TITLE: Bayesian parameter estimation for equation $r_t = Z_t^T\beta$?
QUESTION [3 upvotes]: Consider $r_t = \mathbb{E}[Z_t]^T\beta$; $t= 1, 2, 3, \cdots$, where $r_t \in \mathbb{R}$, $\beta \in \mathbb{R}^d$ and $Z_t$ is a $d$-dimensional random vector.
Let's say $\beta$ is an unknown parameter and we would like to estimate it with a Bayesian approach by observing $r_t, Z_t$ for $t=1, 2, 3, \cdots$ (There is no iid assumption on random variables $Z_t$ and their distribution is unknown).How should we proceed? Any ideas?
My Approach: In the case that $d=1$, we may say $\mathbb{E}[Z_t] = \frac{1}{\beta}r_t$, and then say that assuming $\frac{1}{\beta} \sim f_\theta$ as a prior and $Z_t = \frac{1}{\beta}r_t$ we get a prior on $Z_t$. Then, we can update $\theta$ with posterior based on our observation $Z_t = z_t$. I have two questions here:

First, is my understanding correct in the case that $d=1$?
How can I extend it to $d > 1$?

REPLY [1 votes]: These are just some ideas for a way to approach this.
Assuming that $r_t = E[Z_t]^T\beta = \beta^TE[Z_t]$, we can place $d(\mu_t, \sigma I_p)$ priors with some location and scale parameter on $Z_t$. If we place $N(\mu_t, \sigma I_p)$ priors we get some analytical tractability.
That is if $Z_t \sim \textrm{N}(\mu_t,\sigma^2 I_p)$, then $r_t \sim \textrm{N}(\beta^T\mu_t,\sigma^2\beta^T\beta)$, and we can place a prior on $\mu_t$, $\beta$, and $\sigma$. It's easier to assume that you at least know to what degree you're uncertain about your estimates $Z_t$ of $E[Z_t]$; meaning that $\sigma^2$ is a parameter to be specified.
In that setting, if $\mu_t \overset{iid}{\sim}\textrm{N}(0,\tau^2 I)$, where $\tau$ is a parameter to give a dispersed uninformative prior, and $\beta \sim \textrm{N}(0,\phi^2 I)$.
The likelihood for the model is then:
\begin{equation}
\mathcal{L}(r_t | \mu_t, \beta) = \frac{1}{\sqrt{2\pi\sigma^2\beta^T\beta}}\exp\{-\frac{1}{\sigma^2\beta^T\beta}(r_t - \beta^T\mu_t)\}
\end{equation}
The full posterior is then proportional to:
\begin{equation}
\pi(\mu_t, \beta | r_t) \propto \frac{1}{\sqrt{2\pi\sigma^2\tau^2\phi^2 \beta^T\beta}}\exp\{-\frac{1}{\sigma^2\beta^T\beta}\sum_{t}(r_t - \beta^T\mu_t)^2\}\exp\{-\frac{1}{\tau^2}\beta^T\beta\}\prod_{t}(\exp\{-\frac{1}{\phi^2}\mu_t^T\mu_t\})
\end{equation}
Which seems like a pretty monstrous posterior to analyze, but should be fairly straightforward using Metropolis Hastings.


TITLE: Local minimums and maximums of function of three variables
QUESTION [2 upvotes]: I got such function:
$f(x_1, x_2, x_3) = x_1 x_2 x_3(4-x_1-x_2-x_3)$
I need to find all local minimums and maximums of this function.
I calculated partial derivatives and I got that the only points where extremums of this function can be located are:
$(1,1,1)$, $(0,0,a)$, $(0, a, 0)$, $(a, 0, 0)$, where $a$ is real.
I don't know how to deal with this $a$, when calculating this matrix for determining if an extremum is a minimum/maximum.
How to deal with it? Can someone show please?

REPLY [1 votes]: So the function is
$$
f(x,y,z) = 4xyz - x^2yz - xy^2z - xyz^2
$$
so
$$
\nabla f
 = \begin{pmatrix}
4yz - 2xyz - y^2z - yz^2\\
4xz - 2xyz - x^2z - xz^2\\
4xy - 2xyz - xy^2 - x^2y
\end{pmatrix}
 = \begin{pmatrix}
yz(4 - 2x - y - z)\\
xz(4 - 2y - x - z)\\
xy(4 - 2z - y - x)
\end{pmatrix}
$$
and we must solve $\nabla f = \vec{0}$. The first equation yields 3 cases:
Case I $x \ne 0, y \ne 0, z \ne 0$
Then,
$$
\begin{pmatrix}
2x + y + z = 4\\
2y + x + z = 4\\
2z + y + x = 4
\end{pmatrix}
$$
so $x=y=z=1$.
Case II $y=0$
The only meaningful equation is then the second one, which reduces to $xz(4-x-z)=0$. So we get $(0,0,a), (a,0,0), (4-a,0,a)$ for any real $a$.
All other symmetric cases to the ones we dealt with (which have zeros for one or more of the variables), so you get (1,1,1) and 6 families of solutions: $(a,0,0), (0,a,0), (0,0,a), (4-a,0,a), (0,4-a,a), (a,4-a,0)$.
Plug these back into $f$ to get 0 for all of them except $f(1,1,1)=1$. To classify these, as real minima or saddle points, you can use the 2nd derivative tests. So you possibly have an uncountable infinity of minima which all result in 0 value, and one maximum with value of 1.
Note that the absolute minimum of this surface occurs as $-\infty$, so the minima candidates are only local, not absolute.


text:
\begin{document}

\numberwithin{equation}{section}

\newtheorem{theorem}{Theorem}[section]

\newtheorem{proposition}[theorem]{Proposition}

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\def\theconjecture{\unskip}
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\def\complex{{\mathbb C}\/}
\def\dist{\operatorname{dist}\,}
\def\spec{\operatorname{spec}\,}
\def\interior{\operatorname{int}\,}
\def\trace{\operatorname{tr}\,}
\def\cl{\operatorname{cl}\,}
\def\essspec{\operatorname{esspec}\,}
\def\range{\operatorname{\mathcal R}\,}
\def\kernel{\operatorname{\mathcal N}\,}
\def\dom{\operatorname{\mathcal D}\,}
\def\linearspan{\operatorname{span}\,}
\def\lip{\operatorname{Lip}\,}
\def\sgn{\operatorname{sgn}\,}
\def\Z{ {\mathbb Z} }
\def\e{\varepsilon}
\def\p{\partial}
\def\rp{{ ^{-1} }}
\def\Re{\operatorname{Re\,} }
\def\Im{\operatorname{Im\,} }
\def\dbarb{\bar\partial_b}
\def\eps{\varepsilon}
\def\O{\Omega}
\def\Lip{\operatorname{Lip\,}}

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\author{Siqi Fu}
\thanks
{The author was supported in part by NSF grant DMS-0805852.}
\address{Department of Mathematical Sciences,
Rutgers University-Camden, Camden, NJ 08102}
\email{sfu@camden.rutgers.edu}
\title[]
{Positivity of the $\bar\partial$-Neumann Laplacian}
\maketitle

\leftline{\it Dedicated to Professor Linda Rothschild}

\bigskip

\begin{abstract}
We study the $\bar\partial$-Neumann Laplacian from spectral theoretic perspectives.
In particular, we show how pseudoconvexity of a bounded domain is characterized by positivity of the $\bar\partial$-Neumann Laplacian.

\bigskip
\noindent{{\sc Mathematics Subject Classification} (2000): 32W05.}

\smallskip

\noindent{{\sc Keywords}: Pseudoconvex, $\bar\partial$-Neumann Laplacian, Dolbeault cohomology, $L^2$-cohomology.}
\end{abstract}

\bigskip

\section{Introduction}\label{sec:intro}

Whether or not a given system has positive ground state energy is
a widely studied problem with significant repercussions in physics, particularly in quantum mechanics. It follows from the classical Hardy inequality that the bottom of the spectrum
of the Dirichlet Laplacian on a domain in $\R^n$ that satisfies the outer cone
condition is positive if and only if its inradius is finite (see~\cite{Davies95}).
Whereas spectral behavior of the Dirichlet Laplacian is insensitive to boundary
geometry, the story for the $\bar\partial$-Neumann Laplacian is different.
Since the work of Kohn \cite{Kohn63, Kohn64} and H\"{o}rmander \cite{Hormander65},
it has been known that existence and regularity of the $\bar\partial$-Neumann Laplacian
closely depend on the underlying geometry (see the surveys \cite{BoasStraube99, Christ99, DangeloKohn99, FuStraube01} and the monographs \cite{ChenShaw99, Straube09}).

Let $\Omega$ be a domain in $\C^n$.  It follows from the classical Theorem B of Cartan
that if $\Omega$ is pseudoconvex, then the Dolbeault cohomology groups $H^{0, q}(\Omega)$
vanish for all $q\ge 1$. (More generally, for any coherent analytic sheaf $\scriptf$ over
a Stein manifold, the sheaf cohomology groups $H^q(X, \scriptf)$ vanish for all $q\ge 1$.) The converse is also true (\cite{Serre53}, p.~65). Cartan's Theorem B and its converse were generalized by Laufer \cite{Laufer66} and Siu \cite{Siu67} to a Riemann domain over a Stein manifold.  When $\Omega$ is bounded, it follows from H\"{o}rmander's $L^2$-existence theorem for the $\dbar$-operator that if $\Omega$ is in addition pseudoconvex, then the $L^2$-cohomology groups $\widetilde H^{0, q}(\Omega)$ vanish for $q\ge 1$. The converse of H\"{o}rmander's theorem also holds, under the assumption that the interior of the closure of $\Omega$ is the domain itself.  Sheaf theoretic arguments for the Dolbeault cohomology groups can be modified to give a proof of this fact (cf.~\cite{Serre53, Laufer66, Siu67, Broecker83, Ohsawa88}; see also \cite{Fu05} and Section~\ref{sec:finite} below).

In this expository paper, we study positivity of the $\dbar$-Neumann Laplacian, in connection with the above-mentioned classical results, through the lens of spectral theory.  Our emphasis is on the interplay between spectral behavior of the $\dbar$-Neumann Laplacian and the geometry of the domains. This is evidently motivated by Marc Kac's famous question ``Can one hear the shape of a drum?" \cite{Kac66}.  Here we are interested in determining the geometry of a domain in $\C^n$ from the spectrum of the $\dbar$-Neumann Laplacian. (See \cite{Fu05, Fu08} for related results.) We make an effort to present a more accessible and self-contained treatment, using extensively spectral theoretic language but bypassing sheaf cohomology theory.

\section{Preliminaries}\label{sec:pre}

In this section, we review the spectral theoretic setup for the $\bar\partial$-Neumann
Laplacian. The emphasis here is slightly different from the one in the extant literature (cf.~\cite{FollandKohn72, ChenShaw99}).  The $\dbar$-Neumann Laplacian is defined through its associated quadratic form. As such, the self-adjoint
property and the domain of its square root come out directly from the definition.

Let $Q$ be a non-negative, densely defined, and closed
sesquilinear form on a complex Hilbert space $H$ with domain $\dom(Q)$. Then $Q$
uniquely determines a non-negative and
self-adjoint operator $S$ such that
$\dom(S^{1/2})=\dom(Q)$ and
\[
Q(u, v)=\langle S^{1/2}u, \; S^{1/2}v\rangle
\]
for all $u, v\in \dom(Q)$. (See Theorem 4.4.2 in \cite{Davies95}, to which we refer the reader
for the necessary spectral theoretic background used in this paper.)
For any subspace
$L\subset\dom(Q)$, let $\lambda(L)=\sup\{Q(u, u) \mid  u\in L,
\|u\|=1\}$.  For any positive integer $j$, let
\begin{equation}\label{minmax}
\lambda_{j} (Q)=\inf\{\lambda(L) \mid L\subset \dom(Q),
\dim(L)=j\}.
\end{equation}
The resolvent set $\rho(S)$ of $S$ consists of all $\lambda\in\C$ such that
the operator $S-\lambda I\colon\dom(S)\to H$ is both one-to-one and onto (and hence has
a bounded inverse by the closed graph theorem). The spectrum $\sigma(S)$, the complement of $\rho(S)$ in $\C$, is a non-empty closed subset of $[0, \ \infty)$.
Its bottom $\inf\sigma(S)$ is given by $\lambda_1(Q)$.  The essential spectrum $\sigma_e(S)$ is a closed subset of $\sigma(S)$ that consists of isolated eigenvalues
of infinite multiplicity and accumulation points of the spectrum.
It is empty if and only if $\lambda_{j}(Q)\to\infty$ as $j\to\infty$. In this case, $\lambda_{j}(Q)$ is the $j^{\text
{th}}$ eigenvalue of $S$, arranged in increasing order and repeated according to multiplicity.
The bottom of the essential spectrum $\inf\sigma_e(T)$ is the limit of $\lambda_j(Q)$ as $j\to\infty$. (When $\sigma_e(S)=\emptyset$, we set $\inf\sigma_e(S)=\infty$.)

