text:
\begin{document}
\title{Groups with twisted ${\bf p}$-periodic cohomology}
\author{Guido Mislin}
\author{Olympia Talelli}
\address{Dept. of Mathematics ETH Z\"urich, Switzerland}
\curraddr{Dept. of Mathematics, Ohio State University}
\email{mislin@math.ethz.ch}
\address{Dept. of Mathematics, University of Athens, Greece}
\email{otalelli@cc.uoa.gr}
\date{May 17, 2014}
\begin{abstract}
We give a characterization of groups with twisted $p$-periodic cohomology
in terms of group actions on mod $p$ homology spheres. An
equivalent algebraic characterization of such groups is also presented.
\end{abstract}

\maketitle
\section{Introduction}
We will be considering groups with twisted $p$-periodic cohomology ($p$ a prime) in the following sense.
Write $\hat{\Z}_{p}(\omega)$ for the group of $p$-adic integers, equipped with a $G$-action via a homomorphism $\omega: G\to \hat{\Z}_p^\times$. For $M$ a $\Z G$-module, we write $M_\omega$ for the
$\Z G$-module $M\otimes\hat{\Z}_{p}(\omega)$ with diagonal $G$ action.
\begin{defi}{\label{twisted}}
A group $G$ is said to have twisted $p$-periodic cohomology, if there is a $k>0$, a homomorphism
$\omega:G\to \hat{\Z}_p^\times$ and a cohomology class
$e_\omega\in H^n(G,\hat{\Z}_{p}(\omega))$ for some $n>0$, such that
$$e_\omega\cup - : H^i(G,M)\to H^{i+n}(G,M_\omega)$$
is an isomorphism for all $i\ge k$ and all $p$-torsion $\Z G$-modules $M$
of finite exponent.
In case the twisting $\omega$ can be chosen to be trivial, we say that
$G$ has $p$-periodic cohomology.
\end{defi}

By replacing $e_\omega$ with $e_\omega^2$ we see that for
$G$ with twisted $p$-periodic cohomology one can assume, if one wishes to,
that the degree $n$ of the periodicity
generator is even. In case of a finite group $G$ we infer, by replacing $e_\omega$ by a suitable cup power, that if $G$ has twisted $p$-periodic cohomology, it also has $p$-periodic cohomology.
A classical theorem states that a finite group has $p$-periodic cohomology
if and only if all
its abelian $p$-subgroups are cyclic. Moreover, the finite groups
with $p$-periodic cohomology have the following characterization in terms of actions on $\Z/p\Z$-homology
spheres.

\begin{theo}[Swan \cite{Swan}]\label{Swan} A finite group $G$ has $p$-periodic cohomology if and only if
there exists a finite, simply connected free $G$-$CW$-complex, which has the same $\Z/p\Z$-homology as
some sphere.
\end{theo}

Our goal is to find a similar characterization for arbitrary groups with (twisted) $p$-periodic
cohomology.

\begin{defi} A $CW$-complex $X$ is called a $\Z/p\Z$-homology $n$-sphere, if $H_*(X,\Z/p\Z)\cong H_*(S^n,\Z/p\Z)$.
\end{defi}

In Section \ref{mainsection} we will prove the following generalization of Theorem \ref{Swan}.

\begin{theo}\label{main} A group $G$ has twisted $p$-periodic cohomology if and only
if there exist a simply connected $\Z/p\Z$-homology sphere $X$, which is a free
$G$-$CW$-complex satisfying $cd_{\Z/p\Z}(X/G)<\infty$.
\end{theo}

\smallskip
For the definition of the cohomological dimension $cd_{\Z/p\Z}$ of a space see Section \ref{dim}.

\smallskip
As we will see (cf. Section \ref{last}), there are groups which have twisted $p$-periodic cohomology
but which do not have $p$-periodic cohomology. For
groups with $p$-periodic cohomology we prove the following characterization.

\begin{theo}{\label{main2}} A group $G$ has $p$-periodic cohomology
if and only if
there exist a free $G$-$CW$-complex $X$ with homotopically trivial $G$-action such that
$X$ is a $\Z/p\Z$-homology sphere satisfying $cd_{\Z/p\Z}(X/G)<\infty$.
\end{theo}
\noindent

We will also be considering groups with $\Z/p\Z$-periodic cohomology in the following sense.

\begin{defi}
A group $G$ is said to have $\Z/p\Z$-periodic cohomology, if there is a cohomology class
$e\in H^n(G,\Z/p\Z)$ for some $n>0$ and an integer $k>0$, such that for every $\Z/p\Z[G]$-module $M$ the map
$$e\cup - : H^i(G,M)\to H^{i+n}(G,M)$$
is an isomorphism for all $i\ge k$.
\end{defi}

The following is a simple observation.

\begin{lemm}\label{twistedmodp}
Suppose that $G$ has twisted $p$-periodic cohomology. Then $G$ has $\Z/p\Z$-periodic cohomology.
\end{lemm}

Indeed, if $e_\omega\in H^n(G,\hat{\Z}_p(\omega))$ gives rise to twisted periodicity
as above
and $e_\omega(p)\in H^{n}(G,(\Z/p\Z)_\omega)$ denotes the mod p reduction of $e_\omega$, then
$G$ has $\Z/p\Z$-periodic cohomology with periodicity generator the $(p-1)$-fold
cup product $e:=e_\omega(p)^{p-1}\in H^{n(p-1)}(G,\Z/p\Z)$.

\smallskip
If $M$ is a fixed $\Z G$-module which is $p$-torsion of finite exponent $p^{k+1}$,
then the $p^k(p-1)$-fold twisted module
$$M_{\omega^{p^k(p-1)}}:=((\cdots(M_\omega)\cdots)_\omega)_\omega$$

is naturally isomorphic as a $\Z G$-module to $M$. Therefore, if $G$ has
twisted $p$-periodic cohomology of some period $n$,
its cohomology with $M$ coefficients will actually be periodic in high dimensions $d\ge d_0(M)$, with period $n\cdot p^k(p-1)$. In general, it is not possible to choose the dimensions $d_0(M)$ so that they are bounded
by a number independent of $M$. This observation leads to an example of a group with twisted $p$-periodic cohomology
but not having $p$-periodic cohomology (cf. Example \ref{mainexample}).

\smallskip
It is this
example together with the fundamental paper by Adem and Smith \cite{AS} which inspired our work.
For background on groups acting freely on finite dimensional homology spheres, see \cite{MT} and \cite{T}.

\smallskip

\section{$\Z/p\Z$-dimension for spaces and $\Z/p\Z$-localization}\label{dim}
Similarly to the definition of the $\Z/p\Z$-cohomological dimension of groups,
one defines the $\Z/p\Z$-cohomological dimension for spaces as follows.

\begin{defi}\label{cdim}
Let $X$ be a connected $CW$-complex and $k>0$.
The $\Z/p\Z$-co\-ho\-mo\-logical dimension $cd_{\Z/p\Z}(X)$ of $X$ is the
smallest integer~$n$
such that $H^i(X,M)=0$ for all $\Z/p\Z[\pi_1(X)]$-modules $M$ and all $i>n$; if
there is no such $n$, we write $cd_{\Z/p\Z}(X)=\infty$.
\end{defi}

A simple induction on $k$ shows that if $cd_{\Z/p\Z}X<\infty$ then there exists an $i>0$
such that for all $k$ and all $\Z/p^k\Z[\pi_1(X)]$-modules $M$,
$H^j(X,M)=0$ for all $j>i$.

\bigskip

Bousfield constructed
in \cite{B} on the homotopy category of CW-com\-plexes the localization with respect to $H_*(-,\Z/p\Z)$,
which we call the $\Z/p\Z$-localization and which consists of a functorial map
$$c(X):X\to X_{\Z/p\Z}$$
which is characterized by the following universal property :
for every $H_*(-,\Z/p\Z)$-isomorphism $f:X\to Z$ there is a unique map (up to homotopy)
$g : Z\to X_{\Z/p\Z}$ which is an $H_*(-,\Z/p\Z)$-isomorphism such that $g\circ f\simeq c(X)$.

$\quad\quad\quad\quad\quad\quad\xymatrixrowsep{3pc}\xymatrixcolsep{3pc}\xymatrix{
X\ar[rr]^-{H_*(-,\Z/p\Z)-iso\  f}\ar[d]_{c(X)} && Z\ar[dll]^{\exists !\  g}\\
X_{\Z/p\Z}\,.\\
}$

\noindent
If $X$ is simply connected (or nilpotent) and of finite type, then $X_{\Z/p\Z}$ agrees with Sullivan's
$p$-completion $\hat{X}_p$ (cf.~\cite{S}), and $X\to \hat{X}_p$ is profinite $p$-completion on the level of
homotopy groups.
\smallskip

Note that if $X$ is simply connected, one has $cd_{\Z/p\Z}X=cd_{\Z/p\Z}X_{\Z/p\Z}$, but for instance
$cd_{\Z/p\Z}S^1=1<cd_{\Z/p\Z}S^1_{\Z/p\Z}=\infty$ (because $\pi_1(S^1_{\Z/p\Z})$ contains a free
abelian subgroup of infinite rank).
\smallskip

By the {\sl{standard}} $\Z/p\Z$-homology
$n$-sphere we mean $S^n_{\Z/p\Z}$.

\begin{lemm}\label{Z/pZ-sphere} Let $X$ be a $\Z/p\Z$-homology $n$-sphere.
Then $X_{\Z/p\Z}$ is homotopy equivalent to $S^n_{\Z/p\Z}$.
\end{lemm}

\begin{proof}
Assume that
$H_*(X,\Z/p\Z)\cong H_*(S^n,\Z/p\Z)$. We first consider the case of $n=1$. It follows that
$\pi_1(X)_{ab}\otimes \Z/p\Z$ $\cong$ $\Z/p\Z$. Choose an $f:S^1\to X$ mapping to a generator of
$\pi_1(X)_{ab}\otimes \Z/p\Z$. Then $f$ induces an isomorphism in homology with $\Z/p\Z$-coefficients.
It follows that $f$ induces a homotopy equivalence $S^1_{\Z/p\Z}\to X_{\Z/p\Z}$.
Now assume that $n>1$. Since $H_1(X,\Z/p\Z)\cong H_1(X_{\Z/p\Z}, \Z/p\Z)=0$, we also have
$H_1(\pi_1(X_{\Z/p\Z}),{\Z/p\Z})$=$0$. But $\pi_1(X_{\Z/p\Z})$
is a $H\Z/p\Z$-local group, thus $\pi_1(X_{\Z/pZ})=0$ (see Theorem 5.5 of \cite{B}).
We proceed by showing that $X_{\Z/p\Z}$ is $(n-1)$-connected. Let $\pi_i(X_{\Z/p\Z})$ be the first
non-vanishing homotopy group of $X_{\Z/p\Z}$, $i>1$. Because a $H_*(-,\Z_{(p)})$-isomorphism is also
an $H_*(-,\Z/p\Z)$-isomorphism, $X_{\Z/p\Z}$ is $H\Z_{(p)}$-local and therefore its homology groups with
$\Z$-coefficients are uniquely $q$-divisible for $q$ prime to $p$. Moreover, for $n>i>1$, multiplication by $p$
is bijective on $H_i(X_{\Z/p\Z},\Z)$, because $H_j(X_{\Z/p\Z},\Z/p\Z)=0$ for $j=i-1,i$. Thus $H_i(X_{\Z/p\Z},\Z)$
is a $\Q$-vector space for $1<i<n$. Since the only $\Q$-vector space, which is $H\Z/p\Z$-local as an abelian group,
is the trivial one, and because the homotopy groups of $X_{\Z/p\Z}$ are $H\Z/p\Z$-local, we conclude from the
Hurewicz Theorem that $X_{\Z/p\Z}$ must be $(n-1)$-connected. It follows that the natural maps
$$\pi_n(X_{\Z/p\Z})\to H_n(X_{\Z/p\Z},\Z)\to H_n(X_{\Z/p\Z},\Z,p\Z)\cong \Z/p\Z$$
are both surjective. Choose an $f: S^n\to X_{\Z/p\Z}$ which maps to a generator of $H_n(X_{\Z/p\Z},\Z/p\Z)$ and it
follows that $f$ induces a homotopy equivalence $S^n_{\Z/p\Z}\to X_{\Z/p\Z}$.
\end{proof}
\smallskip
There is also a fiberwise version of $\Z/p\Z$-localization (see \cite{M} for details). If
$$X\to E\to B$$ is a fibration
of connected CW-complexes, one can construct a new fibration
$$X_{\Z/p\Z}\to E^f_{\Z/p\Z}\to B\,,$$ together with a map $E\to E^f_{\Z/p\Z}$ over $B$, which restricts on
the fibers to $\Z/p\Z$-localization $X\to X_{\Z/p\Z}$.
Using the Serre spectral sequence, we conclude the following. If $F\to E\to B$ is a fibration
of connected $CW$-complexes with $F$ simply connected, then $cd_{\Z/p\Z}E=cd_{\Z/p\Z}E^f_{\Z/p\Z}$.
Also, if the fiber $F$ is a $\Z/p\Z$-homology sphere, then fiberwise $\Z/p\Z$-localization
yields a fibration with fiber a standard $\Z/p\Z$-homology sphere $S^n_{\Z/p\Z}\to E^f_{\Z/p\Z}\to B.$

\section{Fibrations, orientation and Euler class}

If $F\to E\to B$ is a fibration of connected $CW$-complexes, then $\pi_1(E)\to \pi_1(B)$
is surjective and lifting of loops defines a natural map $\theta: \pi_1(B)\to [F,F]$, a
{\sl homotopy action} of $\pi_1(B)$ on $F$.

\begin{defi}\label{orientable}
Let $F\to E\to B$ be a fibration of connected CW-complexes. The fibration is
called orientable, if the associated homotopy action $\pi_1(B)\to [F,F]$ is
trivial. We call the fibration $H\Z/p^k\Z$-orientable, if $\pi_1(B)$ acts trivially
on $H_*(F,\Z/p^k\Z)$.
\end{defi}
Clearly, if a fibration is orientable, it is $H\Z/p^k\Z$-orientable for all $k$.

