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I’ve just opened the research thread for the mini-polymath4 project over at the polymath blog to collaboratively solve one of the six questions from this year’s IMO.  This year I have selected Q3, which is a somewhat intricate game-theoretic question.  (The full list of questions this year may be found here.)

This post will serve as the discussion thread of the project, intended to focus all the non-research aspects of the project such as organisational matters or commentary on the progress of the project.    The third component of the project is the wiki page, which is intended to summarise the progress made so far on the problem.

As with the previous mini-polymath projects, I myself will be serving primarily as a moderator, and hope other participants will take the lead in the research and in keeping the wiki up-to-date.

Just a reminder that the mini-polymath4 project will begin in three hours at Thu July 12 2012 UTC 22:00.

Two quick updates with regards to polymath projects.  Firstly, given the poll on starting the mini-polymath4 project, I will start the project at Thu July 12 2012 UTC 22:00.  As usual, the main research thread on this project will be held at the polymath blog, with the discussion thread hosted separately on this blog.

Second, the Polymath7 project, which seeks to establish the “hot spots conjecture” for acute-angled triangles, has made a fair amount of progress so far; for instance, the first part of the conjecture (asserting that the second Neumann eigenfunction of an acute non-equilateral triangle is simple) is now solved, and the second part (asserting that the “hot spots” (i.e. extrema) of that second eigenfunction lie on the boundary of the triangle) has been solved in a number of special cases (such as the isosceles case).  It’s been quite an active discussion in the last week or so, with almost 200 comments across two threads (and a third thread freshly opened up just now).  While the problem is still not completely solved, I feel optimistic that it should fall within the next few weeks (if nothing else, it seems that the problem is now at least amenable to a brute force numerical attack, though personally I would prefer to see a more conceptual solution).

Two polymath related items for this post. Firstly, there is a new polymath proposal over at the polymath blog, proposing to attack the “hot spots conjecture” (concerning a maximum principle for a heat equation) in the case when the domain is an acute-angled triangle (the case of the right and obtuse-angled triangles already being solved). Please feel free to comment on the proposal blog post if you are interested in participating.

Secondly, it is once again time to set up the annual “mini-polymath” project to collaboratively solve one of this year’s International Mathematical Olympiad problems. This year, the Olympiad is being held in Argentina, with the problems given out on July 10-11. As usual, there will be a wiki page, discussion thread, and research thread for the project. As in previous years, the first thing to resolve is the starting date and time, so I am setting up a poll here to fix a time (and also to get a preliminary indication of interest in the project).  (I am using 24-hour Coordinated Universal Time (UTC) for these times.  Here is a link that converts the first time given in the poll (Thu Jul 12 2012 UTC 6:00) into other time zones.) Given that the last three mini-polymaths were reasonably successful, I am not planning any changes to the format, but of course if there are any suggestions for changes, I’d be happy to hear them in the comments.

High school algebra marks a key transition point in one’s early mathematical education, and is a common point at which students feel that mathematics becomes really difficult. One of the reasons for this is that the problem solving process for a high school algebra question is significantly more free-form than the mechanical algorithms one is taught for elementary arithmetic, and a certain amount of planning and strategy now comes into play. For instance, if one wants to, say, write ${\frac{1,572,342}{4,124}}$ as a mixed fraction, there is a clear (albeit lengthy) algorithm to do this: one simply sets up the long division problem, extracts the quotient and remainder, and organises these numbers into the desired mixed fraction. After a suitable amount of drill, this is a task that can be accomplished by a large fraction of students at the middle school level. But if, for instance, one has to solve a system of equations such as

$\displaystyle a^2 + bc = 7$

$\displaystyle 2b - c = 1$

$\displaystyle a + 2 = c$

for ${a,b,c}$, there is no similarly mechanical procedure that can be easily taught to a high school student in order to solve such a problem “mindlessly”. (I doubt, for instance, that any attempt to teach Buchberger’s algorithm to such students will be all that successful.) Instead, one is taught the basic “moves” (e.g. multiplying both sides of an equation by some factor, subtracting one equation from another, substituting an expression for a variable into another equation, and so forth), and some basic principles (e.g. simplifying an expression whenever possible, for instance by gathering terms, or solving for one variable in terms of others in order to eliminate it from the system). It is then up to the student to find a suitable combination of moves that isolates each of the variables in turn, to reveal their identities.

Once one is sufficiently expert in algebraic manipulation, this is not all that difficult to do, but when one is just starting to learn algebra, this type of problem can be quite daunting, in part because of an “embarrasment of riches”; there are several possible “moves” one could try to apply to the equations given, and to the novice it is not always clear in advance which moves will simplify the problem and which ones will make it more complicated. Often, such a student may simply try moves at random, which can lead to a dishearteningly large amount of effort expended without getting any closer to a solution. What is worse, each move introduces the possibility of an arithmetic error (such as a sign error), the effect of which is usually not apparent until the student finally arrives at his or her solution and either checks it against the original equation, or submits the answer to be graded. (My own son can get quite frustrated after performing a lengthy series of computations to solve an algebra problem, only to be told that the answer was wrong due to an arithmetic error; I am sure this experience is common to many other schoolchildren.)

It occurred to me recently, though, that the set of problem-solving skills needed to solve algebra problems (and, to some extent, calculus problems also) is somewhat similar to the set of skills needed to solve puzzle-type computer games, in which a certain limited set of moves must be applied in a certain order to achieve a desired result. (There are countless games of this general type; a typical example is “Factory balls“.) Given that the computer game format is already quite familiar to many schoolchildren, one could then try to teach the strategy component of algebraic problem-solving via such a game, which could automate mechanical tasks such as gathering terms and performing arithmetic in order to reduce some of the more frustrating aspects of algebra. (Note that this is distinct from the type of maths games one often sees on educational web sites, which are usually based on asking the player to correctly answer some maths problem in order to advance within the game, making the game essentially a disguised version of a maths quiz. Here, the focus is not so much on being able to supply the correct answer, but on being able to select an effective problem-solving strategy.)