Let $T_k\colon H_k\to
H_{k+1}$, $k=1, 2$, be densely defined and closed operators on
Hilbert spaces. Assume that $\range(T_1)\subset \kernel(T_2)$,
where $\range$ and $\kernel$ denote the range and kernel of the
operators. Let $T^*_k$ be the Hilbert space adjoint of $T_k$,
defined in the sense of Von Neumann by
\[
\dom(T^*_k)=\{u\in H_{k+1} \mid
\exists C>0, |\langle u, T_k v\rangle|\le C \|v\|,
\forall v\in \dom(T_k)\}
\]
and
\[
\langle T^*_k u, v\rangle=\langle u, T_k v\rangle, \quad \text{for all }
u\in\dom(T^*_k) \text{ and } v\in\dom(T_k).
\]
Then
$T^*_k$ is also densely defined and closed.  Let
\[
Q(u, v)=\langle T^*_1 u, T^*_1 v\rangle+\langle T_2 u, T_2 v\rangle
\]
with its domain given by $\dom(Q)=\dom(T^*_1)\cap\dom(T_2)$. The following proposition elucidates the above approach to the $\dbar$-Neumann Laplacian.

\begin{proposition}\label{prop:spectral1} $Q(u, v)$ is a densely defined, closed, non-negative sesquilinear form. The associated self-adjoint operator $\square$ is given by
\begin{equation}\label{eq:square}
\dom(\square)=\{f\in H_2 \mid f\in\dom(Q), T_2 f\in \dom(T_2^*), T_1^* f\in\dom(T_1)\},
\quad \square=T_1T_1^*+T_2^*T_2.
\end{equation}
\end{proposition}

\begin{proof} The closedness of $Q$ follows easily from that of $T_1$ and $T_2$. The non-negativity is evident. We now prove that $\dom(Q)$ is dense in $H_2$. Since
$\kernel(T_2)^\perp=\overline{\range(T_2^*)}\subset \kernel(T_1^*)$ and
\[
\dom(T_2)=\kernel (T_2)\oplus\big(\dom(T_2)\cap \kernel(T_2)^\perp\big),
\]
we have
\[
\dom(Q)=\dom(T_1^*)\cap\dom(T_2)=\big(\kernel (T_2)\cap\dom(T_1^*)\big)\oplus\big(\dom(T_2)\cap \kernel(T_2)^\perp\big).
\]
Since $\dom(T_1^*)$ and $\dom(T_2)$ are dense in $H_2$, $\dom(Q)$ is dense in $\kernel(T_2)\oplus
\kernel (T_2)^\perp=H_2$.

It follows from the above definition of $\square$ that $f\in\dom(\square)$ if and only if $f\in\dom(Q)$
and there exists a $g\in H_2$ such that
\begin{equation}\label{eq:q}
Q(u, f)=\langle u, g\rangle, \quad \text{for all } u\in\dom(Q)
\end{equation}
(cf.~Lemma 4.4.1 in \cite{Davies95}). Thus
\[
\dom(\square)\supset \{f\in H_2 \mid f\in\dom(Q), T_2 f\in \dom(T_2^*), T_1^* f\in\dom(T_1)\}.
\]
We now prove the opposite containment. Suppose $f\in\dom(\square)$. For any $u\in\dom(T_2)$, we write $u=u_1+u_2\in (\kernel(T_1^*)\cap\dom(T_2))\oplus \kernel(T_1^*)^\perp$. Note that $\kernel(T_1^*)^\perp \subset\range(T_2^*)^\perp=\kernel(T_2)$. It follows from
\eqref{eq:q} that
\[
|\langle T_2 u, T_2 f\rangle|=|\langle T_2 u_1, T_2 f\rangle=|Q(u_1, f)|=|\langle u_1, g\rangle|\le \|u\|\cdot\|g\|.
\]
Hence $T_2 f\in\dom(T_2^*)$. The proof of $T_1^* f\in\dom(T_1)$ is similar. For any $w\in\dom(T_1^*)$, we write $w=w_1+w_2\in (\kernel(T_2)\cap\dom(T_1^*))\oplus\kernel(T_2)^\perp$. Note that $\kernel(T_2)^\perp=\overline{\range(T_2^*)}\subset \kernel(T_1^*)$. Therefore, by \eqref{eq:q},
\[
|\langle T^*_1 w, T^*_1 f\rangle|=|\langle T_1^* w_1, T^*_1 f\rangle=|Q(w_1, f)|=|\langle w_1, g\rangle|\le \|w\|\cdot\|g\|.
\]
Hence $T_1^* f\in\dom(T_1^{**})=\dom(T_1)$. It follows from the definition of $\square$ that for any $f\in\dom(\square)$ and $u\in\dom(Q)$,
\[
\langle\square f, u\rangle=\langle \square^{1/2} f, \square^{1/2} u\rangle=Q(f, u)=\langle T_1^* f, T_1^*u\rangle + \langle T_2 f, T_2 u\rangle
=\langle (T_1T_1^*+T^*_2T_2)f, u\rangle.
\]
Hence $\square=T_1T_1^*+T^*_2T_2$. \end{proof}

The following proposition is well-known (compare
\cite{Hormander65}, Theorem 1.1.2 and Theorem 1.1.4; \cite{Catlin83}, Proposition 3;
and \cite{Shaw92}, Proposition 2.3). We provide a proof here for
completeness.

\begin{proposition}\label{prop:spectral2} $\inf\sigma(\square)>0$ if and only if
$\range(T_1)=\kernel(T_2)$ and $\range(T_2)$ is closed.
\end{proposition}

\begin{proof}
Assume $\inf\sigma(\square)>0$. Then $0$ is in the resolvent set of
$\square$ and hence $\square$ has a bounded inverse $G\colon H_2\to\dom(\square)$.
For any $u\in H_2$, write $u=T_1 T_1^*Gu +T_2^*T_2 Gu$. If $u\in
\kernel(T_2)$, then $0=(T_2u, \ T_2 Gu)=(T_2T_2^*T_2Gu,\ T_2Gu)
=(T_2^* T_2Gu, \ T_2^*T_2Gu)$.  Hence $T_2^*T_2Gu=0$ and
$u=T_1T_1^*Gu$. Therefore, $\range(T_1)=\kernel(T_2)$. Similarly,
$\range(T_2^*)=\kernel(T_1^*)$. Therefore $T_2^*$ and hence $T_2$ have closed
range. To prove the opposite implication,
we write $u=u_1+u_2 \in \kernel(T_2)\oplus\kernel(T_2)^\perp$, for any $u\in\dom(Q)$.
Note that $u_1, u_2\in\dom(Q)$. It follows
from $\kernel(T_2)=\range(T_1)$ and the closed range property of $T_2$
that there exists a positive constant $c$ such that $c\|u_1\|^2\le \|T_1^* u_1\|^2$
and $c\|u_2\|^2\le \|T_2 u_2\|^2$. Thus
\[
c\|u\|^2=c(\|u_1\|^2+\|u_2\|^2)\le
\|T_1^* u_1\|^2+\|T_2 u_2\|^2=Q(u, u).
\]
Hence $\inf\sigma (\square)\ge c>0$ (cf.~Theorem 4.3.1 in \cite{Davies95}).
\end{proof}

Let $\kernel(Q)=\kernel(T_1^*)\cap\kernel(T_2)$. Note that when it is non-trivial,
$\kernel(Q)$ is the eigenspace of the zero eigenvalue of $\square$. When $\range(T_1)$ is closed, $\kernel(T_2)=\range(T_1)\oplus\kernel(Q)$. For a subspace $L\subset H_2$, denote by $P_{L^\perp}$ the orthogonal projection onto $L^\perp$ and $T_2|_{L^\perp}$ the restriction of $T_2$ to $L^\perp$. The next proposition clarifies and strengthens the second part of Lemma 2.1 in \cite{Fu05}.

\begin{proposition}\label{prop:spectral3} The following statements are equivalent:
\begin{enumerate}
\item $\inf\sigma_e(\square)>0$.

\item $\range(T_1)$ and $\range(T_2)$ are closed and $\kernel(Q)$ is
finite dimensional.

\item There exists a finite dimensional subspace $L\subset \dom(T_1^*)\cap\kernel(T_2)$
such that $\kernel(T_2)\cap L^\perp =P_{L^\perp}(\range(T_1))$ and $\range(T_2|_{L^\perp})$ is closed.
\end{enumerate}
\end{proposition}

\begin{proof} We first prove (1) implies (2). Suppose $a=\inf\sigma_e (\square)>0$.  If $\inf\sigma(\square)>0$, then $\kernel(Q)$ is trivial and (2) follows from Proposition~\ref{prop:spectral2}.  Suppose $\inf\sigma(\square)=0$. Then $\sigma(\square)\cap [0, a)$ consists only of isolated points, all of which are eigenvalues of finite multiplicity of
$\square$ (cf. Theorem 4.5.2 in \cite{Davies95}).  Hence $\kernel(Q)$, the eigenspace of the eigenvalue $0$, is
finite dimensional.  Choose a sufficiently small $c>0$ so that $\sigma(\square)\cap [0, c)=\{0\}$. By the spectral theorem for self-adjoint operators (cf. Theorem 2.5.1 in \cite{Davies95}), there
exists a finite regular Borel measure $\mu$ on $\sigma(\square)\times\N$ and a unitary transformation $U\colon H_2\to L^2(\sigma(\square)\times \N, d\mu)$ such that $U\square U^{-1}=M_x$, where $M_x \varphi(x, n)=x \varphi(x, n)$ is the multiplication operator by $x$ on
$L^2(\sigma(\square)\times\N, d\mu)$.  Let $P_{\kernel(Q)}$ be the orthogonal projection onto $\kernel(Q)$. For any $f\in\dom(Q)\cap \kernel(Q)^\perp$,
\[
UP_{\kernel(Q)} f= \chi_{[0, c)} Uf =0,
\]
where $\chi_{[0, c)}$ is the characteristic function of $[0, c)$. Hence $Uf$
is supported on $[c, \infty)$.  Therefore,
\[
Q(f, f)=\int_{\sigma(\square)\times\N} x |Uf|^2\, d\mu\ge c\|Uf\|^2=c\|f\|^2.
\]
It then follows from Theorem 1.1.2 in \cite{Hormander65} that both $T_1$ and $T_2$ have
closed range.

To prove (2) implies (1), we use Theorem 1.1.2 in \cite{Hormander65} in
the opposite direction: There exists a positive constant $c$ such that
\begin{equation}\label{eq:h}
c\|f\|^2\le Q(f, f), \quad \text{for all } f\in\dom(Q)\cap\kernel(Q)^\perp.
\end{equation}
Proving by contradiction, we assume $\inf\sigma_e(\square)=0$. Let $\eps$ be
any positive number less than $c$. Since $L_{[0, \eps)}=\range(\chi_{[0, \eps)}(\square))$
is infinite dimensional (cf. Lemma 4.1.4 in \cite{Davies95}), there exists a non-zero $g\in L_{[0, \eps)}$
such that $g\perp \kernel(Q)$.  However,
\[
Q(g, g)=\int_{\sigma(\square)\times\N} x\chi_{[0, \eps)}|Ug|^2 \, d\mu \le \eps \|Ug\|^2=\eps\|g\|^2,
\]
contradicting \eqref{eq:h}.

We do some preparations before proving the equivalence of (3) with (1) and (2).
Let $L$ be any finite dimensional subspace of $\dom(T_1^*)\cap\kernel(T_2)$.
Let $H_2'=H_2\ominus L$. Let
$T_2'=T_2\big|_{H_2'}$ and let ${T_1^*}'=T_1^*\big|_{H_2'}$.  Then
$T_2'\colon H_2'\to H_3$ and ${T_1^*}'\colon H_2'\to H_1$ are
densely defined, closed operators. Let $T_1'\colon H_1\to H_2'$ be
the adjoint of ${T_1^*}'$.  It follows from the definitions that
$\dom(T_1)\subset \dom(T_1')$.  The finite dimensionality of $L$
implies the opposite containment. Thus, $\dom(T_1)=\dom(T_1')$.
For any $f\in\dom(T_1)$ and $g\in\dom({T_1^*}')=\dom(T_1^*)\cap L^\perp$,
\[
\langle T'_1 f, g\rangle =\langle f, {T_1^*}'g\rangle=\langle f, T_1^* g\rangle
=\langle T_1 f, g\rangle.
\]
Hence $T_1'=P_{L^\perp}\circ T_1$ and $\range(T_1')=P_{L^\perp}(\range(T_1))\subset \kernel(T_2')$.  Let
\[
Q'(f, g)=\langle {T_1'}^* f, {T_1'}^*g\rangle+\langle T_2' f, T_2'g\rangle
\]
be the associated sesquilinear form on $H_2'$ with $\dom(Q')=\dom(Q)\cap L^\perp$.

