text:
\begin{document}

\title[Loop space homotopy types]
         {The loop space homotopy type of simply-connected four-manifolds
         and their generalizations}
\author{Piotr Beben}
\address{School of Mathematics, University of Southampton,
        Southampton SO17 1BJ, United Kingdom}
\email{P.D.Beben@soton.ac.uk}
\author{Stephen Theriault}
\address{School of Mathematics, University
         of Southampton, Southampton SO17 1BJ, United Kingdom}
\email{S.D.Theriault@soton.ac.uk}

\subjclass[2010]{Primary 55P35, 57N65, Secondary 57P10}
\keywords{manifold, loop space, homotopy decomposition}

\begin{abstract}
We determine loop space decompositions of simply-connected four-manifolds,
$(n-1)$-connected $2n$-dimensional manifolds provided $n\notin\{4,8\}$,
and connected sums of products of two spheres. These are obtained as
special cases of a more general loop space decomposition of certain
torsion-free $CW$-complexes with well-behaved skeleta and some
Poincar\'{e} duality features.
\end{abstract}

\maketitle

\section{Introduction}

The topology of simply-connected four-manifolds is a subject
of widespread and enduring interest. They have been classified up to
homotopy type by Milnor~\cite{Mi} and up to homeomorphism type
by Freedman~\cite{F}. Their classification up to diffeomorphism
type is one of the great unsolved questions in modern mathematics,
with significant advances achieved by Donaldson~\cite{D} and
Seiberg and Witten~\cite{Wi}. They have also been studied in view
of their connections to other areas of mathematics, such as
knot theory~\cite{FS} and symplectic geometry~\cite{P}.

The homotopy theory of simply-connected four-manifolds has
continued to attract considerable attention since Milnor's classification.
For example, given simply-connected four-manifolds $M$ and~$N$,
Cochran and Habegger~\cite{CH} calculated the group of
self-homotopy equivalences of $M$; Zhao, Gao and Su~\cite{ZGS}
calculated the homotopy classes of maps $[M,N]$; and Baues~\cite{B}
has written a monograph entirely devoted to investigating the
homotopy theory of $M$, $N$ and the maps between them.

In another direction, Wall~\cite{Wa} initiated the study of $(n-1)$-connected
$2n$-dimensional manifolds as a generalization of simply-connected
four-manifolds. Such manifolds have received considerable recent attention
as certain families of them arise as intersections of quadrics in geometric
topology and moment-angle manifolds in toric topology~\cite{BM,GL}.
Another variation is connected sums of products of two spheres, which
generalizes the sub-collection of simply-connected four-manifolds that are
connected sums of $S^{2}\times S^{2}$. Such connected sums appear
in the classification by McGavran~\cite{Mc} of $n$-torus actions on closed,
compact, simply-connected $(n+2)$-manifolds, and they also appear as
intersections of quadrics and moment-angle manifolds~\cite{BM,GL}.

In this paper we study simply-connected four-manifolds and their
generalizations from a new perspective. Let $M$ be a simply-connected manifold.
Let $\Omega M$ be the space of continuous basepoint preserving maps from
the circle to~$M$, called the \emph{(based) loop space} of $M$. When $M$
is a simply-connected four-manifold, an $(n-1)$-connected $2n$-manifold,
or a connected sum of  products of two spheres, we aim to give an explicit,
integral homotopy decomposition of $\Omega M$ as a product of simpler factors.

Decomposing the loops on large classes of manifolds has long been thought
to be too hard to do. However, the methods used in the paper are relatively
accessible and flexible. Essentially, the starting input is information about
the integral cohomology of $M$ derived from Poincar\'{e} duality. This is then
manipulated by creating appropriate homotopy fibrations involving $M$ which
allow one to apply decomposition methods from homotopy theory, in the
spirt of~\cite{CMN}. It should be the case that the same methods can be
used to investigate the loops on other classes of manifolds.

Such decompositions are useful and to illustrate this, we give three examples.
First, in toric topology one associates to a simplicial complex $K$ a space
called a moment-angle complex. If $K$ is a simple polytope then
this moment-angle complex is a manifold. For example, if $K$ is
an $n$-gon then the moment-angle complex is a connected sum of
products of two spheres~\cite{BM,GL}. The combinatorics of the
polytope and the geometry of the manifold are deeply connected,
but the relationship is not well understood. Decomposing the loops
on such connected sums and relating the factors to the
combinatorics of the polytope should be insightful.
Second, string topology is concerned with
properties of the free loop space $\Lambda M$ of $M$: the space of
continuous unbased maps from the circle to $M$. There is a fibration
\(\seqmm{\Omega M}{}{\Lambda M}{e}{M}\)
where $e$ evaluates a map at the basepoint, and $e$ has a section.
The section implies that
$\pi_{m}(\Lambda M)\cong\pi_{m}(M)\oplus\pi_{m}(\Omega M)$
for $m\geq 2$, so the homotopy groups of~$\Lambda M$ can be
determined to the same extent as those of the factors of $\Omega M$.
This has implications for counting the geodesics on $M$ (see~\cite{SV}),
and the decomposition of $\Omega M$ may help clarify the
homology and cohomology of $\Lambda M$. The third application
is to configuration spaces, which will be discussed in more detail
in Section~\ref{sec:config}. Let $F(M,k)$ be the configuration space of
ordered $k$-tuples of distint points in the product space $M^{k}$.
In certain cases, for example if $M$ is a product of two non-trivial
manifolds, Cohen and Gitler~\cite{CG} showed that $\Omega M$ is a
factor of $\Omega F(M,k)$. A decomposition for $\Omega M$ further
refines this, and allows for the calculation of a significant subgroup of
the homotopy groups of the configuration space.

To present our results, we start with a classification theorem.
Assume that homology is taken with integral
coefficients and use the symbol ``$\simeq$" to denote a homotopy equivalence.
By a connected sum of sphere products, we mean a connected sum of
products of two spheres.

\begin{theorem}
   \label{classify}
   The following hold:
   \begin{letterlist}
      \item if $M$ and $N$ are simply-connected four-manifolds, then
               $\Omega M\simeq\Omega N$ if and only if
               $H^{2}(M)\cong H^{2}(N)$;
      \item if $M$ and $N$ are $(n-1)$-connected $2n$-dimensional manifolds
                and $n\notin\{2,4,8\}$, then $\Omega M\simeq\Omega N$ if and only if
                $H^{n}(M)\cong H^{n}(N)$;
      \item if $M$ and $N$ are $n$-dimensional connected sums of sphere
                products, then $\Omega M\simeq\Omega N$ if and only if
                 $H^{m}(M)\cong H^{m}(N)$ for each $m<n$.
   \end{letterlist}
\end{theorem}

Observe that in each case, the homotopy type of $\Omega M$ depends
only on the cohomology of $M$, regarded as a $\mathbb{Z}$-module,
in degrees strictly less than the dimension of $M$. This contrasts with
the situation before looping. For example, Milnor~\cite{Mi} proved that
two simply-connected four-manifolds $M$ and $N$ are homotopy
equivalent if and only if $M$ and $N$ have isomorphic cohomology
rings. Theorem~\ref{classify} states that after looping the ring structure
in cohomology plays no role, only the rank in degree~$2$ cohomology
does. So looping considerably simplifies the homotopy types. This is
interesting because $\Omega M$ has the same homotopy
groups as $M$, just shifted down one dimension. We therefore
immediately obtain the following corollary.

\begin{corollary}
   \label{htpygroups}
   The following hold:
   \begin{letterlist}
      \item if $M$ and $N$ are simply-connected four-manifolds, then
               $\pi_{\ast}(M)\cong\pi_{\ast}(N)$ if and only if
               $H^{2}(M)\cong H^{2}(N)$;
      \item if $M$ and $N$ are $(n-1)$-connected $2n$-dimensional manifolds
                and $n\notin\{2,4,8\}$, then $\pi_{\ast}(M)\cong\pi_{\ast}(N)$ if and only if
                $H^{n}(M)\cong H^{n}(N)$;
      \item if $M$ and $N$ are $n$-dimensional connected sums of sphere
                products, then $\pi_{\ast}(M)\cong\pi_{\ast}(N)$ if and only if
                 $H^{m}(M)\cong H^{m}(N)$ for each $m<n$.
   \end{letterlist}
   $\qqed$
\end{corollary}

Part~(a) of Corollary~\ref{htpygroups} reproves a theorem of Duan and Liang~\cite{DL}
via more homotopy theoretic methods, while parts~(b) and~(c) generalize it to wider classes of manifolds.
Another proof of Part~(a), again using more geometric techniques, is given in a recent preprint by Basu and Basu~\cite{BB}.
In fact, they show that the result holds for stable homotopy groups in place of homotopy groups.
It would be interesting to see whether this also holds for the generalizations presented here.

Theorem~\ref{classify} is proved by decomposing $\Omega M$ into
a product of spaces, up to homotopy. Explicitly, we have the following.

\begin{theorem}
   \label{1conndecomp}
   Let $M$ be a simply-connected four-manifold and suppose that
   $\mbox{dim}\,H^{2}(M)=k$. If $k=0$ then $M\simeq S^{4}$, if $k=1$
   then $\Omega M\simeq S^{1}\times\Omega S^{5}$, and if $k\geq 2$
   then there is a homotopy equivalence
   \[\Omega M\simeq S^1\times\Omega(S^{2}\times S^{3})
                  \times\Omega(J\vee (J\wedge \Omega(S^{2}\times S^{3})))\]
   where $J=\bigvee_{i=1}^{k-1} (S^{2}\vee S^{3})$ if $k>2$ and $J=\ast$ if $k=2$.
\end{theorem}

\begin{theorem}
   \label{Walldecomp}
   Let $M$ be an $(n-1)$-connected $2n$-dimensional manifold and
   suppose that \mbox{$\mbox{dim}\,H^{n}(M)=k$}. If $k\geq 2$ then there
   is a homotopy equivalence
   \[\Omega M\simeq\Omega(S^{n}\times S^{n})\times
        \Omega (J\vee(J\wedge\Omega(S^{n}\times S^{n})))\]
   where $J=\bigvee_{i=1}^{k-2} S^{n}$.
\end{theorem}

\begin{theorem}
\label{connsum}
Let $M$ and $N$ be closed oriented $(m-1)$-connected $n$-dimensional manifolds,
with $1<m\leq n-m$. Suppose that $H_*(M)$ is torsion-free and there is a ring isomorphism
$H^*(N)\cong H^*(S^m\times S^{n-m})$.
Let $M-\ast$ and $N-\ast$ be the punctured manifolds with a single point $\ast$ removed.
Then the following hold:
\begin{romanlist}
\item there is a homotopy equivalence
$$
\Omega (M\#N)\simeq \Omega(S^{m}\times S^{n-m})\times
    \Omega((M-\ast)\vee ((M-\ast)\wedge \Omega(S^{m}\times S^{n-m})));
$$
\item the looped inclusion
$\seqm{\Omega((M-\ast)\vee\bar N)\simeq\Omega((M-\ast)\vee S^{m}\vee S^{n-m})}
    {\Omega i}{\Omega M}$
has a right homotopy inverse.
\end{romanlist}
Consequently, the homotopy type of $\Omega (M\#N)$ is independent of
the homotopy type of $N$, and depends only on the homotopy type of $M-\ast$.
\end{theorem}

Theorems~\ref{Walldecomp} and~\ref{connsum} are consequences of much more general results presented in
Theorem~\ref{TMain} and Proposition~\ref{C1}, both of which are stated in the context of $CW$-complexes
and Poincar\'e duality spaces.

Observe that in each of these theorems, the decompositions
can be further refined. In each case, $J$ is a wedge of simply-connected
spheres, so $J\simeq\Sigma J^{\prime}$ where $J^{\prime}$ is a wedge of
spheres. Therefore, using the facts that
$\Omega(X\times Y)\simeq\Omega X\times\Omega Y$,
$\Sigma (X\times Y)\simeq\Sigma X\vee\Sigma Y\vee(\Sigma X\wedge Y)$
and by~\cite{J}, $\Sigma\Omega S^{s}\simeq\bigvee_{i=1}^{\infty} S^{(s-1)i+1}$,
we see that $J\wedge\Omega(S^{s}\times S^{t})$ is homotopy equivalent to
a wedge of spheres. Thus the factor
$\Omega(J\vee(J\wedge\Omega(S^{s}\times S^{t})))$
is homotopy equivalent to the loops on a large wedge of spheres,
and the Hilton-Milnor Theorem can be applied to decompose this as a
product of loops on spheres of varying dimensions. In particular,
in each case, $\Omega M$ decomposes as a product of loops on spheres,
and so the homotopy groups of $M$ can be determined to the same
extent as the homotopy groups of spheres.
A similar refinement is possible in Theorem~\ref{connsum} when $M-\ast$
has the homotopy type of a suspension.