It is difficult to explain in words exactly what type of game I have in mind, so I decided to create a quick mockup of what a sample “level” would look like here (note: requires Java). I didn’t want to spend too much time to make this mockup, so I wrote it in Scratch, which is a somewhat limited programming language intended for use by children, but which has the benefit of being able to churn out simple but functional apps very quickly (the mockup took less than half an hour to write). (I would certainly not attempt to write a full game in this language, though.) In this mockup level, the objective is to solve a single linear equation in one variable, such as ${2x+7=11}$, with only two “moves” available: the ability to subtract ${1}$ from both sides of the equation, and the ability to divide both sides of the equation by ${2}$, which one performs by clicking on an appropriate icon.

It seems to me that one could actually teach a fair amount of algebra through a game such as this, with a progressively difficult sequence of levels that gradually introduce more and more types of “moves” that can handle increasingly difficult problems (e.g. simultaneous linear equations in several unknowns, quadratic equations in one or more variables, inequalities, etc.). Even within a single class of problem, such as solving linear equations, one could create additional challenge by placing some restriction on the available moves, for instance by limiting the number of available moves (as was done in the mockup), or requiring that each move cost some amount of game currency (which might possibly be waived if one is willing to perform the move “by hand”, i.e. by entering the transformed equation manually). And of course one could also make the graphics, sound, and gameplay fancier (e.g. by allowing for various competitive gameplay modes). One could also imagine some other types of high-school and lower-division undergraduate mathematics being amenable to this sort of gamification (calculus and ODE comes to mind, and maybe propositional logic), though I doubt that one could use it effectively for advanced undergraduate or graduate topics. (Though I have sometimes wished for an “integrate by parts” or “use Sobolev embedding” button when trying to control solutions to a PDE…)

This would however be a fair amount of work to actually implement, and is not something I could do by myself with the time I have available these days. But perhaps it may be possible to develop such a game (or platform for such a game) collaboratively, somewhat in the spirit of the polymath projects? I have almost no experience in modern software development (other than a summer programming job I had as a teenager, which hardly counts), so I would be curious to know how projects such as this actually get started in practice.

Let ${\alpha \in {\bf R}/{\bf Z}}$ be an element of the unit circle, let ${N \geq 1}$, and let ${\rho > 0}$. We define the (rank one) Bohr set ${B_N(\alpha;\rho)}$ to be the set

$\displaystyle B_N(\alpha;\rho) := \{ n \in {\bf Z}: -N \leq n \leq N; \|n\alpha\|_{{\bf R}/{\bf Z}} \leq \rho \}$

where ${\|x\|_{{\bf R}/{\bf Z}}}$ is the distance to the origin in the unit circle (or equivalently, the distance to the nearest integer, after lifting up to ${{\bf R}}$). These sets play an important role in additive combinatorics and in additive number theory. For instance, they arise naturally when applying the circle method, because Bohr sets describe the oscillation of exponential phases such as ${n \mapsto e^{2\pi i n \alpha}}$.

Observe that Bohr sets enjoy the doubling property

$\displaystyle B_N(\alpha;\rho) + B_N(\alpha;\rho) \subset B_{2N}(\alpha;2\rho),$

thus doubling the Bohr set doubles both the length parameter ${N}$ and the radius parameter ${\rho}$. As such, these Bohr sets resemble two-dimensional balls (or boxes). Indeed, one can view ${B_N(\alpha;\rho)}$ as the preimage of the two-dimensional box ${[-1,1] \times [-\rho,\rho] \subset {\bf R} \times {\bf R}/{\bf Z}}$ under the homomorphism ${n \mapsto (n/N, \alpha n \hbox{ mod } 1)}$.

Another class of finite set with two-dimensional behaviour is the class of (rank two) generalised arithmetic progressions

$\displaystyle P( a_1,a_2; N_1,N_2 ) := \{ n_1 a_1 + n_2 a_2: n_1,n_2 \in {\bf Z}; |n_1| \leq N_1, |n_2| \leq N_2 \}$

with ${a_1,a_2 \in {\bf Z}}$ and ${N_1,N_2 > 0}$ Indeed, we have

$\displaystyle P( a_1,a_2; N_1,N_2 ) + P( a_1,a_2; N_1,N_2 ) \subset P( a_1,a_2; 2N_1, 2N_2 )$

and so we see, as with the Bohr set, that doubling the generalised arithmetic progressions doubles the two defining parameters of that progression.

More generally, there is an analogy between rank ${r}$ Bohr sets

$\displaystyle B_N(\alpha_1,\ldots,\alpha_r; \rho_1,\ldots,\rho_r) := \{ n \in {\bf Z}: -N \leq n \leq N; \|n\alpha_i\|_{{\bf R}/{\bf Z}} \leq \rho_i$

$\displaystyle \hbox{ for all } 1 \leq i \leq r \}$

and the rank ${r+1}$ generalised arithmetic progressions

$\displaystyle P( a_1,\ldots,a_{r+1}; N_1,\ldots,N_{r+1} ) := \{ n_1 a_1 + \ldots + n_{r+1} a_{r+1}:$

$\displaystyle n_1,\ldots,n_{r+1} \in {\bf Z}; |n_i| \leq N_i \hbox{ for all } 1 \leq i \leq r+1 \}.$

One of the aims of additive combinatorics is to formalise analogies such as the one given above. By using some arguments from the geometry of numbers, for instance, one can show that for any rank ${r}$ Bohr set ${B_N(\alpha_1,\ldots,\alpha_r;\rho_1,\ldots,\rho_r)}$, there is a rank ${r+1}$ generalised arithmetic progression ${P(a_1,\ldots,a_{r+1}; N_1,\ldots,N_{r+1})}$ for which one has the containments

$\displaystyle B_N(\alpha_1,\ldots,\alpha_r;\epsilon \rho_1,\ldots,\epsilon \rho_r) \subset P(a_1,\ldots,a_{r+1}; N_1,\ldots,N_{r+1})$

$\displaystyle \subset B_N(\alpha_1,\ldots,\alpha_r;\rho_1,\ldots,\rho_r)$

for some explicit ${\epsilon>0}$ depending only on ${r}$ (in fact one can take ${\epsilon = (r+1)^{-2(r+1)}}$); this is (a slight modification of) Lemma 4.22 of my book with Van Vu.

