TITLE: Norm of an Ideal.
QUESTION [3 upvotes]: Let Let $K = \mathbb{Q}(\sqrt{5})$ and $\alpha = \frac{1+\sqrt{5}}{2}.$
How do we describe the ideal $I$ for ring of integers $\mathcal{O}_K$, given by $I = \{ (2+\alpha) a + (1+ 3 \alpha) b : a, b \in \mathbb{Z}\}$ and how do we calculate norm of $I.$
I tried writing $I = \{ (2a+b) + (a+ 3b)\alpha : a, b \in \mathbb{Z}\}.$ Now in ring $\mathcal{O}_K/I$ we have $I = 0.$ So $(2a+b) + (a+ 3b)\alpha = 0$ and since $1,\alpha$ are linearly independent we have $2a+b = 0$ and $a+ 3b=0.$ Which gives $a=b=0.$ I dont know what to interpret from this or even I am doing right or wrong. Help please.

REPLY [6 votes]: A useful fact here is that $\mathcal O_K$ is a PID. This means that $I$ can be written in terms of one generator.
As you've written it, $I$ consists of integer combinations of $2+\alpha$ and $1+3\alpha$, so as an ideal, it is generated by these elements. Denote this fact using the notation
$$I = \langle 2+\alpha, 1+3\alpha\rangle.$$
Observe that $$\alpha^2-\alpha -1 = 0$$
and that
$$\alpha(2+\alpha) =2\alpha + \alpha^2 = 2\alpha + (\alpha + 1) = 1+3\alpha.$$
So $I = \langle 2+\alpha\rangle$, and hence the norm of the ideal $I$ will just be the $N_{K/\mathbb Q}(2+\alpha)$.


TITLE: Value of $\lim_{n\to\infty}{(1+\frac{2n^2+\cos{n}}{n^3+n})^n}$
QUESTION [11 upvotes]: How should one go about computing $$\lim_{n\to\infty}{\left(1+\frac{2n^2+\cos{n}}{n^3+n}\right)^n}\quad?$$ What surprised me about this is that
$$\lim_{n\to\infty}{\left(1+\frac{2n^2+\cos{n}}{n^3+n}\right)^\frac{n^3+n}{2n^2+\cos{n}}}=1$$(according to wolfram), instead of $e$, which is what I expected. Could someone comment on that also?

REPLY [5 votes]: Lets consider the more general case of an arbitrary function.  Then we have the following theorem:
Theorem: Given a function $g(n)$ such that $g_0=\lim_{n\rightarrow \infty} g(n)$ exists, we have $$\lim_{n\rightarrow \infty }\left(1+\frac{g(n)}{n}\right)^n=e^{g_0}.$$
Example: In particular, for your above question, $g(n)=\frac{2n^2+\cos(n)}{n^2+1}$, so that $g_0 =2$, and hence the value of the original limit is $e^2$.
Proof of theorem:  Let $f(n)=\frac{g(n)}{g_0}$ so that  $f(n)=1+o(1)$.  Then since $\lim_{n\rightarrow \infty} \left(1+\frac{a}{n}\right)^n=e^a$ we see that $$\lim_{n\rightarrow \infty} \left(1+\frac{g(n)}{n}\right)^\frac{n}{f(n)}=e^{g_0}.$$  Then, because  $f(n)=1+o(1)$, it follows that $\frac{n}{f(n)}=n+o(n)$.  But, $$\lim_{n\rightarrow \infty} \left(1+\frac{O(1)}{n}\right)^{o(n)}=1,$$ so we conclude that $$\lim_{n\rightarrow \infty }\left(1+\frac{g(n)}{n}\right)^n=e^{g_0}.$$
I hope that helps,


TITLE: Sequences with divisibility
QUESTION [1 upvotes]: A sequence is such that $a_o =1, a_1= 1, a_{n+1}
=a_{n}a_{n-1}+1$ so we have to comment on divisibilty of $a_{2007} $ by 4.
I found out first few values in sequence as
1,1,2,3,7,22, .... which told me that only $a_{3n} $ is even.
But can there be some other elaborative method?

REPLY [1 votes]: Hint $ $ There are $j<k$ with $\,(a_j,a_{j+1})\equiv (a_k,a_{k+1})\,\pmod{\!4}\ $ since there are only finitely many such pairs $\!\bmod {\!4}$. The recurrence depends only on the pair of prior values so this leads to cyclic behavior.


TITLE: Expectation from Joint Distribution
QUESTION [3 upvotes]: We have
$$f_{X,Y}(x,y) =
    \begin{cases}
      1 & 0 < x < 1,  x < y < x+1; \\
      0 & \text{otherwise}
    \end{cases}
$$
and we're looking for $\operatorname{Var}Y$.
I got two different answers with two different methods and I'm not sure which one is correct.

Method 1:
$$\operatorname E(Y) = \int_0^1 \int_x^{x+1}yf_{X,Y}(x,y)\,dy\,dx = \int_0^1 \left(x+\frac{1}{2}\right) \, dx = 1$$
$$\operatorname E(Y^2) = \int_0^1 \int_x^{x+1}y^2f_{X,Y}(x,y) \, dy \, dx = \int_0^1 \left(x^2 + x + \frac{1}{3}\right) \, dx = \frac{7}{6}$$
So $\operatorname{Var}(Y) = \frac{1}{6}$

Method 2:
Marginal distribution $f_Y(y) =1$
$$\operatorname E(Y) = \int_x^{x+1}yf_Y(y)\,dy = x+\frac{1}{2}$$
$$\operatorname E(Y^2) = \int_x^{x+1}y^2f_Y(y) \, dy = x^2 + x +\frac{1}{3}$$
So $\operatorname{Var}(Y) = \left(x^2 + x +\frac{1}{3}\right) - \left(x+\frac{1}{2} \right)^2 = \frac{1}{12}$

One of these solutions is probably wrong, but I can't really identify the error.

REPLY [1 votes]: Your first method is correct, but your second message can be salvaged. What you calculate in the second method are the conditional expectation and the conditional variance, conditional on the value of $X$:
\begin{eqnarray*}
\mathsf E[Y\mid X=x]&=&\int_x^{x+1}y\,f_{X,Y}(x,y)\,\mathrm dy=x+\frac12\;,\\
\mathsf E[Y^2\mid X=x]&=&\int_x^{x+1}y^2\,f_{X,Y}(x,y)\,\mathrm dy=x^2+x+\frac13\;,\\
\mathsf{Var}(Y\mid X=x)&=&\mathsf E[Y^2\mid X=x]-\mathsf E[Y\mid X=x]^2=\left(x^2+x+\frac13\right)-\left(x+\frac12\right)^2=\frac1{12}\;.
\end{eqnarray*}
You can get the unconditional variance of $Y$ from these results by applying the law of total variance:
$$
\mathsf{Var}(Y)=\mathsf E[\mathsf{Var}(Y\mid X)]+\mathsf{Var}(\mathsf E[Y\mid X])=\frac1{12}+\mathsf{Var}\left(X+\frac12\right)=\frac1{12}+\mathsf{Var}(X)=\frac1{12}+\frac1{12}=\frac16\;.
$$


TITLE: Applications of Cauchy's convergence theorem
QUESTION [0 upvotes]: Cauchy's Convergence Theorem:
Let {$P_n$} be a sequence of points and let $d(P_m,P_n)$ be the distance between $P_m$ and $P_n$. Then for a sequence to be convergent, $d(P_m,P_n)$ should $\to$ 0, as $n$ and $m$ become infinite.
I understand the theorem, but if I'm given a sequence, how can I apply this theorem to check if the sequence is Cauchy? I have a midterm tomorrow and I'm positive this will be a question.
The only thing I can think to do would be to some how prove that the distance is always less than some $\epsilon$. But I'm not sure how to even do that. Also, my book doesn't have any problems which require the use of this theorem, so I have nothing to really check any kind of work against. There is only the proof of the formula.
Any help is greatly appreciated!

