Tarski’s circle squaring problem has a constructive solution

You need to know: Euclidean plane ${\mathbb R}^2$ , notation $A+v$ for the translation of set $A\subset {\mathbb R}^2$ by a vector $v\in {\mathbb R}^2$ , open subset of ${\mathbb R}^2$ , countable union, countable intersection, and set difference ( $B \setminus A = \{x\in B, \, x\not\in A\}$ ) of sets. You also need to know what is the Axiom of choice to fully understand the context.

Background: A subset $A\subset {\mathbb R}^2$ is a Borel set if it can be formed from open sets through the operations of countable union, countable intersection, and set difference. We call two sets $A,B \subset {\mathbb R}^2$ equidecomposable by translations if there are partitions $A = A_1 \cup \dots \cup A_m$ and $B = B_1 \cup \dots \cup B_m$ , such that $B_i = A_i + v_i$ , $i=1,\dots,m$ , for some vectors $v_1, \dots, v_m \in {\mathbb R}^2$ . If, moreover, all $A_i$ (and thus $B_i$ ) are Borel sets, we say that A and B are equidecomposable by translations with Borel parts.

The Theorem: On 17th December 2016, Andrew Marks and Spencer Unger submitted to arxiv a paper in which they proved that a circle and a square of the same area on the plane are equidecomposable by translations with Borel parts.

Short context: In 1990, Laczkovich, answering a 1925 question of Tarski, proved that circle and a square of the same area are equidecomposable by translations. In a paper submitted in 2015, Grabowski, Máthé, and Pikhurko proved that this is possible even using only Lebesgue measurable pieces (that is, those having a well-define area). However, both results use axiom of choice and the resulting pieces $A_i$ are impossible to construct explicitly. The Theorem states that the circle can be squared with only Borel pieces. The proof does not use the axiom of choice. If such a proof is possible, we say that a problem has a constructive solution.

Links: Free arxiv version of the original paper is here, journal version is here.

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The elliptic Harnack inequality on a graph is stable under rough isometries

You need to know: Graph, connected graph, infinite graph, degree of a vertex of a graph, bounded degree graph (for which there is a constant $C<\infty$ such that every vertex has degree at most C), path connecting 2 vertices, length of a path, notation $d_G(x,y)$ for the length of the shortest path connecting vertices $x$ and $y$ in graph G.

Background: Let G be a connected graph with infinite vertex set V. For each pair $x, y \in V$ we define a conductance $C_{xy}\geq 0$ such that $C_{xy} = C_{yx}$ and also $C_{xy} = 0$ unless x and y are connected by edge. The graph G together with the conductances $C_{xy}$ is denoted $(G,C)$ and called a weighted graph. For $x\in V$ , let $\mu_G(x)=\sum\limits_{y\in V}C_{xy}$ . For $A\subset V$ , let $\mu_G(A)=\sum\limits_{x \in A}\mu_G(x)$ . A function $h:A\to {\mathbb R}$ is called harmonic on A if $h(x)=\sum\limits_{y\in V}h(y)C_{xy}$ for all $x\in A$ . For $x\in V$ and $r>0$ , let $B_G(x,r)$ be the set of $y\in V$ with $d_G(x,y)<r$ . We say that the elliptic Harnack inequality holds for $(G,C)$ if there exists $c_1$ such that whenever $x_0\in V$ , $r \geq 1$ , and h is non-negative and harmonic in $B_G(x_0, 2r)$ , then $h(x) \leq c_1 h(y)$ for all $x,y \in B_G(x_0,r)$ . We say that weighed graphs $(G,C)$ and $(H,C')$ with vertex sets $V_G$ and $V_H$ are roughly isometric if there is a function $\phi:V_G\to V_H$ and constants $C_1>0$ and $C_2,C_3>1$ such that (i) for every $y\in V_H$ there exists $x\in V_G$ , such that $d_H(y,\phi(x))\leq C_1$ , (ii) $C_2^{-1}(d_G(x,y)-C_1) \leq d_H(\phi(x), \phi(y)) \leq C_2(d_G(x,y)+C_1)$ for all $x,y \in V_G$ , and (iii) $C_3^{-1} \mu_G(B_G(x,r)) \leq \mu_H(B_H(\phi(x),r)) \leq C_3 \mu_G(B_G(x,r))$ for all $x\in V_G$ and $r>0$ .

