# Relatively projective profinie groups

Pavel A. Zalesskii  P. A. Zalesskiĭ
Department of Mathematics
University of Brasilia
70.910 Brasilia DF
Brazil<EMAIL_ADDRESS>
MSC classification: 20E18

Key-words: profinite groups, prosolvable groups.

##  1\. Introduction

It is well-known that the Schreier theorem does not hold for free profinite
groups, i.e., a subgroup of a free profinite group need not be free. Subgroups
of free profinite groups are groups of cohomological dimension $\leq 1$; they
are called projective, as they satisfy the same universal property as
projective modules.

Similarly, the Kurosh Subgroup Theorem does not hold for free products of
profinite groups. Subgroups of free profinite product are relatively
projective that gave motivation to study relatively projective groups in the
context of profinite groups and Galois theory (see [3, 4, 5, 6, 8]).

Following [6] a profinite group $G$ acting continuously on a profinite space
$T$ will be called a profinite pile (or a pair if one follows [1, Chapter 7])
and denoted by $(G,T)$. We say that $(G,T)$ is projective relatively to its
point stabilizers if for given equivariant morphisms $(\varphi,\alpha)$

$\textstyle{(G,T)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\scriptstyle{\psi}$$\textstyle{(B,Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\textstyle{(A,X)}$
(1.1)

such that $\alpha$ is an epimorphism injective on point stabilizers that
induces a homeomorphism on the spaces of orbits, there exists an equivariant
map $\psi:(G,T)\rightarrow(B,Y)$ such that $\alpha\psi=\varphi$. If $G$ is a
pro-$\mathcal{C}$ group (where $\mathcal{C}$ is a class of finite groups
closed for subgroups, quotients and abelian extensions) then the whole diagram
is restricted to pro-$\mathcal{C}$ groups and the notion is the relative
$\mathcal{C}$-projectivity. This definition is slightly differs from the
definition in [3], [4] and [12]; in [12] the definition is more restrictive
for profinite groups of big cardinality as the author assumes the existence of
a continuous section $\sigma:T/G\longrightarrow T$; in [4] the term is
strongly projective; we refer the reader to [6, Proposition 5.7] for the
interrelation of our definition and the one in [4].

The objective of this note is to give a homological characterization of
relatively prosolvable projective groups in the spirit of Ribes’ cohomological
characterization of free (amalgamated) products [9]. Let $\pi$ be a set of
primes. We say that a pile $(G,T)$ is $\pi$-projective if the embedding
problem (1.1) with $\operatorname{Ker}(\alpha)$ being an elementary abelian
$p$-group is solvable for any $p\in\pi$. If $G$ is prosolvable,
$\pi$-projectivity coincides with $\mathcal{C}$-projectivity, where
$\mathcal{C}$ is the class of all solvable $\pi$-groups.

######  Theorem 1.1.

Let $(G,T)$ be a profinite pile. Then the following statements are equivalent:

  1. (i)

$(G,T)$ is $\pi$-projective;

  2. (ii)

For each prime $p\in\pi$ and every simple $p$-primary $\mathbb{F}_{p}G$-module
$M$ the map $H_{i}(G,\mathbb{F}_{p}[[T]]\otimes M)\to H_{i}(G,M)$, induced
from $t\mapsto 1$ for every $t\in\dot{t}\in T/G$, is surjective for $i=2$ and
injective for $i=1$.

######  Corollary 1.2.

Let $(G,T)$ be a prosolvable pile and $\mathcal{C}$ the class of all finite
solvable groups. Then $(G,T)$ is $\mathcal{C}$-projective if and only if for
each prime $p$ and every simple $p$-primary $\mathbb{F}_{p}G$-module $M$ the
map $H_{i}(G,\mathbb{F}_{p}[[T]]\otimes M)\to H_{i}(G,M)$, induced from
$t\mapsto 1$ for every $t\in\dot{t}$, is surjective for $i=2$ and injective
for $i=1$.

The structure of the paper is the following. Section 2 is dedicated to
auxiliary results on relatively projective groups. Theorem 1.1 and Corollary
1.2 are proved in Section 3. We list open questions about relatively
projective groups in Section 4.

Acknowledgements.

  1. (i)

The idea of the homological characterization of relatively projective groups
arose when the author was working jointly with Dan Haran on the article [6];
Sections 2 and 3 of this paper benefited a lot from numerous discussions with
him.

  2. (ii)

During the work on the manuscript the author learned that Theorem 1.1 was
independetly obtained by Gareth Wilkes [12] using cohomology of profinite
pairs; I am independent to him for sharing his early draft with me.

  3. (iii)

This paper was written when the author visited the Department of Pure
Mathematics and Mathematical Statistics of the University of Cambridge; the
author thanks Henry Wilton and the department for the hospitality.

##  2\. Relatively projective groups

We shall consider here a continuous action on the right of a profinite group
$G$ on a profinite space $T$ and use the notation $t^{g}$ for the translation.

######  Definition 2.1.

A pile $\mathbf{G}=(G,T)$ consists of a profinite group $G$, a profinite space
$T$, and a continuous action of $G$ on $T$ (from the right).

A pile $\mathbf{G}=(G,T)$ is finite if both $G$ and $T$ are finite.

A morphism of group piles $\alpha\colon\mathbf{B}=(B,Y)\to\mathbf{A}=(A,X)$
consists of a group homomorphism $\alpha\colon B\to A$ and a continuous map
$\alpha\colon Y\to X$ such that $\alpha(y^{b})=\alpha(y)^{\alpha(b)}$ for all
$y\in Y$ and $b\in B$.

The kernel $\operatorname{Ker}\alpha$ of $\alpha$ is the kernel of the group
homomorphism $\alpha\colon B\to A$.

The above morphism $\alpha$ is an epimorphism if $\alpha(B)=A$, $\alpha(Y)=X$,
and for every $x\in X$ there is $y\in Y$ such that $\alpha(y)=x$ and
$\alpha(B_{y})=A_{x}$. It is rigid, if $\alpha$ maps $B_{y}$ isomorphically
onto $A_{\alpha(y)}$, for all $y\in Y$, and the induced map of the orbit
spaces $Y/B\to X/A$ is a homeomorphism.

######  Definition 2.2.

An embedding problem for a pile $\mathbf{G}$ is a pair

$(\varphi\colon\mathbf{G}\to\mathbf{A},\ \alpha\colon\mathbf{B}\to\mathbf{A})$
(2.1)

of morphisms of group piles such that $\alpha$ is an epimorphism. It is
finite, if $\mathbf{B}$ is finite and rigid if $\alpha$ is rigid. It is a
$\mathcal{C}$-embedding problem if $G,B,A$ are pro-$\mathcal{C}$. It is a
$\pi$-embedding problem if $\operatorname{Ker}\alpha$ is an abelian
$\pi$-group, where $\pi$ is a set of primes.

A solution of (1.1) is a morphism $\gamma\colon\mathbf{G}\to\mathbf{B}$ such
that $\alpha\circ\gamma=\varphi$.

A pile $\mathbf{G}$ is $\mathcal{C}$-projective (resp. $\pi$-projective), if
every finite rigid $\mathcal{C}$-embedding problem for $\mathbf{G}$ (resp.
every $\pi$-embedding problem) has a solution.

######  Definition 2.3.

A group homomorphism (resp. epimorphism) $\gamma:G\rightarrow B$ such that
$\alpha\circ\gamma=\varphi$ will be called a weak solution (resp. proper weak
solution) of (1.1).

Following [4] we shall say that a pro-$\mathcal{C}$ group $G$ is
$\mathcal{G}$-projective (resp.properly $\mathcal{G}$-projective) if every
finite $\mathcal{C}$-embedding problem has a weak solution (resp. proper weak
solution). If every finite rigid $\pi$-embedding problem has a weak solution
(resp. proper weak solution) then $G$ shall be called $\mathcal{G}$-projective
(resp.properly $\mathcal{G}$-projective) over $\pi$.

######  Remark 2.4.

(a) Of course if $\mathbf{G}$ is projective then $G$ is
$\mathcal{G}$-projective. Moreover, the projectivity of $\mathbf{G}$ implies
strongly $\mathcal{G}$-projectivity of $G$ in the sense of [4] which is called
$\mathcal{G}$-projectivity in [6]. Conversely, if $G$ is strongly
$\mathcal{G}$-projective and $G_{t}\neq G_{s}$ for any pair $t\neq s$ then
$\mathbf{G}$ is projective (see [6, Proposition 5.7]).

(b) $\mathbf{G}$ is $\pi$-projective, if every finite rigid embedding problem
(1.1) for $\mathbf{G}$, in which $\operatorname{Ker}(\alpha)$ is a minimal
normal subgroup of $B$, has a solution. So we can always assume that
$\operatorname{Ker}(\alpha)$ is elementary abelian.

This follows by induction on $|\operatorname{Ker}(\alpha)|$. If
$\operatorname{Ker}(\alpha)$ is not minimal, it has a subgroup $C\triangleleft
B$, such that $1\neq C\neq B$. Then $\alpha$ factors into rigid epimorphisms
$\alpha_{1},\alpha_{2}$ in the diagram below, where
$\mathbf{B}^{\prime}=(B/C,T/C)$. By the induction hypothesis there is a
solution $\gamma_{2}$ of the embedding problem $(\varphi,\alpha_{2})$, that
gives the embedding problem $(\gamma_{2},\alpha_{1})$, whose solution $\gamma$
gives a solution of (1.1):

$\textstyle{\mathbf{G}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\scriptstyle{\gamma_{2}}$$\scriptstyle{\varphi}$$\textstyle{\mathbf{B}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{1}}$$\textstyle{\mathbf{B}^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{2}}$$\textstyle{\mathbf{A}\hbox
to0.0pt{.\hss}}$

(c) A profinite group $G$ is $\mathcal{G}$-projective over $\pi$ if and only
if its free profinite product $G\amalg F_{\omega}$ is properly
$\mathcal{G}$-projective over $\pi$, where $F_{\omega}$ is free profinite of
countable rank (if we are in the class of pro-$\mathcal{C}$ groups, then the
free product and $F_{\omega}$ should be considered in this class). Indeed, let
$\varphi_{\omega}:G\amalg F_{\omega}\longrightarrow A$ be an epimorphism and
$\varphi$ its restriction to $G$. Since any finite embedding problem is
properly solvable for $F_{\omega}$, a weak solution of $(\alpha,\varphi)$ can
be always extended to a proper solution of the induced embedding problem
$(\alpha,\varphi_{\omega})$, and any proper solution of
$(\alpha,\varphi_{\omega})$ can be restricted to $G$ producing a weak solution
of $(\alpha,\varphi)$.

We finish the section with proposition that will be used in Section 4.

######  Proposition 2.5.

Let $\mathcal{C}$ be a class of finite groups closed for subgroups, quotients
and extensions. Let $(G,T)$ be a $\mathcal{C}$-projective pile such that there
exists a continuous section $\sigma:T/G\longrightarrow T$. Then $G$ embeds
into a free pro-$\mathcal{C}$-product $H=\coprod_{s\in Im(\sigma)}G_{s}\amalg
F$, where $F$ is a free pro-$\mathcal{C}$ group of rank $d(G)$, the minimal
number of generators of $G$. Moreover, $\\{G_{t}\mid t\in
T\\}=\\{H_{s}^{h}\cap G\mid h\in H,s\in Im(\sigma)\\}$.

###### Proof.

Let $\alpha:H\longrightarrow G$ be an epimorphism that sends $H_{s}$ on their
isomorphic copies in $G$ and $F$ onto $G$. Put $S=\bigcup_{s\in
Im(\sigma)}H_{s}\backslash H$. By [6, Constriction 4.3] $(H,S)$ is a pile and
so extending $\alpha$ to $S\rightarrow T$ by $\alpha(H_{s}h)=s^{\alpha(h)}$
one gets a rigid epimorphism $\alpha:(H,S)\longrightarrow(G,T)$. Since $(G,T)$
is $\mathcal{C}$-projective the embedding problem $(id,\alpha)$ has a solution
which is clearly an embedding.

∎

##  3\. Homological characterization

Let $G$ be a profinite group and $\\{G_{t}\mathop{|\,}t\in T\\}$ be a family
of subgroups indexed by a profinite space $T$. Following [10, Section 5.2] we
say that it is continuous if for any open subgroup $U$ of $G$ the subset
$\\{t\in T\mathop{|\,}G_{x}\leq U\\}$ is open.

######  Lemma 3.1 ([10, Lemma 5.2.1]).

Let $G$ be a profinite group and let $\\{G_{t}\mathop{|\,}t\in T\\}$ be a
collection of subgroups indexed by a profinite space $T$. Then the following
conditions are equivalent:

  1. (a)

$\\{G_{t}\mathop{|\,}t\in T\\}$ is continuous;

  2. (b)

The set $\hat{\mathcal{G}}=\\{(g,t)\in G\times T\mathop{|\,}t\in T,g\in
G_{t}\\}$ is closed in $G\times T$;

  3. (c)

The map $\varphi\colon T\to\operatorname{Subgr}(G)$, given by
$\varphi(t)=G_{t}$, is continuous, where $\operatorname{Subgr}(G)$ is endowed
with the étale topology;

  4. (d)

$\bigcup_{t\in T}G_{t}$ is closed in $G$.

If $(G,T)$ is a pile then denoting by $G_{t}$ the $G$-stabilizer of $t$, for
every $t\in T$, we note that $\mathcal{G}=\\{G_{t}\\}_{t\in T}$ is a
continuous family of closed subgroups of $G$ ([10, Lemma 5.2.2]) closed under
the conjugation in $G$, such that $G_{t^{g}}=G_{t}^{g}$ for all $t\in T$ and
$g\in G$.

Having a continuous family $\\{G_{t}\mid t\in T\\}$ of subgroups of a
pro-$\mathcal{C}$ group $G$ one can define a free pro-$\mathcal{C}$ product
$\coprod_{t\in T}G_{t}$ (see [10, Chapter 5]). If $\mathcal{C}$ consists of
abelian groups, then the free pro-$\mathcal{C}$ product becomes a profinite
direct sum $\bigoplus_{t\in T}G_{t}$. This also extends to the case of
pro-$\mathcal{C}$ modules.

######  Lemma 3.2.

Let $\mathbf{G}=(G,T)$ be a pile and let $\mathcal{G}=\\{G_{t}\\}_{t\in T}$ be
the family of $G$-stabilizers. Consider the following rigid embedding problem

$\big{(}\operatorname{id}\colon\mathbf{G}\to\mathbf{G},\alpha\colon\mathbf{B}=(B,Y)\to\mathbf{G}\big{)}$
(3.1)

with $M=\operatorname{Ker}(\alpha)$ finite minimal abelian normal subgroup
(i.e. an elementary abelian of order $p^{n}$). Let

$\kappa_{i}\colon H_{i}(G,\mathbb{F}_{p}[[T]]\otimes M^{*})\to H_{i}(G,M^{*})$

be the homology maps induced from $T\mapsto 1$, for $i=1,2$, where $M^{*}$ is
dual of $M$.