REPLY [3 votes]: Note that the theorem refers to a complete metric space (if you haven't done metric spaces, I presume your points are real numbers with the usual distances).
For calculations, your intuition is correct: if you can prove that $d(x_n,x_m)<\epsilon$ eventually for all $\epsilon$, then you can conclude that the sequence is Cauchy. However, this is not always required, as you can just take limits as well! Choose your favourite convergent sequence and try it out.
As an example, take your sequence of points to be $P_n=\frac{1}{n}$ in $\mathbb{R}$ with the usual metric. Then
$d(P_n,P_m)=\left|\frac{1}{n}-\frac{1}{m}\right|\leq\left|\frac{1}{n}\right|+\left|\frac{1}{m}\right|\to0 $ as $m,n\to\infty$
If you really love your $\epsilon's$, you can also write it like so. Fix $\epsilon>0$. There is a positive integer $k>0$ such that $\frac{1}{k}<\epsilon$. Given $m,n>2k$ (so that $\frac{1}{m}+\frac{1}{n}<\frac{1}{k}<\epsilon$), we have
$d(P_n,P_m)=\left|\frac{1}{n}-\frac{1}{m}\right|\leq\left|\frac{1}{n}\right|+\left|\frac{1}{m}\right|<\frac{1}{2k}+\frac{1}{2k}=\frac{1}{k}<\epsilon$
Your friends in such calculations include the triangle and Cauchy-Schwarz inequalities.


TITLE: What should say the isoperimetric inequality for taxicab geometry?
QUESTION [1 upvotes]: It seems that there is literature for this topic,

Question. What should say the isoperimetric inequality for taxicab geometry? Have you got any idea? Then you tell us some easy facts or idea about this. Thanks in advance.

I've tried read about this geometry, and previous topic. In this geometry $4$ is $\pi$!, but still I have no idea about previous question.
The statement of the isoperimetrical inequality that I refer to you is in first paragraph of [1], and a reference for the Taxicab gometry is [2]. Yoo I've found nice sites about conics in this geometry, for example ellipses. Another relates quadrilaterals.
References:
[1] Wikipedia, Isoperimetric inequality  https://en.wikipedia.org/wiki/Isoperimetric_inequality
[2] Wikipedia, Taxicab geometry https://en.wikipedia.org/wiki/Taxicab_geometry

REPLY [0 votes]: The best possible ratio is from the unit square, with perimeter $4$ and area $1$, as adding any other square increases $A$ by $1$ and $P$ by at most $2$, so we have:
$$\dfrac{P}{A}-\dfrac{P+2}{A+1}=\dfrac{P-2A}{A(A+1)}\ge0$$
if we start from  the unit square.
We also only need consider convex regions as concave regions merely increase $P$ and decrease $A$, and we can note that a convex region has the same perimeter as a bounding rectangle.
As $\large{\pi_{\small TC}}=4$, we have $4{\large\pi_{\small TC}} A\le P^2
$, with equality iff the shape is a square.
We can define a concave region as having a larger perimeter than the bounding rectangle, which is equivalent to some row or column having $4+$ edges present.


TITLE: Crossing Circles
QUESTION [1 upvotes]: On a plane, you are only allowed to draw circles. After drawing 1 circle, can you ALWAYS draw another so that the new circle crosses all existing circles at 2 points? Why?

REPLY [2 votes]: This can be seen by taking a circle of any radius, but lets say one, centered at $(\frac{1}{2},0)$.  Then take a copy of that circle, and rotate it $180^ \circ$.

Then you repeat this process, placing as many circles as you want between them, in the top or bottom of the plane.  It could look something like this, which is only 4 circles, but more is harder to tell what is going on.

This can be done as long as they all contain the same interior point and have the same radius. For our example, they all contain the origin.


TITLE: Proof Question - Calculus
QUESTION [1 upvotes]: Proof Question - Calculus
Proof that if $a>1/4$ so to the equation $\sin x = \sqrt{x}+\frac{a}{\sqrt{x}}$ there is no real solutions.
Thanks!

REPLY [7 votes]: Clearly $x > 0$.
By AMGM
$$\sqrt{x} + \frac{a}{\sqrt{x}} \geq 2\sqrt{a} > 2*1/2 =1$$
Since $0 \leq |\sin(x)| \leq 1$ , there are no solutions.


TITLE: Two consecutive numbers are removed from the progression 1, 2, 3..n. The arithmetic mean of the remaining numbers is 105/4. What is the value of n?
QUESTION [3 upvotes]: Two consecutive numbers are removed from the progression 1, 2, 3..n. The arithmetic mean of the remaining numbers is 105/4. What is the smallest value of n?
?What are the two numbers removed?
Now many answers are given but they are given in a hit and trial method.
Is there any other method to solve this problem?

REPLY [0 votes]: Since the given arithmetic mean in the title is $261/4$, we know that $(n-2)\times 261/4$ must be an integer, so $n\equiv 2 \bmod 4$.
The sum of the full set of integers from $1$ to $n$ is $n(n+1)/2$, with average $\color{blue}{(n+1)/2}$. Removing two numbers from that set cannot change the average by more than $1$, since removing the largest two makes it $(n-1)/2 = \color{blue}{(n+1)/2}-1$ and removing the two smallest makes it $(n-1)/2 + 2= \color{blue}{(n+1)/2}+1$.
So we know that $261/4 -1 \le \color{blue}{(n+1)/2} \le 261/4+1$ so $128.5\le n\le 132.5$. Now since only $130\equiv 2 \bmod 4$ in this range, we can be confident that  $n=130$ (assuming a well-posed question).


TITLE: Counter example to unique solvability of Dirichlet Problem
QUESTION [1 upvotes]: In Gilbarg and Trudinger, Theorem 6.14 states that for a uniformly elliptic operator $L=A^{ij}D_{ij}+B^iD_i + c$ the Dirichlet problem
$ Lu=f $
has a unique solution $u$ for every $f$ under some suitable assumptions.
One of these assumptions is that $c \leq 0$, and the book says that there are counter examples where the above does not hold for $c \geq 0$. So, what would be such a counter example, and why is it a counter example?
Thanks in advance.

REPLY [1 votes]: The classic examples are eigenvalues of the Laplacian: if $\Delta v = \lambda v$ for some nontrivial $v$ with zero Dirichlet condition, then any constant multiple $u=Cv$ is a solution of the Dirichlet problem $$\Delta u - \lambda u = 0.$$
Thus for an explicit counterexample you can take $\Omega = (0,\pi),$ $A^{ij}=\delta^{ij},$ $B^i = 0,$ $c = 1,$ $f = 0,$ which admits the family of solutions $$u(x) = C \sin(x),\quad C\in\mathbb R.$$


TITLE: Using the work-energy theorem in thermodynamics
QUESTION [0 upvotes]: If an $n$ moles of an ideal gas (one atomic) is heated at constant volume $V$ from temperature $T_1$ to temperature $T_2$, the amount of energy heat needed can be calculated as
$$ Q = nC_V \Delta T $$
Solving with work-energy theorem
By the work-energy theorem the change in total kinetic energy is equal to the work done by net sum of the external forces ($\Delta K = W_f$). In this case, the work done on the gas is due to adding the heat $Q$ to the system. The total kinetic energy in a one atomic gas is given by $K = \frac{3}{2}pV = \frac{3}{2}nRT$. Therefore
$$ Q = \Delta K = K_2 - K_1 = \frac{3}{2}nR(T_2 - T_1) = \frac{3}{2}nR \Delta T$$
And since $C_V = \frac{3}{2}R$ for one atomic ideal gases, the two equations is identical.
Is my second solution reasoning valid?

