TITLE: Number of non-negative integral solution to $Ax=0$ for an integral $m\times n$ matrix $A$ with rank $m$ ($m\le n$)?
QUESTION [1 upvotes]: Assume $A$ is an $m\times n$ integral matrix (all entries are integers) with rank $m$ and $m\le n$.
I am wondering if the number of solutions to $Ax=0$ in $\{0,1,2,\dots,N\}^n$ can be bounded by $(N+1)^{O(n-m)}$, where we may assume $n(N)-m(N)$ is a constant and the integer $N$ tends to infinity.
(We may also assume all $m\times m$ submatrix of $A$ has rank $m$ and even $m<n$ if it will make the bound holds.)
My feeling is that when we are in a field $F$, then the number of solutions is at most $|F|^{n-m}$. I am wondering if there is any counterpart in integral solution case.

REPLY [2 votes]: Sure the bound holds, and in fact you can prove it the same way as you might prove it for a finite field, i.e., by decomposing $A$ and $x$ into blocks.
Without loss of generality assume the first $m$ columns of $A$ are independent.  Let $A = [ B_{m \times m} \mid C_{m \times (n-m)} ]$ and $x^T = [y^T \mid z^T]$ of matching sizes s.t. $Ax = By + Cz$.  Note that $B$ is invertible.
For convenience define $K = \{0, 1, \dots, N\}$.  Then for any $z \in K^{n-m}$, we have
$$Ax = 0 \iff By + Cz = 0 \iff y = -B^{-1}Cz$$
So, for any $z \in K^{n-m}$, a solution for $x$ exists.  However, this solution uses $y \in \mathbb{R}^m$ (or you can think of it as $\mathbb{Q}^m$, i.e. $m$-vector of rationals), so this solution might not exist for $y \in K^m$.  Therefore the number of solutions for $x \in K^n$ is $\le |K|^{n-m}$.
(In fact, since $K^m$ is so much "smaller" than $\mathbb{Q}^m$, I would not be surprised if a much tighter upper bound can apply - at least, to almost all cases.)


TITLE: Showing that the number of non-null elements of an unconditionally Cauchy family is countable
QUESTION [0 upvotes]: Let $I \neq \emptyset$ set and $(x_i)_{i \in I}$ a family in a normed space $X$. We say that $(x_i)_{i \in I}$ is unconditionally Cauchy if $\forall \epsilon>0$ exists $F_{\epsilon} \subset I$ finite such that for all finite subset $F' \subset I$ with $F' \cap F_{\epsilon} = \emptyset$ we have:
$$||\sum_{i \in F'}x_i|| < \epsilon$$
I'm trying to show that if $(x_i)_{i \in I}$ is unconditionally cauchy then $A=\{ i \in I : x_i \neq0 \}$ is countable. I thought writing $A= \bigcup_{n \in \mathbb{N}}A_n $ where $A_n=\{ i \in I: ||x_i||>\frac{1}{n} \}$. Now the problem is fixed $n_0$ show that $A_{n_0}$ is countable.I tried to show by contradiction, but I did not succeed.

REPLY [1 votes]: Claim: $A=\bigcup_n F_{1/n}$.
Proof: Suppose $x_i\neq 0$.  Then $\lVert x_i\rVert>1/n$ for some $n$.
If $i\notin F_{1/n}$ then choosing $F'=\{i\}$, we have $F'\cap F_{1/n}=\varnothing$ and hence by definition of $F_{1/n}$:
$$
\left\lVert\sum_{j\in F'} x_j\right\rVert<\frac1n
$$
but LHS is $\lVert x_i\rVert>\frac1n$, contradiction.  QED.
Now each $F_{1/n}$ is finite, so $A$ is a countable union of finite, hence countable.


text:
\begin{document}

\title[The Morse Complex on a Manifold With Corners]{The Morse Complex for a
Morse Function on a Manifold with Corners}
\author{David G.C. Handron}
\date{\today}
\address{Department of Mathematical Sciences\\
Carnegie Mellon University \\
	Pittsburgh, PA \\
	(412)268-2545}
\email{handron@andrew.cmu.edu}
\keywords{Morse theory, stratified spaces, manifolds with corners}
\subjclass{57R70, 57R25}
\date{\today}

\begin{abstract}
A Morse function $f$ on a manifold with corners $M$ allows the
characterization of the Morse data for a critical point by the Morse
index. In fact, a modified gradient flow allows a proof of the Morse
theorems in a manner similar to that of classical Morse theory. It
follows that $M$ is homotopy equivalent to a CW-complex with one cell
of dimension $\lambda$ for each essential critical point of index
$\lambda$. The goal of this article is to determine the boundary maps
of this CW-complex, in the case where $M$ is compact and orientable.

First, the boundary maps are defined in terms of the modified gradient
flow.  Then a transversality condition is imposed which insures that
the attaching map is non-degenerate in a neighborhood of each critical
point. The degree of this map is then interpreted as a sum of
trajectories connecting two critical points each counted with a
multiplicity determined by a choice of orientations on the tangent
spaces of the unstable manifold at each critical point.
\end{abstract}

\maketitle

\section{Introduction}
Goresky and MacPherson's {\em Stratified Morse Theory}
(\cite{Gor-Mac}) represented a great step forward in extending the
ideas of Morse, Thom and Smale to spaces more general than smooth manifolds.
The complexities involved in dealing with stratified spaces prompted
Goresky and MacPherson to comment on the nostalgia their Stratified
Morse Theory might inspire for the classical version, where the Morse
data of a critical point is determined by a single number, the Morse
index.

In \cite{Vakhrameev} Vakhremeev proved the Morse theorems in the
setting of manifolds with corners, a setting which strikes a nice
balance between the simplicity of classical Morse theory on one hand,
and the generality of Stratified Morse Theory on the other.  Manifolds
with corners are a class of stratified space that arise naturally in
many applications, yet on these spaces the Morse data for an
(essential) critical point still is determined by a single number.

Others have studied Morse Theory in similar settings, (e.g. Hamm
(\cite{Hamm2},
\cite{Hamm2.5}), Siersma (\cite{Siersma}) and Braess (\cite{Braess}),
but none of these make use of a global flow on manifolds with corners.
The Morse theory for manifolds with corners developed in
\cite{handron} utilizes a modified gradient flow to prove the theorems
in a more classical manner (e.g. as in~\cite{Milnor}).  This allows,
as we shall see, the construction of a Morse complex with boundary
maps determined by the trajectories connecting (essential) critical
points.

\subsection{Setup and Definitions}

Let $\{ {\mathbf e}_1,\dots,{\mathbf e}_\ell \}$ denote the standard
basis vectors in ${\mathbb R}^n$.  Define ${\mathbb H}_\ell^n$ to be the
set
$$
{\mathbb H}_\ell^n=\{ {\mathbf w}\in{\mathbb R}^n: {\mathbf
w}\cdot{\mathbf e}_i\geq 0 \text{ for all }i>n-\ell \},
$$
where $\cdot$ denotes the standard inner product on ${\mathbb R}^n$.

\begin{definition} An {\em $n$-dimensional manifold with corners} is a
topological space $M$ together with an atlas ${\mathcal{A}}$ of charts
${\mathbf x}_a:U_a\to{\mathbb H}^n_{\ell_a}$ such that
$\{U_a\}_a$ is an open cover of $M$.  We say that $(M,\mathcal{A})$ is
a $C^r$-manifold with corners if each transition function
$\mathbf{x}_b^{}\mathbf{x}_a^{-1}:\mathbf{x}_a(U_a\cap
U_b)\to\mathbf{x}_b(U_a\cap U_b)$ is of class $C^r$, i.e. can be
extended to a $C^r$ map between open sets of $\mathbb{R}^n$.
\label{mwc}
\end{definition}

The tangent space $T_pM$ of a manifold with corners $M$ at a point $p$ may be
thought of as equivalence classes of the set
$$
\{ ({\mathbf x},{\mathbf v}):{\mathbf x}\text{ is a coordinate chart at }
p\in M\text{ and }{\mathbf v}\in{\mathbb R}^n \},
$$
where $({\mathbf x},{\mathbf v})\sim({\mathbf y},{\mathbf w})$ if
$D({\mathbf x}\circ{\mathbf y}^{-1})({\mathbf w})={\mathbf v}$.
Then for each $p\in M$, $T_pM$ is an $n$-dimensional vector space.

If $p\in M$, we will say a coordinate chart {\em at $p$} is a chart
${\mathbf x}_p\in{\mathcal{A}}$ such that ${\mathbf
x}_p(p)=\mathbf{0}\in{\mathbb H}^n_\ell$.  In this case, the number
$\ell=\ell(p)$ is uniquely determined by $p$.  This allows
consideration of $M$ as a stratified space.  Each connected component
of ${\mathcal E}_\ell=\{p\in M:\ell(p)=\ell\}$ is a stratum of
dimension $n-\ell$.

\begin{definition}
A tangent vector in $T_pM$ points {\em outward} if some representative
$({\bold x},{\bold v})$ has ${\bold v}\notin{\mathbb H}^n_{\ell(p)}$.  A tangent
vector is $T_pM$ points {\em inward} (or {\em into} M) if some representative
$({\bold x},{\bold v})$ has ${\bold v}\in{\mathbb H}^n_{\ell(p)}$.
\label{in_out_vect}
\end{definition}

These terms are well defined, since for any two coordinate charts at
$p$, the transition functions preserve ${\mathbb H}^n_{\ell(p)}$.
Note that the definition of an inward pointing vector includes those
vectors which are tangent to the boundary of $M$.

If $K$ is a stratum of $M$ then $K$ is a manifold.  We may thus
consider at a point $p\in K$ the tangent space $T_pK$.  In fact,
$T_pK\subseteq T_pM$ as
$$
T_pK=\{\mathbf{v}\in T_pM|\mathbf{v}\mbox{ and }-\mathbf{v}\mbox{ point inward}\}
$$

If $p\in K$ is in the closure of another stratum, $L$, we can define the
generalized tangent space
$$
T_pL=\{ {\bold w}\in T_pM: {\bold w}=\lim_{i\to\infty}{\bold v}_i\in T_{q_i}K
\text{ for some sequence }\{q_i\}\to p \}.
$$
We may also write this as $T_pL=\lim_{q\to p}T_qL$.

\begin{definition}
A {\em Riemannian metric} on a $C^r$-manifold with corners $M$
consists of an inner product $\langle\cdot,\cdot\rangle_p$ on each
tangent space $T_pM$, such that with respect to any coordinate chart
$\mathbf{x}:U\to\mathbb{H}^n_\ell$, the functions
$\langle\frac{\partial}{\partial x_i},\frac{\partial}{\partial
x_j}\rangle:\mathbf{x}(U)\to\mathbb{R}$ are of class $C^r$, i.e. can
be extended to a $C^r$ function on an open set in $\mathbb{R}^n$.
\end{definition}

We say that a coordinate chart ${\mathbf x}:U\to \mathbb{H}^n_\ell$ at a
point $p$ is a {\em standard} coordinate chart if whenever $i\leq n-\ell$
and $j>n-\ell$ the tangent vectors $\frac{\partial}{\partial x_i}$ and
$\frac{\partial}{\partial x_j}$ are orthogonal.

\begin{lemma}
For each point $p\in M$, there is a standard coordinate chart at $p$.
\end{lemma}

\begin{proof}
Let
$\mathbf{y}:U\to\mathbb{H}^n_\ell\cong\mathbb{R}^{n-\ell}\times[0,\infty)^\ell$
be a coordinate chart at $p$.  Then $\mathbf{y}(p)=\mathbf{0}$.
Define an inner product on $\mathbb{R}^n$ by
$$
\langle\mathbf{v}, \mathbf{w}\rangle =
\langle\mathbf{y}_\ast^{-1}(\mathbf{v}),
\mathbf{y}_\ast^{-1}(\mathbf{w})\rangle_p.
$$
Here we have used the natural identification of $\mathbb{R}^n$ with
$T_\mathbf{0}\mathbb{R}^n$.

Using this inner product, define an orthogonal projection onto
$\mathbb{R}^{n-\ell}$.  For each vector $\mathbf{v}\in\mathbb{R}^n$,
$\mbox{proj}(\mathbf{v})\in\mbox{span}(\mathbf{e}_1,\dots,\mathbf{e}_{n-\ell})$
is the unique vector such that
$\langle\mathbf{v}-\mbox{proj}(\mathbf{v}),\mathbf{w}\rangle=0$ for
every
$\mathbf{w}\in\mbox{span}(\mathbf{e}_1,\dots,\mathbf{e}_{n-\ell})$.

Now define a new basis for $\mathbb{R}^n$ by
$$
\begin{array}{rlc}
\overline{\mathbf{e}}_i\negmedspace\negthickspace\negthickspace &=\mathbf{e}_i & \mbox{for }i\leq n-\ell \\
\overline{\mathbf{e}}_j\negmedspace\negthickspace\negthickspace &=\mathbf{e}_j-\mbox{proj}(\mathbf{e}_j) & j>n-\ell.
\end{array}
$$
Then if $i\leq n-\ell$ and $j>n-\ell$,
$\langle\overline{\mathbf{e}}_i,\overline{\mathbf{e}}_j\rangle=0$.

Let $J:\mathbb{R}^n\to\mathbb{R}^n$ be the linear transformation that
carries each $\mathbf{e}_k$ to $\overline{\mathbf{e}}_k$.  Then $J$ is
an isomorphism.  We wish to show that
$J:\mathbb{H}^n_\ell\to\mathbb{H}^n_\ell$.  It is sufficient to show
that if
$$
\mathbf{v}=a_1\mathbf{e}_1+\dots+a_n\mathbf{e}_n
$$
with $a_j\geq0$ whenever $j>n-\ell$, then the expression
$$
\mathbf{v}=\overline{a}_1\overline{\mathbf{e}}_1+\dots+\overline{a}_n\overline{\mathbf{e}}_n
$$
satisfies $\overline{a}_j\geq0$ whenever $j>n-\ell$.  But if
$\mathbf{v}=\overline{a}_1\overline{\mathbf{e}}_1+\dots+\overline{a}_n\overline{\mathbf{e}}_n$, then
$$
\begin{array}{rl}
\mathbf{v}&=\overline{a}_1\mathbf{e}_1+\dots+\overline{a}_{n-\ell}\mathbf{e}_{n-\ell} \\
&\quad\quad+\overline{a}_{n-\ell+1}\left(\mathbf{e}_{n-\ell+1}-\mbox{proj}(\mathbf{e}_{n-\ell+1})\right)+\dots+
\overline{a}_{n}\left(\mathbf{e}_{n}-\mbox{proj}(\mathbf{e}_{n})\right). \\
\end{array}
$$
Then, setting
$\mbox{proj}(\mathbf{e}_j)=\sum_{k=1}^{n-\ell}p_{j,k}\mathbf{e}_k$, we
see that
$$
\begin{array}{rl}
\mathbf{v}&=\left(\overline{a}_1-\sum_{j=n-\ell+1}^{n}p_{j,1}\right)\mathbf{e}_{1}+\dots+\left(\overline{a}_{n-\ell}-\sum_{j=n-\ell+1}^{n}p_{j,n-\ell}\right)\mathbf{e}_{n-\ell}
\\
&\quad\quad +\overline{a}_{n-\ell+1}\mathbf{e}_{n-\ell+1}+\dots+\overline{a}_{n-\ell}\mathbf{e}_{n-\ell}.
\end{array}
$$
So for $j>n-\ell$, $\overline{a}_{j}=a_j$.  It follows that
$J:\mathbb{H}^n_\ell\to\mathbb{H}^n_\ell$.

