# Maximal Haagerup subgroups in
$\mathbb{Z}^{n+1}\rtimes_{\rho_{n}}GL_{2}(\mathbb{Z})$

Alain VALETTE

###### Abstract

For $n\geq 1$, let $\rho_{n}$ denote the standard action of
$GL_{2}(\mathbb{Z})$ on the space $P_{n}(\mathbb{Z})\simeq\mathbb{Z}^{n+1}$ of
homogeneous polynomials of degree $n$ in two variables, with integer
coefficients. For $G$ a non-amenable subgroup of $GL_{2}(\mathbb{Z})$, we
describe the maximal Haagerup subgroups of the semi-direct product
$\mathbb{Z}^{n+1}\rtimes_{\rho_{n}}G$, extending the classification of Jiang-
Skalski [JS21] of the maximal Haagerup subgroups in $\mathbb{Z}^{2}\rtimes
SL_{2}(\mathbb{Z})$. We prove that, for $n$ odd, the group
$P_{n}(\mathbb{Z})\rtimes SL_{2}(\mathbb{Z})$ admits infinitely many pairwise
non-conjugate maximal Haagerup subgroups which are free groups; and that, for
$n$ even, the group $P_{n}(\mathbb{Z})\rtimes GL_{2}(\mathbb{Z})$ admits
infinitely many pairwise non-conjugate maximal Haagerup subgroups which are
isomorphic to $SL_{2}(\mathbb{Z})$.

##  1 Introduction

For discrete countable groups, the Haagerup property is a weak form of
amenability that proved to be useful in many questions in analytical group
theory, ranging from K-theory to dynamical systems (see [CCJJV]). It is not
difficult to see that, in a countable group, every Haagerup subgroup is
contained in a maximal one (see Proposition 1.3 in [JS21] or lemma 2.1 below).
This raises the question, given a group $G$, of describing the maximal
Haagerup subgroups of $G$.

The study of maximal Haagerup subgroups was initiated by Y. Jiang and A.
Skalski [JS21], and we refer to this paper for many interesting and intriguing
examples. We mention here Theorem 2.12 in [JS21], where the authors classify
maximal Haagerup subgroups of the semi-direct product $\mathbb{Z}^{2}\rtimes
SL_{2}(\mathbb{Z})$. This example is especially interesting in view of a
result of Burger (Example 2 following Proposition 7 in [Bur91]): if $G$ is a
non-amenable subgroup of $SL_{2}(\mathbb{Z})$, then the pair
$(\mathbb{Z}^{2}\rtimes G,\mathbb{Z}^{2})$ has the relative property (T); in
particular $\mathbb{Z}^{2}\rtimes G$ is not Haagerup, in spite of the fact
that $\mathbb{Z}^{2}$ and $G$ are both Haagerup.

######  Theorem 1.1 (Theorem 2.12 in [JS21]).

Let $H$ be a maximal Haagerup subgroup of $\mathbb{Z}^{2}\rtimes
SL_{2}(\mathbb{Z})$. Then there is a dichotomy:

  1. 1.

either $H=\mathbb{Z}^{2}\rtimes C$, where $C$ is a maximal amenable111We
classify those subgroups in Proposition 2.6. subgroup of $SL_{2}(\mathbb{Z})$
;

  2. 2.

or $H\cap\mathbb{Z}^{2}$ is trivial; then $H$ is not amenable. If $K$ denotes
the image of $H$ under the quotient map $\mathbb{Z}^{2}\rtimes
SL_{2}(\mathbb{Z})\rightarrow SL_{2}(\mathbb{Z})$ (so that $K$ is isomorphic
to $H$), then $H=\\{(b(g),g):g\in K\\}$ where $b:K\rightarrow\mathbb{Z}^{2}$
is a 1-cocycle that cannot be extended to a larger subgroup of
$SL_{2}(\mathbb{Z})$.

######  Remark 1.2.

Denote by $L(G)$ the group von Neumann algebra of the group $G$. In Theorem
3.1 of [JS21], Jiang and Skalski are able to prove the stronger result that,
if $C$ is a maximal amenable subgroup of $SL_{2}(\mathbb{Z})$, such that
$\mathbb{Z}^{2}\rtimes C$ has infinite conjugacy classes, then
$L(\mathbb{Z}^{2}\rtimes C)$ is a maximal Haagerup von Neumann subalgebra of
$L(\mathbb{Z}^{2}\rtimes SL_{2}(\mathbb{Z}))$, where the Haagerup property for
finite von Neumann algebras was defined in [Jo02]. Subsequently Y. Jiang
(Corollary 4.3 in [Ji21]) showed that $L(SL_{2}(\mathbb{Z}))$ is a maximal
Haagerup subalgebra in $L(\mathbb{Z}^{2}\rtimes SL_{2}(\mathbb{Z}))$. It was
pointed out to us by A. Skalski, that there is no known example of a maximal
Haagerup subgroup $H$ in a group $G$, such that $L(H)$ is not maximal Haagerup
in $L(G)$.

We now come to the present paper. Fix $n\geq 1$ and, for
$A=\mathbb{Z},\mathbb{R},\mathbb{C}$, denote by $P_{n}(A)$ the set of
polynomials in two variables $X,Y$, with coefficients in $A$, which are
homogeneous of degree $n$, so that $P_{n}(A)\simeq A^{n+1}$. It is a classical
fact that $GL_{2}(\mathbb{R})$ admits an irreducible representation $\rho_{n}$
on $P_{n}(\mathbb{R})$ given as follows: for $P\in P_{n}(\mathbb{R})$ and
$A=\left(\begin{array}[]{cc}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{array}\right)\in GL_{2}(\mathbb{R})$, set:

$(\rho_{n}(A)(P))(X,Y)=P((X,Y)\cdot A)=P(a_{11}X+a_{21}Y,a_{12}X+a_{22}Y).$

Since $\rho_{n}(GL_{2}(\mathbb{Z}))$ leaves $P_{n}(\mathbb{Z})$ invariant, we
may form the semi-direct product

$G_{n}=:P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}GL_{2}(\mathbb{Z})$

(observe $G_{1}=\mathbb{Z}^{2}\rtimes GL_{2}(\mathbb{Z})$, that contains
$\mathbb{Z}^{2}\rtimes SL_{2}(\mathbb{Z})$ as a subgroup of index 2). Let $G$
be a non-amenable subgroup of $GL_{2}(\mathbb{Z})$: our goal is to classify
the maximal Haagerup subgroups of $P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}G\subset
G_{n}$. So here is our first main result, extending Theorem 1.1:

######  Theorem 1.3.

Fix $n\geq 1$ and a non-amenable subgroup $G$ of $GL_{2}(\mathbb{Z})$. Let $H$
be a maximal Haagerup subgroup of $P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}G$. Then
there is a dichotomy.

  1. 1.

Either $H$ is amenable, in which case
$H=P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}C$, with $C$ maximal amenable in $G$;

  2. 2.

or $H$ is non amenable and there exists a subgroup $K\subset G$ isomorphic to
$H$, and a 1-cocycle $b\in Z^{1}(K,P_{n}(\mathbb{Z}))$ such that

$H=\\{(b(k),k):k\in K\\}$

and $b$ cannot be extended to a larger subgroup of $G$. (In particular, if $b$
is a 1-coboundary, then $K=G$.)

Conversely, any subgroup of $P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}G$ of one of
the above 2 forms, defines a maximal Haagerup subgroup in
$P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}G$.

Even if the conclusion looks very similar to Theorem 1.1, we emphasize that we
had to come up with a totally different argument to show that the intersection
$H\cap P_{n}(\mathbb{Z})$ is either $P_{n}(\mathbb{Z})$ or trivial.

Theorems 1.1 and 1.3 raise an interesting - and somewhat unusual - question in
the 1-cohomology of groups: to describe maximal Haagerup subgroups that are
non-amenable, we need to describe 1-cocycles that cannot be extended to a
larger subgroup. In the case of $\rho_{1}$, this question is extensively
studied in section 2 of [JS21] (from Lemma 2.14 to Proposition 2.18). We
attack the question by observing that, for $K$ a subgroup in
$SL_{2}(\mathbb{Z})$ or $GL_{2}(\mathbb{Z})$, a cocycle $b$ on $K$ cannot be
extended to a larger subgroup if and only if, for any overgroup $L$ of $K$,
the 1-cohomology class of $b$ is not in the image of the restriction map
$H^{1}(L,P_{n}(\mathbb{Z}))\rightarrow H^{1}(K,P_{n}(\mathbb{Z}))$. This
suggests to look for subgroups $K$ which are maximal, or close to being
maximal, in $SL_{2}(\mathbb{Z})$ or $GL_{2}(\mathbb{Z})$. Using this approach
we prove the following result, which seems to be new even for $n=1$:

######  Theorem 1.4.

Assume that $n\geq 1$ is odd. Then there exists a free subgroup $K$ in
$SL_{2}(\mathbb{Z})$ such that $P_{n}(\mathbb{Z})\rtimes SL_{2}(\mathbb{Z})$
contains infinitely many maximal Haagerup subgroups of the form
$H=\\{(b(k),k):k\in K\\}$, with $b\in Z^{1}(K,P_{n}(\mathbb{Z}))$, which are
pairwise not conjugate under $P_{n}(\mathbb{Z})$. Moreover the subgroup $K$
may be chosen either with infinite index or with arbitrarily large finite
index.

In the even case we get:

######  Theorem 1.5.

Assume that $n\geq 2$ is even. Then the semi-direct product $G_{n}$ contains
infinitely many maximal Haagerup subgroups $H$ of the form $H=\\{(b(g),g):g\in
SL_{2}(\mathbb{Z})\\}$, where $b\in
Z^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$, which moreover are pairwise not
conjugate under $P_{n}(\mathbb{Z})$.

