TITLE: How can I prove this statement about filters?
QUESTION [0 upvotes]: I have the following question:

Let $M$ be a finite set an $F$ a filter on $M$. Show that this filter has to be fixed

My Idea was the following:
Since $M$ is finite also $P(M)$ is. But since $F\subset P(M)$ we have that $$|F|\leq |P(M)|=2^{|M|}$$i.e. $F$ is finite. So let us numerate all the elements in $F$, i.e. $$F=\{A_1,...A_n\}$$ we can indeed do this since $F\neq \emptyset$.  Now let us look at $$A_1\cap A_2\cap...\cap A_n\,\,\,\,\,\,\,\,\,(1)$$ We know by thefinition of a filter that $$U,V\in F\Rightarrow U\cap V\in F$$Using this property we can proceede inductivly on the index in $(1)$ and get that $$\bigcap_{i=1}^n A_i\in F$$But since we know that $\emptyset \notin F$ we have that $\bigcap_{i=1}^n A_i\neq \emptyset$ and thus $F$ is fixed.
Does this work like this or is it wrong?
Thank you for your help

REPLY [1 votes]: Yes, this is true  and your proof valid. A filter $\mathcal F$ on a finite set $X$ is a subset of $\mathscr{P}(X)$ and hence is finite, i.e. $\mathcal{F}$ is a finite set of subsets. One axioms of filters tells us that $\mathcal{F}$ is closed under binary intersection, and indeed a standard induction argument (which is often skipped as too self-evident) shows that in fact $F_0:=\mathcal{F}$ is closed under all finite intersections, and in particular $\bigcap \mathcal{F} \in \mathcal F$. As $F_0 \neq \emptyset$ (a filter does not contain $\emptyset$ by another axiom), this means that $\mathcal F$ is a fixed filter (i.e. has non-empty intersection).
Using the enlargement axiom it is in fact easy to see that then the following holds:
$$\mathcal F = \{A \subseteq X\mid F_0 \subseteq A\}$$


TITLE: What is the spin-orbit force?
QUESTION [1 upvotes]: I am trying to brush up on my physics knowledge, and I have run across a term used to justify some results in a few texts that I don't recognize, the spin-orbit force $\vec s \cdot \vec l$. Quickly googling around has not yielded much, if any information, and it's just brushed over as being trivial in texts (as are most concepts, I'm finding). Would anyone be able to explain this concept to me, either conceptually or mathematically? My apologies if I'm just missing a core concept I should know inherently, my education is not in physics.

REPLY [2 votes]: Magnetic fields repel or attract each other.
An electric charge performing a rotation (angular momentum $\mathbf{L}$) generates a magnetic field $\mathbf{B}_1 \propto \mathbf{L}$.
An electron has an intrinsic magnetic field (as if it had a bar magnetc inside) due to its spin $\mathbf{S}$, $\mathbf{B}_2\propto \mathbf{S}$.
These two magnetic fields have an interaction that goes as $\mathbf{B}_1 \cdot \mathbf{B}_2$, and hence $\mathbf{L} \cdot \mathbf{S}$.


text:
\begin{document}

\maketitle

\begin{abstract}
The independence polynomial of a hypergraph is the generating function for its independent (vertex) sets with respect to their cardinality. This article aims to discuss several recurrence relations for the independence polynomial using some vertex and edge operations. Further, an extension of the well-known recurrence relation for simple graphs to hypergraphs is proven and other novel recurrence relations are also discussed.
\\[+1.6ex]
\noindent\emph{Keywords:} independence polynomial, hypergraph, recurrence relation, independent sets, edge operation, vertex operation, graph, vertex cover polynomial
\\[+1.6ex]
\noindent \emph{Mathematics Subject Classification (2010):} 05C30, 05C31, 05C65, 05C69
\end{abstract}

\input{introduction}
\input{preliminaries}
\input{known_recurrence_relations}
\input{new_recurrence_relations}
\input{vertex_cover_polynomial}
\input{discussion}

\section*{Acknowledgement}
Many thanks are due to Julian A. Allagan for his suggestions improving the presentation of this paper.

