text:
\section{Dynamics of the resonant model}\label{sec:resonantmodel}

The reduced Hamiltonian \eqref{redhamN} for $N=2$ is of the form
\begin{equation}\label{calham0}
\begin{aligned}
\mathcal{H}(\varepsilon; \psi_1, \psi_2, K_1, K_2)=&\mathcal{H}_0(\psi_1, \psi_2, K_1, K_2)+\varepsilon \mathcal{H}_1(\psi_1, \psi_2, K_1, K_2)\\
\mathcal{H}_0(\psi_1, \psi_2, K_1, K_2)=&\,\mathcal{H}_0^{(1)}(\psi_1, K_1)+\mathcal{H}_0^{(2)} (\psi_2, K_2)\\

\mathcal{H}_0^{(1)}(\psi_1, K_1)= &\,K_1(1-K_1)(1+2\cos(\psi_1))\\
\mathcal{H}_0^{(2)}(\psi_2, K_2)=&\,K_2(1-K_2)(1+2\cos(\psi_2))\\
\mathcal{H}_1(\psi_1, \psi_2, K_1, K_2)=&\,a_1K_1+b_1K_1^2+a_2 K_2+b_2 K_2^2\\
&+c_1K_1(1-K_1)\cos(\psi_1)+c_2K_2(1-K_2)\cos(\psi_2) +d_{12}K_1K_2.

\end{aligned}
\end{equation}
Note that the only term which couples the two unperturbed Hamiltonians $\mathcal{H}_0^{(1)}$, $\mathcal{H}_0^{(2)}$ is $d_{12}K_1K_2$. The Hamiltonian $\mathcal{H}$ is reversible with respect to the involution
\begin{equation}\label{anglesymmetry}
\Upsilon(\psi_1, \psi_2, K_1, K_2)=(-\psi_1,-\psi_2, K_1, K_2).
\end{equation}

\subsection{Unperturbed dynamics ($\varepsilon=0$)}\label{sec:unperturbeddynamics}

For $\eps=0$, the Hamiltonian system $\mathcal{H}_0$ is the product of the two uncoupled $1$-d.o.f systems with  Hamiltonian $\mathcal{H}_0^{(i)}$, $i=1, 2$ and therefore it is integrable.

We analyze the dynamics given by $\mathcal{H}_0^{(i)}$. We analyze it only for $\mathcal{H}_0^{(1)}$ since both Hamiltonians are equal.

The associated equations of motion  are given by
\begin{equation*}

\begin{split}
\dot{\Psi}_1=&\,(1-2 K_1) (1+2\cos(\psi_1))\\
\dot{K}_1=&\,2\sin(\psi_1)\,K_1\,(1-K_1).
\end{split}
\end{equation*}
 The sets $\{  K_1=0\}$ and $\{K_1=1\}$ are $\mathcal{H}_0^{(1)}$-invariant $1$-dimensional tori which correspond to the hyperbolic  tori \eqref{def:HypTori} after symplectic reduction  and correspond to saddles in proper ``blow down'' coordinates (see \eqref{def:cartesian}, \eqref{def:saddledown}). The sets $\{  K_1=0\}$ and $\{K_1=1\}$ possess the hyperbolic equilibrium points $\left(\pm \Psi_{*}, 0\right)$, $\left( \pm \Psi_{*}, 1 \right)$

  with
  \begin{equation}\label{def:psi}
  \Psi_*=2\pi/3.
  \end{equation}

  Such equilibria are hyperbolic with eigenvalues  $\pm \sqrt{3}$. Their invariant manifolds outside of  $\{  K_1=0\}$ and $\{K_1=1\}$ correspond to the invariant manifolds of the saddles $P_\pm$ in  \eqref{def:saddledown}.

The tori $\{  K_1=0\}$ and $\{K_1=1\}$ are on the same energy level $\mathcal{H}_0^{(1)}=0$ and the saddles  $\left(\pm \Psi_{*}, 0\right)$ and  $\left( \pm \Psi_{*}, 1 \right)$ are

 connected through the heteroclinic orbits
 \begin{equation*}

 (\psi_1(t), K_1(t))=\left(\pm \Psi_*, \dfrac{1}{1+e^{\mp \sqrt{3} t}}\right)
 \end{equation*}
 (see Figure \ref{fig:PhaseSpaceH0}).

\begin{figure}
\begin{center}
\includegraphics[width=0.7\textwidth]{2rect_unpert.pdf}
\end{center}
\caption{{\small Phase space of the Hamiltonian $\mathcal{H}_0^{(1)}$ in \eqref{calham0}} }
\label{fig:PhaseSpaceH0}
\end{figure}

 For $0<K_1<1$, the dynamics of the Hamiltonian $\mathcal{H}_0^{(1)}$ can be also analyzed easily. Consider the ``half'' of the phase space $ (-2\pi/3,2\pi/3)\times (0, 1)\subset\T\times (0, 1)$ limited by the heteroclinic orbits (the other ``half'' is symmetric). It has an elliptic points at $(\psi_1,K_1)=(0,1/2)$ and the rest is foliated by periodic orbits
 \begin{equation}\label{def:Gammah}
\mathtt{P}_h:=\{\mathcal{H}_0^{(1)}=h\} \qquad \mbox{ with} \qquad h\in (0,3/4).
 \end{equation}
  When $h\to 0$, the periodic orbits ``tend'' to the sequence of heteroclinics and $K_1=0,1$ and therefore their period $\mathtt{T}_h\to+\infty$.

Hence, the dynamics  of the 2-dof Hamiltonian $\mathcal{H}_0$ in \eqref{calham0} has the following features. The invariant tori

 \begin{equation}\label{tori}
\mathbb{T}_0:=\{ K_1=K_2=0\}, \qquad \mathbb{T}_1:=\{ K_1=K_2=1 \}
\end{equation}
are two invariant Lagrangian tori for the system \eqref{calham0}. They possess the equilibrium points
\begin{equation}\label{def:saddles}
\mathfrak{e}_+^{(0)}:=(\Psi_{*}, \Psi_{*} , 0, 0), \quad \mathfrak{e}_+^{(1)}:=(\Psi_{*}, \Psi_{*} , 1, 1), \quad \mathfrak{e}_-^{(1)}:=(-\Psi_{*}, -\Psi_{*} , 1, 1), \quad \mathfrak{e}_-^{(0)}:=(-\Psi_{*}, -\Psi_{*} , 0, 0)
\end{equation} connected by the following heteroclinic manifolds
\begin{equation}\label{gammaplus}
\begin{split}
\gamma_{+}(\tau_1, \tau_2):=(\psi_1^+(\tau_1), \psi_2^+(\tau_2), K_1^+(\tau_1), K_2^+(\tau_2))&=\left(\Psi_{*}, \Psi_{*}, \dfrac{1}{1+e^{-\sqrt{3}\tau_1}},   \dfrac{1}{1+e^{-\sqrt{3} \tau_2}}  \right),\\
\gamma_{-}(\tau_1, \tau_2):=(\psi_1^-(\tau_1), \psi_2^-(\tau_2), K_1^-(\tau_1), K_2^-(\tau_2))&=\left(-\Psi_*, -\Psi_*, \dfrac{1}{1+e^{\sqrt{3}\tau_1}},   \dfrac{1}{1+e^{\sqrt{3}\tau_2}}  \right).
\end{split}
\end{equation}
In particular $\gamma_+$ connects the  points $\mathfrak{e}_+^{(0)}$, $\mathfrak{e}_+^{(1)}$ and $\gamma_-$ connects $\mathfrak{e}_-^{(1)}$ with $\mathfrak{e}_-^{(0)}$. The trajectories in the heteroclinic manifolds are just given by $\gamma_{\pm}(\tau_1+t, \tau_2+t)$, $t\in\mathbb{R}$.

The $4$-dimensional phase space of Hamiltonian $\mathcal{H}$ in \eqref{calham0} with $\eps=0$ has several three-dimensional invariant subspaces setting either $K_1$ or $K_2$ equal to $0$ or $1$, and two dimensional invariant subspaces setting either $(\psi_1,K_1)$ or $(\psi_2,K_2)$ at one of the saddles. Thus, one can define the hyperbolic periodic orbits (recall \eqref{def:Gammah})
\begin{align}\label{def:PO:ext}

\mathtt{P}^{\sigma,s}_h := \{ (\psi_1,\psi_2,K_1,K_2) : (\psi_1,K_1) \in \mathtt{P}_h, \,\psi_2= \sigma \Psi_*,\, K_2=k \}, \; \; \sigma = \pm, \; \; s =0,1,
\end{align}
and one could define analogously the other ones placing them at the other saddles.

For  $\varepsilon=0$, the Hamiltonian system \eqref{calham0} possesses two  $2$-dimensional heteroclinic manifolds $W^{s}(\mathfrak{e}_{+}^{(0)})=W^u(\mathfrak{e}_{+}^{(1)})$, $W^{u}(\mathfrak{e}_{-}^{(0)})=W^s(\mathfrak{e}_{-}^{(1)})$. They are certainly not robust under perturbations. We show that, under a generic non-degeneracy condition, those heteroclinic manifolds break down when $0<\varepsilon\ll 1$ creating transverse intersections between some of the stable and unstable invariant manifolds.

\subsection{Non-integrable dynamics $(\varepsilon>0)$: $2$ resonant tuples}

For $\eps>0$, the tori
$\mathbb{T}_0$ and $\mathbb{T}_1$ in \eqref{tori} are still invariant and they still possess  saddles which are $\eps$--close to the unperturbed saddles $\mathfrak{e}_\pm^{(j)}$, $j=0,1$. These saddles have 2-dimensional stable and unstable invariant manifolds.

\begin{remark}\label{rmk:SaddlesPerturbed}
Abusing notation, we also denote by $\mathfrak{e}_\pm^{(j)}$, $j=0,1$ the saddles of the perturbed Hamiltonian \eqref{calham0} with $0<\eps\ll 1$, which are $\eps$-close to those defined by \eqref{def:saddles}.
\end{remark}

\begin{teor}\label{thm:SplittingHomo2Rect}
Consider the Hamiltonian \eqref{calham0} and assume that
 \begin{equation}\label{tildekappas}
d_{12}\neq 0\qquad \text{(see \eqref{calham0})}.
\end{equation}
Then, there exists $\varepsilon_0>0$ such that for all $\varepsilon\in (0, \varepsilon_0)$,

 the invariant manifolds $W_{\varepsilon}^{u}(\mathfrak{e}_{+}^{(0)})$  and $W_{\varepsilon}^s(\mathfrak{e}_{-}^{(0)})$ of the saddles \eqref{def:saddles} of the Hamiltonian \eqref{calham0} intersect transversally along an orbit (within the energy level).

\end{teor}

Note that this theorem is not a classical perturbative result. Indeed, for $\eps=0$ the saddles $\mathfrak{e}_{\pm}^{(0)}$ did not have any connection since  their invariant manifolds coincided with those of $\mathfrak{e}_{\pm}^{(1)}$ along heteroclinic connections. Therefore, the prove of Theorem \ref{thm:SplittingHomo2Rect} is not a direct consequence of Melnikov Theory (is not a theorem about persistence, is a theorem about \emph{new} heteroclinic connections). Thus, we prove this theorem in two steps. First in Section \ref{sec:HeteroMelnikov} we apply Melnikov Theory to prove the existence of transverse (within the energy level) heteroclinic connections between  $\mathfrak{e}_{+}^{(0)}$ and  $\mathfrak{e}_{+}^{(1)}$ (under certain conditions). Then, in Section \ref{sec:HomoMelnikov}, we use this analysis to prove the existence of the connections given in Theorem \ref{thm:SplittingHomo2Rect} through a suitable modification of Melnikov Theory.

\subsubsection{Transversal heteroclinic orbits to saddles}\label{sec:HeteroMelnikov}

The first step to prove Theorem \ref{thm:SplittingHomo2Rect} is to prove the existence of heteroclinic intersections between the saddles  $\mathfrak{e}_{\pm}^{(0)}$ and $\mathfrak{e}_{\pm}^{(1)}$. This step is certainly not necessary to obtain homoclinic intersections. Nevertheless, it will make considerably easier the computation of the Melnikov function associated to the homoclinic intersections.
To obtain the mentioned heteroclinic intersections, one certainly needs that the saddles belong to the same energy level, that is, $\mathcal{H}(\mathfrak{e}_{\pm}^{(0)})=\mathcal{H}(\mathfrak{e}_{\pm}^{(1)})$. By \eqref{calham0} this condition is equivalent to
\begin{equation}\label{elc}
a_1+b_1+a_2+b_2+d_{12}=0.
\end{equation}

\begin{prop}\label{teoetero}

The Hamiltonian \eqref{calham0} possesses four hyperbolic fixed points $\mathfrak{e}_{\pm}^{(0)}$, $\mathfrak{e}_{\pm}^{(1)}$ such that the following holds. If \eqref{elc} is satisfied and
\begin{equation}

\label{concord}
(a_1+b_1)(a_2+b_2)>0,
\end{equation}
there exists $\varepsilon_0>0$ such that for $\varepsilon\in (0, \varepsilon_0)$ the manifolds $W_{\varepsilon}^{u}(\mathfrak{e}_+^{(0)})$ and $W_{\varepsilon}^{s}(\mathfrak{e}_+^{(1)})$ intersect transversally along orbits (within the energy level). The same happens for  $W_{\varepsilon}^{s}(\mathfrak{e}_-^{(0)})$ and $W_{\varepsilon}^{u}(\mathfrak{e}_-^{(1)})$.