It turns out that the cases of odd and even $n$’s are very different: the
reason is that the $GL_{2}(\mathbb{Z})$-action on $P_{n}(\mathbb{Z})$ factors
through $PGL_{2}(\mathbb{Z})$ for $n$ even, while it doesn’t for $n$ odd.

The paper is organized as follows. Section 2 is devoted to prerequisites, with
the exception of Proposition 2.6, describing precisely the maximal amenable
subgroups in $SL_{2}(\mathbb{Z})$. In section 3 we prove the analogue of
Burger’s aforementioned result, namely that the pair
$(G_{n},P_{n}(\mathbb{Z}))$ has the relative property (T) for $n\geq 1$.
Theorem 1.3 is proved in section 4, while cohomological questions are treated
in sections 5 ($n$ odd) and 6 ($n$ even); in particular Theorem 1.4 is proved
in section 5, and Theorem 1.5 in section 6. Theorem 1.5 is actually proved by
establishing explicit formulae for the ranks 222By the rank of a finitely
generated abelian group, we mean the torsion-free rank. of
$H^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ and
$H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$; those formulae may have their
own interest. A combinatorial application of the methods ends the paper.

Acknowledgements: Thanks are due to Y. Benoist, Y. Cornulier, L. Hayez, P.
Jolissaint, A. Skalski, Y. Stalder and A. Zumbrunnen for useful discussions at
various stages of the project.

##  2 Generalities

###  2.1 Haagerup property

The following results on countable groups will be used freely:

  * •

The group $GL_{2}(\mathbb{Z})$ is Haagerup (see sections 1.2.2 or 1.2.3 in
[CCJJV]).

  * •

An extension of a Haagerup group by an amenable group, is Haagerup; in
particular every amenable group is Haagerup (see Proposition 6.1.5 in
[CCJJV]).

######  Lemma 2.1.

Let $G$ be a countable group. Any Haagerup subgroup of $G$ is contained in a
maximal Haagerup subgroup.

Proof: (compare with Proposition 1.3 in [JS21]) To apply Zorn’s lemma, we must
show that being Haagerup is stable by arbitrary increasing unions of
subgroups. This follows from the fact that, for countable groups, being
Haagerup is a local property, i.e a countable group is Haagerup if and only if
every finitely generated subgroup is Haagerup (see Proposition 6.1.1 in
[CCJJV]). $\blacksquare$

###  2.2 1-cohomology of groups

We recall the following facts from the cohomology of groups. For $G$ a group
and $A$ a $G$-module, define the group of 1-cocycles:

$Z^{1}(G,A)=\\{b:G\rightarrow A:b(gh)=gb(h)+b(g)\,\mbox{for all}\,g,h\in
G\\};$

the group of 1-coboundaries:

$B^{1}(G,A)=\\{b\in Z^{1}(G,A):\mbox{there exists}\,a\in A\,\mbox{such
that}\,b(g)=ga-a\,\mbox{for all}\,g\in G\\};$

and the first cohomology group:

$H^{1}(G,A)=Z^{1}(G,A)/B^{1}(G,A).$

The following is proved e.g. in Proposition 2.3 of [Br82].

######  Proposition 2.2.

Let $G$ be a group and let $A$ be a $G$-module. Splittings of the split
extension

$0\rightarrow A\rightarrow A\rtimes G\rightarrow G\rightarrow 1$

are given by $i_{b}:G\rightarrow A\rtimes G:g\mapsto(b(g),g)$ with $b\in
Z^{1}(G,A)$ and are classified up to $A$-conjugacy by the first cohomology
group $H^{1}(G,A)$. $\blacksquare$

The first part of the following lemma is a variation on Corollaire 7.2, page
39 of [Gu80].

######  Lemma 2.3.

Let $G$ be a group and $A$ be a $G$-module with cancellation by 2 (i.e.
$2x=0\Rightarrow x=0$).

  1. 1.

Assume $G$ contains a central element $z$ that acts on $A$ by $-1$. Then
$H^{1}(G,A)$ is a vector space over the field with 2 elements. Moreover $b\in
B^{1}(G,A)$ if $b(z)$ belongs to $2A$.

  2. 2.

Assume that $G$ contains a central element $c$ of order 2, such that the
module action on $A$ factors through $G/\langle c\rangle$. Then the map
$H^{1}(G/\langle c\rangle,A)\rightarrow H^{1}(G,A)$ (induced by the quotient
map $G\rightarrow G/\langle c\rangle$) is an isomorphism.

Proof:

  1. 1.

We must prove that, for every $b\in Z^{1}(G,A)$, we have $2b\in B^{1}(G,A)$.
But for $g\in G$ we have, using $zg=gz$ in the 3rd equality:

$b(z)-b(g)=b(z)+zb(g)=b(zg)=b(gz)=gb(z)+b(g).$

Re-arranging:

$2b(g)=(1-g)b(z).$

Hence $2b\in B^{1}(G,A)$. If we may write $b(z)=2a$ for some $a\in A$,
cancelling 2 on both sides we get $b(g)=(1-g)a$, so $b\in B^{1}(G,A)$.

  2. 2.

The map $H^{1}(G/\langle c\rangle,A)\rightarrow H^{1}(G,A)$ is clearly
injective. For surjectivity, fix $b\in Z^{1}(G,A)$ and consider
$b(c^{2})=b(1)=0$: using the cocycle relation

$0=(1+c)b(c)=2b(c),$

hence $b(c)=0$ and $b$ factors through $G/\langle c\rangle$. $\blacksquare$

###  2.3 About $SL_{2}(\mathbb{Z})$ and $GL_{2}(\mathbb{Z})$

We denote by $C_{n}$ the cyclic group of order $n$, and by $D_{n}$ the
dihedral group of order $2n$. Set

$s=\left(\begin{array}[]{cc}0&-1\\\ 1&0\end{array}\right),\quad
t=\left(\begin{array}[]{cc}0&-1\\\
1&1\end{array}\right),\quad\varepsilon=\begin{pmatrix}-1&0\\\
0&-1\end{pmatrix},\quad w=\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)\quad.$

It is well-known (see Example 1.5.3 in [Se77]) that $SL_{2}(\mathbb{Z})$
admits a decomposition as an amalgamated product:

$SL_{2}(\mathbb{Z})\simeq C_{4}\ast_{C_{2}}C_{6}.$

with $C_{4}=\langle s\rangle,C_{6}=\langle t\rangle$ and
$C_{2}=\\{1,\varepsilon\\}$. It extends to an amalgamated product
decomposition of $GL_{2}(\mathbb{Z})$ (see Section 6 in [BT16]):

$GL_{2}(\mathbb{Z})\simeq D_{4}\ast_{D_{2}}D_{6},$ (1)

with $D_{4}=\langle s,w\rangle,D_{6}=\langle t,w\rangle$ and
$D_{2}=\langle\varepsilon,w\rangle$.

######  Lemma 2.4.

The amenable radical of $GL_{2}(\mathbb{Z})$ is the subgroup $C_{2}$.

Proof: This is a very particular case of a result of Cornulier (Proposition 7
in [Co09]): let $G=A\ast_{C}B$ be an amalgamated product such that $[A:C]\geq
2$ and $[B:C]\geq 3$. Then the amenable radical of $G$ is the largest normal
subgroup of $C$ which is amenable and normalized both by $A$ and $B$.
$\blacksquare$

Our aim now is to describe the maximal amenable subgroups of
$PSL_{2}(\mathbb{Z})$. Recall that an element $A\in
SL_{2}(\mathbb{R}),A\neq\pm Id$, is elliptic if $|Tr(A)|<2$, parabolic if
$|Tr(A)|=2$, and hyperbolic if $|Tr(A)|>2$. These concepts clearly descend to
$PSL_{2}(\mathbb{R})$. For $A=\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\in SL_{2}(\mathbb{R})$, we denote by
$\left[\begin{array}[]{cc}a&b\\\ c&d\end{array}\right]$ the image of $A$ in
$PSL_{2}(\mathbb{R})$.

######  Lemma 2.5.

  1. 1.

The maximal amenable subgroups ot $PSL_{2}(\mathbb{Z})$ are isomorphic to
$C_{3}$, or $\mathbb{Z}$, or $D_{\infty}$ (the infinite dihedral group).

  2. 2.

If $A\in PSL_{2}(\mathbb{Z})$ is parabolic, then $A$ is contained in a unique
maximal amenable subgroup, isomorphic to $\mathbb{Z}$.

  3. 3.

If $A=\left[\begin{array}[]{cc}a&b\\\ c&d\end{array}\right]\in
PSL_{2}(\mathbb{Z})$ is hyperbolic, then $A$ is contained in a unique maximal
amenable subgroup, isomorphic either to $\mathbb{Z}$ or $D_{\infty}$. The
second case happens if $A$ is conjugate to a symmetric matrix. If the second
case happens, then the integer binary quadratic form

$Q_{A}(x,y)=:bx^{2}+(d-a)xy-cy^{2}$

represents $-b$ over $\mathbb{Z}$.

Proof:

  1. 1.

Since $PSL_{2}(\mathbb{Z})$ is the free product $C_{2}\ast C_{3}$, by the
Kurosh theorem any subgroup of $PSL_{2}(\mathbb{Z})$ is a free product of a
free group with conjugates of the free factors. So a subgroup does not contain
a free group if and only if it is isomorphic to one of the following:
$C_{2},C_{3},\mathbb{Z},C_{2}\ast C_{2}=D_{\infty}$; note that these four
groups are amenable. It remains to show that any copy of $C_{2}$ is contained
in at least one copy of $D_{\infty}$; but any element of order 2 in
$PSL_{2}(\mathbb{Z})$ is conjugate to the image of $s$, and the images of $s$
and $tst^{-1}$ together generate a copy of $D_{\infty}$.