\begin{defi}\label{R-spherical}
Let $
 F\to E\to B$
be a fibration of connected CW-complexes.
We call such a fibration {\sl{$\Z/p\Z$-spherical}}
in case $F$ is a $\Z/p\Z$-homology
sphere (or, equivalently, if $F_{\Z/p\Z}\simeq S^n_{\Z/p\Z}$ for some $n>0$).
\end{defi}

We will make use of the following
observation.

\begin{lemm}\label{spacetofibration}
For a group $G$ the following are equivalent.
\begin{itemize}
\item[a)] There exists a simply connected free $G$-$CW$-complex $X$ which is a
$\Z/p\Z$-homology sphere satisfying $cd_{\Z/p\Z}X/G<\infty$.
\item[b)] There exists a $\Z/p\Z$-spherical fibration
$F\to E\to K(G,1)$ with $F$ simply connected and $cd_{\Z/p\Z}E<\infty$.
\end{itemize}
\end{lemm}

\begin{proof}
Let $X$ be as in $a)$ and $f: X/G\to K(G,1)$ the classifying map for the universal cover $X$
of $X/G$. Then the homotopy fiber of $f$ is $G$-homotopy equivalent to $X$, thus $b)$ holds.
If $F\to E\to K(G,1)$ is as in $b)$, the universal cover of $E$ is $G$-homotopy equivalent to $F$, thus $a)$ holds.
\end{proof}

Note that if $X$ any $\Z/p\Z$-homology sphere, it is also a $\Z/p^k\Z$-homo\-lo\-gy sphere
for $k>1$ as
one easily sees by induction on $k$.
Thus, for a $\Z/p\Z$-spherical fibration
$$F\to E\to B$$
as in Definition \ref{R-spherical},
the $\pi_1(B)$-module $H_n(F,\Z/p^k\Z)$
is isomorphic to a twisted module $(\Z/p^k\Z)_\omega$, where
$\omega: \pi_1(B)\to (\Z/p^k\Z)^\times$ corresponds to the action of $\pi_1(B)$ on $H_n(F,\Z/p^k\Z)$.
(If we need to emphasize the dependence of $\omega$ on $k$, we write $\omega(k)$ in place of $\omega$).
We call the twisted module $(\Z/p^k\Z)_\omega$ the $k$-orientation module.
The fibration is {{H$\Z/p^k\Z$-orientable}}
in the sense of Definition \ref{orientable}, if the $k$-orientation
module is the trivial $\Z/p^k\Z[\pi_1(B)]$-module $\Z/p^k\Z$. We write ${\bar{\omega}}$ for the map
$\pi_1(B)\to (\Z/p^k\Z)^\times$ given by ${\bar{\omega}}(x)=\omega(x^{-1})$,
and more generally $\omega^n$, for the map with $\omega^n(x)=\omega(x^n)$, $n\in\Z$.
For any $\Z/p^k\Z[\pi_1(B)]$-module $M$ we write $M_\omega$ for $M\otimes (\Z/p^k\Z)_\omega$
with diagonal $\pi_1(B)$-action $x\cdot (m\otimes z)=xm\otimes\omega(x)z$. Similarly, we consider
the diagonal action on $Hom_{\Z/p^k\Z}((\Z/p^k\Z)_\omega),M)$ given by $(xf)(z)=x\cdot f(\bar{\omega}(x)z)$.
Therefore, there is a natural isomorphism of $\Z/p^k\Z[\pi_1(B)]$-modules
$$H^n(F,M)\cong Hom(H_n(F,\Z/p^k\Z),M)\cong Hom((\Z/p^k\Z)_\omega,M)
\cong M_{\bar{\omega}}\,.$$

In the case of a $\Z/p\Z$-spherical fibration $F\to E\to B$,
the only possibly non-zero differential in the Serre spectral sequence
with coefficients in a $\Z/p^k\Z[\pi(E)]$-module $K$,

$$E_2^{i,\ell}=H^i(B,H^\ell(F,K))\Longrightarrow H^{i+\ell}(E,K),$$

is the transgression differential
$$d_{n+1}: E^{i,n}_2=E^{i,n}_{n+1}\to E^{i+n+1,0}_{n+1}=E^{i+n+1,0}_2.$$
Taking for $K$ the $k$-orientation module $(\Z/p^k\Z)_\omega$
and choosing $i=0$, this yields
$$d_{n+1}: \Z/p^k\Z=H^0(B,\Z/p^k\Z)\to H^{n+1}(B,(\Z/p^k\Z)_{\omega})\,,$$
and the image of $1\in \Z/p^k\Z$, $d_{n+1}(1):=e(k)_\omega\in H^{n+1}(B, (\Z/p^k\Z)_\omega)$
is called the {\sl{twisted}} $\Z/p^k\Z$-Euler class of the given $\Z/p\Z$-spherical
fibration.

Let now $M$ be an arbitrary $\Z/p^k\Z[\pi_1(B)]$-module and choose $K=M_\omega$. Thus $H^n(F,M_\omega)=M$
and
$$d_{n+1}: E^{i,n}_2=H^i(B,M)\to H^{i+n+1}(B,M_\omega)=E^{i+n+1,0}_2\,$$

is given by the cup-product with $e(k)_\omega$. The kernel and image of $d_{n+1}$ are determined as follows:
$$E^{i,n}_\infty=\ker{d_{n+1}}\subset E^{i,n}_2=H^i(B,M)\stackrel{d_{n+1}}\longrightarrow H^{i+n+1}(B,M_\omega)
\,,$$
respectively
$$
H^i(B,M)\stackrel{d_{n+1}}\longrightarrow H^{i+n+1}(B,M_\omega)=E^{i+n+1,0}_2
\twoheadrightarrow\operatorname{coker}{d_{n+1}}=E^{i+n+1,0}_\infty. $$
The natural surjection $\sigma: H^{i+n}(E,M)\to E^{i,n}_\infty$ has as kernel the subgroup
$E^{i+n,0}_\infty$ and, by splicing things together one gets the {\sl{Gysin-sequence}}
$$\to\! H^{i+n}(E,M)\!\overset{\sigma}\rightarrow\! H^i(B,{M}\,)\!\stackrel{e(k)_\omega\cup -}\longrightarrow\!
H^{i+n+1}(B,M_\omega)\!\to \!H^{i+n+1}(E,M_\omega)\!\to$$
One concludes that for large values of $i$ and all $\Z/p^k\Z[\pi_1(B)]$-modules $M$, cup product with $e(k)_\omega$
induces for all $k$ isomorphisms
$$e(k)_\omega\cup - : H^i(B,M)\overset{\cong}\longrightarrow H^{i+n+1}(B,M_\omega)$$
if and only there exists a $j_0$ such that for all $j>j_0$, $H^j(E,M)=0$ for all $\Z/p^k\Z[\pi_1(B)]$-modules $M$
and all $k$ (here $M$ is viewed as $\pi_1(E)$-modules via $\pi_1(E)\to \pi_1(B)$).
In case $F$ is simply connected, this amounts to $cd_{\Z/p\Z}E<\infty$.

\begin{coro}{\label{c1}}
Let $F\to E\to B$ be a $\Z/p\Z$-spherical fibration of $CW$-complexes with $B$ connected and $F$
simply connected, with twisted $\Z/p^k\Z$-Euler classes
$e(k)_{\omega(k)}\in H^{n}(B,(\Z/p^k\Z)_{\omega(k)})$, $k\ge 1$. Then
the following are equivalent.
\begin{itemize}
\item[1)] $cd_{\Z/p\Z}E<\infty$;
\item[2)] there exists $i_0$ such for all $i>i_0$ and all $k\ge 1$
$$e(k)_{\omega(k)}\cup - : H^i(B,M)\longrightarrow H^{i+n}(B,M_{\omega(k)})$$
is an isomorphism for all $\Z/p^k\Z[\pi_1(B)]$-modules $M$.
\end{itemize}
\end{coro}

In the situation of Corollary \ref{c1}, it follows from the naturality of the Serre
spectral sequence that the twisted $\Z/p^k\Z$-Euler classes $e(k)_{\omega(k)}$
are the reduction mod $p^k$ of a class $e_\omega\in H^n(B,\hat{\Z}_p(\omega))$,
where $\hat{\Z}_p(\omega)$ is isomorphic to $\pi_{n-1}(F_{\Z/p\Z})\cong\pi_{n-1}(S^{n-1}_{\Z/p\Z})$
as a $\pi_1(B)$-module.
Therefore, the following holds.

\begin{coro}\label{fibration to twisted}
If there exists a $\Z/p\Z$-spherical fibration of $CW$-com\-plexes $F\to E\to K(G,1)$ with $F$
simply connected and $cd_{\Z/p\Z}E<\infty$, then $G$ has twisted $p$-periodic cohomology.
\end{coro}

The following lemma permits us to pass from $\Z/p\Z$-spherical fibrations to $H\Z/p\Z$-orientable ones.

\begin{lemm}\label{c2} Let $F_1\to E_1\to B$ be a $\Z/p\Z$-spherical fibration of $CW$-complexes with $B$ connected and $F_1$ simply connected, such that $cd_{\Z/p\Z}E_1<\infty$.
Then the $(p-1)$-fold fiberwise join yields a $H\Z/p\Z$-orientable $\Z/p\Z$-spherical fibration
$F_2\to E_2\to B$ over the same base, with $cd_{\Z/p\Z}E_2<\infty$.
\end{lemm}

\begin{proof}
Let $e_\omega\in H^n(B,(\Z/p\Z)_\omega)$ be the twisted Euler class of the fibration $F_1\to E_1\to B$.
Because $cd_{\Z/p\Z}E_1<\infty$, we infer from Corollary \ref{c1} that
there exists $i_0$ such that
$$e_\omega\cup - : H^i(B, M)\to H^{i+n}(B,M_\omega)$$
is an isomorphism for all $i>i_0$ and all $\Z/p\Z[\pi_1(B)]$-modules $M$. We then perform a fiberwise $(p-1)$-fold join
to obtain a new $\Z/p\Z$-spherical fibration
$F_2\to E_2\to B$ with Euler class $e=e_\omega^{p-1}$. This new
fibration is $H\Z/p\Z$-orientable, because the $(p-1)$-fold tensor product of $(\Z/p\Z)_\omega$
with diagonal action is the
trivial $\Z/p\Z[\pi_1(B)]$-module $\Z/p\Z$. Moreover
$$e\cup -: H^i(B,M)\to H^{i+(p-1)n}(B,M)$$
is an isomorphism for $i>i_0$ and all $\Z/p\Z[\pi_1(B)]$-modules $M$.
Note that $e$ is the reduction mod $p$ of the twisted $\Z/p^k\Z$-Euler class
$e(k)_{\omega(k)}\in H^n(B,(\Z/p^k\Z)_{\omega(k)})$ of the $\Z/p\Z$-spherical fibration
$F_2\to E_2\to B$. Induction on $k$ then shows that
$$e(k)_{\omega(k)}\cup -: H^i(B,L)\to H^i(B,L_{\omega(k)})$$
is an isomorphism for all $\Z/p^k\Z[\pi_1(B)]$-modules $L$.
We infer from
Corollary \ref{c1} that $cd_{\Z/p\Z}E_2<\infty.$
\end{proof}

\section{Partial Euler classes}

For a connected $CW$-complex $X$ we write $P_qX$ for its
$q$-th Postnikov section, with canonical map $X\to P_qX$ such that
\begin{itemize}\item[(1)]
$\pi_i(P_q(X))=0$ for $i>q\,,$
\item[(2)] $\pi_j(X)\stackrel{\cong}\longrightarrow\pi_j(P_qX)$ for $j\le q\,.$
\end{itemize}
\smallskip\noindent
In case that $X$ is a $\Z/p\Z$-homology sphere, we have $X_{\Z/p\Z}\simeq S^n_{\Z/p\Z}$.
Therefore, $P_q(X_{\Z/p\Z})=\{\ast$\} for $q<n$ and $P_n(X_{\Z/p\Z})\simeq K(\hat{\Z}_p,n)$.
Adapting the terminology of \cite{AS} we define {\sl k-partial $\Z/p\Z$-Euler classes} as follows.
\begin{defi}\label{kpartial}
Let $B$ be a connected $CW$-complex and and $k\ge 0$. Then $\epsilon\in H^n(B,\Z/p\Z)$ is
a {\sl k-partial $\Z/p\Z$-Euler class} if there exists a fibration

$$(\Phi):\quad\quad P_{n-1+k}(S^{n-1}_{\Z/p\Z})\to E\to B$$

such that $\pi_1(B)$ acts trivially on $H^{n-1}(P_{n-1+k}(S^{n-1}_{\Z/p\Z}),\Z/p\Z)\cong \Z/p\Z$
and there is a generator of that group which transgresses to $\epsilon$
in the Serre spectral sequence with $\Z/p\Z$-coefficients for the fibration $(\Phi)$.
The $k$-partial $\Z/p\Z$-Euler class $\epsilon$ is called {\sl{orientable}},
if the fibration $(\Phi)$ can be chosen to be orientable in the sense of Definition \ref{orientable}.
\end{defi}

\begin{lemm}\label{powers}
Let $B$ be a connected $CW$-complex and $\epsilon\in H^n(B,\Z/p\Z)$ a $k$-partial $\Z/p\Z$-Euler class.
Then for all $\ell>0$, $\epsilon^\ell$ is a $k$-partial $\Z/p\Z$-Euler class. If $\epsilon$ is orientable
in the sense of Definition \ref{kpartial}, then so is
$\epsilon^\ell$.
\end{lemm}
\begin{proof}
Let
$P:= P_{n-1+k}(S^{n-1}_{\Z/p\Z})\to E\to B$
be a fibration such that $\pi_1(B)$ acts trivially on $H^{n-1}(P,\Z/p\Z)$ and let $\alpha\in
H^{n-1}(P,\Z/p\Z)$ be an element which transgresses to $\epsilon$. By forming fiberwise the $\ell$-fold join
and applying $\Z/p\Z$-localization,
we obtain a new fibration
$(\ast^\ell P)_{\Z/p\Z}\to E(\ell)\to B\,.$
In the Serre spectral sequence with $\Z/p\Z$-coefficients for this fibration, $(\alpha\ast\cdots\ast\alpha)_{\Z/p\Z}$
transgresses to $\epsilon^\ell$. Since $\ast^\ell S^{n-1}\simeq S^{n\ell-1}$,
$$P_{n\ell-1+k}((\ast^\ell P)_{\Z/p\Z})=P_{n\ell-1+k}(S^{n\ell-1}_{\Z/p\Z})$$
and we obtain by taking fiberwise Postnikov sections a fibration
$$P_{n\ell-1+k}(S^{n\ell-1}_{\Z/p\Z})\to E^f(\ell)\to B$$ for which the image of $(\ast^\ell\alpha)_{\Z/p\Z}$ under the natural map
$$ H^{n\ell-1}((\ast^\ell P)_{\Z/p\Z},\Z/p\Z)\stackrel{\cong}\longrightarrow H^{n\ell-1}(P_{n\ell-1+k}(S^{n\ell-1}_{\Z/p\Z}),\Z/p\Z)$$
transgresses to $\epsilon^\ell$. It is obvious that $\epsilon^\ell$ is orientable if $\epsilon$ is.
\end{proof}

\begin{lemm}\label{P10}
Let
$(\Phi_0): S^n_{\Z/p\Z}\to E \to B$
\noindent
be a fibration with $B$ connected and $n>0$. By taking fiberwise Postnikov sections, we obtain fibrations
$$(\Phi_k):\quad P_{n+k}S^n_{\Z/p\Z} \to E_k \to B\,,\quad k\ge 0.$$

