text:
\begin{document}

\maketitle

\begin{abstract}
We prove that, among metrics on a compact quotient of $\mathbb{H}^2 \times\cdots\times \mathbb{H}^2$ (product of hyperbolic planes) of prescribed total volume, the product of hyperbolic metrics has minimal volume entropy.
\end{abstract}

\section{Introduction}

Let $(X,g)$ be a compact Riemannian $n$-manifold and $\widetilde{X}$ be the universal Riemannian cover of $X$. The volume entropy is defined as
\[h(g)=\lim_{R \to \infty}{\frac{1}{R}\log(\Vol(B(x,R)))}\]
where $B(x,R)$ is the ball of radius $R$ in $\widetilde{X}$ centered at any point $x\in\widetilde{X}$. The limit exists and is independent of the choice of $x$ (see \cite{manning} p.568).

In this paper, we are interested in the following problem.

\begin{quest} \label{question}
Let $M$ be a compact locally symmetric space of noncompact type with locally symmetric metric $g_0$. Let $g$ be any other metric on $M$ such that $\Vol(M,g_0)=\Vol(M,g)$. Do we have
\[h(g) \geqslant h(g_0)?\]
\end{quest}

M. Gromov was the first to conjecture such a result in \cite{gromovfilling}. He was only interested in the real hyperbolic case. But the question still makes sense for a general symmetric space.\\

In the case where $\widetilde{M}$ is reducible, there exist a unique locally symmetric metric of minimal entropy among locally symmetric metrics of prescribed volume, which are obtained by scaling the metric in the factors (\cite{connellfarbmer} Chapter 2). This metric is called "the" locally symmetric metric and is denoted by $g_0$.\\

In this work, we give a positive answer to the previous question in the case of compact quotients of products of $n$ hyperbolic planes, that is $M=\Gamma\backslash\left(\mathbb{H}^2\right)^n$ where $\Gamma$ is a uniform lattice in $\left(\PSL_2(\R)\right)^n$. More precisely our main result is the following:

\begin{thm}[Main theorem] \label{main}
Assume that $(M,g_0)$ is a compact quotient of the product of $n$ hyperbolic planes. Then, for any other metric $g$ on $M$,
\[h^{2n}(g)\Vol(M,g) \geqslant h^{2n}(g_0)\Vol(M,g_0).\]
\end{thm}

Let us remark that the above inequality is sharp and no assumption is made on the metric $g$.\\

According to \cite{eberlein} p.260, there exist in PSL$_2(\mathbb{R}) \times $PSL$_2(\mathbb{R})$ two types of uniform lattices: virtual products and irreducible lattices.
Then our main theorem applies for the compact products of hyperbolic surfaces or Riemannian manifolds finitely covered by such a product but also for quotients of $\left(\mathbb{H}^2\right)^n$ which are far from being products. Both compact examples do exist (see \cite{shimizu} chapter 6 or \cite{borel} for arithmetic examples).\\

In the fundamental paper \cite{bcg2}, G. Besson, G. Courtois and S. Gallot dealt with the case where $M$ is a locally symmetric space of rank one ($g_0$ is negatively curved) and obtained a similar statement than the main theorem for such spaces. The same result was obtained before (see \cite{bcg1}) in any rank but for a metric $g$ in the conformal class of the locally symmetric metric $g_0$ (supposed to be irreducible). In the case where $\dim M \geqslant 3$, the method is based on the barycenter map and the inequality in theorem \ref{main} appears as an inequality of calibration (to be described below). In the case where $\dim M=2$, already proved by Katok (see \cite{katokconformal}), one can still ask if the inequality can be seen as an inequality of calibration. Besson, Courtois and Gallot showed that it is indeed true and gave another proof of the conjecture for the hyperbolic surfaces.

The barycenter method was improved by Connell and Farb in \cite{connellfarbmer} and, working factor by factor, they also gave a positive answer to the question \ref{question} in the case where $M$ is locally a product of rank one symmetric spaces with no factor $\mathbb{H}^2$. In fact in that case, there exist a unique locally symmetric metric (when one scales the metric in each factor) of minimal entropy and with volume one.

In both of those papers, the authors pointed out that the case of products of hyperbolic surfaces still remained unknown.

Note also that there is an answer to question \ref{question} in the same setting as \cite{connellfarbmer} (products of rank $1$ symmetric spaces without $\mathbb{H}^2$-factors) in \cite{bcgmilnorwood} using an interesting different point of view. The inequality between volumes and entropies appears as a corollary of a general work on representations of fundamental groups of compact manifolds into Lie groups of noncompact types.\\

We now describe the content of this paper. To prove our main result, we use the general outline introduced in \cite{bcg2}. It consists on an application of a method of calibration. To make this method efficient, we embed the universal cover $\mathbb{H}^2 \times\cdots\times \mathbb{H}^2$ in the unit sphere of $\Le^{2}$-functions of the Furstenberg boundary $\mathbb{T}^n=\mathbb{S}^1 \times\cdots\times \mathbb{S}^1$ by the products of Poisson kernels. The aim is to show that this embedding has minimal volume. In order to detect this minimality property, one may use a differential $2n$-form taking its extremal values over orthonormal frames in tangent frames of the embedding. The method is briefly recalled in sections \ref{spherevol} and \ref{calibration}.

The hardest part is to find the calibrating form. Apart from the barycenter map, which is not efficient in the $2$-dimensional case, Besson, Courtois and Gallot developed an alternative idea.
In Chapter 3 of \cite{bcg1}, following Gromov \cite{gromovvabc}, there is a general process to build suitable differential forms using bounded cocycles. The choice of the appropriate cocycle in $\mathbb{H}^2$ is then discussed in \cite{bcg2} Chapter 6.
We generalize this approach for the compact quotients of $\left(\mathbb{H}^2\right)^n$ using the bounded $2n$-cocycle on $\mathbb{T}^n$ that M. Bucher exploits in \cite{bucherh22} (for $n=2$). We describe this bounded cocycle in paragraph \ref{form} and we check that the derived differential form has the required properties afterwards. The calibrating inequality is finally obtained in paragraph \ref{inequality}.

The last section is devoted to applications. We obtain a (non optimal) estimate for the minimal volume of a compact quotient $\Gamma\backslash\left(\mathbb{H}^2\right)^n$. The most spectacular application is that we are able to give an optimal bound for degrees of maps
$f : Y^{2n}\longrightarrow \Gamma\backslash\left(\mathbb{H}^2\right)^n$ from any Riemannian $2n$-manifold. Precisely,

\begin{cor}
Let $Y$ be a smooth manifold of dimension $2n$ endowed with a Riemannian metric $g$ and let $f$ be a continuous map
\[f : (Y,g) \longrightarrow (M,g_0)\]
Then
\[h(g)^{2n} \Vol(Y,g) \geqslant \abs{\textit{deg }f} h(g_0)^{2n} \Vol(M,g_0)\]
\end{cor}

I would like thank my PhD advisor C. Bavard for many useful discussions and comments. I am also grateful to G. Besson for his encouragements and the remarks he made on a preliminary version of this paper.

\section{Calibration method}\label{method}

\subsection{The spherical volume}\label{spherevol}

In this section, $M$ is a quotient of $\left(\mathbb{H}^2\right)^n$ by a uniform torsion-free lattice $\Gamma=\pi_1(M)$ and then $\widetilde{M}$ is the Riemannian product $\left(\mathbb{H}^2\right)^n$ with the usual metric of curvature $-1$ in each factors (called $g_0$ in both of the manifolds $M$ and $\tilde{M}$). Remark that $g_0$ is the best locally symmetric metric in the sense of \cite{connellfarbmer} Chapter 2. As above, $g$ is any other metric on $M$.\\

Let us start with a few notations. We choose once and for all a basepoint $o \in \mathbb{H}^2$, for instance $o=(0,0)$ in the Poincaré disk model. We will denote by the same letter $o$ the basepoint in $\left(\mathbb{H}^2\right)^n$. There will be no ambiguity resulting of that convention. The basepoint is used to compute the Buseman functions and to identify the boundary at infinity $\pa_{\infty} \mathbb{H}^2$ with the circle $\mathbb{S}^1$ (see below).

In what follows, $\pa_F(\widetilde{M})$ will denote the Furstenberg boundary of $\widetilde{M}$: the space of Weyl chambers at infinity in $\widetilde{M}$ emanating from the same point (see \cite{eberlein} or \cite{gjt} for further discussions).
There will be no conceptual difficulties coming from a general theory of Furstenberg boundary: we just use the fact that $\pa_F(\left(\mathbb{H}^2\right)^n)$ is identified to the $n$-dimensional torus $\mathbb{T}^n=\left(\mathbb{S}^1\right)^n$, by the above choice of a basepoint.
The Furstenberg boundary is better adapted to the case of higher rank symmetric spaces and it is one of the key points in \cite{connellfarbmer}. The Furstenberg boundary and the visual boundary are the same in the rank one case and that's why the distinction doesn't appear in \cite{bcg2}.\\

The Furstenberg boundary turns out to be a probability space in the following way. The circle $\mathbb{S}^1=\R\slash 2\pi\Z$ is endowed with the Lebesgue probability measure $d\theta$ (normalized in such a way that $d\theta(\mathbb{S}^1)=1$). The $n$-torus is the product (in the sense of probability spaces) of $n$ such circles. The spaces of $\Le^2$ functions on $\mathbb{T}^n$ is defined with respect to this measure.\\

The Poisson kernel $p_o$ of the disk is defined by:
\[p_o(x,\theta)=e^{-\mathcal{B}_o(x,\theta)},\]
where $\mathcal{B}_o(x,\theta)$ is the Buseman function.
A classical computation gives the explicit expression in the Poincaré disk
\[p_o(x,\theta)= \frac{1-\abs{x}^2}{\abs{x-e^{2i\pi\theta}}^2}\]
for any $x \in B(0,1)$ and $\theta \in \mathbb{S}^1$.\\

The definition of the spherical volume in \cite{bcg2} extends to $\left(\mathbb{H}^2\right)^n$ in the following way. We consider two representations of $\Gamma$. The first one in Isom$(\tilde{M},g_0)$ is the holonomy representation of $\Gamma$. The second one is the representation in the unit sphere of $\Le^2(\partial_F\widetilde{M})$, the Hilbert space of $\Le^{2}$-functions on the Furstenberg boundary with real values. We denote this unit sphere by $\mathcal{S}^{\infty}$ or $\mathcal{S}^{\infty}(\pa_F\widetilde{M})$ if the universal cover needs to be specified. More precisely, it is a unitary representation restricted to $\mathcal{S}^{\infty}$. It is defined by
\[(\gamma f) (\theta)= f(\gamma^{-1}(\theta))\sqrt{p_0(\gamma o_1,\theta^1)}\cdots \sqrt{p_0(\gamma o_n,\theta^n)} \]
where $o$ is the basepoint of $\left(\mathbb{H}^2\right)^n$, $\gamma o =(\gamma o_1,\cdots,\gamma o_n) \in \left(\mathbb{H}^2\right)^n$ and $\theta=(\theta^1,\cdots,\theta^n) \in \mathbb{T}^n$. This is the change of variables formula (for $\Le^2$ functions). Indeed the Jacobian of an isometry acting on the Furstenberg boundary is given by the product of Poisson kernels.\