\phantomsection
\addcontentsline{toc}{section}{References}

\printbibliography

\end{document}

config: arxiv


TITLE: Quotient Space on Normed Vector Spaces, Notation and Intuition
QUESTION [0 upvotes]: Let $X$ be a normed vector space and $M$ a subspace of $X$.
In Folland's text on Real Analysis, he defines an equivalence relation on $X$ as follows: $x \sim y$ iff $x - y \in M$. From this, he denotes the equivalence class of $x \in X$ by $x + M$. What is the intuition behind this equivalence relation? Why is it important in functional analysis?
Later he states a theorem (here he assumes the scalar field is $\mathbb{C}$:

Let $X$ be a normed vector space. If $M$ is a closed subspace of $X$ and $x \in X \setminus M$, there exists $f \in X^*$ such that $f(x) \neq 0$ and $f|M = 0$. In fact, if $\delta = \inf_{y \in M} \|x - y\|$, $f$ can be taken to satisfy $\|f\| = 1$ and $f(x) = \delta$.

In the proof he begins by defining a function $f$ on $M + \mathbb{C}x$. What exactly is the space $M + \mathbb{C}x$? Is it the set of all equivalence classes for $x \in \mathbb{C}$?

REPLY [1 votes]: To answer the first part of your question, the equivalence relation $x \sim y \Leftrightarrow x - y \in M$ is a preliminary step in defining the quotient space $X/M$. The elements of $X/M$ are precisely the equivalence classes $\{[x]_{\sim} : x \in X\} = \{x + M : x \in X\}$, and you can define addition and scalar multiplication on $X/M$ in the obvious way. The intuition of the equivalence relation is essentially the same as that which underlies quotient spaces in general - you are treating the elements of $M$ collectively as the "zero element" of your newly-defined space. If, furthermore, $M$ is a closed subspace of $X$, then one can define a norm on $X/M$ by $\| [x] \|_{X/M} = \inf\{\|x - m \|_X : m \in M\}$ (if $M$ is not closed, then this will not be a norm in general since you end up with non-zero elements of $X/M$ being mapped to zero by $\| \cdot \|_{X/M}$). And this is how you construct the normed space $(X/M, \|\cdot\|_{X/M})$.
As for the second part of your question, for $x \in X$ (not $x \in \mathbb{C}$ as you presumably unintentionally wrote), you can think of $M + \mathbb{C}x$ as the smallest subspace of $X$ that contains $M \cup \{x\}$. Clearly, if $x \in M$, then $M + \mathbb{C}x = M$. Otherwise, $M + \mathbb{C}x$ is a subspace of $X$ that properly contains the subspace $M$.


TITLE: In a universe where the speed of light is infinite, are relativistic models and Newtonian models equivalent?
QUESTION [10 upvotes]: Take our universe. Observations are consistent with relativity, but not  consistent with Newtonian mechanics. Assume that our current (relativistic) model of gravitation is correct. Now increase $c$ to infinity. Experimental observations in this altered universe are still consistent with relativity (by the assumption that relativity is correct). Are they now consistent with Newtonian mechanics as well?

REPLY [8 votes]: In terms of special relativity versus Newtonian mechanics, the answer is yes.  However in terms of general relativity this would be a much different universe.  Instead of having curvature in a 4 dimensional Minkowki space-time, We would have a 1 dimensional time and a 3 dimensional space universe with no possible connections between space and time. So I don't think the equations of general relativity would result in the kind of gravitational forces we experience in our universe.

REPLY [4 votes]: A better definition of an 'infinite speed of light' is needed. If by that you mean that observers in relative motion will view the same space up to isometry and observe equal passages of time between events, then yes. However, there is no such thing as a 'relatavistic model' in Galilean space-time.
For example, electrodynamics is obviously Lorenz invariant. If you take as axioms Coulomb's law of electrostatic force and the Lorenz invariance of electric charge, you can derive the magnetic field (the Lorenz force may be derived from the Coulomb expression of force by Lorenz transformations since force is not a 4-vector).
However, if you replace Lorenz invariance with Galilean invariance in special relativity, all you have is Coulomb's law for electrostatics equipped with the fact that it also works for moving charges. This sort of physics, however, is really just like Newtonian gravity but without the principle of equivalence.