Finally, let $\mathbf{x}=J\circ\mathbf{y}$.  Then
$\mathbf{x}:U\to\mathbb{H}^n_\ell$ and $\mathbf{x}(p)=\mathbf{0}$, so
$\mathbf{x}$ is a standard coordinate chart at $p$.  Moreover,
$$
\mathbf{x}_\ast^{-1}\left(\frac{\partial}{\partial
x_k}\right)=J\circ\mathbf{y}_\ast^{-1}\left(\frac{\partial}{\partial
y_k}\right).
$$
Since
$$
\mathbf{y}_\ast^{-1}\left(\frac{\partial}{\partial
y_k}\right)=\mathbf{e}_k,
$$
 it follows that
$\mathbf{x}_\ast^{-1}\left(\frac{\partial}{\partial
x_k}\right)=\overline{\mathbf{e}}_k$, and $\mathbf{x}$ is a standard
coordinate chart at $p$.

\end{proof}

Let $M$ be a compact, $n$-dimensional $C^r$-manifold with corners
($r\geq 2$) endowed with a Riemannian metric.  Let $f:M\to\mathbb{R}$
be a $C^r$ function.  For every stratum $K$ of $M$, the function
$f|_K:K\to\mathbb{R}$ is a $C^r$ function on the manifold $K$.  If
$p\in K$, we say $p$ is a {\it critical point} of $f$ whenever $p$ is
a critical point of $f\vert_K$. Recall that a critical point of $f|_K$
is {\em non-degenerate} if the Hessian of $f|_K:K\to\mathbb{R}$ is
non-zero at $p$.

If $p$ is an interior point of $M$, then it is contained in a stratum
$K$ with dimension $n$.  The gradient $\nabla f$ of the function
$f:M\to\mathbb{R}$ has a natural definition.  $K$ is a manifold with a
Riemannian metric induced by the metric on $M$.  We let $\nabla
f(p)=\nabla f|_K(p)\in T_pK=T_pM$.  If $p$ is in the boundary of $M$, we
can define $\nabla f(p)=\lim_{q\to p}\nabla f(q)$ where $q$ is taken
to be in the interior of $M$.

We can now define the class of functions we wish to consider:

\begin{definition}
Let $f:M\to{\mathbb R}$ be a $C^r$ function from a compact,
$n$-dimensional $C^r$-manifold with corners ($r\geq 2$) endowed with a
Riemannian metric into the real numbers.  Then $f$ is a Morse function
on if the following hold:

\begin{enumerate}
	\item If $K$ is a stratum of $M$, and $p\in K$ is a critical point of
		$f\vert_K:K\to {\mathbb R}$, then either
\begin{enumerate}
		\item\label{onea} $p$ is a non-degenerate critical point of
			$f\vert_K:K\to {\mathbb R}$ or
		\item\label{oneb} the vector $-\nabla f(p)$ points into $M$.
\end{enumerate}
	\item If $p\in K$ is a critical point of $f|_K$, then for any stratum
		$L$ with $p$ in the closure of $L$, $-\nabla
		f(p)\notin T_pL^\perp\subseteq T_pM$.

	\item\label{three} If $-\nabla f$ is tangent to a stratum $K$
		at a point $p\in K$, and
		$\mathbf{x}:U\to\mathbb{H}^n_\ell$ is a standard
		coordinate chart at $p$, then for $j>n-\ell$, the
		directional derivative of $\frac{\partial f}{\partial
		x_j}$ in the direction $-\nabla f(p)$ is non-zero.
\end{enumerate}
\end{definition}

This definition is inspired by the definition of Morse function in
\cite{Gor-Mac}.  It differs in condition~(\ref{oneb}), where the
critical point is not required to be non-degenerate, and in the
addition of condition~(\ref{three}), which ensures that the vector
field $G$ defined below produces a continuous flow.

We say that $p$ is an {\em essential critical point} of $f$ if $f$ satisfies
condition~(\ref{onea}) at $p$, but not condition~(\ref{oneb}).  The {\em index} of
an essential critical point $p$ in a stratum $K$ is equal to its index (in
the classical sense) as
a critical point of $f|_K$.

Suppose $f:M\to{\mathbb R}$ is a Morse function such that
$f^{-1}(-\infty,r]$ is compact for each $r\in\mathbb{R}$.  It is shown in
\cite{handron} that $M$ is homotopy equivalent to a CW-complex with one
cell of dimension
$\lambda$ for each essential critical point of
index $\lambda$.

Let $G$ be the {\em modified gradient vector field} of $f$ on $M$,
defined in \cite{handron} by projecting the vector $-\nabla f(p)$ onto
the stratum of highest dimension such that the resulting vector does
not point outward from $M$.  It is shown in~\cite{handron} that $G$
induces a flow $\varphi:[0,\infty)\times M\to M$ and that the
stationary points of this flow are exactly the essential critical
points of $f$.  The material above can be found in greatly expanded
form in \cite{handron}.

We can define the stable and
unstable sets of an essential critical point $p$ by
$$
S(p)=\left\{q\in M:\lim_{t\to\infty}\varphi(t,q)=p\right\},
$$
and
$$
U(p)=\left\{q\in M:\exists \{q_k\}_{k=1}^\infty\subset M \mbox{ such that }
\varphi(k,q_k)=q \mbox{ and }\lim_{k\to\infty}q_k=p\right\}.
$$

\section{Results Proved}

Our goal is to describe the boundary maps of the CW-complex generated
by $f$ in terms of the trajectories of the flow $\varphi$.  If we
assume a Smale-like condition (Definition~\ref{smale_condition}) for
the function $f$ then the number of trajectories connecting a pair of
critical points whose indeces differ by one is finite.  Furthermore,
if $M$ is orientable and the index of $p_2$ is one greater than that
of $p_1$, then we can associate to each trajectory $\tau$ connecting
$p_2$ to $p_1$ a sign $+1$ or $-1$.  This sign is determined by
comparing, via the flow $\varphi$, orientations on $U(p_1)$ and
$U(p_2)$.  The degree of the pair $(p_2,p_1)$ is
$$
\mbox{degree}(p_2,p_1)=\sum_{\tau}\mbox{sign}(\tau),
$$
where the sum is over all trajectories connecting $p_2$ to $p_1$.

If $P_1$ and $P_2$ are the cells in the CW-complex corresponding to the
critical points $p_1$ and $p_2$, then the degree of the attaching map of $P_2$
along $P_1$ is the sum of the signs of the trajectories from $p_2$ to $p_1$.

Now suppose $M$ is a compact oriented manifold with corners and
$f:M\to{\mathbb R}$ is a Morse function satisfying the conditions of
Definition~\ref{smale_condition}.  The goal of this article is to
define a chain complex which is determined by the function $f$ and
whose homology is that of $M$.  Let $C_\ell(f)$ be the set of essential
critical points of $f$ with index $j$, and $V_\ell$ the free Abelian
group of formal ${\mathbb Z}$-linear combinations of the elements of
$C_\ell(f)$.  We can define $\partial_\ell:V_\ell\to V_{j-1}$ by setting
$$
\partial_\ell(p)=\sum_{q\in C_{j-1}(f)}\mbox{degree}(p,q)q
$$
for each $p\in C_\ell(f)$, and extending linearly to all of $V_\ell$.
Finally, we can state the

\begin{main}
\textbf{(Theorem~\ref{main_theorem})}
The free Abelian groups $V_\ell$ and maps $\partial_\ell$ form a chain complex whose
homology groups are identical to the $\mathbb{Z}$-homology groups
 of the topological space $M$.
\end{main}

\section{The Attaching Map of a Cell in the Morse Complex}

\label{attaching}

Let $p_1$ and $p_2$ be essential critical points with indices
$\lambda_1$ and $\lambda_2$ respectively.  Let $c_i=f(p_i)$ for
$i=1,2$.  We can choose $\varepsilon$
sufficiently small that the set
$$
U_\varepsilon(p_i)=U(p_i)-f^{-1}((c_i-\varepsilon])=U(p_i)-M_{c_i-\varepsilon}
$$
is homeomorphic to the interior of $e^{\lambda_i}$, a cell with
dimension $\lambda_i$, for $i=1,2$.  Here, we follow the convention
that $M_{a}=f^{-1}((-\infty,a])$.  In addition,
suppose that $f(p_1)=c_1$ and $f(p_2)=c_2$, where $c_1<c_2$, and that
there are no other critical points in $f^{-1}([c_1,c_2])$.

The
flow $\varphi$ can be viewed as a deformation retraction, and used to
define homotopies
$$
M_{c_2-\varepsilon}\simeq\varphi(t,M_{c_2-\varepsilon})
$$
and
\begin{equation}
\label{first_case}
M_{c_2-\varepsilon}\cup U(p_2)\simeq\varphi(t,M_{c_2-\varepsilon}\cup U(p_2))
\end{equation}

Since $U(p_2)$ is invariant under the flow $\varphi$ , it follows that
\begin{equation}
\varphi(t,M_{c_2-\varepsilon}\cup
U(p_2))=\varphi(t,M_{c_2-\varepsilon})\cup U(p_2).
\label{middle_case}
\end{equation}
The space on the left hand side of (\ref{first_case}) is
$M_{c_2-\varepsilon}$ with a $\lambda_2$-cell attached, but on the
right hand side of (\ref{middle_case}),
$U(p_2)-\varphi(t,M_{c_2-\varepsilon})$ may not be homoemorphic to an
open cell.  This is a notable difference from the classical theory.

To simplify our thinking, we can consider
$M_{c_2-\varepsilon}\cup U(p_2)$ to be $M_{c_2-\varepsilon}$ with a
$\lambda_2$-cell attached via the inclusion map
$$
\sigma:\partial U_\varepsilon(p_2)\hookrightarrow M_{c_2-\varepsilon}.
$$
Then the family of homotopy equivalences
$$
M_{c_2-\varepsilon}\simeq \varphi(t,M_{c_2-\varepsilon}),
$$
gives rise to homotopy equivalences
\begin{equation}
\label{second_case}
M_{c_2-\varepsilon}\cup_\sigma e^{\lambda_2} \simeq
\varphi(t,M_{c_2-\varepsilon})
\cup_{\varphi(t,\cdot)\circ\sigma}e^{\lambda_2}.
\end{equation}
Here we have identified $\overline{U_\varepsilon(p_2)}$ with
$e^{\lambda_2}$.

In the first case (\ref{middle_case}), we allow the whole space to
flow by $\varphi$.  The equality follows from the $\varphi$-invariance
of $U(p_2)$.  In the second case (\ref{second_case}), we simply allow
$M_{c_2-\varepsilon}$ to flow by $\varphi$, and with it the image of
the attaching map $\sigma_2$.  (c.f. Lemma~3.7 in \cite{Milnor}.)

From here it is a short step to prove

\begin{lemma}
Let $p_1$ and $p_2$ be two essential critical points of $f$ such that
$f(p_1)=c_1$, $f(p_2)=c_2$, $c_1<c_2$ and there are no other critical
points in $f^{-1}([c_1-\varepsilon,c_2])$.  Then there is a map
$\Phi_2:M_{c_2-\varepsilon}\to M_{c_1+\varepsilon}$ such that
$$
M_{c_2-\varepsilon}\cup U(p_2)=M_{c_2-\varepsilon}\cup_\sigma
e^{\lambda_2}\simeq M_{c_1+\varepsilon}\cup_{\Phi_2\circ\sigma} e^{\lambda_2},
$$
where $\lambda_2$ is the index of $p_2$.  Moreover, $\Phi_2$ may be
defined in a way that preserves $\varphi$-trajectories in the sense
that for every $q\in M_{c_2-\varepsilon}$, the points $q$ and
$\Phi_2(q)$ lie on the same $\varphi$-trajectory.
\label{lemma_1}
\end{lemma}

\begin{proof}
The equality follows immediately from the discussion above.  To define
the map $\Phi_2$, we use the flow $\varphi$ to construct a homotopy.
For each $q\in M_{c_2-\varepsilon}$, define
$$
t_q^{(1)}=\inf\left\{t\in\mathbb{R}^+:\varphi(t,q)\in M_{c_1+\varepsilon}\right\}
$$
The function $q\mapsto t_q^{(1)}$ is continuous. Since there are no
critical values between $c_1$ and $c_2$, $t_q^{(1)}$ is finite for each
$q\in M_{c_2-\varepsilon}$.  Moreover, since $M_{c_2-\varepsilon}$ is
compact,
$T_1=\sup\{t_q^{(1)}:q\in M_{c_2-\varepsilon}\}$ is finite.

Now we can define a  homotopy
$H_1:[0,1]\times M_{c_2-\varepsilon}\to M_{c_1-\varepsilon}$
by
$$
H_1(s,q)=\left\{
\begin{array}{cc}
\varphi(T_1s,q)	& T_1s\leq t_q^{(1)} 	\\
\varphi(t_q^{(1)},q)	& T_1s\geq t_q^{(1)}
\end{array}
\right.
$$

If we set $\Phi_2(q)=H_1(1,q)$ for $q\in M_{c_2-\varepsilon}$, the
homotopy equivalence in the lemma follows.
\end{proof}

$\Phi_2$ can be interpreted as allowing each point to flow by
$\varphi$ until it reaches $M_{c_1+\varepsilon}$.
Now we must deal with the more complicated situation in the
neighborhood of the critical point $p_1$.

\begin{lemma}
Let $p_1$ be an essential critical point of $f$ such that $f(p_1)=c_1$
and let $\varepsilon$ be small enough that there are no other critical
points in $f^{-1}([c_1-\varepsilon,c_1+\varepsilon])$.  Then there is
a homotopy equivalence
$$
\Psi_2:M_{c_1+\varepsilon}\to M_{c_1-\varepsilon}\cup U(p_1).
$$
Moreover, $\Psi_2$ can be chosen such that $\Psi_2(q)=p_1$ if and only
if $q\in S(p_1)\cap M_{c_1+\varepsilon}$.
\label{lemma_2}
\end{lemma}

\begin{proof}
Since $p_1$ is an essential critical point with index $\lambda_1$, we
can choose a coordinate chart such that
$$
f(\mathbf{x})=c_1-x_1^2-\dots-x_{\lambda_1}^2+x_{\lambda_1+1}^2+\dots+x_{n-j}^2+
x_{n-j+1}+\dots+x_n
$$
We can define, for $\varepsilon$ sufficiently small, a region $H$ and
show that $M_{c_1-\varepsilon}\cup H$ is homotopy equivalent to both
$M_{c_1+\varepsilon}$ and $M_{c_1-\varepsilon}\cup U(p_1)$.  To do so
we first choose a smooth function $\mu:\mathbb{R}\to\mathbb{R}$ so
that $\mu(0)>\varepsilon$, $\mu(r)=0$ for $r>2\varepsilon$, and
$-1<\mu'\leq0$.  Then we set
$$
F(\mathbf{x})=f(\mathbf{x})-\mu(x_1^2+\dots+x_{\lambda_1}^2+2(x_{\lambda_1+1}^2+\dots+x_{n-j}^2)+2(x_{n-j+1}+\dots+x_n))
$$
Finally, we define
$H=\overline{F^{-1}(-\infty,c_1-\varepsilon]-M_{c_1-\varepsilon}}$.
Note that $H$ is used to denote this space, as it is in \cite{Milnor},
despite the possibility of confusion with the homotopies defined above
and below.  In this article, $H_i$ is a homotopy and $H$ is a subset
of $M$.

The proof of the lemma will be in two parts.  First, we construct a
homotopy to show that $M_{c_1+\varepsilon}\simeq
M_{c_1-\varepsilon}\cup H$.  We then construct a different homotopy
to show that $M_{c_1-\varepsilon}\cup H\simeq M_{c_1-\varepsilon}\cup
U(p_1)$.

For each $q\in M_{c_1+\varepsilon}$, define
$$
t_q^{(2)}=\inf\left\{t\in\mathbb{R}^+:\varphi(t,q)\in M_{c_1-\varepsilon}\cup H\right\}
$$
The function $q\mapsto t_q^{(2)}$ is continuous, and $t_q^{(2)}$ is finite for each
$q\in M_{c_2-\varepsilon}$.  Moreover, since $M_{c_2-\varepsilon}$ is compact,
$T_2=\sup\{t_q^{(2)}:q\in M_{c_2-\varepsilon}\}$ is finite.

