text:
\begin{document}
\title{Domain perturbations for elliptic problems with Robin boundary conditions of opposite sign.}
\pagestyle{myheadings}
\maketitle
\centerline{\scshape Catherine Bandle}
\medskip
{\footnotesize
\centerline{Mathematische Institut, Universit\"at Basel,}
\centerline{Rheinsprung 21, CH-4051 Basel, Switzerland}
}
\medskip
\centerline{\scshape Alfred Wagner}
\medskip
{\footnotesize
\centerline{Institut f\"ur Mathematik, RWTH Aachen  }
\centerline{Templergraben 55, D-52062 Aachen, Germany}}

\bigskip

\abstract{We consider the energy of the torsion problem with Robin boundary
conditions in the case where the solution is not a minimizer. Its dependence
on the volume of the domain and the surface area of the boundary is discussed.
In contrast to the case of positive elasticity constants, the ball does not provide
a minimum. For nearly spherical domains and elasticity constants close to zero
the energy is largest for the ball. This result is true for general domains in the
plane under an additional condition on the first non-trivial Steklov eigenvalue.
For more general elasticity constants the situation is more involved and it
is strongly related to the particular domain perturbation. The methods used
in this paper are the series representation of the solution in terms of
Steklov eigenfunctions, the first and second shape derivatives and an
isoperimetric inequality of Payne and Weinberger \cite{PaWe61} for the torsional
rigidity.}
\bigskip

{\bf  Key words}: Robin boundary condition, energy representation, Steklov eigenfunction, extremal domain, first and second domain variation, optimality conditions.
\bigskip

\section{Introduction}
Let $\Omega \subset \mathbb{R}^n$ be a  bounded smooth domain  and let $\nu$ denote its outer normal. In this paper we study the Poisson problem
\begin{align}\label{torsion}
\Delta  u+1 =0 \tx{in} \Omega, \quad \p_{\nu }  {u}= \alpha  u \tx{on} \p \Omega.
\end{align}
It is the Euler-Lagrange equation corresponding to the energy functional
$$
E(V,\Omega):=\int_{\Omega} |\nabla V|^2\:dx -\alpha \oint_{\partial \Omega} V^2\:ds -2\int_{\Omega}V\:ds.
$$
If $\alpha<0$ there exists a unique solution $u(x)$ which minimizes  the  energy among all functions in $W^{1,2}(\Omega)$.
In this case  Bucur and Giacomini \cite{BuGi14} have shown that among all domains of given volume the ball has the smallest energy. This property is well-known if $u$ satisfies Dirichlet conditions and follows immediately by symmetrization. The presence of Robin boundary conditions requires completely new arguments.

In this study we are interesting in the case where $\alpha>0$. The motivation came from the eigenvalue problem $\Delta \varphi +\lambda \varphi =0$ in $\Omega$,  $\p_\nu \varphi= \alpha \varphi$ on $\p\Omega$, considered for the first time by Bareket
\cite{Ba77}. She observed that for nearly circular domains of given area the circle has the largest first eigenvalue. Recently this result was extended to higher dimensions for nearly spherical domains by Ferone, Nitsch and Trombetti \cite{ FeNiTr14} cf. also \cite{BaWa2_14}. The question whether or not the ball is optimal for all domains of the same volume remained opened until recently when Freitas and Krejcirik \cite{FrKr14} showed that annuli have for large $\alpha$ a larger eigenvalue than the ball with the same volume.

If $\alpha >0$ Problem \eqref{torsion} is not always solvable. In fact if $\alpha$
coincides with an eigenvalue $0=\mu_1<\mu_2\leq \dots$ of the {\sl Steklov problem}
\begin{align}\label{steklov}
\Delta \phi =0 \tx{in} \Omega, \quad \p_\nu \phi= \mu \phi \tx{on} \p \Omega.
\end{align}
then problem \eqref{torsion} has a solution  if and only if the compatibility condition
\begin{align}\label{compatibility}
\int_{\Omega} \phi_i\:dy =0
\end{align}
holds for all eigenfunctions corresponding to $\mu_i=\alpha$. If \eqref{compatibility} is satisfied then \eqref{torsion} is solvable but the  the solution is not unique.
\smallskip

If $\alpha \neq \mu_i$ then there exists a unique solution.  It is a critical point of $E(V,\Omega)$ in $ W^{1,2}(\Omega)$ in the sense that the Fr\'echet derivative vanishes. However in contrast to the case $\alpha<0$ the critical point is not a local extremum but a saddle point.
\medskip

The goal of this paper is to investigate  $E(u,\Omega)$ among all domains with given volume. In contrast to the case where $\alpha$ is positive the ball has in general not the smallest energy. By means of the shape derivative and a result of Serrin \cite{Se71}for overdetermined boundary value problems it can be shown that the ball is the only critical domain. The analysis of the second shape derivative reveals that for nearly spherical domains and for $\alpha$ small enough the energy is larger or smaller that the one of the ball, depending on the perturbation. The most surprising result in this context is that for $\alpha$ close to zero the ball has the largest energy for all domains of given volume. This phenomenon is related in a wider sense to anti-maximum principles \cite{ClPe79}. At the end we use an upper bound for the torsion of Payne and Weinberger \cite{PaWe61} and obtain an isoperimetric inequality for $E(u,\Omega)$ for all domains in the plane.
\medskip

This paper is organized as follows. First we use the Steklov eigenfunctions to derive a series representation of the energy which will be useful to derive global estimates. This is the content of Section 3. Then we discuss the first shape derivative for general domains and the second shape derivative for nearly spherical domains. At the end we prove the optimality of the disc in two-dimensions.

\section{Preliminaries}
The Steklov eigenvalues and eigenfunctions will play a crucial role in our considerations. If $\p \Omega$ is Lipschitz continuous then they belong to the Sobolev space $W^{1,2}(\Omega)$ and have a trace in $L^2(\partial \Omega)$. The eigenfunctions can be chosen such that
\begin{align}\label{orthog}
\oint_{\p \Omega} \phi_i \phi_j\:dS = \delta_{ij}, \quad  \int_{\Omega} \nabla \phi_i\cdot \nabla \phi_j \:dx= 0 \tx{if} i\neq j \tx{and} \int_{\Omega} |\nabla \phi_i|^2\:dx= \mu_i.
\end{align}
Moreover every harmonic function $h$ in $\Omega$ with a trace in $L^2(\partial \Omega)$ can be expanded in a series of Steklov eigenfunctions which converges in $W^{1,2}(\Omega)$. It should be mentioned that by a result of Mazya \cite{Ma85} in a Lipschitz domain the norms corresponding to the inner products $<u,v>_\Omega = \int_\Omega \nabla u\cdot \nabla v\:dx +\int_\Omega uv\:dx$ and $<u,v>_{\partial \Omega} = \int_\Omega \nabla u\cdot \nabla v\:dx +\oint_{\partial \Omega} uv\:dS$ are equivalent.

In the next lemma we show how to expand harmonic functions into a series of Steklov eigenfunctions.

\begin{lemma}\label{le:harmonic} (i) Suppose that $\alpha\in \mathbb{R}$ does not coincide with a Steklov eigenvalue $\mu_i$.  Let $h$ be the solution of
\begin{align}\label{eq:harmonic}
\Delta h=0 \tx{in} \Omega, \quad \p_{\nu} h = \alpha h + g(x) \tx{on} \p\Omega.
\end{align}
Then $h= \sum_1^\infty h_i \phi_i$ where
$$
h_i= \frac{\oint_{\p \Omega} \phi_i g \: dS}{\mu_i-\alpha}.
$$
This series converges in $W^{1,2}(\Omega)\cap L^2(\p \Omega)$.
\medskip

(ii) Assume $\alpha= \mu_k$ and denote by ${\mathcal{L}}_k$ the linear space generated by the Steklov eigenfunctions belonging to the eigenvalue $\mu_k$.  A solution exists if and only if the compatibility condition
\begin{align}\label{compatibility:h}
\oint_{\p\Omega} g\phi_k\:dS =0.
\end{align}
is satisfied for all $\phi_k\in {\mathcal{L}}_k$ . In this case  \eqref{eq:harmonic} has infinitely many solutions which are expressed as
$$
h= \sum_{i,i\neq k} h_i \phi_i + {\mathcal{L}}_k,
$$
where $h_i$ is given in (i).
\end{lemma}

{\bf Proof}  Because of the completeness of the Steklov eigenfunctions we can write $h=\sum_1^\infty h_i\phi_i$.
Testing \eqref{eq:harmonic} with $\phi_j$ we get by \eqref{orthog}
$$
0= \int_{\Omega} \phi_j \Delta h \:dy = \oint_{\p \Omega}( \phi_j \p_{\nu} h -h\mu_j\phi_j)\:dS=(\alpha -\mu_j)h_j +\oint_{\p\Omega} \phi_j g\:dS.
$$
This proves the first assertion. The convergence follows from  results by \cite{Ma85,Au04}. The second statement is a consequence of the classical theory on inhomogeneous linear problems.\hfill $\square$
\medskip

For the next considerations we decompose the solution $ u$ of \eqref{torsion} into $ u = h + s$ where $s$ is the solution of the Dirichlet problem
\begin{eqnarray}\label{seq}
\Delta s+1 =0 \:\hbox{in}\: \Omega, \quad s=0 \:\hbox{on}\: \p \Omega.
\end{eqnarray}
Then $h$ is a solution of \eqref{eq:harmonic} with $g=  -\p_{\nu}s$.

A straightforward computation shows that
$$
E(  u,\Omega) = -\int_{\Omega}  u\: dx=  -\int_{\Omega} (h+ s)= - \sum_1^\infty h_i \int_{\Omega}\phi_i\:dx
-\int_{\Omega}s \:dx.
$$
Observe that
$$
-\int_{\Omega} s\:dy= \min_{W^{1,2}_0(\Omega)} \int_{\Omega}\left (|\nabla V|^2 -2V\right)\:dy =:T(\Omega).
$$
Moreover we have
$$
-\sum_1^\infty h_i\int_{\Omega} \phi_i\:dx= \sum_1^\infty\frac{\oint_{\p \Omega}\phi_i\p_{\nu} s\:dS}{\mu_i-\alpha}\int_{\Omega}\phi_i\:dx
$$
and
$$
\int_{\Omega} \phi_i\:dx=  -\int_{\Omega} \phi_i \Delta s\:dx= -\oint_{\p \Omega} \phi_i\p_{\nu}s\:dS.
$$
Hence
\begin{align}\label{id:E}
E(  u,\Omega) =T(\Omega) +\sum_1^\infty \frac{\left (\oint_{\p \Omega} \phi_i\p_{\nu}s\:dS\right)^2}{\alpha -\mu_i}.
\end{align}

\begin{theorem} \label{thm:energy}Assume that $\mu_p<\alpha <\mu_{p+1}$. Set
$$
{\mathcal{E}}^+:= \sum_1^p  \frac{\left (\oint_{\p \Omega} \phi_i\p_{\nu}s\:dS\right)^2}{\alpha-\mu_i}\geq 0 \tx{and} {\mathcal{E}}^-:= \sum_{p+1}^\infty \frac{\left (\oint_{\p \Omega} \phi_i\p_{\nu}s\:dS\right)^2}{\alpha-\mu_i}\leq 0.
$$
Then the following statements hold true.
\begin{enumerate}
\item
$$
E(  u,\Omega) = T(\Omega) + {\mathcal{E}}^+ +{\mathcal{E}}^-.
$$
\item
Let ${\bf L_p}$ be the linear space generated by $\{\phi_j\}_{j=1}^p$ and let ${\bf L^\infty_p}$ be the orthogonal complement spanned by
$\{\phi_j\}_{j=p+1}^\infty$. Set
$$
H(v)=\int_{\Omega }|\nabla v|^2 \:dy-\alpha \oint_{\p \Omega} v^2\:dS + 2\oint_{\p \Omega} v\p_\nu s \:dS.
$$
Then
$$
{\mathcal{E}}^+=\max_{\bf L_p}H(v) \tx{and} {\mathcal{E}}^-=\min_{\bf L_p^\infty}H(v)=\min_vH(v), \tx{where} \oint_{\partial \Omega}v\phi_k\:dS=0 \tx{for} i=1,2..p.
$$
\item If $\alpha = \mu_i$ then \eqref{torsion} has a solution if and only if $\int_{\Omega} \phi_i\:dx= -\oint_{\partial \Omega} \phi_i \p_\nu s\:dS=0$ for all eigenfunctions belonging to $\mu_i$.
\end{enumerate}
\end{theorem}

{\bf Proof} The first assertion follows from \eqref{id:E}.
Replacing $v \in {\bf L_p}$ by its series $\sum_1^p v_i \phi_i$ we find
$$
H(v)= \sum_i^p v_i^2(\mu_i-\alpha)+2\sum_1^p v_i\oint_{\p \Omega} \phi_i \p_\nu s \:dS.
$$
By assumption $\mu_i-\alpha <0$ which implies that the positive  maximum is achieved for $v_i=-\oint_{\p \Omega} \phi_i \p_\nu s \:dS/(\mu_i-\alpha)$ which is the Fourier coefficient $h_i$. Inserting this expression into $H(v)$ we obtain $\mathcal{E}^+$.The same argument yields the result for ${\mathcal{E}}^-$. This establishes the second
statement and the last assertion is the compatibility condition stated in \eqref{compatibility}. \hfill $\square$
\medskip

{\sl In the sequel $h_i$ stands for the Fourier coefficient of $h$ in the decomposition $ u= s+h$.}
\medskip

{\sc Remark} The series development \eqref{id:E} holds also for negative $\alpha$.  In this case $\mathcal{E}^+ =0$ and therefore $E( u,\Omega) = T(\Omega) +{\mathcal{E}}^-.$
\medskip

{\sc examples}
1. Le $\Omega= B_R$ be the ball of radius $R$ centered at the origin. The Steklov eigenfunctions for the ball are of the form $r^kX_k(\theta)$ where $\theta\in \p B_1$ and $X_k(\theta)$ are the spherical harmonics of degree $k$. The eigenvalues are
$\mu=\frac{k}{R}$, $k\in \mathbb{N}$, and their multiplicity is $\frac{(k+n-1)!}{k! (n-1)!}$. By the maximum principle for harmonic functions $\phi_1=$const. is the only radial
eigenfunction. Here $s=-\frac{R^2}{2n}-\frac{r^2}{2n}$ and thus
$$
h_i (\mu_i-\alpha)= -\oint_{\p B_R} \phi_i \p_\nu s \:dS = 0 \quad \forall i>1.
$$
Consequently  \eqref{torsion} has a solution for all $\alpha\neq 0$. It is of the form
$$
 u=
\begin{cases}
\frac{R^2}{2n}- \frac{R}{\alpha n} -\frac{r^2}{2n} &\tx{if} \alpha \neq \mu_j ,\\
\frac{R^2}{2n}- \frac{R}{\alpha n} -\frac{r^2}{2n} + w&\tx{if} \alpha = \mu_j,
\end{cases}
$$
where $w$ is any function in the eigenspace of $\mu_j$.
In both cases we get
\begin{align}\label{e:ball}
E(  u,B_R)= T(B_R) + \frac{|B_R|^2}{\alpha |\p B_R|} =|B_R|\left(-\frac{R^2}{n(n+2)} +\frac{R}{\alpha n}\right).
\end{align}

2. Let $\Omega=\{y:r_0<|y|<R\}$ be an annulus and set $r_0= \kappa R$.
Suppose for simplicity that $\Omega \subset \mathbb{R}^n$, $n>2$. The radial solutions of \eqref{torsion} are of the form
$$
 u=-\frac{r^2}{2n}+c_1 +\frac{c_2}{r^{n-2}}.
$$
The boundary conditions lead to the linear system
\begin{align}\label{sy:linear}
c_1\alpha +c_2\left(\frac{\alpha}{R} + \frac{n-2}{R^{n-1}}\right)&= \frac{\alpha R^2}{2n} -\frac{R}{n},\\
\nonumber c_1\alpha +c_2\left(\frac{\alpha}{\kappa R} -\frac{n-2}{(\kappa R)^{n-1}}\right)&= \frac{\alpha(\kappa R)^2}{2n} +\frac{\kappa R}{n}
\end{align}
This system has a unique solution if the determinant is different from zero. The determinant vanishes if
$$
\alpha =\alpha_1 =0 \tx{and} \alpha=\alpha_2 = \frac{n-2}{R^{n-2}}\left(\frac{\kappa^{1-n}+1}{\kappa^{-1}-1}\right).
$$
The eigenfunctions of the
Steklov problem in annulus are similar to those for the ball, namely  $(c_1r^k +c_2r^{-k})X_k(\theta)$, $k=1,\cdots$. In addition to $\phi_1=$const. there is a radial eigenfunction $\phi_r=c_1+c_2r^{2-n}$ with $c_1\mu_r + c_2\left(\frac{\mu_r}{R^{2-n}}+\frac{n-2}{R^{n-1}}\right)=0$ and $c_1\mu_r + c_2\left(\frac{\mu_r}{(\kappa R)^{2-n}}-\frac{n-2}{(\kappa R)^{n-1}}\right)=0$.
Notice that
$\alpha_1$ and $\alpha_2$ correspond to the Steklov eigenvalues $\mu_1$ and $\mu_r$ of the radial eigenfunctions. For $\kappa \neq 1$ the inhomogeneous linear system \eqref{sy:linear} is not solvable if
 $\alpha = \mu_r$. Hence  the Fourier
coefficient $h_r$ is not defined. From the symmetry of the annulus it follows that $h_k =0$ for all $k\neq 1,r$.