\noindent
The fibrations $(\Phi_k)$, $k\ge 0$, are all orientable
if and only if
$\pi_1(B)$ acts trivially on $\pi_n(S^n_{\Z/p\Z})\cong \hat{\Z}_p$.
\end{lemm}

\begin{proof} This follows from the functoriality of $P_{n+k}$
and the fact that homotopy classes $S^n_{\Z/p\Z}\to S^n_{\Z/p\Z}$
correspond naturally to elements of $\pi_n(S^n_{\Z/p\Z})$.
\end{proof}

\begin{defi}\label{omega-p-integral}
Let $X$ be a connected $CW$-complex with fundamental group $G$.
We call an element $x\in H^n(X,\Z/p\Z)$ $\omega$-p-integral, if there
exists an action $\omega: G\to {\hat{\Z}_p}^\times$ such that $G$ acts
trivially on $\hat{\Z}_p/p\hat{\Z}_p$ and $x$ lies
in the image of the natural coefficient homomorphism
$H^n(X,\hat{\Z}_p(\omega))$ $\to H^n(X,\Z/p\Z)\,.$ In case the action
$\omega$ can be chosen to be trivial, we call $x$ p-integral.
\end{defi}

To deal with non-orientable fibrations, we recall
the following fact. Let
$$\hspace{-2 cm} (F):\hspace{2 cm} K(M,m)\to E \to B$$
be a fibration with connected base $B$, $m>0$ and induced action of $\pi_1(B)=G$ on $M$
corresponding to the homomorphism
$\phi:G\to Aut(M)\,.$
Such fibrations are classified by cohomology elements with local coefficients as follows. There
is a universal fibration

$$K(M,m+1)\to L_\phi(M,m+1)\to K(G,1)$$
such that fibrations of type $(F)$
correspond to homotopy classes of maps $f:B\to L_\phi (M,m+1)$ over $K(G,1)$.
The homotopy class over $K(G,1)$ of such an $f$ corresponds to an element in the cohomology with local coefficients $H^{m+1}(B,M)$,
see \cite{Baues} or \cite{G}.

The following lemma is a variation of Lemma 2.5 of \cite{AS}.
\begin{lemm}\label{beta}
Let
$x\in H^{2n}(X,\Z/p\Z)$ be an $\omega$-p-integral element. Then some cup power of $x$
is a $k$-partial $\Z/p\Z$-Euler class and this $k$-partial $\Z/p\Z$-Euler class
is orientable (in the sense of Definition \ref{kpartial}) in case $x$ is p-integral.
\end{lemm}

\begin{proof} Let $G$ be the fundamental group of $X$.
Since $x$ is $\omega$-$p$-integral,
there exists $\omega: G\to\hat{\Z}_p^\times$ and $\tilde{x}\in H^{2n}(X,\hat{\Z}_p(\omega))$
mapping to $x$ under restriction mod $p$.
Let $\mu: X\to L_\omega(\hat{\Z}_p,2n)$ correspond to $\tilde{x}$. It
classifies a fibration
$$K(\hat{\Z}_p(\omega),2n-1)\to E\to X$$
with $H^{2n-1}(K(\hat{\Z}_p(\omega),2n-1),\Z/p\Z)$ =  $H^{2n-1}(K(\hat{\Z}_p/p\hat{\Z}_p,2n-1),\Z/p\Z)$ $\cong$  $\Z/p\Z$
having trivial $G$-action. This shows that $x$ is a $0$-partial $\Z/p\Z$-Euler class.
Suppose now that $k>0$ is given and that $x^m$ is a $(k-1)$-partial
$\Z/p\Z$-Euler class. Thus there is a fibration
$$P_{2nm-1+k-1}(S^{2nm-1}_{\Z/p\Z})=:P(k-1)\to E(k-1)\to X$$
with a generator of $H^{2nm-1}(P(k-1),\Z/p\Z)^G=\Z/p\Z$ transgressing to $y:=x^m$. By Lemma \ref{powers},
for all $j$, $y^j$ is a $(k-1)$-partial $\Z/p\Z$-Euler class too. Thus there are
fibrations
$$P_{2nmj-1+k-1}(S^{2nmj-1}_{\Z/p\Z})=:Q(k-1)\to F(k-1)\to X$$
with a generator of $H^{2nmj-1}(Q(k-1),\Z/p\Z)^G=\Z/p\Z$ transgressing to $y^j=x^{mj}$. To show that
for a suitable $j$,
the power
$y^j$ gives rise to
a $k$-partial $\Z/p\Z$-Euler class, we need to check that the classifying map $\theta:Q(k-1)\to K(\pi,2nmj+k)$ for the fibration $Q(k)\to Q(k-1)$ factors through $F(k-1)$. Note that
$$\pi:=\pi_{2nmj+k-1}(Q(k))=\pi_{2nmj+k-1}(S^{2nmj-1}_{\Z/p\Z})$$
is a finite $p$-group on which $\pi_1(X)=G$ acts
via
$$\omega^{mj}:G\to\hat{\Z}_p^\times=HoAut(S^{2nmj-1}_{\Z/p\Z})\,.$$
We write $\underline{\pi}$ for $\pi$
with that action.
Because of the naturality of the Postnikov section functor, the homotopy
fibration
$$Q(k)\to Q(k-1)\stackrel{\theta}\longrightarrow K(\underline{\pi},2nmj+k)$$
is compatible with the homotopy $G$-action via $\omega^{mj}$ on these spaces.
Therefore,
$$[\theta]\in H^{2nmj+k}(Q(k-1),\underline{\pi})$$
is $G$-invariant with respect to the diagonal $G$-action on this cohomology group.
In the Serre spectral
sequence for $Q(k-1)\to F(k-1)\to X$ with $\underline{\pi}$ coefficients
$$H^s(X, H^t(Q(k-1),\underline{\pi}))\Rightarrow H^{s+t}(F(k-1),\underline{\pi})$$
the cohomology class $[\theta]$ lies thus in
$$E_2^{0,2nmj+k}=H^{2nmj+k}(Q(k-1),\underline{\pi})^G$$
and to show that it is the restriction of a class in the cohomology of $F(k-1)$
with $\underline{\pi}$-coefficients amounts to show that $[\theta]$ is a permanent cycle.
The same argument as in \cite{AS}*{Lemma 2.5}
shows that this is the case for $j$ a large enough $p$-power.
It follows that some power of $x$ is $k$-partial $\Z/p\Z$-Euler class.
In case $x$ is $p$-integral, the argument shows that the $k$-partial
$\Z/p\Z$-Euler class we obtained is orientable.   \end{proof}

\section{Proof of Theorems \ref{main} and \ref{main2}}\label{mainsection}
We will give the proof of Theorem \ref{main}. The proof of Theorem \ref{main2}
is analogous but simpler.

Suppose that $G$ has twisted $p$-periodic cohomology. Then there exists
an $\omega$-$p$-integral class
$\epsilon\in H^{2n}(G,\Z/p\Z)$ and $\epsilon_\omega
\in H^{2n}(G,\hat{\Z}_p(\omega))$ whose reduction mod $p$ is $\epsilon$. By assumption,
there is and a $\ell_0>0$, such that
cup product with $\epsilon_\omega$ induces isomorphisms $H^i(G,M)\to H^{i+2n}(G,M_\omega)$ for all $i\ge \ell_0$
and all $p$-torsion $\Z G$-modules $M$ of finite exponent.
By Lemma \ref{beta} we can find a cup-power $\epsilon^m$ which
is an $\ell_o$-partial $\Z/p\Z$-Euler class.
Therefore, we
have a fibration

$$F(\ell_0):\quad P_{2nm-1+\ell_0}(S^{2nm-1}_{\Z/p\Z})\to E(\ell_0)\to K(G,1)\;,$$
with the property that a generator of $H^{2nm-1}(P_{2nm-1+\ell_0}(S^{2nm-1}_{\Z/p\Z}),\Z/p\Z)$
$\cong \Z/p\Z$
transgresses to $\epsilon^m$ in the Serre spectral sequence for $F(\ell_0)$.
We want to show inductively that $\epsilon^m$ is a $k$-partial Euler class for all $k\ge\ell_0$.
Write $P(j)$ for $P_{2nm-1+j}(S^{2nm-1}_{\Z/p\Z})$.
We will inductively construct fibrations
$$F(k):\quad P(k)\to E(k)\to K(G,1)$$
for $k>\ell_0$
with the property that a generator of $H^{2nm-1}(P(k),\Z/p\Z)$ transgresses to $\epsilon^m$.
To pass from $F(k-1)$ to $F(k)$ we
argue as follows.
We have a diagram
$${\xymatrix{
F(k-1): &P(k-1)\ar[r]&E(k-1)\ar[r]&K(G,1)\\
F(k): &P(k)\ar[u]\ar@{.>}[r]&E(k)\ar@{.>}[r]\ar@{.>}[u]&K(G,1)\ar[u]^{=}\\
}}
$$
in which the fibration $P(k)\to P(k-1)$ has fiber $K(\pi(\omega),2nm-1+k)$ and
is classified by a map
$$\theta: P(k-1)\to K(\pi(\omega),2nm+k),$$
where $\pi(\omega)$ stands for the finite $p$-group $\pi:=\pi_{2nm-1+k}(S^{2nm-1})\otimes \hat{\Z}_p$
with $G$-action induced by $\omega^m: G\to\hat{\Z}_p^\times\subset Aut(\pi_{2nm-1}(S^{2nm-1}_{\Z/p\Z}))$.
To construct the fibration $F(k)$ and the dotted arrows depicted above, we need to
show that $\theta$ factors through $E(k-1)$. This amounts to showing that
$[\theta]\in H^{2nm+k}(P(k-1),\pi(\omega))$ lies in the image of the restriction map
$$H^{2nm+k}(E(k-1),\pi(\omega))\to H^{2nm+k}(P(k-1),\pi(\omega))\,.$$
As argued in the proof of Lemma \ref{beta}, $[\theta]\in H^{2nm+k}(P(k-1),\pi(\omega))$
is $G$-invariant with respect to the diagonal $G$-action  via $\omega^m$ on this cohomology group.
The restriction map in question
corresponds to an edge homomorphism
in the Serre spectral sequence with $\pi(\omega)$-coefficients for the fibration
$P(k-1)\to E(k-1)\to K(G,1)$,
$$H^{2nm+k}\!(E(k-1),\!\pi(\omega))\!\twoheadrightarrow\! E^{0,2nm+k}_\infty\!\subset\! E^{0,2nm+k}_2\!=\!
H^{2nm+k}\!(P(k-1),\!\pi(\omega))^G\!.$$
We need therefore to check that $[\theta]$ is a permanent cycle in the Serre spectral sequence. The only differentials
on $[\theta]$ which could be non-zero are, for dimension reasons, the differentials
$$d_{k+2}:E^{0,2nm+k}_2=E^{0,2nm+k}_{k+2}\to E^{k+2,2nm-1}_{k+2}$$
which takes values in
$$\ker(\epsilon_\omega^m\cup-: H^{k+2}(G,\pi(\omega)_{\bar{\omega}^m})\to H^{k+2+2nm}(G,\pi(\omega)))\,,$$
respectively
$$d_{2nm+k+1}: E^{0,2nm+k}_{2nm+k+1}\to E^{2nm+k+1,0}_{2nm+k+1}\,,$$ which takes values in
$$\operatorname{coker}(\epsilon^m_\omega\cup-: H^{k+1}(G,\pi(\omega))\to H^{2nm+k+1}(G,\pi(\omega)_{\omega^m}))\,.$$
Because $k>\ell_0$, we know that
$$\epsilon^m_\omega\cup -: H^{s}(G,M)\to H^{s+2nm}(G,M_{\omega^m})$$
is an isomorphism for $s=k+1$, respectively $s=k+2$, and any $p$-torsion module $M$ of bounded exponent.
The differentials $d_{k+2}$, respectively
$d_{2nm+k+1}$ depicted above are therefore equal to 0. We conclude that the fibrations in the diagram above can be constructed as displayed.
Passing to homotopy limits in the towers $\{F(k)\}_{k\ge 0}$ of that diagram, one obtains a fibration
$$F(\infty):\quad \quad S^{2n-1}_{\Z/p\Z}\to E\to K(G,1)\,,$$
as desired. To check that $cd_{\Z/p\Z}(E)<\infty$, one considers the
Serre spectral sequence of the fibration $F(\infty)$ with coefficients in a $\Z/p\Z[G]$-module $L$ and finds that
$H^{j}(E,L)=0$ for $j$ large enough, independent of $L$, finishing the first part of the proof.