From this point of view, Theorems~\ref{1conndecomp}, \ref{Walldecomp}
and~\ref{connsum} should be regarded as analogues of the
Hilton-Milnor Theorem. As such, these theorems are very
practical and should have numerous applications. We have already
mentioned how they can be used to determine the homotopy groups
of $M$. As another application described in detail in
Section~\ref{sec:4manifold}, we consider principle $G$-bundles
\(\seqm{P}{}{M}\),
where $M$ is a simply-connected four-manifold and $G$ is a
simply-connected, simple compact Lie group. It is well known
that there are $[M,BG]\cong\mathbb{Z}$ distinct equivalence
classes of such principle $G$-bundles. However, after looping
the homotopy types of $\Omega P$ all coincide as
$\Omega M\times\Omega G$.

To prove Theorems~\ref{1conndecomp}, \ref{Walldecomp}
and~\ref{connsum}, we consider a more general class
of torsion-free $CW$-complexes which resemble Poincar\'{e}
duality spaces. For such a space $P$ of connectivity $m-1$ and
dimension~$n$, we assume that the $(n-1)$-skeleton $\bar P$
has $S^{m}\vee S^{n-m}$ as a wedge summand, and that there is a
space $Q$ and a map
$\seqm{P}{q}{Q}$
such that there is a ring isomorphism
$H^{\ast}(Q)\cong H^{\ast}(S^{m}\times S^{n-m})$ and the composite
\(\seqmm{S^{m}\vee S^{n-m}}{}{P}{}{Q}\)
is onto in cohomology. Taking $F$ to be the homotopy fibre of $q$,
we analyze the homology of $F$ via the Serre spectral sequence,
and then use this to determine its homotopy type. This is then fed
into a decomposition of $\Omega P$ as $\Omega Q\times F$. The
decompositions in the three theorems above then follow as
special cases of this more general decomposition. All of this goes
through provided there are no cup product squares in cohomology,
which is the reason for the excluded cases $\{2,4,8\}$ in
Theorem~\ref{Walldecomp}. In the case of simply-connected
four-manifolds, these difficulties can be overcome through a
novel modification. If the simply-connected four-manifold~$M$
is mapped into $\mathbb{C}P^{\infty}$ by representing a cohomology
class in degree~$2$, then the homotopy fibre is a simply-connected
$5$-dimensional Poincar\'{e} duality complex $Z$ which fits into the
general class of torsion-free spaces $P$ above. The resulting
decomposition of $\Omega Z$ is then used to determine the homotopy
type of $\Omega M$.

\section{A general homotopy decomposition}

The loop space functor and localization functors both have effect of simplifying
homotopy types while retaining most of the original homotopy theoretic information.
At one extreme a conjecture of Anick~\cite{Anick2}
(for which there is some evidence~\cite{MW,Anick1,Stelzer1})
asserts that the loop space of any simply connected finite $CW$-complex
localized away from a predetermined finite set of primes decomposes
as a weak product of a certain countable list of indecomposable spaces,
while at the other end of the spectrum the loop space homotopy type of a
highly connected $CW$-complex is uniquely determined.
This is not difficult to see, for if $X$ and $Y$ and $(2n-2)$-dimensional $(n-1)$-connected,
then they are the $2n$-skeletons of $\Sigma\Omega X$ and $\Sigma\Omega Y$ respectively,
so a homotopy equivalence $\Omega X\simeq\Omega Y$ would allow one to construct a composite
\seqmmm{X}{incl.}{\Sigma\Omega X}{\simeq}{\Sigma\Omega Y}{eval.}{Y}
that induces an isomorphism on homology, and is therefore a homotopy
equivalence. Recently, a much stronger result of Grbi\'{c} and Wu~\cite{GW}
shows that if $X$ and $Y$ are simply-connected finite dimensional co-$H$ spaces
then $X\simeq Y$ if and only if $\Omega X\simeq\Omega Y$.

This leads to a natural question. Starting with a finite $CW$-complex $\bar P$,
and attaching a cell to $\bar P$ to form a space $P$, which homotopy classes of
attaching maps yield the same loop space homotopy type for $P$?
By the above remarks, distinct homotopy classes of $co$-$H$-maps tend to yield
distinct loop space homotopy types.
Our goal is to provide sufficient cohomological criteria given a few conditions on $\bar P$.
More precisely, we give a loop space decomposition for a certain class
of spaces, which includes certain connected-sums and certain
Poincar\'{e} duality spaces (both examples to be discussed in more detail in the
next section). Looping will have the effect of simplifying homotopy types,
and the homotopy types of the loop spaces will be shown to depend only
on simple data, often obtained from the homology of the original space
in degrees strictly less than the dimension of the space.
We begin by defining the class of spaces we have in mind. Throughout,
homology is taken with integer coefficients.

\begin{definition}
\label{Pdef}
Let $m$ and $n$ be integers such that $1<m\leq n-m$.
Suppose $P$ is a finite $n$-dimensional $(m-1)$-connected $CW$-complex
with torsion-free integral homology given by
$$
H_*(P)\cong \Z\paren{a_1,\ldots,a_\ell,z}
$$
where
$$
1<m=|a_1|\leq |a_2|\leq\cdots\leq |a_\ell|=n-m<|z|=n.
$$
Let $\bar P$ be the $(n-1)$-skeleton of $P$ and let
$i\,\colon\seqm{\bar P}{}{P}$
be the skeletal inclusion. Notice that
the bottom cell of $\bar P$ occurs in dimension $m$ while the
top cell occurs in dimension $n-m$.
\smallskip

Define $\mathcal{P}$ as the collection of all such spaces $P$ which also
satisfy the following two properties:
\begin{itemize}
\item[($1$)] there is a homotopy equivalence $\bar P\simeq J\vee (S^{m}\vee S^{n-m})$
         for some space $J$;
\item[($2$)] if $Q$ is the homotopy cofibre of the composite
         $J\hookrightarrow\seqm{\bar{P}}{i}{P}$, then there is a ring isomorphism
         $H^*(Q)\cong H^*(S^{m}\times S^{n-m})$.
\end{itemize}
\end{definition}

To analyze $\Omega P$ for $P\in\mathcal{P}$, some observations and notation
are required.

\noindent
\textbf{Observations}:
\begin{itemize}
\item[($1$)] If $X$ is a space and $H_*(X)$ is torsion-free, an element
$x\in H_*(X)$ has a dual class in $H^*(X)$ which we label as $x^*$.
In our case, since $H_*(P)$ is torsion-free, whenever $|a_i|+|a_j|=n$,
the cup product $a_{i}^*a_{j}^*$ is some multiple of $z^*$; define
the integer $c_{ij}$ by $a_{i}^*a_{j}^*=c_{ij}z^*$.

\item[($2$)] Observe that the homological description of $P$ implies that there is a
homotopy cofibration
$$
\seqmm{S^{n-1}}{\alpha}{\bar P}{i}{P}
$$
where $\alpha$ attaches the top cell to $P$. A basis for $H_*(\bar P)$ is given
by the elements $\{a_1,\ldots,a_\ell\}$.

\item[($3$)] The homotopy decomposition of $\bar P$ lets us define composites
$$
s\,\colon J\hookrightarrow\seqm{\bar P}{i}{P}
$$
$$
s'\,\colon S^{m}\vee S^{n-m}\hookrightarrow\seqm{\bar P}{i}{P}.
$$
Let $\iota_{t}\in H_t(S^t)$ represent a generator. Without loss of
generality we may assume that the basis for $H_*(P)$ has been chosen
so that $(s')_*(\iota_{m})=a_{1}$ and $(s')_*(\iota_{n-m})=a_{\ell}$.
Then the decomposition $\bar P\simeq J\vee(S^{m}\vee S^{n-m})$ implies
that $s_*$ induces an injection onto $\{a_{2},\ldots,a_{\ell-1}\}$.

\item[($4$)] The definition of $Q$ also lets us define a map $q$ by the homotopy
cofibration
$$
\seqmm{J}{s}{P}{q}{Q}.
$$
As this cofibration induces a long exact sequence in homology, the fact
that $\bar P\simeq J\vee (S^{m}\vee S^{n-m})$ is the $(n-1)$-skeleton of $P$
implies that the composite
$\seqmm{S^m\vee S^{n-m}}{s'}{P}{q}{Q}$
induces an injection in homology.

\item[($5$)] The ring isomorphism $H^*(Q)\cong H^*(S^{m}\times S^{n-m})$
implies that
$$
H_*(Q)\cong \Z\{x,y,e\},
$$
where $|x|=m$, $|y|=n-m$, $|e|=n$ and the generators can be
chosen so that $(x^*)^2=(y^*)^2=0$ and $y^*x^*=e^*$. Further,
since $(q\circ s')_*$ is an injection, we have $q_*(a_{1})=x$,
$q_{*}(a_{2})=y$ and $q_{*}(z)=e$; and as $q\circ s$ is null
homotopic we have $q_{*}(a_{i})=0$ for $2\leq i\leq \ell-1$.

\item[($6$)] The description of $q_{\ast}$ on the generators of $H_*(P)$
implies that $c_{\ell 1}=1$, $c_{1\ell}=(-1)^{m(n-m)}$, and
$c_{11}=c_{\ell\ell}=0$.
\end{itemize}

We are aiming for the homotopy decomposition of $\Omega P$
stated in Theorem~\ref{TMain}. To get started, we begin with an initial
decomposition. Define the space $F$ and the maps $f$ and $\delta$
by the homotopy fibration sequence
$$
\seqmmm{\Omega Q}{\delta}{F}{f}{P}{q}{Q}.
$$
We first calculate the homology of $\Omega Q$ and relate it to the
homology of $\Omega (S^{m}\vee S^{n-m})$. By the Bott-Samelson
theorem, there is an algebra isomorphism
$$
H_*(\Omega(S^{m}\vee S^{n-m}))\cong T(u,v)
$$
where $T(u,v)$ is the free tensor algebra on generators $u$ and $v$
of degrees $m-1$ and $n-m-1$ respectively. Let $\mathbb{Z}[u,v]$
be the polynomial algebra generated by $u$ and $v$.

\begin{lemma}
\label{Qhlgy}
There is a coalgebra isomorphism
$H_*(\Omega Q)\cong\mathbb{Z}[u,v]$
which can be chosen so that the map
\(\llnamedright{\Omega(S^{m}\vee S^{n-m})}{\Omega (q\circ s')}{\Omega Q}\)
induces in homology the abelianization
$\seqm{T(u,v)}{}{\mathbb{Z}[u,v]}$.
\end{lemma}

\begin{proof}
First, consider the homology Serre spectral sequence for the
path-loop homotopy fibration
$\seqmm{\Omega (S^{m}\vee S^{n-m})}{}{\ast}{}{S^{m}\vee S^{n-m}}$.
Let $\iota_{k}\in H_{k}(S^{k})$ represent a generator. Then
the elements $\iota_{m},\iota_{n-m}\in H_*(S^{m}\vee S^{n-m})$
transgress to the elements $u,v\in T(u,v)$, and the element
$[u,v]\in T(u,v)$ arises in the spectral sequence as the element
$u\otimes\iota_{n-m}+(-1)^{\vert u\vert\vert\iota_{n-m}\vert}\iota_{m}\otimes v$.

Next, consider the homology Serre spectral sequence
for the path-loop homotopy fibration
$\seqmm{\Omega Q}{}{\ast}{}{Q}$.
By Observation~(5), $H_*(Q)\cong\mathbb{Z}\{x,y,e\}$ where
$\vert x\vert=m$, $\vert y\vert=n-m$, $\vert e\vert=n$ and
the cohomology duals satisfy $x^*y^*=e^*$. Thus in homology,
the reduced diagonal $\overline{\Delta}(e)$ equals $x\otimes y+y\otimes x$.
Thus in the Serre spectral sequence for the path-loop homotopy fibration, we
have $x$ and $y$ transgressing to elements $a$ and~$b$ respectively, and
$d^{n}(e)=a\otimes y+(-1)^{\vert a\vert\vert y\vert} x\otimes b$. It is
now a straightforward calculation to show that there is an isomorphism of
vector spaces $H_*(\Omega Q)\cong\mathbb{Z}[a,b]$ where $\vert a\vert=m-1$
and $\vert b\vert=n-m-1$.