TITLE: Is the set of real matrices diagonalizable in $\mathcal M_n(C)$ dense in the set of all companion matrices?
QUESTION [2 upvotes]: Let $\mathcal E=\{A \in \mathcal M_n(\mathbb R): A \text{ is of companion form} \}$. That is, $A$ has the following form
\begin{align*}
A=\begin{pmatrix}  0 & \cdots & 0& -a_{0} \\  1 & \cdots & 0 & -a_{1}\\ \vdots &\ddots & \vdots &\vdots \\ 0 &\cdots & 1 & -a_{n-1} \end{pmatrix}.
\end{align*}
Let $\mathcal D = \{A \in \mathcal M_n(\mathbb R): A \text{ is diagonalizable in $\mathcal M_n(\mathbb C)$}\}$. We know $\mathcal D$ is dense in $\mathcal M_n(\mathbb R)$. I am wondering whether $\mathcal E \cap \mathcal D$ is dense in $\mathcal E$. The set $\mathcal E$ is closed. It is not clear to me whether this statement is true.

REPLY [2 votes]: This companion matrix is diagonalisable over $\Bbb C$, iff $f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ has $n$ distinct complex zeros. So, the question
is whether real polynomials with distinct zeros are dense in the set of all real polynomials of a certain degree.
The answer is yes. If $f$ has a repeated zero, then there is a complex number $z_0$ with $f(z_0)=f'(z_0)=0$. For a small nonzero real $\epsilon$, $g(x)=f(x)+\epsilon$ has $g'(z_0)=0$ and $g(z_0)\ne0$. For small enough
$\epsilon$. This holds for all common roots of $f$ and $f'$. But if $f(z_1)\ne0$ and $f'(z_1)=0$ choosing $\epsilon $ small enough will
ensure $g(z_1)\ne0$. Then $g$ and $g'=f'$ have no common roots.
So for any $f$, there are $g$ with no repeated roots close to $f$.


TITLE: How to work out which of these two equations is larger?
QUESTION [0 upvotes]: I have two equations:

$x = 2^n (p+i)^{3n}$
$x = 14^n p^{3n} $

Here $n$, $p$, and $i$ are all integers $\geq0$.
I worked out (using a spreadsheet) that if $i > 1$ then the value of x in expression 1 is larger than the value of x in expression 2.
How can I show this mathematically?

REPLY [1 votes]: Comparing the two equations.
\begin{align}
\require{cancel}
\frac{x_1}{x_2}&=\frac{2^n(p+i)^{3n}}{14^n p^{3n}}\\
&=\frac{2^n(p+i)^{3n}}{(2\cdot7)^n p^{3n}}\\
&=\frac{\cancel{2^n}(p+i)^{3n}}{\cancel{2^n} 7^n p^{3n}}\\
&=\frac{1}{7^n}\left(\frac{p+i}{p}\right)^{3n}\\
&=\left(\frac{\left(\frac{p+i}{p}\right)^{3}}{7}\right)^n\\
&=\left(\frac{\left(1+\frac{i}{p}\right)^{3}}{7}\right)^n\\
\end{align}
$x_1>x_2$ if the value of $i,p,n$ are greater than zero and $i\ge p$. For instance, let $i=p=1$, you get
\begin{equation}
\frac{x_1}{x_2}=\left(\frac{\left(1+\frac{1}{1}\right)^{3}}{7}\right)^n=\left(\frac{8}{7}\right)^n\qquad\text{hence } x_1>x_2
\end{equation}


TITLE: homographic random variable
QUESTION [0 upvotes]: I am working on Galton-Watson processes and I encountered the term "homographic random variables". Although I googled, I could not find anything. Do you have any idea about it? Do you know any book about it?

REPLY [2 votes]: These are nonnegative integer random variables $N$ whose generating function $g_N:s\mapsto\mathrm E(s^N)$ is homographic, that is, such that $g_N(s)=(as+b)/(cs+d)$ for some parameters $(a,b,c,d)$.