The Theorem: On 5th October 2016, Martin Barlow and Mathav Murugan submitted to arxiv a paper in which they proved the following result. Let $(G,C)$ and $(H,C')$ be two connected bounded degree weighed graphs that are roughly isometric. Then the elliptic Harnack inequality holds for $(G,C)$ if and only if it holds for $(H,C')$ .

Short context: The elliptic Harnack inequality was proved by Moser in 1961 for some partial differential equations, but since that turned out to be useful in many other applications, for example for weighted graphs. The Theorem proves that this inequality is stable under rough isometries, resolving a long-standing open question. While we state the Theorem only for weighed graphs here, Barlow and Murugan actually proved it in much more general setting.

Links: Free arxiv version of the original paper is here, journal version is here.

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The Furstenberg’s conjecture on the intersections of invariant sets is true

You need to know: Set ${\mathbb Q}$ of rational numbers, closed subset of $[0,1]$ , multiplication of a set by a constant $s\cdot A = \{s\cdot a: a \in A\}$ , sum of two sets $A + B = \{a + b : a \in A; b \in B\}$ , Hausdorff dimension $\text{dim}(A)$ of set $A \subset [0,1]$ .

Background: For real number $x$ , let $\left \lfloor{x}\right \rfloor$ be the largest integer not exceeding x, and let $\text{frac}(x)=x-\left \lfloor{x}\right \rfloor$ . For integer m, let $T_m:[0,1)\to[0,1)$ be a function given by $T_m(x)=\text{frac}(mx)$ . We say that set $S \subset [0,1]$ is invariant under $T_m$ if $T_m(x) \in S$ for every $x \in S$ .

The Theorem: On 25th September 2016, Pablo Shmerkin submitted to arxiv a paper in which he proved the following result. Let $p,q \geq 2$ be positive integers such that $\frac{\log p}{\log q}\not\in{\mathbb Q}$ . Let $A; B \subseteq [0,1]$ be closed sets which are invariant under $T_p$ and $T_q$ , respectively. Then for all real numbers $u$ and $v$ , $\text{dim}((u\cdot A + v) \cap B) \leq \max\{0, \text{dim}(A) + \text{dim}(B) - 1\}$ .

Short context: The Theorem confirms a long-standing conjecture of Furstenberg made in 1969. In fact, the Furstenberg’s conjecture corresponds to the case $u=1$ and $v=0$ of the Theorem. On 26th September 2016, Meng Wu submitted to arxiv a paper with independent and different proof of the same result. Also, see here for an earlier theorem resolving another related conjecture of Furstenberg.

Links: Free arxiv version of the original paper is here, journal version is here.

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The Nadirashvili’s conjecture on the volume of the zero sets of harmonic functions is true

You need to know: Eucledean space ${\mathbb R}^n$ , origin $0$ in ${\mathbb R}^n$ , norm $||x||=\sqrt{\sum_{i=1}^nx_i^2}$ of $x=(x_1,\dots,x_n)\in{\mathbb R}^n$ , unit ball $B_n(0,1) =\{ x \in {\mathbb R}^n\,:\,||x||<1\}$ , open subset V of ${\mathbb R}^n$ , twice continuously differentiable function f on V, notation $\frac{\partial^2 f}{\partial x_i^2}$ for second-order partial derivatives.