  * (a)

$\kappa_{2}$ is surjective if and only if for any embedding problem (3.1) the
group homomorphism $\alpha\colon B\to G$ splits and hence $B\cong M\rtimes G$.

  * (b)

$\kappa_{1}$ is injective if and only if every split embedding problem (3.1)
has a solution.

###### Proof.

(a) We just need to show that the extension $\alpha\colon B\to G$, as an
element $\beta$ of $H^{2}(G,M)$, is trivial if and only if $\kappa_{2}$ is
surjective.

Since $\alpha$ is rigid, $\alpha_{|G_{y}}$ is injective, for all $y\in Y$
which means that for every $G_{t}$ there exists $y\in Y$ and an isomorphism
$\gamma_{t}:G_{t}\longrightarrow B_{y}$ such that
$\alpha\circ\gamma_{t}=\operatorname{id}$. Note that the $G$-orbit $\dot{t}$
of $t$ is homeomorphic to $G_{t}\backslash G$, hence
$\mathbb{F}_{p}[[\dot{t}]]\cong F_{p}[[G_{t}\backslash G]]$. Observe also that
$Tor_{n}^{\mathbb{F}_{p}[[G]]}(M^{*},\mathbb{F}_{p}[[\dot{t}]])$ is naturally
isomorphic to $H_{n}(G,\mathbb{F}_{p}[[T]]\otimes M^{*})$. Therefore, by [11,
Lemma 6.10.10] for each $\dot{t}\in T/G$ one has the following commutative
diagram

$\textstyle{H_{n}(G,\mathbb{F}_{p}[[\dot{t}]]\otimes
M^{*})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\epsilon_{n}}$$\scriptstyle{\varphi_{n}}$$\textstyle{H_{n}(G,M^{*})}$$\textstyle{H_{n}(G_{t},M^{*})\hbox
to0.0pt{,\hss}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{Cor}_{t}}$
(3.2)

where $\varphi_{n}$ is an isomorphism and $\epsilon_{n}$ is induced from the
augmentation map
$\epsilon\colon\mathbb{F}_{p}[[\dot{t}]]\rightarrow\mathbb{F}_{p}$, for every
$n\geq 0$.

By [10, Proposition 5.5.4] $\mathbb{F}_{p}[[T]]=\bigoplus_{\dot{t}\in
T/G}\mathbb{F}_{p}[[\dot{t}]])$ and so

$H_{i}(G,\mathbb{F}_{p}[[T]]\otimes M^{*})=\oplus_{\dot{t}\in
T/G}H_{i}(G,\mathbb{F}_{p}[[\dot{t}]]\otimes M^{*})$

by [10, Theorem 9.1.3 (b)]. Thus by (3.2) the surjectivity of

$\bigoplus_{\dot{t}\in T/G}H_{2}(G,\mathbb{F}_{p}[[\dot{t}]]\otimes M^{*})\to
H_{2}(G,M^{*})$

is equivalent to generation of $H_{2}(G,M)$ by
$\operatorname{Cor}_{t}(H_{2}(G_{t},M^{*}))$, $t\in T$. Dually this means that
$\bigcap_{t\in T}\operatorname{Ker}(\operatorname{Res}^{2}_{t})=0$, where

$\operatorname{Res}_{t}^{2}\colon H^{2}(G,M)\longrightarrow H^{2}(G_{t},M)$

is the restriction map (cf. [10, Proposition 6.3.6]).

But, for every $t\in T$, the restriction $\operatorname{Res}_{t}^{2}(\beta)$
of $\beta$ to $G_{t}$ is the $0$-map, because $\alpha|_{B_{t}}$ is an
isomorphism. Hence $\bigcap_{t\in
T}\operatorname{Ker}(\operatorname{Res}^{2}_{t})=0$ implies $\beta=0$ and
conversely $\beta\neq 0$ implies $\bigcap_{t\in
T}\operatorname{Ker}(\operatorname{Res}^{2}_{t})\neq 0$ and so non-
surjectivity of $\kappa_{2}$.

(b) As $B=M\rtimes G$, the group $H^{1}(G,\mathbb{F}_{p})$ is isomorphic to
the group of continuous monomorphisms of $\gamma:G\longrightarrow B$ with
$\alpha\gamma=id$ modulo the conjugation by $M$ (cf. [11, Exercise 6.8.2
(b)]). Similarly, for every $t\in T$, $H^{1}(G_{t},M)$ is isomorphic to the
group of continuous monomorphisms $\gamma:G_{t}\to M\rtimes G_{t}$ with
$\alpha\gamma_{t}=id$ modulo the conjugation by $M$. Thus $B_{y}\leq M\rtimes
G_{t}$ gives an element $\beta_{t}\in H^{1}(G_{t},M)$. Moreover, we can view
$\beta_{t}$ as a homomorphism
$H_{1}(G_{t},M^{*})\longrightarrow\mathbb{F}_{p}$ and its composition with
$\varphi_{1}^{-1}$ from (3.2) gives the continuous homomorphism

$\delta_{\dot{t}}:H_{n}(G,\mathbb{F}_{p}[[\dot{t}]]\otimes
M^{*})\longrightarrow\mathbb{F}_{p}.$

Besides since $\\{G_{y}\mid y\in Y\\}$ is a continuous family ([10, Lemma
5.2.2]) closed for conjugation, the maps

$\delta_{\dot{t}}:H_{1}(G,\mathbb{F}_{p}[[\dot{t}]]\otimes
M^{*})\longrightarrow\mathbb{F}_{p}$

are well defined (i.e. $\beta_{t}$ and $\beta_{t^{g}}$ define the same
$\delta_{\dot{t}}$).

Claim. The maps $\delta_{\dot{t}}$ determine a continuous homomorphism

$\beta:\bigoplus_{\dot{t}\in T/G}H_{1}(G,\mathbb{F}_{p}[[\dot{t}]]\otimes
M^{*})\longrightarrow\mathbb{F}_{p}.$

Indeed, let $U_{i}$ be the family of all open normal subgroups of $B$ such
that $M\cap U_{i}=1$ and $\alpha(U_{i})$ acts trivially on $M$. Then there
exist decompositions $(B,Y)=\varprojlim_{i\in I}(B/U_{i},Y_{i})$,
$(G,T)=\varprojlim_{i\in I}(G_{i},T_{i})$ as inverse limits of finite piles
(see [10, Lemma 5.6.4]), where $G_{i}=G/\alpha(U_{i})$,
$T_{i}=T/\alpha(U_{i})$, and we have $B_{i}=B/U_{i}=M\rtimes G_{i}$. Put
$T_{i}=Y_{i}/M$. Then we have a splitting homomorphism $\alpha_{i}:M\rtimes
G_{i}\rightarrow G_{i}$ and moreover
$\alpha_{i}:(B_{i},Y_{i})\rightarrow(G_{i},T_{i})$ is rigid, for each $i$.
This means that we have a family of $\beta_{i,t_{i}}\in
H^{1}({G_{i}}_{t_{i}},M)$ which determine homomorphisms
$\delta_{i,\dot{t}_{i}}:H_{1}({G_{i}},\mathbb{F}_{p}[\dot{t}_{i}]\otimes
M^{*})\longrightarrow\mathbb{F}_{p}$. This defines uniquely the homomorphism

$\beta_{i}:\oplus_{t_{i}\in
T_{i}}H_{1}({G_{i}},\mathbb{F}_{p}[\dot{t}_{i}]\otimes
M^{*})\rightarrow\mathbb{F}_{p}.$

Now since a projective limit commutes with profinite direct sums (cf. [10,
Theorem 5.3.4]) $\beta=\varprojlim_{i\in I}\beta_{i}$ defines the desired
homomorphism

$\beta:\bigoplus_{\dot{t}\in
T/G}H_{1}({G_{i}},\mathbb{F}_{p}[\dot{t}_{i}]\otimes
M^{*})\longrightarrow\mathbb{F}_{p}.$

Recall that $H_{1}(G,\mathbb{F}_{p}[[T]]\otimes M^{*})=\oplus_{\dot{t}\in
T/G}H_{1}(G,\mathbb{F}_{p}[[\dot{t}]]\otimes M^{*})$. Hence $\kappa_{1}$ can
be rewritten as

$\kappa_{1}\colon\oplus_{\dot{t}\in
T/G}H_{1}(G,\mathbb{F}_{p}[[\dot{t}]]\otimes M^{*})\to H_{1}(G,M^{*})$

Then the injectivity of $\kappa_{1}$ allows to extend
$\delta_{\dot{t}},\dot{t}\in T/G$ to

$\beta\colon H_{1}(G,M^{*})\rightarrow\mathbb{F}_{p},$

whose restriction to $G_{t}$ is $\beta_{t}$, for every $t\in T$.

Conversely, if $\kappa_{1}$ is not injective and $0\neq k$ is in the kernel of
$\kappa_{1}$, then there exists a choice of $\delta_{\dot{t}},\dot{t}\in T/G$
that determine a homomorphism

$\beta:\bigoplus_{t\in T}H_{1}(G,\mathbb{F}_{p}[[\dot{t}]]\otimes
M^{*})\longrightarrow\mathbb{F}_{p}$

with $\beta(k)\neq 0$. Then there exists $i$ such that for
$\beta_{i}=\oplus_{i}\delta_{i,\dot{t}_{i}}$ one has $\beta_{i}(k_{i})\neq 0$,
where $k_{i}$ is the image of $k$ under the projection

$\bigoplus_{\dot{t}\in T/G}H_{1}(G,\mathbb{F}_{p}[[\dot{t}]]\otimes
M^{*})\longrightarrow\oplus_{t_{i}\in
T_{i}}H_{1}(G_{i},\mathbb{F}_{p}[[\dot{t}_{i}]]\otimes M^{*}).$

For each $t_{i}$ choose a homomorphism
$\gamma_{t_{i}}:{G_{i}}_{t_{i}}\longrightarrow B_{i}$ with
$\gamma_{t_{i}^{g}}(g_{i})=\gamma_{t_{i}}(g_{i})^{g}$ for all $g_{i}\in
G_{t_{i}},g\in G_{i}$ that represents $\beta_{i,t_{i}}$ determined by
$\delta_{\dot{t}_{i}}$. Define
$Y_{i}=\bigcup_{t_{i}}B_{i}/\gamma_{t_{i}}({G_{i}}_{t_{i}})$. Then we have a
morphism of piles

$\alpha_{i}:\mathbf{B}_{i}=(B_{i},Y_{i})\longrightarrow\mathbf{G}_{i}=(G_{i},T_{i})$

and the embedding problem $(\alpha_{i},\operatorname{id})$ does not have
solution, since $\beta_{i}$ can not be an element of
$H^{1}(G_{i},M^{*})=Hom(H_{1}(G_{i},M),\mathbb{F}_{p})$. Consider the pull-
back

$\textstyle{\hat{\mathbf{B}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\hat{\alpha}}$$\scriptstyle{\pi}$$\textstyle{\mathbf{G}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\textstyle{\mathbf{B}_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\textstyle{\mathbf{G}_{i}}$

Then the embedding problem $(\hat{\alpha},\operatorname{id})$ does not have a
solution, since otherwise $(\alpha,\operatorname{id})$ does, contradicting the
above. ∎

###### Proof of Theorem 1.1.

Suppose for each $p\in\pi$ and every simple $p$-primary module $M$ the map
$\kappa_{2}$ is surjective. Consider a finite rigid embedding problem

$\big{(}\varphi\colon\mathbf{G}\to\mathbf{A},\
\alpha\colon\mathbf{B}\to\mathbf{A})\big{)}$ (1.1)

with $M=|\operatorname{Ker}(\alpha)|$ is minimal normal elementary abelian
$p$-group for $p\in\pi$ (cf. Remark 2.4(b)).

Let $\hat{B}=B\times_{A}G$, $\hat{Y}=Y\times_{X}T$ be the pull-backs. This
gives a pull-back $\hat{\mathbf{B}}=\mathbf{B}\times_{\mathbf{A}}\mathbf{G}$
of piles and let $\pi\colon\hat{\mathbf{B}}\to\mathbf{B}$,
$\hat{\alpha}\colon\hat{\mathbf{B}}\to\mathbf{G}$ be the coordinate
projections. Thus we have a commutative diagram

$\textstyle{\hat{\mathbf{B}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\hat{\alpha}}$$\scriptstyle{\pi}$$\textstyle{\mathbf{G}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\textstyle{\mathbf{B}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\textstyle{\mathbf{A}}$

By [6, Lemma 4.8] $\hat{\alpha}$ is rigid.

By Lemma 3.2 (a) if $\kappa_{2}$ is surjective then $\hat{\alpha}$ splits, i.e
there is a morphism $\psi\colon\mathbf{G}\to\mathbf{B}$ such that
$\hat{\alpha}\circ\psi=\operatorname{id}$.

By Lemma 3.2 (b) if $\kappa_{1}$ is injective then the embedding problem
$(\operatorname{id},\hat{\alpha})$ has a solution $\hat{\gamma}$ in which case
$\pi\circ\hat{\gamma}$ is the required solution.

Conversely, if $\mathbf{G}$ is not $\pi$-projective then (1.1) does not have
any solution and so $(\operatorname{id},\hat{\alpha})$ does not have solution
so by Lemma 3.2 either $\kappa_{2}$ is not surjective or $\kappa_{1}$ is not
injective. ∎

###### Proof of Corollary 1.2.

By Remark 2.4(b) we may consider only $\mathcal{C}$-embedding problems
$(\alpha,\varphi)$ with $\ker(\alpha)$ minimal normal. But if $\mathcal{C}$ is
the class of all finite solvable groups then such $\ker(\alpha)$ is elementary
abelian. Thus the result follows from Theorem 1.1. ∎

######  Corollary 3.3.

Let $(G,T)$ be a pro-$p$ pile. Then $(G,T)$ is projective if and only if
$H_{i}(G,\mathbb{F}_{p}[[T]])\to H_{i}(G,\mathbb{F}_{p})$, induced from
$t\mapsto 1$ for every $t\in\dot{t}$, is surjective for $i=2$ and injective
for $i=1$.

##  4\. open problems

1\. Let $(G,T)$ be a profinite projective pile and $T^{\prime}=\\{t\in T\mid
G_{t}\neq 1\\}$. Suppose that $G$ is finitely generated.

(a) Is $T^{\prime}/G$ finite?

(b) Is $G_{t}$ a retract of $G$?

2\. Can one embed a projective profinite pile $(G,T)$ into a free profinite
product $H=\coprod_{x\in X}H_{x}\amalg F$ with $F$ free profinite such that
for every $t\in T$ one has $G_{t}=H_{x}^{h}\cap G$ for some $x\in X$?