REPLY [0 votes]: Let us try to understand the basics of First  Law of Thermodynamics-
Heat (Q) and work (W) are the two ways to add or remove energy from a physical system.

n moles of an ideal gas (one atomic) is heated at constant volume V from temperature T1 to temperature T2, the amount of energy heat needed can be calculated as

Q = n C(v)ΔT
The processes are very different.
Heat is driven by temperature differences, while work involves a force exerted through a distance.
The heat and work can produce identical results.
For example, both can cause a temperature increase.
Heat transfers energy into a system, such as when the sun warms the air in a bicycle tire and increases the air’s temperature.
Similarly, work can be done on the system, as when someone pumps air into the tire. Heat and work are both energy in transit—neither is stored as such in a system. However, both can change the internal energy, U, of a system.
Internal energy is the sum of the kinetic and potential energies of a system’s atoms and molecules.
It can be divided into many subcategories, such as thermal and chemical energy, and depends only on the state of a system (that is, P, V, and T), not on how the energy enters or leaves the system.
In order to understand the relationship between heat, work, and internal energy, we use the first law of thermodynamics. The first law of thermodynamics applies the conservation of energy principle to systems where heat and work are the methods of transferring energy into and out of the systems.
It can also be used to describe how energy transferred by heat is converted and transferred again by work.
so  use
ΔU = Q – W
and the process given by you says that no p.dv   work is being done ,therefore the  alternate calculation does  not  stand correct ,though it may be relating to change in temperature  in the initial and final  state of the system and trying to relate K.E. with work done, which  is not happening..
Reference- https://www.texasgateway.org/resource/122-first-law-thermodynamics-thermal-energy-and-work


TITLE: Sobolev spaces embeddings and conjugate exponents
QUESTION [1 upvotes]: Let $N\geq 1$ and $1<p<N$ with conjugate exponenr $p^{\prime}$. I would like to understand if there exists some inclusion relation between $W^{1, p}(\mathbb{R}^N)$ and $W^{1, p^{\prime}}(\mathbb{R}^N)$, maybe something
$$W^{1, p}(\mathbb{R}^N)\hookrightarrow W^{1, p^{\prime}}(\mathbb{R}^N).$$
If not, it makes sense to consider
$$W^{1, p}(\mathbb{R}^N)\cap W^{1, p^{\prime}}(\mathbb{R}^N)?$$
Could someone please help?
Thank you in advance!

REPLY [2 votes]: Sketch. In the case $p=2=p'$ there is nothing to prove. Otherwise, say $p<2$ so that $p'>2$. Let $\chi$ be a bump function equal to $1$ near the origin. Then it is a standard exercise to check that $f(x)=\frac1{|x|^\theta}\in W^{1,p}(B_1(0))$ and $f\notin W^{1,p'}(B_1(0))$ for a certain $\theta$ depending on $N$ and $p$. (The basic calculation needed is that $\frac1{|x|^r}\in L^1_{\text{loc}}$ if $r<n$). Then $\chi f$ is a counterexample on $\mathbb R^n$.  The $p>2$ case is also false, and for this you instead consider $(1-\chi)f$. Since $p''=p$ this also covers the reverse inclusion.
I don't know what it means to "consider" it; you can certainly write down the set.


text:
(*  Title:       Verification components with relational MKA
    Author:      Jonathan Julián Huerta y Munive, 2020
    Maintainer:  Jonathan Julián Huerta y Munive <jonjulian23@gmail.com>
*)

subsection \<open> State transformer model \<close>

text \<open> We show that Kleene algebras have a state transformer model. For this we use the type of
non-deterministic functions of the Transformer\_Semantics.Kleisli\_Quantale theory. Below we
prove some auxiliary lemmas for them and show this instantiation. \<close>

theory HS_VC_KA_ndfun
  imports
    Kleene_Algebra.Kleene_Algebra
    "Transformer_Semantics.Kleisli_Quantale"

begin

notation Abs_nd_fun ("_\<^sup>\<bullet>" [101] 100)
     and Rep_nd_fun ("_\<^sub>\<bullet>" [101] 100)

declare Abs_nd_fun_inverse [simp]

lemma nd_fun_ext: "(\<And>x. (f\<^sub>\<bullet>) x = (g\<^sub>\<bullet>) x) \<Longrightarrow> f = g"
  apply(subgoal_tac "Rep_nd_fun f = Rep_nd_fun g")
  using Rep_nd_fun_inject
   apply blast
  by(rule ext, simp)

lemma nd_fun_eq_iff: "(f = g) = (\<forall>x. (f\<^sub>\<bullet>) x = (g\<^sub>\<bullet>) x)"
  by (auto simp: nd_fun_ext)

instantiation nd_fun :: (type) kleene_algebra
begin

definition "0 = \<zeta>\<^sup>\<bullet>"

definition "star_nd_fun f = qstar f" for f::"'a nd_fun"

definition "f + g = ((f\<^sub>\<bullet>) \<squnion> (g\<^sub>\<bullet>))\<^sup>\<bullet>"

thm sup_nd_fun_def sup_fun_def

named_theorems nd_fun_ka "kleene algebra properties for nondeterministic functions."

lemma nd_fun_plus_assoc[nd_fun_ka]: "x + y + z = x + (y + z)"
  and nd_fun_plus_comm[nd_fun_ka]: "x + y = y + x"
  and nd_fun_plus_idem[nd_fun_ka]: "x + x = x" for x::"'a nd_fun"
  unfolding plus_nd_fun_def by (simp add: ksup_assoc, simp_all add: ksup_comm)

lemma nd_fun_distr[nd_fun_ka]: "(x + y) \<cdot> z = x \<cdot> z + y \<cdot> z"
  and nd_fun_distl[nd_fun_ka]: "x \<cdot> (y + z) = x \<cdot> y + x \<cdot> z" for x::"'a nd_fun"
  unfolding plus_nd_fun_def times_nd_fun_def by (simp_all add: kcomp_distr kcomp_distl)

lemma nd_fun_plus_zerol[nd_fun_ka]: "0 + x = x"
  and nd_fun_mult_zerol[nd_fun_ka]: "0 \<cdot> x = 0"
  and nd_fun_mult_zeror[nd_fun_ka]: "x \<cdot> 0 = 0" for x::"'a nd_fun"
  unfolding plus_nd_fun_def zero_nd_fun_def times_nd_fun_def by auto

lemma nd_fun_leq[nd_fun_ka]: "(x \<le> y) = (x + y = y)"
  and nd_fun_less[nd_fun_ka]: "(x < y) = (x + y = y \<and> x \<noteq> y)"
  and nd_fun_leq_add[nd_fun_ka]: "z \<cdot> x \<le> z \<cdot> (x + y)" for x::"'a nd_fun"
  unfolding less_eq_nd_fun_def less_nd_fun_def plus_nd_fun_def times_nd_fun_def sup_fun_def
  by (unfold nd_fun_eq_iff le_fun_def, auto simp: kcomp_def)

lemma nd_star_one[nd_fun_ka]: "1 + x \<cdot> x\<^sup>\<star> \<le> x\<^sup>\<star>"
  and nd_star_unfoldl[nd_fun_ka]: "z + x \<cdot> y \<le> y \<Longrightarrow> x\<^sup>\<star> \<cdot> z \<le> y"
  and nd_star_unfoldr[nd_fun_ka]: "z + y \<cdot> x \<le> y \<Longrightarrow> z \<cdot> x\<^sup>\<star> \<le> y" for x::"'a nd_fun"
  unfolding plus_nd_fun_def star_nd_fun_def
    apply(simp_all add: fun_star_inductl sup_nd_fun.rep_eq fun_star_inductr)
  by (metis order_refl sup_nd_fun.rep_eq uwqlka.conway.dagger_unfoldl_eq)

instance
  apply intro_classes
  using nd_fun_ka by simp_all

end

end

subset_name: curated


TITLE: Setting up null and alternate hypotheses
QUESTION [1 upvotes]: I'm a bit confused when it comes to setting up the null and alternate hypotheses. Should we try to support the value of the sample mean when setting up the alternate hypothesis, or verify the claim that is given in the question?
E.g.
The manager would like to use automatic cashiers, however, she believes that at least 50% of customers prefer human cashiers. In the sample of 75 customers, 28 said they prefer automatic cashiers. Is there sufficient evidence that the manager's belief is wrong?
p̂ = 47/75 = 0.63
p0 = 0.50

If we try to prove that the sample mean is true, then the null and alternate hypothesis would be:
Ho: p0<=0.5
Ha: p0>0.5

If we use the claim in the question, then null and alternate hypothesis would be:
Ho: p0>=0.5
Ha: p0<0.5

Which one is correct and why?

