text:
\begin{document}

\author{Laurent Bartholdi}
\thanks{Y. de Cornulier and D. Kochloukova thank the ``Courant
  Research Centre'' of the Georg-August-Universit\"at zu G\"ottingen
  for supporting their visit to G\"ottingen in June 2010 and for
  hospitality during a visit in July 2014.}
\address{
L.B.: Mathematisches Institut\\
Georg-August-Universit\"at zu G\"ottingen\\
D-37073 G\"ottingen, Germany}

\author{Yves de Cornulier}
\thanks{Y. de Cornulier was supported by the
  Brazilian-French Network in Mathematics for the travel funding of
  his visit in Campinas in February, 2010}

\address{
Y.C.: D\'epartement de Math\'ematiques\\
Universit\'e Paris-Sud\\
F-91405 Orsay Cedex, France}

\author{Dessislava H. Kochloukova}
\thanks{D. Kochloukova is partially supported by ``bolsa de
  produtividade em pesquisa'' from CNPq, Brazil.}
\address{
D.K.: Department of Mathematics\\
State University of Campinas (UNICAMP)\\
CEP 13083-859 Campinas, SP, Brazil}

\date{November 24, 2014}

\title{Homological finiteness properties of wreath products}

\begin{abstract}
  We study the homological finiteness properties $FP_m$ of wreath
  products $\Gamma = H \wr_X G$. We show that, when $H$ has infinite
  abelianization, $\Gamma$ has type $FP_m$ if and only if both $G$ and
  $H$ have type $FP_m$ and $G$ acts (diagonally) on $X^i$
with  stabilizers of type $FP_{m-i}$ and with finitely many orbits for all $1
  \leq i \leq m$.

  If furthermore $H$ is torsion-free we give a criterion for $\Gamma$
  to be Bredon-$FP_m$ with respect to the class of finite subgroups of
  $\Gamma$.

  Finally, when $H$ has infinite abelianization and $\chi : \Gamma \to \mathbb{R}$ is a non-zero homomorphism with $\chi(H) = 0$, we classify when $[\chi]$ belongs to
  the Bieri-Neumann-Strebel-Renz invariant $\Sigma^m(\Gamma, \mathbb{Z})$.
\end{abstract}

\maketitle

\section{Introduction}

In this paper we consider the homological finiteness
property $FP_m$ of a group. By definition a group $G$ is of
type $FP_m$ if the trivial $\mathbb{Z} G$-module $\mathbb{Z}$
has a projective resolution with all projectives finitely
generated in dimensions up to $m$. If $G$ is
finitely presented and of type $FP_m$ for some $m \geq 2$
then $G$ is of homotopical type $F_m$ i.e. there is a
$K(G,1)$ with finite $m$-skeleton.

In general it is hard to determine the homological type $FP_m$ of
a group $G$. Even in the case of metabelian groups $G$ there is
no complete classification of the ones of type $FP_m$ for $m > 2$ though
there is an open conjecture, the $FP_m$-Conjecture, that relates the homological type $FP_m$ with the $m$-element subsets of the complement of the Bieri-Strebel invariant $\Sigma_A(Q)$ in the character sphere $S(Q)$ \cite{BieriGroves}. Later on the Bieri-Strebel invariant was generalised for any finitely generated group $G$ \cite{BieriRenz},\cite{BNS} and is widely  referred to as the Bieri-Neumann-Strebel-Renz invariant $\Sigma^1(G, \mathbb{Z})$.  Both properties $FP_2$ and $F_2$ (i.e. finite presentability)  coincide for a metabelian group $G$ and the $FP_2$-Conjecture holds \cite{BieriStrebel}. It is worth mentioning that for a general group $G$ the properties $FP_2$ and $F_2$
do not coincide \cite{BestvinaBrady}. Though there are some sufficient conditions, see \cite{Groves},
there is no
complete classification of finite presentability or type $FP_2$ even in the
class of  nilpotent-by-abelian groups.

Let $G$, $H$ be groups and $X$ be a $G$-set. The
(permutational) wreath product $\Gamma = H \wr_X G$
is defined to be the semidirect product $H^{(X)} \rtimes G$, where $H^{(X)}$ is the direct sum of $|X|$ copies of $H$ and $G$ acts on $H^{(X)}$  by permuting the summands.
The classification of the
finitely presented wreath products $\Gamma = H \wr_X G$ was
established in \cite{Yves}. It was shown that $\Gamma$ is finitely
presented if and only if both $H$ and $G$ are finitely
presented, $G$ acts  on $X$ with finitely generated stabilizers, and $G$ acts (diagonally) on
$X^2$ with finitely many orbits. We generalize this result by showing in Lemma \ref{m=2} when $G$ is of type $FP_2$. If $H$ has
infinite abelianization we give a
criterion for $\Gamma$ to be of type $FP_m$ for $m \geq 3$. The sufficiency of the conditions of
the criterion do not require that $H$ has infinite
abelianization, see Proposition \ref{easydirection}, but our proof of the necessity of the
conditions uses significantly the fact that $H$ has infinite
abelianization.

 Our main results are the following theorems. The second is
 a homotopy version of the first one. The proof of Theorem A is homological and Theorem B is an easy corollary of Theorem A and the fact that for $m \geq 2$
a group is of type $F_m$ if and only if it is of type $FP_m$ and is finitely presented.

\bigskip
{\bf Theorem A} {\it
Let $\Gamma = H \wr_X G$ be a wreath product, where  $X \not= \emptyset$ and $H$ has infinite abelianization.
Then the following are equivalent :

1. $\Gamma$  is of type
  $FP_m$;

2. $H$ is of type $FP_m$, $G$ is of type $FP_m$,
$G$ acts on $X^i$ with
  stabilizers of type $FP_{m-i}$ and with finitely many
orbits   for all $1 \leq i \leq m$.}

\bigskip

{\bf Theorem B}
{\it Let $\Gamma = H \wr_X G$ be a wreath product, where $ X \not= \emptyset$ and $H$ has infinite abelianization.
Then the following are equivalent :

1. $\Gamma$  is of type
  $F_m$;

2. $H$ is of type $F_m$, $G$ is of type $F_m$,
$G$ acts on $X^i$ with
  stabilizers of type $FP_{m-i}$ and with finitely many
orbits  for all $1 \leq i \leq m$.}

\bigskip
As a corollary we obtain some results about the Bredon-$FP_m$ type with respect to the class of finite subgroups, denoted  $\underline{FP}_m$. This requires that $H$ is torsion-free in order to control the conjugacy classes of finite subgroups in $\Gamma = H \wr_X G$. In general a group is of type $\underline{FP}_m$ if it has finitely many conjugacy classes of finite subgroups and the centralizer of every finite subgroup is of type $FP_m$ \cite{KropNucPer}. The homotopical counterpart of the property $\underline{FP}_{\infty}$ was studied earlier in \cite{Luck}.

\bigskip
{\bf Theorem C}
{\it Let  $\Gamma=H\wr_X G$  be a wreath
  product, where $ X \not= \emptyset$ and  $H$ be torsion-free, with
  infinite abelianization.   Then $\Gamma$ has type
  $\underline{FP}_m$ if and only if  the following
  conditions hold :

1. $G$ has type $\underline{FP}_m$;

2. $H$ has type $FP_m$;

3. for every finite subgroup $K$ of $G$ and every $ 1 \leq i
\leq m$ the centralizer
$C_G(K)$ acts on $(K \backslash X )^i$ with
stabilizers of type $FP_{m-i}$ and with finitely many orbits.}

\bigskip
{\it Remark.} If $\Gamma=H\wr_X G$  has type
  $\underline{FP}_m$, $X \not= \emptyset$ and $H$ has a non-trivial finite subgroup $K$ then $X$ is finite. Indeed since in $\Gamma$ there are only finitely many conjugacy classes of finite subgroups there is an upper bound of the order of the finite subgroups. On other hand  for every finite subset $X_i$ of $X$ there is a finite subgroup of $\Gamma$ isomorphic to $K^{X_i}$.

  \bigskip
Finally a monoid version of Theorem A is given in Theorem
\ref{sigmamain} for the monoid $\Gamma_{\chi} = \{ g \in
\Gamma | \chi(g) \geq 0 \}$, where $\chi : \Gamma \to
\mathbb{R}$ is a non-trivial character of $\Gamma$
such that $\chi(H) = 0$. This describes the  points $[\chi]$
of the Bieri-Neumann-Strebel-Renz invariant $\Sigma^m(\Gamma,
\mathbb{Z})$ with $\chi(H) = 0$ in terms of the invariant $\Sigma^m(G,
\mathbb{Z})$ and the action of $G_{\chi}$ on $X^i$ for $i
\leq m$.

For a finitely generated group $G$ the Bieri-Neumann-Strebel-Renz
invariants $\{ \Sigma^m(G, \mathbb{Z}) \}_{m \geq 1}$ were first defined in the case $m = 1$
 in \cite{BNS} and for general $m \geq 1$ were considered in \cite{BieriRenz}.  In general
the homological invariant
$\Sigma^m(G, \mathbb{Z})$ is an open subset of the unit sphere $S(G)$  and
$\Sigma^m(G, \mathbb{Z})$ determines which subgroups of $G$
above the commutator are of homological type $FP_m$ \cite{BieriRenz}.
The homological and homotopical $\Sigma^m$-invariants of a group are quite difficult to calculate but they are known for right-angled Artin groups \cite{MMV}, the R. Thompson group $F$ \cite{BGK}, metabelian groups of finite Pr\"ufer rank \cite{Meinert} and for split extensions metabelian groups if $m = 3$ \cite{H-K}.

\subsection*{Structure of the paper} In Section \ref{prel} we collect some basic properties of the homological type $FP_m$ for groups and modules. In section \ref{sectionsemiinduced} we classify the homological type of exterior and tensor powers of induced modules. In Section \ref{split} we classify when a group $G = M \rtimes G$ has type $FP_m$ provided $M$ is a  finitely generated induced $\mathbb{Z} G$-module. The results from Section \ref{sectionsemiinduced} and Section \ref{split} are applied in Section \ref{sectionwreath}, where we prove Theorem A and Theorem B. In Section \ref{sectionbredon} we prove Theorem C and in Section \ref{sigmasection} we prove a $\Sigma$-version of Theorem A.

\section{\boldmath Preliminaries on the homological type $FP_m$}  \label{prel}

Let $R$ be an associative ring with unity.
 We recall that an $R$-module $M$ is of type $FP_m$ if $M$ has a projective resolution with all projectives finitely generated up to dimension $m$. If not otherwise stated $\otimes$ is the tensor product over $\z$ and the modules and the group actions considered are left ones.

\begin{lemma}[{\cite[Prop.~1.4]{BieriQMWbook}}] \label{dimensionshifting1} Let $ 0 \to M_1 \to M  \to M_2 \to 0$ be a short exact sequence of $R$-modules.

(a) if $M$ and $M_2$ are of type $FP_s$ and $s \geq 1$ then $M_1$ is of type $FP_{s-1}$;

(b) if $M_1$ and $M_2$ are of type $FP_s$ then $M$ is of type $FP_s$;

(c) if $M_1$ is of type $FP_{s-1}$ and $M$ is of type $FP_s$ then $M_2$ is of type $FP_s$.
\end{lemma}

The following results follow easily from Lemma \ref{dimensionshifting1}. For completeness we include proofs.

\begin{lemma}   \label{dimensionshifting2} Let $$ \cdots \to Q_i \mapnew{d_i} Q_{i-1} \to \cdots \to Q_0 \mapnew{d_0} V \to 0$$ be an exact complex of $R$-modules with $Q_i$ of type $FP_{m-i}$ for all $i \leq m$. Then $V$ has type $FP_m$.
\end{lemma}

\begin{proof} Apply Lemma \ref{dimensionshifting1}(c) for the short exact sequences
$$
0 \to \im (d_{i+1}) = \ker (d_i) \to Q_i \to \im (d_i) \to 0
$$
for $0 \leq i \leq m-1$ to prove by inverse induction on $i$ that $\im (d_i)$ is $FP_{m-i}$.
\end{proof}

\begin{lemma}   \label{dimensionshifting3} Let $ 0 \leq k \leq m$ be integers and let $$ \cdots \to Q_i \mapnew{d_i} Q_{i-1} \to \cdots \to Q_0 \mapnew{d_0} V \to 0$$ be an exact complex of $R$-modules with $Q_i$ of type $FP_{m-i}$ for all $i < k$. Suppose further that $V$ has type $FP_m$. Then $\im (d_k)$ has type $FP_{m-k}$.
\end{lemma}

\begin{proof} Consider the exact complex
$$
0 \to \im (d_k) \to Q_{k-1} \mapnew{d_{k-1}} Q_{k-2} \to \cdots \to Q_0 \mapnew{d_0} V \to 0
$$
and apply Lemma 1(a) for the short exact sequences associated to the above complex.
\end{proof}

\begin{lemma}   \label{dimensionshifting4} Let
$${\mathcal Q} :  \  \  \cdots \to Q_i \mapnew{d_i} Q_{i-1} \to \cdots \to Q_0 \mapnew{d_0} V \to 0$$
be a complex of $R$-modules with $Q_i$ of type $FP_{m-i}$ for all $0 \leq i \leq m$ and $H_i({\mathcal Q})$ of type $FP_{m-i-1}$ for $ 0 \leq i \leq m-1$. Then $V$ has type $FP_m$.
\end{lemma}

\begin{proof} Apply Lemma \ref{dimensionshifting1} for the short exact sequences
$$
0 \to \ker (d_i) \to Q_i \to \im (d_i) \to 0
$$
and
$$
0 \to \im (d_{i+1}) \to \ker (d_i) \to H_i({\mathcal Q}) \to 0
$$ to prove by inverse induction on $i$ that $\im (d_i)$ is $FP_{m-i}$ for $0 \leq i \leq m$.
\end{proof}

The following lemma appeared as a footnote in \cite{Aberg} and was later on explained with more details in \cite[Prop. 4.1]{Bux}.

\begin{lemma} \label{retract} A retract of a group of type $FP_m$ is a group of type $FP_m$ i.e. if the split extension $\Gamma = M \rtimes G$ is of type $FP_m$ then $G$ is of type $FP_m$.
\end{lemma}

Let $\Gamma = M \rtimes G$ be a group. Then $\mathbb{Z} M$ is a left $\mathbb{Z} \Gamma$-module,
where $G$ acts via conjugation on $M$ and $M$ acts on $\z M$ via left multiplication. Note that the augmentation ideal $\Aug(\z M)$ is a $\mathbb{Z} \Gamma$-submodule of $\mathbb{Z} M$.

\begin{lemma} \label{condition} Let $\Gamma = M \rtimes G$
  be a group. Then $\Gamma$ is $FP_m$ if and only if $G$ is $FP_m$ and $\Aug (\mathbb{Z} M)$ is $FP_{m-1}$ as $\mathbb{Z} \Gamma$-module.
\end{lemma}

\begin{proof} By Lemma \ref{dimensionshifting1} applied to the short exact sequence $0 \to \Aug (\z \Gamma) \to \z \Gamma \to \z \to 0$  the group $\Gamma$ is of type $FP_m$ if and only if $\Aug(\z \Gamma)$ is $FP_{m-1}$ as $\z \Gamma$-module.
Consider the short exact sequence of $\z \Gamma$-modules
\begin{equation} \label{ses1}
0 \to \z \Gamma \Aug (\z G) \mapnew{\alpha} \Aug(\z \Gamma) \to \Aug(\z M) \to 0,
\end{equation}
where $\alpha$ is the inclusion map.