The same argument as for the ball shows that in an annulus \eqref{torsion}
\begin{itemize}
\item has a unique solution if $\alpha \neq \mu_i$,
\item  no solution if $\alpha = \mu_r$,
\item  a family of solutions of the form $-\frac{r^2}{2n}+c_1 +\frac{c_2}{r^{n-2}} +w$ where is in the eigenspace of $\mu_i$  if $\alpha = \mu_i$.
\end{itemize}
Therefore by Theorem \ref{thm:energy} we obtain for the annulus
\begin{align}\label{Eannulus}
E(  u,\Omega) = T(\Omega) +\frac{|\Omega|^2}{\alpha |\p \Omega|} +\frac{h_r^2 }{\alpha-\mu_r}\tx{for all} \alpha \neq 0\tx{and} \alpha\neq \mu_r=\frac{n-2}{R^{n-2}}\left(\frac{\kappa^{1-n}+1}{\kappa^{-1}-1}\right).
\end{align}

 \section{Global estimates}
3.1. General estimates.
\bigskip

From Theorem \ref{thm:energy} we have  for all $\alpha \neq \mu_j$ that $E(u,\Omega)=T(\Omega) + \mathcal{E}^+ + \mathcal{E}^-$. Many estimates are known for $T(\Omega)$ which is related to the torsion. Less known and more difficult to estimate are the expressions $\mathcal{E}^\pm$. We first start with the observation that
 $$
 a_i:= \oint_{\partial \Omega} \phi_i\partial_\nu s\:dS
 $$
 is the Fourier coefficient of $\partial_\nu s$ with respect to the Steklov eigenfunction $\phi_i$. Let us write
 $$
 \partial_\nu s= \underbrace{\sum_1^p a_i \phi_i}_{\partial_\nu s^+} +\underbrace{\sum_{p+1}^\infty a_i \phi_i}_{\partial_\nu s^-}.
 $$
 Furthermore set for short $\|v\|:=\|v\|_{L^2(\partial \Omega)}$. Then $\|\partial_\nu s^+\|^2= \sum_1^p a_i^2$
 and $\|\partial_\nu s^-\|^2=\sum_{p+1}^\infty a_i^2$.
 Under the assumption $0\leq \mu_p<\alpha <\mu_{p+1}$ it follows immediately that
 \begin{align} \label{e:E+E-}
 \alpha^{-1} \|\partial_\nu s^+\|^2 \leq \mathcal{E}^+\leq (\alpha -\mu_p)^{-1} \|\partial_\nu s^+\|^2,\\
 \nonumber (\alpha-\mu_{p+1})^{-1} \|\partial_\nu s^-\|^2 \leq \mathcal{E}^- \leq (\alpha-\mu_m)^{-1} \sum_{p+1}^m a_i^2.
\end{align}
\medskip

{\sc Application} If $\alpha=-c^2$ is negative we have $\mathcal{E}^+=0$  and therefore \ $\mathcal{E}^-\geq \alpha^{-1}\| \partial_\nu s\|^2_{L^2(\partial \Omega)}$. Hence
$$
E( u,\Omega) \geq T(\Omega)-c^{-2} \|\p_\nu s\|^2.
$$
Equality holds for the balls.  From Schwarz symmetrization it follows immediately that $T(\Omega)\geq T(B_R)$  where $B_R$ is the ball with the same volume as $\Omega$. Also $\int_\Omega |\nabla s_\Omega|^2\:dx \leq \int_{B_R}|\nabla s_{B_R}|^2\:dx$. However it is not clear that
$\|\p_\nu s\|_{L^2(\p\Omega)}\leq \|\p_\nu s\|_{L^2(\p B_R)}$ which would prove that the ball has the smallest energy. Pointwise estimates for $|\nabla s|^2$ are well-known in the literature cf. \cite {GiTr83}, \cite{SpeSch76}.
\bigskip

3.2. Let $0<\alpha<\mu_2(\Omega)$.
\bigskip

 In this case
 $$
 \mathcal{E}^+= \alpha^{-1} \left(\oint_{\partial \Omega} \phi_1\partial_\nu s\:dS\right)^2.
 $$
 Since $\phi_1=\frac{1}{\sqrt{|\partial \Omega|}}.$ we find
 $$
 \mathcal{E}^+= \frac{|\Omega|^2}{\alpha |\partial \Omega|}.
 $$
 This together with Theorem \ref{thm:energy} leads to

\begin{lemma} \label{le:E} Assume $0<\alpha <\mu_2(\Omega)$. Then
$$
E( u,\Omega) \leq T(\Omega) +\frac{|\Omega|^2}{\alpha |\p\Omega|}.
$$
Equality holds for the ball.
\end{lemma}

If $a_i=0$ for $i=1,\cdots r$, like for instance in the annulus, then the estimate above holds for $0<\alpha<\mu_r(\Omega)$.

 An interesting question is to find an isoperimetric upper bound for
 $$
 \mathcal{J}(\Omega):=T(\Omega) + \frac{|\Omega|^2}{\alpha |\p \Omega|}.
 $$
 If the volume $|\Omega|=|B_R|$ is fixed then -as mentioned before- Schwarz symmetrization implies that $T(\Omega) \geq T(B_R)$,
whereas $\frac{|\Omega|^2}{\alpha |\p\Omega|}\leq \frac{|B_R|^2}{\alpha |\p B_R|}$. The question arises which inequality prevails.

\begin{proposition}\label{le:comp1}
Let $\Omega\neq B_R$ be a fixed domain in $\mathbb{R}^n$ such that $|\Omega|=|B_R|$. Then there exists a positive number $\alpha_0(\Omega)>0$ such that
$$
\mathcal{J}(\Omega)\begin{cases}
<\mathcal{J}(B_R)&\tx{if} \alpha <\alpha_0,\\
>\mathcal{J}(B_R)&\tx{if} \alpha >\alpha_0.
\end{cases}
$$
\end{proposition}

{\bf Proof}  It is well-known that for any domain different from a a ball $T(\Omega)-T(B_R)= \epsilon_0>0$. Define $\alpha_0= \frac{|B_R|^2}{\epsilon_0}\left( \frac{1}{|\partial B_R|}-\frac{1}{|\partial \Omega|}\right)$. Then
the assertion follows.
\hfill $\square$
\medskip

{\sc remarks}
\begin{enumerate}

\item  A sharper estimate than in Lemma \ref{le:E} can be derived from Theorem \ref{thm:energy} (2). In fact
$$
E( u,\Omega)\leq T(\Omega) +H(V),
$$
where $V$ is any trial function such that $\oint_{\partial \Omega} V\:dS=0$. Observe that if $V$ is a admissible trial function so is $tV$ for any $t\in \mathbb{R}$. Thus
$$
\mathcal{E}^- \leq \min_{\mathbb{R}} H(tV)= -\frac{\left(\oint_{\partial \Omega} V\partial_\nu s \:dS\right)^2}{ \int_\Omega |\nabla V|^2\:dy -\alpha \oint_{\partial \Omega} V^2\:dS}.
$$
Suppose that the origin is the barycenter with respect to $\partial \Omega$, {\sl i.e.} $\oint_{\partial \Omega} x_i\:dS=0$ for $i=1,\cdots n$. Then $x_i$ is admissible for the variational characterization of $\mathcal{E}^-$.  By our assumption $ \int_\Omega |\nabla x_i|^2 \:dy \geq \mu_2 \oint_{\partial \Omega} x_i^2\:dS >\alpha \oint_{\partial \Omega} x_i^2\:dS $.
Consequently
$$
\mathcal{E}^- \leq - \frac{\sum_1^n\left(\int_ \Omega x_i\:dy\right)^2}{n|\Omega| -\alpha \oint_{\partial \Omega} |x|^2|\:dS}.
$$
\item By the Brock-Weinstock inequality \cite {We54}, \cite{Br01}, $\mu_2(\Omega)\leq \mu_2(B_\rho)$ where $B_\rho$ is the ball of the same boundary measure as $\Omega$, {\sl i.e} $|\partial B_\rho|= |\partial \Omega|$.
Thus if $|\partial \Omega|$ is  large, $\mu_2(\Omega)$ is small.
\end{enumerate}

3.3. Let $\mu_p(\Omega)<\alpha <\mu_{p+1}(\Omega)$
\bigskip

This case is more involved. From Theorem \ref{thm:energy} it follows that
$E( u,\Omega) =T(\Omega) +\mathcal{E}^++\mathcal{E}^-$.  Rough  estimates are obtained from \eqref{e:E+E-}.

Observe that if the Fourier coefficient $a_p=\oint_{\partial \Omega} \phi_p \partial_\nu s \:dS \neq 0$ then $\mathcal{E}^+ $ is positive and becomes arbitrarily large as $\alpha$ tends to $\mu_p$ from above. Similarly if the Fourier coefficient $\oint_{\partial \Omega} \phi_{p+1} \partial_\nu s \:dS \neq 0$ then $\mathcal{E}^-\neq 0$ then $\mathcal{E}^-$ is negative and becomes arbitraryly small if $\alpha$ tends to $\mu_{p+1}$ from below.
\medskip

{\sc Examples}

1. In a ball $E( u,\Omega;\alpha)$ has only one pole $\alpha =\mu_1=0$. Hence
$\lim_{\alpha\searrow 0}E( u,\Omega;\alpha)= \infty$ and $\lim_{\alpha\nearrow 0}E( u,\Omega;\alpha)= -\infty$.

2. In an annulus $E( u,\Omega;\alpha)$  has two poles $\alpha =\mu_1=0$ and $\alpha=\mu_r$
see \eqref{Eannulus}.

\section{Domain variations}

\subsection{First domain variation}

\subsubsection{General remarks}
Let $\Omega$ be a family of perturbations of the domain $\Omega$ given by
\begin{equation}\label{not1}
\overline{\Omega}_t=\left\{y:y=x+tv(x)+\frac{t^2}{2}w(x)+o(t^2)\::\:
x\in\overline{\Omega}\right\},
\end{equation}
where  $v$ and $w$ are smooth vector fields $v,w:\overline{\Omega}\to\R$ belonging to $C^{2,\epsilon}(\overline{\Omega}).$

We assume that  on $\p \Omega$, $v$ points in the normal direction, i.e. $v=(\nu\cdot v)\nu$.
The parameter $t$ belongs to $(-t_0,t_0)$ where $t_0$ is chosen so small that $y:\Omega \to \Omega_t$ is a diffeomorphism.
We consider the family of problems
$$
\Delta_y  u(y,t) +1=0 \tx{in} \Omega_t, \quad  \partial_{\nu_t} u(y,t) =\alpha u(y,t) \tx{on} \partial \Omega_t,
$$
where $\nu_t$ is the outer unit normal at $\Omega_t$. For short we set
\begin{eqnarray*}
\tilde{u}(t):=u(y(x,t),t)\qquad\hbox{for}\:x\in\Omega,\:\:\vert t\vert< t_0.
\end{eqnarray*}
We now map this problem by means of $y(x,t)$ into
$\Omega$ and obtain after the change of variable $y \to  x$
\begin{align}\label{eq:torsion2}
\p_j \left(A_{ij}(x,t)\p_j\tilde u(t)\right) +J(t)=0 \tx{in} \Omega, \quad \p_{\nu_A}\tilde u(t)=\alpha m(x,t)\tilde u(t)\tx{on}\p\Omega,
\end{align}
where
\begin{eqnarray*}
\partial_i = \frac{\p}{\p x_i},\quad
dy=J(t)dx,\quad
dS_y=m(t)dS_x,\quad
A_{ij}(t):=\frac{\partial x_i}{\partial y_k}\frac{\partial x_j}{\partial y_k}\:J(t),
\quad\partial_{\nu_A} =\nu_iA_{ij} \partial_j.
\end{eqnarray*}
In \cite{BaWa14} it was shown that for small $|t|$
\begin{eqnarray}\label{jform}
J(t)&:=&\tx{det}(I+tD_v+\frac{t^2}{2}D_w)\\\nonumber&=&1+t\:\Div v+\frac{t^2}{2}\left((\Div v)^2-D_v : D_v+\Div\: w\right)+o(t^2).
\end{eqnarray}
Here we used the notation
\begin{eqnarray*}
D_v : D_v:=\partial_{i}v_j\partial_{j}v_{i},
\end{eqnarray*}
where summation over repeated indices is undestood. Furthermore
$$
m(t)=1 + t (n-1)(v\cdot \nu) H+o(t)
$$
where $H$ is the mean curvature of $\p\Omega$ and
\begin{eqnarray*}
\Div_{\partial\Omega}v=\Div v-\nu\cdot D_v \nu:=\partial_{i}v_{i}-\nu_{j}\partial_{j}v_{i}\nu_{i}.
\end{eqnarray*}
We also showed that
\begin{eqnarray*}
A_{ij}(0)&=&\delta_{ij};\\
\dot{A}_{ij}(0)&=&\Div v\:\delta_{ij}-\partial_{j}v_{i}-\partial_{i}v_{j};\\
\ddot{A}_{ij}(0)&=&\left((\Div v)^2-D_v:D_v\right)\:\delta_{ij} +2\left(\partial_{k}v_i\:\partial_{j}v_k+\partial_{k}v_j\:\partial_{i}v_k\right)\\
&&
+
2\:\partial_{k}v_i\:\partial_{k}v_j
-
2\:\Div v\left(\partial_{j}v_{i}+\partial_{i}v_{j}\right)
+\Div\:w\: \delta_{ij} -\p_iw_j-\p_j w_i.
\end{eqnarray*}
Similarly we can transform the Steklov problem. In terms of the x-coordinates it reads as
\begin{align}\label{eq:Steklov2}
L_A \phi(t) =0 \tx{in} \Omega, \quad \p_{\nu_A} \phi(t) = \mu(t) m(t) \phi(t) \tx{on} \p \Omega, \quad L_A:= \p_j(A_{ij} \p_i).
\end{align}
The next Lemma is well-known, s. for instance \cite[IV, Sec. 3.5]{Ka} or \cite[VI, Sec. 6]{CoHi65}.

\begin{lemma} Suppose that $\mu_p(\Omega)<\alpha <\mu_{p+1}(\Omega)$. Then there exists $t_0>0$ sich that
$\mu_p(\Omega_t)<\alpha  <\mu_{p+1}(\Omega_t)$ for all $t\in(-t_0,t_0)$..
\end{lemma}

{\bf Proof} By the min-max principle
$$
\mu_p(\Omega_t) =\rm{min}_{L_p}\rm{max}_{V\in L_p} \frac{\int_{\Omega}\nabla V\cdot A(t)\nabla V\:dx}{\oint_{\p\Omega}V^2 m(t)\:dS_x},
$$
where $L_p$ is an $n-$dimensional linear space in $W^{1,2}(\Omega)$. Since $|\nabla V|^2 (1-c_1|t|)<\nabla V\cdot A(t) \nabla V \leq |\nabla V|^2 (1+c_1|t|)$
and $V^2(1-c_2 |t|)<m(t)V^2 \leq V^2(1+c_2 |t|)$. From the min-max principle we obtain that $|\mu_p(\Omega)-\mu_p(\Omega_t)| \leq t C$
where $C$ depends on $v$ and $w$. \hfill $\square$.
\medskip

{\sl In the sequel se shall always assume that $\alpha$ does not coincide with an eigenvalue of $\Omega_t$ for all $t\in (-t_0,t_0)$.}

Suppose that $\Omega \in C^{2,\epsilon}$, $A_{ij}(t)\in C^{1,\epsilon}$, $J(t)\in C^{0,\epsilon}$ and $m(t)\in C^{1,\epsilon} $. We also assume that all the data are at least twice continuously differentiable in $t$. Then by Schauder's regularity theory \cite{GiTr83} it follows that $\tilde u(t) \to \tilde u(0)=:u(x)$ in $C^{2,\epsilon'}$, $\epsilon '<\epsilon$. Our assumptions imply that $|\partial \Omega_t| \to |\partial \Omega|$ which is crucial for the convergence of the eigenvalues.
A  general study of domain perturbations for elliptic problems with Robin boundary conditions is carried out by Dancer
and Daners in \cite{DaDa97}.