REPLY [2 votes]: Usually the alternative hypothesis is what we wish to show. In this case we wish to show that the manager's claim is incorrect. Letting $p$ denote the proportion of customers who prefer human cashiers, the manager being incorrect would mean $p<0.5$
We have
$$H_0: p\geq0.5$$
$$H_a: p\lt0.5$$
Since $\hat{p}=\frac{47}{75}\approx0.627$ we certainly don't have evidence to reject the null hypothesis.
The more interesting question would be to test whether the manager's claim is correct. However, the wording of the problem indicates to me that these are the hypotheses.
Note: You cannot accept a null hypothesis. You can only fail to reject it or reject it.


text:
\section{Homology cobordism}\label{s3}
In this section we prove that the composition $SH_+^{*,\le 1+\delta, \le \epsilon}(W) \to SH_+^{*,\le 1+\epsilon+2\delta}(W)\to H^{*+1}(W)\mapsto H^{*+1}(Y)$ is independent of symplectically aspherical/Calabi-Yau fillings, which,  combined with the case for the standard filling $V\times \D$, will yield the proof of Theorem \ref{thm:main}. Here the first is map is the continuation map \eqref{eqn:continuation}. We separate the proof into the symplectically aspherical case and the Calabi-Yau case. The symplectically aspherical case is more involved due to the missing of a global $\Z$ grading. But the action filtration $F_k$ will serve as a substitute of the grading.
\subsection{ The symplectically aspherical case}\label{ss31}
Let $\widehat{W\backslash Y_{\epsilon}}$ denote the completion in the negative direction of the domain in $\widehat{W}$ outside $Y_{\epsilon}$, i.e.\ $\widehat{X}$ in \S \ref{ss:NS}. Then $\widetilde{H}_{1+\epsilon+2\delta}$ is well-defined on  $\widehat{W\backslash Y_{\epsilon}}$. We consider the moduli space $\cM_{\overline{\gamma}_p,\gamma_q}(\widetilde{H}_{1+\epsilon+2\delta})$, which is the compactification of the following
$$\left\{u:\C\backslash \{*\}\to \widehat{W\backslash Y_{\epsilon}}|\partial_s u+J(\partial_tu-X_{\widetilde{H}_{1+\epsilon+2\delta}})=0, \lim_{s\to \infty} u(t) = \overline{\gamma}_p(t+\theta), \lim_{\to *} u = (\gamma_q,-\infty) \right\}/\R$$
where $\gamma_q$ is a Reeb orbit on $Y_{\epsilon}$ which is the asymptotic of the {\em free} negative puncture $*$ and $\R$ is the translation (which moves the puncture) action on $\C$. Note that $\overline{\gamma}_p,\gamma_q$ are both contractible in $Y_{\epsilon}$ with a standard bounding disk in the standard filling $V\times \D$, which can be pushed into the boundary. For $u\in \cM_{\overline{\gamma}_p,\gamma_q}(\widetilde{H}_{1+\epsilon+2\delta})$, we use $[u]$ to denote the class in $H_2(Y)$ given by capping off $u$ with the two standard disks. $u$ is called homologically trivial iff $[u]=0$. The following is based on the compactness results in \cite{bourgeois2002morse,bourgeois2009symplectic}.

\begin{proposition}\label{prop:homology}
	For $\epsilon$ sufficiently small,  all curves in $\cM_{\overline{\gamma}_p,\gamma_q}(\widetilde{H}_{1+\epsilon+2\delta})$ for any $p,q$ must be homologically trivial.
\end{proposition}
\begin{proof}
	Assume otherwise, we have $u_{\epsilon} \in \cM_{\overline{\gamma}_p,\gamma_q}(\widetilde{H}_{1+\epsilon+2\delta})$ which are not homologically trivial for $\epsilon \to 0$. Then $u_{\epsilon}$ converges to a cascade as a hybrid of \cite{bourgeois2002morse} (for the symplectization end) and \cite{bourgeois2009symplectic} (for the Hamiltonian end). The only place which can contribute nontrivial homology is the middle Floer cylinder. But in the case when $\epsilon=0$, the contact energy $\int u^*\circ \pi^*(\lambda|_{Y_0})$ must be zero, where $\pi$ is the projection from the positive symplectization $\widehat{W\backslash Y_0}Y_{0}\times [1,\infty)$ to $Y_{0}$. Hence the middle Floer cylinder is a reparametrization of a trivial cylinder, which is homologically trivial, contradiction.
\end{proof}

Although $c_1(Y)$ is not zero as long as $c_1(V)\ne 0$, as we will see below, Proposition \ref{prop:homology} implies that the relevant moduli spaces of holomorphic  curves do not pick up nontrivial first Chern classes from $V$, which allows us to compute the dimension after neck-stretching. We are interested in the cochain map $\delta_{\partial}:C_+^*(\widetilde{H}_{1+\epsilon+2\delta})\to C_0(Y)$, which computes the map $SH^{*,\le 1+\epsilon+2\delta}_+(W)\to H^{*+1}(W)\to H^{*+1}(Y)$. For this, we pick a Morse function $h$ on $Y_{\epsilon}$ with a generic metric. Then following \cite[\S 3]{zhou2019symplectic}, we know that cochain map is define by counting the following configuration.

\begin{figure}[H]\label{fig:2}
	\begin{tikzpicture}
	\draw (0,0) to [out=90, in = 180] (0.5, 0.25) to [out=0, in=90] (1,0) to [out=270, in=0] (0.5,-0.25)
	to [out = 180, in=270] (0,0) to (0,-1);
	\draw[dotted] (0,-1) to  [out=90, in = 180] (0.5, -0.75) to [out=0, in=90] (1,-1);
	\draw (1,-1) to [out=270, in=0] (0.5,-1.25) to [out = 180, in=270] (0,-1);
	\draw (1,0) to (1,-1);
	\draw[->] (1,-1) to (1.25,-1);
	\draw (1.25,-1) to (1.5,-1);
	\draw (1.5,-1) to [out=90, in = 180] (2, -0.75) to [out=0, in=90] (2.5,-1) to [out=270, in=0] (2,-1.25)
	to [out = 180, in=270] (1.5,-1) to (1.5,-2);
	\draw (2.5,-2) to [out=270, in=0] (2,-2.5) to [out = 180, in=270] (1.5,-2);
	\draw (2.5,-1) to (2.5,-2);
	\draw[->] (-0.5,0) to (-0.25,0);
	\draw (-0.25,0) to (0,0);
	\draw[->] (2,-2.5) to (2.5,-2.5);
	\draw (2.5,-2.5) to (3,-2.5);
	\node at (2.5, -2.8) {$\nabla h$};
	\node at (0.5,-0.5) {$u_1$};
	\node at (2, -1.5) {$u_2$};
	\end{tikzpicture}
	\caption{$\delta_{\partial}$ from $2$ level cascades}
\end{figure}
By the same $S^1$ equivariant transversality argument as before, any solution from a hat orbit is never rigid.  For $p\in \Crit(f),q\in \Crit(h)$, we use $\cM^{\partial}_{\check{\gamma}_p,q}$ to denote the compactified moduli space.
\begin{proposition}\label{prop:stretched}
Assume $\epsilon$ is sufficiently small. Let $W$ be a symplectic aspherical filling of $Y_\epsilon$, then the cochain morphism $\delta_{\partial}:C_+^*(\widetilde{H}_{1+\epsilon+2\delta})\to C_0(h)$ has the following property for some choice of $J$.
\begin{enumerate}
	\item $\delta_{\partial}(\hat{\gamma}_p)=0$.
	\item $\delta_{\partial}(\check{\gamma}_p) = a+b$ with $\ind(a)=\ind(p)$ and $\ind(b)>\ind(a)=\ind(p)$\footnote{Here $b$ may not have pure degree, then that $\ind(b)>\ind(a)$ means that any component of $b$ has larger Morse index compared to $a$.}, moreover $a$ does not depend on the filling,
\end{enumerate}
\end{proposition}
\begin{proof}
The first property follows from $S^1$-equivariant transversality. In order to prove the second claim, we need to prove the following two properties.
\begin{enumerate}
	\item $\la \delta_{\partial} \check{\gamma}_p, q \ra = 0$ if $\ind(p) > \ind(q)$.
	\item $\la \delta_{\partial} \check {\gamma}_p,q\ra$ is independent of the filling if $\ind(p)=\ind(q)$.
\end{enumerate}
Note that the first property holds for the standard filling $V\times \D$. For both claims, it is equivalent to prove that
$\la \delta_{\partial} \check {\gamma}_p,q\ra$ is independent of the filling if $\ind(p)\ge \ind(q)$.  Note that $\la \delta_{\partial} \check{\gamma}_p, q \ra = \# \cM^{\partial}_{\check{\gamma}_p,q}$, we claim $\cM^{\partial}_{\check{\gamma}_p,q}$ is contained outside $Y_{\epsilon}$ for sufficiently stretched almost complex structure as long as $\ind(p)\ge \ind(q)$. Assume otherwise, in the fully stretched situation, the top curve will have multiple negative punctures asymptotic to Reeb orbits on $Y_{\epsilon}$. Then by the action constraint \eqref{eqn:positive2}, there is exactly one negative puncture with the asymptotic Reeb orbit $\gamma_w$ for a critical point $w$ of $f$. The Conley-Zehnder index of $\gamma_{w}$ using the obvious disk is $n-\ind(w)+2$ following \cite[Theorem 6.3]{zhou2019symplectic}. Therefore the Floer part is a curve in $\cM_{\overline{\gamma}_p,\gamma_w}(\widetilde{H}_{1+\epsilon+2\delta})$ in Proposition \ref{prop:homology}, which has trivial homology class. Then by Proposition \ref{prop:homology}, the virtual dimension of such configuration is
$$\ind(q)-\ind(p) - (2n-\ind (w))<0, \text{ when } \ind(p)\ge \ind (q).$$
As a consequence, there is no such curve. We reach at a contradiction.
\end{proof}