  2. 2.

If $A$ is parabolic then, up to conjugacy in $PSL_{2}(\mathbb{Z})$, we may
assume that $A=\left[\begin{array}[]{cc}1&k\\\ 0&1\end{array}\right]$ for some
$k\in\mathbb{Z},\,k\neq 0$. Let $H$ be a maximal amenable subgroup of
$PSL_{2}(\mathbb{Z})$ containing $A$; by the previous point $H$ is isomorphic
either to $\mathbb{Z}$ or to $D_{\infty}$. In both cases $H$ normalizes the
subgroup $\langle A\rangle$. But by direct computation the normalizer of
$\langle A\rangle$ is $\langle\left[\begin{array}[]{cc}1&1\\\
0&1\end{array}\right]\rangle$, which is therefore the unique maximal amenable
subgroup contaning $A$.

  3. 3.

Let $H$ be a maximal amenable subgroup of $PSL_{2}(\mathbb{Z})$ containing
$A$. Again, $H$ normalizes $\langle A\rangle$. But $A$ is conjugate in
$PSL_{2}(\mathbb{R})$ to $\left[\begin{array}[]{cc}\lambda&0\\\
0&1/\lambda\end{array}\right]$ (with $\lambda>1$) whose normalizer in
$PSL_{2}(\mathbb{R})$ is

$\mathbb{R}\rtimes C_{2}=\\{\left[\begin{array}[]{cc}e^{t}&0\\\
0&e^{-t}\end{array}\right]:t\in\mathbb{R}\\}\rtimes\langle\left[\begin{array}[]{cc}0&-1\\\
1&0\end{array}\right]\rangle.$

So $H$ is the intersection of $PSL_{2}(\mathbb{Z})$ with a given conjugate of
$\mathbb{R}\rtimes C_{2}$, which proves uniqueness of $H$ (and re-proves that
$H$ is isomorphic either to $\mathbb{Z}$ or to $D_{\infty}$).

For the second statement, we may assume that $A$ is symmetric. Then, as the
eigenspaces of $A$ are orthogonal in $\mathbb{R}^{2}$, we see without
computation that $\left[\begin{array}[]{cc}0&-1\\\ 1&0\end{array}\right]$
conjugates $A$ to $A^{-1}$, so that $A$ and $\left[\begin{array}[]{cc}0&-1\\\
1&0\end{array}\right]$ together generate a copy of $D_{\infty}$ (then
contained in a maximal one).

For the final statement, observe that the maximal amenable subgroup containing
$A$ is isomorphic to $D_{\infty}$ if and only if there exists an involution
$B\in PSL_{2}(\mathbb{Z})$ that conjugates $A$ into $A^{-1}$. Assume such a
$B$ exists. Then, denoting by $\overline{A},\overline{B}$ lifts of $A,B$ in
$SL_{2}(\mathbb{Z})$, we will have
$\overline{B}\,\overline{A}\,\overline{B}^{-1}=\pm\overline{A}^{-1}$. Taking
the trace of both sides, we see that the minus sign leads to
$Tr(\overline{A})=0$, in contradiction with $|Tr(\overline{A})|>2$. So
$\overline{B}\,\overline{A}\,\overline{B}^{-1}=\overline{A}^{-1}$ or
$\overline{B}\,\overline{A}=\overline{A}^{-1}\,\overline{B}$. Now
$\overline{B}$ has order 4 in $SL_{2}(\mathbb{Z})$, and therefore
$Tr(\overline{B})=0$. So we may write
$\overline{B}=\left(\begin{array}[]{cc}x&y\\\ z&-x\end{array}\right)$ and we
get

$\left(\begin{array}[]{cc}x&y\\\
z&-x\end{array}\right)\cdot\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)=\left(\begin{array}[]{cc}d&-b\\\
-c&a\end{array}\right)\cdot\left(\begin{array}[]{cc}x&y\\\
z&-x\end{array}\right).$

Equating the (1,1)-coefficients on both sides yields:

$ax+cy=dx-bz\Longleftrightarrow z=\frac{(d-a)x-cy}{b}.$

Taking into account the condition $\det(\overline{B})=1$, i.e. $-1=x^{2}+yz$,
and inserting the previous value of $z$:

$-1=x^{2}+y(\frac{(d-a)x-cy}{b})\Longleftrightarrow Q_{A}(x,y)=-b,$

which concludes the proof. Note that if $A$ is symmetric, i.e. $b=c$, then
$Q_{A}(0,1)=-b$. $\blacksquare$

######  Example 1.

We claim that the maximal amenable subgroup of $PSL_{2}(\mathbb{Z})$
containing $A=\left[\begin{array}[]{cc}3&1\\\ 2&1\end{array}\right]$ is
infinite cyclic. To see it, we check that the quadratic form
$Q_{A}(x,y)=x^{2}-2xy-2y^{2}$ does not represent -1. But the equation
$x^{2}-2xy-2y^{2}=-1$ is equivalent to $(x-y)^{2}-3y^{2}=-1$, which leads to
Pell’s equation $X^{2}-3Y^{2}=-1$; as the fundamental unit $2+\sqrt{3}$ in
$\mathbb{Z}[\sqrt{3}]$ has norm 1, this equation has no solution.

We now lift the results of Lemma 2.5 to $SL_{2}(\mathbb{Z})$. Since a maximal
amenable subgroup of $SL_{2}(\mathbb{Z})$ must clearly contain the center, we
see that maximal amenable subgroups of $SL_{2}(\mathbb{Z})$ are exactly
pullbacks of maximal amenable subgroups of $PSL_{2}(\mathbb{Z})$ by the
quotient map $SL_{2}(\mathbb{Z})\rightarrow PSL_{2}(\mathbb{Z})$. Observe that
the inverse image of an involution in $PSL_{2}(\mathbb{Z})$ has order 4 in
$SL_{2}(\mathbb{Z})$; consequently the inverse image in $SL_{2}(\mathbb{Z})$
of a copy of $D_{\infty}$ in $PSL_{2}(\mathbb{Z})$, will be isomorphic to the
semi-direct product $\mathbb{Z}\rtimes C_{4}$ where a generator of $C_{4}$
acts on $\mathbb{Z}$ by $n\mapsto-n$. So we get immediately the following,
that improves on Proposition 2.13 in [JS21]:

######  Proposition 2.6.

  1. 1.

The maximal amenable subgroups ot $SL_{2}(\mathbb{Z})$ are isomorphic to
$C_{6}$, or $\mathbb{Z}\times C_{2}$, or $\mathbb{Z}\rtimes C_{4}$.

  2. 2.

For parabolic (resp. hyperbolic) matrices in $SL_{2}(\mathbb{Z})$, item (2)
(resp. item (3)) of lemma 2.5 applies with the obvious changes. $\blacksquare$

##  3 Relative property (T)

######  Proposition 3.1.

Let $G$ be a non-amenable subgroup of $GL_{2}(\mathbb{Z})$.

  1. 1.

The restriction of the representation $\rho_{n}$ to $G$ is irreducible.

  2. 2.

The pair $(P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}G,P_{n}(\mathbb{Z}))$ has the
relative property (T). In particular $P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}G$ is
not Haagerup.

Proof:

  1. 1.

Since $G\cap SL_{2}(\mathbb{Z})$ has index at most 2 in $G$, replacing $G$ by
$G\cap SL_{2}(\mathbb{Z})$ we may assume that $G\subset SL_{2}(\mathbb{Z})$.
Let $L$ be the Zariski closure of $G$ in $SL_{2}(\mathbb{R})$, so that $L$ is
a Lie subgroup of $SL_{2}(\mathbb{R})$, hence of dimension 0, 1, 2 or 3. As
Lie subgroups of dimension 0, 1, 2 are virtually solvable hence amenable, $L$
has dimension 3, i.e. $G$ is Zariski dense in $SL_{2}(\mathbb{R})$. Since the
representation $\rho_{n}$ is algebraic, irreducibility is preserved by passing
to a Zariski dense subgroup.333We recall the argument: if $W$ is a
$\rho_{n}(G)$-invariant subspace, and $(f_{i})_{i\in I}$ is a set of linear
forms such that $W=\cap_{i\in I}\text{Ker}(f_{i})$, then
$\rho_{n}(G)$-invariance of $W$ is equivalent to $f_{i}(\rho_{n}(g)w)=0$ for
every $g\in G,w\in W,i\in I$. View this as a system of polynomial equations in
the matrix coefficients of $g$: it vanishes on $G$, hence also vanishes on
$SL_{2}(\mathbb{R})$ by Zariski density.

  2. 2.

Set $V_{n}=P_{n}(\mathbb{R})$. By Proposition 7 in [Bur91] (see especially
Example (2) on p. 62 of [Bur91]), if $G$ does not fix any probability measure
on the projective space $P(V_{n}^{*})$, then the pair
$(P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}G,P_{n}(\mathbb{Z}))$ has the relative
property (T). Since the representation $\rho_{n}$ of $SL_{2}(\mathbb{R})$ is
equivalent to its contragredient $\rho_{n}^{*}$, it is enough to check that
there is no $G$-fixed probability measure on the projective space $P(V_{n})$.
So assume by contradiction that there is such a probability measure $\mu$.
Then, by Corollary 3.2.2 in [Zim], there are exactly two cases:

     * •

The measure $\mu$ is not supported on a finite union of proper projective
subspaces. Then the stabilizer $PGL(V_{n})_{\mu}$ is compact, which
contradicts the fact that the image of $G$ in $PGL(V_{n})$ is infinite
discrete.

