text:
\begin{document}

\title{Torsion cohomology for solvable groups of finite rank}

\author{\sc Karl Lorensen}

\maketitle

\vspace{-20pt}

\centerline{\footnotesize Department of Mathematics and Statistics}
\begin{spacing}{.5}

\centerline{\footnotesize Pennsylvania State University, Altoona College}

\centerline{\footnotesize Altoona, PA 16601}

\centerline{\footnotesize USA}
\vspace{5pt}

\centerline{\footnotesize E-mail address: {\tt kql3@psu.edu}}

\end{spacing}
\begin{abstract}  We define a class $\mathcal{U}$ of solvable groups of finite abelian section rank which includes all such groups that are virtually torsion-free as well as those that are finitely generated.  Assume that $G$ is a group in $\mathcal{U}$ and $A$ a $\mathbb ZG$-module. If $A$ is $\mathbb Z$-torsion-free and has finite $\mathbb Z$-rank,  we stipulate a condition on $A$ that guarantees that $H^n(G,A)$ and $H_n(G,A)$ must be finite for $n\geq 0$.  Moreover, if the underlying abelian group of $A$ is a \v{C}ernikov group,  we identify a similar condition on $A$ that ensures that $H^n(G,A)$ must be a \v{C}ernikov group for all $n\geq 0$.
\vspace{10pt}

\noindent {\bf Mathematics Subject Classification (2010)}:  20F16, 20J06
\end{abstract}
\vspace{20pt}

\section{Introduction}

This paper continues the study of the homology and cohomology of solvable groups of finite abelian section rank begun by P. H. Kropholler and the author in \cite{kroploren}. As in the previous paper, we will use the shorter term {\it solvable group of finite rank} to refer to such a group.
For a virtually torsion-free solvable group $G$ of finite rank and a $\mathbb ZG$-module $A$ that is torsion-free and has finite rank as an abelian group, the article \cite{kroploren} identifies a condition on $A$ that ensures that $H^n(G,A)$ and $H_n(G,A)$ are finite. The condition stipulates that $A$ must not possess a nonzero submodule whose underlying abelian group has a spectrum contained in that of $G$.  We remind the reader that the {\it spectrum} of a group is the set of primes for which the group has a quasicyclic section. Moreover, if $\pi$ is a set of primes, then a group is said to be {\it $\pi$-spectral} if its spectrum is contained in $\pi$.

Rather than just focusing on groups that are virtually torsion-free, the present paper considers the following, larger class of solvable groups of finite rank.

\begin{definition}{\rm  Let $\mathcal{U}$ be the class of solvable groups $G$ of finite rank such that $G$ contains a nilpotent normal subgroup $N$ with the following two properties:
\vspace{5pt}

\noindent (i) The quotient $Q=G/N$ is finitely generated and virtually abelian.
\vspace{5pt}

\noindent (ii) If $T$ is the torsion subgroup of $N$ and $\pi$ the spectrum of $G$, then there is a $G$-invariant, $N$-central series in $T$  of finite length in which each infinite factor, viewed as a $\mathbb ZQ$-module, can be expressed as a quotient of a $\mathbb ZQ$-module that is torsion-free and $\pi$-spectral as an abelian group.}

\end{definition}

\noindent Prominent examples of groups that belong to $\mathcal{U}$ are all finitely generated solvable groups of finite rank (Proposition 3.9).

Our first aim is to establish the following generalization of the main result in \cite{kroploren}.

\begin{thmA} Let $G$ be a group in $\mathcal{U}$ with spectrum $\pi$. Assume that $A$ is a $\mathbb ZG$-module whose underlying abelian group is torsion-free and has finite rank. Suppose further that $A$ does not have any nontrivial $\mathbb ZG$-submodules that are $\pi$-spectral as abelian groups. Then $H^n(G,A)$ and $H_n(G,A)$ are finite
for all $n\geq 0$.

\end{thmA}

To prove Theorem A, we employ the same reasoning that is used for the virtually torsion-free case in \cite{kroploren}. The argument is based on the Lyndon-Hochschild-Serre (LHS) spectral sequences for cohomology and homology for the group extension $1\rightarrow N\rightarrow G\rightarrow Q\rightarrow 1$, where $N$ and $Q$ are as related in the definition of $\mathcal{U}$.  Moreover, in order to study the initial pages of these spectral sequences, we apply [{\bf 3}, Proposition A], describing a property of the functors ${\rm Ext}^n_{\mathbb Z\Gamma}$ and ${\rm Tor}_n^{\mathbb Z\Gamma}$ when $\Gamma$ is an abelian group. The success of this approach hinges on the $\mathbb ZQ$-module structure of $H_nN$. What is required is that $H_nN$ belongs to $\mathcal{C}(Q,\pi)$, where $\mathcal{C}(Q,\pi)$ is the class of $\mathbb ZQ$-modules that can be constructed from modules that are virtually torsion-free of finite rank and $\pi$-spectral by forming finitely many quotients and extensions. The main theorem in \cite{kroploren} is established by showing that $H_nN$ belongs to $\mathcal{C}(Q,\pi)$ if $N$ is torsion-free.  In the present article, we invoke L. Breen's results from \cite{breen} concerning the integral homology of abelian groups in order to show that $H_nN$ always lies in $\mathcal{C}(Q,\pi)$ if $G$ belongs to $\mathcal{U}$ (Corollary 3.8).

The second, and primary, purpose of this article is to examine the situation where the underlying additive group of $A$ is torsion; more specifically, we will suppose that $A$ is a {\it \v{C}ernikov group}, meaning that it is a finite extension of a direct sum of finitely many quasicyclic groups. We wish to specify a condition, analogous to the one above for torsion-free modules, that guarantees that $H^n(G,A)$ is a \v{C}ernikov group for all $n\geq 0$.
The precise result that we prove is Theorem B below.

\begin{thmB}  Let $G$ be a group in $\mathcal{U}$ with spectrum $\pi$. Assume that $A$ is a $\mathbb ZG$-module whose underlying abelian group is a \v{C}ernikov $\omega$-group for a set of primes $\omega$. Suppose further that $A$ fails to have an infinite $\mathbb ZG$-submodule that is a quotient of a $\mathbb ZG$-module that is torsion-free of finite rank and $\pi$-spectral as an abelian group. Then, for all $n\geq 0$, $H^n(G,A)$ is a \v{C}ernikov $\omega$-group.
\end{thmB}

\noindent Observe that the conclusion that $H^n(G,A)$ is a \v{C}ernikov group is the best possible outcome under the hypotheses of Theorem B: very easy examples reveal that the cohomology groups may not be finite. Also, there is no mention of homology in the theorem because $H_n(G,A)$ will always be a \v{C}ernikov $\omega$-group if $A$ is a \v{C}ernikov $\omega$-group and $G$ is solvable of finite rank.

The proof of Theorem B, like that of its predecessor, draws extensively upon the techniques and findings of \cite{kroploren}. In addition, a key role is played by a theorem of J.C. Lennox and D. J. S. Robinson on the cohomology of nilpotent groups. The latter result enables us to reduce Theorem B to the case where $A$ lacks any infinite proper submodules and the subgroup $N$ from the definition of $\mathcal{U}$ acts trivially on $A$. It is shown in \cite{kroploren} that, provided $A$ is infinite, the triviality of the action of $N$ allows us to express the $E_2$-page of the LHS cohomology spectral sequence for $1\rightarrow N\rightarrow G\rightarrow Q\rightarrow 1$  as $E_2^{pq}={\rm Ext}^p_{\mathbb ZQ}(H_qN,A)$. As a result, we can obtain the conclusion of Theorem B by establishing that ${\rm Ext}^p_{\mathbb ZQ}(H_qN,A)$ is \v{C}ernikov. To accomplish this, we identify the following property of modules over integral group rings of finitely generated abelian groups.

\begin{propA} Let $G$ be a finitely generated abelian group. Assume that $A$ is a $\mathbb ZG$-module that has finite rank {\it qua} abelian group. Suppose further that $B$ is a $\mathbb ZG$-module whose underlying additive group is a \v{C}ernikov $\omega$-group for a set of primes $\omega$. If $A$ fails to have an infinite $\mathbb ZG$-module quotient that is isomorphic to a quotient of $B$, then, for $n\geq 0$,  ${\rm Ext}_{\mathbb ZG}^n(A,B)$ is a \v{C}ernikov $\omega$-group.
\end{propA}

\noindent Similar to the use of [{\bf 3}, Proposition A] in the proof of Theorem A, the application of Proposition A to the $E_2$-page of the spectral sequence described above requires that $H_nN$ lies in the class $\mathcal{C}(Q,\pi)$.

The plan of the paper is as follows. Proposition A is established  in \S 2; the proof relies on a generalization of [{\bf 3}, Proposition A], as well as an elementary property of Pontryagin duals of modules (Lemma 2.8). In \S 3 we discuss the class $\mathcal{C}(G,\pi)$ for a group $G$ and set of primes $\pi$, proving the pivotal Corollary 3.8.
 The last section of the article is devoted to the proofs of Theorems A and B, with the former established right at the outset.  The proof of Theorem B demands some further preparation: most importantly, we must show that the condition on the $\mathbb ZG$-submodules of $A$ in the theorem also holds for $\mathbb ZG_0$-submodules, where $G_0$ is any subgroup of finite index (Lemma 4.2). After proving Theorem B and enunciating a corollary, we conclude the paper by presenting two examples that demonstrate the necessity of the hypotheses of the theorem.

 Below we describe the notation and terminology that we will employ throughout the paper. In addition, we state a proposition summarizing the fundamental structural properties of solvable groups of finite rank. For the proofs of these properties, we refer the reader to \cite{robinson-lennox2}.
\vspace{10pt}

\begin{notation}{\rm

\vspace{5pt}

A {\it section} of a group is a quotient of one of its subgroups.
\vspace{5pt}

Let $R$ be a ring and $A$ an $R$-module. An {\it $R$-module section} of $A$ is an $R$-module quotient of one of its $R$-submodules.
\vspace{5pt}

Let $R$ be a ring. An $R$-module $A$ is said to be {\it bounded} if there exists $r\in R$ such that $rA=0$.
\vspace{5pt}

Suppose that $R$ is a principal ideal domain and $A$ an $R$-module. Let $F$ be the field of fractions of $R$. The {\it torsion-free rank} of $A$, denoted $r_0(A)$, is defined by $$r_0(A)={\rm dim}_F(A\otimes_RF).$$ Moreover, for each prime $p$ in $R$, the {\it $p$-rank}, $r_p(A)$, is defined by $$r_p(A)={\rm dim}_{R/pR}({\rm Hom}(R/pR,A)).$$ We say that $A$ has finite $R$-rank if $r_0(A)$ is finite and $r_p(A)$ is finite for every prime $p\in R$.
\vspace{5pt}

If $p$ is a prime, then $\hat{\mathbb Z}_p$ is the ring of $p$-adic integers and $\mathbb Z_{p^\infty}$ the quasicyclic $p$-group.
\vspace{5pt}

Let $G$ be a group. If $g, x\in G$, then $x^g=g^{-1}xg$. Moreover, for $H<G$ and $g\in G$, $H^g=g^{-1}Hg$.
\vspace{5pt}

The join of all the nilpotent normal subgroups of a group $G$, known as the {\it Fitting subgroup}, is denoted ${\rm Fitt}(G)$.
\vspace{5pt}

We designate the join of all the torsion normal subgroups of a group $G$ by $\tau(G)$.
\vspace{5pt}

A solvable group is said to have {\it finite rank} if its abelian sections all have finite $\mathbb Z$-rank.
\vspace{5pt}

If $G$ is a solvable group such that $G/\tau(G)$ has finite rank, then $h(G)$ denotes the {\it Hirsch rank} of $G$, namely, the number of infinite cyclic factors in any series of finite length whose factors are all either infinite cyclic or torsion. By the Schreier refinement theorem, this number is independent of the particular series selected.
\vspace{5pt}

A solvable group is {\it minimax} if it has a series of finite length in which each factor is either cyclic or quasicyclic. If $\pi$ is a set of primes, then a {\it $\pi$-minimax group} is a solvable minimax group that is $\pi$-spectral.
\vspace{5pt}

If $G$ is a \v{C}ernikov group, then the number of quasicyclic summands in the divisible subgroup of finite index is called the {\it \v{C}ernikov rank} of $G$, written $c(G)$.
\vspace{5pt}

Let $G$ be a group and $A$ a $\mathbb ZG$-module. If $A\otimes_{\mathbb Z}\mathbb Q$ is a simple $\mathbb QG$-module, then we will refer to $A$ as {\it rationally irreducible}. In other words, $A$ is rationally irreducible if and only if the additive group of $A$ is not torsion and, for every submodule $B$ of $A$, either $B$ or $A/B$ is torsion as an abelian group.
\vspace{5pt}

Assume that $G$ is a group. Let $A$ be a $\mathbb ZG$-module and $B$ a subgroup of the underlying additive group of $A$. We define

$$C_G(A)=\{g\in G: ga=a \  \ \forall\,  a\in A\}; \ \ \  N_G(B)=\{g\in G: gb\in B\  \ \forall\,  b\in B\}.$$}
\end{notation}
\vspace{10pt}

\begin{proposition} Let $G$ be a solvable group of finite rank.
\vspace{5pt}

\noindent (i) $G$ is virtually torsion-free if and only if $\tau(G)$ is finite.
\vspace{5pt}

\noindent (ii) If $\tau(G)$ is a \v{C}ernikov group, then ${\rm Fitt}(G)$ is nilpotent and $G/{\rm Fitt}(G)$ virtually abelian.
\vspace{5pt}

\noindent (iii) If $\tau(G)$ is finite, then $G/{\rm Fitt}(G)$ is finitely generated.
\vspace{5pt}

\noindent (iv) If $G$ is finitely generated, then $G$ is minimax.  \nolinebreak \hfill\(\square\)

\end{proposition}

\section{Proposition A}

Proposition A is based on the following more general formulation of [{\bf 3}, Proposition A]. The original result was stated for the case $R=\mathbb Z$, but the proof applies to any principal ideal domain satisfying the finiteness condition given.