In the special case when ${r=1}$, one can make a significantly more detailed description of the link between rank one Bohr sets and rank two generalised arithmetic progressions, by using the classical theory of continued fractions, which among other things gives a fairly precise formula for the generators ${a_1,a_2}$ and lengths ${N_1,N_2}$ of the generalised arithmetic progression associated to a rank one Bohr set ${B_N(\alpha;\rho)}$. While this connection is already implicit in the continued fraction literature (for instance, in the classic text of Hardy and Wright), I thought it would be a good exercise to work it out explicitly and write it up, which I will do below the fold.

It is unfortunate that the theory of continued fractions is restricted to the rank one setting (it relies very heavily on the total ordering of one-dimensional sets such as ${{\bf Z}}$ or ${{\bf R}}$). A higher rank version of the theory could potentially help with questions such as the Littlewood conjecture, which remains open despite a substantial amount of effort and partial progress on the problem. At the end of this post I discuss how one can use the rank one theory to rephrase the Littlewood conjecture as a conjecture about a doubly indexed family of rank four progressions, which can be used to heuristically justify why this conjecture should be true, but does not otherwise seem to shed much light on the problem.

Let ${G = (G,+)}$ be a finite additive group. A tiling pair is a pair of non-empty subsets ${A, B}$ such that every element of ${G}$ can be written in exactly one way as a sum of an element ${a}$ of ${A}$ and an element ${b}$ of ${B}$, in which case we write ${G = A \oplus B}$. The sets ${A, B}$ are then called tiles, with ${B}$ being a complementary tile to ${A}$ and vice versa. For instance, every subgroup ${H}$ of ${G}$ is a tile, as one can pick one representative from each coset of ${H}$ to form the complementary tile. Conversely, any set formed by taking one representative from each coset of ${H}$ is also a tile.

Tiles can be quite complicated, particularly when the group ${G}$ is “high-dimensional”. We will therefore restrict to the simple case of a cyclic group ${G = {\bf Z}/N{\bf Z}}$, and restrict even further to the special case when the modulus ${N}$ is square-free. Here, the situation should be much simpler. In particular, we have the following conjecture of Coven and Meyerowitz, which asserts that the previous construction of a tile is, in fact, the only such construction:

Conjecture 1 (Coven-Meyerowitz conjecture, square-free case) Let ${N}$ be square-free, and let ${A}$ be a tile of ${{\bf Z}/N{\bf Z}}$. Then there exists a subgroup ${H}$ of ${{\bf Z}/N{\bf Z}}$ such that ${A}$ consists of a single representative from each coset of ${H}$.

Note that in the square-free case, every subgroup ${H}$ of ${{\bf Z}/N{\bf Z}}$ has a complementary subgroup ${H^\perp}$ (thus ${{\bf Z}/N{\bf Z} = H \oplus H^\perp}$). In particular, ${H}$ consists of a single representative from each coset of ${H^\perp}$, and so the examples of subgroups of ${{\bf Z}/N{\bf Z}}$ are covered by the above conjecture in the square-free case.

In the non-square free case, the above assertion is not true; for instance, if ${p}$ is a prime, then the multiples of ${p}$ in ${{\bf Z}/p^2{\bf Z}}$ are a tile, but cannot be formed from taking a single representative from all the cosets of a given subgroup. There is a more general conjecture of Coven and Meyerowitz to handle this more general case, although it is more difficult to state:

Conjecture 2 (Coven-Meyerowitz conjecture, general case) Let ${N}$ be a natural number, and let ${A}$ be a tile of ${{\bf Z}/N{\bf Z}}$. Then there exists a set ${S_A}$ of prime powers with ${|A| = \prod_{p^j \in S_A} p}$ such that the Fourier transform

$\displaystyle \hat 1_A(k) := \sum_{n \in A} e^{2\pi i kn / N}$

vanishes whenever ${k}$ is a non-zero element of ${{\bf Z}/N{\bf Z}}$ whose order is the product of elements of ${S_A}$ that are powers of distinct primes. Equivalently, the generating polynomial ${\sum_{n \in A} x^n}$ is divisible by the cyclotomic polynomials ${\phi_m}$ whenever ${m}$ is the product of elements of ${S_A}$ that are powers of distinct primes.

It can be shown (with a modest amount of effort) that Conjecture 2 implies Conjecture 1, but we will not do so here, focusing instead exclusively on the square-free case for simplicity.

It was observed by Laba that Conjecture 2 is connected to the following well-known conjecture of Fuglede:

Conjecture 3 (One-dimensional Fuglede conjecture, tiling to spectral direction) Let ${E}$ be a compact subset of ${{\bf R}}$ of positive measure which is a tile (thus ${{\bf R} = E \oplus \Lambda}$ for some set ${\Lambda \subset {\bf R}}$). Then ${L^2(E)}$ (with Lebesgue measure) has a spectrum, that is to say an orthogonal set of plane waves ${x \mapsto e^{2\pi i \xi x}}$.

Indeed, it was shown by Laba that Conjecture 2 implies Conjecture 3 in the case when ${E}$ is the finite union of unit intervals. Actually, thanks to the more recent work of Farkas, Matolcsi, and Mora we know that Conjecture 2 in fact implies the universal spectrum conjecture of Lagarias and Wang, which in turn was known to imply Conjecture 3 in full generality. (On the other hand, the conjecture fails in four and higher dimensions; see the papers of Kolountzakis-Matolcsi and of Farkas-Revesz.)