Suppose now conversely that $X$ is a simply connected free $G$-$CW$-complex which is
a $\Z/p\Z$-homology sphere satisfying $cd_{\Z/p\Z}X/G<\infty$.
By Lemma \ref{spacetofibration} there exists a $\Z/p\Z$-spherical fibration
$F\to E\to K(G,1)$ with $F$ simply connected and $cd_{\Z/p\Z}E<\infty$.
Corollary \ref{fibration to twisted} then implies that $G$ has twisted
$p$-periodic cohomology,
completing the proof of Theorem \ref{main}.\hfill$\square$

\section{Algebraic characterization}
Let $x\in H^n(G,\Z/p\Z)$ and consider a $\Z/p\Z [G]$-projective resolution
$$\mathcal{P}_\ast:\quad \cdots\longrightarrow P_n\stackrel{\partial_n}\longrightarrow P_{n-1}
\stackrel{\partial_{n-1}}\longrightarrow\cdots\longrightarrow P_0\longrightarrow\Z/p\Z\,.$$
Put $K_i:=\partial_i{P_i}$. A cocycle representative of $x$ corresponds to a map
$\theta: K_n\to\Z/p\Z$. Form the following diagram

$${\xymatrix{
 &K_n\ar[r]\ar[d]^\theta&P_{n-1}\ar[r]\ar[d]&P_{n-2}\ar[r]\ar[d]^=&\ar@{.}&\ar[r]&P_0\ar[r]\ar[d]^=&\mathbb{Z}/p\mathbb{Z}
 \ar[d]^=\\
 \tilde{x}:&\Z/p\Z\ar[r]&A\ar[r]&P_{n-2}\ar[r]&\ar@{.}[r]&\ar[r]&P_0\ar[r]&\Z/p\Z
}}
$$
where the square on the left is a push-out square. Then the class of the $n$-fold extension
$\tilde{x}$, $[\tilde{x}]\in Ext^n_{\Z/p\Z G}(\Z/p\Z,\Z/p\Z)$, corresponds
to $x\in H^n(G,\Z/p\Z)$.

\begin{lemm}{\label{proj.dim}} Let $e\in H^n(G,\Z/p\Z)$ and consider the $n$-extension
$$\tilde{e}:\quad \Z/p\Z\to A\to P_{n-2}\to\cdots\to P_{0}\to \Z/p\Z$$
as above. Then the following are equivalent.
\begin{itemize}
\item[1)] $G$ has $\Z/p\Z$-periodic cohomology via cup-product with $e$;
\item[2)] the $\Z/p\Z[G]$-projective dimension of $A$ is finite.
\end{itemize}
\end{lemm}
\begin{proof}
Let $\mathcal{P}_\ast$ be a $\Z/p\Z[G]$-projective resolution of $\Z/p\Z$ and choose
$\theta: K_n\to \Z/p\Z$ to represent $e$ as above, giving rise to the $n$-extension $\tilde{e}$.
It is known that cup product with $e$ is induced by a chain map $\Theta:\mathcal{P}_\ast\to
\mathcal{P}_\ast$ of degree $-n$ which extends $\theta$. Consider the following commutative diagram
with exact rows

$${\xymatrix{
 0\ar[r]&K_n\ar[r]\ar[d]^\theta&P_{n-1}\ar[r]\ar[d]^\mu&K_{n-1}\ar[r]\ar[d]^=&0\\
 0\ar[r]&\Z/p\Z\ar[r]&A\ar[r]&K_{n-1}\ar[r]&0\,.
 }}
$$
From the corresponding commutative diagram of long exact $Ext$-sequen\-ces
$${\xymatrix@C-12pt{
\ar[r]& Ext^i_{\Z/p\Z[G]}(K_{n-1},-)\ar[r]& Ext^i_{\Z/p\Z[G]}(P_{n-1},-)\ar[r]&
 Ext^i_{\Z/p\Z[G]}(K_n,-)\ar[r]&\\
 \ar[r]& Ext^i_{\Z/p\Z[G]}(K_{n-1},-)\ar[r]\ar[u]^=&Ext^i_{\Z/p\Z[G]}(A,-)
 \ar[r]\ar[u]&Ext^i_{\Z/p\Z[G]}(\Z/p\Z,-)\ar[r]\ar[u]^{Ext^i_{\Z/p\Z[G]}(\theta,-)}&
}}
$$
follows that $Ext^i_{\Z/p\Z[G]}(\theta,-)$ is an isomorphism for large $i$ if and only
if $proj.dim_{\Z/p\Z[G]}A <\infty$.
\end{proof}

\begin{coro}\label{zeroproj}
Let $G$ be a group with $\Z/p\Z$-periodic cohomology. There exist $k>0$ such
that for all $i\ge k$ and all projective $\Z/p\Z[G]$-modules $P$,
$H^i(G,P)=0$.
\end{coro}
\begin{proof}
By Lemma \ref{proj.dim}, there is a monomorphism $\iota:\Z/p\Z \to A$ with $A$ a
$\Z/p\Z[G]$-module of finite projective dimension $d$ over $\Z/p\Z[G]$. Let $I$ be an injective $\Z/p\Z[G]$-module.
$I$ injects into $A\otimes_{\Z/p\Z}I$
via $x\mapsto \iota(1)\otimes x$ and, as $I$ is injective,
$I$ is a retract of $A\otimes_{\Z/p\Z}I$. For any projective
$\Z/p\Z[G]$-module $P$, $P\otimes_{\Z/p\Z} I$ is projective. Thus $proj.dim_{\Z/p\Z[G]}A\otimes_{\Z/p\Z}I\le d$
and, because $I$ is a retract of that module, $proj.dim_{\Z/p\Z[G]}I\le d$ too. We conclude
that the supremum of the the projective length of injective $\Z/p\Z[G]$-modules, $spli\, {\Z/p\Z}[G]$, is $\le d$.
It follows then that $silp\, {\Z/p\Z}[G]$, the supremum of the injective length of projective $\Z/p\Z[G]$-modules,
is $\le d$  too (see \cite{GG}*{Theorem 2.4}) and we infer that $H^i(G,P)=0$ for $i>d$ and all projective $\Z/p\Z$-modules $P$.
\end{proof}
The following is an algebraic characterization of groups with twisted $p$-periodic cohomology.

\begin{lemm}
A group $G$ has twisted $p$-periodic cohomology if and only if
there exists an $n>1$ and an exact sequence
of $\Z/p\Z [G]$-modules
$$\epsilon:\quad 0\to\Z/p\Z\to A \to P_{n-2}\to\cdots P_0\to \Z/p\Z\to 0$$
with $P_i$ projective for $0\le i\le n-2$ and $proj.dim_{\Z/pZ [G]}(A)<\infty$,
such that $[\epsilon]\in Ext^n_{\Z/p\Z G}(\Z/p\Z,\Z/p\Z)= H^n(G,\Z/p\Z)$ is $\omega$-$p$-integral
(in the sense of Definition \ref{omega-p-integral})
for some $\omega:G\to\hat{\Z}_p^\times$. $G$ has $p$-periodic cohomology if and only if
there is an $\epsilon$ as above with $[\epsilon]$ $p$-integral (for the definition of $\omega$-$p$-integral
cohomology elements see Definition \ref{omega-p-integral}).
\end{lemm}
\begin{proof}
Suppose $G$ has twisted $p$-periodic cohomology. By
definition, there exists
$\sigma: G\to \hat{\Z}^\times_p$ and an an $\sigma$-$p$-integral cohomology class $e_\sigma\in H^m(G,\hat{\Z}_p(\sigma))$
and a $k>0$ such that cup product with $e_\sigma$ induces isomorphisms $H^i(G,M)\to H^{i+m}(G,M_\sigma)$
for all $p$-torsion $\Z[G]$-modules $M$
of finite exponent (we may assume without loss of generality that $m>1$). It follows
(cf.~Lemma \ref{twistedmodp}) that $G$ has $\Z/p\Z$-periodic cohomology via cup product with
$e:=e_{\sigma}(p)^{p-1}$, where $e_\sigma(p)$ denotes the mod $p$ reduction of $e_\sigma$. Putting $n=m(p-1)$,
it follows that $e$ is $\omega$-$p$-integral with respect to $\omega =\sigma^{p-1}$
and can be represented by an $n$-extension
$$\tilde{e}:\quad 0\to\Z/p\Z\to A \to P_{n-2}\to\cdots \to P_0\to \Z/p\Z\to 0$$
with $P_i$ projective for $0\le i\le n-2$ and $proj.dim_{\Z/pZ [G]}(A)<\infty$.
Conversely, if we are given an $n$-extension
$$\epsilon:\quad 0\to\Z/p\Z\to A \to P_{n-2}\to\cdots \to P_0\to \Z/p\Z\to 0$$
with $P_i$ projective for $0\le i\le n-2$ and $proj.dim_{\Z/pZ [G]}(A)<\infty$
representing an $\omega$-$p$-integral class $e\in H^n(G,\Z/p\Z)$, then $G$ is twisted
$p$-periodic via cup product with $e$. The untwisted version of the lemma corresponds
to the case where we can choose for $\omega$ the trivial homomorphism.
\end{proof}

\section{Some remarks and examples}{\label{last}}

One cannot expect a group $G$ to have $\Z/p\Z$-periodic cohomology if all its finite subgroups
do (for instance, torsion-free groups do not have $\Z/p\Z$-periodic cohomology in general).
We will display below a class of groups, for which this assertion holds. For the proofs, we will make use
of Tate cohomology $\hat{H}^*(G,-)$ for arbitrary groups, as defined in \cite {GM}.
In case $G$ admits a finite dimensional classifying space for proper actions
$\underbar{EG}$, there is a finitely
convergent stabilizer spectral sequence
$$E_1^{m,n}=\prod_{\sigma\in \Sigma_m}\hat{H}^n(G_\sigma,M)\Longrightarrow \hat{H}^{m+n}(G,M)$$
where $\Sigma_m$ is a set of representatives of $m$-cells of $\underbar{EG}$ and $M$ a $\Z G$-module.
For $G$ a group, $M$ a $\Z G$-module and $\mathcal{F}$ the set of finite
subgroups of $G$, we write
$$\mathcal{H}^q(G,M)\subset \prod_{H\in \mathcal F}\hat{H}^q(H,M)$$
for the set of {\sl{compatible}} families $(u_H)_{H\in\mathcal F}$ with respect to
restriction maps of finite subgroups of $G$, induced by embeddings given by conjugation
by elements of $G$.

There are many results on groups $G$ which imply the existence of a finite dimensional $\underbar{EG}$.
For instance, groups of cohomological dimension $1$ over $\Q$ do: they act on a tree with finite stabilizers.
Also, if there is a short exact sequence $H\to G\to Q$ of groups and $H$ as well as $Q$ admit a finite dimensional $\underbar{E}$ and
there is a bound on the order of the finite subgroups of $Q$, then there exist a finite dimensional
model for $\underbar{EG}$
(cf. L\"uck \cite{L}*{Theorem 3.1}).

\begin{lemm}\label{help}
Suppose $G$ admits a finite dimensional $\underbar{EG}$. Then the following holds.
\begin{itemize}
\item[i)] The natural map induced by restricting to finite subgroups
$$\rho: \hat{H}^\ast(G,\Z/p\Z)\to \mathcal{H}^\ast(G,\Z/p\Z)$$
has the property that every element in the kernel of $\rho$ is nilpotent,
and that for every $u\in\mathcal{H}^\ast(G,\Z/p\Z)$ there is a $k$
such that $u^{p^k}$ lies in the image of $\rho$.
\item[ii)] If $\dim\underbar{EG}=t$ and the order of every finite $p$-subgroup of $G$ divides $p^s$
then for every $\Z_{(p)} G$-module $M$ and all $i$
$$p^{s(t+1)}\cdot\hat{H}^i(G,M)=0\,.$$
\item[iii)] If there is a bound on the order of the finite $p$-subgroups of $G$ then
the natural map
$$\alpha: \hat{H}^\ast(G,\Z_{(p)})\to \hat{H}^\ast(G,\Z/p\Z)$$
has the property that every element in the kernel of $\alpha$ is nilpotent and for any
$u\in \hat{H}^\ast(G,\Z/p\Z)$ there exist $k$ such that $u^{p^k}$ lies
in the image of $\alpha$.

\end{itemize}
\end{lemm}
\begin{proof} i) is Corollary 3.3 of \cite{MT}.
For ii) we observe that for every $\Z_{(p)}G$-module $M$, the $E_1$-term of the stabilizer
spectral sequence is annihilated by $p^s$. Since $\underbar{EG}$ has dimension $t$, this
implies that $p^{s(t+1)}$ annihilates all groups $\hat{H}^\ast(G,M)$.
For iii) we first use ii) to conclude that $p^{s(t+1)}$ annihilates the
groups $\hat{H}^*(G,\Z_{(p)})$. One then
argues as in the proof of Lemma 6.6 , Ch.~X in \cite{Br} that for any $\ell>0$ and $x\in \hat{H}^\ast(G,\Z/p\Z),$
$x^{p^\ell}$ lies in the image
$I_\ell$ of
$$\hat{H}^\ast(G,\Z/p^{\ell+1}\Z)\to \hat{H}^\ast(G,\Z/p\Z)$$
and that for $\ell$ large enough $I_\ell$ equals the image of the natural map
$$\alpha: \hat{H}^\ast(G,\Z_{(p)})\to \hat{H}^\ast(G,\Z/p\Z)\,,$$ implying one part of iii).
If $y$ lies in the kernel of $\alpha$, the long exact coefficient sequence
associated with the short exact sequence $\Z_{(p)}\stackrel{p}\to \Z_{(p)}\to \Z/p\Z$
shows that $y=pz$ for some $z$ and therefore $y^{s(t+1)}=p^{s(t+1)}z^{s(t+1)}=0$, finishing the proof of iii).
\end{proof}