Then, as in \cite{bcg2}, we introduce the family $\mathcal{N}$ of Lipschitz immersions
\[\Phi: \widetilde{M} \longrightarrow \Le^2(\pa_F\widetilde{M})\]
which are $\Gamma$-equivariant, that is satisfying the following equation
\[\Phi(\gamma x)=\gamma \Phi(x)\]
for all $\gamma\in\Gamma$ and all $x\in \widetilde{M}$.
We also require that
\[\forall x \in \left(\mathbb{H}^2\right)^n,\;\;\; \norm{\Phi(x)}_{\Le^2}=1, \]
that is $\Phi(x) \in \mathcal{S}^\infty$ and that, for every $x\in\widetilde{M}$, $\Phi(x)$ is positive almost everywhere.
We can also consider those immersions as functions of two variables $\Phi : (x,\theta) \mapsto \Phi(x)(\theta)$.
The product of square roots of Poisson kernels
\[
\begin{array}{c c c c}
\Phi_0 = : & \left(\mathbb{H}^2\right)^n & \longrightarrow & \rons^{\infty} \\
& (x^1,\cdots,x^n) & \longmapsto & \sqrt{p_o(x^1,\cdot)}\times\cdots\times \sqrt{p_o(x^n,\cdot)}\\
\end{array}
\]
is an example of such an immersion (see Lemma \ref{poisson} below). Moreover it is an embedding. We will think of $\left(\mathbb{H}^2\right)^n$ as embedded in $\mathcal{S}^{\infty}$ by the product of Poisson kernels.
The spherical volume is then defined by
\[
\SphereVol(M)  =  \inf_{\Phi \in \mathcal{N}} \{\Vol (\Phi )\}\\
\]
where
\[
\Vol(\Phi)= \int_M{\sqrt{\abs{\det\,_{g_0}(g_{\Phi}(x))}} dv_{g_0}(x)} \}
,\]
where $g_{\Phi}$ denotes the almost everywhere defined pull-back of the usual Hilbertian metric on $\Le^2$ by the Lipschitz immersion $\Phi$ and $\det_g(g_{\Phi})$ is computed in any $g$-orthonormal basis. The integral on $M$ means that we integrate on a fundamental domain in $\left(\mathbb{H}^2\right)^n$ for the $\Gamma$-action and the equivariance relation satisfied by $\Phi$ shows that this does not depend on the fundamental domain.

The spherical volume is a transitional object. We use it to make a link between entropies and volumes. We want to prove the following inequalities.
\[\Vol(M,g_0) \left(\frac{h(g_0)^2}{8n}\right)^{n}=\SphereVol(M) \leqslant \Vol(M,g) \left(\frac{h(g)^2}{8n}\right)^{n}.\]

First we recall the second inequality.
\begin{prop}[\cite{bcg2} Chapter 3]\label{boundedabove}
We have
\[\SphereVol(M) \leqslant \left(\frac{h(g)^2}{8n}\right)^{n} \Vol (M,g).\]
\end{prop}
\begin{proof}
We refer to Chapter 3 of \cite{bcg2} for the same proof in the rank one case.
There are only very few modifications to make in our setting.
Let's first reintroduce a family of immersions which satisfy the conditions above. For a real parameter $c>h(g)$, we consider :
\[\Psi_c(x,\theta)=\left(\int_{\tilde{M}}e^{-cd(x,y)} p_0(y^1,\theta^1)\cdots p_0(y^n,\theta^n)dv_g(y)\right)^{1/2}.\]
The condition on $c$ ensures that the integral converges. Indeed for uniform lattices volume entropy and critical exponent are the same. Then we define an element of $\mathcal{N}$ by
\[\Phi_c(x,\theta)=\frac{\Psi_c(x,\theta)}{\left(\int_{\mathbb{T}^n}\Psi_c^2(x,\theta)d\theta\right)^{1/2}}.\]
We just have replaced the boundary sphere by the Furstenberg boundary. We can now perform the very same computation as \cite{bcg2} (p.742 for a proof of the Lipschitz regularity and p.746 for the volume computation). We get the required estimate for the spherical volume.
\end{proof}

It remains to check now, if $M$ is a compact quotient of $\left(\mathbb{H}^2\right)^n$, that
\[\SphereVol(M) = \left(\frac{h(g_0)^2}{8n}\right)^{n} \Vol (M,g_0).\]
In fact, we can find an immersion $\Phi$ of $\mathcal{N}$ which has precisely the needed volume.

\begin{lem}\label{poisson}
Let $M$ be a compact quotient of $\left(\mathbb{H}^2\right)^n$. Let $\Phi_0 : \left(\mathbb{H}^2\right)^n \rightarrow \Le^2(\mathbb{T}^n)$ be defined by
\[\Phi_0(x^1,\cdots,x^n,\theta^1,\cdots,\theta^n)=\prod_{i=1}^n\sqrt{\frac{1-\abs{x^i}^2}{\abs{x^i-e^{2i\pi\theta^i}}^2}}.\]
Then
\begin{enumerate}
\item $\Phi_0$ is a smooth embedding, it belongs to $\mathcal{N}$ and
\[\Vol(\Phi_0)=\left(\frac{h(g_0)^2}{8n}\right)^n\Vol(g_0)=\left(\frac{1}{8}\right)^n\Vol(g_0).\]
\item The tangent space at the basepoint $\mathds{1}\in\mathcal{S}^\infty$ of the image of $\Phi_0$, $T_{\mathds{1}}\Phi_0(\left(\mathbb{H}^2\right)^n)$ is generated by the $2n$ functions
\[\fonction{f_i}{\mathbb{T}^n}{\R}{\theta=(\theta^1,\cdots,\theta^n)}{\sqrt{2}\cos\theta^i}\]
and
\[\fonction{f_{i+1}}{\mathbb{T}^n}{\R}{\theta=(\theta^1,\cdots,\theta^n)}{\sqrt{2}\sin\theta^i},\]
for $i=1,\cdots,n$.
\end{enumerate}
\end{lem}

\begin{proof}
It is enough to handle the same situation with only one factor $\mathbb{H}^2$, the situation appearing completely as a product. The family of measures $(\nu_z)_{z\in\mathbb{H}^2}$ on $\mathbb{S}^1$ which are in the Lebesgue class and satisfy
\[\frac{d\nu_z}{d\theta}=\frac{1-\abs{z}^2}{\abs{z-e^{2i\pi\theta}}^2} \]
are in fact the so-called Patterson-Sullivan measures of the hyperbolic plane which were constructed in \cite{patterson}. We refer to this original paper for the equivariance relation (remark this family of measures is even SL$_2(\R)$-equivariant).\\

Then it is enough to show that $\Phi_0$ is an immersion at the basepoint $o$, SL$_2(\R)$ acting (transitively on $\mathbb{H}^2$) by diffeomorphisms. The differential of $\Phi_0$ is easy to compute and we get the basis we claimed for the tangent.

This shows in particular the point 2 of the above Lemma which shall be used later. As the $2$ functions $\cos$ and $\sin$ are a free family in $\Le^2(\mathbb{S}^1)$, we obtain that $\Phi_0$ is an immersion at $o$.\\

Finally the volume of $\Phi_0$ has been computed in \cite{bcg2} at page 744.
\end{proof}

\subsection{Calibration theory}\label{calibration}

In order to show that the Spherical Volume is achieved by the map $\Phi_0$, we use a classical method of calibration, following \cite{bcg2} chapter 4. Let us make a brief review on how we implement this method.\\

Let $\Omega$ be a differential $2n$-form which is $\Gamma$-invariant.

\begin{defi}
\begin{enumerate}
\item The comass of $\Omega$ is the quantity
\[\comass(\Omega)=\sup \abs{\Omega_{\varphi}(f_1,\cdots,f_{2n})},\]
where the supremum is taken over all functions $\varphi\in\mathcal{S}^\infty$ and every orthonormal family $(f_1,\cdots,f_{2n})$ where each $f_i$ belongs to $T_{\varphi}\mathcal{S}^\infty$.\
\item One says that the differential form $\Omega$ calibrates some immersion $\Phi_0\in\mathcal{N}$ if
\begin{enumerate}
\item The form $\Omega$ is closed,\
\item its comass is finite and nonzero and \
\item when we restrict $\Omega$ to orthonormal families, it is maximal on the tangent space $T\Phi_0(\left(\mathbb{H}^2\right)^n)$, that is,
\[\frac{\abs{\Omega_{\Phi_0(x)}(d_x\Phi_0(u_1), \cdots,d_x\Phi_0(u_{2n}))}}{\norm{d_x\Phi_0(u_1)\wedge \cdots\wedge d_x\Phi_0(u_{2n})}} =\comass(\Omega)\]
for every $x\in\left(\mathbb{H}^2\right)^n$ and every orthonormal family $(u_1,\cdots,u_{2n})$
\end{enumerate}
\end{enumerate}
\end{defi}

Here is the way we shall exploit a calibrating differential form. The following proposition follows readily from Stokes theorem. Here, we emphasize the fact that we use in a decisive way the compactness hypothesis for $M$ (see \cite{stormnonuniform}).

\begin{prop}[\cite{bcg2} proposition 4.3 p.748]
Assume there exists a differential $2n$ form which calibrates an immersion $\Phi_0\in\mathcal{N}$. Then
\[\SphereVol(M)=\Vol (\Phi_0). \]
\end{prop}

From now on it remains to find such a calibrating form for the Poisson kernel.

\section{A calibrating form for \texorpdfstring{$\left(\mathbb{H}^2\right)^n$}{h22}}\label{proof}

\subsection{Definition of the form}\label{form}

We will denote by $e$ the Euler class of the circle,
\[
   \left \{
   \begin{array}{r c l}
      e(\theta_0,\theta_1,\theta_2)  & =  & 1 \mbox{ if the points are cyclically ordered on $\mathbb{S}^1$  } \\
      e(\theta_0,\theta_1,\theta_2)  & =  & -1 \mbox{ if not} \\
   \end{array}
   \right .
\]
Then we consider the application $C : \left(\mathbb{T}^n\right)^{2n+1}\rightarrow \R$ given by the following formula,
\[C(\theta_0,\cdots,\theta_{2n})=\frac{1}{(2n+1)!} \sum_{\sigma\in\mathfrak{S}_{2n+1}}\sign (\sigma) \prod_{i=1}^ne(\theta^i_{\sigma(2i-2)},\theta^i_{\sigma(2i-1)},\theta^i_{\sigma(2i)}). \]
For example if $n=1$ then $C=e$, the Euler class and if $n=2$, this is the application used in \cite{bucherh22}. This map $C$ could be seen as the alternation of the cup product of $n$ Euler classes. In particular $C$ is alternate, that is,
\[C(\theta_{\sigma(0)},\cdots,\theta_{\sigma(2n)})=\sign (\sigma) C(\theta_0,\cdots,\theta_{2n}).\]
Then the following formula defines a differential $2n$-form on $\mathcal{S}^\infty$:
\[\Omega_{\varphi}(f_1,\cdots,f_{2n})=\int_{\left(\mathbb{T}^n\right)^{2n+1}} C(\theta_0,\cdots,\theta_{2n})\varphi^2(\theta_0) \varphi f(\theta_1)\cdots \varphi f (\theta_{2n}) d\theta_0\cdots d\theta_{2n}.
\]
We conclude this paragraph by stating some properties of $C$ that we shall use later on.