We are now in position to prove that (2) implies (3). In this case, we can take
$L=\kernel(Q)$.  By Theorem 1.1.2 in \cite{Hormander65}, there exists a
positive constant $c$ such that
\[
Q(f, f)=Q'(f, f)\ge c\|f\|^2, \quad \text{for all } f\in\dom(Q').
\]
We then obtain (3) by applying Proposition~\ref{prop:spectral2} to
$T_1'$, $T_2'$, and $Q'(f, g)$.

Finally, we prove (3) implies (1). Applying Proposition~\ref{prop:spectral2} in
the opposite direction, we know that there exists a positive constant $c$ such that
\[
Q(f, f)\ge c\|f\|^2, \quad\text{for all } f\in\dom(Q)\cap L^\perp.
\]
The rest of the proof follows the same lines of the above proof of the implication $(2)\Rightarrow (1)$, with $\kernel(Q)$ there replaced by $L$.
\end{proof}

We now recall the definition of the $\dbar$-Neumann Laplacian
on a complex manifold. Let $X$ be a complex hermitian manifold of dimension $n$.
Let $C^\infty_{(0, q)}(X)=C^\infty(X, \Lambda^{0, q}T^*X)$ be
the space of smooth $(0, q)$-forms on $X$.
Let $\dbar_q\colon C^\infty_{(0, q)}(X)\to C^\infty_{(0, q+1)}(X)$
be the composition of the exterior differential
operator and the projection onto $C^\infty_{(0, q+1)}(X)$.

Let $\Omega$ be a domain in $X$. For $u, v\in C^\infty_{(0, q)}(X)$,
let $\langle u, v\rangle$ be the
point-wise inner product of $u$ and $v$, and let
\[
\langle\langle u, v\rangle\rangle_\Omega=\int_\Omega \langle u,
v\rangle dV
\]
be the inner product of $u$ and $v$ over $\Omega$.  Let $L^2_{(0,
q)}(\Omega)$ be the completion of the space of compactly supported forms in
$C^\infty_{(0, q)}(\Omega)$ with respect to the above inner
product.  The operator $\dbar_q$ has a closed extension on
$L^2_{(0, q)}(\Omega)$. We also denote the closure by $\dbar_q$.
Thus $\dbar_q\colon L^2_{(0, q)}(\Omega)\to L^2_{(0, q+1)}(\Omega)$
is densely defined and closed. Let $\dbarstar_q$ be its adjoint.
For $1\le q\le n-1$, let
\[
Q_q(u, v)=\langle\langle\dbar_q u, \dbar_q
v\rangle\rangle_\Omega+\langle\langle\dbarstar_{q-1} u,
\dbarstar_{q-1} v\rangle\rangle_\Omega
\]
be the sesquilinear form on $L^2_{(0, q)}(\Omega)$ with domain
$\dom(Q_{q})=\dom(\dbar_q)\cap \dom(\dbarstar_{q-1})$.
The self-adjoint operator $\square_{q}$ associated with
$Q_{q}$ is  called {\it the $\dbar$-Neumann Laplacian} on $L^2_{(0, q)}(\Omega)$.  It is an
elliptic operator with non-coercive boundary conditions \cite{KohnNirenberg65}.

The Dolbeault and $L^2$-cohomology groups on $\Omega$
are defined respectively by
\[
H^{0, q}(\Omega)=\frac{\{f\in C^\infty_{(0, q)}(\Omega)\mid
\dbar_q f=0\}}{\{\dbar_{q-1} g \mid g\in C^\infty_{(0, q-1)}(\Omega)\}} \ \text{ and }\
\widetilde H^{0, q}(\Omega)=\frac{\{f\in L^2_{(0,
q)}(\Omega)\mid \dbar_q f=0\}}{\{\dbar_{q-1} g \mid g\in L^2_{(0, q-1)}(\Omega)\}}.
\]
These cohomology groups are in general not isomorphic. For example, when a complex variety
is deleted from $\Omega$, the $L^2$-cohomology group remains the same but the Dolbeault cohomology group could change from trivial to infinite dimensional.
As noted in the paragraph preceding Proposition~\ref{prop:spectral3}, when $\range(\dbar_{q-1})$ is closed in  $L^2_{(0, q)}(\Omega)$, $\widetilde H^{0, q}(\Omega)\cong \kernel(\square_q)$.
We refer the reader to \cite{Demailly} for an extensive treatise on the subject and to \cite{Hormander65} and \cite{Ohsawa82} for results relating these cohomology groups.

\section{Positivity of the spectrum and essential spectrum}\label{sec:finite}

Laufer proved in \cite{Laufer75} that for any open subset of a Stein manifold, if a Dolbeault cohomology group is finite dimensional, then it is trivial. In this section, we establish the
following $L^2$-analogue of this result on a bounded domain in a Stein mainfold:

\begin{theorem}\label{th:finite} Let $\Omega\subset\subset X$ be a domain in a Stein manifold $X$ with $C^1$ boundary. Let $\square_q$, $1\le q\le n-1$, be the $\dbar$-Neumann Laplacian on $L^2_{(0, q)}(\Omega)$. Assume that $\scriptn(\square_q)\subset W^1(\Omega)$. Then $\inf\sigma(\square_q)>0$ if and only if $\inf\sigma_e(\square_q)>0$.
\end{theorem}

The proof of Theorem~\ref{th:finite} follows the same line of arguments as Laufer's. We provide the details below.

Let $H^\infty(\Omega)$ be the space of bounded holomorphic functions on $\Omega$. For any $f\in H^\infty(\Omega)$, let $M_f$ be the multiplication operator by $f$:
\[
M_f\colon L^2_{(0, q)}(\Omega)\to L^2_{(0, q)}(\Omega), \qquad M_f(u)=fu.
\]
Then $M_f$ induces an endomorphism on $\widetilde H^{0, q}(\Omega)$.  Let $\scripti$ be
set of all holomorphic functions $f\in H^\infty(\Omega)$ such that $M_f=0$ on $\widetilde H^{0, q}(\Omega)$.  Evidently, $\scripti$ is an ideal of $H^\infty(\Omega)$. Assume $\inf\sigma_e(\square_q)>0$. To show that $\widetilde H^{0, q}(\Omega)$ is trivial, it suffices to show that $1\in\scripti$.

\begin{lemma}\label{lm:vector} Let $\xi$ be a holomorphic vector field on $X$ and let $f\in\scripti$. Then
$\xi(f)\in \scripti$.
\end{lemma}

\begin{proof} Let $D=\xi\lrcorner \partial\colon C^\infty_{(0, q)}(\Omega)\to
C^\infty_{(0, q)}(\Omega)$, where $\lrcorner$ denotes the contraction operator.
It is easy to check that $D$ commute with the
$\dbar$ operator. Therefore, $D$ induces an endomorphism on $\widetilde H^{0, q}(\Omega)$.
(Recall that under
the assumption, $\widetilde H^{0, q}(\Omega)\cong \kernel(\square_q)\subset W^1(\Omega)$.) For any $u\in \kernel(\square_q)$,
\[
D(fu)-fD(u)=\xi\lrcorner\partial(fu)-f\xi\lrcorner\partial u=\xi(f) u.
\]
Notice that $\Omega$ is locally starlike near the boundary. Using partition of unity and the Friedrichs Lemma, we obtain $[D(fu)]=0$. Therefore, $\left[\xi(f) u\right]=
\left[D(fu)\right]-\left[fD(u)\right]=\left[0\right]$. \end{proof}

We now return to the proof of the theorem. Let $F=(f_1, \ldots, f_{n+1})\colon X\to \C^{2n+1}$ be a proper embedding of $X$ into $\C^{2n+1}$ (cf.~Theorem 5.3.9 in \cite{Hormander91}). Since $\Omega$ is relatively compact in $X$, $f_j\in H^\infty(\Omega)$. For any $f_j$,  let $P_j(\lambda)$ be the characteristic polynomial of $M_{f_j}\colon \widetilde H^{0, q}(\Omega)\to \widetilde H^{0, q}(\Omega)$. By the Cayley-Hamilton theorem, $P_j(M_{f_j})=0$ (cf.~Theorem~2.4.2 in \cite{HornJohnson}). Thus $P_j(f_j)\in \scripti$.

The number of points in the set $\{(\lambda_1, \lambda_2, \ldots, \lambda_{2n+1})\in\C^{2n+1} \mid P_j(\lambda_j)=0, 1\le j\le 2n+1 \}$ is finite.  Since $F\colon X\to \C^{2n+1}$ is one-to-one, the number of common zeroes of $P_j(f_j(z))$, $1\le j\le 2n+1$, on $X$ is also finite. Denote these zeroes by $z^k$, $1\le k\le N$. For each  $z^k$, let $g_k$ be a function in $\scripti$ whose vanishing order at $z^k$ is minimal. (Since $P_j(f_j)\in\scripti$, $g_k\not\equiv0$.) We claim that $g_k(z^k)\not=0$.  Suppose otherwise $g_k(z^k)=0$.  Since there exists a holomorphic vector field $\xi$ on $X$ with any prescribed holomorphic tangent vector at any given point (cf.~Corollary 5.6.3 in \cite{Hormander91}), one can find an appropriate choice of
$\xi$ so that $\xi(g_j)$ vanishes to lower order at $z^k$. According to Lemma~\ref{lm:vector}, $\xi(g_j)\in\scripti$. We thus arrive at a contradiction.

Now we know that there are holomorphic functions, $P_j(f_j)$, $1\le j\le 2n+1$, and  $g_k$, $1\le k\le N$, that have no common zeroes on $X$. It then follows that there exist holomorphic functions
$h_j$ on $X$ such that
\[
\sum P_j(f_j) h_j +\sum g_k h_k =1.
\]
(See, for example, Corollary 16 on p.~244 in \cite{GunningRossi65}, Theorem 7.2.9 in \cite{Hormander91}, and Theorem~7.2.5 in \cite{Krantz01}. Compare also Theorem 2 in \cite{Skoda72}.) Since $P_j(f_j)\in\scripti, g_k\in \scripti$, and $h_j\in H^\infty(\Omega)$, we have $1\in \scripti$.  We thus conclude the proof of Theorem~\ref{th:finite}.

\begin{remark}
(1) Unlike the above-mentioned result of Laufer on the Dolbeault cohomology groups \cite{Laufer75}, Theorem~\ref{th:finite} is not expected to hold if the boundedness condition on $\Omega$ is removed (compare \cite{Wiegerinck84}). It would be interesting to know whether
Theorem~\ref{th:finite} remains true if the assumption $\scriptn(\square_q)\subset W^1(\Omega)$ is dropped and whether it remains true for unbounded pseudoconvex domains.

(2) Notice that in the above proof, we use the fact that $\range(\dbar_{q-1})$ is closed, as a consequence of the assumption $\inf\sigma_e(\square_q)>0$ by Proposition~\ref{prop:spectral3}. It is well known that for any infinite dimensional Hilbert space $H$, there exists a subspace $R$ of $H$ such that $H/R$ is finite dimensional but $R$ is not closed. However, the construction of such a subspace usually involves Zorn's lemma (equivalently, the axiom of choice). It would be of interest to know whether there exists a domain $\Omega$ in a Stein
 manifold such that $\widetilde H^{0, q}(\Omega)$ is finite dimensional but $\range(\dbar_{q-1})$ is not closed.

