分类: Analytic number theory

  • From Sum-Free Sets to Strongly 2-Primitive Sets: Localization, Patching, and the Search for Stability

    Two extremal problems can share a proof architecture without sharing a proof. Benjamin Bedert’s 2025 breakthrough on large sum-free subsets is additive and Fourier-analytic. A July 2026 manuscript on strongly $2$-primitive sets is multiplicative and hypergraph-theoretic. Their common language is not a transferable sieve, but a four-step design: localize, build a strong object in each local block, prevent leakage between blocks, and add the gains. That comparison points toward the right stability question – and also reveals why the most naive version of stability is false.

    Status and scope. Bedert’s theorem is the February 2025 arXiv preprint cited below. The $27/2$ theorem is contained in a five-page manuscript by Przemek Chojecki supplied for this post in July 2026; no public preprint or peer-reviewed version was found, and the Erdős Problems page for #793 still labels the problem open at the time of writing. The written proof was checked here for internal coherence, but that is not independent peer review. The factor-graph and normalized-stability results later in this article are the additional analysis developed in the accompanying long note. Conditional and open statements are labelled explicitly.

    Here is the headline before the details. The supplied manuscript claims

    $$
    F(n)=\pi(n)+\left(\frac{27}{2}+o(1)\right)
    \frac{n^{2/3}}{(\log n)^2},
    $$

    where $F(n)$ is the largest size of a strongly $2$-primitive subset of $[1,n]$. One might expect every near-extremizer to be close to the construction behind this formula: almost all primes, plus products of three primes near $n^{1/3}$ arranged as a nearly saturated linear $3$-uniform hypergraph. That literal statement is false. There is, however, an exact and useful replacement: after contracting harmless private-prime lifts, every near-extremizer has a canonical prime layer and a small exceptional hub carrying the entire second-order excess. If the residual hub is mostly made of squarefree triples, then pair saturation and $n^{1/3}$-scale stability follow. Proving – or refuting – that triple-core hypothesis is the remaining inverse problem.

    1. Two problems that look similar only from far away

    1.1 The sum-free extraction problem

    A set $B\subset\mathbb Z$ is sum-free if it contains no $x,y,z$, not necessarily distinct, with

    $$x+y=z.$$

    For a finite set $A$ of integers, write

    $$
    S(A)=\max\{|B|:B\subseteq A,\ B\text{ is sum-free}\},
    $$

    and, for $N$-element sets of positive integers,

    $$
    S(N)=\min_{|A|=N}S(A).
    $$

    This is an extraction problem. An adversary gives us an arbitrary host set $A$, and we must find a large structured subset inside it. Erdős’s middle-third argument gives $S(A)\ge |A|/3$. Alon and Kleitman improved this to $(N+1)/3$, and Jean Bourgain proved $S(N)\ge (N+2)/3$ in 1997. The long-standing qualitative question was whether the additive improvement can tend to infinity:

    $$S(N)\ge \frac N3+\omega(N),\qquad \omega(N)\longrightarrow\infty.$$

    Bedert answered yes, proving that an absolute $c\gt0$ exists such that

    $$
    S(A)\ge \frac{|A|}{3}+c\log\log |A|.
    $$

    1.2 The strongly $2$-primitive packing problem

    A set $A\subseteq[1,n]$ is strongly $2$-primitive when

    $$
    a\nmid bc
    \qquad
    (a,b,c\in A,\ a\notin\{b,c\}),
    $$

    where $b=c$ is allowed. The word strongly matters. Under a more recent convention, a $2$-primitive set only forbids witnesses $b,c$ that are distinct. For example, $\{4,5,6\}$ passes that weaker test but fails the strong one because $4\mid6^2$.

    Now define

    $$
    F(n)=\max\{|A|:A\subseteq[1,n],\ A\text{ is strongly }2\text{-primitive}\}.
    $$

    This is a packing problem. The host interval is fixed, and we directly construct the largest possible forbidden-divisibility family. The all-primes set gives the leading term $\pi(n)$. In 1938 Erdős proved upper and lower bounds of the form

    $$
    \pi(n)+c_1\frac{n^{2/3}}{(\log n)^2}
    \le F(n)\le
    \pi(n)+c_2\frac{n^{2/3}}{(\log n)^2},
    $$

    and later asked whether the second-order term has an asymptotic constant. This is the modern Erdős Problem #793. The supplied 2026 manuscript proposes that the constant is $27/2$; it is important not to say that Erdős himself conjectured this numerical value.

    Comparison of the sum-free extraction problem and the strongly 2-primitive packing problem
    Figure 1. The quantifiers already separate the two questions. Sum-free theory extracts a subset from an arbitrary host; the multiplicative problem packs a forbidden-divisibility family into a fixed interval.

    1.3 Equality versus order

    The relation $a\nmid bc$ is not ordinary product-freeness. The set $\{6,10,15\}$, for instance, has no internal equality $xy=z$, yet $6\mid10\cdot15$. Prime valuations expose the real geometry:

    $$
    a\mid bc
    \quad\Longleftrightarrow\quad
    v_p(a)\le v_p(b)+v_p(c)
    \quad\text{for every prime }p.
    $$

    Thus the forbidden relation is a coordinatewise domination inequality in the divisor lattice. For squarefree integers, if $E_a=\{p:p\mid a\}$, it becomes

    $$E_a\subseteq E_b\cup E_c.$$

    So the natural combinatorial object is a $2$-cover-free family, not the solution set of a linear equation. Fourier characters are superb at detecting equations such as $x+y=z$. They do not come with an evident contractive projection that detects the partial order $\nu(a)\le\nu(b)+\nu(c)$. This is the first reason Bedert’s proof cannot simply be copied into the multiplicative setting.

    2. The $27/2$ upper bound: every element needs a private factor

    Set

    $$
    y=n^{1/3},\qquad
    M=\frac{y}{\log n},\qquad
    \Sigma_n=M^2=\frac{n^{2/3}}{(\log n)^2}.
    $$

    The upper bound begins with a small lemma that contains more information than the inequality it proves.

    Private-factor lemma. Let $\mathcal B$ be a set of positive integers, and choose for every $a\in A$ a factorization $a=u_av_a$ with $u_a,v_a\in\mathcal B$. If $A$ is strongly $2$-primitive, then $|A|\le|\mathcal B|$.

    Think of $E_a=\{u_a,v_a\}$ as a two-element multiset. If $u_a\ne v_a$ and neither coordinate is private to $a$, another chosen pair contains $u_a$ and another contains $v_a$; the corresponding two elements have product divisible by $a$. If $u_a=v_a=x$, failure of privacy would give another pair containing $x$ twice, hence another element equal to $x^2=a$. Therefore every $a$ has a private coordinate, and these coordinates are distinct. That is the injection $A\hookrightarrow\mathcal B$.

    The manuscript chooses the multiplicative $2$-basis

    $$
    \begin{aligned}
    \mathcal B_0&=[1,n^{3/5}],\\
    \mathcal B_1&=\{p\text{ prime}:n^{3/5}\lt p\le n\},\\
    \mathcal B_2&=\{pq:p,q\le y\text{ prime}\},\\
    \mathcal B_3&=\{qr:y\lt q\le n^{2/5},\ r\le n/q^2,\ q,r\text{ prime}\}.
    \end{aligned}
    $$

    Every $m\le n$ can be factored into two elements of $\mathcal B=\mathcal B_0\cup\mathcal B_1\cup\mathcal B_2\cup\mathcal B_3$. The case split is elementary but carefully tuned. Small $m$ can be balanced into two factors below $n^{3/5}$, a prime factor above $n^{2/5}$ can be split off, and the remaining difficult case groups two of the three largest prime factors into an element of $\mathcal B_2$ or $\mathcal B_3$.

    The main prime layer comes from $\mathcal B_1$. The genuinely second-order counts are

    $$
    |\mathcal B_2|
    =\binom{\pi(y)+1}{2}
    =\left(\frac92+o(1)\right)\Sigma_n,
    $$

    because $\pi(n^{1/3})\sim3n^{1/3}/\log n=3M$, and

    $$
    |\mathcal B_3|
    =\sum_{y\lt q\le n^{2/5}}\pi\!\left(\frac n{q^2}\right)
    =(9+o(1))\Sigma_n.
    $$

    The contribution of $\mathcal B_0$ and all relevant overlaps is $o(\Sigma_n)$. Hence

    $$
    |\mathcal B|
    =\pi(n)+\left(\frac92+9+o(1)\right)\Sigma_n
    =\pi(n)+\left(\frac{27}{2}+o(1)\right)\Sigma_n.
    $$

    The private-factor injection then gives the upper bound. A caution that becomes crucial for stability: $\mathcal B_2$ and $\mathcal B_3$ are labels in an upper-bound certificate. Near-saturation of these labels does not immediately say that the original elements of $A$ are themselves products of two or three primes.

    3. The lower bound: turn properly coloured edges into prime triples

    The lower bound lives in a linear $3$-uniform hypergraph. Let $\mathcal H$ be a family of triples of distinct primes such that any two triples share at most one prime and every edge product is at most $n$. Define

    $$
    A_{\mathcal H}
    =\{p\le n:p\text{ prime and }p\notin V(\mathcal H)\}
    \cup
    \left\{\prod_{p\in E}p:E\in\mathcal H\right\}.
    $$

    Then

    $$|A_{\mathcal H}|=\pi(n)-|V(\mathcal H)|+|\mathcal H|.$$

    Why is this strongly $2$-primitive? A target triple product has three distinct prime coordinates. Each other hyperedge supplies at most one of them, so two other elements supply at most two. The same argument still works when the two witnesses coincide. Primes outside the vertex set remain singleton elements and cannot divide any other chosen element.

    3.1 Logarithmic prime bins

    Fix a small mesh $h\gt0$ and divide primes near $y=n^{1/3}$ into bins

    $$
    P_i=\{p\text{ prime}:ye^{ih}\lt p\le ye^{(i+1)h}\},
    \qquad
    \Delta_i=e^{(i+1)h}-e^{ih}.
    $$

    For a cell satisfying

    $$i\le j,\qquad i+2j\le-4,$$

    put $k=-i-j-3$. Then $i\le j\lt k$, and any $p\in P_i$, $q\in P_j$, $r\in P_k$ obeys $pqr\le n$.

    If $i\lt j$, properly edge-colour the complete bipartite graph between $P_i$ and $P_j$, using primes of $P_k$ as colours. When $i=j$, do the same with the complete graph on $P_i$. The lower pair $\{p,q\}$, coloured by $r$, becomes the hyperedge $\{p,q,r\}$.

    A complete bipartite graph between two prime bins properly edge-coloured by a third prime bin
    Figure 2. Within a cell, every colour class is a matching, so two produced triples never share a lower pair. Across cells, the sorted signature $(i,j,k)$ has fixed sum $-3$; two shared bin indices force the third and hence force the same cell.

    The proper colouring is the local no-interference mechanism. The constant-sum signature is the global one. Together they make the union over all cells a linear hypergraph.

    3.2 The cell weight

    For any fixed finite collection of bins, the prime number theorem gives

    $$|P_i|=(3+o(1))M\Delta_i.$$

    Off the diagonal, a cell contributes approximately $9M^2\Delta_i\Delta_j$ edges. A diagonal cell contributes half as many unordered pairs. The exact geometric-series identity is

    $$
    \sum_{\substack{i\lt j\\i+2j\le-4}}\Delta_i\Delta_j
    +\frac12\sum_{i\le-2}\Delta_i^2
    =e^{-h}+\frac12e^{-2h}.
    $$

    Letting $h\to0$, the weight tends to $3/2$, so

    $$
    |\mathcal H|
    =\left(9\cdot\frac32-o(1)\right)\Sigma_n
    =\left(\frac{27}{2}-o(1)\right)\Sigma_n.
    $$

    The number of vertices used is only $o(\Sigma_n)$. Replacing those primes by the triple products therefore yields the matching lower bound.

    The logarithmic feasible pair region split into the 9 over 2 and 9 counting regimes
    Figure 3. The same constant is visible in the canonical feasible pair space. The labels $9/2$ and $9$ are prime-counting contributions, not Euclidean areas of the drawing.

    4. From Erdős’s middle third to Bedert’s $c\log\log N$

    We now return to the additive problem. Let $\mathbb T=\mathbb R/\mathbb Z$, and let $\phi=\mathbf1_{(1/3,2/3)}$. The middle third of the circle is sum-free: two points in it cannot add, modulo $1$, to another point in it. Therefore, for every $x\in\mathbb T$,

    $$A_x=\{a\in A:ax\pmod1\in(1/3,2/3)\}$$

    is sum-free. Averaging $|A_x|$ over $x$ gives $|A|/3$. The problem is to force a positive fluctuation above that mean.

    After a harmless normalization, the relevant Fourier series has the form

    $$
    F_A(x)=\sum_{a\in A}\sum_{m\ge1}
    \frac{\chi(m)}m\cos(2\pi max),
    $$

    where $\chi$ is the non-principal real character modulo $3$. A lower bound for $\|F_A\|_1$ gives a one-sided large value and hence a large sum-free subset.

    4.1 Where Bourgain’s sieve loses a logarithm

    The historical correction is worth making explicit. The relevant predecessor is Jean Bourgain’s 1997 paper, not “John Bogan, 1993.” Also, the Littlewood $L^1$ conjecture had already been proved in 1981, independently by McGehee-Pigno-Smith and Konyagin. Bourgain’s argument combines Fourier analysis with a Möbius operation that filters harmonic indices.

    Schematically, if $R_Q$ denotes the positive $Q$-rough integers, then

    $$
    \sum_{d\mid\prod_{p\le Q}p}
    \frac{\mu(d)\chi(d)}d F_A(dx)
    =\sum_{a\in A}\cos(2\pi ax)+\mathcal R_Q(x).
    $$

    The left side costs

    $$
    \prod_{p\le Q}\left(1+\frac1p\right)\asymp\log Q
    $$

    under the triangle inequality. If $Q$ is large enough to make the rough remainder negligible directly, this factor consumes the logarithm delivered by a Littlewood-type lower bound. This explains why merely “using the Littlewood theorem harder” does not produce the desired unbounded gain.

    4.2 Bedert’s split: medium primes are sieved, small primes are projected

    Bedert chooses

    $$Q_1=(\log N)^{1/2},\qquad Q=(\log N)^{20}.$$

    He Möbius-sieves only the medium primes $Q_1\le p\le Q$. Their reciprocal sum is $O(1)$, so the norm loss is a constant rather than a logarithm. The small primes $p\le Q_1$ are handled by a different operation.

    For every small prime, record the exact valuation $\nu_p(a)$ and the unit residue after removing that prime power. Chinese remaindering packages all this data into a joint fibre $A(r,\nu)$. Fourier projection to a residue class,

    $$
    \operatorname{Proj}(H;\rho\bmod q)(x)
    =\sum_{m\equiv\rho\ (q)}\widehat H(m)e(mx),
    $$

    is an $L^1$-contraction:

    $$
    \|\operatorname{Proj}(H;\rho\bmod q)\|_1\le\|H\|_1.
    $$

    At the main valuation level, the projection removes small-prime harmonic contamination without paying the full Möbius product. After truncation, one sees a large exponential sum plus an $L^2$-small error, to which a robust McGehee-Pigno-Smith test function applies.

    At a general valuation level, lower fibres can still leak into the projected channel. Bedert proves an isolation-or-descent alternative: either the desired $L^1$ lower bound is already present, or a large fibre has a strictly lower fibre losing at most a controlled polylogarithmic factor. Iterating and reversing this descent produces a chain

    $$
    \nu^{(1)}\prec\nu^{(2)}\prec\cdots\prec\nu^{(J)},
    \qquad
    J\gg\frac{\log N}{\log\log N},
    $$

    whose fibre sizes grow geometrically.