\begin{prop}\label{invariancecocycle}
Let $G$ be the group $\left(\Diff^+(\mathbb{S}^1)\right)^n$ embedded diagonally in $\Diff^+(\mathbb{T}^n)$. The map $C$ is $G$-invariant, that is
\[\forall g\in G,\;\; C(g\theta_0,\cdots,g\theta_{2n})=C(\theta_0,\cdots,\theta_{2n}). \]
\end{prop}

\begin{proof}
The group $\Diff^+(\mathbb{S}^1)$ preserves the cyclic order on $\mathbb{S}^1$ then, for $\gamma\in \Diff^+(\mathbb{S}^1)$,
\[e(\gamma\theta_0,\gamma\theta_1,\gamma\theta_2)=e(\theta_0,\theta_1,\theta_2).\]
Taking $g=(\gamma^1,\cdots,\gamma^n)\in G$, one has
\begin{eqnarray*}
C(g\theta_0,\cdots,g\theta_{2n}) & = & \frac{1}{(2n+1)!} \sum_{\sigma\in\mathfrak{S}_{2n+1}}\sign (\sigma) \prod_{i=1}^n e(\gamma^i\theta^i_{\sigma(2i-2)},\gamma^i\theta^i_{\sigma(2i-1)},\gamma^i\theta^i_{\sigma(2i)})\\
 & = & \frac{1}{(2n+1)!} \sum_{\sigma\in\mathfrak{S}_{2n+1}}\sign (\sigma) \prod_{i=1}^ne(\theta^i_{\sigma(2i-2)},\theta^i_{\sigma(2i-1)},\theta^i_{\sigma(2i)})\\
 & = & C(\theta_0,\cdots,\theta_{2n}).
\end{eqnarray*}
\end{proof}

\begin{prop}\label{closedcocycle}
The map $C$ is closed as a combinatorial cochain, that is
\[\sum_{i=0}^{2n+1}(-1)^i C(\theta_0,\cdots,\hat{\theta_i},\cdots,\theta_{2n+1})=0. \]
\end{prop}

\begin{proof}
We refer to \cite{bucherpolygons} p.331.
\end{proof}

\subsection{An invariance relation satisfied by \texorpdfstring{$\Omega$}{w}}\label{invarianceform}

Let us show now that the form $\Omega$ is invariant under the action of the group $G=\left(\Diff^+(\mathbb{S}^1)\right)^n$. This group contains $\left(\PSL_2(\R)\right)^n$ and extends its action on $\Le^2(\mathbb{T}^n)$ by the change of variables formula,
\[(gf)(\theta)=\sqrt{\Jac g^{-1}(\theta)}f\circ g^{-1}(\theta). \]
The action being unitary, we let act $G$ on the unit sphere $\mathcal{S}^\infty$ by restriction and on the tangent space of this sphere. Then we have, for $g\in G$,
\begin{eqnarray*}
(g^*\Omega)_{\varphi}(f_1,\cdots,f_{2n}) &  = &  \Omega_{g\varphi}(gf_1,\cdots,gf_{2n})\\
& =  & \int_{\left(\mathbb{T}^n\right)^{2n+1}} C(\theta_0,\cdots,\theta_{2n})(g\varphi)^2(\theta_0)\prod_{i=1}^{2n} (g\varphi) (gf_i)(\theta_i)d\theta_{i}
\\
& =  & \int_{\left(\mathbb{T}^n\right)^{2n+1}} C(\theta_0,\cdots,\theta_{2n})\Jac g^{-1}(\theta_0)\varphi^2(\theta_0) \prod_{i=1}^{2n}\Jac g^{-1}(\theta_i)\varphi f_i(\theta_i)d\theta_{i}.
\end{eqnarray*}
We now perform the change of variables formula $\theta_i'=g^{-1}(\theta_i)$. We get
\[(g^*\Omega)_{\varphi}(f_1,\cdots,f_{2n}) = \int_{\left(\mathbb{T}^n\right)^{2n+1}} C(g\theta_0,\cdots,g\theta_{2n})\varphi^2(\theta_0) \varphi f(\theta_1)\cdots \varphi f (\theta_{2n}) d\theta_0\cdots d\theta_{2n}.\]
The conclusion now follows from proposition \ref{invariancecocycle}.

\subsection{Closure of \texorpdfstring{$\Omega$}{w}}\label{closure}

In order to use the method of calibration, we have to deal with a closed form. This is the aim of this paragraph.

\begin{prop}
The differential $2n$-form $\Omega$ is closed.
\end{prop}

\begin{proof}

To differentiate $\Omega$, it is easier to have an expression on a space of measures instead of the unit $\Le^2$-sphere. Let us begin by a quick review on the structure of the space of measures we will deal with.\\

Let $\ronm$ be the Banach vector space dual to $\mathcal{C}^0(\mathbb{T}^n)$, the Banach space of continuous functions on $\mathbb{T}^n$. The space $\ronm$ is also the space of Radon measures on $\mathbb{T}^n$. We consider the affine space
\[\ronm_1=\set{\mu\in\ronm |  \mu(\mathbb{T}^n)=1}.\]
We think as this affine space as a infinite dimensional manifold shaped on a Banach space. The tangent space in each point is the vector space of zero-mass measures.\\

The differential form $\Omega$ comes from a form on $\ronm_1$ pulled-back by the smooth map
\[\fonction{\mathcal{A}}{\mathcal{S}^\infty}{\ronm_1}{\varphi}{B\mapsto\int_B\varphi^2(\theta)d\theta}. \]
The image of a function $\varphi$ is the measure with density $\varphi^2$ with respect to the Lebesgue measure.
Let us define a differential  $2n$-form $\Omega'$ on $\ronm_1$,
\[
\Omega'_{\mu}(\alpha_1,\cdots,\alpha_{2n})  = \int_{\left(\mathbb{T}^n\right)^{2n+1}}C(\theta_0,\cdots,\theta_{2n})d\mu(\theta_0)d\alpha_1(\theta_1)\cdots d\alpha_{2n}(\theta_{2n}).
\]
Clearly
\[\Omega=2^{2n}\mathcal{A}^*\Omega'.\]
Then it is enough to show that $\Omega'$ is closed. But now, $\Omega'$ is a linear map with respect to $\mu$. Then, we have (see \cite{lang} p.84 for the differential formula of a form on a Banach manifold)
\[d\Omega_{\mu}'(\alpha_0,\cdots,\alpha_{2n})=\sum_{i=0}^{2n}(-1)^i\partial_{\mu}\Omega_{\mu}(\alpha_0,\cdots,\hat{\alpha_i},\cdots,\alpha_{2n})\cdot\alpha_i .\]
By linearity this expression is also
\[d\Omega_{\mu}'(\alpha_0,\cdots,\alpha_{2n})=\sum_{i=0}^{2n}(-1)^i\Omega_{\alpha_i}(\alpha_0,\cdots,\hat{\alpha_i},\cdots,\alpha_{2n}).\]
With skew-symmetry, this is again
\[d\Omega_{\mu}'(\alpha_0,\cdots,\alpha_{2n}) = (2n+1)\int_{\left(\mathbb{T}^n\right)^{2n+1}} C(\theta_0,\cdots,\theta_{2n})d\alpha_0(\theta_0)\cdots d\alpha_{2n}(\theta_{2n}).\]
Each measure has a vanishing total mass. Let us show that the above expression is in fact
\[d\Omega_{\mu}'(\alpha_0,\cdots,\alpha_{2n}) = -(2n+1)\int_{\left(\mathbb{T}^n\right)^{2n+2}}\pa C(\theta_0,\cdots,\theta_{2n+1})d\alpha_0(\theta_0)\cdots d\alpha_{2n}(\theta_{2n})d\mu(\theta_{2n+1}).\]
\label{vanishing}
Let us recall the definition of the combinatorial boundary of a cochain,
\[\pa C(\theta_0,\cdots,\theta_{2n+1})=\sum_{i=0}^{2n+1} (-1)^i C(\theta_0,\cdots,\hat{\theta_i},\cdots,\theta_{2n+1}). \]
First we have,
\[\int_{\left(\mathbb{T}^n\right)^{2n+1}}C(\theta_0,\cdots,\theta_{2n})d\alpha_0(\theta_0)\cdots d\alpha_{2n}(\theta_{2n})\]
\[= \int_{\left(\mathbb{T}^n\right)^{2n+2}} C(\theta_0,\cdots,\theta_{2n})d\alpha_0(\theta_0)\cdots d\alpha_{2n}(\theta_{2n})d\mu(\theta_{2n+1})\]
because $\mu(\mathbb{T}^n)=1$.
We recognize the opposite of the last term in the expression of $\pa C$ (corresponding to the index $i=2n+1$).
We now prove that the other terms in the expression of $\pa C$ vanish. Each of this remaining terms are in the form of
\[\int_{\left(\mathbb{T}^n\right)^{2n+2}}C(\theta_0,\cdots,\hat{\theta_i},\cdots,\theta_{2n+1})d\alpha_0(\theta_0)\cdots d\alpha_{2n}(\theta_{2n})d\mu(\theta_{2n+1})\]
with $i\neq 2n+1$. Integrating with respect to the variable $\theta_i$ which is not involved in the cocycle, make appear a multiplicative term, the total mass of one of a tangent measure, supposed to be zero.\\

So finally the closure of $C$ as a combinatorial cocycle (see proposition \ref{closedcocycle}) proves the result.
\end{proof}

\subsection{The calibrating inequality}\label{inequality}

Let us first remark that we only have to establish the inequality at the basepoint $\varphi=\mathds{1}$. Indeed one can show exactly as in paragraph \ref{closure} that
\[\int_{\left(\mathbb{T}^n\right)^{2n+1}} C(g\theta_0,\cdots,g\theta_{2n})\left(\varphi^2(\theta_0)-1\right) \varphi f(\theta_1)\cdots \varphi f (\theta_{2n}) d\theta_0\cdots d\theta_{2n}=0.\]
Replacing the functions $f_i\in T_{\varphi}\mathcal{S}^\infty$ by the functions $\varphi f_i\in T_{\mathds{1}}\mathcal{S}^\infty$, we can make the assumption $\varphi=\mathds{1}$.\\

Besson, Courtois and Gallot studied the $2$-dimensional case in \cite{bcg2} chapter 6. They used the Euler class to build a differential form
\[\omega_{\varphi}(f_1,f_2)= \int_{\left(\mathbb{S}^1\right)^3}e(\theta_0,\theta_1,\theta_2)\varphi^2(\theta_0) \varphi f_1(\theta_1)\varphi f_2(\theta_2)d\theta_0 d\theta_1 d\theta_2.\]
Here is what they proved

\begin{prop}[\cite{bcg2} chapter 6]\label{bcgcalibration}
\begin{enumerate}
\item One may use the following alternative expression for $\omega$,
\[\omega_{\varphi}(f_1,f_2)=2\int_{\mathbb{S}^1}F_1dF_2,\]
where $F_1$ and $F_2$ are the primitives with vanishing integrals for $\varphi f_1$ and $\varphi f_2$.\
\item The comass of $\omega$ equals $\frac{1}{\pi}$.
\item Moreover $\omega$ is a calibrating form for the Poisson kernel, that is, at the basepoint $\mathds{1}$, $\omega$ is maximal over orthonormal families $(f_1,f_2)$ if
\[f_1(\theta)=\sqrt{2}\cos \theta \;\;\mbox{ and }\;\; f_2(\theta)=\sqrt{2}\sin \theta.\]
\end{enumerate}
\end{prop}

Back to the $2n$-dimensional situation, we first show that the differential form $\Omega$ is nonzero, evaluating it on $T_{\mathds{1}}\Phi_0(\left(\mathbb{H}^2\right)^n)$. Remember the family of functions given by
\[\fonction{f_i}{\mathbb{T}^n}{\R}{\theta=(\theta^1,\cdots,\theta^n)}{\sqrt{2}\cos\theta^i}\]
and
\[\fonction{f_{i+1}}{\mathbb{T}^n}{\R}{\theta=(\theta^1,\cdots,\theta^n)}{\sqrt{2}\sin\theta^i},\]
for $i=1,\cdots,n$, is an orthonormal basis of $T_{\mathds{1}}\Phi_0(\left(\mathbb{H}^2\right)^n)$.

