text:
\section{Introduction}

	In the last few years, local stability, as developed by Hrushovski and Pillay \cite{HP94}, has
	found several applications \cite{TW19, CPT20, gC21,
	MPPW21} to questions on asymptotic
	behaviour in additive combinatorics. Recall that a  subset $A$ of a
	group
	$G$ is
	$k$-\emph{stable} if its Cayley graph
	 \[ \mathrm{Cay}(G,A)=\{ (g,h) \in G\times G \ | \
	g^{-1} \cdot h \in A\}\] induces no half-graph of height $k$, that is,
	there is no sequence $(a_1,b_1,\ldots,a_k,b_k)$ in $G^{2k}$ such that the
	pair $a_i\inv\cdot b_j$ belongs to $A$ if and only if $i\le j$.

	Although the notion of stability has played and continues to play a crucial role in
	the development of model theory, from a combinatorial perspective it is
	exceptionally sensitive: the slightest perturbation of the set has the
	potential to destroy it (by adding a single instance of a half-graph).
	Motivated by this, we introduce a relaxation of stability which takes such
	perturbations into account. A relation $S$ in two finite
	sets $X$ and $Y$ is \emph{robustly $k$-stable} if there are few (in the
	sense of the counting measure associated to $X\times Y$) half-graphs of
	height $k$ induced by $S$ (see Definition \ref{D:robust}).  Similar
	attempts to quantify stability have appeared in recent work of
	Coregliano and Malliaris \cite[Section 3]{CM22}, as well as in work
	of Terry and the third author \cite[Section 5]{TW21}.

	Our purpose for this work is twofold: on the one hand, we aim to develop the notion of robust stability from a model-theoretic point of view. On the other hand, we hope to further cement the relevance of stability to classical questions in arithmetic
	combinatorics (both in the abelian and the non-commutative case). The paper thus focuses on the existence of certain
	two-dimensional shapes, such as corners, squares, and $L$-shapes, in dense subsets of groups.

    Given an abelian group $G$, a (non-trivial) \emph{corner} in a subset
    $S\subseteq G\times G$ is a triple $(x, y, d)$ in $G^3$, with $d\neq 0_G$,
    such that all three pairs $(x,y)$, $(x+d, y)$, and $(x,y+d)$
lie in $S$. It is well-known that the existence of
non-trivial corners for dense subsets in $G\times G$ implies Roth's theorem
\cite{kR53} on
$3$-term arithmetic progressions in dense subsets $A$ of $G$. Indeed, a non-trivial
corner $(x, y, d)$ in
$\mathrm{Cay}(G,A)$ gives rise, by a simple projection, to the $3$-arithmetic progression $x-y-d, x-y,
x-y+d$ in $A$.

 The existence of corners in subsets $S$ of $G\times G$ of positive
density  follows straight from the multidimensional
Szemer\'edi theorem. The latter was first proved qualitatively in
\cite{FK78}, and quantitative versions (albeit with poor bounds)
follow from the hypergraph regularity lemmas of Gowers
\cite{G07} and Nagle, R\"odl, Schacht and Skokan
\cite{NRS06,RS04}. However, finding corners is not quite as difficult as
resolving the full multidimensional Szemer\'edi theorem, and indeed
several direct attempts have been made at a strongly quantitative resolution \cite{ASz74,S05,S06,G05,LM07}.

	In the case of a non-abelian group $G$, we distinguish between
so-called
\emph{naive corners}, that is, configurations of the form
$(x, y), (gx, y), (x, gy)$ with $g\neq 1_G$, and \emph{BMZ
	corners} (for Bergelson, McCutcheon and Zhang \cite{BMZ97}), which are of
	the
form $(x, y),
(gx, y), (gx, gy)$ with $g\neq 1_G$. The latter configuration can also be written
as $(x, y),
(xg, y), (x, gy)$ with $g\neq 1_G$. In abelian groups, naive and BMZ
corners are easily seen to be equivalent. Some abelian techniques can be adapted to
the non-abelian context for BMZ corners, but do not seem to apply to
naive corners \cite{BMZ97}, and an extensive literature now exists \cite{S13,A16,BRZK17}.

Various other 2-dimensional shapes have been studied for abelian groups, such as \emph{squares} (configurations of the form $(x, y, d)$ in $G^3$, with $d\neq 0_G$ and all of $(x,y)$, $(x+d, y)$, $(x,y+d)$, $(x+d,y+d)$ in $S$),  and
$L$-shapes (where $(x,y)$, $(x+d, y)$, $(x,y+d)$, $(x,y+2d)$ in $S$).
In a recent preprint, Peluse
\cite{sP22} obtained the first reasonable bound for the existence of
$L$-shapes in dense subsets of $\mathbb{F}_p^n\times \mathbb{F}_p^n$.

In the presence of robust stability, that is, when
few half-graphs of height $k$ are induced by the subset $S$ of $G\times G$, we are able to
obtain qualitative asymptotic bounds on the density of $S$ that guarantees the existence of a square in $S$. In fact, our methods work in both abelian and non-abelian finite groups
and produce a subgroup of bounded index almost all of whose elements
appear as the side length of many squares.

Concretely, we will prove the following result (see Corollary
\ref{C:corners_bddexp}).

\begin{teoA}
	Given integers $k, r\ge 1$ and real numbers $\delta, \epsilon>0$ there exists
	an integer
	$N=N(k, r, \delta, \epsilon)\ge 1$ and a real number $\theta=\theta(k, r,
	\delta, \epsilon)>0$ with the following property. Given a finite
	group $G$ of
	exponent at most $r$ as well as subsets
	$X\subseteq G^n$ and $Y\subseteq G^m$ for some $n,m\geq 1$,
	let $S\subseteq
	X\times Y$ be a relation such that
\begin{itemize}
\item the size of $S$ is $|S|\ge \delta |G|^{n+m}$, and
\item the collection
$\HH_k(S)$ of all
half-graphs of height $k$ induced by $S$ on  $G$ has size $|\HH_k(S)|\le
\theta  |G|^{k(n+m)}$.
	\end{itemize} Then there exists a subgroup $H$ of $G$ of index at most
	$N$ with 	\[  |\left\{  g \in H \ |\ \exists \, (a,
	b)\in S \text{ s.t } (a, b\cdot g), (a\cdot g, b), (a\cdot g,
	b\cdot g) \in S \right\} | > (1-\epsilon) |H|.\]
\end{teoA}
    We obtain similar results for the existence of $L$-shapes (see Theorem
    \ref{T:Lshape} and Corollary \ref{C:Lshape}) as well as for the existence
    of arbitrarily long arithmetic progressions (see Corollary
    \ref{C:AP_robust_bddexp}). The techniques used in this work are of
   a model-theoretic flavor, and rely heavily on a stationarity result for
    robustly stable relations (see Theorem \ref{T:robuststable_main}). For
    reasons we will comment on towards the end of the paper, they do not extend
    to more sophisticated 2-dimensional shapes such as $3 \times 3$ grids.

    In forthcoming
	work, we establish some of the results presented here in the language
	of additive combinatorics. While the techniques used will be different, the
	structure of the proofs will follow a similar pattern. The present paper should therefore be viewed as an effort to highlight the close interactions and foster
	stronger synergies between model theory and additive combinatorics.

	In order to render the presentation of this first work more accessible to
	an audience who may not be versed with the language of model theory, we
	have adapted the results using some of the (perhaps more familiar) machinery of ultraproducts,
	with the aim of keeping the presentation as self-contained as possible. This choice
	should not present an obstacle for the model-theoretic reader, who will easily translate the terminology and techniques to the more general setting.

	\section{Non-standard finite groups and Loeb
	measures}\label{S:measure}

	Recall that a non-principal ultrafilter $\Ult$ on $\N$ is a
	non-empty collection of infinite sets closed under finite
	intersections and
	with the property that either a subset of $\N$ or its complement
	belongs to
	$\Ult$.  Such ultrafilters exist and each one induces a finitely
	additive probability measure
	on all subsets of $\N$ with only possible values $0$ and $1$ such
	that no
	finite subset has measure $1$.

	\begin{definition}\label{D:pseudofte}A \emph{non-standard finite
	group} has as underlying set $\prod_{n\to \Ult} G_n$ for  some
		collection $(G_n)_{n\in \N}$ of finite groups of strictly
		increasing size (that is $|G_n|<|G_{n+1}|$ for all $n$ in $\N$)
		and some non-principal
		ultrafilter  $\Ult$ on $\N$,  where the set
		$\prod_{n\to \Ult} G_n$ consists of the infinite Cartesian
		product  $\prod_{n\in \N} G_n$ modulo the equivalence relation
		\[ (g_n)_{n\in\N}
		\sim_\Ult (h_n)_{n\in\N}\ \Longleftrightarrow \{n\in \N \ | \
		g_n=h_n\} \in
		\Ult.\] That is, we identify two sequences if they are equal
		$\Ult$-almost everywhere (with respect to the measure
		induced by $\Ult$ on all subsets of $\N$).
	\end{definition}
	If we denote the class of $(g_n)_{n\in\N}$ modulo $\sim_\Ult$ by
	$[(g_n)_{n\in\N}]_\Ult$, it follows immediately from the definition
	of the
	non-principal ultrafilter $\Ult$ that the quotient $\prod_{n\to
	\Ult} G_n$ is a
	group, where \[
	[(g_n)_{n\in\N}]_\Ult \cdot [(h_n)_{n\in\N}]_\Ult = [(g_n\cdot
	h_n)_{n\in\N}]_\Ult,\] is well-defined and has neutral element
	$1_{\prod_{\Ult}
	G_n} =
	[(1_{G_n})_{n\in\N}]_\Ult$. Furthermore, the inverse
	$[(g_n)_{n\in\N}]_\Ult\inv=
	[(g_n\inv)_{n\in\N}]_\Ult$.

	Note that a finite Cartesian product of non-standard finite groups is
	again
	a non-standard finite group.

	\begin{definition}\label{D:internal}
		A subset $X$ of a non-standard finite group $\prod_{n\to \Ult}
		G_n$ is \emph{internal} if
		there exists a collection $(X_n)_{n\in\N}$ such that each $X_n$
		is a subset of
		$G_n$ and  \[X= \left\{ [(g_n)_{n\in\N}]_\Ult \in \prod_{n\to
		\Ult} G_n \ | \ g_n
		\text{ belongs to } X_n \text{ for $\Ult$-almost all $n$  in } \N
		\right\}.\]
		The internal set $X$ is \emph{defined over} the parameter
		subset $A$ of
		$G=\prod_{n\to\Ult} G_n$ if there is a finite tuple $\bar a$ in
		$A$ and an internal
		subset $Y$ of the Cartesian product \[G\times
		(\underbrace{G\times \cdots \times
			G}_{|\bar
			a|})\] such that for all  $g=[(g_n)_{n\in\N}]_\Ult$ in $G$, \[
			\text{$g$
			belongs to $X$ if and only if $(g, \bar a)$ belongs to $Y$.}\]
	\end{definition}
	Given an internal subset $X$ of $G=\prod_{n\to \Ult} G_n$
	induced from the
	collection $(X_n)_{n\in\N}$ as above, we will denote the subset
	$X_n$ of $G_n$
	by $X(G_n)$. Likewise, we will sometimes denote the internal set
	$X$ by $\prod_{n\to \Ult} X_n$.
	\begin{remark}\label{R:internal}~
		\begin{enumerate}[(a)]
			\item Every finite subset of a non-standard finite group is
			internal and defined with parameters.
			\item The collection of internal sets is closed under Boolean
			combinations and
			projections. Given a (group) word $w=w(u, v_1\ldots, v_n)$
			and a tuple $\bar
			a=(a_1,\ldots, a_n)$ in a non-standard finite group $G$, the set
			\[X=\left\{ g\in G \ | \
			w(g, \bar a)=1_G\right\} \]   is internal and defined over $\bar
			a$.
			\item Given internal subsets $X_1,\ldots,X_n$ of a
			non-standard finite group $G$, the set $X_1\times\dots\times
			X_n$ is an internal subset of the $n$th Cartesian product
			$G^n=G\times\dots\times G$. It is defined over the same set of
			parameters as $X_1,\ldots,X_n$.
			\item Given an internal set $X$ of $G$, the subset
			\[
			\left\{ (x,y) \in G\times G \ | \ x\cdot y\in X \right\}
			\]
			is an internal subset of $G\times G$ defined over the same
			parameters. In particular, every translate (left or right) of an
			internal set is internal.
			\item In this particular set-up, \L o\'s's Theorem becomes
			tautologically immediate: given an internal set $X$ in a
			non-standard finite group $G=
			\prod_{n\to\Ult} G_n$,  we have that \[ X\ne \emptyset \
			\Longleftrightarrow \
			X(G_n) \ne \emptyset \text{ for $\Ult$-almost all $n$  in } \N.
			\]
		\end{enumerate}
	\end{remark}

	Every non-standard finite group arising as a limit of finite groups
 has
cardinality continuum
	and so it has continuum many internal sets. However, whenever we
	restrict our attention to only countably many internal sets
	non-standard finite groups have the following remarkable property,
	which in model-theoretic terms is called {\em
	$\aleph_1$-saturation}.

	\begin{fact}[$\aleph_1$-saturation]\label{F:Saturation}
		Let $G$ be a non-standard finite group. Every internal countable
		cover of an internal set admits a finite sub-covering: given an
		internal subset $X$ of $G$ and a countable family $(Y_n)_{n\in
		\N}$ of internal subsets of $G$ such that $X=\bigcup_{n\in\N}
		Y_n$, there is some natural number $k$ such that
		\[
		X=Y_{n_1} \cup \dots \cup Y_{n_k}.
		\]
		Equivalently, a countable intersection of internal set is
		non-empty whenever every finite sub-intersection is.
	\end{fact}
	In particular, every infinite internal subset of a
	non-standard
	finite group must be uncountable.