Now we can define the first of our two homotopies
$H_2:[0,1]\times M_{c_1+\varepsilon}\to M_{c_1-\varepsilon}\cup H$
by
$$
H_2(s,q)=\left\{
\begin{array}{cc}
\varphi(T_2s,q)	& T_2s\leq t_q^{(2)} 	\\
\varphi(t_q^{(2)},q)	& T_2s\geq t_q^{(2)}
\end{array}
\right.
$$
This shows that  $M_{c_1+\varepsilon}= H_2(0,M_{c_1+\varepsilon})\simeq
H_2(1,M_{c_1+\varepsilon})= M_{c_1-\varepsilon}\cup H$.

Now we show that $M_{c_1-\varepsilon}\cup H\simeq
M_{c_1-\varepsilon}\cup U(p_1)$.  In our chosen coordinate system the
stable set of $p_1$ corresponds to
$$
\{(x_1,\dots,x_n)|x_1=\dots=x_\lambda=0\}.
$$
  The unstable set of
$p_1$ corresponds to
$$
\{(x_1,\dots,x_n)|x_{\lambda+1}=\dots=x_n=0\}.
$$

Consider the function

$g(\mathbf{x})=f(p_1)+x_{\lambda+1}^2+\dots+x_{n-j}^2+x_{n-j+1}+\dots+x_n$.
This function is constant on the unstable set, with a value of
$f(p_1)$.  For $q$ outside the unstable set, $f(q)>f(p_1)$.  Moreover,
on the stable set the $-\nabla g$ agrees with $-\nabla f$.

Let $X$ denote the modified (negative) gradient vector field of the function
$g$.  Then $X(q)=G(q)$ for $q\in S(p_1)$, and $X(q)=0$ for $q\in U(p_1)$.
The flow of $X$ is directed toward $U(p_1)$.  Let $\psi$ denote the flow of
the vector field $\frac{X}{\|X\|}$, and note that $\psi$ preserves the
$\varphi$-trajectories that have $p_1$ as a lower endpoint.

For each $q$ in $H$, we can set
$t_q^{(3)}=\inf\{t\in\mathbb{R}^+:\psi(t,q)\in M_{c_1-\varepsilon}\cup U(p_1)\}$, and for $q\in M_{c_1-\varepsilon}$ set $t_q^{(3)}=0$.
Then define $T_3=\sup\{t_q^{(3)}:q\in H\}$.
Then the homotopy
$H_3:[0,1]\times M_{c_1-\varepsilon}\cup H\to
M_{c_1-\varepsilon}\cup U(p_1)$
given by
$$
H_3(s,q)=\left\{
\begin{array}{cc}
\psi(T_3s,q)	& T_3s\leq t_q^{(3)}	\\
\psi(t_q^{(3)},q)	& T_3s\geq t_q^{(3)}
\end{array}
\right.
$$
gives a homotopy equivalence
$H_3(1,\cdot):M_{c_1-\varepsilon}\cup H\to M_{c_1-\varepsilon}\cup U(p_1)$.

Then the composition $\Psi_2=H_3(1,\cdot)\circ H_2(1,\cdot)$ is a
homotopy equivalence $\Psi_2:M_{c_1+\varepsilon}\to
M_{c_1-\varepsilon}\cup U(p_1)$.

\end{proof}

There is no reason, in principle, that the $\varepsilon$ used to
define $U_\varepsilon(p_2)$ must be the same as the $\varepsilon$ in
Lemma~\ref{lemma_1} and Lemma~\ref{lemma_2}.  Nor must the
$\varepsilon$'s from the two Lemmas agree with each other.  They can
be chosen to be the same, however, and this serves to simplify the
notation in the remainder of our argument.

Now lets view this in terms of attaching maps.  Let $\sigma_2$ denote
the inclusion map $\sigma_2:\partial U_\varepsilon(p_2)\to
M_{c_2-\varepsilon}$.  (Previously this was simply $\sigma$.)  Then we
can use the homotopy equivalence $\Phi_2$ to define an attaching map
$$
\Phi_2\circ\sigma_2:\partial U_\varepsilon(p_2)\to M_{c_1+\varepsilon}.
$$
Now, identifying $U_\varepsilon(p_2)$ with $e^{\lambda_2}$, we find that
$$
M_{c_2+\varepsilon}\simeq M_{c_2-\varepsilon}\cup U(p_2)\simeq
M_{c_1+\varepsilon}\cup_{\Phi_2\circ\sigma_2}e^{\lambda_2}.
$$
The first homotopy equivalence follows from Lemma~\ref{lemma_2}.

Using the homotopy equivalence $\Psi_2$, we can define an attaching map
$$
\Psi_2\circ\Phi_2\circ\sigma_2:\partial U_\varepsilon(p_2)\to
M_{c_1-\varepsilon}\cup U(p_1).
$$
From this, it follows that
$$
M_{c_2+\varepsilon}\simeq M_{c_2-\varepsilon}\cup U(p_2)\simeq
M_{c_1-\varepsilon}\cup
U(p_1)\cup_{\Psi_2\circ\Phi_2\circ\sigma_2}e^{\lambda_2}.
$$

Let $\Sigma_2=\Psi_2\circ\Phi_2\circ\sigma_2$.  Because $\Phi_2$
preserves all the $\varphi$-trajectories and $\Psi_2$ preserves those
in $S(p_1)$, it follows that $\Sigma_2(q)=p_1$ if and only if $q\in
\partial U_\varepsilon(p_2)$ lies on a trajectory connecting $p_2$ to $p_1$.

Using these ideas, we can construct a cellular complex (which will
later prove to be a CW-complex) that is homotopy equivalent to $M$.
Let $p_1,p_2,\dots,p_m$ be the essential critical points of $M$, each
$p_i$ having index $\lambda_i$.  Assume the values $f(p_i)$ are
distinct and the essential critical points are labeled such that
$f(p_i)<f(p_j)$ whenever $i<j$.  Then $c_1=f(p_1)$ must be an absolute
minimum, and $M_{c_1+\varepsilon}$ is homotopy equivalent to the
0-cell, $U_\varepsilon(p_1)$.  In fact the map $\Psi_2$ is a homotopy
equivalence.

Now, as in Lemma~\ref{lemma_1} and Lemma~\ref{lemma_2}, we consider the
inclusion map
$\sigma_2:\partial U_\varepsilon(p_2)\hookrightarrow M_{c_2-\varepsilon}$.
Viewing $U_\varepsilon(p_2)$ as a $\lambda_2$-cell,
and using the attaching map
$\Sigma_2=\Psi_2\circ\Phi_2\circ\sigma_2$, of
$U_\varepsilon(p_2)$ onto $U_\varepsilon(p_1)=\{p_1\}$
we see that
$$
\begin{array}{rl}
M_{c_2+\varepsilon} &\simeq
U_\varepsilon(p_1)\cup_{\Sigma_2}U_\varepsilon(p_2) \\
& \simeq e^{\lambda_1}\cup_{\Sigma_2}e^{\lambda_2}.
\end{array}
$$

If $\varepsilon$ is sufficiently small, then $U_\varepsilon(p_3)$ is
homeomorphic to a $\lambda_3$-cell.  It takes only a slight extension
to define the attaching map for this cell.  Let $\sigma_3$ be the
inclusion map $\partial U_\varepsilon(p_3)\hookrightarrow
M_{c_3-\varepsilon}$, and define $\Phi_3$, $\Psi_3$ for the critical
points $p_2$ and $p_3$ using Lemma~\ref{lemma_1} and
Lemma~\ref{lemma_2}.  Then
$\widetilde{\Sigma}_3=\Psi_3\circ\Phi_3\circ\sigma_3$ gives the
appropriate attaching map of $\partial U_\varepsilon(p_3)$ onto
$M_{c_2-\varepsilon}\cup U(p_2)\simeq
M_{c_2-\varepsilon}\cup_{\sigma_2}e^{\lambda_2}$.

Since $\Psi_2\circ\Phi_2$ is a homotopy equivalence from
$M_{c_2-\varepsilon}\cup_{\sigma_2}e^{\lambda_2}$ to
$e^{\lambda_1}\cup_{\Sigma_2}e^{\lambda_2}$, the desired attaching map
is given by
$$
\Sigma_3=\Psi_2\circ\Phi_2\circ\Psi_3\circ\Phi_3\circ\sigma_3.
$$
It follows that
$$
\begin{array}{rl}
M_{c_3+\varepsilon} &\simeq
U_\varepsilon(p_1)\cup_{\Sigma_2}U_\varepsilon(p_2)\cup_{\Sigma_3}U_\varepsilon(p_3) \\
& \simeq e^{\lambda_1}\cup_{\Sigma_2}e^{\lambda_2}\cup_{\Sigma_3}e^{\lambda_3}.
\end{array}
$$

Because of the construction of $\Phi_3$ and $\Psi_3$,
$\Sigma_3(q)=p_2$ if and only if $q\in\partial U_\varepsilon(p_3)$
lies on a $\varphi$-trajectory connecting $p_3$ to $p_2$.  We would
like to be able to say that $\Sigma_3(q)=p_1$ if and only if
$q\in\partial U_\varepsilon(p_3)$ lies on a $\varphi$-trajectory
connecting $p_3$ to $p_1$.  This will be true provided that $\Psi_3$
does not disrupt these trajectories.

This can not be true in every case.  With care, though, we can be sure
that trajectories connecting critical points whose indices differ by
one are not disturbed.  This issue is resolved by
Lemmas~\ref{critical_point_lemma} and \ref{cw-complex_lemma}.

In general, for each critical point $p_k$ we define
$$
\Sigma_k=\Psi_2\circ\Phi_2\circ\dots\Psi_k\circ\Phi_k\circ\sigma_k.
$$
These maps gives a homotopy equivalence
$$
\begin{array}{rl}
M & \simeq
U_\varepsilon(p_1)\cup_{\Sigma_2}U_\varepsilon(p_2)\cup\dots\cup_{\Sigma_m}U_\varepsilon(p_m)
\\
& \simeq e^{\lambda_1}\cup_{\Sigma_2}e^{\lambda_2}\cup\dots\cup_{\Sigma_m}e^{\lambda_m}.
\end{array}
$$

We shall see that under appropriate conditions
(Definition~\ref{smale_condition}), this cellular complex is in fact
a CW-complex.

\section{The Degree of an Attaching Map}
\label{degree}

Our goal is to show that, as in classical Morse theory, the degree of
this attaching map can be determined by counting the trajectories
connecting essential critical points.  One difficulty is that,
although these attaching maps are continuous, they are in general not
smooth.  Lemmas~\ref{lemma2.1}, \ref{lemma2.2} and \ref{lemma2.3} show
that the maps possess enough regularity that we can establish a
Smale-like transversality condition
(Definition~\ref{smale_condition}).  Using this definition,
Lemmas~\ref{transversality_lemma} and \ref{finite_trajectories} show
that a finite number of trajectories connect critical points whose
indices differ by one.  Finally a after a careful consideration of
orientations along each trajectory connecting critical points, we
define a $\mbox{degree}$ to each such pair.

\begin{lemma}
Let $K$ be a stratum with dimension $k$ and $L$ be a stratum in the closure of
$K$ with dimension $k-1$.  Let $p\in K$ and suppose that
$\varphi(s,p)\in K$ for
$s<t_p$ and $\varphi(s,p)\in L$ for $t_p< s\leq T$.  Then there is a
neighborhood $U\subset K$ of $p$ such that $\varphi(T,\cdot):U\to M$ is
smooth.
\label{lemma2.1}
\end{lemma}

\begin{proof}
Choose a coordinate system near $\varphi(t_p,p)$ such that for $q\in K$,
$x_{k+1}(q)=\dots=x_n(q)=0$ and for $q\in L$, $x_{k}(q)=\dots=x_n(q)=0$.
We know that the $k$th coordinate satisfies $\varphi_k(t_p,p)=0$, and in
addition
$$
\left.\frac{\partial \varphi_k}{\partial t}\right|_{(t_p,p)}\neq0,
$$
by the definition of Morse function.

From the Implicit Function Theorem, it follows that there is an open set
$U\subset K$ containing $p$ and an open set $V\subset \mathbb{R}$ such that
for each $q\in U$ there is a unique $g(q)\in V$ such that
$\varphi_k(g(q),q)=0$.  Moreover, the function $g$ is smooth.

We can use this function $g$ to define two additional functions:
$\gamma:U\to{\mathbb R}\times U$ given by $\gamma(q)=(g(q),q)$ and
$\tau:\gamma(U)\to{\mathbb R}\times M$ given by $\tau(t,q)=(T-t,\varphi(t,q))$.
Since $\varphi$ is smooth whenever it remains in a single stratum, these are
themselves smooth functions.
We can write
$$
\varphi(T,q)=(\varphi\circ\tau\circ\gamma)(q),
$$
a composition
of smooth functions, so $\varphi(T,\cdot):U\to M$ is smooth.

\end{proof}

\begin{lemma}
Let $K$ be a stratum with dimension $k-1$ in the closure of a stratum $L$
with dimension $k$.  Let $p\in K$ and suppose that
$\varphi(s,p)\in K$ for
$s\leq t_p$ and $\varphi(s,p)\in L$ for $t_p< s\leq T$.  Then there is a
neighborhood $U\subset K$ of $p$ such that $\varphi(T,\cdot):U\to M$ is
smooth.
\label{lemma2.2}
\end{lemma}

\begin{proof}
Choose a coordinate system near $\varphi(t_p,p)$ such that for $q\in K$,
$x_{k}(q)=\dots=x_n(q)=0$ and for $q\in L$, $x_{k+1}(q)=\dots=x_n(q)=0$.
Let $h(s,q)=(-\nabla f)_k(\varphi(s,q))$. Then $h$ is a smooth real valued
function.  Since the trajectory $\varphi(\cdot,p)$ moves from $K$ to $L$ at
time  $t_p$, we must have $h(t_p,p)=0$.

By the definition of a Morse function, we know that
$$
\left.\frac{\partial h}{\partial t}\right|_{(t_p,p)}\neq 0.
$$
It then follows from the Implicit Function Theorem, that there is an open set
$U\subset K$ containing $p$ and an open subset $V\subset \mathbb{R}$ such that
for each $q\in U$ there is a unique $g(q)\in V$ such that
$h(g(q),q)=0$.  Moreover, the function $g$ is smooth.

As before, we can define smooth functions
$\gamma:U\to{\mathbb R}\times U$ given by $\gamma(q)=(g(q),q)$ and
$\tau:\gamma(U)\to{\mathbb R}\times M$ given by $\tau(t,q)=(T-t,\varphi(t,q))$
and write
$$
\varphi(T,q)=(\varphi\circ\tau\circ\gamma)(q),
$$
Again, this is a composition of smooth functions, and so
$\varphi(T,\cdot):U\to M$ is smooth.

\end{proof}

Now let $p_2$ be an essential critical point with index $\lambda_2=j$ and
$p_1$ an essential critical point with index $\lambda_1=j-1$.  Suppose that
$S(p_1)\cap U(p_2)\neq \emptyset$.  Choose a
$$
q\in S(p_1)\cap U(p_2)\cap M_{c_1+\varepsilon}.
$$
Then there is a $q_1\in U_\varepsilon(p_2)$ and a time $T$ such that
$q=\varphi(T,q_1)$.

\begin{lemma}
If the trajectory $\varphi(s,q_1)$ only ever changes between strata whose
difference in dimension is 1, then there is a neighborhood $V$ in the stratum
containing $q_1$ such that $\varphi(T,\cdot):V\to M$ is smooth.
\label{lemma2.3}
\end{lemma}

\begin{proof}
The result follows from repeated application of Lemma~\ref{lemma2.1} and
Lemma~\ref{lemma2.2}.  Since $f^{-1}(-\infty,r]$ is compact for each choice
of $r$, there can be only
finitely many stratum changes~(\cite{handron}).  For each stratum
change, $i=1,\dots,N$, we get a corresponding neighborhood $V_i$.  Since
there are finitely
many, we may choose
$$
V=V_1\cap\dots\cap V_N.
$$
\end{proof}

\begin{definition}
We say that the Morse function $f$ is a {\em Morse-Smale function} if for each
pair of critical points, $p$ and $p'$ with $f(p)>f(p')$, and for
$\varepsilon$ sufficiently small
the parametrized space $U(p)$ and the set
$S_\varepsilon(p')=S(p')\cap M_{f(p')+\varepsilon}$
intersect transversely in the following sense: if $q\in
S_\varepsilon(p')\cap U(p)$, $K$ is the stratum containing $q$, and $q$ has a
neighborhood $N\subset M$ such that $S_\varepsilon(p')\cap U(p)\cap
N=S_\varepsilon(p')\cap U(p)\cap K$, then $U(p)$ and
$S_\varepsilon(p')\cap K$ intersect transversely in $K$.
\label{smale_condition}
\end{definition}

Let $p_i$ and $p_j$ be two critical points with indices $\lambda_i$ and
$\lambda_j$, respectively, satisfying  $\lambda_j-\lambda_i=1$.  Let
$\mathfrak{T}$ be the set of trajectories from $p_j$ to $p_i$.  We wish
to show that the
transversality condition in Definition~\ref{smale_condition} together
with the compactness of $M$ ensure that $\mathfrak{T}$
is a finite set.