\begin{proposition}{\rm (Kropholler, Lorensen, and Robinson)} Let $G$ be an abelian group and $R$ a principal ideal domain such that $R/Ra$ is finite for every  nonzero element $a$ of $R$. Assume that $A$ and $B$ are $RG$-modules that are $R$-torsion-free and have finite $R$-rank. Suppose further that $A$ fails to contain a nontrivial $RG$-submodule that is isomorphic to a submodule of $B$. Then  ${\rm Ext}_{RG}^n(A,B)$ and ${\rm Tor}_n^{RG}(A,B)$ are bounded $R$-modules
for all $n\geq 0$.  \nolinebreak \hfill\(\square\)
\end{proposition}

In proving Proposition 2.1, one requires the observation below from \cite{kroploren} concerning near splittings of modules. We will invoke this property in proving Proposition A, too.

\begin{proposition} Let $R$ be a ring. Suppose that $A$ and $B$ are $R$-modules such that, for every positive integer $m$, both the kernel and the cokernel of the map $b\mapsto mb$ from $B$ to $B$ are finite.  Let $0\rightarrow B\rightarrow E\rightarrow A\rightarrow 0$  be an $R$-module extension and $\xi$ the element of ${\rm Ext}_R^1(A,B)$ that corresponds to this extension. Then $\xi$ has finite order if and only if $E$ has a submodule $X$ such that $B\cap X$ and $E/B+X$ are finite. \nolinebreak \hfill\(\square\)
\end{proposition}

Two other results from \cite{kroploren} that we will employ in this section, and in \S 4, are the following proposition and its corollary.

\begin{proposition} Let $G$ be a group and $R$ a commutative ring. Suppose that $A$ and $B $ are $RG$-modules and regard ${\rm Ext}^n_R(A,B)$ and ${\rm Tor}_n^R(A,B)$ as $\mathbb ZG$-modules via the diagonal action for $n\geq 0$.
Then the following two statements hold.
\vspace{5pt}

\noindent (i) There is a cohomology spectral sequence whose $E_2$-page is given by
$$E_2^{pq}=H^p(G,{\rm Ext}^q_R(A,B)),$$
\noindent and that converges to ${\rm Ext}^n_{RG}(A,B)$.
\vspace{5pt}

\noindent (ii) There is a homology spectral sequence whose $E^2$-page is given by
$$E^2_{pq}=H_p(G,{\rm Tor}_q^R(A,B)),$$
\noindent and that converges to ${\rm Tor}_n^{RG}(A,B)$. \hfill\(\square\)
\end{proposition}

\begin{corollary} Let $G$ be a group and $R$ a commutative ring. Suppose that $A$ and $B $ are $RG$-modules and view both ${\rm Hom}_R(A,B)$ and $A\otimes_R B$ as $\mathbb ZG$-modules under the diagonal actions.

\noindent (i) If either $A$ is projective or $B$ is injective as an $R$-module, then, for $n\geq 0$,

$${\rm Ext}^n_{RG}(A,B)\cong H^n(G,{\rm Hom}_R(A,B)).$$

\noindent (ii) If $B$ is flat as an $R$-module, then, for $n\geq 0$,

$${\rm Tor}_n^{RG}(A,B)\cong H_n(G, A\otimes_R B).$$
\hfill\(\square\)
\end{corollary}

The argument for Proposition A  also makes use of the properties of modules described in the following two lemmas. The proofs of these are entirely elementary and therefore left to the reader.

\begin{lemma} Let $G$ be a group. Suppose that $D$ is a $\mathbb ZG$-module whose underlying abelian group is a divisible \v{C}ernikov group. Then any $\mathbb ZG$-module extension of $D$ by a finite module can be expressed as a $\mathbb ZG$-module extension of a finite module by a quotient of $D$.  \nolinebreak \hfill\(\square\)
\end{lemma}

\begin{lemma} Let $R$ be a principal ideal domain and $G$ a group. Assume that $A$ and $B$ are $RG$-modules such that $A$ is finitely generated as an $RG$-module.
\vspace{5pt}

\noindent (i) If $B$ is finitely generated as an $R$-module, then ${\rm Hom}_{RG}(A,B)$ is finitely generated as an $R$-module.
\vspace{5pt}

\noindent (ii) If $B$ has finite rank as an $R$-module, then ${\rm Hom}_{RG}(A,B)$ has finite rank as an $R$-module.
\nolinebreak \hfill\(\square\)

\end{lemma}

Before establishing Proposition A, we discuss a key tool employed in the proof, namely, the Pontryagin dual of a module, defined for the convenience of the reader below.

\begin{definition}{\rm Assume that $p$ is a prime. If $A$ is an abelian $p$-group with finite rank, then the {\it Pontryagin dual} $A^\ast$ of $A$ is the group ${\rm Hom}_{\mathbb Z}(A,\mathbb Z_{p^\infty})$. The dual $A^\ast$ is a finitely generated $\hat{\mathbb Z}_p$-module. If $A$ happens to be endowed with a $\mathbb ZG$-module structure for a group $G$, then we equip $A^\ast$ with a $G$-action by letting $(g \phi)(x)=\phi(g^{-1}x)$ for every $g\in G$, $\phi\in A^\ast$, and  $x\in A$. In this way, we may view $A^\ast$ as a $\hat{\mathbb Z}_pG$-module.

For any finitely generated $\hat{\mathbb Z}_p$-module $M$, the {\it Pontryagin dual} $M^\ast$ of $M$ is the group ${\rm Hom}_{\mathbb Z}(M,\mathbb Z_{p^\infty})$.  In this case, the dual $M^\ast$ is an abelian $p$-group with finite rank. If $M$ is a $\mathbb ZG$-module for a group $G$, then we regard $M^\ast$ as a $\mathbb ZG$-module using the same $G$-action as above.
}

\end{definition}

Some elementary properties of Pontryagin duals are described in the following two lemmas. For proofs of these, we refer the reader to \cite{wilson}.

\begin{lemma} Let $p$ be a prime.
\vspace{5pt}

\noindent (i) If $A$ and $B$ are abelian $p$-groups of finite rank, then

$${\rm Hom}_{\mathbb Z}(A, B)\cong {\rm Hom}_{\hat{\mathbb Z}_p}(B^\ast, A^\ast).$$
\vspace{5pt}

\noindent (ii) If $M$ and $N$ are finitely generated $\hat{\mathbb Z}_p$-modules,  then

$${\rm Hom}_{\hat{\mathbb Z}_p}(M, N)\cong {\rm Hom}_{\mathbb Z}(N^\ast, M^\ast).$$ \nolinebreak \hfill\(\square\)

\end{lemma}

\begin{lemma} Let $p$ be a prime and $G$ a group.
\vspace{5pt}

\noindent (i) If $A$ is a $\mathbb ZG$-module whose additive group is a $p$-group of finite rank, then $A$ is isomorphic to its double dual $A^{\ast \ast}$.
\vspace{5pt}

\noindent (ii) If $M$ is a $\hat{\mathbb Z}_pG$-module that is finitely generated as a $\hat{\mathbb Z}_p$-module, then $M$ is isomorphic to its double dual $M^{\ast \ast}$. \nolinebreak \hfill\(\square\)
\end{lemma}

The crucial property of Pontryagin duals for our purposes is described in Lemma 2.9 below, which appeared in a preliminary version of \cite{kroploren}. This fact is undoubtedly well known; however, as we are unaware of any reference to it in the published literature, we include a proof.

\begin{lemma} Assume that $p$ is a prime and $G$ a group. Let $A$ and $B$ be $\mathbb ZG$-modules whose additive groups are $p$-groups of finite rank. Suppose further that, as an abelian group, $B$ is divisible. Then

$${\rm Ext}_{\mathbb ZG}^n(A,B)\cong {\rm Ext}_{\hat{\mathbb Z}_pG}^n(B^\ast,A^\ast)$$
for $n\geq 0$.
\end{lemma}

\begin{proof} We have the following chain of isomorphisms.

$${\rm Ext}_{\mathbb ZG}^n(A,B)\cong H^n(G,{\rm Hom}_{\mathbb Z}(A,B))\cong H^n(G,{\rm Hom}_{\hat{\mathbb Z}_p}(B^\ast,A^\ast))\cong {\rm Ext}_{\hat{\mathbb Z}_pG}^n(B^\ast,A^\ast).$$
The first and last isomorphisms in this chain are consequences of Corollary 2.4(i). The first arises from the divisibility of the underlying abelian group of $B$ and the last from the resulting projectivity of $B^\ast$ as a $\hat{\mathbb Z}_p$-module.
\end{proof}

We are now prepared to prove Proposition A.

\begin{proof}[{\bf Proof of Proposition A}] Without incurring any significant loss of generality, we can assume that $\omega$ contains just a single prime $p$. We begin the proof with the observation that, if $B$ is finite, then ${\rm Ext}_{\mathbb ZG}^n(A,B)$ must be a finite $p$-group for all $n\geq 0$.
This follows immediately from Proposition 2.3(i) and the fact that ${\rm Ext}_{\mathbb Z}^n(A,B)$ is a finite $p$-group for all $n\geq 0$. Combining this observation with Lemma 2.5 permits a further reduction to the case where $B$ is divisible.

In the next two paragraphs, we consider the case where either $A$ is rationally irreducible or the additive group of $A$ is torsion. Under either of these assumptions, there is a finitely generated submodule $A_0$ of $A$ such that the additive group of $A/A_0$ is torsion. Since $\mathbb ZG$ is Noetherian, $A_0$ is of type ${\rm FP}_\infty$. Invoking Lemma 2.6(ii), we deduce that ${\rm Ext}^n_{\mathbb ZG}(A_0,B)$ is a \v{C}ernikov $p$-group for $n\geq 0$.  Now let $P$ be the $p$-torsion subgroup of $A/A_0$.
 Our intention is to show that  ${\rm Ext}_{\mathbb ZG}^n(P,B)$ is a finite $p$-group for $n\geq 0$, which will yield the conclusion of the proposition.

To establish that ${\rm Ext}_{\mathbb ZG}^n(P,B)$ is a finite $p$-group, we use the fact that ${\rm Ext}_{\mathbb ZG}^n(P, B)$ and ${\rm Ext}_{\hat{\mathbb Z}_pG}^n(B^\ast,P^\ast)$ are isomorphic for all $n\geq 0$. Since $\hat{\mathbb Z}_pG$ is Noetherian and $B^\ast$ is a finitely generated $\hat{\mathbb Z}_pG$-module,  $B^\ast$ has type ${\rm FP}_{\infty}$ as a $\hat{\mathbb Z}_pG$-module. Thus, by Lemma 2.6(i),  ${\rm Ext}_{\hat{\mathbb Z}_pG}^n(B^\ast,P^\ast)$ is finitely generated as a $\hat{\mathbb Z}_p$-module. Suppose that ${\rm Ext}_{\hat{\mathbb Z}_pG}^n(B^\ast,P^\ast)$ is not a finite $p$-group for some $n\geq 0$. Then ${\rm Ext}_{\hat{\mathbb Z}_pG}^n(B^\ast,P^\ast)$ is not bounded as a $\hat{\mathbb Z}_p$-module. According to Proposition 2.1, this means that $P^\ast$ and $B^\ast$ have isomorphic nontrivial $\hat{\mathbb Z}_pG$-submodules. Dualizing a second time, we deduce that $P$ and $B$ have isomorphic infinite $\mathbb ZG$-module quotients, thereby contradicting our hypothesis. Therefore, for each $n\geq 0$, ${\rm Ext}_{\hat{\mathbb Z}_pG}^n(B^\ast,P^\ast)$, and hence ${\rm Ext}_{\mathbb ZG}^n(P, B)$, must be a finite $p$-group. This concludes the argument for the case where $A$ is either rationally irreducible as a module or torsion {\it qua} abelian group.

We handle the general case by inducting on $h(A)$, the case $h(A)=0$ having already been established above. Suppose $h(A)\geq 1$. Let $U$ be a submodule of $A$ with $h(U)$ as large as possible while still remaining less than $h(A)$. It follows from the case proved in the first paragraph that ${\rm Ext}^n_{\mathbb ZG}(A/U, B)$ is a \v{C}ernikov $p$-group for $n\geq 0$. Consequently, the conclusion of the theorem will follow from the inductive hypothesis if we succeed in showing that $U$ does not have a quotient that is isomorphic to a nontrivial quotient of $B$. To accomplish this, suppose that $U$ possesses such a quotient and take $V$ to be a submodule of $U$  such that $U/V\cong W$, where $W$ is a nontrivial quotient of $B$. By the case of the proposition established above, ${\rm Ext}^1_{\mathbb ZG}(A/U,U/V)$ is torsion. Appealing to Proposition~2.2, we deduce that $A$ has a quotient that is isomorphic to a finite extension of a nontrivial quotient of $W$. Lemma 2.5 implies, then, that there is a nontrivial quotient of $A$ that is isomorphic to a quotient of $W$. But the presence of such a quotient contradicts our hypothesis. Therefore, $U$ cannot possess a nontrivial quotient that is isomorphic to a quotient of $B$. This completes the proof of Proposition A.

