Amir’s conjecture for random walks on groups is true

You need to know: Group, set of generators, finitely generated group, basic probability theory, independent identically distributed (i.i.d.) random variables.

Background: Let $G$ be an infinite group with finite generating set $S$ . Let $S^{-1}=\{g \in G\,|\,g^{-1}\in S\}$ , and let $|g|$ be the smallest integer $k\geq 0$ such that $g\in G$ has representation $g=\prod_{i=1}^k s_i$ with each $s_i \in S \cup S^{-1}$ . A symmetric finitely supported probability measure on $G$ is a function $\mu:G\to[0,1]$ such that set $\{g \in G\,|\,\mu(g)>0\}$ is finite, $\sum_{g\in G} \mu(g)=1$ , and $\mu(g^{-1})=\mu(g)$ for all $g \in G$ . Let $X_1, X_2, \dots$ be a sequence of i.i.d. random variables with distribution $\mu$ , that is, such that ${\mathbb P}[X_i=g]=\mu(g)$ for all $g\in G$ and all $i$ . A random walk on $G$ with step distribution $\mu$ is the sequence $W_1,W_2,\dots$ given by $W_n=\prod_{i=1}^n X_i$ . A speed function of random walk $W_n$ is $L_\mu(n) = {\mathbb E}[|W_n|] = \sum_{g\in G}|g|{\mathbb P}[W_n=g]$ . Its entropy is $H_\mu(n) = - \sum_{g\in G}{\mathbb P}[W_n=g]\log{\mathbb P}[W_n=g]$ . The exponent of a function $f$ is $\lim_{n\to\infty} \frac{\log f(n)}{\log n}$ , provided that the limit exists.

The Theorem: On 27th October 2015, Jérémie Brieussel and Tianyi Zheng submitted to arxiv a paper in which they proved that for any $\gamma\in[1/2,1]$ and $\theta\in[1/2,1]$ satisfying $\theta \leq \gamma \leq \frac{1}{2}(\theta+1)$ , there exist a finitely generated group $G$ and a symmetric probability measure $\mu$ of finite support on $G$ , such that the random walk on $G$ with step distribution $\mu$ has speed exponent $\gamma$ and entropy exponent $\theta$ .

Short context: The main research directions in studying random walks on groups are (i) given a group, establish properties of random walk on it, and, conversely (ii) given properties of a random walk, find if there exists a group with such random walk. The Theorem contributes to direction (ii). It confirms a conjecture of Amir who proved the same result for $\gamma,\theta\in[3/4,1]$ .

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

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There exist full-sized groups that are profinitely rigid in the absolute sense

You need to know: Group, finite and infinite group, abelian and non-abelian group, subgroup, presentation of a group in the form $<S|R>$ , where S is a set of generators and R is a set of relations.

Background: A subgroup N of group G is called normal, if $g\cdot a\cdot g^{-1}\in N$ for any $a \in N$ and $g \in G$ . A (left) coset of N in G with respect to $g\in G$ , denoted $gN$ , is the set $\{g \cdot a: \, a\in N\}$ . If N is normal, cosets form a group with operation $(gN)\cdot (hN)=(g \cdot h)N$ , which is called the quotient group and denoted $G/N$ . If $G/N$ is a finite group, we say that N has a finite index. A group G is called residually finite if the intersection of all its normal subgroups of finite index is an identity element. Let $c(G)$ be the set of all finite quotients of G.

A group G is called finitely generated if it has a presentation $<S|R>$ with finite number of generators. A group G is free if R is an empty set. A group G is called full-sized if it has a non-abelian free subgroup. A finitely generated, residually finite group G is called profinitely rigid in the absolute sense, if, for any finitely generated residually finite group H, $c(G)=c(H)$ (up to isomorphism) implies that H is isomorphic to G.

The Theorem: On 11th November 2018, Martin Bridson, David McReynolds, Alan Reid and Ryan Spitler submitted to arxiv a paper in which they provided the first examples of full-sized groups that are profinitely rigid in the absolute sense.