\end{prop}

See in Figure \ref{fig:Heteros} an example of heteroclinic connections. We devote the rest of the section to prove Proposition \ref{teoetero}.

\begin{figure}
\begin{center}
\includegraphics[width=0.7\textwidth]{transv_intersection.pdf}
\end{center}
\caption{Transverse heteroclinic orbits for $\eps$ small enough.}
\label{fig:Heteros}
\end{figure}

\begin{proof}[Proof of Proposition \ref{teoetero}]
Thanks to the symmetry \eqref{anglesymmetry} of the system \eqref{calham0}, one of the intersections implies the other one.

We just deal with the first one.

Consider a compact subset $\mathbf{K}$ of $\mathbb{R}^2$.

Let $\vec{\tau}=(\tau_1, \tau_2)\in \mathbf{K}$ and $m>0$. We consider the line
\begin{equation*}

\Sigma(\vec{\tau})=\{\gamma_+(\vec{\tau})+r\,\nabla \mathcal{H}^{(1)}_0 (\gamma_+(\vec{\tau})),\,\,\,r\in (-m, m)\},
\end{equation*}
which passes through $\gamma_+(\vec{\tau})$ and it is orthogonal to $\{\mathcal{H}^{(1)}_0=\mathcal{H}^{(1)}_0(\gamma_+(\vec{\tau}))\}$ at $\gamma_+(\vec{\tau})$. Since the system has two degrees of freedom and energy conservation, it is enough to measure the distance along this line. It would be equivalent to consider $\mathcal{H}^{(2)}_0$.
Since $\gamma_+(\vec{\tau})\in W_{0}^s(\mathfrak{e}_+^{(1)})=W_{0}^u(\mathfrak{e}_+^{(0)})$,
if we consider $\varepsilon$ small enough we can ensure that $\Sigma(\vec{\tau})$ intersects transversally $W_{\varepsilon}^s(\mathfrak{e}_+^{(1)})$ and $W_{\varepsilon}^u(\mathfrak{e}_+^{(0)})$ at just one point, $q_{\varepsilon}^s=q_{\varepsilon}^s(\vec{\tau})$ and $q_{\varepsilon}^u=q_{\varepsilon}^u(\vec{\tau})$ respectively.

Then,  the distance between the invariant manifolds in $\Sigma(\vec{\tau})$ is given by
\begin{equation}\label{distsign}
d(\vec{\tau}):=\left\langle \dfrac{ \nabla \mathcal{H}^{(1)}_0 (\gamma_+(\vec{\tau}))}{\lVert    \nabla \mathcal{H}^{(1)}_0 (\gamma_+(\vec{\tau}))    \rVert}, q_{\varepsilon}^s(\vec{\tau})-q_{\varepsilon}^u(\vec{\tau}) \right\rangle.
\end{equation}

Application of the classical Melnikov Theory gives the following result.

\begin{lem}\label{melnikovlemma}
The function $d(\vec{\tau})$ introduced in \eqref{distsign} satisfies
\begin{equation*}

d(\vec{\tau})=\dfrac{\varepsilon}{\lVert    \nabla \mathcal{H}^{(1)}_0 (\gamma_+(\vec{\tau}))    \rVert} \,\mathcal{M_+}(\vec{\tau})+\mathcal{O}_{C^1(\mathbf{K})}(\varepsilon^2), \qquad \vec{\tau}\in\mathbf{K},
\end{equation*}
where
\begin{equation}\label{melint}
\begin{aligned}
&\mathcal{M_+}(\vec{\tau}):= \int_{\mathbb{R}} \{ \mathcal{H}^{(1)}_0 ,  \mathcal{H}_1\} \circ \Phi^t_{\mathcal{H}_0}(\gamma_{+}(\vec{\tau}))\,dt = \int_{\mathbb{R}} \{ \mathcal{H}^{(1)}_0 ,  \mathcal{H}_1\} \circ (\gamma_{+}(\tau_1+t,\tau_2+t))\,dt
\end{aligned}
\end{equation}
is the so-called \emph{Melnikov function} (see \cite{Melnikov63}).
\end{lem}

Since the Hamiltonian system \eqref{calham0} is autonomous, the Melnikov function  $\mathcal{M_+}$ depends just on the one-dimensional variable $\tau_1-\tau_2$. That is, there exists a function $\MM_+^{(0)}:\R\to\R$ such that
\[
 \MM_+(\tau_1,\tau_2)=\MM^{(0)}_+(\tau_1-\tau_2).
\]
By Lemma \ref{melnikovlemma}, we will deduce Theorem \ref{teoetero} by proving that there exists a non-degenerate zero of the function $\mathcal{M}^{(0)}_+$ in \eqref{melint}.

It is convenient to introduce the Melnikov potential $\mathcal{L}_{+}\colon\mathbb{R}^2\to\mathbb{R}$, since it is usually easier to compute. It is defined, up to constants, as a primitive of the Melnikov function, namely
\begin{equation*}

\partial_{\tau_1} \mathcal{L}_+ (\vec{\tau})=\mathcal{M}_+ (\vec{\tau}).
\end{equation*}

We have
\begin{equation*}

\mathcal{L}_+(\vec{\tau})=\int_{\mathbb{R}}\mathcal{H}_1\circ \Phi^t_{\mathcal{H}_0}\left(\gamma_+(\vec{\tau})\right)\,dt=\int_{\mathbb{R}}\mathcal{H}_1\left(\gamma_+(\tau_1+t,\tau_2+t)\right)\,dt.
\end{equation*}
Recall that we are assuming \eqref{elc}, which implies $\mathcal{H}_1(\mathfrak{e}_+^{(0)})=\mathcal{H}_1(\mathfrak{e}_+^{(1)})=0$. Therefore, the integrand decays exponentially to zero as $t\to\pm\infty$.

The Melnikov potential satisfies  $\mathcal{L}_+(\vec{\tau})= \mathcal{L}^{(0)}_+(\tau_0)$ where $\tau_0:=\tau_1-\tau_2$ and $ \mathcal{L}^{(0)}_+$ is called \emph{reduced Melnikov potential}.

Then,
\[
\partial_{\tau_0} \mathcal{L}_{+}^{(0)}(\tau_0)=\mathcal{M}^{(0)}_+(\tau_0).
\]

Hence we shall look for non-degenerate critical points of $\mathcal{L}_{+}^{(0)}$, which correspond to non-degenerate zeros of $\mathcal{M}_{+}^{(0)}$.

The following lemma concludes the proof of Proposition \ref{teoetero}.

\begin{lem}\label{lemma:HeteroMelnPot}
There exists a constant $\tilde{\eta}\in\mathbb{R}$ such that  the reduced Melnikov potential  $\mathcal{L}^{(0)}_+$ is given by
\begin{equation}\label{nice}
 \mathcal{L}^{(0)}_+(\tau_0) =\tau_0\,\,\dfrac{(a_1+b_1)\,e^{-\sqrt{3}\tau_0}+(a_2+b_2)}{ 1-e^{-\sqrt{3} \tau_0} }+\tilde{\eta}.
\end{equation}
Therefore, provided \eqref{concord} is satisfied, it possesses a non-degenerate critical point.

\end{lem}

\begin{remark}\label{rem}
Note that $\Upsilon\gamma_+(\vec{\tau})=\gamma_-(-\vec{\tau})$, $i=1, 2$ (see \eqref{gammaplus}) where $\Upsilon$ is the involution introduced in \eqref{anglesymmetry}.Then $\mathcal{L}_+(\vec{\tau})=\mathcal{L}_-(-\vec{\tau})$ and $\mathcal{L}^{(0)}_+(\tau_0)=\mathcal{L}^{(0)}_-(-\tau_0)$ . Therefore, if \eqref{concord} holds, $\mathcal{L}_-^{(0)}$ has a non-degenerate critical point.
\end{remark}

\begin{proof}[Proof of Lemma \ref{lemma:HeteroMelnPot}]
Using the definition of $\mathcal{H}_0$ in \eqref{calham0}, \eqref{elc},

one can write $\mathcal{L}_+$ as
\begin{equation*}

\begin{aligned}
\mathcal{L}_+(\vec{\tau})&= (a_2+b_2)\int_{\mathbb{R}} K_1^+(t+\tau_1)(1-K_2^+(t+\tau_2))\,dt +(a_1+b_1)\int_{\mathbb{R}} K_2^+(t+\tau_2)(1-K_1^+(t+\tau_1))\,dt+\tilde{\eta}\\
&=(a_2+b_2)\int_{\mathbb{R}} K_1^+(s+\tau_0)(1-K_2^+(s))\,ds +(a_1+b_1)\,\int_{\mathbb{R}} K_2^+(s)(1-K_1^+(s+\tau_0))\,ds+\tilde{\eta}\\
&=:\mathcal{L}_+^{(0)}(\tau_0),
\end{aligned}
\end{equation*}
where $\tau_0=\tau_1-\tau_2$ and the constant $\tilde{\eta}\in\R$ is given by
\begin{equation*}

\begin{split}
\tilde{ \eta}:=&\,\int_\R \left(b_1 K_1(t)(K_1(t)-1)+b_2 K_2(t)(K_2(t)-1)\right)\,dt\\
 &+\,\int_\R \left( c_1 K_1(t)(1-K_1(t))\cos\Psi_*+c_2 K_2(t)(1-K_2(t))\cos\Psi_* \right)\,dt.
\end{split}
\end{equation*}
For $i, j=1, 2$ we have (recall \eqref{gammaplus})
 \begin{equation}\label{def:IntegralForMelnikov}
\begin{split}
 \int_{\mathbb{R}} K^+_i(t+\tau_i) (1-K^+_j(t+\tau_j))\,dt&= \int_{\mathbb{R}} \dfrac{e^{-\sqrt{3}(t+\tau_j)}}{(1+e^{-\sqrt{3}(t+\tau_i)})(1+e^{-\sqrt{3}(t+\tau_j)}) }  \,dt\\
 &=(\tau_i-\tau_j) \dfrac{1}{1-e^{-\sqrt{3}(\tau_i-\tau_j)}},
\end{split}\end{equation}
which gives \eqref{nice}.
Therefore,  we have that

\begin{align*}
\lim_{\tau_0\to +\infty} \partial_{\tau_0} \mathcal{L}^{(0)}_+(\tau_0)=a_2+b_2, \qquad \lim_{\tau_0\to -\infty} \partial_{\tau_0}\mathcal{L}^{(0)}_+(\tau_0)=-(a_1+b_1) \label{limelle+}.

\end{align*}
If $(a_1+b_1)(a_2+ b_2)>0$ (see \eqref{concord}) the reduced Melnikov potential $\mathcal{L}^{(0)}_{+}$ has at least one critical point.

Moreover,
\begin{equation*}

\,\partial_{\tau_0}^2\mathcal{L}^{(0)}_+(\tau_0)=-(a_1+b_1+a_2+b_2)
\dfrac{\sqrt{3}}{4}\left( 2-\sqrt{3} \tau_0\coth\left(\dfrac{\sqrt{3}\tau_0}{2} \right)\right) {\mathrm{csch}^2\left(  \dfrac{\sqrt{3} \tau_0}{2} \right) } .
\end{equation*}
By \eqref{concord} this function has constant sign since
\begin{align*}
 2-\sqrt{3}\tau_0\,\coth\left(\dfrac{\sqrt{3}\tau_0}{2} \right)< 0 \qquad \forall \tau_0\neq 0,\\
 \lim_{\tau_0\to 0} \left( 2-\sqrt{3}\tau_0\coth\left(\dfrac{\sqrt{3}\tau_0}{2} \right)\right) {         \mathrm{csch}^2\left(  \dfrac{\sqrt{3} \tau_0}{2} \right) }  =-\frac{2}{3}.
\end{align*}
Therefore $\mathcal{L}_{+}^{(0)}$ is either convex or concave (depending on the sign of $a_1+b_1+a_2+b_2$) and its critical points are non-degenerate.
\end{proof}

\end{proof}
\subsubsection{Transversal homoclinic orbits to saddles: Proof of Theorem \ref{thm:SplittingHomo2Rect}}\label{sec:HomoMelnikov}

We use the computation of the heteroclinic Melnikov potential in Lemma \ref{lemma:HeteroMelnPot} to prove the existence of  homoclinic transversal intersections given by Theorem \ref{thm:SplittingHomo2Rect}.