\end{proof}

\section{The Class $\mathfrak{C}(G, \pi)$}

To prove Theorems A and B, we need to understand the following class of modules. This class is a slight extension of the class with the same designation discussed in \cite{kroploren}.

\begin{definition} {\rm Assume that $G$ is a group and $\pi$ a set of primes. Let $\mathfrak{C}(G, \pi)$ be the smallest class of $\mathbb ZG$-modules with the following two properties.
\vspace{5pt}

\noindent (i) The class $\mathfrak{C}(G, \pi)$ contains every finite $\mathbb ZG$-module as well as every $\mathbb ZG$-module whose underlying abelian group is torsion-free of finite rank and $\pi$-spectral.
\vspace{5pt}

\noindent (ii) The class $\mathfrak{C}(G, \pi)$ is closed under forming $\mathbb ZG$-module quotients and extensions.}
\end{definition}

The class $\mathfrak{C}(G, \pi)$ has the following two additional closure properties, which can be established by arguments in the same vein as those for [{\bf 3}, Lemmas 2.4, 2.5]. The basic strategy is to induct on the number of closure operations from (ii) that are necessary to construct the module from ones that are either finite or torsion-free of finite rank and $\pi$-spectral. We leave the small task of supplying complete proofs to the reader.

\begin{lemma} For any group $G$ and set of primes $\pi$, the class $\mathfrak{C}(G, \pi)$ is closed under forming $\mathbb ZG$-module sections. \nolinebreak \hfill\(\square\)
\end{lemma}

\begin{lemma} Assume that $G$ is a group and $\pi$ a set of primes.
Let $A$ and $B$ be $\mathbb ZG$-modules  and, for each $n\geq 0$, regard ${\rm Tor}^{\mathbb Z}_n(A,B)$ as a $\mathbb ZG$-module under the diagonal action. If $A$ and $B$ belong to $\mathfrak{C}(G, \pi)$, then ${\rm Tor}^{\mathbb Z}_n(A,B)$ lies in $\mathfrak{C}(G, \pi)$ for $n\geq 0$. \nolinebreak \hfill\(\square\)
\end{lemma}

Of central importance for our purposes are the following two homological properties of the class $\mathfrak{C}(G, \pi)$. Both of these can be deduced in a manner similar to [{\bf 3}, Lemma 2.6], although for Lemma 3.4 one must invoke Proposition A of the present paper instead of [{\bf 3}, Proposition A].

\begin{lemma} Let $\pi$ be a set of primes and $G$ an abelian group. Assume that $B$ is a $\mathbb ZG$-module whose additive group is torsion-free with finite rank. Suppose further that there are no nontrivial $\mathbb ZG$-submodules of $B$ that are $\pi$-spectral as abelian groups. If $A$ is a $\mathbb ZG$-module in $\mathfrak{C}(G, \pi)$ and $\bar{B}$ a $\mathbb ZG$-module quotient of $B$, then ${\rm Ext}_{\mathbb ZG}^n(A,\bar{B})$ and ${\rm Tor}^{\mathbb ZG}_n(A,\bar{B})$ are bounded $\mathbb Z$-modules for $n\geq 0$. \nolinebreak \hfill\(\square\)
\end{lemma}

\begin{lemma} Let $\pi$ be a set of primes and $G$ a finitely generated abelian group. Assume that $B$ is a $\mathbb ZG$-module whose additive group is a \v{C}ernikov $\omega$-group for some set of primes $\omega$. Suppose further that there is no infinite $\mathbb ZG$-module quotent of $B$ that is also a quotient of a $\mathbb ZG$-module that is torsion-free of finite rank and $\pi$-spectral as an abelian group. If $A$ is a $\mathbb ZG$-module in $\mathfrak{C}(G, \pi)$, then, for each $n\geq 0$, ${\rm Ext}_{\mathbb ZG}^n(A, B)$ is a \v{C}ernikov $\omega$-group. \nolinebreak \hfill\(\square\)
\end{lemma}

Our main goal in this section is to venture beyond what was shown in \cite{kroploren} about the contents of $\mathfrak{C}(G, \pi)$ by showing, first, that the class is closed under the integral homology functor. This will then permit us to identify a large class of nilpotent normal subgroups of solvable groups of finite rank  whose integral homology lies within the eponymous class with respect to the ambient group. In order to establish the former property, we will avail ourselves of the results of Breen in \cite{breen} concerning the integral homology of abelian groups. We begin by defining the $m$-fold torsion product of abelian groups, following Breen's notation and terminology.

\begin{definition}{\rm  Assume that $A_1, \dots, A_m$ are abelian groups. For each $i=1, \dots, m$, let ${\bf F}^i$ be a $\mathbb Z$-flat resolution of $A_i$. Then, for $n\geq 0$, we define

$${\rm Tor}^{\mathbb Z}_n(A_1, \dots, A_m)=H_n({\bf F}^1\otimes_{\mathbb Z} \dots \otimes_{\mathbb Z}{\bf F}^m).$$

\noindent If $m=2$, this group is isomorphic to ${\rm Tor}^{\mathbb Z}_n(A_1, A_2)$ as it is usually defined.}

\end{definition}

Lemma 3.2 can be generalized to these higher order torsion products.

\begin{lemma} Assume that $G$ is a group and $\pi$ a set of primes.
Let  $A_1, \dots, A_m$ be $\mathbb ZG$-modules  and, for each $n\geq 0$, regard ${\rm Tor}^{\mathbb Z}_n(A_1, \dots, A_m)$ as a $\mathbb ZG$-module under the diagonal action. If $A_1, \dots , A_m$ belong to $\mathfrak{C}(G, \pi)$, then ${\rm Tor}^{\mathbb Z}_n(A_1, \dots, A_m)$ lies in $\mathfrak{C}(G, \pi)$ for $n\geq 0$.
\end{lemma}

\begin{proof}  We proceed by induction on $m$, the case $m=2$ having been disposed of in Lemma 3.2. Suppose $m>2$. By the K\"unneth formula ([{\bf 2}, Theorem V.2.1]), there is a short exact sequence

$$0\longrightarrow {\rm Tor}^\mathbb Z_n(A_1,\cdots A_{m-1})\otimes_{\mathbb Z} A_m\longrightarrow {\rm Tor}^{\mathbb Z}_n(A_1, \cdots, A_m)\longrightarrow {\rm Tor}_1^{\mathbb Z}({\rm Tor}^\mathbb Z_{n-1}(A_1,\cdots A_{m-1}), A_m)\longrightarrow 0$$

\noindent of $\mathbb ZG$-modules for $n\geq 0$. In view of the inductive hypothesis, ${\rm Tor}^\mathbb Z_i(A_1,\cdots A_{m-1})$ belongs to $\mathfrak{C}(G,\pi)$ for all $i\geq 0$.  Invoking Lemma 3.2 and the closure of $\mathfrak{C}(G,\pi)$ under extensions, we deduce that ${\rm Tor}^{\mathbb Z}_n(A_1, \dots, A_m)$ must reside in $\mathfrak{C}(G,\pi)$ for $n\geq 0$.
\end{proof}

Equipped with the preceding lemma, we can apply two of the results in \cite{breen} to obtain our most important closure property for $\mathfrak{C}(G, \pi)$.

\begin{lemma} Assume that $G$ is a group and $\pi$ a set of primes. If $A$ is a $\mathbb ZG$-module that belongs to $\mathfrak{C}(G,\pi)$, then $H_nA$ lies in $\mathfrak{C}(G,\pi)$ for $n\geq 0$.
\end{lemma}

\begin{proof} According to [{\bf 1}, (1.10)], there is a natural spectral sequence $\{E^r_{pq}\}$ converging to $H_nA$ such that $E^2_{pq}=L_p\Lambda^qA$ for $p, \ q\geq 0$. Here $L_p\Lambda^qA$ represents the $p$th derived functor of the $q$th exterior power of $A$. Moreover, by [{\bf 1}, Theorem 4.7], there is a $\mathbb ZG$-module monomorphism from $L_p\Lambda^qA$ to ${\rm Tor}^{\mathbb Z}_p(\underbrace{A, \dots, A)}_q$ for $q>0$. It follows, then, from Lemmas 3.5 and 3.1 that $E^2_{pq}$ lies in $\mathfrak{C}(G,\pi)$ for $q>0$. In addition, $E^2_{p0}$, being trivial, belongs to $\mathfrak{C}(G,\pi)$. Since the differentials in Breen's spectral sequence are $\mathbb ZG$-module homomorphisms, $E^\infty_{pq}$ can be regarded as a $\mathbb ZG$-module section of $E^2_{pq}$. Hence $E^\infty_{pq}$ must be a member of $\mathfrak{C}(G,\pi)$ for $p,\ q\geq 0$.  Furthermore, $H_nA$ has a $\mathbb ZG$-module series whose factors are the modules $E^\infty_{pq}$ for $p+q=n$. Therefore, $H_nA$ must belong to $\mathfrak{C}(G,\pi)$ for $n\geq 0$.
\end{proof}

Now we prove the desired result concerning the integral homology of a nilpotent normal subgroup.

\begin{proposition} Assume that $G$ is a group with a nilpotent normal subgroup $N$ of finite rank. Let $\pi={\rm spec}(N)$ and $Q=G/N$. Suppose further that $N$ has a $G$-invariant central series in which each infinite factor can be expressed as a quotient of a $\mathbb ZQ$-module whose underlying abelian group is torsion-free of finite rank and $\pi$-spectral. Then $H_nN$ belongs to $\mathfrak{C}(Q, \pi)$ for $n\geq 0$.

\end{proposition}

\begin{proof} Let $1=N_0<N_1<\dots <N_r=N$ be a central series with the properties described. We prove the proposition by induction on $r$, the case $r=0$ being trivial. Assume $r>0$. By the inductive hypothesis, $H_n(N/N_1)$ is a member of $\mathfrak{C}(Q, \pi)$ for $n\geq 0$.  Moreover, according to Lemma 3.6, $H_nN_1$ lies in  $\mathfrak{C}(Q, \pi)$ for $n\geq 0$.  We now consider the LHS homology spectral sequence associated to the central extension $1\rightarrow N_1\rightarrow N\rightarrow N/N_1\rightarrow 1$. In this spectral sequence, $E_{pq}^2=H_p(N/N_1, H_qN_1)$, and so there is an exact sequence

$$0\longrightarrow H_p(N/N_1)\otimes_{\mathbb Z} H_qN_1\longrightarrow E_{pq}^2\longrightarrow {\rm Tor}_1^{\mathbb Z}(H_{p-1}(N/N_1), H_qN_1)\longrightarrow 0.$$

\noindent An appeal to Lemma 3.2 allows us to deduce that $E_{pq}^2$ lies in $\mathfrak{C}(Q, \pi)$.  Since the differentials in the spectral sequence are $\mathbb ZQ$-module homomorphisms, it follows from Lemma 3.1 that $E_{pq}^\infty$ belongs to $\mathfrak{C}(Q, \pi)$.  As a result, $H_nN$ is a member of $\mathfrak{C}(Q, \pi)$ for $n\geq 0$.

\end{proof}

As an immediate consequence of Proposition 3.7, we have

\begin{corollary} Assume that $G$ is a group in the class $\mathcal{U}$ and $N$ and $Q$ are as described in the definition of $\mathcal{U}$. Then $H_nN$ belongs to $\mathfrak{C}(Q, \pi)$ for $n\geq 0$.  \nolinebreak \hfill\(\square\)
\end{corollary}

We conclude this section by highlighting an important type of group that belongs to $\mathcal{U}$.

\begin{proposition}  Every finitely generated solvable minimax group lies in $\mathcal{U}$.
\end{proposition}

\begin{proof} Let $G$ be a finitely generated solvable minimax group with spectrum $\pi$. Write $N={\rm Fitt}(G)$ and $Q=G/N$. Let $i$ be a positive integer and consider the canonical $\mathbb ZQ$-module epimorphism $\theta_i: \underbrace{N_{\rm ab}\otimes_{\mathbb Z} \dots \otimes_{\mathbb Z} N_{\rm ab}}_{i}\rightarrow \gamma_iN/\gamma_{i+1}N$. Since $G$ is finitely generated as a group and $Q$ polycyclic, $N_{\rm ab}$ is finitely generated as a $\mathbb ZQ$-module. As a result,  $N_{\rm ab}$ is a Noetherian $\mathbb ZQ$-module and hence virtually torsion-free as an abelian group. Therefore, each factor in the lower central series of $N$ is a homomorphic image of a $\mathbb ZQ$-module whose underlying additive group is virtually torsion-free of finite rank and $\pi$-spectral. This, then, implies that $G$ belongs to $\mathcal{U}$ .

\end{proof}

\section{Theorems A and B}

This section is devoted to the proofs of the main results of the paper, Theorems A and B.  We begin with the former, as it presents the least difficulty.

\begin{proof}[{\bf Proof of Theorem A}]

As pointed out at the beginning of the proof of [{\bf 3}, Theorem A], the conclusion will follow if we can show that the cohomology and homology groups are bounded. Let $N$ be a nilpotent normal subgroup of $G$ with the two properties described in the definition of $\mathcal{U}$. Put $Q=G/N$.
First we treat the case where $A$ is rationally irreducible. Under this assumption, $G/C_G(A)$ is virtually abelian due to [{\bf 5}, 3.1.6]. Furthermore, as argued in the proof of [{\bf 5}, 5.2.3], the structure of the multiplicative group of an algebraic number field allows us to conclude that $G/C_G(A)$ is finitely generated. Hence, by replacing $N$ with $N\cap C_G(A)$, we can assume $N<C_G(A)$. Let $G_0$ be a subgroup of $G$ with finite index such that $Q_0=G_0/N$ is abelian. Since $H_nN$ lies in $\mathfrak{C}(Q_0,\pi)$ by Corollary 3.8, we can argue exactly as in the second and third paragraphs of the proof of [{\bf 3}, Theorem A], except that now we invoke Lemma 3.3 rather than [{\bf 3}, Lemma 2.6]. This reasoning allows us to deduce that  $H^n(G_0,A)$ and $H_n(G_0,A)$ are bounded for $n\geq 0$. Thus the same holds for $H^n(G,A)$ and $H_n(G,A)$.