    4.3 The non-Archimedean no-leakage lemma

    Associate to these levels the nested moduli

    $$
    q_i=\prod_{p\le Q_1}p^{\nu_p^{(i)}+1},
    \qquad q_1\mid q_2\mid\cdots\mid q_J.
    $$

    Let $g_i$ be the normalized indicator of the $i$-th residue block and put $Q_i(x)=\exp(-|\widehat g_i(x)|)$. The non-Archimedean MPS test function is assembled from

    $$
    \Phi_J=
    \widehat g_J+\widehat g_{J-1}Q_J+\cdots+
    \widehat g_1Q_2\cdots Q_J.
    $$

    The decisive observation is that $|\widehat g_i|$ is $1/q_i$-periodic, hence $\widehat Q_i$ is supported on $q_i\mathbb Z$. Multiplying an earlier block by a later $Q_k$ shifts its frequencies only by multiples of $q_k$. Since $q_i\mid q_k$, the earlier block cannot escape its residue class modulo $q_i$. Main inner products contribute one controlled unit per block, while geometric fibre growth makes the cross terms summable.

    That is Bedert’s genuine no-leakage mechanism. It is more precise than the slogan “use many congruence classes and patch them.”

    One further qualification prevents a common misstatement. Bedert obtains a suitable $F_4$-isomorphic model $B$ of the original set and proves the required $L^1$ bound there. The norm $\|F_A\|_1$ is not claimed to be invariant under an $F_4$-isomorphism. What is preserved is the four-term additive information needed for sum-freeness, so $S(A)=S(B)$.

    Bedert’s inverse output is also stronger than the numerical lower bound. If $S(A)\le N/3+C$, his results force low additive dimension, a dense $F_4$-model, large additive energy in every substantial subset, and a “99% Structure Theorem” decomposing all but a small exceptional set into large small-doubling pieces. This is a genuine inverse theorem for host sets resisting sum-free extraction, but not an edit-distance classification by one canonical extremizer.

    5. The real bridge: localize, build, prevent leakage, sum

    Parallel flowcharts for Bedert's p-adic Fourier proof and the logarithmic hypergraph construction
    Figure 4. The analogy is architectural. Bedert preserves Fourier residue lanes through nested moduli; the multiplicative construction preserves low codegrees through matching colours and unique scale signatures.

    The dictionary is now clean:

    Role Bedert’s sum-free proof Strongly $2$-primitive proof
    Underlying relation Linear equation in an abelian group Coordinatewise domination of prime valuations
    Local coordinates Exact small-prime valuations and unit residues Logarithmic sizes of prime factors
    Local block Joint $p$-adic fibre and MPS block Scale cell and properly coloured graph
    No interference Nested moduli preserve Fourier support Colour matchings and unique cell signatures preserve linearity
    Accumulation One controlled inner product per fibre One triple per admissible lower pair
    Inverse output Additive dimension, energy, small-doubling pieces Private factors and a cover-free exponent core

    This is a substantial connection, but it is a connection of proof design. Bedert’s Fourier projection has no direct analogue for the divisor partial order. The most plausible transfer is therefore not a line-by-line proof but a research program: find a multiplicative localization, an isolation-or-descent alternative, and a no-leakage invariant that survives repeated prime powers and larger supports.

    6. The first stability guess is false

    The lower construction suggests a tempting statement: every set with

    $$
    |A|\ge
    \pi(n)+\left(\frac{27}{2}-o(1)\right)\Sigma_n
    $$

    should differ in only $o(\Sigma_n)$ places from almost all primes plus a near-optimal linear family of triple products. Two examples show why this is too rigid.

    6.1 A private-prime lift

    Start with a near-optimal linear triple family $\mathcal H_n$ using odd primes below $n/2$. Keep primes above $n/2$, replace every unused prime $p\le n/2$ by $2p$, and retain the triple products. Each $2p$ has the private prime divisor $p$; no other chosen element contains that prime. The triple products are still protected by linearity. The new set remains strongly $2$-primitive and has the same second-order asymptotic size.

    But essentially every prime below $n/2$ has been replaced by a composite. The symmetric-difference distance from the all-primes model is

    $$
    \Theta(\pi(n/2))=\Theta\!\left(\frac n{\log n}\right),
    $$

    and

    $$
    \frac{\pi(n/2)}{\Sigma_n}
    \asymp n^{1/3}\log n\longrightarrow\infty.
    $$

    So literal edit-distance stability fails by much more than the scale of the second-order term.

    6.2 A nonlinear sunflower at almost no cost

    Even if we insist on squarefree triple products, linearity need not hold in the original representation. Add a sunflower

    $$\{\{2,3,r\}:r\in R\}$$

    for many private petals $r$. Remove the singleton primes $2,3,r$ and add the products $6r$. The net cardinality loss is only $2$, while the support family contains $\asymp n/\log n$ edges sharing the pair $\{2,3\}$. It is $2$-cover-free because every target has its own private petal, but any linear subfamily contains at most one sunflower edge.

    The two examples have the same cause: degree-one prime coordinates carry a huge amount of neutral decoration. Stability can only become true after quotienting out this freedom.

    7. Near equality in the upper bound is an exact star forest

    Fix the chosen factorizations $a=u_av_a$ from the multiplicative basis proof. Make a graph $G_A$ with vertex set $\mathcal B$, one edge $\{u_a,v_a\}$ for each $a\in A$, and allow loops.

    Exact factor-graph theorem. Every nonloop edge has a degree-one endpoint, and every loop is an isolated component. Thus $G_A$ is a disjoint union of stars and isolated loops. If $U$ is the number of unused vertices and $c(G_A)$ is the number of nonloop star components, then

    $$|\mathcal B|-|A|=U+c(G_A).$$

    The proof is the private-factor lemma read without discarding information. If a nonloop edge $\{u,v\}$ had both endpoints in other edges, the corresponding two elements would have product divisible by $uv=a$. If a loop at $u$ met another edge, then $u^2$ would divide the square of the other element.

    A factor graph decomposed into stars, an isolated loop and unused vertices
    Figure 5. The upper-bound defect is counted exactly: an unused basis coordinate costs one, and each nonloop star component costs one.

    More quantitatively,

    $$
    \bigl|\{b\in\mathcal B\setminus\mathcal B_0:
    \deg_{G_A}(b)\ne1\}\bigr|
    \le |\mathcal B|-|A|.
    $$

    If

    $$
    |A|\ge
    \pi(n)+\left(\frac{27}{2}-\delta_n\right)\Sigma_n,
    $$

    then the defect $D_n=|\mathcal B|-|A|$ is at most $(\delta_n+o(1))\Sigma_n$. Hence almost every secondary coordinate in both $\mathcal B_2$ and $\mathcal B_3$ occurs in exactly one chosen factor pair. This is a rigorous stability theorem for the upper-bound certificate. It is not yet a classification of the original integers.

    8. Private-prime compression gives the correct normal form

    Suppose a prime $p$ divides $a\in A$ and divides no other element of $A$. Replacing $a$ by $p$ preserves the cardinality and strong $2$-primitivity. The new target $p$ cannot divide a product of two other elements because neither contains $p$; for any unchanged target, replacing $a$ by a divisor only decreases the product available to cover it.

    For primes $p\gt n^{3/5}$, the basis factorization can be chosen to expose $p$ in every multiple. Thus a degree-one large-prime coordinate is genuinely a globally private prime and can be contracted. Contract all such coordinates simultaneously.

    Normalized stability modulo private-prime lifts. From every strongly $2$-primitive $A\subseteq[1,n]$ satisfying the near-extremal bound above, private-prime contractions produce a strongly $2$-primitive set $\widetilde A$ of the same size with

    $$\widetilde A=(\mathbb P\cap[1,n]\setminus V)\cup C,\qquad \operatorname{supp}(c)\subseteq V\quad(c\in C).$$

    where

    $$|V|\le(\delta_n+o(1))\Sigma_n,\qquad |C|-|V|=|A|-\pi(n).$$

    In particular, if $\delta_n=o(1)$, then

    $$
    |V|=o(\Sigma_n),
    \qquad
    |C|=\left(\frac{27}{2}+o(1)\right)\Sigma_n.
    $$

    The geometry is now transparent. Before compression, the leading $\pi(n)$ coordinates may be represented by arbitrary private multiples. After compression, almost every prime appears literally. Every residual composite is supported entirely on the small exceptional prime hub $V$, and its excess over the missing primes is exactly the second-order gain.

    Private-prime lifts contracted to a normal form with singleton primes and a small exceptional hub
    Figure 6. Literal stability fails on the left. The quotient by private-prime contractions produces the exact normal form on the right.

    There is one more unconditional consequence. At most $|V|$ members of $C$ have support of size at most two. Indeed, any one- or two-prime residual element must have a valuation coordinate in which it is a strict global maximum; assign the element to such a prime. Two elements cannot receive the same prime. Therefore, in a normalized near-extremizer, all but $o(\Sigma_n)$ residual composites have at least three distinct prime divisors.

    This is close to the hoped-for triple picture, but it does not prove that the typical element has exactly three prime factors, that it is squarefree, or that its prime factors lie near $n^{1/3}$.

    9. Conditional stability when the residual core is made of triples

    Now impose an additional hypothesis:

    Squarefree-triple hypothesis. All but $o(\Sigma_n)$ elements of $C$ are products of three distinct primes.

    Let $\mathcal H$ be the support family of those triples. Strong $2$-primitivity says precisely that it is $2$-cover-free:

    $$
    E\nsubseteq F\cup G
    \qquad(E,F,G\in\mathcal H,\ E\notin\{F,G\}),
    $$

    with $F=G$ allowed.

    9.1 Pruning to a linear core

    If two triples $E,F$ share a pair and $x$ is the third vertex of $E$, then $x$ has degree one. Otherwise another edge $G$ containing $x$, together with $F$, would cover $E$. Delete such a petal edge whenever a repeated pair occurs. Every deletion removes at least one vertex as well as one edge, so the excess $|\mathcal H|-|V(\mathcal H)|$ does not decrease.

    The result is a linear subfamily $\mathcal L$ with

    $$
    |\mathcal L|-|V(\mathcal L)|
    \ge |\mathcal H|-|V(\mathcal H)|,
    $$

    and only $o(\Sigma_n)$ edges are lost in the normalized near-extremal setting. Notice the nuance: linearity is recovered after pruning private petals; it is not asserted for every cover-free representation.

    9.2 Saturating the feasible pair space

    Define

    $$
    \mathcal D_n=
    \{(p,q):p\lt q\text{ prime and }pq^2\le n\}.
    $$

    Sort an edge of $\mathcal L$ as $p\lt q\lt r$ and map it to $(p,q)$. Linearity makes this map injective. Since $r\gt q$ and $pqr\le n$, the image lies in $\mathcal D_n$. Direct counting gives

    $$
    |\mathcal D_n|
    =\left(\frac{27}{2}+o(1)\right)\Sigma_n.
    $$

    The range $q\le n^{1/3}$ contributes $(9/2+o(1))\Sigma_n$; the range $n^{1/3}\lt q\le n^{2/5}$ contributes $(9+o(1))\Sigma_n$; the tail is negligible. Since the linear core already has $(27/2-o(1))\Sigma_n$ edges, its lower-pair map misses only $o(\Sigma_n)$ feasible pairs.

    Moreover, for every fixed $\varepsilon\gt0$, all but $o(\Sigma_n)$ edges satisfy

    $$
    n^{1/3-\varepsilon}
    \le p,q,r\le
    n^{1/3+\varepsilon}.
    $$

    This is the desired scale stability. It does not imply uniqueness. Different one-factorizations or different proper edge-colourings can change $\Theta(\Sigma_n)$ triples while preserving the same pair occupancy and extremal count. The stable object is the saturated pair space, not an individual colouring.

    10. The remaining inverse problem is weighted and cover-free

    After normalization, write every $c\in C$ as an exponent vector

    $$
    \nu(c)=(v_p(c))_{p\in V}\in\mathbb Z_{\ge0}^{V}.
    $$

    Strong $2$-primitivity becomes the weighted cover-free condition

    $$
    \nu(c)\nleq\nu(c_1)+\nu(c_2)
    \quad\text{coordinatewise}
    $$

    whenever $c\notin\{c_1,c_2\}$. We know that

    $$
    |C|=\left(\frac{27}{2}+o(1)\right)\Sigma_n,
    \qquad |V|=o(\Sigma_n),
    $$

    and that almost every vector has support at least three. What remains unknown is the sharper assertion

    $$
    \#\{c\in C:c\text{ is not a squarefree product of three primes}\}
    =o(\Sigma_n).
    $$

    This gap may conceal genuine alternative near-extremizers. If a triple $pqr$ has product slack, one can contemplate replacing it by $2pqr$ while deleting the singleton prime $2$. For a linear support family, the underlying three private coordinates still prevent coverage. It is not known whether a positive density of such bounded-hub multiplier layers can coexist globally near the $27/2$ threshold, or whether cross-layer collisions force them to be negligible.

    Open normalized stability problem. Classify weighted $2$-cover-free exponent families under the product constraint $c\le n$ whose excess is $(27/2-o(1))\Sigma_n$. Decide whether the residual core is predominantly squarefree of support three, or whether bounded-hub multiplier layers produce genuinely different normalized near-extremizers.

    10.1 What a Bedert-style descent would need

    The additive proof suggests three design requirements, not three ready-made lemmas.

    1. A local order projection. One needs to isolate a valuation or support layer while controlling contamination from lower exponent vectors. A divisor-lattice zeta or Möbius transform is a possible language, but no analogue of Bedert’s $L^1$-contractive residue projection is presently available.
    2. An isolation-or-descent alternative. If the secondary basis slots do not have predominantly prime cofactors, the argument should descend to a structured hub of repeated factors with a quantitative gain or a summable loss.
    3. A no-leakage invariant. Low codegree and scale signatures work for squarefree triples. A general invariant must survive repeated exponents and supports of size at least four.

    The star-forest theorem is already a first inverse statement of this kind: near equality forces almost every basis coordinate to be a leaf attached to a small collection of hubs. The hard step is to turn that certificate-level hub structure into a classification of the original weighted exponent vectors.

    11. What is proved, conditional, and open

    • Published/preprint additive result: Bedert proves $S(A)\ge |A|/3+c\log\log|A|$, together with inverse information involving additive dimension, dense $F_4$-models, energy and a $99\%$ decomposition into large small-doubling pieces.
    • Supplied 2026 manuscript: the claimed $27/2$ asymptotic follows from the multiplicative-basis upper bound and the logarithmic-cell hypergraph construction described above. The status caveat at the beginning remains in force.
    • Unconditional stability developed in the accompanying note: literal edit stability is false; the upper-bound factor graph is an exact star forest; private-prime compression gives the normal form $(\mathbb P\setminus V)\cup C$; and almost every residual composite has at least three distinct prime divisors.
    • Conditional stability: if almost all residual composites are squarefree triples, pruning gives a near-maximal linear core whose lower pairs saturate $\mathcal D_n$, and almost every prime factor lies at scale $n^{1/3+o(1)}$.
    • Open: prove that the normalized residual core is predominantly squarefree of support three, or construct a different near-extremal weighted cover-free core.

    The conceptual moral is simple but useful. Bedert’s $p$-adic fibres and the multiplicative logarithmic cells are not the same mathematical object. Yet both proofs win by choosing coordinates in which local constructions can be made strong and then finding an exact invariant that prevents those constructions from interfering. In the stability problem, the same philosophy says to quotient the neutral directions first. Once private-prime lifts are removed, the true obstruction becomes visible: a weighted cover-free family on a tiny prime hub. That is the right object for the next theorem.