\subsubsection{First variation of the energy}

Consider problem \eqref{torsion} in a class of domains $\Omega$ described in \eqref{not1}. As before the solutions of \eqref{torsion} in $\Omega$ will be denoted by $u(x)$. We shall use the abbreviation ${\mathcal{E}}(t)$ for $E(\tilde u(t),\Omega_t)$. Under the conditions stated above the solution of \eqref{torsion} $\tilde u(t)=\tilde u(y,t)$ is continuous and continuously differentiable in
$t$.

\medskip

It was shown in \cite{BaWa14} that the first domain variation  $\frac{d}{dt}{\mathcal{E}}(t)\large |_{t=0}$ is given by
\begin{align*}
\dot{{\mathcal{E}}}(0)=\int_{\p \Omega} (v\cdot \nu)\left[\vert\nabla u\vert^2-2u-2\alpha^2 u^2-\alpha(n-1)u^2H\right]\:dS.
\end{align*}
{\sc Example}
\smallskip

If $\Omega=B_R$ then
\begin{align}\label{edotball}
\dot{\tilde{\mathcal{E}}}(0)=\left(\frac{(n+1)R}{\alpha n^2}-\frac{R^2}{n^2}\right)\int_{\p B_R} (v\cdot \nu)\:dS.
\end{align}

This leads to the following

\begin{corollary} \label{co:critical}Let $\Omega_t$ be a family of nearly spherical domains with prescribed volume $|\Omega_t|=|B_R|$.
Then $\dot{{\mathcal{E}}}(0)=0$.
\end{corollary}

{\bf Proof} From \eqref{jform} it follows that for volume preserving transformations
\begin{align}\label{volume1}
\oint_{\p \Omega} (v\cdot \nu)\:dS=0.
\end{align}
This together with \eqref{edotball} establishes the assertion. \hfill $\square$
\newline
\newline

A further consequence of \eqref{edotball} is the local monotonicity property.

\begin{corollary}
 If $0<\alpha R<n+1$ and $|\Omega_t|>|B_R|$ then $\dot{\cal{E}}(0)>0$, otherwise if $\alpha R>n+1$ then $\dot{\cal{E}}(0)<0$.
 \end{corollary}
 {\bf Proof} By our assumption we have $\int_{\p B_R}(v\cdot \nu)\:dS>0$.  The sign of  $\dot{\cal{E}}(0)$ depends in view of \eqref{edotball} on the sign of $(n+1)\alpha R -(\alpha R)^2$.
 \hfill  $\square$

\subsubsection{First variation of $\mathcal{J}(\Omega_t)$}

In the case $0<\alpha<\mu_2(\Omega)$ (see chapter 3.2) the energy $E(\tilde u,\Omega_t)$ is bounded from above by  ${\cal{J}}(\Omega)=T(\Omega) + \frac{|\Omega_t|^2}{\alpha |\p \Omega_t|}$.
Let ${\cal{S}}(t)=\vert\partial\Omega_t\vert$. If $|\Omega_t|=|\Omega|$ then the first variation is given by
\begin{eqnarray}
\label{var1}\dot{\J}(0)&=&\dot{T}(0)-\frac{\vert\Omega\vert^2}{\alpha\vert\partial\Omega\vert^2}\dot{{\cal{S}}}(0)
\end{eqnarray}
where
\begin{eqnarray}
\label{var2}\dot{T}(0)&=&-\int\limits_{\partial\Omega}Ê\vert\nabla s\vert^2(v\cdot\nu)\:dSÊ\\
\label{var3}\dot{{\cal{S}}}(0)&=&(n-1)Ê\int\limits_{\partial\Omega}Ê(v\cdot\nu)H\:dSÊ
\end{eqnarray}
Thus for all critical domains the solution $s$ of \eqref{seq} solves the additional boundary condition
\begin{eqnarray}\label{overdet}
\frac{\vert\Omega\vert^2}{\alpha\vert\partial\Omega\vert^2}(n-1)H+\vert\nabla s\vert^2=const.\quad\hbox{on}\:\partial\Omega.
\end{eqnarray}
This is a direct consequence of \eqref{volume1}. By Theorem 3 in \cite{Se71} concerning overdetermined boundary value problems, the ball is the only domain for which on $\p\Omega$, $s$ is constant and $|\nabla s|= c(H)$ for a non-increasing function $c$.
Consequently

\begin{lemma}\label{serrin}
For $\alpha>0$ the ball is the only critical domain for the functional
${\cal{J}}(\Omega)$ among all domains of equal volume.
\end{lemma}

\subsection{Second domain variation for nearly spherical domains}

\subsubsection{Second variation for the energy}

Corollary \ref{co:critical} gives rise to the following question: {\sl  is $E(u, B_R)$ a local extremum among
the family
 $\Omega_t$, $t\in(-t_0,t_0)$,  of perturbed domains with the same volume as $B_R$?} The answer will be obtained from the second variation.

Consider the family of nearly spherical domains $\Omega_t:=\{y=x+tv(x)+\frac{t^2}{2}w(x): x\in \overline{B_R}\}$. Let $\tilde{u}(t):=u(y(x),t)$ be the solution of $\Delta u+1=0$ in $\Omega_t$, $\p_\nu u=\alpha u$ on $\p\Omega_t$ transformed onto $\overline{\Omega}$.
If $\tilde{u}(t)$ is differentiable - this is the case when the data are H\"older continuous as described in the previous section and $\alpha \neq \mu_i(\Omega_t)$ for all $t\in (-t_0,t_0)$ - then
\begin{eqnarray*}
\frac{d}{dt}\tilde{u}(t)\vert_{t=0}=u'(x)+v\cdot\nabla u_0,
\end{eqnarray*}
where $u=u_0$ is the solution of \eqref{torsion} in $B_R$.

It was shown in \cite{BaWa14} that  the shape derivative $u'$ solves the inhomogeneous boundary value problem
\begin{eqnarray}
\label{uprtball1}\Delta u'&=&0\qquad\hbox{in}\:B_R\\
\label{uprtball2}\partial_{\nu}u'&=&\alpha u'+\left(\frac{1-\alpha\:R}{n}\right)v
\cdot\nu\qquad\hbox{on}\:\partial B_R.
\end{eqnarray}
Let us assume that such a solution $u'$ exists. This is certainly the case if $\alpha$ does not coincide with a Steklov eigenvalue $\mu_i(B_R)$.

For the next result we consider perturbations which, in addition to the condition \eqref{volume1}, satisfy the volume preservation of the second order, namely
\begin{align}\label{volume2}
\int_{B_R}((\Div v)^2-D_v : D_v+\Div\:w)\:dx=0.
\end{align}
This formula can be simplified if $v$ points into normal direction only. It takes the form
\begin{eqnarray}\label{vol2}
(n-1)
\int\limits_{\partial\Omega}H(v\cdot\nu)^2\:dS+\int\limits_{\partial\Omega}(w\cdot\nu)\:dS=0.
\end{eqnarray}
Set
\begin{eqnarray*}
Q(u'):= \int\limits_{B_R} |\nabla u'|^2\:dx -\alpha\: \int\limits_{\partial B_R} u'^2\:dS.
\end{eqnarray*}
The following formula was derived in \cite{BaWa14}. Remember that for nearly spherical domains $\mathcal{E}(0)= E(u, B_R)$. Moreover if $\alpha \neq \mu_i(B_R)$,  Lemma 2 implies that for $t$ sufficiently small $\alpha$ never coincides with an eigenvalue $\mu_j(\Omega_t)$.

\begin{lemma}\label{secvar}
Assume $\alpha \neq \mu_i(B_R)$ and let the volume preservation conditions \eqref{volume1} and \eqref{volume2} be satisfied. Put ${\cal{S}}(t):=|\p\Omega|$. Then
\begin{eqnarray}\label{Eradial}
\ddot{{\cal{E}}}(0)=-2 Q(u')
+
\frac{2R}{n^2}(1-\alpha R) \int_{\p B_R}(v\cdot \nu)^2\:dS
-
\frac{R^2}{\alpha n^2}\ddot{\cal{S}}(0) .
\end{eqnarray}
For a ball the second variation of the surface area is of the form
\begin{eqnarray*}
\ddot{\cal{S}}(0) =\oint_{\partial B_R}\left(|\nabla^{*} (v\cdot\nu)|^2 -\frac{(n-1)}{R^2} (v\cdot \nu)^2\right)\:dS,
\end{eqnarray*}
where $\nabla^{*}$ stands for the tangential gradient on $\p B_R$.
\end{lemma}

\subsubsection{Discussion of the sign of $\ddot{{\cal{E}}}(0)$}

We write for short
\begin{eqnarray}\label{feq}
{\cal{F}}:=-2Q(u')
+
\frac{2R}{n^2}(1-\alpha R) \int_{\p B_R}(v\cdot \nu)^2\:dS.
\end{eqnarray}
In order to estimate ${\cal{F}}$  we consider the Steklov eigenvalue problem \eqref{steklov}
An elementary computation yields $\mu_1=0$,  and $\mu_k =\frac{k-1}{R}$ (for $k\geq 2$ and counted without multiplicity). The second eigenvalue $ \mu_2=1/R$ has multiplicity $n$ and its  eigenfunctions are
$\frac{x_1}{R}, \hdots, \frac{x_n}{R}$.
\medskip

{\sl From now on we shall count the eigenvalues $\mu_i$ with their multiplicity, i.e. $\mu_2=\mu_3=\mu_{n+1}=1/R$ and $\mu_{n+2}=2/R$ etc.}

Let $\{\phi_i\}_{i\geq 1}$ be system of Steklov eigenfunctions introduced in Section 2. The function $u'$ solves \eqref{eq:harmonic} with $g=\left(\frac{1-\alpha\:R}{n}\right)v
\cdot\nu$. Hence by Lemma \ref{le:harmonic}
\begin{align*}
u'(x)=\sum_{i=1}^\infty c_i \phi_i \quad\tx{and} \quad(v\cdot \nu)=\sum_{i=1}^\infty b_i\phi_i.
\end{align*}
Note that since the first eigenfunction $\phi_1$ is a constant constant.
the condition
$$
0=\oint_{\p B_R}(v\cdot \nu)\:dS=\oint_{\p B_R} \phi_1 (v\cdot \nu)\:dS
$$
implies that $b_1=0$. From \eqref{uprtball2} we have also $c_1=0$.
The coefficients $b_i$ for  $i\geq 2$ are determined from the boundary value
problem \eqref{uprtball1}, \eqref{uprtball2}. In fact
\begin{eqnarray}\label{coeff}
b_i= \frac{n\:c_i\:(\mu_i-\alpha)}{1-\alpha\:R}\qquad\hbox{for} \quad i=2,3,\hdots .
\end{eqnarray}
From the orthonormality conditions \eqref{orthog} of the eigenfunctions it follows that
\begin{align*}
Q(u')&=\sum_{i=2}^\infty c_i^2(\mu_i-\alpha).
\end{align*}
Inserting this into \eqref{feq} we get
\begin{align*}
{\cal{F}}= 2\sum_2^\infty c_i^2\:(\mu_i-\alpha)^2\left[\frac{R}{1-\alpha\:R}-\frac{1}{\mu_i-\alpha}\right].
\end{align*}
Since $\mu_2=\cdots=\mu_{n+1}= \frac{1}{R}$ it follows that
\begin{eqnarray}\label{F}
{\cal{F}}&=& 2\sum_{n+2}^\infty c_i^2\:(\mu_i-\alpha)^2\left[\frac{R}{1-\alpha\:R}-\frac{1}{\mu_i-\alpha}\right]\\
\nonumber&=&2\frac{(1-\alpha R)^2}{n^2}\sum_{n+2}^\infty b_i^2\left[\frac{R}{1-\alpha\:R}-\frac{1}{\mu_i-\alpha}\right].
\end{eqnarray}
\smallskip

Next we shall discuss the sign of $\ddot{\cal{S}}(0)$. Observe that
$$
{\cal{R}}[\chi]=\frac{\oint_{\partial B_R}|\nabla^{*} \chi |^2\:dS}{\oint_{\p B_R} \chi^2\:dS}
$$
is the Rayleigh quotient of the Laplace- Beltrami operator on $\p B_R$. Its eigenvalues $\Lambda_{i_k}$ are $k(n-2+k)/R^2$,
$k\in \mathbb{N}^+$. Observe that the multiplicity of this eigenvalue is the same as for the Steklov eigenvalue corresponding to $k/R$. Remember that for volume preserving perturbations of the first order we  have
 $\oint_{\p B_R} (v\cdot \nu)\:dS=0$ and therefore $(v\cdot \nu)$ is orthogonal to the first eigenfunctionwhich is a constant. Thus
$$
{\cal{R}}[(v\cdot \nu)]\geq \frac{n-1}{R^2}.
$$
Equality holds if and only if $(v\cdot \nu)$ belongs to the eigenspace spanned by $\{\frac{x_i}{R}\}_{i=1}^{n}$. This does not occur if we exclude small translations. Consequently $\ddot{\cal{S}}(0)>0$. This is consistent  with the isoperimetric inequality.
\smallskip

If we replace in $\ddot{\cal{S}}(0)$, $(v\cdot \nu)$ by
$\sum_1^\infty b_i\phi_i$ we obtain
\begin{align}\label{ddS}
\ddot{\cal{S}}(0)= \sum_2^\infty b_i^2(\Lambda_i- \frac{n-1}{R^2}).
\end{align}
From \eqref{F} and \eqref{ddS} we then get
\begin{align}\label{dee}
\ddot{\mathcal{E}}(0) = \sum_{n+2}^\infty\frac{b_i^2}{\alpha n^2}\underbrace{\left\{2(1-\alpha R)^2(\frac{\alpha R}{1-\alpha R}-\frac{\alpha}{\mu_i-\alpha})-R^2\Lambda_i+n-1)\right\}}_{d_i}.
\end{align}
Since the multiplicity of the Steklov eigenvalues and $\Lambda_i$ depending on $k$ is the same we can replace $\mu_i$ by $k_i/R$ for a suitable integer $k_i$ and $\Lambda_i$ by $k_i(k_i+n-2)/R^2$.
Consequently
\begin{align}\label{d}
d_i=\frac{2\xi(1-\xi)(k_i-1)}{k_i-\xi}-k_i(k_i+n-2)+n-1,
\end{align}
where $ \xi:=\alpha R$ and $k_i=2,3,4 ...$.
\smallskip

Next we shall discuss the sign of $d_i$. Suppose that $k_p<\xi<k_{p+1}$, $k_p\geq 2$. It is easy to see that
\begin{align*}
d_i<0   \tx{if} k_ i>k_p \tx{and} i\geq n+2 ,
\end{align*}
If  $\xi =k_p+\epsilon$, $(0<\epsilon<1)$ and $k_p\geq 2$,  then by \eqref{d}
$$
d_p= \frac{2(k_p+\epsilon)(k_p+\epsilon -1)(k_p-1)}{\epsilon} -k_p(k_p+n-2)+n-1.
$$
For given $k_p\geq 2$ and $n$ we can always find $0<\epsilon$ sufficiently small such that $d_p>0$.
Observe that for $k_p\geq 2$,
$d_p-k_p(k_p+n-2)+n-1$ is a monotone decreasing function of $\epsilon$.
 A lower bound is obtained for $\epsilon=1$, namely
$$
d_p > 2k_p^3-k_p^2-nk_p +n-1.
$$
For $n=2,3,4$ this expression is positive. However in general the sign varies.
\smallskip

\noindent If $k_p=0$, i.e. $0<\xi <1$ or $k_p=1$, i.e. $1<\xi<2$, no positive terms appear in the expression
of $\ddot{\mathcal{E}}(0)$.
\smallskip

\noindent The same situation as for $d_p$ holds for $ d_i<d_p$. Since
$d_i >k_i^2 +(4-n)k_i +n-1$ we have $d_i>0$ if $n\leq 4$.

$k_i<k_p$ the sign of $d_i$ depends on $k_i$ and $n$. If $k_i$ is large compared to $n$ it is positive, otherwise negative.