     * •

There exists a proper linear subspace $W$ in $V_{n}$ such that $\mu([W])>0$
(where $[W]$ denotes the image of $W$ in $P(V_{n})$), and moreover the orbit
of $[W]$ under the stabilizer $PGL(V_{n})_{\mu}$ is finite. In particular
there is a finite index subgroup of $PGL(V_{n})_{\mu}$ that leaves $[W]$
invariant. So there is a finite index subgroup $G_{0}$ of $G$ that leaves the
linear subspace $W$ invariant, in contradiction with the first part of the
Proposition. $\blacksquare$

##  4 Maximal Haagerup subgroups

An interesting question raised in [JS21] is whether every countable group
admits a Haagerup radical, i.e. a unique maximal normal subgroup with the
Haagerup property. We first show that $G_{n}$ admits such a Haagerup radical.

######  Proposition 4.1.

For every $n\geq 1$, the Haagerup radical of $G_{n}$ is
$P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}C_{2}$.

Proof: Set $U=:P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}C_{2}$. Let $N\triangleleft
G_{n}$ be a normal Haagerup subgroup, we want to prove that $N$ is contained
in $U$. We proceed as in the proof of Proposition 2.10 in [JS21]: the subgroup
$UN$ is normal and since $UN/N\simeq U/(U\cap N)$ is amenable, $UN$ is an
amenable extension of a Haagerup group, hence $UN$ is Haagerup. Since $UN$
contains $U$ we have in particular $P_{n}(\mathbb{Z})\subset UN$ so
$UN=\mathbb{Z}^{n+1}\rtimes_{\rho_{n}}K$, for some normal subgroup
$K\triangleleft GL_{2}(\mathbb{Z})$. By Proposition 3.1(2), the subgroup $K$
must be amenable, i.e. $K\subset C_{2}$ by lemma 2.4. So $UN\subset U$ and
therefore $N\subset U$. $\blacksquare$

Since $P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}C_{2}$ is actually amenable, we
immediately have:

######  Corollary 4.2.

The amenable radical of $G_{n}$ is $P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}C_{2}$.
$\blacksquare$

We now come to the proof of our first main result.

Proof of Theorem 1.3: Let

$q_{n}:P_{n}(\mathbb{Z})\rtimes GL_{2}(\mathbb{Z})\rightarrow
GL_{2}(\mathbb{Z}):(v,S)\mapsto S$

be the quotient map. Observe that, as $\text{Ker}(q_{n}|_{H})=H\cap
P_{n}(\mathbb{Z})$ is abelian, $H$ is amenable if and only if $q_{n}(H)$ is
amenable. We separate the two cases.

  1. 1.

If $H$ is amenable, set $C=q_{n}(H)$, so that $H$ is contained in the amenable
subgroup $P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}C$. By maximality, we have
$H=P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}C$, and $C$ is maximal amenable in $G$.

  2. 2.

Assume now that $H$ is not amenable, and set $K=q_{n}(H)$.

Claim 1: the subgroup $H\cap P_{n}(\mathbb{Z})$ is invariant by $\rho_{n}(K)$.
Indeed, fix $h\in H$ and write $h=(v_{h},q_{n}(h))$ as an element of the semi-
direct product $G_{n}$. For $(w,1)\in H\cap P_{n}(\mathbb{Z})$, we have,
because $P_{n}(\mathbb{Z})$ is abelian:

$(\rho_{n}(q_{n}(h^{-1}))w,1)=(0,q_{n}(h))^{-1}(w,1)(0,q_{n}(h))$
$=(0,q_{n}(h))^{-1}(-v_{h},1)(w,1)(v_{h},1)(0,q_{n}(h))$
$=((v_{h},1)(0,q_{n}(h)))^{-1}(w,1)((v_{h},1)(0,q_{n}(h)))$
$=(v_{h},q_{n}(h))^{-1}(w,1)(v_{h},q_{n}(h))=h^{-1}(w,1)h$

which belongs to $H\cap P_{n}(\mathbb{Z})$ because the latter is normal in
$H$. This proves Claim 1.

Claim 2: We have $H\cap P_{n}(\mathbb{Z})=\\{0\\}$. To see it, let $k$ be the
rank of the free abelian group $H\cap P_{n}(\mathbb{Z})$, so that $0\leq k\leq
n+1$, we must show that $k=0$.

If $k=n+1$, then $H\cap P_{n}(\mathbb{Z})$ has finite index in
$P_{n}(\mathbb{Z})$, so that $H$ has finite index in $H\cdot
P_{n}(\mathbb{Z})$. By maximality, we must have $H\cdot P_{n}(\mathbb{Z})=H$,
i.e. $P_{n}(\mathbb{Z})\subset H$ and
$H=P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}K$; as $K$ is not amenable, this
contradicts part (2) of Proposition 3.1.

If $1\leq k\leq n$, we denote by $W$ the linear subspace of
$P_{n}(\mathbb{R})$ generated by $H\cap\mathbb{Z}^{n+1}$. By Claim 1, the
subspace $W$ is invariant by $\rho_{n}(K)$, contradicting Part 1 of
Proposition 3.1. This proves Claim 2.

At this point we know that $q_{n}|_{H}$ induces an isomorphism from $H$ onto
$K$, so that by Proposition 2.2 there exists a 1-cocycle $b\in
Z^{1}(K,P_{n}(\mathbb{Z}))$ so that $H=\\{(b(k),k):k\in K\\}$. By maximality
of $H$, the 1-cocycle cannot be extended to a larger subgroup of $G$. This
proves the direct implication of the Theorem.

For the converse, if $C$ is a maximal amenable subgroup of $G$, then
$P_{n}(\mathbb{Z})\rtimes C$ is Haagerup, and maximality follows immediately
from Part 2 of Proposition 3.1. If $K$ is a non-amenable subgroup of $G$ and
$b\in Z^{1}(K,P_{n}(\mathbb{Z}))$ is a 1-cocycle that cannot be extended to a
larger subgroup, then $H=\\{(b(k),k):k\in K\\}$ is a Haagerup subgroup of
$P_{n}(\mathbb{Z})\rtimes_{\rho_{n}}G$, and maximality follows the dichotomy
in the direct implication of the Theorem. $\blacksquare$

##  5 Cohomological matters: $n$ odd

###  5.1 Maximal subgroups in $PSL_{2}(\mathbb{Z})$

Recall that a subgroup $H$ in a group $G$ is said to be maximal if $H$ is
maximal among proper subgroups of $G$.

######  Proposition 5.1.

The group $PSL_{2}(\mathbb{Z})$ admits free maximal subgroups of arbitrary
large finite index; and also of infinite index.

Proof: Since $PSL_{2}(\mathbb{Z})$ is isomorphic to the free product
$C_{2}\ast C_{3}$, a subgroup $H$ of $PSL_{2}(\mathbb{Z})$ is free if and only
it is torsion-free, if and only if it has no element of order 2 or 3, if and
only if it does not meet the conjugacy classes of (the images of) $s$ and $t$
in $PSL_{2}(\mathbb{Z})$.

  1. 1.

Finite index. Let $p$ be a prime, with $p\equiv 11\mod 12$. Consider the
congruence subgroup

$\overline{\Gamma}_{0}(p)=\\{\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\in PSL_{2}(\mathbb{Z}):c\equiv 0\mod p\\}.$

So $\overline{\Gamma}_{0}(p)$ is the inverse image under reduction modulo $p$,
of the upper triangular subgroup in $PSL_{2}(p)$. Hence
$\overline{\Gamma}_{0}(p)$ is a maximal subgroup of index $p+1$ in
$PSL_{2}(\mathbb{Z})$. For $g=\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\in PSL_{2}(\mathbb{Z})$, by direct computation we have
that the (2,1)-entry of $gsg^{-1}$ is $c^{2}+d^{2}$, which is not divisible by
$p$ as $p\equiv 3\mod 4$; so $\overline{\Gamma}_{0}(p)$ has no element of
order 2. Similarly, the (2,1)-entry of $gtg^{-1}$ is $c^{2}-cd+d^{2}$. Assume
by contradiction that $p$ divides $c^{2}-cd+d^{2}$. Then it also divides
$4(c^{2}-cd+d^{2})=(2c-d)^{2}+3d^{2}$, from which it follows that the Legendre
symbol $\left(\begin{array}[]{c}-3\\\ \hline\cr p\end{array}\right)=-1$, as
one sees by inverting $d$ modulo $p$. Using quadratic reciprocity, this
contradicts $p\equiv 2\mod 3$. So $\overline{\Gamma}_{0}(p)$ has no element of
order 3, and therefore is a free group.

  2. 2.

Infinite index. We will appeal to Theorem E in [FLMS22] together with its
proof. Consider the following two permutations of $\mathbf{N}=\\{1,2,...\\}$:

$a=(12)(34)(56)(78)...$ $b=(123)(456)(789)...$

View the symmetric group $Sym(\mathbf{N})$ as a Polish group with the topology
of pointwise convergence. Let $T$ be the subset of $Sym(\mathbf{N})$
consisting of those $\vartheta$’s in $Sym(\mathbf{N})$ such that the action of
$PSL_{2}(\mathbb{Z})$ on $\mathbf{N}$ defined by mapping $s$ to $a$ and $t$ to
$\vartheta b\vartheta^{-1}$, is transitive on $\mathbf{N}$. By lemma 6.2 in
[FLMS22], the set $T$ is a non-empty $G_{\delta}$ in $Sym(\mathbf{N})$, hence
$T$ is itself Polish. Now Theorem E together with its proof in section 6.2 of
[FLMS22], shows that for a generic choice of $\vartheta\in T$, the
corresponding action of $PSL_{2}(\mathbb{Z})$ will be 2-transitive (in fact
highly transitive, i.e. $n$-transitive for every $n\geq 1$). By
2-transitivity, the stabilizer of any point in $\mathbf{N}$ will be a maximal
subgroup of infinite index in $PSL_{2}(\mathbb{Z})$. Since $a$ and $\vartheta
b\vartheta^{-1}$ are permutations without fixed points, that stabilizer avoids
the conjugacy classes of $s$ and $t$, and therefore it is free. $\blacksquare$

###  5.2 The case $n=1$

The next result is specific to $n=1$; it applies in particular to
$G=SL_{2}(\mathbb{Z})$ and $G=GL_{2}(\mathbb{Z})$. For $SL_{2}(\mathbb{Z})$, a
different proof appears in lemma 2.14 in [JS21].