    References and source trail

    1. P. Erdős, On sequences of integers no one of which divides the product of two others and on some related problems, 1938.
    2. P. Erdős, On some applications of graph theory to number-theoretic problems, 1969; see also Erdős Problem #793.
    3. O. Carruth McGehee, L. Pigno and B. Smith, Hardy’s inequality and the $L^1$-norm of exponential sums, Annals of Mathematics 113 (1981), 613-618.
    4. J. Bourgain, Estimates related to sumfree subsets of sets of integers, Israel Journal of Mathematics 97 (1997), 71-92.
    5. B. Bedert, Large sum-free subsets of sets of integers via $L^1$-estimates for trigonometric series, arXiv:2502.08624v1, 12 February 2025.
    6. P. Chojecki, The Second Term for Strongly 2-Primitive Sets, user-supplied five-page manuscript, July 2026; no public identifier located at the time of writing.
  • Dirichlet hyperbola method:用双曲线拆分除数和

    旧博客原文

    原题:Dirichlet hyperbola method

    A pdf version is Dirichlet hyperbola method.

    1. Introduction

    Theorem 1

    \displaystyle \sum_{1\leq n\leq x}d(n)=\sum_{1\leq n\leq x}[\frac{x}{n}]=xlogx+(2\gamma-1) x+O(\sqrt{x}) \ \ \ \ \ (1)

     

    Remark 1 I thought this problem initial 5 years ago, cost me several days to find a answer, I definitely get something without the argument of Dirchlet hyperbola method and which is weaker but morally the same camparable with the result get by Dirichlet hyperbola method.

    Remark 2 How to get the formula:

    \displaystyle \sum_{1\leq n\leq x}d(x)=\sum_{1\leq n\leq x}[\frac{x}{n}]? \ \ \ \ \ (2)

    In fact,

    \displaystyle \sum_{1\leq n\leq x}d(x)=\sum_{1\leq ab\leq x}1=\sum_{1\leq n\leq x}[\frac{x}{n}]\ \ \ \ \ (3)

    Which is the integer lattices under or lying on the hyperbola {\{(a,b)|ab=x\}}.

    Remark 3 By trivial argument, we can bound the quantity as following way,

    \displaystyle \begin{array}{rcl} \sum_{1\leq n\leq x}[\frac{x}{n}] & = & \sum_{1\leq ab\leq x}1\\ & = & x\sum_{i=1}^x\frac{1}{i}-\sum_{i=1}^x\{\frac{x}{i}\}\\ & = &xlnx+\gamma x+O(x) \end{array}

    The error term is {O(x)}, which is too big. But fortunately we can use the symmetry of hyperbola to improve the error term.

    Proof:

    \displaystyle \begin{array}{rcl} \sum_{1\leq n\leq x}d(n) & = & \sum_{ab\leq x}1\\ & = & \sum_{a\geq \sqrt{x}}[\frac{x}{b}]+\sum_{b\geq \sqrt{x}}[\frac{x}{a}]-\sum_{1\leq a,b\leq \sqrt{x}}1\\ & = & xlogx+(2\gamma-1)x+O(\sqrt{x}) \end{array}

    \Box

    Theorem 2

    Given a natural number k, use the hyperbola method together
    with induction and partial summation to show that

    \displaystyle \sum_{n\leq x}d_k(n) = xP_k(log x) + O(x^{1-\frac{1}{k}+\epsilon}), n\leq x \ \ \ \ \ (4)

    where {P_k(t)} denotes a polynomial of degree {k-1} with leading term {\frac{t^{k-1}}{(k-1)!}}.

    Remark 4 {P_k(x)} is the residue of {\zeta(s)^kx^ss^{-1}} at {s=1}.

    Proof:

    We can establish the dimension 3 case directly, which is the following asymptotic formula,

    \displaystyle \sum_{1\leq xy\leq n}[\frac{n}{xy}]=xP_2(logx)+O(x^{1-\frac{1}{3}+\epsilon}) \ \ \ \ \ (5)

    The approach is following, we first observe that

    \displaystyle \sum_{1\leq xy\leq n}[\frac{n}{xy}]=\sum_{xyz\leq n}1 \ \ \ \ \ (6)

    The problem transform to get a asymptotic formula for the lattices under 3 dimension hyperbola. The first key point is, morally {([n^{\frac{1}{3}}],[n^{\frac{1}{3}}],[n^{\frac{1}{3}}])} is the central point under the hyperbola.

    Then we can divide the range into 3 parts, and try to get a asymptotic formula for each part then add them together. Assume we have:

    1. {A_x=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq yz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}.
    2. {A_y=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq xz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}.
    3. {A_z=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq xy\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}.

    Then the task transform to get a asymptotic formula,

    \displaystyle A_x=A_y=A_z=xQ_2(logx)+O(x^{1-\frac{1}{3}+\epsilon}) \ \ \ \ \ (7)

    But we can do the same thing for {\sum_{1\leq yz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]} and then integral it. This end the proof. For general {k\in {\mathbb N}}, the story is the same, by induction.

    Induction on {k} and use the Fubini theorem to calculate {\sum_{x_1...x_r\leq n}\frac{n}{x_1...x_r},\forall 1\leq r\leq k}. \Box

    There is a major unsolved problem called Dirichlet divisor problem.

    \displaystyle \sum_{n\leq x}d(n) \ \ \ \ \ (8)

    What is the error term? The conjecture is the error term is {O(x^{\theta}), \forall \theta>\frac{1}{4}}, it is known that {\theta=\frac{1}{4}} is not right.

    Remark 5

    To beats this problem, need some tools in algebraic geometry.

    2. Several problems

    {\forall k\in {\mathbb N}}, is there a asymptotic formula for {\sum_{t=1}^n\{\frac{kn}{t}\}} ?

    {\forall k\in {\mathbb N}}, {f(n)} is a polynomial with degree {k}, is there a asymptotic formula for {\sum_{t=1}^n\{\frac{f(n)}{t}\}} ?

    {\forall k\in {\mathbb N}}, {g(n)} is a polynomial with degree {k}, is there a asymptotic formula for {\sum_{t=1}^n\{\frac{n}{g(t)}\}} ?

    Theorem 3 {k\in {\mathbb N}}, then we have

    \displaystyle \lim_{n\rightarrow \infty}\frac{\{\frac{kn}{1}\}+\{\frac{kn}{2}\}+...+\{\frac{kn}{n}\}}{n}=k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma) \ \ \ \ \ (9)

    Proof:

    \displaystyle \begin{array}{rcl} \frac{\{\frac{kn}{1}\}+\{\frac{kn}{2}\}+...+\{\frac{kn}{n}\}}{n} & = & \frac{\sum_{i=1}^k\frac{kn}{i}-\sum_{i=1}^n[\frac{kn}{i}]}{n}\\ & = & k(lnn+\gamma +\epsilon_n)-\frac{\sum_{i=1}^{kn}[\frac{kn}{i}]-\sum_{i=n+1}^{kn}[\frac{kn}{i}]}{n} \end{array}

    \Box

    Now we try to estimate

    \displaystyle S_k(n)=\sum_{i=1}^{kn}[\frac{kn}{i}]-\sum_{i=n+1}^{kn}[\frac{kn}{i}] \ \ \ \ \ (10)

    In fact, we have,

    \displaystyle \begin{array}{rcl} S_k(n) & = & (2\sum_{i=1}^{[\sqrt{kn}]}[\frac{kn}{i}]-[\sqrt{kn}]^2)-(\sum_{i=1}^k[\frac{kn}{i}]-kn)\\ & = & 2\sum_{i=1}^{[\sqrt{kn}]}\frac{kn}{i}-\sum_{i=1}^k\frac{kn}{i}+2\{\sqrt{kn}\}[\sqrt{kn}]+\{\sqrt{kn}\}^2-2\sum_{i=1}^{[\sqrt{kn}]}\{\frac{kn}{i}\}+\sum_{i=1}^k\{\frac{kn}{i}\}\\ & = & 2kn(ln[\sqrt{kn}]+\gamma+\epsilon_{[\sqrt{kn}]})-kn\sum_{i=1}^k\frac{1}{i}+r(n)\\ & = & knln(kn)+kn(2\gamma-\sum_{i=1}^k\frac{1}{i})+r'(n)\\ & = & knln n+kn(2\gamma+lnk-\sum_{i=1}^k\frac{1}{i})+r'(n) \end{array}

    Where {-3\sqrt{n}<r(n)<3\sqrt{n}}, {-3\sqrt{n}<r'(n)<3\sqrt{n}}.

    So by 1 we know,

    \displaystyle \begin{array}{rcl} \frac{\{\frac{kn}{1}\}+...+\{\frac{kn}{n}\}}{n} & = & k(lnn+\gamma+\epsilon_n)-klnn-k(2\gamma+lnk-\sum_{i=1}^k\frac{1}{i})+\frac{r'(n)}{n}\\ & = & k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma)+\frac{r'(n)}{n}+k\epsilon_n \end{array}

    So we have,

    \displaystyle \lim_{n\rightarrow \infty}\frac{\{\frac{kn}{1}\}+...+\{\frac{kn}{n}\}}{n} =k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma)=k\epsilon_k \ \ \ \ \ (11)

    Remark 6 In fact we can get {0<k\epsilon_k<\frac{1}{2}, \forall k\in {\mathbb N}}, by combining the theorem 3 and 1.

    3. Lattice points in ball

    Gauss use the cube packing circle get a rough estimate,

    \displaystyle \sum_{n\leq x}r_2(n)=\pi x+O(\sqrt{x}) \ \ \ \ \ (12)

     

    In the same way one can obtain,

    \displaystyle \sum_{n\leq x}r_k(n)=\rho_kx^{\frac{k}{2}}+O(x^{\frac{k-1}{2}}) \ \ \ \ \ (13)

    Remark 7 Where {\rho_k=\frac{\pi^{\frac{k}{2}}}{\Gamma(\frac{k}{2}+1)}} is the volume of the unit ball in {k} dimension.

    Dirchlet’s hyperbola method works nicely for the lattic points in a ball of dimension {k\geq 4}. Langrange proved that every natural number can be represented as the sum of four squares, i.e. {r_4(n)>0}, and Jacobi established the exact formula for the number of representations

    \displaystyle r_4(n)=8(2+(-1)^n)\sum_{d|n,d\ odd}d. \ \ \ \ \ (14)

    Hence we derive,

    \displaystyle \begin{array}{rcl} \sum_{n\leq x}r_4(n) & = & 8\sum_{m\leq x}(2+(-1)^m)\sum_{dm\leq x, d\ odd}d\\ & = & 8\sum_{m\leq x}(2+(-1)^m)(\frac{x^2}{4m^2}+O(\frac{x}{m}))\\ & = & 2x^2\sum_1^{\infty}(2+(-1)^m)m^{-2}+O(xlogx)\\ & = & 3\zeta(2)x^2+O(xlogx) = \frac{1}{2}(\pi x)^2+O(xlogx) \end{array}

    This result extend easily for any {k\geq 4}, write {r_k} as the additive convolution of {r_4} and {r_{k-4}}, i.e.

    \displaystyle r_k(n)=\sum_{0\leq t\leq n}r_4(t)r_{k-4}(n-t) \ \ \ \ \ (15)

    Apply the above result for {r_4} and execute the summation over the remaining {k-4} squares by integration.

    \displaystyle \sum_{n\leq x}r_k(n)=\frac{(\pi x)^{\frac{k}{2}}}{\Gamma(\frac{k}{2}+1)}+O(x^{\frac{k}{2}-1}logx) \ \ \ \ \ (16)

     

    Remark 8 Notice that this improve the formula 12 which was obtained by the method of packing with a unit square. The exponent {\frac{k}{2}-1} in 16 is the best possible because the individual terms of summation can be as large as the error term (apart from {logx}), indeed for {k=4} we have {r_4(n)\geq 16n} if {n} is odd by the Jacobi formula. The only case of the lattice point problem for a ball which is not yet solved (i.e. the best possible error terms are not yet established) are for the circle({k=2}) and the sphere ({k=3}).

    Theorem 4

    \displaystyle \sum_{n\leq x}\tau(n^2+1)=\frac{3}{\pi}xlogx+O(x) \ \ \ \ \ (17)

    4. Application in finite fields

    Suppose {f(x)\in {\mathbb Z}[x]} is a irreducible polynomial. And for each prime {p}, let

    \displaystyle \rho_f(p)=\# \ of \ solutions\ of f(x)\equiv 0(mod\ p) \ \ \ \ \ (18)

    By Langrange theorem we know {\rho_f(p)\leq deg(f)}. Is there a asymptotic formula for

    \displaystyle \sum_{p\leq x}\rho_f(p)? \ \ \ \ \ (19)

    A general version, we can naturally generated it to algebraic variety.

    \displaystyle \rho_{f_1,...,f_k}(p)=\#\ of \ solutions\ of f_i(x)\equiv 0(mod\ p) ,\ \forall 1\leq i\leq k \ \ \ \ \ (20)

    Is there a asymptotic formula for

    \displaystyle \sum_{p\leq x}\rho_{f_1,...,f_k}(p)? \ \ \ \ \ (21)

    Example 1 We give an example to observe what is involved. {f(x)=x^2+1}. We know {x^2+1\equiv 0 (mod \ p)} is solvable iff {p\equiv 1 (mod\ 4)} or {p=2}. One side is easy, just by Fermat little theorem, the other hand need Fermat descent procedure, which of course could be done by Willson theorem. In this case,

    \displaystyle \sum_{p\leq n}\rho_f(p)=\# \ of \{primes \ of \ type\ 4k+1 \ in \ 1,2,...,n\} \ \ \ \ \ (22)

    Which is a special case of Dirichlet prime theorem.

    Let {K} be an algebraic number field, i.e. the finite field extension of rational numbers, let

    \displaystyle \mathcal{O}_K=\{\alpha\in K, \alpha \ satisfied \ a\ monic \ polynomial\ in\ {\mathbb Z}[x]\} \ \ \ \ \ (23)

     

    Dedekind proved that,

    Theorem 5

    1. {\mathcal{O}_K} is a ring, we call it the ring of integer of {K}.
    2. He showed further every non-zero ideal of {\mathcal{O}_K} could write as the product of prime ideal in {\mathcal{O}_k} uniquely.
    3. the index of every non-zero ideal {I} in {\mathcal{O}_K} is finite, i.e. {[\mathcal{O}_K:I]<\infty}, and we can define the norm induce by index.

      \displaystyle N(I):=[\mathcal{O}_K:I] \ \ \ \ \ (24)

      Then the norm is a multiplication function in the space of ideal, i.e. {N(IJ)=N(I)N(J), \forall I,J \in \ ideal\ class\ group\ of\ \mathcal{O}_K}.

    4. Now he construct the Dedekind Riemann zeta function,

      \displaystyle \zeta_K(s)=\sum_{N(I)\neq 0}\frac{1}{N(I)^s}=\prod_{J\ prime \ ideal\ }\frac{1}{1-\frac{1}{N(J)^s}},\ \forall Re(s)>1 \ \ \ \ \ (25)

     

    Now we consider the analog of the prime number theorem. Let {\pi_K(x)=\{I,N(I)<x\}}, does the exist a asymptotic formula,

    \displaystyle \pi_K(x)\sim \frac{x}{ln x}\ as\ x\rightarrow \infty? \ \ \ \ \ (26)

    Given a prime {p}, we may consider the prime ideal

    \displaystyle p\mathcal{O}_K=\mathfrak{P}_1^{e_1}\mathfrak{P}_2^{e_2}...\mathfrak{P}_k^{e_k} \ \ \ \ \ (27)

    Where {\mathfrak{P}_i } is different prime ideal in {\mathcal{O}_K}. But the question is how to find these {\mathfrak{P}_i}? For the question, there is a satisfied answer.