Since the Hamiltonian \eqref{calham0} with $\eps=0$ does not have connections between $\mathfrak{e}_{\pm}^{(0)}$, we  cannot apply directly Melnikov Theory to obtain such connections for $\eps>0$. Instead, we exploit the usual technique of considering a modified unperturbed Hamiltonian and using two parameters $\eps$ and $\delta$.

We consider  the Hamiltonian
\begin{equation}\label{bfham0}
\begin{aligned}
\mathcal{H} =\mathbf{H}_0+\varepsilon\mathbf{H}_1, \quad \mathbf{H}_0(\psi_1, \psi_2, K_1, K_2)& =\mathbf{H}_0^{(1)}+\mathbf{H}_0^{(2)},\\
\mathbf{H}_0^{(1)}(\psi_1, \psi_2, K_1, K_2)&=K_1(1-K_1)(1+2\, \cos(\psi_1))-\delta K_1^2,\\
\mathbf{H}_0^{(2)}(\psi_1, \psi_2, K_1, K_2)&=K_2(1-K_2)(1+2\, \cos(\psi_2))-\delta K_2^2
\end{aligned}
\end{equation}
and

\begin{equation}\label{trueperturbation}
\begin{aligned}
\mathbf{H}_1(\psi_1, \psi_2, K_1, K_2):=&\,d_{12}\, K_1 \,K_2\\
&+ a_1 K_1+(b_1+1) K_1^2+c_1\,K_1(1-K_1) \cos(\psi_1) \\
&+a_2 K_2+(b_2+1) K_2^2+c_2\,K_2(1-K_2) \cos(\psi_2) .
\end{aligned}
\end{equation}
If one takes $\de=\eps$, this Hamiltonian coincides with \eqref{calham0}. Nevertheless, for now we consider $\delta$ and $\eps$ independent parameters. Later one we will take $\de=\eps$.

If $\delta=0$, then the dynamics of $\mathbf{H}_0$ is the same described in Section \ref{sec:unperturbeddynamics}.

If $\delta\neq 0$, the tori defined in \eqref{tori} are $\mathbf{H}_0$-invariant; moreover, they belong to different energy levels, since

\[
\mathbf{H}_{0_{| \mathbb{T}_0 }}=0, \qquad \mathbf{H}_{0_{| \mathbb{T}_1 }}=-2\delta.
\]

The equilibrium points contained in $\mathbb{T}_0$ are the saddles $\mathfrak{e}_{\pm}^{(0)}$ defined in \eqref{def:saddles}.

 Now we compute the heteroclinic manifold that connects (forward in time) $\mathfrak{e}^{(0)}_+$ with $\mathfrak{e}^{(0)}_-$ (see Figure \ref{fig:PhaseSpaceHomoclinic}). Such orbit corresponds to a homoclinic to the saddle $P_-$ in \eqref{def:saddledown} (expressed in  the ``blow down'' coordinates \eqref{def:cartesian}).

\begin{lem}\label{lemma:UnperturedHomo2Rect}
 The saddles $\mathfrak{e}^{(0)}_+$ with $\mathfrak{e}^{(0)}_-$ of Hamiltonian $\mathbf{H}_0$ in \eqref{bfham0} are connected by a two-dimensional heteroclinic manifold parameterized as
\begin{equation}\label{gammazero}
\begin{aligned}
\gamma_0(\vec{\tau}):&=(\gamma_0^{(1)}(\tau_1), \gamma_0^{(2)}(\tau_2))=(\psi_1^{(0)}(\tau_1), \psi_2^{(0)}(\tau_2), K_1^{(0)}(\tau_1), K_2^{(0)}(\tau_2)),\\
\psi_j^{(0)}(\tau_j):&=2\,\arctan (\Lambda (\tau_j)), \quad  K_j^{(0)}(\tau_j)=\frac{1}{1-\frac{\delta}{3}(1-2\cosh(\sqrt{3}\tau_j))}\quad j=1, 2,

\end{aligned}
\end{equation}
where $\Lambda(t):=-\sqrt{3}\,\, \tanh \left( \dfrac{\sqrt{3}}{2} t \right)$.
\end{lem}

\begin{proof}
Using that $\mathbf{H}_0^{(1)}$ is zero when restricted to $\mathbb{T}_0$ we get
\begin{equation}\label{def:kappa1}
K_1=\frac{1+2 \cos(\psi_1)}{1+2 \cos(\psi_1)+\delta}.
\end{equation}
When the angle $\psi_1\in [-\Psi_*, \Psi_*]$ the numerator in \eqref{def:kappa1} is positive. Hence $K_1\in (0, 1)$ if $\delta>0$.
Plugging \eqref{def:kappa1} in the equation for $\psi_1$ we have
\[
\dot{\psi}_1=-(1+2\cos(\psi_1)),
\]
which leads to
\begin{equation}\label{psi1}
\psi_1(t)=2\,\arctan(\Lambda (t)), \quad \Lambda(t)=-\sqrt{3}\,\, \tanh \left( \dfrac{\sqrt{3}}{2} t \right).
\end{equation}

By using \eqref{def:kappa1} and the trigonometric identity  $\cos(2 \arctan (x))=(1-x^2)/(1+x^2)$ we have
\begin{equation}\label{def:kappasimple}
K_j^{(0)}(t)=\frac{1}{1-\frac{\delta}{3}(1-2\cosh(\sqrt{3}t))}.
\end{equation}
Reasoning in the same way for $(\psi_2, K_2)$ we get that
the homoclinic orbit to $\mathbb{T}_0$ is given by \eqref{gammazero}.
\end{proof}

\begin{figure}
\begin{center}
\includegraphics[width=0.7\textwidth]{2rect_pert_hom.pdf}
\end{center}
\caption{{\small Phase space restricted to the $(\psi_1,K_1)$-coordinates for the Hamiltonian $\mathcal{H}$ in \eqref{bfham0}-\eqref{trueperturbation}.} }\label{fig:PhaseSpaceHomoclinic}
\end{figure}

By reasoning as in the proof of Theorem \ref{teoetero} we have that the distance between the manifolds in a suitable section is given by
\begin{equation}\label{dist2}
d(\vec{\tau})=\dfrac{\varepsilon}{\lVert    \nabla \mathbf{H}^{(1)}_0 (\gamma_0(\vec{\tau}))    \rVert} \,\mathcal{M}_0(\vec{\tau})+\mathcal{O}_{C^1(\mathbf{K})}(\varepsilon^2), \qquad \vec{\tau}\in\mathbf{K},
\end{equation}
where the Melnikov function is given by
\begin{equation}\label{melfunz}
\mathcal{M}_0(\vec{\tau})=\int_{\mathbb{R}} \{ \mathbf{H}^{(1)}_0, \mathbf{H}_1\}\circ \Phi^t_{\mathbf{H}_0}(\gamma_0(\vec{\tau}))\,dt.
\end{equation}
It can be easily checked that the $\mathcal{O}_{C^1(\mathbf{K})}(\varepsilon^2)$ are uniform for $\de$ small enough.

The associated  Melnikov potential is
\begin{align*}
\mathcal{L}_0(\vec{\tau})&=\int_{\mathbb{R}}  \mathbf{H}_1\circ \Phi^t_{\mathbf{H}_0}(\gamma_0(\vec{\tau}))\,dt    \label{melpot}
={d_{12}} \int_{\mathbb{R}} K^{(0)}_1(t+\tau_1) \,K^{(0)}_2(t+\tau_2)\,dt+\eta_*
\end{align*}
where
\begin{equation*}

\begin{split}
\eta_*=&\int_{\mathbb{R}} \Big(a_1 K^{(0)}_1(t)+(b_1+1) (K^{(0)}_1(t))^2+c_1\,K^{(0)}_1(t)(1-K^{(0)}_1(t)) \cos\psi^{(0)}_1(t)\Big)\,dt\\
&+\int_{\mathbb{R}} \Big(a_2 K^{(0)}_2(t)+(b_2+1) (K^{(0)}_2(t))^2+c_2\,K^{(0)}_2(t)(1-K^{(0)}_2(t)) \cos\psi^{(0)}_2(t) \Big)\,dt.
\end{split}
\end{equation*}
As before we consider the reduced Melnikov potential
\begin{equation}\label{def:melpotred}
\mathcal{L}_0^{(0)}(\tau_0)={d_{12}} \int_{\mathbb{R}} K^{(0)}_1(s+\tau_0) \,K^{(0)}_2(s)\,ds+\eta_*.
\end{equation}

We want to deduce that $\mathcal{L}_0^{(0)}$ has non-degenerate critical points by using the information on the Melnikov potentials \eqref{nice} of the heteroclinic case.

\begin{prop}\label{propfond}

Fix an interval $\mathcal{I}\subset\mathbb{R}$. There exists $\de_0>0$ such that $\forall \de\in (0, \de_0)$ there exists  a real number $\eta$ and a constant $\nu_0>0$ such that, for $\tau_0\in \mathcal{I}$,

\begin{equation*}

 \mathcal{L}^{(0)}_0(\tau_0)=\eta+d_{1 2}\,\tau_0\,\coth\left( \frac{\sqrt{3}\tau_0}{2} \right)+\mathcal{O}_{C^2(\mathcal{I})}(\de^{\nu_0}).
\end{equation*}
\end{prop}

The proof of this proposition is deferred to Section \ref{def:prop59}.

To complete the proof of Theorem \ref{thm:SplittingHomo2Rect} it is enough  use \eqref{dist2} and Proposition \ref{propfond} and take  $\de=\eps$. Indeed, the transverse homoclinic  points are $\eps$-close to the non-degenerate critical point for $\mathcal{L}_0^{(0)}$. By Proposition \ref{propfond}, $\mathcal{L}_0^{(0)}$ has a non-degenerate critical point $\eps^{\nu_0}$-close to $\tau_0=0$.

\subsubsection{Proof of Proposition \ref{propfond}}\label{def:prop59}

Thanks to the exponential convergence of the homoclinic orbit to the equilibrium points $\mathfrak{e}_\pm^{(0)}$ we have that (recall \eqref{gammazero})
\[
\int_{\mathbb{R}} K_j^{(0)}(t)\,dt<\infty \qquad i=1, 2.
\]
We write $\mathcal{L}^{(0)}_0$ as
\begin{equation}\label{ellezero}
\begin{aligned}
\mathcal{L}^{(0)}_0(\tau_0)=-d_{12}\int_{\mathbb{R}} K^{(0)}_1(s+\tau_0) \,(1-K^{(0)}_2(s))\,ds+\eta_1,

\end{aligned}
\end{equation}
where
\[
 \eta_1=\eta_*+d_{12}\int_\mathbb{R}K^{(0)}_1(s)ds.
\]
Define the function
\[
 F(\psi_1,\psi_2,K_1,K_2)=K_1 (1-K_2).
\]
By \eqref{def:IntegralForMelnikov}, we have that
\[
 \begin{split}
\mathcal{F}_+(\tau_1,\tau_2)&= \int_\mathbb{R}F(\gamma_+(\tau_1+t,\tau_2+t))dt=(\tau_1-\tau_2)\frac{1}{1-e^{-\sqrt{3}(\tau_1-\tau_2)}},\\
\mathcal{F}_-(\tau_1,\tau_2)&  =  \int_\mathbb{R}F(\gamma_-(\tau_1+t,\tau_2+t))dt=-(\tau_1-\tau_2)\frac{1}{1-e^{\sqrt{3}(\tau_1-\tau_2)}},
 \end{split}
\]
which is just the integral of the function $F$ along the heteroclinic orbits $\gamma_\pm$ introduced in \eqref{gammaplus}. These functions satisfies $\mathcal{F}_\pm(\tau_1,\tau_2)=\mathcal{F}_\pm(0,\tau_2-\tau_1)$.

Since the homoclinic orbit \eqref{gammazero} is ``close'' to the concatenation of $\gamma_+$ and $\gamma_-$ in \eqref{gammaplus}, we show that there exists $\nu_0>0$ such that
 the integral in \eqref{ellezero} satisfies
\[
\begin{split}
\int_{\mathbb{R}} K^{(0)}_1(s+\tau_0) \,(1-K^{(0)}_2(s))\,ds&=\mathcal{F}_+(0,\tau_0)+\mathcal{F}_-(0,\tau_0)+\OO\left(\delta^{\nu_0}\right)\\
&=\tau_0 \mathrm{coth} \left( \frac{\sqrt{3}\tau_0}{2} \right)+\OO\left(\delta^{\nu_0}\right).
\end{split}
\]

The estimate for the error is proved in the following lemma. To state it, we define
\begin{equation}\label{def:oeffe}
\mathfrak{O}_F(\vec{\tau}):=\int_{\mathbb{R}}\left[ F(\gamma_0(\tau_1+t,\tau_2+ t))-F(\gamma_+(\tau_1+t,\tau_2+ t))-F(\gamma_-(\tau_1+t,\tau_2+ t))\right]\,dt.
\end{equation}

 \begin{lem}\label{propdiff}
 Let $\mathbf{K}$ be a compact subset of $\mathbb{R}^2$.

