text:
\begin{document}
\maketitle

\begin{abstract}
Recently, M. Mitkovski \cite{mitkovski2017basis} gave a criterion for the basicity of a sequence of complex exponentials in terms of the invertibility properties of a certain naturally associated Toeplitz operator, in the spirit of the celebrated work of Khrushch{\"e}v--Nikolski--Pavlov \cite{khruscev1981unconditional}. In our paper, we extend the results of Mitkovski to model spaces associated with meromorphic inner functions and we also give an analogue for the property of being an asymptotically orthonormal basis.
\end{abstract}

\section{Introduction}
The classical theory of Fourier series says that the trigonometric system $(e^{int})_{n\in\mathbb Z}$ forms an orthonormal basis of $L^2(0,2\pi)$. It is natural to ask what happens if we replace the trigonometric system by another systems of complex exponential $(e^{i\lambda_n t})_{n\in\mathbb Z}$, $\Lambda=(\lambda_n)_{n\in\mathbb Z}\subset\mathbb R$. Indeed this problem has a very long and rich history and finds its roots in the work of Paley--Wiener and Levinson. There at least two directions of approach to this problem. First, we can try to find small perturbations of the trigonometric systems which retains the desired basis properties of expansion. This problem of perturbation in fact gave rise to a whole direction of research, culminating with the beautiful theorem of Ingham--Kadets. See \cite{Young} for more details. The second direction is to find characterization of discrete sequences $\Lambda=(\lambda_n)_{n\in\mathbb Z}\subset\mathbb R$ such that $(e^{i\lambda_n t})_{n\in\mathbb Z}$ forms a Riesz basis/asymptotically orthonormal basis of $L^2(0,2\pi)$. In the seventies, an approach using Toeplitz operators was developed by Douglas, Sarason and Clark. Not only did it allow the recapture of all the classical results but also permitted Khrushch{\"e}v--Nikolski--Pavlov \cite{khruscev1981unconditional} to give the solution of the Riesz basis problem for complex exponentials. We briefly explain the basic idea of the Toeplitz approach. Since the map $f(t)\longmapsto f(t)e^{-t}$ is an isomorphism on $L^2(0,2\pi)$, it is clear that $(e^{i\lambda_n t})_{n\in\mathbb Z}$ forms a Riesz basis of $L^2(0,2\pi)$ if and only if $(e^{i\lambda'_n t})_{n\in\mathbb Z}$ does, where $\lambda'_n=\lambda_n+i$, $n\in\mathbb Z$. If we apply the inverse of the Fourier transform, it is easy to see that $(e^{i\lambda'_n t})_{n\in\mathbb Z}$ is a Riesz basis for $L^2(0,2\pi)$ if and only if $(k_{\lambda'_n}^{\Theta_{2\pi}})_{n\in\mathbb Z}$ is a Riesz basis for $\mathcal K_{\Theta_{2\pi}}$. Here $\mathcal K_{\Theta_{2\pi}}$ is the model space associated to $\Theta_{2\pi}(z)=e^{2i\pi z}$ and $k_{\lambda'_n}^{\Theta_{2\pi}}$ is its reproducing kernel. Then, the crucial idea is to observe that the basis properties of the reproducing kernels in $K_{\Theta_{2\pi}}$ are encoded in the invertibility property of the Toeplitz operator $T_{\overline{B_{\Lambda'}}\Theta_{2\pi}}$, where $B_{\Lambda'}$ is the Blaschke product associated to $\Lambda'=(\lambda_n')_{n\in\mathbb Z}$. If we want to apply the same method to study the asymptotically orthonormal basis property, we are faced with a difficulty because the latter property is not preserved by isomorphism, and consequently we cannot shift $\Lambda$ to $\Lambda'$. But recently, Mitkovski \cite{mitkovski2017basis} has shown that we can adapt the approach of Khrushch{\"e}v--Nikolski--Pavlov avoiding the use of the translated sequence but instead exploiting the ideas of Makarov--Poltoratski's Toeplitz approach to the completeness problem of exponential systems. In this paper, we will show that one can generalize Mitkovski's results to more general model spaces and we also give an analogous result for the asymptotically orthonormal basis property. \\