TITLE: Given an adjunction $<F,G,\varphi>$ , $G$ is faithful $G$ is faithful if and only if $\varphi^{-1}$ carries epis to epis
QUESTION [2 upvotes]: I have the next doubt about the solution of this problem
Given an adjunction $<F,G,\varphi>:X\rightarrow A$ Prove that $G$ is faithful if and only if $\varphi^{-1}$ carries epimorphism to epimorphism.
$(\Rightarrow)$ Let it be $f:x\rightarrow Ga$ an epimorphism we want to show that $\varphi^{-1}(f)$ is and epimorphism, so let it be $g_{1}, g_{1}: a\rightarrow b$ morphisms such that:
$g_{1}\circ \varphi^{-1}(f)=g_{2}\circ \varphi^{-1}(f)$
by definition of $\varphi^{-1}(f)$ we have:
$g_{1}\circ \epsilon_{a}\circ F(f)=g_{2}\circ \epsilon_{a}\circ F(f)$
we have that $g_{1}\circ \epsilon_{a}\circ F(f):F(x)\rightarrow b$ if we apply $\varphi$ to $g_{1}\circ \epsilon_{a}\circ F(f)$ by definition of $\varphi$ we have
$\begin{eqnarray}
\varphi(g_{1}\circ \epsilon_{a}\circ F(f))&=&G(g_{1}\circ \epsilon_{a}\circ F(f))\circ \eta_{x}\\
&=&G(g_{1})\circ G(\epsilon_{a})\circ GF(f)\circ \eta_{x}\\
&=&G(g_{1})\circ G(\epsilon_{a})\circ \eta_{Ga}\circ f\\
&=&G(g_{1})\circ id\circ f\\
&=&G(g_{1})\circ f
\end{eqnarray}$
The third line  by naturality of $\eta$ it follows  and the fourth one by the triangular identity. In the same way for $g_{2}\circ \epsilon_{a}\circ F(f)$ we will have the same. Therefore
$G(g_{1})\circ f=G(g_{2})\circ f$
Because $f$ is an epimorphism and $G$ is faithful, it follows that $g_{1}=g_{2}$, therefore $\varphi^{-1}$ carries epimorphism to epimorphism.
I hope the idea is correct for this part, for  the $(\Leftarrow)$ part i do not know how to use the hipothesis because if $G(g_{1})=G(g_{1})$ for morphisms $g_{1}, g_{2}:a\rightarrow b$ the only thing i have is if we apply $\varphi^{-1}$ an we use the naturality of $\epsilon$, i will have:
$g_{1}\circ \epsilon_{a}=g_{2}\circ \epsilon_{a}$
Thank you for your time.

REPLY [1 votes]: Consider the composition $[a,b]\overset{G}{\rightarrow}[Ga,Gb]\overset{\varphi ^{-1}}{\rightarrow}[FGa,b]$.
By Yoneda, it is determined by $(\varphi ^{-1}\circ G)(id _a)=\varepsilon _a$ and since, $\varphi ^{-1}$ carries epis to epis, and $G(id_a)$ is epic, $\varepsilon _a$ is an epi. But then, using the fact that
$f:b\rightarrow a$ is monic (epic) $\Leftrightarrow f^{*}: [a,-]\rightarrow [b,-]$ is epic (monic), we have
$\varphi ^{-1}\circ G$ is a monic, and this only happens when $G$ is faithful.
Remark: we can use this idea to get a fast proof of the forward direction: note that since $F$ is a left adjoint, is preserves epimorphisms (because it preserves colimits and hence pushouts). Now if $G$ is faithful, then $\varphi ^{-1}\circ G$ is injective and so by the above argument, $\varepsilon _a$ is epi. Then the result is immediate since if $f$ is epi, then $$g_{1}\circ \varphi^{-1}(f)=g_{2}\circ \varphi^{-1}(f)\Rightarrow g_{1}\circ \epsilon_{a}\circ F(f)=g_{2}\circ \epsilon_{a}\circ F(f)\Rightarrow g_1\circ \varepsilon _a=g_2\circ \varepsilon _a\Rightarrow g_1=g_2$$.


