TITLE: Convergence of conditional measures for a convergent sequence of probabilities whose projection is constant
QUESTION [4 upvotes]: Setting
Suppose $\mu_n$ is a sequence of probability measures on $[0,1]\times [0,1]$ converging to a limit probability $\mu$ meaning that
$$ \lim_{n\to+\infty}\int f(x,y)d\mu_n(x,y) = \int f(x,y)d\mu(x,y)$$
for all continuous $f:[0,1]\times [0,1] \to \mathbb{R}$.
Suppose furthermore that all these probabilities project to the uniform measure on the first coordinate.  This implies there are Borel mappings (conditional probabilities) $x \mapsto \mu_{n,x}$ and $x \mapsto \mu_{x}$ from $[0,1]$ to the space of probabilities on $[0,1]$ satisfying
$$\int f(x,y) d\mu_n(x,y) = \int_0^1 \int_0^1 f(x,y)d\mu_{n,x}(y) dx,$$
and
$$\int f(x,y) d\mu(x,y) = \int_0^1 \int_0^1 f(x,y)d\mu_{x}(y) dx.$$
Question
I'm looking for a reference for the fact that
$\lim_{n\to+\infty}\mu_{n,x} = \mu_x$
for almost every $x \in [0,1]$.
More generally, I'm looking for some reference covering the situation when $\mu_n$ are probabilities on some compact space with constant push-forward under some continuous mapping of that space.
Proof
Here's a proof of the claim (I would still love to have a reference).
Take $f(x,y) = h(x)g(y)$ with $h$ and $g$ continuous and notice that
$$\lim_{n \to +\infty}\int_0^1  h(x)(\mu_{n,x}(g) - \mu_x(g)) dx = 0,$$
where we use $m(g)$ for the integral of $g$ with respect to the measure $m$.
Using $h$ to approximate the set $A_{\epsilon} = \lbrace x \in [0,1]: \liminf_{n \to +\infty} \mu_{n,x}(g) - \mu_x(g) > \epsilon\rbrace$ and Fatou's lemma (all functions are bounded)
$$\frac{\epsilon}{2}|A_{\epsilon}| \le \int_0^1 h(x)\liminf_{n \to +\infty}(\mu_{n,x}(g)-\mu_x(g)) dx \le \liminf_{n \to +\infty} \int_0^1 h(x)(\mu_{n,x}(g)-\mu_x(g)) dx  = 0,$$
where $|A|$ denotes the Lebesgue measure of $A$.   This shows that $A_\epsilon$ has measure $0$.
Since this holds for all $\epsilon > 0$ and also for the function $-g$ we get
$$\lim_{n \to +\infty}\mu_{n,x}(g) = \mu_x(g),$$
for almost every $x$.
Intersecting the full measure sets where this holds, over all $g$ in a countable dense set of continuous functions on $[0,1]$, the claim follows.

REPLY [4 votes]: A simple special case of Dan's answer above: Define  $f_n:[0,1] \to [0,1]$ by $f_n(x)= nx \mod 1$  and define $g_n:[0,1] \to [0,1]^2$ by $g_n(x)=(x,f_n(x))$. The pushforward $\mu_n=\lambda g_n^{-1}$ of Lebesgue measure $\lambda$ on $[0,1]$ is the uniform measure on the graph of $f_n$. The sequence $\mu_n$ converges weakly to Lebesgue measure $\mu$ on $[0,1]^2$
but for each $x$ in the unit interval, $\mu_{n,x}$ are Dirac measures that cannot converge weakly to $\mu_x=\lambda$.


text:
\begin{document}

\title{On the  Reidemeister torsion of rational homology spheres}

\author{Liviu I. Nicolaescu \\University of Notre Dame\\Notre Dame, IN 46556\\
http://www.nd.edu/ $\tilde{}$ lnicolae/ }

\date{June 2000}

\maketitle

\section*{Introduction}

In   the paper \cite{Tu5} V. Turaev  has   proved   a certain identity involving the Reidemeister torsion  of a rational homology   sphere. In this very short note  we will suitably interpret  this identity  as a  second order finite difference equation satisfied by the torsion which will allow  us to prove   a general structure result for the $\mod {\bZ}$ reduction of the torsion.   More precisely  we prove that the mod ${\bf Z}$ reduction of the torsion is completely determined by three data.

\medskip

\noindent $\bullet$ a certain canonical $spin^c$ structure,

\noindent $\bullet$ the linking form of the rational homology sphere and

\noindent $\bullet$ a constant $c\in {\bQ}/{\bZ}$.

\medskip

As a consequence, the constant $c$ is  a  ${\bQ}/{\bZ}$-valued invariant of the rational homology sphere. Experimentations with lens spaces suggest this invariant is as powerful as the torsion itself.

\tableofcontents

\section{The Reidemeister torsion}
\setcounter{equation}{0}

We review briefly a few basic facts about the Reidemeister torsion a  rational homology $3$-sphere. For more details and examples we refer to \cite{Nicol, Tu5}.

 Suppose $M$ is a rational homology sphere. We set $H:= H_1(M, {\bZ})$  and use the multiplicative notation to denote the  group operation on $H$. Denote $Spin^c(M)$ the  $H$-torsor of isomorphism classes of  $spin^c$ structure on $M$.          We denote by  ${\gF}$  the space of functions
\[
\phi: H\ra {\bQ}.
\]
The group $H$ acts on  ${\gF}_M$  by
\[
H\times {\gF}\ni (g,\phi)\mapsto g\cdot \phi
\]
where
\[
(g\cdot\phi)(h)= \phi(hg).
\]
We denote by $\int_H$ the augmentation map
\[
{\gF}_M\ra {\bQ},\;\;\int_H \phi:=\sum_{h\in H}\phi(h).
\]
According to \cite{Tu5} Reidemeister torsion  is a $H$-equivariant map
\[
\tau: Spin^c(M)\ra {\gF}_M,\;\;Spin^c(M)\si\mapsto \tau_\si\in {\gF}_M
\]
such that
\[
\int_H\tau_\si=0
\]
Denote by ${\bf lk}_M$ the linking form of $M$,
\[
{\bf lk}_M: H\times H\ra {\bQ}/{\bZ}.
\]
V. Turaev has proved in \cite{Tu5}  that  $\tau_\si$ satisfies the identity
\begin{equation}
\tau_\si(g_1g_2h)-\tau_\si(g_1h)-\tau_\si(g_2h)+\tau_\si(h)=-{\bf lk}_M(g_1,g_2)\mod {\bZ}
\label{eq: tur}
\end{equation}
$\forall g_1,g_2,h\in H,\;\si\in Spin^c(M)$.

\section{A second order ``differential equation''}
\setcounter{equation}{0}

The identity (\ref{eq: tur}) admits a more suggestive interpretation. To describe it we need a few more notation.

Denote by $\s$ the   space of functions $H\ra {\bQ}/{\bZ}$.   Each $g\in H$ defines a  first order differential operator
\[
\Delta_g:\s\ra\s,\;\;(\Delta_g u)(h):=u(gh)-u(h),\;\;\forall u\in \s,\;\;h\in H.
\]
If $\Xi=\Xi_\si$ denotes the mod ${\bZ}$ reduction of  $\tau_\si$ then  we can rewrite  (\ref{eq: tur}) as \begin{equation}
(\Delta_{g_1}\Delta_{g_2}\Xi)(h)=-{\bf lk}_M(g_1,g_2)
\label{eq: second}
\end{equation}

We will prove  uniqueness and existence results for this equation. We begin with the (almost) uniqueness part.

\begin{lemma}
The  second order linear  differential equation  (\ref{eq: second})  determines $\Xi$ up to an ``affine''  function.
\end{lemma}

\noindent {\bf Proof}\hspace{.3cm} Suppose $\Xi_1$, $\Xi_2$ are two solutions of  the above equation.  Set $\Psi:=\Xi_1-\Xi_2$.   $\Psi$ satisfies the  equation
\[
\Delta_{g_1}\Delta_{g_2}\Psi=0.
\]
Now observe  that any function $F\in \s$ satisfying the second order equation
\[
\Delta_u\Delta_vF=0,\;\;\forall u,v\in H
\]
is affine, i.e. it has the form
\[
F=c + \lambda
\]
where $c\in {\bQ}/{\bH}$ is a constant and $\lambda:H \ra {\bQ}/{\bZ}$ is a character. Indeed, the condition
\[
\Delta_u(\Delta_vF)=0,\;\;\forall u
\]
implies $\Delta_vF$ is a constant depending on $v$, $c(v)$. Thus
\[
F(vh)-F(h)=c(v),\;\;\forall h.
\]
The function $G= F-F(1)$ satisfies the same differential equation
\[
G(vh)-G(h)=c(v)
\]
and the additional condition $G(1)=0$.   If we set $h=1$ in the above equation we deduce
\[
G(v)=c(v).
\]
Hence
\[
G(vh)= G(h)+ G(v),\;\;\forall v, h
\]
 so that $G$ is a character and $F= F(1)+ G$.  Thus, the  differential equation (\ref{eq: second}) determines $\Xi$ up to a constant and a character. $\Box$

 \vspace{1cm}

\begin{lemma}  Suppose $b:H\times H\ra {\bQ}/{\bZ}$ is a nonsigular, symmetric bilinear form on $H$. Then  there exists a quadratic form $q: H\ra {\bQ}/{\bZ}$ such that
\[
\Delta q=b
\]
where
\[
(\Delta q)(uv):=q(uv)-q(u)-q(v).
\]
\end{lemma}

\noindent{\bf Proof}\footnote{We are indebted to Andrew Ranicki for suggesting this approach.}\hspace{.3cm}   Let us briefly recall the terminology in this lemma.   $b$ is nonsingular if the induced map $G\ra G^\sharp$ is an isomorphism. A quadratic map form is a function $q: H\ra {\bQ}/{\bZ}$ such that
\[
q(0)=0,\;\;q(u^k)=k^2q(u),\;\;\forall u\in H, \;k\in {\bZ}
\]
and $\Delta q$ is a bilinear  form.

Suppose $b$ is a nonsingular, symmetric , bilinear  form $H\times H\ra {\bQ}/{\Z}$.  Then, according to \cite[\S 7]{Wall}, $b$ admits a resolution. This  is a  nondegenerate,  symmetric,  bilinear form
\[
B: \Lambda\times \Lambda \ra {\bZ}
\]
on a free abelian group  $\Lambda$ such that,  the induced monomorphism $J_B:\Lambda\ra \Lambda^*$   is a resolution of $H$
\[
0\hra \Lambda\stackrel{J_B}{\Lra}\Lambda^*\stackrel{\pi}{\twoheadrightarrow }H\ra 0
\]
and  $b$ coincides with the induced bilinear form on $\Lambda^*/(J_B\Lambda)$ ($n:=\# H$)
\[
b(\pi(u), \pi(v))= \frac{1}{n^2}B(J_B^{-1}(nu), J_B^{-1}(nv))\mod {\bZ},\;\;\forall u, v\in \Lambda^*.
\]
 Now set
\[
q(\pi(u))=\frac{1}{2n^2}B(J_B^{-1}(nu),J_B^{-1}(nu))\;\;\mod{\bZ}
\]
It is clear that this quantity is well defined i.e.
\[
\frac{1}{2n^2}B(J_B^{-1}(nu),J_B^{-1}(nu))=\frac{1}{2n^2}B(J_B^{-1}(nv),J_B^{-1}(nv))\;\mod{\bZ}
\]
if $v= u+ J_B\lambda, \;\lambda\in \Lambda$. Clearly
\[
\Delta q= b.\;\;\;\Box
\]

\vspace{.7cm}

We deduce that  there exists a constant $c$, a character $\lambda: H\ra {\bQ}/{\bZ}$ and a quadratic form $q$ such that
\[
\Xi(h)=\Xi_\si (h)= c+\lambda(h) + q(h),\;\;\Delta q= {\bf lk}_M.
\]
In the above discussion the choice of the $spin^c$ structure $\si$ is tantamount to a choice of  an origin of $H$ which allowed us to identify the torsion of $M$ as a function $H\ra {\bQ}$. Once we make such a non-canonical choice, we have to replace  $\Xi$  with the family of translates
\[
\{\Xi_g(\bullet):=\Xi(g\bullet);\;\;g\in H\}
\]
In particular
\[
\Xi_g(h):=\Xi(gh)= c+\lambda(gh) + q(gh)= \underbrace{\Bigl(c+ \lambda(g) + q(g)\Bigr)}_{c(g)} + \underbrace{\Bigl(\lambda(h) + (\Delta q)(g,h)\Bigr)}_{\lambda_g(h)} + q(h)
\]
where $\lambda_g(\bullet)= \lambda(\bullet)+ {\bf lk}_M(g, \bullet)$. Since the linking from is nondegenerate  we can find  an {\em unique} $g$ such that $\lambda_g=0$.

We have proved the following  result.

\begin{proposition} Suppose $M$ is a  rational homology sphere. Then there exists an unique $spin^c$-structure $\si$ on $M$ so that, with respect to this  choice  the mod ${\bZ}$ reduction of $\tau_{M,\si}$
\[
\Xi(h):=\tau_{\si}(h)\;\mod {\bZ}
\]
has the form
\[
\Xi(h)= c+ q(h)
\]
where $c\in {\bQ}/{\bZ}$ is a constant  while  $q(u)$ is the unique quadratic form such that
\[
\ \Delta q =-{\bf lk}_M.
\]
In particular,
\[
\Xi(h)=\Xi(h^{-1})\;\mod{\bZ},
\]
and the constant $c\in {\bQ}/{\bZ}$ is a topological invariant of $M$.
\label{prop: new}
\end{proposition}

\section{Examples}
\setcounter{section}{0}

We want to show on  some simple examples that the invariant $c$ is nontrivial.