We begin with the following

\begin{lemma}
\label{transversality_lemma}
Suppose that $\widetilde{M}$ is an $m$-manifold, $\widetilde{S}$ a $k$-dimensional regular
submanifold.  Let $\widetilde{U}$ be an $(m-k)$-manifold, and
$\Phi:\widetilde{U}\to\widetilde{M}$ a smooth map such that $\widetilde{S}\pitchfork\Phi(\widetilde{U})$.  Then
$\Phi^{-1}(\widetilde{S}\cap\Phi(\widetilde{U}))$ consists of isolated points in $\widetilde{U}$.
\end{lemma}

\begin{proof}
First, since $\dim(\widetilde{S})+\dim(\Phi(\widetilde{U}))=k+(m-k)=m$, the transverse intersection
of $\widetilde{S}$ and $\Phi(\widetilde{U})$ is zero dimensional.  Let $q\in \widetilde{U}$ such that
$\Phi(q)\in \widetilde{S}$.  We will show that there is a
neighborhood $N_q$ of $q$ such that $\Phi^{-1}(\widetilde{S})\cap N_q=\{q\}$.

let $\mathbf{x}=(x^1,\dots,x^m)$ be a coordinate chart on $\widetilde{M}$
such that
$\mathbf{x}(\Phi(q))=(0,\dots,0)$, and $(x^1,\dots,x^k)$ gives a coordinate
chart on $\widetilde{S}$.  Then for $p\in \widetilde{S}$
$$
x^i(p)=0,\quad\mbox{for }i=k+1,\dots,m.
$$

Let $\bar{\mathbf{x}}=(x^{k+1},\dots,x^m)$.  We want to show that
$\tilde{\mathbf{x}}=\bar{\mathbf{x}}\circ\Phi$ is a coordinate chart on $\widetilde{U}$.
It suffices to show that if $\mathbf{y}$ is a coordinate chart on $\widetilde{U}$, then
$D(\tilde{\mathbf{x}}\circ\mathbf{y}^{-1})$ is non-singular.

Since $\widetilde{S}\pitchfork\Phi(\widetilde{U})$, the vectors
$$
\left\{
\frac{\partial}{\partial x^1}|_{\Phi(q)},
\dots,
\frac{\partial}{\partial x^k}|_{\Phi(q)},
\Phi_\ast\left(\frac{\partial}{\partial y^{k+1}|_q}\right),
\dots,
\Phi_\ast\left(\frac{\partial}{\partial y^{n}|_q}\right)
\right\}
$$
form a basis for $T_{\Phi(q)}(\widetilde{M})$, as do
$\left\{
\frac{\partial}{\partial x^1}|_{\Phi(q)},
\dots,
\frac{\partial}{\partial x^m}|_{\Phi(q)}
\right\}$.
We can thus write
$$
\Phi_\ast\left(\frac{\partial}{\partial y^{i}}\right)=\sum_{j=1}^n
\alpha_i^j\frac{\partial}{\partial x^j}\quad\mbox{ for }i=k+1,\dots,n.
$$

Using this expression, it is clear that the matrix
$$
\begin{bmatrix}
1              & 0 & \cdots & 0              \\
0              & 1 & \cdots & 0              \\
\vdots         &   &        & \vdots         \\
\alpha_{k+1}^1 &   & \cdots & \alpha_{k+1}^m \\
\vdots         &   &        & \vdots         \\
\alpha_{m}^1 &   & \cdots & \alpha_{m}^m
\end{bmatrix}
=
\left[
\begin{array}{c|c}
I_k  & 0 \\ \hline
\ast &
\begin{matrix}
\alpha_{k+1}^{k+1} & \cdots & \alpha_{k+1}^m \\
\vdots             & \ddots & \vdots         \\
\alpha_m^{k+1}     & \cdots & \alpha_m^m
\end{matrix}
\end{array}
\right]
=
\left[
\begin{array}{c|c}
I_k  & 0 \\ \hline
\ast & M
\end{array}
\right]
$$
is non-singular (having linearly independent rows).  It follows that the
matrix $M$ is non-singular.  But $M$ is the transpose of the matrix for
$$
D(\bar{\mathbf{x}}\circ(\Phi\circ\mathbf{y}^{-1}))
=D((\bar{\mathbf{x}}\circ\Phi)\circ\mathbf{y}^{-1})
=D(\tilde{\mathbf{x}}\circ\mathbf{y}^{-1}),
$$
evaluated at $\mathbf{y}(q)$, so
$D(\tilde{\mathbf{x}}\circ\mathbf{y}^{-1})$ is non-singular in a neighborhood
of $\mathbf{y}(q)$.

Now, since $\tilde{\mathbf{x}}$ is a coordinate chart in a neighborhood of
$q$, there is an open set
$N_q$ of $q$ such that for $q'\in N_q$ with $q'\neq q$,
$\tilde{\mathbf{x}}(q')\neq(0,\dots,0)$.  So $\Phi(q')\notin S$.

\end{proof}

\begin{lemma}
Let $f:M\to\mathbb{R}$  be a Morse-Smale function on a manifold with corners
$M$.
Let $p_i$ and $p_j$ be two critical points of $f$ with indices $\lambda_i$ and
$\lambda_j$ respectively, satisfying $\lambda_j-\lambda_i=1$.
Then the set
$\mathfrak{T}$ of trajectories from $p_j$ to $p_i$ is finite.
\label{finite_trajectories}
\end{lemma}

\begin{proof}
Let $c_i=f(p_i)$ and $c_j=f(p_j)$.  Then $\partial U_\varepsilon(p_j)$
is diffeomorphic to $S^{\lambda_j-1}$.  For each element of
$\tau\in\mathfrak{T}$ there is exactly one $q_\tau\in \partial
U_\varepsilon(p_j)\cap\tau$.

Suppose that $\mathfrak{T}$ is an infinite set.  Then, since $\partial U_\varepsilon(p_j)$ is
compact, there is a convergent sequence $\{q_i\}_{i=1}^\infty$ such that
$q_i\in \partial U_\varepsilon(p_j)\cap\tau_i$ for some $\tau_i\in\mathfrak{T}$.  Let
$q_0=\lim_{i\to\infty}q_i$.

We wish to show that $q_0=\partial U_\varepsilon(p_j)\cap\tau_0$ for some
$\tau_0\in\mathfrak{T}$.  Choose $T$ such that for $p= \varphi(T,q_0)$,
$f(p)<c_i+\varepsilon$.  Then for $i$ sufficiently large
$f(\varphi(T,q_i))<c_i+\varepsilon$, and since for each $i$
$\lim_{t\to\infty}\varphi(t,q_i)=p_i$, $\varphi(T,q_i)\in S_\varepsilon(p_i)$
for $i$ sufficiently large.  Now, $\varphi$ is continuous, and
$S_\varepsilon(p_i)$ is closed, so
$p=\lim_{i\to\infty}\varphi(T,q_i)\in S_\varepsilon(p_i)$.
It follows that $q_0$ lies on some trajectory connecting $p_j$ to $p_i$.

Now we will show that $p_0$ is isolated in $\partial
U_\varepsilon\cap\mathfrak{T}$, contradicting the choice of $p_0$.
Let $K$ be the stratum containing $p=\varphi(T,q_0)$ and set $c=f(p)$.
Let $\widetilde{M}=f^{-1}(c)\cap K$,
$\widetilde{S}=S_\varepsilon(p_i)\cap\widetilde{M}$.  Assume there is
a neighborhood $U_0$ of $q_0$ in $M$ on which $\varphi(T,\cdot)$ is
smooth.  (If there is not, replace $T$ with a slightly larger $T'$.)
For each $q\in U_0$ there is a trajectory of $\varphi$ containing $q$.
We will denote this trajectory by $\gamma_q$.  For a given $q\in U_0$,
let $\tilde{q}=\gamma_q\cap\partial U_\varepsilon(p_j)$,
$\bar{q}=\gamma_q\cap\varphi(T,\cdot)^{-1}(\widetilde{S})$, and let
$t_q\in\mathbb{R}$ such that
$$
\begin{array}{c}
\varphi(t_q,\tilde{q})=\bar{q}\quad\mbox{ if }f(\tilde{q})\geq f(\bar{q})  \\
\varphi(-t_q,\bar{q})=\tilde{q}\quad\mbox{ if }f(\tilde{q})\leq f(\bar{q}).
\end{array}
$$

The maps $q\mapsto\tilde{q}$, $q\mapsto\bar{q}$ and $q\mapsto t_q$ are
all smooth, and so the function $\Phi_0:U_0\to M$ given by
$\Phi_0(q)=\varphi(T+t_q,q)$ is smooth.  Let $\widetilde{U}=\partial
U_\varepsilon(p_j)\cap U_0$, and $\Phi=\Phi_0\vert_{\widetilde{U}}:\widetilde{U}\to\widetilde{M}$.

We can let $k=\dim(\widetilde{S})=\dim(K)-\lambda_i-1$ and
$m=\dim(\widetilde{M})=\dim(K)-1$.  Then $\dim(\partial U_\varepsilon(p_j))=\lambda_j-1=m-k$.
Choose a coordinate chart at $p\in\widetilde{M}$, $\mathbf{x}=(x^1,\dots,x^m)$,
such that $(x^1,\dots,x^k)$ is a coordinate chart on $\widetilde{S}$ and
$$
x^i(p')\neq0\quad\mbox{ whenever }p'\notin\widetilde{S}\mbox{ and }i>k.
$$
This can be done in such a way that
$\frac{\partial}{\partial x^{k}}\vert_p$ is tangent to the trajectory
$\tau_0$.

Now choose a coordinate chart $\mathbf{y}=(y^{k+1},\dots,y^{m+1})$
on $U_0$, such that $\mathbf{y}=(y^{k+1},\dots,y^m)$ is a
coordinate chart on $\widetilde{U}$ and
$\frac{\partial}{\partial y^{m+1}}\vert_{q_0}$ is tangent to the trajectory
$\tau_0$.  Then
$$
\Phi_\ast\left(\frac{\partial}{\partial y^i}\vert_{q_0}\right)={\Phi_0}_\ast\left(\frac{\partial}{\partial y^i}\vert_{q_0}\right)\in T_p\widetilde{M}
\mbox{ for }i=k,\dots,m.
$$
In addition, ${\Phi_0}_\ast(\frac{\partial}{\partial y^n}\vert_{q_0})$ will be
tangent (at $p$) to the trajectory $\tau_0$.

Since $f$ is a Morse-Smale function, the vectors
$$
\left\{
\frac{\partial}{\partial x^{1}}\vert_p,
\dots,
\frac{\partial}{\partial x^{k+1}}\vert_p,
{\Phi_0}_\ast\left(\frac{\partial}{\partial y^{k+1}}\vert_{q_0}\right),
\dots,
{\Phi_0}_\ast\left(\frac{\partial}{\partial y^{m+1}}\vert_{q_0}\right)
\right\}
$$
span $T_pK$.  Since the vectors $\frac{\partial}{\partial x^{k+1}}\vert_p$
and ${\Phi_0}_\ast(\frac{\partial}{\partial y^{m+1}}\vert_{q_0})$ are tangent
to $T_p\tau_0$, and the remaining $m=\dim(\widetilde{M})$ vectors are all in
$T_p\widetilde{M}$,
$$
\left\{
\frac{\partial}{\partial x^{1}}\vert_p,
\dots,
\frac{\partial}{\partial x^{k}}\vert_p,
\Phi_\ast\left(\frac{\partial}{\partial y^{k+1}}\vert_{q_0}\right),
\dots,
\Phi_\ast\left(\frac{\partial}{\partial y^{m}}\vert_{q_0}\right)
\right\}
$$
form a basis for $T_p\widetilde{M}$.  The first $k$ vectors of this basis
are a basis of $T_p\widetilde{S}$, and the remaining $m-k$ vectors are a
basis for $T_p\Phi(\partial U_\varepsilon(p_j))$.  It follows that
$$
\widetilde{S}\pitchfork\Phi(\partial U_\varepsilon(p_j)).
$$

By Lemma~\ref{transversality_lemma},
$\Phi^{-1}(\widetilde{S}\cap\Phi(\widetilde{U}))$ consists of isolated points.
Note that for sufficiently large $i$,
$q_i\in\Phi^{-1}(\widetilde{S}\cap\Phi(\widetilde{U}))$ and in addition
$q_0\in\Phi^{-1}(\widetilde{S}\cap\Phi(\widetilde{U}))$.  This, however,
contradicts the fact that $\lim_{i\to\infty}q_i=q_0$.  Consequently, the
number of trajectories connecting $p_j$ to $p_i$ must be finite.
\end{proof}

Now assume that $M$ is an oriented manifold.
At each essential critical point $p$, the
tangent space $T_pM$ can be decomposed into a stable space and an unstable
space, $T_pM=E^+(p)\oplus E^-(p)$.  Choose an orientation for the
unstable space $E^-(p)$.  This induces an orientation on $E^+(p)$ in the
following way:

Let $\lambda_p$ be the index of $p$, and choose a basis
$v_1,\dots,v_{\lambda_p}\in E^-(p)$ which represents the orientation
on $E^-(p)$.  Then choose $v_{\lambda_p+1},\dots,v_n\in E^+(p)$ such
that $v_1,\dots,v_n$ is a basis which represents the orientation on
$T_pM$.  The vectors $v_{\lambda_p+1},\dots,v_n$ determine an
orientation for $E^+(p)$.

Now for critical points $p_i$ and $p_j$, the orientation on $E^+(p_i)$
determines an orientation on
$S_\varepsilon(p_i)$, which
in turn determines an orientation on $T_qS(p_i)$ for $q\in S(p_i)$.  In a
similar way, the
choice of orientation on $E^-(p_j)$ induces an orientation on
$T_{q'}U(p_j)$ for $q'\in U_\varepsilon(p_j)$.

Note that if the function $f:M\to{\mathbb R}$ is a Morse-Smale function
and $\varphi(T,q')=q$, then
the derivative $\varphi(T,\cdot)_\ast$ restricted to $U(p_j)$ has full rank at
$\varphi(T,\cdot)^{-1}(q)$.

Now the map $\varphi(T,\cdot)_\ast$ pushes forward the orientation on
$T_{q'}U(p_j)$ to give an
orientation on $T_q\varphi(T,U(p_j))$.  Since
$\dim(\varphi(T,U(p_j)))=j$ and
$\dim(S_\varepsilon(p_i))=n-j+1$, these two sets intersect in the
one dimensional trajectory through $q$.  We can choose a basis for
$T_q\varphi(T,U(p_j))$ which represents the induced orientation and such that
the last vector is $\frac{d}{dt}|_T\varphi(t,q_1)$.  We also may choose
a basis representing the orientation on $T_qS(p_i)$, with
$\frac{d}{dt}|_T\varphi(t,q_1)$ as its first
vector.  When combined, these bases give an orientation for $T_qM$.
If this orientation agrees with the chosen orientation on $M$ then we say
the sign of the trajectory from $p_j$ to $p_i$ through $q_1$ is $+1$.
Otherwise, the sign is $-1$.