TITLE: Different ways to write a number given digit constraints
QUESTION [2 upvotes]: Originally I was just going to ask the problem on my practice math contest, which is asking how many ways there are to write a nine-digit number containing each digit $1-9$ so that the first digit is twice as big as the second and the last two digits are odd. (e.g. $846527913$).
I understood that in this case, the number would be in the form of $EE(EEEOO)OO$ or $EO(EEOOO)OO$ where $E$ is an even, $O$ is an odd, the numbers in the parentheses can be ordered in any way, and the only four cases for the first two are $21, 42, 63,$ and $84.$ Calculating this gave me $240,$ which is definitely wrong (I made a program to check, and it says $7680$).
So here are the questions:

How do I solve this particular problem correctly? Am I on the right track?
How are these solved in general? Given that it's a $9$-digit number and each digit $1-9$ is used once, how does one handle constraints such as these? This has always confused me (I'm not very strong in combinatorics)

REPLY [2 votes]: You're on the right track.  I'd break the problem up into four cases: where the number starts with 21, 42, 63 or 84.
In the case where you start with 21, you must have 21(EEEOO)OO in your notation. (There are four even digits, so three of them must be in the middle part of the number.)
So we can ask the questions:
- in how many different places can 4 go?
- in how many different places can  6 go (given that we've already placed 4?)
- in how many different places can  8 go (given that we've already placed 4 and 6)?
The answers are 5, 4, and 3 respectively.
Then we need to place the four remaining odd digits among the four remaining places; there are of course $4! = 24$ ways to do this.
So there are a total of $5 \times 4 \times 3 \times 24 = 1440$ such numbers beginning with 21.  The same is true if we start with 63 instead of 21. When we start with 42 or 84 the result is different but the method is much the same, and adding up the results from the four cases gives 7680.
As for (2), there aren't really "general" methods. But one thing that is good to keep in mind is trying to place the digits one at a time (as I've done above) instead of trying to do everything in one fell swoop.


TITLE: Kernel of a conjugation map
QUESTION [1 upvotes]: Let $$X=\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}$$
If $M$ is any orthogonal matrix on a vector space $V$ of dimension $2$. I noted that $M$ can be written as $$M= \frac{M-XMX}{2} + \frac{M+XMX}{2},$$ the first summand is denoted by $P_M$ and the second is denoted by $Q_M$. I also saw that $P_M X= XP_M$ and $Q_M X=-XQ_M$, i.e. $ $ conmutes (resp. anticonmutes) with $P_M$ (resp. $Q_M$).
How can I find $\ker P_M$ and $\ker Q_M$? I believe that the rank-nullity theorem can help me, but I’m not so sure.

REPLY [1 votes]: I will assume we are working over $\Bbb R$.
$O(2)$ consists in rotations and rotations multiplied by  $\begin {pmatrix}-1\quad 0\\0\quad 1\end {pmatrix}$ or $\begin {pmatrix}1\quad 0\\0\quad -1\end {pmatrix}$.
Recall that rotations are the $\begin{pmatrix} \cos\theta\quad -\sin\theta \\ \sin\theta\quad \cos\theta\end{pmatrix}$.
Take a rotation, call it $R$.  Then $XRX=2R$.
Thus $P_R=-\frac R2$.
Similarly $Q_R=\frac {3R}2$.
Since $R$ has trivial kernel,  so do $P_R$ and $Q_R$.
I leave the case $\det M=-1$ to you.


text:
\subsection{Automatic transversality.}\label{sec:trans}
If the space $\mathcal{M}_{\operatorname{hyb}}(\Lambda_0 ,\Lambda)$ contains no stationary solutions, transversality can be achieved for generic choices of almost complex structures. This is a standard -- albeit, difficult -- argument, which involves no new ideas not already present in the proof that the moduli spaces defining the Rabinowitz Floer complex are generically transverse (see for instance \cite[Thm.\ 4.11]{AbbonMerry2014} or \cite[Thm.\ 2]{Wisniewska2017} for the non-compact case). We therefore omit the proof.

The case where  $\mathcal{M}_{\operatorname{hyb}}(\Lambda_0 ,\Lambda)$ admits stationary solutions is somewhat less standard, however. In the presence of stationary solutions it is not possible to obtain regularity by perturbing $J$, and one must therefore prove transversality ``by hand''. This is the content of the following section.