If there exists a continuous section $T/G\longrightarrow T$ this follows from
Proposition 2.5. The pro-$p$ case is the subject of [6].

3\. Let $F$ be an abstract free group and $H$ a finitely generated subgroup of
$F$. Suppose $(\widehat{F},T)$ is a projective pile with $|T|>1$ and
$\overline{H}=\widehat{F}_{t}$ for some $t\in T$. Is $H$ a free factor of $F$?

This is the case if $\overline{H}$ is a free factor of $\widehat{F}$, see [7,
W, 2] for 3 different proofs.

4\. Let $F$ be a finitely generated free group and $1\neq g\in F$. Suppose
$(\widehat{F},T)$ is a projective pile with $|T|>1$ and $g\in\widehat{F}_{t}$
for some $t\in T$. Does $g$ belong to a non-trivial free factor of $F$?

5\. Is it true that $(G,T)$ is a projective prosolvable pile if and only if
$G$ acts on a profinite tree with trivial edge stabilizers and vertex
stabilizers being point stabilizers $G_{t}$, $t\in T$?

Theorem below states that this is true if $T\longrightarrow T/G$ admits a
continuous section.

######  Theorem 4.1.

  1. (i)

Let $G$ be a prosolvable group acting on a profinite tree $D$ with trivial
edge stabilizers. Then $(G,V(D))$ is $\mathcal{C}$-projective, where
$\mathcal{C}$ is the class of all finite solvable groups.

  2. (ii)

Let $\mathcal{C}$ be a class of finite groups closed for subgroups, quotients
and extensions. Let $(G,T)$ be a projective pro-$\mathcal{C}$ pile. Suppose
there exists a continuous section $\sigma:T/G\longrightarrow T$. Then $G$ acts
on a pro-$\mathcal{C}$ tree with trivial edge stabilizers and non-trivial
vertex stabilizers being stabilizers of points in $T$.

###### Proof.

(i). Suppose $G$ acts on a profinite tree $D$ with trivial edge stabilizers.
This means that we have the short exact sequence of $F_{p}[[G]]$-modules for
any prime $p$

$0\to\mathbb{F}_{p}[[E^{*},*]]\to\mathbb{F}_{p}[[V]]\to\mathbb{F}_{p}\to 0$

where $\mathbb{F}_{p}[[E^{*}]]$ is free $\mathbb{F}_{p}[[G]]$-module and $V$
is the set of vertices of $D$ (see [10, Section 2.4]).

Then, after tensoring it with $\widehat{\otimes}_{\mathbb{F}_{p}[[G]]}M$, we
get a long exact sequence in homology

$Tor_{2}^{G}(M,\mathbb{F}_{p}[[E^{*},*]])\to
Tor_{2}^{G}(M,\mathbb{F}_{p}[[V]])\to H_{2}(G,M)\to
Tor_{1}^{G}(M,\mathbb{F}_{p}[[E^{*},*]]).$

Since $F_{p}[[E^{*},*]]$ is a free $\mathbb{F}_{p}[[G]]$-module, the left and
right terms are $0$. It remains to observe that
$V=\mathop{\vphantom{\bigcup}\mathchoice{\leavevmode\vtop{\halign{\hfil$\m@th\displaystyle#$\hfil\cr\bigcup\cr\cdot\crcr}}}{\leavevmode\vtop{\halign{\hfil$\m@th\textstyle#$\hfil\cr\bigcup\cr\cdot\crcr}}}{\leavevmode\vtop{\halign{\hfil$\m@th\scriptstyle#$\hfil\cr\bigcup\cr\cdot\crcr}}}{\leavevmode\vtop{\halign{\hfil$\m@th\scriptscriptstyle#$\hfil\cr\bigcup\cr\cdot\crcr}}}}\displaylimits_{\dot{v}\in
V/G}\dot{v}$ is the disjoint union of orbits, so
$\mathbb{F}_{p}[[V]]=\bigoplus_{\dot{v}}\mathbb{F}_{p}[[\dot{v}]]$, and the
pro-$p$ direct sum commutes with homology ([10, Theorem 9.1.3(b)]). Since
$Tor_{2}^{G}(M,\mathbb{F}_{p}[[V]])=H_{2}(G,\mathbb{F}_{p}[[V]]\otimes M)$ the
result follows from Theorem 1.1.

(ii) By Proposition 2.5 $G$ embeds into a free pro-$\mathcal{C}$ product of
$H=\coprod_{x\in X}H_{x}$ such that $\\{G_{t}\mid t\in T\\}=\\{H_{x}^{h}\cap
G\mid h\in H,x\in X\\}$. Let $S$ be the standard pro-$\mathcal{C}$ tree for
$H=\coprod_{x\in X}H_{x}$ ([10, Section 6.3]) By [10, Lemma 6.3.2(b)] $H$ acts
on $S$, with trivial edge stabilizers and vertex stabilizers being conjugates
of $H_{x}$. Then the restriction of this action to $G$ gives the required
action.

∎

## References

  * [1] W. Dicks and D. J. Dunwoody, Groups acting on Graphs. Cambridge University Press, 1989.
  * [2] A. Garrido and A. Jaikin-Zapirain. Free factors and profinite completions. Preprint.
  * [3] D. Haran. On closed subgroups of free products of profinite groups. Proc. London Math Soc. 55 (1987), 147–206.
  * [4] D. Haran and M. Jarden, The absolute Galois group of a semi-local field, Springer Monographs in Mathematics, Springer, 2021.
  * [5] D. Haran, M. Jarden, and F. Pop, Projective group structures as absolute Galois structures with block approximation, Memoirs of AMS 189 (2007) 1-56.
  * [6] D. Haran and P.A. Zalesskii. Relatively projective pro-$p$ groups. arXiv:2208.12982.
  * [7] D. Puder, O. Parzanchevski, Measure preserving words are primitive. J. Amer. Math. Soc. 28 (2015), 63–97.
  * [8] F. Pop, On prosolvable subgroups of profinite free products and some applications, Manuscripta mathematica 86 (1995) 125-135.
  * [9] Ribes, L., Cohomological characterization of amalgamated products of groups. J. Pure Appl. Algebra 4, 309–317 (1974).
  * [10] L. Ribes, Profinite graphs and groups, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics, volume 66, Springer, 2017.
  * [11] L.Ribes and P.A. Zalesskii, Profinite Groups, Second Edition, Ergebnisse der Mathematik und ihrer Grenzgebiete (3) 40, Springer-Verlag Berlin Heidelberg 2010.
  * [12] Gareth Wilkes, Relative extensions and cohomology of profinite groups. arXiv:2209.06492.


# Polynomial convexity with degree bounds

Marko Slapar  Faculty of mathematics and physics
University of Ljubljana
Jadranska 19
1000 Ljubljana, Slovenia

Faculty of education
University of Ljubljana
Kardeljeva ploščad 16
1000 Ljubljana, Slovenia

Institute of Mathematics, Physics and Mechanics
Jadranska 19
1000 Ljubljana, Slovenia<EMAIL_ADDRESS>
###### Abstract.

We introduce different notions of polynomial convexity with bounds on degrees
of polynomials in $\mathbb{C}^{n}$. We provide some examples in higher
dimensions and show necessary and sufficient conditions for polynomial
convexity with degree bounds for certain sets of points in $\mathbb{C}$ and
for certain arcs in the unit circle.

###### Key words and phrases:

polynomial convexity, lemniscate, configurations of points

###### 2020 Mathematics Subject Classification:

32E20, 30E10, 30C10

Supported by the research program P1-0291 and research projects J1-3005 and
N1-0237 at the Slovenian Research and Innovation Agency.

##  1\. Introduction

For a compact set $K\subset\mathbb{C}^{n}$, its polynomial hull is defined as

$\mathcal{P}(K)=\\{z\in\mathbb{C}^{n};|P(z)|\leq\|P\|_{K}\text{ for all
polynomials }P\\}.$

The set is called polynomially convex, if $P(K)=K$. We can characterise
polynomial convexity in the following way. Let
$F:[s,\infty)\times\mathbb{C}^{n}\to\mathbb{C}$ be a continuous map, such that
for every $t\in[s,\infty)$, the map

$F_{t}=F(t,\cdot)\\!:\mathbb{C}^{n}\to\mathbb{C}$

is a polynomial, and let for every $t\in[s,\infty)$ the

$H_{t}=\\{z\in\mathbb{C}^{n};F_{t}(z)=0\\}$

be the zero set of $F_{t}$. We call the map $t\mapsto H_{t}$ the curve of
algebraic hypersurfaces in $\mathbb{C}^{n}$. If

$\lim_{t\to\infty}\mathrm{dist}(0,H_{t})=\infty,$

we say that $H_{t}$ tends to infinity. The following result was proved by
Stolzenberg ([2]) and is often referred as the Oka characterisation of
polynomial hulls.

######  Proposition 1 ([2], &1).

A compact set $K\subset\mathbb{C}^{n}$ is polynomially convex, if and only if,
for every $z_{0}\in\mathbb{C}^{n}\backslash K$, there exists a curve of
algebraic surfaces $H_{t}$, $t\geq s$, that tends to infinity, so that
$z_{0}\in H_{s}$ and $H_{t}\cap K=\emptyset$ for every $t\geq s$.

The proof in one direction is immediate. If
$z_{0}\in\mathbb{C}^{n}\backslash\mathcal{P}(K)$ and
$P:\mathbb{C}^{n}\to\mathbb{C}$ is such that $P(z_{0})=1$ and $\|P\|_{K}<1$,
then $H_{t}=\\{P(z)=t\\}$, $t\geq 1$, gives a curve of algebraic surfaces in
the complement of $K$, that tends to infinity, and $z_{0}\in H_{0}$. The other
direction requires approximating $1/F_{t}$ on $\mathcal{P}(K)$ by polynomials
and does not give such a clear correspondence between $F_{t}$ and the
polynomial $P$ that separates $z_{0}$ from $\mathcal{P}(K)$. Notice that one
can already take the curve of algebraic surfaces to be the real level sets of
a single polynomial.

By restricting the degrees of the polynomial, this motivates us to define the
following polynomial hulls with restricted degrees.

######  Definition 2.

Let $K\subset\mathbb{C}^{n}$ be a compact set and let
$d\in\mathbb{N}\cup\\{\infty\\}$.

  * •

The d-polynomial hull of $K$ is the set

$\mathcal{P}_{d}(K)=\\{z\in\mathbb{C}^{n};|P(z)|\leq\|P\|_{K}\text{ for all
polynomials }p\text{ with }\deg P\leq d\\}.$

A compact set $K\subset\mathbb{C}^{n}$ is d-polynomially convex if
$\mathcal{P}_{d}(K)=K$.

  * •

The Oka d-polynomial hull of $K$ is the set $\mathcal{P}^{g}_{d}(K)$
containing all the points $z_{0}\in\mathbb{C}^{n}$, such that every curve of
algebraic surfaces $H_{t}$, $t\geq s$, that tends to infinity, with $z_{0}\in
V_{0}$ and $\deg F_{t}\leq d$, where
$F:[s,\infty)\times\mathbb{C}^{n}\to\mathbb{C}$ is the function defining the
curve of algebraic surfaces, has a nonempty intersection with $K$. A compact
set $K\subset\mathbb{C}^{n}$ is Oka d-polynomially convex if
$\mathcal{P}^{Oka}_{d}(K)=K$.

  * •

The geometric d-polynomial hull of $K$ is the set $\mathcal{P}^{g}_{d}(K)$
containing all the points $z_{0}\in\mathbb{C}^{n}$, such that for every
polynomial $P$, $\deg P\leq d$ and $P(z_{0})=0$, there exists a $\lambda\geq
0$, so that $\\{P(z)=\lambda\\}\cap K\neq\emptyset$. A compact set
$K\subset\mathbb{C}^{n}$ is geometrically d-polynomially convex if
$\mathcal{P}^{g}_{d}(K)=K$

Stolzenberg’s result shows that

$\mathcal{P}(K)=\mathcal{P}_{\infty}(K)=\mathcal{P}^{g}_{\infty}(K)=\mathcal{P}^{Oka}_{\infty}(K),$

and from the proof of his result, we clearly see that

$\mathcal{P}^{Oka}(K)\subseteq\mathcal{P}_{d}^{g}(K)\subseteq\mathcal{P}_{d}(K)$

for every $d\in\mathbb{N}$. It is also obvious that for $d_{1}\leq d_{2}$, we
have inclusions

$\mathcal{P}_{d_{2}}(K)\subseteq\mathcal{P}_{d_{1}}(K),\mathcal{P}_{d_{2}}^{g}(K)\subseteq\mathcal{P}^{g}_{d_{1}}(K),\mathcal{P}_{d_{2}}^{Oka}(K)\subseteq\mathcal{P}^{Oka}_{d_{1}}(K)$

and that

$\mathcal{P}(K)=\bigcap_{d\in\mathbb{N}}P_{d}(K)=\bigcap_{p\in\mathbb{N}}\mathcal{P}_{d}^{g}(K)=\bigcap_{d\in\mathbb{N}}\mathcal{P}_{d}^{Oka}(K)$

holds. We will see from examples below, that in general, inclusions are
strict.

######  Remark 3.

In any dimension, polynomial $1$-convexity is just the standard convexity,
while Oka $1$-polynomial convexity is harder to characterise. A set is
$K\subset\mathbb{C}^{n}\subset\mathbb{CP}^{n}$ is Oka $1$-polynomially convex,
if it is linearly convex and the set $K^{*}\subset\mathbb{CP}^{n*}$, defined
as the set of hyperplanes in $\mathbb{CP}^{n}$ not intersecting $K$, is the
convex hull of a connected set [3, Remark 2.1.10].

######  Remark 4.

When $n=1$, it is clear that if $K\in\mathbb{C}$ is already polynomially
convex, then $P^{Oka}_{1}(K)=K$, since polynomial convexity in dimension one
is equivalent to the complement of $K$ being path connected, and that for
general compact sets $K\subset\mathbb{C}$,
$P^{Oka}_{1}(K)=\mathbb{C}\backslash V$, where $V$ is the unbounded connected
component of $\mathbb{C}\backslash K$. One usually needs polynomials of much
higher degrees to test the polynomial convexity with the other two
definitions.

The paper is organised as follows. In the next section, we give some examples
to illustrate the problem in dimension $2$. In the remaining of the paper, we
focus on dimension one, where characterising $d$-polynomial convexity already
turns out to be a difficult problem. The arguments in the paper are mostly
elementary.

##  2\. Results in $\mathbb{C}^{2}$

Compact subsets in a totally real subspace of $\mathbb{C}^{2}$ are always
polynomially convex and the following proposition shows that one can already
conclude that by using polynomials of degree $2$.