    Lemma 6 (existence of primitive element) There always exist a primetive elements in {K}, such that,

    \displaystyle K={\mathbb Q}(\theta) \ \ \ \ \ (28)

    Where {\theta} is some algebraic number, which’s minor polynomial {f(x)\in {\mathbb Z}[x]}.

    Theorem 7 (Dedekind recipe) Take the polynomial {f(x)}, factorize it in the polynomial ring {{\mathbb Z}_p[x]},

    \displaystyle f(x)\equiv f_1(x)^{e_1}...f_{r}(x)^{e_r}(mod \ p) \ \ \ \ \ (29)

    Consider {\mathfrak{P}_i=(p, f_i(\theta)) \subset \mathcal{O}_K}. Then apart from finite many primes, we have,

    \displaystyle p\mathcal{O}_K=\mathfrak{P}_1^{e_1}\mathfrak{P}_2^{e_2}...\mathfrak{P}_k^{e_k} \ \ \ \ \ (30)

    Where {N(\mathfrak{P}_i)=p^{deg{f_i}}}.

    Remark 9 The apart primes are those divide the discriminant.

    Now we can argue that 4 is morally the same as counting the ideals whose norm is divide by {p} in a certain algebraic number theory.

    And we have following, which is just the version in algebraic number fields of 2.

    Theorem 8 (Weber) {\#} of ideals of {\mathcal{O}_K} with norm {\leq x} equal to,

    \displaystyle \rho_k(X)+O(x^{1-\frac{1}{d}}), where \ d=[K:Q] \ \ \ \ \ (31)

     


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Dirichlet hyperbola method 是解析数论中处理卷积和的基本工具。它的几何图像非常直接:把求和区域看成双曲线 $ab\le x$ 下方的格点,再利用双曲线关于 $\sqrt x$ 的对称性减少误差。

    Dirichlet hyperbola method:用双曲线拆分除数和
    Dirichlet hyperbola method 用双曲线 $ab=x$ 的对称性改进卷积和的误差。

    1. 除数函数的例子

    除数函数满足

    $$d(n)=\sum_{ab=n}1.$$

    因此

    $$\sum_{n\le x}d(n)=\sum_{ab\le x}1.$$

    这就是双曲线 $ab=x$ 下方的整数格点数。

    2. 直接估计的问题

    若对每个 $a$ 求 $\lfloor x/a\rfloor$,得到

    $$\sum_{a\le x}\left\lfloor\frac xa\right\rfloor.$$

    平凡地把 floor 换成 $x/a$ 会产生太大的误差,因为项数有 $x$ 个。hyperbola method 的关键是只在 $a\le\sqrt x$ 和 $b\le\sqrt x$ 的短范围内精确处理。

    3. 基本公式

    由对称性可得

    $$\sum_{n\le x}d(n)=2\sum_{a\le\sqrt x}\left\lfloor\frac xa\right\rfloor-\lfloor\sqrt x\rfloor^2.$$

    于是

    $$\sum_{n\le x}d(n)=x\log x+(2\gamma-1)x+O(\sqrt x).$$

    这个误差已经比直接方法好很多。

    4. 一般卷积

    若 $h=f*g$,则

    $$\sum_{n\le x}h(n)=\sum_{ab\le x}f(a)g(b).$$

    可以选择参数 $Y$,把区域分成 $a\le Y$、$b\le x/Y$ 和重叠部分。合适的 $Y$ 取决于 $f,g$ 的平均阶和可用误差估计。

    5. 高维推广

    对 $k$ 重除数函数 $d_k(n)$,问题变成

    $$a_1a_2\cdots a_k\le x$$

    下方的格点计数。通过归纳、partial summation 和多维双曲面拆分,可以得到

    $$\sum_{n\le x}d_k(n)=xP_{k-1}(\log x)+\text{error},$$

    其中 $P_{k-1}$ 是次数 $k-1$ 的多项式。这个方法的力量在于:它把乘法卷积的求和问题变成可视化的几何区域拆分。

  • 对数平均 Sarnak 猜想:从 BSZ 准则到熵下降

    旧博客原文

    原题:Log average sarnak conjecture

     

    This is a note concentrate on the log average Sarnak conjecture, after the work of Matomaki and Raziwill on the estimate of multiplication function of short interval. Given a overview of the presented tools and method dealing with this conjectue.

     

    1. Introduction

    Sarnak conjecture \cite{Sarnak} assert that for any obersevable {\{f(T^n(x_0))\}_{n=1}^{\infty}} come from a determination systems {(T,X),T:X\rightarrow X}, where {h(T)=0}, {x_0\in X, f\in C(X)}. The correlation of it and the Liuvillou function is 0, i.e. they are orthongonal to each other, more preseicesly it is just to say,

    \displaystyle \sum_{n<x}\mu(x)f(T^n(x_0))=o(x) \ \ \ \ \ (1)

     

    This is a very natural raised conjecture, Liuville function is the presentation of primes, due to we always believe the distribution of primes in {\mathbb N} should be randomness.

    It has been known as observed by Landau \cite{Laudau} that the simplest case,

    \displaystyle \sum_{n<x}\mu(n)=o(x)

    already equivalent to the prime number theorem. It is not difficult to deduce the spetial case of Sarnak conjecture when with the obersevation in $latex {(1)}&fg=000000$ come from finite dynamic system is equivalent to the prime number theorem in athremetic progress by the similar argument. Besides this two classical result, may be the first new result was established by Davenport,

    Theorem 1 Let {T:S_1\rightarrow S_1, T(x)=x+\alpha}, {\alpha} is a inrational, then the obersevation come from {(T,S_1)} is orthogonal to Mobius function. due to {\{e^{2\pi ikx}\}_{k\in \mathbb Z}} is a basis of {C(S_1)}, suffice to proof,

    \displaystyle \sum_{n<x}e^{2\pi ikn\alpha}\mu(n)=o(x), \forall k\in \mathbb N

    There is a lots of spetial situations of Sarnak’s conjecture have been established, The parts I mainly cared is the following:

    1. Interval exchange map.
    2. Skew product flow.
    3. Obersevable come from One dimensional zero entropy flow.
    4. Nilsequences.

    But in this note, I do not want to explain the tecnical and tools to establish this result, but considering an equivalent conjecture of Sarnak conjecture, named Chowla conjecture, and explain the underlying insight of the suitable weak statement, i.e. the log average Chowla conjecture and the underlying insight of it.

    The note is organized as following way, in the next section $latex {(2)}&fg=000000$, we give a self-contained introduction on the tools called Bourgain-Sarnak-Ziegler critation, explain the relationship of this critation and the sum-product phenomenon, also given some more general critation along the philosephy use in establish the Bourgain-Sarnak-Ziegler critation, which maybe useful in following development combine with some other tools. The key point is transform the sum from linear sum to bilinear sum and decomposition the bilinear sum into diagonal part and off-diagonal part, use the assume in the critation to argue the off-diagonal part is small and on the orther hand the diagonal part is also small by the trivial estimate and the volume of diogonal is small, this is very similar to a suitable Caderon-Zugmund decomposition.

    In section $latex {(4)}&fg=000000$, I try to give a proof sketch of the result of Matomaki and Raziwill, which is also a key tools to understanding the Sarnak conjecture, or equivalent the Chowla conjecture. The key points of the proof contains following:

    1. Find a suitable fourier indentity
    2. Construct a multiplication-addition dense subset {S}, and proof that the theorem MR hold we need only to proof it hold for {S\cap [1,2,...,n]} instead of {[1,2,...,n]}
    3. Involve the power of euler product formula. divide the whole interval into a lot of small interval with smaller and smaller scale and a residue part. We look the part come from every small scale as a major term and look the residue part as minor term.
    4. Deal with the major term at every scale, by a combitorios identity and second moments method.
    5. find a enough decay estimate from a scale to the next smaller scale.
    6. Deal with the minor term by the H… lemma.

    Due to the theorem of MR do not exausted the method they developed, we trying to make some more result with their method, Tao and Matomaki attain the average version of Chowla conjecture is true by this way, and combine this argument and the entropy decresment argument they established the 2 partten of the log average Chowla conjecture is true. Very recently Tao and his coperator proved the odd partten case of log average chowla conjecture is true, combine an argument of frustenberg crresponding principle and entopy decresment argument. But it seems the even and large than 2 case is much difficult and seems need something new to combine with the method of MR and entropy decresment and frunstenberg corresponfing principle to make some progress.

    So, in section $latex {(5)}&fg=000000$, we give a self-contain introduction to the entropy decresment argument of Tao, and combine with the frustenberg corresponding principle.

    In the last section $latex {(6)}&fg=000000$, I state some result and method and phylosphy of them I get on nilsequences and wish to combine them with the previous method to make some progress on log average Chowla conjecture on the even partten case.

    \newpage

    2. Bourgain-Sarnak-Zieglar creation

    We begin with the easiest one, this is the main result established in \cite{BSZ}, I try to give the main ideal under the proof, but with a no quantitative version is the following,

    Theorem 2 (Bourgain-Sarnak-Zieglar creation, not quantitative version) if for all primes {p,q>>1} we have:

    \displaystyle \sum_{n=1}^Nf(T^{pn}(x))\overline{f(T^{qn}(x))}=o(N) \ \ \ \ \ (2)

     

    Then for multiplication function {g(n)} we have

    \displaystyle \sum_{n=1}^Ng(n)\overline{ f(T^n(x))}=o(N) \ \ \ \ \ (3)

     

    Remark 1 For simplify we identify {f(T^n(x)):=F(n)}.

    Remark 2

    The idea is following, break the sum into a bilinear one, so, of course, we multiplication it with itself. i.e. we consider to control,

    \displaystyle |\sum_{i=1}^Ng(n)\overline{ F(n)}|^2=\sum_{n=1}^N\sum_{m=1}^Ng(n)g(m)\overline{F(n)F(m)} \ \ \ \ \ (4)

     

    To control 4, we need exhausted the mutiplication property of {g(n)}, we have {g(mn)=g(n)g(m),\forall\ m,n\in {\mathbb N}}. We can not get good estimate for all term,

    \displaystyle g(n)g(m)\overline{F(n)F(m)} \ \ \ \ \ (5)

    The condition in our hand if following,

    \displaystyle \sum_{n=1}^NF(pn)\overline{F(qn)}=o(N), \forall \ p,q\in \mathop{\mathbb P} \ \ \ \ \ (6)

    So, just like the situation of Cotlar-Stein lemma \cite{Cotlar-Stein lemma}, we wish to estimate like following:

    \displaystyle \begin{array}{rcl} |\sum_{p\in W}\sum_{n\in V}F(pn)g(pn)| & \leq & \sum_{n\in V}|g(n)|\cdot |\sum_{p\in W}F(pn)g(p)| \\ & \leq &\sum_{n\in V}|\sum_{p \in W}F(pn)g(p)|\\ & \overset{Cauchy-Schwarz}\leq & |V|^{\frac{1}{2}}[\sum_{n\in V}|\sum_{p\in W}F(pn)g(p)|^2]^{\frac{1}{2}}\\ & = & |V|^{\frac{1}{2}}[\sum_{p_1,p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}}\\ \end{array}

    Then we consider divide the sum into diagonal part and non-diagonal part, as following,

    \displaystyle |V|^{\frac{1}{2}}[\sum_{p_1\neq p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}}+|V|^{\frac{1}{2}}[\sum_{p\in W}\sum_{n\in V}|F(pn)|^2]^{\frac{1}{2}} \ \ \ \ \ (7)

    But the first part is small, i.e.

    \displaystyle |V|^{\frac{1}{2}}[\sum_{p_1\neq p_2\in W}\sum_{n\in V}F(p_1n)\overline{F(p_2n)}g(p_1)\overline{g(p_2)}]^{\frac{1}{2}} =o(|W||V|) \ \ \ \ \ (8)

    Because of

    \displaystyle \sum_{n\in V}F(p_1n)\overline{F(p_2n)}=o(V), \forall p_1\neq p_2\in W \ \ \ \ \ (9)

    and the second part is small, i.e.

    \displaystyle |V|^{\frac{1}{2}}[\sum_{p\in W}\sum_{n\in V}|F(pn)|^2]^{\frac{1}{2}}=o(|W||V|) \ \ \ \ \ (10)

    Because diagonal part is small in {W\times W} and trivial inequality

    \displaystyle \sqrt{\sum_{n\in V}|F(pn)|^2}\leq |V|^{\frac{1}{2}} \ \ \ \ \ (11)

    But the method in remark 2 is not always make sense in any situation, we need to construct two suitable sets {W,V} and then break up {\{1,2,...,n{\mathbb N}\}} into {W\times V}, this mean,

    \displaystyle \{1,2,...,N\}\sim W\times V+o(N) \ \ \ \ \ (12)

    But this {W,V} could be construct in this situation, thanks to the prime number theorem,

    Theorem 3 (Prime number theorem)

    \displaystyle \pi(n)\sim \frac{n}{ln(n)} \ \ \ \ \ (13)

    Morally speaking, this is the statement that the primes, which is the generator of multiplication function, is not very sparse.

    3. Van der curpurt trick

    There is the statement of Van der carport theorem:

    Theorem 4 (Van der curpurt trick) Given a sequences { \{x_n\}_{n=1}^{\infty}} in { S_1}, if { \forall k\in N^*}, { \{x_{n+k}-x_n\}} is uniformly distributed, then { \{x_n\}_{n=1}^{\infty}} is uniformly distributed.

    I do not know how to establish this theorem with no extra condition, but this result is true at least for polynomial flow. \newpage Proof:

    \displaystyle \begin{array}{rcl} |\sum_{n=1}^Ne^{2\pi imQ(n)}|& = &\sqrt{(\sum_{n=1}^Ne^{2\pi imQ(n)})(\overline{\sum_{n=1}^Ne^{2\pi imQ(n)}})}\\ & = &\sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi imQ(n+h_1)-Q(n)}}\\ & = &\sqrt{\sum_{h_1=1}^N\sum_{n=1}^{N-h_1}e^{2\pi im \partial^1_{h_1}Q(n)}}\\ & \leq & \sqrt{\sum_{h_1=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)}|}\\ & = &\sqrt{\sum_{h_1=1}^N\sqrt{ (\sum_{n=1}^{N-h_1}e^{2\pi \partial^1_{h_1}Q(n)} )(\overline{\sum_{n=1}^{N-h}e^{2\pi \partial^1_{h_1}Q(n)})}}}\leq\sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N|\sum_{n=1}^{N-h_1}e^{2\pi\partial^1_{h_2} \partial^1_hQ(n)} |}}\\ & \leq ....\leq & \\ & = & \sqrt{\sum_{h_1=1}^N\sqrt{ \sum_{h_2=1}^N \sqrt{....\sqrt{\sum_{h_{k-1}=1}^{N-h_{k-2}}|\sum_{n=1}^{N-h_{k-1}}e^{2\pi\partial_{h_1h_2...h_{k-1}Q(n)}}|}}}} =o(1) \end{array}

    \Box

    This type of trick could also establish the following result, which could be understand as a discretization of the Vinegradov lemma.

    Remark 3

    Uniformly distribution result of { F_p}: Given {Q(n)=a_kn^k+...+a_1n+a_0}, {\{Q(0),Q(1),...,Q(p-1)\}} coverages to a uniformly distribution in {\{0,1,...,p-1\}} as {p \rightarrow \infty}.