\begin{theo}\label{bound}
Let $G$ be a group which admits a finite dimensional $\underbar{EG}$. Then the following holds.
\begin{itemize}
\item[a)]
$G$ has $\Z/p\Z$-periodic cohomology if and only if all its finite subgroups do;
\item[b)] $G$ has $p$-periodic cohomology if all its finite subgroups do and there is a bound
on the order of the finite $p$-subgroups of $G$.
\end{itemize}
\end{theo}
\begin{proof}
We first prove a). If $G$ has $\Z/p\Z$-periodic cohomology and $e\in H^n(G,\Z/p\Z)$ is a periodicity
generator, then every subgroup $H<G$ has $\Z/p\Z$-periodic cohomology, with periodicity generator
the restriction $e_H\in H^n(H,\Z/p\Z)$. This follows from the natural isomorphism
$H^*(H,M)\cong H^*(G, \operatorname{Coind}_H^G M)$ (Shapiro Lemma). If all finite subgroups
of $G$ have $\Z/p\Z$-periodic cohomology, there exist a unit in $e\in \hat{H}^n(G,\Z/p\Z)$ for some
$n>0$ (cf.~\cite{MT}*{Theorem 4.4}). Since $\dim{\underbar{EG}}$ is finite, there is a $k>0$ such that
the natural map $\theta: H^j(G,M)\to \hat{H}^j(G,M)$ is an isomorphism for all $j\ge k$ and all $\Z G$-modules $M$.
Choose $\ell$ such that the degree of $e^\ell$ is larger than $k$ and choose $\epsilon\in H^{n\ell}(G,\Z/p\Z)$
such that $\theta(\epsilon)=e^\ell$. Then $G$ has $\Z/p\Z$-periodic cohomology with periodicity generator $\epsilon$,
finishing the proof of a).
For b) we assume that all finite subgroups of $G$ have $p$-periodic cohomology and that there is a
bound on the order of the finite $p$-subgroups. From Theorem 4.4 of \cite{MT} we conclude that there exist
a unit $u\in \hat{H}^n(G,\Z/p\Z)$ for some $n>0$. Let $v$ an inverse for $u$. By Lemma \ref{help} we can find $k>0$
and $\tilde{u},\tilde{v}\in \hat{H}^\ast(G,\Z_{(p)})$ such that $\alpha(\tilde{u})=u^{p^k}$ and
$\alpha(\tilde{v})=v^{p^k}$, where $\alpha: \hat{H}^\ast(G,\Z_{(p)})\to \hat{H}^\ast(G,\Z/p\Z)$ is the natural
map. From Lemma \ref{help} we conclude that $1-\tilde{u}\tilde{v}$ is nilpotent, thus $\tilde{u}\tilde{v}$ is
invertible, and we conclude that $\tilde{u}\in \hat{H}^{np^k}(G,\Z_{(p)})$ is a unit. Since $G$ admits a
finite dimensional $\underbar{EG}$, the supremum of the injective length of projective
$\Z G$-modules, $silp\, \Z G$, is finite. Therefore,
there is an $n_0$ such that $H^n(G,P)=0$ for all $n>n_0$ and all projective
$\Z G$-modules $P$. By a basic property of Tate cohomology, this implies that there exist $m>0$ such that the canonical map
$\lambda: H^i(G,L)\to \hat{H}^i(G,L)$ is an isomorphism for all $i>m$ and all $\Z G$-modules $L$.
It follows that by choosing an $r>0$ such that $\tilde{u}^r$ has degree larger than $m$, that there is an $\epsilon\in
H^{np^kr}(G,\Z_{(p)})$ with $\lambda(\epsilon)=\tilde{u}^r$. Let
$$\beta: H^{np^kr}(G,\Z_{(p)})\to H^{np^kr}(G,\hat{\Z}_p)$$ be the canonical map and
put $e=\beta(\epsilon)$. Then cup-product with $e$ induces isomorphisms
$$e\cup - : H^j(G,M)\to H^{j+np^kr}(G,\hat{\Z}_p\otimes M)=H^{j+np^kr}(G,M)$$
for all $j>m$ and all $p$-torsion $\Z G$-modules $M$ of bounded exponent, proving that $G$ has $p$-periodic
cohomology.
\end{proof}

Note that we made use of the bound condition in b) of
Theorem \ref{bound} to prove the result, but that bound is not a necessary condition.
For instance, $\Z_{p^\infty}$ has $p$-periodic cohomology, but no bound on the order
of its finite $p$-subgroups. On the other hand, the following is an example
of a group $G$ which admits a finite dimensional $\underbar{EG}$
and with all finite subgroups having $p$-periodic cohomology, but $G$ not having $p$-periodic
cohomology.
Let $\alpha\in \hat{\Z}_p^\times$ be a $p$-adic
unit and define $G(\alpha)$ to be the semi-direct product  $\Z_{p^\infty}\rtimes_\alpha \Z$,
where we have identified $Aut(\Z_{p^\infty})$ with $\hat{\Z}_p^\times$.

\begin{exam}\label{mainexample}
Let p be an odd prime and put $G(1+p)=\Z_{p^\infty}\rtimes_{1+p}\Z$.
\begin{itemize}
\item[a)]
$G(1+p)$ has $\Z/p\Z$-periodic cohomology
of period 2.
\item[b)] $G(1+p)$ does not have $p$-periodic cohomology.
\item[c)] $G(1+p)$ has twisted $p$-periodic cohomology.
\item[d)] $G(1+p)$ acts freely on a simply connected 7-dimensional\\ $G(1+p)$-$CW$-complex
which is a $\Z/p\Z$-homology $3$-sphere.
\end{itemize}
\end{exam}

\begin{proof}
a): Let $H=\Q\rtimes_{1+p}\Z$. There is a natural surjective map $H\to G(1+p)$
with kernel isomorphic to $\Z_{(p)}$. Note that $H$ has cohomological dimension 3. Choose $Y$ to be a
3-dimensionl model for $K(H,1)$ and $X$ the covering space corresponding to $\Z_{(p)}<H$.
$X$ is a free $G(1+p)$-$CW$-complex and $X\simeq K(\Z_{(p)},1)$, thus $X$ is a $\Z/p\Z$-homology $1$-sphere.
We then have homotopy fibration
$$X\to X/G(1+p)\to BG(1+p)\,,\quad X_{\Z/p\Z}= S^1_{\Z/p\Z}$$
which is $H\Z/p\Z$-orientable, because multiplication by $1+p$
is the identity on $\Z/p\Z$. It follows that the associated $\Z/p\Z$-Euler class $e\in H^2(G(1+p),\Z/p\Z)$ induces
via cup product isomorphisms
$$H^i(G(1+p),M)\stackrel{e\cup -}\longrightarrow H^{i+2}(G(1+p),M)$$
for all $i>3$ and all $\Z/p\Z[G]$-modules
$M$, which proves a).
For b) we consider the subgroups
$$G_n=\Z/p^n\Z\rtimes_{1+p}\Z<G(1+p)$$
and observe that the minimal $p$-period
for $H^*(G_n,\Z/p^n\Z)$ is $\ge 2p^{n-1}$, because multiplication by $1+p$ on $H^2(\Z/p^n\Z,\Z/p^n\Z)=\Z/p^n\Z$ is an
automorphism of order $p^{n-1}$ for odd $p$. Thus, the minimal $p$-period for $G_n$ goes to $\infty$ as $n$ tends to $\infty$ and, therefore, $G(1+p)$ does not have $p$-periodic
cohomology. For c) we observe that the twisted $\hat{\Z}_p$-Euler class $\tilde{e}\in H^2(G(1+p),\hat{\Z}_p(\omega))$
of the homotopy $\Z/p\Z$-spherical fibration constructed in a), with $\omega: G(1+p)\to\hat{\Z}_p^\times$
given by $(x,y)\mapsto (1+p)^y$ for $(x,y)\in\Z_{p^\infty}\rtimes \Z$, has reduction mod $p$ equal to the
$\Z/p\Z$-Euler class $e$ of a). It follows that $G(1+p)$ has twisted
$p$-periodic cohomology of period $2$, with twisted $p$-periodicity induced by cup product with $\tilde{e}$.
For d) we again look at the free $G(1+p)$-$CW$-complex $X$ as constructed in a). The join $X\ast X$ is a simply connected free $G(1+p)$-$CW$-complex of dimension $7$, which is a
$\Z/p\Z$-homology $3$-sphere, completing the proof.
\end{proof}

\end{document}

config: arxiv


TITLE: Halting problem within a finite time interval?
QUESTION [2 upvotes]: Let the finite time halting problem to be one where the program is counted as "not halting" if the Turing machine takes more than a given time interval to run the program.
Is this finite time halting problem undecidable?

REPLY [1 votes]: In addition to the excellent answer already posted, I'd like to point out some intuition for why it must be computable.
If you wanted to compute the characteristic function for the finite time halting problem we could run the machine for s steps on the given input. If we assign the value of 1 to problems that have halted in s steps and the value of 0 otherwise, then we have an algorithm that halts on every input.


TITLE: Calculating limit of $\sum_{k = 0}^\infty \frac{n^k}{k!} \vert k - n \vert$
QUESTION [1 upvotes]: I want to show that for each $n \in \mathbb N$ it holds that
$$\sum_{k = 0}^\infty \frac{n^k}{k!} \vert k - n \vert = 2n \frac{n^n}{n!}.$$
But I have absolutely no idea how to do that. I tried to split the series like this:
$$\sum_{k = 0}^\infty \frac{n^k}{k!} \vert k - n \vert = \sum_{k = 0}^{n - 1} \frac{n^k}{k!}(n - k) + \sum_{k = n + 1}^\infty \frac{n^k}{k!} (k - n)$$
Now I want to show that $\sum_{k = 0}^{n - 1} \frac{n^k}{k!}(n - k) = \sum_{k = n + 1}^\infty \frac{n^k}{k!} (k - n) = n \frac{n^n}{n!}$ holds. But I got no idea how to achieve that. I tried to use induction but things became pretty messy and I don't think it will work like that. On the other hand I got the idea that it might have to do with $\exp$ or binomial coefficients but I don't see where I can facilitate that. I would appreciate some hints :)

REPLY [4 votes]: Split further:
\begin{align}
\sum_{k = 0}^{n-1} \frac{n^k}{k!}(n-k) + \sum_{k = n+1}^{\infty} \frac{n^k}{k!}(k-n) &= \sum_{k = 0}^{n-1} \frac{n^{k+1}}{k!} - \sum_{k = 1}^{n-1} \frac{n^k}{(k-1)!} + \sum_{k = n+1}^{\infty} \frac{n^k}{(k-1)!} - \sum_{k = n+1}^{\infty} \frac{n^{k+1}}{k!} \\
&= \sum_{k = 0}^{n-1} \frac{n^{k+1}}{k!} - \sum_{k = 0}^{n-2} \frac{n^{k+1}}{k!} + \sum_{k = n}^{\infty} \frac{n^{k+1}}{k!} - \sum_{k = n+1}^{\infty} \frac{n^{k+1}}{k!} \\
&= \frac{n^n}{(n-1)!} + \frac{n^{n+1}}{n!} \\
&= 2n\frac{n^n}{n!}.
\end{align}


TITLE: Transformations Linear Algebra Question
QUESTION [0 upvotes]: Let $T : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the linear transformation defined by
$T(x,y) = (-6x -3y; -4x -2y)$
Find a vector $w$ that is not in the image of $T$.
Cannot work this out please help me.

REPLY [4 votes]: $$T\binom xy=\binom{-6x-3y}{-4x-2y}=\binom{\frac32t}t$$
with $\;t=-4x-2y\;$, thus any vector not of this form won't be in the image of $\;T\;$, for example
$$\binom 11\;,\;\;\binom1 3\;,\;\;\binom{\sqrt2}5\;,\ldots$$


text:
\begin{document}

\title[On the Davis-Wielandt shell and the Davis-Wielandt index]{On the Davis-Wielandt shell of an operator and the Davis-Wielandt index of a normed linear space}

\author[Pintu Bhunia, Debmalya Sain and  Kallol Paul]{ Pintu Bhunia, Debmalya Sain and Kallol Paul}

\address[Bhunia] { Department of Mathematics, Jadavpur University, Kolkata 700032, West Bengal, India}
\email{pintubhunia5206@gmail.com}

\address{(Sain) Department of Mathematics, Indian Institute of Science, Bengaluru 560012, Karnataka, India}
\email{saindebmalya@gmail.com}

\address[Paul] {Department of Mathematics, Jadavpur University, Kolkata 700032, West Bengal, India}
\email{kalloldada@gmail.com}

\thanks{Pintu Bhunia would like to thank UGC, Govt. of India for the financial support in the form of SRF. Dr. Debmalya Sain feels elated  to acknowledge the joyful presence of his beloved junior Mr. Debabrata Dey in every sphere of his life. Prof. Paul would like to thank RUSA 2.0, Jadavpur University for  partial support. }
\thanks{}

\subjclass[2010]{Primary 47A12, Secondary 46B20,47L05}
\keywords{Davis-Wielandt shell, Davis-Wielandt radius, Davis-Wielandt index, Inequalities in normed spaces, Polyhedral Banach spaces, Linear operators}

\date{}
\maketitle

\begin{abstract}
We study the Davis-Wielandt shell and the Davis-Wielandt radius of an operator on a normed linear space $\mathcal{X}$. We show that after a suitable modification, the modified Davis-Wielandt radius defines a norm on  $\mathcal{L}(\mathcal{X})$ which is equivalent to the usual operator norm on $\mathcal{L}(\mathcal{X})$. We introduce the Davis-Wielandt index of a normed linear space and compute its value explicitly in case of some particular polyhedral Banach spaces. We also present a general method to estimate the Davis-Wielandt index of any polyhedral Banach space.
\end{abstract}

\section{Introduction}

\smallskip

\noindent The purpose of this article is to study the Davis-Wielandt shell of an operator between normed linear spaces and some related geometric and analytic properties. Let us first establish the relevant notations and terminologies to be used throughout the article.

\noindent For a complex number $c\in \mathbb{C}$, we write $\mathcal{R}(c)$ and $\mathcal{I}(c)$ for the real part of $c$ and the imaginary part of $c$,  respectively. Let $\mathcal{X}$ be a normed linear space, real or complex and let $\mathcal{X}^*$ denote the dual space of $\mathcal{X}.$  Let $ B_{\mathcal{X}}= \{ x \in \mathcal{X}~:\|x\| \leq 1 \} $ and $  S_{\mathcal{X}}= \{ x \in \mathcal{X}~:\|x\| = 1 \} $ be the unit ball and the unit sphere of $ \mathcal{X}$, respectively. Given a convex subset $K$ of $ \mathcal{X}, $  let $\textit{ext}~K$ denote the set of all extreme points of $ K. $ The Krein-Milman theorem asserts that $\textit{ext}~ K \neq \emptyset$, if $K$ is non-empty compact.  For each $x\in S_{\mathcal{X}}$, let us define $\mathcal{J}(x)=\left\{ f\in \mathcal{X}^*: f(x)=1=\|f\| \right\}.$ We denote by $ \Pi$ the set $\{(x,f): x\in S_{\mathcal{X}}, f\in \mathcal{J}(x)\}$.
Let $\mathcal{L}(\mathcal{X})$ be the normed linear space of all bounded linear operators on $ \mathcal{X} $, endowed with the usual operator norm.

