithmetic progressions by the number of representations of the common difference. Now this weight, giving a positive density to the set of sums of two squares, which is sparse, of density , just as the von Mangoldt function is a weight (mostly) on primes giving them a density, cannot be easily eliminated afterwards, in contrast to the von Mangoldt function (one can write for that ).

More precisely, the result that naturally comes out concerning three term arithmetic progressions with common difference a sum of two squares is

where is the representation function and are some explicit constant which I don’t reproduce here. Moreover, we can generalise to other positive definite binary quadratic forms than this one, and there’s nothing special about length three: an asymptotic is available for any length. Here we notice that in some sense, the result is only seemingly polynomial, and truly rather linear: the polynomial nature of the pattern is enclosed in a *linear* input into the representation function of a quadratic form.

In fact, my article contains a more general result of which the one above is only a particular case. My work consisted in mingling the von Mangoldt function with the representation functions of quadratic forms, whose behaviour on linear systems have been already analysed respectively in by Green and Tao and Matthiesen. The idea is to consider sums of the form

where can be the von Mangoldt function or a representation function, and the $\psi_i$ are linear forms. The cohabitation of both types of functions went quite well. One delicate point was to eliminate biases modulo small primes of both types functions, an operation known as the -trick. The difficulty is that while the value of the von Mangoldt function is more or less determined by the coprimality to small primes, it is not so for the representation function, which is also sensitive to the residue modulo large powers of small primes. Once this issue is adressed carefully, it is possible to majorize them by one and the same pseudorandom majorant, which opens the way to the application of the transference principle.

Similarly, the cohabitation between the von Mangoldt function and the divisor function is quite natural, yielding asymptotics for expressions such as . This is reminiscent of the Titchmarsh divisor problem, the evaluation of or (almost equivalently) of , but the latter expression involves a linear system of infinite complexity, and is thus altogether out of reach of my method, just as the twin primes or the basic Hardy-Littlewood conjecture.

]]>We will rather try to get an asymptotic of the form

where is basically the number of points with positive integer coordinates in , so hopefully , and the local factors take into account the obstructions or the absence of obstructions to primality modulo . Recall that classically denotes the von Mangoldt function. There are only very few non-linear polynomials for which an asymptotic for the number of prime values is available. The easiest one is obviously . Indeed, the primes which are sum of two squares are the primes congruent to 1 modulo 4 (and also the prime 2, but it’s a single prime, so we don’t have to care), and they are represented 8 times each as a value of . So

Now let’s check what the conjecture would say. Here . What about the ? They are supposed to be . Now it easy to check that if , and otherwise. So in the former case, and in the latter case. Thus, doesn’t converge absolutely, in contrast with the traditional Green-Tao situation, where … However, if we imagine that to each prime corresponds a prime with , we could compute the product of the by grouping the factors into pairs . A bit more precisely,

and

which implies that

so this product is convergent, although not absolutely. I guess the constant is , see Mertens theorem, and that . I don’t really know how to compute this convergent product. We quickly notice that . If the product could be equal to , which it can’t unfortunately, being smaller than 1, we would apparently get the correct result, but this remains a quite dodgy case, which should make one cautious about stating ambitious generalisations of the Hardy-Littlewood conjecture.

]]>Thus has normal order : almost all numbers have about ; this is the theorem of Hardy-Ramanujan. It’s even better than this : for almost all numbers, is within of its model . In fact the values of follow a Gaussian distribution, as the Erdös-Kac theorem reveals.

In contrary the divisor function doesn’t have any normal order, but it does have an average order which is quite good-looking, namely (while has a normal order which, surprisingly isn’t but a bit a less, namely , showing that a few exceptionally highly divisble numbers are able to make the average deviate substantially).

Now of course on the primes and collapse violently, being stuck at respectively 2 and 1. The question is whether just after such a shocking value, at , one recovers more normal arithmetic indices, whether there are traces of the past catastrophe (or signs of the coming castrophe just before it).

Of course isn’t absolutely any number, for instance it’s surely even; but then is just any number? It must also be said that it has higher chances to be mutiple of 3 than a generic number, in fact one in two chance instead of one in three, because whith equal probability.