	Every non-standard finite group becomes equipped with a finitely
	additively probability measure, on the Boolean algebra of internal
	sets, induced by the normalised counting measure on every finite
	group.
	\begin{definition}\label{D:Measure}
		The {\em Loeb} or {\em non-standard counting measure} of the
		non-standard finite group $G=\prod_{n\to\Ult} G_n$ is defined
		for every internal subset $X$ of $G$ as
		\[
		\mu_G (X) = \lim_{n\to \Ult} \frac{|X(G_n)|}{|G_n|}.
		\]
		In an abuse of notation, we call the value $\mu_G(X)$ the {\em
		density of $X$}.
	\end{definition}
	Note that for every real number $r$ in the interval $[0,1]$,
	the measure $\mu_G(X)\ge r$ whenever  $|X(G_n)|\ge r|G_n|$
	for
	$\Ult$-almost all $n$ in $\N$.

	\begin{remark}\label{R:fibreDef}
		Given an internal subset $Z$ of  the non-standard finite group
		$G^{k+m}$ and a real number $r$ in $[0,1]$, there is an internal
		subset $Y$ defined over the same set of parameters as $Z$ such
		that
		\[
		\left\{ y \in G^m \ | \ \mu_{G^k} (Z_y) > r  \right\} \subseteq Y
		\subseteq \left\{ y \in G^m \ | \ \mu_{G^k} (Z_y)\ge r  \right\},
		\]
		where $Z_y=\{x )\in G^k \ | \ (x,y)\in Z\}$ denotes the
		fibre of $Z$ over $y$. Namely, set
		\[
		Y = \prod_{n\to \Ult} \left\{ y\in G_n^m \ | \ |Z_y(G_n^k)| \ge r
		|G_n^k| \right\}.
		\]
	\end{remark}

	We include the following easy result to emphasise the relevance of
	using non-standard finite objects in order to conclude the existence
	of non-quantitative asymptotic bounds.

	\begin{lemma}\label{L:Transfer}
		Consider a family $(G_n,X_n)_{n\in\N}$, where $X_n$ is a
		subset of the finite group $G_n$. The following are equivalent for
		every real number $r$ in $[0,1]$.
		\begin{enumerate}[(a)]
			\item For every $\epsilon>0$ there is some $n_0=n_0(r, \epsilon)$
			in $\N$  such that $|X_n| \le
			(r+\epsilon) |G_n|$ for all $n > n_0$.
			\item For every non-principal ultrafilter $\Ult$ on $\N$, we
			have that $\mu_G(X) \le r$ for the internal set
			$X=\prod_{n\to\Ult} X_n$ in the non-standard finite group
			$\prod_{n\to\Ult}G_n$.
		\end{enumerate}
	\end{lemma}
	\begin{proof}
		$(a)\Rightarrow(b):$ No finite set lies in a non-principal
		ultrafilter $\Ult$. Hence, by the way the non-standard counting
		measure on the non-standard finite group $G=\prod_{n\to\Ult}
		G_n$ has been defined, it follows that $\mu_G(X)\le
		r+\epsilon$ for every $\epsilon>0$, so $\mu_G(X)\le r$, as desired.

		\noindent $(b)\Rightarrow(a):$ Otherwise, negating quantifiers,
		there is some value $\epsilon>0$ such that for every natural number
		$n_0\ge 1$, we find some $n > n_0$ with  $|X_n| > (r+ \epsilon) \cdot
		|G_n|$. Hence the subset
		\[
		Q_{\epsilon} = \left\{ n\in \N \ | \  |X_n| >
		(r+\epsilon)\cdot |G_n|  \right\}
		\]
		is infinite. Choose a non-principal ultrafilter $\Ult$ containing
		each $Q_\epsilon$ (Note that there exist such ultrafilters!). By
		definition of the non-standard counting measure
		on the non-standard finite group $G=\prod_{n\to\Ult} G_n$ we
		deduce
		\[
		\mu_G(X) = \lim_{n\to\Ult} \frac{|X(G_n)|}{|G_n|} \ge
		\lim_{n\to\Ult} r+\epsilon = r +\epsilon.
		\]
	 Thus $\mu_G(X)> r$,
		which yields the desired contradiction.
	\end{proof}
\begin{remark}\label{R:Transfer}
Taking set-theoretic complements, it follows from Lemma \ref{L:Transfer} that
the following two conditions are equivalent.
\begin{enumerate}[(a)]
	\item For every $\epsilon>0$ there is some $n_0=n_0(r, \epsilon)$
in $\N$  such that $|X_n| \ge
(r-\epsilon)|G_n|$ for all $n > n_0$.
\item For every non-principal ultrafilter $\Ult$ on $\N$, we
have that $\mu_G(X) \ge r$ for the internal set
$X=\prod_{n\to\Ult} X_n$ in the non-standard finite group
$\prod_{n\to\Ult}G_n$.
\end{enumerate}
\end{remark}

	\begin{remark}\textup{(c.f. }\cite[Section
	2]{BT14}\textup{)}\label{R:Fubini}
		Carath\'eodory's  Extension Criterion is satisfied by the Loeb
		measure by $\aleph_1$-saturation (Fact \ref{F:Saturation}).
		Therefore, the Loeb measure $\mu_G$ extends to a
		unique $\sigma$-additive measure on the $\sigma$-algebra
		generated by the internal subsets
		of  $G$. Abusing notation, we will not distinguish between
		$\mu_G$ and its unique extension.

		Furthermore, the family of measures $\{\mu_{G^k}\}_{k\ge 1}$
		satisfies the Fubini-Tonelli Theorem \cite[Theorem 19]{BT14}, meaning that for any
 internal subset $Z$ of $G^{n+m}$ the following holds.
		\begin{itemize}
			\item The function $y\mapsto \mu_{G^n}(Z_y)$, resp.
			$x\mapsto \mu_{G^m}(Z_x)$,  is $\mu_{G^m}$-measurable,
			resp. $\mu_{G^n}$-measurable.
			\item We have the equality
			\[
			\mu_{G^{n+m}}(Z) = \int_{G^n} \mu_{G^m}(Z_x) \,
			\mathrm{d}\mu_{G^n} =\int_{G^m} \mu_{G^n}(Z_y) \,
			\mathrm{d}\mu_{G^m}.
			\]
		\end{itemize}

	\end{remark}

	\section{Tame families and dense subsets}\label{S:tame}

	In view of $\aleph_1$-saturation (Fact \ref{F:Saturation}), we
	introduce the following notion,
	which appears in \cite{BB21} as $\bigwedge$-internal and in
	\cite{lvD15} as $\Pi$-definable. For readers familiar with model
	theory this corresponds to type-definable sets over a  countable
	parameter set in a countable language.

	\begin{definition}\label{D:InfInt}
		A subset $X$ of $G$ is {\em $\omega$-internal} if it is a countable
		intersection of internal subsets of $G$. If each one of the internal
		sets is defined over the parameter set $A$, then we say that $X$
		is {\em defined over $A$}.
	\end{definition}

	\begin{remark}\label{R:InfInt}~
		\begin{enumerate}[(a)]
			\item Internal sets are $\omega$-internal.
			\item Every countable decreasing chain $(Y_n)_{n\in \N}$ of
			infinite internal subsets of a non-standard finite group $G$
			yields a non-empty $\omega$-internal set. By
			$\aleph_1$-saturation (Fact \ref{F:Saturation}), this
			$\omega$-internal set is uncountable.
		\end{enumerate}
	\end{remark}

	\begin{example}
		Let $G$ be the non-standard finite group $\prod_{n\to \Ult}
		\Z/2^n\Z$, for some non-principal ultrafilter $\Ult$. For every
		natural number $k\ge 1$, the function $f_k:x\mapsto 2^k\cdot
		x$ from $G$ to $G$ is an internal function, that is, its graph is an
		internal set. Each internal subgroup $\mathrm{Im}(f_k)$ has
		finite index in $G$ and they form a strictly decreasing chain,
		which yields an $\omega$-internal uncountable subgroup of $G$
		which is $2$-divisible.
	\end{example}

	In view of Fact \ref{F:Saturation} we will need to restrict our
	attention to suitable countable families of internal sets.
	Model-theoretically this is done by fixing a suitable countable
	first-order language and considering the corresponding definable
	sets, which are
	always internal, see for example \cite{eH12,lvD15,dP20,BB21}.
	For the sake of the presentation, we will avoid introducing the
	terminology of first-order formulae and provide an approach
	tailored to our purposes.

	\begin{definition}\label{D:Tame}
		Let $G=\prod_{n\to \Ult} G_n$ be a non-standard finite group. A
		family $\mathcal F$ of internal sets in all finite Cartesian
		products of $G$ is {\em tame} if it satisfies the following
		conditions:
		\begin{enumerate}[(I)]
			\item Every internal set of the form
			\[
			X_w=\left\{ (g_1,\ldots,g_n)\in G^n \ | \
			w(g_1,\ldots,g_n)=1_G\right\}, \]
			where $w=w(u_1,\ldots,u_n)$ is a (group) word, belongs to
			$\mathcal F$.
			\item It is closed under finite Cartesian product, Boolean
			combinations, projections and permutations of the coordinates.
			\item Given an internal subset $Z$ of $G^n$ in $\mathcal F$,
			the internal set
			\[
			\left\{ (x,y)\in G^n\times G^n \ | \ x\cdot y\in Z \right\}
			\]
			also belongs to $\mathcal F$.
			\item Given an internal subset $Z$ of $G^{k+m}$ in $\mathcal
			F$ and a rational number $q$ in $[0,1]$, there is an internal
			set $Y_q$ in $\mathcal F$ (defined over the same parameters
			as
			$Z$) such that
			\[
			\left\{ y \in G^m \ | \ \mu_{G^k} (Z_y) > q  \right\} \subseteq
			Y _q \subseteq \left\{ y \in G^m \ | \ \mu_{G^k} (Z_y)\ge q
			\right\},
			\]
			as in Remark \ref{R:fibreDef}. In particular, for every internal
			subset $Z$ of $G^{k+m}$ in $\mathcal F$, the subset  \[
			\left\{ y \in G^m \ | \ \mu_{G^k} (Z_y) =0
			\right\}=\bigcap\limits_{n\ge 1} \left(G^m\setminus
			Y_{\frac{1}{n}} \right)\] is $\omega$-internal and given  by a
			countable intersection of internal subsets in $\mathcal F$.
		\end{enumerate}
	\end{definition}

	\begin{remark}\label{R:tame_vs_language}
		The readers familiar with the notion of first-order formulae in a
		given language $\LL$ containing the language of groups will
		notice that every such language $\LL$ induces a tame family of
		internal
		sets, setting \[X_\psi=\{ (a_1,\ldots, a_n) \in G^n \ |  \
		\psi(a_1,\ldots, a_n) \text{ holds in the $\LL$-structure } G \} \]
		for every $\LL$-formula $\psi(x_1,\ldots, x_n)$.

		Similarly, every tame family $\mathcal F$ induces a first-order
		language $\LL$ containing the language of groups after adding,
		for every internal subset $X$ of $G^n$ in $\mathcal F$ a
		distinguished predicate $R_X$ such that its interpretation in
		$G$ equals $X$.

		With this translation in mind, we see that
		certain internal sets are automatically in a distinguished tame
		family $\mathcal F$, without having to write it explicitly as a
		Boolean combination of suitable projections. For example, if the
		subset $S$ of $G\times G$ is in $\mathcal F$ (and thus
		\emph{definable}), then so is the set \[ \{ (a,b) \in G^2 \ | \text{
		for some $g$ in $G\setminus\{1_G\}$ the pairs }  (a,b), (g\cdot a,
		b), (a, g\cdot b) \text{ all lie in $S$}   \}.\]
	\end{remark}

	To obtain an example of a countable tame family of
	internal sets it suffices to close
	the family of internal sets given by group words under countably many instances of properties II)-IV). More generally, the following holds.

	\begin{lemma}\label{L:TameNum}
		Given any countable family $\mathcal F_0$ of internal sets, there is
		a countable tame family $\mathcal F$ of internal sets extending
		$\mathcal F_0$.
	\end{lemma}

	\begin{remark}
		By Lemma \ref{L:TameNum}, whenever we want to apply
		Lemma \ref{L:Transfer} to a distinguished internal set $X$ in a
		nonstandard finite group, we may always assume that $X$ belongs to a
		countable tame family.
	\end{remark}

	If the tame family $\mathcal F$ arises as the family of definable
	sets with respect to a fix language $\LL$ as explained in Remark
	\ref{R:tame_vs_language}, the following notion of richness
	corresponds to the classical model-theoretic notion of an elementary
	substructure in that particular language.

	\begin{definition}\label{D:Rich}
		A subset $M$ of a non-standard finite group  $G$ is {\em rich with
		respect to the tame family} $\Fc$ if
		every non-empty fibre $X_{a}$ of $G^n$, where $a$ is tuple in
		$M$ and $X$ is an internal subset of $G^{n+|a|}$ in $\Fc$,
		contains an $n$-tuple whose coordinates all lie in $M$, {\it i.e.}
		$X_a(M)=X_a\cap M^n\neq \emptyset$.
	\end{definition}

	A straightforward chain argument, mimicking the proof of
	Downward
	L\"owenheim-Skolem for first-order languages, yields the following
	result.

	\begin{remark}[Downwards L\"owenheim-Skolem]\label{R:LS}
		Given a countable tame family $\Fc$ and a countable subset $A$ of  a
		non-standard finite group $G$,
		there is a countable subset $M$ of $G$ containing $A$ which is rich
		with respect to $\Fc$.
	\end{remark}

	Notice that every rich subset is in particular an infinite subgroup of
	$G$.
	However, a countable rich subset $M$ does no longer satisfy the
	$\aleph_1$-saturation condition for coverings of internal subsets,
	all defined
	over $M$: Indeed, the $M$-points of the internal set given by
	$x=x$
	can be covered by countably many singletons (running through
	every point in
	$M$), yet it does not admit a finite covering.
	\medskip

	\noindent \textbf{Henceforth, we fix a countable tame
	family $\Fc$ of internal sets. All internal and $\omega$-internal
	sets  we shall consider belong to this particular family. }
	\medskip

	Given a subset $A$ of a non-standard finite group $G$, we will
	denote by $\Fc(A)$ the collection of all fibres $Y_{a}$ of $G^{n}$,
	with $n$ running over all possible natural numbers, where $a$ is a
	tuple in $A$ and $Y$ is an internal subset of $G^{n+|a|}$ in
	$\Fc$.

	\begin{definition}\label{D:Type}
		Consider a non-standard finite group $G$ and a countable subset
		$A$ of $G$. Given a tuple $b$ of elements of $G$,
		its {\em type  over $A$} is the $\omega$-internal set
		\[
		\tp(b/A) = \bigcap_{\mathclap{\substack{b\in X  \\ X\in
		\Fc(A)}}}  X.
		\]
	\end{definition}
	Since the above set $\Fc(A)$ is countable, by Remark
	\ref{R:InfInt} (b) the type $\tp(b/A)$ is an uncountable
	$\omega$-internal set whenever $b$ does not lie in a finite fibre
	defined over $A$ of some internal set in $\Fc$. Note that $\tp(b'/A)
	= \tp(b/A)$ if and only if $b'$ belongs to $\tp(b/A)$.