    Remark 4 But I definitely do not know how to establish the similar result when {Q(n)=n^{-1}}.

    Remark 5

    This trick could also help to establish estimate of correlation of low complexity sequences and multiplicative function, such as result:

    \displaystyle S(x)=\sum_{n\le x}\left(\frac{n}{p}\right)\mu(n)=o(n)

    Maybe with the help of B-Z-S theorem.

    \newpage

    4. Matomaki and Raziwill’s work

    In this section we explain the main idea underlying the paper \cite{KAISA MATOMA 虉KI AND MAKSYM RADZIWILL}. But play with a toy model, i.e. the corresponding corollary of the original result on Liouville鈥檚 function.

    Definition 5 (Lioville’s function)

    \displaystyle \lambda(n)=(-1)^{\alpha_1+\alpha_2+...+\alpha_k}, \forall \ n=p_1^{\alpha_1}...p_k^{\alpha_k}. \ \ \ \ \ (14)

    Remark 6

    \displaystyle |\int_{X}^{2X}\lambda(n)dx|=o(x) \ \ \ \ \ (15)

    is equivalent to the prime number theorem 3.

    The most important beakgrouth of analytic number theory is the new understanding of multiplication function on share interval, this result is established by Kaisa Matom盲ki and Maksym Radziwill. Two very young and intelligent superstars.

    The main theorem in them article is :

    Theorem 6 (Matomaki,Radziwill) As soon as {H\rightarrow \infty} when {x\rightarrow \infty}, one has:

    \displaystyle \sum_{x\leq n\leq x+H}\lambda(n)= o(H) \ \ \ \ \ (16)

    for almost all {1\leq x\leq X} .

    In my understanding of the result, the main strategy is:

    1. Parseval indetity, transform to Dirchelet polynomial.
    2. Involved by multiplication property, spectral decomposition.
    3. From linear to multilinear , Cauchy schwarz inequality.
    4. major term estimate.
    5. Estimate the contribution of area which is not filled.

    4.1. Parseval indetity, transform to Dirchelet polynomial

    We wish to establish the equality,

    \displaystyle \frac{1}{X}\int_{X}^{2X}|\sum_{x\leq n\leq x+H}\lambda(n)|dx=o(H) \ \ \ \ \ (17)

    This is the {L^1} norm, by Chebyschev inequality, this could be control by {L^2} norm, so we only need to establish the following,

    \displaystyle \frac{1}{X}\int_X^{2 X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx=o(H^2) \ \ \ \ \ (18)

     

    We wish to transform from the discretization sum to a continue sum, that is,

    \displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x} \ \ \ \ \ (19)

     

    Remark 7 There are two points to understand why 19 and 18 are the same.

    1. {[xe^{-\frac{1}{T}},xe^{\frac{1}{T}}]\sim [x-H,x+H]}.
    2. {1_{x\leq n\leq 2x}} and {\frac{1}{x}} is to make that {x=O(X)}.

    So the Magnitude of 18 and 19 are the same. i.e.

    \displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x}\sim \frac{1}{X}\int_X^{2 X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx \ \ \ \ \ (20)

    Now we try to transform 19 by Parseval indetity, this is something about the {L^2} norms of the quality we wish to charge. It is just trying to understanding 19 as a quantity in physical space by a more chargeable quality in frequency space. Image,

    \displaystyle \int_{{\mathbb R}}|\sum_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(n)1_{X\leq n\leq 2X}|^2\frac{dx}{x}:=\int_{{\mathbb R}}|f_X(x)|^2dx \ \ \ \ \ (21)

    Then {f_X(x)=\int_{xe^{-\frac{1}{T}}\leq n\leq xe^{\frac{1}{T}}}\lambda(x)1_{X\leq n\leq 2X}}. Note that,

    \displaystyle \begin{array}{rcl} \widehat{f_X(\xi)} & = & \int_{{\mathbb R}}f_X(x)e^{2\pi ix\xi}dx\\ & = & \sum_{x\leq n\leq 2x}\lambda(x)\int_{logn-\frac{1}{T}}^{logn+\frac{1}{T}}e^{2\pi ix\xi}dx, \ T=\frac{X}{H}\\ & = & \sum_{X\leq n\leq 2X}\lambda(x)e^{2\pi ilog(n)\cdot \xi}\cdot\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i-\frac{\xi}{T}}}{2\pi i\xi}\\ \end{array}

    So by Parseval identity, we have,

    \displaystyle \begin{array}{rcl} \int_{{\mathbb R}}|f_X(x)|^2dx & = & \int_{{\mathbb R}}|\widehat{f_X(\xi)}|^2d\xi \\ & = & \int_{{\mathbb R}}|\sum_{X\leq n\leq 2X}\lambda(n)\cdot n^{2\pi i\xi}|^2(\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i\frac{-\xi}{T}}}{2\pi i\xi})^2d\xi\\ & \sim & \int_{{\mathbb R}}|\sum_{X\leq n\leq 2X}\lambda(n)\cdot n^{2\pi i\xi}|^2\frac{1}{T^2}1_{|\xi|^2\leq T}\\ \end{array}

    Remark 8 We know the Fejer kernel satisfied,

    \displaystyle (\frac{e^{2\pi i\frac{\xi}{T}}-e^{2\pi i\frac{-\xi}{T}}}{2\pi i\xi})^2\sim \frac{1}{T^2}1_{|\xi|\leq T} \ \ \ \ \ (22)

    So morally speaking, we get the following identity.

    \displaystyle \frac{1}{X}\int_{X}^{2X}|\sum_{x\leq n\leq x+H}\lambda(n)|^2dx\sim \frac{1}{(x/H)^2}\int_{0}^{\frac{X}{H}}|\sum_{x\leq n\leq 2x}\lambda(x)x^{2\pi i\xi}|^2d\xi \ \ \ \ \ (23)

    In fact we do a cutoff, the quality we really consider is just:

    \displaystyle \frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt \ \ \ \ \ (24)

    established the monotonically inequality:

    Theorem 7 (Paserval type identity)

    \displaystyle \frac{1}{X}\int_{X}^{2X}|\frac{1}{H}\sum_{x\leq n\leq x+H}\lambda(n)|^2dx \sim聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt \ \ \ \ \ (25)

     

    Remark 9

    In my understanding, This is a perspective of the quality, due to the quality is a multiplicative function integral on a domain { \mathbb N^*} with additive structure, it could be looked as a lots of wave with the periodic given by primes, so we could do a orthogonal decomposition in the fractional space, try to prove the cutoff is a error term and we get such a monotonically inequality.

    But at once we get the monotonically inequality, we could look it as a聽compactification process and this process still carry most of the information so lead to the inequality.

    It seems something similar occur in the attack of the moments estimate of zeta function by the second author. And it is also could be looked as something similar to the 聽spectral decomposition with some basis come from multiplication generators, i.e. primes.

    4.2. Involved by multiplication property, spectral decomposition

    I called it is “spectral decomposition”, but this is not very exact. Anyway, the thing I want to say is that for multiplication function {\lambda(n)}, we have Euler-product formula:

    \displaystyle \Pi_{p,prime}(\frac{1}{1-\frac{\lambda(p)}{p^s}})=\sum_{n=1}^{\infty} \frac{\lambda(n)}{n^s} \ \ \ \ \ (26)

     

    But anyway, we do not use the whole power of multiplication just use it on primes, i.e. {\lambda(pn)=\lambda(p)\lambda(n)} leads to following result:

    \displaystyle \lambda(n)=\sum_{n=pm,p\in I}\frac{\lambda(p)\lambda(m)}{\# \{p|m, p\in I\}+1}+\lambda(n)1_{p|n;p\notin I} \ \ \ \ \ (27)

    This is a identity about the function {\lambda(n)}, the point is it is not just use the multiplication at a point,i.e. {\lambda(mn)=\lambda(m)\lambda(n)}, but take average at a area which is natural generated and compatible with multiplication, this identity carry a lot of information of the multiplicative property. Which is crucial to get a good estimate for the quality we consider about.

    4.3. From linear to multilinear , Cauchy schwarz

    Now, we do not use one sets {I}, but use several sets {I_1,...,I_n } which is carefully chosen. And we do not consider [X,2X] with linear structure anymore , instead reconsider the decomposition:

    {[X,2X]=\amalg_{i=1}^n (I_i\times J_i) \amalg U}

    On every {I_i\times J_i} it equipped with a bilinear structure. And {U} is a very small set, {|U|=o(X)} which is in fact have much better estimate.

    {\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt =\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}

    Now we just use a Cauchy-Schwarz:

    {\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}

    4.4. major term estimate

    {=\sum_{i=1}^n\int_{I_i\times J_i}聽聽\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt}

    {\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt}

    4.5. estimate the contribution of area which is not filled

    \newpage

    5. Entropy dcrement argument

    \newpage

    6. Correlation with nilsequences

    I wish to establish the following estimate: {\lambda(n)} is the liouville function we wish the following estimate is true.

    \displaystyle \int_{0\leq x\leq X}|\sup_{f\in \Omega^m}\sum_{x\leq n\leq x+H}\lambda(n)e^{2\pi if(x)}|dx =o(XH). \ \ \ \ \ (28)

    Where we have { H\rightarrow \infty} as { x\rightarrow \infty},

    \displaystyle \Omega^m=\{a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0 | a_m,...,a_1,a_0\in [0,1]\}

    is a compact space.

    I do not know how to prove this but this is result is valuable to consider, because by a Fourier identity we could transform the difficulty of (log average) Chowla conjecture to this type of result.

    There is some clue to show this type of result could be true, the first one is the result established by Matomaki and Raziwill in 2015:

    Theorem 8 (multiplication function in short interval)

    {f(n): \mathbb N\rightarrow \mathbb C} is a multiplicative function, i.e. { f(mn)=f(n)f(m), \forall m,n\in \mathbb N}. {H\rightarrow \infty} as {x\rightarrow \infty}, then we have the following result,

    \displaystyle \int_{1\leq x\leq X}|\sum_{x\leq n\leq x+H}f(n)|=o(XH). \ \ \ \ \ (29)

    And there also exists the result which could be established by Vinagrodov estimate and B-S-Z critation :

    Theorem 9 (correlation of multiplication function and nil-sequences in long interval)

    {f(n): \mathbb N\rightarrow \mathbb C} is a multiplicative function, i.e. { f(mn)=f(n)f(m), \forall m,n\in \mathbb N}. {g(n)=a_n^m+...+a_1n+a_0} is a polynomial function then we have the following result,

    \displaystyle \int_{1\leq n \leq X}|f(n)e^{2\pi i g(n)}|=o(X) \ \ \ \ \ (30)

    \newpage {9} \bibitem{Sarnak} Peter Sarnak, Mobius Randomness and Dynamics.

    \texttt{https://publications.ias.edu/sites/default/files/Mahler }. \bibitem{Laudau} JA 虂NOS PINTZ (BUDAPEST). LANDAU鈥橲 PROBLEMS ON PRIMES.

    \texttt{https://users.renyi.hu/~pintz/pjapr.pdf} \bibitem{BSZ} Knuth: Computers and Typesetting,

    \texttt{http://www-cs-faculty.stanford.edu/\~{}uno/abcde.html}

    \bibitem{Cotlar-Stein lemma} Almost orthogonality

    \texttt{https://hxypqr.wordpress.com/2017/12/18/almost-orthogonality/}

    \bibitem{KAISA MATOMA 虉KI AND MAKSYM RADZIWILL} KAISA MATOMA 虉KI AND MAKSYM RADZIWIL, MULTIPLICATIVE FUNCTIONS IN SHORT INTERVALS.

    \texttt{https://arxiv.org/abs/1501.04585v4/}.

     


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    这篇笔记想整理的是 Sarnak 猜想的一条现代路线:从莫比乌斯函数和零熵动力系统的正交性出发,经过 Bourgain-Sarnak-Ziegler 准则,把线性相关和转成双线性相关;再借助 Matomaki-Radziwill 的短区间乘法函数估计,以及 Tao 的 entropy decrement,把问题推向对数平均 Chowla 猜想。

    对数平均 Sarnak 猜想:从 BSZ 准则到熵下降
    Sarnak/Chowla 的对数平均路线:线性相关先转成双线性相关,再用短区间估计和熵下降选择合适尺度。

    1. Sarnak 猜想的基本形状

    Sarnak 猜想说,如果 $(X,T)$ 是零拓扑熵动力系统,$f\in C(X)$,那么对任意 $x\in X$,莫比乌斯函数与观测序列 $f(T^n x)$ 应该正交:

    $$\frac1N\sum_{n\le N}\mu(n)f(T^n x)\to 0.$$

    这里的哲学是:$\mu(n)$ 携带素数分布中的振荡,而零熵系统产生的是低复杂度的确定序列;二者不应该长期对齐。

    最简单的有限动力系统情形已经包含素数定理和算术级数中的素数定理。圆周无理旋转情形则接近 Davenport 型估计:对任意无理数 $\alpha$,指数序列 $e(n\alpha)$ 与 $\mu(n)$ 的相关和具有消失。

    2. 从 Sarnak 到 Chowla

    Chowla 猜想更像是对 $\mu$ 或 Liouville 函数自身随机性的表述。一个典型的 $k$ 点相关形式是

    $$\frac1N\sum_{n\le N}\lambda(n+h_1)\cdots \lambda(n+h_k)\to 0,$$

    其中 $h_i$ 两两不同。Sarnak 关心的是乘法函数与低复杂度确定序列的相关,Chowla 关心的是乘法函数自身不同平移之间的相关。二者之间可以用 Furstenberg correspondence principle 和动力系统模型联系起来。

    对数平均版本把普通平均换成

    $$\frac1{\log N}\sum_{n\le N}\frac{a(n)}n.$$

    这个权重让尺度选择更加稳定,也更适合短区间分析。很多情况下,对数平均结论比普通平均结论先被证明,因为它容许把不同尺度上的误差以更柔和的方式叠加。

    3. Bourgain-Sarnak-Ziegler 准则

    BSZ 准则的核心是一个线性到双线性的转换。设 $a_n$ 是有界序列,如果对很多不同素数 $p\ne q$,都有

    $$\frac1N\sum_{n\le N}a_{pn}\overline{a_{qn}}\to 0,$$

    那么可以推出

    $$\frac1N\sum_{n\le N}\mu(n)a_n\to 0.$$

    直观上,$\mu$ 的乘法结构允许我们把原来的线性相关拆成不同素数伸缩后的相关。对角部分 $p=q$ 用平凡估计和体积小来处理;非对角部分则由上面的假设控制。这一点和 Calderon-Zygmund 分解的精神很像:把难对象拆成一个小的对角坏集和一个可估计的非对角主体。

    4. Matomaki-Radziwill 的短区间输入

    Matomaki-Radziwill 的工作说明,乘法函数在大多数短区间里的平均行为可以被控制。粗略地说,对许多短区间 $[x,x+H]$,有

    $$\frac1H\sum_{x

    接近它在长区间中的平均。这使得我们可以把一个全局相关和切成许多短尺度,再用欧拉乘积、二阶矩和组合恒等式逐层比较。

    短区间估计真正有用的地方在于:它让乘法函数的局部随机性可以被拿来攻击动力系统中的相关问题,而不是只停留在平均阶的数论命题。

    5. Entropy decrement 的作用

    entropy decrement 的思想是寻找一个尺度,使得随机变量 $n$ 与它的素数倍 $pn$ 之间的条件信息变少。换句话说,在合适尺度上,系统看到的结构不会因为乘一个小素数而增加太多复杂度。

    这一步的意义是把“乘法平移”转成“动力系统里可比较的两个观测”。当这个信息损失足够小的时候,短区间估计、BSZ 双线性结构和 correspondence principle 就可以接上。

    6. Nilsystem 方向

    nilsequence 是零熵系统中非常重要的一类模型。它既有足够丰富的几何结构,又保留了可计算的 Fourier 分析。把 Sarnak、Chowla、短区间乘法函数和 nilsequence 放在一起看,真正的问题是:乘法函数的随机性如何穿过 nilmanifold 上的低复杂度轨道。

    这条路线目前最有价值的地方,不是把所有情形一次性解决,而是提供了一张方法地图:线性相关转双线性相关;短区间估计提供局部随机性;entropy decrement 选择尺度;Furstenberg 原理把数论相关放回动力系统。

  • Large sieve 与 Bombieri-Vinogradov theorem:几乎正交性的数论形态

    旧博客原文

    原题:The large sieve and the Bombieri-Vinogradov theorem

    -1.Motivation-

    Large sieve a philosophy reflect as a large group of inequalities which is very effective on controlling some linear sum or square sum of some correlation of arithmetic function, some idea of which could have originated in harmonic analysis, merely rely on almost orthogonality.