If $\Gamma$ is of type $FP_m$ by Lemma \ref{retract} $G$ is of type $FP_m$, so $\Aug(\z G)$ is $FP_{m-1}$ as $\z G$-module and the induced $\z \Gamma$-module  $\z \Gamma \Aug (\z G)$ is $FP_{m-1}$. Then by Lemma 1(c) applied to the short exact sequence
(\ref{ses1}) $\Aug(\z M)$ is $FP_{m-1}$ as $\z \Gamma$-module.

If $G$ is of type $FP_m$ and $\Aug(\z M)$ is $FP_{m-1}$ as $\z \Gamma$-module then by Lemma \ref{dimensionshifting1}(b) applied to the short exact sequence (\ref{ses1}) $\Aug (\z \Gamma)$ is $FP_{m-1}$ as $\z \Gamma$-module.
\end{proof}

\begin{lemma} \label{tensor9} Let $S$ be a subring of $R$ such that $R$ is
  flat as right $S$-module and $M$ be a left $S$-module. Furthermore assume that the
  inclusion map $S \to R$ of right $S$-modules splits.  Then $M$ is of type
  $FP_m$ as $S$-module  if and only if $R \otimes_S M$ is
  $FP_m$ as $R$-module.
\end{lemma}

\begin{proof}

1. Suppose that $M$ is of type
  $FP_m$ as $S$-module.
Let ${\mathcal F}$ be  a projective
  resolution of $M$ with projectives  finitely generated
   in dimensions $\leq m$.
Since $R \otimes_S - $ is an exact
  functor
$R \otimes_S {\mathcal F}$ is a projective resolution of the $R$-module $R
\otimes_S M$ with projectives finitely generated in
dimensions $\leq m$, so  $R \otimes_S M$ is
  $FP_m$ as $R$-module.

2. Suppose that  $R \otimes_S M$ is
  $FP_m$ as $R$-module. We will prove by induction on $m$ that $M$
  is $FP_m$ as $S$-module.

Suppose first that $m = 0$. Let $X = \{ a_i = \sum_{j}
r_{i,j} \otimes m_{i,j} \}_{i}$
be a finite generating set of  $R \otimes_S M$ as $R$-module,  where $r_{i,j} \in R, m_{i,j} \in M$. Let
$M_0$ be the $S$-submodule of $M$ generated by the finite set $Y = \{
m_{i,j} \}_{i,j}$. Since $R$ is flat as $S$-module
there is a short exact sequence $0 \to R \otimes_S M_0 \mapnew{\alpha}
R \otimes_S M \to  R \otimes_S (M/M_0) \to 0$
and by the choice of $Y$ the map $\alpha$ is surjective, so
$\alpha$ is an
isomorphism. Then $R \otimes_S (M/M_0) = 0$ has a direct
summand $S \otimes_S (M/M_0) = M / M_0$, so $M_0 = M$.

Suppose now that $M$ is $FP_{m-1}$ as $S$-module. Let
$$
{\mathcal F} : \cdots \to F_i \mapnew{d_i} F_{i-1} \to \cdots \to
F_0 \to M \to 0
$$
be a projective resolution of $S$-modules with $F_i$
finitely generated for $i \leq m-1$. Consider the exact
complex induced by ${\mathcal F}$
$$
{\mathcal R} : 0 \to A = \ker (d_{m-1}) \to F_{m-1}
\mapnew{d_{m-1}}  F_{m-2} \to \cdots \to F_0 \to M \to 0.
$$
Since $R$ is flat as right $S$-module
$$R \otimes_S {\mathcal R}  :    0 \to R \otimes_S A
\to R \otimes_S F_{m-1}
\mapnew{id_R \otimes d_{m-1}}  R \otimes_S F_{m-2} \to \cdots \to R
\otimes_S F_0 \to R \otimes_S M \to 0
$$ is an
exact complex and $R \otimes_S F_i$ is a  finitely generated
projective $R$-module for $i \leq m-1$. By Lemma \ref{dimensionshifting3} since $R \otimes_S M$ is $FP_m$ as
$R$-module we have that $R \otimes_S A $ is finitely
generated as $R$-module. Then by the case $m = 0$ we deduce
that $A$ is finitely generated as $S$-module, so $M$ is
$FP_m$ as $S$-module.
\end{proof}

\begin{corollary} \label{inductionFPm} Let $L$ be a subgroup
  of a group $G$ and $A$ be a (left) $\z L$-module. Then
 $\z G \otimes_{\z L} A$ is $FP_m$ as $\z G$-module if and only if $A$ is $FP_m$ as $\z L$-module.
\end{corollary}

\section{Exterior and tensor powers of induced modules} \label{sectionsemiinduced}

Let $G$ be a group and $S$ be a free $\mathbb{Z}$-module with basis $X \not= \emptyset$. If $S$ is a $\mathbb{Z} G$-module via an action of  $G$ on the set
$X$, we call $S$ an induced module. If $S$ is a $\mathbb{Z} G$-module via an action of  $G$ on the set
$X \cup - X$ we say that $S$ is a semi-induced module, in this case we write $S = \mathbb{Z}_t X$ with index $t$ standing for twisted action of $G$.

\begin{lemma} \label{twisting} Let $S= \mathbb{Z}_t X$ be a finitely generated semi-induced $\mathbb{Z} G$-module. Then

(a) $S \simeq \bigoplus_{i \in I} \mathbb{Z} G \otimes_{\mathbb{Z} K_i} V_i$ as $\mathbb{Z} G$-modules where $I$ is finite, for each $i \in I$ there is an element $x_i \in X$ with stabilizer $G_i$ in $G$ such that $K_i$ is a subgroup of $G$ containing $G_i$ such that $[K_i : G_i] = d_i \in \{ 1,2\}$ and each $V_i$ as an abelian group is $\mathbb{Z}$, $G_i$ acts trivially on $V_i$ and $K_i$ acts non-trivially on $V_i$ precisely when $d_i = 2$.

(b) $S$ is of type $FP_m$ over $\mathbb{Z} G$ if and only if all the groups $G_i$ are of type $FP_m$ for $i \in I$.
\end{lemma}

\begin{proof} (a) Observe first that $S$ splits as a finite direct sum of cyclic semi-induced $\mathbb{Z} G$-modules, each generated by an element of $X$. Thus from the very beginning we may assume that $S$ is a cyclic semi-induced $\mathbb{Z} G$-module with a generator $w$ an element from $X$. If $S$ is an induced $\mathbb{Z} G$-module we are done, so assume that
 $S$  is not an induced $\mathbb{Z} G$-module.
Then there are $g_1, g_2$ in $G$ such that $g_1 w = - g_2 w$, so for  $t = g_2^{-1} g_1$ we have $t w = - w$. Then $t^2 w = t(-w) = - t w = -(-w) = w$, so $t^2 \in G_0$, where $G_0$ is the stabilizer of $w$ in $G$. For $g_0 \in G_0$ we have $g_0 t w = - g_0 w = - w = t w$, so $t^{-1} g_0 t \in G_0$. Then for the subgroup $K_0$ of $G$ generated by $G_0$ and $t$ we have that $G_0$ is a normal subgroup of $K_0$ of index 2.

Observe that if $h_1, h_2 \in G$ and $h_1 w = - h_2 w$ then for $t_1 = h_2^{-1} h_1$ we have $t_1 w = -w = t w$, so $t_1^{-1} t \in G_0$. Thus $h_1 w = - h_2 w$ is a corollary of $t w = - w$ so $S \simeq (\mathbb{Z} G \otimes_{\mathbb{Z} G_0} \mathbb{Z}) / \mathbb{Z} G (t G_0 + G_0) \simeq \mathbb{Z} G \otimes_{\mathbb{Z} K_0} V_0$, where $V_0 \simeq (\mathbb{Z} K_0 \otimes_{\mathbb{Z} G_0} \mathbb{Z}) / \mathbb{Z} K_0 (G_0 + t G_0)$ has underlying abelian group $\mathbb{Z}$ and $G_0$ acts trivially and $K_0/ G_0$ changes the sign. This completes the proof of item (a).

(b) Observe that since $S \simeq \bigoplus_{i \in I} \mathbb{Z} G \otimes_{\mathbb{Z} K_i} V_i$  we have that $S$ is of type $FP_m$ if and only if
$ \mathbb{Z} G \otimes_{\mathbb{Z} K_i} V_i$ is $FP_m$ as $\mathbb{Z} G$-module for every $i \in I$. By Corollary \ref{inductionFPm} $ \mathbb{Z} G \otimes_{\mathbb{Z} K_i} V_i$ is $FP_m$ as $\mathbb{Z} G$-module precisely when $V_i$ is of type $FP_m$ over $\mathbb{Z} K_i$. If $K_i = G_i$ we have that $V_i$ is the trivial $\mathbb{Z} G_i$-module, so $V_i$ is of type $FP_m$ over $\mathbb{Z} K_i$ precisely when $G_i = K_i$ is $FP_m$ as a group. If $K_i / G_i$ is the cyclic group of order 2 then $V_i$ is the sign module i.e.\ $G_i$ acts trivially on $V_i = \mathbb{Z}$ and $K_i/ G_i$ acts via sign change.  Since $G_i$ has finite index in $K_i$ we have that $V_i$ is $FP_m$ as $\mathbb{Z} K_i$-module precisely when $V_i$ is $FP_m$ as $\mathbb{Z} G_i$-module, but as $\mathbb{Z} G_i$-module $V_i$ is the trivial one, so $V_i$ is $FP_m$ as $\mathbb{Z} G_i$-module is the same as $G_i$ is $FP_m$.
\end{proof}

 The following proposition is the main result of this section. It classifies the homological type of tensor and exterior powers of induced modules.

\begin{proposition} \label{wedgeproducts} Let $M= \mathbb{Z} X$ be an induced $\mathbb{Z} G$-module. Then the following are equivalent :

1.  $\bigwedge^i M$ is of type $FP_{m-i}$ for all $1 \leq i \leq m$;

2.   $\bigotimes^i M$ is of type $FP_{m-i}$ for all $1 \leq i \leq m$;

3. all stabilizers of the diagonal action of $G$ on $X^i$   are of type $FP_{m-i}$  and
 $\otimes^i M$ is finitely generated as  $\mathbb{Z} G$-module for all $1 \leq i \leq m$;

4. all stabilizers  of $i$ element subsets of $X$  are of type $FP_{m-i}$  and
 $\wedge^i M$ is finitely generated as  $\mathbb{Z} G$-module for all $1 \leq i \leq m$;

5.  all stabilizers of the diagonal action of $G$ on $X^i$  are of type $FP_{m-i}$  and $G \backslash X^i $ is finite for all $1 \leq i \leq m$.

\end{proposition}

\begin{proof}
Observe that $\wedge^i M = \z_t Y_i$ is an semi-induced $\mathbb{Z} G$-module, where $Y_i$ is the set of $i$-element subsets of $X$ and $\otimes^i M = \z Z_i$ is an induced $\mathbb{Z} G$-module, where $Z_i = X^i$.

It is obvious that conditions 2,3 and 5  are equivalent.