Now consider the homotopy commutative diagram of path-loop homotopy fibrations
\[\diagram
         \Omega (S^{m}\vee S^{n-m})\rto\dto^{\Omega(q\circ s')}
               & \ast\rto\dto & S^{m}\vee S^{n-m}\dto^{q\circ s'} \\
         \Omega Q\rto & \ast\rto & Q.
  \enddiagram\]
This induces a morphism of Serre spectral sequences between the
two path-loop homotopy fibrations. By Observation~(5), the map $(q\circ s')_{\ast}$
is an isomorphism in degrees~$<n$. Therefore, comparing Serre spectral
sequences, $(\Omega(q\circ s'))_{\ast}$ is an isomorphism in degrees~$<n-1$.
In particular, $(\Omega(q\circ s'))_{\ast}$ is an isomorphism in degrees
$m-1$ and $n-m-1$. Thus, up to sign, $(\Omega(q\circ s'))_{\ast}$ sends
$u,v\in T(u,v)$ to $a,b\in\mathbb{Z}[u,v]$. Comparing spectral sequences,
we also have the element
$u\otimes\iota_{n-m}+(-1)^{\vert u\vert\vert\iota_{n-m}\vert}\iota_{m}\otimes v$
sent to $a\otimes y+(-1)^{\vert a\vert\vert y\vert} x\otimes b$, which is
the image of the differential $d^{n}(e)$. That is,
$[u,v]\in T(u,v)$ is sent to $0\in\mathbb{Z}[a,b]$. Further, it is
straightforward to see that once the $d^{n}$ differential is taken
into account and we move to $E^{n+1}$, that the $E^{n+1}$ page
for the fibration
$\seqmm{\Omega (S^{m}\vee S^{n-m})}{}{\ast}{}{S^{m}\vee S^{n-m}}$
maps onto the $E^{n+1}$ page for the fibration
$\seqmm{\Omega Q}{}{\ast}{}{Q}$. As there are no more non-trivial
differentials, the same is true of the $E^{\infty}$ pages, and so
$(\Omega(q\circ s'))_{\ast}$ is onto.

Finally, since $(\Omega(q\circ s'))_{\ast}$ is an algebra map and
$(\Omega(q\circ s'))_{\ast}([u,v])=0$, there is a factorization
\[\diagram
         T(u,v)\rrto^-{(\Omega(q\circ s'))_{\ast}}\dto^{\pi}
                & & H_*(\Omega Q)\cong\mathbb{Z}[a,b] \\
         \mathbb{Z}[u,v]\urrto_{g} & &
  \enddiagram\]
for some map $g$, where $\pi$ is the abelianization map.
Since $(\Omega(q\circ s'))_{\ast}$ is onto and both
$\mathbb{Z}[u,v]$ and $\mathbb{Z}[a,b]$ have the same
Poincar\'{e} series, $g$ must be an isomorphism. The statement
of the lemma now follows.
\end{proof}

By the Hilton-Milnor Theorem, the inclusion of the
wedge into the product
$\seqm{S^{m}\vee S^{n-m}}{j}{S^{n}\times S^{n-m}}$
has a right homotopy inverse after looping. That is, there is a map
$$
\phi\,\colon\seqm{\Omega S^{n}\times\Omega S^{n-m}}{}
     {\Omega(S^{m}\vee S^{n-m})}
$$
which is a right homotopy inverse of $\Omega j$.

\begin{lemma}
\label{loopQsplit}
The composite
$\seqmmm{\Omega S^{m}\times\Omega S^{n-m}}{\phi}{\Omega(S^{m}\vee S^{n-m})}
   {\Omega s'}{\Omega P}{\Omega q}{\Omega Q}$
is a homotopy equivalence. Consequently, in the homotopy fibration sequence
\seqmmm{\Omega Q}{\delta}{F}{f}{P}{q}{Q},
the map $\delta$ is null homotopic,
implying that there are homotopy equivalences
$$
\Omega P\simeq\Omega Q\times\Omega F\simeq
    \Omega S^{m}\times\Omega S^{n-m}\times\Omega F.
$$
\end{lemma}

\begin{proof}
The fact that $\phi$ is a right homotopy inverse of $\Omega i$ implies
that $\phi_{\ast}$ is a coalgebra map which maps onto the sub-coalgebra
$\mathbb{Z}[u,v]$ of $T(u,v)\cong H_*(\Omega(S^{m}\vee S^{n-m}))$.
By Lemma~\ref{Qhlgy}, $(\Omega(q\circ s'))_{\ast}$ maps this sub-coalgebra
isomorphically onto $H_{\ast}(Q)$. Thus $\Omega q\circ\Omega s'\circ\phi$
induces an isomorphism in homology and so is a homotopy equivalence.

For the consequences, consider the homotopy fibration sequence
$\seqmmm{\Omega F}{}{\Omega P}{\Omega q}{\Omega Q}{\delta}{F}$.
We have just shown that $\phi\circ\Omega s'$ is a right homotopy inverse
for~$\Omega q$. Therefore, the map $\delta$
is null homotopic, and this immediately implies that there is a
homotopy equivalence $\Omega P\simeq\Omega Q\times\Omega F$.
\end{proof}

Next, we wish to give an explicit homotopy decomposition
of the space $\Omega F$. The first step is to calculate its homology.
By Observation~(4), the composite \seqmm{J}{s}{P}{q}{Q} is a homotopy
cofibration, so it is null homotopic. Therefore, $s$ lifts through
\seqm{F}{f}{P}
to a map
$$
\bar s\colon\seqm{J}{}{F}.
$$
By Observation~(3), $s_{\ast}$ induces an injection onto $\{a_{2},\ldots,a_{\ell-1}\}$.
So its lift $\bar s$ has the property that~$(\bar s)_{\ast}$ is an
injection, and we will also label a basis for the image of $(\bar s)_{\ast}$ by
$\{a_{2},\ldots,a_{\ell-1}\}$.

As the homotopy fibration
\begin{equation}
\label{EPrincipalFib}
\seqmm{\Omega Q}{\delta}{F}{f}{P}.
\end{equation}
is principal, there exists a left action
$$
\theta\,\colon\seqm{\Omega Q\times F}{}{F}
$$
such that the following diagram commutes up to homotopy
\begin{equation}
\label{EAction}
\diagram
\Omega Q\times \Omega Q\rto^{\ID\times\delta}\dto^{\mu}
    & \Omega Q\times F\dto^{\theta}\\
\Omega Q\rto^{\delta} & F
\enddiagram
\end{equation}
where $\ID$ is the identity map and $\mu$ is the standard loop space
multiplication.

\begin{proposition}
\label{P4}
There is an isomorphism of left $H_{*}(\Omega Q)$-modules
$$
H_*(F)\cong \Z\paren{a_2,\ldots,a_{\ell-1}}\otimes H_*(\Omega Q),
$$
where $\mathbb{Z}\{a_{2},\ldots, a_{\ell-1}\}$ is the image of
$\bar{s}_{\ast}$ and the left action of $H_{*}(\Omega Q)$ given by $\theta_*$.
\end{proposition}

\begin{proof}
By a result of Moore~\cite{Mo}, the homology Serre spectral sequence $E$
for the principal homotopy fibration sequence \seqmm{\Omega Q}{\delta}{F}{f}{P}
is a spectral sequence of left $H_*(\Omega Q)$-modules, with
\begin{equation}
\label{ESerre}
E^{2}_{*,*}\cong H_*(P)\otimes H_*(\Omega Q).
\end{equation}
Here, the left action is induced by $\theta_{\ast}$ and the
differentials respect the left action of $H_*(\Omega Q)$.
That is, up to sign, $d^n(f\otimes gh)=(1\otimes g)d^n(f\otimes h)$ whenever
the differential $d^n$ is defined. We now proceed to calculate the spectral
sequence. In doing so, it will be helpful to rewrite~(\ref{ESerre}) as
\begin{equation}
\label{ESerre2}
E^{2}_{\ast,\ast}\cong\mathbb{Z}\{1,a_{1},\ldots,a_{\ell},z\}\otimes H_{\ast}(\Omega Q).
\end{equation}
\medskip

\noindent
\textit{Initial information on the differentials}.
Consider the composite
$\seqmm{S^{m}\vee S^{n-m}}{s'}{P}{q}{Q}$.
By Observation~(4), $(q\circ s')_{\ast}$ is an injection in homology.
The composite induces a homotopy fibration diagram
\[\diagram
     \Omega Q\rto\ddouble & Z\rto\dto
          & S^{m}\vee S^{n-m}\rto^-{q\circ s'}\dto^{s'} & Q\ddouble \\
     \Omega Q\rto^-{\delta} & F\rto^-{f} & P\rto^-{q} & Q
  \enddiagram\]
which defines the space $Z$. Since $(q\circ s')_{\ast}$ is an injection
in homology and there is a coalgebra isomorphism
$H_{\ast}(Q)\cong H_{\ast}(S^{m}\times S^{n-m})$,
in the homology Serre spectral sequence for the fibration
$\seqmm{\Omega Q}{}{Z}{}{S^{m}\vee S^{n-m}}$ the generators
$\iota_{m},\iota_{n-m}\in H_{\ast}(S^{m}\vee S^{n-m})$ transgress
to the elements $u,v\in H_{\ast}(\Omega Q)$ respectively, where $u,v$
are as in Lemma~\ref{Qhlgy}. Now consider the homology Serre spectral
sequence for the fibration
$\seqmm{\Omega Q}{\delta}{F}{f}{P}$.
By Observation~(3), we may assume that $(s')_{\ast}(\iota_{m})=a_{1}$
and $(s')_{\ast}(\iota_{n-m})=a_{l}$,
so a comparison of spectral sequences implies that the elements
$a_{1},a_{l}$ transgress to $u,v\in H_{\ast}(\Omega Q)$. That is,
in terms of $E^{2}_{\ast,\ast}$, we have
$$
d^{m}(a_1\otimes 1)=1\otimes u,\,\, d^{n-m}(a_\ell\otimes 1)=1\otimes v.
$$
Further, by Observation~(3), the map
$\seqm{J}{s}{P}$
induces an injection in homology onto $\{a_{2},\ldots,a_{\ell-1}\}$,
and it was observed before the statement of the proposition that
the map $s$ lifts through $f$ to $F$. Therefore the elements $\{a_{2},\ldots,a_{\ell-1}\}$
survive the spectral sequence. Consequently,
\begin{equation}
  \label{dtinfo}
  d^{t}(a_{i})=0\ \mbox{for all $t\geq 2$ and $2\leq i\leq\ell-1$}.
\end{equation}
\medskip

\noindent
\textit{Case 1: $m<n-m$}. For degree reasons, the differentials
$d^{2},\ldots, d^{m-1}$ are all zero on the elements $a_{1},\ldots,a_{\ell}$,
so the left action of $H_{\ast}(\Omega Q)$ implies that these differentials
are identially zero. Therefore
$$
E^{2}_{\ast,\ast}\cong E^{m}_{\ast,\ast}.
$$

For $d^{m}$ we have $d^{m}(a_{1}\otimes 1)=1\otimes u$. The left
action of $\theta_{\ast}$ implies that for any element $g\in H_{\ast}(\Omega Q)$,
we have (up to sign),
$$
d^{m}(a_1\otimes g)= (1\otimes g)d^{m}(a_1\otimes 1)=
     (1\otimes g)(1\otimes u) = 1\otimes gu.
$$
By~(\ref{dtinfo}), $d^{m}(a_{i})=0$ for $2\leq i\leq\ell$. So
the left action of $\theta_{\ast}$ implies that for
$d^{m}(a_{i}\otimes g)=0$ for any $2\leq i\leq\ell$ and any $g\in H_*(\Omega Q)$.
Next, consider the element $z\otimes 1$.
Dualizing to the cohomology spectral sequence associated with $E$,
we have for each $i$ such that $|a_i|=n-m$,
\begin{align*}
d_{m}(a_i^*\otimes u^*)&=(d_{m}(a_i^*\otimes 1))(1\otimes u^*)+
    (-1)^{|a_i|}(a_i^*\otimes 1)d_{m}(1\otimes u^*)\\
&=(-1)^{\vert a_{i}\vert} (a_i^*\otimes 1)(a_1^*\otimes 1) =
     (-1)^{\vert a_{i}\vert} c_{i1}(z^*\otimes 1).
\end{align*}
This implies that in the homology Serre spectral sequence $E$ we have
$$
d^{m}(z\otimes 1)=\csum{|a_i|=n-m}{}{(-1)^{\vert a_{i}\vert} c_{i1}(a_i\otimes u)}.
$$
The left action of $\theta_{\ast}$ therefore implies that
$$
d^{m}(z\otimes g)=\csum{|a_i|=n-m}{}{(-1)^{\vert a_{i}\vert} c_{i1}(a_i\otimes gu)}
$$
for each $g\in H_*(\Omega Q)$. Therefore, as $c_{\ell 1}=1$ by Observation~(6),
$c_{\ell 1}(a_\ell\otimes gu)=(a_\ell\otimes gu)$ is identified in $E^{m+1}_{n-m,*}$
with a linear combination of elements $a_i\otimes gu$ for $|a_i|=n-m$. Note
that $a_{1}$ is excluded here since $\vert a_{1}\vert=m$ and in this case we
have assumed that $m<n-m$. Collectively, we have determined the differential $d^{m}$,
and obtain an isomorphism of left $H_{\ast}(\Omega Q)$-modules
$$
E^{m+1}_{\ast,\ast}\cong\mathbb{Z}\{a_{2},\ldots,a_{\ell}\}\otimes H_{\ast}(\Omega Q).
$$