These observations are summarized in the following

\begin{lemma}\label{thm1}
(i) Let $0<\alpha R<2$, $\alpha \neq \frac{1}{R}$. Then $\ddot{{\cal{E}}}(0)\leq 0$. Equality holds if and only if $b_i =0$ for all $i\geq n+2$.
\smallskip

(ii) If $k_p<\alpha R<k_{p+1}$, $k_p\geq 2$, then $\ddot{{\cal{E}}}(0)\leq 0$ if $b_i=0$ for $i=n+2,..,k_{p-1}$.
\medskip

(iii) Assume $\alpha R=k_p+\epsilon$, $k_p\geq 2$, $\epsilon\in (0,1)$. Then for every $n$ there exists $\epsilon$ sufficiently small such that $\ddot{{\cal{E}}}(0)\geq 0$ for $b_p\neq 0$ and $b_i=0$ for $i=n+2,..,p-1$ and $b_i=0$ for $i>p$.
\medskip

(iv) If $n\leq 4$ and $k_p<\alpha R<k_{p+1}$ then $\ddot{{\cal{E}}}(0)\geq 0$ if $b_i=0$ for all $i>p$ and arbitrary
$b_i$, $i\leq p$, and $\ddot{{\cal{E}}}(0)\leq 0$ if $b_i=0$ for $i\leq p$ and arbitrary $b_i$, $i>p$.
\end{lemma}

{\sc Example}
\medskip

Let $\Omega\subset \mathbb{R}^2$ be the ellipse whose boundary $\partial \Omega$ is given by
$$
\left\{\frac{R\cos(\theta)}{1+t},(1+t)R\sin(\theta)\right\},
$$
where $(r,\theta)$
 are the polar coordinates in the plane. This ellipse has the same area as the circle $B_R$ and can be interpreted as a perturbation described in \eqref{not1}. We have $y= x+t(-x_1,x_2) + \frac{t^2}{2}(x_1,0) +o(t^2)$. The eigenvalues and eigenfunctions of the Steklov eigenvalue problem \eqref{steklov} in $B_R$ are
$$
\mu = \frac{k}{R} \tx{and} \phi=r^k\begin{cases}
a\cos(k\theta) \\
a\sin(k\theta)
\end{cases}
$$
where $a=\frac{1}{\sqrt{\pi R^{2k+1}}}$ is the normalization constant.
We have
$$
(v\cdot \nu)=-R\cos(2\theta)=b_4\phi_3,
$$
and
\begin{eqnarray*}
 \ddot{S}(0)=\oint_{\partial B_R}\left(|\nabla^{*} (v\cdot \nu)|^2 -\frac{(v\cdot \nu)^2}{R^2}\right)\:dS=3\pi R.
\end{eqnarray*}
A straightforward computation
yields
\begin{align*}
\ddot{\mathcal{E}}(0)= \left[ -\frac{3}{4\alpha} +\frac{R(1-\alpha R)}{2(2-\alpha R)}\right]\oint_{\partial B_R}(v\cdot\nu)^2\:dS.
\end{align*}
with
$\oint_{\partial B_R}(v\cdot\nu)^2\:dS=\pi R^3$.
From this expression it follows immediately that
\begin{align*}
\ddot{\mathcal{E}}(0)
\begin{cases}
>0 \tx{if} \alpha R>2\\
<0 \tx{if} \alpha R<2.
\end{cases}
\end{align*}
This result is in accordance with Lemma \ref{thm1} (i). In this example $b_i=0$ for all $i\neq 4$. The sign of $\ddot{\mathcal{E}}(0)$ depends therefore on $d_4$. It changes sign at $\alpha R=2$
\bigskip

As we have already mentioned $b_1=0$ for all volume preserving perturbations. The coefficients $b_2,..,b_{n+1}$ belong all to the Steklov eigenvalue $\mu_2=\dots=\mu_{n+1}=1/R$ and give no contribution to $\ddot{\mathcal{E}}(0)$. This is due to the fact that on $\partial B_R$
$$
\sum_2^{n+1} b_i\phi_i =\sum_1^n b_{i+1} c x_i= \vec{b}\cdot \nu,
$$
where $\vec{b}$ is a constant vector. The presence of $b_i$ for $i=2,..,n+1$ means that the perturbed domain
$\Omega$ has been shifted by a vector $t\vec{b}$. Notice that such a shift does not affect the higher coefficients
$b_k$, $k\geq n+2$. Obviously it  leaves the energy invariant. There is therefore no loss in generality to assume that
\begin{align}\label{ barycenter}
b_2=b_3=\dots=b_{n+1}=0.
\end{align}
This condition also implies that $c_2=\dots =c_{n+1}=0$. Hence Problem \eqref{uprtball1}, \eqref{uprtball2} is solvable for $\alpha R=1$. This observation together with \eqref{dee} implies that for perturbations which are not pure translations or rotations the following result holds true.

\begin{theorem}
\begin{enumerate}
\item Assume $0<\alpha R<1$. Then
$$
\ddot{\mathcal{E}}(0)\leq - \frac{n-0.5}{\alpha n^2} \oint_{\partial B_R} (v\cdot \nu)^2\:dS<0.
$$
\item Assume $1<\alpha R<2$. Then
$$
\ddot{\mathcal{E}}(0)\leq\frac{1}{\alpha n^2} \left(\frac{2\alpha R(1-\alpha R)}{2-\alpha R}-n-1\right)\oint_{\partial B_R} (v\cdot \nu)^2\:dS<0.
$$
\end{enumerate}
In both cases the energy is  maximal for the ball among all nearly spherical domains of given volume.
\end{theorem}

 In general if $\alpha R>2$ the energy $\mathcal{E}(t)$ has a saddle in $t=0$.

\begin{theorem}
Assume $n=2,3,4$ and $k_p<\alpha R<k_{p+1}$. Let $\mathcal{L}_p$ be the linear space generated by the eigenfunctions $\phi_i$ belonging to the eigenvalues $\mu_i=1/R,..,k_p/R$ and $\mathcal{L}_p^\perp$ be its complement generated by $\phi_i$ belonging to the remaining eigenvalues $\mu_i= k_{p+1}/R,..., \infty$. Then
$$
\ddot{\mathcal{E}}(0)\begin{cases}
>0 \tx{if} (v\cdot \nu) \in \mathcal{L}_p\\
<0 \tx{if} (v\cdot \nu) \in \mathcal{L}_p^\perp.
\end{cases}
$$
 \end{theorem}

\subsubsection{The second variation of ${\J}(\Omega_t)$}

As for $\ddot{\E}(0)$ we can derive a formula for the second volume preserving domain variation for the functional $\J$. Applying the rules of differentiation we get
\begin{eqnarray}
\label{var4}\ddot{\J}(0)&=& Ê\ddot{\T}(0)+\frac{2\vert\Omega\vert^2}{\alpha{\cal{S}}(\partial\Omega)^3}\dot{{\cal{S}}}^2(0)-\frac{\vert\Omega\vert^2}{\alpha\vert\partial\Omega\vert^2}Ê\ddot{{\cal{S}}}(0)
\end{eqnarray}
In analogy to formulas \eqref{var1} - \eqref{var3} and with the help of \eqref{vol2}
we get
\begin{eqnarray}\label{var5}\ddot{\T}(0)
&=Ê
\int\limits_{\partial\Omega}\vert\nabla s\vert^2
\left((n-1)H(v\cdot\nu)^2-(w\cdot\nu)\right)
\:dSÊ\\
\nonumber &+
2\int\limits_{\Omega}\vert\nabla s'\vert^2\:dx +2\int\limits_{\partial\Omega}(v\cdot\nu)^2\partial_{\nu}s\:dS,
\end{eqnarray}
where  the shape derivative $s'$ satisfies
\begin{eqnarray}\label{shapes}
\Delta s'=0\quad\hbox{in}\:\Omega,\qquad\qquad s'=-v\cdot\nabla s=v\cdot\nu\vert\nabla s\vert\quad\hbox{in}\:\partial\Omega.
\end{eqnarray}
Moreover by formula (2.20) in \cite{BaWa14}
\begin{eqnarray}
\label{var6}\ddot {S}(0)&=&\int\limits_{\partial\Omega}\vert\nabla^{*}(v\cdot\nu)\vert^2\:dS-
\int\limits_{\partial\Omega}\left(\vert A\vert^2-(n-1)^2H^2\right)(v\cdot\nu)^2\:dS\\
\nonumber&&+
(n-1)\int\limits_{\partial\Omega}Ê(w\cdot\nu)HÊ\:dS,
\end{eqnarray}
where
\begin{eqnarray*}
\vert A\vert^2=\sum_{i,j}^{n-1}(\partial_{i}^{*}\nu\cdot x_{\xi_k})(\partial_{k}^{*}\nu\cdot x_{\xi_i}).
\end{eqnarray*}
denotes the socond fundamental form of $\partial\Omega$.
\newline
\newline
From Section 4.1.3 we know that the ball is the only critical point of $\J$. For the ball $B_R$ we have
\begin{eqnarray*}
\dot{S}(0)&=&0,\\
\ddot{S}(0)&=&\oint_{\partial B_R}\left(|\nabla^{*} (v\cdot \nu)|^2 -\frac{n-1}{R^2}(v\cdot \nu)^2\right)\:dS\geq 0,
\end{eqnarray*}
and
\begin{eqnarray*}
s(x)=\frac{1}{2n}\left(R^2-\vert x\vert^2\right).
\end{eqnarray*}
If $R$ is chosen such that $\vert\Omega_t\vert=\vert B_R\vert$ for all $t\in (-t_0.t_0)$  we get
\begin{eqnarray*}
\ddot{\J}(0)&=&\frac{R^2}{n^2}\int\limits_{\partial B_R}
\left((n-1)H(v\cdot\nu)^2-(w\cdot\nu)\right)
\:dSÊ
+
2\int\limits_{B_R}\vert\nabla s'\vert^2\:dx
-
\frac{2R}{n}\int\limits_{\partial B_R}(v\cdot\nu)^2\:dS\\
&&-
\frac{R^2}{\alpha\: n^2}Ê\:\ddot{{\cal{S}}}(0).
\end{eqnarray*}
The volume constraint \eqref{vol2} then implies
\begin{eqnarray*}
\ddot{\J}(0)=-2\frac{R}{n^2}\int\limits_{\partial B_R}(v\cdot\nu)^2
\:dSÊ
+
2\int\limits_{B_R}\vert\nabla s'\vert^2\:dx
-
\frac{R^2}{\alpha\: n^2}Ê\:\ddot{{\cal{S}}}(0).
\end{eqnarray*}
If we use \eqref{shapes} to eliminate $(v \cdot \nu)$ we can write $\ddot{\J}(0)$ as a functional in $s'$ alone.
\begin{eqnarray*}
\ddot{\J}(0)=\I(s')
:=
2\int\limits_{B_R}\vert\nabla s'\vert^2\:dx
-
\frac{2}{R}\int\limits_{\partial B_R}s'^2\:dS
-
\frac{1}{\alpha}\int\limits_{\partial B_R}\left(\vert\nabla^{*} s'\vert^2 -\frac{n-1}{R^2}s'^2\right)\:dS.
\end{eqnarray*}

\subsubsection{Sign of $\ddot{\J}(0)$}

We like to find the sign of $\I$. For the ball it follows from the volume constraint  that $\oint_{\p B_R}
s' \:dS=0$. Hence
$$
\int_{B_R} |\nabla s'|^2\:dx \geq \mu_2(B_R) \oint_{\p B_R} s'^2\:dS.
$$
Since $\mu_2=1/R$ we get the lower estimate
\begin{eqnarray}\label{lowbd}
\I(s')\geq-\frac{1}{\alpha}\ddot{S}(0).
\end{eqnarray}
Keeping in mind that $s'$ is harmonic we get
\begin{eqnarray}\label{ests'}
\int_{B_R} |\nabla s'|^2 \:dx =\oint_{\p B_R} s'\partial_{\nu} s'\:dS \leq \frac{1}{2R} \oint_{\p B_R} s'^2\:dS + \frac{R}{2} \oint_{\p B_R}(\p_\nu s')^2\:dS.
\end{eqnarray}
Next we multiply $-\Delta s=1$ with $x\cdot\nabla s$ and integrate over $\Omega$. Since $s=0$ on $\partial\Omega$ this gives
\begin{eqnarray*}
\int\limits_{B_R}\vert\nabla s'\vert^2\:dx=
\frac{R}{n-2}\int\limits_{\partial B_R}\vert\nabla^{*} s'\vert^2\:dS
-
\frac{R}{n-2}\int\limits_{\partial B_R}(\partial_{\nu} s')^2\:dS
\end{eqnarray*}
If we put this together with the estimate \eqref{ests'} we get
\begin{align*}
\int\limits_{B_R}\vert\nabla s'\vert^2\:dx \leq \frac{1}{nR}\oint_{\p B_R} s'^2\:dS +\frac{R}{2}\oint_{\p B_R} |\nabla ^*s'|^2\:dS.
\end{align*}
This results in the following upper bound.
$$
\mathcal{I}(s') \leq \left( \frac{2R}{n} -\frac{1}{\alpha}\right) \ddot{S}(0).
$$
Thus we have proved

\begin{lemma}\label{maximizer}
For $\alpha<\frac{n}{2R}$ the ball is a local maximizer of $\J(\Omega)$ among nearly circular domains of equal volume.
\end{lemma}

\subsection{Optimality of the ball in two dimensions}
From Proposition \ref{le:comp1} it follows that among all domains of given area $A:=\vert B_R\vert$ the functional $\mathcal{J}(\Omega)$ is smaller than the corresponding expression for the circle provided $\alpha <\alpha_0$ where
$$
\alpha_0=\frac{A^2}{T(\Omega)-T(B_R)}\left(|\p B_R|^{-1}-L^{-1}\right) \tx{where} L=|\p \Omega|.
$$
If we replace $T(\Omega)$ by an upper bound $T^{*}$ then
$$
\alpha_0\geq \frac{A^2}{T^{*}-T(B_R)}\left(|\p B_R|^{-1}-L^{-1}\right).
$$
Observe that
$$
T(B_R)=-\frac{A^2}{8\pi}\tx{and} |\p B_R|=\sqrt{4\pi A}.
$$
We are interested in estimates for $T(\Omega)$ which depend only on $L$ and $A$.
in \cite{PaWe61} Payne and Weinberger derived by means of the method of parallel lines such an inequality.
\medskip

Let us introduce the notation
\begin{align*}
2\pi \tilde R:=L, \quad A=:\pi(\tilde R^2- \tilde r^2) \tx{and} \tilde r=y\tilde R.
\end{align*}
Then
$$
y^2=1-\frac{4\pi A}{L^2}, \quad L^2=\frac{4 \pi A}{1-y^2}=\frac{4\pi^2 R^2}{1-y^2} \tx{and} \tilde R^2=\frac{A}{\pi(1-y^2)}=\frac{R^2}{1-y^2}.
$$
Payne-Weinberger's inequality says that
$$
T(\Omega)\leq \frac{\pi}{2}\left(\tilde r^4\log \frac{\tilde r}{\tilde R} -\frac{3}{4}\tilde r^4 +\tilde R^2\tilde r^2-\frac{\tilde R^4}{4}\right).
$$
The expression at the right-hand side is the energy corresponding to the boundary value problem $\Delta U+1=0$ in $B_{\tilde R}\setminus B_{\tilde r}$ with $U=0$ on $\p B_{\tilde R}$ and $\p_\nu U=0$ on $\p B_{\tilde r}$. Consequently equality holds for the disc.