######  Proposition 5.2.

Let $G$ be a subgroup of $GL_{2}(\mathbb{Z})$ containing an element $t_{0}$ of
order 6. Let $\rho:G\rightarrow GL_{N}(\mathbb{Z})$ be a homomorphism such
that $\rho(t_{0}^{3})=-1$, and $\rho(t_{0})$ does not admit $-1$ as an
eigenvalue. Then $H^{1}(G,\mathbb{Z}^{N})=0$. This applies in particular to
$\rho=\rho_{1}|_{G}$.

Proof: In an amalgamated product of groups, any element of finite order is
conjugate to a finite order element in one of the factors (see e.g. Cor. 1 in
section 3 of [Se77]). So in $GL_{2}(\mathbb{Z})=D_{4}\ast_{D_{2}}D_{6}$,
conjugating $G$ if necessary we may assume that $t_{0}=t$, so that
$\varepsilon=t_{0}^{3}$ belongs to $G$.

Fix $b\in Z^{1}(G,\mathbb{Z}^{N})$, we want to prove that $b$ is a
1-coboundary. By lemma 2.3 (applied with $z=\varepsilon$) it is enough to
prove that $b(\varepsilon)\in 2\mathbb{Z}^{N}$.

Set $T=:\rho(t_{0})$, then:

$0=T^{3}+1=(T+1)(T^{2}-T+1).$

Since by assumption $T+1$ is invertible in $M_{N}(\mathbb{C})$, we have
$0=T^{2}-T+1$, i.e. $2T=T^{2}+T+1$. Now expanding $t_{0}^{3}=\varepsilon$ by
the 1-cocycle relation we get

$b(\varepsilon)=(T^{2}+T+1)b(t_{0})=2Tb(t_{0}).$

i.e. $b(\varepsilon)\in 2\mathbb{Z}^{N}$. For the final statement about
$\rho_{1}$, it is enough to observe that $-1$ is not an eigenvalue of
$t=\left(\begin{array}[]{cc}0&-1\\\ 1&1\end{array}\right)$. $\blacksquare$

We now revisit an example from Jiang-Skalski (Proposition 2.16 in [JS21]): for
$N\geq 2$, define

$\Gamma_{1}(N)=\\{\left(\begin{array}[]{cc}a&b\\\ c&d\end{array}\right)\in
SL_{2}(\mathbb{Z}):c\equiv 0\mod N,\,a\equiv d\equiv 1\mod N\\}.$

Define also the vector $v_{N}=\left(\begin{array}[]{c}1/N\\\
0\end{array}\right)$ in $\mathbb{Q}^{2}$. Then Jiang-Skalski prove that
$\Gamma_{1}(N)$ is exactly the set of elements of $SL_{2}(\mathbb{Z})$ such
that $b_{N}(g)=gv_{N}-v_{N}$ belongs to $\mathbb{Z}^{2}$, so that $b_{N}$ is a
1-cocycle in $Z^{1}(\Gamma_{1}(N),\mathbb{Z}^{2})$ that does not extend to an
overgroup, and $H_{N}=\\{(b_{N}(k),k):k\in\Gamma_{1}(N)\\}$ is maximal
Haagerup in $\mathbb{Z}^{2}\rtimes SL_{2}(\mathbb{Z})$. We make the result
more precise by observing that $H_{N}$ is actually a free group for infinitely
many values of $N$.

######  Proposition 5.3.

If $N$ is divisible by some prime $p$ with $p\equiv 11\mod 12$, then $H_{N}$
is a free group.

Proof: We will use the fact that any subgroup of a free group is free. It is
enough to show that $\Gamma_{1}(N)$ is free. Since
$\Gamma_{1}(N)\subset\Gamma_{1}(p)$, we reduce to the case of $\Gamma_{1}(p)$.
Clearly $\varepsilon\notin\Gamma_{1}(p)$, so $\Gamma_{1}(p)$ maps injectively
to its image in $PSL_{2}(\mathbb{Z})$. Clearly the latter is contained in
$\overline{\Gamma}_{0}(p)$, which is free by the proof of Proposition 5.1.
Hence the conclusion. $\blacksquare$

###  5.3 $n$ odd, general case

For general odd $n$, we have the following:

######  Proposition 5.4.

Let $G$ be a finitely generated subgroup of $GL_{2}(\mathbb{Z})$. Assume that
$G$ is generated by $k$ elements $g_{1},g_{2},...,g_{k}$ and contains
$\varepsilon=\begin{pmatrix}-1&0\\\ 0&-1\end{pmatrix}$. Then, for $n$ odd,
$H^{1}(G,P_{n}(\mathbb{Z}))$ is a vector space over the field $\mathbf{F}_{2}$
of two elements, with dimension at most $(k-1)(n+1)$.

Proof: By the first part of lemma 2.3, we know that
$H^{1}(G,P_{n}(\mathbb{Z}))$ is a vector space over $\mathbf{F}_{2}$. Now any
cocycle $b\in Z^{1}(G,P_{n}(\mathbb{Z}))$ is completely determined by the
vectors $b(g_{1}),...,b(g_{k})$. On the other hand the boundary map
$\partial:P_{n}(\mathbb{Z})\rightarrow
Z^{1}(G,P_{n}(\mathbb{Z})):P\mapsto\partial P$, with $(\partial
P)(g)=\rho_{n}(g)P-P$, is injective. So $H^{1}(G,P_{n}(\mathbb{Z}))$ appears
as a sub-quotient of $(P_{n}(\mathbb{Z}))^{k-1}$, which gives the desired
bound on the dimension. $\blacksquare$

Proof of Theorem 1.4: Let $M$ be a a maximal free subgroup of
$PSL_{2}(\mathbb{Z})$ as in Proposition 5.1. Lifting arbitrarily a free basis
of $M$ to $SL_{2}(\mathbb{Z})$, we get a free subgroup $K$ in
$SL_{2}(\mathbb{Z})$, say of rank $k$ (possibly $k=\infty$). In particular,
the rank of $H^{1}(K,P_{n}(\mathbb{Z}))$ as an abelian group is $(k-1)(n+1)$.
Now by maximality of $M$ in $PSL_{2}(\mathbb{Z})$, together with the fact that
the central extension

$1\rightarrow\langle\varepsilon\rangle\rightarrow
SL_{2}(\mathbb{Z})\rightarrow PSL_{2}(\mathbb{Z})\rightarrow 1$

does not split, we see that the only overgroups of $K$ in $SL_{2}(\mathbb{Z})$
are $K\times\langle\varepsilon\rangle$ and $SL_{2}(\mathbb{Z})$. For both of
them, $H^{1}$ is a torsion group (by Proposition 5.4), so any 1-cocycle $b$ of
infinite order in $H^{1}(K,P_{n}(\mathbb{Z}))$ will not extend to an overgroup
of $K$, and the subgroup $H=\\{(b(k),k):k\in K\\}$ is maximal Haagerup in
$P_{n}(\mathbb{Z})\rtimes SL_{2}(\mathbb{Z})$. $\blacksquare$

##  6 Cohomological matters: $n$ even

We will see that for $n$ even the groups
$H^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ and
$H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ are infinite, so that the
situation is completely different from the case $n$ odd.

Since $\rho_{n}(\varepsilon)=1$ as $n$ is even, by the second part of Lemma
2.3 we may replace $H^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ by
$H^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$, and similarly
$H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ by
$H^{1}(PGL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$.

###  6.1 The case of $SL_{2}(\mathbb{Z})$

Set $S=\rho_{n}(s),T=\rho_{n}(t)$, and recall that $S,T$ generate
$\rho_{n}(SL_{2}(\mathbb{Z}))\simeq PSL_{2}(\mathbb{Z})$.

######  Lemma 6.1.

An assignment $S\mapsto b(S),T\mapsto b(T)$ of two vectors in
$P_{n}(\mathbb{Z})$, extends to $b\in
Z^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ if and only if $(1+S)b(S)=0$ and
$(1+T+T^{2})b(T)=0$.

Proof: View $PSL_{2}(\mathbb{Z})$ as the free product $\langle
S\rangle\ast\langle T\rangle=\langle S,T|S^{2}=1,T^{3}=1\rangle$. So the
assignment $S\mapsto b(S),T\mapsto b(T)$ extends to a 1-cocycle if and only it
satisfies $(S+1)b(S)=0$ (obtained by expanding $S^{2}=1$ by the 1-cocycle
relation) and $(1+T+T^{2})b(T)=0$ (obtained by expanding $T^{3}=1$ by the
cocycle relation). $\blacksquare$

Define, for $n$ even:

$\eta(n)=\left\\{\begin{array}[]{ccc}0&if&n\equiv 2\mod 3\\\ 1&if&n\equiv
0\mod 3\\\ -1&if&n\equiv 1\mod 3\end{array}\right.$

######  Theorem 6.2.

For $n$ even:

$rk(H^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))=\frac{n+1+3(-1)^{\frac{n}{2}+1}-4\eta(n)}{6}>0.$

In particular
$rk(H^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))\geq\frac{n-6}{6}$.