 (3) We refer the reader to \cite{Shaw09} for related results on the relationship between triviality and finite dimensionality of the $L^2$-cohomology gourps using the $\dbar$-Cauchy problem. We also refer the reader to \cite{Brinkschulte02} for a related result on embedded $CR$ manifolds.
\end{remark}

\section{Hearing pseudoconvexity}\label{sec:psc}

The following theorem illustrates that one can easily determine pseudoconvexity from
the spectrum of the $\dbar$-Neumann Laplacian.

\begin{theorem}\label{th:psc} Let $\Omega$ be a bounded domain in $\C^n$ such that
$\interior(\cl(\Omega))=\Omega$. Then the following statements are equivalent:
\begin{enumerate}
\item $\Omega$ is pseudoconvex.
\item $\inf\sigma(\square_q)>0$, for all $1\le q\le n-1$.
\item $\inf\sigma_e(\square_q)>0$, for all $1\le q\le n-1$.
\end{enumerate}
\end{theorem}

The implication $(1)\Rightarrow (2)$ is a consequence of H\"{o}rmander's fundamental
$L^2$-estimates of the $\dbar$-operator \cite{Hormander65}, in light of Proposition~\ref{prop:spectral2}, and it holds without the assumption $\interior(\cl(\Omega))=\Omega$. The implications $(2)\Rightarrow (1)$ and $(3)\Rightarrow (1)$ are consequences of the sheaf cohomology theory dated back to Oka and Cartan (cf.~\cite{Serre53, Laufer66, Siu67, Broecker83, Ohsawa88}).  A elementary proof of (2) implying (1), as explained in \cite{Fu05}, is given below. The proof uses sheaf cohomology arguments in \cite{Laufer66}. When adapting Laufer's method to study the $L^2$-cohomology groups, one encounters a difficulty: While the restriction to the complex hyperplane of the smooth function resulting from the sheaf cohomology arguements for the Dolbeault cohomology groups is well-defined, the restriction of the corresponding $L^2$ function is not. This difficulty
was overcome in \cite{Fu05} by appropriately modifying the construction of auxiliary $(0, q)$-forms (see the remark after the proof for more elaborations on this point).

We now show that (2) implies (1). Proving by contradiction, we assume that $\Omega$ is not pseudoconvex.
Then there exists a domain $\widetilde\Omega\supsetneqq\Omega$ such that every holomorphic
function on $\Omega$ extends to $\widetilde\Omega$. Since $\interior(\cl(\Omega))=\Omega$,
$\widetilde\Omega\setminus\cl(\Omega)$ is non-empty. After a
translation and a unitary transformation, we may assume that the
origin is in $\widetilde\Omega\setminus\cl(\Omega)$ and there is
a point $z^0$ in the intersection of
$z_n$-plane with $\Omega$ that is in the same connected component
of the intersection of the $z_n$-plane with $\widetilde\Omega$.

Let $m$ be a positive integer (to be specified later). Let $k_q=n$. For any
$\{k_1, \ldots, k_{q-1}\}\subset \{1, 2, \ldots, n-1\}$, we define
\begin{equation}\label{eq:rm}
u(k_1, \ldots, k_q)=\frac{(q-1)!
(\bar z_{k_1}\cdots\bar z_{k_q})^{m-1}}{ r_m^q}\sum_{j=1}^q (-1)^j \bar{z}_{k_j} d\bar{z}_{k_1}\wedge
\ldots\wedge\widehat{d\bar{z}_{k_j}}\wedge\ldots\wedge
d\bar{z}_{k_q},
\end{equation}
where $r_m=|z_1|^{2m}+\ldots +|z_n|^{2m}$. As usual,
$\widehat{d\bar{z}_{k_j}}$ indicates the deletion of
$d\bar z_{k_j}$ from the wedge product. Evidently,
$u(k_1, \ldots, k_q)\in L^2_{(0, q-1)}(\Omega)$ is a smooth form on
$\C^n\setminus\{ 0\}$.
Moreover, $u(k_1, \ldots, k_q)$ is skew-symmetric with respect to the
indices $(k_1, \ldots, k_{q-1})$.  In particular, $u(k_1, \ldots, k_q)=0$ when two $k_j$'s are identical.

We now fix some notional conventions. Let $K=(k_1,\ldots, k_q)$ and
$J$ a collection of indices from $\{k_1, \ldots, k_q\}$.  Write $d\bar z_K=d\bar z_{k_1}\wedge \ldots\wedge
d\bar z_{k_q}$, $\bar z_K^{m-1}= (\bar z_{k_1}\cdots\bar
z_{k_q})^{m-1}$, and $\widetilde{d\bar
z_{k_j}}=d\bar{z}_{k_1}\wedge
\ldots\wedge\widehat{d\bar{z}_{k_j}}\wedge\ldots\wedge
d\bar{z}_{k_q}$.  Denote by
$(k_1, \ldots, k_q \mid J)$ the tuple of remaining indices after deleting those
in $J$ from $(k_1, \ldots, k_q)$. For example, $(2, 5, 3, 1 \mid (4, 1, 6 \mid 4, 6))=(2, 5, 3)$.

It follows from a straightforward calculation that
\begin{align}
\dbar u(k_1, \ldots, k_q)&=-\frac{q! m
\bar{z}_K^{m-1}} {r_m^{q+1}}\big(r_m d\bar z_K
+ \big(\sum_{\ell =1}^n \bar z_\ell^{m-1} z_\ell^m d\bar
z_\ell\big)\wedge \big(\sum_{j=1}^q (-1)^j \bar
z_{k_j}\widetilde{d\bar z_{k_j}}\big)\big)\notag\\
&=-\frac{q! m\bar
z_K^{m-1}}{r_m^{q+1}} \sum_{\begin{subarray}{c}\ell=1 \\
\ell\not=k_1, \ldots, k_q\end{subarray}}^n z^m_\ell \bar
z_\ell^{m-1}\big(\bar z_\ell d\bar z_K+d\bar
z_\ell\wedge\sum_{j=1}^q(-1)^j \bar z_{k_j}
\widetilde{d\bar z_{k_j}}\big)\notag\\
&=m\sum_{\ell=1}^{n-1} z^m_\ell u(\ell, k_1, \ldots,
k_q).\label{eq:um}
\end{align}
It follows that $u(1, \ldots, n)$ is a $\dbar$-closed $(0, n-1)$-form.

By Proposition~\ref{prop:spectral2}, we have $\range(\dbar_{q-1})=\kernel(\dbar_q)$
for all $1\le q\le n-1$. We now solve
the $\dbar$-equations inductively, using $u(1, \ldots, n)$ as initial
data. Let $v\in L^2_{(0, n-2)}(\Omega)$ be a solution to
$\dbar v=u (1, \ldots, n)$.  For any $k_1\in \{1, \ldots, n-1\}$, define
\[
w(k_1)=-mz^m_{k_1} v+(-1)^{1+k_1} u(1, \ldots, n \mid k_1).
\]
Then it follows from \eqref{eq:um} that $\dbar w(k_1)=0$.  Let
$v(k_1)\in L^2_{(0, n-3)}(\Omega)$ be a solution of $\dbar v(k_1)=w(k_1)$.

Suppose for any $(q-1)$-tuple $K'=(k_1, \ldots, k_{q-1})$ of integers from $\{1, \ldots, n-1\}$, $q\ge 2$, we have constructed $v(K')\in L^2_{(0, n-q-1)}(\Omega)$ such that it is skew-symmetric with respect to the indices and satisfies
\begin{equation}\label{eq:v}
\dbar v(K')=m\sum_{j=1}^{q-1} (-1)^j z^m_{k_j} v(K' \mid
{k_j}) + (-1)^{q+|K'|} u(1, \ldots, n \mid {K'})
\end{equation}
where $|K'|=k_1+ \ldots +k_{q-1}$ as usual.  We now construct a $(0, n-q-2)$-forms $v(K)$
satisfying \eqref{eq:v} for any $q$-tuple $K=(k_1, \ldots, k_q)$ of integers from $\{1, \ldots, n-1\}$ (with $K'$ replaced by $K$).  Let
\[
w(K)=m\sum_{j=1}^q (-1)^j z^m_{k_j} v(K \mid {k_j})
+(-1)^{q+|K|} u(1, \ldots, n \mid  K) .
\]
Then it follows from \eqref{eq:um} that
\begin{align*}
\dbar w (K)&=m\sum_{j=1}^q (-1)^j z^m_{k_j}\dbar v(K \mid
{k_j})
+(-1)^{q+|K|} \dbar u(1, \ldots, n \mid {K})\\
&=m\sum_{j=1}^q (-1)^j z^m_{k_j} \Big( m\sum_{1\le i <j} (-1)^i
z^m_{k_i} v(K \mid  k_j,  k_i) +m\sum_{j<i\le q} (-1)^{i-1}
z^m_{k_i}  v(K \mid  k_j,
 k_i) \\
&\qquad -(-1)^{q+|K|-k_j} u(1, \ldots, n \mid {(K \mid  k_j)})
\Big) +(-1)^{q +|K|} \dbar u(1, \ldots, n \mid  K) \\
&=(-1)^{q+|K|}\big( -m\sum_{j=1}^q (-1)^{j-k_j} z^m_{k_j} u(1,
\ldots, n \mid {(K \mid  k_j)}) +\dbar
u(1, \ldots, n \mid  K)\big) \\
&=(-1)^{q+|K|}\big( -m\sum_{j=1}^q z^m_{k_j} u(k_j, (1, \ldots, n \mid
 K)) + \dbar u(1, \ldots, n \mid  K)\big)=0
\end{align*}
Therefore, by the hypothesis, there exists a $v(K)\in L^2_{(0, n-q-2)}(\Omega)$ such that
$\dbar v(K)=w(K)$. Since $w(K)$ is skew-symmetric with respect to indices $K$, we
may also choose a likewise $v(K)$. This then concludes the inductive step.

Now let
\[
F=w(1, \ldots, n-1)= m\sum_{j=1}^{n-1} z^m_j v(1, \ldots,
j, \ldots, n-1)- (-1)^{n+\frac{n(n-1)}2} u(n),
\]
where $u(n)=-\bar{z}_n^m/r_m$, as given by \eqref{eq:rm}.
Then $F(z)\in L^2(\Omega)$ and $\dbar F(z) =0$.  By the hypothesis, $F(z)$ has a
holomorphic extension to $\widetilde{\Omega}$.  We now restrict $F(z)$ to
the coordinate hyperplane $z'=(z_1, \ldots, z_{n-1})=0$. Notice that
so far we only choose the $v(K)$'s and $w(K)$'s from  $L^2$-spaces. The
restriction to the coordinate hyperplane $z'=0$ is not well-defined.
To overcome this difficulty, we choose $m>2(n-1)$. For sufficiently
small $\e>0$ and $\delta>0$,
\begin{align*}
&\left\{\int_{\{|z'|<\e\}\cap\Omega}
\big|\big(F+(-1)^{n+\frac{n(n-1)}2}u(n)\big)(\delta z', z_n))\big|^2dV(z)\right\}^{1/2}\\
&\qquad\le m \delta^m\e^m \sum_{j=1}^{n-1}\left\{\int_{\{|z'|<\e\}\cap\Omega}
|v(1, \ldots, \hat j, \ldots, n-1)(\delta z', z_n)|^2
dV(z)\right\}^{1/2} \\
&\qquad\le m\delta^{m-2(n-1)}\e^m\sum_{j=1}^{n-1} \|v(1, \ldots,
\hat j, \ldots, n-1)\|_{L^2(\Omega)}.
\end{align*}
Letting $\delta\to 0$, we then obtain
\[
F(0, z_n)=-(-1)^{n+\frac{n(n-1)}2} u(n)(0, z_n)=(-1)^{n+\frac{n(n-1)}2}z_n^{-m}.
\]
for $z_n$ near $z^0_n$.  (Recall that $z^0\in\Omega$ is in the same
connected component of $\{z'=0\}\cap\widetilde\Omega$ as the origin.)
This contradicts the analyticity of $F$ near the origin. We
therefore conclude the proof of Theorem~\ref{th:psc}.

\begin{remark} (1) The above proof of the implication $(2)\Rightarrow (1)$ uses only the fact that
the $L^2$-cohomology groups $\widetilde H^{0, q}(\Omega)$ are trivial for all $1\le q\le n-1$. Under the (possibly) stronger assumption $\inf\sigma(\square_q)>0$, $1\le q\le n-1$, the difficulty regarding the restriction of the $L^2$ function to the complex hyperplane in the proof becomes superficial. In this case, the $\dbar$-Neumannn Laplacian $\square_q$ has a bounded inverse. The interior ellipticity of the $\dbar$-complex implies that one can in fact choose
the forms $v(K)$ and $w(K)$ to be smooth inside $\Omega$, using the canonical solution operator to the $\dbar$-equation. Therefore, in this case, the restriction to $\{z'=0\}\cap\Omega$ is well-defined.  Hence one can choose $m=1$. This was indeed the choice in \cite{Laufer66}, where the forms involved are smooth and the restriction posts no problem.  It is interesting to note that by having the freedom to choose $m$ sufficiently large, one can leave out the use of interior ellipticity. Also, the freedom to choose $m$ becomes
crucial when one proves an analogue of Theorem~\ref{th:psc} for the Kohn Laplacian because the $\dbar_b$-complex is no longer elliptic. The construction of $u(k_1, \ldots, k_q)$ in
\eqref{eq:rm} with the exponent $m$ was introduced in \cite{Fu05} to handle this
difficulty.