 There exists $\delta_0>0$ small, such that $\forall \delta\in (0, \delta_0)$ and $\tau\in\mathbf{K}$ there exists a positive constant $\nu_0\in (0, 1)$ such that the following holds
 \begin{equation}\label{integrand}
  \lVert \mathfrak{O}_F \rVert_{C^0(\mathbf{K})}+ \lVert \partial_{\tau_1} \mathfrak{O}_F \rVert_{C^0(\mathbf{K})}+\lVert \partial^2_{\tau_1} \mathfrak{O}_F \rVert_{C^0(\mathbf{K})}  \lesssim \delta^{\nu_0}.
 \end{equation}
 \end{lem}

This lemma implies Proposition \ref{propfond}. We devote the rest of this section to prove Lemma \ref{propdiff}.

\begin{proof}[Proof of Lemma \ref{propdiff}]
We write the function $\mathfrak{O}_F$ in \eqref{def:oeffe} as
\[
\mathfrak{O}_F(\vec{\tau}):=\int_{\mathbb{R}} F(\gamma_0(\tau_1-p+t, \tau_2-p+t)-F(\gamma_+(\tau_1+t, \tau_2+t)-F(\gamma_-(\tau_1-2 p+t, \tau_2- 2 p+t))\,dt
\]
where
\begin{equation}\label{delta}
p:=\dfrac{1}{\sqrt{3}}\left|\ln\frac{\delta}{3}\right|.
\end{equation}
Note that the shifts by the vector $(p, p)$ do not alter the value of the integral. These shifts are  useful to bound the integrand.
To obtain such estimates, we need the following lemmas.

\begin{lem}\label{difforb}
Let $\sigma_1\in (0, 1)$.
Consider $\gamma_\pm, \gamma_0$ in  \eqref{gammazero}, \eqref{gammaplus} and
\begin{equation}\label{def:interval}
\mathcal{I}:=\left(-\frac{\sigma_1}{\sqrt{3}}\left|\ln\frac{\delta}{3}\right|, \frac{\sigma_1}{\sqrt{3}} \left|\ln\frac{\delta}{3}\right|\right).
\end{equation}
 There exists a constant $\nu\in(0, 1)$ such that
\begin{equation}\label{differenceaction}
\left\lVert K_i^{(0)}\left(\tau_i\mp p+ t\right)-K_i^{\pm}(\tau_i+ t) \right\rVert_{C^0(\mathcal{I}\times\mathbf{K})}\lesssim  \delta^{\nu}, \qquad i=1, 2,
\end{equation}
\begin{equation}\label{differenceangle}
\left\lVert \sin\left(\psi_i^{(0)}\left(\tau_i\mp p+ t\right)\right)-\sin (\psi_i^{\pm}(\tau+ t)) \right\rVert_{C^0(\mathcal{I}\times\mathbf{K})}\lesssim  \delta^{\nu}\qquad i=1, 2.
\end{equation}

\end{lem}

\begin{proof}
To simplify the notation let us consider $i=1$.
By \eqref{def:kappasimple} we have that
\begin{equation*}

K_1^{(0)}\left(t\pm p\right)=\frac{1}{1+e^{\pm\sqrt{3}t}-\frac{\delta}{3}+\frac{\delta^2}{9} e^{\mp\sqrt{3}t}}.
\end{equation*}
Thus for $\sigma\in [-\sigma_1, \sigma_1]$,
\[
 \left\lvert K_1^{(0)}\left(\frac{\sigma}{\sqrt{3}} \left|\ln\frac{\delta}{3}\right|\pm p\right)-K_1^\mp\left( \frac{\sigma}{\sqrt{3}}\left|\ln\frac{\delta}{3}\right|\right)\right\rvert \lesssim \max\{ \delta, \delta^{2-\sigma_1} \}.
 \]
 This gives the bounds \eqref{differenceaction}.

By using \eqref{psi1} and the trigonometric identity $\sin(2\arctan(x))=2 x/(1+x^2)$ we have

\[
 \sin(\psi^{(0)}_1(t))=-\frac{\sqrt{3}\sinh\left(\sqrt{3}t\right)}{2\cosh\left(\sqrt{3}t\right)-1}=-\frac{\sqrt{3}}{2} \tanh(\sqrt{3} t)\Big( 1+\frac{1}{2 \cosh(\sqrt{3} t)-1} \Big).
\]
Then, for $t_\pm=\displaystyle \pm p+\frac{\sigma}{\sqrt{3}}\left|\ln\frac{\delta}{3}\right|$ with $\sigma\in [-\sigma_1, \sigma_1]$, we have
\[

\tanh(\sqrt{3} t_{\pm})=

\pm1+\OO\left(\delta^{2(1-\sigma_1)}\right),\qquad
\frac{1}{2 \cosh(\sqrt{3} t_{\pm})-1}=\OO\left(\delta^{1-\sigma_1}\right).

\]
To prove \eqref{differenceangle} it is enough to use these estimates and \eqref{gammaplus}, to obtain
\[
 \sin(\psi^{(0)}_1(t_\pm))=\mp \frac{\sqrt{3}}{2}+\OO\left(\de^{1-\sigma_1}\right)=\sin (\psi_i^{\mp}(\tau_i))+\OO\left(\de^{1-\sigma_1}\right).
 \]

\end{proof}

\begin{lem}\label{convexp}
There exists $T>0$ independent of $\varepsilon$ such that
(see \eqref{gammaplus},  \eqref{gammazero})
\[
\begin{aligned}
\lVert  \gamma_+(\tau_1+t, \tau_2+t)-\mathfrak{e}_+^{(0)} \rVert_{C^0(\mathbf{K})}&\lesssim  \,e^{\sqrt{3} t} \qquad &\text{for}\,\, t<0,&\\
\lVert  \gamma_+(\tau_1+t, \tau_2+t)-\mathfrak{e}_+^{(1)} \rVert_{C^0(\mathbf{K})}&\lesssim \,e^{-\sqrt{3}  t} \qquad &\text{for}\,\, t>0,&\\
\lVert  \gamma_-(\tau_1+t, \tau_2+t)-\mathfrak{e}_-^{(1)} \rVert_{C^0(\mathbf{K})}&\lesssim \,e^{\sqrt{3} t}  \qquad &\text{for}\,\, t<0,&\\
\lVert  \gamma_-(\tau_1+t, \tau_2+t)-\mathfrak{e}_-^{(0)} \rVert_{C^0(\mathbf{K})}&\lesssim \,e^{-\sqrt{3} t}  \qquad &\text{for}\,\, t>0,&\\

\lVert  \gamma_0(\tau_1-p+t, \tau_2-p+t)-\mathfrak{e}^{(0)}_{+} \rVert_{C^0(\mathbf{K})}&\lesssim \,e^{\sqrt{3}t}\qquad &\text{for}\,\, t<-T,&\\
\lVert  \gamma_0(\tau_1+p+t, \tau_2+p+t)-\mathfrak{e}^{(0)}_{-} \rVert_{C^0(\mathbf{K})}&\lesssim \,e^{-\sqrt{3}t}\qquad &\text{for}\,\, t>T&.
\end{aligned}
\]

\end{lem}

\begin{proof}
The lemma follows by straightforward estimates and the hyperbolicity of the equilibria.

\end{proof}

We split $\mathbb{R}=\bigcup_{k=1}^5 \mathcal{I}_k$
\[
\mathcal{I}_1:=(-\infty, \mathtt{a}), \quad \mathcal{I}_2:=[\mathtt{a}, \mathtt{b}], \quad \mathcal{I}_3:=[\mathtt{b}, \mathtt{c}], \quad \mathcal{I}_4:=[\mathtt{c}, \mathtt{d}], \quad \mathcal{I}_5:=(\mathtt{d}, +\infty)
\]
with
\begin{equation*}

\mathtt{a}=-\frac{3}{4\sqrt{3}} \left| \ln\delta\right|,\quad \mathtt{b}=\frac{3}{4\sqrt{3}}\left| \ln\delta\right|, \quad \mathtt{c}=2p-\frac{3}{4\sqrt{3}}\left| \ln\delta\right|, \quad \mathtt{d}=2p+\frac{3}{4\sqrt{3}}\left| \ln\delta\right|.
\end{equation*}

We have
\[
\int_{\mathbb{R}}\left[ F(\gamma( \tau_1-p +t, \tau_2-p+t))-F(\gamma_+(\tau_1+t, \tau_2+t))-F(\gamma_-( \tau_1-2p +t, \tau_2-2p+t))\right]\,dt=\sum_{j=1}^5\mathtt{T}_j
\]
where
\begin{align}
\mathtt{T}_1=&\,\int_{\mathcal{I}_1} \left[ F(\gamma_0( \tau_1-p +t, \tau_2-p+t))-F(\gamma_+(\tau_1+t, \tau_2+t))\right]\,dt   \label{t1} \\
\mathtt{T}_2=&\,\int_{\mathcal{I}_2}\left[  F(\gamma_0( \tau_1-p +t, \tau_2-p+t))-F(\gamma_+(\tau_1+t, \tau_2+t))\right]\,dt     \label{t2}\\
\mathtt{T}_3=&\,\int_{\mathcal{I}_3} F(\gamma_0( \tau_1-p +t, \tau_2-p+t))\,dt-\int_{\mathtt{b}}^{+\infty} F(\gamma_+(\tau_1+t, \tau_2+t))\,dt\label{t3}\\
&- \int^{\mathtt{c}}_{-\infty} F(\gamma_-(\tau_1-2 p+t, \tau_2-2 p+t))\,dt    \notag\\
\mathtt{T}_4=&\,\int_{\mathcal{I}_4} \left[  F(\gamma_0( \tau_1-p +t, \tau_2-p+t))-F(\gamma_-(\tau_1-2 p+t, \tau_2-2 p+t))\right]\,dt \label{t4}\\
\mathtt{T}_5=&\,\int_{\mathcal{I}_5}\left[  F(\gamma_0( \tau_1-p +t, \tau_2-p+t))-F(\gamma_-(\tau_1-2 p+t, \tau_2-2 p+t))\right]\,dt. \label{t5}
\end{align}
By the symmetry of the problem it is sufficient to provide bounds for $\lvert \mathtt{T}_i \rvert$ with $i=1, 2, 3$. The idea is to use the exponentially fast convergence of the orbits $\gamma_0$, $\gamma_{\pm}$ to the saddles (see Lemma \ref{convexp}) to get bounds on the integrals over the unbounded intervals and to exploit the closeness of such orbits on the compact intervals using Lemma \ref{difforb}.

\begin{itemize}
\item \textbf{Bound for $\mathtt{T}_1$} (see \eqref{t1}): Let us call $S:=\{ \gamma_0( \tau_1-p +t, \tau_2-p+t) \}_{t\in\mathcal{I}_1}$. $S$ is a compact subset of $\mathbb{R}^2$.
Recalling that $F(\mathfrak{e}^{(0)}_+)=0$, by Lemma \ref{convexp} and the mean value theorem we have that (recall \eqref{gammazero})
\begin{equation}\label{stima1}
\begin{split}
\int_{\mathcal{I}_1} \lvert F(\gamma_0( \tau_1-p +t, \tau_2-p+t)) \rvert\,dt &=\int_{\mathcal{I}_1} \lvert F(\gamma_0( \tau_1-p +t, \tau_2-p+t))-F(\mathfrak{e}^{(0)}_+) \rvert\,dt\\
&\le \lVert F \rVert_{C^1(S)} \int_{\mathcal{I}_1} \lVert \gamma_0( \tau_1-p +t, \tau_2-p+t)-\mathfrak{e}_+^{(0)}\rVert\,dt\\
&\lesssim \delta^{3/4}.
\end{split}
\end{equation}
By Lemma \ref{convexp}, the compactness of the orbit $\{\gamma_+(\tau_1+t, \tau_2+t)\}_{t\in\mathcal{I}_1}$, $F(\mathfrak{e}_+^{(0)})=0$ one can reason in the same way to obtain the same bound for the term involving $\gamma_+$.

\item \textbf{Bound for $\mathtt{T}_2$} (see \eqref{t2}): We note that $\mathcal{I}_2$ is of the form \eqref{def:interval}. By using the compactness of the orbits and the mean value theorem as in the previous step, we can apply Lemma \ref{difforb} and obtain
\begin{equation}\label{stima2}
\lvert \mathtt{T}_2 \rvert\lesssim \delta^{\nu}\,\lvert \mathcal{I}_2 \rvert\lesssim \delta^{\nu}\,\lvert \ln\delta \rvert
\end{equation}
where $\nu\in(0, 1)$ is given by Lemma \ref{difforb}.