**From the point of view of **

There the answer is: yes, perfectly. Indeed, Heini Halberstam established that the Erdös-Kac theorem holds when is constrained to range in the shifted primes . That is, lies within of for a positive proportion of primes, the density being again gaussian.

**From the point of view of **

Not quite. For this consider the Titchmarsh divisor problem, consisting in estimating . If was replaceable by as we suppose, then this sum would be asymptotic to by Mertens formula. It turns out that it is in fact asymptotic to , the constant prefactor being well over 1,9, so that it can be said that has almost twice as many divisors as banal numbers of his size. Now remember it’s always even; now there are good heuristic reasons to believe that , so that in average is close to , but 1,9 is still much larger sensibly larger than 4/3. Here the higher chances of to be divisible by 3, 5 etc. in comparison to a banal number weigh enough to make the average deviate.

It would be interesting to determine whether has a normal order, in the same vein as Halberstam’s result.

**Another criterion of banality: the probability of being squarefree**

As we know, the density of squarefree numbers is . It is then reasonable to wonder if among numbers of the forme , the proportion of square free is the same. It’s clear to see that it can’t quite be so: indeed, has one on two chance of being divisible by the square 4 (while a generic number has of course one on four chance), one on six chances of being divisible by 9, one on chance of being divisibly by the prime . So one guesses that is a little less likely to be squarefree than any number; indeed, it’s been proven by Mirsky that the proportion of squarefree numbers among numbers of the form is , to compare with .

The property of being squarefree can be replaced by the property of being -free for instance, the density among shifted primes being then , to compare with .

**Appendix: average of the divisor function on odd and even numbers**

We know that

And let us suppose that and .

Now the sum on even numbers can be decomposed into a sum on numbers of 2-adic valuation 1,2,3… But numbers of 2-adic valuation k are numbers which are of the form 2^km with odd, so

for fixed . Pretending this asymptotic also holds when is allowed to grow with , we infer from that . Hence and .

It might be that this intuition has been proven more rigorously, I’d like to see where!

]]>**First proof**

This one requires a good deal of knowledge of the theory finite fields.

We prove that is the product of all monic irreducible polynomials whose degree divide , which immediately provides the above formula by comparison of the degrees. To prove this, notice that the roots of are all the elements of the extension of , so that , as an element of can be factored as . On the other hand, if is an irreducible polynomial of degree , it splits into linear factors in . So has a root and all its other roots are images of under the Frobenius element, more precisely,

so that (at first sight in , but also in because they are both in – it is well known that the gcd is preserved under field extensions). Moreover, all monic irreducible are coprimes, so that their product still divides . Finally, we need to prove the converse divisibility, either in or in , which is equivalent. So let’s do it in the latter; take a factor with . Let be the order of the Frobenius element on , so that and no smaller integer has this property; then which is indeed an irreducible of degree , hence a factor of . Given the factorization of and the fact that the factors are coprime, their product still divides

**Second proof**

This one proves exactly what we want and not more and uses only power series. We need the notation .

Let

be the zeta function (for s of real part >1). It is easy to see that by looking at the sum expression, and by looking at the product expression. Putting , both sides become nice power series so that we get the equality

and doing classical operations (logarithmic derivation and expansion in power series followed by termwise identification in particular) on both sides yield the result.

This proof seems to use about zero input of algebra, so its success looks like a miracle. The only step where a (tiny) bit of knowledge seems to be used is the transformation of the sum into an Euler product – it needs the existence of unicity of the decomposition of a monic polynomial into product of irreducibles.

**A kind of sieve method ?**

It could be reasonable to hope that the sieve of Eratosthenes would work better in the realm of polynomial rings. We’ll try to adapt it here. Notice that in this case we have count primes up to a certain large degree, and not of exactly certain large degree (both quantities are far from equivalent in fact !).

Let’s define to be the number of monic irreducible (let’s say primes) polynomials of degree at most n, and . We also use Moebius’ function for monic polynomials, also defined in terms of the number of prime factors. The Moebius inversion formula still holds in the polynomial world. We also write

Thus

and if we forget about we obtain exactly

this looks quite nice, but the error term, just like in the integer case, can’t easily be bounded by anything better than which we know will be vastly bigger than the main term. By the way this product can very easily be shown to be but the exact asymptotic is not any easier than the PNT itself… So this sieve seems to have very little interest, given the convenience of the other methods.

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