	A classical Ramsey argument and $\aleph_1$-saturation allows us to
	produce, out of a given infinite sequence and a countable set of
	parameters, a new sequence with a remarkable property known as
	\emph{indiscernibility}.

	\begin{definition}
		We say that a sequence $(a_i)_{i\in \mathbb N}$ is
		\emph{indiscernible} over the countable subset $A$ (or
		\emph{$A$-indiscernible}) if for every natural number $n$ and
		every increasing enumeration $i_0<\dots<i_{n-1}$ we have that
		$\tp(a_0\ldots a_{n-1} / A )$ equals $\tp(a_{i_0}\ldots
		a_{i_{n-1}} / A )$,  that is, \[ (a_0,\ldots, a_{n-1}) \in X \
		\Leftrightarrow \ (a_{i_0},\ldots, a_{i_{n-1}}) \in X\] for every
		internal subset $X$ in $\Fc(A)$ of the appropriate arity.
	\end{definition}

	\begin{definition}\label{D:Dense}
		Let $G$ be a non-standard finite group.
		\begin{itemize}
			\item
			An $\omega$-internal subset is {\em dense} if it is not contained
			in any internal set of density $0$ (see Definition
			\ref{D:Measure}).
			\item
			Given a countable subset $A$ of $G$ and a tuple $b$ of
			elements of $G$, we say that $b$ is {\em dense over $A$} if
			$\tp(b/A)$ is dense.
		\end{itemize}
	\end{definition}

	\begin{remark}\label{R:Dense}~
		\begin{enumerate}[(a)]
			\item An internal subset $X$ is dense if and only if it intersects
			every internal subset of density $1$. Notice that $X$ is dense if
			and only if it has positive density. However, an
			$\omega$-internal set can be dense and yet  of density $0$ (with
			respect to the extension of the Loeb measure to the
			$\sigma$-algebra generated by all internal sets, see
			\cite[Remark 16]{BT14}).
			\item If an $\omega$-internal set $X=\bigcap_{n\in \N} X_n$ is
			given by a decreasing chain, then the set $X$ is dense if and
			only if each internal set $X_n$ is. Indeed, if $X$ were not
			dense witnessed by an internal set $Y$ of density $0$, then the
			decreasing intersection \[\bigcap_{n\in \N} (X_n \setminus Y)
			= \emptyset,\] which yields by $\aleph_1$-saturation that
			some $X_n\setminus Y$ must be empty.
			\item If $b$ is dense over $A$, then so is every element in
			$\tp(b/A)$.
			\item If $b$ is dense over $A\cup \{g\}$, then so are $b\cdot
			g$ and $g\cdot b$.
		\end{enumerate}
	\end{remark}

	\begin{lemma}\label{L:ExtensionDense}
		Let $X$ be a non-empty $\omega$-internal dense subset of a
		non-standard finite group $G$ defined over a countable subset
		$A$. If  $B$ is a subset of parameters extending $A$, then there
		exists some element in $X$ which is dense over $B$.

		In particular, if $c$ is dense over $A$, then there is some $c'$ in
		$\tp(c/A)$ which is dense over $B$.
	\end{lemma}
	\begin{proof}
		Write $X=\bigcap_{n\in \N} X_n$. By $\aleph_1$-saturation, it
		suffices to show that
		\[
		X_1\cap \dots\cap X_m  \cap (G\setminus Y_1) \cap \dots \cap
		(G\setminus Y_m) \neq\emptyset
		\]
		for every $m\ge 1$, where $Y_1, \dots , Y_m$ are fibres of
		internal sets in $\Fc$ defined over $B$ and of density
		$0$. Otherwise,  if  the above intersection were empty,
		then $ 	X_1\cap \dots\cap X_m$ would have density $0$, so
		$X$
		would not be dense.
	\end{proof}

Model-theoretically, it is convenient to capture a global
behaviour of an internal (or definable) set in terms of a suitable
(dense) element in the set. An example of this translation is the
following result, which follows from the Fubini-Tonelli Theorem and
 Caratheodory's Extension Theorem for the Loeb measure.

	\begin{fact} \cite[Lemma 1.9 \& Remark 1.12]{MPP21}
	\label{F:UdiExercise}
		Let $G$ be a non-standard finite group and $X$ be an internal
		subset of $G^n$ defined over a countable subset $A$. The
		following are equivalent.
		\begin{enumerate}[(a)]
			\item There exists some $(b_1,\ldots,b_n)$ in $X$ \emph{in
				good position} over $A$, that is, each $b_i$ is dense over
			$A\cup\{b_j:j<i\}$.
			\item The set $X$ is dense.

		\end{enumerate}

		Furthermore, every tuple as in (a) is dense over $A$
		with respect to the Loeb measure on $G^n$.

	\end{fact}
	Note that if a tuple is in good position, then so is every subtuple.
	\begin{definition}\label{D:Comp}
		Two internal subsets $X$ and $Y$ of $G^n$  are {\em
		comparable} if the set $X\triangle Y$ has Loeb measure $0$.
	\end{definition}

	\begin{remark}\label{R:Comp}
		Note that by Lemma \ref{L:ExtensionDense} two internal sets
		$X$ and $Y$ both defined over $A$ are comparable if and only if
		they contain the same dense elements over $A$.

		By Fubini-Tonelli Theorem, two internal subsets $X$ and $Y$ of
		$G^{m+n}$ are comparable if and only if for
		$\mu_{G^n}$-almost all elements $b$ in $G^n$, the fibres $X_b$
		and $Y_b$ are comparable.
	\end{remark}

	\section{Stability and robust stability}\label{S:stab}

	Let $k\ge 1$ be a natural number. A \emph{half-graph of height
	$k$ induced by} the relation $S \subseteq  X\times Y$ consists of a
	sequence $(a_1,b_1,\ldots,a_k,b_k)$ with $a_i$ in $X$
	and $b_i$ in $Y$ for each $i=1,\dots, k$, such that the pair $(a_i,b_j)$ belongs to $S$
	if and only if $i\le j$. We denote by $\HH_k(S)$ the collection of all
	half-graphs of height $k$ induced by $S$. Note that for each
	sequence $(a_1,b_1,\ldots,a_k,b_k)$ in $\HH_k(S)$, the elements
	$a_i$ must be pairwise distinct and similarly for the elements $b_i$.

	\begin{definition}\label{D:stab}
		The relation $S \subseteq  X\times Y$ is \emph{$k$-stable} if it
		induces no half-graph of height $k$, or equivalently, if
		$\HH_k(S)=\emptyset$. A subset $A$ of a group $G$ is
		\emph{$k$-stable} if the relation given by the Cayley graph of
		$A$ in $G$ \[ \mathrm{Cay}(G,A)=\{ (g,h) \in G\times G \ | \
		g^{-1} \cdot h \in A\}\] is $k$-stable, or equivalently, if the
		relation
		\[ \Gamma(G,A)=\{ (g,h) \in G\times G \ | \ h\inv \cdot g \in
		A\}\] is $k$-stable.
	\end{definition}

	\begin{example}\label{E:stable_examples}
		A non-empty subset $A$ of a group $G$ is $2$-stable if and only
		if it is a coset of a subgroup of $G$. Sidon subsets of abelian
		groups, such as $2^{\mathbb N}$ in $\Z$, are easily seen to be
		$3$-stable \cite[Lemma 1.3]{tS20}.
	\end{example}

	\begin{remark}\label{R:stab_ultrap}
		Consider a non-standard finite group $G=\prod_{n\to \Ult} G_n$
		and  an internal relation $S$ on $G\times G$ defined over the
		countable subset $A$.  By \L o\'s' Theorem (see Remark
		\ref{R:internal} (e)), the relation $S$ is $k$-stable if and only if $S(G_n)$ is $k$-stable in $G_n$ for $\Ult$-almost all $n$ in $\N$.
		Furthermore, a straightforward $\aleph_1$-saturation argument
		yields that the internal relation $S$ is $k$-stable for some $k\ge
		2$ if and only if there is no $A$-indiscernible sequence
		$(a_i,b_i)_{i\in\N}$ such that the pair $(a_i,b_j)$ belongs to $S$
		if and only if $i\le j$.
	\end{remark}

	The previous remark motivates the following definition.
	\begin{definition}\label{D:stab_infinite}
		Consider a non-standard finite group $G=\prod_{n\to \Ult} G_n$.
		An $\omega$-internal relation  $S$  on $G\times G$ defined over
		the countable subset $A$ is \emph{stable} if there is no infinite
		$A$-indiscernible sequence $(a_i,b_i)_{i\in\N}$ such that the
		pair $(a_i,b_j)$ belongs to $S$ if and only if $i\le j$.
	\end{definition}
	Using the Krein-Milman theorem on the
	locally compact Hausdorff topological real vector space of all
	$\sigma$-additive probability measures, Hrushovski  \cite[Proposition 2.25]{eH12}
	proved the following result. Roughly speaking, it asserts that the
	relation given by the measure of an intersection of two internal sets
	exceeding a threshold is stable. In the particular case of the
	intersection
	having measure $0$, stability is an easy consequence of  the
	so-called
	\emph{S1-property} of the
	null ideal of a measure. For the
	purpose of this work, we will state an adapted version of
	\cite[Lemma 2.10]{eH12}, extracting it from the formulation in
	\cite[Fact 2.2 \& Corollary 2.3]{MPP20}.
	\begin{fact}\label{F:Measure_eq}
		Let $G$ be a non-standard finite group and let $\alpha$ be a real
		number in $[0,1]$.
		\begin{enumerate}[(a)]
			\item Given two internal subsets $X$ of $G^{n+r}$ and $Y$ of
			$G^{n+s}$ in $\Fc$, the relation $R^\alpha_{X,Y}$ of
			$G^r\times G^s$ defined by  \[ R^\alpha_{X,Y}(a,b) \
			\Leftrightarrow \ \mu_{G^n}(X_a \cap Y_b) \le \alpha \] is
			stable. Notice that  $R^\alpha_{X,Y}$ is $\omega$-internal by
			Definition \ref{D:Tame} (IV).
			\item If the  two  subsets $X$ of $G^{n+r}$ and $Y$ of
			$G^{n+s}$ are $\omega$-internal and defined over $A$, the
			relation  $R_{X,Y}$ of $G^r\times G^s$ defined by \[
			R_{X,Y}(a,b) \ \Leftrightarrow \ \text{ the $\omega$-internal
			subset $X_a \cap Y_b$ of $G^n$  is not dense}\] is
			\emph{equational}, that is, there is no  $A$-indiscernible sequence
			$(a_i, b_i)_{i\in \N}$ such that $X_{a_0} \cap Y_{b_1}$ is not
			dense yet $X_{a_0} \cap Y_{b_0}$ is dense.