In Section 2, we start with some general preliminaries on Riesz basis, asymptotically orthonormal basis and meromorphic inner functions. As we will see, our characterization of asymptotically orthonormal basis involves the condition that a certain Toeplitz operator is unitary modulo compact (meaning that it can be written as the sum of a unitary operator and a compact one). In Section 3, we thus revisit some classical results on  Toeplitz operators. In particular, we explain how to derive a result of Douglas (characterizing the Toeplitz operators which are unitary modulo compact) from Hartman's theorem on compactness of Hankel operators and Devinatz--Widom's theorem on invertibility of Toeplitz operators. These results in the context of Hardy space of the unit disc can be found for instance in Peller \cite{Peller} and Nikolski \cite{nikolski2009functions}. Here we need the analogue for the upper-half plane and we provide the details with an aim to make our presentation complete. In Section 4, we give a characterization for Riesz basis property and in the last section, we discuss the asymptotically orthonormal basis property for systems of reproducing kernels in the model spaces.

\section{Preliminaries}

\subsection{Definition of Riesz sequences and AOB}
Let $\HH$ be a Hilbert space, $\mathfrak X=(x_n)_{n\geq 1}$ be a sequence of vectors in $\HH$. We recall that $\mathfrak X$ is said to be:
\begin{enumerate}
\item[(a)] \emph{minimal} if for every $n\geq 1$,
$$
x_n\not\in\mbox{span}(x_\ell:\ell\not=n),
$$
where $\mbox{span}(\dots)$ denotes the closure of the finite linear combination of $(\dots)$;
\item[(b)] A \emph{Riesz sequence} (abbreviated RS) if there exists two positive constants $c,C$ such that
\begin{equation}\label{eq:defn-RS}
c\sum_{n\geq 1}|a_n|^2\leq \|\sum_{n\geq 1}a_n x_n\|^2_{\HH}\leq C\sum_{n\geq 1}|a_n|^2,
\end{equation}
for every finitely supported sequence of complex numbers $(a_n)_n$;
\item[(c)] An \emph{asymptotically orthonormal sequence} (abbreviated AOS) if there exists $N_0\in\mathbb{N}$ such that for all $N\geq N_0$ there are positive constants $c_N,C_N$ verifying
\begin{equation}\label{eq:AOB}
c_N\sum_{n\geq N}|a_n|^2\leq \|\sum_{n\geq N}a_n x_n\|^2_{\HH}\leq C_N\sum_{n\geq N}|a_n|^2,
\end{equation}
for every finitely supported sequence of complex numbers $(a_n)_n$ and $\lim_{N\to \infty}c_N=1=\lim_{N\to \infty}C_N$;
\item[(d)] An \emph{asymptotically orthonormal basic sequence} (abbreviated AOB) if it is an AOS with $N_0=1$;
\item[(e)] A \emph{Riesz basis} for $\mathcal H$ (abbreviated RB) if it is a complete Riesz sequence, that is a Riesz sequence satisfying
$$
\mbox{span}(x_n:n\geq 1)=\HH.
$$
 \end{enumerate}
It is easy to see that $(x_n)_{n\geq 1}$ is an AOB if and only if it is an AOS as well as a RS. Also, $(x_n)_{n\geq 1}$ is an AOB if and only if it is minimal and an AOS. The reader should pay attention to the fact that AOB does not imply completeness; an AOB is a basis for its span but not necessarily for the whole space.