\begin{lem}\label{comass}
We have the relation
\[\Omega_{\mathds{1}}(f_1,f_2,\cdots,f_{2n})=\frac{2^n}{\pi^n(2n)!}.\]
\end{lem}

\begin{proof}
Let us look for permutations $\sigma\in\mathfrak{S}_{2n+1}$ for which the corresponding term
\[2^n \sign(\sigma)\int_{\left(\mathbb{T}^n\right)^{2n+1}}\prod_{i=1}^n\left(e(\theta^i_{\sigma(2i-2)},\theta^i_{\sigma(2i-1)},\theta^i_{\sigma(2i)}) \cos \theta_{2i-1}^i \sin\theta_{2i-2}^i\right) d\theta_0\cdots d\theta_{2n}\]
is nonzero. In order to get such a permutation, the variables $\theta_{2i-1}^i$ and $\theta_{2i}^i$ which are involved in the expression
\[  \cos \theta_1^1 \sin\theta_2^1 \cdots \cos \theta_{2n-1}^n \sin\theta_{2n}^n\]
have to appear also in the expression
\[\prod_{i=1}^ne(\theta^i_{\sigma(2i-2)},\theta^i_{\sigma(2i-1)},\theta^i_{\sigma(2i)}).\]
If not we integrate with respect to one of the variables $\theta_{2i-1}^i$ or $\theta_{2i}^i$ missing in the term of the cocycle and we get a multiplicative vanishing term which is the integral of one of the functions $f_i$.
Hence, we need that, for each $i$,
\[ \set{\theta_{2i-1},\theta_{2i}}\subset \set{\theta_{\sigma(2i-2)},\theta_{\sigma(2i-1)},\theta_{\sigma(2i)}}. \]
Let us now compute the number of suitable permutations. The first image $\sigma(0)$ could be any of the $2n+1$ elements. Let us say $\sigma(0)$ falls in the $i$\textsuperscript{th} Euler class, $\sigma(0)\in \set{2i-2,2i-1,2i}$ (e.g $\sigma(0)=2i-2$, other cases are similar). From the condition above, it remains only two possibilities for $\sigma(2i-1)$ and $\sigma(2i)$, namely,
\[\sigma(2i-1)=\sigma(2i-1) \;\;\mbox{ and }\;\; \sigma(2i)=\sigma(2i)\]
or
\[\sigma(2i-1)=\sigma(2i) \;\;\mbox{ and }\;\; \sigma(2i)=\sigma(2i-1).\]
Once we fill the $i$\textsuperscript{th} Euler class, the adjacent one(s) already have one element imposed (remember $\theta_{2i}$ appears in both the $i$\textsuperscript{th} Euler class and the $(i+1)$\textsuperscript{th} Euler class). Hence we have again two possibilities for the two remaining arguments and inductively for each Euler class.

Finally we find
\[2^n(2n+1)\]
suitable permutations.\\

We now play with the skew-symmetry relation to show that, for each suitable permutation $\sigma$, the corresponding term
\[2^n \sign(\sigma)\int_{\left(\mathbb{T}^n\right)^{2n+1}}\prod_{i=1}^n\left(e(\theta^i_{\sigma(2i-2)},\theta^i_{\sigma(2i-1)},\theta^i_{\sigma(2i)}) \cos \theta_{2i-1}^i \sin\theta_{2i-2}^i\right) d\theta_0\cdots d\theta_{2n}\]
equals $\frac{1}{\pi^n}$. From now on, we assume that $\sigma$ is some suitable permutation characterized above.

The variable $\theta_0$ appears in one or two Euler classes, let us say for example that
\[\sigma(0)=2i\]
for some $i<n$ (other cases are simpler). Then the variable $\theta_0$ appears in both the $i$\textsuperscript{th} and the $(i+1)$\textsuperscript{th} Euler class. We compute the term
\[T_{\sigma}:= \sign(\sigma)\int_{\left(\mathbb{T}^n\right)^{2n+1}}\prod_{i=1}^n\left(e(\theta^i_{\sigma(2i-2)},\theta^i_{\sigma(2i-1)},\theta^i_{\sigma(2i)}) \cos \theta_{2i-1}^i \sin\theta_{2i-2}^i\right) d\theta_0\cdots d\theta_{2n}\]
by first performing the change of variables
\[\theta_{\sigma(2i-2)}=\theta'_1,\;\; \theta_{\sigma(2i-1)}=\theta'_2,\;\; \theta_{\sigma(2i+1)}=\theta'_3, \;\; \theta_{\sigma(2i+2)}=\theta'_4 \;\; \mbox{ and } \;\; \theta_k=\theta'_k,\]
for $k\neq 2i-2, 2i-1, 2i+1, 2i+2$.
Since the Jacobian of this change of variable is $1$, we do not change the value of $T_{\sigma}$. Remember we took $\sigma$ such that
\[\set{\sigma(2i-2),\sigma(2i-1)}=\set{2i-2,2i-1}\; \mbox{ and } \set{\sigma(2i+1),\sigma(2i+2)}=\set{2i+1,2i+2}.\]
So we can assume that $\theta_0$ appears in the first two Euler classes.

Now the support disjoint cycles of the permutation $\sigma$ are all contained in sets of the form
\[\set{2i-2,2i-1,2i}.\]
Hence, when one reorders the variables in each Euler class such that
$\theta_{\sigma(2i-1)}$ sits in the $(2i-1)$\textsuperscript{th} place and $\theta_{\sigma(2i)}$ sits in the $(2i)$\textsuperscript{th} place, the sign by which we changed $T_{\sigma}$ is precisely $\sign(\sigma)$.

Therefore, because the first variables in each Euler class don't play any role, each term $T_{\sigma}$ equals $T_{id}$. The computation of $T_{id}$ is easy because the situation appears now as a product and one can use Lemma \ref{bcgcalibration}. Indeed,
\begin{eqnarray*}
T_{id} & = & 2^n \prod_{i=1}^n\int_{\left(\mathbb{S}^1\right)^3}e(\theta^i_{2i-2},\theta^i_{2i-1},\theta_{2i}^i) \cos \theta_{2i-1}^i \sin_{2i}^id\theta_{2i-2}^id\theta_{2i-1}^id\theta_{2i}^i\\
 & = & \prod_{i=1}^n \comass(\omega)\\
 & = & \frac{1}{\pi^n}
\end{eqnarray*}
\end{proof}

We now proceed to the calibrating inequality. We introduce the $(2n-1)$-cocycle,
\[D(\theta_1,\cdots,\theta_{2n})=\int_{\mathbb{T}^n}C(\theta_0,\cdots,\theta_{2n})d\theta_0.\]
The reasoning involves the Fourier coefficient of $D$. A function $f\in T_{\mathds{1}}\mathcal{S}^\infty$ is written as
\[f(\theta^1,\cdots,\theta^n)=\sum_{k=(k^1,\cdots,k^n)\in I}a_k\cos (k^1\theta^1+\cdots+k^n\theta^n) + b_k\sin (k^1\theta^1+\cdots+k^n\theta^n),\]
where $\theta=(\theta^1,\cdots,\theta^n)\in\mathbb{T}^n$ and $k\in I$, the subset of $\mathbb{Z}^n$ given by
\[I=\N\backslash\set{0}\times \Z^{n-1} \cup \set{0}\times \N\backslash \set{0}\times\Z^{n-2}\cup \cdots \cup \set{0,\cdots,0}\times\N\backslash \set{0}.\]
The convergence of the sum has to be understood for the $\Le^2$ topology.

We denote by $\gamma : \mathbb{S}^1 \rightarrow \R$ a function which can be either $\sqrt{2}\cos$ or $\sqrt{2}\sin$. For some $K=(k^1,\cdots,k^n)\in I$ we denote by $\gamma_K : \mathbb{T}^n\rightarrow \R$ a function which can be either
\[\theta=(\theta^1,\cdots,\theta^n)\longmapsto \sqrt{2}\cos(k^1\theta^1+\cdots +k^n\theta^n)\]
or
\[\theta=(\theta^1,\cdots,\theta^n)\longmapsto \sqrt{2}\sin(k^1\theta^1+\cdots +k^n\theta^n)\]

\begin{defi}[Fourier transform and Fourier coefficients]
Let $F$ be a map from $\left(\mathbb{T}^n\right)^{2n}$ to $\R$. The Fourier transform of $F$ is the map
\[\fonction{\hat{F}}{I^{2n}}{\R^{2^{2n}}}{(K_1,\cdots,K_{2n})}{\int_{\left(\mathbb{T}^n\right)^2n} D(\theta_1,\cdots,\theta_{2n}\gamma_{K_1}(\theta_1)\cdots \gamma_{K_{2n}}(\theta_{2n})d\theta_1\cdots d\theta_{2n}}.\]
We called $\hat{F}(K_1,\cdots,K_{2n})$ a Fourier coefficient. The $2^{2n}$ coordinates of a Fourier coefficient correspond to the possible choices of functions $\gamma$.
\end{defi}

\begin{rmq}
Fourier coefficients of the cocycle $D$ contain values of $\Omega$. In particular
\[\hat{D}\begin{pmatrix}
1 & 1 & 0 & 0 & \cdots & 0\\
0 & 0 & 1 & 1 & \cdots & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & 0 & &\cdots & 1 & 1
\end{pmatrix}\]
contains the value of $\Omega$ on the orthonormal tangent of the image of the Poisson kernel. That is why we investigate the combinatorial properties of $\hat{D}$
\end{rmq}

Here is a characterization of the nonzero Fourier coefficients of $D$

\begin{lem}\label{nonzerocoeff}
\begin{enumerate}
\item The Fourier coefficient $\hat{D}(K_1,\cdots,K_{2n})$ is nonzero if and only if there exist some nonzero integers $k^1,\cdots,k^n$ such that
\[(K_1,\cdots,K_{2n})=\begin{pmatrix}
k^1 & k^1 & 0 & 0 & \cdots & 0\\
0 & 0 & k^2 & k^2 & \cdots & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & 0 & &\cdots & k^n & k^n
\end{pmatrix}\]
up to permutation of lines and rows. Moreover if $(K_1,\cdots,K_{2n})$ is as above, the nonzero coordinates of $\hat{D}(K_1,\cdots,K_{2n})$ (among the $2^{2n}$ ones) is constructed by taking
\[\gamma_{K_{2i-1}}(\theta)=\sqrt{2}\cos(k^i\theta^1)\; \mbox{and} \; \gamma_{K_{2i}}=\sqrt{2}\sin(k^i\theta^n)\]
for all $i$ and up to permutations of the functions.\
\item A nonzero coordinate of some Fourier coefficient, that is,
\[2^{n}\int_{\left(\mathbb{T}^n \right)^{2n+1}}C(\theta_0,\cdots,\theta_{2n})\left(\prod_{i=1}^n\cos(k^i\theta_{2i-1}^i) \sin(k^i\theta_{2i}^i)\right) d\theta_0\cdots d\theta_{2n} \]
equals
\[\frac{2^n}{(2n)!\pi^n\prod_{i=1}^nk^i}.\]
In particular, any nonzero value is smaller than the value of $\Omega$ on the tangent of the image of the Poisson kernel.
\end{enumerate}
\end{lem}