Proof: Since we are only interested in the torsion free part of
$H^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ we may work rationally and
compute the $\mathbb{Q}$-dimension of
$H^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q}))$.

We define a subgroup of $Z^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q}))$ by

$Z_{0}^{1}=:\\{b\in Z^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q})):b(T)=0\\},$

We first observe that every cocycle $b\in
H^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q}))$ is cohomologous to a cocycle in
$Z^{1}_{0}$. This is clear in terms of affine actions: if $\alpha(g)v=gv+b(g)$
(with $v\in P_{n}(\mathbb{Q})$) is the affine action associated with $b$, then
for any $w\in P_{n}(\mathbb{Q})$ the vector
$w_{0}=:\frac{1}{3}(w+\alpha(T)w+\alpha(T^{2})w)$ is $\alpha(T)$-fixed, so the
coboundary $b^{\prime}(g)=b(g)+gw_{0}-w_{0}$ vanishes on $T$.

Consequently, setting $m_{0}=:\dim Z_{0}^{1}$ and $n_{0}=:\dim(Z_{0}^{1}\cap
B^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q})))$, we have

$\dim(H^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q})))=\dim(Z_{0}^{1}/(Z_{0}^{1}\cap
B^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q}))))=m_{0}-n_{0},$

and it is enough to compute $m_{0}$ and $n_{0}$ separately. This will be done
in 3 steps. Note that, by lemma 6.1, the space $Z^{1}_{0}$ can be identified
with $\ker(S+1)$

  1. 1.

Computation of $m_{0}=\dim(\ker(S+1))$. Observe that $(SP)(X,Y)=P(Y,-X)$ for
$P\in P_{n}(\mathbb{Z})$. Expanding $P$ as a sum of monomials
$P(X,Y)=\sum_{k=0}^{n}a_{k}X^{n-k}Y^{k}$, we get

$(SP)(X,Y)=\sum_{k=0}^{n}(-1)^{k}a_{k}Y^{n-k}X^{k}=\sum_{k=0}^{n}(-1)^{k}a_{n-k}X^{n-k}Y^{k},$

so that $SP=-P$ if and only if

$a_{k}=(-1)^{k+1}a_{n-k}\>\mbox{for every}\;k=0,1,...,n.$ (2)

So we may choose $a_{0},a_{1},...,a_{n/2-1}$ arbitrarily, while $a_{n/2}=0$ if
$n\equiv 0\mod 4$, and $a_{n/2}$ can be chosen arbitrarily if $n\equiv 2\mod
4$. So

$m_{0}=\frac{n+1+(-1)^{\frac{n}{2}+1}}{2}.$ (3)

  2. 2.

We claim that $Z_{0}^{1}\cap B^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q}))$ can
be identified with $\ker(T-1)$. Indeed $b\in Z^{1}_{0}$ is a 1-coboundary if
and only if there exists $P\in P_{n}(\mathbb{Q})$ such that $b(S)=(S-1)(P)$
and $0=(T-1)(P)$. This is still equivalent to $b(S)\in(S-1)(\ker(T-1))$. This
already identifies $Z_{0}^{1}\cap
B^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q}))$ with $(S-1)(\ker(T-1))$.

To prove the claim it remains to show that $(S-1)|_{\ker(T-1)}$ is injective,
i.e. $\ker(S-1)\cap\ker(T-1)=\\{0\\}$. But, as $S,T$ generate
$PSL_{2}(\mathbb{Z})$, the space $\ker(S-1)\cap\ker(T-1)$ is exactly the space
of $\rho_{n}(SL_{2}(\mathbb{Z}))$-fixed vectors in $P_{n}(\mathbb{Q})$, which
is 0 by the first part of Proposition 3.1.

  3. 3.

Computation of $n_{0}=\dim(\ker(T-1))$. Since $T^{3}=1$, the operator $T$
defines a representation $\sigma$ of the cyclic group $C_{3}$, and
$n_{0}=\dim(\ker(T-1))$ is the multiplicity of the trivial representation in
this representation. Recall that $C_{3}$ has 3 irreducible representations,
all of dimension 1: the trivial representation $\chi_{0}$ and the characters
$\chi_{\pm}$ defined by $\chi_{\pm}(T)=e^{\pm\frac{2\pi i}{3}}$. Now $\sigma$
is equivalent to the direct sum of $n_{0}$ copies of $\chi_{0}$ with $n_{+}$
copies of $\chi_{+}$ and $n_{-}$ copies of $\chi_{-}$. Note that $n_{+}=n_{-}$
because $\sigma$ is a real representation. Computing the character of $\sigma$
we get

$n+1=Tr(1)=n_{0}\chi_{0}(1)+n_{+}\chi_{+}(1)+n_{-}\chi_{-}(1)=n_{0}+2n_{+};$
$Tr(T)=n_{0}\chi_{0}(T)+n_{+}\chi_{+}(T)+n_{-}\chi_{-}(T)=n_{0}-n_{+}.$

Solving this system for $n_{0}$ we get:

$rk(\ker(T-1))=n_{0}=\frac{n+1+2Tr(T)}{3}.$ (4)

It remains to compute $Tr(T)$ to get the exact value of $n_{0}$. We observe
that $\rho_{n}$ extends to a representation of $SL_{2}(\mathbb{C})$ on the
space $P_{n}(\mathbb{C})$. Restricting to the compact torus
$T=\\{a_{\vartheta}=\left(\begin{array}[]{cc}e^{i\vartheta}&0\\\
0&e^{-i\vartheta}\end{array}\right):\vartheta\in\mathbb{R}\\}$ we have the
classical formula (see e.g. formula (7.26) in [Ha03]):

$Tr(\rho_{n}(a_{\vartheta}))=\frac{\sin((n+1)\vartheta)}{\sin(\vartheta)}.$

Since $t$ has order 6 in $SL_{2}(\mathbb{Z})$, it is conjugate to $a_{\pi/3}$
in $SL_{2}(\mathbb{C})$, and therefore:

$Tr(T)=Tr(\rho_{n}(a_{\pi/3}))=\frac{\sin((n+1)\pi/3)}{\sin(\pi/3)}=\eta(n).$
(5)

Using equations (3), (4) and (5), we get the desired result. Note that for
$n=2,4,6$ we get $m_{0}>n_{0}$ from the following table:

$\left.\begin{array}[]{c|c|c}n&n_{0}&m_{0}\\\ \hline\cr 2&1&2\\\ 4&1&2\\\
6&3&4\end{array}\right.$ (6)

$\blacksquare$

######  Corollary 6.3.

For $n\geq 2$ even, let $M$ be a free maximal subgroup of
$PSL_{2}(\mathbb{Z})$ as in Proposition 5.1; let $K\simeq
M\times\langle\varepsilon\rangle$ denote the inverse image of $M$ in
$SL_{2}(\mathbb{Z})$. Then the semi-direct product $P_{n}(\mathbb{Z})\rtimes
SL_{2}(\mathbb{Z})$ contains infinitely many maximal Haagerup subgroups $H$ of
the form $H=\\{(b(g),g):g\in K\\}$, where $b\in
Z^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$, which moreover are pairwise not
conjugate under $P_{n}(\mathbb{Z})$.

Proof: Say that the rank of $M$ is $k$ (possibly $k=\infty$). By lemma 2.3, we
have $rk(H^{1}(K,P_{n}(\mathbb{Z}))=rk(M,P_{n}(\mathbb{Z}))=(k-1)(n+1)$ as $M$
is free, So by Theorem 6.2:

$rk(H^{1}(K,P_{n}(\mathbb{Z}))\geq
n+1>\frac{n+1+3(-1)^{\frac{n}{2}+1}-4\eta(n)}{6}=rk(H^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z})).$

So taking for $b$ cocycles on $K$ whose classes have infinite order in the co-
kernel of the restriction map
$H^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))\rightarrow
H^{1}(K,P_{n}(\mathbb{Z}))$, we may construct maximal Haagerup subgroups of
the desired form. $\blacksquare$

###  6.2 The case of $GL_{2}(\mathbb{Z})$

We recall the notations $S=\rho_{n}(s),T=\rho_{n}(t)$, to which we add
$W=:\rho_{n}(w)$. Note that, for $P\in P_{n}(\mathbb{Z})$, we have
$(WP)(X,Y)=P(Y,X)$; in particular $WP=P$ (resp. $WP=-P$) means that $P$ is
symmetric (resp. antisymmetric).

To estimate the rank of $H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$, we will
need the following representation-theoretic lemma.

######  Lemma 6.4.

The $\mathbb{Q}$-dimension of the space $P_{n}^{0}=:\\{P\in
P_{n}(\mathbb{Q}):P\,\mbox{symmetric},TP=P\\}$ is $\frac{n+4+2\eta(n)}{6}$.