######  Proposition 5.

Let $K\subset\mathbb{R}^{2}\subset\mathbb{C}^{2}$ be a compact subset. Then
for every $(z_{0},w_{0})\in\mathbb{C}^{2}\backslash K$, there exist a
polynomial $P(z,w)$ of degree two, so that $P(z_{0},w_{0})=1$ and
$\|P\|_{K}<1.$ So $K=\mathcal{P}(K)=\mathcal{P}_{2}(K)$.

###### Proof.

Using the linear map $F(z,w)=(z+iw,z-iw)$, we can assume that $K\subset V$,
where $V$ is the totally real plane
$V=\\{(\zeta,\overline{\zeta});\zeta\in\mathbb{C}\\}$. Let
$(z_{0},w_{0})\in\mathbb{C}\backslash K$ and let

$M=\max\\{|(z-z_{0})(w-w_{0})|;(z,w)\in K\\}.$

Let first $(z_{0},w_{0})\in\mathbb{C}\backslash V$. Let $z_{0}w_{0}=a+ib$. If
$b\neq 0$, we set

$P(z,w)=1+\frac{i}{m}(zw-z_{0}w_{0}),$

with $m\in\mathbb{R}$, $m\neq 0$, to be determined. Then $P(z_{0},w_{0})=1$
and

$|p(\zeta,\overline{\zeta})|^{2}=(1+\frac{b}{m})^{2}+\frac{1}{m^{2}}(|\zeta|^{2}-a)^{2}.$

On $K$, $|P(z,w)|<1$, as long as

$-2bm>b^{2}+(|\zeta|^{2}-a)^{2})$

for $(\zeta,\overline{\zeta})\in K$, which is equivalent to $-2bm>M$.

If $b=0$, we can set

$P(z,w)=1+\frac{i}{m}(zw-iz_{0}w_{0}).$

Again $P(z_{0},w_{0})=1$ and

$|p(\zeta,\overline{\zeta})|^{2}=(1+\frac{a}{m})^{2}+\frac{1}{m^{2}}|\zeta|^{4}.$

On $K$, $|P(z,w)|<1$, as long as

$-2am>a^{2}+|\zeta|^{4}$

for $(\zeta,\overline{\zeta})\in K$.

If $(z_{0},w_{0})\in V\backslash K$, we can set

$P(z,w)=1-\frac{1}{2M}(z-z_{0})(w-w_{0}).$

∎

As expected, the same does not hold for general compact sets in
$\mathbb{C}^{2}$, even if we restrict to simple curves in the standard
Lagrangian torus.

######  Proposition 6.

Let $p,q\in\mathbb{N}$, $D(p,q)=1$, and
$K=\\{(e^{ip\theta},e^{-iq\theta};0\leq\theta\leq
2\pi\\}\subset\mathbb{C}^{2}$ be the $(p,-q)-$torus knot. Then every
polynomial $P(z,w)$, with $P(0,0)=1$ and $\|p\|_{K}<1$, must be of degree at
least $p+q$. Furthermore $\mathcal{P}_{p+q}(K)=K$ and for $d<p+q$,
$\mathcal{P}_{d}(K)$ strictly contains $K$.

###### Proof.

We can assume that $p>q$ and let $P(z,w)=1+Q(z,w,)$ be a polynomial of degree
less then $p+q$, $Q(0,0)=0$, with $|1+Q(\zeta^{p},\overline{\zeta}^{q})|<1$
for $|\zeta|=1$. By substituting $\overline{\zeta}=\zeta^{-1}$, we get

$\left|1+Q(\zeta^{p},\zeta^{-q})\right|<1.$

Since $p$ and $q$ are coprime and the degree of $Q$ is less then $p+q$,
$Q(\zeta^{p},\zeta^{-q})$ does not contain a nonzero constant term. Let $j\geq
0$ be the smallest integer, so that $\zeta^{j}Q(\zeta^{p},\zeta^{-q})$ does
not contain any terms of negative degree in $\zeta$. Multiplying the
inequality with $\zeta^{j}$, the inequality is equivalent to

$\left|\zeta^{j}+R(\zeta)\right|<1,\quad|\zeta|=1,$

where $R$ is a polynomial in $\zeta$, not containing the term $\zeta^{j}$. Let

$a=\frac{1}{\max_{|\zeta|=1}{|\zeta^{j}+R(\zeta)|}}.$

Using Cauchy formula for the $j-$th derivative of $f$ at $0$, we find that
this is not possible.

Strict containment for $d<p+q$ obviously follows from above. To show that that
$\mathcal{P}(K)=\mathcal{P}_{p+q}(K)$, we take $(a,b)\not\in K$. If
$(a,b)\notin S=\\{z^{q}w^{p}=1\\}$, we can take

$Q(z)=z^{q}w^{p}-1.$

Then $Q(K)=\\{0\\}$ and $|Q(a,b)|>0$. If $(a,b)\in S\backslash K$, then the
point already lies outside the (convex) standard closed polydisc, and we can
separate the point by affine functions. ∎

While we see from the above proposition, that one cannot demonstrate
polynomial convexity (using the standard definition) of torus knots
$K_{p.-q}=\\{(e^{ip\theta},e^{-iq\theta};0\leq\theta\leq 2\pi\\}$ with
polynomials of degree less than $p+q$, it might be possible to find curves of
surfaces of lower degree for this purpose (using Oka or geometric
characterisation).

######  Example 7.

Let $K_{p}=\\{(e^{pi\theta},e^{-i\theta});0\leq\theta\leq 2\pi\\}\subset
S^{1}\times S^{1}\subset\mathbb{C}^{2}.$

  * •

Let $H_{\lambda}=\\{(z,w)\in\mathbb{C}^{2};zw-z-w=\lambda\\}$. Then $(0,0)\in
H_{0}$ and $K_{2}\cap H_{\lambda}=\emptyset$ for all $\lambda\in[0,\infty)$.

  * •

Let $H_{\lambda}=\\{(z,w)\in\mathbb{C}^{2};-2zw+z-2w=\lambda\\}$. Then
$(0,0)\in H_{0}$ and $K_{3}\cap H_{\lambda}=\emptyset$ for all
$\lambda\in[0,\infty)$.

  * •

Let $H_{\lambda}=\\{(z,w)\in\mathbb{C}^{2};3zw-3z-5w=\lambda\\}$. Then
$(0,0)\in H_{0}$ and $K_{4}\cap H_{\lambda}=\emptyset$ for all
$\lambda\in[0,\infty)$.

So, at least for $p\leq 4$, one can test that $(0,0)$ is not in the geometric
polynomial hull of $K_{p}$, using polynomials of degree $2$. We do not believe
that this holds for larger $p$, but we are not able to find the lowest degree
of polynomials for even such a simple case.

##  3\. Points in $\mathbb{C}$

The first already interesting example is understanding $n$-polynomial
convexity of points in $\mathbb{C}$. If a set of points has cardinality less
or equal to $n$, than it is trivially $n$ polynomially convex, since we can
always interpolate $n+1$ points with a polynomial of degree $n$. We will see
that for $n>2$, a generic configuration of $n+1$ points, turns out to be
$n$-polynomially convex.

######  Lemma 8.

Let $K=\\{z_{0},z_{1},\ldots,z_{n}\\}$ be a set of distinct points in
$\mathbb{C}$ and let $w\not\in K$. The point $w$ is in the $n$-polynomial hull
of $K$ if and only if all the values

$(z_{i}-w)\prod_{j\neq i}(z_{i}-z_{j})$

lie on the same real half-line from the origin.

###### Proof.

We first assume that $w=0$ and let $K=\\{z_{0},z_{1},\ldots,z_{n}\\}$,
$0\not\in K$. The point $w$ is in $n$-polynomial hull of $K$, if for every
polynomial

$Q(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+\ldots+a_{0},$

such that $\|Q\|_{K}=1$, we also have $\|Q(0)\|\leq 1$. Let us write

$\begin{bmatrix}1&z_{0}&\cdots&z_{0}^{n}\\\ 1&z_{1}&\cdots&z_{n}^{n}\\\
\vdots&\vdots&\ddots&\vdots\\\
1&z_{n}&\cdots&z_{n}^{n}\end{bmatrix}\begin{bmatrix}a_{0}\\\ a_{1}\\\
\vdots\\\ a_{n}\end{bmatrix}=\begin{bmatrix}\alpha_{0}\\\ \alpha_{1}\\\
\vdots\\\ \alpha_{n}\end{bmatrix},$

where
$\alpha=\begin{bmatrix}\alpha_{0},&\alpha_{1},&\cdots,&\alpha_{n}\end{bmatrix}^{T}\in\mathbb{C}^{n+1}$
is some point with $\|\alpha\|_{\infty}=1$. If we denote the Vandermonde
matrix above by $V$, we must have

$\left|\left<\begin{bmatrix}1,&0,\cdots,&0\end{bmatrix},V^{-1}\alpha\right>\right|\leq
1.$

This is true for all $|\alpha|_{\infty}\leq 1$, if and only if, the first
column of $V^{-T}$ has $l_{1}$ norm less or equal to $1$. Using Cramer’s rule,
this gives the inequality

(3.1) $\frac{1}{|\det V|}\sum_{i=0}^{n}\left|\frac{z_{0}z_{1}\cdots
z_{n}}{z_{i}}\det V_{i}\right|\leq 1,$

where $V_{i}$ is the Vandermonde determinant of the of points
$z_{0},\ldots,\hat{z_{i}},,\ldots,z_{n}.$ Since the determinant of a
Vandermonde matrix is the product of deferences of the points,

$\det V=\prod_{0\leq i<j\leq n}(z_{j}-z_{i}),$

we get

$\det V=(-1)^{i}\det V_{i}\prod_{j\neq i}(z_{i}-z_{j}).$

By expanding the determinant of $V$ by the first column, we also get

$\det V=z_{0}z_{1}\cdots z_{n}\sum_{i}(-1)^{i}\frac{\det
V_{i}}{z_{i}}=z_{0}z_{1}\cdots z_{n}\det V\sum_{i}\frac{1}{z_{i}\prod_{j\neq
i}(z_{i}-z_{j})}.$

The inequality (3.1) can be rewritten as

$\sum_{i=0}^{n}\left|\frac{\det V}{z_{i}\prod_{j\neq
i}(z_{i}-z_{j})}\right|\leq\left|\sum_{i=0}^{n}\frac{\det V}{z_{i}\prod_{j\neq
i}(z_{i}-z_{j})}\right|.$

This happens, if and only if, all the values

$z_{i}\prod_{j\neq i}(z_{i}-z_{j})$

lie on the same real half-line from the origin in $\mathbb{C}$. If $w\neq 0$,
we get the result by using a simple affine transformation $z\mapsto z-w$. ∎

Using the above lemma, we can geometrically describe the $n$-polynomial hull
of $n+1$ points in $\mathbb{C}$.

######  Proposition 9.

Let $n>1$ and $K=\\{z_{0},z_{1},\ldots,z_{n}\\}$, where the points
$z_{0},z_{1},\ldots,z_{n}$ are (in a counter-clockwise order) vertices of a
convex $n+1$-sided polygon. Let $w$ be a point in the convex hull of $K$. For
every $i=0,\ldots,n$, let $\alpha_{i,j}$ be the angle at the point $z_{j}$,
$j\neq i-1,i,i+1$ ($n+1:=0$), of the triangle $\Delta z_{i}z_{i+1}z_{j}$, and
let $\beta_{i}$ be the angle at $w$ of the triangle $\Delta z_{i}z_{i+1}w$.
Then $w$ is in the $n$-polynomial hull of $K$, if and only if, for every
$i=0,\ldots,n$,

$\beta_{i}+\sum_{j}\alpha_{i,j}=\pi.$

If such a point $w$ exists, it is unique.

###### Proof.

Let $i=0,1,\ldots,n$. From Lemma 8, we see that $w$ is in the $n$-polynomial
hull of $K$ if and only all the equalities

$\frac{z_{i}-w}{|z_{i}-w|}\prod_{j\neq
i}\frac{z_{j}-z_{i}}{|z_{j}-z_{i}|}=\frac{z_{i+1}-w}{|z_{i+1}-w|}\prod_{j\neq
i+1}\frac{z_{j}-z_{i+1}}{|z_{j}-z_{i+1}|}$

hold, which is equivalent to

$\frac{(z_{i+1}-w)/|z_{i+1}-w|}{(z_{i}-w)/|z_{i}-w|}\prod_{j\neq
i,i+1}\frac{(z_{j}-z_{i+1})/|z_{j}-z_{i+1}|}{(z_{j}-z_{i})/|z_{j}-z_{i}|}=-1.$

This is equivalent to

$e^{i(\beta_{i}+\sum_{j}\alpha_{i,j})}=-1.$

Since the sum of all $\alpha_{i,j}$ over all $i,j$ is just the sum of the
interior angles of the polygon, while the sum of all $\beta_{i}$ is $2\pi$, we
get

$\sum_{i}\left(\beta_{i}+\sum_{j}\alpha_{i,j}\right)=2\pi+(n-1)\pi=(n+1)\pi,$

and all the summands must be equal to $\pi$.

If such a point $w$ exists, it must therefore lie on the intersection of $n$
distinct circular arcs over the edges of the polygon, and must be inside the
polygon (all the point with a given angle with respect to a line segment lie
on a circle having that line segment as a cord, Figure 1).

$z_{0}$$z_{1}$$z_{2}$$z_{3}$$w$ Figure 1.

Since two distinct circles only intersect at most two points, and one of them
is either a vertex, or lies outside the polygon, such a point $w$ must be
unique. ∎

######  Remark 10.

We can write a similar condition for angles of $w$ for arbitrary line segments
with end points in the set $K$. Let $z_{i}z_{j}$ be a line segment. The
formula we get is

$\beta+\sum\epsilon_{k}\alpha_{k}=\pi,$

where $\epsilon_{k}$ equals to $1$ if the point $z_{k}$ lies in the same half
plane defined by the line through $z_{i}$ and $z_{j}$ as $w$, and $-1$ if the
point lies in the opposite half plane. The convexity condition in the
proposition was only used to ensure that all points in the same half plane
defined by the line through $z_{i}$ and $z_{i+1}$.

######  Example 11.

For $3$ points $z_{0},z_{1},z_{2}\in\mathbb{C}$, it is easy to check that the
condition in Proposition 9 is equivalent to $w$ being the orthocenter of the
triangle $\Delta z_{0}z_{1}z_{2}$, which has to be acute. So three points are
$2$-polynomially convex if and only if they are the vertices of an obtuse or
right triangle (any number of collinear points are $2$-polynomially convex).
If we take the vertices of an acute triangle and add the orthocenter, the four
point set is also $2$-polynomially convex.

For convex dependent points, we have a simple result.

######  Proposition 12.

If $K$ is a set of $n+1$ distinct points in $K$ that is not convex
independent, then $K$ is $n$-polynomially convex.