Now if $\mathfrak{T}$ is the set of trajectories from $p_j$ to $p_i$,
Lemma~\ref{finite_trajectories} ensures that $\mathfrak{T}$
is a finite set.  So, we can define
$$
\mbox{degree}(p_j,p_i)=\sum_{\tau\in\mathfrak{T}}\mbox{sign}(\tau).
$$
This degree, we shall see, is the degree of the attaching map
$\Sigma_j$ at $p_i$.

\section{The Morse Complex for a Morse Function on a Manifold with Corners}
\label{complex}

\subsection{The Geometric Complex of a Morse Function}

Let $P_k$ denote the cell $U_\varepsilon(p_k)$.
We know that $M$ is homotopy equivalent to the cellular complex
$$
X=P_1\cup_{\Sigma_2}P_2\cup\dots\cup_{\Sigma_m}P_m.
$$

Recall that $\Sigma_j$ was defined by
$$
\Sigma_j=\Psi_2(1,\cdot)\circ\Phi_2(1,\cdot)\circ\dots\circ\Psi_j(1,\cdot)\circ\Phi_j(1,\cdot)\circ\sigma_j,
$$
which attaches $P_j$ to
$P_1\cup_{\Sigma_2}P_2\cup\dots\cup_{\Sigma_{j-1}}P_{j-1}$.  The map
$\Sigma_{j}$ is defined by a sequence of maps derived from flows.
They alternate between a $\Phi_k$ which preserves the
$\varphi$-trajectories, and a $\Psi_k$ which does not.

We wish to show that this cellular complex is, in fact, a CW-complex.
In order to do so we need the following result regarding the behavior
of the trajectories of $\varphi$.

\begin{lemma}
Let $p_i$ and $p_j$ be essential critical points of a Morse-Smale
function $f$ on a manifold with corners $M$.  If $p_j\in\overline{U(p_i)}$
then $\mbox{index}(p_j)<\mbox{index}(p_i)$.  If $p_i\in\overline{S(p_j)}$
then $\mbox{index}(p_i)>\mbox{index}(p_j)$.
\label{critical_point_lemma}
\end{lemma}

This lemma allows the maps $\Phi_k$ and $\Psi_k$ to be defined in a
way that preserves $\varphi$-trajectories connecting critical points
whose indices differ by one.

\begin{proof}
We will show that the boundaries of $U(p_i)$ and of $S(p_j)$ are
composed of a union of trajectories.  Suppose, in this case, that
$p_j\in\overline{U(p_i)}$.  Then there is a trajectory $\tau\in
S(p_j)$, with $\tau\subset\overline{U(p_i)}$.  The upper endpoint of
$\tau$ must also be a critical point $p_k$ in $\overline{U(p_i)}$.
Since $f$ is a Morse-Smale function, it follows that
$\mbox{index}(p_k)>\mbox{index}(p_j)$.

Traveling up from $p_j$ in this
way, there is a sequence of essential critical points with increasing
indices.  Since the topmost critical point in this sequence must be
$p_i$, the result follows.  The second half of the lemma is
established in a similar way.

To show that the boundary of $U(p_i)$ and of $S(p_j)$ is composed of a
union of trajectories, we will assume that $q\in\overline{U(p_i)}$ and
we will show two things.  First, that for any $t_0>0$,
$\varphi(t_0,q)\in\overline{U(p_i)}$ and second, that that there exist
$t_0$ and $q'$ such that $q'\in\overline{U(p_i)}$ and
$\varphi(t_0,q')=q$.

If $q\in\overline{U(p_i)}$, then there is a sequence $\{q_n\}\subset
U(p_i)$ such that $\lim_{n\to\infty}q_n=q$.  Since $\varphi(t_0,\cdot)$ is
continuous, it follows that
$\lim_{n\to\infty}\varphi(t_0,q_n)=\varphi(t_0,q)$.

The second assertion is more of a challenge, since $\varphi$ provides
a flow in only one direction.  Again, let $\{q_n\}\subset
U(p_i)$ be a sequence such that $\lim_{n\to\infty}q_n=q$. Let $r=f(q)$.

If $q$ is an essential critical point, then let $q'=q$ and $t_0$ be
any real number greater than zero.  Then $\varphi(t_0,q')=q$.

If $q$ is not an essential critical point, let then for each $n$, let
$q_n'$ denote the intersection of the trajectory containing $q_n$ with
$f^{-1}(r)$.  Then $\lim_{n\to\infty}q_n'=q$ also.

Let $r'>r=f(q)$ be such that $f^{-1}([r,r'])$ contains no critical
points (at least in a neighborhood of $q$).  Then let $q_n''$ denote
the intersection of the trajectory containing $q_n$ with $f^{-1}(r')$.
Since $f^{-1}(r')$ is a closed submanifold, it is compact.  There is a
subsequence of $\{q_n''\}$ that converges to a point $q'$.  Relabel,
if necessary, so that $\lim_{n\to\infty}q_n''=q'$.  Note that
$q'\in\overline{U(p_i)}$.

The map from $f^{-1}(r')$ to $f^{-1}(r)$ given by allowing each point
of $f^{-1}(r')$ to flow by $\varphi$ until it hits $f^{-1}(r)$ is both
well defined and continuous (at least in a neighborhood of
$q'$). Since this map carries each $q_n''$ to $q_n'$, it follows that
the limit $q'$ is carried to the corresponding limit $q$.  Thus there
is some $t_0$ such that $\varphi(t_0,q')=q$.

It follows that $\overline{U(p_i)}$ contains the entirety of some
trajectory passing through $q$.

The above argument uses only the fact that $\overline{U(p_i)}$ is a
union of trajectories.  The argument applies equally well to
$\overline{S(p_j)}$

\end{proof}

Note that this lemma does not imply that if $\tau$ is a trajectory
that intersects $\overline{U(p_i)}$, then
$\tau\subset\overline{U(p_i)}$.  Because trajectories combine wherever
$-\nabla f$ points outward, a point may lie in more than one
trajectory.  Of course, all trajectories containing a particular point
must have the same lower limit.

\begin{lemma}
The cellular complex $X$ of the Morse-Smale function $f:M\to\mathbb{R}$
is a CW-complex.
\label{cw-complex_lemma}
\end{lemma}

\begin{proof}
By construction, $\Sigma_i$ maps $\partial P_i=\partial
U_\varepsilon(p_i)$ onto a union of cells.  The important point here
is to show that these cells have dimension smaller than that of
$P_i$.  Ideally, a cell $P_j$ will be in the image of $\Sigma_i$ if
and only if there is a $\varphi$-trajectory connecting $p_i$ to $p_j$.

By Lemma~\ref{critical_point_lemma}, for each critical point $p_i$
there is an open set $W$ containing $p_i$, such that $U(p_j)\cap
W=\varnothing$ whenever $\mbox{index}(p_j)\leq\mbox{index}(p_i)$ and
$S(p_j)\cap W=\varnothing$ whenever
$\mbox{index}(p_j)\geq\mbox{index}(p_i)$.  We can then choose
$\varepsilon$ small enough that $\Psi_{i+1}$ is the identity outside
$W$.

Now suppose that $S(p_j)\cap U(p_i)=\varnothing$, but
$P_j\subset\mbox{image}(\Sigma_i)$.  The sets $S(p_j)$ and $U(p_i)$
must have been ``confused'' in a neighborhood of some essential
critical point $p_k$ by $\Psi_{k+1}$.  It follows that
$\mbox{index}(p_k)<\mbox{index}(p_i)$ and
$\mbox{index}(p_j)<\mbox{index}(p_k)$. It follows that
$$
\mbox{index}(p_j)\leq\mbox{index}(p_i)-2$$.

Hence, the boundary of each cell is mapped to a union of lower
dimensional cells, and the geometric complex is a CW-complex.

\end{proof}

The proof gives us some other information about the attaching maps of
this CW-complex.  Suppose that $\Sigma$ is the attaching map for
$P=U_\varepsilon(p)$ and $Q=U_\varepsilon(q)$ is a cell with dimension
one less than that of $P$.  Then $\Sigma^{-1}(q)$ is exactly the set
of points in $\partial U_\varepsilon(q)$ that lie on trajectories
connecting $p$ to $q$.

This CW-complex gives rise to a chain complex
$$
0 \longleftarrow  W_0  \overset{\partial_1}{\longleftarrow}  W_1  \overset{\partial_2}{\longleftarrow} \dots \overset{\partial_n}{\longleftarrow}  W_n.
$$
The free Abelian groups $W_i$ are generated by the $i$-cells of the
above CW-complex.  We will refer to this chain complex as the {\em
geometric complex} of $f:M\to\mathbb{R}$.

\subsection{The Algebraic Complex of a Morse Function}
Let $C_j(f)$ be the set of essential critical points of $f$ with index
$j$, and $V_j$ the free Abelian group of formal ${\mathbb Z}$-linear
combinations of the elements of $C_j(f)$.  For each
$j=1,\dots,n=\dim(M)$ define a map $\partial'_j:V_j\to V_{j-1}$ by
setting
$$
\partial'_j(p)=\sum_{q\in C_{j-1}(f)}\mbox{degree}(p,q)q
$$
for each $p\in C_j(f)$, and extending linearly to all of $V_j$.  For
$j=0$, set $\partial'_0(p)=0$

In the next section, we will see that the free Abelian groups $V_j$
and maps $\partial'_j$ form a chain complex
$$
0 \longleftarrow  V_0  \overset{\partial'_1}{\longleftarrow}  V_1  \overset{\partial'_2}{\longleftarrow} \dots \overset{\partial'_n}{\longleftarrow}  V_n
$$
We will refer to this chain complex as the {\em algebraic complex}
of $f:M\to\mathbb{R}$.

\subsection{Main Theorem}

We are now ready to state and prove the main result of this article.

\begin{theorem}
Let $M$ be a compact orientable $n$-manifold with corners, and
$f:M\to\mathbb{R}$ be a Morse-Smale function satisfying the conditions of
Lemma~\ref{lemma2.3}.  Then the algebraic complex of
$f:M\to\mathbb{R}$ is a chain complex.  Moreover, the algebraic
complex and geometric complex have the same homology groups.
\label{main_theorem}
\end{theorem}

\begin{proof}
Let $C_k(X)$ be the set of cells in $X$ having dimension $k$.  As
above, $C_k(f)$ is the set of critical points of $f$ having index
$k$.  These sets are in one to one correspondence;  each $P\in C_k(X)$
satisfies $P=U_\varepsilon(p)$ for some $p\in C_k(f)$ and for each
$p\in C_k(f)$, $P=U_\varepsilon(p)\in C_k(X)$.

The bijection $h_k:C_k(X)\to C_k(f)$ by $U_\varepsilon(p)\mapsto p$
can be extended to an isomorphism $h_k:W_k\to V_k$ by linearity.
Thus we have a chain map $h$ given by
$$
\begin{CD}
0    @<<<  W_0   @<{\partial_1}<<   W_1   @<{\partial_2}<<   \dots   @<{\partial_n}<<  W_n   @<<< 0\\
@VVV       @V{h_0}VV                @V{h_1}VV                @VVV                        @V{h_n}VV \\
0    @<<<  V_0   @<{\partial'_1}<<  V_1   @<{\partial'_2}<<  \dots   @<{\partial'_n}<<  V_n  @<<< 0
\end{CD}
$$
Where each $h_i$ is an isomorphism.  It suffices to show that
$\partial_k=h_{k-1}^{-1}\circ\partial'_k\circ h_k$.

The map $\partial_k$ is defined by
$$
\partial_k(P_j)=\sum_{P_i\in C_{k-1}(X)}\deg(\Sigma_j,P_i)P_i
$$
for each $P_j\in C_k(X)$, where $\Sigma_j$ is the attaching map for the
cell $P_j$.
The degree $\deg(\Sigma_j,P_i)$ is determined by choosing a $q\in P_i$ such that
$\Sigma_j(\partial P)$ intersects $P_i$ transversely at $q$, and then adding the
number of points in $\Sigma_j^{-1}(q)$ at which $\Sigma_j$ is orientation
preserving, and subtracting the number of points in $\Sigma_j^{-1}(q)$ at
which $\Sigma_j$ is orientation reversing.  Given $P_j\in C_k(X)$ and
$P_i\in C_{k-1}(X)$ we must determine this degree.

We have the relationship
$$
\deg(\Sigma_j,P_i)=\deg(\Sigma_{ij},U_\varepsilon(p_i)).
$$

Consider a point $q\in\partial U_\varepsilon(p_j)\cap\tau$  for some
trajectory connecting $p_j$ to $p_i$.  The map $\Sigma_{ij}$ is orientation
preserving at $q$ if $\mbox{sign}(\tau)=+1$, and orientation reversing if
$\mbox{sign}(\tau)=-1$.  It follows that
$$
\deg(\Sigma_j,P_i)=\deg(\Sigma_{ij},p_i)=\mbox{degree}(p_j,p_i).
$$
and so
$$
\deg(\Sigma_j,P_i)=\mbox{degree}(h_k(p_j),h_{k-1}(p_i)).
$$
It follows that
$$
\partial_k(P)=\sum_{P_i\in
C_{k-1}(X)}\mbox{degree}(h_k(p_j),h_{k-1}(p_i))h_{k-1}^{-1}(p_i).
$$
and so $\partial_k=h_{k-1}^{-1}\circ\partial'_k\circ h_k$ as claimed.
This proves the theorem.

\end{proof}

It should be pointed out that in the the descending set of an
essential critical point may not be homeomorphic to an open disk.
Consequently, the descending sets may not be interpreted directly as
cells in a CW-complex, as they may in the classical theory.
Nonetheless, this theorem provides a useful parallel with the classical
Morse theory in the setting of manifolds with corners.

config: arxiv


TITLE: Limits and Continuity of a Complex Function
QUESTION [0 upvotes]: Prove that $f(z)= (\frac{(Im(z))^3}{|z|^2})  \;\;\;\;\;\ z\neq0\\
                  \qquad \;\;\;\;\;\;0\qquad \quad  z=0\\$
$\quad$ is continuous for all $z_0\in \mathbb{C}$.
I have done this by checking that the limit is the same over two paths, real and imaginary. I think that both paths equal 0. Is this enough to confirm continuity?
I have done the same for
Prove that $f(z)= \frac{\bar{z}}{z} \qquad z\neq0 \\
\qquad \; \; 0 \qquad z=0 \\$
is not continuous at $z_0=0$, but continuous for all other $z_0$,
and have gotten that the real and imaginary paths have different limits when tending to $0$ but i don't know how to show that it is continuous for all other $z_0$.

REPLY [1 votes]: For the first one, we have $|f(z)|=\dfrac{\left|\text{Im}(z)\right|^{3}}{|z|^{2}}\leq\dfrac{|z|^{3}}{|z|^{2}}=|z|$ for $z\ne 0$. Of course this inequality also satisfies for $z=0$. Anyway we have $\lim_{z\rightarrow 0}|f(z)|=0$ by Squeeze Theorem, then $\lim_{z\rightarrow 0}f(z)=f(0)$.
Considering only two paths to prove existence is not conclusive.


TITLE: existing limit and differentiability
QUESTION [2 upvotes]: Let $f: (a, b) \to \mathbb R$ continuous function that is differentiable in every $x \in (a, b)$ except $x_0$. Show that if there exists a finite limit
$L =  \lim_{x \to x_0} f´(x)$
then $f$ is differentiable in $x_0$ and $f´(x_0) = L$
I think I have a kind of a hunch here, maybe by using MVT? But I'm not sure how to put it.

REPLY [3 votes]: HINT:
From the mean value theorem, for $a<x_0<x<b$, there exists a number $\bar x\in (x_0,x)$ such that $\frac{f(x)-f(x_0)}{x-x_0}=f'(\bar x)$.
Take the limit as $x\to x_0$ from the right.  The limit is $L$.  Now, show that the limit from the left exists.