\medskip

(a) Suppose $M= L(8,3)$. Then its torsion is  (see \cite{Nico2})
 \[
{T_{8, \,3}} \sim  - {\displaystyle \frac {9}{32}} \,x^{7} -
{\displaystyle \frac {3}{32}} \,x^{6} - {\displaystyle \frac {9}{
32}} \,x^{5} + {\displaystyle \frac {5}{32}} \,x^{4} +
{\displaystyle \frac {7}{32}} \,x^{3} - {\displaystyle \frac {3}{
32}} \,x^{2} + {\displaystyle \frac {7}{32}} \,x +
{\displaystyle \frac {5}{32}}
\]
where $x^8=1$ is a generator of ${\bZ}^8$.  Then
\[
q(x^n) = \frac{-3k^2n^2}{16}
\]
The set of  possible  values $\frac{-3m^2}{16}$  mod ${\bZ}$ is
\[
A:=\{0,\;\; \frac{-3}{16},\;\;\frac{4}{16},\;\;\frac{5}{16}\}
\]
The set  possible values of $\Xi(h)$ is
\[
B:=\{-\frac{9}{32},\;\;-\frac{3}{32},\;\;\frac{5}{32},\;\;\frac{7}{32}\}.
\]
We need to find a constant $c\in {\bQ}/{\bZ}$ such that
\[
B-c=A.
\]
Equivalently, we need to figure out orderings $\{a_1, a_2,a_3,a_4\}$    and $\{b_1, b_2, b_3, b_4\}$ of $A$ and $B$ such that $b_i-a_i$ mod ${\bZ}$ is a constant independent of $i$. A little trial and error shows that
\[
\vec{A}=(0, -\frac{3}{16},\frac{4}{16},\frac{5}{16}),\;\; \vec{B}=(-\frac{3}{32},-\frac{9}{32}, \frac{5}{32},\frac{7}{32})
\]
and the constant is $c=-\frac{3}{32}$. This is the coefficient of $x^2$. We deduce that (modulo ${\bZ}$)
\[
F:=T_{8,3}(x)+\frac{3}{32}\sim - {\displaystyle \frac {3}{16}} \,x^{7} -
0\cdot x^{6} - {\displaystyle \frac {3}{16}} \,x^{5} + {\displaystyle \frac {1}{4}} \,x^{4} +
{\displaystyle \frac {1}{4}} \,x^{3} - 0\cdot x^{2} + {\displaystyle \frac {1}{4}} \,x +
{\displaystyle \frac {1}{4}}
\]
The translation of $F$ by $x^{-2}$ is
\[
x^{-2}(T_{8,3}+\frac{3}{32})=\frac{1}{4}x^7+\frac{1}{4}x^6-\frac{3}{16}x^5-\frac{3}{16}x^3 +\frac{1}{4}x^2 +\frac{1}{4}x.
\]

(b) Suppose $M=L(7,2)$. Then, its torsion is (see \cite{Nico2})
\[
{T_{7, \,2}} \sim  - {\displaystyle \frac {2}{7}} \,x^{6} +
{\displaystyle \frac {1}{7}} \,x^{5} + {\displaystyle \frac {2}{7
}} \,x^{3} + {\displaystyle \frac {1}{7}} \,x - {\displaystyle
\frac {2}{7}}
\]
where $x^7=1$ is a generator of ${\bZ}_7$. We see that  in this form $T_{7,2}$ is symmetric, i.e. the coefficient of $x^k$ is equal to the coefficient of $x^{6-k}$. The constant $c$ in this case must be the   coefficient of the middle  monomial $x^3$,  which is $\frac{2}{7}$.

(c) Suppose $M=L(7,1)$. Then
\[
{T_{7, \,1}} \sim {\displaystyle \frac {2}{7}} \,x^{6} +
{\displaystyle \frac {1}{7}} \,x^{5} - {\displaystyle \frac {1}{7
}} \,x^{4} - {\displaystyle \frac {4}{7}} \,x^{3} -
{\displaystyle \frac {1}{7}} \,x^{2} + {\displaystyle \frac {1}{7
}} \,x + {\displaystyle \frac {2}{7}}
\]
This is again a symmetric polynomial and   the coefficient of the middle monomial is $-4/7$.  We see that this invariant distinguishes the lens spaces $L(7,1)$, $L(7,2)$.

(d) For $M= L(9,2)$ we have
\[
 {T_{9, \,2}} \sim  - {\displaystyle \frac {10}{27}} \,x^{8} +
{\displaystyle \frac {2}{27}} \,x^{7} - {\displaystyle \frac {1}{
27}} \,x^{6} + {\displaystyle \frac {8}{27}} \,x^{5} +
{\displaystyle \frac {2}{27}} \,x^{4} + {\displaystyle \frac {8}{
27}} \,x^{3} - {\displaystyle \frac {1}{27}} \,x^{2} +
{\displaystyle \frac {2}{27}} \,x - {\displaystyle \frac {10}{27}}
\]
Again, this is a symmetric function, i.e the coefficient of $x^k$  is equal to the coefficient of $x^{8-k}$, $x^9=1$. The constant  is the coefficient of $x^5$, which is $2/27$. We deduce that, mod ${\bZ}$, we have
\[
T_{9,2}=-\frac{2}{3}x^8-\frac{2}{9}x^7-\frac{1}{3}x^6 -\frac{2}{9}x^7
\]

(e) Finally  when $M= L(9,7)$ we have
\[
{T_{9, \,7}} \sim  - {\displaystyle \frac {8}{27}} \,x^{8} -
{\displaystyle \frac {2}{27}} \,x^{7} + {\displaystyle \frac {10
}{27}} \,x^{6} + {\displaystyle \frac {1}{27}} \,x^{5} -
{\displaystyle \frac {2}{27}} \,x^{4} + {\displaystyle \frac {1}{
27}} \,x^{3} + {\displaystyle \frac {10}{27}} \,x^{2} -
{\displaystyle \frac {2}{27}} \,x - {\displaystyle \frac {8}{27}
}
\]
the  polynomial is again symmetric so that the constant $c$ is the coefficient of $x^4$ which is $-2/7$.

\bigskip

It would be very interesting to know whether the invariant $c$ satisfies  any surgery properties.  This is not a trivial issue because     we cannot  relate  the potential surgery properties of $c$ to the surgery properties of the torsion.  In the case of torsion the surgery formula involve finite difference operators which  kill the constants so $c$ will not appear in any of them.

\addcontentsline{toc}{section}{Bibliography}

config: arxiv


TITLE: $1$D Wave equation complete solution confusion
QUESTION [1 upvotes]: I have a question regarding the $1$D wave equation:
$$ \cfrac{\partial^2y}{\partial x^2} = \cfrac{1}{c^2} \cfrac{\partial^2y}{\partial t^2}$$
I have seen in several physics books that its complete solution is:
$$
y(x,t)=f_1(x-ct)+f_2(x+ct)
$$

I do not understand what $f_1$ and $f_2$ exactly are
i) Since, given some boundary conditions, the solution is an infinite sum of sine and cosine functions, shouldn't each one of those functions ($f_1$ and $f_2$) be an infinite sum of $g_n(x \pm ct)$ functions, for instance, accompanied by arbitrary $c_n$ constants ?
ii) I understand that $f_1$ and $f_2$ satisfy the PDE, but how do we know there should not be another $f_3( ... )$ term with another argument in the $y(x,t)$ expression?

REPLY [1 votes]: $f_1$ and $f_2$ are general expressions.  The point of the representation is that the two argument function of $(x,t)$ is the sum of two one argument functions of $(x-ct)$ and $(x+ct)$.


TITLE: Condition of closed embedding
QUESTION [2 upvotes]: One condition, (2) of Definition 25.4.1, for a morphism of ringed spaces $i:Z\rightarrow X$ to be a closed immersion is that
$$O_X \rightarrow i_*O_Z$$
is surjective.

I have two confusions

(a)
  $i^*O_X \rightarrow O_Z$ is surjective does this show $O_X \rightarrow i_*O_Z$ is surjective?  This doesn't seem to be the case for me. The counit of $i^*,i_*$ doens't seem to be special.
(b) I could prove $(i^*O_X)_x=O_{X,i(x)}$ but is it the case $(i_*O_Z)_{i(z)}=(O_Z)_{z}$?

I do not see why both cases have to be true, it would be nice if a counter example is provided too.

REPLY [1 votes]: First, as the pullback of the structure sheaf is the structure sheaf, the answer to (a) is that the natural map $i^*\mathcal{O}_X\to \mathcal{O}_Z$ is always an isomorphism for any morphism of schemes. So you're right to be suspicious here.
Second, $(i_*\mathcal{O}_Z)_{i(z)} = \mathcal{O}_{Z,z}$ is true. You can find this result on wikipedia or in most introductory algebraic geometry books. The point is that we can combine the definition $\Gamma(U,i_*\mathcal{F})=\Gamma(U\cap Z,\mathcal{F})$ for any open $U\subset X$ and any sheaf $\mathcal{F}$ on $Z$ with the fact that $Z$ has the subspace topology and every open set in $Z$ comes from the intersection of some open set in $X$ with $Z$.


TITLE: Calculate the sum of all the six-digit numbers you can build using $1,2,3,4,5,6$
QUESTION [0 upvotes]: Consider all the six-digit numbers you can form using the numbers $1,2,3,4,5,6 $(Each digit is included exactly once in each number). Calculate the sum of these numbers.
b) Calculate the sum of these numbers where each digit can be included several times in each number.
I haven't found a good way to calculate this. Especially b, I know there are $720$ numbers and $120$ of them have $1$ in unit digit, and I think $120$ has $6$ in unit digit? (I'm not sure if it's correct tho)

REPLY [0 votes]: For part b) again, the mean number is $\frac{111 \ 111+666 \ 666}{2}$. Since this is not an integer, and there are an even number of integers, the pairs of numbers will cancel out ($111\ 111$ and $666 \ 666$; $111\ 112$ and $666 \ 665$ etc.) and sum to twice the mean.
Therefore, the sum of all these numbers is $6^6 \cdot \frac{111 \ 111+666 \ 666}{2}$, which gives the same number as before.


TITLE: Solving surds without compairing
QUESTION [2 upvotes]: Question: Let $a + \sqrt{2b}  = 3 - 2\sqrt{2}$ .Find the value of $a - \sqrt{2b}$
What I did: I compared the whole numbers and the irrational numbers in both sides and calculated the answer $3 + 2\sqrt{2}$.
However, I am not very satisfied to do it this way. I could not manage to calculate it with the  algebraic formulas.
How to do it?

REPLY [4 votes]: Assuming that $a$ and $b$ are integer (or rational), we have that
$$a-3=2\sqrt2-\sqrt{2b}=\sqrt2(2-\sqrt b)$$
or
$$\sqrt b=2-\frac{a-3}{\sqrt2}$$
Squaring,
$$b=4-2\sqrt2(a-3)+\frac{(a-3)^2}2$$
That is,
$$2\sqrt2(a-3)=4-b+\frac{(a-3)^2}2$$
R.H.S. is rational. Thus, $a-3=0$.


TITLE: Proving matrix equation: $GM^\top(MGM^\top)^{-1}MG=G$
QUESTION [0 upvotes]: I want to show $$GM^\top(MGM^\top)^{-1}MG=G$$ where $MGM^\top$ is invertible, $G$ is a symmetric square matrix. $M$ does not have to be a square matrix. I am not 100% sure this is true (90% sure it is) so a contradictory proof will be great as well.

Thanks Justingpassby for the contradictory proof. I came into above from the following but I'm not sure where it is wrong?
Let $u = Mv$, $\text{var}(v)=G$, $u$ and $v$ be both Normal random variables with mean 0 so $\text{var}(u)=MGM^\top$.
Then $v=GM^\top(MGM^\top)^{-1}u$ is one possible solution of $u = Mv$ since
$$ Mv=MGM^\top(MGM^\top)^{-1}u = u.$$
Now $$\begin{array}\text{var}(v) &=& GM^\top(MGM^\top)^{-1}MGM^\top(MGM^\top)^{-1}MG\\
&=& GM^\top(MGM^\top)^{-1}MG
\end{array}$$
Therefore $GM^\top(MGM^\top)^{-1}MG=G$. But as shown this it not true for some non-square $M$ so there must be a hole in my logic somewhere above. Any idea anyone?

REPLY [0 votes]: Formally,
$$(MGM^\top)^{-1}= M^{\top(-1)}G^{-1}M^{-1}$$
So
$$\begin{align}
GM^\top(MGM^\top)^{-1}MG &= GM^\top M^{\top(-1)}G^{-1}M^{-1} MG \\
& = G
\end{align}$$
since everything else cancels.
This assumes that both $G$ and $M$ are invertible.


TITLE: Convergence of the series for the Dedekind zeta function
QUESTION [5 upvotes]: The Dedekind zeta function of an algebraic number field $K$ is defined as
$\zeta_K(s)\mathrel{\stackrel{\rm def}=} \sum\limits_{I \subset\mathcal O_K} \frac{1}{(N_{K/\mathbb Q}(I))^s}$, where $N_{K/\mathbb Q} (I)$ is the norm of the ideal $I$, i.e., the cardinality of $\mathcal O_K/I$. The first question I have about this function is why the series converges for $s \in \mathbb C$ such that ${\rm Re}(s)>1$ and diverges otherwise. To answer this question should I somehow describe ideals in algebraic number fields? Is there a general way to do this?