Proof: We observe that $P_{n}^{0}$ is invariant under $\rho_{n}|_{D_{3}}$,
which suggests to use representation theory of the dihedral group
$D_{3}=\langle T,W\rangle$. This group has 3 irreducible representations
defined over $\mathbb{Q}$: the trivial character $\chi_{0}$, the non-trivial
character $\chi_{1}$ defined by $\chi_{1}(T)=1$ and $\chi_{1}(W)=-1$, and the
2-dimensional irreducible representation on vectors in $\mathbb{R}^{3}$ whose
3 coordinates sum up to 0. The character table of $D_{3}$ is:

$\begin{array}[]{c|ccc}&e&T&W\\\ \hline\cr\chi_{0}&1&1&1\\\ \chi_{1}&1&1&-1\\\
\pi&2&-1&0\end{array}$

Write $\rho_{n}|_{D_{3}}=n_{0}\chi_{0}\oplus n_{1}\chi_{1}\oplus n_{\pi}\pi$.
From the character table, it follows that the dimension of $P_{n}^{0}$ is
exactly the multiplicity $n_{0}$ of $\chi_{0}$ in $\rho_{n}|_{D_{3}}$. To
compute $n_{0}$ , we will compute the character of $\rho_{n}|_{D_{3}}$. Denote
by $P_{n}^{s}$ (resp. $P_{n}^{a}$) the subspace of symmetric (resp.
antisymmetric) polynomials in $P_{n}(\mathbb{Q})$. Then

$Tr(W)=\dim(P_{n}^{s})-\dim(P_{n}^{a})=(\frac{n}{2}+1)-\frac{n}{2}=1,$

so that the character of $\rho_{n}|_{D_{3}}$ is given by:

$n+1=Tr(1)=n_{0}+n_{1}+2n_{\pi};$ $Tr(T)=n_{0}+n_{1}-n_{\pi};$
$1=Tr(W)=n_{0}-n_{1}.$

Solving for $n_{0}$, and using formula (5), gives the desired result.
$\blacksquare$

######  Theorem 6.5.

For even $n$, we have

$rk(H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))=\frac{n-5+3(-1)^{\frac{n}{2}+1}-4\eta(n)}{12}.$

In particular
$rk(H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z})))\geq\frac{n-12}{12}$.

Proof: As for $SL_{2}(\mathbb{Z})$, we will work rationally. We will appeal to
a part of the Hochschild-Serre long exact sequence in group cohomology,
namely: if $G$ is a group, $N\triangleleft G$ a normal subgroup, and $V$ is a
$G$-module with $V^{G}=0$, then the restriction map $H^{1}(G,V)\rightarrow
H^{1}(N,V)^{G/N}$ is an isomorphism (see section 8.1 in [Gu80]).

We apply this to $V=P_{n}(\mathbb{Q}),\,G=GL_{2}(\mathbb{Z})$ and
$N=SL_{2}(\mathbb{Z})$, so that $G/N=\langle W\rangle$ has order 2. We must
therefore work out the $W$-invariants on
$H^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q}))$. This amounts to finding the
$W$-invariants in $Z^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q})$ and dividing
by the $W$-invariants in $B^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q}))$. Of
course we will enjoy the fact that the $W$-invariants in a subspace of
$P_{n}(\mathbb{Q})$ are just the symmetric polynomials in that subspace.

By the beginning of the proof of Theorem 6.2, every cocycle in
$Z^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q})$ is cohomologous to a cocycle in
$Z^{1}_{0}$, which identifies with $\ker(S+1)$. By formula (2), a polynomial
$P(X,Y)=\sum_{k=0}^{n}a_{k}X^{n-k}Y^{k}$ is in $\ker(S+1)$ if and only if
$a_{k}=(-1)^{k+1}a_{n-k}$ for every $k=0,1,...,n$. So this polynomial $P(X,Y)$
is symmetric if and only if $a_{k}=0$ for even $k$, and $a_{k}=a_{n-k}$ for
odd $k$. Hence we get:

$\dim(Z^{1}_{0})^{W}=\frac{n+1+(-1)^{\frac{n}{2}+1}}{4}.$ (7)

Now we turn to the computation of the dimension of $(Z^{1}_{0}\cap
B^{1}(PGL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q})))^{W}$. As in the third step of
the proof of Theorem 6.2, using $SW=WS$ we identify this space first with
$(S-1)(\ker(T-1)\cap\ker(W-1))$, second with $\ker(T-1)\cap\ker(W-1)$, which
is the subgroup $P_{n}^{0}$ from lemma 6.4: its dimension is
$\frac{n+4+2\eta(n)}{6}$.

Summarizing: we have by (7):

$\dim(H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))=\dim(Z^{1}_{0})^{W}-\dim(Z^{1}_{0}\cap
B^{1}(PGL_{2}(\mathbb{Z}),P_{n}(\mathbb{Q})))^{W}$
$=\frac{n+1+(-1)^{\frac{n}{2}+1}}{4}-\frac{n+4+2\eta(n)}{6}$
$=\frac{n-5+3(-1)^{\frac{n}{2}+1}-4\eta(n)}{12}.$

It is clear that this quantity is bounded below by $\frac{n-12}{12}$.
$\blacksquare$

With this we reach the proof of Theorem 1.5.

Proof of Theorem 1.5: It is enough to show that the co-kernel of the
restriction map $H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))\rightarrow
H^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ has rank $>0$. This follows by
comparing the formulae in Theorems 6.2 and 6.5. Indeed we have

$\frac{n+1+3(-1)^{\frac{n}{2}+1}-4\eta(n)}{6}-\frac{n-5+3(-1)^{\frac{n}{2}+1}-4\eta(n)}{12}=\frac{n+7+3(-1)^{\frac{n}{2}+1}-4\eta(n)}{12}>0$

where the inequality follows from $7+3(-1)^{\frac{n}{2}+1}-4\eta(n)\geq 0$.
$\square$

######  Example 2.

For $n=2,4$, we can give explicit cocycles witnessing the fact that the image
of the restriction map
$Rest:H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))\rightarrow
H^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ has infinite index.

Recall from the proof of Theorem 6.2 that we defined $Z_{0}^{1}=:\\{b\in
Z^{1}(PSL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z})):b(T)=0\\}$ and identified it
naturally with $\ker(1+S)$. For $a\in\mathbb{Z}$, define $b_{a}\in
Z^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ by prescribing:

$(b_{a}(S))(X,Y)=a(X^{n}-Y^{n});\;b_{a}(T)=0;$

it follows from (2) that $b_{a}(S)\in\ker(1+S)$, hence $b_{a}\in Z_{0}^{1}$.
We first show ihat, for $a\neq 0$, the cocycle $b_{a}$ is non-zero in $H^{1}$.

Assume that $b_{a}$ is a coboundary. By the 3rd step of the proof of Theorem
6.2, the space $Z^{1}_{0}\cap B^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$
identifies naturally with $(S-1)(\ker(T-1))$. For $n=2,4$, we have
$rk(\ker(T-1))=1$ by table (6). For $n=2$, the space $\ker(T-1)$ is generated
by $X^{2}-XY+Y^{2}$ (a $t$-invariant quadratic form on $\mathbb{Z}^{2}$), and
by direct computation one gets $(S-1)(X^{2}-XY+Y^{2})=2XY$, forcing $a=0$. For
$n=4$, we replace $X^{2}-XY+Y^{2}$ by $(X^{2}-XY+Y^{2})^{2}$ and proceed
analogously.

Finally, as $b_{a}$ is anti-symmetric, for $a\neq 0$ its class in $H^{1}$ is
certainly not in $H^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))^{W}$ which is
the image of the restriction map
$H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))\rightarrow
H^{1}(SL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$.

Coming back to Theorem 6.5, we observe that
$rk(H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z})))=0$ for $n=2,4,6,8,12$. This
means that $H^{1}(GL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ is a finite group for
these values. We will show that it is not zero.

######  Corollary 6.6.

Fix $n\geq 2$ even. Define

$m=\left\\{\begin{array}[]{ccc}n/4&if&n\equiv 0\mod 4\\\ (n+2)/4&if&n\equiv
2\mod 4.\end{array}\right.$

Then $H^{1}(GL_{2}(Z),P_{n}(\mathbb{Z}))$ has at least order $2^{m}$.

Proof: We start with two observations.

  1. 1.

A 1-cocycle $b\in Z^{1}_{0}$ such that $b(S)$ is symmetric, extends from
$PSL_{2}(\mathbb{Z})$ to $PGL_{2}(\mathbb{Z})$ by the prescription $b(W)=0$.
Indeed $PGL_{2}(\mathbb{Z})\simeq D_{2}\ast_{C_{2}}D_{3}$, with $C_{2}=\langle
W\rangle,D_{2}=\langle S,W\rangle,D_{3}=\langle T,W\rangle$. So by the cocycle
relation $b$ extends to $D_{3}$ by $b|_{D_{3}}=0$, and to $D_{2}$ by defining
$b(SW)=b(S)$, which is indeed equal to $b(WS)=Wb(S)$ as $b(S)$ is symmetric.
So $b$ extends to the amalgamated product.

  2. 2.

If a cocycle $b\in Z^{1}(PGL_{2}(\mathbb{Z}),P_{n}(\mathbb{Z}))$ with $b(W)=0$
is a coboundary, then there exists $P\in P_{n}(\mathbb{Z})$ such that
$(W-1)P=0$ (i.e. $P$ is symmetric) and $b(S)=(1-S)P$. But
$SP(X,Y)=P(Y,-X)=P(-X,Y)$ (by symmetry), so writing
$P(X,Y)=\sum_{k=0}^{\frac{n}{2}-1}a_{k}(X^{n-k}Y^{k}+X^{k}Y^{n-k})+a_{n/2}X^{\frac{n}{2}}Y^{\frac{n}{2}}$
we get

$((1-S)P)(X,Y)=\sum_{k=0}^{\frac{n}{2}-1}a_{k}(1-(-1)^{k})(X^{n-k}Y^{k}+X^{k}Y^{n-k})+a_{n/2}(1-(-1)^{\frac{n}{2}})X^{\frac{n}{2}}Y^{\frac{n}{2}},$

so that $(1-S)P$ only has even coefficients.