We handle the general case by inducting on $h(A)$, the case $h(A)=0$ being trivial. Assume $h(A)\geq 1$. Choose $B$ to be a submodule of $A$ with $h(B)$ as large as possible subject to the conditions that $h(B)<h(A)$ and the additive group of $A/B$ is torsion-free. By the inductive hypothesis, $H^n(G,B)$ and $H_n(G,B)$ are finite for $n\geq 0$. Since $A/B$ is rationally irreducible, the conclusion will then follow from the case proved above if we can show that $A/B$ fails to contain a nonzero $\mathbb ZG$-submodule that is torsion-free and $\pi$-spectral as an abelian group. Suppose that $A/B$ contains such a submodule, and let $C$ be a submodule of $A$ properly containing $B$ such that $C/B$ is torsion-free and $\pi$-spectral {\it qua} abelian group. Because $G/C_G(C/B)$ is finitely generated and virtually abelian, we may assume $N<C_G(C/B)$. Now, as in the proof of [{\bf 3}, Theorem A], we put $\Gamma=C\rtimes G$. It is straightforward to deduce that $\Gamma/B$ belongs to the class $\mathcal{U}$.  As a result, $H^2(\Gamma/B,B)$ is finite. This allows us to derive a contradiction by adducing the argument employed at the end of the proof of [{\bf 3}, Theorem A]. Therefore, we are compelled to conclude that $A/B$ fails to possess a nonzero $\mathbb ZG$-submodule that is torsion-free and $\pi$-spectral as an abelian group. This completes the proof of the theorem.
\end{proof}

Combining Theorem A with Proposition 3.9, we obtain

\begin{corollary} Let $G$ be a finitely generated solvable minimax group with spectrum $\pi$. Assume that $A$ is a $\mathbb ZG$-module whose underlying abelian group is torsion-free and has finite rank. Suppose further that $A$ does not have any nontrivial $\mathbb ZG$-submodules that are $\pi$-spectral as abelian groups. Then $H^n(G,A)$ and $H_n(G,A)$ are finite
for all $n\geq 0$.  \nolinebreak \hfill\(\square\)

\end{corollary}

The necessity of the condition on the submodules of $A$ in Theorem A is immediately apparent if we take $G=C_{\infty}\times C_{\infty}$ and $A=\mathbb Z$, with $G$ acting trivially. In this situation, $H^0(G,A)=\mathbb Z$, $H^1(G,A)=\mathbb Z\oplus \mathbb Z$, and $H^2(G,A)=\mathbb Z$. Rather less obvious is that the hypothesis that $G$ belongs to $\mathcal{U}$ can also not be discarded.  To verify that, we present the example below, which offers a cohomological interpretation of  [{\bf 3}, Example 3.3].

\begin{example} {\rm Let $p$ and $q$ be distinct primes.  Take $Q$ to be an infinite cyclic group, and let $B$ be the $\mathbb ZQ$-module obtained by letting the generator of $Q$ act on $\mathbb Z_{p^\infty}$ by multiplication by $q$. Put $G=B\rtimes Q$.  Let $A$ be the underlying additive group of the ring $\mathbb Z[1/q]$ and endow $A$ with a $\mathbb ZG$-module structure by letting the generator of $Q$ act  by multiplication by $q$ and $B$ centralize $A$.  We will employ the LHS spectral sequence associated to the extension $1\rightarrow B\rightarrow G\rightarrow Q\rightarrow 1$ to compute $H^2(G,A)$. In this spectral sequence,  $E^{11}_2=0$, and the universal coefficient theorem yields $E^{02}_2\cong {\rm Ext}_{\mathbb Z}^1(B,A)^Q$.  It is straightforward to ascertain ${\rm Ext}^1_{\mathbb Z}(B,A)\cong \hat{\mathbb Z}_p$. Therefore, since the diagonal action of $Q$ on ${\rm Ext}^1_{\mathbb Z}(B,A)$ is trivial, we deduce $H^2(G,A)\cong \hat{\mathbb Z}_p$. Thus we have described an example where all the hypotheses of Theorem A are satisfied save the stipulation that $G$ is a member of \ $\mathcal{U}$, and yet the cohomology in dimension two is torsion-free and nontrivial.
}

\end{example}

To lay the groundwork for Theorem B, we prove three lemmas concerning modules that can be expressed as quotients of modules that are torsion-free of finite rank.

\begin{lemma} Suppose that $\pi$ is a set of primes. Let $G$ be a group and $G_0<G$ with $[G:G_0]<\infty$. Assume that $A$ is a $\mathbb ZG$-module that can be expressed as a $\mathbb ZG_0$-module quotient of a $\mathbb ZG_0$-module whose underlying abelian group is torson-free, has finite rank, and is $\pi$-spectral.  Then $A$ can also be realized as a $\mathbb ZG$-module quotient of a $\mathbb ZG$-module whose additive group possesses the same three properties.
\end{lemma}

\begin{proof} Assume that  $M_0$ is a $\mathbb ZG_0$-module that is torsion-free of finite rank and $\pi$-spectral as an abelian group, and that there is a $\mathbb ZG_0$-module epimorphism $\phi_0: M_0\to A$. Set $\Gamma=A\rtimes G$, $\Gamma_0=A\rtimes G_0$, and $\Omega_0=M_0\rtimes G_0$. Moreover, let $\phi_0^\ast: \Omega_0\to \Gamma_0$ be the epimorphism induced by $\phi_0$. In addition, suppose that $\theta:F\to \Gamma$ is a group epimorphism such that $F$ is a free group, and put $F_0=\theta^{-1}(\Gamma_0)$ and $N=\theta^{-1}(A)$. Since $F_0$ is free, there is a homomorphism $\psi_0: F_0\to \Omega_0$ that makes the diagram

\begin{displaymath}\xymatrix{
&\Omega_0  \ar[r]^{\phi_0^\ast} &\Gamma_0\\
& F_0 \ar[u]_{\psi_0}  \ar[ur]_{\theta_0} &}
\end{displaymath}
commute, where $\theta_0:F_0\to \Gamma_0$ is the epimorphism induced by $\theta$. Setting $K={\rm Ker}\ \theta$ and $K_0={\rm Ker}\ \psi_0$, we have the chain of subgroups $N'< K_0<K<N<F_0<F$. Now let $L=\bigcap_{f\in F} K_0^f$. Then $L\lhd F$ and $N'<L<K_0$.  Because $[F:F_0]$ is finite, there are only finitely many distinct conjugates of $K_0$ in $F$. Furthermore, for each $f\in F$, $N/K_0^f$ is a torsion-free abelian group of finite rank that is $\pi$-spectral. As a result, $N/L$ must enjoy the same properties. In addition, $K$ acts trivially on $N/L$ by conjugation. This means that we can regard $N/L$ as a $\mathbb Z\Gamma$-module, and hence as a $\mathbb ZG$-module via the map $G\to \Gamma$. Denoting this $\mathbb ZG$-module by $M$, we then have a $\mathbb ZG$-module epimorphism $\phi:M\to A$ induced by $\theta$. The lemma has now been proven.
\end{proof}

\begin{lemma} Suppose that $\pi$ is a set of primes. Let $G$ be a group and $G_0<G$ with $[G:G_0]<\infty$. Assume that $A$ is a $\mathbb ZG$-module and $B$ a $\mathbb ZG_0$-submodule of $A$. Suppose further that $B$ can be expressed as a quotient of a $\mathbb ZG_0$-module whose underlying abelian group is torson-free, has finite rank, and is $\pi$-spectral.  Then the $\mathbb ZG$-submodule of $A$ generated by $B$ can be realized as a $\mathbb ZG$-module quotient of a $\mathbb ZG$-module whose additive group enjoys the same three properties.
\end{lemma}

\begin{proof}  Since $G_0$ contains a $G$-invariant subgroup of finite index, it suffices to consider the case where $G_0\lhd G$. Let $g_1,\dots, g_r$ be a complete list of coset representatives of $G_0$ in $G$.  Then $\bar{B}=g_1B+\dots +g_r B$ is the $\mathbb ZG$-submodule of $A$ generated by $B$. According to our hypotheses, there is a $\mathbb ZG_0$-module $M$ and a $\mathbb ZG_0$-module epimorphism $\phi:M\to B$. Now, for each $i=1, \dots, r$, we construct a new $\mathbb ZG_0$-module $M_i$ with the same underlying abelian group as $M$. The new action, denoted $\circ_i$ for each $i$, is defined in terms of the old action as follows: $g_0\circ_i x=g_0^{g_i}x$ for all $x\in M_i$ and $g_0\in G_0$. Next set $\bar{M}=  M_1\oplus \dots \oplus M_r$, and define the map $\bar{\phi}:\bar{M}\to \bar{B}$ by
$$\bar{\phi}(x_1,\dots, x_r)=g_1\phi (x_1)+\dots + g_r\phi(x_r)$$
for all $(x_1, \dots, x_r)\in \bar{M}$. It is easily checked that $\bar{\phi}$ is an $\mathbb ZG_0$-module epimorphism. Therefore, the conclusion of the lemma follows by applying Lemma 4.1.

\end{proof}

\begin{lemma} Assume that $\pi$ is a set of primes.  Let $G$ be a group and $A$ a $\mathbb ZG$-module that is divisible as an abelian group. Suppose that $A$ has a finite submodule $B$, and that there is an epimorphism $\theta: U\to A/B$, where $U$ is a $\mathbb ZG$-module whose additive group is torsion-free of finite rank and $\pi$-spectral. Then the module $A$, too, can be realized as a quotient of a $\mathbb ZG$-module whose additive group is torsion-free of finite rank and $\pi$-spectral.

\end{lemma}

\begin{proof}  From the map $\theta: U\to A/B$, we can construct a $\mathbb ZG$-module extension $0\rightarrow B\rightarrow M\rightarrow U\rightarrow 0$ that fits into a commutative diagram of the form

\begin{displaymath} \begin{CD}
0 @>>> B@>>> M @>>> U @>>>0\\
&& @| @VVV @VV{\theta}V &&\\
0 @>>> B @>>> A@>>> A/B @>>>0.
\end{CD} \end{displaymath}
Since $M$ is virtually torsion-free, we can select a torsion-free submodule $M_0$ of $M$ with finite index such that $M_0\cap B=0$. The submodule $M_0$, then, is mapped surjectively to $A$, thereby witnessing the property desired.
\end{proof}

In addition to the preceding three lemmas, the proof of Theorem B depends upon three homological properties. The first two are contained in Lemma 3.4 and Corollary 3.8, respectively; the third is articulated in the following special case of a theorem due to
Lennox and Robinson (see \cite{robinson-lennox1} and [{\bf 5}, 10.3.6]).

\begin{theorem}{\rm (Lennox and Robinson)} Assume that $\omega$ is a set of primes. Let $N$ be a nilpotent group with finite rank and $A$ a $\mathbb ZN$-module whose underlying additive group is a \v{C}ernikov $\omega$-group. If $A^N$ is finite, then $H^n(N,A)$ is a finite $\omega$-group for $n\geq 0$. \nolinebreak \hfill\(\square\)
\end{theorem}

In the original statement of Lennox and Robinson's theorem, $N$ is not assumed to have finite rank, which leads to the weaker conclusion that $H^n(N,A)$ is merely bounded as a $\mathbb Z$-module. However, as argued in the first paragraph in the proof of [{\bf 3}, Theorem A], $N$ and $A$ having finite rank means that, if $H^n(N,A)$ is bounded, then it must be finite.

Armed with the results above, we proceed with the proof of our second theorem.

\begin{proof} [{\bf Proof of Theorem B}]

Assume that $N$ is the normal subgroup of $G$ described in the definition of $\mathcal{U}$ and write $Q=G/N$.
Without any real loss of generality, we may assume that $A$ is divisible {\it qua} abelian group. In this paragraph, we treat the case where, for every subgroup $H$ of $G$ of finite index,
each proper $\mathbb ZH$-submodule of $A$ is finite. Consider first the situation where $A^N\neq A$. Under this assumption, Lennox and Robinson's result yields that $H^n(N,A)$ is a finite $\omega$-group for $n\geq 0$. It follows, then, from the LHS spectral sequence that $H^n(G,A)$ is a finite $\omega$-group for $n\geq 0$.  Next we suppose $N<C_G(A)$.  Let $G_0$ be a normal subgroup of $G$ with finite index such that $N<G_0$ and $Q_0=G_0/N$ is a finitely generated abelian group.
We will employ the LHS spectral sequence for the group extension $1\to N\to G_0\to Q_0\to 1$ to investigate $H^n(G_0, A)$. The universal coefficient theorem and Corollary 2.4 give rise to the following chain of isomorphisms for $p, \ q\geq 0$.
\begin{displaymath} H^p(Q_0, H^q(N, A)) \cong H^p(Q_0 , {\rm Hom}_{\mathbb Z}(H_qN, A))\cong {\rm Ext}^p_{\mathbb ZQ_0}(H_qN, A).\end{displaymath}
According to Lemmas 4.4 and 4.5, $A$ fails to have a quotient that is also a quotient of a $\mathbb ZQ_0$-module that is torsion-free of finite rank and $\pi$-spectral as an abelian group. Moreover, Corollary 3.8 implies that $H_qN$ belongs to $\mathfrak{C}(Q_0, \pi)$ for $q\geq 0$. Therefore, by Lemma 3.4, ${\rm Ext}^p_{\mathbb ZQ_0}(H_qN, A)$ is a \v{C}ernikov $\omega$-group for $p,\ q\geq 0$. It follows, then, that $H^n(G_0, A)$ is a \v{C}ernikov $\omega$-group for $n\geq 0$.  Finally, appeals to the LHS spectral sequence for the extension $1\rightarrow G_0\rightarrow G\rightarrow G/G_0\rightarrow 1$ and [{\bf 6}, Lemma 4.3] permit the deduction that $H^n(G,A)$ is a \v{C}ernikov $\omega$-group for $n\geq 0$.

Now we tackle the general case, proceeding by induction on $c(A)$. As the case $c(A)=0$ is trivial, we suppose $c(A)>0$. Select an additive subgroup $B$ of $A$ with $c(B)$ as large as possible subject to the conditions that $c(B)<c(A)$ and $N_G(B)$ has finite index in $G$. Let $G_1$ be a normal subgroup of $G$ contained in $N_G(B)$ such that $[G:G_1]$ is finite. Notice that our choice of $B$ guarantees that, for any subgroup $H$ of $G_1$ with finite index, every proper $\mathbb ZH$-submodule of $A/B$ is finite. Furthermore, in view of Lemma 4.4 and the inductive hypothesis, $H^n(G_1,B)$ is a \v{C}ernikov $\omega$-group for $n\geq 0$. Hence the conclusion of the theorem will follow  if we manage to establish that $H^n(G_1,A/B)$ is a \v{C}ernikov $\omega$-group for $n\geq 0$. Set $N_1=G_1\cap N$. If $N_1$ fails to act trivially on $A/B$, then we deduce that $H^n(G_1,A/B)$ is a finite $\omega$-group from Lennox and Robinson's theorem.  Suppose that $N_1$ acts trivially on $A/B$.
Our plan is to show that, in this case, $A/B$ cannot be a $\mathbb ZG_1$-module quotient of a $\mathbb ZG_1$-module whose additive group is torsion-free of finite rank and $\pi$-spectral; then it will follow from the case proved above that $H^n(G_1,A/B)$ is a \v{C}ernikov $\omega$-group for $n\geq 0$, thus yielding the desired conclusion.