(2) One can similarly give a proof of the implication $(3)\Rightarrow (1)$. Indeed, the above proof can be easily modified to show that the finite dimensionality of $\widetilde H^{0, q}(\Omega)$, $1\le q\le n-1$, implies the pseudoconvexity of $\Omega$. In this case, the $u(K)$'s are defined by
\[
u(k_1, \ldots, k_q)=\frac{(\alpha+q-1)! \bar
z_n^{m\alpha}(\bar z_{k_1}\cdots\bar z_{k_q})^{m-1}}{ r_m^{\alpha+
q}}\sum_{j=1}^q (-1)^j \bar{z}_{k_j} d\bar{z}_{k_1}\wedge
\ldots\wedge\widehat{d\bar{z}_{k_j}}\wedge\ldots\wedge
d\bar{z}_{k_q},
\]
where $\alpha$ is any non-negative integers. One now
fixes a choice of $m>2(n-1)$ and let $\alpha$ runs from $0$ to $N$ for a sufficiently
large $N$, depending on the dimensions of the $L^2$-cohomology groups. We refer the reader to \cite{Fu05} for details.

(3) As noted in Sections~\ref{sec:pre} and \ref{sec:finite}, unlike the Dolbeault cohomology case, one cannot remove the assumption $\interior(\cl(\Omega))=\Omega$ or the boundedness
condition on $\Omega$ from Theorem~\ref{th:psc}.  For example, a
bounded pseudoconvex domain in $\C^n$ with a complex analytic variety removed still satisfies condition (2) in Theorem~\ref{th:finite}.

(4) As in \cite{Laufer66}, Theorem~\ref{th:psc} remains true for a Stein manifold. More generally, as a consequence of Andreotti-Grauert's theory \cite{AndreottiGrauert62}, the $q$-convexity of a bounded domain $\Omega$ in a Stein manifold such that $\interior(\cl(\Omega))=\Omega$ is characterized by $\inf\sigma(\square_k)>0$ or $\inf\sigma_e(\square_k)>0$ for all $q\le k\le n-1$.

(5) It follows from Theorem~3.1 in \cite{Hormander04} that for a domain $\Omega$ in a complex hermitian manifold of dimension $n$, if $\inf\sigma_e(\square_q)>0$ for some $q$ between $1$ and $n-1$, then wherever the boundary is $C^3$-smooth, its Levi-form cannot have exactly $n-q-1$ positive and $q$ negative eigenvalues.  A complete characterization of a domain in a complex hermitian manifold, in fact, even in $\C^n$, that has $\inf\sigma_e(\square_q)>0$ or $\inf\sigma(\square_q)>0$ is unknown.

\end{remark}

\bigskip

\noindent{\bf Acknowledgments.} We thank Professor Yum-Tong Siu for drawing our attention to the work of Laufer \cite{Laufer66}, by which our work was inspired. We also thank Professors Mei-Chi Shaw and Emil Straube, and the referee for stimulating discussions and constructive suggestions.

\bibliography{survey}
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config: arxiv


TITLE: What is a concrete example to demonstrate that $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ is NOT a norm-Euclidean domain?
QUESTION [7 upvotes]: I'm talking about the ring of algebraic integers of the form $\frac{a}{2} + \frac{b \sqrt{-19}}{2}$, where $a, b \in \mathbb{Z}$.
This ring is said to be a principal ideal domain but not a Euclidean domain to any suitable function.
Since its not norm-Euclidean, it should be possible to find specific numbers $n, b, q, r \in \mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ such that $n = qb + r$ yet $N(r) > N(b)$. Or would the failure of Euclideanness manifest itself in some other way?
I've looked at a few different books and papers, and alot of them give a concrete example (usually $6$) for $\mathbb{Z}[\sqrt{-5}]$ not being UFD, but no example to show $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ is PID but not Euclidean with respect to some function.

REPLY [5 votes]: I think it is important to remember that Euclid's GCD algorithm is what motivates the definition of a Euclidean domain. For example, in $\mathbb{Z}$, if I ask you to compute $\gcd(7, 30)$, your mind automatically comes up with the answer, but if you have any doubt you can always use Euclid's GCD algorithm to assure yourself the answer is correct. Along the way you might think about how $30 = 4 \times 7 + 2$. But you probably won't think about how $30 = 10 \times 7 - 40$, which is true, but irrelevant to the task at hand because $N(-40) > N(7)$ (here I'm using $N(n) = \sqrt{n^2}$, where $\sqrt{n}$ gives the principal square root, just in case you're curious).
Before we tackle $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$, let's look at a UFD that is a Euclidean domain, $\mathbb{Z}[\sqrt{-2}]$. What is $\gcd(\sqrt{-2}, 2 + 2\sqrt{-2})$? My gut feeling is that the answer is $\sqrt{-2}$. But I'm not sure this is correct, so I use Euclid's algorithm: $2 + 2\sqrt{-2} = (2 - \sqrt{-2}) \times \sqrt{-2} + 0$. Clearly $N(0) < N(\sqrt{-2})$ (now I'm using $N(a + b\sqrt{-2}) = a^2 + 2b^2$). This confirms my hunch that $\gcd(\sqrt{-2}, 2 + 2\sqrt{-2}) = \sqrt{-2}$. Of course this is insufficient to prove that $\mathbb{Z}[\sqrt{-2}]$ is norm-Euclidean, but that was not my purpose here.
But it does suggest that to find an example where "Euclideanness" fails, we need to choose numbers in $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ such that one number is not a divisor of the other, because in that case we can just do $r = 0$ and determine $q$ accordingly. Our norm function is now $N\left(\frac{a}{2} + \frac{b\sqrt{-19}}{2}\right) = \frac{a^2}{4} + \frac{19b^2}{4}$. Let's try $\gcd\left(\frac{3}{2} + \frac{\sqrt{-19}}{2}, 10\right)$. How about $\left(\frac{3}{2} - \frac{\sqrt{-19}}{2}\right) \times \left(\frac{3}{2} + \frac{\sqrt{-19}}{2}\right) + 3$? But $N(3) > N\left(\frac{3}{2} + \frac{\sqrt{-19}}{2}\right)$. And if we try $2\left(\frac{3}{2} - \frac{\sqrt{-19}}{2}\right) \times \left(\frac{3}{2} + \frac{\sqrt{-19}}{2}\right) - 4$ we run into a similar problem: $N(-4) >  N\left(\frac{3}{2} + \frac{\sqrt{-19}}{2}\right)$. Hmm, let's see about $\left(\frac{1}{2} - \frac{\sqrt{-19}}{2}\right) \times \left(\frac{3}{2} + \frac{\sqrt{-19}}{2}\right) + \left(\frac{9}{2} + \frac{\sqrt{-19}}{2}\right)$. This is much worse, for now we have $25 > 7$ rather than just $9 > 7$.
I'm going to try one more thing: $\left(\frac{5}{2} - \frac{\sqrt{-19}}{2}\right) \times \left(\frac{3}{2} + \frac{\sqrt{-19}}{2}\right) + \left(\frac{3}{2} - \frac{\sqrt{-19}}{2}\right)$. This is the worst attempt so far, because $7 = 7$ exclamation mark! If we were working in $\mathbb{Z}$, running into b == r would probably prompt us to q++ (by which I mean q := q + 1) and r := 0. Suffice it to say that trying this here would lead us to $r = \sqrt{-19}$ and clearly $19 > 7$.
It's up to you if you want to continue the wild goose chase or read R. A. Wilson's 2011 proof that $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ is a principal ideal domain but not Euclidean. I will close by remarking that Euclid's GCD algorithm is not the only one for finding the GCD. If you can factorize into primes, you can figure out the GCD that way.


text:
\begin{document}

\title{Persistent current in a thin superconducting wire}

\author{Ilya Vilkoviskiy \\ email: \href{mailto:reminguk@gmail.com}{reminguk@gmail.com}}
\affil{Moscow Institute of Physics and Technology, 141700 Dolgoprudny, Russia}

\affil{I.E. Tamm Department of Theoretical Physics, P.N. Lebedev Physical
Institute,\\ Leninsky ave. 53, 119991 Moscow, Russia}

\affil{Center for Advanced Studies,
Skolkovo Institute of Science and Technology,\\
143026 Moscow, Russia}

\maketitle
\begin{abstract}{In this paper, we explore the persistent current in thin superconducting wires and accurately examine the effects of the phase slips on that current. The main result of the paper is the formula for persistent current in terms of the solutions of certain (nonlinear) integral equation. This equation allows to find asymptotics of the current at long(small) length of the wire, in that paper, we interested in the region in which the system becomes strongly interacting and very few amounts of information can be extracted by perturbation theory. Nevertheless, due to the integrability, exact results for the current can be obtained. We observe that at the limit of a long wire, the current becomes exponentially small, we believe that it is the signal that phase slips may destroy superconductivity for long wires, below BKT phase transition.}
\end{abstract}
\newpage
\tableofcontents
\newpage

\section{Introduction}

    The quantum effects in strongly interacting systems is a very interesting theme. It is known, that thin superconducting wires may lost two general features of superconductors : zero resistivity and existence of dissipationless (persistent) current in an external magnetic flux. This effects was observed at the whole range of temperatures below the critical $T_c$, even at $T \to 0$ \cite{Arutyun}. Which means that effect of superconductivity breaking, is of a quantum nature because thermal fluctuations is suppressed at small  temperatures. It was realised \cite{ZGOZ} that these effects caused by the one particular process - quantum phase slip (QPS). To understand what is the QPS, let us assume that low energy behaviour of the system could be described by the two fields : module and phase of the order parameter $\Delta(x,t)=|\Delta(x,t)| e^{i\phi(x,t)}$ , note that as $\phi$ is a phase, it is defined only modulo $2\pi$ : $\phi \sim \phi +2\pi$. If we look at the two dimensional Euclidean plane $(x,\tau=i t)$ , the phase slip could be thought as the following field $\phi$ configuration
\beq
\phi(x,\tau)=\arctan(\frac{x}{\tau})
\eeq
While $|\Delta|$ is nontrivial only in a small region near the point $x=0, \tau =0$ : $|\Delta| \sim const $ for $x^2+\tau^2 \gg 1$ . Though such solutions have a singularity at $\tau=0$ for small $\tau$ , they looks like a step function, and due to the fact that $\phi$ is a phase very close to the configurations $\phi = const$.
\begin{figure}[h]
\begin{minipage}[h]{1 \linewidth}
\center{\includegraphics[width=0.7\linewidth]{BKT.png}}
\end{minipage}
\caption{BKT phase diagram, $m^2$ - is a phase slips rate. $\alpha^2\sim \sqrt{S}$  , where S is the wire cross section}
\label{wire1}
\end{figure}
    In this paper, we examine the phase and wire length dependence of the persistent current at zero temperature, in order to separate the quantum and thermal fluctuations. It is known that under varying the thickness of the wire the system undergoes a BKT phase transition see fig 1. The phase $\alpha^2>2 $ is a superconducting one and has been investigated in \cite{AS}, the opposite phase $ \alpha^2<2 $ , which as we will argue is insulating one, will be the subject of our interest. This phase hasn't been investigated yet, both because of lack of experiments and also because there is no simple theoretical model for that region. In that paper, we try to fill this gap.

        Generally, in the superconducting phase, we have some macroscopic theory at high energies, then, if the initial coupling is small enough we can use Renormalisation group to scale to the lower energies (and larger distances) and arrive at the weakly interacting system at the limit of large distances. That means that at the large distances we can treat the system perturbatively, effects of interaction becomes stronger when one goes to the smallest distances, here, in order to calculate current one can use quantum mechanical approximation - which neglects coordinate dependence of fields: $\phi(x,t) \to \phi (t)$ \cite{AZS}. Instead in the insulating phase, system is asymptotically free at the small distances and strong interacting at large ones, physically it means that we expect that current will  have simple sawtooth-like behaviour for the rings of small circumference and experience some unknown behaviour at the limit of large (compared to the only dimensional parameter of the system - the phase slips rate) circumference. This fact makes the insulating region interesting both for theoretical and experimental physicists.

            It is very hard to analyze the current in the whole range of lengths because there is no any simple model for wires of a large circumference, instead, one can use underlying integrability in order to derive an exact equation for the current, which then can be solved numerically or analysed exactly in the limits of large or small wires, see appendix D for details.