Note that the stabilizer of $(x_1, \ldots, x_i) \in X^i$ in $G$ (via the diagonal action)  is  commensurable with the stabilizer of the set $\{ x_1, \ldots, x_i \}$ in $G$, in particular one of the stabilizers is $FP_{m-i}$ precisely when the other is $FP_{m-i}$.
 Furthermore $\wedge^i M$ is finitely generated as $\mathbb{Z}
 G$-module for all $i \leq m$ if and only if $G$ acts on the set of $i$ element subsets
 of $X$ with finitely many orbits for all $i \leq m$. The last is equivalent with  $G$ acts on $X^i$ with finitely many orbits for all $i \leq m$.
Thus conditions 5 and 4 are equivalent, and by Lemma \ref{twisting}(b) conditions 1 and 2 are equivalent too.
\end{proof}

\begin{lemma} \label{wedgeproducts3}
Let $X$ be a (left) $G$-set such that  $G$ acts  with finitely many orbits on $X^i$ and
with stabilizers  of type $FP_{m-i}$ for all $1 \leq i \leq m$.
Let $i_1, i_2, \ldots, i_s$ be non-negative integers such that $i_1 + \cdots + i_s = r \leq m$
 and
$X^{(i_1,i_2, \ldots,i_s)}
$ be the set of $s$-tuples $(X_1,
\ldots, X_s)$ of
pairwise disjoint subsets $X_1, \ldots , X_s$ of $X$ such
that $X_j$ has cardinality $i_j$ for all $1 \leq j \leq s$.
Then $\mathbb{Z} X^{(i_1,i_2, \ldots,i_s)}$ is of type $FP_{m-r}$ as $\mathbb{Z} G$-module.
\end{lemma}

\begin{proof}
Let $B$ be the stabilizer in $G$ of the $s$-tuple $(X_1,
\ldots, X_s)$ i.e. $B$ is the set of $g \in G$ such that $g X_i = X_i $ for all $ 1 \leq i \leq s$. Then since all $X_i$ are finite $B$ is a subgroup of finite index in the group $D$, where $D$ is  the stabilizer  in $G$ of the set $\bigcup_{1 \leq i \leq s} X_i$. Note that $| \bigcup_i X_i | = i_1 + \cdots + i_s = r$. Then by Proposition \ref{wedgeproducts} (i.e. item 5 implies item 4), $D$ is of type $FP_{m-r}$, hence $B$ is of type $FP_{m-r}$.
Note that $G \backslash X^r $ finite implies that $G \backslash X^{(i_1,i_2, \ldots,i_s)}$ is finite. Then by Lemma \ref{twisting} the  $\mathbb{Z} G$-module $\mathbb{Z} X^{(i_1,i_2, \ldots,i_s)}$ is of type $FP_{m-r}$.
\end{proof}

\section{\boldmath Type $FP_m$ for split extensions of induced modules} \label{split}

Let $\Gamma = M \rtimes G$ be a group, where $M$ is abelian.
Note we do
not impose restrictions on $G$. We consider $M$ as a left $\z G$-module, where $G$ acts by conjugation. In this section $M$ is an induced finitely generated (left) $\mathbb{Z} G$-module i.e. \begin{equation} \label{directsum} M = \bigoplus_{i \in I} \mathbb{Z} G  \otimes_{\mathbb{Z} H_i}\mathbb{Z} \simeq \bigoplus_{i \in I} \mathbb{Z} [G/ H_i] = \mathbb{Z} X,
 \end{equation}
where $X = \bigcup_{i \in I} G/ H_i$,
 $I$ is a finite set and $\{ H_i \}_{i \in I}$ are the stabilizers of the action of $G$ on $X$. In this section we classify when a group $\Gamma = M \rtimes G$ is of type $FP_m$ provided $M$  is an induced finitely generated $\mathbb{Z} G$-module.

\begin{proposition} \label{free1}  Let $G$ be a group of
  type $FP_m$ and $M$ be a finitely generated induced $\mathbb{Z}[G]$-module such that  $\wedge^i M$ be of type $FP_{m-i}$ as $\mathbb{Z} G$-module via the diagonal $G$-action for all $1 \leq i \leq m$.  Then $\Gamma = M \rtimes G$ is of type $FP_m$.
\end{proposition}

\begin{proof}
Let $X$ be the disjoint union $\bigcup_i G/ H_i$, so $X$ is a
basis of $M$ as a free $\mathbb{Z}$-module.
Since $M$ is a torsion-free abelian group, there is a Koszul complex \cite{Weibel}
\begin{equation} \label{abelian1}
\cdots \to \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^k M) \mapnew{d_k}  \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^{k-1} M) \to \cdots \to \mathbb{Z} M \otimes_{\mathbb{Z}} M \to \mathbb{Z} M \to \mathbb{Z} \to 0
\end{equation}
with differential given by
$$
d_k(m_1 \wedge \cdots \wedge m_k) = \sum_{1 \leq i \leq k} (-1)^i \epsilon(m_i) \otimes m_1 \wedge \cdots \wedge m_{i-1} \wedge m_{i+1} \wedge \cdots \wedge m_k,
$$
where $\epsilon(m_i) = m_i - 1 \in \z M$ and $m_1, \ldots,
m_k \in X$.
Thus (\ref{abelian1}) gives an exact complex
\begin{equation} \label{abelian1new}
{\mathcal Q} : \cdots \to Q_{k-1} = \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^k M) \mapnew{d_k}  Q_{k-2} = \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^{k-1} M) \to \cdots $$ $$ \to Q_0 = \mathbb{Z} M \otimes_{\mathbb{Z}} M \to Q_{-1} = \Aug(\mathbb{Z} M) \to 0.
\end{equation}
Note that for a left $\mathbb{Z} G$-module $V$ there is an isomorphism of abelian groups
$$
\mathbb{Z} \Gamma \otimes_{\mathbb{Z} G } V \simeq \mathbb{Z} M \otimes_{\mathbb{Z}} V
$$
and thus the action of $\Gamma$ on the induced module $\mathbb{Z} \Gamma \otimes_{\mathbb{Z} G } V $ gives an action of $\Gamma$ on $\mathbb{Z} M \otimes_{\mathbb{Z}} V$.
We apply this for $V = \wedge^k M$ and since $\wedge^k M$ is of type $FP_{m-k}$ as $\mathbb{Z} G$-module and
$\mathbb{Z} \Gamma \otimes_{\mathbb{Z} G }$ is an exact functor
$$
\mathbb{Z} \Gamma \otimes_{\mathbb{Z} G } (\wedge^k  M) \simeq \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^k M)
$$
is of type $FP_{m-k}$ as $\mathbb{Z} \Gamma$-module.
Note that since $X$ is a $G$-invariant set the differentials in (\ref{abelian1new}) are homomorphisms of $\z G$-modules, hence are homomorphisms of $\z \Gamma$-modules.
Then (\ref{abelian1new}) is an exact complex with $Q_i$ of homological type $FP_{m-i-1}$ as $\mathbb{Z} \Gamma$-module for $0 \leq  i \leq m-1$. Then by Lemma \ref{dimensionshifting2} applied to the complex  (\ref{abelian1new}) $\Aug (\mathbb{Z} M)$ is of type $FP_{m-1}$ as $\mathbb{Z} \Gamma$-module, hence by Lemma \ref{condition} $\Gamma$ is of type $FP_m$.
\end{proof}

\begin{lemma} \label{free2} Suppose that $\Gamma = M \rtimes
  G$ is a group of type $FP_m$ and $M$ is a finitely generated induced  $\mathbb{Z}[G]$-module such that $\wedge^i M$ is of type $FP_{m-i-1}$ as $\mathbb{Z} G$-module (via the diagonal $G$-action) for all $1 \leq  i \leq m -1$. Then  $M$ is of type $FP_{m-1}$ as $\mathbb{Z} G$-module.
\end{lemma}

\begin{proof}  By Lemma \ref{condition} $\Omega = \Aug(\mathbb{Z} M)$ is of type $FP_{m-1}$ as $\mathbb{Z} \Gamma$-module. Note that $M \simeq \Omega / \Omega^2 \simeq \mathbb{Z} \otimes_{\mathbb{Z} M} \Aug (\mathbb{Z} M)$. Let
$$
{\mathcal F} : \cdots \to F_i \to F_{i-1} \to \cdots \to F_0 \to \Omega \to 0
$$
be a free resolution of $\Omega$ as $\mathbb{Z} \Gamma$-module with $F_i$ finitely generated for $i \leq m-1$.
Let ${\mathcal Q}$ be the complex $\mathbb{Z} \otimes_{\mathbb{Z} M} {\mathcal F}$ i.e.
 \begin{equation} \label{complexQ}
{\mathcal Q} : \cdots \to Q_i \to Q_{i-1} \to \cdots \to Q_0 \to M \to 0
\end{equation}
is a complex of free $\mathbb{Z} G$-modules with $Q_i$ finitely generated for $i \leq m-1$ and its homology groups
are $$
H_i({\mathcal Q}) \simeq \Tor_{i}^{\mathbb{Z} M}(\mathbb{Z}, \Omega) \hbox{ for } i \geq 1.
$$
Note that the long exact sequence in $\Tor$ applied to the short exact sequence $\Omega \to \mathbb{Z} M \to \mathbb{Z}$ gives
$$
\Tor_{i}^{\mathbb{Z} M}(\mathbb{Z}, \Omega) \simeq \Tor_{i+1}^{\mathbb{Z} M}(\mathbb{Z}, \mathbb{Z}) = H_{i+1}(M, \mathbb{Z}) \simeq \wedge^{i+1} M \hbox{ for } i \geq 1.
$$
The last isomorphism is \cite[Ch.~V,~Thm.~6.4]{Brownbook}.
Then $H_i({\mathcal Q})$ is of type $FP_{m-i-2}$ as $\mathbb{Z} G$-module for all $ i \leq m-2$. Then by Lemma \ref{dimensionshifting4} applied to the complex (\ref{complexQ})
$M$ is of type $FP_{m-1}$ as $\mathbb{Z} G$-module.
\end{proof}

\begin{proposition} \label{free3} Let $\Gamma = M \rtimes G$
  be a group of type $FP_m$ and
$M$ be a finitely generated induced $\mathbb{Z}[G]$-module. Then $\wedge^i M$ is of type $FP_{m-i}$ as $\mathbb{Z} G$-module via the diagonal $G$-action for all $1 \leq i \leq m$.
\end{proposition}

\begin{proof} We induct on $m$ and assume that the proposition holds for smaller values of $m$, in particular $\wedge^i M$ is $FP_{m-i-1}$ as $\mathbb{Z} G$-module. By Lemma \ref{free2} $M$ is of type $FP_{m-1}$ as $\mathbb{Z} G$-module, thus the proposition holds for $i = 1$. We proceed by induction on $i$ i.e. assume that we have proved  that $\wedge^j M$ is of type $FP_{m-j}$ as $\mathbb{Z} G$-module via the diagonal $G$-action for all $1 \leq j \leq i-1$ and will  show that
$\wedge^i M$ is of type $FP_{m-i}$ as $\mathbb{Z} G$-module.

Consider the Koszul complex
\begin{equation} \label{abelian2}
\cdots \to \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^k M) \mapnew{d_k}  \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^{k-1} M) \to \cdots \to \mathbb{Z} M \otimes_{\mathbb{Z}} M \mapnew{d_1} \mathbb{Z} M \to \mathbb{Z} \to 0
\end{equation}
and its modified version
\begin{equation} \label{abelian*new}
{\mathcal Q} : \cdots \to Q_{k-1} = \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^k M) \mapnew{d_k}  Q_{k-2} = \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^{k-1} M) \to \cdots $$ $$\to Q_0 = \mathbb{Z} M \otimes_{\mathbb{Z}} M \mapnew{d_1} Q_{-1} = \Aug(\mathbb{Z} M) \to 0
\end{equation}
Since $\wedge^k M$ is $FP_{m-k}$ as $\mathbb{Z} G$-module for $k \leq i-1$ the induced module
$Q_{k-1} = \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^k M)$ is $FP_{m-k}$ as $\mathbb{Z} \Gamma$-module for $k \leq i-1$. Note that by Lemma \ref{condition} $\Aug (\z M)$ is $FP_{m-1}$ as $\z \Gamma$-module.
Then by Lemma \ref{dimensionshifting3} applied to the complex (\ref{abelian*new}) $\im (d_i)$ is of type $FP_{m-1-(i-1)}$ as $\mathbb{Z} \Gamma$-module.
Note that
$$\im (d_{i}) \simeq (\mathbb{Z} M \otimes (\wedge^i M)) / \ker (d_{i}) \simeq
(\mathbb{Z} M \otimes (\wedge^i M)) / \im (d_{i+1}).$$
Denote $(\mathbb{Z} M \otimes (\wedge^i M)) / \im (d_{i+1})$ by $V$ i.e. $V$ has type $FP_{m-i}$ as $\z \Gamma$-module.
Let $W$ be $\mathbb{Z} \otimes_{\mathbb{Z} M} V$. Since $ \mathbb{Z} \otimes_{\mathbb{Z} M} -$ is right exact there is
an exact sequence
$$
\mathbb{Z} \otimes_{\mathbb{Z} M} \im (d_{i+1}) \mapnew{\beta} \mathbb{Z} \otimes_{\mathbb{Z} M} (\mathbb{Z} M \otimes (\wedge^{i} M)) \to W \to 0.$$
By the definition of $d_{i+1}$ we have $\beta = 0$, furthermore the module in the middle is isomorphic to $\wedge^i M$, so
$$
W \simeq \wedge^i M.$$ If $W$ is of type $FP_{m-i}$ as $\mathbb{Z} G$-module the inductive step is completed.