###### Proof.

We number the points in $K$ so that
${\mathrm{Conv}}K={\mathrm{Conv}}\\{z_{0},\ldots,z_{k}\\}$ and
$z_{0},\ldots,z_{k}$ are convex independent. The rest of the points
$z_{k+1},\ldots,z_{n}$ lie in the interior of the convex hull of $K$. Let
$w\not\in K$ be a point in the $n$-polynomial hull of $K$. Then $w$ lies in
the interior of ${\mathrm{Conv}}K$, and for every $i=0,1,\ldots,k$, we have by
Lemma 8

$\beta_{i}+\sum_{j\leq k}\alpha_{i,j}+\sum_{j>k}\alpha_{i,j}=\pi,$

where $\beta_{i}$ is the angle of a triangle $\Delta z_{i}z_{i+1}w$ at $w$ and
$\alpha_{i,j}$ the angle at $z_{j}$ of the triangle $\Delta
z_{i}z_{i+1}z_{j}$, $j\neq i,i+1$ (we use $k+1=0$). Note that, since
$z_{0},\ldots,z_{k}$ are extremal points, of ${\mathrm{Conv}}K$, all the other
points are in the same side of line segments
$z_{0}z_{1},z_{1}z_{2},\ldots,z_{k}z_{0}$. We have

$\sum_{i=0}^{k}\sum_{j\leq k}\alpha_{i,j}=(k-1)\pi$

since this is precisely the sum of interior angles of convex polygon
$z_{0}z_{1}\ldots z_{k}$. Since

$\sum_{i=0}^{k}\sum_{j>k}\alpha_{i,j}=2(n-k)\pi\ \text{and}\
\sum_{i=1}^{k}\beta_{i}=2\pi,$

we have

$2\pi+(k-1)\pi+2(n-k)\pi=(k+1)\pi,$

which is only possible if $n-k=0$. ∎

We can easily see from Lemma 8 that most configurations of $n+1$ point are
$n$-polynomially convex. The set of all sets of not necessarily distinct $n+1$
points in $\mathbb{C}$ equals

$C_{n+1}=\\{\\{z_{0},z_{1},\ldots,z_{n}\\},z_{0},z_{1},\ldots,z_{n}\in\mathbb{C}\\}=\mathbb{C}^{n+1}/{\mathrm{Symm}}_{n+1},$

where ${\mathrm{Symm}}_{n+1}$ is the symmetric group on $n+1$ elements, and
for the set of all sets of distinct $n+1$ points in $\mathbb{C}$, the set of
unordered $n+1$-configurations in $\mathbb{C}$, we have

${\mathrm{UConf}}_{n+1}(\mathbb{C})={\\{z=(z_{0},z_{1},\ldots,z_{n})\in\mathbb{C}^{n+1},z_{i}\neq
z_{j}\\}}/{{\mathrm{Symm}}_{n+1}}.$

It is well known that the map $(z_{0},z_{1},\ldots
z_{n})\mapsto(z-z_{0})(z-z_{1})\cdots(z-z_{n})$ descents to a map from
$\mathbb{C}^{n+1}/{\mathrm{Symm}}_{n+1}$ to monic polynomials of degree $n+1$,
and gives a structure of an $n+1$ dimensional algebraic manifold on both
$C_{n+1}$ and ${\mathrm{UConf}}_{n+1}(\mathbb{C})$. Also, $C_{n+1}$ is
algebraically isomorphic to $\mathbb{C}^{n+1}$.

######  Proposition 13.

Let $n>1$. The set $\\{c\in C_{n+1},c\text{\ is $n$-polynomially convex}\\}$
is the complement of an $n+4$ dimensional real algebraic subset of $C_{n+1}$.
The set $\\{c\in{\mathrm{UConf}}_{n+1}(\mathbb{C}),c\text{\ is
$n$-polynomially convex}\\}$ is open in ${\mathrm{Conf}}_{n+1}(\mathbb{C})$,
and is the complement of an $n+4$ dimensional real algebraic submanifold.

###### Proof.

Let

$S=\\{(z_{0},z_{1},\ldots,z_{n})\in\mathbb{C}^{n+1};\\{z_{0},z_{1},\ldots,z_{n}\\}\
\text{is not polynomially convex}\\}.$

For every $Z\in S$, the coordinates of $Z$ are all distinct and there exists a
unique $w_{Z}\in\mathbb{C}\backslash\\{z_{0},z_{1},\ldots,z_{n}\\}$, that lies
in the polynomial hull of $\\{z_{0},z_{1},\ldots,z_{n}\\}$. In a small
neighbourhood $U$ of any point in $S$, the map
$\mapsto\frac{1}{z_{0}-w_{Z}}(z_{1}-w_{Z},\cdots,z_{n}-w_{Z})$ gives us a
smooth fibration from $U$ onto an open subset of the set

$S^{\prime}=\\{(z_{1},z_{2},\ldots,z_{n})\in\mathbb{C}^{n};0\ \text{is in the
polynomial hull of}\ \\{1,z_{1},\ldots,z_{n}\\}.$

The fibers are $4$-dimensional. It is thus enough to check that $S^{\prime}$
is an $n$ dimensional real algebraic submanifold in $\mathbb{C}^{n}$. Let
every $k=1,2,\ldots,n$,

$F_{k}=\frac{(z_{k}-z_{1})\cdots(z_{k}-z_{k-1})(z_{k}-z_{k+1})\cdots(z_{k}-z_{n})z_{k}}{(1-z_{1})\cdots(1-z_{n})}$

and let $F=(F_{1},\ldots,F_{n})$. From the condition in Lemma 8, we get that
near $Z^{\prime}\in S^{\prime}$, the point in $S^{\prime}$ are solutions of
$\mathrm{Im\ }F=0$. We thus need to check that the real derivative
$D(F-\overline{F})$ is of maximal rank for any $Z^{\prime}\in S^{\prime}$. If
we denote by $\partial F$ the complex derivative of $F$, this will be true, if
$\det(\partial F)\neq 0$. From

$\displaystyle\frac{\partial F_{k}}{\partial z_{k}}$
$\displaystyle=F_{k}\left(\frac{1}{z_{k}-z_{1}}+\cdots+\frac{1}{z_{k}-z_{n}}+\frac{1}{z_{k}}\right)$
$\displaystyle\frac{\partial F_{k}}{\partial z_{j}}$
$\displaystyle=F_{k}\left(\frac{1}{z_{k}-z_{j}}-\frac{1}{z_{k}-1}\right)$

we get

$\det(\partial F)=\frac{F_{1}\cdots F_{n}}{z_{1}\cdots
z_{n}}\begin{vmatrix}\frac{z_{1}}{z_{1}-z_{2}}+\cdots+\frac{z_{1}}{z_{1}-z_{n}}+1&\cdots&\frac{z_{1}}{1-z_{1}}-\frac{z_{1}}{z_{n}-z_{1}}\\\
\dots&\ddots&\vdots\\\
\frac{z_{n}}{1-z_{n}}-\frac{z_{n}}{z_{1}-z_{n}}&\cdots&\frac{z_{n}}{z_{n}-z_{1}}+\cdots+\frac{z_{n}}{z_{n}-z_{n-1}}+1\end{vmatrix}.$

The last determinant is a rational function in $z_{1}$ and we can see that it
does not have a singularity at (only possible singularities) $z_{1}=z_{2}$ and
$z_{1}=1$ and has a finite value at $z_{1}=\infty$. So the determinant does
not depend on $z_{1}$. From symmetry, the determinant does not depend on other
variables as well, so it is constant. To calculate the constant, we send the
variables successively to infinity to see that it is nonzero ($n!$). ∎

##  4\. Subsets on the unit circle

Let us first look at sets of $n+1$ distinct points $z_{j}=e^{i\phi_{j}}$,
$0\leq\phi_{0}<\ldots<\phi_{n}<2\pi$, $j=0,\ldots,n-1$, on the unit circle.
For every $j$, the angle $\angle z_{j}z_{k}z_{j+1}$ equals half of the angle
$\angle z_{j}0z_{j+1}$, the condition for $w$ in the unit disk from
Proposition 9 to be in the $n$-polynomial hull has an obvious simplification
using the Thales’s theorem:

$\beta_{j}=\pi-(n-1)\frac{\phi_{j+1}-\phi_{j}}{2},$

where $\beta_{j}$ is the angle $\angle\phi_{j}w\phi_{j+1}$ and the point $w$
most lie on the intersection of $n$ circles with arcs $z_{j}z_{j+1}$,
$j=0,\ldots,n-1$, with centers

$c_{j}=\frac{\sin(n(\phi_{j+1}-\phi_{j})/2)}{\sin((n-1)(\phi_{j+1}-\phi_{j})/2)}\left(\cos\frac{\phi_{j}+\phi_{j+1}}{2},\sin\frac{\phi_{j}+\phi_{j+1}}{2}\right)$

and radia

$r_{j}=\frac{\sin\frac{\phi_{j+1}-\phi_{j}}{2}}{\sin\frac{(n-1)(\phi_{j+1}-\phi_{j})}{2}}.$

Notice that for points on the unit circle, the condition in Proposition 9 only
has to be checked for $n$ of the angles, since it then automatically holds for
the last one. With this notation, we get the following observation.

######  Proposition 14.

Let $z_{j}=e^{i\phi_{j}}$, $0\leq\phi_{0}<\ldots<\phi_{n}<2\pi$ be $n+1$
distinct points o the unit circle.

  * (i)

If any of the angles $\phi_{j+1}-\phi_{j}$ is greater or equal $2\pi/n$, the
set of points is $n$-polynomially convex.

  * (ii)

If for any $i,j$, $(\phi_{j+1}-\phi_{j})+(\phi_{i+1}-\phi_{i})$ is less or
equal $2\pi/n$, the set of points is $n$-polynomially convex.

  * (iii)

If the set of points is not $n$-polynomially convex, then the extra point $w$
in the $n$-polynomial hull is contained in the triangle $\Delta z_{j}0z_{j+1}$
with the smallest angle $\angle z_{j}0z_{j+1}$.

###### Proof.

Let us assume that for some $j$, $\phi_{j+1}-\phi_{j}>\frac{2\pi}{n}$ and let
$w$ be the extra point in the $n$-polynomial hull. The angle at $w$ of the
triangle $z_{j}wz_{j+1}$ would have to be smaller than
$\frac{\pi}{n}<\frac{\phi_{j+1}-\phi_{j}}{2}$ and so $w$ would need lie
outside of the unit disk and thus outside of the convex hull of the points.
For (ii), we only need to observe that if
$(\phi_{j+1}-\phi_{j})+(\phi_{i+1}-\phi_{i})\leq 2\pi/n$, at least one of the
the other angles $\phi_{k+1}-\phi_{k}$ has to be greater or equal to $2\pi/n$.

To prove (iii), let us assume that $\phi_{0}=0$ and that $\phi_{1}-\phi_{0}$
is the smallest of the angles. The two circles with centers $c_{0}$ ad $c_{1}$
with appropriate radii $r_{0}$ and $r_{1}$ already intersect at the point
$z_{1}$. The other point of intersection is the reflection of $z_{1}$ across
the line segment, connecting $c_{0}$ and $c_{1}$.

$z_{0}$$z_{1}$$z_{2}$$c_{0}$$c_{1}$$z_{2}$$w$ Figure 2.

The point of intersection necessarily lies below (or on) the line segment
$0z_{1}$, since $\phi_{2}-\phi_{1}\geq\phi_{1}-\phi_{0}$. Analogously, the
point $w$ must be above the segment $0z_{0}$, since
$2\pi-\phi_{n}\geq\phi_{1}-\phi_{0}$. So $w$ must lie in the triangle
$z_{0}0z_{1}$. ∎

######  Example 15.

One obvious example of not $n$-polynomially convex set of $n+1$ points is a
set $S$ of $n+1$ equally distributed points on the unit circle. Then its
$n$-polynomial hull equals $S\cup\\{0\\}$. Furthermore, if the smallest angle
$\phi_{j+1}-\phi_{j}$ goes to $0$, then all the other angles
$\phi_{k+1}-\phi_{k}$ must approach $2\pi/n$. To see this, let
$\phi_{j+1}-\phi_{j}<\epsilon$ and let
$\phi_{k+1}-\phi_{k}=2\pi/n-\epsilon_{k}$. Since the sum of all angles equals
$2\pi$, we have $\epsilon_{k}<\epsilon$ for all $k\neq j$. So, if $w$ in the
interior of the unit disk is close to the boundary, there exists a set of $S$
of $n+1$ points on the unit circle, so that $w$ is contained in the
$n$-polynomial hull of $S$ with the angle at $0$ of the triangle $\Delta
z_{j}0z_{j+1}$, that contains $w$, is small, while all the other angles are
almost $2\pi/n$.

######  Proposition 16.

Let $A_{\alpha}=\\{z\in S^{1};\ -\alpha\leq\arg z\leq\alpha\\}$. If
$\alpha>\frac{n-1}{n}\pi$, then $A_{\alpha}$ is not $n$-polynomially convex.

###### Proof.

We see from the above example, that if $\alpha<2\pi/n$, we can always find a
set of $n+1$ points $S\subset A_{\alpha}$ so that some point near the unit
circle is in the $n$-polynomial hull of $S$ and thus of $A_{\alpha}$. ∎

######  Remark 17.

We are not able to show that $A_{\alpha}=\\{z\in S^{1};\ -\alpha\leq\arg
z\leq\alpha\\}$, $\alpha\leq\frac{n-1}{n}\pi$, then $A_{\alpha}$ is actually
$n$-polynomially convex. It is not hard to see that $A_{\alpha}=\\{z\in
S^{1};\ -\alpha\leq\arg z\leq\alpha\\}$, $\alpha<\frac{\pi}{4}$, is
$2$-polynomially convex. To see this, let $|\phi|\leq\alpha$ and we take the
polynomial $p(x)=z^{2}-2e^{i\phi}z+e^{2i\phi}(a^{2}+1)$. We can see that
$|p(z)|_{A}<|p(re^{i\phi})|$ for $0\leq r<1$, as long as $a>0$ is large
enough. So any point in the angular sector $\\{|z|<1;\ -\alpha\leq\arg
z\leq\alpha\\}$ can be separated from $A$ by a quadratic polynomial.

It would be nice to be able to characterise those compact sets $A$ in
$\mathbb{C}$ (or $\mathbb{C}^{n}$), for which there exists an $n$, so that
$\mathcal{P}(A)=\mathcal{P}_{n}(A)$. We call such sets _finitely polynomially
convex_. For example.

######  Question 18.

Let $\gamma$ be a smooth simple Jordan arc in $\mathbb{C}$. Is $\gamma$
finitely polynomially convex?

We are certainly not able to show finite polynomial convexity of sets by
finding lemniscates of fixed degree, uniformly approximating the arc. To
achieve this, we would need polynomials of larger and larger degrees. In this
direction, there is a large body of research on the improvements of Hilbert
lemniscate theorem. See for example [1].