Continuing, by~(\ref{dtinfo}), $d^{m+1},\ldots,d^{n-m-1}$ are all identically
zero on the elements $a_{2},\ldots,a_{\ell-1}$ and for degree reasons,
$d^{m+1},\ldots,d^{n-m-1}$ are all identically zero on $a_{\ell}$. So the left
action of $\theta_{\ast}$ implies that these differentials are identically zero on
all elements. Therefore there is an isomorphism
$$
E^{m+1}_{\ast,\ast}\cong E^{n-m}_{\ast,\ast}.
$$

For $d^{n-m}$, by~(\ref{dtinfo}), $d^{n-m}(a_{i})=0$ for
$2\leq i\leq\ell -1$, so the left action of $\theta_{\ast}$ implies that
$d^{n-m}(a_{i}\otimes g)=0$ for any $2\leq i\leq\ell-1$ and for
any $g\in H_{\ast}(\Omega Q)$. From the initial calculation of differentials,
we obtained $d^{n-m}(a_{\ell}\otimes 1)=1\otimes v$. The left
action of $\theta_{\ast}$ therefore implies that for any element
$g\in H_{\ast}(\Omega Q)$ we have (up to sign),
$$
d^{n-m}(a_{\ell}\otimes g)= (1\otimes g)d^{n-m}(a_{\ell}\otimes 1)=
     (1\otimes g)(1\otimes v) = 1\otimes gv.
$$
Thus we have determined the differential $d^{n-m}$, and obtain an isomorphism
of left $H_{\ast}(\Omega Q)$-modules
$$
E^{n-m+1}_{\ast,\ast}\cong\mathbb{Z}\{a_{2},\ldots, a_{\ell-1}\}\otimes H_{*}(\Omega Q).
$$

Finally, by~(\ref{dtinfo}), the differentials $d^{t}$ for $t>n-m$ are all
identically zero on $a_{2},\ldots,a_{\ell-1}$, so the left acton of $\theta_{\ast}$
implies that these differentials are identically zero on all elements. Hence
$$
E^{\infty}_{\ast,\ast}\cong E^{n-m+1}_{\ast,\ast}.
$$
Since there is no torsion in $E^{\infty}_{*,*}$, there is no extension problem,
and we have
\begin{equation}
\label{EIso}
H_{\ast}(F)\cong \cplus{i+j=*}{}{E^{\infty}_{i,j}}
\cong\mathbb{Z}\{a_{2},\ldots,a_{\ell-1}\}\otimes H_{\ast}(\Omega Q).
\end{equation}
To see that this is an isomorphism of left $H_{*}(\Omega Q)$-modules,
recall that the left action
\seqm{H_{\ast}(\Omega Q)\otimes E^{\infty}_{i,j}}{}{E^{\infty}_{i,j+\ast}}
coincides with the left action of associated graded objects
$$
\seqm{H_{\ast}(\Omega Q)\otimes \frac{\mc F_{i,i+j}}{\mc F_{i-1,i+j}}}{}
{\frac{\mc F_{i,i+j+\ast}}{\mc F_{i-1,i+j+\ast}}\cong E^{\infty}_{i,j+\ast}}
$$
induced by the action \seqm{H_{*}(\Omega Q)\otimes H_{i+j}(F)}{\mu_*}{H_{i+j+\ast}(F)},
where $\mc F_{i,j}=\mc F_i H_j(F)\subseteq H_j(F)$ is the increasing filtration
associated with our spectral sequence.
Observe from the calculations above that the action on the $E^{\infty}_{\ast,\ast}$ is free,
so the action on the associated graded objects is free.
Therefore the action $\mu_*$ must also be free,
and so the isomorphism~(\ref{EIso}) is one of left $H_{*}(\Omega Q)$-modules.

\medskip

\noindent
\textit{Case 2: $m=n-m$}. This case is simpler.
We have $n=2m$ and $|u|=|v|=m-1$. So the only differential which
comes into play is $d^{m}$. This time $d^{m}(z\otimes g)$ is the sum
of linear combinations of the elements
$c_{i1}(a_i\otimes gu)$ and $c_{i\ell}(a_i\otimes gv)$ for all $i$,
where
$c_{\ell 1}=1$, $c_{1\ell}=(-1)^{m(n-m)}=-(-1)^{|u||v|}$ and $c_{11}=c_{\ell\ell}=0$.
Therefore the elements $a_\ell\otimes gu-(-1)^{|u||v|}a_1\otimes gv$
are identified in $E^{m+1}_{m,*}$ with a linear combination of
elements of the form $a_i\otimes gu$ or $a_i\otimes gv$ for $2\leq i\leq \ell-1$,
and the calculation goes through as before.
\end{proof}

Now we refine the homotopy decomposition
$\Omega P\simeq\Omega Q\times\Omega F$
of Lemma~\ref{loopQsplit} by identifying the homotopy type of $F$.
For spaces $X$ and $Y$, the \emph{left half-smash} of $X$ and $Y$ is defined by
$$
X\ltimes Y=(X\times Y)/(\ast\times Y).
$$
It is well-known that if $Y$ is a suspension then there is a homotopy equivalence
$$
X\ltimes Y\simeq Y\vee (X\wedge Y).
$$

\begin{proposition}
\label{Ftype}
There is a homotopy equivalence
$$
F\simeq \Omega Q\ltimes J.
$$
\end{proposition}

\begin{proof}
Using the lift $\seqm{J}{\bar s}{F}$ of $\seqm{J}{s}{P}$ and the
homotopy action $\seqm{\Omega Q\times F}{\theta}{F}$, define
$\lambda$ as the composite
$$
\lambda\,\colon\seqmm{\Omega Q\times J}{\ID\times \bar s}
    {\Omega Q\times F}{\theta}{F}.
$$
By~(\ref{EAction}), the restriction of $\theta$ to $\Omega Q$ is
homotopic to $\delta$, which by Lemma~\ref{loopQsplit} is null homotopic.
Therefore the composite
$$
\seqmm{\Omega Q\times *}{\ID\times *}{\Omega Q\times J}{\lambda}{F}
$$
is null homotopic. Since the homotopy cofibre of $\ID\times\ast$ is
$\Omega Q\ltimes F$, the map $\lambda$ extends to a map $\hat\lambda$
that makes the following diagram homotopy commute
\[\diagram
\Omega Q\times *\rto^-{\ID\times *}
& \Omega Q\times J\rto^{}\dto^{\lambda}
& \Omega Q\ltimes J\ar@{.>}[dl]^{\hat\lambda}\\
& F.
\enddiagram\]
By definition, $\lambda=\theta\circ(\ID\times\bar s)$, so Proposition~\ref{P4}
implies that $\hat\lambda_{\ast}$ is an isomorphism. Thus $\hat\lambda$
is a homotopy equivalence.
\end{proof}

\begin{theorem}
\label{TMain}
Let $P\in\mathcal{P}$ and suppose that $P$ is $(m-1)$-connected
and $n$-dimensional. Then the following hold:
\begin{romanlist}
\item there is a homotopy equivalence
$$
\Omega P\simeq \Omega(S^{m}\times S^{n-m})\times
\Omega(\Omega(S^{m}\times S^{n-m})\ltimes J),
$$
which, if $J$ is a suspension, refines to a homotopy equivalence
$$
\Omega P\simeq \Omega(S^{m}\times S^{n-m})\times \Omega(J\vee (J\wedge \Omega(S^{m}\times S^{n-m})));
$$
\item the map $\seqm{\Omega\bar P}{\Omega i}{\Omega P}$ has a right
homotopy inverse.
\end{romanlist}
\end{theorem}

\begin{proof}
For part~(i), by Lemma~\ref{loopQsplit}, $\Omega P\simeq\Omega Q\times\Omega F$
and $\Omega Q\simeq\Omega S^{m}\times\Omega S^{n-m}$, and by
Proposition~\ref{Ftype}, $F\simeq \Omega Q\ltimes J$. Thus
$$
\Omega P\simeq\Omega S^{m}\times\Omega S^{n-m}\times
    \Omega((\Omega S^{m}\times\Omega S^{n-m})\ltimes J).
$$
If $J$ is a suspension, this decomposition refines due to the fact
that $\Omega Q\ltimes J\simeq J\vee (J\wedge\Omega Q)$.

For part~(ii), define $\bar q$ as the composite
$$
\bar q\,\colon\seqmm{\bar P}{i}{P}{q}{Q}.
$$
From this composite we obtain a homotopy pullback diagram
\begin{equation}
\label{FbarFdgrm}
\diagram
\Omega Q\rto^{\bar\delta}\ddouble & \bar F\rto^{\bar f}\dto^{\tau}
     & \bar P\rto^{\bar q}\dto^{i} & Q\ddouble\\
\Omega Q\rto^{\delta} & F\rto^{f} & P\rto^{q} & Q
\enddiagram
\end{equation}
which defines the space $\bar F$ and the maps $\bar f$, $\bar\delta$ and $\tau$.
In particular, this is a homotopy commutative diagram of principal fibration
sequences, so if
$\bar\theta\,\colon\seqm{\Omega Q\times\bar F}{}{\bar F}$
is the homotopy action for the top fibration sequence, then there is a
homotopy commutative diagram of actions
\[\diagram
         \Omega Q\times\bar F\rto^-{\bar\theta}\dto^{\ID\times\tau}
                & \bar F\dto^{\tau} \\
        \Omega Q\times F\rto^-{\theta} & F.
  \enddiagram\]

By definition, the map $\seqm{J}{s}{P}$ factors as the composite
$\seqmm{J}{r}{\bar P}{i}{P}$,
where $r$ is the inclusion of the wedge summand in
$\bar P\simeq J\vee (S^{m}\vee S^{n-m})$. Since $s$ lifts
through $f$ to the map $\seqm{J}{\bar s}{F}$, the definition
of $\bar F$ as a homotopy pullback in~(\ref{FbarFdgrm}) implies that there
is a pullback map $\bar r\,\colon\seqm{J}{}{\bar F}$ such that
$\bar f\circ\bar r\simeq r$ and $\tau\circ\bar r\simeq\bar s$.
Combining this with the preceding diagram, we obtain a homotopy
commutative diagram
\begin{equation}
\label{actiondgrm}
  \diagram
         \Omega Q\times J\rto^-{\ID\times\bar r}\ddouble
                & \Omega Q\times\bar F\rto^-{\bar\theta}\dto^{\ID\times\tau}
                & \bar F\dto^{\tau} \\
        \Omega Q\times J\rto^-{\ID\times\bar s} & \Omega Q\times F\rto^-{\theta} & F.
  \enddiagram
\end{equation}

By definition, the map $\seqm{S^{m}\vee S^{n-m}}{s'}{P}$ factors as
the composite
$\seqmm{S^{m}\vee S^{n-m}}{j}{\bar P}{i}{P}$,
where $j$ is the inclusion of the wedge summand in
$\bar P\simeq J\vee (S^{m}\vee S^{n-m})$. By Lemma~\ref{loopQsplit},
$\Omega(q\circ s')$ has a right homotopy inverse. As
$\bar q=q\circ i$, we have $q\circ s'=q\circ i\circ j=\bar q\circ j$,
so $\Omega(\bar q\circ j)$ has a right homotopy inverse. Consequently,
$\Omega\bar q$ has a right homotopy inverse, which implies that in
the homotopy fibration
$\seqmm{\Omega P}{\Omega\bar q}{\Omega Q}{\bar\delta}{\bar F}$,
the map $\bar\delta$ is null homotopic.

Let $\bar\lambda$ be the composite along the top row of~(\ref{actiondgrm}),
$$
\bar\lambda\,\colon\seqmm{\Omega Q\times J}{\ID\times\bar r}
     {\Omega Q\times F}{\bar\theta}{\bar F}.
$$
Since $\bar{\theta}$ is a homotopy action, its restriction to $\Omega Q$
is $\bar\delta$. Therefore the restriction of $\bar\lambda$ to $\Omega Q$
is $\bar\delta$, which is null homotopic. Thus there is a homotopy
commutative diagram
\[\diagram
       \Omega Q\times\ast\rto^-{\ID\times\ast}
          & \Omega Q\times J\rto\dto^{\bar\lambda}
          & \Omega Q\ltimes J\ar@{.>}[dl]^{\tilde\lambda} \\
       & F &
  \enddiagram\]
where the top row is a homotopy cofibration and $\tilde\lambda$
is an extension of $\bar\lambda$. Now let $\gamma$ be the composite
$$
\gamma\,\colon\seqmm{\Omega Q\ltimes\bar F}{\tilde\lambda}{\bar F}{\tau}{F}.
$$
Observe that $\gamma$ is a choice of the extension $\hat\lambda$ in
the proof of Proposition~\ref{Ftype}. Thus $\gamma$ induces an isomorphism
in homology and so is a homotopy equivalence. Consequently,
the map $\tau$ has a right homotopy inverse
$\sigma\,\colon\seqm{F}{}{\bar F}$.