    One fundamental example is the estimate of the quality,

    \sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|

    One naive idea of control this quality is using Cauchy-schwarz inequality. But stupid use this we gain something even worse than trivial estimate. In fact by triangle inequality and trivial estimate we gain trivial bound: \sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|\leq x. But by stupid use Cauchy we get following,

    \sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|\leq ((\sum_{n\leq x}|\Lambda(n)|^2)(\sum_{n\leq x}|\chi(n)|^2))^{\frac{1}{2}}\leq xlog^{\frac{1}{2}}x

    But this does not mean Cauchy-Schwarz is useless on charge this quality, we careful look at the inequality and try to understand why the bound will be even worse. Every time we successful use Cauchy-Schwarz there are two main phenomenon, first, we lower down the complexity of the quantity we wish to bound, second we almost do not loss any thing at all. So we just reformulate the quantity and find it lower down the complexity and the change is compatible with the equivalent condition of Cauchy-Schwarz. For example we have following identity,

    \sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|=\sqrt{ \sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}| \sum_{m\leq x}|\Lambda(m)\overline{\chi(m)}|}=\sqrt{ \sum_{k_1,k_2\in \mathbb F_p^{\times}}\sum_{n',m'\leq \frac{x}{p}}|\Lambda(n')\Lambda(m')\overline{\chi(k_1)\chi(k_2)}| }

    So we could understand this quality as the Variation of primes in arithmetic profession constructed by \{pn+b| b\in\{1,2,...,p-1\}\}. But this is still difficult to estimate, merely because of we need to control the variation of convolution of \Lambda with itself on \mathbb F_p^{\times}\simeq \{pn+b| b\in\{1,2,...,p-1\}\}.

    Now we change our perspective, recall a variant of Cauchy-Schwarz inequality, which called Bessel inequality, as following,

    Bessel inequality

    Let {g_1,\dots,g_J: {\bf N} \rightarrow {\bf C}} be finitely supported functions obeying the orthonormality relationship,

    \displaystyle \sum_n g_j(n) \overline{g_{j'}(n)} = 1_{j=j'}

    for all {1 \leq j,j' \leq J}. Then for any function {f: {\bf N} \rightarrow {\bf C}}, we have,

    \displaystyle (\sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}|^2)^{1/2} \leq (\sum_n |f(n)|^2)^{1/2}.

    Pf: The proof is not very difficult, we just need to keep an orthogonal picture in our mind, consider \{g_{j}(n)\}, 1\leq j\leq J to be a orthogonal basis on l^2(\mathbb N), then this inequality is a natural corollary.

    Have this inequality in mind, by the standard argument given by transform from version of orthogonal to almost orthogonal which was merely explained in the previous note.  We could image the following corresponding almost orthogonal variate of “Bessel inequality” is true:

    Generalised Bessel inequality

    Let {g_1,\dots,g_J: {\bf N} \rightarrow {\bf C}} be finitely supported functions, and let {\nu: {\bf N} \rightarrow {\bf R}^+} be a non-negative function. Let {f: {\bf N} \rightarrow {\bf C}} be such that {f} vanishes whenever {\nu} vanishes, we have

    \displaystyle (\sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}|^2)^{1/2} \leq (\sum_n |f(n)|^2 / \nu(n))^{1/2} \times ( \sum_{j=1}^J \sum_{j'=1}^J c_j \overline{c_{j'}} \sum_n \nu(n) g_j(n) \overline{g_{j'}(n)} )^{1/2}

    for some sequence {c_1,\dots,c_J} of complex numbers with {\sum_{j=1}^J |c_j|^2 = 1}, with the convention that {|f(n)|^2/\nu(n)} vanishes whenever {f(n), \nu(n)} both vanish.

     


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Large sieve 是一族控制算术序列在线性相位或剩余类中分布的强大不等式。它的核心精神和调和分析中的 almost orthogonality 非常接近。

    Large sieve 与 Bombieri-Vinogradov theorem:几乎正交性的数论形态
    large sieve 用几乎正交性控制许多模数和 residue phases 上的平方和。

    1. 基本形状

    典型 large sieve inequality 形如

    $$\sum_{q\le Q}\sum_{\substack{a\bmod q\\(a,q)=1}}
    \left|\sum_{n\le N}a_n e(an/q)\right|^2
    \le (N+Q^2)\sum_{n\le N}|a_n|^2.$$

    它说明不同模数和不同 residue phases 之间几乎正交。

    2. 为什么 Cauchy-Schwarz 要小心用

    粗暴使用 Cauchy-Schwarz 可能比平凡估计还差。成功的关键是先把要求的量重写成具有正交结构的平方和,再用 Cauchy-Schwarz 或 Parseval。

    3. 算术级数中的素数

    Bombieri-Vinogradov theorem 控制素数在平均模数意义下的分布:

    $$\sum_{q\le Q}\max_{(a,q)=1}\left|\psi(x;q,a)-\frac{x}{\varphi(q)}\right|$$

    在 $Q\le x^{1/2}$ 附近仍有强估计。它可以看作广义 Riemann hypothesis 在平均意义下的替代。

    4. Large sieve 的作用

    large sieve 给出对字符和指数和的均方控制。结合 Vaughan identity、零点密度估计或双线性分解,可以处理素数在很多模数下的平均误差。

    5. 哲学

    Large sieve 的力量在于:我们不逐个模数证明最优分布,而是在整个模数族上利用几乎正交性。这个思想和调和分析中 square function、wave packet 的平均控制是一脉相承的。

  • 短区间上的乘法函数:Matomaki-Radziwill 定理的分析图像

    旧博客原文

    原题:Multiplication function on short interval

    The most important beakgrouth of analytic number theory is the new understanding of multiplication function on share interval, this result is established by Kaisa Matomäki & Maksym Radziwill. Two very young and intelligent superstars.

    The main theorem in them article is :

    Theorem(Matomaki,Radziwill)
    As soon as H\to \infty when x\to \infty, one has:
    
                        \sum_{x\leq n\leq x+H}\lambda(n)= o(H)
    
    for almost all x\sim X .

     

    In my understanding of the result, the main strategy is:

    Step 1:Parseval indetity, monotonically inequality

    Parseval indetity, monotonically inequality, this is something about the L^2 norms of the quality we wish to charge. It is just trying to understanding

    \frac{1}{X}\int_{X}^{2X}|\frac{1}{H}\sum_{x\leq n\leq x+H}\lambda(n)|^2dx

    as a fuzzy thing by a more chargeable quality:

      \frac{1}{X^2}\int_{0}^{\infty}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt

    In fact we do a cutoff, the quality we really consider is just:

    \frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt

    established the monotonically inequality:

    \frac{1}{X}\int_{X}^{2X}|\frac{1}{H}\sum_{x\leq n\leq x+H}\lambda(n)|^2dx << \frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt

    In my understanding, This is a perspective of the quality, due to the quality is a multiplicative function integral on a domain (\mathbb N^*) with additive structure, it could be looked as a lots of wave with the periodic given by primes, so we could do a orthogonal decomposition in the fractional space, try to prove the cutoff is a error term and we get such a monotonically inequality.

    But at once we get the monotonically inequality, we could look it as a compactification process and this process still carry most of the information so lead to the inequality.

    It seems something similar occur in the attack of the moments estimate of zeta function by the second author. And it is also could be looked as something similar to the  spectral decomposition with some basis come from multiplication unclear, i.e. primes.

     

    Step 2: Involved by multiplication property, spectral decomposition 

    I called it is “spectral decomposition”, but this is not very exact. Anyway, the thing I want to say is that for multiplication function \lambda(n), we have Euler-product formula:

    Euler-product formula:
                          \Pi_{p,prime}(\frac{1}{1-\frac{\lambda(p)}{p^s}})=\sum_{n=1}^{\infty} \frac{\lambda(n)}{n^s}

    But anyway, we do not use the whole power of multiplication just use it on primes, i.e. \lambda(pn)=\lambda(p)\lambda(n) leads to following result:

    \lambda(n)=\sum_{n=pm,p\in I}\frac{\lambda(p)\lambda(m)}{\# \{p|n, p\in I\}+1}+\lambda(n)1_{p|n;p\notin I}

    This is a identity about the function \lambda(n), the point is it is not just use the multiplication at a point,i.e. \lambda(mn)=\lambda(m)\lambda(n), but take average at a area which is natural generated and compatible with multiplication, this identity carry a lot of information of the multiplicative property. Which is crucial to get a good estimate for the quality we consider about.

     

    Step 3:from linear to multilinear , Cauchy schwarz

    Now, we do not use one sets I, but use several sets I_1,...,I_n which is carefully chosen. And we do not consider [X,2X] with linear structure anymore , instead reconsider the decomposition:

    [X,2X]=\amalg_{i=1}^n (I_i\times J_i) \amalg U

    On every I_i\times J_i it equipped with a bilinear structure. And U is a very small set, $|U|=o(X)$ which is in fact have much better estimate.

    \int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt =\sum_{i=1}^n\int_{I_i\times J_i}  \frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt

    Now we just use a Cauchy-Schwarz:

    \sum_{i=1}^n\int_{I_i\times J_i}  \frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt$

     

    Step 4: major term estimate

     

    step 5:minor term estimate

     

    step 6: estimate the contribution of area which is not filled

     


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Matomaki-Radziwill 的工作改变了我们对乘法函数短区间平均的理解。它说明,有界乘法函数在几乎所有短区间中的平均,通常接近其长区间平均。

    短区间上的乘法函数:Matomaki-Radziwill 定理的分析图像
    短区间乘法函数问题把局部平均转化为 Dirichlet polynomial 的频率估计。

    1. 基本问题

    设 $f$ 是有界乘法函数。我们关心

    $$\frac1H\sum_{x

    对多数 $x$ 的行为。传统解析数论更擅长长区间平均,而短区间要求理解局部波动。

    2. Matomaki-Radziwill 定理

    粗略地说,只要 $H\to\infty$ 不太慢,对几乎所有 $x\le X$,短区间平均可以由长区间信息控制。这一结果为 Chowla、Sarnak 和 pretentious multiplicative functions 提供了关键输入。

    3. Dirichlet polynomial 视角

    把乘法函数平均转化成 Dirichlet polynomial:

    $$\sum_{n\le X}\frac{f(n)}{n^{1+it}}.$$

    Parseval 型恒等式把短区间均方问题变成 $t$-空间上的积分估计。这是从 additive intervals 进入 multiplicative Fourier analysis 的桥。

    4. Cut-off 与单调性

    证明中需要去掉某些坏尺度,并建立类似单调性的控制:截断后的对象仍然保留主要信息。这个过程有点像 compactification,把原来粗糙的短区间平均换成更可估的频率对象。

    5. 为什么它重要

    短区间乘法函数估计让“乘法随机性”可以在局部尺度上使用。Sarnak 和 Chowla 的许多对数平均进展,都依赖这种把局部平均、Dirichlet polynomial 和 entropy decrement 结合起来的能力。

  • Vinogradov 估计的一条思路:环面均匀分布与连分数尺度

    旧博客原文

    原题:An approach to Vinogradov estimate

    Vinogradov estimate is:

    |\sum_{n=1}^{N}e^{2\pi i\alpha P(n)}|\leq c_A\frac{N}{log^A N}

    For fix \alpha is irrational and \forall A>0 ... (*).

    Assume deg(P)=n, this could view as a effective uniformly distribute result of dynamic system:  ([0,1]^n,T), where T: x\to (A+B)x, b is a nilpotent matrix, matrix A is identity but with a irrational number \alpha in the (n, n) elements.

    First approach

    we could easily to get a “uniform distribute on fiber” result without very much tough estimate to attach the theorem. That is just a application by my  “rigid trick” that is describe in my early note. But this approach is according to the understanding of the result as a uniformly distribute result on Torus T^n, we could do this approach with the last S^1, which will corresponding to \partial^{n-1}x_k, i.e. we could apply the “rigid trick” to prove sequences (x_k,\partial^1 x_k,..., \partial^{n-1} x_k) is uniformly distribute according to \partial^{n-1} x_k\in S^1 .

    Graph

    But this approach seems difficult to generate. The difficulty is come from both there is no  similar uniformly distribute of the other perimeter use the rigid trick (At least as I know, I try to prove there could be one but I failed) and if in the best case we have the similar uniformly distribute result for other perimeter there is still some thing more need to be established. See this graph for a counterexample that the uniformly distribute for all fiberation could not derive a uniformly distribute for the original space.

     

    Second approach 

    In this approach we need use the information of continue fractional to get some information (Which is of course critical to get some information about the estimate). But I do not know if it is necessary, maybe this could be a interesting question weather the information come from continue fractional must involve to get such a estimate in the future, but not today.

    Any way, there is two different type of continue fractional:

    1.\alpha=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+...}}}.

    2.\alpha=q_0+\frac{1}{q_1}+\frac{1}{q_1q_2}+\frac{1}{q_1q_2q_3}+\frac{1}{q_1q_2q_3q_4}+....

    Anyway, these could be understand as a same thing more or less (if fact we can calculate some quantitive with a_i,q_i which is roughly the same). That is just the orbits \{e^{2\pi i\alpha}\} have quasi-period property, that is to say, under certain norms, it could be understand as the limits of periodic sequences. So it is natural to approximation \{e^{2\pi i\alpha}\} by periodic sequences and will lead to a very good point-wise coverage result:

    T_k^{n}(x) \longrightarrow T^{n}(x)

    Where T_k^{n}(x)=e^{2\pi i\sum_{i=1}^k\frac{1}{p_1...p_i}} is just the periodic approximation sequence which come from the best approximation (critical point of ||\frac{q}{p}-\alpha||), which natural occur in continue fractional. And by this we already arrive a non qualitative form result of (*) with deg(P)=1.

    But unfortunately this approximation is too good to be true for deg(P)\geq 2 case. The reason of this result could be true is just because the natural estimate for the best approximation of \alpha; i.e. Dirichlet approximation theorem.

    But for higher degree case, although we could not expect this thing to be true, we still could image a weaker but enough result to be true:

    \{T_k^{n}(x)\} \longrightarrow \{T^{n}(x)\}

    in the Gromov Hausdorff metric sense, and the $T_k^n(x)$ is carefully chose, which have a finite torsion structure(which could be view as a multilinear structure which will play a central role in the estimate). Here is a graph for deg(P)=2:

     

    Roughly speaking,  in general deg(P)=n case, there is a cube structure in the orbits e^{2\pi iP(n\alpha)} and is critical to observe that the progression of difference structure in it. The goal of this approach is to establish some result from the finite torsion structure(multilinear structure). That is to say, the boundary is high order thing in all direction but there is only one direction attend to infinity the other is just a finite torsion, and we wish to get more information from the extra structure.