Short context: One of the fundamental questions in the theory of finitely generated groups is to what extend such groups are determined by their set of finite quotients. To have a hope to recover group G from its finite quotients, we must assume that G is residually finite. It is conjectured that a broad class of residually finite groups are recoverable, but, before 2018, this was known only for some special groups like abelian. In particular, no examples of full-sized groups with this property was known. The Theorem provides first such examples.

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

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There exist finitely generated coarsely non-amenable groups that are coarsely embeddable into l^2

You need to know: Symmetric difference $A \Delta B = (A \setminus B) \cup (B \setminus A)$ of sets $A$ and $B$ , notation $|A|$ for the number of elements in finite set $A$ , notation $\times$ for set product, set ${\mathbb N}$ of natural numbers, metric space $(X; d)$ where $X$ is a set and $d$ is a metric, infinite dimensional Hilbert space $l^2$ , group, generating set of a group, finitely generated group, word metric of a group with respect to a generating set.

Background: We say that metric space $(X; d_X)$ is coarsely embeddable into metric space $(Y; d_Y)$ is there exists a map $f:(X; d_X)\to(Y; d_Y )$ such that $d_Y(f(x_n); f(y_n)) \to \infty$ if and only if $d_X(x_n; y_n) \to \infty$ for all sequences $(x_n)_{n\in {\mathbb N}}$ and $(y_n)_{n\in {\mathbb N}}$ in $X$ . We say that a finitely generated group $G$ is coarsely embeddable into $(Y; d_Y )$ if it is so for the word metric with respect to a finite generating set $S$ . This property does not depend on the choice of $S$ .

A metric space $(X; d)$ is called uniformly discrete if there exists a constant $r>0$ such that, for any $x,y \in X$ , we have either $x=y$ or $d(x,y)>r$ . A uniformly discrete metric space $(X; d)$ is coarsely amenable if for every $\epsilon>0$ and $R>0$ , there exist a constant $S>0$ and a collection of finite subsets $\{A_x\}_{x \in X}$ , $A_x \subseteq X \times {\mathbb N}$ for every $x\in X$ , such that (a) $|A_x \Delta A_y|/|A_x \cap A_y|\leq \epsilon$ when $d(x,y) \leq R$ , and (b) $A_x \subseteq B(x,S) \times N$ . A finitely generated group $G$ is called coarsely amenable if it is so for the word metric with respect to a finite generating set $S$ , and is called coarsely non-amenable otherwise.

The Theorem: On 19th June 2014, Damian Osajda submitted to arxiv a paper in which he proved the existence of finitely generated coarsely non-amenable groups that are coarsely embeddable into the infinite dimensional Hilbert space $l^2$ .

Short context: The notion of coarse amenability is a week version of amenability, and has many equivalent formulations and applications. In particular, it is known that any finitely generated coarsely amenable groups is coarsely embeddable into $l^2$ . The question whether the converse is true (Are groups coarsely embeddable into $l^2$ coarsely amenable?) is a natural question raised by a number of researchers. The Theorem provides a negative answer.

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

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A universally measurable homomorphism between Polish groups is automatically continuous

You need to know: Groups and homomopshisms between them, topological spaces and homeomopshisms between them, continuous function on a topological space, probability measure on a topological space, metric space, complete metric space, dense subset of a metric space.

Background: A topological group G is a topological space that is also a group such that the group operations of product $G\times G \to G: (x,y) \to xy$ and taking inverses $G \to G: x \to x^{-1}$ are continuous. A topological space is called Polish space if it is homeomorphic to a complete metric space that has a countable dense subset. A Polish group is a topological group G that is also a Polish space. A Borel probability measure on a topological space is a probability measure that is defined on all open sets. A subset of a Polish group G is called universally measurable if it is measurable with respect to
every Borel probability measure on G. A homomorphism $\phi: G\to H$ between Polish groups G and H is called universally measurable if $\phi^{-1}(U)$ is a universally measurable set in G for every open set $U \subseteq H$ .