\begin{remark}
In the case when $V$ is Weinstein and $c_1(V)=0$, the SFT degree ($\mu_{CZ}+n-3$) of $\gamma_p$ is bounded below by $n$. However, for general $V$ with $H^{2n-1}(V) \ne 0$ and $c_1(V)=0$, the SFT degree of $\gamma_p$ is bounded below by $1$. From the proof of Proposition \ref{prop:stretched}, we see that $H^{2n}(V)\ne 0$ is exactly the borderline case for the argument fails. The proof of Proposition \ref{prop:stretched} shows that even though there might be interesting augmentations, the augmentation does not affect the part we are interested in. The situation changes dramatically when $V$ becomes closed, i.e.\ if we consider negative line bundles over a symplectically aspherical manifold $V$. Then by \cite{oancea2008fibered}, the symplectic cohomology is zero. But now the augmentation to the Reeb orbit corresponding to $H^{2n}(V)$ plays an essential role. And the elimination pattern is completely different, in particular, $1$ is only killed after we include the $n$th-multiple covers of the simple Reeb orbits, see \cite{ritter2014floer}.
\end{remark}

\begin{proof}[Proof of Theorem \ref{thm:main} for the symplectically apherical case] We first assume $W$ is exact for simplicity. Combining Proposition \ref{prop:filtration}, \ref{prop:better} and \ref{prop:stretched} together, we know that $\Phi:C_+^*(K_{1+\delta}\oplus\epsilon(f-1))\to C_+^*(H_{1+\epsilon+2\delta})\to C_+^*(\widetilde{H}_{1+\epsilon+2\delta})\to C^{*+1}(Y)$ preserves the index filtration, and the map on the  associated graded group is independent of fillings. Then on the associated graded group of cohomology, the induced map $\oplus \Phi_k$ is also independent of fillings. Since for the standard filling $V\times \D$, $\oplus \Phi_k$ is injective on the check component (the quotient of hat component). This implies that $\Phi$ must be injective for any filling on the check component. On the other hand, note that $1\in \Ima \Phi_0$ for $V\times \D$, therefore $1+A\in \Ima \Phi$ for some $A\in \oplus_{i>0}H^{2i}(Y)$ by the $\Z/2$ grading. By Proposition \ref{prop:kill}, we have that $\delta:SH_+^{*,\le 1+\delta,\le\epsilon}(W)\to H^{*+1}(W)$ is surjective and $SH^*(W)=0$.  Moreover, by Proposition \ref{prop:checkhat}, $\delta$ factors through the projection to the check component. Then the injectivity of $\Phi$ on the check component implies that $\delta$ is an isomorphism on the check component. Hence $H^*(W)\to H^*(Y)$ is also injective. Therefore, to finish the proof, it is sufficient to show that the image of $H^*(W)\to H^*(Y)$ is also independent of filling.  The injectivity of $\oplus \Phi_k$ implies that  $\Ima \oplus \Phi_k=\oplus \left((\Ima \Phi\cap F_kH^*(Y))/(\Ima \Phi\cap F_{k+1}H^*(Y))\right)$. On the other hand, the filtration on $H^*(Y)$ is the natural filtration by grading and $\Ima \Phi$ is the image of  $H^*(W)\to H^*(Y)$, where the associated graded groups are naturally isomorphic to the original groups. Hence $\Ima \oplus \Phi_k$ is naturally isomorphic to $\Ima \Phi$. The invariance of the former implies that $\Ima \Phi$ is independent of the filling, the claim follows. The claim on homology cobordism is from Proposition \ref{prop:homologycob} below.

When $W$ is only symplectically spherical, the symplectic action is well-defined for contractible orbits but not necessarily in the form on \eqref{eqn:action}. But since all the relevant orbits $\overline{\gamma}_p$ are contractible inside the cylindrical end of the boundary, the symplectic actions of those orbits are indeed given by \eqref{eqn:action}.  Therefore the same argument above goes through for symplectically spherical fillings. Note that $V\times \D$ is built from handles with indices at most $2n-1$. As a consequence, we have that $H^1(V\times \D) \to H^1(Y)$ is an isomorphism. Combining with the fact that $H^2(V\times \D)\to H^2(Y)$ is injective, we know that the symplectic form $\omega$ on a symplectically aspherical filling $W$ is necessarily exact and has a primitive whose restriction on the boundary is the original contact form. This proves that $W$ is an exact filling.
\end{proof}

\subsection{The Calabi-Yau case}
First of all, the symplectic cohomology and positive symplectic cohomology are defined for Calabi-Yau fillings using the Novikov coefficient $\Lambda$ over $\Q$, see \cite[\S 8]{zhou2019symplectic}. In particular, the reason that positive symplectic cohomology is defined is no longer for action restrictions but because of the asymptotic behavior lemma \cite[Lemma 2.3]{cieliebak2018symplectic}. Similar to Proposition \ref{prop:kill}, we have the following analogue for Calabi-Yau fillings (i.e.\ a strong filling $W$ such that $c_1(W)$ is torsion) due to the fact that $1$ is a unit in $QH^*(W;\Lambda)$, which is $H^*(W;\Lambda)$ as a group.
\begin{proposition}\label{prop:kill2}
	Let $W$ be a Calabi-Yau filling and if $1$ is in the image of $SH^{*,\le a, \le b}_+(W;\Lambda) \to QH^{*+1}(W) $ is zero. Then $SH^*(W;\Lambda)=0$ and $SH^{*,\le a, \le b}_+(W;\Lambda)\to QH^{*+1}(W;\Lambda)$ is surjective.
\end{proposition}
\begin{proposition}\label{prop:exact}
	Let $W$ be a Calabi-Yau filling of $Y$, then $W$ is symplectically aspherical.
\end{proposition}
\begin{proof}
	In the Calabi-Yau case, we have a $\Z$ grading on symplectic cohomology. In particular,  we do not need Proposition \ref{prop:homology} to control the homology class and all generators of $C^*(K_{1+\delta}\oplus \epsilon(f-1))$ have a well-defined grading, since all of them are contractible. More explicitly, the grading is given by $|p\otimes \check{\gamma}_0|=\ind(p)-1, |p\otimes \hat{\gamma}_0|=\ind(p)-2$ and the Conley-Zehnder index of $\gamma_p$ is given by $n+2-\ind(p)$, in particular, the SFT degree $\mu_{CZ}(\gamma_p)+(n+1)-3$ is positive. There is also no need to use $H_{1+\epsilon+2\delta}$ or $\widetilde{H}_{1+\epsilon+2\delta}$. We can apply the same argument in Proposition \ref{prop:stretched} to $K_{1+\delta}\oplus \epsilon(f-1)$ directly, which shows that $\delta_{\partial}:H^*(C_+(K_{1+\delta}\oplus\epsilon(f-1)))\to H^*(Y;\Lambda)$ is independent of the Calabi-Yau filling. Then by Proposition \ref{prop:kill2}, we have $H^*(W;\Lambda)\to H^*(Y;\Lambda)$ is isomorphic to $H^*(V\times \D;\Lambda) \to H^*(Y;\Lambda)$ similar to the symplectically aspherical case. In particular, we have that $H^*(W;\Q)\to H^*(Y;\Q)$ is injective. As a consequence, $\omega$ is an exact form, i.e.\ $W$ is symplectically aspherical.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm:main} for the Calabi-Yau case]
	It follows from Proposition \ref{prop:exact} and the symplectically aspherical case of Theorem \ref{thm:main}.
\end{proof}

\begin{remark}
	When $c_1(V)=0$, it was shown in \cite{zhou2019symplectic} that $\partial (V\times \D)$ is asymptotically dynamically convex. For Calabi-Yau fillings of $\partial(V\times \D)$,  the index neck-stretching argument in \cite{zhou2019symplectic} requires that $\partial V$ has a Reeb dynamics with Conley-Zehnder indices bounded from below. But the index neck-stretching argument is still applicable if we attach additional flexible handles, while the action neck-stretching in this paper breaks down.
\end{remark}