Let
$$
{\mathcal F} : \cdots \to F_j \to F_{j-1} \to \cdots \to F_0 \to V \to 0
$$
be a free resolution of $V$ as $\mathbb{Z} \Gamma$-module with $F_j$ finitely generated for $j \leq m-i$.
  Let ${\mathcal R}$ be the complex $\mathbb{Z} \otimes_{\mathbb{Z} M} {\mathcal F}$ i.e.
 $$
{\mathcal R} : \cdots \to R_j \to R_{j-1} \to \cdots \to R_0 \to W \to 0
$$
is a complex of free $\mathbb{Z} G$-modules with $R_j$ finitely generated for $j \leq m-i$ and its homology groups are
$$
H_j({\mathcal R}) \simeq \Tor_{j}^{\mathbb{Z} M}(\mathbb{Z}, V) \hbox{ for } j \geq 1.
$$
Note that for the $\z M$-module  $V = \im (d_i)$ the Koszul complex (\ref{abelian2}) gives a free resolution
\begin{equation}
{\mathcal C} : \cdots \to \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^k M) \mapnew{d_k}  \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^{k-1} M) \to \cdots \to $$ $$\mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^{i+1} M) \to  \mathbb{Z} M \otimes_{\mathbb{Z}} (\wedge^{i} M) \mapnew{d_i} V \to 0
\end{equation}
and by the definition of the differentials $d_k$ the complex
\begin{equation}  \mathbb{Z} \otimes_{\mathbb{Z} M} {\mathcal C} :
 \cdots \to  \wedge^{i+2} M \to  \wedge^{i+1} M \to  \wedge^{i} M \mapnew{\alpha} \mathbb{Z} \otimes_{\mathbb{Z} M} \im (d_i) = W \to 0
\end{equation}
has all zero differentials except possibly $\alpha$.
Thus
$$
H_j({\mathcal R}) \simeq \Tor_{j}^{\mathbb{Z} M}(\mathbb{Z}, V) \simeq H_j(\mathbb{Z} \otimes_{\mathbb{Z} M} {\mathcal C}) \simeq \wedge^{j+i} M  \hbox{ for } j \geq 1.
$$
By Lemma \ref{dimensionshifting4} applied to the complex ${\mathcal R}$, for $W$ to be of type $FP_{m-i}$ as $\z G$-module it is sufficient that
$$H_j({\mathcal R}) \hbox{ is of type } FP_{m-i-j-1} \hbox{ as } \mathbb{Z} G\hbox{-module for } 1 \leq j \leq m-i-1.$$ Thus we need for $k = i+j$ that
$$\wedge^k M \hbox{ is of type } FP_{m-k-1} \hbox{ as } \mathbb{Z} G\hbox{-module for } 1 \leq k \leq m-1$$
but this follows from the fact that the proposition holds for $m-1$.
\end{proof}

The following theorem is the main result of this section and classifies when  a group $\Gamma = M \rtimes G$ is of type $FP_m$ provided $M$  is an induced finitely generated $\mathbb{Z} G$-module.

\begin{theorem} \label{mainthm1}  Let $\Gamma = M \rtimes G$
  be a group and $M$ be a finitely generated induced $\mathbb{Z}[G]$-module. Then
  $\Gamma$ is of type $FP_{m}$ if and only if $G$ is of type
  $FP_m$ and $\wedge^i M$ is of type $FP_{m-i}$ as $\z
  G$-module via the diagonal action for all $1 \leq i \leq m$.
\end{theorem}

\begin{proof} Follows directly from Lemma \ref{retract}, Proposition \ref{free1} and Proposition \ref{free3}.
\end{proof}

\section{\boldmath Homological type $FP_m$ for wreath products} \label{sectionwreath}

\subsection{Preliminaries on tensor products of complexes} \label{sectionKunneth}
Let $({\mathcal A}, \partial)$ be a non-negative complex of free
$\z$-modules i.e. all non-zero modules are in
dimension $\geq 0$. Let ${\mathcal A}_i = {\mathcal A}$ for all $1 \leq i \leq s$.
Consider the tensor product ${\mathcal A}_1 \otimes
\cdots \otimes {\mathcal A}_s$ with
differential  given by
\begin{equation} \label{differential01}
d(a_1 \otimes \cdots \otimes a_s) =
\sum_{1 \leq j \leq s}
(-1)^{\deg(a_1) + \cdots + \deg(a_j)} a_1 \otimes \cdots
\otimes \partial(a_j)\otimes \cdots \otimes a_s.
\end{equation}
For the transposition $\pi = (i, i+1) \in S_s$ and each
$a_i$ an element of one of the free modules in ${\mathcal A}_i$ we define
\begin{equation} \label{symmetricgroup}
\pi(a_1 \otimes \cdots \otimes a_s) := (-1)^{\deg(a_i) \deg(a_{i+1})} (a_1 \otimes \cdots \otimes  a_{i+1} \otimes a_i \otimes \cdots \otimes a_s).
\end{equation}
Since the symmetric group $S_s$ is generated by $\{ (i, i+1) \}_{1 \leq i \leq
  s-1}$ we get an action of $S_s$ on
the tensor product ${\mathcal A}_1 \otimes
\cdots \otimes {\mathcal A}_s$. As the action of the symmetric
group $S_s$ commutes with the differential from (\ref{differential01}), this induces an
action of $S_s$ on the homology groups of ${\mathcal A}_1 \otimes
\cdots \otimes {\mathcal A}_s$.

\subsection{Wreath products : proofs of Theorem A and Theorem B}

In this section we classify when a wreath product  $\Gamma = H \wr_X G$  is of type $FP_m$. First we show  in Proposition
\ref{easydirection} some sufficient conditions for $\Gamma$ to be of type $FP_m$. The difficult part of the classification is to show that these sufficient conditions are necessary. We establish this in Theorem \ref{2direction} under the extra condition that $H$ has infinite abelianization.

\begin{proposition} \label{easydirection} Let $\Gamma = H \wr_X G$ be a wreath
  product such that both $H$ and $G$ are of type $FP_m$,
  $G$ acts (diagonally) on $X^i$  with
  stabilizers of type $FP_{m-i}$ and with finitely many orbits for $1 \leq i \leq m$. Then
  $\Gamma$ has type $FP_m$.
\end{proposition}

\begin{proof}
Let
$$
{\mathcal H} :  \cdots \to \mathbb{Z} H \otimes \mathbb{Z} Y_i \mapnew{\partial_i}
\mathbb{Z} H \otimes \mathbb{Z} Y_{i-1} \to \cdots \to
\mathbb{Z} H \otimes \mathbb{Z} Y_1 \mapnew{\partial_1} \mathbb{Z} H \to
\mathbb{Z} \to 0
$$
be a free resolution of the trivial $\mathbb{Z} H$-module
$\mathbb{Z}$ with all $Y_i$ finite  for $i \leq
m$.

Consider $H_x$ an isomorphic copy of the group $H$ and let
${\mathcal H}_x$  be the  complex obtained from ${\mathcal H}$ by
substituting $H$ with $H_x$. We write $\partial_{x}$ for the
differential of ${\mathcal H}_x$ and if we want to stress its
degree, say $i$, we write $\partial_{i,x}$. Let $M$ be the normal closure of $H$ in $\Gamma$. Thus $M$
is the subgroup of the direct product $\prod_{x \in X} H_x$ that
contains the elements with all but finitely
many trivial coordinates.

We can take the tensor product over $\mathbb{Z}$ of the deleted complexes ${\mathcal H}_x^{del}$ (i.e. in ${\mathcal H}_x$ delete $\mathbb{Z}$) for $x \in X$
and obtain the deleted complex ${\mathcal F}^{del}$.
This makes sense for infinite $X$ as the direct limit of the
tensor products of any finite number of the complexes ${\mathcal
  H}_x$. Note that we have fixed a linear order $\leq$ on $X$
and the tensor product ${\mathcal H}^{del}_{x_1} \otimes \cdots
\otimes {\mathcal H}^{del}_{x_t}$ is made for $x_1 < \cdots <
x_t$ and $t \geq 1$.
By the K\"unneth formula
 we get a free resolution of the trivial $\mathbb{Z}
 M$-module $\mathbb{Z}$
$$
{\mathcal F} : \cdots \to \mathbb{Z} M \otimes W_i \mapnew{d_i}
\mathbb{Z} M \otimes W_{i-1} \to \cdots
\to \mathbb{Z} M \otimes W_1 \mapnew{d_1} \mathbb{Z} M \to
\mathbb{Z} \to 0,
$$
where $W_i$ is a free abelian group with a basis $Z_i$, $Z_i$ is the disjoint union
$$
Z_i = \bigcup (Y_1^{i_1}) \times
(Y_2^{i_2} ) \times \cdots \times (Y_s^{i_s}) \times X^{(i_1,i_2, \ldots,i_s)},
$$
over all possible $s \geq 1$ such that  $i_1 +2i_2
+ \cdots+s i_s = i$, $i_j \geq 0$ and
$
X^{(i_1,i_2, \ldots,i_s)}$ is the set of $s$-tuples $(X_1,
\ldots, X_s)$ of
pairwise disjoint subsets $X_1, \ldots , X_s$ of $X$ such
that $X_j$ has cardinality $i_j$ for all $1 \leq j \leq s$.
We write the elements of $Z_i$ as formal products $y_1
\cdots y_j x_1 \cdots x_j$, which indicates that in the tensor product $y_i \in Y
= \bigcup_{i \geq 1} Y_i$ is taken from the complex ${\mathcal
  H}_{x_i}$ and $x_1 < \cdots < x_j$ are elements of $X$,
recall we have fixed some linear order $\leq$ on $X$. Indeed the element $y_1
\cdots y_j x_1 \cdots x_j$ corresponds to the element  $B_1 \times \cdots \times B_s \times (X_1, \ldots , X_s)$ of $Z_i$, where $X_r$ is the set of those elements $x_t$ of $\{ x_1, \ldots , x_j \}$ for which $y_t \in Y_r$, we write $X_r = \{ x_{j_1} , \ldots , x_{j_{i_r}} \}$ where $x_{j_1} < \cdots < x_{j_{i_r}}$ and set $B_r = (y_{j_1}, \ldots , y_{j_{i_r}}) \in Y_r^{i_r}$.

Sometimes it will be convenient to use general products $y_1
\cdots y_j x_1 \cdots x_j$, where $x_1, \ldots , x_j$ are pairwise different elements of $X$ but we do not assume that $x_1 < \cdots < x_j$.  We follow (\ref{symmetricgroup}) and define for all $ 1 \leq i \leq j-1$
$$ y_1
\cdots y_j x_1 \cdots x_j = (-1)^{\deg(y_i) \deg (y_{i+1})} y_1
\cdots y_{i+1} y_i \cdots y_j x_1 \cdots x_{i+1} x_i \cdots x_j.
$$
This completes the definition of the general products $y_1
\cdots y_j x_1 \cdots x_j$ and shows that they belong to $Z_i \cup - Z_i$.

As $X$ is a (left) $G$-set we get an action of $G$ on the general products given by
\begin{equation} \label{actionaction}
g (y_1
\cdots y_j x_1 \cdots x_j) = y_1
\cdots y_j (g x_1) \cdots (g x_j).\end{equation}
Thus $W_i$ is a semi-induced $\mathbb{Z} G$-module i.e. $W_i = \mathbb{Z}_t Z_i$. Furthermore since $Y_i$ is a finite set for $i \leq m$ and by Lemma \ref{wedgeproducts3} we deduce that $W_i$ is finitely generated as $\mathbb{Z} G$-module.
Using the other notation for the elements of $Z_i$ we have for
$B_1 \times \cdots \times B_s \times (X_1, \ldots , X_s) \in Z_i$ and $g \in G$ that
$$
g (B_1 \times \cdots \times B_s \times (X_1, \ldots , X_s)) = \pm
(g B_1 \times \cdots \times g B_s \times (g X_1, \ldots , g X_s))
$$
where $g B_k$ is obtained from $B_k \in Y_k^{i_k}$ by some permutation (possible the trivial one) of the $i_k$ coordinates. By Lemma
  \ref{twisting}(b) $W_i$ is of type $FP_{m-i}$ as $\mathbb{Z} G$-module if the stabilizer in $G$ of the element $B_1 \times \cdots \times B_s \times (X_1, \ldots , X_s) \in Z_i$ is of type $FP_{m-i}$ for all  $B_1 \times \cdots \times B_s \times (X_1, \ldots , X_s) \in Z_i$ . Any such stabilizer is commensurable with the stabilizer in $G$ of  $ (X_1, \ldots , X_s) \in X^{(i_1,i_2, \ldots,i_s)}$ and by the proof of Lemma \ref{wedgeproducts3}  the last stabilizer is of type $FP_{m-i}$.
Note we have proved that $W_i$ is of type $FP_{m-i}$ as $\mathbb{Z} G$-module for $1
\leq i \leq m$.
Then $\mathbb{Z} \Gamma \otimes_{\mathbb{Z} G} W_i \simeq  \mathbb{Z} M \otimes W_i$ is of type $FP_{m-i}$ as $\mathbb{Z} \Gamma$-module.