\item \textbf{Bound for $\mathtt{T}_3$} (see \eqref{t3}):  We use that $F( 1, 1)=0$. Let us denote $\mathtt{m}:=-(\mathtt{b}-p)=\mathtt{c}-p>0$ (see \eqref{delta} for the definition of $p$). By translating the variable $t$ we obtain
\begin{align*}
\int_{\mathcal{I}_3} \lvert F(\gamma_0(\tau_1- p+t, \tau_2- p+t)) \rvert\,dt& =\int_{\mathcal{I}_3} \lvert F(\gamma_0(\tau_1- p+t, \tau_2- p+t))-F( 1, 1) \rvert\,dt\\
&=\int_{-\mathtt{m}}^{\mathtt{m}}  \lvert F(\gamma_0(\tau_1+t, \tau_2+t))-F(1, 1) \rvert\,dt\\
&\lesssim \lVert F \rVert_{C^1(S)} \sum_{i=1}^2 \int_{-\mathtt{m}}^{\mathtt{m}} | K_i^{(0)} (t)-1 |\,dt.
\end{align*}
Now, using \eqref{def:kappasimple}, one can see that on the  interval $[-\mathtt{m}, \mathtt{m}]$, one has that $| K_i^{(0)} (t)-1 |\lesssim \delta^{3/4}$, which implies
\[
\int_{\mathcal{I}_3} \lvert F(\gamma_0(\tau_1- p+t, \tau_2- p+t)) \rvert\,dt\lesssim \delta^{3/4} \,\lvert \ln\delta\rvert.
\]
By Lemma \ref{convexp} we have
\[
\begin{split}
\int_{\mathtt{b}}^{\infty} \lvert F(\gamma_+(\tau_1+t, \tau_2+t))\rvert\,dt&\lesssim  \delta^{3/4}\\
\int^{\mathtt{c}}_{-\infty} \lvert F(\gamma_-(\tau_1- 2p+t, \tau_2- 2p+t)) \rvert\,dt& = \int^{\mathtt{a}}_{-\infty} \lvert F(\gamma_-(\tau_1+t, \tau_2+t)) \rvert\,dt\lesssim  \delta^{3/4}.
\end{split}
\]
Hence,
\begin{equation}\label{stima3}
\lvert \mathtt{T}_3\rvert\lesssim  \delta^{3/4} \,\lvert \ln \delta \rvert.
\end{equation}

\end{itemize}

By \eqref{stima1}, \eqref{stima2}, \eqref{stima3}
we have that
\[
\lVert \mathfrak{O}_F \rVert_{C^0(\mathcal{A})}\lesssim \delta^{\nu}\lvert \ln \delta \rvert
\]
(changing $\nu$ if necessary). Now we observe that (recall \eqref{calham0}, \eqref{bfham0})
\begin{align*}
\partial_{\tau_1} F(\gamma_0(\tau_1- p+t, \tau_2- p+t))&=\{ F, \mathbf{H}^{(1)}_0\} (\gamma_0(\tau_1- p+t, \tau_2- p+t))\\
\partial_{\tau_1} F(\gamma_+(\tau_1+t, \tau_2+t))&=\{ F, \mathcal{H}^{(1)}_0 \}(\gamma_+(\tau_1+t, \tau_2+t)), \\
\partial_{\tau_1} F(\gamma_-(\tau_1-2 p+t, \tau_2-2 p+t))&=\{ F, \mathcal{H}^{(1)}_0\}(\gamma_-(\tau_1-2 p+t, \tau_2-2 p+t)).
\end{align*}
By the particular form of the Hamiltonians $\mathcal{H}^{(1)}_0$ and $\mathbf{H}^{(1)}_0$ in  \eqref{calham0}, \eqref{bfham0} one can check that $\{ F, \mathcal{H}^{(1)}_0\}=\{ F, \mathbf{H}^{(1)}_0\}$. Let us call $G:=\{ F, \mathcal{H}^{(1)}_0\}$. Clearly $G(\mathfrak{e}_\pm^{(0)})=G(\mathfrak{e}_\pm^{(1)})=0$. Now we can repeat the same strategy to get the bounds for the associated $\mathtt{T}_i$. The only difference is that when we compare the orbits $\gamma_0, \gamma_{\pm}$ on compact intervals we need to use also \eqref{differenceangle}. Then we obtain $\lVert \mathfrak{O}_{G} \rVert_{C^0(\mathbf{K})}\lesssim \delta ^{\nu}\lvert \ln \delta \rvert$ (recall \eqref{def:oeffe}).

Regarding the second derivatives in $\tau_1$ we have
\begin{align*}
\partial_{\tau_1}^2 F(\gamma_0(\tau_1- p+t, \tau_2- p+t))&=\{ G, \mathbf{H}^{(1)}_0\} (\gamma_0(\tau_1- p+t, \tau_2- p+t))\\
\partial_{\tau_1}^2 F(\gamma_+(\tau_1+t, \tau_2+t))&=\{ G, \mathcal{H}^{(1)}_0 \}(\gamma_+(\tau_1+t, \tau_2+t)), \\
\partial_{\tau_1}^2 F(\gamma_-(\tau_1-2 p+t, \tau_2-2 p+t))&=\{ G, \mathcal{H}^{(1)}_0\}(\gamma_-(\tau_1-2 p+t, \tau_2-2 p+t)).
\end{align*}
We observe that on a compact set $\lvert \{G, \mathbf{H}^{(1)}_0\}-\{ G, \mathcal{H}^{(1)}_0\}\rvert\lesssim \delta$. Then we can consider the function $E:=\{G, \mathcal{H}^{(1)}_0\}$ and repeat the same arguments above to prove that $\lVert \mathfrak{O}_{E} \rVert_{C^0(\mathbf{K})}$ has a bound like \eqref{integrand}. We conclude by noting that
\[
 \lVert \mathfrak{O}_F \rVert_{C^0(\mathbf{K})}+ \lVert \partial_{\tau_1} \mathfrak{O}_F \rVert_{C^0(\mathbf{K})}+\lVert \partial^2_{\tau_1} \mathfrak{O}_F \rVert_{C^0(\mathbf{K})}    \lesssim \Vert \mathfrak{O}_F \rVert_{C^0(\mathbf{K})}+\lVert \mathfrak{O}_{G} \rVert_{C^0(\mathbf{K})}+\lVert \mathfrak{O}_{E} \rVert_{C^0(\mathbf{K})}.
\]
 \end{proof}

\subsection{Transversal homoclinic orbits to saddles: $N$ resonant tuples}

In this section we prove the generalization of Theorem \ref{thm:SplittingHomo2Rect} for the case of multiple resonant tuples. To break integrability we need to impose a non-degeneracy condition on the coefficients $d_{ij}$ in \eqref{coeffN}. To state it we introduce the matrix
\begin{equation}\label{def:MatrixD}

\mathcal{D}
=
\begin{pmatrix}
d_{1,N} + \sum_{j \neq 1} d_{1,j} & -d_{1,2} & \ldots & -d_{1,N-1} \\
-d_{2,1} & \ddots & \vdots & \vdots \\
\vdots & \vdots & \ddots &  -d_{N-2,N-1} \\
-d_{N-1,1} & \ldots & -d_{N-1,N-2} & d_{N-1,N} + \sum_{j \neq N-1} d_{N-1,j}
\end{pmatrix}.
\end{equation}

\begin{prop}\label{prop:homointersaddlesN}
Assume that the matrix $\mathcal{D}$ satisfies
\begin{equation}\label{cond:HessD}
\det \mathcal{D}\neq 0.
\end{equation}
Then, there exists $\varepsilon_0>0$ such that for all $\varepsilon\in(0, \varepsilon_0)$  the invariant manifolds $W_{\varepsilon}^-(\mathfrak{e}^{(0)}_-)$ and $W_{\varepsilon}^+(\mathfrak{e}^{(0)}_+)$ of the saddles \eqref{saddlesN} of the Hamiltonian \eqref{def:hamN} intersect transversally along an orbit (within the energy level).

\end{prop}

\begin{remark}
Note that condition \eqref{cond:HessD} is satisfied for a generic choice of coefficients $d_{ij}$. Indeed, the determinant of such matrix is a polynomial in the variables $d_{i j}$. Then, it is enough to  show that such polynomial is not identically zero. If one consider
\[
d_{i j}=\begin{cases}
1 \quad \text{if}\,\,i=1, \dots, N-1,\,\,j=N,\\
0 \quad \text{otherwise}
\end{cases}
\]
the matrix \eqref{def:MatrixD} is a multiple of the identity. This means that at some point the polynomial is not zero and therefore it is not-zero for almost every choice of $d_{ij}$.
In Section \ref{sec:ApplicationMultiple}, we prove that condition \eqref{cond:HessD} is satisfied for the resonant models associated to the Wave, Beam and Hartree equations that we consider.
\end{remark}

\begin{proof}
We proceed as for the case $N=2$ in Section \ref{sec:HomoMelnikov}. That is, we introduce a second parameter $\de$ and we define the Hamiltonian $\mathcal{H}=\sum_{j=1}^N\mathbf{H}_0^{(j)}+\varepsilon \mathbf{H}_1$ given by (recall \eqref{coeffN})
\begin{equation}\label{def:hamN}
\begin{aligned}
\mathbf{H}_0^{(j)}(\psi_j, K_j; \delta) :=&\,K_j (1-K_j)(1 +2  \,\cos(\psi_j))-\delta\, K^2_j ,\\
\mathbf{H}_1(\psi_1, \dots, \psi_N, K_1, \dots, K_N) :=&\,\sum_{j=1}^N \big(a_{j} K_j + (b_{j}+1) K^2_j +c_j\,K_j (1-K_j)\,\cos(\psi_j)\big)\\
&\,+ \sum_{i, j=1, i< j}^N d_{i j} K_i\,K_j.
\end{aligned}
\end{equation}
If $\de=\eps$, it coincides with \eqref{redhamN}.

We proceed as in the proof of Theorem \ref{thm:SplittingHomo2Rect}. For $\varepsilon=0$ the dynamics is the same described in Section \ref{sec:HomoMelnikov}.
In particular it is easy to see that, when $\varepsilon=0$, one can consider the two saddle points (recall that $\Psi_*:=2\pi/3$)
\begin{equation}\label{saddlesN}
\mathfrak{e}^{(0)}_{\pm}:=(\pm \Psi_*, \dots, \pm \Psi_*, 0, \dots, 0)
\end{equation}
connected by the $\de$-dependent homoclinic manifolds (recall \eqref{psi1}, \eqref{gammazero})
\begin{equation}\label{gammazeroN}

\gamma_0(\vec{\tau}):=(\psi_1^{(0)}(\tau_1), \ldots, \psi_N^{(0)}(\tau_N), K_1^{(0)}(\tau_1), \ldots, K_N^{(0)}(\tau_N)),\qquad \vec{\tau}=(\tau_1,\ldots,\tau_N).\nonumber \\

\end{equation}
We define the associated Melnikov potential
\begin{equation}\label{def:malnikovpotcaseN}
\mathcal{L}_{0, N}(\vec{\tau}):=\int_{\mathbb{R}} \mathbf{H}_1\circ \Phi^t_{\mathbf{H}_0} (\gamma_0(\vec{\tau}))\,dt= \sum_{i, j=1, i< j}^N d_{i j} \int_{\mathbb{R}}  K^{(0)}_i(\tau_i+t)\,K^{(0)}_j(\tau_j+t) \,dt+\eta_*,
\end{equation}
where
\[
\eta_*:=\sum_{i=1}^N\int_{\mathbb{R}} a_{i} K^{(0)}_i(t) + (b_{i}+1) (K^{(0)}_i)^2(t) +c_i\,K^{(0)}_i (1-K^{(0)}_i(t))\,\cos\psi^{(0)}_i(t)\,dt.
\]
We note that such function is the sum of terms of the form \eqref{def:melpotred}.
Thanks to the autonomous nature of the system the potential, $\mathcal{L}_{0, N}$ depends just on $\tau_1-\tau_N, \dots, \tau_{N-1}-\tau_N$. Thus one can consider the reduced Melnikov potential $\mathcal{L}_{0, N}^{(0)}$, which satisfies
\[
 \mathcal{L}_{0, N}^{(0)}\left(\tau_1-\tau_N, \dots, \tau_{N-1}-\tau_N\right)=\mathcal{L}_{0, N}(\tau_1,\ldots,\tau_N).
\]
Classical Melnikov Theory ensures that  non-degenerate critical points of this reduced Melnikov potential gives rise to transversal (within the energy level) intersections between $W_{\varepsilon}^-(\mathfrak{e}^{(0)}_-)$ and $W_{\varepsilon}^+(\mathfrak{e}^{(0)}_+)$.