REPLY [3 votes]: The convergence for ${\rm Re}(s) > 1$ is discussed in textbooks on algebraic number theory, e.g., Borevich and Shafarevich's "Number Theory" or Lang's "Algebraic Number Theory." Have you looked in any books that discuss zeta-functions of number fields?  If not, where are you finding the definition of $\zeta_K(s)$?
To show the series converges when ${\rm Re}(s) > 1$, we can show first that it converges for real $s > 1$. Then it will converge absolutely for $s \in \mathbf C$ with real part greater than $1$, since $|1/{\rm N}(I)^s| = 1/{\rm N}(I)^{{\rm Re}(s)}$, and thus the series converges for $s$ with ${\rm Re}(s) > 1$.
When $s$ is real and greater than $1$, consider the Euler product $\prod_{{\mathfrak p}} 1/(1 - 1/{\rm N}(\mathfrak p)^s)$. Convergence of this product is equivalent to convergence of $\prod_{{\mathfrak p}} (1 - 1/{\rm N}(\mathfrak p)^s)$. A theorem about infinite products is that a product $\prod_{n \geq 1} (1 + x_n)$ with $|x_n| < 1$ converges absolutely (implying it can be expanded out into a convergent series by multiplying out all the terms) if the series $\sum_{n \geq 1} x_n$ converges.  Therefore we consider the series $\sum_{\mathfrak p} 1/{\rm N}(\mathfrak p)^s$. If $\mathfrak p$ divides $p\mathcal O_K$ then ${\rm N}(\mathfrak p) \geq p$ and there are at most $[K:\mathbf Q]$ primes $\mathfrak p$ dividing each $p\mathcal O_K$. Therefore
$$
\sum_{\mathfrak p} \frac{1}{{\rm N}(\mathfrak p)^s} \leq \sum_p \frac{[K:\mathbf Q]}{p^s} < \infty
$$
by convergence of the Dirichlet series for the Riemann zeta-function. This proves
$\prod_{\mathfrak p} 1/(1 - 1/{\rm N}(\mathfrak p)^s)$ converges and is equal to
$\sum_I 1/{\rm N}(I)^s$.
To show failure of convergence of $\sum 1/{\rm N}(I)^s$ with ${\rm Re}(s) \leq 1$, the case ${\rm Re}(s) < 1$ follows from divergence at $s = 1$ because a basic theorem about Dirichlet series says convergence at a point $s_0$ implies convergence at all $s$ with ${\rm Re}(s) > {\rm Re}(s_0)$. Therefore divergence at $s = 1$ implies divergence if ${\rm Re}(s) < 1$. The divergence at $s = 1$ is related to the existence of a pole at $s = 1$; see any textbook discussing the zeta-function of a number field.
Divergence if ${\rm Re}(s) = 1$ with $s \not= 1$ is more complicated. Have you ever seen a proof that $\sum_{n \geq 1} 1/n^s$ does not converge if ${\rm Re}(s) = 1$ and $s \not= 1$?


text:
\begin{document}

\title{Symplectic quasi-states on the quadric surface and Lagrangian submanifolds\\
}

\renewcommand{\thefootnote}{\alph{footnote}}

\author{\textsc Yakov Eliashberg

and \ Leonid
Polterovich

}

\date{June 12, 2010}

\maketitle

\begin{abstract} The quantum homology of the monotone complex
quadric surface splits into the sum of two fields. We outline a
proof of the following statement: The unities of these fields give
rise to distinct symplectic quasi-states defined by asymptotic
spectral invariants. In fact, these quasi-states turn out to be
``supported" on disjoint Lagrangian submanifolds. Our method
involves a spectral sequence which starts at homology of the loop
space of the 2-sphere and whose higher differentials are computed
via symplectic field theory, in particular with the help of the
Bourgeois-Oancea exact sequence.
\end{abstract}

\renewcommand{\thefootnote}{\arabic{footnote}}

\section{Introduction and main result}
\label{sec-intro}

The quantum homology of the monotone complex
quadric surface $S^2 \times S^2$ splits into the sum of two fields. The
unities of these fields give rise to symplectic
quasi-states defined by asymptotic spectral invariants (see \cite{EP-qst}).
One of these quasi-states is ``supported" on a Lagrangian sphere, the anti-diagonal \cite{EP-rigid}.
Our main finding (see Theorem \ref{thm-main} below) is that the second quasi-state
is ``supported" on the exotic monotone Lagrangian torus described in \cite{EP-rigid}  which is disjoint from the anti-diagonal. Thus these quasi-states are distinct, and the exotic torus
has strong symplectic rigidity properties. Let us pass to precise formulations.

Let $(W,\omega)$ be the standard symplectic quadric surface $S^2
\times S^2$, where both $S^2$ factors have equal areas $1$. Consider
the field $\cK = \C[[t^{-1},t]$ of Laurent series and the quantum
homology algebra $QH(W,\omega)= H_*(W,\C) \otimes \cK$ which is
graded by
$$\text{deg}(at^N) = \text{deg}(a) + 4N\;.$$ The graded component
$QH_4$ is an algebra over $\C$ with respect to the quantum product.
It splits into the sum of two fields generated by idempotents
$e_{\pm}= (1 \pm Pt)/2$, where $1=[W]$ is the fundamental class and
$P$ stands for the class of the point. Define the functionals
$\zeta_\pm : C^{\infty}(W) \to \R$ by
$$\zeta_{\pm}(H) = \lim_{E \to \infty} c(e_\pm,EH)/E\;,$$
where $c(\cdot,\cdot)$ is a spectral invariant
\cite{Oh-spectral,MS}.

It was shown in \cite{EP-qst} that these functionals are {\it
symplectic quasi-states}, that is they are monotone, linear on all
Poisson-commutative subspaces and normalized by $\zeta_\pm(1)=1$.
 Recall that a {\it quasi-measure} associated
to a quasi-state $\zeta$ is a set-function whose value on a closed
subset $X$  equals, roughly speaking, $\zeta(\chi_X)$ where $\chi_X$
is the indicator function of $X$. We write $\tau_\pm$ for the
quasi-measures associated to $\zeta_\pm$. The reader is referred to
\cite{EP-qmm,EP-qst} for further preliminaries.

We view $W$ as the symplectic cut \cite{Lerman} of the unit
cotangent bundle $\{|p| \leq 1\} \subset T^*S^2$, where the zero
section is identified with the Lagrangian anti-diagonal $L \subset
W$, and the length $|p|$ is understood with respect to the standard
spherical metric so that the length of the equator equals $1$. The
level $\{|p|=1/2\}$ contains unique (up to a Hamiltonian isotopy)
monotone Lagrangian torus denoted by $K$. Write $T \subset W$ for
the Lagrangian Clifford torus (the product of equators).

It has been proved in \cite[Example 1.20]{EP-rigid} that
$\tau_- (L)=1$ and $\tau_-(K)=0$. Together with the equality $\tau_-(T)=\tau_+(T)=1$
(see \cite{EP-qst}) this yields that $K$ and $T$ are not Hamiltonian isotopic.

\medskip
\noindent
\begin{thm}\label{thm-main} $\tau_+(L)=0, \tau_+(K) = 1$. In
particular, $\zeta_- \neq \zeta_+$.
\end{thm}

\medskip
\noindent Equality $\tau_+(K) =1$ yields that $K$ is
$e_+$-superheavy in the terminology of \cite{EP-rigid}. In
particular, it is non-displaceable and intersects every image of the
Clifford torus under a symplectomorphism.

The rest of the note contains a detailed outline of the proof of
Theorem \ref{thm-main}. Our method involves a spectral sequence
which starts at homology of the loop space of the 2-sphere and whose
higher differentials are computed via symplectic field theory. The
main technical ingredient comes from a paper by Bourgeois and Oancea
\cite{BO}.

Non-displaceability of the exotic torus $K$ has been recently
established via Lagrangian Floer homology by several independent
groups of researchers:  Fukaya, Oh, Ohta and Ono \cite{FO3},
Chekanov and Schlenk \cite{CH} and Wehrheim (unpublished). The paper
\cite{CH} presents new constructions of exotic Lagrangian tori in
the product of spheres. A related construction of exotic tori is
given by Biran and Cornea in \cite{BC} in the context of their study
of narrow Lagrangian submanifolds. It would be interesting to
understand whether these tori can be distinguished by appropriate
symplectic quasi-measures.

The paper \cite{FO3} contains a more general version of Theorem
\ref{thm-main}. According to \cite{FO3},  the exotic torus $K$ lies
in an infinite family of non-displaceable Lagrangian tori whose
Liouville class varies. Though our approach is quite different from
the one in \cite{FO3}, there are some similarities which deserve
further exploration. Let us mention also that the tori of the
above-mentioned family appear as invariant sets of a semitoric
integrable system (see \cite{PV} and Section \ref{sec-reduction}
below). In dimension four, semitoric means that one of the integrals
generates a Hamiltonian circle action. The study of this class of
integrable systems was initiated recently in \cite{PV}. Semitoric
systems have some amusing properties and appear in meaningful
physical models. It would be interesting to detect ``symplectically
rigid" invariant Lagrangian tori in other examples of semitoric
systems.

\section{Reduction to a Floer-homological
calculation}\label{sec-reduction}

We work with Floer homology with $\C$-coefficients. In our
conventions on the Conley-Zehnder indices, the PSS-isomorphism
identifies $FH_k$ with $QH_{k+2}$ (see \cite{MS} for preliminaries).

Throughout that paper, we denote by $\Sigma$ the diagonal in $W=S^2
\times S^2$. In our picture, $\Sigma$ is obtained from the
hypersurface $\{|p|=1\}$ by the symplectic cut \cite{Lerman}.

Fix $r \in (0;1/2)$, $E>0$ large enough, and $\epsilon >0$ small
enough. We assume that the data is ''non-resonant": $1/r \notin \Z$
and $(E+\epsilon)/\epsilon \notin \Z$.

Consider a Hamiltonian $H_E(|p|)$ on $W$ given by a piece-wise
linear function which equals $E$ on $U:= \{|p| \leq r-\epsilon\}$
and equals $-\epsilon$ on  $V:= \{|p|\geq r\}$.

We
refer e.g. to \cite{EKP} for a discussion on Floer-homological
calculations with piece-wise linear Hamiltonians.

Orbits of period $1$ of $H_E$ form several critical submanifolds,
which we are going to describe now. Each of these submanifolds is
equipped with a Morse function which is used for a small
perturbation of the action functional associated to $H_E$. In
addition, we fix capping discs for orbits from these submanifolds.
Critical points of these Morse functions together with the capping
discs give rise to generators of the Floer complex. Let us pass to
the precise description of this data.

\medskip
\noindent {\sc The maximum set $U$:} Here we have constant orbits
capped with the constant discs. We choose an exhausting Morse
function $f_U$ on $U$ with two critical points: a saddle point $x_0$
of Morse index $2$ and a maximum point $x_2$. Their Conley-Zehnder
indices are $0$ and $2$ respectively, and their actions equal $E$.
We refer to the elements of the Floer complex of the form $\gamma
t^{-N}$ for $\gamma \in \{x_0,x_2\}$ as {\bf $U$-generators}, and
call the number $N$ {\it the $t$-degree}.

\medskip
\noindent{\sc The minimum set $V$:} Here we have constant orbits
capped by the constant discs. We choose an exhausting Morse function
$f_V$ on $V$ with two critical points: a saddle point $y_0$ of Morse
index $2$ and a minimum point $y_{-2}$. Their Conley-Zehnder indices
are $0$ and $-2$ respectively, and their actions equal $-\epsilon$.
We refer to the elements of the Floer complex of the form $\gamma
t^{-N}$ for $\gamma \in \{y_0,y_{-2}\}$ as {\bf $V$-generators}.

\medskip
\noindent {\sc Non-constant orbits:} They form two series of
submanifolds diffeomorphic to $\R P^3$. We denote these submanifolds
by $Z^{\pm}_k$. Here the lower index $k$ corresponds to
$k$-times-covered simple closed geodesics on $L$, $Z^+$ stands for
the orbits on the submanifold $\{|p|= r-\epsilon\}$ and $Z^-$ stands
for the orbits on the submanifold $\{|p|= r\}$. Note that the
multiplicity $k$ satisfies inequalities $k \geq 1$ and
\begin{equation}\label{eq-k-bound}
k \leq \frac{E+\epsilon}{\epsilon}\;.
\end{equation}
In the discussion below we assume that $k \geq 1$ is arbitrary, and
that inequality \eqref{eq-k-bound} is an extra restriction which
selects orbits relevant for the Floer complex corresponding to the
fixed value of $\epsilon>0$. Next, we fix a Morse function
$f^{\pm}_k$ on $Z^{\pm}_k$ with critical points $\check{m}^{\pm}_k$,
$\hat{m}^{\pm}_k$, $\check{M}^{\pm}_k$ and $\hat{M}^{\pm}_k$ of
Morse indices $0,1,2,3$ respectively. It will be convenient for us
to choose $f_k$ in such a way that the orbits in each pair of orbits
$(\hat m_k^\pm,\check m_k^\pm)$ and $(\hat M_k^\pm,\check M_k^\pm)$
represent the same unparameterized orbit. The orbits from
$Z^{\pm}_k$ are capped by discs lying in $W \setminus \Sigma$. We
refer to the elements of the Floer complex of the form $\gamma
t^{-N}$ for $\gamma \in
\check{m}^{+}_k,\hat{m}^{+}_k,\check{M}^{+}_k, \hat{M}^{+}_k$ as
{\bf upper generators} and for $\gamma \in
\check{m}^{-}_k,\hat{m}^{-}_k,\check{M}^{-}_k, \hat{M}^{-}_k$ as
{\bf lower generators}, and call the number $N$ {\it the
$t$-degree}.

The Conley-Zehnder indices of the generators corresponding to the
non-constant orbits are given in Table 1 below.

\begin{table}[h]\label{tab:hresult}
\caption{Indices of upper/lower generators} \centering
\begin{tabular}{c rrrr}
\hline \hline & & & &\\
& $\check{m}_k$ & $\hat{m}_k$ & $\check{M}_k$ & $\hat{M}_k$  \\
upper &  $2k-1$ & $2k$ & $2k+1$ & $2k+2$ \\
lower & $2k-2$  & $2k-1$ & $2k$ & $2k+1$  \\
\hline
\end{tabular}
\end{table}

The actions of the lower generators equal $-\epsilon + kr$ and of
the upper generators $E+k(r-\epsilon)$.

\medskip\noindent In what follows we write $CZ(\gamma)$ for the Conley-Zehnder index of a
capped orbit $\gamma$ and $A(\gamma)$ for its action. We have
$$CZ(\gamma t^{-N}) = CZ(\gamma) -4N, \; A(\gamma t^{-N}) = A(\gamma) -N\;.$$

A direct calculation (crucially based on the fact that $r < 1/2$)
yields the following lemma which will be used throughout the paper.

\begin{lemma}\label{lem-0}
\begin{itemize}
\item[{(i)}] All lower generators and $V$-generators of the Conley-Zehnder indices $1,2,3$ have action $< 1$.
\item[{(ii)}]All upper generators and $U$-generators of the Conley-Zehnder
indices $1,2,3$ have action $< E+1$ and non-negative $t$-degree.
\item[{(iii}] Let $\gamma$ be an upper generator of
the Conley-Zehnder index $1,2,3$ and action $A(\gamma) > 1$. Then
inequality \eqref{eq-k-bound} holds automatically: $k <
\frac{E+\epsilon}{\epsilon}$.
\item[{(iv)}] There exist numbers $\mu_-(E) < \mu_+(E)$,
$\mu_\pm(E) \to \infty$ as $E \to \infty$ with the following
property: Let $\gamma$ be an upper generator or a $U$-generator of
the Conley-Zehnder index $1,2,3$ and $t$-degree $N$. Then $A(\gamma)
>1$ for  $N \leq \mu_-$ and $A(\gamma) < 1$ for $N > \mu_+ (E)$.
\end{itemize}
\end{lemma}

\begin{proof}

\medskip\noindent{\sc Lower generators:}
Take $\gamma_k \in
\{\check{m}_k^-,\hat{m}_k^-,\check{M}_k^-,\hat{M}_k^-\}$ and put
$\gamma = \gamma_k t^{-N}$. We have that $$CZ(\gamma) = 2k+j-4N \in
[1;3],\;\;j= -2,-1,0,1\;.$$ Thus $0 \leq 2k-4N \leq 4$. Note that
$$A(\gamma) = kr-\epsilon - N\;.$$ Since $r < 1/2$ we have that
$$A(\gamma) < (k/2-N) -\epsilon < 1\;.$$

\medskip\noindent{\sc Upper generators:}
Take $\gamma_k \in
\{\check{m}_k^+,\hat{m}_k^+,\check{M}_k^+,\hat{M}_k^+\}$ and put
$\gamma = \gamma_k t^{-N}$. We have that $$CZ(\gamma) = 2k+j-4N \in
[1;3],\;\;j= -1,0,1,2\;.$$ Thus
\begin{equation} \label{eq-kN}
0 \leq 2k-4N \leq 4\;.
\end{equation}
 In particular, $N \geq 0$ since $k \geq 1$.
Note that
\begin{equation}\label{eq-action}
A(\gamma) = E + kr-k\epsilon -N\;.
\end{equation}
Since $r < 1/2$ we have that
$$A(\gamma) < E + (k/2-N) -k\epsilon \leq E+1\;.$$

Put $\kappa =1-2r+2\epsilon$. Observe that by  \eqref{eq-kN} and
\eqref{eq-action} $A(\gamma) >1$ yields $k < 2E/\kappa <
(E+\epsilon)/\epsilon$ which proves statement (iii) of the lemma.