            The paper is organized as follows: at the section 2 we discuss the physical system and make a review of the effective model for 1-dimensional superconducting rings. At the section 3 we argue what interacting terms could be added to the Lagrangian, and argue that all physics is dominated by the one term - which bring us to the Sine-Gordon theory.  At the section 3 we present the results of the paper. The ground state energy and the current could be written in terms of solutions of Destri-De Vega equation (DDV) , (\ref{DDV1}), the details explained  in the Appendix. In the Appendix A we review dualities between bosons and fermions and relations between their statistical sums (\ref{bosonisation}) .In Appendix B we sketchy explain the relations between XXZ spin chains and Thirring model. Relations between XXZ Bethe Ansatz (\ref{BAE}) and nonlinear equations (\ref{DDV}) which we finally use as the main computational tool explained in the Appendix C. Finally in the Appendix D , we show how to calculate persistent current , using the DDV equation.
\section{Physical system}
 \begin{figure}[h]
\begin{minipage}[h]{1 \linewidth}
\center{\includegraphics[width=0.5\linewidth]{wire.png} }
\end{minipage}
\caption{Superconducting ring of the circumference L, in the external magnetic flux}
\label{wire1}
\end{figure}
We consider a superconducting ring of the circumference L, at the limit of small temperature $T\to 0$. Low energy physics of the wire is described by the order parameter
$\Delta=|\Delta(x,t)|e^{i\phi(x,t)}$ , moreover as fluctuations of the $ |\Delta(x,t)|$ described by a massive field we assume that at low energies $ |\Delta(x,t)|$ is an independent of
$\{x,t\}$ constant. \\
Effective theory described be the Lagrangian , see \cite{AS} for details:\\
\beq
L=\frac{\alpha^2}{4
\pi } \left(\frac{1}{2}(v \partial_t \phi(x,t))^2 -\frac{1}{2}(v^{-1}\partial_x \phi(x,t))^2\right) +m^2 V_{slips} \label{MS}
\eeq

As was argued in \cite{AS} : for the standard dirty metals the only dimensionless parameter which control the physics of the system is $\frac{\alpha^2}{4\pi}=\frac{1}{2e^2}\sqrt{\frac{C_w}{L_{kin}}}$ , $v^2=\frac{\pi \sigma S}{C_w}$ where $C_w$-is the geometric wire capacitance per length and  $L_{kin}=\frac{1}{\pi \sigma_N S}$- is the kinetic wire inductance  , $\sigma_N$ -  is the normal state Drude conductance of the wire and $S$ is the wire cross section.

The term $M V_{slips}$ is introduced to take phase slips into account, and we will investigate it carefully below.
As usual , the current is the gradient of an order parameter phase $J=\frac{\sigma}{2e}\partial_x \phi$, and the charge density $\rho=V C_w=-\frac{C}{e}\partial_t \phi$, so that the continuity equation $\partial_t \rho+\partial_x J=0$ turns to an equation of motion for the field $\phi$
\beq
v^2 \partial_t^2\phi -\partial_x^2\phi=0
\eeq
In order to couple the external electro-magnetic field we should analyze the transformation law of field $\phi$ under the gauge transformations
\begin{eqnarray}
A_{\mu}\to A_{\mu}+\partial_{\mu}f \\
\Delta\to \Delta e^{i f}
\end{eqnarray}
The gauge invariant Lagrangian is
\beq
L=\frac{\alpha^2}{2
\pi } \left(\frac{1}{2}(\partial_t \phi(x,t)-A_{t})^2 -\frac{1}{2}(\partial_x \phi(x,t)-A_x)^2\right) \label{L}
\eeq
In order to introduce external magnetic flux, one can choose $A_{\mu}$ in the form
\begin{eqnarray}
A_t=0 \\
A_x=\frac{\Phi}{L\alpha^2}
\end{eqnarray}
Here we choose the normalization of the flux $\Phi$ in such a way that it is $2\pi$ periodic: statistical sum is periodic function of $\Phi$ with the period $2\pi$.
\\
To the end of this section, let us note that velocity in the Lagrangian [\ref{L}] can be scaled out by redefinition of coordinate $x \to v x$ and time $t\to v t$ , later, for simplicity, we will use this convention.

\subsection{Mathematical model}
We will study the Sine-Gordon theory
\beq
S=\int\limits_0 ^{\beta \sim \infty} \int\limits_{0}^L[\frac{1}{2\pi \alpha^2} \left(\frac{1}{2}(\partial_t \chi(x,t))^2 -\frac{1}{2}(\partial_x \chi(x,t))^2+m^2 \cos(\chi(x,t))\right)- \frac{\Phi}{2\pi L} \partial_t \chi(x,t)] dx dt \label{A}
\eeq
There are several ways to arrive to this effective theory, which takes phase slips into account, in  that paper we try to write it down from general principles, but it also can be derived as an effective theory of phase slips - which are vortices in the imaginary time formalism \cite{AS}
.\\

Starting from the free theory, it is natural to ask in which ways this theory can be deformed to an interacting one and what is the physical meaning of that deformation.
\\

As $\phi$ is  a phase, it is natural to consider following operators
\beq
V_{m}(x,t)=e^{i(m \phi(x,t)} , m\in \mathbb{Z}
\eeq
This operators has fixed conformal dimension $\Delta=\frac{m^2}{2\alpha^2}$, and mutual local to each other, that means that the product
\beq
V_{m_1}(x,t)V_{m_2}(y,\tau)=((x-y)^2+(t-\tau)^2)^{\frac{m_1m_2}{2\alpha^2}}:V_{m_1}(x,t)V_{m_2}(y,\tau):
\eeq
is single valued operator in the complex plane : $z=x+it$ , $\bar{z}=x-it$, here we assume that $:V_{m_1}(x,t)V_{m_2}(y,\tau):$ means the normal ordered product, which is meromorphic in $z$, $\bar{z}$. It is possible to enlarge the algebra of local operators, introducing the dual field
\begin{gather}
\partial_x \chi(x,t)=\partial_t \phi(x,t)\\
\partial_t \chi(x,t)=-\partial_x \phi(x,t)
\end{gather}
Consider the following operators \cite{ZZ} :
\beq
V_{m,n}(z,\bar{z})=e^{i(m \phi(z,\bar{z})+n\alpha^2\chi(z,\bar{z}))}
\eeq
\beq
V_{m_1,n_1}(z,\bar{z})V_{m_2,n_2}(0,0)=|z|^{(\frac{m_1m_2}{\alpha^2}+n_1n_2\alpha^2)}z^{\frac{m_1n_2+n_1m_2}{2}}{\bar{z}}^{-\frac{m_1n_2+n_1m_2}{2}}:V_{m_1,n_1}(z,\bar{z})V_{m_2,n_2}(0,0):
\eeq
Which are mutual local to each other at the integers n, it is imply that $\chi$ is compactifyed on a circle of radius $\frac{1}{\alpha^{2}}$ ,  $\chi\sim \chi+\frac{2\pi}{\alpha^2}$\\

We also want our Lagrangian to be invariant under the global gauge transformations: $\phi \to \phi+f, \chi\to \chi$, general Lagrangian satisfying this restrictions contains only operators $V_{0,n}$, obviously it can be written down in terms of field $\chi$ only (also it is useful to make a little redefinition $\chi\to \frac{1}{\alpha^{2}} \chi$:
\beq
L=\frac{1}{2\pi \alpha^2} \left(\frac{1}{2}(\partial_t \chi(x,t))^2+\frac{1}{2}(\partial_x \chi(x,t))^2+\sum\limits_{n}m^2_n \cos(n \chi(x,t))\right)
\eeq
Note that operators $e^{\pm i(n\chi(z,\bar{z})}$ can be identified with operators of creation/annihilation of vortexes (phase slips) with vorticity n, see \cite{AZS} for more details and physical interpretation.\\
Here the masses $m_n$ is the dimensional parameters, with the dimension
\beq
[m_n]=2(1-\frac{(n\alpha)^2}{4})
\eeq

Easy to understand that at the low energies (high distances L)the ratio $\frac{m_n}{m_1}$ scales to zero as $L^{(1-n^2)\frac{\alpha^2}{4}}$, so, it is natural to leave only first of them. This is a standard way to get a Sine-Gordon Lagrangian.
\beq
L=\frac{1}{2\pi \alpha^2} \left(\frac{1}{2}(\partial_t \chi(x,t))^2 +\frac{1}{2}(\partial_x \chi(x,t))^2+ M^2 \cos(\chi(x,t))\right)
\eeq
It is important to note that this theory have two qualitatively different regimes :$\alpha>2 , \alpha<2$ in the first regime mass M scales to zero at high distances and perturbation theory works perfectly, in the second regime mass M scales to infinity which breaks perturbation theory down. The transition between these two regimes known an BKT transition.\\
In this paper we will work at the region $ \alpha<2$ at which Sine-Gordon theory is integrable. Natural question to be interested in is how that mass parameter affect the superconductivity, the easiest one example is the computation of the ground state energy and persistent current of the Sine-Gordon theory  in the presence of external magnetic flux $\Phi$, at zero temperature $T=\frac{1}{\beta} \to 0$
\beq
S=\int\limits_0 ^{\beta \sim \infty} \int\limits_{0}^L[\frac{1}{2\pi \alpha^2} \left(\frac{1}{2}(\partial_t \chi(x,t))^2 -\frac{1}{2}(\partial_x \chi(x,t))^2+m^2 \cos(\chi(x,t))\right)- \frac{\Phi}{2\pi L} \partial_t \chi(x,t)] dx dt \label{A}
\eeq
\beq
J=\frac{L}{2\pi \beta}\frac{\partial \log(Z(\Phi))}{\partial \Phi}=\frac{L}{2\pi}\frac{\partial E_{vac}(\Phi)}{\partial \Phi}
\eeq

The best way to solve this problem was provided by Destri and De-Vega \cite{Light-Cone}.
Below, I review their results, and apply them to our problem:

\section{Results}
Our aim is to compute statistical sum, ground state energy and current at the limit of zero temperature, here and later we will work in the imaginary time formalism:
\beq
Z=\int D\phi \ \ e^{-S_{cl}[\phi]} \sim e^{-\beta E_{vac}}
\eeq
Where $S_{cl}$ is given by the formula (\ref{A})
\beq
J=\frac{L}{2\pi \beta}\frac{\partial \log(Z(\Phi))}{\partial \Phi}=\frac{L}{2\pi}\frac{\partial E_{vac}(\Phi)}{\partial \Phi}
\eeq

Classical result (at $m=0$) tells that
\beq
E_{vac}(\Phi)=-\frac{\pi}{6L}(1-3\frac{\Phi^2 \alpha^2}{2\pi }) \label{Evac<}
\eeq
\beq
J(\Phi)=\frac{\alpha^2 \Phi}{4\pi^2} \label{Jphi}
\eeq
For the case $m^2\ne0$,  we will write down an intregral equation for a ground state energy (\ref{DDV}) , \cite{DDV}, see appendix for the details.
\beq
\epsilon(\theta)=-M L \cosh(\theta)-i\Phi-G\star\log(1+e^{\epsilon})+G_{c}\star \log(1+e^{\bar{\epsilon}}) \label{DDV1}
\eeq
Where $\star$ is a convolution, and $G_c$ ,   $G$ are known functions (\ref{G})
\beq
G(\theta)=\int\limits_{0}^{\infty} \frac{dk}{4\pi}\frac{\sinh(\pi k(\frac{\pi}{2\gamma}-1))}{\sinh(\frac{\pi k}{2} (\frac{\pi}{\gamma}-1))\cosh(\frac{\pi k}{2})}\cos(k \theta)
\eeq
\beq
G_c(\lambda)=G(\lambda+i\pi)
\eeq
\beq
E=-M \int\limits_{-\infty}^{\infty} \cosh(\theta)\left(\log(1+e^{\epsilon})+\log(1+e^{\bar {\epsilon}})\right) \frac{d\theta}{2\pi} \label{E}
\eeq

It is simple to get from this equation that in the limit of large $M L\to \infty$
\beq
E=-M \left(\frac{2}{\pi}\right)^{3/2}e^{-(M L-\frac{3}{4}\pi)}\cos(\Phi) +o(e^{-2M L}) \label{Evac>}
\eeq
\beq
J=\frac{M L\sqrt{2}}{\pi^{\frac{5}{2}}} e^{-(M L-\frac{3}{4}\pi)}\sin(\Phi) +o(e^{-2M L})
\eeq
Here $M $ is the soliton mass, it is related to the bare mass $m$ through the formula \cite{ZM}
\beq
 M=\left(\frac{\pi m \Gamma\left(\frac{2-2\alpha^2}{2}\right)}{\Gamma\left(\frac{\alpha^2}{2}\right)}\right)^{\alpha^2-2}\frac{2\Gamma\left(\frac{\alpha^2}{2(2-\alpha^2)}\right)}{\sqrt{\pi}\Gamma\left(\frac{1}{2-\alpha^2}\right)}
\eeq
Higher corrections could be also computed order by order. The technique allows one to compute the current numerically, here is the plots of the current $J({\Phi})$, for different values of $M L$
 \begin{figure}[h]
\begin{minipage}[h]{1 \linewidth}
\center{\includegraphics[width=0.75\linewidth]{J12.png} }
\end{minipage}
\caption{For different values of $ML$ , Current $J({\Phi})$ - interpolates between Sawtooth and Sine}
\label{ris:image1}
\end{figure}
\newpage
\section{Discussion}
In this paper we started investigation of the so-called `insulating' region of thin superconducting nanowires. Following the paper of A.Semenov and A.Zaikin we used Sine-Gordon theory as an effective low energy theory for our system. In order to explore superconducting properties of the ring we computed the simplest quantity of superconductors : the persistent current dependence on the external magnetic flux and length of the ring. We found that for the ring of large lengths the current is exponentially suppresed $J\sim e^{-M  L}$. This result can not be obtained in pertrubation theory, instead we used integrability of the theory in order to find asymptotics of the current exactly. Our results agrees with the results of $\cite{AS}$ , which was obtained with the help of renorm group argument, which are not very confident at the regime of strong coupling. Although, our exact computation shows that heuristic argument of renorm group leads to a qualitatively correct results. Also our method reproduce exponential and pre-exponential factors of current dependence as well as systematical computation of higher order of pertrubation series in ($e^{-ML})$.