Denoting  $\,\,{\tilde{\tau}}:= (\tau_1-\tau_N, \dots, \tau_{N-1}-\tau_N)$, Proposition \ref{propfond} implies that there exists a constant $\eta\in\mathbb{R}$ such that
\[
\mathcal{L}_{0, N}^{(0)}(\tilde{\tau})=\eta+\sum_{i, j=1, i< j}^{N-1} d_{i j}\, (\tilde{\tau}_i-\tilde{\tau}_j)\coth\left(\frac{\sqrt{3}}{2} (\tilde{\tau}_i-\tilde{\tau}_j) \right)+\sum_{j=1}^{N-1} d_{j N} \tilde{\tau}_j \coth\left(\frac{\sqrt{3}}{2} \tilde{\tau}_j \right)+\mathcal{O}_{C^2}(\delta^{\nu_0})
\]
for some $\nu_0>0$.
Since $x\,\coth((\sqrt{3}/2)x)$ is an even function, the origin
 $(0, \dots, 0)\in\mathbb{R}^{N-1}$ is a critical point of the first order of  $\mathcal{L}_{0, N}^{(0)}$ (that is, dropping the errors $\mathcal{O}_{C^2}(\de^{\nu_0})$). The  Hessian matrix of the first order of $\mathcal{L}_{0, N}^{(0)}$ at the origin is

\[\mathrm{Hess}=  \frac{1}{\sqrt{3}}\mathcal{D}\]

where $\mathcal{D}$ is the matrix introduced in \eqref{def:MatrixD}. Then, condition \eqref{cond:HessD} implies $\det\mathrm{Hess}\neq 0$.

The non-degeneracy of the Hessian implies that the reduced Melnikov potential $\mathcal{L}_{0, N}^{(0)}$ has a non-degenerate critical point $\de^{\nu_0}$--close to $\tilde\tau=0$. Then, taking $\de=\eps$ one can use  classical Melnikov Theory to ensure the existence of the transverse intersection between invariant manifolds stated in Proposition \ref{prop:homointersaddlesN}.
\end{proof}

config: arxiv


TITLE: Vector Calculus Proof of Integration
QUESTION [1 upvotes]: Let R be a solid region bounded by an oriented closed suface $\partial{R}$. Let f and g be $c^2$ scalar functions. Let the $\hat{n}$ be the outward normal to $\partial{R}$. Prove that
$$\iiint \nabla{f} \cdot{} \nabla{g} \ dV = \iint_{\partial{R}} f\nabla{g}\cdot{}d\vec{S} - \iiint_{R} f \nabla{}^2g \ dV$$
I am aware that the process to prove this is to use Gauss theorem on the RHS and that should somehow allow me to show the LHS, but I am not sure how to even begin this problem since f and g aren't given to me.

REPLY [0 votes]: Just remember that
$$
\nabla\cdot(f {\bf F}) = \nabla f \cdot {\bf F} + f\nabla\cdot{\bf F} \tag{1}
$$
Now use ${\bf F} = \nabla g$ to find
$$
\nabla\cdot(f \nabla g) = \nabla f \cdot {\nabla g} + f\nabla\cdot{\nabla g} = \nabla f \cdot \nabla g + f \nabla^2 g \tag{2}
$$
Integrating at both sides
$$
\iiint_R \nabla\cdot(f\nabla g) {\rm d}V = \iiint_R \nabla f\cdot \nabla g {\rm d}V + \iiint_R f\nabla^2 g {\rm d}V \tag{3}
$$
Now apply the divergence theorem to the L.H.S
$$
\iint_{\partial R} f\nabla g \cdot{\rm d}{\bf S} = \iiint_R \nabla f\cdot \nabla g {\rm d}V + \iiint_R f\nabla^2 g {\rm d}V \tag{4}
$$


TITLE: Do order isomorphisms of sub-pospaces extend to the whole space?
QUESTION [1 upvotes]: Let $(X,\leq)$ be a pospace: a topological poset such that uppersets and downsets are closed in the topology. Let $A\subseteq X$ be a closed (connected) subspace and suppose $\phi: A\rightarrow A$ is an order isomorphism: a monotone bijection whose inverse is also monotone. Does $\phi$ extend to an order isomorphism of $X$? If not, what kind of conditions on $X$ do guarantee this property?
I'm specifically interested in the case where $X=[0,1]^n$ in which case things like Tietze's extension theorem might be helpful (or just the general fact that it is a compact metric space).

REPLY [0 votes]: I found a counter-example to the case of $X=[0,1]^n$:
Pick the subspace $A=\frac{1}{2}+[0,\frac{1}{2}]^n$. So it is an upper corner that is order-isomorphic to $X$. One of the corners of $A$ is an interior point of $X$, while the other corners of $A$ are also corners of $X$. If we permute the coordinates of $A$ such that one of the corner points of $A$ is mapped to the "interior" corner point, then we get an order isomorphism of $A$ that can't be extended to an order isomorphism of $X$, since any order isomorphism of $X$ must map corner points to corner points.
Since this applies to a connected compact subset of a compact metric space, it is unlikely to hold in general for any other general class of pospaces.


TITLE: For a given $\lambda$, what are its corresponding eigenvectors if $A - \lambda I = 0$?
QUESTION [0 upvotes]: Given any square matrix $A$ and its eigenvalue $\lambda$, if $A - \lambda I = 0$, then $(A - \lambda I) \mathbf{x} = \mathbf{0}$ becomes
$$(A - \lambda I)\mathbf{x} =
\begin{bmatrix}
0 & \dots & 0 \\
\vdots & \ddots & \vdots \\
0 & \dots & 0
\end{bmatrix}
\begin{bmatrix}
x_1 \\ \vdots \\ x_n
\end{bmatrix}
=
\begin{bmatrix}
0 \\ \vdots \\ 0
\end{bmatrix} \ .
$$
What are the corresponding eigenvectors of $\lambda$? Is it all the possible $n \times 1$ vectors in $\mathbb{R}^n$? If so, is this possible?

REPLY [0 votes]: The relation between an eigenvector ($v$) and the correspondent eigenvalue (\lambda) is
$$Av=\lambda v\Leftrightarrow (A-\lambda I)v=0 \quad (1)$$
So, once you find the solution, $\lambda$, of $\det (A-\lambda I)=0$ you have to place it back in the linear system $(1)$ and find the correspondent solution, $v$, of it.


TITLE: Does such a Markov chain exist?
QUESTION [0 upvotes]: Suppose it has finite state space $S$, and $\lim\limits_{n\to \infty}p_{ij}^{(n)}=0$ for all $i,j\in S$. But guess is there isn't, since for a finite transition matrix, it is unlikely to have $P^n\to0$ for each of its entry. I have tried some inequalities and summed over $n$ or state space, trying to find contradiction by recurrence theorem, but got nothing.
Any hint would be appreciated!

REPLY [1 votes]: The sum of each column in transition matrix has to equal 1 (consider a perfectly located state in the beginning - after the matrix multiplication the probability vector will represent the corresponding matrix column). In your case it would be equal to zero.
You can also think of it this way - you would converge to a state where the imaginary particle is nowhere. It is always "somewhere".


TITLE: In 1st law of Thermodynamics, $W=P_{ext}dV$ or $W=P_{int}dV$?
QUESTION [0 upvotes]: Work done by the system of the piston = (Force applied on the system on the piston)×(displacement of the piston) = $F_{int}\cdot dS$=$P_{int}\cdot dV$
But, in my textbook, it is given $W=P_{ext}dV$
For a reversible process, this is okay, since $P_{int}=P_{ext}$ at all points, but what if it is not reversible? If $P_{int}\neq P_{ext}$, then which one is correct? Why?

REPLY [1 votes]: It is always $dW = - P_\text{ext} dV$.

If the process is quasistatic, then by definition $P_\text{ext}=P_\text{int}$ and in this special case

$$ dW = dW_\text{rev} = - P_\text{int} dV$$


TITLE: Measure and Lebesgue Integral
QUESTION [3 upvotes]: I got this exercise as homework and I found some problems in solving it. So I hope that someone can help me.
Let $f:[0,1] \rightarrow R$ Lebesgue measurable and $S=\{x \in [0,1]:f(x) \in Z\}$.
Show that:
$\lambda(S)=\lim_{n \rightarrow \infty} \int_0^1 \lvert \cos(\pi f(x))\rvert^n \,d\lambda$

REPLY [0 votes]: I'm going to sum up a complete answer:
$\lambda(S)={x \in [0,1]:f(x) \in Z}$
considering : $\lim_{n \rightarrow \infty} \int_0^1 \vert \cos(\pi f(x))\vert^n d\lambda$
 so $0 \leq \vert \cos(\pi f(x))\vert^n \leq 1$.
If $\vert \cos(\pi f(x))\vert=1$ it means that $f(x)=2k,2k+1\qquad k \in N \forall x \in [0,1]$ so $\lim_{n \rightarrow \infty} \int_0^1 \vert \cos(\pi f(x))\vert^n d\lambda=\int_0^1 1 d\lambda=1$
At the same time if $f(x)=2k,2k+1\qquad k \in N \forall x \in [0,1]$ it means that $2k,2k+1 \in Z$ and so $\lambda(S)=\lambda([0,1])=1$.
If $\vert\cos(\pi f(x))\vert \neq 1$ you get $\lim_{n \rightarrow \infty} \int_0^1 \vert \cos(\pi f(x))\vert^n d\lambda=\int_0^1  d\lambda=0$ and in this case S has cardinality as maximum numerable so it could be considered as union of singletons of null measure since we get that S has measure zero too.
So I can use this convergence to apply the monotone convergenze theorem to $g-f_n$ as suggested by Davide Giraudo.


TITLE: Show that the union of two equivalence relations G and H on A is transitive if G ◦ H ⊆ G ∪ H and H ◦ G ⊆ G ∪ H
QUESTION [1 upvotes]: The original question is: "Suppose G and H are equivalence relations on A. Then G ∪ H is an equivalence relation on A if and only if G ◦ H ⊆ G ∪ H and H ◦ G ⊆ G ∪ H"
At first, I assumed that G ∪ H is an equivalence relation and proved that G ◦ H ⊆ G ∪ H and H ◦ G ⊆ G ∪ H
The second assumption is that G ◦ H ⊆ G ∪ H and H ◦ G ⊆ G ∪ H. So I have to prove that G ∪ H is reflexive, symmetric and transitive. I have done the first two, but I'm stuck in the third one. I'm trying to solve it like this:
Suppose $(a,b) \in  G \circ H$ and $(b,c) \in  G \circ H$. (This means $(a,b)\in G \cup H$ and $(b, c) \in G \cup H$)
Then by the definition of composition, $(a,s) \in H \wedge (s,b)\in G$ and $(b, t) \in H \wedge (t,c) \in G$
As $((s,b)\in G \wedge (b, t)\in H)$, $(s, t) \in H\circ G$
I don't know where to go from here. If I could somehow show that $(s, t) \in G\circ H$, then I could write
$(s, d)\in H$ and $(d, t) \in G$
$(a, s) \in H \wedge (s, d) \in H$ means $(a, d) \in H$
Similarly, $(d, t) \in G \wedge (t, c) \in G$ means $(d, c) \in G$
Thus, $(a, d) \in H \wedge (d, c) \in G$ means $(a,c) \in G \circ H$
Therefore, $(a,c) \in G \cup H$
Repeat the same thing for $H\circ G$
Any help would be appreciated. Thanks in advance :)

REPLY [1 votes]: You have already proven that $G\cup H$ is reflexive and symmetric, so it remains to prove that it is transitive.
Suppose $(a,b),(b,c) \in G\cup H$.
You want to prove that $(a,c) \in G\cup H$.
If $(a,b),(b,c) \in G$, then $(a,c) \in G \subseteq G\cup H$;
likewise if $(a,b),(b,c) \in H$.
If $(a,b) \in G$ and $(b,c) \in H$, then $(a,c) \in G\circ H \subseteq G\cup H$;
likewise if $(a,b) \in H$ and $(b,c) \in G$.


TITLE: Partial sums and growth
QUESTION [1 upvotes]: Define $f \sim g$ to mean $\lim_{x \to \infty}\frac{f(x)}{g(x)}=1$.
I am trying to either prove or disprove that $f \sim g$ $\iff$ $\sum_{k=1}^{x}f(k) \sim \sum_{k=1}^{x} g(k)$ provided $f$ and $g$ are positive and unbounded for large $x$.
I showed that $\leftarrow$ does not hold, but dropped the assumption that $f$ and $g$ remain unbounded (I took $f(x)=\frac{1}{x(x+1)}$ and $g(x)=\frac{1}{x(x+1)(x+2)}$, in which case the partial sums are $\sim$, but $f$ and $g$ aren't.) For $\rightarrow$, I tried to use Abel's summation formula and attempted to show that $\sum_{k=1}^{x}f(k) -\sum_{k=1}^{x} g(k) \rightarrow 0$, but got nowhere.