\subsection{Meromorphic inner functions and model spaces.}
A function $\Theta:\mathbb C_+\longrightarrow \mathbb C$ is said to be {\em inner} on the upper-half plane $\mathbb C_+=\{z\in\mathbb C:\Im(z)>0\}$ if $\Theta$ is analytic and bounded on $\mathbb C_+$ and if its radial limits are of modulus one almost everywhere on $\mathbb R$. We say that the inner function $\Theta$ is a meromorphic inner function (abbreviated by MIF) if $\Theta$ admits a meromorphic extension to the whole complex plane. A well-known theorem by Riesz and V. Smirnov says that all MIF functions have the form
$$
\Theta(z)=B(z)e^{iaz}
$$
where $a\geq 0$ and $B$ is the Blaschke product formed with the zeros of the function $\Theta$ given by $Z=(z_n)_{n\geq 1}$, where $|z_n|\to\infty$ and satisfy the convergence criterion
$$
\sum_{n\geq 1}\frac{\Im z_n}{1+|z_n|^2}<\infty.
$$
In particular, using this representation, it is not difficult to see that $\Theta$ will admit an analytic extension through the real line.

To each inner function $\Theta$, we associate a {\em model space}
$$
\mathcal K_\Theta=\mathcal H^2\ominus\Theta\mathcal H^2=\mathcal H^2\cap \Theta\overline{\mathcal H^2},
$$
where $\mathcal H^2$ is the Hardy space on $\mathbb C_+$. It is well known that if $\Theta$ is a MIF, then every function $f\in\mathcal K_\Theta$ can be extended analytically through the real line. In the case when $\Theta$ is a MIF with $\Theta'\in L^\infty(\mathbb R)$, we will need some uniform $L^2$--estimate on this extension. We start with the following result appearing in \cite[Lemma 2]{dyakonov2002differentiation}. For completeness, we provide a proof.

\begin{lem}\label{Lem:mean-value-thm}
Let $\Theta$ be a MIF inner function such that $\Theta' \in L^\infty(\R)$. Then,  for every $\varepsilon>0$, we have
$$\displaystyle\inf_{0\leq \Im z <\varepsilon} |\Theta(z)|\geq 1-\varepsilon \|\Theta'\|_\infty.
$$
\end{lem}
\begin{proof}
Since $\Theta$ is analytic on the real line, we can apply the mean value theorem. Then, for all $x,h \in \R$,
$$|\Theta(x+h)-\Theta(x)|\leq M|h|,$$
where $M=\|\Theta'\|_\infty$.
Consider the function $z \rightarrow \Theta(z+h)-\Theta(z)$. This function is the Poisson integral of the boundary function $x\rightarrow \Theta(x+h)-\Theta(x),$ which permits to extend the latter estimate in $\mathbb C_+$. In other words, for all $h\in\mathbb R_+$ and $z\in\mathbb C_+$, we have
$$
|\Theta(z+h)-\Theta(z)|\leq M|h|.
$$
But then by the definition of the derivative,
$$|\Theta'(z)|\leq M, $$
for all $z \in \mathbb C_+$. Note that we have proved that $\Theta' \in H^\infty(\mathbb C_+)$. Now letting $z=x+iy$, $0<y<\varepsilon$, we have
$$
|\Theta(z)-\Theta(x)|\leq |z-x|\displaystyle\sup_{0\leq \Im z <\varepsilon} |\Theta'(z)|\leq \varepsilon M,
$$
which gives
\begin{equation}\label{eq2-Theta-minoration1}
|\Theta(z)|\geq 1-\varepsilon M.
\end{equation}

\end{proof}
Now, let $f\in \mathcal K_\Theta$ and $0<\varepsilon<\|\Theta'\|_\infty^{-1}$. We can write $f(z)=\Theta(z)\overline{h(z)}$, $z\in\mathbb C_+$, for some $h\in\mathcal H^2$.
Then, if we define for $-\varepsilon<\Im(z)<0$,
\begin{equation}\label{eq1-analytic-continuation}
f(z)=\frac{1}{\overline{\Theta(\bar z)}} \overline{h(\bar z)},
\end{equation}
we obtain the analytic continuation of $f$ on $\{z\in\mathbb C: \Im(z)>-\varepsilon\}$.