			\item Equationality implies the following (see \cite[Remark 2.1
			\& Corollary 2.3]{MPP20}). For every $\omega$-internal subset
			$X$ of $G^{n+r}$ defined over a countable rich subset $M$, if
			the fibre $X_a$ is dense for some $a$ which is itself dense over $M$, then so
			is the intersection $\bigcap_{i=0}^m X_{b_i}$ for every $m$ in
			$\N$ and every tuple $(b_0, \ldots, b_m)$ of elements in
			$\tp(a/M)$ in good position over $M$, as in Fact
			\ref{F:UdiExercise} (a), that is, the element
			$b_i$ is dense over $M\cup\{b_j:j<i\}$ for $i\le m$.
		\end{enumerate}
	\end{fact}

	While the notion of stability has proven highly profitable in model theory, from a more combinatorial perspective it is open to the criticism that whether or not a relation is stable is very sensitive to minor
	perturbations. In particular, modifications by an internal set of
	measure $0$ can
	destroy stability, as the following example illustrates.
	\begin{example}\label{E:destroy__stab}
		For some non-principal ultrafilter $\Ult$, consider the
		non-standard finite group $G=\prod_{n\to \Ult} \Z/n^2\Z$  as
		well as the internal set $X=\prod_{n\to \Ult} \{\overline
		0,\ldots, \overline{n-1}\}$ and the internal relation $S$ on
		$X\times X$ arising from the standard linear order of $\Z$
		restricted to the set of representatives $\{0,\ldots, n-1\}$. By \L
		o\'s's theorem (see Remark \ref{R:internal} (e)), the collection
		$\HH_k(S)$ of induced half-graphs of height $k$ is non-empty for
		every natural number $k\ge 1$, so the internal relation $S$ is never
		stable. However, for every index $k\ge 1$, the internal set
		$\HH_k(S)$ has $\mu_{G^{2k}}$-measure $0$, so it is
		comparable to the empty relation, which in turn is $k$-stable.
	\end{example}
	Motivated by the above example, we introduce a weakening of
	stability which is combinatorially robust with respect to
	perturbations by sets of measure $0$.

	\begin{definition}\label{D:robust}
		Consider a non-standard finite group $G$ as well as two internal
		subsets $X$ of $G^{n}$ and $Y$ of $G^{m}$. Given a natural number $k\geq 1$, an internal
		relation $S \subseteq  X\times Y$ is \emph{robustly $k$-stable},
		with $k\ge 1$ in $\N$, if $\mu_{G^{(n+m)k}}(\HH_k(S))=0$,
		that is,  the set all half-graphs of height $k$ induced by $S$ is
		negligible in $G^{(n+m)k}$ with respect to the non-standard
		counting measure.
	\end{definition}

	\begin{remark}\label{R:robust}~
		\begin{enumerate}[(a)]
			\item 	It is immediate that robust stability is preserved
			by
			finite
			Boolean combinations as well as by permutation of the
			variables:
			if $S\subseteq X\times Y$ is robustly $k$-stable, then so is the
			inverse relation \[S^{opp}=\{(y,x)\in Y\times X \ | \ (x,y) \in
			S\}.\]

			\item 	Fact \ref{F:UdiExercise} yields that the internal
			relation
			$S$ defined over the countable subset $A$ is not robustly
			$k$-stable if and only if it induces a half-graph of height $k$
			witnessed by a dense tuple $(a_1, b_1, \ldots, a_k, b_k)$ in
			good
			position over $A$.

			As a consequence, whenever a relation $S$ is comparable to an
			internal $k$-stable or even a robustly $k$-stable relation $S'$,
			Remark \ref{R:Comp} yields
			that $S$ is  robustly $k$-stable.
		\end{enumerate}
	\end{remark}

	We will see in Proposition \ref{P:robuststable_boxes} below that every
	robustly stable relation $S\subseteq X\times Y$ can be
	approximated for every $\epsilon>0$ by a finite union of  internal
	boxes of the form $X'\times Y'$. Hence, the robustly $k$-stable
	relation is $\epsilon$-comparable to a stable relation $S'$, although
	our methods do not allow us to compute the degree of stability of
	$S'$.

	 In fact,  the (bipartite version of the) induced removal lemma
	\cite{EFR86} implies that every robustly $k$-stable must be
	comparable to a $k$-stable relation, but we do not know how to give
	a model-theoretic proof of this result. So we ask the following:

	\begin{question}
		Is there a model-theoretic account that every robustly $k$-stable
		relation is comparable (or $\epsilon$-comparable) to an internal
		$k$-stable relation?
	\end{question}
A straightforward application of Lemma
\ref{L:Transfer} (with $r=0$) yields an analogue of robust stability for families
of
finite groups
in terms of  their asymptotic behaviour:
\begin{remark}\label{R:robust_asympt}
Consider a family $(G_\ell, X_\ell, Y_\ell, S_\ell)_{\ell\in\N}$,
 where
$X_\ell\subseteq G_\ell^n$ and $Y_\ell\subseteq G_\ell^m$ are
subsets of the finite group $G_\ell$ and $S_\ell\subseteq X_\ell\times
Y_\ell$ for every $\ell$ in $\N$.  The following are equivalent for
every natural number $k\ge 1$.
\begin{enumerate}[(a)]
	\item For every  $\theta>0$ there is some
	$\ell_0=\ell_0(k,\theta)$ in
	$\N$ such that $|\HH_k(S_\ell)| \le \theta |G_\ell|^{k(n+m)}$ for all
	$\ell\ge \ell_0$, where $\HH_k(S_\ell)$ is the collection of all
	half-graphs
	of height $k$ induced by $S_\ell$ on the finite group $G_\ell$.
	\item For every non-principal ultrafilter $\Ult$ on $\N$, the
	internal relation $S=\prod_{n\to\Ult} S_\ell$ is robustly $k$-stable.
\end{enumerate}
If  (a) holds, we say that  $(G_\ell, X_\ell, Y_\ell, S_\ell)_{\ell\in\N}$
is a \emph{robustly $k$-stable family}.
\end{remark}

	A key feature of stable relations is that they are \emph{stationary} \cite[Lemma 2.3]{eH12}, in the sense that their
	truth value is constant along
	the set of pairs of realisations which are \emph{non-forking independent}
	(non-forking independence is a fundamental notion in model
	theory, originally due to Shelah, defined in combinatorial terms for
	any structure \cite[Chapter III.1, Definition 1.4]{sSbook}). We
	will now show that the truth value of a robust
	stable relation remains constant along the set of pairs of realisations for
	which one coordinate is dense over the other.

	Before stating the corresponding result, we introduce some
	notation.
	\begin{definition}\label{D:goodpos}
		Fix two types $\tp(a/M)$ and $\tp(b/M)$ over a countable rich
		subset $M$. We denote by $\GP(\tp(a/M),\tp(b/M))$  the set of
		all
		pairs $(a',b')$ with $a'$
		in $\tp(a/M)$ and $b'$ in $\tp(b/M)$ such that $(a',b')$ or
		$(b',a')$ is in good position over $M$ (see Fact
		\ref{F:UdiExercise}).
	\end{definition}
	Note that the set $\GP(\tp(a/M),\tp(b/M))$ is empty exactly if
	one of the types $\tp(a/M)$ or $\tp(b/M)$ is not dense, by Lemma
	\ref{L:ExtensionDense}. Moreover:
	\[
	(a',b')\in \GP(\tp(a/M),\tp(b/M)) \ \Leftrightarrow \ (b',a')\in
	\GP(\tp(b/M),\tp(a/M)).
	\]

	\begin{theorem}\label{T:robuststable_main}
		Consider a non-standard finite group $G$ as well as two internal
		subsets $X$ of $G^{n}$ and $Y$ of $G^{m}$ and an internal
		relation $S$ on $X\times Y$,  all defined over a countable rich
		subset $M$.

		If  $S$ is robustly $k$-stable for some $k\ge 1$, then for every two (dense)
		types $\tp(a/M)$ and $\tp(b/M)$, the set
		$\GP(\tp(a/M),\tp(b/M))$ is either
		fully contained in $S$ or disjoint from $S$.
	\end{theorem}

	\begin{proof}
		If $a$ does not lie in $X$, neither does any element of
		$\tp(a/M)$, and likewise if $b$ does not lie in $Y$. In this case
		$X\times Y$ (and thus $S$) is clearly disjoint from
		$\GP(\tp(a/M),\tp(b/M))$. So we
		may assume that $\GP(\tp(a/M),\tp(b/M))$ is a subset of
		$X\times Y$.

		Suppose now that $S$ is not  disjoint from
		$\GP(\tp(a/M),\tp(b/M))$.  Up to relabelling, we may assume by
		Remark \ref{R:robust} (a) that $(a,b)$ lies in $S$, with $(a, b)$
		in
		good position, that is, the element $b$ is dense over
		$M\cup\{a\}$ (keeping in mind that $a$ is already dense over
		$M$).

		\begin{claim*}
			All pairs $(a', b')$ in $\GP(\tp(a/M),\tp(b/M))$ with $(b', a')$
			in good position over $M$ are contained in $S$.
		\end{claim*}
		\begin{claimproof*}
			Suppose to the contrary that there is a pair $(a',b')$ with $a'$ dense
			over
			$M\cup\{b'\}$ not contained in $S$. By $\aleph_1$-saturation
			(using the fact that $b'$ belongs to the type $\tp(b/M)$), there is such
			a pair
			of the form $(a',b)$. So we may assume that $a'$ belongs to
			$\tp(a/M)$ and is dense over $M\cup\{b\}$ but $(a', b)$ does
			not lie
			in $S$.
			In particular, we have two pairs $(a, b)$ and $(a', b)$ satisfying
			the following:
			\begin{itemize}
				\item the type $\tp(a/M)=\tp(a'/M)$ is dense;
				\item the element $b$ is dense over $M \cup\{a\}$ and
				$a'$ is dense
				over $M \cup\{b\}$;
				\item the pair $(a, b)$ belongs to $S$ but $(a', b)$ does not.
			\end{itemize}
			We will now construct inductively, for every $n$ in $\N$, a
			dense
			sequence $(a_0, b_0, \ldots, a_n, b_n)_{0\le i\le n}$ in good
			position
			over $M$ witnessing that $S$ induces a half-graph of height
			$n$
			such that  $a_i$ belongs to $\tp(a/M)$ and $b_i$ to
			$\tp(b/M)$ for
			each $i\le n$. In particular, for $n=k$, we contradict the robust
			stability of $S$ by Remark \ref{R:robust} (b).

			For $n=0$, set $a_0=a$ and $b_0=b$, so $(a_0, b_0)$ is in good
			position over $M$ and lies in $S$, as desired. Suppose now that
			the
			sequence $(a_0, b_0, \ldots, a_r, b_r)$ has already been
			constructed
			for some $r< n$. Consider the $\omega$-internal set \[Z =\{
			(x,y) \in
			X\times Y \ |\  x\in \tp(a/M) \ \text{and} \ (x,y) \notin S \},\]
			which is
			defined
			over $M$. Clearly $(a', b)$ belongs to $Z$, so the fibre
			$Z_b\subseteq
			X$ is dense.
			By Fact \ref{F:Measure_eq} (c), so is the intersection $
			Z_{b_0}\cap\cdots \cap Z_{b_r}$, since the subtuple
			$(b_0,\ldots,
			b_r)$ is also in good position over $M$ and all  coordinates lie
			in
			$\tp(b/M)$. Hence, there is an element $a_{r+1}$ in the above
			intersection which is dense over $M\cup\{a_i, b_i\}_{0\le i\le
			r}$. In
			particular, the element $a_{r+1}$ lies in $\tp(a/M)$ but
			$(a_{r+1},
			b_i)$ does not lie in $S$ for $i\le r$.

			Similarly, the $\omega$-internal set \[W =\{ (x,y) \in
			X\times Y \ |\  y\in \tp(b/M) \ \text{and}\ (x,y) \in S \}\]  is
			also defined
			over $M$ and the fibre $W_a\subseteq Y$ is dense, since $(a,
			b)$ lies
			in
			$W$. Thus, there is an element $b_{r+1}$ in $\tp(b/M)$
			contained in
			$W_{a_0}\cap\cdots\cap W_{a_{r+1}}$ dense over $M\cup
			\{a_i,
			b_i\}_{0\le i\le r} \cup \{a_{r+1}\}$, which yields that the
			tuple \[
			(a_0, b_0, \ldots, a_{r+1}, b_{r+1})\] is in good position over
			$M$ by Fact
			\ref{F:UdiExercise} and satisfies the desired properties.
		\end{claimproof*}

		By the previous claim, choose any pair $(a',b')$ in $S\cap
		\GP(\tp(a/M), \tp(b/M))$ such that $a'$ is dense over
		$M\cup\{b'\}$. The pair $(b',a')$ is in good position over $M$
		and it belongs to the inverse relation $S^{\rm opp}$, which is
		again robustly $k$-stable, by Remark \ref{R:robust} (a).
		We conclude by the Claim (applied to $S^{\rm opp}$ inverting
		the roles
		of $X$ and $Y$) that all pairs $(b'', a'')$ in $\GP(\tp(b/M),
		\tp(a/M))$
		with $b''$ dense over $M\cup\{a''\}$ must be contained
		in $S^{\rm opp}$. Hence, all pairs $(a'', b'')$ in $\GP(\tp(a/M),
		\tp(b/M))$ in good position over $M$ belong to $S$, so the set
		$\GP(\tp(a/M), \tp(b/M))$  is fully contained in $S$, as desired.
	\end{proof}

	\begin{cor}\label{C:keylemma}
		Consider a non-standard finite group $G$ as well as two internal
		subsets $X$ of $G^{n}$ and $Y$ of $G^{m}$ and an internal
		relation $S$ on $X\times Y$,  all defined over a countable rich
		subset $M$.  If  $S$ is robustly $k$-stable for some $k\ge 1$,
		then the fibres $S_a$ and $S_{a'}$ are comparable
		whenever $a'$ belongs to the dense type $\tp(a/M)$.
	\end{cor}
	\begin{proof}
		If the set-theoretic difference $S_a \setminus S_{a'}$ were
		dense for some choice $a$ and $a'$, both in the same dense type
		over $M$, then it would contain some element $b$ which is dense over
		$M\cup\{a, a '\}$.  In particular, both $(a, b)$ and $(a', b)$ lie in
		$\GP(\tp(a/M), \tp(b/M))$, yet $(a,b)$ belongs to $S$ but $(a',
		b)$ does not, which contradicts Theorem
		\ref{T:robuststable_main}.
	\end{proof}

	\begin{prop}\label{P:robuststable_boxes}
		Consider a non-standard finite group $G$ as well as two internal
		subsets $X$ of $G^{n}$ and $Y$ of $G^{m}$ and an internal
		relation $S$ on $X\times Y$,  all defined over a countable rich
		subset $M$.  If  $S$ is robustly $k$-stable for some $k\ge 1$,
		then for every $\epsilon >0$ there is a finite (possibly empty)
		union
		of boxes \[ U=\bigcup\limits_{i=1}^\ell  X_i\times Y_i,\] with all
		$X_i\times Y_i \subseteq X\times Y$,
		defined over $M$, such that $\mu(U\triangle S) < \epsilon$.
	\end{prop}
	Note that every box is $2$-stable and thus the finite union $U$ is
	$r$-stable for some $r$ which can be explicitly computed in
	terms of the number $\ell$ of boxes occurring in $U$. However, at the time of writing we
	are unable to give an explicit bound on $\ell$
	or show that $U$ and $S$ are actually comparable.
	\begin{proof}
		If $S$ is not dense in $G^{n+m}$, then it suffices to consider for
		$U$ the empty union of boxes. Thus, we may assume that the
		internal subset
		$S$ of $G^{n+m}$ is dense.