This inequality implies that
\begin{align*}
\epsilon_0= T(\Omega)-T(B_R)\leq  \frac{\pi}{4}\tilde R^4y^2\left[1+y^2\log y^2 -y^2\right]=\frac{\pi R^4}{4(1-y^2)^2}y^2\left[1+y^2\log y^2 -y^2\right].
\end{align*}
Moreover
$$
|\p B_R|^{-1}- L^{-1}= \frac{1}{\sqrt{4\pi A}}(1-\frac{\sqrt{4\pi A}}{L})=  \frac{y^2}{\sqrt{4\pi A}\left(1+\frac{\sqrt{4\pi A}}{L}\right)}=\frac{y^2}{2\pi R(1+\sqrt{1-y^2})}.
$$
Collecting all the terms we obtain the estimate
\begin{align}\label{A0}
\alpha_0\geq \frac{2(1-y^2)^2}{R(1+\sqrt{1-y^2})(1+y^2\log y^2 -y^2)}=:\frac{2}{R}g(y^2).
\end{align}
The function $g(t)$ is monotone increasing for $t\in (0,1)$ with $\lim_{t\to 1}g(t)= 2$ and $g(0)=1/2$. The number $y^2$ measures the defect of $\Omega$ with respect to the circle. The estimate \eqref{A0} together with the monotonicity of $g$ implies

\begin{theorem} \label{J}
(i) Let $\Omega\subset \mathbb{R}^2$ be a domain with fixed area $A$ and let $B_R$ be a disc with the same area. Then
$$
\mathcal{J}(\Omega)\leq \mathcal{J}(B_R) \tx{for all} \alpha \leq \frac{2}{R}g(y^2).
$$
(ii) In particular $\mathcal{J}(\Omega)$ is smaller than the corresponding quantity for the disc if $\alpha\leq 1/R$.
\end{theorem}
Observe that the second statement is consistent with Lemma \ref{maximizer}. As a consequence we have

\begin{corollary}\label{last} Under the same assumptions we have $E(\tilde u,\Omega)$ achieves its maximum for the disc provided $\alpha <\mu_2(\Omega)$.
\end{corollary}

{\bf Proof} From Lemma \ref{le:E} and Theorem \ref{J} (ii) it follows that $E(\tilde u,\Omega) <\min\{\frac{1}{R},\mu_2(\Omega)\}$.
Note that by Weinstock's result $\mu_2(\Omega)\leq \mu_2(B_\rho)\leq\frac{1}{R}$, thus $\min\{\mu_2(\Omega), \frac{1}{R}\}=\mu_2(\Omega)$. \hfill $\square$
\medskip

{\bf Open problem} In order to extend Corollary \ref{last} a generalization of Payne-Weinberger's inequality to higher dimensions would be helpful. This inequality is based on estimates for the length of parallel curves which to our knowledge are not available in higher dimensions.
\bigskip

{\bf Acknowledgement} This paper was initiated during a visit at the Newton Institute in Cambridge. Both authors would like to thank this Institute for the excellent working atmosphere.


text:
\begin{document}
\title{On the deformation groupoid of the inhomogeneous pseudo-differential Calculus}
\author{Omar Mohsen}
\date{}
\maketitle
\abstract{In 1974, Folland and Stein constructed an inhomogeneous pseudo-differential calculus based on analysis on the Heisenberg group. This Heisenberg calculus was generalized by several authors, to any subbundle of the tangent bundle.

van Erp and Yuncken, following Debord and Skandalis showed that this calculus can be recovered using a deformation groupoid \textit{alla} tangent groupoid of Connes.

Using functoriality of the deformation to the normal cone construction, we give an elementary construction of this groupoid. We then extend it to the general case of a filtration of the tangent bundle by an iterated deformation.}
\section*{Introduction} In order to construct a parametrix for Hörmander's \cite{MR0222474} subelliptic operators on a contact manifold, Folland and Stein \cite{MR0344699,MR0367477} defined a noncommutative pseudo-differential calculus where the principal cosymbol is a function on a bundle of Heisenberg groups. A fundamental characteristic of this pseudo-differential calculus is that a vector field defines a differential operator of order $1$ if it is everywhere tangent to the contact subbundle and of order $2$ if not. Later on, this was generalised to an arbitrary subbundle of the tangent bundle, and even further to a filtration of the tangent bundle under conditions on the Lie bracket (see \cite{MR0493005,MR0370271,MR660648,MR973272,MR1185817,MR764508,MR0423432,MR953082,MR0442149,MR0436223}) To such a structure one associates a bundle of graded nilpotent Lie groups over which the cosymbols are functions. Let us remark that the general situation is more involved because the bundle of graded nilpotent Lie groups doesn't need to be locally trivial and hence the analogue of the theorem of Darboux doesn't hold in general.

This calculus was later used by many authors, for instance by Connes and  Moscovici \cite{MR1657389,MR1334867} to define a transversal signature operator on foliated manifolds and do computations in cyclic cohomology, following a construction of Hilsum and Skandalis \cite{MR925720}, by Julg and Kasparov \cite{MR1332908} to compute the $SU(n,1)$ equivariant $\KK$-theory following the work of Rumin  \cite{MR1267892}.

In \cite{MR3187645}, Debord and Skandalis showed how to recover the classical pseudo-differential calculus thanks to the tangent groupoid. In \cite{MR2247877,MR2680395}, Ponge and van-Erp independentaly define a deformation groupoid for a contact manifold. van Erp and Yuncken \cite{Erp:2015aa} used this groupoid to give an alternate presentation of the pseudo-differential calculus mentioned above. This groupoid was also used by van Erp \cite{MR2680396,MR2680395} and later (with Baum \cite{MR3261009}) to formulate and prove an index formula in the same spirit as that of Atiyah-Singer. Their index theorem is for differential operators whose cosymbol is invertible in the above calculus associated to a contact structure. These operators are necessarily hypoelliptic, hence their analytic index is well defined but they are rarely elliptic.

The groupoid defined by Ponge and van-Erp was later extended by Choi and Ponge \cite{Choi:2015aa,Choi:2017aa,Ponge:2017aa}, and independently by van Erp and Yuncken \cite{Erp:2016aa} following work by Julg and van Erp \cite{MR3694098}. This extension was also used by van Erp \cite{MR2746652} to formulate and prove an index theorem for hypoelliptic operators on foliated manifolds.

In this article, we prove that the deformation groupoids defined in \cite{MR2247877,MR2680395,MR2417549,Choi:2015aa,Choi:2017aa,Ponge:2017aa,Erp:2016aa} are special cases of the deformation to the normal cone construction. Let $H\subseteq TM$ be a vector bundle. Recall the tangent groupoid $$\dnc(M\times M,M)=M\times M\times \R^*\coprod TM\times \{0\}\rightrightarrows M\times \R$$ defined by Connes. The space $H\times \{0\}\subseteq TM\times \{0\}\subseteq\dnc(M\times M,M)$ is a Lie subgroupoid. Hence by the naturality of the $\dnc$ construction, the space $$\dnc(\dnc(M\times M,M),H\times\{0\})\rightrightarrows \dnc(M\times \R,M\times \{0\})=M\times \R^2$$ is a Lie groupoid. We prove that the fiber over $M\times \{1\}\times \R$ is the Heisenberg Lie groupoid. Furthermore the groupoid  $\dnc(\dnc(M\times M,M),H\times\{0\})\rightrightarrows M\times \R^2$ is a quite natural object to study because it contains `the deformations in all the directions'. In \Cref{Heisenberg groupoid sect}, we show that the general case (replacing $H$ by a filtartion of $TM$) is just an iterated deformation to the normal cone construction. Our approach gives rise to noncommutative Lie groupoids/symbols precisely because we deform Lie groupoids with respect to subgroupoids and not with respect to spaces, and contrary to the methods used in \cite{MR2247877,MR2680395,MR2417549,Choi:2015aa,Choi:2017aa,Ponge:2017aa,Erp:2016aa} no analysis on local coordinates is needed to construct the Lie groupoid, only functoriality of the $\dnc$ construction.

The methods developed here can be used to give a variety of examples of Lie groupoids. In particular we extend the Heisenberg Lie groupoid to cover the case of transverse (to a foliation) hypoelliptic pseudo-differential calculus without any difficulty (examples \ref{example H foliation single} and \ref{examp foliation multipleHeisen}).

\bigskip

This article is organised as follows;  in \Cref{dnc section}, we recall the notion of the deformation to the normal cone following \cite{Debord:2017aa}. \Cref{prop exact seq} is  also proved which will be used in \Cref{M single H examp}, to prove that our construction gives the Lie groupoid defined in \cite{MR2417549}. Quotient of Lie groupoids is also recalled which will be used in \Cref{Heisenberg groupoid sect}.

 In \Cref{Dnc iterated sect}, the iterated deformation to the normal cone construction is introduced.

In \Cref{M single H examp}, we prove that the the fiber over $M\times \{1\}\times\R$ of the Lie groupoid $$\dnc(\dnc(M\times M,M),H\times\{0\})\rightrightarrows M\times \R^2$$ have the same features as the Heisenberg groupoid.

  In \Cref{sect H bundle 1 case}, an alternate description (in a more general situation) of the horizontal fiber regarded in \Cref{M single H examp} is given, also local charts of our space are given which shows that our space coincides with the space in  \cite{Choi:2015aa,Choi:2017aa,Ponge:2017aa} as a Lie groupoid.

  In \Cref{Heisenberg groupoid sect}, we generalize the construction given in \Cref{M single H examp} but for a filtration of the tangent bundle proving that iterated deformation to the normal cone gives rise to the Carnot groupoid\footnote{Following Ponge's recommendation, the deformation groupoid in the case of a filtration of $TM$ should be called the Carnot groupoid.} in the general case. This section is independent of \Cref{M single H examp} and provides another proof of \Cref{main thm H single}. Finally in \Cref{examp Higson}, we show that the deformation constructed in \cite{Sadegh:2016aa} is just the quotient (see \Cref{examp quotient dnc}) of Carnot Lie groupoids.
\section*{Acknowledgments}
I wish to express my gratitude to my thesis director, G. Skandalis, for his support and his numerous insights and remarks into this work. I would also like to thank Prof. Higson and Prof. Ponge for their kind remarks and recommendations. This work was supported by grants from Région Ile-de-France.

\section{Deformation to the normal cone}\label{dnc section}
In this section, we recall the deformation to the normal cone construction following \cite{Debord:2017aa}. \textit{The deformation to the normal cone}\index{deformation to the normal cone} of a manifold $M$ along an immersed submanifold $V$ is a manifold whose underlying set is $$\dnc(M,V):=M\times \R^*\sqcup N^M_V\times\{0\},$$ where $N^M_V$ is the normal bundle of $V$ inside $M$. The smooth structure is defined by covering the manifold with two sets; the first is $M\times \R^*$ and the second is $\phi(N^M_V)\times\R^*\sqcup N^M_V\times\{0\}$ where $\phi:N^M_V\to M$ is a tubular embedding. To simplify the exposition, we will always assume that tubular neighbourhoods are diffeomorphisms on $N^M_V$. In the case of immersed manifolds, the tubular neighbourhoods are only local in $V$.. The smooth structure on $\phi(N^M_V)\times\R^*\sqcup N^M_V\times\{0\}$ is given by declaring the following map a diffeomorphism \begin{align*}
&\tilde{\phi}:N^M_V\times \R\to \phi(N^M_V)\times\R^*\sqcup N^M_V\times\{0\}\\
 &\tilde{\phi}(x,X,t)=(\phi(x,tX),t)\in M\times \R^*,\quad t\neq 0\\
 &\tilde{\phi}(x,X,0)=(x,X,0)\in N^M_V\times\{0\}.
\end{align*}
\begin{prop}[cf. chapter IV from \cite{MR1074006}]\label{smooth structure dnc construction} The above charts are compatible and the smooth structure is independent of $\phi$.
\end{prop}
Compatibility is clear. Independence of $\phi$ follows by noticing that the following functions are smooth functions that generate the smooth structure:\begin{enumerate}
\item the function\begin{align*}
(\pi_M,\pi_\R):\dnc(M,V)&\to M\times \R\\
(x,t)&\to (x,t),\quad t\neq 0\\
(x,X,0)&\to (x,0)
\end{align*}
\item Let $f\in C^\infty(M)$ be a smooth function which vanishes on $V$. Therefore $df:N^M_V\to \R$ is well defined. The following function is smooth
\begin{align*}
\dnc(f):\dnc(M,V)&\to \R\\
(x,t)&\to \frac{f(x)}{t},\quad t\neq 0\\
(x,X,0)&\to df_x(X)
\end{align*}
\end{enumerate}

The group $\R^*$ acts smoothly on $\dnc(M,V)$. The action is given by $\lambda_{u}(x,t)=(x,ut)$ and $\lambda_u(x,X,0)=(x,\frac{X}{u},0)$ for $u\in \R^*$.

\bigskip

\begin{prop}[Functoriality of DNC]\label{Functoriality of DNC}
Let  $M,M'$ be smooth manifolds, $V\subseteq M$, $V'\subseteq M'$ submanifolds, $f:M\to M'$ a smooth map such that $f(V)\subseteq V'$. Then the map defined by
\begin{align*}
\dnc(M,V)&\to \dnc(M',V')\\
(x,t)&\to (f(x),t),\quad t\neq 0\\
(x,X,0)&\to (f(x),df_x(X),0)
\end{align*}
is a smooth map\footnote{In the case where $V'$ is an immersed submanifold, one must also suppose that $f_{|V}:V\to V'$ is continuous.} that will be denoted by $\dnc(f)$.
 Furthermore the map  $\dnc(f)$ is \begin{itemize}
\item  a submersion if and only if $f$ is a submersion and $f_{|V}:V\to V'$ is also a submersion.
 \item an immersion if and only if $f$ is an immersion and for every $v\in V$, $T_vV=df_v^{-1}(TV').$
 \end{itemize}
\end{prop}
\begin{proof}
Smoothness of $\dnc(f)$ follows from the description of smooth maps given above. For statements concerning submersions and immersions. Let $U\subseteq \dnc(M,V)$ be the set where the differential of $\dnc(f)$ is onto (respectively injective). It is clear that $U$ is an open set that is invariant under the $\R^*$ action and contains $M\times\R^*$. To prove that $U=\dnc(M,V)$, it suffices to prove that $V\times \{0\}\subseteq U.$ If $v\in V$, then one sees directly that $$T_{(v,0)}\dnc(M,V)=\R\oplus T_vV\oplus T_vM/T_vV$$ The differential of $\dnc(f)$ is then $d\dnc(f)_{(v,0)}(t,X,Y)=(t,df_v(X),df_v(Y)).$ The proposition is then clear.
\end{proof}
The map \begin{align*}
 N^M_V\to N^{M'}_{V'},\quad (x,X)&\to (f(x),df_x(X))
\end{align*} will be denoted by $Nf$.

\begin{rem}\label{action dnc}It follows from \Cref{Functoriality of DNC}, that if $G$ is a Lie group acting smoothly on a manifold $M$ that leaves a submanifold $V$ invariant, then $G$ acts smoothly on $\dnc(M,V)$. This action commutes with the $\R^*$ action $\lambda$. In particular the group $G\times \R^*$ acts on $\dnc(M,V).$
\end{rem}
\begin{prop}\label{DNC fibered product}
Let $M_1, M_2, M$ be manifolds, $V_i\subseteq M_i,V\subseteq M$ submanifolds, $f_i:M_i\to M$ smooth maps such that \begin{enumerate}
\item $f_i(V_i)\subseteq V$ for $i\in \{1,2\}$
\item the maps $f_i$ are transverse
\item the maps $f_{i}|V_i:V_i\to V$ are transverse
\end{enumerate} Then \begin{enumerate}
\item\begin{enumerate}[label=(\alph*)]
\item the maps $Nf_i:N^{M_i}_{V_i}\to N^M_V$ are transverse.
\item the natural map $$N^{M_1\times_MM_2}_{V_1\times_VV_2}\to N^{M_1}_{V_1}\times_{N^{M}_{V}}N^{M_1}_{V_2}$$ is a diffeomorphism.