\begin{lem}\label{lem:analytic-continuation}
Let $\Theta$ be a meromorphic inner function such that $\Theta'\in L^\infty(\mathbb R)$. Let $\varepsilon>0$ satisfying $\varepsilon<\|\Theta'\|_\infty^{-1}$. Then, for every $f\in \mathcal K_\Theta$, we have
$$
\sup_{|y|<\varepsilon}\int_{-\infty}^{+\infty}|f(x+iy)|^2\,dx\leq \frac{\|f\|_2^2}{(1-\varepsilon \|\Theta'\|_\infty)^2}.
$$
\end{lem}
\begin{proof}
It is clear that
$$
\sup_{0\leq y<\varepsilon}\int_{-\infty}^{+\infty}|f(x+iy)|^2\,dx\leq \|f\|_2^2.
$$
Now assume that $-\varepsilon<y<0$. According to \eqref{eq2-Theta-minoration1} and \eqref{eq1-analytic-continuation}, we have
\begin{eqnarray*}
|f(x+iy)|^2&=&\frac{|h(x-iy)|^2}{|\Theta(x-iy)|^2}\\
&\leq & \frac{|h(x-iy)|^2}{(1-\varepsilon \|\Theta'\|_\infty)^2}.
\end{eqnarray*}
Hence,
\begin{eqnarray*}
\int_{-\infty}^{+\infty}|f(x+iy)|^2\,dx&\leq & \frac{1}{(1-\varepsilon \|\Theta'\|_\infty)^2} \int_{-\infty}^{+\infty}|h(x-iy)|^2\,dx \\
&\leq & \frac{\|h\|_2^2}{(1-\varepsilon \|\Theta'\|_\infty)^2}\\
&=&  \frac{\|f\|_2^2}{(1-\varepsilon \|\Theta'\|_\infty)^2}.
\end{eqnarray*}
\end{proof}

Recall that for a MIF $U$ and $\lambda\in\mathbb R$, the function
$$
K_\lambda^U(z)=\frac{i}{2\pi}\frac{1-\overline{U(\lambda)}U(z)}{z-\bar\lambda}
$$
is the reproducing kernel of the space $\mathcal K_U$ and $2\pi \|K_\lambda^U\|_2^2=|U'(\lambda)|$. Hence, the function
$$
k_\lambda^U(z)=\frac{i}{\sqrt{2\pi |U'(\lambda)|}} \frac{1-\overline{U(\lambda)}U(z)}{z-\bar\lambda}
$$
is the normalized reproducing kernel of $\mathcal K_U$.

\begin{lem}\label{Lem:identite-clee}
Let $I$ and $\Theta$ be two MIF. Assume that $\{I=1\}=\{\lambda_n:n\geq 1\}$ and denote by $\eta_n=|I'(\lambda_n)|^{1/2} |\Theta'(\lambda_n)|^{-1/2}$, $n\geq 1$. Then, for every finitely supported sequence   of complex numbers $(a_n)_{n\geq 1}$, we have
\begin{equation}
(1-I)\sum_{n\geq 1}a_n k_{\lambda_n}^\Theta=\sum_{n\geq 1}a_n \eta_n k_{\lambda_n}^I -\Theta \sum_{n\geq 1}a_n \eta_n \overline{\Theta(\lambda_n)} k_{\lambda_n}^I.
\end{equation}
\end{lem}