If we restrict ourselves to rectifiable simple Jordan arcs, the answer to
above question is no, as it follows from the next simple example.

######  Example 19.

Let $K_{n}$, $n=1,2,\ldots$, be a circle with radius $\frac{1}{2n^{2}}$,
centred at $(1/n,0)$, with open circular arcs of angle $\pi/(n+1)$ removed at
each side. Let

$K=\bigcup_{n}K_{n}\cup\bigcup_{n}(\gamma^{-}_{n}\cup\gamma_{n}^{+})\cup\\{(0,0)\\},$

where $\gamma_{n}^{\pm}$ are two arcs connecting sets $K_{n}$ and $K_{n+1}$ at
their boundary points (see Figure 3).

Figure 3.

The arcs $\gamma_{n}^{\pm}$ can be made so that $K$ is smooth except at the
point $(0,0)$. Since $K_{n}$ contains equally spread $2n+2$ points, the point
$(1/n,0)$ is in the $(2n+1)$-polynomial hull of $K$, and so $K$ is not
finitely polynomially convex.

######  Remark 20.

Let $A\subset S^{1}$ be a compact set such that $\overline{A}=A$ ($A$ is
symmetric across $x$ axis). If $0$ is in the $n$-polynomial hull of $A$, then
for any polynomial $p(z)=a_{n}z^{n}+\cdots+a_{1}z+1$, we have $||p||_{A}\geq
1=|p(0)|$ and so
$||z^{n}p(1/z)||_{A}=||z^{n}+a_{1}z^{n-1}+\cdots+a_{n}||_{A}\geq 1.$ So among
all the monic polynomials of degree $n$, the polynomial $z^{n}$ has the
smallest sup norm on $A$ and is thus the (unique) Chebyshev polynomial for
$A$. An example of such a set is $A_{\alpha}=\\{z\in S^{1};\ -\alpha\leq\arg
z\leq\alpha\\}$, with $\alpha>\frac{n}{n+1}\pi,$ but we could take any compact
$A=\overline{A}\subset S^{1}$ that contains a set of $n+1$ equally spread
points. With a different proof, this is a well known result from the theory of
Chebyshev polynomials [4, Theorem 3].

Acknowledgments. I express my gratitude to Josip Globevnik for engaging in
insightful discussions on the topic, and to Thomas Mark for not only
identifying the problem of $2$-polynomial convexity of three points in
$\mathbb{C}$, but also providing a solution to the problem (personal
communication).

## References

  * [1] Andrievskii, V., On the approximation of a continuum by lemniscates. J. Approx. Theory, 105, no.2, 292–304, 2000.
  * [2] Stolzenberg, G., Polynomially and rationally convex sets. Acta Math. 109, 259-289, 1963.
  * [3] Andersson, M., Passare M., Sigurdsson R., Complex Convexity and Analytic Functionals, Progress in Mathematics, Vol. 225, Birkhüser Basel, 2004.
  * [4] Thiran, J.P., Detaille C., Chebyshev polynomials on circular arcs in the complex plane. Progress in approximation theory, Academic Press, 771–786, 1991.


# On the tensor degree of finite groups

Peyman Niroomand  School of Mathematics and Computer Science
Damghan University of Basic Sciences
Damghan, Iran<EMAIL_ADDRESS>and  Francesco G. Russo  DIEETCAM
Universitá Degli Studi di Palermo
Viale Delle Scienze, Edificio 8, 90128, Palermo, Italy
<EMAIL_ADDRESS>
###### Abstract.

We study the number of elements $x$ and $y$ of a finite group $G$ such that
$x\otimes y=1_{{}_{G\otimes G}}$ in the nonabelian tensor square $G\otimes G$
of $G$. This number, divided by $|G|^{2}$, is called the tensor degree of $G$
and has connection with the exterior degree, introduced few years ago in [P.
Niroomand and R. Rezaei, On the exterior degree of finite groups, Comm.
Algebra 39 (2011), 335–343]. The analysis of upper and lower bounds of the
tensor degree allows us to find interesting structural restrictions for the
whole group.

###### Key words and phrases:

Tensor degree, commutativity degree, exterior degree, Schur multiplier,
dihedral groups, $p$–groups.

###### 2010 Mathematics Subject Classification:

Primary: 20J99, 20D15; Secondary: 20D60; 20C25.

##  1\. Commutativity, exterior and tensor degrees

In the present paper all the groups are supposed to be finite. Brown and
others [4] wrote an influential contribution on a generalization of the usual
abelian tensor product of abelian groups. Following their terminology, the
nonabelian tensor product $G\otimes H$ of two groups $G$ and $H$ is the group
generated by the symbols $g\otimes h$ with defining relations $xy\otimes
h=(y^{x}\otimes h^{x})(x\otimes h)$ and $x\otimes hk=(x\otimes h)(x^{h}\otimes
k^{h})$ for all $x,y\in G$ and $h,k\in H$ (as usual, $y^{x}=x^{-1}yx$). In
case $G=H$ and all actions are by conjugation, $G\otimes G$ is called the
nonabelian tensor square of $G$. [4, Propositions 1,2,3] describe the main
calculus rules in $G\otimes H$, which are:

(1.1) $(x^{-1}\otimes h)^{x}=(x\otimes h)^{-1}=(x\otimes h^{-1})^{h};\ \ \ \
(y\otimes k)^{xh}(x\otimes h)=(x\otimes h)(y\otimes k)^{hx};$
$y\otimes(h^{x}h^{-1})=(x\otimes h)^{y}(x\otimes h)^{-1};\ \ \ \
(x(x^{-1})^{h})\otimes y=(x\otimes h)((x\otimes h)^{-1})^{k};$ $(y\otimes
k)~{}^{(x\otimes h)}=(y\otimes k)^{[x,h]};\ \ \ \ [x\otimes h,y\otimes
k]=(x(x^{-1})^{h})(k^{y}k^{-1}).$

They allow us to conclude that

$\kappa:x\otimes y\in G\otimes G\mapsto\kappa(x\otimes y)=[x,y]\in G^{\prime}$

is an epimorphism of groups such that $\ker\kappa=J_{2}(G)$ is a central
subgroup of $G\otimes G$. Furthermore, $\nabla(G)=\langle x\otimes x\ |\ x\in
G\rangle\subseteq J_{2}(G)$ and $J_{2}(G)$ is important from the point of view
of the algebraic topology (see [4]). Defining the nonabelian exterior square
$G\wedge G=(G\otimes G)/\nabla(G)=\langle(x\otimes y)\nabla(G)\ |\ x,y\in
G\rangle=\langle x\wedge y\ |\ x,y\in G\rangle$ of $G$, we can see easily that

$\kappa^{\prime}:x\wedge y\in G\wedge G\mapsto\kappa^{\prime}(x\wedge
y)=[x,y]\in G^{\prime}$

is an epimorphism of groups such that
$\ker\kappa^{\prime}=M(G)=H_{2}(G,\mathbb{Z})$ is the Schur multiplier of $G$,
that is, the second integral homology group of $G$. The following diagram
involves the third integral homology group of $G$ and the Whitehead functor
$\Gamma$. It has exact rows and central extensions as columns (see [4] for
details).

(1.2) $\setcounter{MaxMatrixCols}{11}\begin{CD}00\\\ @V{}V{}V@V{}V{}V\\\
H_{3}(G)@>{}>{}>\Gamma(G/G^{\prime})@>{}>{}>J_{2}(G)@>{}>{}>H_{2}(G)@>{}>{}>0\\\
\Big{\|}\Big{\|}@V{}V{}V@V{}V{}V\\\
H_{3}(G)@>{}>{}>\Gamma(G/G^{\prime})@>{}>{}>G\otimes G@>{}>{}>G\wedge
G@>{}>{}>1\\\ @V{\kappa}V{}V@V{\kappa^{\prime}}V{}V\\\
G^{\prime}@=G^{\prime}\\\ @V{}V{}V@V{}V{}V\\\ 11\\\ \end{CD}$

The exterior centralizer of $x\in G$ is the set

$C_{G}^{\wedge}(x)=\\{a\in G\ |\ a\wedge x=1_{{}_{G\wedge G}}\\},$

which turns out to be a subgroup of $G$ (see [12]), and the exterior center of
$G$ is the set

$Z^{\wedge}(G)=\\{g\in G\ |\ 1_{{}_{G\wedge G}}=g\wedge y\in G\wedge G,\forall
y\in G\\}={\underset{x\in G}{\bigcap}}C_{G}^{\wedge}(x)$

which is a subgroup of the center $Z(G)$ of $G$ (see [12]). We mention that
the interest in studying $C_{G}^{\wedge}(x)$ and $Z^{\wedge}(G)$ is due to the
fact that they allow us to decide whether $G$ is a capable group or not, that
is, whether $G$ is isomorphic to $E/Z(E)$ for some group $E$ or not. In
analogy, we may consider the tensor centralizer

$C_{G}^{\otimes}(x)=\\{a\in G\ |\ a\otimes x=1_{{}_{G\otimes G}}\\},$

of $x\in G$ and it turns out to be a subgroup of $G$ (see [3, Theorem 3.1]),
and the tensor center

$Z^{\otimes}(G)=\\{g\in G\ |\ 1_{{}_{G\otimes G}}=g\otimes y\in G\otimes
G,\forall y\in G\\}={\underset{x\in G}{\bigcap}}C_{G}^{\otimes}(x),$

which is a subgroup of $Z(G)$ (see [3, Corollary 3.3]).

A combinatorial approach, in order to measure how far we are from
$Z^{\wedge}(G)$, has been investigated in [13], introducing the exterior
degree of $G$

$d^{\wedge}(G)=\frac{|\\{(x,y)\in G\times G\ |\ x\wedge y=1_{{}_{G\wedge
G}}\\}|}{|G|^{2}}=\frac{1}{|G|}\sum^{k(G)}_{i=1}\frac{|C^{\wedge}_{G}(x_{i})|}{|C_{G}(x_{i})|},$

where $k(G)$ is the number of conjugacy classes of $G$ and the second equality
is [13, Lemma 2.2]. It is in fact clear that $d^{\wedge}(G)=1$ if and only if
$G=Z^{\wedge}(G)$ so that the exterior degree represents the probability that
two random chosen elements commute with respect to the operator $\wedge$. On
the other hand, $d^{\wedge}(G)$ has been connected with the commutativity
degree

$d(G)=\frac{|\\{(x,y)\in G\times G\ |\
[x,y]=1\\}|}{|G|^{2}}=\frac{1}{|G|^{2}}\sum^{k(G)}_{i=1}|C_{G}(x_{i})|$

of $G$, studied by Gustafson and others [6, 7, 9, 10, 11, 16]. The reader may
refer also to [8, 14] for recent studies on the topic. Of course, $d(G)=1$ if
and only if $G$ is abelian. On the other hand, among groups with exterior
degree equal to 1, we find all cyclic groups, but not necessarily all abelian
groups (see [1, 3, 13]). Then, roughly speaking, $d^{\wedge}(G)$ gives a
measure of how far is $G$ from being cyclic. It is interesting to note that

$d^{\wedge}(G)\leq d(G),$

by [13, Theorem 2.3]. Here we introduce the tensor degree

$d^{\otimes}(G)=\frac{|\\{(x,y)\in G\times G\ |\ x\otimes y=1_{{}_{G\otimes
G}}\\}|}{|G|^{2}},$

evaluating the distance of $G$ from being equal to $Z^{\otimes}(G)$, since
$d^{\otimes}(G)=1$ if and only if $Z^{\otimes}(G)=G$. On the other hand, one
may easily check that $d^{\otimes}(G)=1$ if and only if $G$ is trivial, by
using [5, Proposition 18].

We note that there are several examples of abelian groups with exterior degree
different from 1 and here we show, not only examples of abelian groups with
tensor degree different from 1, but also examples of abelian groups with
different values of tensor and exterior degrees. In fact, the study of
$d^{\otimes}(G)$ will present a new perspective of investigation of specific
subclasses of nilpotent groups. We will prove restrictions on $d^{\otimes}(G)$
of numerical nature, which will influence the structure of $G$, and relations
among $d^{\otimes}(G)$, $d^{\wedge}(G)$ and $d(G)$.

##  2\. Fundamental inequalities for the tensor degree

In a group $G$, $C^{\wedge}_{G}(x)$ and $C^{\otimes}_{G}(x)$ are normal
subgroups of $C_{G}(x)$ (see [3, 12]).

######  Lemma 2.1.

Let $x$ be an element of a group $G$. Then

$|C_{G}(x):C^{\otimes}_{G}(x)|\leq|J_{2}(G)|.$

Furthermore, $|C_{G}(x):C^{\otimes}_{G}(x)|\leq|M(G)|\ |\nabla(G)|.$

###### Proof.

The map

$\varphi:yC^{\otimes}_{G}(x)\in C_{G}(x)/C^{\otimes}_{G}(x)\longmapsto
y\otimes x\in J_{2}(G)$

satisfies the condition $\varphi(abC^{\otimes}_{G}(x))=ab\otimes x=(a\otimes
x)^{b}\ (b\otimes x)=(a\otimes x)\ (b\otimes x)=\varphi(aC^{\otimes}_{G}(x))\
\varphi(bC^{\otimes}_{G}(x))$ for all $a,b\in C_{G}(x)$, where (1.1) have been
applied. Furthermore, $\ker\varphi=\\{yC^{\otimes}_{G}(x)\ |\ y\otimes
x=1_{G\otimes G}\\}=C^{\otimes}_{G}(x)$. Then $\varphi$ is a monomorphism and
$|C_{G}(x):C^{\otimes}_{G}(x)|\leq|J_{2}(G)|$. On the other hand, we know from
[4] that the map $\pi:a\otimes b\in J_{2}(G)\mapsto a\wedge b\in M(G)$ is an
epimorphism of groups such that $J_{2}(G)/\ker\pi=J_{2}(G)/\nabla(G)\simeq
M(G)$. Thus $|C_{G}(x):C^{\otimes}_{G}(x)|\leq|M(G)|\ |\nabla(G)|$. ∎

The above bound shows that the section $C_{G}(x)/C^{\otimes}_{G}(x)$ depends
on the size of $J_{2}(G)$, or, equivalently, from that of $\nabla(G)$ and
$M(G)$. This agrees with the homological sequences of (1.2). The following
lemma deals with different aspects and correlates the tensor centralizers with
the tensor degree.