Suppose that $A/B$ is a quotient of a $\mathbb ZG_1$-module whose additive group is torsion-free of finite rank and $\pi$-spectral.
 Let $\Gamma=A\rtimes G_1$. Then the group $\Gamma/B$ plainly belongs to $\mathcal{U}$.  Consequently, by the inductive hypothesis, $H^n(\Gamma/B,B)$ is torsion for $n\geq 0$. That this holds for $n=2$ implies the existence of a subgroup $X$ of $\Gamma$ such that $B\cap X$ is finite and $[\Gamma:BX]<\infty$. Writing $G_2=BX$, we have that $A\cap X$ and $B\cap X$ are $\mathbb ZG_2$-submodules of $A$. Moreover, $A\cap X/B\cap X$ and $A/B$
are isomorphic as $\mathbb ZG_2$-modules. Thus Lemma 4.5 yields that $A\cap X$ is a $\mathbb ZG_2$-module quotient of a $\mathbb ZG_2$-module whose underlying additive group is torsion-free of finite rank and $\pi$-spectral.  In view of Lemma 4.4, this contradicts our hypothesis concerning $A$. Therefore, $A/B$ cannot be a $\mathbb ZG_1$-module quotient of a $\mathbb ZG_1$-module whose additive group is torsion-free of finite rank and $\pi$-spectral. This concludes the proof of the theorem.

\end{proof}

Theorem B has the following corollary.

\begin{corollary}  Let $G$ be a finitely generated solvable minimax solvable group with spectrum $\pi$. Assume that $A$ is a $\mathbb ZG$-module whose underlying abelian group is a \v{C}ernikov $\omega$-group for a set of primes $\omega$. Suppose further that $A$ fails to have an infinite $\mathbb ZG$-submodule that is a quotient of a $\mathbb ZG$-module that is torsion-free of finite rank and $\pi$-spectral as an abelian group. Then, for all $n\geq 0$, $H^n(G,A)$ is a \v{C}ernikov $\omega$-group.  \nolinebreak \hfill\(\square\)
\end{corollary}

In a second corollary below, we identify a common situation in which the condition on the submodules of $A$ in Theorem B applies, namely, where the action of the group on the module is transcendental.

\begin{corollary} Let $G$ be a group in the class $\mathcal{U}$ with spectrum $\pi$. Assume that $A$ is a $\mathbb ZG$-module whose additive group is a \v{C}ernikov $\omega$-group for a set of primes $\omega$, and whose proper submodules are all finite. Suppose further that $\phi: G\to {\rm Aut}(A)$ is the homomorphism arising from the action of $G$ on $A$. If, for some $g\in G$, $\phi(g)$ fails to satisfy any polynomial over $\mathbb Z$, then, for $n\geq 0$, $H^n(G,A)$ must be a \v{C}ernikov $\omega$-group.
\end{corollary}

\begin{proof} If $A$ could be realized as a quotient of a $\mathbb ZG$-module whose additive group were torsion-free of finite rank, then $\phi(g)$ would plainly satisfy a polynomial over $\mathbb Z$ for every $g\in G$. Hence it must be impossible to express $A$ as such a quotient. Theorem A, therefore, yields that $H^n(G,A)$ is a \v{C}ernikov $\omega$-group for $n\geq 0$.
\end{proof}

We conclude the paper with two examples illustrating that, in Theorem B, we cannot dispense with either the restriction on $G$ or the condition on the submodules of $A$.

\begin{example} {\rm Let $p$ and $q$ be distinct primes.  Take $Q$ to be an infinite cyclic group, and let $A$ be the $\mathbb ZQ$-module obtained by letting the generator of $Q$ act on $\mathbb Z_{p^\infty}$ by multiplication by $q$. Set $G=A\rtimes Q$. We may also regard $A$ as a $\mathbb ZG$-module via the epimorphism $G\to Q$. First we claim that $A$ fails to be a quotient of a torsion-free $\mathbb ZG$-module that has finite rank and is $\{p\}$-spectral. To show this, we let $B$ be the underlying additive group of the ring $\mathbb Z[1/q]$ and endow $B$ with a $\mathbb ZQ$-module structure by letting the generator of $Q$ act again by multiplication by $q$. By Proposition 2.3(i),
$${\rm Ext}_{\mathbb ZQ}^1(A,B)\cong {\rm Ext}^1_{\mathbb Z}(A,B)^Q.$$
Moreover, a simple calculation reveals  ${\rm Ext}^1_{\mathbb Z}(A,B)\cong \hat{\mathbb Z}_p$. Since the diagonal action of $Q$ on ${\rm Ext}^1_{\mathbb Z}(A,B)$ is trivial, this means
${\rm Ext}_{\mathbb ZQ}^1(A,B)\cong \hat{\mathbb Z}_p$. However, according to Lemma 3.3, if $A$ were a quotient of a torsion-free $\{p\}$-minimax $\mathbb ZG$-module, then ${\rm Ext}_{\mathbb ZQ}^1(A,B)$ would have to be bounded. Hence $A$ cannot be a quotient of such a module.

Now we apply the LHS spectral sequence $\{E^{ij}_r\}$ associated to the group extension $1\rightarrow A\rightarrow G\rightarrow Q\rightarrow 1$ to calculate $H^n(G,A)$ for $n=1,\ 2$.  Since $E^{10}_2=H^1(Q,A)=0$, we deduce $H^1(G,A)\cong {\rm Hom}_{\mathbb Z}(A,A)^Q$. With the diagonal action being trivial, this yields $H^1(G,A)\cong \hat{\mathbb Z}_p$. Next we observe
$$E^{11}_2\cong H^1(Q, {\rm Hom}_{\mathbb Z}(A,A))\cong \hat{\mathbb Z}_p.$$ Furthermore, invoking the universal coefficient theorem, we obtain $$E^{02}_2\cong {\rm Hom}_{\mathbb Z}(A,A)^Q\cong \hat{\mathbb Z}_p.$$
Hence $H^2(G,A)\cong \hat{\mathbb Z}_p\oplus \hat{\mathbb Z}_p$.
This example, then, fulfills all  the hypotheses of Theorem B except the condition that $G$ is a member of $\mathcal{U}$, and the cohomology is torsion-free and uncountable in dimensions one and two.

}

\end{example}

\begin{example}{\rm Let $p$ be a prime and $Q=\langle t\rangle$ an infinite cyclic group. Form a $\mathbb ZQ$-module $B_1$ by letting $t$ act on $\mathbb Z[1/p]$ by multiplication by $p$. Moreover, let $B_2$ be the $\mathbb ZQ$-module with the same underlying additive group but where $t$ acts instead by multiplication by $1/p$. Next set $G=(B_1\oplus B_2)\rtimes Q$. Then $G$ is a finitely generated torsion-free solvable minimax group. Take $A$ to be the trivial $\mathbb ZG$-module with underlying abelian group $\mathbb Z_{p^\infty}$. We will employ the LHS spectral sequence $\{E^{ij}_r\}$ associated to the extension  $1\rightarrow B_1\oplus B_2\rightarrow G\rightarrow Q\rightarrow 1$ to compute $H^2(G,A)$. For this purpose, we let $V={\rm Hom}_{\mathbb Z}(\mathbb Z[1/p], A)$, so that $V$ is a vector space over $\mathbb Q$ with dimension equal to the cardinality of the continuum. We thus have $H^1(B_1\oplus B_2,A)\cong V\oplus V$, where $Q$ acts on the first copy of $V$ via  multiplication by $p$ and on the second by multiplication by $1/p$. Under this action, every derivation from $Q$ to $V\oplus V$ is inner, rendering $E^{11}_2=0$. Also, the universal coefficient theorem and K\"unneth formula yield $H^2(B_1\oplus B_2,A)\cong V$, where the action of $Q$ on $V$ is trivial. This means $E^{02}_2\cong V$. Consequently, $H^2(G,A)\cong V$. This example, then, demonstrates the necessity of the condition in Theorem B  on the submodules of $A$. }

\end{example}

\begin{acknowledgement} {\rm The author is greatly indebted to Peter Kropholler for many stimulating and enlightening exchanges concerning solvable groups. In addition, the author would like to express his gratitude to the anonymous referee for a number of salutary suggestions.}
\end{acknowledgement}


TITLE: calculating the area of a circle by integrating over straight lines
QUESTION [0 upvotes]: The Problem
Imagine having a circle of radius $R$. We all know the area of this circle is $A=\pi R^2$ and the circumference is $2\pi R$. I would like to know why this approach for calculating the area is wrong, and why it gives me the circumference instead:
If I have a line of length $R$, I can rotate it around its edge a full rotation ($2\pi$ rad) and it would cover the whole circle, meaning it should give me the area, so we have the area be equal to $A=2\pi R$ or alternatively $A=\int\limits _{0}^{2\pi}Rd\theta =2\pi R$.
Note: By rotating the line I mean taking the original line (purple in the image) and then placing its end (which isn't in the center of the circle) on some other point on the circumference of the circle, as seen in the picture (the red lines), doing so for every possible point on the circle would look like the entire area of the circle is colored red, since its all covered by the rotations of this one purple line.

Note: I tried to look at other similar SE questions such as this one, but they didn't answer my question well.
Context
The reason I thought of this is because a few days ago I tried to calculate the electric field of a ring centered at $\left(0,0,0\right)$ with some inner radius $R_1$  and outer radius $R_2$ and charge density $\sigma = \frac{\lambda}{r^{3}}$, at the point $\left(0,0,z\right)$ for some $z$. since the ring is symmetrical in the $x,y$ directions the field has affect in the $z$ direction at that point. I thought about finding the field one "line" in the ring causes, and multiply that by $2 \pi$, but I found this approach to be incorrect.

REPLY [1 votes]: Consider in your illustration, an outer ring of some width $h$. The "spokes" covering it have an area of $9hw$, where $w$ is the width of each spoke. But near the center, spokes covering another ring of width $h$ would also have an area of $9hw$, even though the area of that ring is less. So it ends up weighing the area near the center disproportionately more than the area close to the edge of the circle.
You can do what Yves Daoust suggested of adding up triangles, or you could add up rings. For the latter, you can add up the areas of small rings (which is the circumference of the ring times $dr$) as $$\int_0^R 2\pi r dr$$


TITLE: Smart way to deal with factoring quadratics?
QUESTION [1 upvotes]: For some quadratics such as:
$$10x^2+89x-9$$
I factoring by making the observations $10=(1)(10)=(5)(2)$ and $9=(3)(3)=9(1)$, then I try:
$$(10x ? 1)(x ? 9)$$
$$(10x ? 9)(x ? 1)$$
$$(10 x ? 3)(x ? 3)$$
$$(5x ? 1)(2x ? 10)$$
$$(5x ? 10)(2x ? 1)$$
$$(5x ? 3)(2x ? 3)$$
We got lucky , by our first guess,  we can get;
$$(10x-1)(x+9)$$
This is the way my Algebra $2$ teacher made me switch to because he said it's less magic and more factoring. This method is a bit hard to teach, or is it not in your opinion?
But in many cases I find myself going through a lot of possibilities with this method, sometimes to find that my quadratic cannot be factored.
Of course there is the method what multiplies to $-90$ and adds to $89$...this may take a little less thought then we proceed as:
$$10x^2+90x-x-9$$
And of course we may go straight to the quadratic formula...(if the question is to factor then, using the roots we can find a factored form).
If I'm taking an sat or some other timed test, what method would you suggest. And do you have a better method?

REPLY [0 votes]: Broadly, unless the factors "jump out at you", factoring is a pretty horrible approach to this. Best reserved for folks who simply cannot "get" any other method, and like most academic (and life things), a group of them does exist.
To factor, just solve it using the quadratic formula. You end up with:
x = root 1 and x = root 2
Not factored, eh? Let's say the roots are x = 3½ and x = 2¾. Using simple algebra, get expressions equal to 0, then multiply by the denominators in the fractions. For example:
x = 3½    move the 3½ to the left side
x - 3½ = 0    multiply each side by 2 (½'s denominator)
2x - 7 = 0

That's the first factor candidate. One step left regarding it and its brother, (4x - 11).
Mutliply the "x" terms in the potential factors. You get 8x in this example. Divide "a" of "ax^2" fame in the quadratic you were given to factor by the result of multiplying them, 8 in this case.
If the result is 1, you are done. You have your factors, (2x - 7) and (4x - 11) in the example.
If it is anything else, 3.24 say, you would simply show it leading the two factors you have:
3.24 * (2x - 7) * (4x - 11)

That's the most proper expression of it. Also, if the quadratic went down, not up (a < 0), this is when you catch that, putting a negative out front.
So, QUICK and easy, and for the SAT your work is not graded, only your answer. If handing it in to a teacher, do the work leaving some space "to the left and above" it and after you KNOW the result, fill the empty space with some maundering about with potential factors and such, don't be careless and write things that couldn't've led to your answer(!!!), be on point, but do it quickly and incompletely... you... didn't have all the room in the world to write so you had to do some of it in your head, eh?
Factoring, for people who can understand other methods and use them, is like the person who spends three minutes deciding what calculational "trick" to use to (cleverly?) solve some problem when 12 seconds was all it would have taken to solve it in the usual, standard way. Awesome if the factors leap out at you, boom, you're done. But a nightmare when you have 20 sets of possible factors, all of which have negative and positive possibilities, and a million combinations. Teachers use them to separate the men from the boys, the A's from the B's. Well, you want the A, right?
Finally, if you have to show every bit of work, well, you have to grunt through it. But doing this you at least know the right answer, without having to check each halfway likely possibility. You're just listing them as if you really evaluated each along the way but NOT spending all that time evaluating. So... faster.
All that said, like anything, there is a huge gain in gradually building a list of common quadratics that you know the factors for, you know, like 4x^2 + 4x + 1 = 0 is (2x + 1)^2 and that kind of thing. Then there's no need for smart ways, kludgy, hard to work ways, or any ways at all. You just write the answer and move on.