\begin{proof}
We have
\begin{eqnarray*}
(1-I(z))k_{\lambda_n}^\Theta(z)&=&\frac{i}{\sqrt{2\pi}}\frac{(1-I(z))}{|\Theta'(\lambda_n)|^{1/2}}\frac{1-\overline{\Theta(\lambda_n)}\Theta(z)}{z-\bar\lambda_n}\\
&=&Ê\frac{|I'(\lambda_n)|^{1/2}}{|\Theta'(\lambda_n)|^{1/2}} (1-\overline{\Theta(\lambda_n)}\Theta(z)) \frac{i}{\sqrt{2\pi |I'(\lambda_n)|}}\frac{1-\overline{I(\lambda_n)}I(z)}{z-\bar\lambda_n}\\
&=& \eta_n (1-\overline{\Theta(\lambda_n)}\Theta(z)) k_{\lambda_n}^I(z).
\end{eqnarray*}
Hence
$$
(1-I)k_{\lambda_n}^\Theta=\eta_n k_{\lambda_n}^I -\Theta \eta_n  \overline{\Theta(\lambda_n)} k_{\lambda_n}^I.
$$
\end{proof}
Assume that the real sequence $(\lambda_n)_{n\geq 1}$ satisfies the following two conditions:
\begin{equation}\label{eq:lambda_tecn1}
\delta:=\inf_{n\geq 1}(\lambda_{n+1}-\lambda_n)>0,
\end{equation}
and
\begin{equation}\label{eq:lambda_tecn2}
\sup_{n}\left|\sum_{k\neq n}\left(\frac{1}{\lambda_n-\lambda_k}+\frac{\lambda_k}{\lambda_k^2+1}\right)\right|<\infty.
\end{equation}
Define the measure $\mu=\sum_{k=1}^\infty \nu_k \delta_{\lambda_k}$, where $(\nu_k)_{k\geq 1}$ is a sequence of positive real numbers satisfying $\sum_k \frac{\nu_k}{1+\lambda^2_k}<\infty$ and $\sup_k |\nu_k|<\infty$.
Let $G$ be the function defined by
$$
G(z)=\sum_{k\geq 1}\nu_k \left(\frac{1}{\lambda_k-z}-\frac{\lambda_k}{\lambda_k^2+1}\right).
$$
Clearly, $G$ is analytic in $\mathbb C_+$ and $\Im(G(z))>0$, $z\in\mathbb C_+$. Therefore, the function
\begin{equation}\label{eq:definition-I}
I=\frac{G-i}{G+i}
\end{equation}
is a meromorphic inner function in the upper half-plane and it is easy to see that $\{I=1\}=\{\lambda_n:n\geq 1\}$. The measure $\mu$ is the so-called Clark measure for $I$ and the sequence $(k_{\lambda_n}^I)_{n\geq 1}$ is an orthonormal basis for $\mathcal K_I$, see \cite{Clark}. Moreover, according to \cite[Lemma 5.2]{baranov-2006}, we have $I'\in L^\infty(\mathbb R)$ and $|I'(\lambda_n)|=2/\nu_n$. Recall that if $I$ is given by $I(z)=e^{iaz} B_Z(z) $, where $a\geq 0$ and $Z=(z_n)_n$ is the zero sequence of $I$, and $B_Z$ is the Blaschke product associated to $Z$, then for $t\in\mathbb R$, we have
$$
|I'(t)|=a+2\sum_{k\geq 1}\frac{\Im(z_k)}{|t-z_k|^{2}}.
$$
Since $I'\in L^\infty(\mathbb R)$, Lemma~\ref{Lem:mean-value-thm} implies $\delta_1:=\inf_n(\Im(z_n))>0$. Now, it is not difficult to see that if $|t-\lambda_n|\leq \delta$ (where $\delta$ is the constant in \eqref{eq:lambda_tecn1}), then
$$
\frac{1}{2(1+\frac{\delta^2}{\delta_1^2})}\leq \frac{|I'(t)|}{|I'(\lambda_n)|}=\frac{\nu_n}{2}|I'(t)|.
$$
Since $\mathbb R=\bigcup_n [\lambda_n-\delta,\lambda_n+\delta]$ and $\sup_n \nu_n<\infty$, we finally get
\begin{equation}\label{eq:bounded-below-above}
|I'(t)|\asymp 1,\qquad t\in\mathbb R.
\end{equation}