    This also have a physics explaining, for which see the graph:

     

    Third approach

    For P(n)=an^2+bn+c case:

    P(k+\Delta)=P(k)+(ka+b)\Delta+\Delta^2


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Vinogradov 型估计可以看成带有有效误差的均匀分布定理。若相位含有无理参数,问题往往在两个语言之间切换:动力系统上是环面轨道的均匀分布,解析数论上是指数和的消失。

    Vinogradov 估计的一条思路:环面均匀分布与连分数尺度
    Vinogradov 型估计可以看成环面轨道均匀分布的定量版本,核心是控制指数和。

    1. 动力系统图像

    考虑环面上的序列

    $$n\mapsto (n\alpha,n^2\alpha,\ldots)\pmod1.$$

    当 $\alpha$ 无理时,这类序列常常均匀分布。Vinogradov 估计要求更强:不仅要知道平均极限,还要给出可用的误差项。

    2. 指数和形式

    Weyl criterion 把均匀分布转成指数和:

    $$\frac1N\sum_{n\le N}e(P(n))\to0.$$

    有效估计则要控制

    $$\left|\sum_{n\le N}e(P(n))\right|.$$

    当 $P$ 的系数包含无理数时,连分数近似会决定哪些尺度上相位最接近有理、哪些尺度上可以得到抵消。

    3. Fiber 均匀分布的限制

    一种诱人的想法是先证明每个 fiber 上的均匀分布,再合成整个空间的均匀分布。但这并不总是成立:所有纤维方向看起来平均,不代表整体分布没有隐藏相关性。这个失败提示我们必须直接控制整体指数和。

    4. 连分数尺度

    设 $\alpha$ 的 convergents 是 $p_k/q_k$。在长度接近 $q_k$ 的区间上,$n\alpha$ 的分布有特别好的结构。把 $[1,N]$ 拆成这些标准尺度,可以把任意长度的问题化成一族可估的块。

    5. 证明路线

    一个合理策略是:先用 Weyl differencing 降低多项式相位阶数;再用连分数控制主要尺度;最后把误差在不同块上求和。动力系统语言提供几何直觉,真正的定量估计则来自指数和技术。

  • Sarnak 猜想的标准模型:skew product 与 interval exchange

    旧博客原文

    原题:Sarnak conjecture, understand with standard model

    Sarnak conjecture is a conjecture lie in the overlap of dynamic system and number theory. It is mainly focus on understanding the behavior of entropy zero dynamic system by look at the correlation of an observable and the Mobius function .

    We state it in a rigorous way:

    let (X,T) be a entropy zero topological dynamic system. Let Mobius function be defined as \mu(n)=(-1)^t, where $latex$ is the number of different primes occur in the decomposition of n.

    Then for any continuous function f:X\to R and x\in X, observable \xi(n)=f(T^n(x)) is orthogonal to the Mobius function; i.e. ,

    \lim_{N\to \infty}\frac{1}{N}\sum_{n=0}^{N-1}\mu(n)\xi(n)=o(N).

    I mainly focus on the special cases when dynamic system X is the skew product on T^2 and when the dynamic system which is a interval exchange in [0,1].

    Skew product

    For the first one, \Theta=(T,T^2),T:T^2\longrightarrow T^2 :
    T(x)=x+\alpha,T(y)=cx+y+h(x)
    y_1(n)=T^{n}(x)=x+n\alpha,y_2(n)=T^n(y)=nx+\frac{n(n-1)}{2}\alpha+y+\sum_{n=1}^{N-1}h(x+i\alpha) , where c=1,-1.

    by Bourgain-Ziegelar-Sarnak theorem we know the difficulties is focus on deal with the exponent

    S_{p,q}(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)(\frac{e(npm\alpha)-1}{e(m\alpha)-1}- \frac{e(nqm\alpha)-1}{e(m\alpha)-1})}

    for all p,q is suffice large primes pair.

    and a much simper case is the affine map:T:(x,y)\to (x+\alpha,cx+y+\beta) on \mathbb T^2 and the general case T:(x_1,...,x_n)\to A(x_1,...,x_n) where A is a upper-triangle matrix with diagonal 1; i.e. A=I+B, B is nilpotent. So the sarnak conjecture in this case is reduce to the Davenport estimate on exponent by B-Z-S theorem:

    |\sum_{n=0}^{N}e^{2\pi if(n)}|\leq c_A\frac{N}{(log N)^A}, \forall A>0.

    Interval exchange map

    For the interval exchange map, we can explain it by a composition of rotation of some part of S_1 step by step and with a renormalization process to glue the neighbor rotations.

    Now let us explain a little with this interesting dynamic system. We focus in the simplest nontrivial case, which is the 3-interval exchange map. In this case, just consider the permutation of intervals I_1,I_2,I_3, and it is easy to see there is only one case is nontrivial that is permutation: I_1\to I_3,I_2\to I_2,I_3\to I_1. We explain a little more with other trivial case:

    When  I_1\to I_2,I_2\to I_3,I_3\to I_1, the interval exchange map is just a rotation and for which the sarnak conjecture is just come from:

    |\sum_{n=0}^{N}e^{2\pi in\alpha}\mu(n)|=o(N), \forall \alpha\in R.

    Which is trivial because \sum_{n=0}^{N}e^{2\pi in\alpha}\mu(n)=\frac{1-e^{2\pi iN\alpha}}{1-e^{2\pi i\alpha}}.

    For the case $I_1\to I_2, I_2\to I_1, I_3\to i_3$ the map T is a rotation on I_1\cap I_2 but it is a identity map on I_3 and the orbits of point only lying one of $I_1\cap I_2, I_3$, lying in which one depend on the original point x we take is lying in which one.

    Now we focus on the most difficult situation. It is annoying but it is the obstacle we must get over to go far. Fortunately it could be explained as in the following picture.

    img_0069.jpg
    3-Interval exchange map as two rotation map glue with a renormalization map.

     

    Now we explain what happen in the picture, it is mainly say one identity, which explain how to look 3-interval exchange map as a composition of rotation map with a renormalization map to glue them. Rotation is a kind of map we have good understanding but we do not understand very well with the renormalization map which is glue the two endpoints of I_2,I_3 which are not the common endpoint of them. Then you get two circle glue like a “8” , and T_2 is just rotate one of it and make the other one to be invariance.

    Now we roughly could think about what is the thing we need to charge with, it is just:

    \sum_{n=0}^{N}f((T_1\circ R\circ T_1)^n(x))\mu(n)=o(N).

    Now we do some calculate with this geometric explain of interval exchange map.

    Let A=I_1, B=I_2\cap I_3, then A\cap B=\emptyset, A\cup B=[0,1]. And |A|=\alpha, 0<\beta<|B|. the rotation T_1:x\to x-\alpha, T_2:x\to x+\beta.

     

     

    Standard model

    Is there a standard model of entropy zero dynamic system?

    This problem seems to be too ambitious. But it occur naturally when I an trying to have a global understand of the Sarnak conjecture.

     


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Sarnak 猜想位于动力系统和解析数论的交界处。它说零熵动力系统产生的确定序列,应该和莫比乌斯函数这样的算术随机序列正交。

    Sarnak 猜想的标准模型:skew product 与 interval exchange
    Sarnak 猜想的标准模型包括 skew product、unipotent affine maps 和 interval exchange maps。

    1. 基本陈述

    设 $(X,T)$ 是零拓扑熵系统,$f\in C(X)$。Sarnak 猜想断言

    $$\frac1N\sum_{n\le N}\mu(n)f(T^n x)\to0.$$

    这里 $\mu(n)$ 是 Mobius function。零熵表示轨道复杂度低,而 $\mu(n)$ 预期具有强随机性。

    2. Skew product 模型

    典型例子是

    $$T(x,y)=(x+\alpha,y+h(x))\pmod1.$$

    对 Fourier character 展开后,问题会变成

    $$\sum_{n\le N}\mu(n)e(P(n))$$

    或更一般的旋转 Birkhoff sum 相位。Bourgain-Sarnak-Ziegler 准则可以把莫比乌斯相关转为不同素数伸缩下的双线性相关。

    3. Affine nilsystem 情形

    若环面自同态由上三角 unipotent 矩阵给出,例如 $A=I+B$ 且 $B$ nilpotent,那么 $T^n$ 的坐标是 $n$ 的多项式。因此 Sarnak 猜想可归约到 Davenport 型多项式指数和估计。

    4. Interval exchange maps

    interval exchange map 可以看作把区间切成有限段后重排。它通常是零熵,但没有简单的光滑结构。它的 renormalization 来自 Rauzy induction,类似连续分数在旋转中的作用。

    这里的困难是:相位不再是一个光滑多项式,而是经过多次 induction 拼接出来的低复杂度序列。

    5. 标准模型的意义

    skew product 展示了“低熵加光滑结构”如何导出指数和;interval exchange 展示了“低熵但不光滑”的困难。理解这两个模型,就能看清 Sarnak 猜想里动力系统复杂度与数论随机性之间的真正接口。

  • Vinogradov mean value theorem:矩估计、Weyl sums 与 decoupling

    旧博客原文

    原题:Note on Vinogradov main theorem

    1.Introduction

     

    Question:
    Vinogradov mean value
    Let k,s\in \mathbb N,x\in R^k .

    J_{s,k}(N)=|\{(n_1,...,n_s,n_{s+1},...,n_{2s})|n_1^j+...+n_s^j=n_{s+1}^j+...+n_{2s}^j) \forall 1\leq j\leq k,1\leq n_i\leq N(1\leq i\leq s) \}|
    How to estimate J_{s,k}(N)?

    We assume f_{k}(x,N)=\sum_{1\leq n\leq N}e(nx_1+n^2x_2+...+n^kx_k), then by following clear calculate:

    \int_{[0,1]^k}|f_k(x,N)|^{2s}dx_1...dx_k =\int_{[0,1]^k}|\sum_{1\leq n\leq N}e(nx_1+n^2x_2+...+n^kx_k)|^{2s}dx_1dx_2...dx_k &=\int_{[0,1]^k}\sum_{1\leq n_1,...,n_{2s}\leq N}e^{2\pi i[(n_1+...+n_{2s})x_1+...+(n_1^k+...+n_{2s}^k)x_k-(n_{s+1}+...+n_{2s})x_1-...-(n_{s+1}^k+...+n_{2s}^k)x_k]} &=|\{(n_1,...,n_{2s})| n_1^j+...+n_s^j=n_{s+1}^j+...+n_{2s}^j,\forall 1\leq j\leq k\}|

    we have:
    J_{s,k}(N)=\int_{[0,1]^k}|f_k(x,N)|^{2s}dx_1...dx_k
    main conjecture:
    \forall \epsilon >0,we have:
    J_{s,k}(N)<<N^{\epsilon}(N^s+N^{2s-\frac{1}{2}k(k+1)})

    theorem(Bourgain-Demeter-Guth)
    Main conjecture hold in general.

    2.Application

    We have following directly application:

    1.Waring problem

    2.Bound Weyl sums.\

     

    3.Zero-free region for Riemann-zeta function.

    3.Relate to the decoupling theorem

    Now we discuss the decoupling theorem. This theorem describe the phenomenon when we are considering the “expension” operator E_{[0,1]}(g) cut off $E_{[0,1]}(g)$ into a lot of small boxes E_{J}(g), then the $L_{d(d+1)})$ norms of the operator could be bounded very well, in fact it is near orthonagonal.

    [B-D-G]
    Let d\geq 2,$0<\delta\leq 1$. Then for each ball B\subset R^d of radious at least \delta^{-d}.
    ||E_{[0,1]}g||_{L^{d(d+1)}(w_B)}<< \delta^{-\epsilon}(\sum_{J\subset [0,1],|J|=\delta}||E_Jg||^2_{L^{d(d+1)(w_B)}})^{\frac{1}{2}}
    (J runs over a partition of [0,1] in \delta-intervals)

    Discretized version:
    Now we discuss the discretization of decoupling type result. We could establish a relationship between the decoupling theorem and Vinogradov mean theorem. look at the sum:
    \int_{[0,1]^k}|\sum_{1\leq n\leq N}e(nx_1+n^2x_2+...+n^kx_k)|^{2s}dx_1dx_2...dx_n
    This could be view as a 2s norm of a constant function h=1, with a lebergue measure d\sigma on curve \Gamma=\{(t,t^2,...,t^d):0\leq t\leq 1\}. this curve \Gamma could be view as a canonical curve with non-vanish guess curvature.
    ||\widehat {hd\sigma}||_{2s}^{2s}=\int_{R^{k}}|\int_{\Gamma}h(t,t^2...,t^n)e(tx_1+...+t^kx_k)|^{2s}d\sigma

    this is very similar with the restriction theorem:

    [restriction theorem]
    let \Gamma be $n-1$ dimension parabolic in R^n, then guess curvature of \Gamma is non-vanish.\sigma is a natural induced lebergue measure on \Gamma, we have, for suitable exponents p,p' come from rescaling arument.
    ||\widehat{gd\sigma}||_{p'}\lesssim ||g||_p

    So it seems like these are the same thing, but unfortunately they are not,there are two things distinct them:
    1.the density is defferent, it is a discrete sum in:
    \int_{[0,1]^k}|\sum_{1\leq n\leq N}e(nx_1+n^2x_2+...+n^kx_k)|^{2s}dx_1dx_2...dx_n
    but a continue integral in:
    ||\widehat {hd\sigma}||_{2s}^{2s}=\int_{R^k}|\int_{\Gamma}h(t,t^2...,t^n)e(tx_1+...+t^kx_k)|^{2s}d\sigma
    so we need to construct a rescaling way to make the discretization one coverage to the continue one.a suitable fexiable function seems like (w_B,B_{r}(c_B)),w_B(x)=(1-\frac{|x-c_B|}{R})^{-100k}
    2.there

     

     


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    Vinogradov mean value theorem 控制 Weyl sums 的高阶矩,是 Waring problem、指数和估计和 zeta 函数零点区域中的基础工具。Bourgain-Demeter-Guth 用 decoupling theorem 证明了主猜想。

    Vinogradov mean value theorem:矩估计、Weyl sums 与 decoupling
    Vinogradov mean value theorem 把 Weyl sums 的高阶矩与 moment curve decoupling 联系起来。

    1. Mean value

    $$S(\alpha)=\sum_{n\le N}e(\alpha_1n+\alpha_2n^2+\cdots+\alpha_kn^k).$$

    Vinogradov mean value 研究

    $$J_{s,k}(N)=\int_{[0,1]^k}|S(\alpha)|^{2s}\,d\alpha.$$

    它也等于某个 Diophantine system 解的个数。

    2. 主猜想

    主猜想断言

    $$J_{s,k}(N)\lesssim_\varepsilon N^\varepsilon\left(N^s+N^{2s-k(k+1)/2}\right).$$

    两个项分别对应 diagonal solutions 和维数计数给出的主项。

    3. 应用

    这个估计直接用于 Waring problem,也给出 Weyl sums 的强上界。通过指数和控制,可以进一步进入 zeta 函数零点区域和等分布问题。

    4. Decoupling 视角

    考虑 moment curve

    $$\gamma(t)=(t,t^2,\ldots,t^k).$$

    decoupling theorem 描述 extension operator 在小区间分解后的 $L^p$ 几乎正交性。离散化后,它与 Vinogradov mean value theorem 精确相连。

    5. 思想总结

    Vinogradov mean value 把数论中的方程计数、调和分析中的 Fourier extension、以及几何中的曲率结构放到同一个问题里。这是现代解析数论和 decoupling 理论交汇的代表。

  • Heat flow 与多项式零点:从变形思想到 Riemann Hypothesis 的 toy model

    旧博客原文

    原题:Heat flow and the zero of polynomial-a approach to Riemann Hypesis

    this is a note after reading the blog:Heat flow and the zero of polynomial.