\subsection{Homology cobordism}
A cobordism $W$ from $\partial_0W$ to $\partial_1W$ is called a homology cobordism iff $\partial_0W \to W$ and $\partial_1 W \to W$ both induce isomorphism on homology.
\begin{proposition}\label{prop:homologycob}
	Under the assumption in Theorem \ref{thm:main}, the filling $W$ can be obtained from $V\times \D$ by attaching a homology cobordism from $Y$ to $Y$.
\end{proposition}
\begin{proof}
	Let $W_0$ be a copy of $V\times \D(\epsilon) \subset V\times \D$ placed near $Y=\partial (V\times \D)$ for $\epsilon$ small. Then for any symplectically aspherical/Calabi-Yau filling $W$, we can assume $W_0$ is also contained in $W$. Let $X$ denote the cobordism from $\partial W_0$ to $\partial W$. We can assume $W_0$ is inside the strip $Y \times (1-3\epsilon,1)$ near the boundary. In particular, $H^*(Y \times (1-3\epsilon,1)) \to H^*(W_0)$ is an isomorphism when restricted to the image of $H^*(V\times \D )\hookrightarrow H^*(Y)$. Since $H^*(W)\to H^*(Y)$ is independent of filling, we have that $H^*(W)\to H^*(W_0)$ is an isomorphism. Therefore $H^*(X,\partial W_0)=0$ by excision. By Lefschetz duality and the universal coefficient theorem, we have that $H_*(X,\partial W_0)$ and $H_*(X,\partial W)$ are both zero. Hence $X$ is a homology cobordism.
\end{proof}

\subsection{General strong fillings}
The obstruction of applying Proposition \ref{prop:kill} for general strong fillings is that we may have a zero divisor $1+A$ in $QH^*(W;\Lambda)$ for $A\in \oplus_{i>0}H^{2i}(W;\Lambda)$, since $(1+A)\cup \cdot :QH^*(W;\Lambda)\to QH^*(W;\Lambda)$ is a linear map between finite dimensional $\Lambda$-spaces. For general strong fillings, symplectic cohomology and positive symplectic cohomology can be defined as usual if one applies a suitable virtual technique to overcome the transversality issue. For simplicity, we assume the strong filling is semi-positive as in \cite[Definition 6.4.1]{mcduff2012j}, so that the theory can be defined using generic almost complex structures.

\begin{proposition}\label{prop:sphere}
	Assume $W^{2n+2}$ is a strong (semi-positive) filling, such that there is no embedded symplectic sphere $S$, with $2-n\le c_1(S) \le 2n-1$, then there is no zero divisor of $QH^*(W;\Lambda)$ in the form of $1+A$ for $A\in \oplus_{i>0}H^{2i}(W;\Lambda)$.
\end{proposition}
\begin{proof}
	If there is a such zero divisor, we claim there exists $B\in \oplus_{i>0} H^{2i}(W;\Lambda)$ such that $(1+A)\cup B=0$. First of all, there is a $\Z/2$ grading, hence there exist $B\in \oplus_{i>0} H^{2i}(W;\Lambda)$ and $b\in \Q$ such that $(1+A)\cup (b+B)=0$.  Note that $\langle C\cup D, 1\rangle=0$ for any $C,D\in \oplus_{i>0} H^{2i}(W;\Lambda)$, since the corresponding moduli space counts curves with a point constraint. However, such a moduli space must be empty as we can choose the point constraint near the boundary, and the maximal principle will obstruct such curve. Since $0=(1+A)\cup (b+B)=b+bA+B+A\cup B$, we must have $b=0$, i.e.\ the claim holds.
	Then to have $(1+A)\cup B=0$, the quantum product $\oplus_{i>0} H^{2i}(W;\Lambda)\otimes \oplus_{i>0} H^{2i}(W;\Lambda) \to \oplus_{i>0} H^{2i}(W;\Lambda)$ must be deformed, hence there must be some holomorphic sphere (possibly nodal) $S$, such that $6\le 2c_1(S)+2n+2 \le 6n$, i.e.\ $2-n\le c_1(S)\le 2n-1$. When $n\ge 2$, a (nodal) holomorphic sphere can be perturbed into an embedded symplectic sphere with the same first Chern class, hence a contradiction. When $n=1$, it is always semi-positive, and the curve contributing to the deformation of the product is necessarily somewhere injective by \cite[\S 6.6]{mcduff2012j}, hence the curve can be assumed to an embedded symplectic sphere, which is a contradiction.
\end{proof}

\begin{proof}[Proof of Corollary \ref{cor:strong}]
	The proof follows from the same argument for Theorem \ref{thm:main}. Although we do not have a well-defined symplectic action for strong fillings, but the continuation maps used in the proof of Theorem \ref{thm:main} can be described by moduli spaces contained outside the boundary by neck-stretching, where the symplectic action is well-defined and can be used to restrict Floer trajectories. Then by Proposition \ref{prop:sphere}, we still have $SH^*(W)=0$ and $H^*(W;\Lambda)\to H^*(Y;\Lambda)$ is always injective. Hence the symplectic form is exact.
\end{proof}

\section{$h$-cobordisms}
In this section, we will upgrade the homology cobordism $X$ in Proposition \ref{prop:homologycob} to an $h$-cobordism assuming $\pi_1(Y)$ is abelian. Unlike the cohomology information, we can not quite get the full information on the fundamental group or more generally higher homotopy groups. But in the case of $\pi_1(Y)$ abelian, we do have enough ingredients to get some information on $\pi_1$ and conclude an $h$-cobordism.
\subsection{Symplectic cohomology of covering spaces}

Recall from \cite[\S 3.3]{zhou2019symplectic}, for every covering space $\widetilde{W}\to W$, we can define the symplectic cohomology of the covering space. A cochain is a sum of formal sums of different lifts of periodic orbits on $W$. The differential is defined by lifting the differential on $W$ according to the (unique) parallel transportation. In particular, we have the following commutative long exact sequences,
\begin{equation}\label{eqn:dia}
\xymatrix{
\ldots \ar[r] & H^*(W) \ar[r]\ar[d] & SH^*(W) \ar[r]\ar[d] & SH^*_+(W)\ar[r]\ar[d] & H^{*+1}(W) \ar[r]\ar[d] & \ldots\\
\ldots \ar[r] & H^*(\widetilde{W}) \ar[r] & SH^*(\widetilde{W}) \ar[r] & SH^*_+(\widetilde{W})\ar[r] & H^{*+1}(\widetilde{W}) \ar[r] & \ldots
}
\end{equation}
Similarly for the filtered version. Note that we use only contractible orbits to define $SH^*(W)$. A map from the thrice punctured sphere (i.e.\ a pair of pants) can be completed (as a topological map) to a map from sphere, as all asymptotics are contractible orbits. As a consequence, we can lift the map to the universal cover and gives $SH^*(\widetilde{W})$ a unital ring structure. Then the first square in \eqref{eqn:dia} is a commutative square of unital rings.