Background: Let V be an open subset of ${\mathbb R}^n$ . A function $f: V \to {\mathbb R}$ is called harmonic if $\frac{\partial^2 f}{\partial x_1^2} + \dots + \frac{\partial^2 f}{\partial x_n^2} = 0$ for every $x\in V$ . A diameter of set $U\subset {\mathbb R}^n$ is $\text{diam} U=\sup\limits_{x,y \in U}||x-y||$ . For $\delta>0$ , a $\delta$ -cover of set $S \subset {\mathbb R}^n$ is a sequence of sets $U_1, U_2, \dots, U_i, \dots$ of diameters $\text{diam}U_i\leq \delta$ for all i such that $S \subset \bigcup\limits_{i=1}^\infty U_i$ . For $d\geq 0$ , let $H_\delta^d(S) = \inf \sum\limits_{i=1}^{\infty}(\text{diam}U_i)^d$ , where the infimum is taken over all $\delta$ -covers of S. The number $H^d(S)=\lim\limits_{\delta\to 0} H_\delta^d(S)$ is called the d-dimensional Hausdorff measure of S.

The Theorem: On 9th May 2016, Alexander Logunov submitted to arxiv a paper in which he proved that for every dimension $n\geq 3$ , there exists a constant $c=c(n)>0$ , depending only on n, such that inequality $H^{n-1}(\{f = 0\} \cap B(0,1)) \geq c$ holds for every harmonic function $f: B(0,1)\to {\mathbb R}$ such that $f(0)=0$ .

Short context: Harmonic functions are studied in many areas of mathematics, such as mathematical physics and the theory of stochastic processes. The Hausdorff measure $H^{n-1}$ is (up to a constant factor) just the usual $n-1$ -dimensional volume, e.g. $H^2$ is the area. The Theorem confirms a conjecture of Nadirashvili made in 1997. In particular, it implies that the zero set of any non-constant harmonic function $h:{\mathbb R}^3 \to {\mathbb R}$ has an infinite area Also, the Theorem is an important step towards proving a more general conjecture of Yau, which predicts a similar result on n-dimensional curved spaces (called “ $C^\infty$ -smooth Riemannian manifolds”) in place of ${\mathbb R}^n$ .

Links: Free arxiv version of the original paper is here, journal version is here.

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The family {x, xy, x+y} is Ramsey

You need to know: Set ${\mathbb N}$ of positive integers

Background: By finite colouring of ${\mathbb N}$ we mean partition ${\mathbb N}=C_1\cup\dots\cup C_r$ into a finite number r of disjoint subsets $C_i$ . We say that set $S \subset {\mathbb N}$ is monochromatic if $S \subset C_i$ for some i.

The Theorem: On 4th May 2016, Joel Moreira submitted to the Annals of Mathematics a paper in which he proved that for any finite colouring of ${\mathbb N}$ there exist (infinitely many) pairs $x,y \in {\mathbb N}$ such that the set $\{x, xy, x + y\}$ is monochromatic.

Short context: A set of k polynomials $P_1,\dots, P_k$ in s variables $x_1, \dots, x_s$ with integer coefficients is called a Ramsey family if for any finite colouring of ${\mathbb N}$ there exist $x_1, \dots, x_s \in {\mathbb N}$ such that the set $P_1(x_1, \dots, x_s),\dots, P_k(x_1, \dots, x_s)$ is monochromatic. A classical problem asks to develop necessary and sufficient conditions on the polynomials $P_1,\dots, P_k$ that guarantee that they form a Ramsey family. However, before 2016, it was not even known if a simple family $\{xy, x + y\}$ is Ramsey. The Theorem establishes this, even for a lager family $\{x, xy, x + y\}$ . In fact, the authors developed a general methodology and established that many other families are Ramsey as well. See here and here for related recent results.

Links: Free arxiv version of the original paper is here, journal version is here.

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The restriction conjecture for paraboloids is true for p>2(3n+1)/(3n-3), n>=2

You need to know: Notations: ${\mathbb C}$ for the set of complex numbers, $|z|$ for the absolute value of complex number z, $i = \sqrt{-1}$ , $e^{ix} = \cos(x)+i\sin(x)$ , ${\mathbb R}^n$ for n-dimensional Euclidean space, $B^{n-1}$ for the unit ball in ${\mathbb R}^{n-1}$ , $|w|^2=\sum\limits_{i=1}^{n-1}w_i^2$ for $w=(w_1, \dots, w_{n-1}) \in {\mathbb R}^{n-1}$ , $||f||_{L^p(\Omega)} = \left(\int_\Omega |f(\omega)|^p d\omega\right)^{1/p}$ , where $p\geq 1$ , $\Omega\subseteq {\mathbb R}^n$ , and $f:\Omega \to {\mathbb C}$ .