Finally, consider the diagram
\[\diagram
        (\Omega S^{m}\times\Omega S^{n-m})\times\Omega F
                    \rto^-{\ID\times\Omega\sigma}\drdouble
             & (\Omega S^{m}\times\Omega S^{n-m})\times\Omega\bar F
                    \rto^-{\Omega j\times\Omega\bar f}\dto^{\ID\times\Omega\tau}
             & \Omega\bar P\times\Omega\bar P\rto^-{\mu}\dto^{\Omega i\times\Omega i}
             & \Omega\bar P\dto^{\Omega i} \\
        & (\Omega S^{m}\times\Omega S^{n-m})\times\Omega F
                   \rto^-{\Omega s'\times\Omega f}
              & \Omega P\times\Omega P\rto^-{\mu} & \Omega P
  \enddiagram\]
where $\mu$ is the standard loop multiplication. The left triangle homotopy
commutes since $\sigma$ is a right homotopy inverse of $\tau$. The
middle square homotopy commutes since, by definition, $s'=i\circ j$,
and by~(\ref{FbarFdgrm}), $f\simeq i\circ\bar f$. The right square homotopy
commutes since $\Omega i$ is a loop map. By part~(i), the bottom row is a
homotopy equivalence, so the homotopy commutativity of the diagram
implies that~$\Omega i$ has a right homotopy inverse.
\end{proof}

\section{Consequences}

In this section we apply Theorem~\ref{TMain} to two classes of examples,
first to certain connected sums, and then to certain Poincar\'{e} duality
complexes, and prove Theorems~\ref{Walldecomp}.

If $M$ is a closed oriented $n$-dimensional manifold, let $\bar M$
be the $(n-1)$-skeleton of $M$.
In particular, $M$ is homotopy equivalent to $M-\ast$,
and is obtained from $\bar M$ by attaching a single $n$-cell.
Observe that if $N$ is a closed oriented
$n$-dimensional manifold and there is a ring isomorphism
$H^*(N)\cong H^*(S^{m}\times S^{n-m})$, then $\bar N\simeq S^{m}\vee S^{n-m}$.
Denote the connected sum of two closed oriented $n$-dimensional
manifolds $M$ and $N$ by $M\#N$. Observe that the $(n-1)$-skeleton
of $M\#N$ is homotopy equivalent to $\bar M\vee\bar N$. Let
$$
i\,\colon\seqm{\bar M\vee\bar N}{}{M\#N}
$$
be the skeletal inclusion.

\begin{proof}[Proof of Theorem~\ref{connsum}]
We will show that $M\#N\in\mathcal{P}$. Let
$P=M\#N$, let $\bar P$ be the $(n-1)$-skeleton of $P$, and let
$\seqm{\bar P}{i}{P}$ be the skeletal inclusion. By the definitions
of $M$ and $N$, $P$ is an $(m-1)$-connected, $n$-dimensional
$CW$-complex. Since $P=M\#N$ is a closed oriented manifold, it satisfies
Poincar\'{e} duality,  which implies that $\bar P\simeq\bar M\vee\bar N$
is actually $(n-m)$-dimensional. Note that as $m>1$ we have $n-m<n-1$,
so $H_*(P)$ is torsion-free if and only if $H_*(\bar P)$ is torsion-free.
But as $H_*(M)$ is torsion-free, so is $H_*(\bar M)$, which implies that
$\bar P\simeq\bar M\vee\bar N\simeq\bar M\vee (S^{m}\vee S^{n-m})$
also has $H_*(\bar P)$ torsion-free. Thus $H_*(P)$ is torsion-free.

Now if $J=\bar M$ then as $\bar N\simeq S^{m}\vee S^{n-m}$, we have
$\bar P\simeq J\vee(S^{m}\vee S^{n-m})$. Let $Q$ be the cofibre of the
composite $\seqmm{J}{}{\bar P}{i}{P}$, that is, $Q$ is the cofibre of the
composite $\seqmm{\bar M}{}{\bar M\vee\bar N}{i}{M\#N}$. Then
$Q\simeq N$, which implies that
$H^*(Q)\cong H^*(N)\cong H^*(S^{m}\times S^{n-m})$.
Thus $P=M\#N$ satisfies all the conditions of Definition~\ref{Pdef},
so $P\in\mathcal{P}$. The assertions of the proposition are now all
direct applications of Theorem~\ref{TMain}.
\end{proof}

\begin{example}
As an example of Theorem~\ref{connsum} in action, recall that
an $n$-dimensional manifold $M$ is a connected sum of sphere products if
$$
M\cong (S^{m_{1}}\times S^{n-m_{1}})\#\cdots\#(S^{m_{k}}\times S^{n-m_{k}})
$$
for some integers $m_{1},\ldots,m_{k}$.
Let $M_{1}=(S^{m_{1}}\times S^{n-m_{1}})\#\cdots\#(S^{m_{k-1}}\times S^{n-m_{k-1}})$
and $N=S^{m_{k}}\times S^{n-m_{k}}$ so that $M=M_{1}\# N$. Observe that
$\bar M_{1}=\bigvee_{i=1}^{k-1}(S^{m_{i}}\vee S^{n-m_{i}})$.
So by Theorem~\ref{connsum}, there is a homotopy equivalence
\[\Omega M\simeq\Omega(M_{1}\#N)\simeq\Omega(S^{m_{k}}\times S^{n-m_{k}})\times
      \Omega(\bar M_{1}\vee(\bar M_{1}\wedge\Omega(S^{m_{k}}\times S^{n-m_{k}}))).\]

\end{example}

Recall that $P$ is a \emph{Poincar\'e duality complex} if it has the homotopy type of
a finite $CW$-complex and its cohomology ring $H^*(P;R)$ satisfies
Poincar\'e duality for all coefficient rings $R$. In particular every
oriented simply-connected manifold is a Poincar\'e duality complex.

\begin{proposition}
\label{C1}
Fix $1<m\leq n$. If $m=n-m$, assume that $m\notin\{2,4,8\}$.
Let $P$ be an $(m-1)$-connected $n$-dimensional Poincar\'e duality complex
such that $(n-1)$-skeleton $\bar P$ of $P$ has the homotopy type of a wedge
of spheres. Let $i\,\colon\seqm{\bar P}{}{P}$ be the skeletal inclusion. Then
the following hold:
\begin{romanlist}
\item there is a homotopy equivalence
$$
\Omega P\simeq \Omega(S^{m}\times S^{n-m})\times \Omega(J\vee (J\wedge \Omega(S^{m}\times S^{n-m})))
$$
where $J$ is obtained from $\bar P$ by quotienting out a copy of $S^{m}\vee S^{n-m}$;
\item the map
$\seqm{\Omega\bar P}{\Omega i}{\Omega P}$
has a right homotopy inverse.
\end{romanlist}
Consequently, the homotopy type of $\Omega P$ depends only on the
homotopy type of~$\bar P$.
\end{proposition}

We will need a preliminary lemma about the cohomology ring of Poincar\'e duality
complexes before we can prove this.

\begin{lemma}
\label{LPD}
Let $P$ be an $n$-dimensional Poincar\'e duality complex such that $H_*(P)$
is torsion-free, and let $e^*$ be a generator of $H^n(P)\cong\mathbb{Z}$.
Then for any positive integer $i\leq n$ and basis element $x^*$ in $H^i(P)$,
there exists a choice of basis for $H^{n-i}(P)$ such that $x^*y^* = e^*$ for
some $y^*$ in this basis.
\end{lemma}

\begin{proof}
Let $x$ and $e$ be the homology duals of $x^*$ and $e^*$.
Since $H^*(P)$ satisfies Poincar\'e duality, the cap product homomorphism
$$
\seqm{e\cap H^i(P)}{}{H_{n-i}(P)}
$$
is an isomorphism, so it maps a basis of $H^i(P)$ to a basis of $H_{n-i}(P)$.
Therefore
$$
y = e\cap x^*
$$
is an element in a basis for $H_{n-i}(P)$.

Since $H_*(P)$ is torsion-free, the cup product is dual to the cap product.
That is, there is a commutative diagram
\[\diagram
H^{n-i}(P)\rto^-{\cong}\dto^{\cup x^*} & Hom(H_{n-i}(P),Z)\dto^{(\cap x^*)^*}\\
H^n(P)\rto^-{\cong} & Hom(H_n(P),Z).
\enddiagram\]
In particular, since the homomorphism $(\cap x^*)$ sends $e$ to $y$
and $e$ generates $H^n(P)$, its dual $(\cap x^*)^* = (\cup x^*)$ sends $y^*$ to $e^*$,
so we have
$$
y^*\cup x^* = e^*.
$$
Since $y$ is an element in a basis for $H_{n-i}(P)$,
$y^*$ is an element in the dual basis for $H^{n-i}(P)$, and we are done.
\end{proof}

Note that if $m=n$ and $m\notin\{2,4,8\}$ then the element $y^*$
in Lemma~\ref{LPD} is \emph{not} equal to $\pm x^*$. But if $m\in\{2,4,8\}$
then we may have $y^*=\pm x^*$. This is the reason for the exclusion of
this case in the statement of Proposition~\ref{C1}.
\medskip

\begin{proof}[Proof of Proposition~\ref{C1}]
We will check that $P\in\mathcal{P}$. By Poincar\'e
duality, $\bar P$ is $(n-m)$-dimensional. So as $m>1$, we have
$n-m<n-1$, implying that $H_*(P)$ is torsion-free if and only if
$H_*(\bar P)$ is torsion-free. But as $\bar P$ is homotopy equivalent
to a wedge of spheres, $H_*(\bar P)$ is torsion-free and therefore
$H_*(P)$ is torsion-free.

Fix $e^*$ as a generator of $H^n(P)\cong\mathbb{Z}$.
Let $x^*\in H^m(P)$ be a basis element. By Lemma~\ref{LPD} there
exists a basis element $y^*\in H^{n-i}(P)$ such that $x^*y^*=e^*$.
Since $\bar P$ is homotopy equivalent to a wedge of spheres,
$x^*$ and $y^*$ are spherical classes represented by maps
\seqm{S^{m}}{\alpha}{\bar P} and \seqm{S^{n-m}}{\beta}{\bar P}
and the wedge sum
$\seqm{S^{m}\vee S^{n-m}}{\alpha+\beta}{\bar P}$
has a left homotopy inverse. Thus $\bar P\simeq J\vee (S^{m}\vee S^{n-m})$
where $J$ is the homotopy cofibre of $\alpha+\beta$.
Let $Q$ be the homotopy cofibre of the composite
$\seqmm{J}{}{\bar P}{i}{P}$. The homotopy equivalence for $\bar P$
and the fact that, as a $CW$-complex, $P=\bar P\cup e^{n}$ implies
that $Q$ is a three-cell complex, $Q=(S^{m}\vee S^{n-m})\cup e^{n}$,
and the map to the cofibre, $\seqm{P}{q}{Q}$, is onto in homology.
Dualizing, $q^*$ is an injection. Suppose that $u*\in H^m(Q)$,
$v^*\in H^{n-m}(Q)$ and $z^*\in H^{n}(Q)$ satisfy  $q^*(u^*)=x^*$,
$q^*(v^*)=y^*$ and $q^*(z^*)=e^*$. Then the fact that $x^*y^*=e^*$
implies that $u^*v^*=z^*$. Thus there is a ring isomorphism
$H^*(Q)\cong H^*(S^{m}\times S^{n-m})$. Thus $P$ satisfies all
the conditions of Definition~\ref{Pdef}, so $P\in\mathcal{P}$. The assertions
of the proposition are now all direct applications of Theorem~\ref{TMain}.
\end{proof}

As an example of Proposition~\ref{LPD} in action, we prove
Theorem~\ref{Walldecomp}. Let $M$ be an $(n-1)$-connected
$2n$-dimensional manifold. Observe that the $(2n-1)$-skeleton
of $M$ is homotopy equivalent to $\bigvee_{i=1}^{k} S^{n}$, where
$k=\mbox{dim}\,H^{n}(M)$. We aim to decompose $\Omega M$.

\begin{proof}[Proof of Theorem~\ref{Walldecomp}]
If $n\notin\{2,4,8\}$ and $k\geq 2$, then by
Proposition~\ref{LPD} there is a homotopy equivalence
\[\Omega M\simeq\Omega(S^{n}\times S^{n})\times
     \Omega (J\vee(J\wedge\Omega(S^{n}\times S^{n})))\]
where $J=\bigvee_{i=1}^{k-2} S^{n}$.
\end{proof}

\section{The case of simply-connected $4$-manifolds}
\label{sec:4manifold}

Proposition~\ref{C1} does not cover the cases of simply-connected $4$-manifolds,
$3$-connected $8$-manifolds, or $7$-connected $16$-manifolds,
due to the potential presence of nonzero cup product squares. To handle
the case of simply-connected $4$-manifolds and prove
Theorem~\ref{1conndecomp}, we use the fact that such
spaces appear as the base space in a certain $S^1$ homotopy fibration
whose total space is a Poincar\'e duality complex. These homotopy fibrations
generalize the fiber bundle
\seqmm{S^1}{}{S^5}{}{\mb CP^2}.