\begin{remark}
	One can also define a symplectic cohomology with local coefficients, i.e.\ the underlying cochain complex is the free $\Z[\pi_1]$ module generated by periodic orbits. The differential is again the lifting of the ordinary differential and it respects the $\Z[\pi_1]$ module structure. The corresponding cohomology is the same as the symplectic cohomology of the universal cover if $\pi_1$ is finite. If $\pi_1$ is infinite, the symplectic cohomology of the universal cover allows generators which can be viewed as an infinite sum in the group ring. On the regular cohomology level, i.e.\ the Morse theory level, the cohomology with local coefficient $H^*(W;\Z[\pi_1])$ is the compactly supported cohomology of the universal cover $H_c^*(\widetilde{W})\ne H^*(\widetilde{W})$, see \cite[Proposition 3H.5]{hatcher2002algebraic}. It still carries a product structure (the pair of pants construction holds), but it is not unital ($1\in H^*(\widetilde{W})$ is represented by an infinite sum in the group ring).
\end{remark}

\begin{proposition}\label{prop:fundamental}
		If $\pi_1(Y)$ is abelian, then we have $\pi_1(V)$ is abelian  and $\pi_1(Y)\to \pi_1(V\times \D)$ is an isomorphism. Then for any exact/symplectically aspherical  filling $W$, we have $\pi_1(Y) \to \pi_1(W)$ is an isomorphism. Moreover, $H^*(\widetilde{W}) \to H^*(\widetilde{Y})$ is injective and independent of fillings for the universal covers.
\end{proposition}
\begin{proof}
	Since $V\times \D$ can be built from handles with indices at most $2n-1$ (i.e.\ co-indices at least $3$), we have $\pi_1(Y)\to \pi_1(V\times \D)$ is an isomorphism. In particular, $H_1(Y)\to H_1(V\times \D)$ is an isomorphism. Then by universal coefficient theorem, we have $H^1(V\times \D) \to H^1(Y)$ and $\tor H^2(V\times \D) \to \tor H^2(Y)$ are isomorphisms.  As a consequence, \eqref{t1} of Theorem \ref{thm:main} implies that $H_1(Y)\to H_1(W)$ is an isomorphism. Therefore $\pi_1(Y)\to \pi_1(W)$ is at least injective.  Then $\pi_1(Y)\to \pi_1(W)$ is surjective by the same argument in \cite[Theorem 3.16]{zhou2019symplectic} by considering the symplectic cohomology of the universal cover. The only difference is replacing the grading in \cite{zhou2019symplectic}  with  the associated graded group from the filtration. Hence $\pi_1(Y)\to \pi_1(W)$ is an isomorphism. The independence of $H^*(\widetilde{W})\to H^*(\widetilde{Y})$ then follows from the same proof of Theorem \ref{thm:main}.
\end{proof}

\begin{lemma}\label{lemma:ext}
	For a $\Z$-module $A$, if $\Hom(A,\Z)=\Ext(A,\Z)=0$, then $A=0$.\footnote{This is from Eric Wofsey's solution to \href{https://math.stackexchange.com/questions/1734222/does-trivial-cohomology-imply-trivial-homology-does-operatornamehoma-math}{https://math.stackexchange.com/questions/1734222/does-trivial-cohomology-imply-trivial-homology-does-operatornamehoma-math}}
\end{lemma}
\begin{proof}
	Since $\Ext(\cdot,\Z)$ turns injective maps into surjective maps by $\Ext^2(\cdot,\Z)=0$, we have $\Ext(B,\Z)=0$ for any $B\subset A$. Therefore any finitely generated subgroup of $A$ is free. Hence $A$ is torsion free. Next we fix a prime $p$. Since $A$ is torsion free, we have short exact sequence $0\to A \stackrel{p\times }{\to} A \to A/pA \to 0$, which induces exact sequence
	$$\Hom(A,\Z)\to \Ext(A/pA, \Z) \to \Ext(A,\Z).$$
	Hence $\Ext(A/pA,\Z)=0$. But $A/pA$ is a direct sum of copies of $\Z/p$. Therefore $A/pA=0$ for any $p$. Then $A$ is a divisible torsion free group, hence a $\Q$-vector space. Since $\Ext(\Q,\Z)\ne 0$, we have $A=0$.
\end{proof}

We also need the following form of the universal coefficient theorem.
\begin{lemma}\label{lemma:univ}
Let $R$ be a ring and  $(C_*,\partial)$ be a cochain complex of $R$ modules such that $H_*(C_*)=0$, then $H^*(\hom_R(C_*,R))=0$.
\end{lemma}
\begin{proof}
    We use $B_n\subset C_n$ to denote the image of $\partial$ and $Z_n\subset C_n$ to denote the kernel of $\partial$. By assumption, we have $B_n=Z_n$. Note that we have  a tautological short exact sequence,
    $$0\to Z_n\to C_n\to B_{n-1}\to 0.$$
    Then the tautological short exact sequence
    $$0\to Z_n/B_n \to C_n/B_n \to C_n/Z_n \to 0$$
    is
    $$0\to 0 \to C_n/B_n \to B_{n-1}\to 0.$$
    If we use $M^*$ to denote $\hom_R(M,R)$ for a $R$-module $M$, then we have the following
    $$
    \xymatrix{
    & & 0 & & \\
    B_n^* & C_n^* \ar[l] & \left(\displaystyle\frac{C_n}{B_n}\right)^* \ar[l]\ar[u] & 0\ar[l] & 0\\
    & & B_{n-1}^*\ar[u] & & Z_{n-1}^*\ar[ll]^{\simeq} \ar[u]  \\
    & & 0\ar[u] & & C_{n-1}^*\ar[ull]\ar[u]}
    $$
    where all vertical and horizontal lines are exact.

    Since the coboundary $\partial^*$ on $C^*$ is defined as the map from $C_{n-1}^*$ to $C_n^*$ in the diagram above. As a consequence, we have $\ker \partial^*$ is $\ker [C_{n-1}^*\to B_{n-1}^*]$. On the other hand, we have $\Ima \partial^*=\Ima (Z_{n-2}^*\to C^*_{n-1})$. Since $B_{n-2}^*\simeq Z_{n-2}^*$ and $(C_{n-1}/B_{n-1})^*\simeq B_{n-2}^*$, we have $\Ima \partial^*=\Ima ((C_{n-1}/B_{n-1})^*\to C^*_{n-1})=\ker [C_{n-1}^*\to B_{n-1}^*] = \ker \partial^*$ by the exactness of the topmost row. That is $H^*(\hom_R(C_*,R))=0$.
\end{proof}
\begin{remark}
    The general universal coefficient theorem (which computes cohomology from homology) needs some assumptions on $R$ or $C_*$, see \cite[\S 3.6.5]{MR1269324}. The above version without any assumption works because $C_*$ is acyclic, which is sufficient for our purpose in Proposition \ref{prop:cob}.
\end{remark}