		\begin{claim}\label{Claim:family}
			Consider a subset $Z$ of $S$ in the $\sigma$-algebra of
			internal sets defined over $M$. (Note that $Z$ need not be
			$\omega$-internal, but it is measurable with respect to the
			extension of the Loeb measure.) If $Z$ has positive measure,
			then there exists a dense element $a$ in $G^n$ such that the fibre $Z_a$ is dense and satisfies the following property ($\star$): for every internal
			subset $W$ of $Y$ defined over $M$, if
			$\mu_{G^m}(W\setminus
			S_a)=0$,
			then \[ \mu_{G^n}\big( \{x\in G^n \ | \
			\mu_{G^m}(W\setminus S_x)=0\} \big) > 0.\]
		\end{claim}
		\begin{claimproof}

			By Fubini-Tonelli (Remark \ref{R:Fubini}), the measurable set
			\[Z_1=\{ x\in G^n \ | \ \mu_{G^m}(Z_x)>0\}\] has positive
			measure (since $\mu_{G^{n+m}} (Z)$ is not zero). For every
			internal subset $W$ of $Y$ defined over $M$, set \[ D_W=\{
			x\in G^n
			\ | \ \mu_{G^m}(W \setminus S_x)=0 \}.\]  Notice that there
			are only
			countably many such subsets $D_W$. Consider the
			subfamily
			$\mathcal D$ of all such subsets $D_W$ of measure $0$.
			Similarly, let $\mathcal E$ be the countable family of all
			internal subsets $X'$ of $G^n$ defined over $M$ of measure
			$0$. By $\sigma$-additivity of the (extension of the) Loeb
			measure, the set
			\[
			Z_1 \setminus \Big( \bigcup_{D_W\in\mathcal D} D_W \ \cup \
			\bigcup_{X'\in \mathcal E} X' \Big)
			\]  has positive measure, so it is non-empty. Now choose some
			element $a$ in the above set-theoretic difference. By
			the choice of $\mathcal E$, the type $\tp(a/M)$ is dense and so is
			the fibre $Z_a$ (since $a$ lies in $Z_1$).  Furthermore, the
			element $a$ satisfies ($\star$)
			by our choice of the family $\mathcal D$.
		\end{claimproof}
	As the internal set $S$ is dense, it has positive measure. By the previous
	claim, there is an element $a$ which is dense over $M$  satisfying
	$(\star)$ with respect to $Z=S$ such that the fibre $S_a$ is dense. Thus,
	the fibre $S_a$ contains an element $b$ which is dense over $M\cup\{a\}$.
	By Fact \ref{F:UdiExercise}, the dense pair $(a,b)$ is in
	good position over $M$ and clearly lies in $S$.

		\begin{claim}\label{Claim:specialpt}
			For every pair $(a, b)$ in $S$ in good position over $M$ such that
			$a$ satisfies
			($\star$), there are an internal subset $Y'$ of $Y$ and an
			$\omega$-internal subset $X'$ of $X$,  both defined
			over $M$, such $(a,b)$ lies in $X'\times Y'$ and $\mu((X'\times
			Y')\setminus S)=0$.
		\end{claim}
		\begin{claimproof}
			By Theorem \ref{T:robuststable_main},  there is no element in
			$\tp(b/M)$ dense over $M\cup\{a\}$ outside of $S_a$, so
			the $\omega$-internal subset $\tp(b/M) \setminus S_a$ of $Y$
			is not dense. By $\aleph_1$-saturation, there is some internal
			subset $Y'$ of $Y$ defined over $M$ containing $b$ with
			$\mu_{G^m}(Y'\setminus S_a)=0$. We
			deduce  by $(\star)$ that $\mu_{G^n}(X')>0$, where
			\[ X'= \{x \in X \ |
			\ \mu_{G^m}(Y'\setminus S_x)=0\}\] is an $\omega$-internal subset
			(by Definition \ref{D:Tame} (IV)) containing $a$ (this set is
			$D_{Y'}$ with $Z=S$ in the notation of the Claim
			\ref{Claim:family}).

			A
			straightforward computation using Fubini-Tonelli yields
			\[
			\mu_{G^{n+m}} ((X'\times Y')\setminus S) = \int_{X'}
			\mu_{G^m}(Y'\setminus S_x) \, \mathrm{d}\mu_{G^n} = 0, \]
			as desired.
		\end{claimproof}

		Consider now the countable collection $\mathcal B_S$ of all
		subsets
		$X'\times Y'$ as in Claim \ref{Claim:specialpt}, that is, the
		set $Y'$ is internal whilst $X'$ is $\omega$-internal, both are
		defined over $M$ and 	\[ X'= \{x \in X \ |
		\ \mu_{G^m}(Y'\setminus S_x)=0\},\] so
		$\mu_{G^{n+m}}((X'\times Y')\setminus
		S)=0$. By
		$\sigma$-additivity, we have that \[ \Big(\bigcup_{X'\times
		Y'\in \mathcal B_S} X'\times Y' \Big)\setminus S \] has measure
		$0$ (with respect to the extension of the Loeb measure).

		Observe that the set \[ Z=S\setminus \bigcup_{X'\times Y'\in \mathcal
		B_S} X'\times Y' \]
		belongs to the $\sigma$-algebra of internal sets defined over
		$M$.  If $Z$ has positive measure, we deduce as in the discussion right
		after Claim \ref{Claim:family} that $Z$ contains a pair $(a,b)$ in good
		position over $M$ with $a$ satisfying
		$(\star)$. It follows from Claim
		\ref{Claim:specialpt} that $(a,b)$ lies in some
		box $X'\times Y'$ in $\mathcal B_S$, contradicting the choice of
		$Z$.  Thus, the sets $S$ and $\bigcup_{X'\times
		Y'\in \mathcal B_S} X'\times Y'$ are comparable  with respect to
		the extension of the Loeb measure. We need only show that we
		can replace the latter union, modulo an $\epsilon$-error, by
		an actual finite union of internal 	boxes defined over $M$.

		Given $\epsilon>0$, by continuity of the Loeb measure there is some
		integer $\ell$
		and subsets $X'_1\times
		Y_1,\ldots, X'_\ell\times Y_\ell$ in $\mathcal B_S$ such that \[
		\mu_{G^{n+m}}(
		S\setminus
		\bigcup\limits_{i=1}^\ell X'_i\times Y_i)<\frac{\epsilon}{2}.\] Note
		that $\ell>0$, since $\mathcal B_S$ is non-empty, for $S$ is assumed to
		be dense.
		For each $1\le i\le \ell$, the $\omega$-internal subset
		$(X'_i\times Y_i)\setminus
		S$ has measure $0$, so we can find a finite intersection $X_i$ of
		some internal subsets occurring in $X'_i$ such that
		\[\mu_{G^{n+m}}\big((X_i\times Y_i)\setminus S\big)
		<\frac{\epsilon}{2\ell}.\]
		Note that $X_i$ is again internal and defined over $M$. Now set
		$U=\bigcup_{i=1}^\ell X_i\times Y_i$, so $U$ is
		contained in $\bigcup_{i=1}^\ell X'_i\times Y_i$, and it follows
		that
		 \begin{align*}
		\mu_{G^{n+m}}\left(S \triangle U\right) & = \mu_{G^{n+m}} (S
		\setminus U) + \mu_{G^{n+m}}(U\setminus S)  \\ & \le \mu_{G^{n+m}}
		\big(S
		\setminus \bigcup_{i=1}^\ell X_i'\times Y_i \big)  + \sum_{i=1}^{\ell}
		\mu_{G^{n+m}}\left((X_i\times Y_i)\setminus S \right) \\ & <
		\frac{\epsilon}{2} +
		\ell\cdot \frac{\epsilon}{2\ell} = \epsilon.
		\end{align*}
		This completes the proof of Proposition \ref{P:robuststable_boxes}.
	\end{proof}
	\section{Corners and related shapes}\label{S:corner}

The results in Section \ref{S:stab} did not use the underlying
	group structure. In this section, the group structure will play a
	fundamental role in guaranteeing the existence of certain 2-dimensional patterns for a dense
	robustly stable relation.

	Given a countable rich substructure $M$,  the group $G(M)=G\cap
	M$ of
	$M$-rational points of $G$ naturally acts on the collection of  types
	over $M$ by \[  \tp(a/M) \star g = \tp(a\cdot g/M) .\] This is a right
	action, but there is also a natural left action, defined analogously.
	In the presence of stability, the stabilizer of a type becomes $\omega$-internal
	(or type-definable).

	After the work of Kim and
	Pillay \cite{KP97}, many notions and techniques from
	geometric stability
	have   been adapted successfully to other contexts, in particular to
	simple theories. Pushing beyond simplicity, in seminal work \cite{eH12}
	Hrushovski established the existence of an $\omega$-internal
	subgroup of a nonstandard finite group that plays
	the role of the
	\emph{stabilizer} of every dense type over the countable rich substructure $M$.
	This subgroup is known in model theory as the \emph{connected
	component} of $G$ over $M$ and denoted by $\GO M$.

	The next fact summarises the content of Hrushovski's \emph{stabiliser theorem}, tailored to our particular context.

	\begin{fact}\textup{(}\cite[Theorem 3.5]{eH12}\textup{)}
	\label{F:G00}
		Let  $M$ be a countable rich subset of a non-standard finite group
		$G$.
		There exists a $\omega$-internal normal subgroup $\GO M$
		defined over $M$ with the following properties:
		\begin{enumerate}[(a)]
			\item  The $\omega$-internal subset $\GO M$ equals a countable
			intersection of internal \emph{generic symmetric neighborhoods}
			$Z_n$ of
			the
			identity, each defined over $M$, that is,
			\begin{itemize}
	\item  each $Z_n=Z_n\inv$ is symmetric and contains the identity element
	$1_G$;
	\item finitely many translates of each $Z_n$ cover the group $G$;
	\item $Z_{n+1}\cdot Z_{n+1}\subseteq Z_n$ for all $n$ in $\N$.
			\end{itemize}

			\item Whenever
			$(a, b)$ is in good position over $M$ with $b$ in $\tp(a/M)$,
			the dense element $b\cdot
			a\inv$ over $M$ belongs to $\GO M$. In particular, the
			$\omega$-internal set $\GO M$ is dense.
			\item  Whenever $a$ is dense over $M$ and $g$ in $\GO M$ is
			dense over $M$,
			the $\omega$-internal set \[ \tp(a/M)\cap  \tp(a/M)\cdot g\] is
			again dense,  that
			is, there exists some $a'$  in $\tp(a/M)$ which is dense over
			$M\cup\{g\}$
			such that $a'\cdot g$ belongs to $\tp(a/M)$.

			Similarly, the $\omega$-internal set  $\tp(a/M)\cap  g\cdot
			\tp(a/M)$ is 		also dense,  so there exists some $a''$  in
			$\tp(a/M)$ which is dense over 		$M\cup\{g\}$
			such that $g\cdot a''$ belongs to $\tp(a/M)$.
			\item If $N$ is a countable rich subset of $G$ containing $M$,
			then $\GO
			N$ is a
			subgroup of $\GO M$.
			\end{enumerate}
		Furthermore, it follows from \cite[Theorem 4.5]{rI68} that if
		$G$ has finite exponent, then $\GO M$ is a countable
		intersection of internal subgroups, each defined over $M$ and
		of finite index. The reason for this is that the quotient group $G/ \GO
		M$ admits a topology (called \emph{the logic topology}) which makes it a compact Hausdorff topological group.
	\end{fact}

Since the $\omega$-internal normal subgroup $\GO M$ equals a countable
intersection of internal generic symmetric neighborhoods $Z_n$, each defined
over $M$, the elements $a'$ of any given type $\tp(a/M)$ all lie in the
same
coset $a\cdot \GO M$. Indeed, we need only show that $a\inv \cdot a'$ belongs
to $Z_n$ for every $n$ in $\N$. Since $M$ is a rich subset, finitely many
translates of each $Z_n$ by elements of $G(M)$ cover $G$. In particular, the
element $a$ belongs to $g \cdot Z_{n+1}$ for some $g$ in $G(M)$ and therefore
so does $a'$. Hence, we deduce that $a\inv \cdot a'$ belongs to $Z_{n+1}\cdot
Z_{n+1}\subset Z_n$, as desired.