The Theorem: On 8th December 2018, Christian Rosendal submitted to the Forum of mathematics, Pi a paper in which he proved that every universally measurable homomorphism between Polish groups is automatically continuous.

Short context: The Theorem resolves a longstanding problem posed by Christensen in 1971. It can be viewed as a generalisation of old classical theorem stating that any Lebesgue measurable function $f:{\mathbb R}\to{\mathbb R}$ satisfying the functional equation $f(x+y)=f(x)+f(y), \, x,y \in {\mathbb R}$ must be continuous.

Links: The original paper is available here.

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Any finite subgroup of GL_n(C) has a semi-invariant of degree at most Cn

You need to know: Set ${\mathbb C}$ of complex numbers, set ${\mathbb C}^n$ of vectors $z=(z_1,\dots,z_n)$ with complex entries, homoheneous polynomial in n variables, degree of a polynomial, group, subgroup, matrix, determinant of a matrix, group $\text{GL}_n({\mathbb C})$ (group of $n\times n$ matrices with complex entries and non-zero determinant).

Background: Let $H_{n}({\mathbb C})$ be the set of homogeneous polynomials in n complex variables with complex coefficients. Let G be a finite subgroup of $\text{GL}_n({\mathbb C})$ . A polynomial $P \in H_{n}({\mathbb C})$ is called semi-invariant of G if for every matrix $A\in G$ we have $P(Az)=B_A P(z),\, \forall z \in {\mathbb C}^n$ , where $B_A\in {\mathbb C}$ is a constant depending on A. Let $d(G)$ be the minimum integer k such that G has a semi-invariant of degree k.

The Theorem: On 8th January 2016, Pham Huu Tiep submitted to Forum of Mathematics, Pi a paper in which he proved the existence of constant $C<\infty$ such that $d(G) \leq Cn$ for every finite subgroup G of $\text{GL}_n({\mathbb C})$ . In fact, $C=1,184,036$ works.

Short context: In 1981, Thompson proved that, if $n>1$ is any integer and G is any finite subgroup of $\text{GL}_n({\mathbb C})$ , then $d(G)\leq 4n^2$ . He further conjectured that this quadratic upper bound can be improved to a linear one. The Theorem confirms this conjecture.

Links: The original paper is available here.

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The Grothendieck constant is strictly smaller than Krivine’s bound

You need to know: Euclidean space ${\mathbb R}^n$ , inner product $\langle x,y\rangle=\sum\limits_{i=1}^n x_i y_i$ in ${\mathbb R}^n$ , unit vector in ${\mathbb R}^n$ , matrix, big O notation.

Background: In 1953, Grothendieck proved the existence of a universal constant $K<\infty$ such that, given any $m\times n$ matrix with real entries $a_{ij}$ and arbitrary unit vectors $x_1,\dots, x_m,y_1,\dots, y_n$ in ${\mathbb R}^{m+n}$ , there are signs $\epsilon_1,\dots,\epsilon_m,\delta_1,\dots,\delta_n \in \{-1, +1\}$ such that inequality $\sum\limits_{i=1}^m \sum\limits_{j=1}^n a_{ij}\langle x_i, y_j \rangle \leq K \sum\limits_{i=1}^m \sum\limits_{j=1}^n a_{ij}\epsilon_i \delta_j$ holds. This is called the Grothendieck’s inequality. The minimal constant K for which it holds is called the Grothendieck constant, and is denoted $K_G$ . See here for a version of the Grothendieck constant of a graph.

The Theorem: On 31st March 2011, Mark Braverman, Konstantin Makarychev, Yury Makarychev, and Assaf Naor submitted to arxiv a paper in which they proved the existence of a constant $\epsilon_0>0$ such that $K_G < \frac{\pi}{2\log(1+\sqrt{2})}-\epsilon_0$ .