Given the simple statement of Conjecture 1, it is perhaps somewhat surprising that it remains open, even in simple cases such as when ${N}$ is the product of just four primes. One reason for this is that some plausible strengthenings of this conjecture (such as the Tijdeman-Sands conjecture) are known to be false, as we will review below. On the other hand, as we shall see, tiling sets have a lot of combinatorial structure, and in principle one should be able to work out a lot of special cases of the conjecture. Given the combinatorial nature of this problem, it may well be quite suitable for a polymath project in fact, if there is sufficient interest.

One of the most notorious problems in elementary mathematics that remains unsolved is the Collatz conjecture, concerning the function ${f_0: {\bf N} \rightarrow {\bf N}}$ defined by setting ${f_0(n) := 3n+1}$ when ${n}$ is odd, and ${f_0(n) := n/2}$ when ${n}$ is even. (Here, ${{\bf N}}$ is understood to be the positive natural numbers ${\{1,2,3,\ldots\}}$.)

Conjecture 1 (Collatz conjecture) For any given natural number ${n}$, the orbit ${n, f_0(n), f^2_0(n), f^3_0(n), \ldots}$ passes through ${1}$ (i.e. ${f^k_0(n)=1}$ for some ${k}$).

Open questions with this level of notoriety can lead to what Richard Lipton calls “mathematical diseases” (and what I termed an unhealthy amount of obsession on a single famous problem). (See also this xkcd comic regarding the Collatz conjecture.) As such, most practicing mathematicians tend to spend the majority of their time on more productive research areas that are only just beyond the range of current techniques. Nevertheless, it can still be diverting to spend a day or two each year on these sorts of questions, before returning to other matters; so I recently had a go at the problem. Needless to say, I didn’t solve the problem, but I have a better appreciation of why the conjecture is (a) plausible, and (b) unlikely be proven by current technology, and I thought I would share what I had found out here on this blog.

Let me begin with some very well known facts. If ${n}$ is odd, then ${f_0(n) = 3n+1}$ is even, and so ${f_0^2(n) = \frac{3n+1}{2}}$. Because of this, one could replace ${f_0}$ by the function ${f_1: {\bf N} \rightarrow {\bf N}}$, defined by ${f_1(n) = \frac{3n+1}{2}}$ when ${n}$ is odd, and ${f_1(n)=n/2}$ when ${n}$ is even, and obtain an equivalent conjecture. Now we see that if one chooses ${n}$ “at random”, in the sense that it is odd with probability ${1/2}$ and even with probability ${1/2}$, then ${f_1}$ increases ${n}$ by a factor of roughly ${3/2}$ half the time, and decreases it by a factor of ${1/2}$ half the time. Furthermore, if ${n}$ is uniformly distributed modulo ${4}$, one easily verifies that ${f_1(n)}$ is uniformly distributed modulo ${2}$, and so ${f_1^2(n)}$ should be roughly ${3/2}$ times as large as ${f_1(n)}$ half the time, and roughly ${1/2}$ times as large as ${f_1(n)}$ the other half of the time. Continuing this at a heuristic level, we expect generically that ${f_1^{k+1}(n) \approx \frac{3}{2} f_1^k(n)}$ half the time, and ${f_1^{k+1}(n) \approx \frac{1}{2} f_1^k(n)}$ the other half of the time. The logarithm ${\log f_1^k(n)}$ of this orbit can then be modeled heuristically by a random walk with steps ${\log \frac{3}{2}}$ and ${\log \frac{1}{2}}$ occuring with equal probability. The expectation

$\displaystyle \frac{1}{2} \log \frac{3}{2} + \frac{1}{2} \log \frac{1}{2} = \frac{1}{2} \log \frac{3}{4}$

is negative, and so (by the classic gambler’s ruin) we expect the orbit to decrease over the long term. This can be viewed as heuristic justification of the Collatz conjecture, at least in the “average case” scenario in which ${n}$ is chosen uniform at random (e.g. in some large interval ${\{1,\ldots,N\}}$). (It also suggests that if one modifies the problem, e.g. by replacing ${3n+1}$ to ${5n+1}$, then one can obtain orbits that tend to increase over time, and indeed numerically for this variant one sees orbits that appear to escape to infinity.) Unfortunately, one can only rigorously keep the orbit uniformly distributed modulo ${2}$ for time about ${O(\log N)}$ or so; after that, the system is too complicated for naive methods to control at anything other than a heuristic level.

Remark 1 One can obtain a rigorous analogue of the above arguments by extending ${f_1}$ from the integers ${{\bf Z}}$ to the ${2}$-adics ${{\bf Z}_2}$. This compact abelian group comes with a Haar probability measure, and one can verify that this measure is invariant with respect to ${f_1}$; with a bit more effort one can verify that it is ergodic. This suggests the introduction of ergodic theory methods. For instance, using the pointwise ergodic theorem, we see that if ${n}$ is a random ${2}$-adic integer, then almost surely the orbit ${n, f_1(n), f_1^2(n), \ldots}$ will be even half the time and odd half the time asymptotically, thus supporting the above heuristics. Unfortunately, this does not directly tell us much about the dynamics on ${{\bf Z}}$, as this is a measure zero subset of ${{\bf Z}_2}$. More generally, unless a dynamical system is somehow “polynomial”, “nilpotent”, or “unipotent” in nature, the current state of ergodic theory is usually only able to say something meaningful about generic orbits, but not about all orbits. For instance, the very simple system ${x \rightarrow 10x}$ on the unit circle ${{\bf R}/{\bf Z}}$ is well understood from ergodic theory (in particular, almost all orbits will be uniformly distributed), but the orbit of a specific point, e.g. ${\pi\hbox{ mod } 1}$, is still nearly impossible to understand (this particular problem being equivalent to the notorious unsolved question of whether the digits of ${\pi}$ are uniformly distributed).