REPLY [2 votes]: Under the given assumptions, the implication
$$f(x) \sim g(x) \implies \sum_{k = 1}^x f(k) \sim \sum_{k = 1}^x g(k)$$
holds, but the implication in the other direction does not hold.
To see the latter, consider $f(x) = (2 + \cos (\pi x))\cdot x$ and $g(x) = (2 - \cos (\pi x))\cdot x$. Then $\frac{f(k)}{g(k)}$ alternates between $3$ and $1/3$, so $f(x) \nsim g(x)$, but
$$\sum_{k = 1}^x f(k) = x^2 + O(x) \sim \sum_{k = 1}^x g(k) = x^2 + O(x).$$
To see that the implication above holds, consider an arbitrary $\varepsilon > 0$. By the assumption $f(x) \sim g(x)$ there is an $x_{\varepsilon} \in \mathbb{N}$ such that
$$(1-\varepsilon) f(x) \leqslant g(x) \leqslant (1 + \varepsilon)f(x)$$
for all $x \geqslant x_{\varepsilon}$. Then for $x > x_{\varepsilon}$ we have
$$(1 - \varepsilon)\sum_{k = 1}^x f(k) - (1 - \varepsilon)\sum_{k = 1}^{x_{\varepsilon}} f(k) + \sum_{k = 1}^{x_{\varepsilon}} g(k) \leqslant \sum_{k = 1}^x g(k) \leqslant (1 + \varepsilon)\sum_{k = 1}^x f(k) - (1 + \varepsilon) \sum_{k = 1}^{x_{\varepsilon}} f(k) + \sum_{k = 1}^{x_{\varepsilon}} g(k).$$
Since the sums up to $x_{\varepsilon}$ are independent of $x$, and $\sum_{k = 1}^x f(k) \to \infty$ (that is what we need, $f$ need not be unbounded itself) it follows that
$$1 - 2\varepsilon < \frac{\sum_{k = 1}^x f(k)}{\sum_{k = 1}^x g(k)} < 1 + 2\varepsilon$$
for sufficiently large $x$.


TITLE: Basic Electrostatics Problem Modified
QUESTION [1 upvotes]: I was going through lecture notes of a class(this is not a homework question), and I have made a problem I'm myself confused at:
Two parallel infinite conducting planes lying in the xy-plane are separated by a distance $d$. Each plane is uniformly charged with equal but opposite surface charge densities. Find the electric field everywhere in space.
Now, if the planes were non-conducting, it was pretty trivial. But I want to know what happens if these are replaced by conducting planes. Does Polarization takes place? How do we solve it then and what are the changes that can be noticed?

REPLY [0 votes]: CASE1: - BOTH GROUNDED : No plate should have a charge as a result there would be no electric field.
CASE2: ONE CHARGED OTHER GROUNDED: now whatever charge you give to one plate opposite charge induces to the other plate and the same charge goes to ground so so we have two plates with exactly same charge density (say $\sigma$) but of opposite charges . So the electric field is 2$\times$$\sigma/2\epsilon_0\hat{i}$.
CASE3 BOTH ARE GIVEN CHARGES AND ARE ISOLATED: You will find by demanding that field at the centre of plates 0 . You will find charge distribution like $(Q_i+Q_2/2)$,$(Q_1-Q_2)/2$,$(Q_2-Q_1)/2$,$Q_1+Q_2/2$ if you go to the sides of plates from left to right. And electric  field would be $(Q_1-Q_2)/2A\epsilon_0\hat{i}$.


TITLE: How to measure the wavelengths of ultraviolet lines in the hydrogen spectrum?
QUESTION [1 upvotes]: how to measure the wavelengths of ultraviolet radiation isn't it invisible?
it's possible to measure visible radiation but how to measure invisible radiation?

REPLY [1 votes]: Three different ways (at least).
1) Use a detector, like a photodiode or photomultiplier, which responds to UV by producing electrons when hit by photons even if those photons are not visible,
2) Use a detector, like a bolometer, which measures the temperature rise produced when UV energy is absorbed, or
3) Coat a visible-only detector with a phosphor which emits light when hit with UV, then detect the emitted visible light.


TITLE: Difference between rows and columns of a matrix
QUESTION [0 upvotes]: Say I have three vectors $(2,-1,0), (1,-1,0), (1,0,1)$ and I want to check for linear independence.
At first I thought I need to put the vectors as rows, and then use Gaussian elimination and see.
Then I thought maybe I need to put them as columns rather than as rows, and use Gaussian elimination and see.
After trying both ways, I got to the conclusion there is no linear dependence between the vectors, but I don't understand exactly what I did.

REPLY [1 votes]: Recall the definition of linear independence:

A collection of vectors $\{v_1,v_2,\ldots,v_n\}$ of a vector space $V$ over a field $K$ is linearly independent if $a_1v_1+a_2v_2+\dots+a_nv_n=0$ implies $a_1=a_2=\dots=a_n=0$, where the $a_i$ are elements of the underlying field $K$.

Assuming that you are working in the space $\mathbb R^3$ over $\mathbb R$, what we need to show is that
$$a_1\begin{bmatrix}2\\-1\\0\end{bmatrix}+a_2\begin{bmatrix}1\\-1\\0\end{bmatrix}+a_3\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$
for real numbers $a_1,a_2,a_3$ implies that $a_1=a_2=a_3=0$. This reduces to a $3\times 3$ system of linear equations by just checking each component. This should be relatively easy and can be done by elementary methods, which I will leave to you.

Note: it does not really matter if you consider your vectors as row vectors or column vectors. If you considered them as row vectors, for example, all we need to show is
$$a_1\begin{bmatrix}2&-1&0\end{bmatrix}+a_2\begin{bmatrix}1&-1&0\end{bmatrix}+a_3\begin{bmatrix}1&0&1\end{bmatrix}=\begin{bmatrix}0&0&0\end{bmatrix}$$
implies $a_1=a_2=a_3=0$.
But the way to do this is exactly the same: by considering each component! And of course, the equations that we have and the resulting $3\times 3$ linear system are exactly the same; so this is in fact equivalent to the previous view as column vectors.


TITLE: Joules to do something
QUESTION [1 upvotes]: If my very limited understanding is correct, then, not accounting for gravity:
1 Newton can move 1 kilogram 1m
But, can 2 Newtons move 1 kilogram 2 meters?
Is this because 1 Newton = acceleration of 1/ms, so, 2 Newtons would result in a speed of 2 m/s? Or would it be 1 m/s lasting twice as long?
So what does a joule do: accelerate or just move?

REPLY [2 votes]: 1 Newton can move 1 kilogram 1m
But, can 2 Newtons move 1 kilogram 2 meters?
Is this because 1 Newton = acceleration of 1/ms, so, 2 Newtons would
  result in a speed of 2m/s? Or would it be 1m/s lasting twice as long?
So what does a joule do: accelerate or just move?

A joule does not accelerate, it is the unit, the measure both of

kinetic energy (that a body has when it has been accelerated and has velocity) and the measure of
work done against a force

1 N (newton) does not move but accelerates 1 Kg by 1 m/s every second, using the correct units: $1 m/s^2$. 2 N (newtons) give 1Kg double acceleration:  $2 m/s^2$. Gravity is a Force of ca.

10 N and accelerates 1 Kg by $10m/s^2$ (and not, as you write, 1(0) /ms, I edited your post by I left that, as I cannot know if it is a typo or what units you really wanted to write). That means that every second its velocity increases by 10 m/s. If it lasts for a second, then speed will be 10 m/s and KE 50 joules and it will travel a distance of 5m, if it lasts two seconds its speed will be 20 m/s and its KE 200 joules, and so on.

The energy these body acquires is measured in Joules and depends on the time the force acts on the bodies. This is the case of 1 (or more) Kg of mass that falls to the ground because of gravity for 1 (or more seconds)

If on the contrary, you want to lift up a weight of 1Kg, then you do not accelerate the body but you move it or, better displace it, it gains no speed because you are doing work, you are working against a force (gravity), and with the same amount of energy (joules) you can lift a
k-Kg body by 1m, or a
.5k-kg-body by 2m, or a
2k-Kg body by .5m

That is because joules are the product of force by displacement: (Ke) = Work = $\rightarrow = F . d$, so  $\rightarrow k *1 = .5k *2 = 2k *.5$
If you want to lift that body back to the height of 5m, you must apply a force of 10N by 5 metres, and spend the 50 joules of energy it gained falling down, and that because you fight against a force of 10N over the distance of 5m-
In conclusion: gravity accelerates a body, accelerates it  and gives it energy (joules), if you want to lift a body, you spend energy (joules) to displace/ lift it up, in this case you might say, in approximate language, that: 'joules make a body move*


TITLE: Where did I go wrong in showing $\|x\|_1>\sqrt n\|x\|_2$
QUESTION [1 upvotes]: To show $\|x\|_1>\sqrt n\|x\|_2$ where the norm $\|\cdot||_1,\|\cdot\|_2$ are defined over $\mathbb R^n$
Let $x=(x_1,x_2,\ldots,x_n),y=(1,1,\ldots,1)\in\mathbb R^n$
Using  Hölder's Inequality
$$\sum_{i=1}^n|x_i y_i|\le\left(\sum_{i=1}^n|x_i|^2\right)^{1/2} \left(\sum_{i=1}^n|y_i|^2\right)^{1/2} \\ \implies \sum_{i=1}^n|x_i|\le  \left(\sum_{i=1}^n|x_i|^2\right)^{1/2}\sqrt n\implies \|x\|_1\le\sqrt n\|x\|_2$$
Where did I go wrong?

REPLY [1 votes]: Your argument correctly shows that $\|x\|_1\le \sqrt{n}\|x\|_2$, and thus disproves the opposite inequality stated in the title.
Bonus content: in the opposite direction, $\|x\|_2\le \|x_1\|$ holds. Indeed, squaring both sides we get $\sum_i x_i^2$ on the left and $\sum_i x_i^2 +\sum_i\sum_j |x_i|\,|x_j|$ on the right.


TITLE: Geometry vectorial
QUESTION [0 upvotes]: Consider the lines $L_{1} = \{(2t, 3t - 1, t) \mid t\in\mathbb{R}\}$ and $L_{2} = \{(-t, 2t + 1,- 2t + 3) \mid t\in\mathbb{R}\}$.
Calculate the Cartesian equations of two parallel planes $\pi_{1}$ and $\pi_{2}$ such that $L_{1}\subset \pi_{1}$ and $L_{2} \subset \pi_{2}$.
I used the cross product, but that don't find the plans equations

REPLY [3 votes]: $\ell$ is a line with direction $\langle 2,3,1\rangle$ and passing $(0, -1, 0)$;
$m$  is a line with direction $\langle-1, 2, -2\rangle$ and passing $(0, 1, 3)$.
So the vertical normal vector of the two parallel planes is:
$$\langle2, 3, 1\rangle \times \langle-1, 2, -2\rangle = \langle-8, 3, 7\rangle$$
so the two planes:
$$ \pi_1: -8x + 3y + 7z + 3 = 0;$$
$$ \pi_2: -8x + 3y + 7z - 24 = 0$$


TITLE: Existence of uniform circular motion
QUESTION [3 upvotes]: We know that if any body has some mass and is moving with a velocity v at an instant $t = 0$ , then it is not going to have same velocity after some time if any net force acts on it. It is because the force causes acceleration and acceleration causes change in velocity.

Consider a ball of mass $M$ describing a uniform circular motion in a circle of radius $R$ . This ball is having constant speed. The particle at any instant is having only force (as well as acceleration) towards the centre.
Suppose it had a velocity of magnitude $x$ at the instant it starts the uniform circular motion. From this instant only, the centripetal force works because before this (when it was not doing circular motion) no centripetal force was there.
Now, magnitude of velocity towards the centre is increasing as time is passing because there is a positive acceleration towards the centre. Let it be denoted as $v(t)$ .
The magnitude of resultant velocity is given as :
$$R = \sqrt{(x)^2 + { v(t) } ^2  }$$
as from the Pythagoras theorem.
As $v(t)$ is increasing, thus the resultant's magnitude is also. So, it's not uniform circular motion?
Where did I go wrong ?
I would be glad if the actual description of circular motion is provided. I cannot get any intuitive description about circular motions and it feels very hard.

REPLY [4 votes]: Let $\mathbf{\hat r}$ be a unit vector from the ball to the centre of the circle. If the ball's velocity is $\mathbf{v}$ then the magnitude of component of component of $\mathbf{v}$ directed towards the centre of the circle is
$$v_r=\mathbf{v}.\mathbf{\hat r}$$
so when we differentiate with respect to time we have
$$\displaystyle \frac {d(v_r)}{dt} = \frac {d \mathbf{v}}{dt} . \mathbf {\hat r} + \mathbf v . \frac {d \mathbf{\hat r}}{dt}$$
I think you are assuming that
$$\displaystyle \frac {d(v_r)}{dt} = \frac {d \mathbf{v}}{dt} . \mathbf {\hat r}$$
but this would only be true if the second term were zero, which requires $\mathbf {\hat r}$ to be a constant vector. But $\mathbf {\hat r}$ is not a constant vector because it is changing direction. In fact, for uniform circular motion, the second term exactly cancels the first and we have $\frac {d (v_r)}{dt}=0$, so $v_r$ is constant - and, furthermore, this constant is zero.