TITLE: Continuous function calculus
QUESTION [4 upvotes]: Are there any examples satisfy the continuous function $f:\Bbb R\rightarrow \Bbb R$ such that the image of the closed interval $[0,\infty)$ under $f$ is the open interval $(-1,1)$..like $f([0,\infty))=(-1,1)$?
A picture is also ok.

REPLY [3 votes]: Note that the function
$$g(x) = \frac{2}{\pi} \arctan{x}$$
is monotonically increasing, continuous, and tends to $1$ at infinity (the constant $2/\pi$ is included solely to make the range $[0, 1)$). It follows that something like
$$f(x) = \left(\frac 2 {\pi} \arctan{x}\right) \sin x$$
will have the desired property. It gets arbitrarily close to $\pm 1$, is continuous, but never reaches either.


text:
\begin{document}
\title{On the Thurston-Bennequin invariant of graph divide links}
\author{Masaharu Ishikawa}
\footnote[0]{
  This work is supported by the Sumitomo Foundation,
  Grant for Basic Japanese Research Project.
  It is also supported by MEXT,
  Grant-in-Aid for Young Scientists (B) (No. 16740031).
}
\address{Department of Mathematics, Tokyo Institute of Technology,\\
2-12-1, Oh-okayama, Meguro-ku, Tokyo, 152-8551, Japan}
\email{ishikawa@math.titech.ac.jp}
\keywords{Thurston-Bennequin invariant, divide, $4$-genus, tight contact structure}
\subjclass{57M25 (Primary) 57R17, 57M50 (Secondary)}

\begin{abstract}
In the present paper we determine the Thurston-Bennequin invariant
of graph divide links,
which include all closed positive braids, all divide links
and certain negative twist knots.
As a corollary of this and a result of
P.~Lisca and A.I.~Stipsicz, we prove that
the $3$-manifold obtained from $S^3$ by Dehn surgery along
a non-trivial graph divide knot $K$ with coefficient $r$
carries positive, tight contact structures for every $r$
except the Thurston-Bennequin invariant of $K$.
\end{abstract}

\maketitle

\section{Introduction}

A contact form on an oriented, closed, smooth $3$-manifold $M$
is an $1$-form $\alpha$ satisfying $\alpha\land d\alpha\ne 0$.
The $2$-plane field $\xi$ on $M$ determined by the kernel of $\alpha$
is called a {\it contact structure}.
A contact structure is called {\it positive}
if the orientation determined by the contact structure
agrees with the given one on $M$.

Let $(M,\xi)$ be a contact manifold, i.e.,
an oriented, closed, smooth $3$-manifold $M$ equipped with
a contact structure $\xi$.
A {\it Legendrian curve} $K$ in $M$ is a simple closed curve in $M$
such that at each point in $K$ the tangent vector to $K$
is included in the $2$-plane of $\xi$.
An {\it overtwisted disk} is a disk in $M$ whose boundary
is a Legendrian curve $K$ and at each point in $K$
the disk is transverse to the $2$-plane of $\xi$.
A contact structure $\xi$ is called {\it tight}
if there are no overtwisted disks in $M$.

It is well-known that $S^3$ admits a unique, positive,
tight contact structure $\xi_0$,
so-called the {\it standard contact structure}.
Let $K$ be a Legendrian link in $(S^3,\xi_0)$, which is an oriented link
consisting of Legendrian curves in $(S^3,\xi_0)$,
and denote by $K^\perp$ an oriented link obtained by pushing-off $K$
in the direction normal to the $2$-plane field $\xi_0$.
The linking number $lk(K,K^\perp)$ of $K$ and $K^\perp$
is called the {\it Thurston-Bennequin invariant} of $K$
and we denote it by $\tb(K)$.
It is known by D.~Bennequin that $\tb(K)$ has an upper bound
when $K$ runs in the same ambient isotopy type~\cite{bennequin:83}.
The maximal number of $\tb(K)$ is called
the {\it maximal Thurston-Bennequin invariant} of $K$
and denoted by $\TB(K)$.

In the present paper we will determine
the maximal Thurston-Bennequin invariant
for a class of links, called {\it graph divide links}.
A {\it graph divide}, introduced by T.~Kawamura in~\cite{kawamura},
is the image of a generic immersion $\varphi:G\to D$
of a disjoint sum $G$ of intervals, circles and finite graphs
into the unit disk $D$.
We call the image of a vertex of $G$ a {\it vertex} of $\varphi(G)$
and a transversal crossing point produced by the immersion
a {\it double point} of $\varphi(G)$.

\begin{figure}[htbp]
   \centerline{\input{tb3.pstex_t}}
   \caption{An example of a graph divide and its knot. This knot is $8_{21}$.
(see \cite[Example 3.4]{kawamura}.)
\label{tb3}}
\end{figure}

Let $P=\varphi(G)\subset D$ be a graph divide.
Suppose that each vertex of $P$ has a sign $+$ or $-$.
We then define an oriented link using $P$ and its signs, see Section~2.
We call it the link of a graph divide $P$ with signs, or shortly
a graph divide link of $P$ (with signs).
It is worth noting that the class of graph divide links includes,
for example,
\begin{itemize}
   \item[(i)] all closed positive braid (see~\cite[Remark 5.4]{kawamura}),
   \item[(ii)] all divide links (see~\cite{acampo:99,acampo:98a}
   and the list in~\cite{gi2}), and
   \item[(iii)] twist knots with $q<0$ twists
                in the notation in~\cite[p.112]{rolfsen}
               (see~\cite{gibson2} and \cite[Example 3.3]{kawamura},
                and cf.~the first footnote in~\cite{ls}).
\end{itemize}
Note that the links of complex plane curve singularities
are included in both (i) and (ii).

The slice Euler characteristic $\chi_s(K)$
of an oriented link $K$ in $S^3$ is the maximal number of
Euler characteristics of 2-dimensional manifolds in $B^4$,
bounded by $S^3$, whose boundary is $K$.
The integer $g_s(K)=(2-\mu-\chi_s(K))/2$ is called the {\it $4$-genus} of $K$,
where $\mu$ is the number of components of the link $K$.

The main theorem in the present paper is the following:

\begin{thm}\label{thm1}
Let $K$ be an oriented link obtained as
the link of a graph divide $P=\varphi(G)$ with signs $\ve$.
Then we have $\TB(K)=-\chi_s(K)$.
\end{thm}

Before stating a corollary, we introduce a result of
P.~Lisca and A.I.~Stipsicz.

\begin{thm}[Lisca and Stipsicz~\cite{ls}]\label{thmls}
Let $K$ be a knot in $S^3$ such that $g_s(K)>0$ and $\TB(K)=2g_s(K)-1$.
Then the oriented $3$-manifold obtained from $S^3$
by Dehn surgery along $K$ with coefficient $r$ carries
positive, tight contact structures for every $r\ne\TB(K)$.
\end{thm}

Since $\TB(K)=-\chi_s(K)=2g_s(K)-1$ for every graph divide knot $K$,
we have the following corollary:

\begin{cor}\label{cor1}
Let $K$ be a non-trivial knot obtained as
the link of a graph divide $P=\varphi(G)$ with signs $\ve$.
Then the oriented $3$-manifold obtained from $S^3$
by Dehn surgery along $K$ with coefficient $r$ carries
positive, tight contact structures for every $r\ne\TB(K)$.
\end{cor}

We will recall the definition of the link of a graph divide in Section~2
and prove Theorem~\ref{thm1} and Corollary~\ref{cor1} in Section~3.

The author would like to thank Tomomi Kawamura for her useful suggestions.

\section{Graph divides}

Let $P=\varphi(G)\subset D$ be a graph divide.
Suppose that each vertex of $P$ has a sign $+$ or $-$.
We denote this distribution of signs by $\ve$.
To each pair $(P,\ve)$ of a graph divide $P$
and a distribution $\ve$ of signs,
we define oriented, immersed circles in $D$ according to
the following {\it doubling method}:
\begin{itemize}
   \item[(i)] outside a small neighborhood of the vertices of $P$
we draw two curves parallel to $P$;
   \item[(ii)] inside the small neighborhood of each vertex of $P$
we draw the curves as shown in Figure~\ref{tb2}
according to the sign at the vertex;
   \item[(iii)] assign an orientation to the curves obtained
so that it is ``keeping left'' almost everywhere.
\end{itemize}
We denote the obtained oriented, immersed circles by $d(P;\ve)$.

\begin{figure}[htbp]
   \centerline{\input{tb2.pstex_t}}
   \caption{Doubling method around vertices of graph divides.\label{tb2}}
\end{figure}

An {\it oriented divide} is the image of a generic immersion of
oriented circles into $D$~\cite{gi}.
So, $d(P;\ve)$ is an oriented divide.
Let $T(\R^2)$ be the tangent bundle to $\R^2$,
on which the unit disk $D$ lies,
and $ST(\R^2)$ the unit sphere in $T(\R^2)$.
We denote by $x=(x_1,x_2)$ the coordinates of $\R^2$
and by $u=(u_1,u_2)$ the coordinates of tangent space to $\R^2$.
The {\it link} of an oriented divide $\vec P$ is the set
\[
   L(\vec P)=\{(x,u)\in ST(\R^2)\mid x\in \vec P,\,u\in T_x(\vec P)\}
   \subset S^3,
\]
where $x\in \vec P$ means that $x$ lies on the curve of $\vec P$
and $T_x(\vec P)$ is the set of tangent vectors to $\vec P$
whose direction is consistent with the orientation assigned to $\vec P$.

\begin{dfn}
The {\it link} $L(P;\ve)$ of a graph divide $P$ with signs $\ve$
is the set given by
\[
   L(P;\ve)=\{(x,u)\in ST(\R^2)\mid x\in d(P;\ve),\,u\in T_x(d(P;\ve))\}
   \subset S^3,
\]
i.e., it is the link of the oriented divide $d(P;\ve)$.
\end{dfn}

\begin{rem}
We here note a brief history. $P=\varphi(G)$ is called a {\it divide}
if $G$ is a disjoint sum of intervals and circles
and the image of the endpoints of intervals lies on the boundary of $D$.
The divide is introduced by N.~A'Campo
as a generalization of real morsified curves of
complex plane curve singularities.
In particular, their links are fibered and the fibrations correspond
to the Milnor fibrations of the complex
singularities.
He also determined the unknotting numbers and the slice Euler
characteristics of all divide knots~\cite{acampo:99,acampo:98a}.

Next W.~Gibson and the author introduced {\it free divides}
by forgetting the assumption about the endpoints of intervals~\cite{gi2}.
After that, W.~Gibson introduced {\it tree divides}
as the image of trees and
determined their unknotting numbers, slice Euler characteristics, and
Thurston-Bennequin invariant~\cite{gibson2}.
Finally, T.~Kawamura introduced graph divides, proved
their quasipositivity and determined their
slice Euler characteristics~\cite{kawamura}.
\end{rem}

\section{Proofs}

Let $P=\varphi(G)$ be a graph divide with signs $\ve$
and consider the $\pi/2$-rotation map $R_{\pi/2}:ST(\R^2)\to ST(\R^2)$
given by $(x_1,x_2,u_1,u_2)\mapsto (x_1,x_2,u_2,-u_1)$.
This map corresponds to the $\pi/2$-rotation of tangent vectors
to the unit disk $D$
and the image of the $\pi/2$-rotation of tangent vectors to $d(P;\ve)$
constitutes co-oriented, immersed circles $w$ in $D$, i.e.,
a wave front on $D$.
Since $R_{\pi/2}(L(P;\ve))$ is the preimage of the wave front $w$,
it is Legendrian with respect to the standard contact structure in $S^3$.
In what follows, we regard $L(P;\ve)$ as a Legendrian link
up to the $\pi/2$-rotation map $R_{\pi/2}$.

Let $\delta(P)$ denote the number of double points of $P=\varphi(G)$
and $\chi(G)$ the Euler characteristic of $G$.
We will prove the following:

\begin{thm}\label{thm1a}
For a graph divide $P=\varphi(G)$ with signs $\ve$,
we have $\tb(L(P;\ve))=2\delta(P)-\chi(G)$.
\end{thm}

Before proving Theorem~\ref{thm1a} we give a proof of the main theorem.

\vspace{3mm}

\noindent
{\it Proof of Theorem~\ref{thm1}.}\,
Let $K$ denote the link $L(P;\ve)$ without information about Legendrian.
It is known in~\cite[Proposition 6.2]{kawamura} that
$\chi_s(K)=\chi(G)-2\delta(P)$.
Hence we have $\tb(L(P;\ve))=-\chi_s(K)$.
By the slice Bennequin inequality~\cite{rudolph},
we have $\tb(K^{Leg})\leq -\chi_s(K)$
for any Legendrian link $K^{Leg}$ ambient isotopic to $K$
and hence $\TB(K)=\tb(L(P;\ve))=-\chi_s(K)$.
\qed

\vspace{3mm}

In order to prove Theorem~\ref{thm1a}
we introduce a modified tangle presentation of a graph divide.
First prepare seven kinds of tangles as shown in Figure~\ref{tb1}.
We name them as ``a left endpoint tangle'',
``a right endpoint tangle'', and so on.
The left or right endpoint tangle has two cases depending on
the sign $+$ or $-$ at the endpoint vertex.
A {\it tangle product} is a product of these tangles with well-defined
connections.

\begin{figure}[htbp]
   \centerline{\input{tb1.pstex_t}}
   \caption{Tangles.\label{tb1}}
\end{figure}

\begin{dfn}
A {\it modified tangle presentation} of a graph divide $P=\varphi(G)$
with signs $\ve$ is a tangle product whose immersed graph has
the same configuration as the graph divide $P$ with signs $\ve$.
\end{dfn}

It is easy to see that any graph divide $P$ with signs $\ve$
has a modified tangle presentation.