\end{enumerate}
Similarly for $\dnc$, we have \item\begin{enumerate}[label=(\alph*)]
 \item the maps $\dnc(f_i):\dnc(M_i,V_i)\to \dnc(M,V)$ are transverse.
 \item the natural map $$\dnc(M_1\times_MM_2,V_1\times_VV_2)\to \dnc(M_1,V_1)\times_{\dnc(M,V)}\dnc(M_2,V_2)$$ is a diffeomorphism.
\end{enumerate}

\end{enumerate}
\end{prop}
\begin{proof}
Statements $1.$ $(a)$ and $1.$ $(b)$ are clear. Bijectivity of the natural map $\dnc(M_1\times_MM_2,V_1\times_VV_2)\to \dnc(M_1,V_1)\times_{\dnc(M,V)}\dnc(M_2,V_2)$ is clear as well. To prove that it is a diffeomorphism and that the maps $\dnc(f_i)$ are transverse, we use the same argument as in \Cref{Functoriality of DNC}. The two conditions are open conditions which are $\R^*$-invariant. Hence it suffices to check that them at $V_1\times_V V_2$ which follows directly from $1.$ $(a)$ and $1.$ $(b)$.
\end{proof}

\begin{theorem}\label{Construction of dnc grouopids}Let $G$ be a Lie groupoid, $H$ a Lie subgroupoid. Then
\begin{enumerate}
\item the space $N^G_H\rightrightarrows N^{G^0}_{H^0}$ is a Lie groupoid whose structure maps are $Ns$, $Nr$ and whose Lie algebroid is equal to $N^{\mathfrak{A}G}_{\mathfrak{A}H}.$ Furthermore, $N^G_H$ is a $\mathcal{VB}$-groupoid over $H$.
\item  the manifold $\dnc(G,H)\rightrightarrows\dnc(G^0,H^0)$ is a Lie groupoid whose structure maps are $\dnc(s)$, $\dnc(r)$ and Lie algebroid is equal to $\dnc(\mathfrak{A}G,\mathfrak{A}H)$.
\item if $K\subseteq H$ is a Lie subgroupoid, then the restriction of the normal bundle $N^G_H\big|_{K}\rightrightarrows N^{G^0}_{H^0}\big|_{K^0}$ is a Lie subgroupoid of $N^G_H\rightrightarrows N^{G^0}_{H^0}$ whose Lie algebroid is $N^{\mathfrak{A}G}_{\mathfrak{A}H}\big|_{\mathfrak{A}K}.$ Furthermore $N^G_{H}\big|_{K}$ is a $\mathcal{VB}$-groupoid over $K$.
\end{enumerate}
\end{theorem}
\begin{proof}
Statements $1$ and $2$ are direct consequences of propositions \ref{smooth structure dnc construction}, \ref{Functoriality of DNC} and  \ref{DNC fibered product}. The third statement follows from the first and because the projection map onto the base
$$ \begin{tikzcd}N^G_H\arrow[d,shift right=0.85]\arrow[d, shift left=0.85]\arrow[r]& H\arrow[d,shift right=0.85]\arrow[d, shift left=0.85]\\ N^{G^0}_{H^0}\arrow[r]&H^0
\end{tikzcd}$$ is a submersive morphism of groupoids, hence the inverse image of the Lie subgroupoid $K$ is a Lie groupoid.
\end{proof}
From now on, for a Lie groupoid $G$ and a Lie subgroupoid $H$, we will use $\mathcal{N}^G_H$ to denote the space $N^G_H$ equipped with the structure of a Lie groupoid given by \Cref{Construction of dnc grouopids}.
\begin{rems}\label{sections of dnc algebr}\label{trivializing dnc Rn}\label{metric dnc E}\begin{enumerate}
 \item Let $E\to M$ be a vector bundle, $V\subseteq M$ a submanifold, $F\to V$ a subbundle of the restriction of $E$ to $V$. By \Cref{Construction of dnc grouopids}, the space $\dnc(E,F)$ is a vector bundle over $\dnc(M,V).$ Since a section of $\dnc(E,F)$ is determined by its values on the dense set $M\times \R^*$. It follows that $$\Gamma(\dnc(E,F))=\{X\in \Gamma(E\times \R):X_{|V\times\{0\}}\in \Gamma(F)\},$$ where $\Gamma$ denotes the set of global sections (continuous or smooth).
 \item Let $V=V_0+a\subseteq \R^n$ be an affine subspace where $V_0$ is the underlying vector space, $a\in \R^n$. Let $L$ be the orthogonal of $V_0$, $\pi_{V_0},\pi_L$ the orthogonal projections. The space $\dnc(\R^n,V)$ will be identified with $\R^{n+1}$ by the following map \begin{align*}
&\dnc(\R^n,V)\to \R^{n+1}\\
&(x,t)\to (a+\pi_{V_0}(x-a)+\frac{\pi_L(x-a)}{t},t),\quad t\neq 0\\
&(x,X,0)\to (x+X,0),
\end{align*}
where in the last identity we identified $N^{\R^n}_V$ with $L$.
\end{enumerate}
\end{rems}
\begin{examps}\label{examps dnc}\begin{enumerate}
\item If $M$ is a smooth manifold, then $$\dnc(M\times M,M)=M\times M\times \R^*\sqcup TM\times \{0\}\rightrightarrows M\times \R$$ is the first deformation groupoid constructed by A. Connes. He used it to give a short elegant proof of Atiyah Singer index theorem \cite{MR1303779}. The product law is given by $$(x,y,t)\cdot (y,z,t)=(x,z,t),\; (x,X,0)\cdot (x,Y,0)=(x,X+Y,0).$$
\item Let $L\subseteq G^0$ be a submanifold. Since $N^G_{L}$ is equal to $N^{G^0}_L\oplus \ker(ds)_{|L}$. It follows that the groupoid $\mathcal{N}^G_{L}\rightrightarrows N^{G^0}_L$ is equal to $$\{(X,Y,Z):X,Z\in N^{G^0}_L,Y\in \mathfrak{A}G, Z=X+\natural(Y)\},$$ with the obvious structural maps.
\end{enumerate}
\end{examps}
\begin{prop}\label{prop exact seq}Let $G\rightrightarrows G^0$ be a Lie groupoid, $H\subseteq G$ a Lie subgroupoid which is a bundle of connected Lie groups such that $$(dr-ds)(T_hG)\subseteq T_{s(h)}H^0,\quad \forall h\in H.$$ Then \begin{enumerate}
\item the Lie groupoid $\mathcal{N}^G_H\rightrightarrows N^{G^0}_{H^0}$ is a bundle of Lie groups.
\item the Lie groupoid $\mathcal{N}^G_H\big|_{H^0}\rightrightarrows N^{G^0}_{H^0}$ is a bundle of abelian Lie groups which is isomorphic (as a bundle of Lie groups) to $\mathfrak{A}G/\mathfrak{A}H\times_{H^0}N^{G^0}_{H^0}.$
\item In this way the Lie groupoid $\mathcal{N}^G_H\rightrightarrows N^{G^0}_{H^0}$ sits in an exact sequence of a bundle of Lie groups over $N^{G^0}_{H^0}$ whose fiber at $(x_0,X_0)\in N^{G^0}_{H^0}$ is $$1\to\mathfrak{A}G_{x_0}/\mathfrak{A}H_{x_0}\to\left(\mathcal{N}^G_H\right)_{(x_0,X_0)}\to  H_{x_0}\to 1.$$
\end{enumerate}
Furthermore the action associated to this exact sequence of the Lie algebra $\mathfrak{A}H_{x_0}$ on the abelian group $\mathfrak{A}G_{x_0}/\mathfrak{A}H_{x_0}$ is as follows; if $X,Y\in \Gamma^\infty(\mathfrak{A}G)$ such that $X_{|H^0}\in \Gamma^\infty(\mathfrak{A}H)$, then by our assumption, $$[X,Y](x_0)\mod \mathfrak{A}H_{x_0}$$ only depends on $X(x_0)\in \mathfrak{A}H_{x_0}$ and $Y(x_0)\mod \mathfrak{A}H_{x_0}\in \mathfrak{A}G_{x_0}/\mathfrak{A}H_{x_0}$. In particular the above exact sequence is central if and only if this action is trivial.
\end{prop}
\begin{proof}
\begin{enumerate}
\item The condition $(dr-ds)(T_hG)\subseteq T_{s(h)}H^0$ can be restated as the equality of the maps $Ns,Nr:T_hG/T_hH\to T_{s(h)}G^0/T_{s(h)}H^0$. Those two maps are the source and the target maps of  the Lie groupoid  $\mathcal{N}^G_H=\sqcup_{h\in H}T_hG/T_hH\rightrightarrows N^{G^0}_{H^0}$. By assumption, they coincide which means that $\mathcal{N}^G_H\rightrightarrows N^{G^0}_{H^0}$ is a bundle of Lie groups.

\item If $X\subseteq Y\subseteq Z$ are manifolds, then $N^Z_Y\big|_X=N^Z_X/N^Y_X$. It follows that $\mathcal{N}^G_{H}\big|_{H^0}$ is the surjective image by a groupoid morphism of the Lie groupoid $\mathcal{N}^G_{H^0}.$ One has $$\mathcal{N}^G_{H^0}=\{(X,Y,Z):X,Z\in N^{G^0}_{H^0},Y\in \mathfrak{A}G/\mathfrak{A}H, Z=X+\natural(Y)\}.$$ By assumption, the map $\natural:\mathfrak{A}G/\mathfrak{A}H\to \mathcal{N}^{G^0}_{H^0}$ is the zero map. Hence $\mathcal{N}^G_{H^0}\rightrightarrows N^{G^0}_{H^0}$ is a bundle of abelian Lie groups, hence $\mathcal{N}^G_{H}\big|_{H^0}\rightrightarrows N^{G^0}_{H^0}$ as well.
\item \begin{enumerate}
\item exactness at $\mathfrak{A}G_{x_0}/\mathfrak{A}H_{x_0}$
 is clear, because $\mathcal{N}^G_{H}\big|_{H^0}$ is a subgroupoid of $\mathcal{N}^G_H$
 \item  exactness at $\left(\mathcal{N}^G_H\right)_{(x_0,X_0)}$ follows directly from the definitions.
 \item the map $s:G\to G^0$ is a submersion, hence exactness at $H_{x_0}$.
 \end{enumerate}
 Let us prove that $[X,Y]$ only depends on $X(x_0)$ and $Y(x_0)$, where $X,Y\in \Gamma^\infty(\mathfrak{A}G)$  such that $X_{|H^0}\in \Gamma^\infty(\mathfrak{A}H).$\begin{itemize}
 \item If $Y$ vanishes at $x_0$, then locally it can be written as the sum of sections of the form $fZ$, where $f:M\to \R$ vanishes at $x_0$ and $Z\in \Gamma^\infty(\mathfrak{A}G)$. One has $$[X,fZ]=f(x_0)[X,Z](x_0)+df_{x_0}(\natural(X(x_0)))Z(x_0)=0,$$ because $X(x_0)\in \mathfrak{A}H_{x_0}$ and $H$ is a bundle of Lie groups, hence $\natural (X(x_0))=0$.
 \item If $Y_{|H^0}\in \Gamma^\infty(\mathfrak{A}H)$, then $[X,Y](x_0)\in \mathfrak{A}H_{x_0}$ because the Lie bracket computation could be carried out inside $\mathfrak{A}H$.
 \item If $X$ vanishes at $x_0$, then $dX_{x_0}:T_{x_0}G^0\to \mathfrak{A}_{x_0}G$ is well defined. It is well known that $[X,Y](x_0)=-dX_{x_0}(\natural(Y(x_0))).$ This formula can be proved locally by writing $X$ as sum of $fZ$. The condition $X_{|H^0}\in \Gamma^\infty(\mathfrak{A}H)$ implies that $dX_{x_0}(T_{x_0}H^0)\subseteq \mathfrak{A}H^0$. The assumption on $dr-ds$ implies that $\natural (Y(x_0))\in T_{x_0}H^0$, hence $[X,Y](x_0)\in \mathfrak{A}H_{x_0}.$
 \end{itemize}
\end{enumerate}
That this is the action associated to the abelian extension of $\left(\mathcal{N}^{G}_{H}\right)_{(x_0,X_0)}$ is then clear.
\end{proof}
\paragraph{Quotient of Lie groupoids.}\label{para dfn quotient of Lie groupoids} Let $G\rightrightarrows G^0$ be a Lie groupoid, $H\subseteq G$ a Lie subgroupoid. The Lie groupoid $H$ acts on the smooth manifold $G_{H^0}$ by right translation. This action is clearly free and proper. Hence by \cite[section 5.9.5]{MR0219078}, the quotient space $G_{H^0}/H$ is a smooth manifold, that will be denoted by $G/H$.

\begin{examp}\label{examp quotient dnc}If $V$ is a submanifold of $M$, then $\dnc(V\times V,V)$ is a Lie subgroupoid of $\dnc(M\times M,M)$. It is clear that the quotient space is equal to $$\dnc(M\times M,M)/\dnc(V\times V,V)=\dnc(M,V).$$
\end{examp}
\section{DNC iterated}\label{Dnc iterated sect}
Let $M$ be a smooth manifold, $V_0\subseteq M$ a submanifold, $V_{1}\subseteq \dnc(M,V_0)$ a submanifold. One defines $$\dnc^2(M,V_0,V_{1}):=\dnc(\dnc(M,V_0),V_{1}).$$
This space being a deformation space admits an $\R^*$-action that will be denoted by $\lambda^{(1)},$ and a projection map $\pi_\R^{(1)}:\dnc^2(M,V_{0},V_{1})\to \R$.

\bigskip

If $V_{1}$ is $\R^*$-invariant, then by \Cref{action dnc}, the group $\R^*$ acts on $\dnc^2(M,V_{0},V_{1})$. This action will be denoted by $\lambda^{(0)}$, furthermore the group $\left(\R^*\right)^2$ acts on $\dnc^2(M,V_0,V_1)$ by $\lambda^{(0)}\times \lambda^{(1)}.$

\bigskip

Let $\pi_\R:\dnc(M,V)\to \R$ be the projection constructed in \Cref{dnc section}. If $\pi_\R(V_1)$ is an affine subspace of $\R$ and $\pi_\R|V_{1}:V_1\to \pi_\R(V_1)$ is a submersion, then the map $$\pi_\R^{(0,1)}:=\dnc(\pi_\R):\dnc^2(M,V_{0},V_{1})\to \dnc(\R,\pi_\R(V_{1}))=\R^2$$ is a smooth submersion, where  we identified $\dnc(\R,\pi_\R(V_1))$ with $\R^2$ using \Cref{trivializing dnc Rn}.

If $V_{1}$ is furthermore $\R^*$-invariant, then one has for all $u,t\in\R^*$
\begin{equation}\label{relation projection and action}
\pi^{(0,1)}_\R\lambda^{(1)}_u=(\frac{\pi^{(0)}_\R}{u},u\pi^{(1)}_\R),\; \pi_\R^{(0,1)}\lambda^{(0)}_u=(u\pi^{(0)}_\R,\pi_\R^{(1)}),
\end{equation} where $\pi^{(0,1)}_\R=(\pi^{(0)}_{\R},\pi^{(1)}_\R)$.

\bigskip

By induction, given a sequence of submanifolds $$V_0\subseteq M,\;V_{1}\subseteq \dnc(M,V_0),\;V_{2}\subseteq \dnc^2(M,V_0,V_{1}),\cdots ,\;V_{k}\subseteq \dnc^{k}(M,V_0,\dots ,V_{k-1}).$$ We define the space $$\dnc^{k+1}(M,V_0,\cdots,V_k):=\dnc(\dnc^{k}(M,V_0,\cdots,V_{k-1}),V_{k}).$$

If for each $1\leq i\leq k$, $\pi^{(0,\dots,i-1)}_\R(V_i)$ is an affine subspace of $\R^i$ and $\pi^{(0,\dots,i-1)}_\R:V_i\to \pi^{(0,\dots,i-1)}_\R(V_i)$ is a submersion, then by \Cref{Functoriality of DNC}, the map $$\pi_\R^{(0,\cdots,k)}:=\dnc(\pi_\R^{(0,\cdots,k-1)}):\dnc^{k+1}(M,V_0,\cdots,V_k)\to \dnc(\R^k,\pi^{(0,\dots,k-1)}_\R(V_k))=\R^{k+1}$$ is a smooth submersion, where we identified $\dnc(\R^k,\pi_\R^{(0,\dots,k-1)}(V_k))$ with $\R^{k+1}$ using \Cref{trivializing dnc Rn}.

If each $V_i$ is $(\R^*)^i$ invariant, then the space $\dnc^{k+1}(M,V_0,\cdots,V_k)$ admits $k+1$ pairwise commuting actions of $\R^*$-denoted $\lambda^{(k)},\dots,\lambda^{(0)}$.

\bigskip

Propositions \ref{Functoriality of DNC},\;\ref{DNC fibered product} and \Cref{Construction of dnc grouopids} have obvious extensions to $\dnc^k$. In particular we have by induction if $G\rightrightarrows G^0$ is a Lie groupoid, $H_0\subseteq G$, $H_{1}\subseteq \dnc(G,H_0)$, $\dots ,\;H_{k}\subseteq \dnc^{k}(G,H_0,\dots ,H_{k-1}).$ are Lie subgroupoids, then $$\dnc^{k+1}(G,H_0,H_{1},\cdots,H_k)\rightrightarrows \dnc^{k+1}(G^0,H_0^0,\cdots,H_k^0)$$ is a Lie groupoid.
\section{Computations in the case of a single subbundle}\label{M single H examp}
Let $M$ be a smooth manifold, $H\subseteq \mathcal{N}^{M\times M}_M=TM$ a subbundle. In this section we prove \Cref{main thm H single}, which proves the claim made in the introduction (at least on the algebraic level) that the fiber of the groupoid $\dnc^2(M\times M,M,H\times \{0\})\rightrightarrows \dnc^2(M,M,M\times\{0\})=M\times \R^2$ over $M\times \{1\}\times \R$ is equal to the groupoid constructed in \cite{Choi:2015aa,Choi:2017aa,Ponge:2017aa,Erp:2016aa}. In \Cref{sect H bundle 1 case}, we will write local charts which will prove that in fact the fiber is equal as a smooth manifold to the one constructed in  \cite{Choi:2015aa,Choi:2017aa,Ponge:2017aa,Erp:2016aa}.