Further, pick $$\mu_-(E) <(E-1)/\kappa\;, \mu_+(E) >
(E-\kappa)/\kappa\;.$$ Statement (iv) of the lemma readily follows
from  \eqref{eq-kN} and \eqref{eq-action}.

\medskip
\noindent Finally, the only $U$-generator of the index $1,2,3$ is
$x_2$, and its action equals $E$ and its $t$-degree equals $0$. The
only $V$-generator of index  $2$ is $y_{-2}t$ and its action equals
$1-\epsilon$. This completes the proof of statements (i) and (ii) of
the lemma.
\end{proof}

\medskip \noindent \begin{lemma} [Main lemma] \label{lem-1}
$HF_2^{(1;E+1)}(H_E)=\C$.
\end{lemma}

\medskip
\noindent {\bf Proof of Theorem \ref{thm-main} modulo Main Lemma:}

\medskip
\noindent {\sc Step 1:}  Look at the diagram

\[\xymatrix { & HF_3^{(E+1,+\infty)}(H_E) \ar[d]^-{j} &  \\
 HF_2^{(-\infty,1)}(H_E) \ar[r]^-{k} \ar[rd]^-{m} &
 HF_2^{(-\infty,E+1)}(H_E) \ar[d]^-{i} \ar[r]^-{l} & HF_2^{(1,E+1)}(H_E)
 = \C \\ & HF_2^{(-\infty, \infty)} = QH_4(M) = \C^2 &
 }\]

Here the horizontal and the vertical lines are exact, and the
triangle is commutative. Since $e_{\pm}=(1 \pm Pt) /2$ and $\max H_E
= E$  the spectral invariants $c(e_{\pm},H_E)$ do not exceed $E+1$.
Thus, since $QH_4$ is generated by $e_-,e_+$ the map $i$ is onto. By
Lemma \ref{lem-0}(i),(ii) $HF_3^{(E+1,+\infty)}(H_E)=0$. This yields
that $j=0$ so $i$ is an isomorphism, and in particular $l$ has a
non-trivial kernel. Thus $k \neq 0$ and we conclude that $m\neq 0$.

Assume that some non-zero quantum homology class $a=\alpha e_- +
\beta e_+$, $\alpha,\beta \in \C$, lies in the image of $m$.  This
yields $c(a,H_E) \leq 1$. Since $\tau_-(L)=1$ (see \cite{EP-rigid}),
we have that $c(e_-,H_E) \geq E$, and therefore $\beta \neq 0$.
Observe that the quantum product $e_+ * e_-$ equals $0$, while $e_+
* e_+=e_+$, $ e_-
* e_- = e_-$. Thus, by the triangle inequality for spectral
invariants,
$$c(e_+,H_E) = c(a*e_+,H_E) \leq c(a,H_E) + c(e_+,0) \leq 2\;.$$
Since this holds for every $E$ and $r < 1/2$, we conclude that
\begin{equation}\label{eq-tube}
\tau_+(\{|p| < 1/2\})=0\;.
\end{equation}

\medskip
\noindent {\sc Step 2:} Observe now that the Hamiltonian $|p|^2$ on
$W\setminus \Sigma$ extends to a {\it smooth} Hamiltonian on the
whole $W$. This Hamiltonian is integrable and yields a foliation of
$W \setminus (L \cup \Sigma)$ by Lagrangian tori. Look at these tori
in the tube $\{|p| \geq 1/2\}$. One readily checks by an argument in
the spirit of \cite{M}, that all these tori besides the monotone
exotic torus $K$ are displaceable.

One can prove the displaceability directly in the following way.
Write $W=S^2 \times S^2$ as
$$\{x_1^2 +y_1^2 +z_1^2 =1\} \times  \{x_2^2 +y_2^2 +z_2^2 =1\} \subset \R^3 \times \R^3\;.$$
Introduce functions $F$ and $G$ on $W $ by
$$F(x_1,y_1,z_1,x_2,y_2,z_2)= z_1 + z_2\;,$$
$$G(x_1,y_1,z_1,x_2,y_2,z_2)= x_1x_2+y_1y_2+z_1z_2\;,$$
and a map
\begin{equation}
\label{eq-Phi} \Phi= (F,G) : W  \to \R^2\;.\end{equation} One can
check directly that within this model the Hamiltonian $|p|$
corresponds to $\sqrt{(1+G)/2}$. It defines an integrable
Hamiltonian system with an integral $F$ (since $F$ generates a
circle action, such an integrable system is {\it semitoric}
\cite{PV}). The above-mentioned Lagrangian tori are given by
$$N_{a,b}:= \Phi^{-1}(a,b)= \{z_1 +z_2 =a, x_1x_2+y_1y_2+z_1z_2=b\}\;.$$
The monotone torus $K$ is given by $N_{0,-1/2}$.

Note that for $a\neq 0$,  $N_{a,b}$ is displaceable
by
$$(x_1,y_1,z_1,x_2,y_2,z_2) \to (-x_1,y_1,-z_1, -x_2,y_2,-z_2)\;.$$
For $a=0$, the torus $N_{0,b}$, $b \in (-1/2,1)$ can be displaced
inside the hypersurface $\Pi := \{z_1 +z_2 =0\}$. Indeed, let
$\phi_i$ be the polar angle in the $(x_i,y_i)$-plane. Consider a
fibration $\tau: \Pi \to C:=(-1;1) \times S^1$ given by
$$(x_1,y_1,z_1,x_2,y_2,z_2) \to (z_1, \phi_1 -\phi_2)\;.$$
One readily checks that for every $(z,\theta) \in C$ the preimage
$\tau^{-1}(z,\theta)$ consists of a closed orbit of the Hamiltonian
$z_1 +z_2$. Thus for every simple closed curve $\alpha \subset C$,
the preimage $\tau^{-1}(\alpha)$ is a Lagrangian torus in $\Pi$.
Denote by $\sigma$ the push-forward to $C$ of the symplectic form
restricted to $\Pi$. Since the symplectic form on $W$ is given by
$$\frac{1}{4\pi} (dz_1 \wedge d\phi_1 + dz_2 \wedge d\phi_2)\;,$$
we have that $\sigma=(4\pi)^{-1} dz \wedge d\theta$. Furthermore,
$N_{0,b} = \tau^{-1}(\alpha_b)$ with $$\alpha_b = \Big{\{}z^2 =
\frac{\cos \theta -b}{cos \theta +1}\Big{\}}\;.$$ Observe that
$\alpha_b$ is a contractible simple closed curve in $C$. Integration
yields
$$\frac{1}{4\pi}\int_{\alpha_{-1/2}} z d\theta = \frac{1}{2} =  \frac{1}{2} \text{Area}_\sigma (C)\;.$$
For $b > -1/2$ the curve $\alpha_b$ lies inside the disc bounded by
$\alpha_{-1/2}$ in $C$. Thus $\alpha_b$  is displaceable in $C$, and
therefore, by lifting the displacing isotopy to $\Pi$, we get that
$N_{0,b}$ is displaceable in $\Pi$. This completes the proof of the
displaceablility.

\medskip
\noindent {\sc Step 3:} Consider the push-forward $ \Phi_* \tau_+$
of the quasi-measure $\tau_+$ to the plane $\R^2$ by the map $\Phi$
given by \eqref{eq-Phi}. Let $(u,v)$ be Euclidean coordinates on
$\R^2$. Since $F$ and $G$ Poisson commute, $ \Phi_* \tau_+$ extends
to a measure, say $\sigma$ on $\R^2$. Recall that in our model of
$W$ the function $|p|$ corresponds to $\sqrt{(1+G)/2}$, and hence
the tube $\{|p| < 1/2\}$ is given by $\{G < -1/2\}$. Formula
\eqref{eq-tube} above implies that the support of $\sigma$ lies in
$\{v \geq -1/2\}$.

Further, by Step 2, every non-empty fiber $\Phi^{-1}(a,b)$ with
$$ b \geq -1/2, (a,b) \neq (0,-1/2)$$
is displaceable in $W$. Recall \cite{EP-qst} that every
Floer-homological symplectic quasi-measure vanishes on displaceable
subsets, and hence a point $(a,b) \in \R^2$ cannot lie in the
support of $\sigma$ provided the set $\Phi^{-1}(a,b)$ is
displaceable. Therefore the support of $\sigma$ consists of the
single point $(0,-1/2)$, so that $\sigma$ is the Dirac
$\delta$-measure concentrated in this point. Since the torus $K$ is
given by $\Phi^{-1}(0,-1/2)$, we get that
$$\tau_+(K) = \sigma((0,-1/2))=1\;.$$
This completes the proof of the theorem. \qed

\section{Proof of the Main Lemma}

\medskip
\noindent {\sc The strategy of calculation:}

\noindent By Lemma \ref{lem-0}  The Floer complex
$CF_i^{(1;E+1)}(H_E)$, $i=1,2,3$ is generated by upper generators
and $U$-generators in the action window $(1;E+1)$ satisfying the
multiplicity bound \eqref{eq-k-bound}. The lower generators and
$V$-generators leave the stage. {\it Thus we shall suppress the
upper index $+$ and set $Z_k = Z_k^+$, $f_k:=f_k^+,
\check{m}_k=\check{m}_k^+$, etc.}

Denote by $B_n$, $n \geq 0$ the span over $\C$ of generators
$$\gamma \in \{x_0 ,x_2,\check{m}_{k}, \hat{m}_{k}, \check{M}_{k} ,
\hat{M}_{k}, k \geq 1\}$$ of the Conley-Zehnder index $n$.

Put $C_{i,s} = t^{-s}B_{i+4s}$, where $i\geq 0$ and $s \geq 0$.
Write $C_i = \oplus_s C_{i,s}$ and $C = C_1 \oplus C_2 \oplus C_3$.
Denote by $D \subset C$ the subspace consisting of the elements of
action $< 1$ and set $D_{i,s} = D \cap C_{i,s}$. By Lemma
\ref{lem-0}(i)-(iii) the Floer complex of $H_E$ in the action window
$(1;E+1)$ and in degrees $1,2,3$ is given by $(C/D, \delta)$, where
$\delta: C/D \to C/D$ is the Floer differential. The differential
$\delta$ has the form
\begin{equation}\label{eq-delta}
\delta = \delta_0 + t^{-1}\delta_1 + t^{-2} \delta_2+... \; \mod D
\end{equation}
with $\delta_l : B_* \to B_{*+4l-1}$. Let us emphasize that only
negative powers of $t$ appear in this expression: this follows from
the fact that Floer trajectories of $H_E$ are holomorphic near
$\Sigma$ and intersect it positively. With this notation,
\begin{equation}\label{eq-BO-0}
HF_2^{(1;E+1)}(H_E)  = H_2(C/D,\delta)\;.
\end{equation}

\medskip\noindent To motivate the strategy of calculation of this homology group,
identify
\begin{equation}\label{eq-filt}
CF_i^{(1;E+1)}(H_E)= CF_i^{(1-E;1)}(H_E-E)\;
\end{equation}
and look at  $C_{i,0}/D_{i,0}$ considered as a subspace of
$CF_i^{(1-E;1)}(H_E-E)$. With this identification homology of the
complex $(\oplus C_{i,0}/D_{i,0}, \delta_0)$ converge to symplectic
homology $SH(U')$ of the domain $U'=\{|p| < r\}$ in the action
window $(-\infty; 1)$ provided $E \to \infty$ and $\epsilon \to 0$.
Indeed, functions $H_E-E$ restricted to $U'$ form an exhausting
sequence used in the definition of symplectic homology, the complex
$\oplus C_{i,0}/D_{i,0}$ is generated by closed orbits of $H_E-E$
capped inside $U'$ and the differential $\delta_0$ counts the Floer
trajectories lying inside $U'$. The contribution of the lower
generators disappears in this limit.

Now we can formulate the intuitive idea behind our calculation: The
complex $(C/D,\delta)$ can be considered as a properly understood
deformation of $(\oplus C_{i,0}/D_{i,0}, \delta_0)$ which involves
capping discs and Floer trajectories intersecting $\Sigma$.
Eventually, the required homology $H(C/D,\delta)$ can be calculated
by an appropriate spectral sequence which starts at $SH(U')$.

To make this precise, we use the technology developed by Bourgeois
and Oancea \cite{BO} (and extended further in \cite {BEE})  who
identified symplectic homology of the Liouville domain $\{|p| < r\}$
with the homology of the complex $B: = \oplus B_i = \oplus_i
C_{i,0}$ equipped with certain differential $d_0$ which will be
described below. In fact we shall introduce an appropriate
deformation of their construction which takes into account the fact
that Floer cylinders can intersect $\Sigma$ and which will enable us
to calculate homology of the deformed complex $(C/D,\delta)$.