Let $e_1$ (resp. $e_2$) be the number of left (resp. right) endpoint tangles
and $f_1$ (resp. $f_2$) the number of left (resp. right) fold tangles.
We denote by $v_1,\cdots,v_m$ the vertices in the ($+$) branch tangles
and by $b(v_i)$ the number of edges connected to the vertex $v_i$.

\begin{lemma}\label{lemma1}
$-\chi(G)=\sum_{i=1}^m(b(v_i)-2)-e_2+f_1-f_2$.
\end{lemma}

\begin{proof}
By checking $\chi(G)$ from the right to the left
in the modified tangle presentation,
we have $\chi(G)=e_2+f_2-\sum_{i=1}^m(b(v_i)-2)-f_1$.
\end{proof}

Next we calculate $\tb(L(P;\ve))=lk(L(P;\ve),L^\perp(P;\ve))$.
The link $R_{\pi/2}(L^\perp(P;\ve))$ is the Legendrian
link of a wave front $w^\perp$ obtained by pushing-off the wave front $w$
in the direction of the assigned co-orientation.
Since $R_{\pi/2}$ is an isomorphism, the linking of two links does not change.
So, $lk(L(P;\ve),L^\perp(P;\ve))$ is equal to
the linking number of the Legendrian link of $w$ and that of $w^\perp$.
By the $-\pi/2$-rotation map $R^{-1}_{\pi/2}$, we conclude that
$lk(L(P;\ve),L^\perp(P;\ve))$ is equal to the linking number
of $L(P;\ve)$ and the link of an oriented divide obtained
by parallel-shifting $d(P;\ve)$ in one of the possible directions,
see Figure~\ref{tb4}.

\begin{figure}[htbp]
   \centerline{\input{tb4.pstex_t}}
   \caption{A parallel shift of an oriented divide $\vec P$.\label{tb4}}
\end{figure}

A method of obtaining a link diagram of the link of an oriented divide
has been made in~\cite{gi}.
Let $\vec P$ be an oriented divide in $D$.
Each point $(x,u)$ in $L(\vec P)$ corresponds to a point $x\in D$
and a vector $u\in T_x(\vec P)$.
Fix a parametrization $x(t)$ of $\vec P$ with parameter
$t\in \sqcup_{i=1}^\mu S^1$, where $\mu$ is the number of
components of $\vec P$.
This induces a parametrization $(x(t),u(t))$ of the link $L(\vec P)$.
By setting $\vec P$ in general position,
we assume that $y=\arg u(t)$ is transversal to $y\equiv -\pi/2 \mod 2\pi$
and if $x(t_1)=x(t_2)$ for $t_1\ne t_2$ then
$\arg u(t_i)\not\equiv -\pi/2 \mod 2\pi$ for $i=1,2$.

We regard the curve of oriented divide $\vec P$
as the strands of the link $L(\vec P)$
by applying the following rules:

\begin{dfn}[over/under rule]
Suppose that $x(t_1)=x(t_2)$ for $t_1\ne t_2$.
Let $\theta_i$,\,$i=1,2$,\,be real numbers such that
$-\pi/2 < \theta_i < 3\pi/2$ and $\theta_i\equiv \arg u(t_i)\mod 2\pi$.
If $\theta_1 < \theta_2$ then the strand corresponding to $t_1$ passes
over the strand corresponding to $t_2$.
\end{dfn}

We are drawing the link diagram on $\R^2\supset D$ with
coordinates $(x_1,x_2)$.
Let $x_3$ denote the third coordinate for describing
over/under at crossing points of the link diagram.

\begin{dfn}[winding rule]
If $\arg u(t_0)\equiv -\pi/2 \mod 2\pi$ and $\arg u(t)$ is locally increasing
(resp. decreasing) at $t=t_0$ then we cut the strand at that point, bring it
to $x_3=-\infty$ (resp. $x_3=\infty$)
and take it back from $x_3=\infty$ (resp. $x_3=-\infty$).
\end{dfn}

We fix the coordinates $(x_1,x_2)\in\R^2$ in such a way that
$x_1$ is in the right direction in Figure~\ref{tb1}
and $x_2$ is in the top.
In order to calculate $lk(L(P;\ve),L^\perp(P;\ve))$,
we draw the link diagram of the oriented divide $d(P;\ve)$
and its parallel-shift, and check the algebraic number of crossings
of these links.

\begin{lemma}\label{lemma2}
The contribution to $lk(L(P;\ve),L^\perp(P;\ve))$ in each tangle
is as follows:
\begin{itemize}
   \item[(i)] $0$ for each left endpoint tangle;
   \item[(ii)] $-1$ for each right endpoint tangle;
   \item[(iii)] $+1$ for each left fold tangle;
   \item[(iv)] $-1$ for each right fold tangle;
   \item[(v)] $b(v_i)-2$ for each {\rm ($+$)} branch tangle with vertex $v_i$;
   \item[(vi)] $0$ for each {\rm ($-$)} branch tangle;
   \item[(vii)] $+2$ for each double point tangle.
\end{itemize}
\end{lemma}

\begin{proof}
The cases (i) $\sim$ (iv) and (vii) have been proved
by W.~Gibson in~\cite{gibson}.
We here write their proofs again for completeness of the proof.

Let $\vec P$ denote the oriented divide consisting of
$d(P;\ve)$ and its parallel-shift.
First we consider the case of a left endpoint tangle with $+$ sign.
The corresponding link diagram is described in Figure~\ref{tb5}.
In the left endpoint tangle, two positive crossings and
two negative crossings appear and hence their algebraic sum is $0$.
Outside the tangle the strand drawn by the winding rule may produce
crossings with other strands. However since $\vec P$ is
obtained by the doubling method
the algebraic sum of these signs becomes always $0$,
see the crossings around the four parallel strands in Figure~\ref{tb5}.
Hence the contribution of this tangle to $lk(L(P;\ve),L^\perp(P;\ve))$ is $0$.

\begin{figure}[htbp]
   \centerline{\input{tb5.pstex_t}}
   \caption{A link diagram for a left endpoint tangle with sign $+$.\label{tb5}}
\end{figure}

If a left endpoint tangle has $-$ sign then the oriented divide $\vec P$
has no double points and no point $x(t)$ at
which $\arg u(t)\equiv -\pi/2 \mod 2\pi$.
Hence the corresponding link diagram has no crossings.
This completes the proof of case (i).

The case of a right endpoint tangle with sign $+$
is described in Figure~\ref{tb6}. Hence the contribution is $-1$.
In the case of $-$ sign,
the winding rule produces two negative crossings
and the contribution becomes also $-1$.
This completes the proof of case (ii).

\begin{figure}[htbp]
   \centerline{\input{tb6.pstex_t}}
   \caption{A link diagram for a right endpoint tangle with sign $+$.\label{tb6}}
\end{figure}

The cases of left/right fold tangles and a double point tangle
are described in Figure~\ref{tb7}.
The windings outside the fold tangles do not contribute to the calculation
of the linking number because of the same reason as we mentioned
in case (i).
The contributions are $+1$, $-1$ and $+2$ respectively
and these complete the proofs of cases (iii), (iv) and (vii).

\begin{figure}[htbp]
   \centerline{\input{tb7.pstex_t}}
   \caption{Link diagrams for left/right fold tangles and a double point tangle.\label{tb7}}
\end{figure}

The cases of branch tangles are described in Figure~\ref{tb8}.
The figures are in the case where the numbers of branches
at the vertices are $5$.
Thus the contribution is $b(v_i)-2$ if it is a ($+$) branch tangle
and $0$ if it is ($-$).
This completes the proof.
\end{proof}

\begin{figure}[htbp]
   \centerline{\input{tb8.pstex_t}}
   \caption{Link diagrams for ($+$) and ($-$) branch tangles.\label{tb8}}
\end{figure}

\noindent
{\it Proof of Theorem~\ref{thm1a}}.\,
From Lemma~\ref{lemma2} we have
$lk(L(P;\ve),L^\perp(P;\ve))=\sum_{i=1}^m(b(v_i)-2)-e_2+f_1-f_2+2\delta(P)$
and this coincides with $2\delta(P)-\chi(G)$ by Lemma~\ref{lemma1}.
\qed

\vspace{3mm}

\noindent
{\it Proof of Corollary~\ref{cor1}}.\,
Since we already have $\TB(K)=-\chi_s(K)$,
it is enough to show that $g_s(K)>0$.
Assume $g_s(K)=0$. Then, by~\cite[Proposition~6.2]{kawamura},
we have $\delta(P)=0$ and $\chi(G)=1$.
But in this case $P$ is an embedded tree and therefore
the link $L(P;\ve)$ is a trivial knot (see~\cite[Lemma~6.8]{kawamura}).
This contradicts the assumption.
\qed

\vspace{3mm}

\begin{rem}
A {\it positive link} is an oriented link which has a positive diagram,
i.e. a diagram with only positive crossings.
Using the results in~\cite{tanaka} and~\cite{nakamura},
we can show the equality $TB(K)=-\chi_s(K)$ for every positive link $K$.
Hence the oriented $3$-manifold obtained from $S^3$
by Dehn surgery along a positive knot $K$ with coefficient $r$
carries positive, tight contact structures for every $r\ne\TB(K)$.
\end{rem}

\begin{rem}
The knot $8_{20}$ is slice and quasipositive
(cf.~\cite[Proof of Theorem 6.7]{kawamura})
and satisfies $\TB(8_{20})=-2$ (see~\cite{ng}).
Hence the equality $\TB(K)=2g_s(K)-1$ does not hold.
This means that the equality is a property of graph divide knots and
positive knots, but is not a property of quasipositive knots.
We don't know if it is a property of strongly quasipositive knots or not.
We remark that the graph divide knot $8_{21}$ shown in Figure~\ref{tb3}
is not strongly quasipositive, see~\cite[Remark 6.9]{kawamura}.
\end{rem}

config: arxiv


TITLE: Prove that O is the middle of segment IJ
QUESTION [0 upvotes]: Excuse me if I get some names wrong, as I primarily study Maths in Arabic/French.
I'll try to translate the task:
Let ABCD be a parallelogram of which center (Where the two diameters cross) is point O. I is the center of the circle surrounding the triangle OBC and J is the center of the circle surrounding the triangle OAD.

Construct the figure
Prove that O is the middle of segment IJ

I already have the figure drawn here, but I couldn't find a way to effectively prove that O is the middle of [IJ]?

REPLY [1 votes]: A reflection in the point $O$ maps $A$ to $C$, $D$ to $B$ and $O$ to itself, so the circumcenter of $\triangle OAD$ is mapped to the circumcenter of $\triangle OBC$.

REPLY [0 votes]: The homothethy centered at $O$ with dilatation ratio $-1$ sends $I$ to $J$.  Also, you should use "circumscribing" rather than "surrounding."


TITLE: Does existence of non-trivial solution of $S(x,y,z) = 0, \; S(y,z,x) = 0, \; S(z,x,y) = 0$ implies existence of trivial solution at $x=y=z$ axis?
QUESTION [3 upvotes]: My question is following. Suppose that you have an implicit surface given by equation $S(x,y,z) = 0$ (if it matters, now $S(x,y,z)$ is a polynomial function). I'm interested only in $\mathbb{R}^3_{+}$ part of $\mathbb{R}^3$ (you may think that $S(x,y,z)$ is symmetric under $(x,y,z) \mapsto (\pm x, \pm y, \pm z)$ ).
In my case the intersection of $S(x,y,z)$ with coordinate planes will look like on the picture:

Let the cyclic group of 3rd order act on $\mathbb{R}^3$ via cyclic shift $(x,y,z) \mapsto (y,z,x)$.

It's easy to see that if $\lbrace S(x,y,z) = 0 \rbrace \cap \mathbb{R}^3_+$ is fully contained in some fundamental domain of group action then system of equations $S(x,y,z) = 0, \; S(y,z,x) = 0 \; S(z,x,y) = 0 $ has no solution. If $S(x,y,z)$ intersects with only one face of fundamental domain, then the same result holds.
Now suppose that we have a non-trivial solution of system $S(x,y,z) = 0, \; S(y,z,x) = 0 \; S(z,x,y) = 0 $. By non-trivial I mean solution that is inner point of group action's fundamental domain. Now the question goes here:
My hypothesis is that such surface $S(x,y,z)$ necessarily will have intersection with line $x=y=z$. Are there any counterexamples that immediately jump to your mind or is there any proof sketch?
UPDATE So, the most important feature of $S(x,y,z)$ is that $S(x,y,0) = 0$ has solution. Solution set consists of two compact connectivity components (see the picture 1 for example). The same for $S(x,0,z)$ and $S(0,y,z)$

REPLY [1 votes]: How about the following: Consider a torus
$$\bigl(\sqrt{\xi^2+\eta^2}-a\bigr)^2 +\zeta^2=b^2,\qquad 0<b<a,$$
in $(\xi,\eta,\zeta)$-coordinates. After a suitable squaring this can be written in the form
$$\Phi(\xi,\eta,\zeta)=0\ ,\tag{1}$$ where $\Phi$ is some polynomial of degree $4$ in the variables $\xi$, $\eta$, $\zeta$.
Then let the $\zeta$-axis point in direction $(1,1,1)$ of $(x,y,z)$-space and choose the $\xi$- and the $\eta$-axis orthogonal in the plane $x+y+z=0$. The greek and the latin variables will then be related by an orthogonal linear transformation. It follows that the torus is given in the latin variables by an equation of the form
$$\Psi(x,y,z)=0\ ,$$
where  $\Psi$ is a symmetric polynomial of degree $4$ in the latin variables. Choosing the parameters $a$ and $b$ appropriately you can make sure that the torus intersects the coordinate planes. But it will not intersect the line $x=y=z$.
If symmetry is important, here is another idea: Consider the edge-skeleton of the octahedron $|x|+|y|+|z|\leq1$ and replace the edges by tubes of radius $r\ll1$ soldered together in a smooth way. The resultant surface has the symmetries you want and does not intersect the line $x=y=z$. Some people will even find a polynomial equation $\Omega(x,y,z)=0$ that defines such a surface.