We now come to the proof itself. For $\varepsilon\in\\{0,1\\}^{m}$, define
$b_{\varepsilon}(T)=0$ and

$b_{\varepsilon}(S)(X,Y)=\left\\{\begin{array}[]{ccc}\sum_{k=1}^{m}\varepsilon_{k}(X^{n-2k+1}Y^{2k-1}+X^{2k-1}Y^{n-2k+1})&if&n\equiv
0\mod 4\\\
\sum_{k=1}^{m-1}\varepsilon_{k}(X^{n-2k+1}Y^{2k-1}+X^{2k-1}Y^{n-2k+1})+\varepsilon_{m}X^{\frac{n}{2}}Y^{\frac{n}{2}}&if&n\equiv
2\mod 4.\end{array}\right.$

Observe that $(S+1)b_{\varepsilon}(S)=0$ by (2), so that by lemma 6.1 we get a
cocycle $b_{\varepsilon}\in Z^{1}_{0}$ with $b_{\varepsilon}(S)$ symmetric, so
by the first observation in the proof we may extend it to
$PGL_{2}(\mathbb{Z})$ by $b_{\varepsilon}(W)=0$.

On the other hand, if $\varepsilon_{1},\varepsilon_{2}$ are distinct elements
in $\\{0,1\\}^{m}$, then $(b_{\varepsilon_{1}}-b_{\varepsilon_{2}})(S)$ has at
least one odd coefficient, so $b_{\varepsilon_{1}}-b_{\varepsilon_{2}}$ cannot
be a coboundary by the 2nd observation above. $\blacksquare$

###  6.3 A combinatorial application

We end the paper with an amusing combinatorial consequence of our proof of
Theorem 6.2. For $n$ even, the sum

$\sum_{k=0}^{n/2}(-1)^{k}\left(\begin{array}[]{c}n-k\\\ k\end{array}\right),$

which is an alternating sum on a diagonal in Pascal’s triangle, can be
computed by combinatorial means (see [BQ08]). We give a representation-
theoretic derivation.

######  Corollary 6.7.

For even $n$:

$\sum_{k=0}^{n/2}(-1)^{k}\left(\begin{array}[]{c}n-k\\\
k\end{array}\right)=\eta(n)$

. Proof: In view of formula (5), it is enough to prove that the LHS is equal
to $Tr(T)=Tr(T^{-1})$. For this we observe that $T^{-1}(P)(X,Y)=P(X-Y,X)$ for
$P\in P_{n}(\mathbb{R})$. So in the canonical basis
$X^{n},X^{n-1}Y,...,XY^{n-1},Y^{n}$ we have, for $k=0,1,...,n$:

$(T^{-1})(X^{n-k}Y^{k})=X^{k}(X-Y)^{n-k}=\sum_{\ell=0}^{n-k}(-1)^{\ell}\left(\begin{array}[]{c}n-k\\\
\ell\end{array}\right)X^{n-\ell}Y^{\ell}.$

In the last sum, the term $X^{n-k}Y^{k}$ does not appear if $k>\frac{n}{2}$,
and it appears with coefficient $(-1)^{k}\left(\begin{array}[]{c}n-k\\\
k\end{array}\right)$ if $0\leq k\leq\frac{n}{2}$. In other words, the $k$-th
diagonal coefficient of the matrix of $T^{-1}$ in the canonical basis is

$\left\\{\begin{array}[]{ccc}0&if&k>\frac{n}{2}\\\
(-1)^{k}\left(\begin{array}[]{c}n-k\\\
k\end{array}\right)&if&k\leq\frac{n}{2}\end{array}\right.$

This yields the trace of $T^{-1}$:

$Tr(T^{-1})=\sum_{k=0}^{n/2}(-1)^{k}\left(\begin{array}[]{c}n-k\\\
k\end{array}\right).$

$\blacksquare$

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Author’s address:

Institut de mathématiques
Université de Neuchâtel
11 Rue Emile Argand - Unimail
CH-2000 Neuchâtel - SUISSE
<EMAIL_ADDRESS>


\eps^{\beta'p-(n-1)\beta''} +\eps^{\beta''(1-s_0)}\le \eps^{2\alpha(1+\beta'')}
\]

which gives the condition

\begin{equation}\label{thirdcondalpha}
2\alpha \le \min\lt( \frac{\beta'p -(n-1)\beta''}{1+\beta''},\frac{\beta''}{1+\beta''} (1-s_0)\rt).
\end{equation}

This constraint implies <ref> (which itself implies <ref>). Since the first term is decreasing in $\beta''$ and the second one is increasing, we see that the optimal choice of $\beta''$ is obtained when both terms are equal, that is

\[
\beta''=\frac{\beta p}{(n-s_0) (n+p)}
\]

which gives

\[
2\alpha\le \frac{\beta p}{(n-s_0)(n+p)+\beta p}(1-s_0).
\]

Since $p$ can be  taken arbitrarily close to $p_0$ and $\beta$ arbitrarily close to $1$. This concludes the proof.

M. Goldman was partially supported by the ANR grant SHAPO.
B. Merlet and M. Pegon are partially supported by the Labex CEMPI (ANR-11-LABX-0007-01).

Missing 'biblatex' package
The bibliography requires the 'biblatex' package.

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Oxford ; New York

Clarendon Press

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Calculus of variations,Functions of bounded variation

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abstractAbstract We consider functionals given by the sum of the perimeter and the double integral of some kernel g : ℝ N × ℝ N → ℝ + {g:\mathbb{R}{̂N}\times\mathbb{R}{̂N}\to\mathbb{R}{̂+}} , multiplied by a “mass parameter” ε. We show that, whenever g is admissible, radial and decreasing, the unique minimizer of this functional among sets of given volume is the ball as soon as ε ≪ 1 {\varepsilon\ll 1} .
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Ecole Polytechnique, Centre de Mathématiques

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abstractWe investigate properties of minimizers of a variational model describing the shape of charged liquid droplets. The model, proposed by Muratov and Novaga, takes into account the regularizing effect due to the screening of free counterionions in the droplet. In particular we prove partial regularity of minimizers, a first step toward the understanding of further properties of minimizers.
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ematics - Analysis of PDEs

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abstractAbstract We revisit the liquid drop model with a general Riesz potential. Our new result is the existence of minimizers for the conjectured optimal range of parameters. We also prove a conditional uniqueness of minimizers and a nonexistence result for heavy nuclei.
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abstractWe study the optimal design problem of finding the minimal energy configuration for a mixture of two conducting materials with a perimeter penalization of the unknown domain, when a volume constraint is added. We prove that minimizers of the constrained problem also minimize an unconstrained problem with a volume penalization for which regualarity results are available in the literature.
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Mathematics - Analysis of PDEs

abstractWe consider a variational problem involving competition between surface tension and charge repulsion. We show that, as opposed to the case of weak (short-range) interactions where we proved ill-posedness of the problem in a previous paper, when the repulsion is stronger the perimeter dominates the capacitary term at small scales. In particular we prove existence of minimizers for small charges as well as their regularity. Combining this with the stability of the ball under small $C{̂1,\gamma}$ perturbations, this ultimately leads to the minimality of the ball for small charges. We cover in particular the borderline case of the $1-$capacity where both terms in the energy are of the same order.
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mpeg/snap/zotero-snap/common/Zotero/storage/T2ES6VLK/Goldman et al. - 2022 - Rigidity of the ball for an isoperimetric problem .pdf
ematical Physics,Mathematics - Analysis of PDEs

journaltitleComm. Pure Appl. Math.
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abstractThe seminal results of Bourgain, Brezis, Mironescu and D\'avila show that the classical perimeter can be approximated by a family of nonlocal perimeter functionals. We consider a corresponding second order expansion for the nonlocal perimeter functional. In a special case, the considered family of energies is also relevant for a variational model for thin ferromagnetic films. We derive the Gamma–limit of these functionals. We also show existence for minimizers with prescribed volume fraction. For small volume fraction, the unique, up to translation, minimizer of the limit energy is given by the ball. The analysis is based on a systematic exploitation of the associated symmetrized autocorrelation function.
titleSecond order expansion for the nonlocal perimeter functional
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ematics - Analysis of PDEs

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Cambridge University Press

shorttitleSets of Finite Perimeter and Geometric Variational Problems
titleSets of Finite Perimeter and Geometric Variational Problems: An Introduction to Geometric Measure Theory
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journaltitleJournal de l’École polytechnique — Mathématiques
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ematical Physics,Mathematics - Analysis of PDEs,Mathematics - Optimization and Control

issn0010-3616, 1432-0916
journaltitleCommun. Math. Phys.
titleA nonlocal isoperimetric problem with dipolar repulsion
mpeg/snap/zotero-snap/common/Zotero/storage/RPI4M7JH/Muratov et Simon - 2019 - A Nonlocal Isoperimetric Problem with Dipolar Repu.pdf

journaltitleNonlinear Analysis
titleLarge mass minimizers for isoperimetric problems with integrable nonlocal potentials
020 - Large mass minimizers for isoperimetric problems w.pdf;/home/mpeg/snap/zotero-snap/common/Zotero/storage/SAFCPQJP/Pegon - 2021 - Large mass minimizers for isoperimetric problems w.pdf
Mathematical Physics,Mathematics - Analysis of PDEs,Mathematics - Optimization and Control

issn0249-633X, 2275-3230
journaltitleMémoires Soc. Math. France
titleEnsembles quasi-minimaux avec contrainte de volume et rectifiabilité uniforme
mpeg/snap/zotero-snap/common/Zotero/storage/IJ5LJXF4/Rigot - 2000 - Ensembles quasi-minimaux avec contrainte de volume.pdf

issn0075-4102, 1435-5345
journaltitleJournal für die reine und angewandte Mathematik
titleBoundaries of Caccioppoli sets with Hölder-continuous normal vector
mpeg/snap/zotero-snap/common/Zotero/storage/LNIQGXY8/Tamanini - 1982 - Boundaries of Caccioppoli sets with Hölder-continu.pdf
is is a quick fix for a conflict between todonotes and amsart