The above heuristic argument only suggests decreasing orbits for almost all ${n}$ (though even this remains unproven, the state of the art is that the number of ${n}$ in ${\{1,\ldots,N\}}$ that eventually go to ${1}$ is ${\gg N^{0.84}}$, a result of Krasikov and Lagarias). It leaves open the possibility of some very rare exceptional ${n}$ for which the orbit goes to infinity, or gets trapped in a periodic loop. Since the only loop that ${1}$ lies in is ${1,4,2}$ (for ${f_0}$) or ${1,2}$ (for ${f_1}$), we thus may isolate a weaker consequence of the Collatz conjecture:

Conjecture 2 (Weak Collatz conjecture) Suppose that ${n}$ is a natural number such that ${f^k_0(n)=n}$ for some ${k \geq 1}$. Then ${n}$ is equal to ${1}$, ${2}$, or ${4}$.

Of course, we may replace ${f_0}$ with ${f_1}$ (and delete “${4}$“) and obtain an equivalent conjecture.

This weaker version of the Collatz conjecture is also unproven. However, it was observed by Bohm and Sontacchi that this weak conjecture is equivalent to a divisibility problem involving powers of ${2}$ and ${3}$:

Conjecture 3 (Reformulated weak Collatz conjecture) There does not exist ${k \geq 1}$ and integers

$\displaystyle 0 = a_1 < a_2 < \ldots < a_{k+1}$

such that ${2^{a_{k+1}}-3^k}$ is a positive integer that is a proper divisor of

$\displaystyle 3^{k-1} 2^{a_1} + 3^{k-2} 2^{a_2} + \ldots + 2^{a_k},$

i.e.

$\displaystyle (2^{a_{k+1}} - 3^k) n = 3^{k-1} 2^{a_1} + 3^{k-2} 2^{a_2} + \ldots + 2^{a_k} \ \ \ \ \ (1)$

for some natural number ${n > 1}$.

Proposition 4 Conjecture 2 and Conjecture 3 are equivalent.

Proof: To see this, it is convenient to reformulate Conjecture 2 slightly. Define an equivalence relation ${\sim}$ on ${{\bf N}}$ by declaring ${a \sim b}$ if ${a/b = 2^m}$ for some integer ${m}$, thus giving rise to the quotient space ${{\bf N}/\sim}$ of equivalence classes ${[n]}$ (which can be placed, if one wishes, in one-to-one correspondence with the odd natural numbers). We can then define a function ${f_2: {\bf N}/\sim \rightarrow {\bf N}/\sim}$ by declaring

$\displaystyle f_2( [n] ) := [3n + 2^a] \ \ \ \ \ (2)$

for any ${n \in {\bf N}}$, where ${2^a}$ is the largest power of ${2}$ that divides ${n}$. It is easy to see that ${f_2}$ is well-defined (it is essentially the Syracuse function, after identifying ${{\bf N}/\sim}$ with the odd natural numbers), and that periodic orbits of ${f_2}$ correspond to periodic orbits of ${f_1}$ or ${f_0}$. Thus, Conjecture 2 is equivalent to the conjecture that ${[1]}$ is the only periodic orbit of ${f_2}$.

Now suppose that Conjecture 2 failed, thus there exists ${[n] \neq [1]}$ such that ${f_2^k([n])=[n]}$ for some ${k \geq 1}$. Without loss of generality we may take ${n}$ to be odd, then ${n>1}$. It is easy to see that ${[1]}$ is the only fixed point of ${f_2}$, and so ${k>1}$. An easy induction using (2) shows that

$\displaystyle f_2^k([n]) = [3^k n + 3^{k-1} 2^{a_1} + 3^{k-2} 2^{a_2} + \ldots + 2^{a_k}]$

where, for each ${1 \leq i \leq k}$, ${2^{a_i}}$ is the largest power of ${2}$ that divides

$\displaystyle n_i := 3^{i-1} n + 3^{i-2} 2^{a_1} + \ldots + 2^{a_{i-1}}. \ \ \ \ \ (3)$

In particular, as ${n_1 = n}$ is odd, ${a_1=0}$. Using the recursion

$\displaystyle n_{i+1} = 3n_i + 2^{a_i}, \ \ \ \ \ (4)$

we see from induction that ${2^{a_i+1}}$ divides ${n_{i+1}}$, and thus ${a_{i+1}>a_i}$:

$\displaystyle 0 = a_1 < a_2 < \ldots < a_k.$

Since ${f_2^k([n]) = [n]}$, we have

$\displaystyle 2^{a_{k+1}} n = 3^k n + 3^{k-1} 2^{a_1} + 3^{k-2} 2^{a_2} + \ldots + 2^{a_k} = 3 n_k + 2^{a_k}$

for some integer ${a_{k+1}}$. Since ${3 n_k + 2^{a_k}}$ is divisible by ${2^{a_k+1}}$, and ${n}$ is odd, we conclude ${a_{k+1} > a_k}$; if we rearrange the above equation as (1), then we obtain a counterexample to Conjecture 3.

Conversely, suppose that Conjecture 3 failed. Then we have ${k \geq 1}$, integers

$\displaystyle 0 = a_1 < a_2 < \ldots < a_{k+1}$

and a natural number ${n > 1}$ such that (1) holds. As ${a_1=0}$, we see that the right-hand side of (1) is odd, so ${n}$ is odd also. If we then introduce the natural numbers ${n_i}$ by the formula (3), then an easy induction using (4) shows that

$\displaystyle (2^{a_{k+1}} - 3^k) n_i = 3^{k-1} 2^{a_i} + 3^{k-2} 2^{a_{i+1}} + \ldots + 2^{a_{i+k-1}} \ \ \ \ \ (5)$

with the periodic convention ${a_{k+j} := a_j + a_{k+1}}$ for ${j>1}$. As the ${a_i}$ are increasing in ${i}$ (even for ${i \geq k+1}$), we see that ${2^{a_i}}$ is the largest power of ${2}$ that divides the right-hand side of (5); as ${2^{a_{k+1}}-3^k}$ is odd, we conclude that ${2^{a_i}}$ is also the largest power of ${2}$ that divides ${n_i}$. We conclude that