As a graded and filtered \footnote{ One should shift our filtration
by $E$ to get the standard filtration on symplectic homology used in
\cite{BO}, see \eqref{eq-filt}.} vector space the deformed
Bourgeois-Oancea complex is described as follows.
 Introduce the ring $\Lambda$
consisting of all Laurent series of the form
$$\sum_{s=0}^{+\infty} \lambda_s t^{-s}\;, \lambda_s \in \C \;.$$
With this notation the deformed Bourgeois-Oancea complex is given by
$QB:= B \otimes_{\C} \Lambda$. As before, this complex is graded by
$CZ(\gamma t^{-s}) = CZ(\gamma)-4s$ and filtered by the symplectic
action of $H_E$: $A(\gamma t^{-s}) = A(\gamma)-s$. Its differential
$d$ is $\Lambda$-linear and has the form
$$d = d_0 + t^{-1}d_1 + t^{-2}d_2 +...\;,$$
with $d_l : B_* \to B_{*+4l-1}$. It is instructive to view $(QB,d)$
as a ``quantum" deformation of the complex $(B,d_0)$ where $t$ plays
the role of a deformation parameter. By \cite{BO} the group
$H(B,d_0)$ coincides with symplectic homology of the Liouville
domain $\{|p| < r\} \subset T^*S^2$. The latter, according to
\cite{AS,SW}, equals homology of the free loop space of $S^2 = L$.
Therefore

\begin{equation}
\label{eq-loops} H(B,d_0) = H(\cL S^2)\;.
\end{equation}

In order to describe the differential $d$ we need some
preliminaries.

\medskip
\noindent{\sc Stretching-the-neck:}

\noindent Denote by $\pi: \nu \to \Sigma$ the holomorphic normal
line bundle to $\Sigma$ in $W= \C P^1 \times \C P^1$. Perform a
stretching-the-neck \cite{EGH,B-etc} of $W$ at the hypersurfaces
$\{|p|= r-\epsilon\}$ and $\{|p| = r\}$. The manifold $W$ splits
into three pieces which after gluing in (asymptotically) cylindrical
ends at their boundaries will be identified with $W_{left} := W
\setminus \Sigma$, $W_{middle}: = \nu \setminus \Sigma$ and
$W_{right}:= \nu$.

It would be convenient to view orbits from $Z_k$ as
$k$-times-covered unit circles of the bundle $\nu$. In particular,
the projection $\pi: \nu \to \Sigma$ gives rise to the natural map
$\pi^k: Z_k \to \Sigma$.

We shall assume that the exhausting Morse function $f_U$ is defined
on the whole $W_{left}$.

\medskip\noindent {\sc Matrix coefficient $(d_l\gamma_+,\gamma_-)$ for upper
generators $\gamma_+,\gamma_-$:}

\noindent Denote by $P_{a,b}$   the cylinder $\C \setminus {0}$ with
the set of {\it negative} interior punctures $a=\{a_1,...,a_{l_-}\}$
and the set of {\it positive} interior punctures $b=\{b_1, ...,
b_{l_+}\}$. Here we put $l_+ =l$ (recall that we are defining
$d_l$).

Let $\gamma$ be an orbit from $Z_k$. In our picture it is
interpreted in two different ways. First, it is a point of the
corresponding critical variety $Z_k$ (recall that upper index $+$ is
omitted). We denote this point by $A_\gamma$. Second, $\gamma$ is a
(parameterized, in general multiply covered) unit circle in the
fiber of the bundle $\nu$ over the projection $\pi(A_\gamma)$.

In what follows we work with holomorphic maps $$u : P_{a,b} \to
W_{middle}= \nu \setminus \Sigma \;.$$  We say that $u$ {\it enters}
$\gamma$ at a puncture $\eta \in a \cup \infty$ if $u(z)/|u(z)| \to
\gamma(\text{arg}(z))$ and $|u(z)| \to \infty$  as $z \to \eta$. We
say that $u$ {\it exits } $\gamma$ at a puncture $\eta \in b \cup 0$
if $u(z)/|u(z)| \to \gamma(\text{arg}(z))$ and $|u(z)| \to 0$  as $z
\to \eta$.

Suppose that $u$ exits $\alpha$ at $0$, enters $\beta$ at $\infty$ and
in addition exits (resp. enters) {\it some} orbits at all positive (resp. negative)
interior punctures. We shall denote this by
$$\xymatrix{\alpha \ar[r]^u & \beta}\;.$$ We shall also book-keep the quantity
$l_-$ by putting \begin{equation}\label{eq-weight}\text{weight}(u) =
2^{l_-}\;.\end{equation}

Note that geometrically such $u$'s either are contained in a single
fiber of $\nu$, or correspond to multi-sections of $\nu$ with zeroes
at $\pi(A_\alpha)$ and the projections of the asymptotic images of
the positive punctures, and with poles at $\pi(A_\beta)$ and the
projections of the asymptotic images of the negative punctures.

Suppose now that $\alpha,\beta$ belong to the same critical manifold
$Z_k$. We shall write
\begin{equation}
\label{eq-arrow} \xymatrix{\alpha \ar@{~>}[r]^v & \beta}\;
\end{equation}
 if $v$ is
a parameterized piece of trajectory of the negative gradient flow of
$f_k$ joining the points $A_{\alpha}$ and $A_{\beta}$.
\footnote{We will assume that the gradient vector field for each function $f_k$ satisfies the following    condition:
 the $1$-dimensional stable manifold    of  $\check  M_k$ (resp. the unstable manifold of $\hat m_k$)
 consists of orbits  which differ from $\check M_k$ (resp. from $\hat m_k$) only by their parameterization.}

Let $\gamma_+,\gamma_-$ be two upper generators representing
critical points of $f_{k_+},f_{k_-}$ on $Z_{k_+},Z_{k_-}$
respectively. Assume that $k_+ \neq k_-$. Consider all possible
configurations of the form
\begin{equation}\label{eq-BOconfig}
\xymatrix {\gamma_+ \ar@{~>}[r]^v & \alpha \ar[r]^u &  \beta
\ar@{~>}[r]^w & \gamma_- }\;. \end{equation} We call
$\text{weight}(u)$ the weight of this configuration. Note that the
right and/or the left arrow could be empty. In case $k_+ =k_-=k$, we
consider configurations of the form
\begin{equation}
\label{eq-arrow-1} \xymatrix{\gamma_+ \ar@{~>}[r]^v & \gamma_-}\;,
\end{equation}
and its weight is put to be $1$. Finally, we define
$(d_l\gamma_+,\gamma_-)$ as the sum of weights taken over the
$0$-dimensional components of the moduli spaces of configurations of
the form \eqref{eq-BOconfig} and  \eqref{eq-arrow-1}. Note that in
the definition of the moduli spaces \eqref{eq-BOconfig} the markers
$a,b$ are varying as well. In addition, each weight should be taken
with a sign responsible for the orientation of the moduli space. The
orientation issue will be ignored in this note.

\bigskip
\noindent

\noindent {\sc Matrix coefficients $(d_l x, \gamma )$ and $(d_l
\gamma,x)$ for an upper orbit $\gamma$ and an $U$-orbit $x$:}

\noindent First, note that all the coefficients $(d_l \gamma, x)$
vanish for $l > 0$ by index reasons, so we do not need to describe
here the algorithm for their computing. For the description of
$(d_0\gamma, x)$ we refer to \cite{BEE}.

Let us describe the algorithm for computing of coefficients $(d_l x,
\gamma )$. Suppose that
 the multiplicity of the orbit $\gamma$ is equal to $k$.
 Then the coefficient $(d_lx,\gamma)$ counts rigid configuration
\begin{equation}
\label{eq-guc} (g,u,c)\; ,
\end{equation}
 where
\begin{itemize} \item
$g$ is a  minus gradient trajectory of  the function $f_U$ which
begins
 at $x$ and ends at a point $p\in \p U$; note that $\p U$ can be canonically
  identified with the $S^1$-bundle  associated with the complex line
   bundle $\nu$, and thus $p$ determines a ray $l_p$ in one of the fibers of $\nu$;
\item $u:P_{a,b}\to W_{middle}$ a holomorphic map such that $u$ exits
 (resp. enters)  some orbits at  all positive (resp. negative) punctures,
  enters  an orbit $\wt \gamma$ of multiplicity $k$ at $\infty$
   and $\lim\limits_{z\to 0} u(z) \in l_p$; note that the set of
   positive
    punctures must be non-empty due to the maximum principle;
\item  $c$ is a minus gradient trajectory of the function $f_k$
 connecting $\wt \gamma$ and $\gamma$. \end{itemize}

   A new feature of
     configurations $(g,u,c)$ considered above is the ray $l_p$
which connects the holomorphic curve $u$ with the gradient
trajectory $g$. This requires a justification which will be given
elsewhere.

\bigskip
\noindent {\it This completes the description of the differential
$d$ on $QB$.}

\bigskip

\medskip
\noindent {\sc Comparison of Floer and Bourgeois-Oancea homologies:}

\noindent Recall that $D$ denotes the subspace of $C$ consisting of
elements of action $< 1$. Denote by $QD \subset QB$ the subspace
consisting of elements of action $< 1$. We shall use the following
equality:
\begin{equation}\label{eq-BO}
H_2(C/D, \delta) = H_2(QB/QD,d)\;.
\end{equation}

Note that for $i=1,2,3$ we have $C_i = QB_i = \oplus _s B_{i+4s}
\otimes_\C t^{-s}$, and $D_i = QD_i = QB_i \cap D$. Let us compare
the differentials. The stretching-the-neck procedure described above
has the following effect on the original Floer trajectories of our
Hamiltonian $H_E$: every Floer trajectory joining a pair of upper
generators $\gamma_+$ and $\gamma_-$ splits into three pieces. The
piece lying in $W_{middle} = \nu \setminus \Sigma$ is the Floer
trajectory joining $\gamma_+$ and $\gamma_-$ with positive punctures
(corresponding to the intersection points with $\Sigma$) and
possibly some negative punctures. The orbit $\gamma_+$ lies on the
connected component of the ideal boundary of $W_{middle}$ adjacent
to $W_{right}$, while the orbit $\gamma_-$ lies on the connected
component of the ideal boundary of $W_{middle}$ adjacent to
$W_{left}$. The positive punctures are capped by rigid holomorphic
planes (the fibers of $\nu$) lying in $W_{right}$.  The negative
punctures are capped by rigid holomorphic planes lying in $W_{left}$
corresponding to $\C P^1 \times \text{point}$ and $\text{point}
\times \C P^1$ in $\C P^1 \times \C P^1$.  Thus every negative
puncture is capped by exactly $2$ rigid planes, which yields the
factor $2^{l_-}$ in the definition of the weight in
\eqref{eq-weight}.

Let's focus on the piece lying in $W_{middle}$: one extends
Bourgeois-Oancea theory \cite[Prop. 4]{BO} and finds an isomorphism
between homology whose differential is determined by such punctured
Floer trajectories and the homology whose differential is described
by configurations of the form \eqref{eq-BOconfig} and
\eqref{eq-arrow-1}. This explains equality \eqref{eq-BO}. The formal
proof will be given in a forthcoming paper.

\medskip
\noindent\begin{rem}{\rm From the viewpoint of the Hamiltonian Floer
theory, the differential $\delta: C/D \to C/D$ does not lifts
canonically to a differential $C \to C$: the square of the
expression in the right hand side of formula \eqref{eq-delta}
vanishes modulo the subcomplex $D$. Since $d^2=0$, the argument
above shows that in the limit, ``when the neck is stretched", the
square of this expression vanishes in $C$ itself.}
\end{rem}

\medskip
\noindent {\sc Unperturbed differential:}

\noindent The explicit form of the differential $d_0$ is folkloric
(private communications of F.Bourgeois and T.Ekholm). We shall
present the result right now.

\medskip\noindent
\begin{lemma}\label{lem-folk}
We have
$$d_0\hat{m}_k = 0, d_0\hat{M}_k=0\;, k \geq 1\;,$$
$$ d_0\check{m}_k = 2\hat{m}_{k-1}+2\hat{M}_{k-2}, k \geq 3\;,$$
$$ d_0 \check{M}_k = 2\hat{m}_k + 2\hat{M}_{k-1}, k \geq 2\;,$$
$$ d_0\check{m}_2 = 2\hat{m}_{1}+2 x_2\;,$$
$$ d_0 \check{M}_1 = 2\hat{m}_1 + 2x_2\;,$$
$$d_0 \check{m}_1 =0\;,$$
$$d_0 x_0 =0, d_0 x_2 = 0\;.$$
\end{lemma}

\medskip
\noindent It follows that the homology of the complex $(B,d_0)$ are
given by
$$H_0(B,d_0) = \text{Span}_{\C} ([x_0]), H_2(B,d_0) = \text{Span}_{\C}
([x_2])\;,$$ $$H_{2k+2}(B,d_0) = \text{Span}_{\C} ([\hat{M_k}])\;,
 k \geq 1,$$
$$H_1(B,d_0)= \text{Span}_{\C} ([\check{m}_1])\;,$$
$$H_{2k+1}(B,d_0)= \text{Span}_{\C} ([\check{M}_k-\check{m}_{k+1}])\;,  k \geq 1.$$

\medskip\noindent
It remains to describe the differential $d_l$ for $l \geq 1$.

\medskip
\noindent {\sc Calculation of the quantum corrections:}

\medskip
\noindent \begin{lemma}\label{lem-2}
\begin{itemize}
\item[{(i)}] For all $l \geq 2$ one has $d_l = 0$;
\item[{(ii)}] For all $k \geq 1$,
$$d_1 x_0 = d_1 x_2 = 0$$
$$d_1 \check{m}_k = \hat{m}_{k+1},\;
d_1\check{M}_k=\hat{M}_{k+1},\;d_1 \hat{m}_k=0, d_1 \hat{M}_k=0\;.$$
\end{itemize}
\end{lemma}

\medskip
\noindent {\bf Proof of Main Lemma:}

1) Define a decreasing filtration $\dots \supset QB^{(\mu)} \supset
QB^{(\mu+1)} \supset \dots $ on $QB$ by
$$QB_*^{(\mu)} = \oplus_{s \geq \mu+1}C_{*,s} = \oplus_{s \geq \mu+1}t^{-s}B_{*+4s}\;,$$
and observe that the differential $d$ preserves the filtration.
Furthermore, by Lemma \ref{lem-0}(iv) for $E$ large enough the
subspace $QD_i \subset QB_i$, $i=1,2,3$ consisting of elements of
the $H_E$-action $< 1$ is squeezed between $QB_i^{(\mu_+)}$ and
$QB_i^{(\mu_-)}$ with $\mu_\pm(E) \to \infty$ as $E \to \infty$:
$$QB_i^{(\mu_-)} \supset QD_i \supset QB_i^{(\mu_+)}\;.$$
We shall show in the next step that for $\mu$ large enough
$H_2(QB/QB^{(\mu)},d)$ is isomorphic to $\C$ and is generated by
$x_2 \mod QB^{(\mu)} $. The latter implies that the map
$H_2(QB/QB^{(\mu_+)},d) \to  H_2(QB/QB^{(\mu_-)},d)$ is an
isomorphism. Since it factors through $H_2(QB/QD,d)$ we shall
conclude that $H_2(QB/QD,d)=\C$.