\begin{proof}
Take some functions $\gamma_{K_1},\cdots,\gamma_{K_{2n}}$. One may use the addition formulas for the trigonometric functions $\gamma_{K_i}$. Hence $\hat{D}(K_1,\cdots,K_{2n})$ is a sum of terms of the form
\[\int_{\left(\mathbb{S}^1 \right)^{2n^2}} \prod_{i=1}^ne(\theta^i_{\sigma(2i-2)},\theta^i_{\sigma(2i-1)},\theta^i_{\sigma(2i)}) \prod_{j=1}^{2n}\prod_{l=1}^n \gamma(k_j^l\theta_j^l)d\theta_j^l \]
(if some integer $k_j^l=0$, we set $\gamma(k_j^l\theta_j^l)=1$).
We show that if $(K_1,\cdots,K_{2n})$ is not as in the statement of the Lemma, every such term vanishes.
In order to do that, we will use several times the arguments below.
\begin{enumerate}
\item {\bf Argument 1} : If there is some variable $\theta_j^l\in\mathbb{S}^1$ which is involved in some function $\gamma$ and which is not involved in the term of the cocycle, that is
\[\theta_j^l\notin \bigcup_{i=1}^n\set{\theta^i_{\sigma(2i-2)},\theta^i_{\sigma(2i-1)},\theta^i_{\sigma(2i)}}, \]
then we integrate with respect to this variable $\theta_j^l$ and we get zero as a multiplicative term because the integral of $\theta_j^l\mapsto \gamma(k_j^l\theta_j^l)$ vanishes. \
\item {\bf Argument 2} : One can also show exactly as in paragraph \ref{closure} that
\[\int_{\left(\mathbb{S}^1\right)^3}e(\theta_0,\theta_1,\theta_2)\gamma(k_0\theta_0)\gamma(k_1\theta_1) \gamma(k_2\theta_2)d\theta_0 d\theta_1 d\theta_2=0.\]
Indeed the Euler class $e$ is combinatorially closed.\
\item {\bf Argument 3} : An expression of the form
\[\int_{\left(\mathbb{S}^1 \right)^3}e(\theta_0,\theta_1,\theta_2)\gamma(\theta_0)d\theta_0 d\theta_1 d\theta_2 \]
must vanish. Indeed for a fixed $\theta_0$,
\[\int_{\left(\mathbb{S}^1 \right)^2}e(\theta_0,\theta_1,\theta_2)d\theta_1 d\theta_2=0. \]
\end{enumerate}

In each term of the cocycle there is $3n$ variables involved. Then we can't have more than $3n$ nonzero integers (otherwise we use argument 1 above). We can't have also more than 3 nonzero integers in each row. Indeed,
\begin{itemize}
\item If there is 4 or more, there is at least one variable involved in the functions and not in the term of the cocycle. We use argument 1.
\item If there is 3, there is again 2 possibilities. Either the variables involved in the functions and in the cocycle are not the same and we use argument 1; either they are the same and after we integrate with respect to this 3 variable, we use argument 2.
\end{itemize}

Moreover we can't have more than 2 nonzero integers in each line. Indeed there is at most one repetition of the variables in the different Euler classes. In fact there is exactly two nonzero integers in each line. If there is only one, this means that there is only one of three variables in some Euler class matching with a variable in the functions. We integrate with respect to this three variables and we use argument 3.

Hence there is exactly $2n$ nonzero integers in the $2n^2$ coordinates of a nonzero index $(K_1,\cdots,K_{2n})\in I^{2n}$ and exactly two nonzero integers in each line. So there is exactly one nonzero integer in each row.\\

So far we can assume that
\[(K_1,\cdots,K_{2n})=\begin{pmatrix}
k^1 & k^2 & 0 & 0 & \cdots & 0\\
0 & 0 & k^3 & k^4 & \cdots & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & 0 & &\cdots & k^{2n-1} & k^{2n}
\end{pmatrix}\]
for $2n$ nonzero integers $k^1,k^2,\cdots,k^{2n}$. Indeed, up to sign, the values of $\hat{D}(K_1,\cdots,K_{2n})$ are left unchanged when we permute the $K_i$'s. One can also reorder the lines by performing some change of variables of the form
\[\theta_j^{'\sigma(l)}=\theta_j^l\]
which only changes the sign. From now on the situation appears completely as a product. One can check that the permutations $\sigma\in\mathfrak{S}_{2n+1}$ giving a nonzero contribution are exactly the ones we characterized in Lemma \ref{comass}. Each term gives the same value. A generic coordinate of $\hat{D}(K_1,\cdots,K_{2n})$ can be written as
\[\frac{2^n}{(2n)!}\prod_{i=1}^n \int_{\left(\mathbb{S}^1\right)^3} e(\theta^i_{2i-2},\theta^i_{2i-1},\theta^i_{2i})\gamma(k^{2i-1}\theta^i_{2i-1})  \gamma(k^{2i}\theta^i_{2i})d\theta^i_{2i-2} d\theta^i_{2i-1} d\theta^i_{2i}.\]
We can now conclude with the assertion 1 of Lemma \ref{bcgcalibration}. Indeed we must choose the integers $k^1,k^2,\cdots,k^{2n}$ and the functions $\gamma_{K_1},\gamma_{K_2},\cdots,\gamma_{K_{2n}}$ such that, for all index $i$, $\gamma_{K_{2i}}$ and the primitive of $\gamma_{K_{2i-1}}$ are not $\Le^2$-orthogonal.\\

We can also perform the exact computation of the nonzero values thanks to the same assertion 1 of Lemma \ref{bcgcalibration}. Indeed for each index $i$,
\[
2\int_{\left(\mathbb{S}^1\right)^3} e(\theta^i_{2i-2},\theta^i_{2i-1},\theta^i_{2i})\cos(k^i\theta^i_{2i-1})  \sin(k^i\theta^i_{2i})d\theta^i_{2i-2} d\theta^i_{2i-1} d\theta^i_{2i} \]
\[=  4\int_{\mathbb{S}^1}\frac{1}{k^i}\sin^2(k^i\theta^i)d\theta^i
 = \frac{1}{k^i\pi} \]
\end{proof}

We finally deduce the calibrating inequality. Take $2n$ functions $f_1,\cdots,f_{2n}$ with integrals zero, the family $(f_1,\cdots,f_{2n})$ being orthonormal. For each function $f_i$, we write
\[f_i(\theta)=\sum_{K=(k^1,\cdots,k^n)\in I}a_{K,i}\cos(k^1\theta^1+\cdots + k^n\theta^n) + b_{K,i}\sin(k^1\theta^1+\cdots + k^n\theta^n) .\]
and we denote by $c_{K,i}$ some Fourier coefficient of $f_i$ which can be either $a_{K,i}$ or $b_{K,i}$. Beside, let us set the following convention in order to design the coordinates of some Fourier coefficient. We take $C=(c_{K_1,1},\cdots,c_{K_{2n},2n})\in\prod_{i=1}^{2n}\set{a_{K_i,i},b_{K_i,i}}$ and denote by $\hat{D}(K_1,\cdots,K_{2n})^C$ the coordinate of $\hat{D}(K_1,\cdots,K_{2n})$ corresponding to
\[
   \left \{
   \begin{array}{r c l}
      \gamma_{K_i}(\theta)  & =  & \cos(k^1_i\theta^1+\cdots +k^n_i\theta^n) \mbox{ if } c_{K_i,i} = a_{K_i,i}  \\
      \gamma_{K_i}(\theta)  & =  & \sin(k^1_i\theta^1+\cdots +k^n_i\theta^n) \mbox{ if } c_{K_i,i} = b_{K_i,i} \\
   \end{array}
   \right .
\]
Hence
\[\Omega_{\mathds{1}}(f_1,\cdots,f_{2n})=\sum_{\substack{K_1,\cdots,K_{2n}\\ C\in\prod_{i=1}^{2n}\set{a_{K_i,i},b_{K_i,i}}}}\hat{D}(K_1,\cdots,K_{2n})^C c_{K_1,1}\cdots c_{K_{2n},2n}. \]
We already isolated the vanishing Fourier coefficients. Let $B=(e_1,\cdots,e_n)$ be a basis of $\R^n$. For $\rho\in\mathfrak{S}_n$, we denote by $Q(\rho)$ the $n\times n$ matrix in $B$ of the linear map sending $e_i$ to $e_{\rho (i)}$. We have
\[\Omega_{\mathds{1}}(f_1,\cdots,f_{2n})=\sum\hat{D}\left(Q(\rho)\cdot\begin{pmatrix}
k^1 & k^1 & 0 & 0 & \cdots & 0\\
0 & 0 & k^2 & k^2 & \cdots & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & 0 & &\cdots & k^n & k^n
\end{pmatrix}\right)\prod_{i=1}^nc_{k^{\rho(i)},2i-1}c'_{k^{\rho(i)},2i}  \]
The sum is taken over all possible choices of nonzero integers $k^1,\cdots,k^n$, all permutations $\rho\in \mathfrak{S}_n$ and over all possibilities of $c_{k^{\rho(i)},2i-1}$ and $c'_{k^{\rho(i)},2i}$ such that, for all $i$,
\[ c_{k^{\rho(i)},2i-1}=a_{k^{\rho(i)},2i-1}\;\mbox{ and }\; c'_{k^{\rho(i)},2i}=b_{k^{\rho(i)},2i}\]
or
\[ c_{k^{\rho(i)},2i-1}=b_{k^{\rho(i)},2i-1}\;\mbox{ and }\; c'_{k^{\rho(i)},2i}=a_{k^{\rho(i)},2i}.\]
We now have
\begin{eqnarray*}
\abs{\Omega_{\varphi}(f_1,\cdots,f_{2n})} & \leqslant & \sum\frac{2^n}{(2n)!\pi^n\prod_{i=1}^n k^i}\prod_{i=1}^n\abs{c_{k^{\rho(i)},2i-1}c'_{k^{\rho(i)},2i}}\\
 & \leqslant & \sum \frac{2^n}{(2n)!\pi^n}\prod_{i=1}^n\abs{c_{k^{\rho(i)},2i-1}c'_{k^{\rho(i)},2i}}.
\end{eqnarray*}
It is a standard fact that, using repetitively the Cauchy-Schwartz inequality, one gets
\[\sum_{\alpha\in A}\prod_{\beta=1}^N x_{\alpha,\beta}\leqslant \prod_{\beta=1}^N \sqrt{\sum_{\alpha\in A}x_{\alpha,\beta}^2}, \]
for some countable set $A$, some integer $N$ and some non-negative numbers $x_{\alpha,\beta}$.
Here we obtain
\begin{eqnarray*}
\abs{\Omega_{\mathds{1}}(f_1,\cdots,f_{2n})} & \leqslant & \frac{2^n}{(2n)!\pi^n}\prod_{i=1}^n\sqrt{\sum\abs{c_{k^{\rho(i)},2i-1}}} \prod_{i=1}^n\sqrt{\sum\abs{c'_{k^{\rho(i)},2i}}} \\
 & \leqslant & \frac{2^n}{(2n)!\pi^n} \prod_{i=1}^{2n}\norm{f_i}_{\Le^2}\\
 & = & \frac{2^n}{(2n)!\pi^n}.\\
\end{eqnarray*}
This inequality is an equality when $(f_1,\cdots,f_{2n})$ generates the tangent to the image of the Poisson kernel. The proof of the main theorem is then complete.

\section{Applications}\label{applications}

We finally look for some consequences, suggested by the last chapter of \cite{bcg2}.