    Though the current is suppressed in the bulk of superconductor, we can't confidently claim that superconductivity is broken, for example, it would be interesting to compute resistivity of the wire connected to the resistor, which is yet a nontrivial problem.
    We believe that it will be very interesting to discover the region near, and below, the BKT phase transition experimentally. Not only because it may have applications in nanoelectronics, but because it can boost our understanding of Sine-Gordon theory.
\centerline{\bf Acknowledgements}
I thank Andrew G. Semenov for introducing me to the theme, and for interesting discussions during the work. I also thank Andey Marshakov for permanent support and his interest to this study. Finally I would like to thank Andrei D. Zaikin for his comments on earlier versions of the paper.

\newpage
\appendix
\section{Free field theory and boson-fermion correspondence, from Sine-Gordon to Thirring}
In this section, we rigorously show that the theory of one bosonic field $\phi(x,t)$ - compactified on a circle : $\phi(x,t) \sim \phi(x,t)+2\pi $ is equivalent to a theory of Dirac fermions. \\
Consider , for the moment pure Sine Gordon theory:
\beq
S=\int\limits_0 ^{\beta} \int\limits_{0}^L\frac{1}{2\pi \alpha^2} \left(\frac{1}{2}(\partial_t \chi(x,t))^2 -\frac{1}{2}(\partial_x \chi(x,t))^2+m^2 \cos(\chi(x,t))\right) dx dt
\eeq
Field $\phi$ defined to take values in $\mathbb{R}$ ,  but (at least at the classical level) we should specify a boundary conditions, for the applications to a superconductors we should consider all fields of the form :
\begin{gather}
\chi(x+L,t)=\chi(x,t)+2\pi n \label{con1} \\
\chi(x,t+\beta)=\chi(x,t)+2\pi m \label{con2}
\end{gather}
The best way to deal with different sectors is to introduce chemical potentials $\Phi $ , and $\mu$ in the action
\begin{gather}
S_{m,n}=\int\limits_0 ^{\beta} \int\limits_{0}^L\frac{1}{2\pi \alpha^2} \left(\frac{1}{2}(\partial_t \chi(x,t))^2 -\frac{1}{2}(\partial_x \chi(x,t))^2+m^2 \cos(\chi(x,t))\right) dx dt+m\Phi +n\mu
\end{gather}
\beq
Z(\Phi,\mu)=\sum\limits_{m,n \in \mathbb{Z}} Z_{m,n}=\sum\limits_{m,n \in \mathbb{Z}} \int_{m,n} D\phi e^{iS_{m,n}}
\eeq
Where integration $\int_{m,n}$ performed under the fields satisfying conditions (\ref{con1},\ref{con2}).
It is easy to understand that
\beq
Z(\Phi,\mu)=\int D\phi e^{i S} \   \ e^{i\frac{\Phi}{2\pi L}  \int\limits_0 ^{\beta } \int\limits_{0}^L\partial_t \chi(x,t) dx dt+i\frac{\mu}{2\pi \beta}  \int\limits_0 ^{\beta } \int\limits_{0}^L\partial_x \chi(x,t) dx dt }
\eeq
Where integration performed in all sectors. Alternatively
\beq
Z_{m,n}=\int\limits_{0}^{2\pi}\int\limits_{0}^{2\pi} Z(\Phi,\mu)e^{i n\phi +i m\mu}d\mu d\phi
\eeq
We state that the same statistical sum $Z(\Phi,\mu)$ can be obtained from the dual fermionic model (Thirring model) with the fields $\psi(x,t)=\begin{pmatrix} \psi_L(x,t) \\ \psi_R(x,t) \end{pmatrix}$ and Hamiltonian
\beq
H_{Th}=\int\left(\psi^{\dagger} \sigma^3 (-i \partial_x+\frac{\Phi}{L}) \psi+\frac{\mu}{ \beta} \psi^{\dagger}\psi+M \psi^{\dagger} \sigma^2 \psi +2g\psi^{\dagger}_R\psi_R\psi^{\dagger}_L\psi_L  \right)dx \label{Th}
\eeq

This statement can be easily proved by direct computation at the free (and massles) fermions point $g=M=0$, and than one can follow the Coleman's standard argument to understand it holds for any orders of  $g$ and $M$ in perturbation theory.
\subsection{Free boson}
\beq
S=\int\limits_0 ^{\beta} \int\limits_{0}^L\frac{1}{2\pi \alpha^2} \left(\frac{1}{2}(\partial_t \chi(x,t))^2 -\frac{1}{2}(\partial_x \chi(x,t))^2+\frac{\Phi}{L} \partial_t \chi(x,t) +\frac{\mu}{\beta}  \partial_x \chi(x,t) \right) dx dt
\eeq
Classical solutions have the form
\beq
\chi(x,t)=\frac{2\pi n x}{L}+\frac{2\pi m t}{\beta}+\sum\limits_{k \in  \mathbb{Z}/\{0\}}\chi_k(t)e^{\frac{2\pi i k x}{L}}
\eeq
$\phi_k(t)$ just the oscillators, each of them make a contribution equal to a $\frac{1}{e^{\pi i k \frac{\beta}{L}}-e^{-\pi i k \frac{\beta}{L}}}$ , and we will get
\beq
Z_{\alpha}(\Phi,\mu)=\prod\limits_{k\ne 0}\frac{1}{e^{\pi i k \frac{\beta}{L}}-e^{-\pi i k \frac{\beta}{L}}}\sum_{m,n}e^{i \pi (\frac{\beta}{L} \frac{M^2}{\alpha^2}-\frac{L}{\beta} \frac{n^2}{\alpha^2})+i\Phi n+i\mu m}
\eeq
Or, after Poisson resummation

\beq
Z_{\alpha}(\Phi,\mu)=\prod\limits_{p\ne 0}\frac{1}{e^{\pi i p \frac{\beta}{L}}-e^{-\pi i p \frac{\beta}{L}}}\sum_{m,n}e^{i \pi (\frac{\beta}{\alpha^2 L} M^2+\frac{\alpha^2\beta}{L}( n+\frac{\Phi}{2\pi})^2)+i\mu m} \label{FB}
\eeq
At the limit of zero temperature $ (\beta\to i\frac{1}{T}\to i\infty)$
\beq
Z\sim e^{-\beta E_{vac}(\Phi)}
\eeq
\beq
E_{vac}(\Phi)=-\frac{\pi}{6L}(1-3\frac{\Phi^2 \alpha^2}{2\pi }) \label{Evac}
\eeq
Which reproduce classical answer for the current of the wire, in the presence of external flux
\beq
J(\Phi)=\frac{\alpha^2 \Phi}{4\pi^2} \label{Jphi}
\eeq
\subsection{Free fermions}
The method of Destri and De-Vega is applied to Thirring model, so it is worth to show directly that , at least at the free level, fermionic theory reproduce the same answer as bosonic.\\
We begin with the free and massles theory of Dirac fermions, and show that it's statistical sum is identical to bosonic one at the point $\alpha=\sqrt{2}$
\beq
H_{ff}=\int\left(\psi^{\dagger} \sigma^3 (-i \partial_x+\frac{\Phi}{L}) \psi+\frac{\mu}{ \beta} \psi^{\dagger}\psi \right)dx
\eeq
\beq
\psi_{\alpha,r}=c_{\alpha,r}e^{\frac{2\pi i}{L} r}
\eeq
\beq
H_{ff}=\sum\limits_{r\in \mathbb{Z}+1/2}\frac{2\pi}{L}c^{\dagger}_{R,-r}c_{R,r}(r+\frac{\Phi}{2\pi}+\frac{\mu L}{2\pi \beta})+\sum\limits_{s\in \mathbb{Z}+1/2}\frac{2\pi}{L} c_{L,s}^{\dagger} c_{L,-s}
(s-\frac{\Phi}{2\pi}+\frac{\mu L}{2\pi \beta})
\eeq
\beq
Z_f(\Phi,\mu)=tr\left(e^{i\beta H_{ff}}\right)
\eeq

\begin{gather}
Z_f(\Phi,\mu)=\prod\limits_{p\ne 0}\frac{1}{e^{\pi i p \frac{\beta}{L}}-e^{-\pi i p \frac{\beta}{L}}}(\sum_{m\in 2\mathbb{Z},n \in \mathbb{Z}}e^{i \pi (\frac{\beta}{2 L} m^2+\frac{2\beta}{L}( n+\frac{\Phi}{2\pi})^2)+i\mu m}+ \nonumber \\+ \sum_{m\in 2\mathbb{Z}+1,n \in \mathbb{Z}+1/2}e^{i \pi (\frac{\beta}{2 L} m^2+\frac{2\beta}{L}( n+\frac{\Phi}{2\pi})^2)+i\mu m})
\end{gather}
If we in addition project to a even m (fermion charge) we obtain
\beq
Z_{f}^{even}(\Phi,\mu)=\prod\limits_{p\ne 0}\frac{1}{e^{\pi i p \frac{\beta}{L}}-e^{-\pi i p \frac{\beta}{L}}}\sum_{m\in 2\mathbb{Z},n \in \mathbb{Z}}e^{i \pi (\frac{\beta}{2 L} m^2+\frac{2\beta}{L}( n+\frac{\Phi}{2\pi})^2)+i\mu m}
\eeq
analogically
\beq
Z_{f}^{odd}(\Phi+\pi,\mu)=\prod\limits_{p\ne 0}\frac{1}{e^{\pi i p \frac{\beta}{L}}-e^{-\pi i p \frac{\beta}{L}}}\sum_{m\in 2\mathbb{Z}+1,n \in \mathbb{Z}}e^{i \pi (\frac{\beta}{2 L} m^2+\frac{2\beta}{L}( n+\frac{\Phi}{2\pi})^2)+i\mu m}
\eeq
Finally, we obtain
\beq
Z^{boson}_{\alpha=\frac{1}{\sqrt{2}}}(\Phi,\mu)=Z_{f}^{even}(\Phi,\mu)+Z_{f}^{odd}(\Phi+\pi,\mu) \label{bosonisation}
\eeq
Which simply means that we have to impose antiperiodic boundary conditions in even charge sector and periodic in odd charge sector. Of course, as the bosonic and fermionic statistical sums are identical, fermionic models reproduce the same answers for the ground state energy and current as the bosonic ones (\ref{Evac},\ref{Jphi}).
\subsection{General $\alpha$ and interacting fermions}