2) Lemma \ref{lem-2} yields that $d_1: H_{2k}(B,d_0) \to
H_{2k+3}(B,d_0)$ vanishes while $d_1: H_{2k+1}(B,d_0) \to
H_{2k+4}(B,d_0)$ is onto for all $k \geq 0$. Fix $\mu$ large enough
and identify $\overline{QB}:= QB/QB^{(\mu)}$ with $\oplus_{s \leq
\mu} C_{*,s}$.

We claim that
$$H_2(\overline{QB},d) = H_2(B,d_0) = \C\;.$$
Indeed, consider a filtration $\cF_p \overline{QB}_*:= \oplus_{\mu
\geq s \geq \mu-p} C_{*,s}$ on the complex $\overline{QB}$. Look at
the homology spectral sequence corresponding to this filtration
\cite{Spanier}: we have that
$$E^1_j = \oplus_{s \leq \mu} H_j(C_{*,s}, d_0) = \oplus_{s \leq \mu}
t^{-s}H_{j+4s}(B,d_0)$$ and the differential $d^1: E^1_j \to
E^1_{j-1}$  can be written as
$$d^1= \oplus d^1_s,\;\, d^1_s = [t^{-1}d_1]: H_j( C_{*,s}, d_0) \to H_{j-1}(C_{*,s+1}, d_0)\;, s \leq \mu-1\;.$$
Since $d=d_0 + t^{-1}d_1$, the spectral sequence degenerates at the
second page and converges to $E^2_j := H_j(E^1, d^1)$, and in
particular $H_2(\overline{QB},d) = E^2_2$. In order to calculate
$E^2_2$ look at the piece
\begin{equation} \label{eq-piece}
H_3(C_{*,s-1}, d_0) \to H_2(C_{*,s}, d_0) \to H_1(C_{*,s+1}, d_0)
\end{equation}
of the complex $(E^1, d^1)$. If $\mu-1 \geq s \geq 1$, the left arrow is onto,
while the right arrow is zero, thus the homology vanishes.
If $s=\mu$, the sequence \eqref{eq-piece} degenerates to
$$H_3(C_{*,\mu-1}, d_0) \to H_2(C_{*,\mu}, d_0) \to 0\;,$$
and since the left arrow is onto, the homology vanishes. Finally,
for $s=1$ the sequence \eqref{eq-piece} degenerates to
$$0 \to H_2(C_{*,0}, d_0) \to H_1(C_{*,1}, d_0)\;.$$
Since the right arrow vanishes, the resulting homology is
$$H_2(C_{*,0},d_0) = H_2(B,d_0)= \C\;.$$ We conclude that $H_2(\overline{QB},d) = E^2_2
=\C$, and the generator of $H_2(\overline{QB},d)$ is $x_2 \mod
QB^{(\mu)}$, as required.

\medskip\noindent
It remains to put all the pieces together: The calculation above
together with equalities \eqref{eq-BO-0} and \eqref{eq-BO} yield

$$ HF_2^{(1;E+1)}(H_E)  = H_2(C/D,\delta)=  H_2(QB/QD,d)= H_2(B,d_0)=\C\;.$$
This completes the proof.
\qed

\medskip
\noindent
{\bf Proof of Lemma \ref{lem-2}:}

1) First we explore coefficients of the form
$(d_l \gamma_+,\gamma_-)$ with $l=l_+ \geq 1$,
 $$\gamma_\pm \in \{\check{m}_{k_\pm},
\hat{m}_{k_\pm}, \check{M}_{k_\pm} , \hat{M}_{k_\pm}\}\;.$$
 The index formula reads
$$CZ(\gamma_+)-CZ(\gamma_-) + 4l_+ = 1\;.$$
Recall that $$CZ(\gamma_\pm) = 2k_\pm + j_\pm, \; j_\pm =
-1,0,1,2\;.$$ Put $h=j_+ -j_-$. Thus the index formula yields
\begin{equation}\label{eq-index}
2(k_+-k_-) + h +4l_+ =1\;.
\end{equation}

Consider a configuration of the form
$$\xymatrix {\gamma_+ \ar@{~>}[r]^v & \alpha \ar[r]^u &  \beta
\ar@{~>}[r]^w & \gamma_- }\;.$$ Denote by $\Delta$ the degree of the
projection of $u$ to $\Sigma$. Since the Chern class of $\nu$ equals
$2$ we have
\begin{equation}\label{eq-RR}
(k_+ + l_+) - (k_-+l_-) = 2\Delta\;.
\end{equation}

It follows from \eqref{eq-index} and \eqref{eq-RR} that
$$l_+ + l_- = (1-h)/2-2\Delta\;.$$
Since $l_+ \geq 1$, $\Delta \geq 0$ and $h$ is an integer from
$[-3;3]$ we conclude that $\Delta=0$. Thus our holomorphic curve
lies in the single fiber of the bundle $\nu$.

We claim that $h \neq -3$. Indeed otherwise we have the connecting
trajectory of the form
\begin{equation}\label{eq-mM}
\xymatrix {\check{m}_{k_+}  \ar@{~>}[r]^v & \alpha \ar[r]^u & \beta
\ar@{~>}[r]^w & \hat{M}_{k_-} } \;. \end{equation} Since
$\check{m}_{k_+}$ is the point of minimum of $f_{k_+}$, $v$ is the
constant trajectory. Since $\hat{M}_{k_-}$ is the point of maximum
of $f_{k_-}$, $w$ is the constant trajectory. Since $u$ lies in the
single fiber, the set of limit points of $u(z)/|u(z)|$ as $z \to
\infty$ lie on a circle which is the fiber of $Z_{k_{-}} =\R P^3$
over $\pi^{k_+}(\check{m}_{k_+})$. Generically, as the dimension
count shows, this circle does not pass through $\hat{M}_{k_-}$, and
hence configuration \eqref{eq-mM} does not exist. The claim follows.

Since $h \neq -3$, we get that
$$h=-1, l_+=1, l_- =0\;.$$
This readily yields that $d_l \gamma_+ = 0$ for $l \geq 2$ and the
only possible non-trivial matrix coefficients could be  (with
$k_+:=k$) $(d_1 \check{m}_k,\hat{m}_{k+1})$, $(d_1
\check{M_k},\hat{M}_{k+1})$ and $(d_1\hat{m}_k,\check{M}_{k+1})$. We
claim that $$(d_1 \check{m}_k,\hat{m}_{k+1})=(d_1
\check{M_k},\hat{M}_{k+1})=1\;.$$

Let us present a calculation (modulo orientations) for the
coefficient $(d_1 \check{m}_k,\hat{m}_{k+1})$ (the calculation for
$(d_1 \check{M_k},\hat{M}_{k+1})$ is analogous). Since $\Delta=0$
and $l_+=1$, we work with holomorphic maps $u$ of the cylinder
$P_{\{b\}}$ punctured at a point $b \in \C \setminus 0$ lying in the
single fiber of $\nu$. Choose functions $f_k$ and $f_{k+1}$ on $Z_k$
and $Z_{k+1}$ respectively so that
$$\pi^k(\check{m}_k)=\pi^{k+1}(\hat{m}_{k+1}) := m \in \Sigma$$ and so that the  circle
$(\pi^{k+1})^{-1}(m) \subset Z_{k+1}$ forms the unstable manifold
$\mathcal{U}$ of $\hat{m}_{k+1}$. Identify the fiber of the line
bundle $\nu$ over $m$ with $\C$. Recall that we identified each
$Z_j$ with the unit circle bundle of $\nu$. Assume that
$\check{m}_k$ corresponds to the point $-1 \in \C$ and
$\hat{m}_{k+1}$ corresponds to the point $1 \in \C$. Since
$\check{m}_k$ is the minimum point of $f_k$, the only gradient
trajectories exiting $\check{m}_k$ are the constant ones. Thus we
are counting configurations of the form
\begin{equation}\label{eq-mm}
\xymatrix {\check{m}_{k} \ar[r]^u &  \beta \ar@{~>}[r]^w &
\hat{m}_{k+1} } \;. \end{equation} Note that the point $A_\beta$
lies both on the circle $\mathcal{U}= (\pi^{k+1})^{-1}(m)$ (since
the image of $u$ is contained in the single fiber) and on the stable
manifold of $\hat{m}_{k+1}$. These two submanifolds intersect
transversally at a single point, $\hat{m}_{k+1}$. In particular, $w$
is constant. Therefore it suffices to show that the holomorphic map
$u$ is unique up to a reparameterization and up to multiplication by
constants (these symmetries should be taken into account when one
passes to the moduli space).

The general form of $u$ is $u(z) = \lambda z^k (z-b)$ with the
asymptotic conditions
$$\text{Arg}(u(t)) \to 0 \;,\; t \in \R_+, t \to +\infty$$
and
$$\text{Arg}(u(t)) \to \pi \;,\; t \in \R_+, t \to 0\;.$$
This readily yields $\lambda, b \in \R_+$. The change of variables
$z \to cz$ with $c \in \R_+$ takes $u$ to the form $u(z) = \lambda
c^{k+1}z^k(z-b/c)$. Thus $u$, up to a reparameterization and up to
multiplication by constants coincides with $z^k(z-1)$. Thus we have
the unique configuration of the form \eqref{eq-mm}, which completes
the calculation.

\medskip

Finally, we claim that $(d_1\hat{m}_k,\check{M}_{k+1})=0$. Indeed,
assume that $d_1\hat{m}_k = n\check{M}_{k+1}$. Observe that since
$d^2=0$ we have that $d_1d_0+d_0d_1 =0$. Thus (in view of the
explicit formulas for the unperturbed differential)
$$d_1d_0\hat{m}_k + d_0d_1\hat{m}_k = 0 + n
d_0\check{M}_{k+1}=0\;.$$ Since $d_0\check{M}_{k+1}\neq 0$ we
conclude that $n=0$ as claimed.

\medskip

2) Now we turn to calculation of $d_l x$ where $x \in \{x_0,x_2\}$
and $l \geq 1$. Let us observe  that  if  $(d_lx,\gamma)\neq 0$ then
the orbit $\gamma$ must have an odd grading, and hence
$\gamma=\check m_k$, or $\gamma=\check M_k$.
 But in this case  if there exists a gradient  trajectory $c$ connecting an orbit $\wt \gamma$ with $\gamma$, and  if
$\wt\gamma_\alpha$ differs  from $\wt\gamma$    by a
reparameterization $s\mapsto se^{i\alpha}$,
 then for almost all values $\alpha$ there exists a gradient trajectory $c_\alpha$ connecting  $\wt\gamma_\alpha $
    with $\gamma$.
  This implies that there  are no rigid configurations $(g,u,c)$ (see \eqref{eq-guc} above) which may contribute to
   $(d_lx,\gamma)$, and hence $(d_lx,\gamma)=0$. Indeed, any such configuration
    belongs to a family $(g,u\circ r_\alpha,c_\alpha)$, where $r_\alpha:\C\setminus 0\to\C\setminus 0$ is
     the rotation $z\mapsto ze^{i\alpha}$.

 This  finishes off the proof of the lemma. \qed

\medskip\noindent{\bf Outline of the proof of Lemma \ref{lem-folk}:}
First we discuss coefficients of the form
$(d_0 \gamma_+,\gamma_-)$ with $l_+ =0$,
 $$\gamma_\pm \in \{\check{m}_{k_\pm},
\hat{m}_{k_\pm}, \check{M}_{k_\pm} , \hat{M}_{k_\pm}\}\;.$$ We argue
as in Step 1 of the proof of Lemma \ref{lem-2}, keeping the same
notations and taking into account that $l_+ =0$ This yields three
possibilities:
\begin{itemize}
\item[{(i)}] $h=-3, \Delta=0, l_- =2$;
\item[{(ii)}] $h=-3, \Delta=1, l_-=0$;
\item[{(iii)}] $h=-1, \Delta=0, l_-=1$.
\end{itemize}
Case (i) is ruled out exactly as in the proof of Lemma \ref{lem-2}.
Case (ii) yields $(d_0 \check{m}_k,\hat{M}_{k-2}) =2$,
case (iii) yields
$$(d_0 \check{m}_k, \hat{m}_{k-1})= (d_0 \check{M}_k, \hat{M}_{k-1}) =2\;,$$
while all other coefficients vanish.

Further, we analyze the matrix coefficients involving $x_0$ and $x_2$.
Equality $(d_0\check{m}_2,x_2) = 2$ follows from the count of degree $1$ (properly parameterized) sections of the bundle
$\nu$ passing through a given generic point and having a single zero of order two
at the point $\pi^2(\check{m}_2) \in \Sigma$.
Equality
$(d_0\check{M}_1,x_2) = 2$ corresponds to the fact that there are exactly two rigid spheres
$S^2 \times \text{point}$ and $\text{point} \times S^2$ in $W$ passing through $x_2$.

The only tricky remaining coefficient is $(d_0 \check{m}_1,x_0)=0$:
Seemingly, there are two rigid spheres in $W$ asymptotic to
$\check{m}_1$ which may contribute to this coefficient. We claim
that in fact a cancelation happens. To see this, recall that by
\eqref{eq-loops} and \cite{Cohen} $H_1(B,d_0) = \Z$. If $(d_0
\check{m}_1,x_0) \neq 0$, we would get that $H_1(B,d_0) = 0$, and
thus arrive at a contradiction. \qed

 \bigskip \noindent {\bf Acknowledgements.} We thank Luis Diogo, Misha Entov, Sam Lisi, Dusa
McDuff, Isidora Milin, Yasha Savelyev and Frol Zapolsky for numerous
useful discussions. Preliminary results of this note were announced
at the PRIMA congress in Sydney in July 2009. The second named
author thanks the Simons Foundation for sponsoring his stay at MSRI,
Berkeley, where a part of this paper has been written, and MSRI for
hospitality and a stimulating research atmosphere.