TITLE: Convergence of $\max$ and $\min$ sequences of first $n$ terms of a convergent sequence
QUESTION [0 upvotes]: Let $(x_n)$ be a sequence of real numbers converging to $x$.
Define sequences $y_n:=\max\{x_1,x_2,\ldots,x_n\}$ and $z_n:=\min\{x_1,x_2,\ldots,x_n\}$
Now do the sequences $(y_n),(z_n)$ converge? If $(x_n)$ is monotonic , then both  these sequences are convergent. What can we say in general case? Please help me with this. Thanks!

REPLY [0 votes]: $(y_n)$ is monotically increasing. It is also bounded: There exists $N$ such that $\lvert x_n - x \rvert < 1$ for $n \ge N$. Thus $x_n < x + 1$ for $n \ge N$ and thefore $y_n \le \max(y_N,x+1)$ for all $n$. Thus $(y_n)$ converges to some $y \in \mathbb R$. Similarly $(z_n)$ converges to some $z \in \mathbb R$.
However, $z < y$ unless $(x_n)$ is constant: If $(x_n)$ is not constant, then $x_N \ne x_M$ for suitable $N, M$. Let $R = \max(N,M)$. Then $z_R < y_R$ and thus
$$z \le z_R < y_R \le y .$$


TITLE: Prove a matrix maps to a point that make up the plane perpendicular to a line.
QUESTION [1 upvotes]: I'm having difficulty understanding what the following question is asking and was hoping someone could explain it to me.
$T = \begin{bmatrix}1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{bmatrix}$
Given a point on the line x = y = z, prove that the set of points in $\mathbb{R}^3$ that $T$ maps to that point make up the plane through that point that is perpendicular to the line $x = y = z$.
According to my instructor, I am suppose to solve for $Tx = \begin{bmatrix}a\\a\\a\end{bmatrix}$. Doing so I get
$\begin{bmatrix} 1/3 && 1/3 && 1/3 && a \\1/3 && 1/3 && 1/3 && a \\1/3 && 1/3 && 1/3 && a \end{bmatrix}$
Reducing:
$\begin{bmatrix} 1 && 1 && 1 && 3a \\ 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 \end{bmatrix}$
$ x_1 + x_2 + x_3 = 3a $
$ x_1 = 3a - x_2 - x_3$
So
$x = \begin{bmatrix}  x_1\\ x_2\\x_3 \end{bmatrix} = x_1\begin{bmatrix} 3a \\ 0 \\ 0 \end{bmatrix} + x_2\begin{bmatrix} -1\\1\\0\end{bmatrix} + x_3\begin{bmatrix} -1\\0\\1 \end{bmatrix} $
Since I don't understand what the question is really asking, I'm not sure what to do with this answer.

REPLY [1 votes]: Well, check that the set of points you denoted as $\;x\;$ indeed passes through the point on $\;x=y=z\;$ and is perpendicular to that line:
$$\begin{pmatrix}a\\a\\a\end{pmatrix}=\begin{pmatrix}1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\frac13\begin{pmatrix}x_1+x_2+x_3\\x_1+x_2+x_3\\x_1+x_2+x_3\end{pmatrix}\iff$$
$$x_1+x_2+x_3=3a$$
Now, check the vector $\;u=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\;$ defined by the above relation actually passes throught the point $\;A:=\begin{pmatrix}a\\a\\a\end{pmatrix}\,$ (very easy) ,and that the vector $\;A-u\;$ is perpendicular to the given line there (very easy, too).


TITLE: How to multiply matrices?
QUESTION [0 upvotes]: I've written a java-programm(JOGL) and have my vertices from the object I want to draw on screen. I've read, when I want to rotate my object I must multiply the matrix of my object with the so-called "rotation-matrix". Link -> rotationsmatrix . I don't know what I've done wrong by multiplying the two matrices. So please, can anyone help me?

REPLY [2 votes]: I cannot read German, but I used Google translate.  Please forgive me if I've misunderstood your goal.
Your mistake was in your definition of the rotation matrix.  Because you are rotating the object around the positive $z$ axis, the $z$ coordinate should not change after the rotation.  The rotation matrix should be a $3\times3$ matrix, not a $4\times4$; the result should be a $3\times1$ column vector that describes a point, not a $4\times4$ matrix.
I will call $\vec M$ the original point vector and $\vec M'$ the rotated point vector.
Thus:
$$\vec M' = \mathbf R \vec M$$
We let $\vec M = \pmatrix{x \\ y \\ z}$.
Because we are rotating around the positive $z$-axis, the rotation matrix is:
$$R = \pmatrix{\cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1}$$
Thus:
$$\begin{align}
\vec M' &= \pmatrix{\cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1}\pmatrix{x \\ y \\ z} \\
&= \pmatrix{x\cos\theta -y\sin\theta \\ x\sin\theta + y\cos\theta \\ z}
\end{align}$$


TITLE: Show that there exist $g\in C^{\infty}$ such that $f(x)=g(x^2)$
QUESTION [3 upvotes]: Let $f\in C^{\infty}(\mathbb{R},\mathbb{R})$, even,.
Show that there exist $g\in C^{\infty}(\mathbb{R^+},\mathbb{R})$ such that for all $x\in \mathbb{R}, f(x)=g(x^2)$

My attempt :
Let $g$ the function defined on $\mathbb{R^+}$ by $g(x)=f(\sqrt{x})$, it's clear that $g$ is continuous on $\mathbb{R^+}$ and $C^{\infty}$ on $\mathbb{R^+_*}$.
So the problem is to find the limit at $0$ but I don't know how can I proceed.
EDIT: Here $\mathbb{R^+}=[0,+\infty)$

REPLY [0 votes]: The basic idea is that a smooth even function looks like a parabola sufficiently close to zero, so that its composition with $\sqrt x$ is still smooth near zero.
To make this precise, you want to remember that the derivative is the best linear approximation: that is, for any differentiable function whatsoever we get $f(x)=f(0)+f'(0)x+h(x)x^2$, where $\lim_{x\to 0} h(x)$ is finite. But for $f$ even, $f'(0)=0$, because $$0=f(x)-f(-x)=2f'(0)x+h(x)x^2-h(-x)x^2$$ And for small enough $x$ we can suppose $|h(x)x^2-h(-x)x^2|\leq (|h(x)|+|h(-x)|)x^2< 2f'(0)x$-to see this, divide by $x^2$ on both sides and use that $h$ converges at $0$.
Thus $f(x)=f(0)+h(x)x^2$ and the first derivative of $f(\sqrt x)=f(0)+h(\sqrt x)x$ exists: namely, it's $h(0)$.


TITLE: Regularity of solution of $u_t -\Delta u=f$ where $f$ is a continuous function
QUESTION [3 upvotes]: Let $f \in C(\bar \Omega \times [0,T])$ and consider the following problem:
$\begin{align}
\partial_tu -\Delta u&=f \quad \text{in} \; \Omega \times (0,T)\\
u(\cdot,0)&=u_0 \quad \text{in} \; \Omega
\end{align}$
Assume that $u_0 \in L^2(\Omega)$. My question is: Can we deduce $u\in C^{1+a, \frac{1+a}{2}}(\Omega \times (0,T))$ for some $0<a<1$? Or can we at least obtain a Hoelder estimate only in terms of the space variable?
My thought was that since $f \in C(\bar \Omega \times [0,T])$ implies $f \in L^2(\Omega \times (0,T))$, we could apply the known $L^2-$regularity and obtain $u\in L^2(0,T;H^1_0(\Omega))\cap H^1(0,T;H^{-1}(\Omega)) \cap W^{2,1}_p(\Omega \times (\sigma,T))$ for all $1 \leq p <\infty\;$ and for any $\sigma>0$. Then, by a t-Sobolev embedding we could obtain the desired regularity.
Is the above correct or did I miss anything? I would appreciate any help or hints!
Thanks in advance!

REPLY [2 votes]: The fact that $u \in W^{2,1}_p(\Omega \times (\sigma,T))$ does not follow from $L^2$-theory using energy estimates; you need to use the fact that $f$ lies in $L^p.$ Your idea is correct to show this and deduce the result using Sobolev embedding, but you would need to use a suitable $L^p$ estimates for the form
$$ \lVert u \rVert_{W^{2,1}_p(\Omega \times (\sigma,\rho))} \leq C \left( \lVert f \rVert_{L^p(\Omega \times (0,T))} + \lVert u \rVert_{L^2(\Omega \times (0,T))} \right),$$
where $0 < \sigma < \rho < T.$ Note this uses a global estimate, which requires some regularity of $\partial\Omega,$ but you can work around this since the question only asks for interior regularity. Then you would use the embedding
$$ W^{2,1}_p(\Omega \times (\sigma,\rho)) \hookrightarrow C^{1+\alpha,\frac{1+\alpha}2}(\overline \Omega \times [\sigma,\rho]) $$
for $p > n,$ where $\alpha = 1-\frac{p}n.$


TITLE: Let $T: V→V$ be a linear operator on a vector space $V$. Assume that $T^n = 0v$ for some $n\ge1$. Prove that the linear map $Iv + T$ is an isomorphism
QUESTION [0 upvotes]: Does $T^n$ refer to $T(T(T(\cdots T(v))))$ - $n$ total compositions of $T$?
If so, I'm not exactly sure what information this gives us on the map $T$ aside from the obvious that it slowly converges to $0v$.
And lastly, does $Iv + T$ mean $I(v) + T(v)$ which $I(v) = v$ so it would be $v + T(v)$ is an isomorphism?
$Iv$ is already known to be bijective, but what about $T$? How can we use the fact that $T^n=0v$ for some $n\ge1$ to prove that $Iv + T$ is an isomorphism?

REPLY [1 votes]: Hint: Try to find a map $(I+T)^{-1}$.


TITLE: Solving a stochastic differential equationn
QUESTION [1 upvotes]: Does anyone has ideas on how to solve this equation.
$$dX_{t} = \left(\sqrt{1+X_{t}^{2}} + \frac{1}{2}X_{t}\right)\,dt + \sqrt{1 + X_{t}^{2}} \,dBt$$ where $Bt$ is a standard Brownian Motion.
I have tried to solve $$\frac{\partial f}{\partial x} = \sqrt{1 + f^2}$$
$$\frac{\partial f}{\partial t} + 0.5\frac{\partial^{2} f}{\partial x^{2}} = 0.5f + \sqrt{1 + f^{2}}$$
As for the 1st equation, I can find the solution $f(x) = \sinh(x) + C$, but clearly, it is not the solution for the 2nd equation. Does anyone have ideas on this?

REPLY [1 votes]: You could take the inspiration from the first solution the other way around, it is easier to transform the known equation than to reverse-engineer starting from an unknown quantity. Set
$$
Y_t=g(X_t)=\arg\sinh(X_t),~~g'(x)=\frac1{\sqrt{1+x^2}},~~g''(x)=-\frac{x}{\sqrt{1+x^2}^3}.
$$
Then by the Ito theorem
$$
dY_t=\frac12g''(X_t)\sigma(X_t)^2dt+g'(X_t)dX_t
=dt+dB_t.
$$


TITLE: Windows lightweight Math Software
QUESTION [11 upvotes]: I'm looking for lightweight, free, Windows, Math software.
Something I can put an expression and get an answer, or graph it.
I tried Euler, but it is quiet complicated and HUGE.
Basic needs:

Expression Based
Supports Variables
Support Functions, User defined and auto loaded.
Supports graphs, 2D.  Not really needing 3D.
Supports History.

What do you use?  What do you recommend?

REPLY [0 votes]: You may consider TeXmacs, it is a free wysiwyw editing platform with special features for scientists.
More related to your question, you can interface it with a lot of third party mathematics tools for algebraic calculus, graphing, numeric calculus, etc.

Developed on GNU/Linux, it runs under Windows and Mac OS X as well. I guess the best user experience is with GNU/Linux.
Downside of TeXmacs regarding your question, it is not exactly lightweight, although absolutely free software.


text:
\begin{document}

\maketitle

\begin{abstract}
In this short note we give an elementary combinatorial
argument, showing that the Conjecture of J. Fern\'andez de Bobadilla, I. Luengo, A. Melle-Hern\'andez, A. N\'emethi (see \cite{boro}, Conjecture 1.2) follows from Theorem 5.4 in \cite{boro} in the case of rational cuspidal curves with two critical points.
\end{abstract}

\noindent {\bf 2010 Mathematics Subject Classification.} Primary 14H50; Secondary 14B05, 57M25, 57R58.

\noindent {\bf Key words and phrases.} Rational cuspidal curve, Alexander polynomial, infimum convolution.

\section{Introduction}\label{intro}

In this short note we deal with irreducible algebraic curves $C\subset \mathbb{C}P^2$. Such a curve has a finite set of singular points $\{z_i\}_{i=1}^n$ such that a neighbourhood of each singular point intersects $C$ in a cone on a link $K_i \subset S^3$. We would like to know what possible configurations of links $\{K_i\}_{i=1}^n$ arise in this way. We consider only the case in which each $K_i$ is connected (in this case $K_i$ is a knot), and thus $C$ is a rational curve, meaning that there is a rational surjective map $\mb{C}P^1 \to C$. Such a curve is called rational cuspidal. We refer to \cite{moe} for a survey on rational cuspidal curves.

Suppose that $z$ is a cuspidal singular point of a curve $C$ and $B$ is a sufficiently small ball around $z$. Let $\Psi(t)=(x(t),y(t))$ be a local parametrization of $C \cap B$ near $z$. For any polynomial $G(x,y)$ we look at the order at $0$ of an analytic map $t \mapsto G(x(t),y(t)) \in C$. Let $S$ be the set of integers, which can be realized as the order for some $G$. Then $S$ is a semigroup of $\mb{Z}_{\geq 0}$. We call it the semigroup of the singular point, see \cite{wall} for the details and proofs. The gap sequence, $G = \mb{Z}_{\geq 0} \backslash S$, has
precisely $\mu/2$ elements, where the largest one is $\mu-1$. Here $\mu$ stands for the Milnor number. Assume that $K$ is the link of the singular point $z$. The Alexander polynomial of $K$ is of the form
\[
	\Delta_K(t) = \sum_{i=0}^{2m} (-1)^i t^{n_i},
\]
where $(n_i)_{i=0}^{2m}$ form an increasing sequence with $n_0=0$ and $n_{2m}=2g$, for $g$ being the genus of $K$. Writing $t^{2n_i}-t^{2n_{i-1}}=(t-1)(t^{2n_{i}-1}+t^{2n_{i}-2}+\ldots+t^{2n_{i-1}})$ yields the representation
\begin{equation}
\label{deltak}
\Delta_K(t) = 1+(t-1)\sum_{j=1}^k t^{g_j},
\end{equation}
for some finite sequence $0<g_1<g_2<\ldots<g_k$. We have the following lemma (see \cite{wall}, Exercise 5.7.7), which relates the Alexander polynomial to the gap sequence of a singular point.

\begin{lem} The sequence $g_1,\ldots, g_k$  in (\ref{deltak}) is the gap sequence of the semigroup of the
singular point. In particular, $k = |G| = \mu/2$, where $\mu$ is the Milnor number, so $|G|$
is the genus.
\end{lem}
If we write $t^{g_j}= (t-1)(t^{g_j-1}+t^{g_j-2}+\ldots+1)+1$, we obtain
\[
	\Delta_K(t) = 1+(t-1)g(K) + (t-1)^2 \sum_{j=0}^{\mu-2} k_j t^j,
\]
where $k_j = |\{ m>j: m \notin S \}|$. This motivates the following definition.

\begin{def*} For any finite increasing sequence of positive integers $G$ we define
\[
	I_G(m) = |\{k \in G \cup \mb{Z}_{<0}: \  k \geq m\}|,
\]
where $\mb{Z}_{<0}$ is the set of negative integers. We shall call $I_G$ the gap function, because in most applications $G$ will be a gap sequence of some semigroup.
\end{def*}

\noindent Clearly, for $j=0,1,\ldots, \mu-2$ we have $I_G(j+1)=k_j$.