######  Lemma 2.2.

Let $x_{1},\ldots,x_{k(G)}$ be a system of representatives for the conjugacy
classes of the group $G$. Then

$d^{\otimes}(G)=\frac{1}{|G|}\sum^{k(G)}_{i=1}\frac{|C^{\otimes}_{G}(x_{i})|}{|C_{G}(x_{i})|}.$

###### Proof.

The idea of [13, Proof of Lemma 2.2] may be adapted here. Let
$C_{1},\ldots,C_{k(G)}$ be the conjugacy classes of $G$ and $x_{i}\in C_{i}$
for $i=1,2,\ldots,k(G)$. For every $y\in C_{i}$ there exists a $g\in G$ such
that $y=x^{g}_{i}$. This implies
$|C^{\otimes}_{G}(y)|=|C^{\otimes}_{G}(x_{i})|$, hence

$|G|^{2}d^{\otimes}(G)=\sum_{x\in
G}|C^{\otimes}_{G}(x)|=\sum^{k(G)}_{i=1}\sum_{x\in
C_{i}}|C^{\otimes}_{G}(x_{i})|=\sum^{k(G)}_{i=1}|G:C_{G}(x_{i})||C^{\otimes}_{G}(x_{i})|$
$=|G|\sum^{k(G)}_{i=1}\left|\frac{C^{\otimes}_{G}(x_{i})}{C_{G}(x_{i})}\right|.$

∎

We may compare commutativity and tensor degrees in the following way.

######  Theorem 2.3.

Let $G$ be a group and $p$ be the smallest prime divisor of $|G|$. Then

$\frac{d(G)}{|J_{2}(G)|}+\frac{|Z^{\otimes}(G)|}{|G|}\left(1-\frac{1}{|J_{2}(G)|}\right)\leq
d^{\otimes}(G)\leq
d(G)-\left(1-\frac{1}{p}\right)\left(\frac{|Z(G)|-|Z^{\otimes}(G)|}{|G|}\right).$

###### Proof.

The idea of [13, Proof of Theorem 2.3] may be applied, thanks to the
preliminaries which we have done. Let $x\not\in Z^{\otimes}(G)$. From Lemma
2.1, $|C^{\otimes}_{G}(x)|/|C_{G}(x)|\geq 1/|J_{2}(G)|$. From Lemma 2.2 and
the equality $d(G)=\frac{k(G)}{|G|}$, we deduce

$d^{\otimes}(G)=\frac{1}{|G|}\sum^{k(G)}_{i=1}\left|\frac{C^{\otimes}_{G}(x_{i})}{C_{G}(x_{i})}\right|\geq\frac{1}{|G|}\left(|Z^{\otimes}(G)|+\frac{k(G)-|Z^{\otimes}(G)|}{|J_{2}(G)|}\right)$
$=\frac{k(G)}{|G|\
|J_{2}(G)|}+\frac{|Z^{\otimes}(G)|}{|G|}\left(1-\frac{1}{|J_{2}(G)|}\right)=\frac{d(G)}{|J_{2}(G)|}+\frac{|Z^{\otimes}(G)|}{|G|}\left(1-\frac{1}{|J_{2}(G)|}\right).$

Conversely, $|G:C^{\otimes}_{G}(x)|\geq p$ implies that

$d^{\otimes}(G)=\frac{1}{|G|}\sum^{k(G)}_{i=1}\left|\frac{C^{\otimes}_{G}(x_{i})}{C_{G}(x_{i})}\right|\leq\frac{|Z^{\otimes}(G)|}{|G|}+\frac{1}{p}\left(\frac{|Z(G)|-|Z^{\otimes}(G)|}{|G|}\right)+\frac{k(G)-|Z(G)|}{|G|}$
$=d(G)-\frac{p-1}{p}\left(\frac{|Z(G)|-|Z^{\otimes}(G)|}{|G|}\right).$

∎

Theorem 2.3 has analogies with [13, Theorem 2.3]. On the other hand, the
literature on $M(G)$ is richer than the literature on $J_{2}(G)$ and it may be
useful to rewrite the lower bound of Theorem 2.3 in the following way.

######  Corollary 2.4.

Let $G$ be a group. Then

$\frac{d(G)}{|M(G)||\nabla(G)|}+\frac{|Z^{\otimes}(G)|}{|G|}\left(1-\frac{1}{|M(G)||\nabla(G)|}\right)\leq
d^{\otimes}(G).$

From [5, Proposition 18], an abelian group $G$ has trivial $Z^{\otimes}(G)$.
This means that the term
$\frac{|Z^{\otimes}(G)|}{|G|}\left(1-\frac{1}{|J_{2}(G)|}\right)$ vanishes in
the lower bound of Theorem 2.3; $J_{2}(G)=G\otimes G$; $d(G)=1$; the term
$\frac{|Z(G)|-|Z^{\otimes}(G)|}{|G|}$ becomes equal to $\frac{|Z(G)|-1}{|G|}$
. Hence we have another interesting consequence of Theorem 2.3.

######  Corollary 2.5.

An abelian group $G$ satisfies $\frac{1}{|G|}+\frac{|G|-1}{|G||G\otimes
G|}\leq d^{\otimes}(G)\leq\frac{1}{p}+\frac{p-1}{p|G|}$. Moreover, if
$|G|\rightarrow\infty$, then $0\leq d^{\otimes}(G)\leq\frac{1}{p}$.

The well–known notion of Schur cover, which can be found in [2, 4], gives a
condition of equality among the tensor degree and the commutativity degree.

######  Proposition 2.6.

Let $G$ be a perfect group and $G^{*}$ be a Schur cover of $G$. Then
$d^{\otimes}(G)=d(G^{*})$.

###### Proof.

By the definition of Schur cover, there is an exact sequence

$1\rightarrow A\rightarrow
G^{*}\stackrel{{\scriptstyle\pi}}{{\rightarrow}}G\rightarrow 1,$

in which $A\cong M(G)$. From [4, Proposition 7], we have a homomorphism
$\xi:G\otimes G\rightarrow G^{*}$ given by $\xi(g\otimes h)=[g_{1},h_{1}]$
such that $\pi(g_{1})=g,\pi(h_{1})=h$. Since $G\otimes G\simeq G^{*}$ by using
[4, Corollary 1], we have

$|\\{(g,h)\in G\times G~{}|~{}g\otimes h=1_{G\otimes
G}\\}||M(G)|=|\\{(g_{1},h_{1})\in G^{*}\times
G^{*}~{}|~{}[g_{1},h_{1}]=1\\}|,$

as required. ∎

Another fundamental relation among $d(G)$, $d^{\wedge}(G)$ and
$d^{\otimes}(G)$ is listed below.

######  Theorem 2.7.

Let $G$ be a group. Then $d^{\otimes}(G)\leq d^{\wedge}(G)\leq d(G)$.

###### Proof.

$d^{\wedge}(G)\leq d(G)$ follows from [13, Theorem 2.3]. Let $x\not\in
Z^{\otimes}(G)$. We have $g\in C^{\wedge}_{G}(x)$ if and only if $g\wedge
x=1_{G\wedge G}$ if and only if $(g\otimes x)\nabla(G)=\nabla(G)$ if and only
if $g\otimes x\in\nabla(G)$. This condition is weaker than the condition
$g\otimes x=1_{G\otimes G}$, characterizing the elements of
$C^{\otimes}_{G}(x)$. Then $C^{\otimes}_{G}(x)\subseteq
C^{\wedge}_{G}(x)\subseteq C_{G}(x)$. From this fact, Lemma 2.2 and [13, Lemma
2.2], we conclude

$d^{\otimes}(G)=\frac{1}{|G|}\sum^{k(G)}_{i=1}\left|\frac{C^{\otimes}_{G}(x_{i})}{C_{G}(x_{i})}\right|\leq\frac{1}{|G|}\sum^{k(G)}_{i=1}\left|\frac{C^{\wedge}_{G}(x_{i})}{C_{G}(x_{i})}\right|=d^{\wedge}(G).$

∎

From [13, Theorem 2.3] and Theorem 2.7, $G$ is unidegree, or right unidegree,
if the upper bound $d^{\wedge}(G)=d(G)$ is achieved. A group $G$ is called
unicentral if $Z^{\wedge}(G)=Z(G)$. Unicentral groups have been studied in [2]
and [13, Corollary 2.5] shows that right unidegree groups are unicentral. Now
we say that $G$ is left unidegree if the lower bound $d^{\otimes}(G)=d(G)$ is
achieved. Finally, $G$ is said to be left and right unidegree if it is both
left and right unidegree. By the concept of left unidegree introduced here,
this condition and Theorem 2.7 imply that of right unidegree. A complete
classification of groups, whose center is equal to its tensor center, is not
available, to the best of our knowledge, against the classification of
unicentral groups in [2].

######  Corollary 2.8.

If $G$ is left unidegree, then $Z(G)=Z^{\otimes}(G)$. Furthermore, if $G$ is
left and right unidegree, then $G$ is unicentral and $Z(G)=Z^{\otimes}(G)$.

###### Proof.

If $d^{\otimes}(G)=d(G)$, then $Z(G)=Z^{\otimes}(G)$ by the upper bound in
Theorem 2.3. Furthemore, if $d^{\wedge}(G)=d(G)$, then we apply [13, Corollary
2.5] and the result follows. ∎

##  3\. Sharpening upper and lower bounds for the tensor degree

The present section is devoted to improve the numerical restrictions on the
tensor degree and to find relations among quotients and subgroups. [13,
Proposition 2.6] has analogies in the context of the tensor degree.

######  Proposition 3.1.

If $N$ is a normal subgroup of a group $G$, then $d^{\otimes}(G)\leq
d^{\otimes}(G/N).$ The equality holds, if $N\subseteq Z^{\otimes}(G)$.

###### Proof.

$|G|^{2}\ d^{\otimes}(G)=\sum_{x\in G}|C^{\otimes}_{G}(x)|=\sum_{xN\in
G/N}\sum_{n\in N}|C^{\otimes}_{G}(xn)|$ $=\sum_{xN\in G/N}\sum_{n\in
N}\frac{|C^{\otimes}_{G}(xn)N|}{|N|}\ |C^{\otimes}_{G}(xn)\cap
N|\leq\sum_{xN\in G/N}\sum_{n\in N}|C^{\otimes}_{G/N}(xN)|\
|C^{\otimes}_{G}(xn)\cap N|$ $=\sum_{xN\in G/N}|C^{\otimes}_{G/N}(xN)|\
\sum_{n\in N}|C^{\otimes}_{G}(xn)\cap N|$ $\leq|N|^{2}\sum_{xN\in
G/N}|C^{\otimes}_{G/N}(xN)|=|G|^{2}\ d^{\otimes}(G/N).$

For each central subgroup $N$ of $G$, [4, Proposition 9] ensures the exactness
of the sequence $(G\otimes N)\times(N\otimes
G){\overset{\alpha}{\longrightarrow}}G\otimes
G{\overset{\beta}{\longrightarrow}}(G/N)\otimes(G/N)\longrightarrow 1$ for
suitable homomorphisms $\alpha$ and $\beta$. This is our case. Furthermore, if
$N\subseteq Z^{\otimes}(G)$, then $\mathrm{Im}\ \alpha=1_{{}_{G\otimes G}}$
and $G/N\otimes G/N\simeq G\otimes G$ so that
$d^{\otimes}(G)=d^{\otimes}(G/N)$. ∎

The next corollary explains better the conditions of equality in Proposition
3.1.

######  Corollary 3.2.

Let $N$ be a normal subgroup of a group $G$. Then the following statements are
equivalent:

  * (i)

$N\subseteq Z^{\otimes}(G);$

  * (ii)

$d^{\otimes}(G)=d^{\otimes}(G/N);$

  * (iii)

$G\otimes G\cong G/N\otimes G/N.$

###### Proof.

(i) implies (ii) by Proposition 3.1. Conversely,
$d^{\otimes}(G)=d^{\otimes}(G/N)$ implies that $N\subseteq C^{\otimes}_{G}(x)$
for all $x\in G$, then $N\subseteq Z^{\otimes}(G).$ Finally, the equivalency
of (ii) and (iii) are obtained directly from [5, Proposition 16]. ∎

The next result is going to improve the upper bound in Theorem 2.3.

######  Proposition 3.3.

Let $G$ be a group, $p$ the smallest prime divisor of $|G|$ and $x\not\in
Z(G)$ such that $C^{\otimes}_{G}(x)\not=C_{G}(x)$. Then

$d^{\otimes}(G)\leq
d(G)-\left(1-\frac{1}{p}\right)\left(\frac{|Z(G)|-|Z^{\otimes}(G)|-1}{|G|}\right).$

In particular, if $Z^{\otimes}(G)=1$, then $d^{\otimes}(G)\leq
d(G)-\left(1-\frac{1}{p}\right)\frac{|Z(G)|}{|G|}.$

###### Proof.

We may apply the same argument, which has been used to prove the upper bound
of Theorem 2.3. The first inequality may be specified as

$d^{\otimes}(G)\leq\frac{|Z^{\otimes}(G)|}{|G|}+\frac{1}{p}\left(\frac{|Z(G)|-|Z^{\otimes}(G)|}{|G|}\right)+\frac{1}{p|G|}+\frac{k(G)-|Z(G)|-1}{|G|}$
$=\frac{1}{|G|}\left(\left(1-\frac{1}{p}\right)|Z^{\otimes}(G)|+\left(\frac{1}{p}-1\right)|Z(G)|+\frac{1}{p}+k(G)-1\right)$
$=d(G)-\left(1-\frac{1}{p}\right)\left(\frac{|Z(G)|-|Z^{\otimes}(G)|-1}{|G|}\right).$

The rest is clear and the result follows. ∎

The extremal case of $Z^{\otimes}(G)=1$ is described by the next result and
has analogies with [13, Theorem 2.8]. There are also analogies for the
commutativity degree in [6, 9, 10, 11, 16].

######  Theorem 3.4.

Let $G$ be a nonabelian group with $Z^{\otimes}(G)=1$ and $p$ be the smallest
prime dividing $|G|$. Then $d^{\otimes}(G)\leq\frac{1}{p}.$

###### Proof.

First we claim that, if $Z(G)\cap G^{\prime}\not=1$, then there exists an
$x\not\in Z(G)$ such that $C^{\otimes}_{G}(x)\not=C_{G}(x)$. Assume that
$C^{\otimes}_{G}(x)=C_{G}(x)$ for all $x\not\in Z(G)$. Since
$Z(G)\subseteq{\underset{x\not\in Z(G)}{\bigcap}}C^{\otimes}_{G}(x)$ and
$G^{\prime}\subseteq{\underset{x\in Z(G)}{\bigcap}}C_{G}^{\otimes}(x)$, we
have $G^{\prime}\cap Z(G)\subseteq Z^{\otimes}(G)$. This implies $Z(G)\cap
G^{\prime}=1$, which is a contradiction. The first claim follows.