$\displaystyle f_2([n_i]) = [3n_i + 2^{a_i}] = [n_{i+1}]$

and thus ${[n]}$ is a periodic orbit of ${f_2}$. Since ${n}$ is an odd number larger than ${1}$, this contradicts Conjecture 3. $\Box$

Call a counterexample a tuple ${(k,a_1,\ldots,a_{k+1})}$ that contradicts Conjecture 3, i.e. an integer ${k \geq 1}$ and an increasing set of integers

$\displaystyle 0 = a_1 < a_2 < \ldots < a_{k+1}$

such that (1) holds for some ${n \geq 1}$. We record a simple bound on such counterexamples, due to Terras and to Garner :

Lemma 5 (Exponent bounds) Let ${N \geq 1}$, and suppose that the Collatz conjecture is true for all ${n < N}$. Let ${(k,a_1,\ldots,a_{k+1})}$ be a counterexample. Then

$\displaystyle \frac{\log 3}{\log 2} k < a_{k+1} < \frac{\log(3+\frac{1}{N})}{\log 2} k.$

Proof: The first bound is immediate from the positivity of ${2^{a_{k+1}}-3^k}$. To prove the second bound, observe from the proof of Proposition 4 that the counterexample ${(k,a_1,\ldots,a_{k+1})}$ will generate a counterexample to Conjecture 2, i.e. a non-trivial periodic orbit ${n, f(n), \ldots, f^K(n) = n}$. As the conjecture is true for all ${n < N}$, all terms in this orbit must be at least ${N}$. An inspection of the proof of Proposition 4 reveals that this orbit consists of ${k}$ steps of the form ${x \mapsto 3x+1}$, and ${a_{k+1}}$ steps of the form ${x \mapsto x/2}$. As all terms are at least ${n}$, the former steps can increase magnitude by a multiplicative factor of at most ${3+\frac{1}{N}}$. As the orbit returns to where it started, we conclude that

$\displaystyle 1 \leq (3+\frac{1}{N})^k (\frac{1}{2})^{a_{k+1}}$

whence the claim. $\Box$

The Collatz conjecture has already been verified for many values of ${n}$ (up to at least ${N = 5 \times 10^{18}}$, according to this web site). Inserting this into the above lemma, one can get lower bounds on ${k}$. For instance, by methods such as this, it is known that any non-trivial periodic orbit has length at least ${105{,}000}$, as shown in Garner’s paper (and this bound, which uses the much smaller value ${N = 2 \times 10^9}$ that was available in 1981, can surely be improved using the most recent computational bounds).

Now we can perform a heuristic count on the number of counterexamples. If we fix ${k}$ and ${a := a_{k+1}}$, then ${2^a > 3^k}$, and from basic combinatorics we see that there are ${\binom{a-1}{k-1}}$ different ways to choose the remaining integers

$\displaystyle 0 = a_1 < a_2 < \ldots < a_{k+1}$

to form a potential counterexample ${(k,a_1,\ldots,a_{k+1})}$. As a crude heuristic, one expects that for a “random” such choice of integers, the expression (1) has a probability ${1/q}$ of holding for some integer ${n}$. (Note that ${q}$ is not divisible by ${2}$ or ${3}$, and so one does not expect the special structure of the right-hand side of (1) with respect to those moduli to be relevant. There will be some choices of ${a_1,\ldots,a_k}$ where the right-hand side in (1) is too small to be divisible by ${q}$, but using the estimates in Lemma 5, one expects this to occur very infrequently.) Thus, the total expected number of solutions for this choice of ${a, k}$ is

$\displaystyle \frac{1}{q} \binom{a-1}{k-1}.$

The heuristic number of solutions overall is then expected to be

$\displaystyle \sum_{a,k} \frac{1}{q} \binom{a-1}{k-1}, \ \ \ \ \ (6)$

where, in view of Lemma 5, one should restrict the double summation to the heuristic regime ${a \approx \frac{\log 3}{\log 2} k}$, with the approximation here accurate to many decimal places.

We need a lower bound on ${q}$. Here, we will use Baker’s theorem (as discussed in this previous post), which among other things gives the lower bound

$\displaystyle q = 2^a - 3^k \gg 2^a / a^C \ \ \ \ \ (7)$

for some absolute constant ${C}$. Meanwhile, Stirling’s formula (as discussed in this previous post) combined with the approximation ${k \approx \frac{\log 2}{\log 3} a}$ gives

$\displaystyle \binom{a-1}{k-1} \approx \exp(h(\frac{\log 2}{\log 3}))^a$

where ${h}$ is the entropy function

$\displaystyle h(x) := - x \log x - (1-x) \log (1-x).$

A brief computation shows that

$\displaystyle \exp(h(\frac{\log 2}{\log 3})) \approx 1.9318 \ldots$

and so (ignoring all subexponential terms)

$\displaystyle \frac{1}{q} \binom{a-1}{k-1} \approx (0.9659\ldots)^a$

which makes the series (6) convergent. (Actually, one does not need the full strength of Lemma 5 here; anything that kept ${k}$ well away from ${a/2}$ would suffice. In particular, one does not need an enormous value of ${N}$; even ${N=5}$ (say) would be more than sufficient to obtain the heuristic that there are finitely many counterexamples.) Heuristically applying the Borel-Cantelli lemma, we thus expect that there are only a finite number of counterexamples to the weak Collatz conjecture (and inserting a bound such as ${k \geq 105{,}000}$, one in fact expects it to be extremely likely that there are no counterexamples at all).