Short context: The Grothendieck’s inequality, despite being innocent-looking at the first glance, turned out to be important in many areas of mathematics and applications, such as functional analysis, harmonic analysis, operator theory, quantum mechanics, and computer science. For many applications, it is important to have it with the best possible constant. However, despite substantial efforts, even the second digit of $K_G$ remains unknown. The best lower bound is $K_G\geq 1.67696...$ proved by Davie in 1984. For the upper bound, Krivine proved in 1979 that $K_G \leq \frac{\pi}{2\log(1+\sqrt{2})}=1.7822...$ , and conjectured that $K_G$ is actually equal to this value. This conjecture was well-believed, and many researchers focused on proving the matching lower bound. The Theorem states that Krivine’s conjecture is false.

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

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The group of boundary fixing homeomorphisms of the disc is not left-orderable

You need to know: Euclidean space ${\mathbb R}^2$ , norm $|x|=\sqrt{x_1^2+x_2^2}$ of $x=(x_1,x_2)\in {\mathbb R}^2$ , continuous functions on ${\mathbb R}^2$ , bijection (function which is one-to-one and onto), inverse function, group.

Background: Let $B=\{x\in {\mathbb R}^2 : |x|\leq 1\}$ and $S^1=\{x\in {\mathbb R}^2 : |x|=1\}$ be the unit disk and the unit circle in ${\mathbb R}^2$ , respectively. A function $f:B\to B$ is called a homeomorphism if (i) f is a bijection, (ii) f is continuous, and (iii) the inverse function $f^{-1}$ is continuous. We say that function $f:B\to B$ is boundary fixing if $f(x)=x$ for all $x\in S^1$ . We say that function $h:B\to B$ is the composition of functions $f:B\to B$ and $g:B\to B$ , if $h(x)=f(g(x))$ for all $x\in B$ . It is known that the set of all boundary fixing homeomorphisms $f:B\to B$ with composition operation forms a group which we denote $\text{Homeo}(B,S^1)$ . In general, group G with operation $\cdot$ is called left orderable if there exists a total order $\leq$ on G such that $a\leq b$ implies $c\cdot a \leq c\cdot b$ for all $a,b,c \in G$ .

The Theorem: On 30th October 2018, James Hyde submitted to arxiv a paper in which he proved that the group $\text{Homeo}(B,S^1)$ of boundary fixing homeomorphisms of B is not left-orderable.

Short context: The question whether the group $\text{Homeo}(B,S^1)$ is left-orderable has been asked by many researchers and has been included in several collections of the open problems in group theory. The Theorem resolves this question affirnatively.

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

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There is a one-relator inverse monoid with undecidable word problem

You need to know: Group, presentation of a group in the form $<S|R>$ , where S is a set of generators and R is a set of relations.

Background: Let S be a set and let $\cdot$ be a binary operation $S \times S \to S$ . We say that pair $(S, \cdot)$ form a monoid if (i) $(a\cdot b) \cdot c = a \cdot (b \cdot c)$ holds for all $a,b,c \in S$ , and (ii) there exists an element e in S, called the identity element, such that $e\cdot a = a \cdot e = a$ holds for all $a\in S$ . A monoid is called inverse monoid, if for every $x\in S$ there exists $y\in S$ (called inverse of x) such that $x=x\cdot y \cdot x$ and $y=y\cdot x\cdot y$ . Note that every group if an inverse monoid with extra property $x\cdot y = y \cdot x = e$ whenever y is an inverse of x.

A group G is called finitely presented if it has a presentation $<S|R>$ with finite number of generators and finite set of relations. If, moreover, set R consists on just one relation, group G is called one-relator. A word in group G is any product of generators and their inverse elements. The word problem in G is the problem to decide whether a given word represents an identity element. Same definitions extends to inverse monoids and (with obvious modifications) to monoids.