TITLE: Prove by induction that $\forall n \in \mathbb{N} (I_A \subseteq R^n \circ R^{-n})$
QUESTION [1 upvotes]: Let $R$ be a binary relation and $A = Dom(R) \cup Ran(R)$. Prove by
induction that $I_A \subseteq R^n \circ R^{-n}$ for every natural $n$.
I reached the conclusion in this post, but I think my problem asks for a generalization of what is found there. The book from where I took the problem gives the hint: use the fact that for any relations $R_1$, $R_2$, $S_1$, and $S_2$, if $R_1 \subseteq R_2$ and $S_1 \subseteq S_2$, then $R_1 \circ S_1 \subseteq R_2 \circ S_2$.
Basis. Let $n = 1$ and $(x,x) \in I_A$. Then either $x \in Dom(R)$ or $x \in Ran(R)$. If $x \in Dom(R)$, there is some $y$ such that $(x,y) \in R$, which implies that $(y,x) \in R^{-1}$. Hence, $(x,x) \in R^{-1} \circ R \therefore I_A \subseteq R^{-1} \circ R$. Likewise, if $x \in Ran(R)$, there is some $y$ such that $(y,x) \in R$, which implies that $(x,y) \in R^{-1}$. Hence, $(x,x) \in R \circ R^{-1} \therefore I_A \subseteq R\circ R^{-1} $.
Step. Let $k \geq 1$. Suppose $I_A \subseteq R^k \circ R^{-k}$. To prove $I_A \subseteq R^{k+1} \circ R^{-(k+1)}$, notice that $$R^{k+1} \circ R^{-(k+1)} = (R^k \circ R) \circ (R^{-k} \circ R^{-1}).$$ I haven't thought much further because the base case is bothering me. When $x \in Ran(R)$, the claim was proved. However, since composition is not commutative, when $x \in Dom(R)$ the conclusion was not reached, and I think that is not even possible. As for the inductive step, am I headed towards the right direction? I didn't see how to use the hint yet.

REPLY [1 votes]: As William Elliot said, the problem is incorrect as stated.  What your argument in "Basis" goes to show is not $I_A \subseteq R \circ R^{-1}$, $I_A \subseteq R^{-1} \circ R$, but $I_Y \subseteq R \circ R^{-1}$, $I_X \subseteq R^{-1} \circ R$, where $X$ is the domain of $R$ and $Y$ its range.
To prove inductively that $I_Y \subseteq R^n \circ R^{-n}$, for the inductive step, begin with
\begin{align}
  R^{k + 1} \circ R^{-(k + 1)} &= \left(R^k \circ R\right) \circ \left(R^{-1} \circ R^{-k}\right)\\
 &= R^k \circ \left(R \circ R^{-1}\right) \circ R^{-k}.
\end{align}
[edited to add:]
Some of the things that William Elliot said don't agree with the definition of composition of relations that I'm familiar with, and that I think you're working with; but even with that definition it's true, and along the same general lines as his remarks, that to prove the result for general $n$ you need some further assumption, and that a sufficient condition is that the domain and range of $R$ are equal.


TITLE: How to prove that $0$ sum could be reached
QUESTION [2 upvotes]: I have $n$ random points $u_1, u_2, \ldots, u_n$ on the unit circle. Points are uniformly distributed.
I want to think about the points as associated angles $\theta_1, \theta_2, \ldots, \theta_n$ that are uniformly picked in $(-\pi, \pi]$.
Suppose, we rotate the circle, thus $\theta_i$ is a function of the angle $\theta$ of the new axis in the original frame.
For example, if $\theta_1 = \dfrac{\pi}{2}$ and we rotate the circle $\dfrac{\pi}{6}$ in the positive direction, then $\theta_1\left(\dfrac{\pi}{6}\right) = \dfrac{2}{3}\pi$.
Angles are adjusted to always stay in $(-\pi, \pi]$ range. See the animation below

Denote $S_{\theta} \triangleq \theta_1(\theta) + \theta_2(\theta) + \ldots + \theta_n(\theta)$.

Does there exist a rotation $\theta$ such that $S_{\theta} = 0$ ?

My gut feeling, is that yes. But I do not see a direction to prove the claim.
What  I have done, is that rotating a circle clock-wise creates $S_{\theta}$ monotonically decreasing, except, when a point "crosses a negative X axis", at that point $2\pi$ are added.
But I don't see why $S_{\theta}$ changes the sign. It could stay negative or positive always.

REPLY [0 votes]: Since the sum of the shifted angles differs from the sum of the original angles by $n\theta$ plus some multiple of $2\pi$,  the problem can only be solved by taking  $\theta=\theta(k)=-\frac{\theta_1+\cdots +\theta_n}n+\frac{2k\pi}n$ for suitable $k\in\Bbb Z$. This also means that we may assume wlog. that the angles are "normalized" to begin with, i.e., $$\tag1\theta_1+\cdots+\theta_n\in2\pi\Bbb Z$$
and
$$\tag2\theta_1\le\theta_2\le\ldots\le \theta_n. $$
If for each interval $(\frac {2j-n}n\pi,\frac{2(j+1)-n}n\pi]$, $0\le j<n$, we know the number $N_j$ of angles $\theta_i$ are in that interval, then this already determines $\sum\theta_i$ up to an interval $(\alpha,\alpha+2\pi]$, and so by $(1)$ determines it completely.
Hence for any $n$-tuple $\mathbf N=(N_0, \ldots, N_{n-1})$ of non-negative integers $N_j$ with $\sum N_j=n$, we can let
$$ F(\mathbf N)=\frac1{2\pi}\sum_{i=1}^n\theta_i$$
where $\sum\theta_i$ is a multiple of $2\pi$ and exactly $N_j$ of the $\theta_i$ are in $(\frac {2j-n}n\pi,\frac{2(j+1)-1}n\pi]$; by the remark above, this is well-defined.
If we define the rotation of such a tuple, $$\operatorname{rot}(N_0,\ldots, N_{n-1}):=(N_1,\ldots, N_{n-1},N_0),$$ then our goal is to
 find, for each valid $\mathbf N$, some $k\in\Bbb Z$ with $F(\operatorname{rot}^k\mathbf N)=0$.
Note that by the condition on the $\theta_i$,
$$ \tag3\sum_{j=0}^{n-1}(\tfrac jn-\tfrac12)N_j<F(\mathbf N)\le 1+ \sum_{j=0}^{n-1}(\tfrac jn-\tfrac12)N_j$$
so that
$$ F(\mathbf N)=\left\lfloor1+\sum_{j=0}^{n-1}(\tfrac jn-\tfrac12)N_j\right\rfloor.$$
It follows that
$$\tag4 F(\operatorname{rot}\mathbf N)=F(\mathbf N)-1+N_0\ge F(\mathbf N)-1.$$
If we sum over all  rotations of a period, i.e.,
with $$G(\mathbf N):=\sum_{k=0}^{n-1}F(\operatorname{rot}^k\mathbf N),$$we find from $(3)$ that
$$ \sum_{j=0}^{n-1}(\tfrac jn-\tfrac12)\sum_{k=0}^{n-1}N_k<G(\mathbf N)\le n+\sum_{j=0}^{n-1}(\tfrac jn-\tfrac12)\sum_{k=0}^{n-1}N_k,$$
i.e.,
$$ -\frac n2<G(\mathbf N)\le \frac n2.$$
Therefore it is neither possible that all $F(\operatorname{rot}^k\mathbf N)$ of a period are positive nor that all are negative. As Hence there must exist some $k\in\Bbb N$ with $F(\operatorname{rot}^k\mathbf N)\ge 0$ and $F(\operatorname{rot}^{k+1}\mathbf N)\le 0$. Using $(4)$, we conclude that one of $F(\operatorname{rot}^k\mathbf N)$, $F(\operatorname{rot}^{k+1}\mathbf N)$ must be $=0$. $\square$


TITLE: Eigenvalues and eigenspaces for the map $T(z,w)=(w,-z)$ on $\mathbb R^2$
QUESTION [0 upvotes]: Find all the eigenvalues and corresponding eigenspaces for $T(z,w)=(w,-z)$ for all $(z,w)$ in $\mathbb{R}^2$.

So I would need to find all the values $x$ for which $T(z,w)=x(z,w)=(w,-z).$
So $xz=w$ and $xw=-z$
So $x$ must be zero? Or $z$ or $w$ are zero?
I'm a bit confused.

REPLY [0 votes]: By definition, an eigenvector should be non-zero. So if $x = 0$, then $0 = (w,-z)$, so $w=z=0$, so this gives never rise to an eigenvector. Also, if $w = 0$, then the above equations imply $z = 0$.
So we can assume that they are non-zero. So
$$xw = x^2z = -z \Rightarrow (x^2+1)z = 0 \Rightarrow x^2+1 = 0$$
But $x$ is real, so there are no solutions.
Notice that this method is not particularly nice. A much nicer method is by computing the characteristic polynomial of the linear transformation (are you familiar with that?)


TITLE: Electric field outside of a conductor with charge $q$
QUESTION [0 upvotes]: Suppose that I have a spherical conductor (radius $R$) with a cavity at the center and a charge $q$ inside it. I know that the electric field outside of the conductor will be
$$\vec{E}=\frac{q}{4\pi\epsilon_0}\frac{1}{r^2}\hat{r}$$
My question is what happens when there is another charge of the same magnitude at distance $R+a$, where $a<R$, from the center of the conductor. Does the flux of the electric field outside of the conductor (for example, at $r=2R$) remain
$$\Phi_E=\frac{q}{\epsilon_0}$$
or does it become
$$\Phi_E=\frac{2q}{e_0}$$
due to the other charge?

REPLY [0 votes]: Note that the flux changes as you suggest (assuming the two charges are equal) but it is not then true that the electric field is twice what you wrote in your first equation. That's because with the second charge being off center, the electric field is no longer spherically symmetric so the flux integral is not easily calculated by assuming E is constant.


TITLE: Torsion - subgroup and quotient
QUESTION [9 upvotes]: Let $G$ be a finitely generated pro-$p$ group.
Let $T$ be the set of all torsion elements in $G$.

Is it possible for $T$ to be a non-closed subgroup?

Anyway,

Can $G/\overline{\langle T\rangle}$ have torsion?

For any finitely generated abelian (more generally,  powerful) pro-$p$ group $G$, I know that 1. and 2. have negative answers: $T$ is finite and $G/T$ is torsion-free.
Thanks.

REPLY [2 votes]: It seems to me that the answer on both questions is Yes.
Let $F$ be a finitely generated free non-abelian pro-$p$ group, $N=\overline{F^{\prime\prime}}$ and $K=\overline {[N,F]}$. Then $N/K$ is torsion free and of infinite rank.
Let $a_1,a_2,\ldots$ be a free  $\mathbb Z_p$-generating set of $N/K$. Put $G_1=F/\langle K, a_1^{p},a_2^{p^2},\ldots\rangle$. Let $a=\prod a_i$ and let $\bar a$ be its image in $G_1$. Then $\bar a$ is not a torsion element but lies in the clousure of the subgroup of torsion elements. This answers the first question.
Let  $G_2$ be the quotient of $G_1\times \langle z\rangle $ by the subgroup generated by $z^p\bar a$. Then the image of $z$ in $G_2$ is not  a torsion element.


TITLE: Does the collection $\{ \ [n, \infty) \ \mid \ n \in \Bbb N\}$ generate the Borel $\sigma$-algebra on $\Bbb R$?
QUESTION [1 upvotes]: Does the collection $\mathfrak C = \{ \ [n, \infty) \ \mid \ n \in \Bbb N\}$
  generate the Borel $\sigma$-algebra on $\Bbb R$?

I came across this problem in a paper. It seems like the answer is No. But I'm having trouble proving it. The only potential solution I can think of is to present a $\sigma$-algebra that contains $\mathfrak C$ but does not contain certain Borel sets (possibly singleton sets?). But I cannot seem to come up with such a $\sigma$ algebra.
Any help is appreciated.

REPLY [1 votes]: Consider the $\sigma$-algebra $\mathcal{A}$ consisting of the sets of the form
$$
T\cup B
$$
where $T=\emptyset$ or $T=(-\infty,0]$ and $B$ is a Borel set in $(0,\infty)$.
It is closed under complements, because, if $B$ is a Borel set in $(0,\infty)$,
\begin{gather}
\mathbb{R}\setminus B=(-\infty,0]\cup\bigl((0,\infty)\setminus B\bigr)
\\
\mathbb{R}\setminus\bigl((-\infty,0]\cup B\bigr)=(0,\infty)\setminus B
\end{gather}
Closure under countable unions is easy to prove.
Now $\mathcal{A}$ contains $\mathfrak{C}$, but not the Borel set $(-2,-1)$.