\noindent For a complex Hilbert space $\mathcal{H},$ the Davis-Wielandt shell and the Davis-Wielandt radius of an operator $T\in \mathcal{L}(\mathcal{H})$ (see \cite{D,W}) are defined respectively as:
\begin{eqnarray*}
 DW(T)&= &\left\{ \left(\langle Tx, x \rangle, \|Tx\|^2 \right) : x\in \mathcal{H}, \|x\|=1 \right\},\\
 dw(T)&= &\sup \left\{ \sqrt{|\langle Tx, x \rangle|^2+ \|Tx\|^4} : x\in \mathcal{H}, \|x\|=1 \right \}.
\end{eqnarray*}
The above concepts are closely related to the numerical range of the operator $T$, studied initially by  Bauer \cite{B} and Lumar \cite{L}, in the more general setting of normed linear spaces.
 Given a $ T \in \mathcal{L}(\mathcal{X}), $ the numerical range and the numerical radius of $ T $ are defined respectively as:
\begin{eqnarray*}
W(T)&=& \left\{ x^*(Tx) : (x,x^*)\in \Pi \right\},\\
 w(T)&=& \sup \left\{ |x^*(Tx)| : (x,x^*)\in \Pi \right\}.
\end{eqnarray*}
Computation of the numerical index of a normed linear space involves the interaction between the geometry of the space and linear operators acting on it. We refer the readers to \cite{DMPW,MJ,MMP1,MMP2,M,SPBB} and references therein for more information on this topic of interest.
Given a normed linear space $\mathcal{X}$, the numerical index of $\mathcal{X}$ is defined as:
 $$n(\mathcal{X})=\inf\{w(T):T\in S_{\mathcal{L}(\mathcal{X})}\}.$$
Motivated by the study on the numerical index of a normed linear space, we introduce another numerical constant associated with a given space, which is related to the study of the Davis-Wielandt shell of an operator and is christened the Davis-Wielandt index. To this end, we first introduce the Davis-Wielandt set of an operator  at a point of unit norm. For $ T \in \mathcal{L}(\mathcal{X}), $ the Davis-Wielandt set of $T$ at the point $x\in S_{\mathcal{X}}$ is defined as:
$$DW(T_x)=\left \{\left (x^*(Tx), \|Tx\|^2\right ):x^* \in \mathcal{J}(x) \right \}.$$
Next we recall that the Davis-Wielandt shell and corresponding the Davis-Wielandt radius of an operator $T$ on $\mathcal{X}$. The Davis-Wielandt shell and the Davis-Wielandt radius of $ T\in \mathcal{L}(\mathcal{X}) $ are defined respectively as:
\begin{eqnarray*}
DW(T)&=& \left\{ \left(x^*(Tx), \|Tx\|^2\right) : (x,x^*)\in \Pi \right\},\\
dw(T)&=&\sup \left\{ \sqrt{| x^*(Tx) |^2+\|Tx\|^4}~:(x,x^*)\in \Pi \right\}.
\end{eqnarray*}
\noindent It is easy to observe that the Davis-Wielandt radius $dw(.)$ does not define a norm on $\mathcal{L}(\mathcal{X})$. However, the following modification allows us to define a norm on $\mathcal{L}(\mathcal{X})$. For $ T \in \mathcal{L}(\mathcal{X}), $ the modified Davis-Wielandt shell and the modified Davis-Wielandt radius of $ T $ are defined respectively as:
\begin{eqnarray*}
 DW^*(T)&=& \left\{ \left(x^*(Tx), \|Tx\|\right) : (x,x^*)\in \Pi \right\},\\
dw^*(T)&=&\sup \left\{ \sqrt{| x^*(Tx) |^2+\|Tx\|^2}~:(x,x^*)\in \Pi \right\}.
\end{eqnarray*}
\noindent Now we are in a position to introduce the Davis-Wielandt index and the modified Davis-Wielandt index of a normed linear space.
Given a normed linear space $ \mathcal{X} $, the Davis-Wielandt index and the modified Davis-Wielandt index of $ \mathcal{X} $ are defined respectively as:
\begin{eqnarray*}
\eta_{dw}(\mathcal{X})&=&\inf \left\{ dw(T):T\in S_{\mathcal{L}(\mathcal{X})} \right \},\\
 \eta_{dw^*}(\mathcal{X})&=&\inf \left\{ dw^*(T):T\in S_{\mathcal{L}(\mathcal{X})} \right \}.
\end{eqnarray*}

\smallskip

\noindent We are especially interested in computing the (modified) Davis-Wielandt index of polyhedral normed linear spaces. Let us first recall that a finite-dimensional real Banach space is said to be polyhedral if $\textit{ext}~ B_{\mathcal{X}}$ is finite. The following definition, introduced in \cite{SPBB}, is useful in studying the geometry of polyhedral Banach spaces.

\begin{definition}
Let $ \mathcal{X} $ be a polyhedral Banach space. Let $ F $ be a facet of the unit ball $ B_{\mathcal{X}} $ of $  \mathcal{X}. $ A functional $ f \in S_{\mathcal{X}^{*}} $ is said to be a \emph{supporting functional corresponding to the facet $ F $ of the unit ball $ B_{\mathcal{X}} $} if the following two conditions are satisfied:\\
$ (i) $ $ f $ attains norm at some point $ v $ of $ F, $\\
$ (ii) $ $ F =(v + \ker f)\cap S_{\mathcal{X}}. $
\end{definition}
\noindent Now we recall the basic definitions related to the concept of smoothness in a normed linear space $\mathcal{X}$. A normed linear space $\mathcal{X}$ is said to be smooth at a non-zero point $x\in {\mathcal{X}}$ if $\mathcal{J}\left(\frac{x}{\|x\|}\right )$ is singleton. $\mathcal{X}$ is said to be smooth if $\mathcal{X}$ is smooth at each non-zero point $x\in \mathcal{X}$.
\noindent Next we define the following two notations useful to the present study. For $T\in \mathcal{L}(\mathcal{X})$, let $W_T$ and $M_T$ be denote the numerical radius attainment set and the norm attainment set of $T$, defined respectively as:
\begin{eqnarray*}
W_T&=& \left\{ x\in S_{\mathcal{X}}:  ~\exists~ x^{*}\in \mathcal{J}(x) ~\mbox{such that}~ |x^{*}(Tx)|=w(T) \right \},\\
M_T &=& \left \{ x\in S_{\mathcal{X}}: \|Tx\|=\|T\| \right \}.
\end{eqnarray*}

\smallskip

\noindent This article is demarcated into three section, including the introductory one. In the second section we explore the convexity properties of the Davis-Wielandt shell of an operator, both from the local and the global perspective. We also study the norming properties of the modified Davis-Wielandt radius in $\mathcal{L}(\mathcal{X})$ and its connection with the usual operator norm. The third section is devoted to the investigation of the Davis-Wielandt index of a normed linear space. We estimate the Davis-Wielandt index of polyhedral Banach spaces. We also obtain the exact values of the Davis-Wielandt index in case of some particular polyhedral Banach spaces.

\section{On the Davis-Wielandt shell}
\smallskip

\noindent
We begin this section with the study of the local convexity properties of the Davis-Wielandt set of a bounded linear operator at a point of unit norm. First we prove the following theorem.

\begin{theorem}\label{DWset}
Let $\mathcal{X}$ be a normed linear space. Let $T\in \mathcal{L}(\mathcal{X})$ and let $x\in S_{\mathcal{X}}$. Then $DW(T_x)$ is convex.  If $\mathcal{X}$ is finite-dimensional, then $DW(T_x)$ is the convex hull of $\Big \{ (x^*(Tx), \|Tx\|^2) : x^*\in \mbox{ext}~~B_{\mathcal{X}^*}, x^*(x)=1 \Big \}.$
\end{theorem}

\begin{proof}
Without loss of generality we assume that $ DW(T_x)$ is not singleton. Let $\left (x^*(Tx), \|Tx\|^2 \right )$, $\left (y^*(Tx), \|Tx\|^2 \right ) \in DW(T_x)$, where $x^*,y^*\in \mathcal{J}({{x}})$.  It is easy to see that $\lambda x^*+(1-\lambda)y^* \in S_{\mathcal{X}^*} $ and $(\lambda x^*+(1-\lambda)y^*)(x)=1$, for all $\lambda \in [0,1].$ Therefore, $\lambda \left (x^*(Tx), \|Tx\|^2 \right )+(1-\lambda)\left (y^*(Tx), \|Tx\|^2 \right )= \Big((\lambda x^*+(1-\lambda)y^*)(Tx), \|Tx\|^2 \Big)\in DW(T_x)$, for all $\lambda \in [0,1].$ So, $ DW(T_x)$ is convex.\\
Let dim $(\mathcal{X}) =n$ and let $x^*\in \mathcal{J}({{x}})$. Then $x^*$ can be written as a convex combination of extreme supporting functionals at $x$. Let $x^*_i$ be the extreme supporting functionals at $x$, for $i=1,2,\ldots,m$. Therefore, there exist scalars $ \lambda_1,\lambda_2,\ldots,\lambda_{m}\geq 0$ with $\sum_{i=1}^m\lambda_i=1$ such that $x^*=\sum_{i=1}^m\lambda_ix_i^*$. So, $x^*(Tx)=\sum_{i=1}^m\lambda_i x^*_i(Tx)$. This proves that  $DW(T_x)$ is the convex hull of $\Big \{ (x^*(Tx), \|Tx\|^2) : x^*\in \mbox{ext}~~B_{\mathcal{X}^*}, x^*(x)=1 \Big \}.$
\end{proof}

\begin{remark}
We would like to remark that if $\mathcal{X}$ is a real normed linear space, then it is easy to observe that  $DW(T_x)$ is the line segment joining
$\Big (\inf \{x^*(Tx): x^*\in \mbox{ext}~~B_{\mathcal{X}^*}, x^*(x)=1\}, \|Tx\|^2 \Big )$ and  $\Big (\sup \{x^*(Tx): x^*\in \mbox{ext}~~B_{\mathcal{X}^*}, x^*(x)=1\}, \|Tx\|^2 \Big )$.
\end{remark}

\noindent Next we study the convexity properties of the Davis-Wielandt shell of an operator $T$ defined on a normed linear space $\mathcal{X}$. In the case of Hilbert space, this convexity property is well-known (see \cite[Th. 2.2 and Th. 2.3]{LPS}, as given in the following theorem.

\begin{theorem}\label{th-2dim}
Let $\mathcal{H}$ be a Hilbert space and let $T\in \mathcal{L}(\mathcal{H})$. Then the following hold true:\\
$(i)$ If dim $(\mathcal{H}) =2$, then $DW(T)$ is convex if and only if $T$ is normal.\\
$(ii)$ If dim $(\mathcal{H}) \geq 3$, then $DW(T)$ is convex.
\end{theorem}

\smallskip

\noindent However, the above convexity property may not hold true in the more general setting of normed linear spaces. To illustrate this, we first make note of the following proposition.

\begin{prop}\label{prop1}
Let $\mathcal{X}$ be a normed linear space and let $T\in \mathcal{L}(\mathcal{X})$. If $W(T)$ is not convex then $DW(T)$ is not convex.
\end{prop}
\begin{proof}
Since $W(T)$ is not convex, there exist $x^*(Tx), y^*(Ty)\in W(T)$ and $\lambda_0\in (0,1)$ such that $\lambda_0x^*(Tx)+(1-\lambda_0) y^*(Ty)\notin W(T)$, where $x,y\in S_{\mathcal{X}}$ and $x^*\in \mathcal{J}({{x}})$, $y^*\in \mathcal{J}({{y}})$. Therefore, there exists no $z^*\in \mathcal{J}({{z}})$, $z \in S_{\mathcal{X}}$ such that   $\lambda_0x^*(Tx)+(1-\lambda_0) y^*(Ty)=z^*(Tz)$.
Clearly, $\left (x^*(Tx), \|Tx\|^2 \right )$, $\left (y^*(Ty), \|Ty\|^2 \right ) \in DW(T)$, but $\lambda_0\left (x^*(Tx), \|Tx\|^2 \right )+(1-\lambda_0)\left (y^*(Ty), \|Ty\|^2 \right ) \notin DW(T)$. So, $DW(T)$ is not convex.
\end{proof}

\noindent We now show that $DW(T)$ is not necessarily convex for $T\in \mathcal{L}(\mathcal{X}),$ even if dim $(\mathcal{X}) \geq 3$. This illustrates that Theorem \ref{th-2dim} $(ii)$ cannot be extended to the setting of Banach spaces.

\begin{theorem}\label{notconvex}
Let $\mathcal{X}={{\ell}}^n_p$, $p\neq 1,2,\infty$ and $n\geq 2$. Let $T\in \mathcal{L}(\mathcal{X})$ be such that $$T=\left(\begin{array}{ccc}
    a&b \\
    c&d
	\end{array}\right)\oplus 0_{n-2\times n-2},~~ a,d\in \mathbb{R}\setminus \{0\},~~b,c\in \mathbb{C}\setminus \{0\} $$
	and $(i)$~~$a+d=0, (ii)~~ |b|=|c|, (iii)~~ \mathcal{R}(b) \mathcal{I}(c)+ \mathcal{R}(c) \mathcal{I}(b)=0.$ Then $DW(T)$ is not convex.
\end{theorem}

\begin{proof}
Let $S=\left(\begin{array}{ccc}
    a&b \\
    c&d
	\end{array}\right).$ Mandal et al.  proved in \cite[Th. 2.9]{MBBP} that $W(S)$ is not convex. By similar computations we can show that $W(T)$ is not convex. Therefore, it follows from Proposition \ref{prop1} that $DW(T)$ is not convex.
\end{proof}

\smallskip

\noindent Our next result relates the smoothness of a normed linear space at a point of unit norm with the local convexity properties of the Davis-Wielandt set of operators on the space.