It's possible to extend the main result to the case where $g$ lives on another differentiable manifold related to $M$ by a map of non-zero degree. The result we obtain is an optimal "degree theorem" as in the article \cite{connellfarbdegree} who misses the case we investigate. There is also a similar result in \cite{lohsauer} but with a nonoptimal constant and with an additional hypothesis on $\Ric g$.

We denote again $M=\Gamma\backslash \left(\mathbb{H}^2\right)^n$ a compact quotient of $\left(\mathbb{H}^2\right)^n$ and $g_0$ the sum of hyperbolic metrics in the different factors $\mathbb{H}^2$.

\begin{cor}
Let $Y$ be a differentiable manifold of dimension $2n$ endowed with a Riemannian metric $g$ and let $f$ be a continuous map
\[f : (Y,g) \longrightarrow (M,g_0)\]
Then
\[h(g)^{2n} \Vol(Y,g) \geqslant \abs{\textit{deg }f} h(g_0)^{2n} \Vol(M,g_0)\]
\end{cor}

\begin{proof}
Observe that the inequality is trivial if deg$(f)=0$. So let us assume that deg$(f)$ is non-zero.
First, one can regularize the map $f$ in a homotopic map, still denoted $f$, which is $\mathcal{C}^1$. We call $\tilde{f}$ the map induced by $f$ from $\tilde{Y}$ to $\tilde{M}$. Let us introduce the invariant appropriated to this new situation
\[\SphereVol(f)=\inf{\Vol((U,\Phi^*(\mbox{can})))}\]
where $\Phi$ are Lipschitz continuous equivariant immersions from $\tilde{Y}$ to $\Le^2(\mathbb{T}^n)$. As before, one example is given by the product of Poisson kernels,
\[\Phi_0(y,\theta)=\prod_{i=1}^n\sqrt{p_0(\tilde{f}(y)^i,\theta^i)},\]
where the $\tilde{f}(y)^i$'s are the coordinates in the factors $\mathbb{H}^2$ and the $\theta^i$'s are the coordinate in $\mathbb{T}^n$.
We also consider
\[\Phi_c(y,\theta)=\left(\frac{\int_{\widetilde{Y}}e^{-cd(y,z)}\Phi_0^2(z,\theta)dv_g(z)}{\int_{\mathbb{T}^n}e^{-cd(y,z)}\Phi_0^2(z,\theta)d\theta}\right)^{1/2}.\]
where $d$ is the $g$-distance in $\widetilde{Y}$.
The two arguments above (page \pageref{boundedabove} for the first and section \ref{proof} for the second) give in this context
\begin{enumerate}
\item $\SphereVol\leqslant \left(\frac{h(g)^2}{8n}\right)^n\Vol(Y,g)$ using the computation of $\Vol(\Phi_c)$\
\item The image of $\Phi_0$ is still calibrated because $f$ is surjective and then $\SphereVol=\Vol(\Phi_0)$. Moreover
\[\Vol(\Phi_0)=\abs{\mbox{deg }f}\left(\frac{h(g_0)^2}{8n}\right)^n\Vol(X,g_0)\]
\end{enumerate}
\end{proof}

We also obtain an estimate for the minimal volume. Let $X$ be a compact manifold. The minimal volume is defined as
\[\MinVol=\inf \set{\Vol(g),\,\abs{K(g)}\leqslant 1}\]
(see \cite{gromovvabc}).

\begin{cor}
Let $M=\Gamma\backslash\left( \mathbb{H}^2\right)^n$ be a compact quotient of $\left( \mathbb{H}^2\right)^n$. Then
\[\MinVol(M)\geqslant\left(\frac{\sqrt{n}}{2n-1}\right)^{2n}\Vol(g_0).\]
\end{cor}

In particular, we reprove in a qualitative way a general theorem of \cite{lafontschmidt} stating that $\MinVol$ is nonzero.

\begin{proof}
We still follow chapter 9 of \cite{bcg2}. Take a metric $g$ on $M$ with $\abs{K(g)}\leqslant 1$. We deduce an equality on the Ricci curvature
\[\Ric (g)\geqslant-(2n-1)g\]
We then apply the Bishop's inequality (\cite{ghl} p.144) comparing volumes of balls for $g$ and volumes of balls in the hyperbolic $2n$-space $\mathbb{H}^{2n}$. Taking $\log$ and making the radius go to infinity, we have
\[h(g)\leqslant 2n-1.\]
We conclude introducing this inequality in
\[\Vol(g)\geqslant\left( \frac{\sqrt{n}}{h(g)}\right)^{2n}\Vol(g_0).\]
\end{proof}

Here is a last consequence in dimension 4. P. Suarez-Serrato in \cite{serrato} classified the 4-dimensional Thurston geometries admitting a metric of minimal normalized volume entropy. The only missing case were the case of quotients of $\mathbb{H}^2\times\mathbb{H}^2$. In dimension 4, there exist 19 geometries admitting a compact quotient (see the list and references in \cite{serrato} p. 366). P. Suarez-Serrato was able to decide among the 18 geometries (all but $\mathbb{H}^2\times\mathbb{H}^2$) which ones admit a metric of volume 1 with minimal volume entropy (theorem A). Hence we obtain

\begin{cor}
The 4-dimensional Thurston geometries admitting a metric with minimal normalized volume entropy are only those of "hyperbolic type"
\[\mathbb{H}_{\mathbb{R}}^4,\mathbb{H}_{\mathbb{C}}^2 \mbox{ and } \mathbb{H}_{\mathbb{R}}^2\times\mathbb{H}_{\mathbb{R}}^2. \]
\end{cor}

\bibliographystyle{alpha}
\bibliography{/home/bibou/Documents/4.Recherche/6.Macros/Bibliographie/biblio}

\end{document}

config: arxiv


TITLE: Why is work defined as force dot displacement?
QUESTION [1 upvotes]: Why is work defined as force dot displacement? I know that it is defined like that based on the observational fact - we do more work when we apply greater force or move to a greater distance. But I wanted to know if there was a more physical reason to back this fundamental relation.

REPLY [1 votes]: $dW = F \cdot ds$
Where $dW$ is an infinitesimal amount of Work, and $ds$ is an infinitesimal displacement. With this,
$dW = m\left(\frac{dv}{dt}\right) \cdot ds $
since $F = ma = m\left(\frac{dv}{dt}\right)$. Therefore,
$dW = m\left(\frac{ds}{dt}\right)dv = m \cdot vdv$
Now from the chain rule it follows that
$vdv = \frac12 d(v^2)$
where $d(v^2)$ means a small change in $v^2$. Therefore
$dW = m\left(\frac{ds}{dt}\right)dv = m \cdot \frac12 d(v^2)$
Integrating over $W$ for constant acceleration yields
$W = \frac12 m(v^2 - u^2)$
where $v$ is the final velocity and $u$ is the initial velocity. Therefore,
$W = \frac12 mv^2 - \frac12 mu^2 = $ final KE - initial KE.
So work being defined as Force $\cdot$ Displacement makes work equivalent to the change in kinetic energy.

REPLY [1 votes]: Actually work as you mentioned has meaning only in classical mechanics, in this physics there are classical (vector) forces and meaning for displacement, but in modern Quantum Physics there are no more such things, or at least their meaning is totally different, there are no more force as this simple vectors or meaning to say that our particle has been displaced 5 meters.
So what I want to say, is that actually energy is much more fundamental that "work" , while other answers explained the relation between work and energy in classical sense, they didn't mention why energy is more fundamental, and that because energy is a concept that deeply related with the structure of our space & time, more precisely because our space & time are homogeneous and isotropic (this means that rotating or displacing a box and applying on it the same experiment will not change the results, rather you will do experiment now or after 100 years), this nature according to "First Noeather theorem" makes energy & momentum conserved.


TITLE: If $A\cong B$, then $A\otimes C\cong B\otimes C$.
QUESTION [6 upvotes]: I think this is kind of true, since $\square\otimes C$ is a functor, so it preserves the isomorphism.
But what if we consider the example, $\mathbb{Z}\otimes\mathbb{Z}_2$ is not isomorphic to $2\mathbb{Z}\otimes\mathbb{Z}_2$. Because the latter is trivial while the first is not.
We can also consider the exact sequences:
$$0\to\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}_2\to0,$$
where the first map is simply multiple by 2,
and
$$0\to2\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}_2\to0,$$
where the first map is the inclusion.
If we tensor by $\mathbb{Z}_2$ we will get
$$\mathbb{Z}_2\to\mathbb{Z}_2\to\mathbb{Z}_2\to0$$
and
$$(0\to)X\to\mathbb{Z}_2\to\mathbb{Z}_2\to0$$
respectively.
Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $r\otimes m$ is 0 in $\mathbb{Z}\otimes\mathbb{Z}_2$, then apparently it is 0 in $2\mathbb{Z}\otimes\mathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?

REPLY [7 votes]: $2\mathbb{Z}\otimes\mathbb{Z}_2$ is not trivial. It is trivial inside of $\mathbb Z \otimes \mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be
$$ 2n \otimes 0 = 0 \text{ and } 2n \otimes 1 = n \otimes 2 = 0 $$
but $n \otimes 2$ is only an element of $2\mathbb{Z}\otimes\mathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $\mathbb Z \otimes \mathbb{Z}_2$ to make sense of it.
Secondly, for your exact sequences. Do keep in mind that the isomorphism $2\mathbb{Z} \to \mathbb{Z}$ is a different map than the inclusion $2\mathbb{Z} \to \mathbb{Z}$.
So if we take the exact sequence
$$ 2\mathbb{Z} \to \mathbb{Z}\to\mathbb{Z}_2\to 0 $$
where the leftmost map is inclusion and tensor with $\mathbb{Z}_2$, we get the following commutative diagram
$\require{AMScd}$
\begin{CD}
    2\mathbb{Z} \otimes \mathbb{Z}_2 @>\text{inclusion}>> \mathbb{Z} \otimes \mathbb{Z}_2 @>>> \mathbb{Z}_2\otimes \mathbb{Z}_2 @>>> 0 \\
    @VVV @VVV @VVV @VVV \\
    \mathbb{Z}_2 @>\times 2>> \mathbb{Z}_2 @>>> \mathbb{Z}_2 @>>>0
\end{CD}
where the rows are exact and the vertical arrows are isomorphisms.
That map $\mathbb{Z}_2 \xrightarrow{\times 2} \mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $\to \mathbb{Z}_2$ are
$$ \begin{array}{ccc}
2\mathbb{Z} \otimes \mathbb{Z}_2 & \longrightarrow & \mathbb{Z}_2 \\
2n \otimes 0 &\longmapsto & 0 \\
2n \otimes 1 & \longmapsto & n \bmod 2
\end{array}
\qquad \text{and} \qquad
\begin{array}{ccc}
\mathbb{Z} \otimes \mathbb{Z}_2 & \longrightarrow & \mathbb{Z}_2 \\
n \otimes 0 &\longmapsto & 0 \\
n \otimes 1 & \longmapsto & n \bmod 2
\end{array}  $$
So the map on the bottom we can compute by looking at the composition $\mathbb{Z}_2 \to 2\mathbb{Z} \otimes \mathbb{Z}_2 \to \mathbb{Z} \otimes \mathbb{Z}_2 \to \mathbb{Z}_2$ and this composition takes $1 \mapsto 2 \otimes 1 \mapsto 2 \otimes 1 \mapsto 0$.