In \cite[Proposition
 2.2]{PSW98}, Pillay, Scanlon and Wagner observed  a particular behaviour in simple theories for dense (or
 rather, generic) \emph{principal} types (that is, those contained in the trivial
 coset $\GO M$). In \cite[Theorem 3.3]{MPP21} their result was adapted to the
 case of
 Loeb measures in order to provide a model-theoretic proof of Roth's
 theorem on  arithmetic progressions of length $3$.  Before stating a
 summary of the results obtained in \cite{MPP21},  we
introduce the following terminology.
\begin{definition}\label{D:squaring}
 Squaring
\emph{preserves dense elements} in the non-standard finite group
$G$ if for every element $a$ and
every countable set of parameters $B$, we have that $a$ is dense
over $B$ if and only the element $a^2=a\cdot a$ is.
 \end{definition}

Squaring preserves dense elements whenever squaring has finite
fibres. A nonstandard finite group obtained as an ultraproduct of
finite  groups of odd order has no involutions, so squaring always
preserves dense elements
\cite[Remark 6.2]{MPP21}.  A relevant  fact from \cite{MPP21} is
the following.

 \begin{fact}\label{F:Roth}\textup{(}\cite[Theorem
 6.1]{MPP21}\textup{)}
Whenever squaring preserves dense elements, for every
dense element $a$  over in $M$, there are
$b$ and $c$ in $\tp(a/M)$ with $(a, c)$ in good position over $M$
and $a\cdot c=b^2$.
 \end{fact}
Note that the pair $(a, b)$ is again in good position over $M$, since
$b$ must be dense over $M\cup\{a\}$, for $b^2=a\cdot c$ is.

We now have all the ingredients in order to state our main result.
	\begin{theorem}\label{T:main_corners}
Consider a non-standard finite group $G$ as well as two internal
subsets $X$ of $G^{n}$ and $Y$ of $G^{m}$ and an internal
relation $S$ on $X\times Y$,   all defined over a countable rich
subset $M$.  If  $S$ is dense and robustly $k$-stable for some $k\ge 1$,  then $S$ contains a dense collection of both naive and of BMZ
corners. Specifically, for every $g$ in $\GO M$ which is dense over $M$, there are $a$ in $X$ and
$b$ in $Y$ such that $(g, b, a)$ is in good position over $M$ (and
thus $\tp(a,b, g/M)$ is dense)  with all four points \[ (a,b), (a\cdot g,
b), (a, b\cdot g) \text{ and } (a\cdot g, b\cdot g)\] in $S$.

In particular, if $G$ is abelian, there is a dense collection of triples
which form a square in $S$.
	\end{theorem}
	\begin{proof}
		Fix some element $g$ in  $\GO M$ which is dense over $M$.
		\begin{claim}\label{Claim:prod_inv}
			There is a tuple $(c, d)$ in good position over $M$
			with $c$
			and $d$ in $\tp(g/M)$  such that $g=c\inv\cdot
			d$.
		\end{claim}
		\begin{claimproof}
By Fact \ref{F:G00} applied to  the
dense element $g\inv$ of $\GO M$ and to the dense type $\tp(g/M)$,
we deduce that there is some
element $a'$ in $\tp(g/M)$ dense over $M\cup\{g\}$ such that
$g''=g\inv \cdot a'$ belongs to $\tp(g/M)$. Hence, for the element
$g''$ in $\tp(g/M)$ there are two elements $c'=g$ and $d'=a'$ in
$\tp(g/M)$ with $(c', d')$ in good position over $M$ such that
$g''=(c')\inv \cdot d'$.  By $\aleph_1$-saturation, we deduce that the
same is true for every element in  $\tp(g''/M)=\tp(g/M)$. We thus obtain
the desired tuple $(c, d)$ as in the statement with $g=c\inv\cdot d$.
		\end{claimproof}
	\begin{claim}\label{Claim:translate_dense}
The internal set $S\cap S\cdot  (1_G, g\inv) $ is dense.
	\end{claim}
		\begin{claimproof}
		Recall that $S$ is dense and thus so is $S\cdot(1_G, g\inv)$.
		Hence, Fact \ref{F:Measure_eq} (c) yields that so is the
		intersection $S\cdot(1_G, c\inv)\cap S\cdot(1_G, d\inv)$, since
		the tuple $(c, d)$ as in Claim \ref{Claim:prod_inv} is in good
		position over $M$ and both $c$ and $d$ lie in the dense type
		$\tp(g/M)$.  Multiplying now by $(1_G, c)$, we deduce that $S\cap S
		\cdot (1_G,
		g\inv)$ is dense, as desired.
				\end{claimproof}

		By Claim \ref{Claim:translate_dense} and Fact
		\ref{F:UdiExercise}, there is a pair $(a', b)$ in good position over
		$M\cup \{g\}$ contained in $S\cap S\cdot (1_G,  g\inv)$. That
		is, both pairs $(a', b)$ and $(a', b\cdot g)$ belong to $S$.

		 Note that we have not yet used  that $S$ is robustly stable. The
		 idea is to replace $a'$ by a suitable realisation $a$ such that
		 $ a\cdot g$ lies again in $\tp(a/M)$. Fact \ref{F:G00} yields that
		 there is some $a$ in  $\tp(a'/M)$ dense over $M\cup\{g\}$ such
		 that  $a\cdot g$ lies again in $\tp(a'/M)$. We may assume that
		 $a$ is
		 dense over $M\cup\{b,g\}$, by $\aleph_1$-saturation. Clearly,
		 each of the pairs
		 $(b, a)$, $(b, a\cdot g)$, $(b\cdot g, a)$ and $(b\cdot g, a\cdot g)$
		 is in good  position over $M$.

		 Theorem \ref{T:robuststable_main} implies that
		 \[\GP(\tp(a'/M), \tp(b/M))\subseteq  S \]
		 since the pair $(a', b)$ lies in $S$ and is in good position over
		 $M$. Thus both pairs $(a, b)$ and $(a\cdot g, b)$ lie in $S$.

		 Now the pair $(a', b\cdot g)$ also lies in $S$ and is in good position
		 over $M$. Again by Theorem \ref{T:robuststable_main}, we
		 conclude that the pairs $(a, b\cdot g)$ and $(a\cdot g, b\cdot g)$ of
		 $\GP(\tp(a'/M), \tp(b\cdot g/M))$ must
		 lie in $S$, which yields the desired conclusion.

		 If $G$ is abelian, then the set of triples that form a square in $S$
		 is clearly internal and defined over $M$. Since it contains a dense
		 triple in good position over $M$, we conclude by Fact
		 \ref{F:UdiExercise} that the set of such triples is dense.
	\end{proof}

In recent work \cite{sP22}, Peluse obtained the first reasonable
upper bounds for the density of $2$-dimensional subsets of the finite
model $\mathbb F_p^n$ without $L$-shaped configurations. We will
show how to obtain a qualitative version of her result valid in any
finite group, possibly non-abelian, under the additional assumption of
robust stability.
\begin{theorem}\label{T:Lshape}
Consider a non-standard finite group $G$ in which squaring
preserves dense elements as well as two internal
subsets $X$ of $G^{n}$ and $Y$ of $G^{m}$ and an internal
relation $S$ on $X\times Y$,   all defined over a countable rich
subset $M$.  If  $S$ is dense and robustly $k$-stable for some $k\ge 1$,  then $S$ contains a dense collection of  \emph{generalized
$2\times 3$-rectangles}. Specifically,
there is a tuple $(b, g, a)$ in good position over $M$ with $a$ in
$X$, $b$ in $Y$ and $g$ in $\GO M$ such that all six points \[ (a, b),
(a\cdot g,
b), (a, b\cdot g), (a\cdot g, b\cdot g),  (a, g\cdot b\cdot g) \text{
and }  (a\cdot g, g\cdot b\cdot g) \] are in $S$.

In particular, if $G$ is obtained as an ultraproduct of finite abelian
groups of odd order, every dense subset $Z$ of $G$ whose Cayley
graph $\mathrm{Cay}(G, Z)$ (see Definition \ref{D:stab}) is
robustly $k$-stable contains a $4$-term arithmetic progression $\{z, z+d,
z+2d, z+3d\}$ given by a dense pair $(x, d)$ (with $d\ne 0$).
\end{theorem}
\begin{proof}
Set $\delta=\mu_{G^{n+m}}(S)/2>0$. By Definition \ref{D:Tame},
there is an internal set $X_\delta$ defined over $M$ such that \[
\{x\in X \ | \
\mu_{G^m}(S_x)> \delta \}\subseteq X_\delta\subseteq \{x\in X \ |
\
\mu_{G^m}(S_x)\ge \delta \}.\] By Fubini-Tonelli
(Remark \ref{R:Fubini}), the internal set $X_\delta$
has positive measure, so
it is dense.
Thus, choose some $a'$ with $\mu_{G^m}(S_{a'})\ge \delta$ with
$a'$ dense over $M$. By Remark \ref{R:LS}, there is a countable rich
subset $N$ containing $M\cup\{a\}$. The internal dense subset
$S_{a'}\subseteq Y$ is defined over $N$, so choose some $b'$ in
$S_{a'}$ which is dense over $N$.

Fact \ref{F:Roth} applied to the dense type $\tp(b'/N)$ yields now
three elements $b_1$, $b_2$ and $b_3$ in $\tp(b'/N)$ with
$b_2^2=b_1\cdot b_3$ and $(b_1, b_2)$ in good position over $N$.
Set now  \[ b=b_1 \text{ and } g=b_3\cdot b_2\inv=b_1\inv\cdot b_2.\]  Note
that $g$ is also
dense over $N\cup\{b_1\}$ and
hence is a dense element of $\GO N$, by Fact \ref{F:G00} (b).  Fact
\ref{F:G00} (d) now yields that $g$ is an element of $\GO M$ which is dense
over $M$.

In
particular, the pairs \[ (a', b) , (a', b\cdot g) \text{ and } (a', g\cdot
b\cdot g)\] are all in good position over $M$ and lie in the robustly
$k$-stable internal relation $S$. Furthermore, the pair $(b,g)$ is in
good position
over $M$.

As in the proof of
Theorem \ref{T:main_corners}, since $g$ is dense in $\GO M$, we
find an element $a$ in
$\tp(a'/M)$ which is dense over $M\cup\{g, b\}$ such that $a\cdot g$
belongs to $\tp(a'/M)$. In particular, the triple $(b,g,a)$ is in good position
over $M$.   Theorem \ref{T:robuststable_main} implies that
\[\GP(\tp(a'/M), \tp(b/M))\subseteq  S,\] so $(a, b)$ and $(a\cdot g,
b)$ belong to $S$. Analogously, since  \[\GP(\tp(a'/M), \tp(b\cdot
g/M))\subseteq  S,\] we have that $(a, b\cdot g)$ and $(a\cdot g,
b\cdot g)$ belong to $S$. Finally, the pair $(a', g\cdot b\cdot g)$  is
in good position over
$M$ and belongs to $S$, so \[\GP(\tp(a'/M), \tp(g\cdot
b\cdot g/M))\subseteq  S,\] and thus the pairs $(a, g\cdot b\cdot g)$
and $(a\cdot g,
g\cdot b\cdot g)$ belong to $S$, as desired.
\end{proof}
\begin{remark}\label{R:Lshapes}
We do not currently know how to prove the existence of
	generalised $3\times 3$-rectangles, even in the presence of robust
	stability. Our techniques above rely heavily on Fact \ref{F:Roth}, which
	produces arithmetic progressions of length
	$3$ via a model-theoretic argument. In order to prove the existence of
	generalised $3\times 3$-rectangles, we would need to know that every dense
	element of $\GO M$ appears as a common difference of an
	arithmetic progression of length $3$ in a particular dense set,
	which is false in general.
\end{remark}

In the particular case of a binary relation given by the Cayley graph of
a subset $A$ of $G$, the proof of Theorem \ref{T:main_corners}
yields a stronger result, since every dense element $g$ in $\GO M$ is
a popular difference for many corners.
\begin{prop}\label{P:AP_robustlystable}
Consider a non-standard finite group $G$ and an internal
robustly $k$-stable dense subset $A$ of $G$ defined over a countable
rich subset $M$.  For every $g$ in $\GO M$ which is dense over $M$, the sets
$A$ and $g\cdot A$ are comparable, that is,  \[ \mu_G(A\triangle
(g\cdot A))=0.\]

In particular, whenever the element $a$ in $A$ is dense over $M\cup\{g\}$,
$g^m\cdot a$ belongs to $A$ for all $m$ in $\Z$. Hence, the
 $\omega$-internal set $\bigcap_{m\in \Z} g^m\cdot A$ is dense with measure
 $\mu_G(A)>0$.
\end{prop}
\begin{proof}
As in Claim \ref{Claim:prod_inv} of Theorem \ref{T:main_corners},
write $g=c\inv \cdot d$, where $(c, d)$ are in good position over $M$
and both lie in the same type $\tp(g/M)$. Suppose towards a contradiction
that $\mu_G(A\triangle (g\cdot A))>0$, or  equivalently, that
\[ \mu_G\big((c\cdot A) \triangle  (d\cdot A))>0.\]
Without loss of generality, we may assume that $(c\cdot A)\setminus
(d\cdot A)$ is dense, so it contains some element $b$ which is dense over
$M\cup\{c,d\}$. This means that $c\inv b$ belongs to $A$ but
$d\inv b$ does not, or equivalently, the pair $(c, b)$ belongs to the robustly
$k$-stable relation $\mathrm{Cay}(G,A)$, but $(d, b)$ does not. Since both
pairs belong
to $\GP(\tp(g/M), \tp(b/M) )$, this contradicts Theorem
\ref{T:robuststable_main}.