Fix $\eta\in \operatorname{CritVal}(\cA^{H_0})=\operatorname{CritVal}(\cA^H)$ and let $\Lambda_0^\eta \subseteq \Crit(\cA^{H_0})$ and $\Lambda^\eta \subseteq \Crit(\cA^H)$ be the corresponding connected components of the respective critical sets. Then  $\mathcal{M}_{\operatorname{hyb}}( \Lambda_0^\eta , \Lambda^\eta)$ consists entirely of stationary solutions by Lemma \ref{lem:stationary} and

\begin{equation}\label{eq:LambdaLambda}
\begin{aligned}
\text{for }\eta\neq 0 \colon  \qquad \mathcal{M}_{\operatorname{hyb}}(\Lambda_0^\eta,\Lambda^\eta)&\cong \Lambda_0^\eta\cong \Lambda^\eta,\\
\text{for }\eta= 0 \colon  \qquad \mathcal{M}_{\operatorname{hyb}}(\Lambda_0^\eta,\Lambda^\eta)&\cong \Sigma_0\times\chi_0,
\end{aligned}
\end{equation}
where $\chi_0\subset T^\ast\R^{n-k}$ denotes the 0-section.

\begin{prop}\label{prop:trans}~
\begin{enumerate}[label=\emph{(\roman*)}]
\item
\label{at2}
The problem $ \mathcal{M}_{\operatorname{hyb}}( \Lambda_0^\eta,\Lambda^\eta )$ is always transversly cut out.
\item
\label{at1}
 The kernel of the linearisation $D\bar{\partial}$ of the problem $ \mathcal{M}^{\operatorname{hyb}}( \Lambda_0^{-\eta},\Lambda^\eta)$ (see \eqref{eq:Banach-sec}) at a stationary solution $(w,\xi)\in\mathcal{M}^{\operatorname{hyb}}( \Lambda_0^{-\eta},\Lambda^\eta)$ has dimension $\frac{1}{2}(\dim\Lambda+\dim\Lambda_0)$.
\end{enumerate}
\end{prop}

\begin{proof}[Proof of Proposition \ref{prop:trans}]
 The proof presented below is an adjustment of \cite[Lem. 4.14]{AbbonMerry2014} to our setting. Note that part \emph{\ref{at2}} is an immediate corollary of part \emph{\ref{at1}}. Indeed, by the correspondence \eqref{bijection}, we have that $\mathcal{M}_{\operatorname{hyb}}( \Lambda_0^\eta,\Lambda^\eta)$ is transversely cut out if and only if $\mathcal{M}^{\operatorname{hyb}}( \Lambda_0^{-\eta},\Lambda^\eta)$ is transversely cut out. The latter holds if and only if the operator $D\bar{\partial}$ is surjective at $(w,\xi)$. Now, $D\bar{\partial}$ is a Fredholm operator of index $\frac{1}{2}(\dim\Lambda+\dim\Lambda_0)$ by Theorem \ref{thm-index} (see also \eqref{dimM}), as $\tCZ(\Lambda_0)=\tCZ(\Lambda)$ by Proposition \ref{prop:muCZ}. Part \emph{\ref{at1}} tells us that $ \dim \ker D\bar{\partial} = \frac{1}{2}(\dim\Lambda+\dim\Lambda_0)$. Thus $D\bar{\partial}$ is surjective, as required.