    1.instead of consider the original version:

    \partial_{zz}f(z,t)=\partial_tf(z,t).

    consider the corresponding “equidistribution version” is also interesting:

    \partial_{zz}f(z,t)=\theta(z,t)\partial_tf(z,t),especially \theta(z,t)=e^{2\pi i\alpha t},\alpha\in R-Q.

    2.

    where f(z)=z^n+a_{n-1}z^{n-1}+...+a_1z+a_0.

    f(z,t)=\sum_{k=1}^n\sum_{0\leq m\leq k-2,2|k-m}\frac{k!}{m!(k-m)!}z^mt^{k-m}.

    =\sum_{k=1}^m\sum_{0\leq m\leq k-2,2|k-m}C_k^mt^{k-m})z^mt^{k-m}

    \sum_{m=0}^{n-2}(\sum_{k=m,2|k-m}^nC_k^mt^{k-m})z^m.

    rescaling:

    F_t:(z_1(t),...,z_n(t))\longrightarrow (\frac{z_1(t)}{t},...,\frac{z_n(t)}{t}).

    F_t\cdot f(z,t)=\sum_{m=0}^{n-2}(\sum_{k=m,2|k-m}^nC_{k}^mt^{k-n})z^m.

    \lim_{t\to \infty}F_t\cdot f(z,t)=\sum_{m=0,2|n-m}^{n-2}C_n^mz^m.(*)

    even term \longrightarrow constant.(after renormelization)

    odd term \longrightarrow 0(invariant).so at least the sum zeros of is invarient.

    by the algebraic fundamental theorem,we have n zero \{z_1,...,z_n\}of (*).

    until now,we already now if the n zeros is distinct,then because the energy is the energy is the same and the entropy is increase so \exists T>>0,\forall t_i,t_j>T,\{t>T|z_i(t)\} \cap \{t>T|z_j(t)\}=\emptyset.\lim_{t\to \infty}|z_i(t)|=\infty and \lim_{t\to \infty}arg(z_i(t))=z_i.

    but how to know the information of the change of direction at “blow up” time?

    1.change direction only at blow up.

    2.energy invariant \sum_{1\leq i\neq j\leq n}\frac{1}{|x_i-x_j|^2}.

    3.general philosophy

    deformation some function under some evolution equation, such like heat equation,wave equation,shrodinger equation.and there is some conversion thing under the equation,and some quantity that could calculate directly such like the trace of spectral.

    4.difficultis

    this philosophy could generate to the analytic function case,but to make the limit case(I only know how ti deal with this now)coverage.we need very good control on the coefficient.

    and to investigate the change of direction at blow up point maybe we need some knowledge about the burid group.

     

     

     


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    用 heat flow 研究多项式零点,是理解更复杂解析函数零点问题的一个 toy model。基本思想是:让函数随时间演化,观察零点如何移动,以及哪些量在演化中保持或单调。

    Heat flow 与多项式零点:从变形思想到 Riemann Hypothesis 的 toy model
    heat flow 让多项式零点随时间运动,提供研究零点实性和不变量的 toy model。

    1. 多项式的 heat deformation

    设 $P(x)$ 是多项式,考虑

    $$\partial_t u=\partial_x^2u,\qquad u(0,x)=P(x).$$

    因为 heat operator 保持多项式空间,$u(t,x)$ 仍然是多项式。其零点随 $t$ 移动。

    2. 不变量与单调量

    某些系数组合在演化中保持不变,另一些量具有单调性。例如最高次项不变,低阶偶次项会随 heat flow 改变。零点的质心或某些对称量可能保持。

    3. 零点碰撞

    若零点始终实且互异,运动图像较清楚;真正困难发生在零点碰撞或分裂时。此时需要理解 blow-up 时间附近的方向变化。

    4. 与 Riemann Hypothesis 的类比

    de Bruijn-Newman 常数研究的是 Xi 函数在 heat flow 型变形下零点保持实的临界时间。多项式模型不能证明 RH,但能展示同一种哲学:通过演化方程追踪零点几何。

    5. 需要的估计

    要从多项式推广到整函数,必须控制系数、增长阶和极限过程。多项式情形的代数基本定理给出有限零点;整函数情形需要更强的紧性和零点分布估计。

  • Sarnak 猜想在 skew product 上的情形

    旧博客原文

    原题:Sarnak猜想在skew product上的情形。

    Cylinder map:
    Cylender map:这是一个动力系统\Theta=(T,T^2),T:T^2\longrightarrow T^2 满足:\\
    T(x)=x+\alpha,T(y)=cx+y+h(x)
    因此
    y_1(n)=T^{n}(x)=x+n\alpha,y_2(n)=T^n(y)=nx+\frac{n(n-1)}{2}\alpha+y+\sum_{n=1}^{N-1}h(x+i\alpha)
    来自动力系统\Theta中的可观测量是指\xi(n)=f(T^n(x)),其中x\in T^2,$f\in C(T^2)$.
    由于Cylender map是零熵的,这个情形下Sarnak猜想成立等价于:
    S(N)=\sum_{n=1}^N\mu(n)\xi(n)=\sum{n=1}^N \mu(n)f(T^nx)
    满足S(N)=o(N),由于f_{\lambda_1\lambda_2}=e^{2\pi i(\lambda_1 x+\lambda_2 y)}C(T^2)的一组基,只需对f_{\lambda_1\lambda_2}证明S(N)=o(N)\\
    展开S(N),我们有\\
    S(N)=\sum_{n=1}^N\mu(n)\xi(n)=\sum_{n=1}^N \mu(n)f(T^nx)\\

    =\sum_{n=1}^N\mu(n)e^{2\pi ik(\lambda_1(x+n\alpha)+\lambda_2(nx+\frac{n(n-1)}{2}+y\sum_{i=1}^{n-1}h(x+i\alpha)))}\\

    =\sum_{n=1}^N\mu(n)e^{2\pi i(\phi(n)+\sum_{i=1}^{n-1}h(x+i\alpha))}\\

    =\sum_{n=1}^N\mu(n)e^{2\pi i(\phi(n)+\sum_{i=1}^{n-1}\sum_{m\in Z}\hat h(m)e^{2\pi im(x+i\alpha)})}\\

    =\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}} \\
    其中我们暂时假定h是解析的,实际上我们要求对h的fourior级数有下界控制,总的来说就是\exists \tau_1,\tau_2:
    e^{\tau_1 m}<<\hat h(m)<<e^{\tau_2 m}

    \begin{lemma}
    \forall A>0,\forall \phi(n) 为多项式函数,我们有指数和估计:
    |\sum_{n=1}^{N}\mu(n)e^{\phi(n)}|<<\frac{N}{(logN)^A}
    \end{lemma}

    此引理来自解析数论指数和理论, 那么\alpha \in Q情形是引理的直接推论。接下来处理\alpha \in R-Q情形,这种情形下,我们定义\alpha的连分数展开为:
    \alpha=[q_1,q_2,q_3,....]

    \begin{lemma}
    如果\alpha的连分数展开有一致的上界,即存在C\in N^*,\forall n\in N^*,1\leq q_n\leq C那么:
    sup_{0\leq a<b\leq 1}|\sum_{k=0}^{N-1}\chi_{(a,b)}(\{k\alpha\})-N(b-a)|=O(log N)

    \end{lemma}

    这个引理的证明由三部分组成,第一部分用一个初等的trick加上连分数表示得到一系列长度区间上的更好的估计,第二部分建立一个有效性估计,第三部分将任何区间拆分成第一种区间的并,并使得余项被有效性估计控制。

    我们现在考察最后这个式子:
    S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}

    我们对这个式子建立有效的估计,指的是能够证明:
    S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=o(N)
    那么我们接下来建立这个估计,这个估计主要由三部分组成,我们分成三节处理这三部分,最后一节是总结。\\
    1.带密度的指数和估计。\\
    2.cut-off估计。\\
    3.一致性均匀估计。\\

    \newpage
    \section{带密度的指数和估计}
    S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=o(N)
    令:A_n=\mu(n)e(\phi(n)),B_n=e(\sum_{m\in Z}e(mx)\hat H(m))
    经典的指数和估计是:
    theorem:
    \forall A>0,\forall \phi(n) 为多项式函数,我们有指数和估计:
    |\sum_{n=1}^{N}\mu(n)e^{\phi(n)}|<<\frac{N}{(logN)^A}

    theorem:
    对于P是一个质数,对于P<<N_1<<N:\\定义\chi_{p}(n)=e^{\frac{2\pi in}{p}}=e_p(n), 定义f:N^*\to Im(\chi_p)满足:\\
    对于任何长度为N_1的一段区间$I$,对任意k\in \{0,1,...,p-1\},
    \sharp\{n\in I|f(n)=e_p(k)\}=\frac{N_1}{p}+O(1)

    |\sum_{n=1}^{N}\mu(n)f(n)e^{\phi(n)}|<<_{C}\frac{N}{(logN)^A}
    其中C\sim P,A\\
    \mu是Mobius函数

     

    cut-off 估计
    在式子S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=o(N)
    中,我们希望对m\in Z1\leq n \leq N做cut off来简化问题。\\
    后者是简单的, 我们待定一个常数c,有:
    S(N)=\hat S(n)+\sum_{n=1}^{cN}\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=\hat S(n)+O(cN)
    c可以待定,之后取得任意小,所以这一部分误差不影响我们最后的结果。\\
    对m做cut off会稍微复杂一些,根据Fourior分析我们知道:\\
    1如果h\in C^{\omega}(T),则
    \hat h(m)=O(e^-\tau m).
    2.若h\in C^{d}(T),则根据分部积分公式\hat h(m)=O(m^{-d}).\\
    接下来的结果可能可以用调和分析中的几乎正交性改进到更好的结果,但是至少我们有:\\
    e(\sum_{|m|>\delta}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1})\sim \sum_{|m|>\delta}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}
    =O(\sum_{|m|>\delta}m \cdot m^{-d})=O(\delta^{d-2})
    所以至少当d>2时,我们可以找到\delta \to \infty当$N\to \infty$,使得|m|>\delta的部分可以被cut off.

    连分数与Ostrowoski表示
    我们知道任何一个(0,1)中的数都有连分数表示,并且这个表示是唯一的。
    \alpha=(q_1,q_2,....,q_n,...)
    此时我们定义正整数集N^*关于\alpha的Ostrowoski表示(wangzhiren 2):
    定义:
    每一个正整数n可以唯一的表示为:
    n=\sum_{i=0}^{\infty}(\Pi_{j=0}^{i-1}q_j)r_i
    其中r_i \in [0,q_{i}-1]

    很明显上面表示中只有有限个r_i不为0,为什么要利用Ostrowoski表示,关键在于Ostrowoski表示中的标架\{q_1...q_k\}是最佳逼近下最好的标架。\\
    实际上我们归纳定义\alpha-标准长度\{l_k\}_{k=1}^{\infty}如下:

    l_1=\alpha,l_{k+1}=1-[\frac{1}{l_k}]l_k

    容易知道l_{k+1}<l_{k},做一些微小的计算会发现第k个\alpha-标准长度和连分数展开的前k项系数乘积之间能够相互控制。
    lemma:
    \forall k\in N^*
    \frac{1}{2q_1...q_k}<l_k<\frac{1}{q_1....q_k}

    直接将\alpha的连分数展开代入计算即可证明。\alpha-标准长度的关键性质是\{n\alpha\}在这个区间中的均匀分布性的余项可以得到很好地控制。
    lemma:
    对任意k\in N^*,对任意长度为l_k=(a,b)的区间I_k\subset (0,1),\forall N\in N^*我们有:
    \sum_{n=1}^N\chi_{(a,b)}(\{n\alpha\})=N(b-a)+O(1)

    证明是对k归纳,实际上k等于1的时候将f(n)=\{n\alpha\}提升为g=n\alpha,因为实轴上长度为\alpha的区间中一定会包含一个\{g(1),...,g(n)\}中的元素,有由于长度为n\alpha的区间中有[n\alpha]个整数,所以:
    \sum_{n=1}^N\chi_{(a,b)}(\{n\alpha\})=[N\alpha]=N(b-a)+O(1)
    归纳过渡也是简单的。

    \section{一致性均匀估计}
    最后我们要建立一致性均匀估计,将对
    S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}
    进行多尺度分解,并且说明他和一个多重带密度的指数和的差是$o(N)$
    \newpage

     


    补充说明

    以下是新整理的中文说明;上方旧博客原文保持不变。

    这篇笔记讨论一个很典型的零熵动力系统:圆环或二维环面上的 skew product。Sarnak 猜想在这里会变成一个指数和问题。动力系统给出相位,莫比乌斯函数给出算术权重,最后要证明两者没有长期相关。

    Sarnak 猜想在 skew product 上的情形
    Skew product 的迭代把 Fourier character 转成带有旋转 Birkhoff sum 的指数和。

    1. Skew product 的形式

    考虑二维环面上的映射

    $$T(x,y)=(x+\alpha,\;y+h(x))\pmod 1.$$

    这里 $\alpha$ 是旋转数,$h$ 是足够光滑或解析的函数。若 $f\in C(\mathbb T^2)$,Sarnak 猜想要求

    $$\frac1N\sum_{n\le N}\mu(n)f(T^n(x,y))\to0.$$

    由于 trigonometric polynomials 在 $C(\mathbb T^2)$ 中稠密,可以先检验 Fourier characters $f(x,y)=e(mx+ny)$。

    2. 相位展开

    迭代 $T$ 得到

    $$T^k(x,y)=\left(x+k\alpha,\;y+\sum_{j=0}^{k-1}h(x+j\alpha)\right).$$

    所以相关和变成

    $$\sum_{k\le N}\mu(k)e\left(m(x+k\alpha)+n y+n\sum_{j

    问题的核心是控制这个由旋转 Birkhoff sum 产生的相位。若 $h$ 是多项式或 Fourier 支持很简单,相位可以化成多项式相位,经典解析数论的指数和估计可以直接进入。

    3. 有界型旋转数

    当 $\alpha$ 的连分数展开系数有一致上界时,旋转轨道具有较好的均匀分布余项。Ostrowski 表示把任意长度拆成由分母 $q_k$ 控制的标准块:

    $$N=\sum_k b_k q_k.$$

    这些标准块是处理旋转和的自然尺度。每一块上相位的波动可控,块与块之间再通过 cut-off 和 Fourier 展开拼接。

    4. Cut-off 与 Fourier 级数

    若 $h$ 解析,它的 Fourier 系数指数衰减;若只要求有限光滑性,则系数只有多项式衰减。把高频部分 cut off 后,误差由

    $$\sum_{|r|>R}|\widehat h(r)|$$

    控制。低频部分则给出有限多个可估的指数和。这里调和分析中的 almost orthogonality 可以改进一些粗糙估计,但最重要的是把问题压缩到有限频率。

    5. 证明图像

    整个论证可以理解成三层:先把 observables 化成 Fourier characters;再把 skew product 的迭代化成旋转和;最后用连分数分块、cut-off 和带密度的指数和估计控制莫比乌斯相关。这个模型清楚展示了 Sarnak 猜想在零熵系统中常见的结构:动力系统低复杂度负责给出可分解相位,解析数论负责证明乘法函数无法跟随这些相位。