To prove the final part, observe that since $A$ and $g\cdot A$ are comparable, so are $A$ and
$g\inv \cdot A$. By continuity, we need only show that each
intersection \[
\bigcap_{-m\le i\le m} g^i\cdot A \]
has constant measure $\mu_G(A)$ for all $m$ in $\Z$. Otherwise, the internal
set \[ A \setminus \bigcap_{-m\le i\le m} g^i\cdot A \]
is dense, so choose by Lemma
\ref{L:ExtensionDense} an element $a$ in
$A$ which is dense over $M\cup\{g\}$. Remark \ref{R:Comp} implies that
$a$ must belong to both $g\cdot A$ and  $g\inv \cdot A$, and inductively, we
conclude that $a$
belongs to $g^{i} \cdot A$ for all $-m\le i\le m$, which gives the desired
contradiction.
\end{proof}

Lemma \ref{L:Transfer} and the accompanying Remark \ref{R:Transfer} allow us to deduce finitary (albeit
ineffective) versions of
Theorems \ref{T:main_corners} and \ref{T:Lshape}.

\begin{cor}\label{C:corners}
Given an integer $k\ge 1$ and a real number $\delta>0$, there is an integer
$\ell_0(k, \delta)\ge 1$ and real numbers $\theta=\theta(k, \delta)>0$ and
$\epsilon=\epsilon(k, \delta)>0$ with the following property. Let $G$ be a finite group of order $|G|\ge \ell_0$ and let
$X\subseteq G^n$ and $Y\subseteq G^m$ for some $n, m\geq 1$. Let $S\subseteq
X\times Y$  be a relation of size $|S|\ge \delta |G|^{n+m}$ such that the collection
$\HH_k(S)$ of all
half-graphs of height $k$ induced by $S$ on  $G$ has size $|\HH_k(S)|\le \theta
|G|^{k(n+m)}$. Then the set
\[ \Lambda_\Box(S)= \left\{ (a, b, g) \in G^n\times G^m \times G \ | \
(a,b), (a\cdot g,
b), (a, b\cdot g)  \text{ and } (a\cdot g, b\cdot g) \text{ lie in } S
\right\}\]
has size  $|\Lambda_\Box(S)| \ge \epsilon |G|^{n+m+1}$. In particular, if
$G$
is abelian, then $S$ contains a (non-trivial) square.
\end{cor}
\begin{proof}
The proof proceeds by contradiction. Suppose there are $k\ge 1$ and
$\delta>0$ such that for every $\ell\ge 1$ (setting $\theta=1/\ell$
and
$\epsilon=1/\ell$), we can find a finite group $G_\ell$ of order at least
$\ell$
and a relation $S_\ell\subseteq X_\ell\times Y_\ell$ of density $\delta$, with
$X_\ell\subseteq G_\ell^n$ and $Y_\ell\subseteq G_\ell^m$, such that
$\HH_k(S_\ell)\le \frac{1}{\ell} |G_\ell|^{k(n+m)}$, yet
$|\Lambda_\Box(S_\ell)|<\frac{1}{\ell}
|G_\ell^{n+m+1}|$.

In particular, the family $(G_\ell, X_\ell, Y_\ell, S_\ell)_{\ell\in \N}$ is
robustly $k$-stable, as in Remark \ref{R:robust_asympt}. Take a non-principal
ultrafilter $\Ult$ on $\N$ and consider the nonstandard finite group
$G=\prod_{\ell\to \Ult} G_\ell$ as well as the internal sets \[
X=\prod_{\ell\to \Ult} X_\ell, \ Y=\prod_{\ell\to \Ult} Y_\ell \ \text{ and } \
S=\prod_{\ell\to \Ult}
S_\ell.\] By Remark \ref{R:Transfer}, the set $S$ has Loeb measure
$\mu_{G^{n+m}}(S)\ge \delta$ and is robustly $k$-stable by  Remark
\ref{R:robust_asympt}. Choose any countable rich subset $M$. Theorem
\ref{T:main_corners} yields the existence of a triple $(a, b, g)$ in
$\Lambda_\Box(S)$ which is
dense over $M$, so $\mu_{G^{n+m+1}}(\Lambda_\Box(S))\ge \eta$ for some $\eta>0$
by Fact \ref{F:UdiExercise}. Remark \ref{L:Transfer} applied to $r=\eta$ yields
the desired contradiction, since the set
$\Lambda_\Box(S_\ell)$ has size at most $\frac{\eta}{2} |G_\ell|^{n+m+1}$
for sufficiently large $\ell$.
\end{proof}
A straightforward adaptation of the previous proof yields the following result
for finite groups of odd order, so that squaring is an injective map.
\begin{cor}\label{C:Lshape}
Given an integer $k\ge 1$ and a real number $\delta>0$, there is an integer
$\ell_0(k, \delta)\ge 1$ and real numbers $\theta=\theta(k, \delta)>0$ and
$\epsilon=\epsilon(k, \delta)>0$ with the following property. Let $G$ be a finite group of odd order $|G|\ge \ell_0$ and let $X\subseteq G^n$ and $Y\subseteq G^m$ for some $n,m\geq 1$. Let $S\subseteq
X\times Y$  be a relation of size $|S|\ge \delta |G|^{n+m}$ such that the collection
$\HH_k(S)$ of all
half-graphs of height $k$ induced by $S$ on  $G$ has size $|\HH_k(S)|\le \theta
|G|^{k(n+m)}$. Then the set
\[ \Lambda_{2\times 3}(S)= \left\{ (a, b, g) \in G^n\times G^m \times G \ \Big|
\
\parbox{8cm}{$(a, b),
(a\cdot g,
b), (a, b\cdot g),  (a\cdot g, b\cdot g), \\ (a, g\cdot b\cdot g) \text{
	and }  (a\cdot g, g\cdot b\cdot g)  \text{ all lie in } S$}
\right\}\]

has size  $|\Lambda_{2\times 3}(S)| \ge \epsilon |G|^{n+m+1}$. In
particular, if $G$
is abelian, then $S$ contains an $L$-shape.
\end{cor}

In the particular case that the groups in question have bounded exponent, we
can strengthen Corollary \ref{C:corners} using Fact \ref{F:G00} to obtain a
group of finite index such that almost all of its elements witness the
existence of a square.
\begin{cor}\label{C:corners_bddexp}
	Given integers $k, r\ge 1$ and real numbers $\delta, \epsilon>0$, there is
	an integer
	$\ell=\ell(k, r, \delta, \epsilon)\ge 1$ and a real number
	$\theta=\theta(k, r,
	\delta, \epsilon)>0$ with the following property.
	Let $G$ be a finite group $G$ of exponent bounded by $r$, and let
	$X\subseteq G^n$ and $Y\subseteq G^m$ for some $n,m\geq 1$. Let $S\subseteq
	X\times Y$ be a relation of size $|S|\ge \delta |G|^{n+m}$ such that the collection
	$\HH_k(S)$ of all
	half-graphs of height $k$ induced by $S$ on  $G$ has size $|\HH_k(S)|\le
	\theta
	|G|^{k(n+m)}$. Then there exists a subgroup $H$ of $G$ of index at most
	$\ell$ such that the set
	\[  H \setminus  \pi_3(\Lambda_\Box(S)) = \left\{ g \in H \ | \text{ there
	is no $(a,
	b)$ in $G^n\times G^m$ with $(a, b, g)$ in $\Lambda_\Box(S)$} 	\right\}\]
	has size  $|H\setminus\pi_3(\Lambda_\Box(S))| < \epsilon |H|$.
\end{cor}
\begin{proof}
As in the proof of Corollary \ref{C:corners}, negating quantifiers, we deduce
that for a fixed choice of $k, r, \delta$ and $\epsilon$, for every
$\ell$ in $\N$
(setting $\theta=1/\ell$) there is a finite group $G_\ell$ and a
relation
$S_\ell\subseteq X_\ell\times Y_\ell$ of density $\delta$, with
$X_\ell\subseteq G_\ell^n$ and $Y_\ell\subseteq G_\ell^m$, such
that
\[ |\HH_k(S_\ell)|\le \frac{1}{\ell} |G_\ell|^{k(n+m)},\] yet for
every
subgroup $H$ of index
at most $\ell$, the subset
$H\setminus\pi_3(\Lambda_\Box(S_\ell))$ has size at least
$\epsilon |H|$.

Take a non-principal
ultrafilter $\Ult$ on $\N$ and consider the nonstandard finite group
$G=\prod_{\ell\to \Ult} G_\ell$ as well as the internal sets \[
X=\prod_{\ell\to \Ult} X_\ell, \ Y=\prod_{\ell\to \Ult} Y_\ell \ \text{
and } \
S=\prod_{\ell\to \Ult}
S_\ell.\]   By Remark \ref{R:Transfer}, the set $S$ has Loeb measure
$\mu_{G^{n+m}}(S)\ge \delta$ and is robustly $k$-stable by  Remark
\ref{R:robust_asympt}.  Choose any countable
rich subset $M$. Notice that the nonstandard finite group $G$ has exponent
bounded by $r$ by \L o\'s's Theorem (Theorem \ref{R:internal}). Fact \ref{F:G00} yields
that the subgroup $\GO M$ is a countable intersection of internal subgroups
$H$, each of finite index and defined over $M$.

Theorem \ref{T:main_corners} now implies that the $\omega$-internal set $\GO
M\setminus \pi_3(\Lambda_\Box(S))$ is not dense. By $\aleph_1$-saturation, we
deduce that for some finite-index internal subgroup $H$
defined over $M$, the
internal set $H\setminus\pi_3(\Lambda_\Box(S))$ has $\mu_G$-measure $0$.
Therefore \[ \mu_G(H\setminus\pi_3(\Lambda_\Box(S)))< \frac{\epsilon}{[G:H]}.\]
For sufficiently large $\ell$ in $\N$, we deduce that the subgroup
$H(G_\ell)$ has
index $[G:H]\le \ell$ and \[
|H(G_\ell)\setminus\pi_3(\Lambda_\Box(S_\ell))| <
\frac{\epsilon}{[G:H]} |G_\ell| = \epsilon |H(G_\ell)|.\] This
contradicts our choice
of $G_\ell$.
\end{proof}

Mimicking the proof above, together with Proposition \ref{P:AP_robustlystable},
we deduce the
following finitary version on the existence of arbitrary long arithmetic
progressions in robustly stable dense sets.

\begin{cor}\label{C:AP_robust_bddexp}
	Given integers $k, m, r\ge 1$ and real numbers $\delta, \epsilon,
	\eta>0$, there
	is
	an integer
	$n=n(k, m, r, \delta, \epsilon, \eta)\ge 1$ and a real number
	$\theta=\theta(k, m,
	r,
	\delta, \epsilon, \eta)>0$ with the following property.
	Let $G$ be a finite group of exponent bounded by $r$, and let $A\subseteq G$ be subset of $G$ of size $|A|\geq \delta|G|$ such that the collection
	$\HH_k(\mathrm{Cay}(G,A))$ of all
	half-graphs of height $k$ induced by its Cayley graph on  $G$ has size
	$|\HH_k(\mathrm{Cay}(G,A))|\le
	\theta
	|G|^{2k}$. Then there exists a subgroup $H$ of $G$ of index at most
	$n$ such that
	\[  \left| \{ h \in H \ | \ |\text{m-}{\rm AP}(A,h)|
	<(1-\eta)|A|\} \right| < \epsilon |H|, \]
	where $\text{m-}{\rm AP}(A,h)= \{ a\in A \ | \ \{a,h\cdot
	a,\ldots,h^{m-1}\cdot a\} \subseteq A\}$.
\end{cor}