TITLE: Finding solvable groups with given Fitting subgroup
QUESTION [2 upvotes]: Let $K=\mathbb Z_2 \times \mathbb Z_2$. What finite solvable groups (up to isomorphism) have Fitting subgroup isomorphic to $K$?

So far I've found that $K, S_4, A_4$ satisfy that property. I don't know whether there are others...  What follows is what I know. Let $G$ be a group we want to find. Then there is $n\in \mathbb N$ such that $G^{(n)}={1_G}$ (a derived subgroup is trivial). Its Fitting subgroup is the intersection of all the $p$-cores of $G$, or equivalently, the unique maximal normal nilpotent subgroup.
Since $p$-groups are nilpotent, if $G$ is a $p$-group, then $G=K$. Let's assume $G$ is not a $2$-group. Then there is $q\neq 2$ prime, dividing $|G|$. For every $p\neq 2$ dividing $|G|$, there are more than one Sylow $p$-subgroups, since its core is trivial.
Another important fact about finite solvable groups is that the centralizer of the Fitting subgroup is its center. In our case $C_G(K)=K$. Yet I can't see how this would help.

Is there a general strategy for any group $K$?

REPLY [3 votes]: Since $C_G(K)=K$, the conjugation action of $G/K$ on $K$ maps $G/K$  into a subgroup of ${\rm Aut}(K) \cong S_3$. We can't have $|G/K| = 2$, so $|G/K| = 1,3,$ or $6$ giving the three examples that you have found. So there are no more examples.
The same strategy would work for any abelian $p$-group $K$. Otherwise it would get more complicated but could still be done because $G/Z(K)$ must be a subgroup of ${\rm Aut}(K)$, and hence $|G|$ is bounded.


TITLE: Find Cardinality of the set $X$ defined below
QUESTION [4 upvotes]: I did this question by crude counting. Since $|S|$ was only $4$, counting didn't took much time but there should be a way to do it for bigger sets.
Let $S = \{a, b, c, d\}$ and  $X = \{f : S \to S \mid f\text{ is bijective and }f(x)\ne x\text{ for each }x \in S\}$
Then
$|X| =\text{ ?}$

REPLY [4 votes]: there is also a derivation using exponential generating functions. an arbitrary permutation of a set is made up of a trivial map on the subset of its fixed points, and a derangement on the complement of the fixed set.
the trivial map has only one representative for each cardinality, so its e.g.f. is $e^z$
the number of permutations of a set has the e.g.f. $\sum \frac{n!}{n!}z^n=(1-z)^{-1}$
if $D(z)$ is the e.g.f. for the number of derangements then:
$$
(1-z)^{-1}=D(z)e^z
$$
i.e.
$$
D(z)=\frac{e^{-z}}{1-z}
$$
as series we have
$$
D(z)=\sum_{j=0}^{\infty}\frac{d_n}{n!} =\sum_{j=0}^{\infty}\frac{(-1)^j}{j!}z^j\sum_{k=0}^{\infty}z^k \\
=\sum_{n=0}^{\infty}\sum_{j+k=n}\frac{(-1)^j}{j!}z^n
$$
which gives the number of derangements of $n$ objects as:
$$
d_n = n!\sum_{j=0}^n \frac{(-1)^j}{j!}
$$


TITLE: Where is my error in solving $y'' + y' + y = 0, y(0) = 1, y'(0) = 0$ with Laplace transform?
QUESTION [2 upvotes]: Im trying to solve a laplace transoform question, but i am stuck.
The question is $y′′(t) + 2ζy′(t) + y(t) = 0, y(0) = 1, y′(0) = 0$ and $ζ = 0.5$.
I have so far done:
Laplace transform which gives
$s^2Y(s)-sy(0)-y'(0)+2ζ(sY(s)-y(0))+Y(s)$
When $ζ = 0.5$
$S^2Y(s)-Sy(0)-y'(0)+sY(s)-y(0)+Y(s)$
$S^2Y(s)-s-sY(s)-1+Y(s)$
$s^2Y(s)-sY(s)+Y(s)=s+1$
$Y(s)[s^2-s+1]=(s+1)$
$Y(s)=(s+1)/(s^2-s+1)$
I'm stuck on this bit not sure what to do after this.

REPLY [1 votes]: Since the denominator is not easily factorable, we avoid partial fractions and instead complete the square. Recall the following inverse Laplace transforms:

$$
Y(s) = \frac{b}{(s - a)^2 + b^2} \implies y(t) = e^{at}\sin bt \\
Y(s) = \frac{s - a}{(s - a)^2 + b^2} \implies y(t) = e^{at}\cos bt
$$

Indeed, observe that:
\begin{align*}
\frac{s + 1}{s^2 - s + 1}
&= \frac{s + 1}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} \\
&= \frac{s - \frac{1}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} + \frac{\frac{3}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} \\
&= \frac{s - \frac{1}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} + \frac{3}{\sqrt 3} \cdot \frac{\frac{\sqrt 3}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2}
\end{align*}
Can you take it from here?


TITLE: Freely adding connected limits preserves accessibility?
QUESTION [3 upvotes]: Let $\mathcal{K}$ be a $\lambda$-accessible category and $\hat{\mathcal{K}}$ its free completion under connected limits.

Is $\hat{\mathcal{K}}$  still accessible?
$\mathcal{K}$ can be identified with a subcategory of $\text{Set}^{\text{Pres}_{\lambda}(\mathcal{K}) ^{\text{op}}}$, is it true that $\hat{\mathcal{K}}$  is just the closure under connected limits of this subcategory?

It looks like the answer should be no to both questions, especially the second one. On the other hand, I am interested in some counterexamples. Still I have not properly understood the phenomenon of (co)completion.

REPLY [4 votes]: An easy way to see the failure of (2) is that if $\mathcal{K}$ already has connected limits, then its restricted Yoneda embedding into $\mathrm{Set}^{\mathrm{Pres}_\lambda(\mathcal{K})^{\mathrm{op}}}$ preserves them, so that its closure under such limits therein would be just itself.  But the free completion under connected limits has to add new limits that do not coincide with the ones that already existed.


TITLE: For continuous and differentiable $f$ with $f(0)=\lim\limits_{x\to\infty}f(x)=0$, how to show $f'$ has a zero on $(0,\infty)$?
QUESTION [2 upvotes]: Problem :
let $f:[0,\infty)\to\mathbb{R}$ be continuous on $[0,\infty)$, differentiable on $(0,\infty)$,
Given that $f(0)=0$, and $\lim\limits_{x\to\infty}f(x)=0$. Prove that there is $c\in (0,\infty)$ such that $f'(c)=0$.
I need help on the above problem! Thanks. Also let me know if my proof provided below is true, if not please let me know the right answer.
Here is my first tentative: let $a>0$. Since $f$ is continuous and differentiable on $(0,\infty)$ and hence on $(0, a)$, then using the Mean Value Theorem we conclude that there is $b\in(0,a)$ such that: $f(a)-f(0)= f'(b)(a-0)$. as $b$ tends to $\infty$, $f(a)$ tends to $0$ which makes the left hand side of the equality equal to $0$, and this implies $f'(b)=0$.

REPLY [2 votes]: You can use a proof by contradiction: if there is no such $c$, then the continuous function $f'$ is either always positive or always negative on $(0,\infty)$. Without loss of generality (since you can replace $f$ by $-f$), say $f'$ is always positive. Since $f(0)=0$, you will be able to disprove the hypothesis $\lim_{x\to\infty} f(x)=0$.

REPLY [2 votes]: First suppose $f$ is identically zero on $[0,\infty)$. This case is trivial. Now assume $f$ not constant. Without loss of generality, assume there is an $a>0$ s.t. $f(a)>0$. Now let $0<s<f(a)$. Then by IVT there is a $b$ in $[0,a]$ s.t. $f(b) = s$.
Now $f(x)$ tends to zero as $x$ tends to infinity so can find N s.t for all $x>N$, $f(x)<s$. Pick $c>N$ so $f(c)<s$. Apply IVT to get a $d$ in $[a,c]$ s.t $f(d)=s$. Now apply MVT to $b$ and $d$.


TITLE: Finite difference, deriving polynomials or derivatives?
QUESTION [0 upvotes]: I just watched this video on finite differences, but this is the Wikipedia page on finite differences, which seems to be talking about something entirely different. And then I found in a book a section deriving the derivatives at discrete points:
Suppose you sample a function at the discrete points $n\Delta x,n \in Z$. We sample the function $f(x) = sin x$ on the interval $[−4, 4]$ with $\Delta x = 1/4$ and plot the data points. We wish to approximate the derivative of the function on the grid points using
only the value of the function on those discrete points. From the definition of the
derivative, one is lead to the formula
$$f'(x) = \frac{f(x+ \Delta x) - f(x)}{\Delta x}$$
...and then it talks about Taylor's theorem. Is any of this related? It seems not to this beginner, it's just sharing the name finite difference. Also, I've seen the difference quotient described as a type of finite difference. I'm guessing the video is just a method of figuring out a polynomial from data, while the others are talking about figuring out derivatives from points when no limit can be taken due to the discrete gaps in the points. If they are related, I'd like to know how, i.e., what is the overarching principle at work here.

REPLY [0 votes]: The derivative of function $f(x)$ can be approximated using formula you provided (forward difference). You can also use backward difference approximation.
$$f'(x) = \frac{f(x ) - f(x - \Delta x)}{\Delta x}$$
Both approximations have an error $O(\Delta x)$.
Central difference approximation
$$f'(x) = \frac{f(x + \Delta x) - f(x - \Delta x)}{2\Delta x}$$
with the error $O(\Delta x ^ 2)$.
Calculation of errors of approximations is based on Taylor's theorem.


TITLE: Deriving the sub-differential of the nuclear norm
QUESTION [13 upvotes]: Let $$f(K) = \| K \|_*$$ be the nuclear norm (sum of the singular values) of $K=U\Sigma V^T$. How can one compute the subdifferential $\partial f$?
This may be a basic question, I'm trying to work my way through a paper in which minimizing $f$ over a convex set of matrices plays a central role. For what it's worth, I have found papers that display the end result, but not the derivation.
EDIT: This paper by Tao and Candes derives an expression, but refers the proof to "Characterization of the subdifferential of some matrix norms" which does not prove it as far as I can tell. I also found a class homework assignment posted online that said this was easy to "grind out" with matrix derivatives, but that there was another way via projections. Any guidance would be greatly appreciated.

REPLY [8 votes]: Contrary to my first impression, this question is answered completely and thoroughly by G.A. Watson in Characterization of the Subdifferential of Some Matrix Norms. For the final derivation, see pg. 40.
The conclusion is that
$$\partial \|K\|_* = \left\{ UV^T+W:\ \ \ \|W\|<1, \text{columnspace}(U) \perp W\perp\text{ rowspace(V)} \right\},$$
where $\|\cdot\|$ is the spectral norm, which is dual to the nuclear norm.


TITLE: Extension of Smooth Functions on Embedded Submanifolds
QUESTION [2 upvotes]: In Lee Smooth Manifolds, this problem is given:
if $S \subset M$ is smoothly embedded and every $f \in \mathcal{C}^{\infty}(S)$ extends to a smooth function on $\textit{all}$ of $M$, then $S$ is properly embedded.
Properly embedded means the inclusion $i: S \to M$ is proper.  For this we have to show $S$ is closed or it's a level set of a continuous function.
I think it should be just a few lines, but I am totally stuck for two hour!  Please to help, thanks =)
What I tried: Since $S$ is embedded, it can be covered with slice charts $U_p$.  If $V = \cup U_p$ then with bump functions we can make $S$ closed in $V$ by making it a level set.  But $V$ is open, so it does not mean $S$ is still closed in the whole manifold.  I could not alter this into proof.
I also tried to make an extension of the identity map of $S$, so that then $S$ would be a retract and closed.

REPLY [0 votes]: I tried this way: consider $K\subset S$ compact and $M$ and $L=\iota^{-1}(K)$. Let $\{U_a\}$ be an open cover for $L$. Consider a smooth partition of the unity $\phi_a$ associated to $\{U_a\}\cup \{U_0\}$ where $U_0 = S\setminus L$ is open. For each a, $\phi_a$ admits a smoooth extension to all $M$ that agrees on $S$, namely $\tilde{\phi_a}$, so that $\sum_a \tilde{\phi_a}(x)|_S = \sum_a \phi_a(x) = 1$. Now consider the sets $V_a = supp(\tilde{\phi_a}) = \tilde{\phi_a}^{-1}((0,\infty))$. Clearly the family $\{V_a\}$ is an open cover for $K$ which admits a partition of the unity ($\tilde{\phi_a}$) so we can conclude that it is locally finite. So for all $x\in K$ there exists a neighborhood $V_x$ such that only finitely many $\{V_a\}$ intersect $V_x$. Using compactness of $K$ cover it by finitely many such sets, i.e.
$$
V_{x_i},i=1,\ldots,n,\ K\subset \cup_{i=1}^n V_{x_i}
$$
Now for $i=1,\ldots,n$ pick $\tilde{\phi_a}_k,k=1,\ldots,n_i$ with respect to their support as above. For what we have said it holds $\sum_{k=1}^{n_i} \tilde{\phi_a}_k(x) = 1$ for all $x\in S$. But then if we come back to S we see that the family $\{supp(\phi_a)_k\}_{k=1\ldots,n_i,\ i=1,\ldots,n} \equiv \{ {U_a}_j \}_{j=1,\ldots,m}$ is such that $\sum_{j=1}^m {\phi_a}_j(x) = 1$ for all $x\in S$. We conclude that $\{ {U_a}_j \}_{j=1\ldots,m}$ is a finite subcover and therefore $L$ is compact. This means that $\iota:S\hookrightarrow M$ is proper. The fact that it is an embedding comes from the preceding point in the problem of the book.