Since the action of the symmetric group in (\ref{symmetricgroup}) commutes with the differential
(\ref{differential01}) for a general product $y_1 \cdots y_j x_1 \cdots x_j$
the differential of
 ${\mathcal F}$ is
\begin{equation} \label{maindif}
d_s(y_1 \cdots y_j x_1 \cdots x_j)
= (\sum_{\deg(y_i) \geq 2} (-1)^{\deg(y_1) + \cdots + \deg(y_i)}   y_1 \cdots \partial_{x_i}(y_i) \cdots y_j x_1
\cdots x_i \cdots x_j) +$$ $$ (\sum_{\deg(y_i) =1}   (-1)^{\deg(y_1) + \cdots + \deg(y_i)}    y_1 \cdots \partial_{x_i}(y_i)
\cdots y_j x_1 \cdots \widehat{x_i} \cdots x_j)
\end{equation}
where $s = \sum_{1  \leq i  \leq j}  \deg(y_i) \geq 1$. Note that by (\ref{actionaction}) the above differential commutes with the $G$-action, hence commutes with the $\Gamma$-action.
Then by Lemma \ref{dimensionshifting2}
applied to the complex (obtained from ${\mathcal F}$)
\begin{equation} \label{+}
\widetilde{\mathcal F} : \cdots \to \mathbb{Z} M \otimes W_i \mapnew{d_i}
\mathbb{Z} M \otimes W_{i-1} \to \cdots
\to \mathbb{Z} M \otimes W_1 \mapnew{d_1} \Aug( \mathbb{Z} M) \to 0,
\end{equation}
we deduce that
$\Aug(\mathbb{Z} M)$ is of type $FP_{m-1}$ as $\mathbb{Z} \Gamma$-module, so by Lemma  \ref{condition}
$\Gamma$ is of type $FP_m$.
\end{proof}

\begin{lemma} \label{m=2} Let $\Gamma = H \wr_X G$ be a group with $H \not= 1, X \not= \emptyset$. Then
  $\Gamma$ is of type $FP_2$ if and only if  $H$ is of type
  $FP_2$, $G$ is of type $FP_2$ and
 $G$ acts on $X^i$  with
  stabilizers of type $FP_{2-i}$ and with finitely many orbits  for $1 \leq i \leq 2$.
\end{lemma}

\begin{proof}
By the previous
theorem it remains to show that if $\Gamma$ is $FP_2$ then all stated conditions hold. This is a particular case of \cite[Prop.~4.9]{BCS} when $X$ is $G$-transitive, the general case follows by  the same argument.
\end{proof}

\begin{proposition} \label{dificultdirection} Let $m \geq
  1$, $\Gamma = H \wr_X G$ be a group of type
  $FP_m$ such that $H \not= 1, X \not= \emptyset$, $H$ is of type $FP_{m-1}$ and
 $G$ acts on $X^i$  with
  stabilizers of type $FP_{m-i}$ and with finitely many orbits  for $1 \leq i \leq m$.
Then $H$ is of type $FP_m$.
\end{proposition}

\begin{proof}
As pointed in \cite{Yves} the case $m = 1$ holds, The case $m = 2$ is a particular case of Lemma \ref{m=2}. So
we may assume that $m \geq 3$.

 Let $Z_i$ and $Y_i$ be as in the proof of
  Proposition \ref{easydirection}.
Since $H$ is of type $FP_{m-1}$
  we  can assume that
  $$Y_1,Y_2, \ldots, Y_{m-1} \hbox{ are all finite }$$ and
will prove that $Y_m$ can be chosen finite.
 The description of $Z_{m}$ and $X^{(0, \ldots, 0,1)} = X$ imply
 that
$$\mathbb{Z} Z_{m} = T_{1,m} \oplus T_{2,m},$$ where $T_{1,m} =
\mathbb{Z} [Y_m \times X]$ and $T_{2,m} = \mathbb{Z} [Z_m
\setminus (Y_m \times X)]$ i.e. $T_{1,m} =
  \mathbb{Z} Y_m \otimes \mathbb{Z} X$.
Observe that since $Y_1, \ldots, Y_{m-1}$ are all finite and
 $G$ acts on $X^i$ with finitely many orbits  for $1 \leq i \leq m$ we have that $T_{2,m}$ is finitely generated as
$\mathbb{Z} G$-module.
Note that by the proof of Proposition \ref{easydirection} $\mathbb{Z} M \otimes \mathbb{Z} Z_i$ is $FP_{m-i}$ as $\mathbb{Z} \Gamma$-module for $1 \leq i \leq m-1$.
Since $\Gamma$ is of type $FP_m$ by Lemma \ref{condition} and
Lemma \ref{dimensionshifting3} applied to the complex (\ref{+})   $\im (d_m)$ is finitely
generated as $\mathbb{Z} \Gamma$-module. Consider the
splitting of the domain $\mathbb{Z} M \otimes \mathbb{Z}
Z_m$ of the differential $d_m$ as $$\mathbb{Z} M \otimes \mathbb{Z}
Z_m = W_{1,m} \oplus W_{2,m},$$ where $W_{1,m} = \mathbb{Z} M \otimes T_{1,m}$
and $W_{2,m} = \mathbb{Z} M \otimes T_{2,m}$. Since
$W_{2,m}$ is finitely generated as $\mathbb{Z}
\Gamma$-module the
condition that $\im (d_m)$ is finitely generated is
equivalent with $d_m(W_{1,m}) / d_m(W_{1,m}) \cap
d_m(W_{2,m})$ is finitely generated as $\mathbb{Z} \Gamma$-module.

Let $w_1 \in
W_{1,m}, w_2 \in W_{2,m}$ be such that $d_m(w_1) =
d_m(w_2)$. Then $w_1 - w_2 \in \ker (d_m) = \im (d_{m+1})$.
Let $$p : W_{1,m} \oplus W_{2,m} \to W_{1,m}$$ be the
canonical projection.
By the definition of $W_{1,m}$ and (\ref{maindif}) applied with $s =
m +1$ we get that
$$w_1 = p(w_1 - w_2) \in p( \im (d_{m+1})) = $$ $$  p d_{m+1}((\mathbb{Z} M \otimes
\mathbb{Z} [Y_1 \times  Y_m \times X^{(1,0, \ldots, 0, 1)}])
\oplus (\mathbb{Z} M \otimes
\mathbb{Z} [Y_{m+1} \times X])
).$$
Then since $d_{m+1}(\mathbb{Z} M \otimes
\mathbb{Z} [Y_{m+1} \times X]
) \subseteq W_{1,m}$ we have $p d_{m+1}(\mathbb{Z} M \otimes
\mathbb{Z} [Y_{m+1} \times X]) = d_{m+1}(\mathbb{Z} M \otimes
\mathbb{Z} [Y_{m+1} \times X])$ and so
$$d_m(w_1) \in  d_m p d_{m+1}(\mathbb{Z} M \otimes
\mathbb{Z} [Y_1 \times  Y_m \times X^{(1,0, \ldots, 0, 1)}])
+ d_m p d_{m+1}(\mathbb{Z} M \otimes
\mathbb{Z} [Y_{m+1} \times X]
) = $$ $$d_m p d_{m+1}(\mathbb{Z} M \otimes
\mathbb{Z} [Y_1 \times  Y_m \times X^{(1,0, \ldots, 0, 1)}])
+ d_m  d_{m+1}(\mathbb{Z} M \otimes
\mathbb{Z} [Y_{m+1} \times X]) = $$
\begin{equation} \label{eqnovo5}  d_m p d_{m+1}(\mathbb{Z} M \otimes
\mathbb{Z} [Y_1 \times  Y_m \times X^{(1,0, \ldots, 0, 1)}]).\end{equation}
Recall that $\partial_{j,x}$ denotes the differential of
${\mathcal H}_x$ in dimension $j$ and when we do not want to
stress the dimension we omit the index $j$ and  use $\partial_x$.
Note that for $y_1 \in Y_1, y_m \in Y_m, x_1, x_m \in X, x_1 \not= x_m$
$$
d_m p d_{m+1} (y_1 y_m x_1 x_m) = d_m p(- \partial_{x_1} (y_1) y_m
x_m  + (-1)^{m+1} y_1 \partial_{x_m}(y_m) x_1 x_m) =
$$
\begin{equation} \label{eqwr0} d_m(-  \partial_{x_1} (y_1) y_m
x_m) = (-1)^{m+1} \partial_{x_1} (y_1) \partial_{x_m}(y_m) x_m. \end{equation}
Decompose $M = S_x \times H_x$, where $S_x
\subseteq  \prod_{t \not= x} H_{t}$. Thus in (\ref{eqwr0}) we have
$\partial_{x_m}(y_m) \in \im (\partial_{m,x_m}),  \partial_{x_1} (y_1) \in \Aug (\mathbb{Z} S_{x_m})$.
Then by (\ref{eqnovo5}) and (\ref{eqwr0})
$$
d_m(w_1) \in \bigoplus_{y_m \in Y_m,y_1 \in Y_1, x_1, x_m \in
  X, x_1 \not= x_m} \mathbb{Z}
M\partial_{x_1}(y_1) \partial_{x_m}(y_m) x_m
=$$
\begin{equation} \label{eqwr1} \bigoplus_{x_m \in X} \Aug(\mathbb{Z} S_{x_m}) \mathbb{Z}
H_{x_m} \im (\partial_{m,x_m}) x_m =
\bigoplus_{x\in X} \Aug(\mathbb{Z}
S_{x})
\im (\partial_{m,x}) x =: J
\end{equation}
and
$$ d_m(W_{1,m}) = \bigoplus_{x \in X} \mathbb{Z} M
 \im (\partial_{m,x}) x =$$
\begin{equation} \label{eqwr2}
\bigoplus_{x \in X} (\mathbb{Z} S_x  \mathbb{Z}
H_x) \im (\partial_{m,x}) x = \bigoplus_{x \in X} \mathbb{Z} S_x  \im (\partial_{m,x}) x.
\end{equation}
Furthermore for every $j \in J$ there is $w_1 \in W_{1,m}$ such that there is some $w_2 \in W_{2, m}$ with $d_m(w_1) = d_m(w_2)$ and $d_m(w_1) = j$.
Thus by (\ref{eqwr1}) and (\ref{eqwr2})
$$K := d_m(W_{1,m}) /J \simeq  \bigoplus_{x \in X}  \im
(\partial_{m,x}) x
$$
Thus the condition that $\im (d_m)$
is finitely generated as $\mathbb{Z} \Gamma$-module is
equivalent with
$K$ is finitely generated as $\mathbb{Z}
\Gamma$-module. Take a finite set $D$ of generators of
$K$ such that $D
\subseteq \bigcup_{x \in X} \im (\partial_{m,x}) x$. Write $d
\in D$ as
$ a_x x$ and define $A_x \subseteq \im (\partial_{m,x})$ as the set of all possible
$a_x$. Thus the set of the images of the finite set $\bigcup_{x \in X}
A_x$ in $\im (\partial_m)$, where we send canonically every
$H_x$ to $H$ by just forgetting the index $x$,  generates $\im (\partial_m)$ as $\mathbb{Z} H$-module. Thus $\im
(\partial_m)$ is finitely generated as $\mathbb{Z}
H$-module, so $Y_m$ can be chosen finite.

\end{proof}

\begin{theorem} \label{2direction} Let $m \geq 1$ and $\Gamma = H \wr_X G$ be of type
  $FP_m$, where $ X \not= \emptyset$ and $H$ has infinite abelianization. Then $H$ is of type $FP_m$
and  $G$ acts on $X^i$ with
  stabilizers of type $FP_{m-i}$ and with finitely many
orbits  for all $1 \leq i \leq m$.
\end{theorem}

\begin{proof} We induct on $m$. The case $m = 1$ is
  trivial as the property $FP_1$ is equivalent with finite
  generation and then  \cite[Prop.~2.1]{Yves} applies.

Since $H$ has infinite abelianization $\mathbb{Z}$ is a
retract of $H$, so $\Gamma_0 =\mathbb{Z} \wr_X G$ is a retract of
$\Gamma$. By Lemma \ref{retract} since $\Gamma$ is of type $FP_m$ the group
$\Gamma_0$ is $FP_m$ too. Then by Proposition \ref{wedgeproducts} and Theorem \ref{mainthm1} $G$ acts on $X^i$ with finitely many
orbits  and
  stabilizers of type $FP_{m-i}$ for all $1 \leq i \leq m$.
 Then Proposition \ref{dificultdirection} completes the
inductive step.
\end{proof}

\begin{theorem} \label{mainwreathresult}
Let $\Gamma = H \wr_X G$ be a wreath product, where $ X \not= \emptyset$ and $H$ has infinite abelianization.
Then the following are equivalent :

1. $\Gamma$  is of type
  $FP_m$;

2. $H$ is of type $FP_m$, $G$ is of type $FP_m$,
$G$ acts on $X^i$ with
  stabilizers of type $FP_{m-i}$ and with finitely many
orbits  for all $1 \leq i \leq m$.
\end{theorem}

\begin{proof}
Follows from Lemma \ref{retract}, Proposition
\ref{easydirection} and Theorem \ref{2direction}.
\end{proof}

The following is a homotopy version of Theorem \ref{mainwreathresult}.

\begin{theorem} \label{mainwreathresult2}
Let $\Gamma = H \wr_X G$ be a wreath product, where $X \not= \emptyset$ and $H$ has infinite abelianization.
Then the following are equivalent :

1. $\Gamma$  is of type
  $F_m$;

2. $H$ is of type $F_m$, $G$ is of type $F_m$,
$G$ acts on $X^i$ with
  stabilizers of type $FP_{m-i}$ and with finitely many
orbits  for all $1 \leq i \leq m$.
\end{theorem}
 Remark. Observe that in the condition 2 we have a mixture
 of both properties $F_m$ and $FP_m$ and not just the
 homotopical property $F_m$.

\begin{proof} In the case $m \leq 2$ this is the main result of
  \cite{Yves}. So we can assume that $m > 2$. In general a
  group is of type $F_m$ if and only if it is finitely
  presented (i.e. is $F_2$) and is of type $FP_m$. Then the
  result follows from Theorem \ref{mainwreathresult} and the fact that the case $m = 2$ of Theorem \ref{mainwreathresult2} was already proven  in  \cite{Yves}.
\end{proof}

 \section{An example}

Let $F$ be the Richard Thompson group with infinite presentation $$\langle x_0, x_1, x_2 \ldots | x_j^{x_i} =  x_{j+1}  \hbox{ for } 0 \leq i < j \rangle.$$ Consider its realization as piecewise linear transformations of the interval $[0,1]$, see \cite{essay}. Let $X =(0, 1) \cap \z [\frac{1} {2}]$. Then $F$ acts with finitely many orbits on $X^i$ for every $i \geq 1$ i.e. the $F$ orbit of $x = (x_1, \ldots, x_i) \in X^i$ contains $y = (y_1, \ldots , y_i)$ if and only if there is a permutation $\sigma \in S_i$ such that $x_{\sigma(1)} \leq x_{\sigma(2)} \leq \cdots \leq x_{\sigma(i)}$ and
 $y_{\sigma(1)} \leq y_{\sigma(2)} \leq \cdots \leq y_{\sigma(i)}$ and $x_{\sigma(k)} = x_{\sigma(k+1)}$ if and only if $y_{\sigma(k)} = y_{\sigma(k+1)}$. Furthermore the stabilizer in $F$ of the point  $x = (x_1, \ldots, x_i) \in X^i$ is $F^{j+1}$, where $j$ is the number of different coordinates of $x$. By \cite{GeogheganBrown} $F$ is of type $FP_{\infty}$, so all stabilizers of the action of $F$ on $X^i$ are of type $FP_{\infty}$. The following result is a direct corollary of Theorem \ref{mainwreathresult} and Theorem \ref{mainwreathresult2}.