Before stating the theorem, let us recall the constuction of the Levi form $\mathcal{L}$ : the map \begin{align*}
\Gamma^\infty(H)\times \Gamma^\infty(H)\to \Gamma^\infty(TM/H),\quad (X,Y)\to [X,Y]\mod H
\end{align*}
is $C^\infty(M)$-linear because \begin{align*}
[fX,Y]=f[X,Y]-Xf Y=f[X,Y]\mod H.
\end{align*} Hence it comes from an anti symmetric bilinear bundle map $\mathcal{L}:H\times H\to TM/H$.
\begin{theorem}\label{main thm H single}
The groupoid $\mathcal{N}^{\dnc(M\times M,M)}_{H\times\{0\}}\rightrightarrows N^{M\times \R}_{M\times\{0\}}=M\times \R$ is isomorphic (by an isomorphism which is equal to the identity on the objects) to the bundle of Lie groups $H\oplus TM/H\times\R\rightrightarrows M\times \R$ equipped with the group law $$(h,n,t)\cdot (h',n',t)=\left(h+h',n+n'+\frac{t}{2}\mathcal{L}(h,h')\right).$$
\end{theorem}
\begin{proof}
First we apply \Cref{prop exact seq} to $\dnc(M\times M,M)\rightrightarrows M\times\R$ and $H\times\{0\}\rightrightarrows M\times \{0\}$. Let us check the condition of \Cref{prop exact seq} and the triviality of the action. \begin{itemize}
 \item Since $\pi_\R\circ r=\pi_\R\circ s$, the condition of \Cref{prop exact seq} is satisfied.

\item the triviality of the action is immediate to check. If $X$ is a section of $TM$ over $M\times \R$ which vanishes on $M\times \{0\}$, $Y$ is a section of $TM$ over $M\times \R$ which vanishes on $M\times \{0\}$ and whose $\partial_t$-derivative on $M\times\{0\}$ is in $H$, then the vector field $[X,Y]$ vanishes over $M\times \{0\}$.
\end{itemize}

The central exact sequence of bundles of Lie groups over $(x_0,t_0)\in N^{M\times\R}_{M\times \{0\}}=M\times \R$ given by \Cref{prop exact seq} is then equal to $$1\to T_{x_0}M/H_{x_0}\to \left(\mathcal{N}^{\dnc(M\times M,M)}_{H\times\{0\}}\right)_{(x_0,t_0)}\to H_{x_0}\to 1.$$

There exists a quite natural section of this exact sequence: let $h\in H_{x_0}$, $f:\R\to M$ any smooth function such that $f'(0)=h$ and $f'(t)\in H_{f(t)} \;\forall t$, \begin{align*}
&\sigma_{x_0,t_0}(h,\cdot):\R\to \dnc(M\times M,M)\\
 &\sigma_{x_0,t_0}(h,u)=(f(tu),f(0),tu) &\text{if}\; tu\neq 0\\
 &\sigma_{x_0,t_0}(h,0)=(x_0,h,0)&\text{if}\; tu=0
\end{align*}
One then sees immediately that the map \begin{align*}\mathfrak{S}_{x_0,t_0}:H_{x_0}&\to  \left(\mathcal{N}^{\dnc(M\times M,M)}_{H\times\{0\}}\right)_{(x_0,t_0)}\\
 h&\to\left( \frac{\partial}{\partial u}\big|_{u=0}\sigma_{x,t}(h,u)\mod T_{(x_0,h,0)}(H\times \{0\})\right)
\end{align*} is well defined (i.e, doesn't depend on the choice of $f$) and is a section of the above exact sequence.

\bigskip

The map $\mathfrak{S}_{x_0,t_0}$ is not a group homomorphism. For $h_1,h_2\in H_{x_0}$, we have \begin{equation*}
 \mathfrak{S}_{x_0,t_0}(h_1)\mathfrak{S}_{x_0,t_0}(h_2)\mathfrak{S}_{x_0,t_0}(-h_1-h_2)=\frac{t_0}{2}\mathcal{L}(h_1,h_2)\in T_{x_0}M/H_{x_0}.
\end{equation*}
This follows from the definition of $\mathcal{L}.$
\end{proof}
\begin{cor}
The fiber of the groupoid $$\dnc^2(M\times M,M,H\times \{0\})\rightrightarrows M\times \R^2$$ over $M\times \{1\}\times \R$ is equal to (as an algebraic groupoid) to $$M\times M\times \R^*\sqcup H\oplus TM/H\times \{0\}\rightrightarrows M\times \R,$$ where the groupoid structure on $M\times M\times \R^*$ is the pair groupoid, and on $H\oplus TM/H$ is the bundle of nilpotent Lie groups $$(h,n)\cdot (h',n')=\left(h+h',n+n'+\frac{1}{2}\mathcal{L}(h,h')\right).$$
\end{cor}
Since $H$ is $\R^*$ invariant, by \Cref{Dnc iterated sect} we have two group actions $\lambda^1$, $\lambda^0$ of $\R^*$. Under the above identification the two actions $\lambda^1$ and $\lambda^0$ become $$\lambda^1_s(h,n,t)=(\frac{h}{s},\frac{n}{s},\frac{t}{s}),\quad \lambda^0_s(h,n,t)=(h,\frac{n}{s},ts).$$
\section{Another description of $N^{\dnc(M,V)}_{H\times\{0\}}$}\label{sect H bundle 1 case}
Let $M$ be a smooth manifold, $V$ a submanifold, $H\subseteq N^M_V$ a smooth subbundle, $\mathcal{H}$ the lift of $H$ to $TM$. In other words $\mathcal{H}$ is a subbundle of the restriction of $TM$ to $V$ such that $TV\subseteq \mathcal{H}$ and $H=\mathcal{H}/TV$. In this section we give an alternate description of the fiber $\left(\pi^{(0,1)}\right)^{-1}(\{1\}\times \R)$ of the space $\dnc^2(M,V,H\times \{0\})$.

\begin{dfn}Let $\tilde{N}^M_{V,H}$ the set of smooth functions $f:\R\to M$ such that $ f(0)\in V$ and $f'(0)\in \mathcal{H}_{f(0)}$.

Let $N^M_{V,H}$ be the quotient of $\tilde{N}^M_{V,H}$ by the equivalence relation where $f,g\in \tilde{N}^M_{V,H}$ are equivalent if and only if \begin{enumerate}
\item $f(0)=g(0)$
\item  $f'(0)-g'(0)\in T_{f(0)}V$.
\item  for every smooth function $l:M\to \R$ which vanishes on $V$ and whose derivative $dl$ vanishes on $\mathcal{H}$, one has $(l\circ f)''(0)=(l\circ g)''(0)$.
\end{enumerate}\end{dfn}
Let $\pi_\R:\dnc(M,V)\to \R$ be the projection. Since $\pi_\R(H)=0$, the map $N\pi_\R:N^{\dnc(M,V)}_H\to N^\R_{0}=\R$ is well defined. We claim that the set $N^M_{V,H}$ is in a natural bijection with $\left(N\pi_\R\right)^{-1}(1).$ To see this let $f\in \tilde{{N}}^M_{V,H}$. Since $f(0)\in V$, the function \begin{align*}
\dnc(f):\R\to \dnc(M,V),\quad t\to (f(t),t),\neq 0,\; 0\to (f'(0),0)
\end{align*}
is smooth. And since $f'(0)\in H$ it follows that $
 \dnc^2(f):\R\to \dnc^2(M,V,H)$ is a welll defined smooth map. Its value at zero is an element in $N^{\dnc(M,V)}_H$ which is clearly in $\left(N\pi_\R\right)^{-1}(1)$. \begin{prop}the map \begin{align*}\label{map description}
\beta:N^M_{V,H}\to \left(N\pi_\R\right)^{-1}(1),\quad [f]\to [\dnc(f)]
\end{align*}
is a well defined bijection
 \end{prop}
 Let us remark that the map $\beta$ is not a linear map and in fact the space $N^M_{V,H}$ is not a vector bundle.
 \begin{proof}
   In \Cref{dnc section}, two types of functions on $\dnc(M,V)$ were described which generate the ring of smooth functions on $\dnc(M,V).$ By regarding each type we see that for two functions $f,g\in \tilde{N}^M_{V,H}$, the classes in   $N^{\dnc(M,V)}_{H}$ of $\dnc(f)$ and $\dnc(g)$ are equal if and only if the classes of $f$ and $g$ are equal in $N^M_{V,H}$. Hence $\beta$ is well defined and injective. Surjectivity follows by looking at a local chart as described below.
\end{proof}
    Let $\psi:N^M_V\to M$ be a tubular neighbourhood embedding, $L:H\oplus N^M_V/H\to N^M_V$ a linear isomorphism given by the choice of a complementary subbundle of $H$ inside $N^M_V$, $\phi=\psi\circ L$.

By the local charts descriped in \Cref{dnc section}, the following is a local chart for $\dnc(M,V)$: \begin{align*}
&\tilde{\phi}:H\oplus N^M_V/H\times \R\to \dnc(M,V)\\
&(h,n,t)\to (\phi(th,tn),t),\quad t\neq 0\\
&(h,n,0)\to (L(h,n,0),0).
\end{align*}
Therefore the following is a local chart for $\dnc^2(M,V,H\times\{0\})$
 \begin{align*}
&H\oplus N^M_V/H\times \R\times\R\to \dnc^2(M,V,H\times\{0\})\\
&(h,n,t,u)\to (\phi(uth,u^2tn),ut,u)\in M\times \R^*\times \R^*\quad t\neq 0,u\neq 0\\
&(h,n,0,u)\to (L(h,un),0,u)\in N^M_V\times \{0\}\times \{u\},\quad u\neq 0\\
&(h,n,t,0)\to (h,n,t,0)\in N^{\dnc(M,V)}_H\times\{0\},
\end{align*}
where in the last identity we identified $N^{\dnc(M,V)}_H$ with  $H\oplus N^M_V\oplus \R$ using $\tilde{\phi}.$ In this local picture, $\pi^{(0,1)}$ is the projection $(h,n,t,u)\to (t,u).$

\bigskip

Let $$\dnc_H(M,V):=M\times \R^*\sqcup N^M_{V,H}\times\{0\}.$$  We equip $\dnc_H(M,V)$ with a smooth structure by identifying it with $\left(\pi_\R^{(0,1)}\right)^{-1}(\{1\}\times \R)$ using the map $\beta$. Its local charts are hence given by \begin{align*}
H\oplus N^M_V/H\times \R\to \dnc_H(M,V)\\
(h,n,u)\to (\phi(uh,u^2n),u),\quad u\neq 0\\
(h,n,0)\to ([t\mapsto \phi(L(th,t^2n))],0).
\end{align*}
The space $\dnc_H(M,V)$ is called \textit{the deformation to the normal cone of $M$ along $V$ with weight $H$.}

\begin{rem}\label{rem all fibers are same}All the other fibers $\left(\pi^{(0,1)}_\R\right)^{-1}(\{t\}\times \R)$ for $t\neq 0$ are isomorphic to $\left(\pi^{(0,1)}_\R\right)^{-1}(\{1\}\times \R)$ by a rescaling in the $u$-variable. The fiber $\left(\pi^{(0,1)}_\R\right)^{-1}(\{0\}\times \R)$ is equal to $\dnc(N^M_V,H).$ In particular the space $\dnc^2(M,V,H)$ should be seen as a deformation of the space $\dnc_H(M,V)$ to the simpler space $\dnc(N^M_V,H).$
\end{rem}
Since $H$ is $\R^*$-invariant, by \Cref{Dnc iterated sect} it follows that there is an $(\R^*)^2$ action on $\dnc^2(M,V,H\times\{0\}).$ It follows from \Cref{relation projection and action} in \Cref{Dnc iterated sect} that $\left(\pi_\R^{(0,1)}\right)^{-1}(\{1\}\times \R)$ is invariant under the diagonal $\lambda^{(1)}_u\lambda^{(0)}_u$. This action is described by $u\cdot (x,t)=(x,tu)$ and $u\cdot ([f],0)=([f(\frac{\cdot}{u})],0)$ for $f\in \tilde{N^M_{V,H}}.$
 \begin{cor}\label{naturality dnc single H}Let $(M,V), (M',V')$ be smooth manifold pairs, $H\subseteq N^M_V, H'\subseteq N^{M'}_{V'}$ subbundles, $g:M\to M'$ a smooth map  such that  $g(V)\subseteq V'$
and $dg(H)\subseteq H'$. Then the maps \begin{itemize}
\item $\begin{aligned}[t]
 Ng:N^M_{V,H}\to N^{M'}_{V',H'}\quad [f]\to [g\circ f]
\end{aligned} $
\item $\begin{aligned}[t]
\dnc(g):\dnc_H(M,V)\to \dnc_{H'}(M',V')\\
(x,t)\to (g(x),t)\\
([f],0)\to ([g\circ f],0)
\end{aligned}$
\end{itemize}
are well defined and smooth.
\end{cor}
\begin{proof}
This is a corollary of \Cref{Functoriality of DNC} applied twice and the identification of $\dnc_H(M,V)$ with $\left(\pi^{(0,1)}_\R\right)^{-1}(\{1\}\times \R)\subseteq \dnc^2(M,V,H\times\{0\})$.
\end{proof}
\begin{prop}\label{fibered product dnc commute single H}Let $M_1, M_2, M$ be manifolds, $V_i\subseteq M_i,V\subseteq M$ submanifolds, $H_i\subseteq N^{M_i}_{V_i}$, $H\subseteq N^M_V$ vector subbundles, $f_i:M_i\to M$ smooth maps such that \begin{enumerate}
\item $f_i(V_i)\subseteq V$
\item the maps $f_i:M_i\to M$ are transverse
\item the maps $f_i|V:V_i\to V$ are transverse
\item  $H=df_1(H_1)+df_2(H_2)$,
\end{enumerate} then \begin{enumerate}
\item the maps $\dnc(f_i):\dnc_{H_i}(M_i,V_i)\to \dnc_H(M,V)$ are transverse.
\item  the natural map $$\dnc_{H_1\times_H H_2}(M_1\times_M M_2,V_1\times
_V V_2)\to\dnc_{H_1}(M_1,V_1)\times_{\dnc_H(M,V)}\dnc_{H_2}(M_2,V_2)$$  is a diffeomorphism.
\end{enumerate}
\end{prop}
\begin{proof}
This is a corollary of \Cref{DNC fibered product} applied twice and the identification of $\dnc_H(M,V)$ with $\left(\pi^{(0,1)}_\R\right)^{-1}(\{1\}\times \R)\subseteq \dnc^2(M,V,H\times\{0\})$.
\iffalse
For example one has \begin{align*}
&\dnc(\dnc(M\times_S M,V\times _T V'),H\times_L H')\\&=\dnc (\dnc(M,V)\times_{\dnc(S,T)}\dnc(M',V'),H\times_L H')\\&=\dnc(\dnc(M,V),H)\times_{\dnc(\dnc(S,T),L)}\times \dnc(\dnc(M',V'),H')
\end{align*}the identification of the fibers of $\pi^{(1,0)}$ is then straightforward.
\fi
\end{proof}
\begin{theorem}\label{dnc groupoid is a groupoid single H}
Let $G\rightrightarrows G^0$ be a groupoid, $G'\rightrightarrows G^{'0}$ a subgroupoid, $H\subseteq \mathcal{N}^G_{G'}$ a $\mathcal{VB}$-subgroupoid \cite{MR941624,MR896907}. Then \begin{enumerate}
\item the space $\mathcal{N}^G_{G',H}\rightrightarrows N^{G^0}_{G^{'0},H^0}$ is a Lie groupoid whose algebroid is equal to $N^{\mathfrak{A}G}_{\mathfrak{A}G',\mathfrak{A}H}.$
\item  the space $\dnc_H(G,G')\rightrightarrows\dnc_{H^0}(G^0,G^{'0})$ is a Lie groupoid whose Lie algebroid is equal to $\dnc_{\mathfrak{A}H}(\mathfrak{A}G,\mathfrak{A}G').$
\end{enumerate}
\end{theorem}
\begin{proof}
This is a corollary of \Cref{naturality dnc single H} and \Cref{fibered product dnc commute single H}.
\end{proof}
\begin{examp}\label{example H foliation single}Let $F\subseteq TM$ be an integrable subundle. We regard the foliation groupoid $\mathcal{G}(M,F)\rightrightarrows M$ as an immersed subgroupoid of $M\times M\rightrightarrows M$ by the map $$(x,[\gamma],y)\to (x,y).$$ This map is not injective but the Lie groupoid $\dnc(M\times M,\mathcal{G}(M,F))\rightrightarrows M\times \R$ is still well defined by a remark in \cite{MR925720}.  Its underlying manifold is a second countable locally Hausdorff manifold.