In \cite{1} the following conjecture was proposed.

\begin{con}\label{conjecture}
Suppose that the rational cuspidal curve $C$ of degree $d$ has critical points $z_1, \ldots, z_n$. Let $K_1, \ldots, K_n$ be the corresponding links of singular points and let $\Delta_1, \ldots, \Delta_n$ be their Alexander polynomials. Let $g$ be the genus of $K$. Let $\Delta = \Delta_1, \ldots, \Delta_n$, expanded as
\[
	\Delta(t) = 1 + \frac{(d-1)(d-2)}{2}(t-1) + (t-1)^2 \sum_{j=0}^{2g-2} k_l
\]
Then for any $j=0,\ldots, d-3 $ we have $k_{d(d-j-3)} \leq (j+1)(j+2)/2$, with equality for $n=1$.
\end{con}
\noindent This conjecture was verified in the case $n=1$ by Borodzik and Livingston, see \cite{boro}.

We define the infimum convolution of two functions.

\begin{def*}
Let $I_1,I_2,\ldots,I_n: \mb{Z} \to \mb{Z}_{\geq 0}$. We define
\[
	(I_1 \diamond I_2 \ldots \diamond I_n)(k) = \min_{\begin{array}{cc}
	\scriptstyle{k_1, k_2,\ldots,k_n \in \mb{Z}} \\
	\scriptstyle{k_1+k_2+\ldots+k_n = k}
	\end{array}}\left( I_1(k_1) + I_2(k_2)+\ldots+  I_n(k_n)  \right).
\]
\end{def*}

\noindent In \cite{boro} the authors gave the proof of the following theorem.

\begin{thm} (\cite{boro}, Theorem 5.4) Let $C$ be a rational cuspidal curve of degree $d$. Let $I_1,\ldots,I_n$ be the gap functions associated to each singular point on $C$. Then for any $j \in \{-1,0,\ldots, d-2\}$ we have
\[
	I_1 \diamond I_2 \diamond \ldots \diamond I_n(jd+1) = \frac12 (j - d + 1)(j - d + 2).
\]
\end{thm}

\noindent Note that we have $|G_1|+ |G_2|+\ldots+|G_n|=\frac{(d-1)(d-2)}{2}$. Therefore, one can give an equivalent reformulation of the Conjecture \ref{conjecture}.

\begin{con}\label{conjecture2}
Suppose that the rational cuspidal curve $C$ of degree $d$ has critical points $z_1, \ldots, z_n$. Let $K_1, \ldots, K_n$ be the corresponding links of singular points and let $\Delta_1, \ldots, \Delta_n$ be their Alexander polynomials. Moreover, let $G_1, G_2,\ldots,G_n$ be the gap sequences of these points. Let $g=|G_1|+ |G_2|+\ldots+|G_n|$ be the genus of $K$. Let $\Delta = \Delta_1, \ldots, \Delta_n$, expanded as
\[
	\Delta(t) = 1 + (t-1)g + (t-1)^2 \sum_{j=0}^{2g-2} k_l
\]
and let $I = I_1 \diamond I_2 \diamond \ldots \diamond I_n$.
Then for any $j=0,\ldots, d-3$ we have $k_{d(d-j-3)} \leq I(d(d-j-3)+1)$, with equality for $n=1$.
\end{con}

In this note we give an elementary argument, showing that FLMN conjecture follows from \cite{boro} for $n=2$. The idea of our proof is to forget about the specific structure of the problem coming from theory of singularities and to prove Conjecture \ref{conjecture2} for general sets $G_1, G_2$. Namely, we have the following theorem.

\begin{thm}
Let $G, H$ be two finite sets of positive integers and let $I_G,I_H:\mb{Z} \to \mb{Z}_{\geq 0}$ be their gap functions. Let us define the polynomials
\[
	\begin{array}{l}
		\Delta_G(t)   =  1 + (t-1) \sum_{j=1}^{|G|} t^{g_j}   =  1+(t-1)|G| + (t-1)^2 \sum_{j \geq 0} k_j^G t^j \\
		\Delta_H(t)   = 1 + (t-1) \sum_{j=1}^{|H|} t^{h_j}  = 1+(t-1)|H| + (t-1)^2 \sum_{j \geq 0} k_j^H t^j
	\end{array},
\]
where $k_j^G = I_G(j+1), k_j^H = I_H(j+1)$, $j \geq 0$. Take $\Delta = \Delta_G \cdot \Delta_H$ and $I =  I_G \diamond I_H$. Then
\[
	\Delta(t) = 1+ (t-1)(|G|+|H|) + (t-1)^2 \sum_{j \geq 0} k_j t^j,
\]
where $k_j \leq I(j+1)$ for $j \geq 0$.
\end{thm}

\noindent This gives the proof of Conjecture \ref{conjecture} in the case $n=2$.

It is natural to ask whether the above theorem is valid for arbitrary $n \geq 2$. Recently, after we found our elementary combinatorial argument for $n=2$, J. Bodn\'ar and A. N\'emethi showed that the Conjecture \ref{conjecture} is false for $n \geq 3$, see \cite{bn}. They also found yet another proof of Conjecture \ref{conjecture} in the case of two singularities.

\begin{comment}

\begin{ex*}
Consider semigroups $S_1=\{2k+3l\colon k,l \geq 1\}$, $S_2=\{2k+5l\colon k,l \geq 1\}$, $S_3=\{3k+4l\colon k,l \geq 1\}$. The corresponding gap functions are $G_1=\{1\}$, $G_2=\{1,3\}$, $G_3=\{1,2,5,7\}$. Then,
\[
	\begin{array}{l}
		\Delta_1(t) = 1+(t-1)t = 1+(t-1)+(t-1)^2 \\
		\Delta_2(t) = 1+(t-1)(t+t^3) = 1+2(t-1)+(t-1)^2(2+t+t^2) \\
		\Delta_3(t) = 1+(t-1)(t+t^2+t^5+t^7) \\
		\hspace{1.5cm}= 1+4(t-1)+(t-1)^2(4+3t+2t^2+2t^3+2t^4+t^5+t^6)
	\end{array}.
\]
We write $\Delta=\Delta_1 \cdot \Delta_2 \cdot \Delta_3$ in the form
\[
	\Delta(t)=1+(|G_1|+|G_2|+|G_3|)(t-1)+(t-1)^2 \sum_{j \geq 0} k_j t^j.
\]

\begin{align*}
	\Delta(t) & = 1+ 7(t-1) \\ & \quad + (t-1)^2(t^{12}-t^{11}+3t^{10}-2t^9+4t^8+3t^6+2t^5+4t^4+3t^3+6t^2+4t+7).
\end{align*}

One can check that
\[
	\begin{array}{ll}
		(k_j)_{j=0}^\infty & =    (7,4,6,3,4,2,3,4,-2,3,-1,1,0,0,0,\ldots) \\
		(I(j+1))_{j=0}^\infty & = (7,6,5,5,4,4,3,3,+2,2,+2,1,1,0,0,\ldots)
	\end{array}		.
\]
We see that, for example, $k_3>I(4)$.
\end{ex*}

\end{comment}

\section{Proof of the main result}\label{proof}

In this section we give a proof of our main result.

\begin{proof}
Our goal is to express the numbers $k_j$ in terms of the numbers $k_j^G$ and $k_j^H$. We have
\begin{align*}
& \Delta(t)  = \Delta_G(t) \Delta_H(t) = 1+ (t-1)(|G|+|H|) \\ &  + (t-1)^2 \Big[ |G|\cdot |H| + \sum_{j \geq 0} (k_j^G + k_j^H)t^j + (t-1)(|G| \sum_{j \geq 0} k_j^H t^j + |H| \sum_{j \geq 0} k_j^G t^j  ) \\ & + (t-1)^2 \Big( \sum_{j \geq 0} k_j^G t^j  \Big) \Big( \sum_{j \geq 0} k_j^H t^j  \Big) \Big] =  1+ (t-1)(|G|+|H|) + (t-1)^2 \Theta(t),
\end{align*}
with
\[
	\Theta(t)  = |G|\cdot |H| +k_0^G(1-|H|) + k_0^H(1-|G|) + k_0^G k_0^H
	 +  \sum_{j \geq 1} t^j  k_j,
\]
where
\[
	k_j =  k_j^G(1-|H|) +|H| k_{j-1}^G + k_j^H(1-|G|) +|G| k_{j-1}^H +l_j
\]
and
\[
	 l_j = \sum_{   u+v=j,\  u,v \geq 0 } k_u^G k_v^H - 2\sum_{u+v=j-1,\  u,v \geq 0} k_u^G k_v^H + \sum_{u+v=j-2,\  u,v \geq 0} k_u^G k_v^H.
\]
Note that $k_0^G = |G|$ and $k_0^H = |H|$.  Therefore,
\begin{align*}
	k_0 & = |G|\cdot |H| +k_0^G(1-|H|) + k_0^H(1-|G|) + k_0^G k_0^H \\ & = |G|\cdot |H| + |G|(1-|H|) +|H|(1-|G|) + |G| \cdot |H| = |G|+|H|.
\end{align*}
Moreover, for $k \geq 1$ we have
\[
	I_H(k) \geq |H|-(k-1), \qquad I_G(1-k) = |G|+(k-1).
\]
Thus, for $k \geq 1$ we obtain
\[
	I_G(1-k) + I_H(k) \geq |G|+|H|.
\]
For $k \leq -1$
\[
	I_G(1-k) \geq |G|+k, \qquad I_H(k) = |H|-k.
\]
In this case we obtain
\[
	I_G(1-k) + I_H(k) \geq |G|+|H|
\]
and we arrive at
\begin{align*}
I(1) & = \min_{k \in \mb{Z}}(I_G(1-k)+I_H(k)) \\ & = I_G(1)+I_H(0) = I_G(0)+I_H(1) = |G|+|H|= k_0.
\end{align*}

Note that
\begin{align*}
	 l_j & = \sum_{   u+v=j,\  u,v \geq 0 } k_u^G k_v^H - \sum_{u+v=j,\  u \geq 0,v \geq 1} k_u^G k_{v-1}^H - \sum_{u+v=j,\  u \geq 1,v \geq 0} k_{u-1}^G k_{v}^H \\
	 & \quad + \sum_{u+v=j,\  u,v \geq 1} k_{u-1}^G k_{v-1}^H = \sum_{u+v=j, u,v \geq 1} (k_u^G-k_{u-1}^G)(k_v^H-k_{v-1}^H) \\
	 & \quad  + k_0^G k_j^H + k_j^G k_0^H - k_0^G k_{j-1}^H -k_{j-1}^G k_0^H .
\end{align*}
Thus,
\begin{align*}
	k_j & = \sum_{u+v=j, u,v \geq 1} (k_u^G-k_{u-1}^G)(k_v^H-k_{v-1}^H) \\
	& \qquad + \Big( (k_0^G k_j^H + k_j^G k_0^H - k_0^G k_{j-1}^H -k_{j-1}^G k_0^H) \\
	& \qquad +( k_j^G(1-|H|) +|H| k_{j-1}^G + k_j^H(1-|G|) +|G| k_{j-1}^H )  \Big).
\end{align*}
Observe that we have a miracle,
\begin{align*}
	 k_0^G k_j^H & + k_j^G k_0^H - k_0^G k_{j-1}^H -k_{j-1}^G k_0^H \\ & \hspace{2cm} + k_j^G(1-|H|) +|H| k_{j-1}^G + k_j^H(1-|G|) +|G| k_{j-1}^H  \\
	&  = |G| k_j^H + k_j^G |H| - |G| k_{j-1}^H -k_{j-1}^G |H| \\
	&  \hspace{1cm} + k_j^G(1-|H|) +|H| k_{j-1}^G + k_j^H(1-|G|) +|G| k_{j-1}^H \\
	&  = k_j^G + k_j^H.
\end{align*}
We get
\[
	k_j = k_j^G + k_j^H + \sum_{u+v=j, u,v \geq 1} (k_{u-1}^G-k_{u}^G)(k_{v-1}^H-k_{v}^H).
\]

We are to prove that $k_j \leq (I_G \diamond I_H)(j+1)$. It suffices to prove that $k_j \leq I_G(j+1-l)+I_H(l)$ for every $l \in \mb{Z}$. Thus, we have to deal with the inequality
\[
	 k_j^G + k_j^H + \sum_{u+v=j, u,v \geq 1} (k_u^G-k_{u-1}^G)(k_v^H-k_{v-1}^H) \leq  I_G(j+1-l)+I_H(l), \quad j \geq 1, l \in \mb{Z}.
\]
Note that if $u+v=j$ then we have either $u \geq j-l+1$ or $v \geq l$. Thus,
\[
	\1_{u \in G} \1_{v \in H} \1_{u+v=j} \leq \1_{u \in G \cap [j-l+1,j]} + \1_{v \in H \cap[l,j]}.
\]
In the above expression we have used the convention $[a,b]=\emptyset$ for $a>b$.
We obtain
\begin{align*}
	\sum_{u+v=j, u,v \geq 1} & (k_{u-1}^G-k_{u}^G)(k_{v-1}^H-k_{v}^H)  = \sum_{u+v=j, u,v \geq 0} (k_{u-1}^G-k_{u}^G)(k_{v-1}^H-k_{v}^H) \\
	& = \sum_{u+v=j, u,v \geq 0} \1_{u \in G} \1_{v \in H} \leq \sum_{u+v=j, u,v \geq 0} \left( \1_{u \in G\cap[j-l+1,j]} + \1_{v \in H\cap[l,j]} \right) \\ & = (k_{j-l}^G-k_j^G) + (k_{l-1}^H-k_j^H),
\end{align*}
what finishes our proof.
\end{proof}

\section*{Acknowledgements}

We are grateful to Maciej Borodzik for his valuable comments about the meaning of our result.


TITLE: Is there list of composite Mersenne numbers with their factorization?
QUESTION [2 upvotes]: Here is a list of known Mersenne primes.
http://mathworld.wolfram.com/news/2009-06-07/mersenne-47/
I'm looking for a list of composite Mersenne numbers(when $p$ is prime $2^p-1$ isn't) with their factorizations for larger $p$.($p<43112609 $)

REPLY [3 votes]: The 'cunningham project' at this website lists factors for $2^n-1, \text{ odd }n<1300$
http://homes.cerias.purdue.edu/~ssw/cun/


TITLE: Translations of field operators in QFT
QUESTION [3 upvotes]: A question in the book QFT of Srednicki: This concerns the relativistic QFT generalization
$$\tag{2.21} {{e}^{-i\hat{P}x/\hbar}}\psi (0){{e}^{i\hat{P}x/\hbar}}~=~\psi (x)$$
of the formula
$$\tag{2.20} {{e}^{i\hat{H}t/\hbar}}\psi (\vec{x},0){{e}^{-i\hat{H}t/\hbar}}~=~\psi (\vec{x},t)$$
from non-relativistic QM.
How can we prove this two formula just use Poincare algebra? $\psi (x)$ is an field operator. About the second formula we can use Schrodinger equation to prove. I think there is a proof just using the Poincare algebra. I just have no idea about this.

REPLY [5 votes]: The formula $e^{-i\hat{P}a/\hbar}\hat\psi(x)e^{i\hat{P}a/\hbar}=\hat\psi(x+a)$ is just the finite transformation version of infinitesimal transformation version formula $[\hat{P},\hat{\psi}(x)]=i\hbar\partial_x\hat{\psi}(x)$ as mentioned by DaniH and Argopulos above. $e^{-i\hat{P}a/\hbar}$ is a finite transformation operator acting on the Hilbert space, while $x\rightarrow x+a$ is the finite transformation acting on the field space. The infinitesimal transformations or the generator of the Lie group is given by $\hat P$ and $i\hbar\partial_x$ accordingly.
Finite transformation can be derived from infinitesimal transformation as followed:
$e^{-i\hat{P}a/\hbar}\hat\psi(x)e^{i\hat{P}a/\hbar}=e^{-i(a/\hbar)[\hat{P},\cdot]}\hat\psi(x)=e^{-i(a/\hbar)(i\hbar\partial_x)}\hat\psi(x)=e^{a\partial_x}\hat\psi(x)=\hat\psi(x+a)$.