This, of course, is far short of any rigorous proof of Conjecture 2. In order to make rigorous progress on this conjecture, it seems that one would need to somehow exploit the structural properties of numbers of the form

$\displaystyle 3^{k-1} 2^{a_1} + 3^{k-2} 2^{a_2} + \ldots + 2^{a_k}. \ \ \ \ \ (8)$

In some very special cases, this can be done. For instance, suppose that one had ${a_{i+1}=a_i+1}$ with at most one exception (this is essentially what is called a ${1}$-cycle by Steiner). Then (8) simplifies via the geometric series formula to a combination of just a bounded number of powers of ${2}$ and ${3}$, rather than an unbounded number. In that case, one can start using tools from transcendence theory such as Baker’s theorem to obtain good results; for instance, in the above-referenced paper of Steiner, it was shown that ${1}$-cycles cannot actually occur, and similar methods have been used to show that ${m}$-cycles (in which there are at most ${m}$ exceptions to ${a_{i+1}=a_i+1}$) do not occur for any ${m \leq 63}$, as was shown by Simons and de Weger. However, for general increasing tuples of integers ${a_1,\ldots,a_k}$, there is no such representation by bounded numbers of powers, and it does not seem that methods from transcendence theory will be sufficient to control the expressions (8) to the extent that one can understand their divisibility properties by quantities such as ${2^a-3^k}$.

Amusingly, there is a slight connection to Littlewood-Offord theory in additive combinatorics – the study of the ${2^n}$ random sums

$\displaystyle \pm v_1 \pm v_2 \pm \ldots \pm v_n$

generated by some elements ${v_1,\ldots,v_n}$ of an additive group ${G}$, or equivalently, the vertices of an ${n}$-dimensional parallelepiped inside ${G}$. Here, the relevant group is ${{\bf Z}/q{\bf Z}}$. The point is that if one fixes ${k}$ and ${a_{k+1}}$ (and hence ${q}$), and lets ${a_1,\ldots,a_k}$ vary inside the simplex

$\displaystyle \Delta := \{ (a_1,\ldots,a_k) \in {\bf N}^k: 0 = a_1 < \ldots < a_k < a_{k+1}\}$

then the set ${S}$ of all sums of the form (8) (viewed as an element of ${{\bf Z}/q{\bf Z}}$) contains many large parallelepipeds. (Note, incidentally, that once one fixes ${k}$, all the sums of the form (8) are distinct; because given (8) and ${k}$, one can read off ${2^{a_1}}$ as the largest power of ${2}$ that divides (8), and then subtracting off ${3^{k-1} 2^{a_1}}$ one can then read off ${2^{a_2}}$, and so forth.) This is because the simplex ${\Delta}$ contains many large cubes. Indeed, if one picks a typical element ${(a_1,\ldots,a_k)}$ of ${\Delta}$, then one expects (thanks to Lemma 5) that there there will be ${\gg k}$ indices ${1 \leq i_1 < \ldots < i_m \leq k}$ such that ${a_{i_j+1} > a_{i_j}+1}$ for ${j=1,\ldots,m}$, which allows one to adjust each of the ${a_{i_j}}$ independently by ${1}$ if desired and still remain inside ${\Delta}$. This gives a cube in ${\Delta}$ of dimension ${\gg k}$, which then induces a parallelepiped of the same dimension in ${S}$. A short computation shows that the generators of this parallelepiped consist of products of a power of ${2}$ and a power of ${3}$, and in particular will be coprime to ${q}$.

If the weak Collatz conjecture is true, then the set ${S}$ must avoid the residue class ${0}$ in ${{\bf Z}/q{\bf Z}}$. Let us suppose temporarily that we did not know about Baker’s theorem (and the associated bound (7)), so that ${q}$ could potentially be quite small. Then we would have a large parallelepiped inside a small cyclic group ${{\bf Z}/q{\bf Z}}$ that did not cover all of ${{\bf Z}/q{\bf Z}}$, which would not be possible for ${q}$ small enough. Indeed, an easy induction shows that a ${d}$-dimensional parallelepiped in ${{\bf Z}/q{\bf Z}}$, with all generators coprime to ${q}$, has cardinality at least ${\min(q, d+1)}$. This argument already shows the lower bound ${q \gg k}$. In other words, we have

Proposition 6 Suppose the weak Collatz conjecture is true. Then for any natural numbers ${a, k}$ with ${2^a > 3^k}$, one has ${2^a-3^k \gg k}$.

This bound is very weak when compared against the unconditional bound (7). However, I know of no way to get a nontrivial separation property between powers of ${2}$ and powers of ${3}$ other than via transcendence theory methods. Thus, this result strongly suggests that any proof of the Collatz conjecture must either use existing results in transcendence theory, or else must contribute a new method to give non-trivial results in transcendence theory. (This already rules out a lot of possible approaches to solve the Collatz conjecture.)

By using more sophisticated tools in additive combinatorics, one can improve the above proposition (though it is still well short of the transcendence theory bound (7)):

Proposition 7 Suppose the weak Collatz conjecture is true. Then for any natural numbers ${a, k}$ with ${2^a > 3^k}$, one has ${2^a-3^k \gg (1+\epsilon)^k}$ for some absolute constant ${\epsilon > 0}$.

Proof: (Informal sketch only) Suppose not, then we can find ${a, k}$ with ${q := 2^a-3^k}$ of size ${(1+o(1))^k = \exp(o(k))}$. We form the set ${S}$ as before, which contains parallelepipeds in ${{\bf Z}/q{\bf Z}}$ of large dimension ${d \gg k}$ that avoid ${0}$. We can count the number of times ${0}$ occurs in one of these parallelepipeds by a standard Fourier-analytic computation involving Riesz products (see Chapter 7 of my book with Van Vu, or this recent preprint of Maples). Using this Fourier representation, the fact that this parallelepiped avoids ${0}$ (and the fact that ${q = \exp(o(k)) = \exp(o(d))}$) forces the generators ${v_1,\ldots,v_d}$ to be concentrated in a Bohr set, in that one can find a non-zero frequency ${\xi \in {\bf Z}/q{\bf Z}}$ such that ${(1-o(1))d}$ of the ${d}$ generators lie in the set ${\{ v: \xi v = o(q) \hbox{ mod } q \}}$. However, one can choose the generators to essentially have the structure of a (generalised) geometric progression (up to scaling, it resembles something like ${2^i 3^{\lfloor \alpha i \rfloor}}$ for ${i}$ ranging over a generalised arithmetic progression, and ${\alpha}$ a fixed irrational), and one can show that such progressions cannot be concentrated in Bohr sets (this is similar in spirit to the exponential sum estimates of Bourgain on approximate multiplicative subgroups of ${{\bf Z}/q{\bf Z}}$, though one can use more elementary methods here due to the very strong nature of the Bohr set concentration (being of the “${99\%}$ concentration” variety rather than the “${1\%}$ concentration”).). This furnishes the required contradiction. $\Box$

Thus we see that any proposed proof of the Collatz conjecture must either use transcendence theory, or introduce new techniques that are powerful enough to create exponential separation between powers of ${2}$ and powers of ${3}$.