Let $M$ be a simply-connected oriented $4$-manifold. If $H^2(M)=0$
then $M$ is homotopy equivalent to $S^{4}$, and the homotopy type
of $\Omega S^{4}$ is well known to be $S^{3}\times\Omega S^{7}$.
So we will assume from now on that $H^{2}(M)\neq 0$. Then, up to homotopy
equivalence, there is a homotopy cofibration
$$
\seqmm{S^3}{\alpha}{\bigvee_{i=1}^{k} S^2}{}{M}
$$
for some map $\alpha$. Suppose that there is an isomorphism
of $\mathbb{Z}$-modules
$$
H^{*}(M)\cong\Z\paren{x_1,\ldots,x_k,z}
$$
where $|x_i|=2$ and $|z|=4$.
Let $c_{ij}$ be such that $x_i x_j = c_{ij}z$.
Let $C$ be the $k\times k$ matrix
$$
C=\sbracket{c_{ij}}.
$$
The anti-commutativity of the cup product implies that
$c_{ij}=c_{ji}$, so $C$ is symmetric, and Poincar\'e duality
implies that $C$ is nonsingular.

Focus on the class $x_{k}\in H^{2}(M)$. By Lemma~\ref{LPD},
we may assume the basis of $H^{2}(M)$ has been chosen so that
$c_{k\bar k}=1$ for some $\bar k$. That is,
$$
x_k x_{\bar k} = z.
$$
The cohomology class $x_{k}$ is represented by a map
$$
q\colon \seqm{M}{}{K(\mathbb{Z},2)}.
$$
Note that $K(\mathbb{Z},2)\simeq\mathbb{C}P^{\infty}$, and
$\Omega\mathbb{C}P^{\infty}\simeq S^{1}$. Define the space $Z$
by the homotopy fibration sequence
$$
\seqmmm{S^{1}}{}{Z}{}{M}{q}{\mathbb{C}P^{\infty}}.
$$
A theorem of Quinn~\cite{Q}
says that in a fibration of spaces having the homotopy type of
finite $CW$-complexes, the total space is a Poincar\'e duality complex if
and only if the fiber and base space are Poincar\'e duality complexes.
This, of course, also holds for homotopy fibrations. Therefore, as we
have a homotopy fibration \seqmm{S^{1}}{}{Z}{}{M}
and both $S^{1}$ and $M$ are Poincar\'e duality complexes, then so is $Z$.

\begin{lemma}
\label{Zprops}
The Poincar\'e duality complex $Z$ satisfies the following:
\begin{romanlist}
\item there is a homotopy cofibration
$$
\seqmm{S^4}{\gamma}{\bigvee_{i=1}^{k} (S^2\vee S^3)}{}{Z}
$$
for some map $\gamma$;
\item $H^*(Z)$ is torsion-free.
\end{romanlist}
\end{lemma}

\begin{proof}
Consider the homotopy fibration
$\seqmm{S^{1}}{}{Z}{}{M}$.
We will use a Serre spectral sequence to calculate $H^*(Z)$. We have
$E_{2}^{\ast,\ast}\cong H^*(S^1)\otimes H^*(M)$.
Let $a\in H^{1}(S^{1})$ represent a generator and recall that, as a
$\mathbb{Z}$-module, $H^*(M)\cong\mathbb{Z}\{x_{1},\ldots,x_{k},z\}$.
Thus a $\mathbb{Z}$-module basis for $E_{2}^{\ast,\ast}$ is given by
$$
\{1,1\otimes x_{1},\ldots, 1\otimes x_{k},1\otimes z, a\otimes 1,
      a\otimes x_{1},\ldots,a\otimes x_{k},a\otimes z\}.
$$
The fibration in question is induced by the map
$\seqm{M}{q}{\mathbb{C}P^{\infty}}$
which represents the cohomology class~$x_{k}$. Therefore
$d_{2}(a)=\pm x_{k}$. Changing the basis of $H^1(S^1)$ if need be,
assume that $d_{2}(a)=x_{k}$. As $d_{2}$ is a differential, the fact
that $x_{k}x_{\bar k}=z$ implies that
$d_{2}(a\otimes x_{\bar k})=x_{k}x_{\bar{k}}=z$, while
$d_{2}(a\otimes x_{j})=x_{k}x_{j}=c_{kj}z$. Thus a $\mathbb{Z}$-module
basis for $E_{3}^{\ast,\ast}$ is given by
$$
\{1,1\otimes x_{1},\ldots, 1\otimes x_{k-1}, (a\otimes x_{1}-a\otimes c_{k1}x_{\bar k}),
   \ldots, (a\otimes x_{\bar k-1}-a\otimes c_{k(\bar k-1)}x_{\bar k}),
$$
$$
\hspace{2cm}
   (a\otimes x_{\bar k+1}-a\otimes c_{k(\bar k+1)}x_{\bar k}),
   (a\otimes x_{k}-a\otimes c_{kk}x_{\bar k}), a\otimes z\}.
$$
All other differentials are trivial for degree reasons, so we have
$H^*(Z)\cong E_{\infty}^{\ast,\ast}\cong E_{3}^{\ast,\ast}$.

Notice that the calculation for the rational cohomology Serre spectral
is exactly the same. Thus the rationalization map
$\seqm{H^*(Z;\mathbb{Z})}{}{H^*(Z;\mathbb{Q})}$
preserves the number of basis elements in each dimension. Thus
$H^*(Z)$ is torsion-free, proving part~(ii).

Notice that the description of $H^*(Z)$ implies that $Z$ has $k-1$
cells in dimension $2$ and $k-1$ cells in dimension $3$. The fact
that $H^*(Z)$ is torsion-free therefore implies that the $3$-skeleton
of $Z$ is homotopy equivalent to $\bigvee_{i=1}^{k} (S^{2}\vee S^{3})$.
The one remaining nontrivial cell of $Z$ occurs in dimension~$5$,
so $Z$ is the homotopy cofibre of a map
$\seqm{S^{4}}{}{\bigvee_{i=1}^{k}(S^{2}\vee S^{3})}$,
proving part~(i).
\end{proof}

\begin{remark}
The space $Z$ is in fact a manifold, not just a Poincar\'{e} duality complex, which is diffeomorphic
to the connected sum of $k$ copies of $S^{2}\times S^{3}$~\cite{DL}.
As we only use the much simpler properties of $Z$ listed in Lemma~\ref{Zprops},
it is clarifying to leave the analysis of $Z$ as it stands in the statement and proof of the lemma.
\end{remark}

Before proceeding to decompose the loop space of a simply-connected $4$-manifold,
we first decompose the loop space of the associated Poincar\'{e} duality space $Z$. Let
$$
i\,\colon\seqm{\bigvee_{i=1}^{k-1}(S^{2}\vee S^{3})}{}{Z}
$$
be the skeletal inclusion.

\begin{proposition}
\label{loopZ}
If $k=1$ then $Z\simeq S^{5}$, so $\Omega Z\simeq\Omega S^{5}$.
If $k\geq 2$ then the following hold:
\begin{romanlist}
\item there is a homotopy equivalence
$$
\Omega Z\simeq\Omega(S^{2}\times S^{3})\times
   \Omega(J\vee (J\wedge \Omega(S^{2}\times S^{3})))
$$
where $J=\bigvee_{i=1}^{k-1} (S^{2}\vee S^{3})$ if $k>2$ and $J=\ast$ if $k=2$;
\item the map $\seqm{\Omega(\bigvee_{i=1}^{k-1} (S^{2}\vee S^{3}))}{\Omega i}
     {\Omega Z}$
has a right homotopy inverse.
\end{romanlist}
\end{proposition}

\begin{proof}
Notice that Proposition~\ref{Zprops}~(i) implies that if $k=1$ then
$Z\simeq S^{5}$. Assume from now on that $k\geq 2$. We will show that
the conditions of Proposition~\ref{C1} hold. The result of Quinn already
cited implies that $Z$ is a Poincar\'{e} duality space, and by
Proposition~\ref{Zprops}~(i), $Z$ is $1$-connected and
$5$-dimensional. So with $m=2$ and $n=5$ we have $m=2<n-m=3$.
By Proposition~\ref{Zprops}, the $4$-skeleton of $Z$ is homotopy equivalent
to $\bigvee_{i=1}^{k-1}(S^{2}\vee S^{3})$. Thus $Z$ satisfies the hypotheses
of Proposition~\ref{C1}, and applying the proposition immediately gives
the statements of the proposition.
\end{proof}

We now prove Theorem~\ref{1conndecomp}, restated as follows.

\begin{theorem}
\label{T1}
Let $M$ be a simply-connected $4$-manifold and suppose
$\mbox{dim}\, H^{2}(M)=k$ for $k>0$. If $k=1$ then there
is a homotopy equivalence
$$
\Omega M\simeq S^1\times\Omega S^5
$$
and if $k\geq 2$ then there is a homotopy equivalence
$$
\Omega M\simeq S^{1}\times\Omega Z\simeq S^1\times\Omega(S^{2}\times S^{3})\times
   \Omega(J\vee (J\wedge \Omega(S^{2}\times S^{3})))
$$
where $J=\bigvee_{i=1}^{k-1} (S^{2}\vee S^{3})$ if $k>2$ and $J=\ast$ if $k=2$.
Consequently, the homotopy type of $\Omega M$ depends only on the
integer $k=\dim\, H^{2}(M)$.
\end{theorem}

\begin{proof}
Consider the map
$\seqm{M}{q}{\mathbb{C}P^{\infty}}$
representing the cohomology class $x_{k}$. Since $M$ is simply-connected,
any generator of $H_{2}(M)$ is in the image of the Hurewicz homomorphism.
In our case, the homology class dual to $x_{k}$ is the Hurewicz image of a map
$t\,\colon\seqm{S^{2}}{}{M}$.
Dualizing, $t^{\ast}(x_{k})=\iota^*_{2}$, where $\iota^*_{2}$
is a generator of $H^2(S^2)$. Therefore, the composite
$\seqmm{S^{2}}{t}{M}{q}{\mathbb{C}P^{\infty}}$
is degree one in cohomology. Let
$\bar t\,\colon\seqm{S^{1}}{}{\Omega M}$
be the adjoint of $t$. Then the composite
$\seqmm{S^{1}}{\bar t}{\Omega M}{\Omega q}{S^{1}}$
is degree one in cohomology, implying that it is a homotopy equivalence.
Therefore, in the homotopy fibration
$\seqmm{\Omega Z}{}{\Omega M}{\Omega q}{S^{1}}$,
the map $\Omega q$ has a right homotopy inverse, implying that there
is a homotopy equivalence
$$
\Omega M\simeq S^{1}\times\Omega Z.
$$
The theorem now follows from the decomposition of $\Omega Z$
in Proposition~\ref{loopZ}.
\end{proof}

An analogue of Theorem~\ref{T1} holds for $3$-connected $8$-manifolds $M$,
provided that there is a map
\seqm{M}{}{\mb HP^{2}}
that induces a surjection onto $H^4(\mb HP^{2})\cong \Z$.
In such a case, composing this map with the inclusion
\seqm{\mb HP^{2}}{}{\mb HP^{\infty}}
and then using the fact that $\mb HP^{\infty}\simeq S^3$, one obtains a
principal homotopy fibration
\seqmm{S^3}{}{Z}{}{M} with total space $Z$ an $11$-dimensional Poincar\'e duality complex.
The only nonzero homology groups of $Z$ are in degrees $4$, $7$, and $11$,
and using the associated action of $S^3$ on $Z$, it is not difficult to show that
the $10$-skeleton of $Z$ is homotopy equivalent to a wedge of $4$-spheres
and $11$-spheres. It is not really clear what may happen in the case of
$7$-connected $16$-manifolds, as $S^{7}$ does not have a classifying space.