Background: The extension operator for the paraboloid is the operator E which puts to every function $f:B^{n-1} \to {\mathbb C}$ with $||f||_{L^p(B^{n-1})}<\infty$ into correspondence a function $E_f: {\mathbb R}^{n} \to {\mathbb C}$ given for every $x=(x_1, \dots, x_n)$ by $E_f(x) = \int_{B^{n-1}}e^{i(x_1w_1+\dots+x_{n-1}w_{n-1}+x_n|w|^2)}f(w)dw$ .

The Theorem: On 14th March 2016, Larry Guth submitted to arxiv a paper in which he proved the following result. Let $n\geq 2$ and let $p>2\frac{3n-1}{3n-3}$ if n odd and $p>2\frac{3n+2}{3n-2}$ if n is even. Then there exists a constant $C=C(p)$ such that $||E_f||_{L^p({\mathbb R}^n)} \leq C ||f||_{L^p(B^{n-1})}$ for every f.

Short context: In 1979, Stein conjectured that the conclusion of the Theorem holds for all $p>\frac{2n}{n-1}$ and $n\geq 2$ . This is known as the restriction conjecture for paraboloids, and is an important conjecture in harmonic analysis. The Theorem proves this conjecture for all $n\geq 2$ under assumption $p>2\frac{3n-1}{3n-3}$ (with slightly less restrictive assumption $p>2\frac{3n+2}{3n-2}$ if n is even).

Links: Free arxiv version of the original paper is here, journal version is here.

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If p>q>2, the maximal t for which the t-snowflake of L_q admits a bi-Lipschitz embedding into L_p is t=q/p

You need to know: Metric space.

Background: We say that a metric space $(X,\rho_X)$ admit a bi-Lipschitz embedding (or just “embeds” for short) into metric space $(Y,\rho_Y)$ if there exist constants $m,M>0$ and a function $f:X\to Y$ such that $m \rho_X(x,y) \leq \rho_Y(f(x),f(y)) \leq M \rho_X(x,y), \, \forall x,y \in X$ . For $\theta\in(0,1]$ , a $\theta$ -snowflake of metric space $(X,\rho_X)$ is the metric space on the same set X with distance $\rho(x,y)=(\rho_X(x,y))^\theta, \, \forall x,y \in X$ . By $L^p$ space we mean, for concreteness, space of functions $f:[0,1]\to{\mathbb R}$ with norm $||f||_p=\left(\int_0^1|f(x)|^pdx\right)^{1/p}<\infty$ .

The Theorem: On 13th January 2016, Assaf Naor submitted to arxiv a paper in which he proved that, for every $2<q<p$ , the maximal $\theta\in(0,1]$ for which the $\theta$ -snowflake of $L_q$ admits a bi-Lipschitz embedding into $L_p$ is equal to $\frac{q}{p}$ .

Short context: It is known that, if $2<q<p$ , then $L_q$ does not embed into $L_p$ . Hence, if $\theta^*=\theta^*(p,q)$ denotes the maximal $\theta\in(0,1]$ for which the $\theta$ -snowflake of $L_q$ embeds into $L_p$ , then $\theta^*<1$ . Quantifying by “how much” $\theta^*$ is bounded away from 1 gives an important quantitative refinement of non-embeddability $L_q$ into $L_p$ . In 2004, Mendel and Naor proved that $\frac{q}{p}\leq \theta^*$ . In a paper submitted in 2014, Naor and Schechtman proved that $\theta^* \leq 1-\frac{(p-q)(q-2)}{2p^3}$ and conjectured that $\theta^*=\frac{q}{p}$ . The Theorem confirms this conjecture.

Links: Free arxiv version of the original paper is here, journal version is here.

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If p>q>2, and the t-snowflake of L_q admits a bi-Lipschitz embedding into L_p, then t is bounded away from 1

You need to know: Metric space.