The Theorem: On 18th September 2018, Robert Gray submitted to Inventiones Mathematicae a paper in which he proved the existence of a one-relator inverse monoid with undecidable word problem.

Short context: In the middle of 20th century, Markov and Post independently proved that there are finitely presented monoids for which the word problem is undecidable. Later, Novikov proved that the same is true for groups. This raises the question for which special classes of monoids and groups the word problem is decidable. In particular, it follows from results in 1932 paper of Magnus that the one-relator groups have decidable word problem. Can this result be extended to a (harder) word problem for inverse monoids? The Theorem gives a negative answer to this question.

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

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The probabilistic Waring problem for finite simple groups has positive solution

You need to know: Group, identity element $1$ , finite group, abelian and non-abelian groups, simple group, free group, notation $|S|$ for the number of elements in finite set S.

Background: Let $F_k$ be the free group with k generators $x_1, \dots x_k$ . Any element $w\in F_k$ (which we call word) can be written as $w=\prod\limits_{j=1}^r x_{i_j}^{\epsilon_j}$ for some $1 \leq i_1 < \dots < i_r \leq k$ and $\epsilon_j = \pm 1$ , $j=1,\dots,r$ . For every group G, w induces a word map from $w:G^k \to G$ given by $w(g_1, \dots, g_k)=\prod\limits_{j=1}^r g_{i_j}^{\epsilon_j}$ . If G is a finite group and $g\in G$ , let $N_G^w(g)$ be the number of $(g_1, \dots, g_k)\in G^k$ such that $w(g_1, \dots, g_k)=g$ . Two words $w_1, w_2$ are said to be disjoint if they are words in disjoint sets of variables. For a function $f: G\to{\mathbb R}$ , we write $||f||_{L^1} = \sum\limits_{g \in G}|f(g)|$ .

The Theorem: On 14th August 2018, Michael Larsen, Aner Shalev, and Pham Tiep
submitted to the Annals of Mathematics a paper in which they proved the following result. Let $w_1, w_2 \neq 1$ be disjoint words and let $w = w_1w_2$ . Then $\lim\limits_{|G|\to\infty}\left\|\frac{N_G^w(g)}{|G|^k}-\frac{1}{|G|}\right\|_{L^1}=0$ , where G ranges over all finite non-abelian simple groups.

Short context: In an earlier work, Larsen, Shalev, and Tiep proved that, for every word w as in the Theorem, and for every sufficiently large finite non-abelian simple group G, we have $N_G^w(g)>0$ for every $g\in G$ . In the language of probability theory, this means that if we select $g_1, \dots, g_k$ from G at random, and compute $w(g_1, \dots, g_k)$ , we can get any element $g\in G$ in this way with non-zero probability. In 2013, Shalev further asked whether $w(g_1, \dots, g_k)$ has almost uniform distribution in G with respect to the $L^1$ norm. This question became known as the probabilistic Waring problem for finite simple groups. The Theorem answers this question affirmatively.

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

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There exist finitely generated infinite simple left orderable groups

You need to know: Group, finite and infinite group, simple group, total order $\leq$ on a set.

Background: A generating set of a group G is a subset $S \subset G$ such that every $g\in G$ can be written as a finite product of elements of S and their inverses. Group G is called finitely generated if it has a finite generating set S. A group G with group operation + is called left orderable if there exists a total order $\leq$ on G such that $a\leq b$ implies $c+a \leq c+b$ for all $a,b,c \in G$ .

The Theorem: On 17th July 2018, James Hyde and Yash Lodha submitted to arxiv a paper in which they proved the existence of finitely generated infinite simple left orderable groups.

Short context: The Theorem answered a long-standing open question posed by Rhemtulla in 1980. In fact, the authors proved the existence of continuum many isomorphism types of such groups. The same is true for right orderable groups (groups G with a total order $\leq$ such that $a\leq b$ implies $a+c \leq b+c$ for all $a,b,c \in G$ ).

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

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