\begin{proposition}\label{prop:cob}
	If $\pi_1(Y)$ is abelian, then any exact/symplectically aspherical filling $W$ is $V\times \D$ glued with an $h$-cobordism from $Y$ to $Y$.
\end{proposition}
\begin{proof}
	Let $X$ be the homology cobordism from $\partial W_0$ to $\partial W$ in the proof of Proposition \ref{prop:homologycob}. By Proposition \ref{prop:fundamental}, we know that $\partial W_0\hookrightarrow W_0\hookrightarrow W$ both induce isomorphisms on $\pi_1$. Then the van Kampen theorem implies that the following push-out diagram consists of isomorphisms,
	$$\xymatrix{\pi_1(\partial W_0) \ar[r]\ar[d] & \pi_1(W_0)\ar[d]\\
		\pi_1(X) \ar[r] & \pi_1(W)}$$
	Then we have $\pi_1(\partial W) \to \pi_1(X)$ is an isomorphism since  $\pi_1(\partial W) \to \pi_1(W)$ is an isomorphism. Applying the argument in Theorem \ref{thm:main} to the universal cover, we have $H^*(\widetilde{W})\to H^*(\widetilde{W_0})$ is an isomorphism by Proposition \ref{prop:fundamental}. Hence by excision, we have $H^*(\widetilde{X}, \widetilde{\partial W_0})=0$. Then by universal coefficient and Lemma \ref{lemma:ext}, we have $H_*(\widetilde{X},\widetilde{\partial W_0})=H_*(X,\partial W_0;\Z[\pi_1])=0$.
	Then by Lemma \ref{lemma:univ} for $R=\Z[\pi_1]$, we have $H^*(X,\partial W_0;\Z[\pi_1])=0$. Then by the Lefschetz duality with twisted coefficients, $H_*(X,\partial W;\Z[\pi_1])=H_*(\widetilde{X},\widetilde{\partial W})=0$. Therefore $X$ is an $h$-cobordism by Whitehead's theorem.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:diff}]
	If the Whitehead group of $Y$ is trivial, then the $h$-cobordism is a trivial cobordism by the $s$-cobordism theorem \cite{milnor1966whitehead}. Hence $W$ is diffeomorphic to $V\times \D$. In general, we can apply the Mazur trick, see \cite{milnor1966whitehead}. That is there is an $h$-cobordism $X'$ from $Y$ to itself such that the concatenations $X\circ X'$ and $X'\circ X$ are trivial cobordisms. Note that $\mathring{W}$ is diffeomorphic to $\widehat{W}$, which is diffeomorphic to $\ldots \circ X\circ X'\circ W$, i.e.\ attaching infinite $X\circ X'$ to $W$. On the other hand, it is $\ldots \circ X \circ X'\circ X \circ V\times \D$, which is diffeomorphic to $\widehat{V}\times \C$ or the interior of $V\times \D$.
\end{proof}
\begin{remark}
	The Whitehead torsion can be put into the framework of Floer theories \cite{abouzaid2018simple}. One can prove that the Whitehead torsion of the cochain map underlying the isomorphism $SH_+^*(W) \to H^{*+1}(Y)\to H^{*+1}(V\times \{1\})$ has zero Whitehead torsion assuming $\pi_1(Y)$ is abelian. What we still need is that the Whitehead torsion of $SH_+^*(W) \to H^{*+1}(W)$ is zero.
\end{remark}

It is very likely that the diffeomorphism type of the filling is unique for any Liouville domain $V$, or at least the homotopy type is unique. However, this requires better ways to probe homotopy groups of the filling, hence we end the paper by asking the following question.
\begin{question}
    Is there a Floer theoretic interpretation of homotopy groups? In particular, is it true that $\pi_k(W)\to \pi_k(Y)$ are independent of exact fillings for any Liouville domain $V$ and $k\ge 1$? What can we say about the Whitehead product on $\pi_k(W)$.
\end{question}

config: arxiv


TITLE: Calculate Cov(X,Y) where X is #children with no books and Y is #children with exactly 1 book while distributing r books to n children.
QUESTION [0 upvotes]: Suppose $r\geq 1$ distinct books are distributed at random among $n\geq 3 $ children. Let $X$ be the number of children who do not get any book, and $Y$ be the number of children who get exactly one book. Then find an expression for $\operatorname{Cov}(X,Y)$.
Given $r$, and $n$, $X$ has support $\{0,1,.....,n-1\}$. Calculating its density function is straightforward,
$$ \Pr(X=x) = \frac{{n \choose x}(n-x)^{r}}{n^r}
$$
I'm not really sure how to calculate a general expression for density of $Y$ and their joint density.

REPLY [0 votes]: For each $1\leq j \leq n$ take $N_j\sim \text{Binomial}\left(r,\frac{1}{n}\right)$ as the number of books that child $j$ receives.
Put $X_j=1_{\{N_j=0\}}$ and $Y_j=1_{\{N_j=1\}}$. Notice how $X=\sum_{j=1}^nX_j$ and $Y=\sum_{j-1}^nY_j$.
Let's examine $\text{cov}(X_j,Y_j)$ for $j\in\{1,...,n\}$
fixed.
$$\begin{eqnarray*}\text{cov}(X_j,Y_j)&=& \mathbb{E}(X_jY_j)-\mathbb{E}(X_j)\mathbb{E}(Y_j) \\ &=& \mathbb{P}(X_j=1,Y_j=1) - \mathbb{P}(X_j=1)\mathbb{P}(Y_j=1) \\ &=& \mathbb{P}\left(N_j=0,N_j=1\right)-\mathbb{P}(N_j=0)\mathbb{P}(N_j=1) \\ &=& 0-\left[{r \choose 0} \left(\frac{1}{n}\right)^0 \left(1-\frac{1}{n}\right)^{r}\right]\left[{r \choose 1} \left(\frac{1}{n}\right)^1 \left(1-\frac{1}{n}\right)^{r-1}\right] \\ &=& - \frac{r}{n}\left(1-\frac{1}{n}\right)^{2r-1}\end{eqnarray*}$$
Now let's study $\text{cov}(X_i,Y_j)$ for $i\neq j$:
$$\begin{eqnarray*}\text{cov}(X_i,Y_j) &=& \mathbb{E}(X_iY_j)-\mathbb{E}(X_i)\mathbb{E}(Y_j) \\ &=& \mathbb{P}(X_i=1,Y_j=1)-\mathbb{P}(X_i=1)\mathbb{P}(Y_j=1) \\ &=& \mathbb{P}\left(\text{Child } i \text{ gets no books & child }j\text{ gets one book}\right)- \frac{r}{n}\left(1-\frac{1}{n}\right)^{2r-1} \\ &=& {r \choose 1}\frac{1}{n} \cdot \left(\frac{n-2}{n}\right)^{r-1}- \frac{r}{n}\left(1-\frac{1}{n}\right)^{2r-1}\end{eqnarray*}$$
We get $$\begin{eqnarray*}\text{cov}\left(X,Y\right) &=& \text{cov}\left(\sum_{j-1}^n X_j,\sum_{j=1}^n Y_j\right) \\ &=& \sum_{j=1}^n \text{cov}\left(X_j,Y_j\right)+\sum_{i\neq j}\text{cov}(X_i,Y_j) \\ &=& n\text{cov}(X_1,Y_1)+(n^2-n)\text{cov}(X_1,Y_2)\end{eqnarray*}$$ Plug in the formulas we obtained above into $\text{cov}(X_1,Y_1),\text{cov}(X_1,Y_2)$ to finish the job.


TITLE: Order not preserved by Galois
QUESTION [2 upvotes]: Let $K/\mathbb Q$ be a Galois extension. I'm trying to come up with an example of an order (necessarily nonmaximal) in $K$ which is not preserved by Galois. For simplicity, I've looked at monogenic orders $\mathcal O_K [\alpha]$ where the question boils down to finding a case where $\mathbb Z[\alpha]$ does not contain all the conjugates of $\alpha$.
Working locally, it seems like if the valuation of $\alpha$ is very large, it should be unlikely that a given conjugate $\alpha'$ of $\alpha$ will be in $\mathbb Z_p[\alpha]$. My thinking is that here, $\alpha$ and $\alpha'$ differ by a unit, but if $v(\alpha)$ is large then the order is a lot smaller, and becomes less and less likely to contain the necessary unit. However, I haven't been able to work out an example along these lines.
Edit: the current answer is good, but I would be more interested in one in the local case that I outlined.

REPLY [1 votes]: $$O=\Bbb{Z}+(1+2\zeta_8+2i)\Bbb{Z}[\zeta_8]\qquad \ne \qquad O'=\Bbb{Z}+(1+2\zeta_8^{-1}-2i)\Bbb{Z}[\zeta_8]$$ $(1+2\zeta_8+2i)$ is a prime ideal above $3$ and residue field $\Bbb{F}_9$.
The image of $O$ in $\Bbb{Z}[\zeta_8]/(3,1+2\zeta_8+2i)$ has 3 elements while the image of its complex conjugate order $O'$ has 9 elements.
Something similar will work whenever $K$ is not a quadratic field.


TITLE: Can overdetermined systems have infinitely many solutions?
QUESTION [2 upvotes]: Can overdeterminend systems have infinitely many solutions? IF so, can someone point me to an example of one?

REPLY [5 votes]: A system with more equations than variables is called overdetermined. It can be either consistent or inconsistent. If one or more equations are a linear combination of the other, you can always obtain a consistent system.For example, $$x+2y=4$$$$3x+6y=12$$$$5x+10y=20$$$$......$$