\begin{prop}\label{smothness}
Let $\mathcal{X}$ be a normed linear space and let $x\in S_{\mathcal{X}}.$ Then $x$ is a smooth point of $\mathcal{X}$ if and only if $ DW(T_x)$ is singleton, for every $T\in \mathcal{L}(\mathcal{X})$.
\end{prop}
\begin{proof}
The necessary part follows trivially. Let us prove only the sufficient part. Let $ DW(T_x)$ be singleton, for every $T\in \mathcal{L}(\mathcal{X})$. Suppose on the contrary that $x$ is not a smooth point. Then there exist distinct supporting functionals $x^*, y^*\in \mathcal{J}(x)$. Let $z\in S_{\mathcal{X}}$ be such that $x^*(z)\neq y^*(z).$ Let $T\in \mathcal{L}(\mathcal{X})$ be such that $Tx=z.$ Therefore, $x^*(Tx)\neq y^*(Tx).$ Thus, there is an operator $T\in \mathcal{L}(\mathcal{X})$ such that $ DW(T_x)$ is not singleton, a contradiction. Hence, $x$ is a smooth point.
\end{proof}

\smallskip

\noindent In the next result we show that the modified Davis-Wielandt radius $dw^*(.)$ defines a norm on $\mathcal{L}(\mathcal{X}),$ for any normed linear space $\mathcal{X}.$

\begin{theorem}\label{norm}
Let $\mathcal{X}$ be a normed linear space. Then $dw^*(.)$ defines a norm on $\mathcal{L}(\mathcal{X})$. It is equivalent to the usual operator norm $\|.\|$, satisfying the following inequality: $$\|T\|\leq dw^*(T)\leq \sqrt{2}\|T\|,~~\mbox{for every}~~ T\in \mathcal{L}(\mathcal{X}).$$
\end{theorem}

\begin{proof}
First we prove that $dw^*(.)$ defines a norm on $\mathcal{L}(\mathcal{X})$. Let $T\in \mathcal{L}(\mathcal{X}).$ Clearly,
 $dw^*(T)=0$ if and only if $T=0$ and $dw^*(cT)=|c|dw^*(T)$, for all scalars $c$. Let $T_1, T_2\in \mathcal{L}(\mathcal{X}).$ Then
\begin{eqnarray*}
&& dw^*(T_1+T_2)\\
 = && \sup \left \{\sqrt{ |x^*((T_1+T_2)x)|^2+\|(T_1+T_2)x\|^2 }: (x,x^*)\in \Pi  \right \}\\
  \leq &&  \sup \left \{\sqrt{ |x^*(T_1x)+x^*(T_2x)|^2+(\|T_1x\|+\|T_2x\|)^2 }: (x,x^*)\in \Pi  \right \}\\
	\leq &&  \sup \Big \{ \sqrt{ |x^*(T_1x)|^2+\|T_1x\|^2 }+\sqrt{|x^*(T_2x)|^2+\|T_2x\|^2}: (x,x^*)\in \Pi  \Big \},\\
		 &&  \hspace{6cm}\mbox{by Minkowski inequality}\\
		\leq &&  \sup \left \{ \sqrt{ |x^*(T_1x)|^2+\|T_1x\|^2 }: (x,x^*)\in \Pi  \right \} \\
	&&  + \sup \left \{ \sqrt{ |x^*(T_2x)|^2+\|T_2x\|^2 }: (x,x^*)\in \Pi  \right \}\\
		=&& dw^*(T_1)+dw^*(T_2).
\end{eqnarray*}
Therefore, $dw^*(T_1+T_2)\leq dw^*(T_1)+dw^*(T_2)$ for every $T_1, T_2\in \mathcal{L}(\mathcal{X}).$ Hence, $dw^*(.)$ defines a norm on $\mathcal{L}(\mathcal{X}).$ Next we prove the other part of the theorem. It follows from the definition of the modified Davis-Wielandt radius that
\begin{eqnarray*}
dw^*(T)&=&\sup \left\{ \sqrt{| x^*(Tx) |^2+\|Tx\|^2}~:(x,x^*)\in \Pi \right\}\\
&\leq& \sqrt{  w^2(T)+ \|T\|^2   }\leq \sqrt{  \|T\|^2 + \|T\|^2 }\leq \sqrt{2} \|T\|.
\end{eqnarray*}
Also,
\begin{eqnarray*}
dw^*(T)&=&\sup \left\{ \sqrt{| x^*(Tx) |^2+\|Tx\|^2}~:(x,x^*)\in \Pi \right\}\\
&\geq& \max \left \{ w(T), \|T\|   \right \}=\|T\|.
\end{eqnarray*}
This completes the proof.

\end{proof}

\begin{remark}
We would like to remark that the inequality in Theorem \ref{norm} is sharp. For the identity operator $T$ on $\mathcal{X}$, the second inequality becomes an equality. For a skew-symmetric operator $T$ on a real Hilbert space $\mathcal{H}$, the first inequality becomes an equality.
\end{remark}

\noindent We would like to estimate the Davis-Wielandt radius of an arbitrary operator on a given polyhedral space. First we recall the following lemma from \cite{SPBB}, which is useful in this context.

\begin{lemma} \label{lemma:extreme}
Let $\mathcal{X}$ be a polyhedral Banach space. Then $f \in S_{\mathcal{X}^{*}}$ is an extreme point of $ B_{\mathcal{X}^{*}} $ if and only if $ f $ is a supporting functional corresponding to a facet of $B_{\mathcal{X}}.$
\end{lemma}

\smallskip

\noindent Now we are in a position to prove the following theorem.

\begin{theorem}\label{prop-est}
Let $\mathcal{X}$ be an $n$-dimensional polyhedral Banach space and let $T\in \mathcal{L}(\mathcal{X}).$ Then
\[dw(T)\leq \max \left\{ \sqrt{| x^*(Tx) |^2+\|T\|^4}~:(x,x^*)\in G \right\}, \]
where $G=\{(x,x^*): x\in \textit{ext}~B_{\mathcal{X}}, x^*\in \textit{ext}~B_{\mathcal{X}^*}, x^*(x)=1 \}.$
\end{theorem}

\begin{proof}
It follows from the definition of the Davis-Wielandt radius of $T$ that
\begin{eqnarray*}
dw(T)&= & \sup \left\{ \sqrt{| x^*(Tx) |^2+\|Tx\|^4}~:(x,x^*)\in \Pi \right\}\\
&\leq & \sup \left\{ \sqrt{| x^*(Tx) |^2+\|T\|^4}~:(x,x^*)\in \Pi \right\}.
\end{eqnarray*}
Let $x\in S_{\mathcal{X}}.$ Let $F$ be a facet of $B_{\mathcal{X}}$ containing $x$. Let $\textit{ext}~B_{\mathcal{X}} \bigcap F =\{x_1,x_2,\ldots,x_m\}$. Then there exist scalars $\lambda_1,\lambda_2,\ldots,\lambda_m \geq 0$ with $\sum_{i=1}^m\lambda_i =1$ such that  $x= \sum_{i=1}^m\lambda_ix_i.$ Let $x^*$ be the supporting functional corresponding to the facet $F$. Then by Lemma \ref{lemma:extreme}, we have $x^*\in \textit{ext}~B_{\mathcal{X}^*}$ and  $x^*\in \mathcal{J}(x_i)$ for all $i=1,2,\ldots,m.$ Now,
\begin{eqnarray*}
|x^*(Tx)|&=&\left |\sum_{i=1}^m\lambda_i x^*(Tx_i) \right |\\
&\leq& \sum_{i=1}^m\lambda_i |x^*(Tx_i)|\\
&\leq& \max_{1\leq i\leq m} \left \{ |x^*(Tx_i)|\right\}.
\end{eqnarray*}
Thus, $|x^*(Tx)|^2+\|T\|^4\leq \max\left \{ |x^*(Tx_i)|^2: 1\leq i\leq m \right\}+\|T\|^4.$ This implies that
\begin{eqnarray*}
\sup \left\{ \sqrt{| x^*(Tx) |^2+\|T\|^4}~:(x,x^*)\in \Pi \right\}=\max \left\{ \sqrt{| x^*(Tx) |^2+\|T\|^4}~:(x,x^*)\in G \right\}.
\end{eqnarray*}
This completes the proof.
\end{proof}

\section{On the Davis-Wielandt index}\label{sec3}

\smallskip

\noindent We begin this section with the following inequalities for the Davis-Wielandt index and the numerical index of a normed linear space $\mathcal{X}$.

\begin{prop}\label{prop-bound eta}
Let $\mathcal{X}$ be a normed linear space. Then $$0\leq n(\mathcal{X})\leq 1\leq \eta_{dw}(\mathcal{X})\leq \sqrt{n^2(\mathcal{X})+1} \leq \sqrt{2}.$$
\end{prop}

\begin{proof}
The inequalities $0\leq n(\mathcal{X})\leq 1\leq \eta_{dw}(\mathcal{X})$ and $\sqrt{n^2(\mathcal{X})+1} \leq \sqrt{2}$ follows trivially. We only prove that $\eta_{dw}(\mathcal{X})\leq \sqrt{n^2(\mathcal{X})+1}.$ Let $T\in S_{\mathcal{L}(\mathcal{X})}.$ Then it follows from the definition of the Davis-Wielandt radius of $T$ that
\begin{eqnarray*}
dw(T)=\sup \left\{ \sqrt{| x^*(Tx) |^2+\|Tx\|^4}~:(x,x^*)\in \Pi \right\} \leq\sqrt{  w^2(T)+ 1 }.
\end{eqnarray*}
Taking infimum over all $T\in S_{\mathcal{L}(\mathcal{X})}$, we get $\eta_{dw}(\mathcal{X})\leq \sqrt{n^2(\mathcal{X})+1}.$
\end{proof}

\smallskip

\begin{remark}\label{remneq}
$(i)$ For the inequalities $ 0\leq n(\mathcal{X})\leq 1$ to be equalities (in separate cases), we refer the readers to the study of normed spaces with numerical index $0$ \cite{MMP1,MMP2} and normed spaces with numerical index $1$ \cite{M}, respectively.\\

\noindent $(ii)$ It follows from Proposition \ref{prop-bound eta} that if $n(\mathcal{X})=0$ then $\eta_{dw}(\mathcal{X})=1.$ However, $\eta_{dw}(\mathcal{X})=1$ does not always imply $n(\mathcal{X})=0$. As for example, we consider $\mathcal{H}$ to be the $n$-dimensional complex Hilbert space, where $n\geq 2$. Let $T=\left(\begin{array}{cc}
    0&1 \\
    0&0
	\end{array}\right) \bigoplus 0$, be the usual matrix representation of $T\in \mathcal{L}(\mathcal{H})$.  It is easy to see that $\|T\|=dw(T)=1$. Since $ \eta_{dw}(\mathcal{H})\geq 1$, it is now immediate that $\eta_{dw}(\mathcal{H})=1$. On the other hand, it is well known that $n(\mathcal{H})=\frac{1}{2}.$
\end{remark}

\noindent In our next result we prove that the following equivalent conditions for the Davis-Wielandt index of a finite-dimensional Banach space to be $\sqrt{2}$.

\begin{prop}\label{equivalent}
Let $\mathcal{X}$ be a finite-dimensional Banach space. Then the following conditions are equivalent:\\
$(i)$ $\eta_{dw}(\mathcal{X})=\sqrt{2}$.\\
$(ii)$ $n(\mathcal{X})=1.$\\
$(iii)$ $w(T)=\|T\|$ for all $T\in {\mathcal{L}(\mathcal{X})}.$\\
$(iv)$ $|x^*(x)|=1$ for every $x\in \textit{ext}~ B_{\mathcal{X}}$ and for every $x^*\in \textit{ext}~ B_{\mathcal{X}^*}.$
\end{prop}

\begin{proof}
\noindent The equivalence of $(ii)$ and $(iv)$ was proved by McGregor in \cite[Th. 3.1]{M}. The equivalence of $(ii) $ and $(iii) $ follows trivially. We only prove the equivalence of $(i) $ and $(iii) $.\\

\noindent $(i) \implies (iii)$: Let $\eta_{dw}(\mathcal{X})=\sqrt{2}$. Then $dw(T)\geq \sqrt{2}$ for all $T\in S_{\mathcal{L}(\mathcal{X})}.$ Therefore, $\sqrt{2} \leq dw(T)\leq \sqrt{w^2(T)+\|T\|^4} $ $\leq \sqrt{2}$ for all $T\in S_{\mathcal{L}(\mathcal{X})}.$ This implies that $w(T)=1$ when $\|T\|=1$. Hence, $w(T)=\|T\|$ for all $T\in {\mathcal{L}(\mathcal{X})}.$\\

\noindent $(iii) \implies (i)$: Let $w(T)=\|T\|$ for all $T\in {\mathcal{L}(\mathcal{X})}.$
Since $\mathcal{X}$ is finite-dimensional, $W_T\neq \emptyset$ for all $T\in {\mathcal{L}(\mathcal{X})}.$ Therefore, it follows from \cite[Th. 3.12]{SMBP} that $W_T \bigcap M_T \neq \emptyset.$ Let $x_0\in W_T \bigcap M_T.$ So, if $\|T\|=1$ then there exist $x^*_0\in \mathcal{J}(x_0)$ such that $w(T)=|x^*_0(Tx_0)|=1$.  Therefore, $dw(T)\geq \sqrt{|x_0^*(Tx_0)|^2+\|Tx_0\|^4}=\sqrt{2}$.  Thus, $dw(T)\geq \sqrt{2}$ for all $T\in S_{\mathcal{L}(\mathcal{X})}$. Hence, $\eta_{dw}(\mathcal{X})\geq \sqrt{2}.$ Also, it follows from Proposition \ref{prop-bound eta} that $\eta_{dw}(\mathcal{X})\leq \sqrt{2}.$  Therefore, $\eta_{dw}(\mathcal{X})=\sqrt{2}$.