Fix $(w,\xi) \in \mathcal{M}^{\operatorname{hyb}}( \Lambda_0^{-\eta},\Lambda^\eta)$ and let $(z, \zeta)\in \ker\big(D_{(w,\xi)}\bar{\partial}\big)$ be arbitrary, i.e.\ $(z,\zeta)$ is a solution to the linearised equation $D_{(w,\xi)}\bar{\partial}(z,\zeta)=0$. In the following, we write $w=(w_0,w_1,w_2)$ and $z=(z_0,z_1,z_2)$ with respect to the splitting $T^\ast\R^k\times T^\ast\R^n=T^\ast\R^k\times T^\ast\R^k\times T^\ast\R^{n-k}$.
Denote $(x_0,x_1)\coloneqq \lim_{s\to+\infty}(w, \xi)(s)\in \Lambda_0^{-\eta}\times\Lambda^\eta$.
Then $(z,\zeta)\in\mathcal{W}^{1,r}_{ \delta, L} \oplus W^{1,r}_{\delta, \Delta}\oplus T_{x_0}\Lambda_0^{-\eta} \oplus T_{x_1}\Lambda^\eta$ and satisfies explicitly
\begin{equation}
    \label{b}
    \frac{d}{ds}(z_0,\zeta_0) + \nabla^2_{(w_0,\xi_0)} \mathcal{A}^{H_0}(z_0,\zeta_0) = 0, \qquad \frac{d}{ds}(z_1,z_2,\zeta_1) + \nabla^2_{(w_1,w_2,\xi_1)} \mathcal{A}^{H}(z_1,z_2,\zeta_1)  = 0,
\end{equation}
the coupling conditions
\begin{equation}
\label{c}
  \zeta_1(0)=-\zeta_0(0), \qquad \forall\ t \in S^1 \colon  \quad z_0(0,t)= D\rho[ z_1(0,t)],\quad z_2(0,t) \in T\chi_0,
\end{equation}
and the asymptotic conditions
\begin{equation}
    \label{d}
    \lim_{s \to + \infty } (z, \zeta)(s) =(y_0,y_1)\in T_{x_0}\Lambda_0^{-\eta} \times T_{x_1}\Lambda^\eta.
\end{equation}
To prove part \emph{\ref{at1}}, we show that any solution $(z,\zeta)$ of $D_{(w,\xi)}\bar{\partial}(z,\zeta)=0$ is constant. Then, we have by \eqref{d} that $(z,\zeta)=(y_0,y_1)\in T_{x_0}\Lambda_0^{-\eta} \oplus T_{x_1}\Lambda^\eta$, which implies with the coupling condition \eqref{c} and with \eqref{eq:LambdaLambda} that the space of such $(z,\zeta)$ has dimension $\frac{1}{2}(\dim\Lambda+\dim\Lambda_0)$.