 \begin{corollary} \label{wreathF} Let $X$ and $F$ be as
   above  and $\Gamma = H \wr_X F$ be a wreath product. Then
   $H$ has type $FP_m$ (resp. $F_m$) if and only if $\Gamma$ is of type $FP_{m}$ (resp. $F_m$).
 \end{corollary}

\begin{proof}
Follows directly from Proposition \ref{easydirection} and Proposition \ref{dificultdirection}.
\end{proof}

\section{\boldmath Bredon type $FP_m$ for wreath products} \label{sectionbredon}

\subsection{Preliminaries on Bredon homology}

 In this section we study the homological finiteness Bredon type  $\underline{FP}_m$. Its homotopical $\infty$-version, the homotopical finiteness Bredon property $\underline{F}_{\infty}$ was considered
in \cite{Luck}. Here we need the following homological
version of L\"uck's result from \cite{Luck}.

\begin{theorem} \label{Bredonref}  \cite{KropNucPer}  A group $\Gamma$ is of type
  $\underline{FP}_m$ if and only if $\Gamma$ has finitely
  many conjugacy classes of finite subgroups and for every
  finite subgroup $K$ of $\Gamma$ the
  centralizer $C_{\Gamma}(K)$ is of type $FP_m$.
\end{theorem}

\subsection{\boldmath Bredon type $FP_m$ for wreath products  : a
  proof of Theorem C }

\begin{theorem} \label{finite}  Let $\Gamma=H\wr_X G$ be a wreath product such that $H$ is torsion-free.
Then every finite subgroup of $\Gamma$ is conjugate to a subgroup of
$G$.
\end{theorem}

\begin{proof} In this proof we consider $X$ as a right $G$-set (if
 $X$ is equipped with a left $G$-action it becomes
a right $G$-action by $x g = g^{-1} x$).
Denote by $\pi : \Gamma\to G$ the natural homomorphism. Let $M$ be the normal closure of $H$ in $\Gamma$. Elements
  $\gamma$ of $\Gamma$ may uniquely be written as $\gamma=mg$ for
  $m\in M$ and $g\in G$; we have $\pi(\gamma)=g$, and write
  $\gamma@x:=m_x$, where $m_x$ is the $x$-coordinate of $m
  \in M \subseteq H^X$.
  The assertion of the theorem is that for every
  finite subgroup $K\le\Gamma$ there exists $\delta\in\Gamma$ such
  that $(k^\delta)@x=1$ for all $x\in X$, $k \in K$.

  Consider first two points $x,y\in X$ in the same orbit under
  $\pi(K)$. Then there exists a unique $h_{x,y}\in H$ such that
  $k@x=h_{x,y}$ for all $k\in K$ with $x^{\pi(k)}=y$. Indeed, consider
  two such $k,k'\in K$ i.e. $x^{\pi(k)}=y = x^{\pi(k')}$. Then $\pi(k'k^{-1})$ fixes $x$, and has finite
  order $s$ because $k'k^{-1}$ belongs to $K$; so
  $1 = (k'k^{-1})^s@x=((k'k^{-1})@x)^s$ implies
  $1 = (k'k^{-1})@x=(k'@x)(k@x)^{-1}$. Note the cocycle identity
  $h_{x,y}h_{y,z}=h_{x,z}$.

  In each orbit $\Omega = x^{\pi(K)}$ of $\pi(K)$ on $X$, choose a representative
  $x_\Omega$, and define $\delta\in H^X$ by $\delta_x=h_{x,x_\Omega}$
  whenever $x$ lies in the orbit $\Omega$.
  Note that since $K$ is finite  there are only finitely
  many $x,y\in X$ such that $h_{x,y}\neq1$. Therefore $\delta$ is a
  finitely supported function on $X$
  i.e. $\delta \in M$.

  Now consider $k\in K$, and write $y=x^{\pi(k)}$. Note that
  $x_{\Omega} = y_{\Omega}$. Then
  \[(k^\delta)@x=(\delta^{-1}@x)(k@x)(\delta@y)=h_{x,x_\Omega}^{-1}h_{x,y}h_{y,y_\Omega}=h_{x,x_\Omega}^{-1}h_{x,y_\Omega}
  = 1.\qedhere\]
\end{proof}

\begin{theorem} Let  $\Gamma=H\wr_X G$  be a wreath
  product, $ X \not= \emptyset$  and $H$ be torsion-free, with
  infinite abelianization.   Then $\Gamma$ has type
  $\underline{FP}_m$ if and only if  the following
  conditions hold :

1. $G$ has type $\underline{FP}_m$;

2. $H$ has type $FP_m$;

3. for every finite subgroup $K$ of $G$ and every $ 1 \leq i
\leq m$ the centralizer
$C_G(K)$ acts on $(K \backslash X)^i$ with stabilizers of type $FP_{m-i}$ and with  finitely many orbits.
\end{theorem}

\begin{proof} By Theorem \ref{Bredonref} we have to understand the centralizers and the conjugacy classes of finite subgroups in $\Gamma$. By Theorem \ref{finite}
  every finite subgroup $K$ of $\Gamma$ is conjugated to a finite
  subgroup $K_0$ of $G$. Thus $\Gamma$ has finitely many
  conjugacy classes of finite subgroups if and only if $G$
  has finitely many conjugacy classes of finite
  subgroups. This completes the proof for $m = 0$.

Furthermore for a finite subgroup $K$ of $G$ we have that $$C_{\Gamma}(K) = C_M(K) \rtimes C_G(K) \simeq H
\wr_{K \backslash X} C_G(K),$$
where $M$ is the normal closure of $H$ in $\Gamma$ and we consider $X$ as a left $G$-set. Then by Theorem \ref{mainwreathresult}
the centralizer $C_{\Gamma}(K) $ is of type $FP_m$ if and only if $C_G(K)$ is $FP_m$ and conditions 2 and  3 hold.
Then  Theorem \ref{Bredonref} completes the proof.
\end{proof}

 \section{\boldmath $\Sigma$-invariants} \label{sigmasection}

\subsection{Some basic properties}
 Following
\cite{BieriRenz} we define an equivalence relation $\sim$ on $\Hom(G, \mathbb{R}) \setminus \{ 0 \}$ by $\chi_1 \sim \chi_2$ if there is $r \in \mathbb{R}_{>0}$ such that $\chi_1 = r \chi_2$. Denote by $[\chi]$ the equivalence class of $\chi$. Then the character sphere $S(G)$ is $(\Hom(G, \mathbb{R}) \setminus \{ 0 \}) / \sim$. By definition
$$
\Sigma^m(G, \mathbb{Z} ) = \{ [\chi] \in S(G)  | \  \ \mathbb{Z} \hbox{ is }
FP_m \hbox{ as } \mathbb{Z} G_{\chi}-\hbox{module} \},
$$where
$G_{\chi} = \{ g \in G | \chi(g) \geq 0 \}$.

\begin{lemma} \label{sigmaretract}  Let $m \geq 1$ be an integer, $\Gamma = M \rtimes G$ be a finitely generated group  and $[\chi] \in \Sigma^m(\Gamma, \z)$ such that $\chi(M) = 0$. Then
$[\chi_0] \in \Sigma^m (G, \z)$, where $\chi_0$ is the restriction of $\chi$ to $G$.
\end{lemma}

\begin{proof} This is a monoid version of Lemma \ref{retract} and Aberg's idea  ( i.e. the footnote from \cite{Aberg}) can be easily modified in this context.
Indeed by \cite{BieriQMWbook} we have that $[\chi] \in \Sigma^m(\Gamma, \z)$ is equivalent with
$$
\Tor^{\z \Gamma_{\chi}}_i(\z, \prod \z \Gamma_{\chi}) = 0 \hbox{ for all }1 \leq i \leq m-1
$$
and the trivial $\mathbb{Z} \Gamma_{\chi}$-module $\mathbb{Z}$ is finitely presented. Note that by the functoriality of $\Tor_i^{\z S} ( \z , \prod \z S)$ on $S$ applied to the monoids $S \in \{ G_{\chi_0}, \Gamma_{\chi} \}$ and the fact that $\Gamma_{\chi} = M \rtimes G_{\chi_0}$ we get that  $\Tor^{\z G_{\chi_0}}_i(\z, \prod \z G_{\chi_0}) = 0$ for all $1 \leq i \leq m-1$. Furthermore the trivial $\mathbb{Z} G_{\chi_0}$-module $\mathbb{Z}$
is obtained from the trivial $\mathbb{Z} \Gamma_{\chi}$-module $\mathbb{Z}$  applying the  right exact functor $\mathbb{Z} \otimes_{\mathbb{Z} M} $, hence the trivial $\mathbb{Z} G_{\chi_0}$-module $\mathbb{Z}$ is finitely presented.
Thus $[\chi_0] \in \Sigma^m (G, \z)$.
\end{proof}

\begin{lemma} \label{condition2} Let $m \geq 1$ be an integer, $\Gamma =  M \rtimes G$
be a finitely generated group
and $\chi : \Gamma \to \real$ be a non-zero homomorphism such that $\chi(M) = 0$.
Denote by $\chi_0$ the restriction of $\chi$ to $G$.
Then $[\chi] \in \Sigma^m(\Gamma, \z)$ if and only if $[\chi_0] \in \Sigma^m( G, \z)$ and $\Aug (\mathbb{Z} M)$ is $FP_{m-1}$ as $\mathbb{Z} \Gamma_{\chi}$-module.
\end{lemma}

\begin{proof} In the proof of Lemma \ref{condition} substitute $\Gamma$ with $\Gamma_{\chi}$ and substitute $G$ with $G_{\chi_0}$.
\end{proof}

 The next lemma collects results from \cite{Master}, that can be found in English version in \cite{MMV}.

\begin{lemma} \label{sigmaofactions} \cite{Master}, \cite{MMV} Let $k \geq 0$ be an integer, $\chi : G \to \real$ be a non-zero  homomorphism, $L$ a subgroup of $G$. Denote by $\chi_0$ the restriction of $\chi$ on $L$. Then

(1) $\z G_{\chi}$ is a flat (right or left) $\z L_{\chi_0}$-module and the inclusion map of $\z L_{\chi_0}$-modules $\z L_{\chi_0} \to \z G_{\chi}$ splits;

(2) if $\chi_0 \not= 0$ then for every $\mathbb{Z} L$-module $V$ we have an isomorphism of  left $\mathbb{Z} G_{\chi}$-modules $\mathbb{Z} G \otimes_{\mathbb{Z} L} V \simeq \mathbb{Z} G_{\chi} \otimes_{\mathbb{Z} L_{\chi_0}} V$;

(3) if $\chi_0 \not= 0$ then for every $\mathbb{Z} L$-module $V$ we have that $\mathbb{Z} G \otimes_{\mathbb{Z} L} V$ is $FP_m$ as $\mathbb{Z} G_{\chi}$-module if and only if $V$ is $FP_m$ as $\mathbb{Z} L_{\chi_0}$-module;

(4) if $L$ has finite index in $G$ then for every $\mathbb{Z} G$-module $V$ we have that $V$ is $FP_m$ as $\mathbb{Z} G_{\chi}$-module if and only if
$V$ is $FP_m$ as $\mathbb{Z} L_{\chi_0}$-module.
\end{lemma}

\begin{proof}
Consider
 $T$  a left transversal of $L$ in $G$  such that $1 \in T$. Then
$
\z G_{\chi} = \bigoplus_{t \in T} t \z L_{\chi_0 \geq - \chi(t)},
$
where $ L_{\chi_0 \geq r} = \{ g \in L | \chi_0(g) \geq r \}$.
Since $1 \in T$ the inclusion map of (right) $\z L_{\chi_0}$-modules $\z L_{\chi_0} \to \z G_{\chi}$ splits.

The rest of item 1. is \cite[Lemma~9.1(i)]{MMV}. Item 2 is \cite[Lemma~9.1(ii)]{MMV}, item 3 is \cite[Lemma~9.2]{MMV} and item 4 is \cite[Thm~9.3]{MMV}.
\end{proof}

 \subsection{Semi-induced modules}

 \begin{lemma} \label{twisting2} Let $\chi: G \to \real$ be
   a non-zero homomorphism.
Let $S$ be a finitely generated semi-induced $\mathbb{Z}
G$-module and consider the decomposition  $S \simeq \bigoplus_{i \in I} \mathbb{Z} G \otimes_{\mathbb{Z} K_i} V_i$ given by Lemma \ref{twisting}. Thus there is a subgroup $G_i$ of $K_i$ of index at most 2 that acts trivially on $V_i$. Then

1. $S$ is finitely generated as $\mathbb{Z} G_{\chi}$-module  if and only if for all $i \in I$ for the restriction $\chi_i$ of $\chi$ on $K_i$ we have $\chi_i \not= 0$;

2. $S$ is of type $FP_m$ as $\mathbb{Z} G_{\chi}$-module if and only if for every $i \in I$ for the restriction $\tilde{\chi}_i$ of $\chi$ to $G_i$ we have that $\widetilde{\chi}_i \not= 0$ and the trivial $\mathbb{Z} (G_i)_{\tilde{\chi}_i}$-module $\mathbb{Z}$ has type $FP_m$ i.e.\ $[\tilde{\chi}_i] \in \Sigma^ m(G_i, \mathbb{Z})$.
\end{lemma}

\begin{proof} Note that item 1 is obvious. By item 1 we can assume that $\widetilde{\chi}_i \not= 0$.
By Lemma \ref{sigmaofactions} (3) $\mathbb{Z} G \otimes_{\mathbb{Z} K_i} V_i $ is $FP_m$ as $\mathbb{Z} G_{\chi}$-module if and only if $V_i$ is $FP_m$ as $\mathbb{Z} (K_i)_{\chi_i}$-module.