\end{proof}

\begin{remark}
The equivalence of  $(i)$, $(ii)$ and $(iii)$ hold true even if $\mathcal{X}$ is infinite-dimensional. The implication $(iv) \implies (ii)$ may not hold true if $\mathcal{X}$ is infinite-dimensional. As for example, we note from \cite{VMR}  that  $\textit{ext}~B_{c_0(\ell_2)}=\emptyset$ but $n(c_0(\ell_2))<1.$
\end{remark}

\smallskip

\noindent For the inequality $\eta_{dw}(\mathcal{X})\leq \sqrt{n^2(\mathcal{X})+1}$ to be an equality, we note that a sufficient condition is $n(\mathcal{X})\in \{0,1\}$. However, there exist polyhedral Banach spaces $\mathcal{X}$, for which $n(\mathcal{X})\neq 0,1$, but the above inequality turns out to be an equality. Before discussing such examples, we refer the readers to \cite{SPBB}, where explicit computation of the exact numerical index of certain $3$-dimensional Banach spaces was done. Since the computation of the Davis-Wielandt index resembles the computation of the numerical index to be considerable extent, we will not repeat the calculations. However, to show that  $\eta_{dw}(\mathcal{X})= \sqrt{n^2(\mathcal{X})+1}$ for the following spaces, denoted in each case by $\mathcal{X}$, we follow the same strategy:\\

\noindent $(1)$ The value of $n(\mathcal{X})$ is known from \cite{SPBB}. We show that ${dw}(T)\geq \sqrt{n^2(\mathcal{X})+1},$ for each $T\in S_{\mathcal{L}(\mathcal{X})}.$ In particular, this implies that $\eta_{dw}(\mathcal{X})\geq \sqrt{n^2(\mathcal{X})+1}$.\\
$(2)$ We explicitly demonstrate an operator $T_0\in S_{\mathcal{L}(\mathcal{X})}$ for which ${dw}(T_0)= \sqrt{n^2(\mathcal{X})+1}.$  This combined with $(1)$ gives the desired equality.\\
\noindent In each case, the operator $T_0$ is chosen such that $w(T_0)=n(\mathcal{X})$, as given in \cite{SPBB}.

\smallskip
We state the following theorems that determine the exact value of  the Davis-Wielandt index of some special $3$-dimensional polyhedral Banach spaces, the proof of which is clear from   the above observations coupled with the results obtained in \cite{SPBB}.

\begin{theorem}\label{theorem:pyramid}

Let $\mathcal{X}$  be a $3$-dimensional polyhedral Banach space such that $B_{\mathcal{X}}$ is a polyhedron obtained by gluing two pyramids at the opposite base faces of a right prism having square base, with vertices $ \pm(1,1,1), \pm(-1,1,1), $   $ \pm(-1,-1,1), $ $ \pm(1,-1,1), \pm(0,0,2)$.  Then $$\eta_{dw}(\mathcal{X})= \frac{\sqrt{5}}{2}.$$
\end{theorem}

\smallskip

\begin{theorem}\label{theorem:prism pyramid}
Let $\mathcal{X}$ be a $3$-dimensional polyhedral Banach space such that $B_{\mathcal{X}}$ is a polyhedron with vertices $(\cos(j-1)\frac{\pi}{n}, \sin(j-1)\frac{\pi}{n},\pm{1}), (0,0,\pm 2)$, $ j \in \{1,2,\ldots,2n\}$, $ n\geq 3$. Then \[\eta_{dw}(\mathcal{X})= \begin{cases}
\sqrt{\sin^2\frac{\pi}{2n}+1}, &  \mbox{when n is odd}  \\
\sqrt{\tan^2\frac{\pi}{2n}+1}, &  \mbox{when n is even}.
\end{cases}
\]

\end{theorem}

\smallskip

\begin{theorem}\label{theorem:odd}
Let $\mathcal{X}$ be a $3$-dimensional polyhedral Banach space such that $B_{\mathcal{X}}$ is a right prism whose base is a regular polygon having $2n$ sides.  Then \[\eta_{dw}(\mathcal{X})= \begin{cases}
\sqrt{\sin^2\frac{\pi}{2n}+1}, &  \mbox{when n is odd}  \\
\sqrt{\tan^2\frac{\pi}{2n}+1}, &  \mbox{when n is even}.
\end{cases}
\]
\end{theorem}

\smallskip

\noindent We observe that Theorem \ref{theorem:odd} does not depend on the height $h$ (say) of the prism.  In particular, taking $ h=0$, we obtain the Davis-Wielandt index of the corresponding two-dimensional polyhedral Banach spaces. This fact is recorded in the following theorem.

\begin{theorem}\label{theorem:odd radius2}
Let $\mathcal{X}$  be a $2$-dimensional polyhedral Banach space such that $B_{\mathcal{X}}$ is a  regular $2n$-gon with vertices $\left (\cos(j-1)\frac{\pi}{n}, \sin(j-1)\frac{\pi}{n}\right)$,  $j \in \{1,2,\ldots,2n\}$. Then \[\eta_{dw}(\mathcal{X})= \begin{cases}
\sqrt{\sin^2\frac{\pi}{2n}+1}, &  \mbox{when n is odd}  \\
\sqrt{\tan^2\frac{\pi}{2n}+1}, &  \mbox{when n is even}.
\end{cases}
\]
\end{theorem}

\smallskip

\begin{remark}
Comparing the polyhedral Banach spaces considered in Theorem \ref{theorem:pyramid} and Theorem \ref{theorem:prism pyramid} with $n=3$, we conclude that non-isometric Banach spaces may have the same Davis-Wielandt index.
\end{remark}

\smallskip

\noindent In our next result we obtain a lower bound of the Davis-Wielandt index of polyhedral Banach spaces.

\begin{theorem}\label{theorem:estimate}
Let $\mathcal{X}$ be an $n$-dimensional polyhedral Banach space such that $\textit{ext}~B_{\mathcal{X}}=\{\pm v_i:i\leq i\leq m\}$. let $ F_{i1},F_{i2},$ $ \ldots,F_{in} $ be any $ n $ facets of $ B_{\mathcal{X}} $ meeting at $ v_i. $ Let $ f_{i1},f_{i2}, \ldots, f_{in} $ be  the $ n $ supporting functionals corresponding to the facets $ F_{i1},F_{i2},\ldots,F_{in} $, respectively. For each $ i=1,2,\ldots,m, $ let \[ \xi_i = \min_{x \in S_{\mathcal{X}}}\left\{ \max_{1 \leq r \leq n} \sqrt{|f_{ir}(x)|^2+1 } \right\}. \]
Then
\[\eta_{dw}(\mathcal{X}) \geq \min \{\xi_1,\xi_2,\ldots,\xi_m\}.\]
\end{theorem}

\begin{proof}
It is easy to observe that $ x \longrightarrow \max_{1 \leq r \leq n}\left\{\sqrt{|f_{ir}(x)|^2+1 } \right \} $, for each $ i=1,2,\ldots,m$,  is a continuous function from $S_{\mathbb{X}}$ to $\mathbb{R}$. Since $ S_{\mathbb{X}} $ is compact, $\xi_i= \min_{x \in S_{\mathcal{X}}}\left\{ \max_{1 \leq r \leq n} \sqrt{|f_{ir}(x)|^2+1 } \right\}$ exists. Let $ T \in S_{\mathcal{L}(\mathcal{X})} $ be arbitrary.  $ T $ must attain its norm at an extreme point of $ B_{\mathcal{X}}.$ Thus, there exists $ v_{j} $ such that $ \|Tv_{j}\|=\| T \|=1 $ for some $j \in \{1,2,\ldots,m\}.$ Therefore, we have
\begin{eqnarray*}
 dw(T) &=& \sup \left \{ \sqrt{|x^*(Tx)|^2+\|Tx\|^4}~:(x,x^*)\in \Pi \right \} \\
&\geq & \max_{1 \leq r \leq n} \left \{\sqrt{\vert f_{jr}(Tv_{j})\vert^2+1} \right \} \\
&\geq&\xi_{j}.
\end{eqnarray*}
This gives the desired inequality.
\end{proof}

\smallskip

\noindent As an illustrative application of the above theorem in estimating the Davis-Wielandt index of polyhedral Banach spaces, we consider a family of Banach spaces considered in \cite{MJ}. For each $\gamma \in (0,1),$ define a norm  $ {\Vert .\Vert}_\gamma  $ on $\mathcal{X}_{\gamma}= \{ (x,y,0) :  x,y\in \mathbb{R} \}$ as  \[{\Vert (x,y,0)\Vert}_\gamma=\max \left\{\vert y\vert, \vert x \vert+(1-\gamma)\vert y\vert \right\} ~~\forall~~(x,y,0)\in \mathbb{R}^3.\] The unit ball $B_{\mathcal{X}_{\gamma}}$ is a hexagon with vertices $\pm(\gamma,1,0),\pm(1,0,0),\pm(\gamma,-1,0)$.
We can estimate the Davis-Wielandt index of the $3$-dimensional polyhedral Banach space $ \mathcal{Y}_{\gamma} $ such that $B_{\mathcal{Y}_{\gamma}}$ is a right prism whose base is a hexagon $B_{\mathcal{X}_{\gamma}}.$
Let us remark that the proofs are omitted as they are lengthy and full of detailed calculations. We identify the extreme functionals of the unit ball of the dual space as the functionals corresponding to the facets of the unit ball of the space and then study the action of these functionals on the unit sphere to obtain a lower bound for the Davis-Wielandt index.

\begin{theorem}\label{theorem:hexagon}
Let $\mathcal{Y}_{\gamma}$ be a $3$-dimensional polyhedral Banach space such that $B_{\mathcal{Y}_{\gamma}}$ is a right prism whose base is a hexagon $B_{\mathcal{X}_{\gamma}}$ defined as above.  Then \[\eta_{dw}(\mathcal{Y}_{\gamma})\geq  \begin{cases}
\sqrt{\frac{1}{(3-2\gamma)^2}+1}, & \mbox{when}~~ 0<\gamma\leq\frac{1}{2} \\
\sqrt{(1-\gamma)^2+1}, & \mbox{when}~~ \frac{1}{2}\leq\gamma<1.
\end{cases}
\]
\end{theorem}

\smallskip

 \noindent Similarly, for $\xi \in (0,1)$ we consider $\mathcal{X}_{\xi}= \{ (x,y,0) : x,y \in \mathbb{R} \} $ and define a norm $ {\Vert .\Vert}_\xi$ on $\mathcal{X}_{\xi}$ as \[{\Vert (x,y,0)\Vert}_\xi=\max \left\{\vert x\vert,\vert y\vert,\frac{\vert x\vert+\vert y\vert}{1+\xi}\right\} ~~\forall~~(x,y,0)\in \mathbb{R}^3. \] The unit ball $B_{\mathcal{X}_{\xi}}$ is an octagon with vertices $\pm(1,\xi,0),\pm(1,-\xi,0),\pm(\xi,1,0),$ $ \pm(\xi,-1,0)$. We also state the following theorem without proof.
\begin{theorem}\label{theorem:octagon}
Let $\mathcal{Y}_{\xi}$ be a $3$-dimensional polyhedral Banach space such that $B_{\mathcal{Y}_{\xi}}$ is a right prism whose base is an octagon $B_{\mathcal{X}_{\xi}}$ defined as above. Then
\[\eta_{dw}(\mathcal{Y}_{\xi}) \geq
\begin{cases}
    \sqrt{\frac{1}{(2+\xi)^2}+1},& \text{when } \xi>\frac{1-\xi}{1+\xi} \\
    \sqrt{\left(\frac{1+\xi}{3+\xi}\right)^2+1},& \text{when } \xi\leq\frac{1-\xi}{1+\xi}.
\end{cases}
\]
\end{theorem}

\smallskip

\noindent We note that all the relevant results in Section \ref{sec3} remain unchanged in case of computing the modified Davis-Wielandt index of the corresponding spaces.\\

\noindent Computation of the exact Davis-Wielandt index of a given normed linear space seems to be a non-trivial problem in the general most case. We end the article with the following open question in this regard, that we could not answer:\\

\noindent Open question: Let $\mathcal{X}$ be a normed linear space. Find separate complete characterizations of the following properties:

$(i)$ $\eta_{dw}(\mathcal{X})=1,$

$(ii)$ $\eta_{dw}(\mathcal{X})= \sqrt{n^2(\mathcal{X})+1}.$

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TITLE: How to show that $0 \leq x \leq \frac{\epsilon}{1+\epsilon} \implies 1\leq \left(1+\frac{x}{n} \right)^{n}\leq \epsilon$
QUESTION [1 upvotes]: Show that
$\forall n\geq 1,\quad  \forall \epsilon > 0,\forall x \in \mathbb{R},$
$$0 \leq x \leq \frac{\epsilon}{1+\epsilon} \implies 1\leq \left(1+\frac{x}{n}  \right)^{n}\leq \epsilon$$

This question is related to that one Let $n\geq 1$ and $\alpha \in [0,1)$ show that : $1\le \bigl(1+\frac{\alpha}{n}\bigr)^{n}\le \frac{1}{1-\alpha}$
My thoughts:
Since $0 \leq x \leq \dfrac{\epsilon}{1+\epsilon}< 1$ (because of $\epsilon < \epsilon+1 \quad \forall \epsilon > 1 $) then by that question Let $n\geq 1$ and $\alpha \in [0,1)$ show that : $1\le \bigl(1+\frac{\alpha}{n}\bigr)^{n}\le \frac{1}{1-\alpha}$ we've :
$$1\le \left(1+\dfrac{x}{n}\right)^{n}\le \dfrac{1}{1-x}$$
Still we've to prove that :
$$\dfrac{1}{1-x}\le 1+\epsilon$$
Or we have $0 \leq x \leq \dfrac{\epsilon}{1+\epsilon} \implies 1-\dfrac{\epsilon}{1+\epsilon} \leq 1-x \leq 1 \implies \dfrac{1}{1+\epsilon} \leq 1-x \leq 1 $ which prove that last inequality

Is my reasoning correct or is there another ways to prove that ?

REPLY [1 votes]: Yes, it's correct, assuming the RHS is $1+\varepsilon$. Another way would be to observe that $\left(1+\frac xn\right)^n$ increases when you increase $x$, so you only have to prove it for $x=\varepsilon/(1+\varepsilon):$
$$\left(1+\frac{\varepsilon/(1+\varepsilon)}n\right)\le\frac1{1-\varepsilon/(1+\varepsilon)}=\frac{1+\varepsilon}{1+\varepsilon-\varepsilon}=1+\varepsilon$$