\bibliographystyle{alpha}


TITLE: is $x_{n}\ll \overline{x}_{n}^{2}$?
QUESTION [2 upvotes]: This question is a cross-post from MSE, cause I didn't get any answer there. I hope it is well suited for MO:
Let $(x_{n})_{n\ge 1}$ be an increasing sequence of positive integers and $\displaystyle{\overline{x}_{n}:=\dfrac{1}{n}\sum_{i=1}^{n}x_{i}}$. Suppose furthermore that $\forall\varepsilon\gt 0, \ \ n.\overline{x}_{n}\ll_{\varepsilon}n^{1+\varepsilon}$. Does it entail that $x_{n}\ll \overline{x}_{n}^{2}$?
Thanks in advance.
Edit: I think I must clear up my idea here. Writing $m_{n}:=\dfrac{x_{1}+x_{n}}{2}$ and using the inequality of arithmetic and geometric means, one gets $x_{1}.x_{n}\le m_{n}^{2}$. My idea is that, under the growth condition above, $m_{n}$ is roughly of the same order of magnitude as $\overline{x}_{n}$.

REPLY [7 votes]: No. Let $(y_k)$ be a rapidly increasing sequence of positive integers, and for $y_k\leq n<y_{k+1}$ put $x_n:=y_k$. Note that $(x_n)$ is an increasing sequence of positive integers satisfying $x_n\leq n$. In particular, $\overline{x}_{n}<n/2$. On the other hand, for $n:=y_k$ we have $x_n=y_k$, while $\overline{x}_{n}<1+y_{k-1}$. Hence $x_{n}\ll \overline{x}_{n}^{2}$ would imply that $y_k\ll y_{k-1}^2$ which is not always true. For a concrete counterexample take $y_k:=2^{3^k}$.


TITLE: Equation of line given translation and rotation that makes the line coincide with $x-$axis.
QUESTION [0 upvotes]: Let $L$ be a straight line with equation $y = mx+c$. Refecting across $L$ is an affine transformation and as a result, we can express it as a matrix.
We can translate and rotate and make the straight line coincide with the $x-$axis then go back and get the original points.
The translation and rotation matrices are given as follows:
$$
\begin{pmatrix}
1 & 0 & -2\\
0 & 1 & -15\\
0 & 0 & 1
\end{pmatrix}
$$
and
$$
\begin{pmatrix}
\frac{1}{\sqrt{37}} & \frac{6}{\sqrt{37}} & 0\\
-\frac{6}{\sqrt{37}} & \frac{1}{\sqrt{37}} & 0\\
0&0&1
\end{pmatrix}
$$
$\textbf{We have to find the equation of the straight line.}$

From the rotation matrix we have $\sin \theta = -\frac{6}{\sqrt{37}}$ and $\cos \theta = \frac{1}{\sqrt{37}}$ thus slope $m = \tan \theta = -6$.
Using the translation the line should pass through origin thus the constant terms should be zero.
$$y - 15 = -6(x-2) + c$$ Equating the constant terms we have $-15 = 12+c$. Thus $c = -27$.
Hence equation of line is $y = -6x -27$.

Is the solution correct?

REPLY [0 votes]: It should be $y = 6x + 3$.
If the transformed line is $~Y = 0, $ using rotation matrix, $\sin\theta = - \frac{6}{\sqrt{37}}, \cos\theta = \frac{1}{\sqrt{37}}$.
So, $Y = 0 = x' \sin\theta + y' \cos\theta \implies y' = 6x' ~$ before rotation.
And using translation matrix,
$(y+k) = 6 (x + h) ~ $ before translation, and as $h = -2, k = -15$,
$y - 15 = 6x - 12 \implies y = 6x + 3 ~$ must be the original line.


TITLE: Does staring at a bright LED light damage your eyes?
QUESTION [9 upvotes]: According to this article it seems that it is the UV part of the spectrum from the Sun that causes damage to the eye.
Would it therefore be "safe" to observe directly an equivalent energy density LED lamp, emitting in the visible part of the spectrum?

REPLY [16 votes]: One can give a highly qualified, but definite "yes" in answer to your question.
Contrary to popular belief, if it weren't for the UV, then staring at the Sun would not be a particularly hazardous thing to do for the majority of people. This is why I said "highly qualified" - for people with certain conditions, simple staring at the Sun may be hazardous, even aside from the UV (more below).
There are two ways that light will damage your eye:

Thermal damage;
Photochemical damage;

Fairly obviously from the nomenclature, the first kind (1) of damage is where so much energy is concentrated on the retina, its temperature is raised and the tissue is damaged or destroyed. The second kind (2) is where the light's photons are energetic enough to beget chemical changes by breaking bonds in organic molecules. This can lead both to acute poisioning of and damage / destruction to the tissue by the weird molecules / free radicals that come out of such light-matter interactions and also long term damage, even nuclear (in the sense of cell nucleus) changes, dysplasia (e.g. cataracts) and ultimately neoplasia (cancer).
The retina in the mammalian eye is superbly, densely envasculated. You won't find a better liquid cooling system in our contemporary human primitive technology. This situation has arisen from two evolutionary drivers: (1) the retina is simply adapted brain tissue, and the brain itself needs sophisticated liquid cooling: of all the organs it is the most sensitive to deviations from the warm blooded homeostatic temperature ($37{\rm ^oC}$) and (2) since we are creatures of the Neogene Eastern Africa, accidental looking at the Sun was everyday and commonplace to us.
Therefore, if you stare straight at the Sun at high noon, your pupils will have shrunken to about $1{\rm mm}$ diameter, thus the power incident on the retina is of the order of $\frac{\pi}{4}\times 10^{-3}{\rm m^2} \times 1000{\rm W} \approx 1{\rm mW}$. This is WAY less than the normal, healthy retina's capacity to dump heat, even if concetrated into a diffraction limited spot. The uncomfortable feeling you get from looking at the Sun is mostly a psychological one: if you did it for thirty seconds, your retinal cone and rod cells would be so drained of ATP (energy stores) that you would be totally (albeit altogether temporarily) blind for many, many minutes. For either a predator or prey creature (we are both), this is not good. But the intensity alone, although it causes severe, but temporary blindness, is not a hazard (as long as you don't get eaten by a Neogene lion wearing sunglasses, or fall over a cliff texting on your phone whilst your sight recovers when the receptors finally get their ATP levels topped up). As you rightly point out the danger from the Sun is UV: it is not the intensity with the uncomfortable, squinty-eye feeling that arises from it that is the danger for someone with a healthy retina. It is the low level but constant UV dose one gets mostly from scattered light that is the problem. This is why sunglasses in many countries must by law fulfil stringent UV attenuation performance standards: sunglasses stop the "glare" but this is not the danger; indeed the "comfort" afforded by non-UV attenuating lenses is a very false sense of security.
Three other points about the Sun's power delivered to someone who stares at it.

You have likely heard that looking at a total eclipse can be dangerous. It indeed can be. And this is because the pupil responds to average light. During a total eclipse, the pupil swells to its full size owing to the twilight's low average intensity. Its diameter swells to about $7{\rm mm}$: so the aperture has fifty times the area it has just before the eclipse begins. If you look at the diamond ring just after totality, you can therefore cop a dose of $20{\rm mW}$ or so in the eye. This can be enough to cause thermal damage. Evolution didn't kit us out to look at total eclipses;
Some eye conditions mean that even the thermal loading on the eye from the Sun can be dangerous. Macular degeneration is a major one, as is albinism or even an extremely white complexion. Heart and circulatory diseases are others. Other diseases and defects mean that the pupil cannot respond to high light levels. Many recreational drugs can dilate the pupils severely, particularly hallucinogens like LSD, psylosin or mescaline can lead to eye damage in this way.
Interestingly, if you begin with the assumption that the Sun's intensity into a fully shrunken pupil of someone staring straight at the Sun represents a safe upper limit to power dose (in the absence of UV), then you come up with dose limits that pretty closely match the ISO60825 laser safety standards for visible light.

The last comment is particularly relevant to you. ISO60825 does NOT tell the difference between coherent and incoherent light. You tread an LED exactly as you would a laser: if you apply ISO60825 and determine that the light dose from your LED is intrinsically safe (i.e. class 1), then this is a sound indication, aside from in severe cases of the diseases I mention above. The other factor I haven't mentioned is the blink response, which is also accounted for by ISO80625, but this sets levels for class 2 and class 3A, which are not intrinsically safe, but which are deemed to be low enough that an accidental looking into the beam will not be harmful owing to the shielding afforded by a healthy blink response (assumed to limit the light dose to 0.25 seconds). Again, drugs severely interfere with this reflex.


TITLE: Complex integral with Cauchy integral formula
QUESTION [0 upvotes]: For $k>1$ integer i want to compute the integral $$\int_0^{\infty}e^{-t^k}dt $$
by using Cauchy's integral formula.
Edit: The actual integral im trying to solve is $$\lim_{x\to\infty}\int_0^{x}e^{it^k}dt$$
but apparantly this is equal to $$e^{\pi i/2k}\int_0^{\infty}e^{-t^k}dt$$

REPLY [0 votes]: Begin by letting $$u=t^k \iff u^{\frac{1}{k}}=t \iff \frac{1}{k}u^{\frac{1}{k}-1}du=dt$$
yielding the integral:$$\frac{1}{k}\int_0^{\infty} u^{\frac{1}{k}-1}e^{-u}$$
Remembering that $$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx$$
We can see that our original integral is then nothing but$$\frac{1}{k}\Gamma(\frac{1}{k})$$


TITLE: Need help in pizza counting total number of possibilities problem.
QUESTION [0 upvotes]: Homer’s Pizza is advertising the following deal:
• 2 pizzas
• up to 3 toppings on each pizza
• 7 toppings to choose from
Double or triple toppings are allowed.
Two of same pizzas allowed.
Arrangement of toppings does not matter.
What is the total number of possibilities for a pizza order in this deal?

REPLY [0 votes]: "Up to three toppings" implies that it is permissible to order a pizza with two toppings, or one, or even with no toppings.
We can handle these as separate cases, or we can invent an eighth choice of "topping," called "none," which can be selected as a single, double, or triple choice to reduce the number of actual toppings.
The number of ways to top a pizza is then the number of ways you can make exactly three selections from eight options with repeated selections allowed, but disregarding the sequence in which the selections are made.
This is the number of $8$-tuples of non-negative integers with the sum $3$,
where the number at the $i$th position in the $8$-tuple says how many times we select the $i$ topping option.
This is a well-known problem in combinatorics,
equal to the number of ways you can arrange three identical items in eight labeled bins, solvable by a method often called "stars and bars"
(see Stars and Bars Derivation for example).
The number of $k$-tuples with sum $n$ is $\binom{n+k-1}{k - 1} = \binom{n+k-1}{n}.$
For the toppings of a single pizza we have $k = 8$ and $n = 3,$ so the number of ways to top a pizza is
$$ \binom{3+8-1}{3} = \binom{10}{3}
 = \frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1} = 120.$$
For two pizzas, presumably an order consisting of one triple mushroom pizza and one triple onion pizza is the same as one triple onion and one triple mushroom,
so the answer is not simply $120^2.$
You can consider separate cases for two different pizzas and for both pizzas the same, or you can consider this a problem of the number of ways to make exactly $2$ selections (with repetition allowed but disregarding the sequence of selections)
from $120$ options, to which we can again apply the formula $\binom{n+k-1}{n}$
with $k = 120$ and $n = 2,$
$$ \binom{2 + 120 + 1}{2} = \binom{121}{2} = \frac{121\cdot 120}{2} = 7260.$$


TITLE: Is this how a Universe could be created from nothing?
QUESTION [2 upvotes]: So this is a rather simplistic understanding but I want to know if it's fundamentally correct:
We have nothing which we'll represent as $0$.
Then we need a Universe which we can represent as any number of parts that sum to $0$
$$ 0 = -1 + 1 $$
In this way we can take $0$ and make a "Universe" that is now equal parts $1$ and $-1$. Nothing in this Universe could be created or destroyed because then it wouldn't equal 0 and yet we've now taken $0$ and made "something" of it by expressing it as different parts.
Is this a valid, albeit simplistic understanding? Or am I missing something here?
This would even mean that an infinite number of Universes could exist, $0$ would just be expressed in different ways, or repeatedly in the same way.

$0 = -1 + 1$
$ 0 = -2 + 2$

REPLY [3 votes]: Although I completely agrees that the question itself doesn't physically make sense, here is somewhat of a answer for the question:

Can you interpret the geometry of the universe in terms of a the potential energy cuonterbalancing its kinetic energy?

The Friedmann equations, which is a isotropic and homogeneous solution of the Einstein field equations of general relativity, describes the evolution of the (scale of the $R$ of the) universe.
$$\left(\frac{\dot{R}}{R}\right)^2=\frac{8\pi G\rho_\text{tot}}{3}-\frac{k c^2}{R^2} $$
The first term represents contains the energy density due to the matter, radiation, dark matter and dark energy in the universe. The last term is related to the curvature of the universe. For $k=+1$ you have a closed universe, $k=-1$ an open en $k=0$ a flat one.

From observation, we believe to be living in a flat universe, i.e. $k=0$.
Although this equation is a solution of the (difficult to solve) field equations of Einstein, it can be classically derived.
Imagine a point mass $m$ being accelerated by gravity at the surface of a sphere of radius $R$, density $\rho$ and mass $=4\pi D^3\rho/3$. So the force equation is
$$m\ddot{D} = -\frac{mMG}{D^2}$$
Some simple calculus give you and using that $D=rR$ (that is not really important: $R$ is not the real distance)
$$\frac{m\dot{R}^2}{2} - \frac{mMG}{R} = \text{constant} = \frac{kc^2}{R^2}$$
where you chose the integration constant to match to one in the Friedmann equation.
And here you see it: on the left hand side you have the total energy, kinetic plus potential.
The Friedmann equation can be interpreted (!) as saying the following

a universe that is spatially closed (with k = +1) has negative total ``energy'': the expansion will eventually be halted by gravity, and the universe will recollapse.
In an unbound model, where k = -1, it will expand forever.
However for a given rate of expansion there is a critical density that will bring the expansion asymptotically to a halt. This is the case when the knietic energy is exactly counterbalancing the potential term, yielding $k=0$.

References: D. Perkins, Particle Astrophysics, Chapter 5.3
or on the web you also have this.