First recall that the second derivative of the functional $\mathcal{A}^{H}$ at a loop $(v,\eta)$, is a symmetric, bilinear operator on $W^{1,2}(S^{1},v^*T^*\R^n)\times \mathbb{R})$ given by:
\begin{equation}\label{d2AH}
 d^2_{(v,\eta)}\mathcal{A}^{H} ((\xi, \sigma),(\xi, \sigma)) = \int \omega(\xi,\partial_{t}\xi) - \eta\int \operatorname{Hess}_{v} H(\xi,\xi) -2\sigma\int dH(\xi).
\end{equation}
We define a function $\varphi \colon [0,+\infty)\to \R$ in the following way:
\[
\varphi(s)\coloneqq \big\|\big(z(s, \cdot),\zeta(s)\big)-(y_0,y_1)\big\|^2_{L^2(S^1)}.
\]
By assumption $\varphi \in L^r_\delta([0,+\infty),\R)$. As $\mathcal{A}^H$ and $\mathcal{A}^{H_0}$ are Morse-Bott (cf.\ \eqref{eq:MorseBott}), we have $(y_0,y_1)\in T_{x_0}\Lambda_0^{-\eta} \times T_{x_1}\Lambda^\eta=\ker \nabla^2_{x_0}\mathcal{A}^{H_0}\oplus\ker\nabla^2_{x_1}\mathcal{A}^H$ and hence\footnote{Observe that $(w_0,\xi_0)\equiv x_0$ and $(w_1,w_2,\xi_1)\equiv x_1$ for all $s\in [0,\infty)$.}
\begin{align*}
\varphi'(s) & = 2 \left( d^2_{(w_0,\xi_0)}\mathcal{A}^{H_0} ((z_0, \zeta_0),(z_0, \zeta_0)) + d^2_{(w_1,w_2,\xi_1)}\mathcal{A}^H ((z_1, z_2, \zeta_1),(z_1, z_2, \zeta_1)) \right), \\
\varphi''(s) & = 4 \left(\left\|\nabla^2_{(w_0,\xi_0)} \cA^{H_0}(z_0, \zeta_0)\right\|^2 + \left\|\nabla^2_{(w_1,w_2,\xi_1)} \cA^H(z_1,z_2, \zeta_1)\right\|^2\right)\geq 0.
\end{align*}
Thus on one hand $\varphi''(s)\geq 0$ and $\varphi$ is convex and on the other hand $\lim_{s\rightarrow\infty}\varphi (s)=0$. That implies that either $\varphi'(0)<0$ or $\varphi'(0)=0$ and $\varphi$ is constant, equal to 0 everywhere. We will show that $\varphi'(0)=0$, which implies that $(z,\zeta)$ is constantly equal to $(y_0,y_1)$. First, we show that
\begin{equation}\label{d2AH1=0}
d^2_{(w_2,\xi_1)}\mathcal{A}^{H_1} (( z_2, \zeta_1),(z_2, \zeta_1))\big|_{s=0}=0.
\end{equation}
Recall from the coupling conditions \eqref{eq-w-second} and \eqref{c} that $w_2(0,t)\in\chi_0$ and $z_2(0,t)\in T\chi_0$ for all $t\in S^1$. Writing \eqref{d2AH1=0} in the form \eqref{d2AH} as 3 integrals, we find that the first integral vanishes, as $\chi_0$ is a Lagrangian submanifold of  $\left(T^*\mathbb{R}^{n-k},\omega_{n-k}\right)$. On the other hand, for every $x\in \chi_0$ we can identify $T_x\chi_0$ with $\mathbb{R}^{n-k}\times\{0\}$ and therefore by \eqref{defH} for every $y\in T_x\chi_0$ we have
\[
\operatorname{Hess}_x H_1(y,y)=y^TA_1y=2 H_1(y)=0,
\]
since $\chi_0 \subseteq H_1^{-1}(0)$ by Remark \ref{H2}.
Consequently, the second integral in \eqref{d2AH} vanishes. Analogously, $T\chi_0\subseteq TH_1^{-1}(0)=\ker(dH_1)$, hence the third integral in \eqref{d2AH} also is $0$. This proves \eqref{d2AH1=0}. Next we will show that
\[
\left( d^2_{(w_0,\xi_0)} \mathcal{A}^{H_0} ((z_0, \zeta_0),(z_0, \zeta_0))  +d^2_{(w_1,\xi_1)}\mathcal{A}^H ((z_1, \zeta_1),(z_1, \zeta_1))\right)\Big|_{s=0}=0.
\]
 Observe that by the coupling conditions \eqref{eq-w-second} and \eqref{c} we have for all $t\in S^1$ that $w_0(0,t) = \rho \circ w_1(0,t)$ and $z_0(0,t) = D\rho[ z_1(0,t)]$ and hence
\[
 d^2_{(w_0,\xi_0)} \mathcal{A}^{H_0} ((z_0, \zeta_0),(z_0, \zeta_0))\Big|_{s=0} = d^2_{(\rho \circ w_1,-\xi_1)}\mathcal{A}^H ((D\rho[z_1], -\zeta_1),(D\rho[z_1], -\zeta_1))\Big|_{s=0}
\]
Thus, the first corresponding integrals in \eqref{d2AH} have opposite signs, since $\rho$ is anti-symplectic. The second integrals in \eqref{d2AH} have opposite signs, since $\xi_0(0)=-\xi_1(0)$ and $H_0 \circ \rho = H_0$ and thus $\operatorname{Hess} H_0(D\rho\, \cdot\,, D\rho\, \cdot\, )= \operatorname{Hess} H_0$. The third integrals  in \eqref{d2AH} have opposite signs, since $\zeta_0(0)=-\zeta_1(0)$ and for every $(x,y)\in T\R^k, \ dH_{\rho (x)}(D\rho[y])=dH_x(y)$. Consequently $\varphi'(0)=0$.
\end{proof}


TITLE: Is it a possible of p-adic number to euclidean geometry
QUESTION [1 upvotes]: I was reading about theorem called Monsky's theorem stating that "A square can never be divided into an odd number of traingles of equal area " .
I read that the proof uses something called 2-adic valuation . I don't know anything about p-adic systems but know that "It is a slightly different number system where the proximity of two numbers is obtained by the number of times their gap is divisible by a certain prime number p " and unknowingly I consider p-adic to be advanced mathematical stuff .
So , I was excited to find a possible instance where advanced mathematics being used in complicated problems of school geometry .
Am I right in concluding that these 2-adic valuation is not same as real number systems as we know it  and this is indeed one instance where higher maths is being applied to euclidean geometry ?
P.S: On a different note , how difficult a question would it be in a maths contest if it asks to prove " A square cannot be divided into odd number of triangles of equal area " ?

REPLY [2 votes]: The geometry of $p$-adic numbers is very different from the geometry of real numbers. For instance, any triangle in $\mathbb{Q}_p$( set of $p$-adic numbers) is isosceles, which is clearly not the case for $\mathbb{R}$.