It remains to consider only the case when $K_i \not= G_i$, so $K_i / G_i$ is cyclic of order 2.
Observe that $V_i$ is the sign module with $G_i$ acting trivially. By Lemma \ref{sigmaofactions} (4) $V_i$ is $FP_m$ as $\mathbb{Z} (K_i)_{\chi_i}$-module if and only if $V_i$ is $FP_m$ as $\mathbb{Z} (G_i)_{\widetilde{\chi}_i}$-module. The proof is completed by the fact that $V_i$ is the trivial $\mathbb{Z} G_i$-module $\mathbb{Z}$.
\end{proof}

 \begin{proposition} \label{wedgeproducts2} Let $M= \mathbb{Z} X$ be an induced $\mathbb{Z} G$-module and $\chi : G \to \real$ a non-zero  homomorphism. Then the following are equivalent :

1.  $\wedge^i M$ is of type $FP_{m-i}$ as $\z G_{\chi}$-module for all $1 \leq i \leq m$;

2.   $\otimes^i M$ is of type $FP_{m-i}$ as $\z G_{\chi}$-module for all $1 \leq i \leq m$;

3. $\otimes^i M$ is finitely generated as  $\mathbb{Z} G_{\chi}$-module via the diagonal action  and
 for representatives   $\{ G_1, \ldots , G_{s_1} \}$ of all $G$-orbits of stabilizers of the diagonal action of $G$ on $X^i$,  the restriction $\chi_j$ of $\chi$ on $G_j$ is non-zero and $[\chi_j] \in \Sigma^{m-i} (G_j, \z)$ for all $1 \leq i \leq m$, $1 \leq j \leq s_1$;

4. $\wedge^i M$ is finitely generated as  $\mathbb{Z} G_{\chi}$-module via the diagonal action and
 for representatives   $\{ \widetilde{G}_1, \ldots , \widetilde{G}_{s_2} \}$ of all $G$-orbits of stabilizers  of the action of $G$ on the $i$-element subsets of $X$, the restriction $\widetilde{\chi}_j$ of $\chi$ on $\widetilde{G}_j$ is non-zero and $[\widetilde{\chi}_j] \in \Sigma^{m-i} (\widetilde{G}_j, \z)$ for all $1 \leq i \leq m$, $1 \leq j \leq s_2$;

 5. $G \backslash X^i $ is finite  and for representatives   $\{ G_1, \ldots , G_{s_1} \}$ of all $G$-orbits of stabilizers
 of the diagonal action of $G$ on $X^i$  the restriction $\chi_j$ of $\chi$ on $G_j$ is non-zero and $[\chi_j] \in \Sigma^{m-i} (G_j, \z)$ for all $1 \leq i \leq m$, $1 \leq j \leq s_1$.
\end{proposition}
 \begin{proof} We say that  the monoid $G_{\chi}$ acts with finitely many orbits on a set $Y$ if there is a finite subset $Y_0$ of $Y$ such that $G_{\chi} Y_0 = Y$. Then  to prove Proposition \ref{wedgeproducts2} it is sufficient to repeat the proof of Proposition
   \ref{wedgeproducts} substituting $G$ with $G_{\chi}$,
  and  apply Lemma \ref{sigmaofactions} and Lemma
   \ref{twisting2} instead of Lemma \ref{twisting}.
 \end{proof}

 \begin{theorem} \label{sigmamain}
Let $\Gamma = H \wr_X G$ be a wreath product of type $FP_m$, $X \not= \emptyset$ and $H$ has infinite abelianization.  Let $\chi : \Gamma \to \mathbb{R}$ be a non-zero character such that $\chi(H) = 0$.
Then the following are equivalent :

1. $[\chi]  \in \Sigma^m(\Gamma, \mathbb{Z} )$;

2. $[\chi \mid_G]  \in \Sigma^m(G, \mathbb{Z} )$  and for representatives   $\{ G_1, \ldots , G_{s_1} \}$ of all $G$-orbits of stabilizers
 of the diagonal action of $G$ on $X^i$  the restriction $\chi_j$ of $\chi$ on $G_j$ is non-zero and $[\chi_j] \in \Sigma^{m-i} (G_j, \z)$ for all $1 \leq i \leq m$, $1 \leq j \leq s_1$.
\end{theorem}

\begin{proof} The proof is the same as the proof of Theorem
  \ref{mainwreathresult} substituting $G$ with $G_{\chi}$.

The fact that 2. implies 1.  is a monoid version of    Proposition \ref{easydirection}.
  Note that in the proof of Proposition \ref{easydirection}  we
  use that some stabilizers are commensurable. Thus they have the same homological type $FP_k$.
   This step requires justification in its $\Sigma$-version, but this is the content of Lemma \ref{sigmaofactions} (4).

The fact that item 1 implies item 2 follows from Lemma \ref{sigmaretract}
applied to the retracts $G$ and $\Gamma_1 = \mathbb{Z} \wr_X G$ of
$\Gamma$. Indeed we get that $[\chi |_G] \in \Sigma^m(G,
\mathbb{Z})$ and $[\chi |_{\Gamma_1}] \in \Sigma^m(\Gamma_1,
\mathbb{Z})$. By the
monoid version of Theorem \ref{mainthm1} the result holds for $\Gamma_1$, so
condition 5 from Proposition \ref{wedgeproducts2}  holds.

\end{proof}


TITLE: $(B_X,w)$ metrizable implies $X^\ast$ separable
QUESTION [1 upvotes]: Let X be a normed space and assume that $(B_X,w)$ is metrizable, i.e. the weak topology is metrizable. Show that $X^\ast$ is separable.
My attempt:
Let $d$ a equivalent metric on $B_X$. For fixed $N\in\mathbb{N}$ there exists a finite family of functionals $F_N\subset X^\ast$ such that $U_N:=\{ x \in X: d(x,0) < N^{-1} \} \supset \{x\in X: |f(x)|<\epsilon_N ,f\in F_N\}$. Define $Z:= \overline{\text{span}}\{ F_N: N\in\mathbb{N}\}$. If $Z=X^*$, we are done; just take the rational span. Now assume that $g\in X^\ast$. Then there exists a $n\in \mathbb{N}$ such that $U_n \subset \{x: |g(x)|< 1\}$. If $x\in \ker f$ for all $f\in F_n$, then $\lambda x \in U_n$ for all $\lambda\in \mathbb R$. Then $|g(x)|<\frac{1}{|\lambda|}$ for all $\lambda \in \mathbb{R}$ and hence, $x\in \ker g$. As $\cap_{f\in F_n} \ker f \subset \ker g$, $g\in \text{span} F_n$ and thus $g\in Z$.
Is the last argument okay. In particular, the argument with the $\lambda$ and the kernel.
Edit:
Claim: If $\ker g\supset \cap_{i=1}^n \ker f_i =: Y$ for linear functionals in $X^*$. Then, $g\in \text{span} \{f_1, \cdots f_n\}$.
Proof. Consider the map  $\vec{f}: X\to \mathbb{R}^n, x \mapsto (f_1(x),\dots, f_n(x))$, i.e. $Y=\ker \vec{f}$. As $Y\subset \ker g$, there exists a unique map $\hat g: \mathbb{R}^n \to \mathbb{R}$ such that  $g=\hat g \circ \vec f$. Note $\mathbb{R}^n = X/Y$. Now, $g(x)=\sum_{i=1}^n \lambda_i x_i$ for $x\in\mathbb{R}^n$ and some fixed $\lambda_i\in \mathbb{R}$. Thus, $g(x)=\sum_{i=1}^n \lambda_i f_i(x)$, which proves the claim.
Edit 2: This proof is definitely false as it contradicts the Baire category theory. We had a countable Hamel basis which cannot be true. See the comment.
Edit 3: The Problem is resolved. I assumed that $X$ and not only the ball $B_X$ was metrizable. Thanks to Nik Weaver who pointet that out on MO.

REPLY [1 votes]: A few comments:
1) you don't need "contradiction". Your argument (if right) shows that if $g\in X^*$, then $g$ is in the span of $F_N$.
2) I don't see why you claim that $\cap_{f\in F_N}\ker f\subset\ker g$ implies $g\in \text{span}\,F_N$.
3) You seem to prove that $X=\text{span}\,\{F_N:\ N\}$ without closure, which looks suspicious to me.
4) Here is a way to fix the proof:
If $g\in X^*\setminus Z$, then we can separate them by Hahn-Banach (considering the locally convex space $X^*$ with the weak$^*$-topology), so there exists $x\in X$ and $c\in\mathbb R$ with
$$\tag{1}
\text{Re}\,f(x)<c\leq\text{Re}\,g(x).
$$
As $Z$ is a subspace, it follows that $\text{Re}\,f(x)=0$ for all $f$ in $Z$, and in particular $-\text{Im}\,if(x)=\text{Re}\,f(x)=0$, so $f(x)=0$ for all $f\in Z$.
Now we can argue as in your attempt. There exists $n$ such that $U_n\subset\{y:\ |g(y)|<\|x\|+1\}$. As $\lambda x\in U_n$ for all $\lambda\in\mathbb R$, we get
$$
|g(x)|\leq\frac{\|x\|+1}\lambda,\ \ \lambda\in\mathbb R,
$$
implying that $g(x)=0$, contradicting (1). So $g$ cannot exist, and $X^*=Z$.


TITLE: General formula of partial sum of series (non-geometric)
QUESTION [0 upvotes]: Like user71317 in his question I am struggling to understand how we arrive at the general formula of the partial sums of a series. In my case the following series:
$$\sum_{n=2}^{\infty} \frac{1}{n^2-1}$$
The general formula of the partial sums is:
$$s_n=\sum_{i=2}^{n} \frac{1}{i^2-1} = \frac{3}{4} - \frac{1}{2n}-\frac{1}{2(n+1)}$$
I understand the answer to the linked question where the answer is the formula for the sum of a finite geometric series. Since my series is not geometric; how do I go from the series to the partial sums in this case?
Update:
How is $\sum_{i=2}^{n} \frac{1}{i^2-1}$ equal to $\frac{3}{4} - \frac{1}{2n}-\frac{1}{2(n+1)}$?

REPLY [0 votes]: This series can be converted into a telescoping series. Using partial fraction decomposition, we can see that
$$\frac{1}{n^2-1}=\frac{1}{(n+1)(n-1)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$
Therefore,
$$\begin{align}
2\sum_{r=2}^n\frac{1}{r^2-1}&=\sum_{r=2}^n\left(\frac{1}{r-1}-\frac{1}{r+1}\right)\\
&=\frac{1}{1}-\frac{1}{3}\\
&~+\frac{1}{2}-\frac{1}{4}\\
&~+\frac{1}{3}-\frac{1}{5}\\
&~~~~~~~\vdots~~~~~~~\vdots\\
&~+\frac{1}{n-2}-\frac{1}{n}\\
&~+\frac{1}{n-1}-\frac{1}{n+1}\\
\end{align}$$
And all the terms will cancel apart from $4$ of them, so we are left with
$$2\sum_{r=2}^n\frac{1}{r^2-1}=1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}=\frac{3}{2}-\frac{1}{n}-\frac{1}{n+1}$$
which means that $$\sum_{r=2}^n\frac{1}{r^2-1}=\frac{3}{4}-\frac{1}{2n}-\frac{1}{2(n+1)}$$

We can take the limit as $n\to\infty$ to find that
$$\sum_{r=1}^\infty\frac{1}{r^2-1}=\lim_{n\to\infty}\sum_{r=1}^n\frac{1}{r^2-1}=\frac{3}{4}$$


TITLE: Probability of at least two bits
QUESTION [0 upvotes]: I have a frame of N bits with per-bit error rate as p sent over wire.
Considering this as a binomial distribution and assuming bit error independence, I am finding the probability of at least 2 bits being corrupted as this :
Probability of none bits being corrupt = $ P(0) = (1-p)^N$
Probability of 1 bit being corrupt = $ P(1) = N \cdot p \cdot (1-p)^{N-1}$
Probability of at least 2 bits being corrupt = $ P = 1 - P(0) - P(1) = 1-(1-p)^N-N \cdot p \cdot (1-p)^{N-1}$
Am doing this correctly or missing something?
Thank you

REPLY [1 votes]: Yup, you are correct.
$$P=1-(1-p)^N-Np(1-p)^{N-1}=1-(1-p)^{N-1}(1-p+Np)$$


TITLE: What is the term for the vector product in dimensions greater than 3?
QUESTION [1 upvotes]: As you know, the cross product maps $\mathbb{R}^{3} \times \mathbb{R}^{3}\rightarrow \mathbb{R}^{3} $ and the product vector is orthogonal to the 2 multiplier vectors and is computed using the determinant with the standard basis vectors as the top row and the multiplier vectors a,b in the respective middle and bottom row(it can also be defined in 7 dimensions but I digress). But what about the vector product in higher dimensions? So for n=4 the vector product maps $\mathbb{R}^{4} \times \mathbb{R}^{4} \times \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}$ using the same technique as the cross product but with an extra row and column. So what is the name for this vector product? This could be generalized to $(\mathbb{R}^{n} | \ n\geq 3) $ with n-1 multiplier vectors.