The second claim is that
$d(G)\leq\frac{1}{p}+\left(1-\frac{1}{p}\right)\frac{|Z(G)|}{|G|}$. Arguing as
in Theorem 2.3, we find

$d(G)=\sum_{x\in
G}\frac{|C_{G}(x)|}{|G|}\leq\frac{1}{p}\left(\frac{|Z(G)|}{|G|}+\frac{|G|-|Z(G)|}{|G|}\right)=\frac{1}{p}+\left(1-\frac{1}{p}\right)\frac{|Z(G)|}{|G|}.$

The third claim is that $d(G)\leq\frac{1}{p}$, provided $G$ is nonabelian with
$Z(G)\cap G^{\prime}=1$. This can be found in [13, Proposition 2.7].

Now we may proceed to prove the result and there is no loss of generality to
assume $Z(G)\cap G^{\prime}\not=1$, by the third claim and Theorem 2.7. Now
the first and the second claims, combined with Proposition 3.3, imply

$d^{\otimes}(G)\leq
d(G)-\left(1-\frac{1}{p}\right)\frac{|Z(G)|}{|G|}\leq\frac{1}{p}+\left(1-\frac{1}{p}\right)\frac{|Z(G)|}{|G|}-\left(1-\frac{1}{p}\right)\frac{|Z(G)|}{|G|}=\frac{1}{p}.$

∎

##  4\. Computations for elementary abelian and extraspecial $p$–groups

The present section is devoted to compute the tensor degree for some classes
of $p$–groups, widely studied in literature. We follow the phylosophy of [15],
where the case of the exterior degree has been studied. As usual, $C_{p^{n}}$
denotes the cyclic group of order $p^{n}$. The tensor degree of elementary
abelian $p$-groups is described by the next result.

######  Proposition 4.1.

If $G\simeq C_{p}\times\ldots\times C_{p}=C^{(n)}_{p}$ is the direct product
of $n\geq 1$ copies of $C_{p}$, then $d^{\otimes}(G)=\frac{2p^{n}-1}{p^{2n}}$.

###### Proof.

We claim that $C^{\otimes}_{G}(x)=1$ for all nontrivial $x\in G$. In our case,
$G\otimes G$ is the usual abelian tensor product of two abelian groups. Then

$G\otimes G\simeq C^{(n)}_{p}\otimes C^{(n)}_{p}\simeq C^{(n^{2})}_{p}.$

If we write $G=\langle a_{1}\rangle\times\ldots\langle a_{n}\rangle$, then
$G\otimes G=\prod^{n}_{i=1}\prod^{n}_{j=1}\langle a_{i}\otimes a_{j}\rangle$
so that $a_{i}\otimes a_{j}\not=1$ for all $i\not=j$. Since $a_{i}\otimes
a_{i}\not=1$ for all $i=1,\ldots,n$, we have $C^{\otimes}_{G}(a_{i})=1$ for
all $i=1,\ldots,n$. Then

$d^{\otimes}(G)=\frac{1}{|G|^{2}}\sum_{x\in
G}|C^{\otimes}_{G}(x)|=\frac{1}{p^{2n}}\ \left(p^{n}+(p^{n}-1)\cdot
1\right)=\frac{2p^{n}-1}{p^{2n}}.$

∎

The following $p$–groups have big sections which are elementary abelian and
have interest in several areas of group theory. Let

$E_{1}=\langle a,b,c\ |\ a^{p}=b^{p}=c^{p}=1,[a,c]=[b,c]=1,[a,b]=c\rangle$

be the extra–special $p$–group of order $p^{3}$ and of exponent $p$,

$D_{2^{n}}=C_{2^{n-1}}\rtimes C_{2}=\langle a,b\ |\
b^{2}=a^{2^{n-1}}=1,b^{-1}ab=a^{-1}\rangle$

be the dihedral 2–group of order $2^{n}$ and

$Q_{2^{n}}=\langle a,b\ |\ b^{2}=a^{2^{n-2}},b^{-1}ab=a^{-1}\rangle$

be the generalized quaternion 2–group of order $2^{n}$. We note that
$Q_{2^{n}}/Z(Q_{2^{n}})\simeq D_{2^{n}}/Z(D_{2^{n}})\simeq C^{(n-1)}_{2}$ for
all $n\geq 1$.

######  Lemma 4.2.

Let $H$ be an extraspecial $p$–group. Then $Z^{\otimes}(H)=Z(H)=H^{\prime}$,
when $H\not\simeq E_{1},Q_{8},D_{8}$. Furthermore,
$Z^{\otimes}(E_{1})=Z^{\otimes}(D_{8})=Z^{\otimes}(Q_{8})=1$.

###### Proof.

From GAP [17], we have that
$Z^{\otimes}(E_{1})=Z^{\otimes}(D_{8})=Z^{\otimes}(Q_{8})=1$. On the other
hand, $H$ is extraspecial, then $H^{\prime}=Z(H)\simeq C_{p}$. Necessarily, we
should have $Z^{\otimes}(H)=H^{\prime}=Z(H)$, when $H\not\simeq
E_{1},Q_{8},D_{8}$. ∎

We are going to calculate the tensor degree for the groups in Lemma 4.2.

######  Theorem 4.3.

Let $p$ be a prime and $m\geq 1$.

  * (i)

$d^{\otimes}(Q_{2^{n}})=\frac{2^{n-3}+2^{n-4}+1}{2^{n}}$ for all $n>3$ and
$d^{\otimes}(Q_{8})=\frac{1}{4}$ for $n=3$.

  * (ii)

$d^{\otimes}(D_{2^{n}})=\frac{2^{n-3}+2^{n-4}+1}{2^{n}}$ for all $n>3$ and
$d^{\otimes}(D_{8})=\frac{5}{16}$ for $n=3$.

  * (iii)

$d^{\otimes}(E_{1})=\frac{2p^{2}+p-2}{p^{5}}$.

  * (iv)

Let $H$ be an extra–special $p$–group of $|H|=p^{2m+1}$ and not isomorphic
with $E_{1},Q_{8},D_{8}$. Then $d^{\otimes}(H)=\frac{2p^{m}-1}{p^{4m}}.$

###### Proof.

(i). We need to have in mind the presentation of $Q_{2^{n}}$ and some
computations in [4, Section 4]. Let $a$ and $b$ be two generators of
$Q_{2^{n}}$ and $1\not=a\otimes b\in Q_{2^{n}}\otimes Q_{2^{n}}$. For all
$l\geq 1$, [4, Equations 4.3 and 4.5] imply

$b\otimes a^{l}=(b\otimes a)^{l}\ (a\otimes a)^{l(l-1)},\ \ \ \ a^{l}\otimes
b=(a\otimes b)^{l}\ (a\otimes a)^{l(l-1)}$

so that the condition

$b\otimes a^{l}=(b\otimes a)^{l}\ (a\otimes a)^{l(l-1)}=(a\otimes b)^{l}\
(a\otimes a)^{l(l-1)}=a^{l}\otimes b$

is satisfied if and only if

$(b\otimes a)^{l}=(a\otimes b)^{l}$

for all $l\geq 1$. This cannot happen, because [4, Equation at p.190, line
+10] implies $(a\otimes b)^{l}=1$ for all $l\geq 1$, and, in particular,
$a\otimes b=1$, which is a contradiction. We conclude that $b\not\in
C^{\otimes}_{Q_{2^{n}}}(a^{l})$ for any choice of $l\geq 1$. In a similar way,
[4, Equations 4.7 and 4.8] imply

$ba^{k}\otimes a^{l}=(b\otimes a)^{l}\ (a\otimes a)^{lk+l(l-1)},\ \ \ \
a^{l}\otimes ba^{k}=(a\otimes b)^{l}\ (a\otimes a)^{lk+l(k-1)}$

for all $l,k\geq 1$ and we find again $ba^{k}\otimes a^{l}=a^{l}\otimes
ba^{k}$ if and only if $(b\otimes a)^{l}=(a\otimes b)^{l}$, which is a
contradiction. Then $ba^{k}\not\in C^{\otimes}_{Q_{2^{n}}}(a^{l})$ for all
$l,k\geq 1$. We conclude that $C^{\otimes}_{Q_{2^{n}}}(a^{l})$ is formed only
by powers of $a$. Now an explicit computation (or looking at [4, Equation
4.5]) shows that only even powers of $a$ commute with $a^{2k+1}$ with respect
to $\otimes$. Then $C^{\otimes}_{Q_{2^{n}}}(a^{2k+1})=\langle a^{2}\rangle$
and $C^{\otimes}_{Q_{2^{n}}}(a^{2k})=\langle a\rangle$. Furthermore, for all
$i=0,1,2,\ldots,2^{n-1}$ we may also deduce
$C^{\otimes}_{Q_{2^{n}}}(a^{i}b)=1$, after all we have said. Then

$d^{\otimes}(Q_{2^{n}})=\frac{1}{|Q_{2^{n}}|^{2}}\sum_{x\in
Q_{2^{n}}}|C^{\otimes}_{Q_{2^{n}}}(x)|$ $=\frac{1}{2^{n}}\
(2^{n}+2^{n-2}|\langle a^{2}\rangle|+(2^{n-2}-1)|\langle
a\rangle|+(2^{n-1}-1)\cdot 1+1)$ $=\frac{1}{2^{n}}\
\left(\frac{2^{n}+2^{n-2}\cdot 2^{n-2}+(2^{n-2}-1)\cdot
2^{n-1}+2^{n-1}}{2^{n}}\right)$
$=\frac{4+2^{n-2}+2(2^{n-2}-1)+2}{2^{n+2}}=\frac{2^{n-3}+2^{n-4}+1}{2^{n}}.$

The case $n=3$, that is, $Q_{8}$ can be solved by using GAP [17] and shows
$d^{\otimes}(Q_{8})=\frac{1}{4}$.

(ii). Exactly the same argument of (i) may be applied for $D_{2^{n}}$ when
$n>3$ and we find $d^{\otimes}(D_{2^{n}})=d^{\otimes}(Q_{2^{n}})$ (see [4,
Section 4]). Similarly, GAP [17] allows us to conclude
$d^{\otimes}(D_{8})=\frac{5}{16}$ in case $n=3$.

(iii). From GAP [17], or by a direct computation, one can see that
$E_{1}\otimes E_{1}\simeq C^{(6)}_{p}$. Thus [1, Proposition 3.5] implies

$E_{1}\otimes E_{1}=\langle a\otimes a\rangle\times\langle a\otimes
b\rangle\times\langle a\otimes c\rangle\times\langle b\otimes
c\rangle\times\langle b\otimes a\rangle\times\langle b\otimes b\rangle.$

Applying [1, Proposition 3.5] again, we get for all $1\leq i\leq p$ that
$C^{\otimes}_{E_{1}}(c^{i})=c^{i}$ and for all $x\in
E_{1}-\\{1,c,c^{2},\ldots,c^{p-1}\\}$ that $C^{\otimes}_{E_{1}}(x)=1$. Then

$d^{\otimes}(E_{1})=\frac{1}{|E_{1}|^{2}}\sum_{x\in
E_{1}}|C^{\otimes}_{E_{1}}(x)|$
$=\frac{1}{p^{6}}\left(p^{3}+{\underbrace{|C^{\otimes}_{E_{1}}(c)|+|C^{\otimes}_{E_{1}}(c^{2})|+\ldots+|C^{\otimes}_{E_{1}}(c^{p-1})|}_{(p-1)-\mathrm{times}}}+(p^{3}-p)\right)$
$=\frac{1}{p^{6}}(p^{3}+(p-1)p+p^{3}-p)=\frac{2p^{2}+p-2}{p^{5}}.$

(iv). From Proposition 3.1, we note that the tensor degree of a group is the
same if we factorize through its tensor center. This and Lemma 4.2 imply
$d^{\otimes}(H)=d^{\otimes}(H/Z^{\otimes}(H))=d^{\otimes}(H/H^{\prime})$.
Since $H/H^{\prime}\simeq C^{(2m)}_{p}$,

$d^{\otimes}(H)=d^{\otimes}(H/H^{\prime})=\frac{2p^{m}-1}{p^{4m}}.$

∎

It is instructive to compare the results of the present section with some of
[6, 9, 10, 11, 13, 15]. We will confirm not only Theorems 2.3, 2.7 and 4.3,
but will verify that exterior degree, tensor degree and commutativity degree
are different group invariants. To conveniece of the reader, we have appended
a list below.

######  Corollary 4.4.

The following inequalities are true:

  * (i)

$d^{\otimes}(E_{1})=\frac{2p^{2}+p-2}{p^{5}}$;
$d^{\wedge}(E_{1})=\frac{p^{3}+p^{2}-1}{p^{5}}$;
$d(E_{1})=\frac{p^{2}+p-1}{p^{3}}$. In particular,
$d^{\otimes}(E_{1})<d^{\wedge}(E_{1})<d(E_{1})$ are proper for all primes
$p\geq 2$.

  * (ii)

$d^{\otimes}(Q_{2^{n}})=\frac{2^{n-3}+2^{n-4}+1}{2^{n}}$ and
$\frac{2^{n-2}+3}{2^{n}}=d^{\wedge}(Q_{2^{n}})=d(Q_{2^{n}})$ for all $n>3$.
Moreover, $d^{\otimes}(Q_{8})=\frac{1}{4}$ and
$\frac{5}{8}=d^{\wedge}(Q_{8})=d(Q_{8})$ in case $n=3$. In particular,
$d^{\otimes}(Q_{2^{n}})<d^{\wedge}(Q_{2^{n}})=d(Q_{2^{n}})$ for all $n\geq 3$.

  * (iii)

$d^{\otimes}(D_{2^{n}})=\frac{2^{n-3}+2^{n-4}+1}{2^{n}}$ and
$d^{\wedge}(D_{2^{n}})=d(D_{2^{n}})=\frac{2^{n-2}+3}{2^{n}}$ for all $n>3$.
Moreover, $d^{\otimes}(D_{8})=\frac{5}{16}$ and
$d^{\wedge}(D_{8})=d(D_{8})=\frac{5}{8}$ in case $n=3$. In particular,
$d^{\otimes}(D_{2^{n}})<d^{\wedge}(D_{2^{n}})=d(D_{2^{n}})$ for all $n\geq 3$.

  * (iv)

$d^{\otimes}(C^{(n)}_{p})=\frac{2p^{n}-1}{p^{2n}}$;
$d^{\wedge}(C^{(n)}_{p})=\frac{p^{n}+p^{n-1}-1}{p^{2n-1}}$;
$d(C^{(n)}_{p})=1$. In particular,
$d^{\otimes}(C^{(n)}_{p})<d^{\wedge}(C^{(n)}_{p})<d(C^{(n)}_{p})$ are proper
for all $n\geq 1$ and for all primes $p\geq 2$.

###### Proof.

(i). See Theorem 4.3 (ii), [15, Theorem 2.2], and [7, Theorem A]. (ii) and
(iii). See Theorem 4.3 (i), [13, Examples 3.1 and 3.2], [11, Remark 4.2].
(iv). See Proposition 4.1 and [13, Example 3.3]. ∎

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