redpajama_set_name: RedPajamaArXiv


text:
The Real Governments of Blue America
Officially, a big part of the federal government shut down late last month. In important ways, however, America's government went AWOL almost two years earlier, when Donald Trump was inaugurated.
After all, politicians supposedly seek office in order to get stuff done — to tackle real problems and implement solutions. But neither Trump, who spends his energy inventing crises at the border, nor the Republicans who controlled Congress for two years have done any of that. Their only major legislative achievement was a tax cut that blew up the deficit without, as far as anyone can tell, doing anything to enhance the economy's long-run growth prospects.
Meanwhile, there has been no hint of the infrastructure plan Trump promised to deliver. And after many years of denouncing Obamacare and promising to provide a far better replacement, Republicans turned out to have no idea how to do that, and in particular no plan to protect Americans with pre-existing conditions.
Why can't Republicans govern? It's not just that their party is committed to an ideology that says that government is always the problem, never the solution. Beyond that, they have systematically deprived themselves of the ability to analyze policies and learn from evidence, because hard thinking might lead someone to question received doctrine.
And Republicans still control the Senate and the White House. So even when (if?) the shutdown ends, it will be at least two years before we have a government in Washington that's actually capable of, or even interested in, governing.
Until recently Republicans had a virtual lock on state government. Almost half the population lived in states with Republican "trifectas," that is, G.O.P. control of both houses plus the governorship. Democrats had comparable control in California, and pretty much nowhere else.
But elections since then have transformed the picture. New Jersey and Washington went full Democratic in 2017, and six more states, including Illinois and New York, flipped in November. At this point more than a third of Americans live under full Democratic control, not far short of the Republican total.
These newly empowered majorities are moving quickly to start governing again. And the experience of states that already had Democratic trifectas suggests that they may achieve a lot.
Consider the experience of California, where Democrats took full control in 2011. Conservatives lambasted Jerry Brown's increases in taxes, spending and the minimum wage, declaring that the state was committing "economic suicide." In reality, the economy boomed, while California's enthusiastic implementation of health reform brought the uninsured share of the population down from 18 percent in 2011 to just 7 percent in 2017 — almost twice the reduction in the U.S. as a whole.
Or consider New Jersey, where Democrats took control last year and used that control to implement a series of measures — including reimposing the requirement that individuals buy health insurance — that reversed many of the Trump administration's efforts at health care sabotage. The result was a sharp drop in insurance premiums, which are now among the lowest in the nation.
Now that Democratic control has expanded, we can expect to see more of this kind of activism.
Gavin Newsom, California's new governor, has proposed further action on health care, including a New Jersey-style state-level mandate and expanded subsidies for the middle class. Washington's governor is proposing creation of a public option, a state insurance plan residents can buy into. And New York City's mayor is proposing measures that would, he says, guarantee coverage for all New Yorkers, including undocumented immigrants.
And health care isn't the only front for new action. For example, Newsom is also proposing major new spending on education and housing affordability. The latter is very important: Soaring housing costs are the biggest flaw in California's otherwise impressive success story.
Now, let's be clear: Not all of the new Democratic policy proposals will actually be implemented, and not all of those that do go into effect will live up to expectations. There's no such thing as perfection, in policy or in life, and leaders who never experience failures or setbacks aren't taking enough risks.
The point, however, is that newly empowered state and local politicians do seem willing to take risks and try new things in an effort to make progress against the nation's problems.
And that's a very hopeful sign for America, because their example may prove contagious.
Justice Louis Brandeis famously described the states as the laboratories of democracy; right now they're the places where we're seeing what it looks like when elected officials try to do what they were elected to do, and actually govern. If we're lucky, two years from now that attitude may re-establish itself in the nation's capital.
Tags: bill-de-blasio, california, gavin-newsom, governement, government-is-the-problem, government-solutions, ideology, jerry-brown, louis-brandeis-states-as-laboratories-of-democracy, new-jersey, new-york-city, paul-krugman, republican-anti-government, republican-unified-control, trump-crisis-seeking, unified-governement, washington-state
What Philip Roth Didn't Know About Women Could Fill a Book
Only after becoming a novelist myself did I understand my discomfort. Philip Roth is celebrated for bringing my family's tiny slice of the world into the American pantheon, widening the literary canon to include American Jews. It is hardly news to point out that he accomplished this feat at the expense of Jewish women.
.. Roth's three favorite topics — Jews, women and New Jersey — all remain socially acceptable targets of irrational public mockery, and Roth was a virtuoso at mocking the combination of all three.
.. The problem is literary: these caricatures reveal a lack of not only empathy, but curiosity.
.. Shakespeare bothered to give the hated Jewish moneylender Shylock a point of view; Mark Twain bothered to imagine emotions for the runaway slave Jim. Both portrayals still reflected popular prejudice, often horribly so. But they also included a glimmer of humanity beyond it, revealing the artists' curiosity about lives they could only imagine.
.. His strength lay in those brilliantly rendered characters and voices like his. His weakness was that those voices denigrated just about everyone else.
.. how many of these women are, after all, precisely the people who made Philip Roth's success possible. The Jewish New Jersey women I know are talented professionals in every field, and often in those two thankless professions that Roth quite likely required to thrive: teachers and therapists.
.. literature means little; the shared humanity that great literature inspires matters even less. What endures, sadly, is Roth's lack of imagination, the unempathetic and incurious caricaturing of others that he turned into a virtue — and which now defines much of American public life.
.. we're still talking about Roth, just like his works taught us to do. Yet in the years to come, the real meaning of his work will emerge not in how we judge Roth, but in how we judge ourselves.
Tags: caricature, gender, jewish, new-jersey, novel, philip-roth, self-absorption, teachers, therapists, uncurious, unempathetic, william-shakespeare
Most U.S. States Aren't Prepared For the Next Recession
Just 16 states have adequate backup money on hand, according to a new Moody's Analytics analysis
If the next recession hit the U.S. this year, more than a quarter of states would be financially unprepared to weather even a moderate downturn, according to a new report.
Fifteen states would struggle in the case of a recession-related tax revenue slump and spike in demand for services, such as Medicaid. They are more than 5 percentage points below the share of funds left in their budgets they would need to tap, according to a new Moody's Analytics analysis. Another 19 states narrowly fall short.
Just 16 states have adequate backup money on hand, with Alaska having almost three times as much as the state would need to keep its economy buoyant.
EXTRA CASH ON HAND
NECESSARY BACKUP FUNDS
DIFFERENCE BETWEEN EXTRA CASH AND NECESSARY BACK-UP FUNDS
Louisiana 3.10% 27.20% -24.00%
North Dakota 0.70% 20.10% -19.40%
Oklahoma 4.00% 16.00% -12.10%
New Mexico -1.10% 10.00% -11.10%
Illinois 0.40% 11.10% -10.70%
Colorado 5.30% 15.10% -9.80%
New Jersey 1.40% 11.00% -9.60%
Pennsylvania -1.80% 6.90% -8.80%
Missouri 5.40% 13.80% -8.40%
Kansas 1.60% 9.20% -7.60%
Source: blogs.wsj.com
Tags: budget, colorodo, financial-health, illinoise, kansas, lousiana, missouri, new-jersey, new-mexico, north-dakota, pennsylvania, states
Jared Kushner's Entitlement Is New Jersey Born and Bred
In his initial application for a security clearance, Mr. Kushner failed to list a single meeting with a foreign government official during the campaign and transition. (Amended forms now reportedly list over a hundred contacts.)
.. Anyone who has ever driven on the New Jersey Turnpike knows that, at a certain point in the road, the entire Manhattan skyline appears to rise from the surrounding marshland like a close-yet-so-far Land of Oz, both tempting and terrorizing with its mysterious jutting cutouts. To traverse this roadway, as Mr. Kushner surely did as a young man, was undoubtedly to exist in a constant state of aspiration and alienation. No matter one's personal glories, for those who call New Jersey home, and especially those who reside in Northern New Jersey, it's difficult to forget that one is still not from "the city," as the landmass across the river is known. Overcompensation tends to follow. Blind arrogance is an occasional byproduct.
.. it's hard not to feel that he's making the same mistake that so many Trump voters did regarding wealth and entitlement — namely, confusing money for virtue and accomplishment — and the luck of birth for talent.
Tags: arrogance, confusing-money-for-virtue, entitlement, jared-kushner, new-jersey, overcompensation, wealth

redpajama_set_name: RedPajamaCommonCrawl


Describes the service interface for managing groups.
Add a new group and optionally map/replicate it to an external storage manager (if configured).
Add a hierarchical association between two groups.
Get a group Groupfor a given group name.
Retrieves a detached and modifiable List of group names, finding groups matching the corresponding group name filter.
Retrieves a detached and modifiable Group list matching the corresponding group name filter.
Retrieve all the groups which are associated from the provided group.
Retrieve all the groups which are associated to the provided group.
Retrieves a detached and modifiable Group list of all the groups associated to a specific user.
Retrieves a detached and modifiable Group list of all the groups in a specific role.
Whether or not a group exists.
Whether or not a user is in a group.
Remove a hierarchical association between two groups.
If an external security storage manager is used, the group will be mapped/replicated to it as well.
throws - a security exception.
mapped - if the new Group should be mapped/replicated to an external security storage manager (if used) or not.
throws - security exception if the group does not exist.
throws - security exception if the user does not exist.
throws - a security exception if the role does not exist.
throws - security exception if the user or group does not exist.
nameFilter - The filter used to retrieve matching groups.
Default supported hierarchical associations are JetspeedPrincipalAssociationType.IS_A and JetspeedPrincipalAssociationType.IS_PART_OF, but it will depend on the actual runtime configuration if the required JetspeedPrincipalAssociationType is available.
If the corresponding JetspeedPrincipalAssociationType is not available, this method will simply return a empty list.
For a JetspeedPrincipalAssociationType.IS_PART_OF association, this will return all the nested groups which together represent the provided group.
For a JetspeedPrincipalAssociationType.IS_A association, this will return all the groups which extend the provided group.
Note: this method will only return the directly associated groups, not further derived associations.
For a JetspeedPrincipalAssociationType.IS_PART_OF association, this will return (at most) the single group where the provided group is part of.
For a JetspeedPrincipalAssociationType.IS_A association, this will return all the groups which the provided group extends.
Note: this method will only return the directly associated group(s), not further derived associations.


Pocatello, ID – Sophomores Santiago Sierra and Johan Jonhagen started the Idaho State Invitational off on the right track, rolling into the quarterfinals. Sierra, seeded No. 1, had a bye in the first round, but convincingly beat Landon Barlow (Weber State), 6-4, 6-3 in the 2nd round. He will face Kellen Bates (Montana State) in quarterfinal action on Saturday.
Jonhagen beat Jake Flowers (Idaho State), 6-2, 6-1 in the 1st round, followed by yet another impressive performance beating No. 2 seeded Alexander Flink (Montana State) 6-2, 6-2. Next up for the Ute is Sebastian Edin (Idaho State).
The two sophomores survived a scare in doubles play, but came up on top defeating Weber State's Hintze/Yin, 7-5. They will square off on Saturday with James/Edgecombe (Montana State) to fight for a spot in the finals.
Singles and doubles play continues tomorrow in Pocatello, Idaho.
This entry was posted in Schedule/Results and tagged Idaho State Invitational, Johan Jonhagen, Santiago Sierra. Bookmark the permalink.


text:
Energy Economy
Prices & Trends
Environmental Cleanup
Design & Remodeling
Electricity & Fuel
Sealing Your Home
DOE Announces Nearly $65 Million for Biofuels Research to Reduce Airplane and Ship Emissions
Projects Align with Administration's New Goal of Decarbonizing the Aviation Sector by 2050
WASHINGTON, D.C. – As part of a White House roundtable to launch the Sustainable Aviation Fuels (SAF) Grand Challenge to decarbonize the aviation sector by 2050, the U.S. Department of Energy (DOE) today announced $64.7 million in funding for projects focused on producing cost-effective, low-carbon biofuels. These investments will advance technologies to create replacements for petroleum fuels used in heavy-duty forms of transportation, like airplanes and ships, and accelerate America's path to a net-zero emissions economy by 2050.
"Decarbonizing transportation – particularly planes and ships that are difficult to electrify – is an essential part of the path to a net-zero carbon future," said Secretary of Energy Jennifer M. Granholm. "These investments mobilize industries to join this effort, which will create new, good-paying jobs across the biofuels, chemical, and agricultural supply chains and boost economic activity in rural economies."
As part of the SAF Grand Challenge, DOE also signed a memorandum of understanding with the U.S. Department of Transportation and U.S. Department of Agriculture to collaborate on the needed research, development, and demonstration (RD&D) to reach the goals of supplying at least 3 billion gallons of SAF per year by 2030 and sufficient SAF to meet 100% of aviation fuel demand – currently 35 billion gallons per year – by 2050.
These efforts seek to cut carbon emissions from the aviation and shipping industries, which – because of their size – are more challenging to electrify. Biofuels, which are produced by converting the renewable carbon from recently living organic materials like crop waste, food waste, and algae into a liquid fuel, can serve as a low-carbon equivalent to fossil-based fuels such as gasoline, jet, and diesel fuel.
"I'm thrilled that the Department of Energy has awarded a grant to Lignolix, a company with roots in Professor Thomas Epps' lab at the University of Delaware, to support their work to transform plant waste into clean fuels and sustainable materials – another success story out of Delaware's entrepreneurship community," said U.S. Senator Chris Coons (D-DE).
"We must cut emissions in aviation and commercial shipping if we're going to meet our net-zero goals," said U.S. Senator John Hickenlooper (CO). "This DOE research grant will put Colorado's National Renewable Energy Research Lab at the center of those efforts and at the forefront of biofuel research."
"I am proud that the National Renewable Energy Laboratory (NREL), located in my district, is helping lead the way to reduce emissions from the aviation industry. NREL's work will help lower the price of cellulosic sugars, which can be used to make sustainable aviation fuel or other biofuels and bioproducts," said U.S. Representative Ed Perlmutter (CO-07). "By reducing the price of biofuels, we can incentivize the use of more sustainable options in the aviation industry and help accelerate our path to a cleaner future."
"Biofuels are a critical component of our nation's energy portfolio and our agriculture economy," said U.S. Representative Rodney Davis (IL-13). "They help reduce emissions while promoting American energy independence. This DOE grant will assist ADM and others in developing new innovations in the biofuels sector. I look forward to watching ADM utilize this grant and continuing to partner with industry stakeholders to further promote the production and use of biofuels."
The 22 selected projects target high-impact bioenergy technology RD&D to bolster foundational knowledge and scale up systems to produce low-carbon biofuels at lower costs. Among the projects are:
National Renewable Energy Laboratory (Golden, CO) will lower the cost and carbon intensity of producing a highly fermentable sugar from corn stover. (Award amount: $2,800,000)
Archer Daniels Midland (Decatur, IL) will couple isobutanol (a precursor for sustainable aviation fuel) fermentation with a membrane separations system – reducing energy used in the separation process by 50%. (Award amount: $3,466,844)
Alder Energy (Charleston, SC) will convert miscanthus, a highly promising biomass crop, to SAF through their advanced pyrolysis oil technology, a process that utilizes heat, pressure, and solvents to deconstruct the miscanthus into oils for conversion to SAF. (Award amount: $3,000,000)
D3MAX (Grand Forks, ND) will design a pilot plant to validate technology for first generation ethanol plants to produce ethanol from corn stover, which is then converted to SAF. (Award amount: $499,988)
T2C Energy (Pinellas Park, FL) was selected to design a demonstration scale plant that converts waste landfill gas to SAF or renewable diesel. (Award amount: $533,619)
AVAPCO (Thomaston, GA) will demonstrate a production process for clean, affordable cellulosic sugars, which are derived from agricultural or woody waste residues and can help reduce greenhouse gas emissions for a variety of products including SAF, bioplastics, and biopolymers. (Award amount: $2,800,000)
Quasar Energy Group (Independence, OH) will use an anaerobic digestor to convert food waste to SAF precursors. The microbes that digest the food waste produce a type of chemical called a volatile fatty acid, which can be converted to SAF. (Award amount: $3,500,000)
These investments are administered by DOE's Bioenergy Technologies Office (BETO), which is focused on developing technologies that convert domestic biomass and other waste resources into low-carbon biofuels and bioproducts. BETO is increasing its emphasis on partnering with industry to demonstrate technologies at large scale in recognition of the urgent need to reduce risks and scale-up SAF production.
For more information about the selected projects, visit BETO's website.
STEM at DOE


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