TITLE: What happens when matrix is moved to the other side of the equation?
QUESTION [1 upvotes]: I have been doing several questions related to Matrix.
I have a question ask me to simplify the expression (knowing that A,X,B are squared matrix with the same order):
$AXB=(AB)^2$
$AXB=A^2B^2$
Then I am thinking where should I put A and B when I move them to the other side (As matrix operation is not commutative):
Should it be: $X=A^2A^{-1}B^2B^{-1}$ which equals to $X=AB$ ?
However, the right answer is $X=BA$. Since they are the same order, does $AB=BA$?
I have always been confused by this kind of question. I have no idea how to properly move the matrix to the other side of the equation

REPLY [2 votes]: There are a couple errors here. First note that $(AB)^2 = (AB)\cdot (AB) = ABAB$, not $A^2B^2$.
The bigger problem is with how you think about what you're doing. There is no mathematical notion of "moving one thing from one side of the equation to the other." There are only things which happen to look like they do this.
The fundamental rule here is that you must do exactly the same thing to both sides.
In this example, we have
$$AXB=(AB)^2=ABAB.$$
We can now multiply by $A^{-1}$ on the left on both sides, this gives
$$A^{-1}AXB = A^{-1}ABAB,$$
or $$XB = BAB,$$ since $A^{-1}A=I$.
Now we can multiply by $B^{-1}$ on the right on both sides, to get
$$XBB^{-1} = BABB^{-1},$$
and simplifying, this gives
$$X=BA,$$
as desired.
Further comments:
Matrix Multiplication:
Matrix multiplication is not commutative. I.e., it is not true in general that $AB=BA$ for matrices $A$ and $B$. Indeed, $AB$ and $BA$ need not even both be the same size or both make sense. Consider the following examples:

$$A=\newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}}\bmat 0 & 1 \\ 0 & 0 \emat, \quad B = \bmat 0 & 0 \\ 1 & 0 \emat, $$ then
$$ AB = \bmat 1 & 0 \\ 0 & 0 \emat,\quad BA = \bmat 0 & 0 \\ 0 & 1 \emat. $$
If $A$ is $2\times 3$ and $B$ is $3\times 2$, then $AB$ is $2\times 2$ and $BA$ is $3\times 3$.
If $A$ is $2\times 3$ and $B$ is $3\times 4$, then $AB$ is $2\times 4$, but $BA$ doesn't even make sense, and can't be multiplied because the sizes of the matrices don't line up.

Invertibility of matrices:
Not every matrix has an inverse, i.e., given a $n\times n$ matrix $A$, there isn't always a matrix $A^{-1}$ such that $AA^{-1}=A^{-1}A=I$. I've assumed that $A$ and $B$ are invertible in your question although it wasn't stated, because otherwise $BA$ is not the unique possibility for $X$, which is the given answer. This is important though, otherwise I couldn't have multiplied by $A^{-1}$ and $B^{-1}$ in the solution.
Also even though nonsquare matrices cannot be invertible in the sense given above, they can have one sided inverses. I.e., if $A$ is $n\times m$, with $n\ge m$, there may be a matrix $B$ with $BA = I_m$, and if $m\ge n$, there may be a matrix $C$ with $AC = I_n$. These can also be very useful at times.


TITLE: 1-1 correspondence between nuclei and regular monomorphisms of a locale
QUESTION [1 upvotes]: I am having a little trouble with Theorem 2.3 in Professor Johnstone's book on "Stone Spaces". The theorem depends on a Lemma (which I am not struggling with; I only include it for context) which refers to a situation in which we have a locale A and a nucleus $j : A \rightarrow A$, and we have defined $A_j = \text{Im}\;j = \{ a \in A \;\mid\; j(a) = a \}$, and claim the following in Lemma 2.2:

$A_j$ is a frame, and $j : A \rightarrow A_j$ is a frame homomorphism, whose right adjoint is the inclusion $A_j \rightarrow A$.

Now Professor Johnstone follows this with the Theorem of section 2.3:

For any locale $A$, there is a bijection between nuclei on $A$ and regular subobjects of $A$ in Loc (i.e. isomorphism classes of regular monomorphisms $B \rightarrow A$)

Now in the proof for this Theorem, we are told that "given a nucleus $j$, Lemma 2.2 gives us a regular monomorphism $A_j \rightarrow A$". My issue with this statement is that, although the Lemma clearly shows that we have a homomorphism $j : A_j \rightarrow A$ in Loc, and one can quickly see that it is monic (because its dual in Frm is surjective by construction and so epi) it is not clear that this is a regular monomorphism in Loc (i interpret "$j$ is a regular monomorphism" to mean that $j$ is to be the equaliser of a pair of arrows in Loc).
After  a bit of pondering and a small amount of research on nlab, I guessed that the work needed to show $j$ is a regular monomorphism in Loc is to take its pushout along itself and then to prove that $j$ is the equaliser of those coprojections. I think this can be done in an elementary fashion by looking at the dual constructions in Frm (which seems to have pullbacks at least). Is that the correct / quickest approach or am I missing something obvious or more categorical in style? I'm wondering whether I should have been paying attention to something that Professor Johnstone might have written earlier in the book.

REPLY [0 votes]: Use that $j^2=j$. The diagram below is thus a coequalizer:
$$ A\underset{j}{\overset{{\rm id}}{\rightrightarrows}} A\to A_j\,.$$
Update: This answer is wrong, as commented below because $j:A\to A$ is not necessarily a frame morphism.


text:
\section{Introduction}
\label{sec:intro}

Nonnegative Matrix Factorization (NMF) is a low-rank model widely used for feature extraction in applications such as multispectral imaging, text mining, or blind source separation; see~\cite{gillis_why_2014,fu2019nonnegative} and the references therein.
Given a nonnegative data matrix $M \in \mathbb{R}_+^{m \times n}$ and a factorization rank $r$, NMF consists in finding two nonnegative matrices $W \in \mathbb{R}^{m \times r}_+$ and $H \in \mathbb{R}^{r \times n}_+$ such that $M \approx W H$.
NMF can be formalized as the following optimization problem:
\begin{equation}
    \min_{W \geq 0, H \geq 0} \| M - WH \|_F^2 .
\end{equation}
In this paper, we use the Frobenius norm to measure the quality of the approximation.
Although other measures are possible, the Frobenius norm is by far the most commonly used, because it assumes Gaussian noise (which is reasonable in many real-life applications) and allows for efficient computations~\cite{gillis_why_2014}.

One of the advantages of NMF over similar methods such as principal component analysis (PCA) is that the nonnegativity constraint favors a part-based representation~\cite{lee1999learning}, which is to say that the factors are more easily interpretable, in particular when they have a physical meaning.
If each column of $M$ represents a data point, then each corresponding column of $H$ contains the coefficients to reconstruct it from the $r$ atoms represented by the columns of $W$, since $M(:,j) \approx WH(:,j)$ for all $j$.
Every data point is therefore expressed as a linear combination of atoms.
For example, when using NMF for multispectral unmixing, a data point is a pixel, an atom is a specific material, and each column of $H$ contains the abundance of these materials in the corresponding pixel; see Section~\ref{sec:xp-real} for more details.
Geometrically, the atoms (columns of $W$) can be seen as $r$ vertices whose convex hull contains the data points (columns of $M$), under appropriate scaling.

\subsection{Separability}

In general, computing NMF is NP-hard~\cite{vavasis_complexity_2010}.
However, Arora et al.~\cite{arora_computing_2012} proved that NMF is solvable in polynomial time under the \emph{separability}  assumption on the input matrix.
\begin{definition}\label{def:separability}
A matrix $M$ is $r$-separable if there exists a subset of $r$ columns of $M$, indexed by $\mathcal{J}$, and a nonnegative matrix $H \geq 0$, such that $M = M(:,\mathcal{J}) H$.
\end{definition}
Equivalently, $M$ is $r$-separable if $M$ has the form
$M = W [ I_r , H' ] \Pi$, where $I_r$ is the identity matrix of size $r$, $H'$ is a nonnegative matrix, and $\Pi$ is a permutation.
Separable NMF consists in selecting the right $r$ columns of $M$ such that $M$ can be reconstructed perfectly.
In other words, it consists in finding the atoms (columns of $W$) \emph{among} the data points (columns of $M$).

\begin{problem}[Separable NMF]
Given a $r$-separable matrix $M$, find $W = M(:,\mathcal{J})$ with $|\mathcal{J}| = r$  and $H \geq 0$ such that $M = W H$.
\end{problem}
Note that, if $W$ is known, the computation of $H$ is straightforward: it is a convex problem that can be solved using any nonnegative least squares (NNLS) solver (for example, it can be solved with the Matlab function \texttt{lsqnonneg}).
However, the solution is not necessarily unique, unless $W$ is full rank.

In the presence of noise, which is typically the case in real-life applications, this problem is called near-separable NMF and is also solvable in polynomial time given that the noise level is sufficiently small~\cite{arora_computing_2012}. In this case, we are given a near-separable matrix $M \approx M(:,\mathcal{J}) H$ where $|\mathcal{J}| = r$ and $H \geq 0$.

\subsection{Successive Nonnegative Projection Algorithm}

Various algorithms have been developed to tackle the (near-)separable NMF problem.
Some examples are the successive projections algorithm (SPA)~\cite{araujo_successive_2001}, the fast canonical hull algorithm~\cite{kumar_fast_2013}, or the successive nonnegative projections algorithm (SNPA)~\cite{gillis_successive_2014}.
Such algorithms start with an empty matrix $W$ and a residual matrix $R = M$, and then alternate between two steps: a greedy selection of one column of $R$ to be added to $W$, and an update of $R$ using $M$ and the  columns extracted so far.
As SNPA was shown, both theoretically and empirically, to perform better and to be more robust than its competitors~\cite{gillis_successive_2014}, it is the one we study here in detail. Moreover, SNPA is able to handle the underdetermined case when $\rank(W) < m$ which will be key for our problem setting (see below for more details).

SNPA is presented in \cref{alg:snpa}.
\input{21pseudocode-snpa}
SNPA selects, at each step, the column of $M$ maximizing a function $f$ (which can be any strongly convex function such that $f(0)=0$, and $f = ||.||_2^2$ is the most common choice).
Then, the columns of $M$ are projected onto the convex hull of the origin and the columns extracted so far, see step 8 where we use the notation
\[
\Delta = \Big\{ h \ \big| \ h \geq 0, \sum_i h_i \leq 1 \Big\},
\]
whose dimension is clear from the context.

After $r$ steps, given that the noise is sufficiently small and that the columns of $W$ are vertices of $\conv(W)$, SNPA is guaranteed to identify $W$.
An important point is that SNPA requires the columns of $H$ to satisfy $\|H(:,j)\|_1 \leq 1$ for all $j$, where $\|x\|_1 = \sum_i |x_i|$ is the $\ell_1$ norm. This assumption can be made without loss of generality by properly scaling the columns of the input matrix to have unit $\ell_1$ norm; see the discussion in~\cite{gillis_successive_2014}.

\subsection{Model Limitations}

Unfortunately, some data sets cannot be handled successfully by separable NMF, even when all data points are linear combinations of a subset of the input matrix.
In fact, in some applications, the columns of the basis matrix $W$, that is, the atoms, might not be vertices of $\conv(W)$.
This may happen when one seeks a matrix $W$ which is not full column rank.
For example, in multispectral unmixing, $m$ is the number of spectral bands which can be smaller than $r$, which is the number of materials  present in the image; see Section~\ref{sec:xp-real} for more details.
Therefore, it is possible for some columns of $W$ to be contained in the convex hull of the other columns, that is, to be additive linear combinations of others columns of $W$; see \cref{fig:examples} for illustrations in three dimensions (that is, $m = 3$).

\begin{figure}[t]
    \centering
    \subfloat[\label{fig:allext}]{
        \input{img/allext}
    }\hfil
    \subfloat[\label{fig:intvertex}]{
        \input{img/oneint}
    }
    \caption{On the left (a), all vertices are exterior, and SNPA is assured to identify them all. On the right (b), the data points are 2-sparse combinations of 4 points, one of which (vertex 4) is ``interior'' hence it cannot be identified with separable NMF.}
    \label{fig:examples}
\end{figure}

These difficult cases cannot be handled with separable NMF, because it assumes the data points to be linear combinations of vertices, so an ``interior vertex'' cannot be distinguished from another data point.
However, if we assume the \emph{sparsity} of the mixture matrix $H$, we may be able to identify these interior vertices.
To do so, we introduce a new model, extending the approach of SNPA using additional sparsity constraints.

We introduce in \cref{sec:ssnmf} a proper definition of this new problem, which we coin as sparse separable NMF (SSNMF). Before doing so, let use recall the literature on sparse NMF.

\subsection{Sparse NMF}

A vector or matrix is said to be \emph{sparse} when it has few non-zero entries.
Sparse NMF is one of the most popular variants of NMF, as it helps producing more interpretable factors.
In this model, we usually consider column-wise sparsity of the factor $H$, meaning that a data point is expressed as the combination of only a few atoms.
For example, in multispectral unmixing, the column-wise sparsity of $H$ means that a pixel is composed of fewer materials than the total number of materials present in the image.
When sparsity is an a priori knowledge on the structure of the data, encouraging sparsity while computing NMF is likely to reduce noise and produce better results.

Sparse NMF is usually solved by extending standard NMF algorithms with a regularization such as the $\ell_1$ penalty~\cite{hoyer2002non,kim_sparse_2007}, or constraints on some sparsity measure, like the one introduced in~\cite{hoyer2004non}.
Recently, exact $k$-sparse methods based on the $\ell_0$-``norm'' have been used for NMF, using a brute-force approach~\cite{cohen_nonnegative_2019-1}, or a dedicated branch-and-bound algorithm~\cite{nadisicexact}.
They allow the explicit definition of a maximum number (usually noted $k$) of non-zero entries per column of $H$.
These approaches leverage the fact that, in most NMF problems, the factorization rank $r$ is small, hence it is reasonable to solve the $k$-sparse NNLS subproblems exactly.

\subsection{Contributions and Outline}

In this work, we study the SSNMF model from a theoretical and a  pratical point of view.
Our contributions can be summarized as follows:
\begin{itemize}
    \item In \cref{sec:ssnmf}, we introduce the SSNMF model. We prove that, unlike separable NMF, SSNMF is NP-complete.

    \item In \cref{sec:algo}, we propose an algorithm to tackle SSNMF, based on SNPA and an exact sparse NNLS solver.

    \item In \cref{sec:analysis}, we prove that our algorithm is correct under reasonable assumptions, in the noiseless case.

    \item In \cref{sec:xp}, experiments on both synthetic and real-world data sets illustrate the relevance and efficiency of our algorithm.
\end{itemize}