The Theorem: On 25th August 2014, Assaf Naor and Gideon Schechtman submitted to arxiv a paper in which they, among other results, proved that, for every $2<q<p$ , if $\theta\in(0,1]$ is such that the $\theta$ -snowflake of $L_q$ admits a bi-Lipschitz embedding into $L_p$ , then necessarily $\theta \leq 1-\frac{(p-q)(q-2)}{2p^3}$ .

Short context: It is known that, if $2<q<p$ , then $L_q$ does not embed into $L_p$ . Hence, if $\theta$ -snowflake of $L_q$ embeds into $L_p$ , then $\theta<1$ . Quantifying by “how much” $\theta$ is bounded away from 1 gives an important quantitative refinement of non-embeddability $L_q$ into $L_p$ . However, before 2014, no estimate in the form $\theta \leq 1-\delta(p,q)$ for any explicit function $\delta(p,q)$ has been known. The Theorem provides the first such estimate. In fact, the authors conjectured that the inequality in the Theorem can be improved to $\theta \leq \frac{q}{p}$ , which would be the best possible. In a later work, Naor proved this conjecture.

Links: Free arxiv version of the original paper is here, journal version is here.

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Tarski’s circle squaring problem can be solved with only measurable pieces

You need to know: Euclidean plane ${\mathbb R}^2$ , Lebesgue measurable subsets of ${\mathbb R}^2$ (that is, subsets for which the area is well-defined), notation $A+v$ for the translation of set $A\subset {\mathbb R}^2$ by a vector $v\in {\mathbb R}^2$ .

Background: We call two sets $A,B \subset {\mathbb R}^2$ equidecomposable by translations if there are partitions $A = A_1 \cup \dots \cup A_m$ and $B = B_1 \cup \dots \cup B_m$ , such that $B_i = A_i + v_i$ , $i=1,\dots,m$ , for some vectors $v_1, \dots, v_m \in {\mathbb R}^2$ . If, moreover, all $A_i$ (and thus $B_i$ ) are Lebesgue measurable, we say that A and B are equidecomposable by translations with Lebesgue measurable pieces.

The Theorem: On 25th January 2015, Łukasz Grabowski, András Máthé, and Oleg Pikhurko submitted to arxiv a paper in which they proved that a circle and a square of the same area on the plane are equidecomposable by translations with Lebesgue measurable pieces.

Short context: In 1990, Laczkovich, answering a 1925 question of Tarski, proved that circle and a square of the same area are equidecomposable by translations. The pieces $A_i$ and $B_i$ in Laczkovich’s proof are not Lebesgue measurable, and he left the the problem whether or not the circle can be squared with measurable pieces as an interesting open question. The Theorem resolves this question affirmatively. Of course, there is nothing special about circle and square, and the Theorem holds in all dimensions, and for any pair of bounded sets with non-empty interiors of the same Lebesgue measure subject to technical condition that the boundaries of the sets have dimensions lower that the sets.

Links: Free arxiv version of the original paper is here, journal version is here.

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The set of points in a rational polygon P not illuminated by any x in P is finite

You need to know: Polygon, angles measured in radians, angle of incidence, angle of reflection.

Background: Let P be a rational polygon, that is, polygon whose angles measured in radians are rational multiples of $\pi$ . For any point $x \in P$ (called a light source), consider a light ray starting from x and moving inside P, with the usual rule that the angle of incidence equals the angle of reflection. A point $x \in P$ is called illuminated if there is a light ray starting from x which passes through y.

The Theorem: On 10th July 2014, Samuel Lelievre, Thierry Monteil, and Barak Weiss
submitted to arxiv a paper in which they proved that in any rational polygon P, and for any point $x\in P$ there are at most finitely many points $y\in P$ which are not illuminated from light source x.

Short context: In 1969, Klee asked whether any point x in any polygon P illuminates the whole P. In 1995, Tokarsky answered this question negatively by constructing an example in which one point y is not illuminated. The Theorem states that, at least for rational polygons, only finitely many points can remain unilluminated. It is just one out of many corollaries of deep “Magic Wand Theorem” proved in 2013 by Eskin and Mirzakhani.

Links: Free arxiv version of the original paper is here, journal version is here.

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