REPLY [1 votes]: The cross-product is the dual to the exterior product (or "wedge" product). For this reason, it can only be defined wiht $n-1$ elements, where $n$ is the dimension of the vector space.
If you're not in 3D where the cross product (which is a pretty poor operator) works as a binary operator, you'll want to use the exterior product.
https://en.wikipedia.org/wiki/Exterior_algebra


TITLE: Method to solve this indefinite integral $ \int \frac{x^{4}}{\sqrt{x^{2}+9}}dx $
QUESTION [2 upvotes]: I'm not able to solve this integral. Can anybody help me?
I tried doing it by parts noticing that
$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = \int x^{3} \cdot \frac{x}{\sqrt{x^{2}+9}} .
$$
where we know that $$ D[\frac{x}{\sqrt{x^{2}+9}}] = \sqrt{x^{2}+9} $$
so that we can view the integral as
$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = x^{3} \cdot \sqrt{x^{2}+9} - \int 3 \cdot x^{2} \cdot \sqrt{x^{2}+9} \, dx  .
$$
which led me in another integration by parts and in another one again, so i realized that maybe this wasn't the right way to do it

REPLY [0 votes]: Denominator term is to the power  -1.
Numerator multiplied by 1.
Use LIATE method to select first and 2nd terms for multiplication rule of Integration.
Use multiplication rule
Solve to get answer


TITLE: $\int \sqrt{\frac{x-1}{x+1}}\frac 1{x^2} dx$
QUESTION [0 upvotes]: How can we integrate
  $$
\int \sqrt{\frac{x-1}{x+1}}\frac 1{x^2} \,  dx
$$

I thought about substituting $u=\frac 1x$ to transform the integral into a slightly-simpler looking one:
$$
-\int \sqrt{\frac{1-u}{1+u}} \,  du
$$
Am I on the right track?
Following @Shobhit's comment, I let $u=\cos (2w)=\cos^2 w - \sin^2 w$, so that $du=-2 \sin(2w) \,  dw = -4 \sin w \cos w \,  dw$ and so
\begin{align}
\int \sqrt{\frac{x-1}{x+1}}\frac 1{x^2} dx &= -\int \sqrt{\frac{1-u}{1+u}} \,  du \\
&= 4 \int |\tan w| \sin w \cos w \,  dw \\
&= 4 \int \sin^2 w \, dw
\end{align}
If this is correct, then I can finish the rest of this integral on my own.

REPLY [3 votes]: The faster way to integrate such a function is to let
$$
u=\sqrt{\frac{x-1}{x+1}}
$$
so that
$$
x= \frac{u^2+1}{1-u^2} \implies dx=\frac{4 u}{\left(u^2-1\right)^2}du
$$
the integral thus becomes
$$
\int \sqrt{\frac{x-1}{x+1}}\frac 1{x^2} dx=\int u \frac{(1-u^2)^2}{(1+u^2)^2} \frac{4 u}{\left(u^2-1\right)^2}du=\int \frac{4u^2}{(1+u^2)^2}du
$$
and now you can use Partial Fraction Decomposition, which goes off easy, you must obtain:
$$
\int \frac{4u^2}{(1+u^2)^2}du=2 \arctan(u)-\frac{2 u}{u^2+1}+c=2 \arctan \left(\sqrt{\frac{x-1}{x+1}}\right)-\frac{\sqrt{x^2-1}}{x}+c
$$

REPLY [2 votes]: After your substitution from $\displaystyle x= \frac{1}{u}$ $$I = -\int\sqrt{\frac{1-u}{1+u}}du = \int\frac{u-1}{\sqrt{1-u^2}}du = \int\frac{u}{\sqrt{1-u^2}}du-\int\frac{1}{\sqrt{1-u^2}}du$$
$$I = -\sqrt{1-u^2}-\arcsin(u)+\mathcal{C} = -\frac{\sqrt{x^2-1}}{x}-\arcsin\left(\frac{1}{x}\right)+\mathcal{C}$$


TITLE: How can I solve this equation in $3$ variables for non negative integral solutions where upper limit of these variables is given
QUESTION [1 upvotes]: Solve  for non negative integral solutions $x_1 + x_2 + x_3 = 48 \text{ such that} \; \;
x_1 \leq 17 \; x_2 \leq 6 \; x_3 \leq 35$

Define,
$A: \text{set of all non negative integral solutions that have } x_1 \leq 17$
$B: \text{set of all non negative integral solutions that have } x_2 \leq 6$
$C: \text{set of all non negative integral solutions that have } x_1 \leq 35$
$U: \text{set of all non negative integral solutions}$
We are interested in $|A \cap B \cap C| =| U\setminus [A^C \cup B^C \cup C^C]| =
|U| - |U \cap [A^C \cup B^C \cup C^C]| = |U| - |A^C \cup B^C \cup C^C|  = |U| - [|A^C| + |B^C| + |C^C| - \big( |A^C \cap B^C |+ |A^C \cap C^C |+ |B^C \cap C^C| \big) + |A^C \cap B^C \cap C^C| ] $
Now I know exactly how to calculate this but I think it is too much work. Is there an easier way to get around this?
at this point, I would proceed like this:
for |U|: we solve $x_1 + x_2 + x_3 = 48  \; \; x_1 \geq 0 \; x_2 \geq 0 \; x_3 \geq 0$
so $|U|= C(50,2)$
for $|A^C|$. we solve  $x_1 + x_2 + x_3 = 48  \; \; x_1 \geq 18 \; x_2 \geq 0 \; x_3 \geq 0$
so $|A^C| = C(32,2) $
similarly, $|B^C| = C(43,2) \; \;|C^C| = C(14,2)$
for $|A^C \cap B^C|$ : we solve $x_1 + x_2 + x_3 = 48  \; \; x_1 \geq 18 \; x_2 \geq 7 \; x_3 \geq 0$
$|A^C \cap B^C| = C(25,2)\ldots$ and so on.

REPLY [1 votes]: The idea is fine, but you can reduce the amount of computation. Multiply the equation by $-1$ and rewrite as
$$(17-x_1)+(6-x_2)+(35-x_3)=10$$
Letting $y_1=17-x_1, y_2=6-x_2, y_3=35-x_3$, you want to find the number of non-negative solutions to $$y_1+y_2+y_3=10$$ such that $y_2\leq 6$. Notice that for each solution $(y_1,y_2,y_3)$ you're going to have $0\leq x_1\leq 17, 0\leq x_2\leq 6,0\leq x_3\leq 35$.


TITLE: Example of Non-Compact Closed Set contained in Open Set With Special Property
QUESTION [1 upvotes]: I am looking for an example of an open set $A$ in a metric space $X$ and a closed, non-compact subset $B$ of $A$ such that there is no $\delta > 0$ s.t. $\{x: \textrm{dist}(x,B) < \delta\} \subset A$.
I have tried using the discrete metric and other obvious tricks but have been unable to come up with an example. Any hints would be appreciated.

REPLY [2 votes]: Let $X= \Bbb R^2.$  Let $A=\{(x, y)~|~ y \gt 0 \}.$  Let $B=\{(x, y)~|x \gt 0 \land xy \geq 1 \}.$
Choose $\delta \gt 0$.  Then $(2\delta, \frac{1}{2\delta}) \in B$, so $(2\delta, -\frac{1}{4\delta}) \notin A$ but $\operatorname{dist}((2\delta, -\frac{1}{4\delta}), B) \leq \frac{3\delta}{4} \lt \delta.$


TITLE: If $\mu^{\ast}$ is an outer measure with enveloping property, $A_n \nearrow A$, $\mu^{\ast}(A) = \lim_{n\to \infty}\mu^{\ast}(A_n)$
QUESTION [5 upvotes]: Let $\mu^{\ast} : \mathcal{H} \to [0,\infty]$ an outer measure, where $\mathcal{H}$ is an hereditary $\sigma$-ring. Suppose that for each $A \in \mathcal{H}$ there is an measurable set $E$ such that $A \subset E$ and $\mu^{\ast}(A) = \mu^{\ast}(E).$ We say in this case that $\mu^{\ast}$ has the enveloping property.
Prove that if a sequence $\{A_n\} \subset \mathcal{H}$ is such that $A_n \nearrow A$, then
$$\mu^{\ast}(A) = \lim_{k \to \infty}\mu^{\ast}(A_k). $$
I tried several naive things like:
For each $k$ there is $\tilde A_k$ measurable such that $A_k \subset \tilde A_k$ and $\mu^{\ast}(A_k) = \mu^{\ast}(\tilde A_k).$ Then, I tried to show that $\tilde A_k \subset \tilde A_{k+1}$, but it seems not to be true.
Then, I considered $\tilde A_N := \cup_{k=1}^N\tilde A_k$. It is true that $\tilde A_N \supset \bigcup_{k=1}^N A_k$, but is not true necessarily that $$\mu^{\ast}(\tilde A_N) = \mu^{\ast}(\bigcup_{k=1}^NA_k),$$ so it led to nowhere.
Any hints?
Thanks in advance.

REPLY [6 votes]: First, notice that $\lim_{n\rightarrow \infty}\mu^*(A_k)=L$ exists since $\mu(A_k)$ is a increasing real sequence. Since $\mu^*$ is monotonic, $L\leq \mu^*(A)$.
For each $k \in \mathbb N$, let $\tilde A_k\supseteq A_k$ be a measurable set such that $\mu(\tilde A_k)=\mu(A_k)$.
For each $k\in \mathbb N$, let $B_k=\bigcap_{l\geq k}\tilde A_l$. Each $B_k$ is measurable and $A_k\subseteq B_k\subseteq \tilde A_k$ (notice that if $l\geq k$, then $A_k\subseteq A_l\subseteq \tilde A_l$, so $A_k\subset B_k)$. Since $\mu^*$ is monotonic, $\mu(A_k)=\mu(B_k)$, and $B_k$ is increasing. Then:
$$\mu^*(A)=\mu^*(\bigcup_{k \in \mathbb N}A_n)\leq \mu^*(\bigcup_{k \in \mathbb N}B_n)=\lim_{n \rightarrow\infty}\mu^*(B_n)=\lim_{n \rightarrow\infty}\mu^*(A_n)=L.$$


TITLE: Higher order derivatives: variation of parameters
QUESTION [1 upvotes]: Higher order of derivatives: variation of parameters?
$$y'''+y' = \tan t, \quad 0 < t< \pi.$$
Use variation of parameters to determine the general solution of the given differential equation.
I know the roots of the characteristic equation are $0$, $-i$, and $i$. But I'm really confused about how to approach the question. I tried doing the Wronskian but because it has a double root, it gave me a weird answer. Can someone please help me?
By the way, the answer of the question is
$$y= c_1 + c_2 \cos t + c_3 \sin t - \ln \cos t - \sin t \ln(\sec t + \tan t).$$

REPLY [2 votes]: First, let $f = y'$ to transform the equation into a second order equation
$$ f'' + f = \tan(t). $$
The characteristic equation is $\lambda^2 + 1 = 0$ with roots $\lambda = \pm i$. The Wronskian will be
$$ \left| \begin{array}{cc}
\sin(t) & \cos(t) \\
\cos(t) & -\sin(t) \end{array} \right| = -1. $$
Solve this equation using variation of parameters to find the general form of $f$ and then integrate to find $y$.


TITLE: existence of neighbor of linear transformation T such that...
QUESTION [0 upvotes]: I want to show if a linear transformation $T$ has rank $k$ then there exist $\delta$ such that the open ball centered at $T$ with radius $\delta$ contains only linear transformations that has rank greater than or equal to k

REPLY [1 votes]: Use the following characterisation of the rank of a matrix:
$$\operatorname{rank}A = \max \{n \in \mathbb N | \exists n \times n-\text{minor of } A \text{ with non-zero determinant}\}$$
Lets conclude (highly non-technical, but I guess you are familiar with $\varepsilon-\delta$-methods):
Give a matrix of rank $k$, you have some $k \times k$-minor with non-vanishing determinant (Let its absolute value be $>2\varepsilon$). Changing the matrix with the amount of some $\delta$, the $k \times k$-minor will also change with at most the amount of $\delta$. Provided $\delta$ is sufficiently small, by continuity of the determinant, you get that the determinant will change by only $\varepsilon$, hence is still positive. Hence we again find a $k \times k$-minor with non-vanishing determinant, thus the rank is at least $k$.


TITLE: Morita equivalence of DG algebras? (reference needed)
QUESTION [17 upvotes]: A stupid question: whether ther exists a universally recognized definition of when two differential graded algebras should be Morita equivalent? I mean a sort of equivalence which would incorporate the usual Morita equivalence of algebras (when there is no differential) and the usual homotopy equivalence of differential graded algebras induced by an $A_\infty$/-quasi-isomorphism. In particular, I would like to believe, that the cyclic cohomology of such algebras should be canonically isomorphic via a generalised trace map.
I believe, this should exist, but a short search on Google rendered only papers on derived Morita equivalence, i.e. as much as I can judge, on equivalence between derived categories of algebras (non-differential). If I am mistaken and this is actually what I need, please, do explain it to me! Or if you know a good reference on the question I am asking, please, do share it with me!

REPLY [2 votes]: Just a comment on "For example, every linear invariant that I know of (algebraic K-theory, cyclic (co)homology, Hochschild (co)homology,...) is not only invariant under Morita equivalence but also under derived Morita equivalence." :
Given an algebra $A$ (view it as trivially d.g.), the (say left) global dimension of $A$ is a Morita invariant, but it is not a derived-Morita invariant. On the positive, the global dimension to be finite or infinite is a derived Morita invariant.