TITLE: Repeated eigenvalue = 0
QUESTION [1 upvotes]: I have this quite straightforward system:
$$ \displaystyle
\begin{array}{l}
\frac{d x}{d t}=\left[\begin{array}{cc}
3 & 9 \\
-1 & -3
\end{array}\right] x \\
\text { with } x(0)=\left[\begin{array}{c}
2 \\
4
\end{array}\right] .
\end{array}
$$
My reasoning:
Using the traditional eyeball-method, one can see that this system is not linearly independant. Taking $det(A-\lambda I)$ yield $-(3+\lambda)(3-\lambda)+9 = 0 \longrightarrow \boxed{\lambda^2 = 0} $.
Since it is a repeated root, I have to get a solution in the form:
$$
\mathbf{u}(t)=c_{1} e^{\lambda t}\left(\begin{array}{l}
a_1 \\
a_2
\end{array}\right)+c_{2}\left(t e^{\lambda t}\left(\begin{array}{l}
a_1 \\
a_2
\end{array}\right)+e^{\lambda t}\left(\begin{array}{l}
b_1 \\
b_2
\end{array}\right)\right)
=e^{\lambda t}\left(c_{1}\left(\begin{array}{l}
a_1 \\
a_2
\end{array}\right)+c_{2}\left(\begin{array}{c}
b_1+a_1t \\
b_2+a_2t
\end{array}\right)\right).
$$
With eigenvectors $$
\mathbf{v_1}=\left[\begin{array}{l}
a_1 \\
a_2
\end{array}\right],
\mathbf{v_2}=\left[\begin{array}{l}
b_1 \\
b_2
\end{array}\right],
$$

$v_1$ is then found with the null space of $A-\lambda I$ =
$$\begin{pmatrix} 3 & 9 \\ -1 & -3 \end{pmatrix} \thicksim \text{gaussa-gaussa}\thicksim \begin{pmatrix} 1 & 3 \\ 0 & 0 \end{pmatrix} \longrightarrow \left[\begin{array}{l}
a_1 \\
a_2
\end{array}\right] = \left[\begin{array}{l}
\hfill 3 \\
-1
\end{array}\right]$$

$v2$ must fullfil the condition:
$$
\left(A-\lambda I\right) \mathbf{v}_{2}=\mathbf{v}_{1}
$$

$$\begin{pmatrix} 3 & 9 \\ -1 & -3 \end{pmatrix} \cdot\left[\begin{array}{l}
b_1 \\
b_2
\end{array}\right] = \left[\begin{array}{l}
\hfill 3 \\
-1
\end{array}\right]$$
I find that $v_2 = \left[\begin{array}{l}
\frac{1}{2} \\
\frac{1}{6}
\end{array}\right]$ does the job.
My solution should be:
$$e^{ \lambda t}\left(c_{1}\left(\begin{array}{l}
3 \\
-1
\end{array}\right)+c_{2}\left(\begin{array}{c}
1/2+3t \\
1/6-t
\end{array}\right)\right)$$, with $e^\lambda$ disappearing since $\lambda = 0$
I find $$c_1 = -5/3 \text{ and } c_2 = 14$$
What is wrong in my reasoning?
The problem: When entering solutions, I get wrong answer:
$$
x_1 = 21t - \frac{3}{2} \\
x_2 = -7t - \frac{1}{2}
$$

REPLY [1 votes]: Your solution is correct up until you plug the constants into your general solution:
$$-\frac53\begin{bmatrix}3 \\ -1\end{bmatrix} + 14\begin{bmatrix}\frac12 + 3t \\ \frac16 - t\end{bmatrix} = \begin{bmatrix}-5 \\ \frac53\end{bmatrix} + \begin{bmatrix}7+42t \\ \frac73 - 14t\end{bmatrix} = \begin{bmatrix}2 + 42t \\ 4 - 14t\end{bmatrix}$$
Verifying the solution:
$$x(0) = \begin{bmatrix}2 + 42(0) \\ 4 - 14(0)\end{bmatrix} = \begin{bmatrix}2 \\ 4\end{bmatrix}$$
$$x' = \begin{bmatrix}42 \\ -14\end{bmatrix}$$
$$Ax = \begin{bmatrix}3 & 9 \\ -1 & -3\end{bmatrix}\begin{bmatrix}2 + 42t \\ 4 - 14t\end{bmatrix} = \begin{bmatrix}(6 + 126t) + (36 - 126t) \\ (-2-42t) + (-12 + 42t)\end{bmatrix} = \begin{bmatrix}42 \\ -14\end{bmatrix}$$


TITLE: Evaluate the following integral: $\int_{0}^{\pi/4} \frac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx$
QUESTION [1 upvotes]: This is a question for the 2006 MIT Integration Bee that went unanswered by the contestants. I am not sure how to solve it either. I was only able to use the double angle formula to simplify the integral: $\sin\left({2x}\right) = 2 \sin\left(x\right) \cos\left(x\right)$
The final answer given was: $\frac{1}{20} \ln \left({3}\right)$
$$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx = $$
$$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 32 \sin\left(x\right) \cos\left(x\right)} \, dx $$

REPLY [2 votes]: $$I=\int_{0}^{\pi/4}\frac{\sin x +\cos x}{9+16 \sin 2x} dx.$$
Use $\sin 2x=1-(\sin x- \cos x)^2$ and re-write
$$I=\int_{0}^{\pi/4} \frac{\sin x +\cos x}{25-16(\sin x -\cos x)^2} dx,$$
Now use $\sin x -\cos x=t \implies (\cos x+ \sin x ) dx=dt$, then
$$I=\frac{1}{16}\int_{-1}^{0} \frac{dt}{25/16-t^2}=\left .\frac{1}{16} \frac{4}{10}\log\frac{5/4+t}{5/4-t} \right|_{-1}^{0}= \frac{1}{40} \log 9=\frac{1}{20}\log3.$$


TITLE: Are rotations of $(0,1)$ by $n \arccos(\frac{1}{3})$ dense in the unit circle?
QUESTION [8 upvotes]: Under which conditions will successive rotations of $(0, 1)$ by an angle $\theta$ guarantee that given $\delta > 0$ and some point $p$ on the unit circle, there exists some $n$ such that rotating $(0, 1)$ by $n\theta$ yields a point within $\delta$ radians of $p$?  That is, are there certain angles that will guarantee that the set generated by rotating $(0, 1)$ by integral multiples of $\theta$ is dense in the unit circle?
More specifically, I am working with the angle $\cos^{-1}\left(\frac{1}{3}\right)$ and wish to show that this property is true, but I'm not sure how to go about proving this.  It seems likely I'm missing something simple here...  Does it hold for any angle that is an irrational multiple of $\pi$?  If so, why?

REPLY [14 votes]: Yes, a necessary and sufficient condition is that the angle not be a rational multiple of $\pi$.
The problem is equivalent to considering the subgroup generated by $\theta$ in the additive group $\mathbb{R}/\{2k\pi\}$, viewed as the interval $[0,2\pi)$. Under the map $x\mapsto \frac{x}{2\pi}$, we can consider $\mathbb{R}/\mathbb{Z}$, and the subgroup generated by $r+\mathbb{Z}$. The assertion is that the subgroup is dense if and only if $r$ is irrational.
If $r$ is rational, then $\langle r+\mathbb{Z}\rangle$ is finite, so it is not dense.
That the subgroup is dense when $r$ is irrational is a consequence of the Equidistribution theorem. Added. Or more simply, a consequence of Dirichlet's approximation theorem (as noted by Gerry Myerson).

Here is Dirichlet's Theorem:
Theorem. (Dirichlet) For any real number $\alpha$ and positive integer $N$, there exist integers $p$ and $q$, $1\leq q\leq N$, such that
$$\left|q\alpha - p\right|\lt\frac{1}{N+1}.$$
To use it to derive the desired result, let $r$ be an irrational number in $[0,1]$ (corresponding to an angle which is not a rational multiple of $\pi$), and let $\beta\in[0,1]$ be any real number We want to show that for every $\epsilon\gt0$ there exists an integer $n$ such that $nr - \lfloor nr\rfloor$ is within $\epsilon$ of $\beta$ (that is, there is a multiple of the rotation which will land $(0,1)$ within $\epsilon$, along the circle, of $\beta$). Find an integer $M$ such that $\frac{1}{M+1}\lt\epsilon$. By Dirichlet's Theorem, you can find integers $p$ and $q$ such that $|q r - p|\lt \frac{1}{M+1} \lt \epsilon$. That means that if you take $qr$ modulo $1$, you lie within $\frac{1}{M+1}$ of either $0$ or $1$. Either way, by taking multiples of $qr$ you divide the circle into segments of length at most $\frac{1}{M+1}$, and by the Archimedean property some multiple of $qr\bmod 1$ will be strictly larger than $\beta$ (Note that $qr-p$ cannot be equal to $0$, because we are assuming that $r$ is irrational). The distance from $\beta$ to the smallest multiple which is strictly larger is less than $\frac{1}{M+1}$, which shows that there is some multiple of $r\bmod 1$ which is within $\epsilon$ if $\beta$, as desired.
(That is: use Dirichlet's Theorem to get a multiple of the rotation that is smaller in absolute value than $\frac{2\pi}{M+1}$, and note that since $\theta$ is not a rational multiple of $\pi$, this small angle cannot be $0$; then taking multiples of this rotation you get within $\frac{2\pi}{M+1}$ of any point in the circle; pick $M$ sufficiently large so that $\frac{2\pi}{M+1}\lt \epsilon$, and you are done).

So the answer to your question is "yes": the images of $(0,1)$ under successive rotations by an angle $\theta$ are dense in the unit circle if and only if $\theta$ is not a rational multiple of $\pi$.

REPLY [2 votes]: It's dense if and only if $\frac{\theta}{\pi}$ is irrational


text:
\section{Preliminary Numerical Experiments}

{
In this section we provide some
{elementary} numerical results of different algorithms for solving RPCA. We selected eight different solvers,
five for solving convex problems: IALM \cite{Lin-Chen-Wu-Ma-iadm-2009}, ADM \cite{Yuan-Yang-pjo-2013}, ADMIP \cite{Aybat-Iyengar-2015}, fastRPCA-max \cite{P1C2-Aravkin14_1P} and fastRPCA-sum \cite{P1C2-Aravkin14_1P}, and three for solving nonconvex problems: LMafit \cite{Shen-Wen-Zhang-Lmafit}, R2PCP \cite{hintermuller2015robust}, GD \cite{Yi-Park-Chen-Caramanis-gradient}. We tested their performance on some standard synthetic data used in many RPCA papers. The synthetic data were generated in the following manner:
\begin{enumerate}
\item $L^o=UV^\top$ such that $U\in\mR^{n\times r}$, $V\in\mR^{n\times r}$ for $r=c_rn$ and $c_r\in\{0.05,0.1\}$. Moreover, $U_{ij}\sim\cN(0,1)$, $V_{ij}\sim\cN(0,1)$ for all $i,j$ are independent standard Gaussian variables,
\item $\Omega\subset\{(i,j):1\leq i,j\leq n\}$ was chosen uniformly at random such that its cardinality $|\Omega|=c_pn^2$ and $c_p\in\{0.05,0.1\}$,
\item ${S^o_{ij}}\sim\cU[-\sqrt{8r/\pi},\sqrt{8r/\pi}]$ for all $(i,j)\in\Omega$ are independent uniform random variables,
\item $N_{ij}^o\sim\varrho\cN(0,1)$ for all $i,j$ are independent Gaussian variables, where for given signal-to-noise ratio (SNR) of $M$, $\varrho$ is computed from
    \begin{align*}SNR(M) & = 10\log_{10}\left(\frac{\mE[\|L^o+S^o\|_F^2]}{\mE[\|N^o\|_F^2]}\right) \\ & = 10\log_{10}\left(\frac{c_rn+c_s\frac{8r}{3\pi}}{\varrho^2}\right), \end{align*}
    and $SNR(M)\in\{50dB,100dB\}$,
\item The data matrix $M=L^o+S^o+N^o$.
\end{enumerate}
{Note that the non-zero entries of the sparse component and the entries of the low-rank component have approximately the same magnitude in expectation. Indeed, for $n\gg 1$, $L^o_{ij}\approx\sqrt{r}\cN(0,1)$; hence, $\mathbb{E}[L^o_{ij}]=\sqrt{\frac{2r}{\pi}}$ for all $i,j$ and $\mathbb{E}[S^o_{ij}]=\sqrt{\frac{2r}{\pi}}$ for $(i,j)\in\Omega$.}