We now extend a result from \cite{mitkovski2017basis}. Part 1 is already contained in that form in \cite[Lemma 5.4]{baranov-2006}.
\begin{lem}\label{Lem:bounded-below-T-1-I}
Let $\Lambda=(\lambda_n)_{n\geq 1}$ be a real sequence satisfying \eqref{eq:lambda_tecn1} and \eqref{eq:lambda_tecn2} and let $I$ be the meromorphic inner function defined by \eqref{eq:definition-I}. Let $\Theta$ be a meromorphic inner function such that $\Theta'\in L^\infty(\mathbb R)$.
\begin{enumerate}
\item If $f\in \mathcal K_\Theta$ and $f(\lambda_n)=0$, $n\geq 1$, then $f/(1-I)\in \mathcal K_\Theta$.
\item The Toeplitz operator $T_{1-I}:\mathcal K_\Theta\longrightarrow H^2$ is bounded below.
\end{enumerate}
\end{lem}
Here, since $1-I$ is analytic, the Toeplitz operator $T_{1-I}$ is just the multiplication by $1-I$.
\begin{proof}
As already mentioned (1) is exactly Lemma 5.4 from \cite{baranov-2006}. Let us now prove (2), that is
$$
\inf_{f\in \mathcal K_\Theta,\|f\|_2=1}\|(1-I)f\|_2>0.
$$
Assume on the contrary that there exists a sequence $(f_k)_{k\geq 1}$ in $\mathcal K_\Theta$, $\|f_k\|_2=1$, such that $\|(1-I)f_k\|_2\to 0$ as $k\to \infty$. Denote by $\phi$ an increasing branch of the argument of $I$. Since $\phi'(t)=|I'(t)|$, according to \eqref{eq:bounded-below-above}, there exists $c_1,c_2>0$ such that
$$
c_1\leq \phi'(t)\leq c_2,\qquad t\in\mathbb R.
$$
Let us fix $\varepsilon>0$ such that
$$
\varepsilon<\min\left(\frac{2\pi}{c_2},\frac{\delta}{6},\frac{1}{4\|\Theta'\|_\infty}\right).
$$
Then, for every $n\geq 1$, we have
$$
0<c_1\varepsilon\leq \phi(\lambda_n+\varepsilon)-\phi(\lambda_n)=\int_{\lambda_n}^{\lambda_n+\varepsilon}\phi'(t)\,dt\leq c_2 \varepsilon<2\pi,
$$
and
$$
0<c_1\varepsilon\leq \phi(\lambda_n)-\phi(\lambda_n-\varepsilon)=\int_{\lambda_n-\varepsilon}^{\lambda_n}\phi'(t)\,dt\leq c_2 \varepsilon<2\pi.
$$
Since $\{\lambda_n\}=\{t\in \mathbb R:e^{i\phi(t)}=1\}$, it is now easy to see that there exists $c_3>0$ such that for all $t\in\mathbb R$ satisfying $\mbox{dist}(t,\Lambda)\geq\varepsilon$, we have
$$
|I(t)-1|\geq c_3.
$$
Let $E_\varepsilon=\bigcup_{n\geq 1}(\lambda_n-\varepsilon,\lambda_n+\varepsilon)$. Therefore, we get
$$
\int_{\mathbb R\setminus E_\varepsilon}|1-I(t)|^2|f_k(t)|^2\,dt\geq c_3^2 \int_{\mathbb R\setminus E_\varepsilon}|f_k(t)|^2\,dt.
$$
Choose now $k$ sufficiently large so that
$$
\|(1-I)f_k\|_2\leq \frac{c_3}{\sqrt{2}},
$$
which gives
$$
\int_{\mathbb R\setminus E_\varepsilon}|f_k(t)|^2\,dt\leq \frac{1}{2}.
$$
Using $1=\|f_k\|_2$, we then get
$$
\frac{1}{2}\leq \int_{E_\varepsilon}|f_k(t)|^2\,dt.
$$
According to Lemma \ref{lem:analytic-continuation}, we know that $f_k$ is analytic on $\{z\in\mathbb C:\Im(z)>-4\varepsilon\}$ and
$$
\sup_{|y|<4\varepsilon}\int_{-\infty}^{+\infty}|f_k(x+iy)|^2\,dx\leq \frac{1}{(1-4\varepsilon\|\Theta'\|_{\infty})^2}.
$$
Now let $t\in (-3\varepsilon,3\varepsilon)$. By subharmonicity of $|f_k|^2$, we have
\begin{eqnarray*}
\sum_{n\geq 1}|f_k(\lambda_n+t)|^2 &\leq  &\sum_{n\geq 1} \frac{1}{9\pi\varepsilon^2} \int_{B(\lambda_n+t,3\varepsilon)} |f_k(z)|^2\,dA(z) \\
&\leq & \sum_{n\geq 1} \frac{1}{9\pi\varepsilon^2}  \int_{-3\varepsilon}^{3\varepsilon} \int_{\lambda_n+t-3\varepsilon}^{\lambda_n+t+3\varepsilon}|f_k(x+iy)|^2\,dx\,dy.
\end{eqnarray*}
\end{proof}
Using the fact that $|\lambda_{n+1}-\lambda_n|\geq \delta>6\varepsilon$, it follows that $[\lambda_{n}+t-3\varepsilon,\lambda_{n}+t+3\varepsilon]$ are disjoints intervals which gives
\begin{eqnarray*}
\sum_{n\geq 1}|f_k(\lambda_n+t)|^2 &\leq  & \frac{1}{9\pi\varepsilon^2}  \int_{-3\varepsilon}^{3\varepsilon}  \int_{-\infty}^{+\infty} |f_k(x+iy)|^2\,dx\,dy \\
&\leq & \frac{2}{3\pi\varepsilon} \frac{1}{(1-4\varepsilon \|\Theta'\|_\infty)^2}.
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
\frac{1}{2}\leq \int_{E_\varepsilon}|f_k(t)|^2\,dt&=&\sum_{n\geq 1} \int_{-\varepsilon}^{\varepsilon}|f_k(t+\lambda_n)|^2\,dt \\
&\leq & \frac{4}{3\pi}\frac{1}{(1-4\varepsilon \|\Theta'\|_\infty)^2}.
\end{eqnarray*}
Hence $(1-4\varepsilon \|\Theta'\|_\infty)^2\leq 8/(3\pi)$, and letting $\varepsilon\to 0$, we get a contradiction, which proves that the operator $T_{1-I}$ is bounded below as an operator from $\mathcal K_\Theta$ to $H^2$.