Unfortunately, once one uses the transcendence theory bound (7), the size ${q}$ of the cyclic group ${{\bf Z}/q{\bf Z}}$ becomes larger than the volume of any cube in ${S}$, and Littlewood-Offord techniques are no longer of much use (they can be used to show that ${S}$ is highly equidistributed in ${{\bf Z}/q{\bf Z}}$, but this does not directly give any way to prevent ${S}$ from containing ${0}$).

One possible toy model problem for the (weak) Collatz conjecture is a conjecture of Erdos asserting that for ${n>8}$, the base ${3}$ representation of ${2^n}$ contains at least one ${2}$. (See this paper of Lagarias for some work on this conjecture and on related problems.) To put it another way, the conjecture asserts that there are no integer solutions to

$\displaystyle 2^n = 3^{a_1} + 3^{a_2} + \ldots + 3^{a_k}$

with ${n > 8}$ and ${0 \leq a_1 < \ldots < a_k}$. (When ${n=8}$, of course, one has ${2^8 = 3^0 + 3^1 + 3^2 + 3^5}$.) In this form we see a resemblance to Conjecture 3, but it looks like a simpler problem to attack (though one which is still a fair distance beyond what one can do with current technology). Note that one has a similar heuristic support for this conjecture as one does for Proposition 3; a number of magnitude ${2^n}$ has about ${n \frac{\log 2}{\log 3}}$ base ${3}$ digits, so the heuristic probability that none of these digits are equal to ${2}$ is ${3^{-n\frac{\log 2}{\log 3}} = 2^{-n}}$, which is absolutely summable.

The classical formulation of Hilbert’s fifth problem asks whether topological groups that have the topological structure of a manifold, are necessarily Lie groups. This is indeed, the case, thanks to following theorem of Gleason and Montgomery-Zippin:

Theorem 1 (Hilbert’s fifth problem) Let ${G}$ be a topological group which is locally Euclidean. Then ${G}$ is isomorphic to a Lie group.

We have discussed the proof of this result, and of related results, in previous posts. There is however a generalisation of Hilbert’s fifth problem which remains open, namely the Hilbert-Smith conjecture, in which it is a space acted on by the group which has the manifold structure, rather than the group itself:

Conjecture 2 (Hilbert-Smith conjecture) Let ${G}$ be a locally compact topological group which acts continuously and faithfully (or effectively) on a connected finite-dimensional manifold ${X}$. Then ${G}$ is isomorphic to a Lie group.

Note that Conjecture 2 easily implies Theorem 1 as one can pass to the connected component ${G^\circ}$ of a locally Euclidean group (which is clearly locally compact), and then look at the action of ${G^\circ}$ on itself by left-multiplication.

The hypothesis that the action is faithful (i.e. each non-identity group element ${g \in G \backslash \{\hbox{id}\}}$ acts non-trivially on ${X}$) cannot be completely eliminated, as any group ${G}$ will have a trivial action on any space ${X}$. The requirement that ${G}$ be locally compact is similarly necessary: consider for instance the diffeomorphism group ${\hbox{Diff}(S^1)}$ of, say, the unit circle ${S^1}$, which acts on ${S^1}$ but is infinite dimensional and is not locally compact (with, say, the uniform topology). Finally, the connectedness of ${X}$ is also important: the infinite torus ${G = ({\bf R}/{\bf Z})^{\bf N}}$ (with the product topology) acts faithfully on the disconnected manifold ${X := {\bf R}/{\bf Z} \times {\bf N}}$ by the action

$\displaystyle (g_n)_{n \in {\bf N}} (\theta, m) := (\theta + g_m, m).$

The conjecture in full generality remains open. However, there are a number of partial results. For instance, it was observed by Montgomery and Zippin that the conjecture is true for transitive actions, by a modification of the argument used to establish Theorem 1. This special case of the Hilbert-Smith conjecture (or more precisely, a generalisation thereof in which “finite-dimensional manifold” was replaced by “locally connected locally compact finite-dimensional”) was used in Gromov’s proof of his famous theorem on groups of polynomial growth. I record the argument of Montgomery and Zippin below the fold.

Another partial result is the reduction of the Hilbert-Smith conjecture to the ${p}$-adic case. Indeed, it is known that Conjecture 2 is equivalent to

Conjecture 3 (Hilbert-Smith conjecture for ${p}$-adic actions) It is not possible for a ${p}$-adic group ${{\bf Z}_p}$ to act continuously and effectively on a connected finite-dimensional manifold ${X}$.

The reduction to the ${p}$-adic case follows from the structural theory of locally compact groups (specifically, the Gleason-Yamabe theorem discussed in previous posts) and some results of Newman that sharply restrict the ability of periodic actions on a manifold ${X}$ to be close to the identity. I record this argument (which appears for instance in this paper of Lee) below the fold also.

I’ve just opened the research thread for the mini-polymath3 project over at the polymath blog.  I decided to use Q2 of this year’s IMO, in part to see how the polymath format copes with a geometric problem.  (The full list of questions for this year is available here.)

This post will serve as the discussion thread of the project, intended to focus all the non-research aspects of the project such as organisational matters or commentary on the progress of the project.    The third component of the project is the wiki page, which is intended to summarise the progress made so far on the problem.

As with mini-polymath1 and mini-polymath2, I myself will be serving primarily as a moderator, and hope other participants will take the lead in the research and in keeping the wiki up-to-date.