We now re-organize the information appearing in the decomposition
in Theorem~\ref{T1} when $k\geq 2$ to make it more clear how the
decomposition depends on the $2$-skeleton of the $4$-manifold. Let
$i\colon\seqm{\bigvee_{i=1}^{k} S^{2}}{}{M}$
be the skeletal inclusion.

\begin{theorem}
   \label{T2}
   Let $M$ be a simply-connected $4$-manifold and suppose
   $\mbox{dim}\, H^{2}(M)=k$ for $k\geq 2$. Then the map
   $\seqm{\Omega(\bigvee_{i=1}^{k} S^{2})}{\Omega i}{\Omega M}$
   has a right homotopy inverse.
\end{theorem}

\begin{proof}
Recall that there is a homotopy fibration
$\seqmm{Z}{r}{M}{q}{\mathbb{C}P^{\infty}}$.
In Theorem~\ref{T1} it was shown that~$\Omega q$ has a right homotopy
inverse,
$f\colon\,\seqm{S^{1}}{}{\Omega M}$. Thus the composite
\[\seqmm{S^{1}\times\Omega Z}{f\times\Omega r}{\Omega M\times\Omega M}
     {\mu}{\Omega M}\]
is a homotopy equivalence, where $\mu$ is the loop multiplication.

By Proposition~\ref{loopZ}, the map
$\seqm{\Omega(\bigvee_{s=1}^{k-1}(S^{2}\vee S^{3}))}{\Omega j}{\Omega Z}$
has a right homotopy inverse, where~$j$ is the inclusion of the $4$-skeleton
into the $5$-dimensional space $Z$. Let
$g\colon\,\seqm{\Omega Z}{}{\Omega(\bigvee_{s=1}^{k-1}(S^{2}\vee S^{3}))}$
be a right homotopy inverse of $\Omega j$. Let $h$ be the composite
\[h\,\colon\seqmm{\bigvee_{s=1}^{k-1}(S^{2}\vee S^{3})}{j}{Z}{r}{M}.\]
Then $\Omega h\circ g$ is homotopic to $\Omega r$. Therefore,
by the previous paragraph, the composite
\[\seqmmm{S^{1}\times\Omega Z}{f\times g}
     {\Omega M\times\Omega(\bigvee_{s=1}^{k-1}(S^{2}\vee S^{3}))}
     {\ID\times\Omega h}{\Omega M\times\Omega M}{\mu}{\Omega M}\]
is a homotopy equivalence.

Since $\bigvee_{s=1}^{k-1}(S^{2}\vee S^{3})$ is $3$-dimensional,
the map $h$ factors through the $3$-skeleton of $M$, which is homotopy
equivalent to $\bigvee_{i=1}^{k} S^{2}$. Thus $h$ factors as a composite
$\seqmm{\bigvee_{s=1}^{k-1}(S^{2}\vee S^{3})}{h^{\prime}}{\bigvee_{i=1}^{k} S^{2}}{i}{M}$
for some map $h^{\prime}$. Also, for connectivity and dimension reasons, the map
$\seqm{S^{1}}{f}{M}$
factors as a composite
$\seqmm{S^{1}}{f^{\prime}}{\Omega(\bigvee_{i=1}^{k} S^{2})}{\Omega i}{\Omega M}$
for some map $f^{\prime}$. Therefore, inserting these factorizations into the
homotopy equivalence $\mu\circ(\ID\times\Omega h)\circ(f\times g)$, we
obtain a homotopy equivalence
\[\seqmmm{S^{1}\times\Omega Z}{f^{\prime}\times g}
     {\Omega(\bigvee_{s=1}^{k-1} S^{2})\times\Omega(\bigvee_{s=1}^{k-1}(S^{2}\vee S^{3}))}
     {\ID\times\Omega h^{\prime}}
     {\Omega(\bigvee_{i=1}^{k} S^{2})\times\Omega(\bigvee_{i=1}^{k} S^{2})}
     {\Omega i\times\Omega i}{\Omega M\times\Omega M}\stackrel{\mu}{\longrightarrow}
     \Omega M.\]
Finally, since $\Omega i$ is a loop map, it commutes with the loop multiplication,
so we obtain a homotopy equivalence
\[\llnamedddright{S^{1}\times\Omega Z}{f^{\prime}\times(\Omega h^{\prime}\circ g)}
        {\Omega(\bigvee_{i=1}^{k} S^{2})\times\Omega(\bigvee_{i=1}^{k} S^{2})}{\mu}
        {\Omega(\bigvee_{i=1}^{k} S^{2})}{\Omega i}{\Omega M}.\]
Consequently, the map $\Omega i$ has a right homotopy inverse.
\end{proof}

Theorem~\ref{T2} is useful. For example, we apply it to determine
the homotopy type of the loops on certain principle $G$-bundles.

\begin{corollary}
   \label{bundlecor}
   Let $G$ be a simply-connected, simple compact Lie group. Let $M$
   be a simply-connected $4$-manifold with $\mbox{dim}\, H^{2}(M)\geq 2$.
   Let $\seqm{P}{\pi}{M}$ be a principle $G$-bundle. Then $\Omega\pi$ has
   a right homotopy inverse, implying that there is a homotopy
   equivalence
   \[\Omega P\simeq\Omega M\times\Omega G.\]
\end{corollary}

\begin{proof}
Any principle $G$-bundle
$\seqm{P}{\pi}{M}$
is classified by a map
$\seqm{M}{g}{BG}$,
where $BG$ is the classifying space of $G$ and $P$ is the homotopy
fibre of $g$. In our case, since $G$ is a simply-connected,
compact simple Lie group, $BG$ is $2$-connected (in fact, it is $3$-connected).
Thus the composite
$\seqmm{\bigvee_{i=1}^{k} S^{2}}{i}{M}{g}{BG}$
is null homotopic by connectivity. By Theorem~\ref{T2}, $\Omega i$
has a right homotopy inverse. Therefore $\Omega g$ is null homotopic.
Hence in the homotopy fibration sequence
$\seqmmm{\Omega G}{}{\Omega P}{\Omega\pi}{\Omega M}{\Omega g}{G}$
the null homotopy for $\Omega g$ implies that $\Omega\pi$ has a right
homotopy inverse, and therefore $\Omega P\simeq\Omega M\times\Omega G$.
\end{proof}

Corollary~\ref{bundlecor} says something interesting. While there
are $[M,BG]\cong\mathbb{Z}$ distinct principle $G$-bundles over $M$,
after looping all those bundles become homotopy equivalent.
Further, the decomposition of $\Omega P$ can be refined by
inserting the decomposition of $\Omega M$ in Theorem~\ref{T1},
and - after localizing at a prime $p$ - by the decompositions of
$\Omega G$ that arise from the $p$-local decompostions
of~$G$ due to Mimura, Nishida and Toda~\cite{MNT}.

\section{Looped Configuration Spaces}
\label{sec:config}

We end with a quick application that is in the spirit of our previous results. Let
$$
F_k(X) = \{(x_1,\ldots,x_k)\in X^{\times k}\mathrel{|}x_i\neq x_j\mbox{ if }i\neq j\}
$$
be the \emph{ordered configuration space} of $k$ distinct points in $X$.
The literature on these spaces is substantial, but many basic questions remain unanswered.
For example, their integral homology is not clearly understood in most cases,
and it is now known that their homotopy type generally does not depend only
on the homotopy type of $X$,
even after restricting the input space to compact manifolds~\cite{LS}.

Things do simplify after looping however.
If we were to take $M$ to be a smooth manifold with a nonvanishing tangent vector field,
then the projection map \seqm{F_k(M)}{}{M} onto the first coordinate has a section.
By~\cite{FN,CG} there is a homotopy decomposition
\begin{equation}
\label{EConfigDecomp}
\Omega F_k(M)\simeq \Omega M\times\Omega(M-Q_1)\times\cdots\times\Omega (M-Q_k)
\end{equation}
for any choice of distinct points $q_1,\ldots,q_k$ in $M$, with $Q_{i}=\{q_1,\ldots,q_i\}$.
Thus, not only are the betti numbers $\Omega F_k(M)$ relatively easy to compute,
but the homotopy type of $\Omega F_k(M)$ depends only on the homotopy type of the input manifold $M$ when $M$ is simply connected.
The following takes this a step further:

\begin{corollary}
Let $1<m\leq n-m$, $n$ be odd, and let $M$ be a closed oriented $(m-1)$-connected $n$-dimensional smooth manifold with torsion-free homology.
Then the homotopy type of the looped configuration space $\Omega F_k(M\#(S^m\times S^{n-m}))$ depends only on the homotopy type of $M-\ast$ for each $k\geq 1$.
\end{corollary}

\begin{proof}
Recall that the connected sum of smooth manifolds can be constructed so that the resulting manifold also has a smooth structure.
Then $M\#(S^m\times S^{n-m})$ is a smooth manifold, and moreover it is odd dimensional, so it has a nonvanishing tangent vector field.
Thus, the decomposition~(\ref{EConfigDecomp}) specializes to
$$
\Omega F_k(M\#(S^m\times S^{n-m}))\simeq\Omega(M\#(S^m\times S^{n-m}))\times\Omega(M\#(S^m\times S^{n-m})-Q_1)\times\cdots\times\Omega(M\#(S^m\times S^{n-m})-Q_k)
$$
for any choice of $k$ distinct points $q_1,\ldots,q_k$ in $M\#(S^m\times S^{n-m})$.

Notice that $M\#(S^m\times S^{n-m})-Q_i$ is homotopy equivalent to the wedge sum of $(M-\ast)\vee S^m\vee S^{n-m}$ with $i-1$ copies of the $(n-1)$-sphere.
Thus, the homotopy type of each factor $\Omega(M\#(S^m\times S^{n-m})-Q_i)$ in the  decomposition above depends only on the homotopy type of $M-\ast$.
Likewise, the homotopy type of the remaining factor $\Omega(M\#(S^m\times S^{n-m}))$ depends only on that of $M-\ast$ by Theorem~(\ref{connsum}).
The result follows.
\end{proof}

\bibliographystyle{amsalpha}

config: arxiv


TITLE: Domain of a function quick question
QUESTION [1 upvotes]: $$f(x)=\arcsin\left(\frac{x-1}{2x}\right)$$
We need to find the domain of this function.
My try:
$$-1\leq \frac{x-1}{2x} \leq 1 $$
We can split this into
$$-1\leq \frac{x-1}{2x} \quad\text{and} \quad\frac{x-1}{2x} \leq 1$$
My idea is to solve for this two inequalities and then take the intersection of them both.
$$-2x\leq x-1\\
-3x\leq -1\\
x\geq\frac{1}{3}$$
$$x-1 \leq 2x\\
-1 \leq x$$
By doing the intersection I get the wrong answer which is: $x\geq\frac{1}{3}$
What have I done wrong?

REPLY [1 votes]: Here it is another approach for the sake of curiosity.
I would start with squaring both sides as it is done next:
\begin{align*}
\left|\frac{x-1}{2x}\right| \leq 1 & \Longleftrightarrow \left(\frac{x-1}{2x}\right)^{2} \leq 1\\\\
& \Longleftrightarrow \frac{x^{2} - 2x + 1 - 4x^{2}}{4x^{2}} \leq 0\\\\
& \Longleftrightarrow \frac{3x^{2} + 2x - 1}{4x^{2}} \geq 0
\end{align*}
We may now study the behavior of the function at the numerator.
To begin with, let us rearrange its expression:
\begin{align*}
3x^{2} + 2x - 1 & = \left(3x^{2} + 2x + \frac{1}{3}\right) - \frac{4}{3}\\\\
& = \left(\sqrt{3}x + \frac{1}{\sqrt{3}}\right)^{2} - \frac{4}{3}
\end{align*}
Consequently, the solution set to problem proposed is given by
\begin{align*}
3x^{2} + 2 x - 1 \geq 0 & \Longleftrightarrow \left(\sqrt{3}x + \frac{1}{\sqrt{3}}\right)^{2} \geq \frac{4}{3}\\\\
& \Longleftrightarrow \left(\sqrt{3}x + \frac{1}{\sqrt{3}} \geq \frac{2}{\sqrt{3}}\right)\vee\left(\sqrt{3}x + \frac{1}{\sqrt{3}} \leq -\frac{2}{\sqrt{3}}\right)\\\\
& \Longleftrightarrow \left(x \geq \frac{1}{3}\right)\vee\left(x \leq -1\right)
\end{align*}
Since both intervals does not contain the value $x = 0$, we conclude the sought domain is given by
\begin{align*}
D_{f} = (-\infty,-1]\cup\left[\frac{1}{3},+\infty\right)
\end{align*}
Hopefully this helps!


TITLE: A down-to-earth introduction to the uses of derived categories
QUESTION [43 upvotes]: When I was learning about spectral sequences, one of the most helpful sources I found was Ravi Vakil's notes here.  These notes are very down-to-earth and give a kind of minimum knowledge needed about spectral sequences in order to use them.
Does anyone know of a similar source for derived categories?  Something that concentrates on showing how these things are used, without developing the entire theory, or necessarily even giving complete, rigorous definitions?

REPLY [2 votes]: "Derived Categories summer graduate school" is also a very good introduction.


TITLE: What's the probability that a shuffled standard deck of 52 cards has two top cards of the same rank?
QUESTION [0 upvotes]: I have done many problems involving color and suit, but I am not sure how rank fits in..
For example: 13(13C2/52C2).
You calculate the probability for 1 rank and multiply it by 13?
Is this correct?