\bigskip

 The vector bundle $TM/F$ will be denoted by $\nu(F).$  If $\gamma:[0,1]\to M$ is path tangent to the leaves, then its holonomy defines a map $d\gamma:\nu(F)_{\gamma(0)}\to \nu(F)_{\gamma(1)}$. One then sees that the groupoid $$\mathcal{N}^{M\times M}_{\mathcal{G}(M,F)}=\{(x,[\gamma],y,X):(x,[\gamma],y)\in \mathcal{G}(M,F),X\in \nu(F)_{y}\}\rightrightarrows M.$$The product is then given by $$(x,[\gamma],y,X)\cdot (y,[\gamma'],z,Y)=(x,[\gamma\gamma'],z,d\gamma'(X)+Z).$$ Let  $H\subseteq \nu(F)$ be a holonomy invariant subbundle, i.e such that for any leafwise path $\gamma:[0,1]\to M$, one has $d\gamma(H_{\gamma(0)})=H_{\gamma(1)}$. It follows that $$L:=\{(x,[\gamma],y,X)\in \mathcal{N}^{M\times M}_{\mathcal{G}(M,F)}:X\in H_y\}\subseteq \mathcal{N}^{M\times M}_{\mathcal{G}(M,F)}$$ is a Lie subgroupoid. The groupoid $$\mathcal{N}^{M\times M}_{\mathcal{G}(M,F),L}=\{(x,[\gamma],y,X,Y):X\in H_y,Y\in \nu(F)_{y}\}\rightrightarrows M$$ has then the groupoid law $$(x,[\gamma],y,X,Y)\cdot (y,[\gamma'],z,X',Y')=(x,[\gamma\gamma'],z,d\gamma'(X)+X',d\gamma'(Y)+Y'+\frac{1}{2}\mathcal{L}(d\gamma'(X),X')),$$where $\mathcal{L}:H\times H\to v(F)/H$ is a Levi form defined similarly to the one defined in \Cref{M single H examp}.
\end{examp}

\section{Carnot Groupoid}\label{Heisenberg groupoid sect}
A more general groupoid will be constructed starting from the following data: Let $M$ be a smooth manifold, $0=H^0\subseteq H^1\subseteq \cdots\subseteq H^{k+1}=TM$ be vector bundles such that $$[\Gamma^\infty(H^i),\Gamma^\infty(H^j)]\subseteq \Gamma^\infty(H^{i+j}),$$ where $H^{i}=TM$ for $i>k$. We will calculate the Lie algebroid of this groupoid and hence show that it is equal to the groupoid constructed in \cite{MR2417549,Choi:2015aa,Choi:2017aa,Ponge:2017aa,Erp:2016aa}.

\bigskip

 Since $[\Gamma^\infty(H^i),\Gamma^\infty(H^i)]\subseteq \Gamma^\infty(H^{i+j})$, it follows that the map \begin{align*}
\Gamma^\infty(H^i/H^{i-1})\times \Gamma^\infty(H^{j}/H^{j-1})\to \Gamma^\infty(H^{i+j}/H^{i+j-1})\\
(X,Y)\to [X,Y]\mod \Gamma^\infty(H^{i+j-1})
\end{align*} is a $C^\infty(M)$-bilinear map, hence it comes from an antisymmetric bilinear map $$\mathcal{L}:H^{i}/H^{i-1}\times H^{j}/H^{j-1}\to H^{i+j}/H^{i+j-1}.$$

For each $a\in M$, the map $\mathcal{L}$ defines the structure of a Lie algebra on $\mathcal{G}(H^\cdot)_a:=\oplus_{i}H^i_a/H^{i+1}_a$ by $$[X,Y]=\mathcal{L}(X,Y),\quad \text{for}\; X\in H^i_a/H^{i+1}_a,Y\in H^j_a/H^{j+1}_a.$$ By Baker–Campbell–Hausdorff formula, the vector space $\mathcal{G}(H^\cdot)_a$ admits the structure of a nilpotent Lie group. It is clear that the structure of group is $C^\infty$ in $a$, hence $\mathcal{G}(H^\cdot)$ is a bundle of nilpotent Lie groups. We will define a Lie groupoid denoted by $\dnc_{H^\cdot}(M\times M,M)$ by induction on $k$ whose underlying set is equal to $$M\times M\times \R^*\sqcup \mathcal{G}(H^\cdot)\times \{0\}.$$ and whose Lie algebroid is equal to $$\Gamma^\infty(\mathfrak{A}_{H^\cdot})=\{X\in \Gamma^\infty(TM\times \R):\partial_t^iX_{|t=0}\in \Gamma^\infty(H^i)\forall i\geq 0\}$$For $k=1$, this is just $\dnc_{H^1}(M\times M,M)\rightrightarrows M\times \R$ defined in \Cref{sect H bundle 1 case}. By induction assuming it is defined for $k-1$, that is the Lie groupoid \begin{align*}
 \dnc_{H^1,\dots,H^{k-1}}(M\times M,M)&=M\times M\times \R^*\sqcup \mathcal{G}(H^1,\dots,H^{k-1})\times \{0\}\\&=M\times M\times \R^*\sqcup H^1\oplus H^2/H^1\oplus \dots\oplus TM/H^{k-1}\times\{0\}
\end{align*} is well defined. The subset $H^1\oplus H^2/H^1\oplus \cdots \oplus H^k/H^{k-1}$ is a Lie subgroupoid of $\mathcal{G}(H^1,\dots,H^{k-1})$ precisely because $$[\Gamma^\infty(H^i),\Gamma^\infty(H^j)]\subseteq \Gamma^\infty(H^{k}),\quad i+j=k.$$ Therefore the space \begin{align*}
 \dnc(\dnc_{H^1,\dots,H^{k-1}}(M\times M,M)&,H^1\oplus H^2/H^1\oplus \cdots \oplus H^k/H^{k-1}\times \{0\})\\&\rightrightarrows \dnc(M\times \R,M\times \{0\})=M\times \R^2
\end{align*} is a Lie groupoid, where we used \Cref{trivializing dnc Rn}. The Lie algebroid of this groupoid is then $$\dnc(\mathfrak{A}_{H^1,\dots,H^{k-1}},H^1\oplus \cdots \oplus H^k/H^{k-1})$$

Using \Cref{sections of dnc algebr}, we get that the space of sections of this algebroid is then equal to \begin{align*}
\Gamma^\infty(\dnc(&\mathfrak{A}_{H^1,\dots,H^{k-1}},H^1\oplus \cdots \oplus H^k/H^{k-1})\\&=\{X\in \Gamma^\infty(TM\times \R\times \R):\partial^i_t(X)(0,u)\in \Gamma^\infty(H^i)\;\forall 0\leq i\leq k-1,u\in \R\\ &\&\;\partial^{k}_t(X)(0,0)\in \Gamma^\infty(H^k)\}.
\end{align*}

We define $\dnc_{H^1,\cdots,H^k}(M\times M,M)$ as the fiber of $\dnc(\dnc_{H^1,\dots,H^{k-1}}(M\times M,M),H^1\oplus H^2/H^1\oplus \cdots \oplus H^k/H^{k-1}\times \{0\})$ over $M\times \{1\}\times \R$. This is clearly a Lie groupoid.

\bigskip

It follows from the above description of $\Gamma^\infty(\dnc(\mathfrak{A}_{H^1,\dots,H^{k-1}},H^1\oplus \cdots \oplus H^k/H^{k-1}))$ by restricting to the diagonal we get that if $$X\in \Gamma^\infty(\dnc(\mathfrak{A}_{H^1,\dots,H^{k-1}},H^1\oplus \cdots \oplus H^k/H^{k-1}))$$ then $\partial_t^iX(0,0)\in\Gamma^\infty(H^i)$ for all $0\leq i\leq k$, where we used that $X(0,u)=0$. This fnishes the induction, and proves that Lie algebroid  of $\dnc_{H^\cdot}(M\times M,M)$ is equal to $\mathfrak{A}_{H^\cdot}$. Hence we proved the following\begin{theorem}
The groupoid $\dnc_{H^\cdot}(M\times M,M)$ is the same as the groupoid constructed in  \cite{Choi:2015aa,Choi:2017aa,Ponge:2017aa,Erp:2016aa}.
\end{theorem}

\begin{rems}\begin{enumerate}
\item In \cite{Erp:2016aa}, a more general case is reagrded where starting from a groupoid $G$, subbundles $H^1\subseteq \dots\subseteq H^r=\mathfrak{A}G$ such that $[\Gamma^\infty(H^i),\Gamma^\infty(H^j)]\subseteq \Gamma^\infty(H^{i+j})$ they construct a groupoid $\dnc_{H^\cdot}(G,G^0)$. It is clear that the above construction works equally well for this case with only notational changes. The advantage of our approach is that we can do the more general case of a groupoid inside another without any extra difficulty.
\item The groupoid \begin{align*}
 \dnc^{k+1}(M\times &M,M,H^1\times\{0\},\dots, H^1\oplus\dots \oplus H^{k}/H^{k-1}\times \R^{k-1}\times \{0\})\\&\rightrightarrows \dnc^{k+1}(M,M\times \{0\},\dots,M\times \R^{k-1}\times \{0\})=M\times \R^{k+1}.
\end{align*}is a Lie groupoid which contains the `deformations in all the directions'. This groupoid admits an $\left(\R^{*}\right)^{k+1}$ action as in \Cref{Dnc iterated sect}. The fiber over $(1,\dots,1,0)$ is then equal to $\dnc_{H^1,\dots, H^k}(M\times M,M)$. The action $\R^*$ defined on $\dnc_{H^1,\dots,H^k}(M\times M,M)$ defined in \cite{Erp:2016aa} is then just the diagonal action of $\R^{k+1}$ which by induction is easily seen to preserve the fiber $(1,\dots ,1,0).$

For example, in the case $k=2$, this gives \begin{align*}
 \dnc^3&(M\times M,H^1\times\{0\},H^1\oplus H^2/H^1\times \R)=M\times M\times \R^*\times \R^*\times \R^*\\&\sqcup TM\times \{0\}\times \R^*\times \R^*\sqcup H^1\oplus TM/H^1\times \R\times\{0\}\times \R^*\\&\sqcup H^1\oplus H^2/H^1\oplus TM/H^2\times\R\times\R\times\{0\}
\end{align*}
Let us remark that the subgroupoid $H^1\oplus H^2/H^1\oplus TM/H^2\times\R\times\R\times\{0\}$ is not trivial as a groupoid, it has a structure \begin{align*}
(h_1,h_2,h_3,t,u,0)\cdot (k_1,k_2,k_3,t,u,0)=\bigg(h_1+k_1,h_2+k_2+\frac{t}{2}[h_1,k_1],\\h_3+k_3+\frac{tu}{2}\left([h_1,k_2]+[h_2,k_1]\right)+\frac{t^2u}{12}\left([h_1,[h_1,k_1]]+[k_1,[k_1,h_1]]\right),t,u,0\bigg)
\end{align*}
Similarly for $M\times \{0\}\times \R^*\times \R^*$ and $H^1\oplus TM/H^1\times \R\times\{0\}\times \R^*$.
\item The existence of the Lie groupoid $\dnc_{H^\cdot}(M\times M,M)$ follows from Debord's result on integrability of Lie algebroids \cite{MR1751671}.
\end{enumerate}
\end{rems}
 \begin{examp}\label{examp Higson}
 Let $V\subseteq M$ a smooth submanifold such that $H^i\cap TV$ is of locally of finite rank. It is then clear that $[\Gamma^\infty(H^i\cap TV),\Gamma^\infty(H^j\cap TV)]\subseteq \Gamma^\infty(H^{i+j}\cap TV)$. Let $G(H^\cdot)$ the bundle of nilpotent Lie groups $\oplus_{i}Hî/H^{i-1}$, $G(H^\cdot\cap TV)$ be the bundle of nilpotent Lie groups $\oplus (H^i\cap TV)/(H^{i-1}\cap TV).$ In \cite{Sadegh:2016aa}, the authors define a smooth manifold whose underlying set is equal to $M\times \R^*\sqcup G(H^\cdot)_{|V}/G(H^\cdot\cap TV),$ where $G(H^\cdot)_{|V}$ is the restriction of $G(H^\cdot)$ to $V$. Similarly to the description in \Cref{examp quotient dnc} of the classical deformation to the normal as a quotient space, the space defined in \cite{Sadegh:2016aa} can also be written as $\dnc_{H^\cdot}(M\times M,M)/\dnc_{H^\cdot\cap TV}(V\times V,V)$.  \end{examp}
\begin{examp}\label{examp foliation multipleHeisen}
Following the notation of \Cref{example H foliation single}. Let $F$ be a foliation, $H^1\subseteq\cdots\subseteq H^{k+1}=\nu(F)$ subbundles such that if $X\in \Gamma^\infty(H^i)$ and $Y\in \Gamma^\infty(H^j)$, then $$[X,Y]\in \Gamma^\infty(H^{i+j}).$$ with the convention $H^{s}=\nu(F)$ for $s>k$ and such that if $i\in\{1,\dots,k\}$, $\gamma:[0,1]\to M$ a path tangent to the leaves, then $d\gamma H^i_{\gamma(0)}=H^i_{\gamma(1)}$. In \Cref{example H foliation single}, we defined the groupoid $\dnc_{L^1}(M\times M,\mathcal{G}(M,F))$. We can by an induction, similar to the above, construct the groupoid $\dnc_{H^\cdot}(M\times M,\mathcal{G}(M,F))$.
 \end{examp}

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TITLE: Value of limit with integral involved in it
QUESTION [1 upvotes]: The question is given by my friend. The problem  is $$L=\lim _{n\to \infty} \int _0 ^1  x^{10}\sin (nx)  dx $$ so what i did was  using by parts method. The relation I get for same thing with $x^8$ is in place of $10$ is that same integral can be written as $\frac {\sin (n)}{n}-\frac {9}{n}(\int _0^1 x^8\sin (nx) $....t. Then continuing we go to an integral of $\sin (nx)=\frac {1-\cos (n)}{n} $After going back with it we have limit as $0$ . But my friend says answer isnt $0$ . Even if we think logically we can see it to be $0$ Wheres the mistake?.Thanks

REPLY [2 votes]: You are right the limit is 0, but your calculation should be
$$\int_0^1 x^{10}\sin(nx)\,dx = \left[x^{10}\left(-\frac{\cos nx}{n}\right)\right]_0^1+ \frac{10}{n}\int_0^1 x^{9}\cos(nx)\,dx$$
which tends to 0 as $n\to\infty$.


TITLE: Let $[0,\omega_1)$ be the set of all countable ordinals with the order topology. Prove that $[0,\omega_1)$ is star-Rothberger.
QUESTION [1 upvotes]: A space $X$ is said to be star-Rothberger if for each sequence $(\mathcal{U}_n)$ of open covers of $X$ there exists a sequence $(U_n)$ such that for each $n$ $U_n\in\mathcal{U}_n$ and $\{St(U_n,\mathcal{U}_n) : n\in\mathbb{N}\}$ is an open cover of $X$.

REPLY [4 votes]: HINT: Let $X=[0,\omega_1)$. Use the pressing-down lemma to show that if $\mathscr{U}$ is an open cover of $[0,\omega_1)$, then there is an $\alpha\in X$ such that
$$\bigcup\{U\in\mathscr{U}:\alpha\in U\}\supseteq[\alpha,\omega_1)\,,$$
and conclude that there is a $U\in\mathscr{U}$ such that $\operatorname{st}(U,\mathscr{U})\supseteq(\alpha,\omega_1]$ and hence $X\setminus\operatorname{st}(U,\mathscr{U})$ is compact.


TITLE: Recurrence relation of sum with binomial coefficients
QUESTION [0 upvotes]: Is there any way of solving for S(n) to obtain a formula that isn't a recurrence relation for: $$S(n)=\frac{ 1}{4 }\sum_{m=0}^n\binom{ n}{m }S(m)$$
The initial condition is: $S(0)=\frac{\pi^2}{6 }$

REPLY [1 votes]: Hint: Take $S(x)=\sum _{m=0}^{\infty}S(m)\frac{x^m}{m!},$ then $$S(x)e^x=\sum _{n=0}^{\infty}\frac{x^m}{n!}\sum _{m+k=n}\binom{n}{m}S(m)=4(S(x)-S(0))+S(0)=4S(x)-3S(0),$$ so $$S(x)=\frac{3S(0)}{4-e^x},$$
can you conclude? If not, take a look at Eulerian Polynomials.