We assme that for a general $\alpha$ ,  formula analogical to (\ref{bosonisation}) holds
\beq
Z^{boson}_{\alpha}(\Phi,\mu)=Z_{Th,g}^{even}(\Phi,\mu)+Z_{Th,g}^{odd}(\Phi+\pi,\mu)
\eeq
Where $Z_{Th,g}^{even/odd}$ is the statistical sum of the Thirring Hamiltonian with the  antiperiodic/periodic boundary conditions in even/odd fermionic charge sector
\beq
H_{Th,g}=\int\left(\psi^{\dagger} \sigma^3 (-i \partial_x+\frac{\Phi}{L}) \psi+\frac{\mu}{ \beta} \psi^{\dagger}\psi+2g\psi^{\dagger}_R\psi_R\psi^{\dagger}_L\psi_L  \right)dx
\eeq
\section{From Thirring to XXZ spin chain}
This section based on the work of Destri and De-Vega \cite{Light-Cone}, see also \cite{LI}\\
Here I just briefly review their results: \\
1) Sine-Gordon can be realised as a certain limit of XXZ spin chain with a number of inhomogeneities\\
2) Spectrum of energy can be analysed with the help of Bethe-Ansatz equations (BAE)\\
\beq
\prod\limits_{n=1}^N\frac{\sinh(\lambda+i\theta_n+i\gamma/2)}{\sinh(\lambda+i\theta_n-i\gamma/2)}=-e^{-2 i \Phi}\prod\limits_{j=1}^{N}\frac{\sinh(\lambda-\lambda_j+i\gamma)}{\sinh(\lambda-\lambda_j-i\gamma)}
\eeq
With the limit $\theta_i =(-1)^{i-1}\Theta/2 , \ \frac{\pi}{\gamma}\Theta=\log(\frac{2N}{M L}) \to \infty $ as $N    \to \infty $, and $\gamma$ relates to the Sine-Gordon coupling constant as \\ $\gamma=\pi(1-\alpha^2) \in [0,\pi]$ , the external magnetic flux corresponds to a twist $e^{-2 i \Phi}$ in the BAE .\\
The energy and momentum of the $\{\lambda_j \}$, configuration given by the formula:
\beq
e^{-i\frac{E\pm P}{2N}}=\prod\limits_{n=1}^N\frac{\sinh(\pm\lambda_j+\Theta/2+i\gamma/2)}{\sinh(\mp \lambda_j-\Theta/2-i\gamma/2)}
\eeq
3) This XXZ regularisation of Sine-Gordon gives the answer for the ground state energy in the presence of external magnetic flux, which (in the massless limit) agrees with the conformal one (instead of coordinate Bethe Ansatz and cut-off regularisation).
\section{From XXZ Bethe Ansatz to DDV equation}
Here we review connection between the XXZ Bethe Ansatz and nonlinear integral equation. \cite{DDV}. \\
Our starting point will be the Bethe-Ansatz equations for XXZ model with inhomogeneities
\beq
\prod\limits_{n=1}^N\frac{\sinh(\lambda+i\theta_n+i\gamma/2)}{\sinh(\lambda+i\theta_n-i\gamma/2)}=-e^{-2 i \Phi}\prod\limits_{j=1}^{N}\frac{\sinh(\lambda-\lambda_j+i\gamma)}{\sinh(\lambda-\lambda_j-i\gamma)} \label{BAE}
\eeq

Taking the logarithm of both sides , and introducing new functions
\beq
Z(\lambda)=\log\left(-e^{-2 i \Phi}\prod\limits_{j=1}^{N}\frac{\sinh(\lambda-\lambda_j+i\gamma)}{\sinh(\lambda-\lambda_j-i\gamma)}\right)-\log \left(\prod\limits_{n=1}^N\frac{\sinh(\lambda+i\theta_n+i\gamma/2)}{\sinh(\lambda+i\theta_n-i\gamma/2)}\right)
\eeq
\beq
\phi(\lambda,x)=i\log\left(\frac{\sinh(ix+\lambda)}{\sinh(ix-\lambda)}\right)
\eeq
We arrive to the following equations
\beq
Z(\lambda)=N[\phi(\lambda+\Theta,\gamma/2)+\phi(\lambda -\Theta,\gamma/2)]-\sum\limits_{j=1}^{N/2}\phi(\lambda-\lambda_k,\gamma)-2\Phi
\eeq
As usual vacuum state characterises by the condition
\beq
Z_{vac}(\lambda_j)=(-N-1+2j)\pi
\eeq
Using it, we can rewrite the sum over roots -$\lambda_k$ as a contour integral
\beq
\sum\limits_{j=1}^{N/2}\phi(\lambda-\lambda_k,\gamma)=\oint\limits_{\Gamma}  \phi(\lambda-\mu,\gamma) \frac{d}{d\mu}\log(1+e^{iZ(\mu)}) \frac{d\mu}{2\pi i} \label{sum}
\eeq
Where contour $\Gamma$ - encircles all roots $\lambda_k$, this is the way how nonlinear integral equation appears
\beq
Z(\lambda)=N[\phi(\lambda+\Theta,\gamma/2)+\phi(\lambda -\Theta,\gamma/2)]-2\Phi-\oint\limits_{\Gamma}  \phi(\lambda-\mu,\gamma) \frac{d}{d\mu}\log(1+e^{iZ(\mu)}) \frac{d\mu}{2\pi i} \label{int}
\eeq
After some manipulations \cite{DDV} this equation could be transformed to a more convenient form:
\begin{gather}
Z(\lambda)=ML \sinh(\frac{\pi \lambda}{\gamma})-\frac{\pi \Phi}{\pi -\gamma}-i  \int\limits_{-\infty}^{\infty} G(\lambda-\mu-i\eta_+) \log(1+e^{iZ(\mu+i\eta_+)}) d\mu+
\nonumber \\
+i  \int\limits_{-\infty}^{\infty}G(\lambda-\mu+i\eta_-,\gamma) \log(1+e^{-iZ(\mu-i\eta_-)}) d\mu
\end{gather}
\beq
G(\mu)=\frac{K}{1+K}=\int\limits_{0}^{\infty} \frac{dk}{4\pi}\frac{\sinh(k(\frac{\pi}{2}-\gamma))}{\sinh(k\frac{\pi-\gamma}{2})\cosh(\frac{\gamma k}{2})}\cos(k \mu)
\eeq
This equation (or its simple modification) is called Destri-De Vega (DDV) equation.

\section{From DDV to asymptotics of the current in the limit of big(small) radius of the wire}
For now we will restrict ourself to a repulsive regime only ($\gamma<\frac{\pi}{2}$ and no breathers).\\ For this case Destri and De Vega suggest a prescription $\eta_+=\eta_-=\eta=\frac{1}{2}min(\gamma ; \pi -\gamma)$.  and the change of an argument : $Z(\frac{\gamma\lambda}{\pi}+i\eta)=i\epsilon(\lambda)$ , $G(\lambda) \to \frac{\gamma}{\pi}G(\frac{\gamma}{\pi}\lambda)$ we will get
\beq
G(\theta)=\int\limits_{0}^{\infty} \frac{dk}{4\pi}\frac{\sinh(\pi k(\frac{\pi}{2\gamma}-1))}{\sinh(\frac{\pi k}{2} (\frac{\pi}{\gamma}-1))\cosh(\frac{\pi k}{2})}\cos(k \theta) \label{G}
\eeq
\beq
\epsilon(\theta)=-ML \cosh(\theta)-i\Phi-G\star\log(1+e^{\epsilon})+G_{c}\star \log(1+e^{\bar{\epsilon}}) \label{DDV}
\eeq
Where $\star$ is a convolution, and $G_c$ is crossed $G$ : $G_c(\lambda)=G(\lambda+i\pi)$ , remind that energy is eqal to
\beq
E=-M\int\limits_{-\infty}^{\infty} \cosh(\theta)\left(\log(1+e^{\epsilon})+\log(1+e^{\bar {\epsilon}})\right) \frac{d\theta}{2\pi} \label{E}
\eeq

    Surprisingly, G coincides with the logarithmic derivative of soliton-soliton scattering matrix $G(\theta)=\frac{1}{2\pi i}\partial_{\theta}\log(S(\theta))$, and $G_c$ coincides with matrix of crossed process. So DDV equation looks similar to a TBA equation, despite the fact that it is much more simple, because in general TBA for Sine-Gordon is a system of a number of coupled nonlinear equations.
\subsection{Large    $ML$ limit}
    Recall that equations \ref{DDV}, \ref{E} contain full information about the energy of the ground state of Sine-Gordon (on finite ring with length $L$) in the external magnetic flux $\Phi$.\\
Having this equation it is easily to analyze the large $ML$ limit, namely we can solve this equation recursively
\beq
\epsilon_0=-ML \cosh(\lambda)-i\Phi
\eeq
\beq
\epsilon_1=G\star\log(1+e^{\epsilon_0})+G_{c}\star \log(1+e^{\bar{\epsilon_0}})
\eeq
\beq
... \nonumber
\eeq
Let us make just zero order approximation,
\beq
E=-M \int\limits_{-\infty}^{\infty} \cosh(\theta) Re\left(\log(1+e^{-ML \cosh(\frac{\pi \lambda}{\gamma})-i\Phi})\right) \frac{d\theta}{2\pi}
\eeq
Expanding logarithm in Taylor series
\beq
E=-\frac{2M}{\pi}\sum\limits_{n \ge 1} \frac{(-1)^{n-1}}{n}K_1(n ML)\cos(n \Phi)
\eeq
For the free fermions point $(\gamma=\frac{\pi}{2})$, nonlinear part just vanishes, and it becomes an exact answer, for example we can make a conformal limit $ML\to 0$, and reproduce the classical answer (\ref{Evac},\ref{Jphi})
\beq
E=-\frac{2}{L\pi}\sum\limits_{n \ge 1} \frac{(-1)^{n-1}}{n^2}\cos(n \Phi)=-\frac{\pi}{6 L}\left(1-\frac{3}{\pi^2 } \Phi^2\right)
\eeq

Away from the free fermions point  , we can leave only the firs term from the infinite sum
\beq
E=-\frac{2M}{\pi} K_1(M L)\cos(\Phi) +o(e^{-2mL})
\eeq
\subsection{Small $ML$ limit (conformal regime)}
The analisys of small $ML$ regime is more complicated, it was done in \cite{DDV}, with the results in agreement with the conformal ones (\ref{Evac},\ref{Jphi}).
\beq
E=-\frac{\pi}{6 L}\left(1-\frac{3(\pi-\gamma)}{2\pi^2 } \Phi^2\right)+...
\eeq


TITLE: For any two sets , the power sets : $\mathcal P(A\setminus B) = \mathcal P(A) \setminus \mathcal P(B)$?
QUESTION [4 upvotes]: For any two sets , the power sets : $\mathcal P(A\setminus B) = \mathcal P(A) \setminus \mathcal P(B)$?
Is this proof correct?
$$\begin{split}x \in \mathcal P(A\setminus B) &\iff x \subseteq A\setminus B  \\ &\iff x \subseteq A \land x \nsubseteq B \\ & \iff x \in\mathcal P(A) \land x \notin \mathcal P(B) \\ & \iff x \in \mathcal P(A) \setminus \mathcal P(B)\end{split}$$

REPLY [1 votes]: Inclusion does not hold in either direction.
$\mathcal P(A\setminus B)\subseteq\mathcal P(A)\setminus\mathcal P(B)$ is always false, because $\emptyset\in\mathcal P(A\setminus B)$ while $\emptyset\notin\mathcal P(A)\setminus\mathcal P(B).$
$\mathcal P(A)\setminus\mathcal P(B)\subseteq\mathcal P(
A\setminus B)$ is sometimes false: if $x\in A\cap B$ and $y\in A\setminus B,$ then $\{x,y\}\in\mathcal P(A)\setminus\mathcal P(B)$ while $\{x,y\}\notin\mathcal P(A\setminus B);$
in fact, $\mathcal P(A)\setminus\mathcal P(B)\subseteq\mathcal P(
A\setminus B)$ holds only when $A\subseteq B$ or $A\cap B=\emptyset.$


TITLE: Show that $\varphi$ is of class $C^{\infty}$ and compute $\varphi'$
QUESTION [0 upvotes]: I am trying to solve the following question, arising from Implicit Function Theorem.
 $$\varphi :L(\Bbb{R}^n,\Bbb{R}^n)\to L(\Bbb{R}^n,\Bbb{R}^n)$$
 $$A\mapsto\varphi(A)=A^2.$$
I want to show that $\varphi$ is of class $C^{\infty}$ and compute $\varphi'$
I need some help in doing these. I would have loved to define $C^\infty$ but based on the arguments about its definition in Why don't we define the class of $C^{\infty}$ in this way?, I think I don't even know which of them I should follow.
Anyway, I would greatly rely on your definition and guidance.

REPLY [1 votes]: Each component of $\phi$ is a polynomial. Therefore $\phi\in C^{\infty}(L(\mathbb{R}^n),\mathbb{R}^n)$.
We need $\phi'(A_0)$ to satisfy $$\begin{align}0&=\lim_{A\to 0}\frac{\|\phi(A+A_0)-\phi(A_0)-\phi'(A_0)(A)\|}{\|A\|}\\&=\lim_{A\to 0}\frac{\|A^2+AA_0+A_0A-\phi'(A_0)(A)\|}{\|A\|}\\&=\lim_{A\to 0}\frac{\|AA_0+A_0A-\phi'(A_0)(A)\|}{\|A\|}\end{align}$$
Therefore, $\phi'(A_0)(A)=AA_0+A_0A$