We created 10 random problems of size $n\in\{500, 1500\}$, i.e., $M\in\mR^{n\times n}$, for each of the two choices of $SNR(M)$, $c_r$ and $c_p$ using the procedure described above. We plot the figures showing the averaged relative errors of the {iterates} over 10 runs versus cpu times in Figure \ref{fig:synthetic}, where the relative error of {$(L,S)$} is defined as
{\[{\rm Error}(L,S) := \frac{\|L-L^o\|_F}{\|L^o\|_F} + \frac{\|S-S^o\|_F}{\|S^o\|_F}.\]}

For all the eight algorithms, we used their default stopping criteria and default parameters setting if the output is of good quality; {otherwise, we tuned some parameters so that the algorithm becomes competitive for our experimental setting. It is worth emphasizing that fastRPCA-sum and fastRPCA-max solve \eqref{eq:aravkin_value_function} with $\psi=\psi_{\rm sum}$ and $\psi=\psi_{\max}$, respectively; and $\rho_{\max}$, the trade-off parameter for $\psi_{\max}$, is set to $\norm{L^o}_*/\norm{S^o}_1$, i.e., this model needs an oracle that provides an ideal separation to tune the parameter.}
\begin{remark}\label{rem:comparison}
We remark that comparing different algorithms for solving RPCA is not an easy task for the following reasons. (i) The algorithms are designed for solving {related but different formulations.}
For example, IALM solves \eqref{rpca-nuclear-L1} and LMafit solves \eqref{Lmafit-problem-rewrite}, so it is difficult to compare which algorithm is better. (ii) The performance of all the algorithms depends on the parameter settings, initial points, and data structures. For example, from Figure \ref{fig:synthetic} we see that LMafit consistently outperforms fastRPCA-sum. However, this is only based on the data and parameters we tested. For other data sets, initial points, and parameter settings, fastRPCA-sum may be better than LMafit.
\end{remark}
}

\begin{figure*}[t]
  \caption{Comparison of different algorithms. The first row shows the comparison results for $n=500$, $SNR=50dB$, and $(c_r,c_p) = (0.05,0.05)$, $(0.05,0.1)$, $(0.1,0.05)$, $(0.1,0.1)$, respectively from left to right. The second row corresponds to $n=500$, $SNR=100dB$, the third row corresponds to $n=1500$, $SNR=50dB$, and the fourth row corresponds to $n=1500$, $SNR=100dB$, all with $(c_r,c_p)$ set the same as the first row from left to right. We remark again that these comparison results only reflect the performance of the algorithms under the current setting of parameters, stopping criteria, initial points, input data and so on. For other settings and data, the performance can be very different. See Remark \ref{rem:comparison}.}\label{fig:synthetic}
  \centering
  \includegraphics[width=0.24\textwidth]{fig_n500_snr50_cr5_cp5}
  \includegraphics[width=0.24\textwidth]{fig_n500_snr50_cr5_cp10}
  \includegraphics[width=0.24\textwidth]{fig_n500_snr50_cr10_cp5}
  \includegraphics[width=0.24\textwidth]{fig_n500_snr50_cr10_cp10}
  \includegraphics[width=0.24\textwidth]{fig_n500_snr100_cr5_cp5}
  \includegraphics[width=0.24\textwidth]{fig_n500_snr100_cr5_cp10}
  \includegraphics[width=0.24\textwidth]{fig_n500_snr100_cr10_cp5}
  \includegraphics[width=0.24\textwidth]{fig_n500_snr100_cr10_cp10}
  \includegraphics[width=0.24\textwidth]{fig_n1500_snr50_cr5_cp5}
  \includegraphics[width=0.24\textwidth]{fig_n1500_snr50_cr5_cp10}
  \includegraphics[width=0.24\textwidth]{fig_n1500_snr50_cr10_cp5}
  \includegraphics[width=0.24\textwidth]{fig_n1500_snr50_cr10_cp10}
  \includegraphics[width=0.24\textwidth]{fig_n1500_snr100_cr5_cp5}
  \includegraphics[width=0.24\textwidth]{fig_n1500_snr100_cr5_cp10}
  \includegraphics[width=0.24\textwidth]{fig_n1500_snr100_cr10_cp5}
  \includegraphics[width=0.24\textwidth]{fig_n1500_snr100_cr10_cp10}
\end{figure*}


TITLE: How can quantum tunnelling lead to spontaneous decay?
QUESTION [9 upvotes]: I have never understood what measuring process (if any) is supposed to be continuously polling the quantum state of an unstable bound system subjected to decay via quantum tunnelling. The reason I reckon some kind of polling process should exist in the first place is the following:
According to the QM postulates, the unitary evolution of such a system should by definition keep it reversible, so it is only when measured that a decay can be observed or not. But this would make the decay rate dependent on the measurement rate, while we well know that the decay probability is constant, and the decay deemed "spontaneaous".
What am I missing?
Edit:
From the discussion in the comment section I gather I have not been clear about what I am asking here exactly. Let me try to reformulate the question.
It is about how quantum tunnelling is supposed to explain the exponential dynamics of a decay process. I am not asking about the Zeno effect; quite the opposite, actually: why, in the absence of any measurement, do we have an exponential decay at all? I just do not understand at what point in the unitary evolution of the unstable system the tunnelling effect is spontaneously happening.
The polling process I imagined is just a way to ask "why does the tunnelling effect manifest iself ?" because I cannot see how it can manifest without a measurement. Please do not infer that I am making up my pet theory here. I am only looking for a way to picture the situation, which at this time I don't get at all.

REPLY [1 votes]: I am adding a few points that Lawrence's answer didn't yet address, but should help clarify the nature of the quantum tunneling process:

According to the QM postulates, the unitary evolution of such a system should by definition keep it reversible, so it is only when measured that a decay can be observed or not. But this would make the decay rate dependent on the measurement rate, while we well know that the decay probability is constant, and the decay deemed "spontaneaous".
  What am I missing?

The reversibility of the unitary evolution does not preclude the irreversibility of single tunneling events. This is because "reversibility" means different things in the two contexts:
1) A unitary evolution $\hat U$ is said to be time-reversible if whenever $|\psi(0)\rangle \rightarrow |\psi(t)\rangle = {\hat U}(t)|\psi(0)\rangle$ is a valid dynamics, then for $|{\bar \psi}(0)\rangle = {\hat T} |\psi(t_0)\rangle$ at some time $t_0$ and with ${\hat T}$ the antilinear time-reversal operator, $ |{\bar \psi}(0)\rangle \rightarrow |{\bar \psi}(t)\rangle =  {\hat U}(t)|{\bar \psi}(0)\rangle \equiv {\hat T} |\psi(t_0-t)\rangle$ gives the exact time-reversed dynamics. This implies that $[{\hat T}, {\hat U}] = 0$, and similarly for the generator of ${\hat U}$, the Hamiltonian ${\hat H}$. For a spin-0 particle the time-reversal operator ${\hat T}$ amounts to complex conjugation in the position representation, but for spin-1/2 and higher it also includes a unitary component acting on the spin degrees of freedom.
2) A tunneling event is irreversible simply because once the particle tunnels through the barrier, the probability that it will tunnel back at some later time is practically null. The actual tunneling corresponds to the $|\psi(0)\rangle \rightarrow |\psi(t)\rangle = {\hat U}(t)|\psi(0)\rangle$ evolution. But let $D$ denote the spatial domain "inside" the barrier, and let
$$
P_D(t) = \int_D{dV\;|\langle {\bf x} |\psi(t)\rangle|^2}
$$
be the probability to localize the particle within $D$. In general it is assumed that the particle is initially localized entirely within D, such that $P_D(0) = \int_D{dV\;|\langle {\bf x} |\psi(0)\rangle|^2} = 1$. The statement that the probability of reverse tunneling is null then amounts to saying that $P_D(t)$ vanishes asymptotically,
$$
\lim_{t \rightarrow \infty}{P_D(t)} = \lim_{t \rightarrow \infty}{\int_D{dV\;|\langle {\bf x} |\psi(t)\rangle|^2}} \;\rightarrow \;0
$$
The fact that the tunneling evolution ${\hat U}(t)$ is still time reversible means that if we now take as initial state the time reversed of the state at some asymptotically large time $t_0 \rightarrow \infty$, $|{\bar \psi}(0)\rangle = {\hat T} |\psi(t_0)\rangle$, when the particle is definitely localized outside of $D$, $\int_D{dV\;|\langle {\bf x} |{\bar \psi}(0)\rangle|^2} \rightarrow 0$, then the evolution of $|{\bar \psi}\rangle$ will be the exact time-reversed of the original one for $|\psi\rangle$, and will asymptotically drive the particle back into domain $D$ through reverse tunneling,
$$
\lim_{t \rightarrow \infty}{\int_D{dV\;|\langle {\bf x} |{\bar \psi}(t)\rangle|^2}} \;\rightarrow \;1
$$
In the sense of an individual event, this reverse tunneling into $D$ is just as irreversible as the original tunneling out of $D$, despite the fact that ${\hat U}(t)$ is a time-reversible evolution.
If you prefer, think of an analogy with a free particle wave-packet of well-defined average momentum ${\bf p}$: it always travels in the direction of $\bf p$ and statistically speaking never turns around, despite inherent zitterbewegung and dispersion. But the free particle evolution is time-reversible, and the time-reversed of this wave packet will travel in the $-\bf p$ direction, etc.

[…] why, in the absence of any measurement, do we have an exponential decay at all? I just do not understand at what point in the unitary evolution of the unstable system the tunneling effect is spontaneously happening.

It is not so much at what point in time tunneling takes place, since this is essentially a statistical process, but what is the probability that at time $t$ the particle is located outside the inner domain $D$. To determine this probability we do not really need to follow the dynamics of the same system through multiple queries. In general we cannot even do this without altering the entire dynamics. What we must do is to query ensembles of identically prepared systems at different times $t$, ideally using a different ensemble for each such query. In principle, this would ensure that the measurement procedure does not interfere with the dynamics up to that point in time, and that the measured probability is indeed reliable.
Exponential decay then means that the fraction of particles detected outside domain $D$ in each such query diminishes exponentially with the time elapsed since the preparation of the respective ensemble. Note that the average tunneling time is statistically well defined, although we do not need to consider an average over exact individual tunneling times for each of the particles in an ensemble.

(From a comment to Lawrence's answer) to get a neat exponential law, it seems to me that some quasi-continuous interaction is needed. I have not seen that point discussed.

Actually it has been extensively researched in relation to a multitude of distinct fields, from nuclear physics to chemical kinetics and biophysics. See these "Lectures on Dissipative Tunneling", and google the Leggett-Caldeira model.