TITLE: Showing that $ \int_0^1 \frac{t^{x}-1}{\ln(t)}dt$ exists for all $x$
QUESTION [1 upvotes]: How do I show that $\displaystyle  \int_0^1 \frac{t^{x}-1}{\ln(t)}\,dt $ exists for every $ x \in \mathbb{R} $?
I have tried bounding it but without success...

REPLY [2 votes]: Notice that the function $f:t\mapsto \frac{t^x-1}{\ln t}$ can be extended by continuity on 1 since by the Hospital rule
$$
\lim_{t\to1}f(t)=\lim_{t\to1}\frac{xt^{x-1}}{1/t}=x,
$$
hence the integral $\int_{1/2}^1f(t)dt$ exists for all $x$. Now for the problem in 0 and by change of variable $u=-\ln t$ the problem becomes at $\infty$
$$\int_0^1 f(t)dt=\int_{0}^\infty\frac{e^{-u}-e^{-(x+1)u}}{u}du$$
so because the integral $\int_1^\infty \frac{e^{-u}}{u}du$ is convergent we see that the given integral converges iff $x+1>0$ i.e. $x>-1$.


TITLE: Rules for factoring out a and b from $a^k - b^k$? Not sure why two expressions are equivalent.
QUESTION [1 upvotes]: I'm working on an inductive proof, and I came across this line but I'm not sure how these are equivalent. Here is the link if more detail is needed. It's in box 2, first line.
$a^k - b^k = (a-b) \sum_{i=0}^{k-1}(a^{k-i-1}b^{i})$

REPLY [1 votes]: The other way (left to right):

First prove by induction the special case:
$$1-x^k=(1-x)\bigl(1+x+x^2+\dots+x^{k-1}\bigr).$$
The initial case ($k=1$) reduces to $\; 1-x=1-x$.
Suppose now the formula is established for some $k\ge 1$, and consider

So the inductive step is proved.

Generalisation: we may suppose $a\ne 0$, and set $x=\dfrac ba$. Note that fot each $i$, $\;a^ix^i = b^i$, so

\begin{align}
a^k-b^k=a^k\bigl(1-x^k\bigr)&=a^k(1-x)\bigl(1+x+x^2+\dots+x^{k-1}\bigr)\\
&=a(1-x)\,a^{k-1}\bigl(1+x+x^2+\dots+x^{k-1}\bigr) \\
&=(a-b)\bigl(a^{k-1}+a^{k-2}(ax)+a^{k-3}(a^2x^2)+\dots+a^{k-1}x^{k-1}\bigr)\\
&=(a-b)\bigl(a^{k-1}+a^{k-2}b+a^{k-3}b^2+\dots+b^{k-1}\bigr).\\
\end{align}


TITLE: Distributive nearring
QUESTION [8 upvotes]: A nearring is a ring-like structure $R$ such that

$(R, +, 0)$ is a group (possibly non-abelian)
$(R, \cdot)$ is a semigroup
$(x+y)\,z=xz+yz$ (right-distributivity)

In other words, three conditions that may fail differ it from an actual ring:

Addition may be non-commutative
There may be no multiplicative identity
Left-distributivity may fail

It's easy enough to give examples of nearrings for which arbitrary subsets of these 3 conditions nevertheless hold, except for (2)+(3) and pure (3). It is well-known that existence of identity and two-sided distributivity implies (1), so (2)+(3) without (1) is impossible. What about pure (3)?
Question: are there fully distributive nearrings with non-abelian additive subgroups?
Some thoughts:
Such nearrings are necessarily pretty weird. Adapting the usual proof that (2)+(3) implies (1), we can argue that
$$
a\,((x+y)-(y+x))=0
$$
for all $x,y,z\in R$. This seems a pretty strong condition, even though lack of multiplicative identity does not allow us to conclude immediately $x+y=y+x$.
Definition Let left-distributing element be $a\in R$ such that $a(x+y)=ax+ay$ for all $x, y\in R$
All nearrings are isomprhic to subsets of $M(G)$ - nearring of functions from group $G$ to itself with pointwise addition and composition as multiplication. Left-distributing elements are precisely endomorphisms. Still, set of endomorphisms does contain identity (likely minor obstacle), and for nonabelian $G$ is not a sub-nearring of $R$ (major problem). Product of left-distributing elements is left-distributing, but sum of left-distributing elements may fail to be, so the sub-nearring generated by endomorphisms in general most likely does not satisfy (3). Sub-nearring generated by endomorphisms mapping $G$ to $Z(G)$ would do the trick, but addition would be abelian.

REPLY [4 votes]: Consider the dihedral group $D_4=\langle a,b :a^4, b^2, (ab)^2\rangle$. Define $x+y=xy$ and $x\cdot y=[x,y]$. Claim: $(D_4,+,\cdot)$ is a distributive near-ring.
Recall commutator identities $[xy,z]=[x,z]^y[y,z]$ and $[z,xy]=[z,y][z,x]^y$. Note that the commutator subgroup coincides with the center. So we have $[xy,z]=[x,z][y,z]$ and $[z,xy]=[z,y][z,x]=[z,x][z,y]$. This translates to $(x+y)\cdot z=x\cdot z+y\cdot z$ and $z\cdot(x+y)=z\cdot x+z\cdot y$.
More generally, this works for any group of nilpotence class 2. For more info see this paper:  Heatherly, H. E. Distributive near-rings. Quart. J. Math. Oxford Ser. (2) 24 (1973), 63–70.


TITLE: Proof regarding complex roots.
QUESTION [0 upvotes]: The question is as follows:
Let $C \in\mathbb {Z^+.}$ Prove that the complex roots of the equation $X^2 + X + \sqrt{C}=0$ are quadratic $\iff $ $C$ is a square in $\mathbb {Z}$.
I have attempted this question for quite some time without getting anywhere, my notes do not seem to be assisting me. Any help would be greatly appreciated!

REPLY [0 votes]: Quadratic integers are roots of an equation $x^2+ax +b=0$ where $a$ and $b$ are integers.
If $C$ is an integer square the condition is automatically satisfied.
Conversely suppose that $X^2 + X + \sqrt{C}=0$ and $X^2+AX +b=0$. Subtracting we have $(A-1)X=\sqrt C-b $ and then $X$ would be real unless $A=1$ and $\sqrt C-b=0. $
Therefore $C=b^2$ as required.


TITLE: Let $(E,d)$ be a metric space. Then $X$ is second-countable if and only if $X$ is Lindelöf if and only if $X$ is separable
QUESTION [1 upvotes]: In proving every subspace of a separable metric space is separable, I need below result. Could you check if my proof is fine?

Theorem: Let $(E,d)$ be a metric space. Then $X$ is second-countable if and only if $X$ is Lindelöf if and only if $X$ is separable.

Proof: Let $X$ be second-countable, i.e., $X$ has a countable base $(O_i)_{i \in \mathbb N}$. Let $(C_j)_{j \in J}$ be an open cover of $X$. Then for each $j \in J$, there is $I_j \subseteq \mathbb N$ such that $C_j = \bigcup_{i \in I_j} O_i$. Let $I := \bigcup_{j \in J} I_j \subseteq \mathbb N$. For each $i \in I$, let $\varphi(i)$ be one of those $j$ such that $i \in I_j$. Then $(C_{\varphi(i)})_{i \in I}$ is a countable subcover of $(C_j)_{j \in J}$.
Let $X$ be Lindelöf. For each $n \in \mathbb N^+$, the collection of open balls $\mathcal U_n :=\{\mathbb B(x, 1/n) \mid x \in X\}$ is an open cover of $X$. Then for each $n \in \mathbb N^+$, there exists a countable subcover $\mathcal U_n' \subseteq \mathcal U_n$. Clearly, $\mathcal U := \bigcup_n \mathcal U'_n$ is a countable union of countable sets and thus countable. It's easy to verify that $\mathcal U$ is a dense countable subset of $X$.
Let $X$ be separable, i.e., $X$ has a countable dense subset $E \subseteq X$. For each $x \in X$ there is a sequence $(e_{x,n}, 1/n)_n$ in $E \times \mathbb N^*$ such that $x \in \mathbb B(e_{x, n}, 1/n)$. Let $\mathcal U$ be the collection of all balls of the form $\mathbb B(e_{x, n}, 1/n)$ where $x \in E$ and $n \in \mathbb N$. Clearly, $\mathcal U$ is countable. Let $O \subseteq X$ be open. For each $x \in O$, there is $r_x >0$ such that $\mathbb B(x, r_x) \subseteq O$. Let $n_x$ be the least $n$ such that $1/n < r_x / 2$. It follows that $x \in \mathbb B(e_{x, n_x}, 1/n_x) \subseteq O$. Hence $O = \bigcup_{x \in O} \mathbb B(e_{x, n_x}, 1/n_x)$. Hence $\mathcal U$ is a countable base of $X$.

REPLY [1 votes]: FYI. You can show directly that if $(X,d)$ is a separable metric space and $Y\subset X,$ then $Y$ is separable, as follows:
Let $E$ be a countable dense subset of $X$. For $n\in \Bbb N$ let $E_n=\{e\in E: B_d(e,1/n)\cap Y\ne \emptyset\}.$ And for $e\in E_n$ take $\psi_n(e)\in  B_d(e,1/n)\cap Y.$
Now let $D=\cup_{n\in \Bbb N}\{\psi_n(e):e\in E_n\}.$ Then $D$ is a countable subset of $Y$.
For $y\in Y$ and $r\in \Bbb R^+,$ we show that $B_d(y,r)\cap D\ne\emptyset$ as follows:
Take $n\in \Bbb N$ with $1/n<r/2.$  Now $E$ is dense in $X$ so take $e\in E\cap B_d(y,1/n).$. Then $e\in E_n$ because $y\in  B_d(e,1/n)\cap Y.$ So $$d(y,\psi_n(e))\le d(y,e)+d(e,\psi_n(e))<2/n<r.$$ So $\psi_n(e)\in B_d(y,r)\cap D$.


TITLE: Determinant of a matrix when its diagonal elements have a certain property
QUESTION [0 upvotes]: Let $A=\begin{pmatrix}
l_1&-1 &-1 &-1&-1 \\
-1&l_2 &-1 & -1 & -1\\
-1 & -1 &l_3 & -1 &-1\\
-1 & -1 & -1 & l_4 & -1\\
-1 & -1 &-1 & -1 &l_5
\end{pmatrix}$
Is there any delicate way to show that det$(A)<0$ when one of the $l_i$ equals to $0$ whereas others are positive?
Thank you for your help!

REPLY [1 votes]: I have got a very easy way to prove that with only the elementary transformation.
First every column $i$ minus column $1$ $(i=2,3,4,5)$, and we got
$\begin{pmatrix}
l1 & -1-l1 & -1-l1 & -1-l1 & -1-l1 \\
-1 & 1+l2 & 0 & 0 & 0\\
-1 & 0 & 1+l3& 0 & 0 \\
-1 & 0 & 0 & 1+l4 & 0 \\
-1 & 0 & 0 & 0 & 1+l5 \\
\end{pmatrix}$
Then column $1$ plus $\frac{1}{l_i-1}$ column $i$ $(i=2,3,4,5)$.
$\begin{pmatrix}
*& -1-l1 & -1-l1 & -1-l1 & -1-l1 \\
0 & 1+l2 & 0 & 0 & 0\\
0 & 0 & 1+l3& 0 & 0 \\
0 & 0 & 0 & 1+l4 & 0 \\
0 & 0 & 0 & 0 & 1+l5 \\
\end{pmatrix}$
Calculate the"$*$", we know that $*<0$.  $\Box$


text:
{\bf Problem.} What is the sum of all of the odd divisors of $180$?
{\bf Level.} Level 4
{\bf Type.} Number Theory
{\bf Solution.} First, we find the prime factorization of $180$ to be $2^2 \cdot 3^2 \cdot 5$. Note that the odd divisors of 180 are precisely the integers of the form $3^a5^b$ where $0\leq a \leq 2$ and $0\leq b\leq 1$. Note also that distributing $(1+3+9)(1+5)$ yields 6 terms, with each integer of the form $3^a5^b$ appearing exactly once.  It follows that the sum of the odd divisors of 180 is $(1+3+9)(1+5)=13 \cdot 6 = \boxed{78}$.

set_name: MATH


TITLE: property of the joint probability of two discrete random variables
QUESTION [1 upvotes]: I have two discrete random variables :
$X$ with probabilities $p_i$ with  $1 \le i \le N_X$
$Y$ with probabilities $q_j$ with  $1 \le j \le N_Y$
let $r_{ij}$ stands for the joint probability $Pr(X=i,Y=j)$
I want to compare the following :
$A=\sum_{ij}(r_{ij}^2)$
$B=\sum_{i}(p_{i}^2).\sum_{j}(q_{j}^2)$
From many tests with $N_X=N_Y=3$, I have only got $A \ge B$. But, I could not devise a proof.
I tried to use convexity of $f(x,y)=xy$ or $g(x)=x^2$

REPLY [1 votes]: It is provable using the convexity of $f(x,y)=xy$.
just write $r_{ij}^2=p_i h(j/i) q_j h(i/j)$ where h are conditional probabilities.
Then use Jensen's inequality with weights $p_iq_j$ (sums to 1)
$A \ge \sum_{ij}(p_iq_jh(i/j))\sum_{ij}(p_iq_jh(j/i))=B$


TITLE: Why a decaying dark matter?
QUESTION [0 upvotes]: I'm reading some papers (model, exp and model) that are dealing with decaying models of DM, why is it interesting? What are the signatures that this type of model can give to observe DM?
I am interested in general on the reasoning behind the choice of a decaying dark matter. Why is that interesting to study? Is there any smoking gun experiment that could detect decaying DM?

REPLY [1 votes]: It's also interesting because it alleviates the $\sigma_8$ tension in cosmology


TITLE: "independent trials" notation in Ross' book
QUESTION [3 upvotes]: In the book "Introduction to Probability Models", while explaining "independent trials", author Ross uses some numerical subscripts that I can not understand. I am quoting the paragraph, can you explain?

Suppose that a sequence of experiments, each of which results in either a "success" or a "failure," is to be performed. Let $E_i$, $i \geq 1$, denote the event that the $i$th experiment results in a success. If, for all $i_1,i_2,\ldots,i_n$ $$P(E_{i_1}E_{i_2} \cdots E_{i_n}) = \prod^n_{j=1} P(E_{i_j})$$

In other words, let $i$ be $2$. $E_2$ is the event second experiment results in a success. What is $2_1$, and what is $E_{2_1}$?
Thanks a lot.

REPLY [0 votes]: The $i$ and the $i_j$ are just dummy variables and have no relations to each other. In less ambiguous but more pedantic language, suppose we have events $E_1, E_2, \dots$. (I'm not sure if the author intends to have only finitely many events here, but it doesn't matter.) For any distinct indices $i_1, \dots, i_n$, we have
\begin{align*}
P(E_{i_1} \cdots E_{i_n}) = P(E_{i_1}) \cdots P(E_{i_n})
\end{align*}
In the other words, the author is saying that any finite subset of the $E_1, E_2, \dots$ is independent.

REPLY [0 votes]: This is really about notation, not about probability or independence.
Notice that it says for all.  Thus: "For all $i_1,\ldots,i_n$".
As a concrete example, one could have $n=6$ and
$$
i_1 = 23, \quad i_2 = 10, \quad i_3=5, \quad i_4=971, \quad i_5 = 154, \quad i_6=3.
$$
Then one would be asserting that
$$
P(E_{23} \, E_{10} \, E_{5} \, E_{971} \, E_{154} \, E_3) = P(E_{23})\,P(E_{10})\,P(E_5)\, P(E_{971}) \, P(E_{154}) \, P(E_3).
$$
The words for all mean that if you had picked any other sequence of values of the indices, or any other number besides $6$, the corresponding thing would still be true.