REPLY [0 votes]: The total number of possible orderings is $52!$.
Now, to get an order we're interested in, we first choose one out of $13$ ranks in $\binom{13}1=13$ ways, then choose $2$ out of the $4$ cards of that rank in $\binom{4}2=6$ ways, and finally choose one of the $2!$ orders for these two. For each such choice, we may yet order the remaining $50$ cards in $50!$ ways. The probability is hence
$$\frac{13\cdot 6 \cdot 2 \cdot 50!}{52!}=\frac{13\cdot 12}{52\cdot 51}=\frac1{17}\simeq 5.88\%$$


TITLE: Let $\sim$ be a relation on $\mathbb R^2$
QUESTION [0 upvotes]: Let $\sim$ be the relation on $\mathbb R^2$ defined by:
$$(x_1, y_1) ∼ (x_2, y_2) \iff x_1^2 + y_1^2=x_2^2+y_2^2$$ Show that $\sim$ is an equivalence relation on $\mathbb R^2$.
Consider the function $f \colon [0, \infty) \to \mathbb R^2/ \sim$ defined by $f(x) = [(x, 0)]$ $\forall x \in [0, \infty)$. Show that $f$ is bijective.
I have shown it is an equivalence relation and that $f$ is injective but I am now stuck on proving it is surjective. I really am not sure how to begin because it seems obvious it is true? Any hints appreciated

REPLY [0 votes]: Surjectivity is, indeed, the most "obvious" of the three tasks here. You need to show that every equivalence class is achieved by $f(x)$ for some $x$. To do this, it suffices to show that every element of $\Bbb R^2$ is equivalent (in the $\sim$ sense) to $(x,0)$, where $x\in[0,\infty)$. [Also, you should convince yourself that this task really is sufficient...]
This isn't quite trivial: since you already know the function is injective, you know that you can't make the domain of $f$ any smaller and still get a surjective function. So you might ask why $[1,\infty)$ isn't good enough, and even $(0,\infty)$ isn't good enough.


TITLE: Is my textbook solution wrong or am I missing something?
QUESTION [0 upvotes]: This is from my textbook:

Example 2.16: Find a root of the equation $x \sin x + \cos x = 0$.
We have
  $$
 f(x) = x \sin x + \cos x
 \quad\text{and}\quad
 f'(x) = x \cos x.
$$
  The iteration formula is, therefore,
  $$
 x_{n+1} = x_n - \frac{x_n \sin x_n + \cos x_n}{x_n \cos x_n}.
$$
  With $x_0 = \pi$, the successive iterates are given below:
  $$
 \begin{array}{cccc}
  n &  x_n   &            f(x_n)           & x_{n+1} \\
  \hline
  0 & 3.1416 &           -1.0\phantom{000} & 2.8233 \\
  1 & 2.8233 &           -0.0662           & 2.7986 \\
  2 & 2.7986 &           -0.0006           & 2.7984 \\
  3 & 2.7984 & \phantom{-}0.0\phantom{000} & 2.7984
 \end{array}
$$

My doubt here is that in last row of table where $x_n = 2.7984$ is $f(2.7984) \approx 1$, which is not near zero. So is this a misprint or am I missing something here? (This has happened to me in earlier questions in which I used the Newton-Raphson method.)

REPLY [2 votes]: Calculators usually are by default in "degree mode", that is, they interpret arguments to trigonometric functions as given in degrees. In mathematical contexts, the default is that the arguments to trigonometric functions are given in radians. Using the wrong mode leads to wrong results, since e.g. $\pi°$ is a small angle whose cosine is close to $1$, while $\pi$ is a straight angle with $\cos \pi = -1$.
Setting the calculator to radians mode avoids the discrepancy.


TITLE: pipette waterfountain
QUESTION [1 upvotes]: Place the tip of a pipette in some hot water.  Cover the other end with you thumb.  Invert the pipette and water sprays out nearly two metres.
Is the cool air in the pipette being heated by the hot water and thus expanding to force some of the water to spray out?  Does the ideal gas law apply?  Is this as fun for other people as it is for me?

REPLY [0 votes]: Theoretical Background
The pipette's ability to hold water with a seal covering the top, occurs due to the balance of forces. One of these forces is surface tension, a combination of the cohesive forces holding together the liquid molecules and the adhesive electrostatic attraction between the liquid and pipette wall (Perlman, 2016). This force coupled with upwards air pressure is equal to a force of gravity, resulting in the water remaining stationary.
As heat is directly proportional to internal energy, when the experiment is performed with hot water the molecules have more kinetic energy and weaker intermolecular bonds between molecules. This results in a decrease in both adhesive and cohesive forces, therefore lowering the surface tension (Nave, 2016).
The hot water fountain phenomenon occurs when a pipette with hot water trapped in the lower half is flipped. This movement momentarily disturbs the surface tension, allowing a small amount of water to run down the side of the pipette. The subsequent gas expansion is explained by Fourier’s Law (Mechanical Engineering, 2016):
dQ/dt=h .A .∆T
dQ/dt=heat transfer
h=heat transfer coefficient
∆T=temperature change
A=surface area
dQ/dt∝A
∴heat transfer∝surface area
As the water runs down the side, there is an increased surface area of water molecules working to transfer heat to air particles. Hence the increase in heat transfer causes an increased rate of gas expansion, further explaining its sudden acceleration.
Additionally, Fourier’s Law also highlights:
dQ/dt∝∆T
∴heat transfer∝temperature change
This showcases that an increased temperature differential between the liquid and the air particles in the pipette would significantly increase the height of the fountain. Additionally, the gas expansion pushing a smaller volume of water with a lower weight would be certainly propelled by the gas expansion, while larger column of water with a greater weight would be effected less by the same gas expansion.
Furthermore, an increase of the liquid temperature results in weakened cohesive and adhesive forces, therefore showing that water at a higher temperature would be affected to a larger extent by the momentary disturbance in the surface tension, caused by the flipping motion. As explained above, this allows more water to slide down the sides of the pipette, hence increasing gas expansion and projection height. This effect can be emphasized by adding detergent to water to further lower its surface tension.
This expansion of gas pushes the water through the tapered end, which further increases its velocity, as per Boyle's Law and The Continuity Principle. Firstly, Boyle's Law explains the inverse relationship between pressure and volume, in which a decreased volume results in particles hitting the sides of the container more often, thus increasing the pressure (Hall, 2015). This increase in pressure results in more force pushing the water molecules, and hence an increased velocity.
P ∝  1/V
P=Pressure
V=Volume
Further, the Continuity Principle states that the mass flow rate must be equal inside the pipette, compared to the water that leaves the tapered tip. As the density of water remains unchanged, this principle can be simplified to show the discharge is equal, and thus (Bryant, 2015):
∴Q_1= Q_2
∴A_1 u_1= A_2 u_2
A=cross sectional area of pipette
u=velocity
u_2=  A_1/A_2  u_1=  (π .  r_1^2  )/(π  .  r_2^2  ) u_1=  (r_1^2)/(r_2^2 ) u_1
∴u_2= (r_1/r_2 )^2 u_1
∴Exit velocity dependent on ratio of diameters (inside of pipette vs tapered end)
∴u_2= (r_1/r_2 )^2 u_1
Substituting the known values of the radius, we can obtain an expression relating the velocity of the liquid before and after its interaction with the tapered end.
u_2=(0.0025/0.0005)^2 u_1
u_2=25 ×u_1
The tapered pipette further accelerates the water, adding to the expanding gas's transferred energy and velocity.


TITLE: Show the space of $(\alpha,\mu)$-Hölder continuous functions is not closed under scalar multiplication.
QUESTION [1 upvotes]: CONTEXT: I want to show $F_{\alpha,\mu}$ is not a linear subspace of $C([0,1],\mathbb{R})$.
We call $f:[0,1]\to\mathbb{R}$ an "$(\alpha,\mu)$-Hölder continuous function" if
$$
\exists_{\alpha,\mu>0} \forall_{x,y\in[0,1]}: |f(x)-f(y)|\leq\mu\cdot|x-y|^{\alpha}.
$$
Let $F_{\alpha,\mu}$ be the set of $(\alpha,\mu)$-Hölder continuous functions.
I have previously shown that if $f\in F_{\alpha,\mu}$, then $p f\in F_{\alpha,|p|\mu}$, for all $p\in\mathbb{R}$. (1)
I now want to show that $F_{\alpha,\mu}$ is not closed under scalar multiplication. To do this, I assume it is closed under scalar multiplication. Proof by contradiction.
Suppose $F_{\alpha,\mu}$ is closed under scalar multiplication for all $\alpha,\mu$. Let $f\in F_{\alpha,\mu}$. Then also $pf\in F_{\alpha,\mu}$ (because of closure). In the same way $\frac{1}{p}pf\in F_{\alpha,\mu}$, but because $pf\in F_{\alpha,\mu}$ we have by our previous result (1) that $f=\frac{1}{p}pf \in F_{\alpha,|\frac{1}{p}|\mu}$.
So we conclude that when $F_{\alpha,\mu}$ is closed under scalar multiplication, we have that for all $f\in F{\alpha,\mu}$ that also $f\in F{\alpha,p\mu}$, for all $p\in\mathbb{R}$.
Returning to the definition I think this means that we now have
$$
\exists_{\alpha>0} \forall_{\mu\in\mathbb{R}\backslash\{0\}}\forall_{x,y\in[0,1]}:|f(x)-f(y)|\leq \mu\cdot|x-y|^{\alpha}.
$$
Because this is true for all $\mu$, also for $\mu$ approaching to zero, we have that $|f(x)-f(y)|=0$ and thus that $f$ is constant. This is the line I cannot prove rigorously. I think it is true because $\alpha$ is fixed.
We conclude that all Hölder continuous functions must be constant. But $f(x)=x$ is (1,1) Hölder continuous. Therefore $F_{\alpha,\mu}$ cannot be closed under scalar multiplication, this was my false assumption. So $F_{\alpha,\mu}$ is not a linear subspace of $C([0,1],\mathbb{R})$.
Is this line of reasoning correct? And how do I fix the vagueness that I put in bold?

REPLY [2 votes]: Your argument works to show that it is not true that $F_{\alpha,\mu}$ is closed under scalar multiplication for all $\alpha$ and $\mu$.  For the step you asked about, there really isn't more you need to say: $|x-y|^\alpha$ is just some number $C$ and you know $|f(x)-f(y)|\leq \mu C$ for all $\mu> 0$, which implies $|f(x)-f(y)|=0$.  To be very explicit, if $|f(x)-f(y)|$ were positive, you could take $\mu=\frac{|f(x)-f(y)|}{2C}$ to get a contradiction (assuming $C\neq 0$, since the $C=0$ case is trivial).
Note though that this does not prove what you wanted to prove, which is that $F_{\alpha,\mu}$ is not closed under scalar multiplication for all $\alpha$ and $\mu$.  All your argument shows is that there is some $\alpha$ and $\mu$ for which it is not closed under scalar multiplication.  To be more precise, what your argument really shows is that if $F_{\alpha,\mu}$ is closed under scalar multiplication for some particular $\alpha$ and $\mu$ then $F_{\alpha,\mu}$ consists only of constant functions.  So, your example $f(x)=x$ being in $F_{1,1}$ only shows that $F_{1,1}$ is not closed under scalar multiplication.  To show it in general, you need to consider arbitrary $\alpha$ and $\mu$ and find a nonconstant function in $F_{\alpha,\mu}$.


TITLE: binary strings and counting sequence problem
QUESTION [1 upvotes]: Hello i'm working on these questions and I have few questions
1) A binary string is a finite sequence of 0 and 1. Ex. 001101 is a string of length 6 a) List all binary strings of length 4 (so I assume 2^4 = 16.)
2) List all binary strings of size at most 6 which do not have two consecutive 0s. Using this list give the beginning of the counting sequence for binary strings with no
consecutive 0s. Can you see the pattern? ( Im not sure how binary string can be size 6. Maybe for example 000011   like this?

REPLY [2 votes]: If $s$ is an admissible string (i.e., has no two consecutive zeros) then any initial segement $s'$ of $s$ is admissible as well.
Denote by $A_n$ the set of admissible strings of length $n$, and put $a_n:=\#A_n$ $(n\geq0)$. One has $A_0=\{\Lambda\}$ (the empty string), $A_1=\{0,1\}$. Now let an $n\geq2$ be given, and consider an $s\in A_n$. If $s=s'1$ then $s'\in A_{n-1}$, with no further conditions on $s'$. If $s=s'0$ then necessarily $s=s''10$ with $s''\in A_{n-2}$, and no further conditions on $s''$.
We therefore have
$$a_n=a_{n-1}+a_{n-2}\qquad(n\geq2)\ .$$
This allows to conclude that for all $n\geq0$ one has  $a_n=F_{n+2}$ (Fibonnacci number, whereby $F_0=0$).