\section{Toeplitz operators of the form unitary plus compact}
In this section we revisit some classical results concerning the Toeplitz operators. As we will see our characterization of AOB we get in Section 5 will involve the condition that a certain Toeplitz operator is of the form unitary plus compact. It turns out that this property of Toeplitz operators has been characterized by Douglas \cite{douglas1973banach}. In this section, we revisit his result and explain how to derive Douglas's result from Hartman's theorem on the compactness of Hankel operators \cite{hartman1958on} and the Devinatz-Widom criterion for the invertibility of Toeplitz operators \cite{nikolski1986treatise,nikolski2009functions}. Most of this can be found in \cite{Peller} or \cite{nikolski1986treatise} in the context of the disc but we need the analogues for the upper half-plane. So we give some details for completeness.

First, let us recall that if $u\in L^\infty(\mathbb R)$, then the \emph{Toeplitz operator} $T_u$ on the Hardy space of the upper half-plane $\mathcal H^2$ is defined by $T_u(f)=P_+(uf)$, $f\in\mathcal H^2$, where $P_+$ is the Riesz projection from $L^2(\mathbb R)$ onto $\mathcal H^2$. Its cousin, the \emph{Hankel operator}, $H_u$ is defined as an operator from $\mathcal H^2$ into $\mathcal H^2_-=L^2(\mathbb R)\ominus\mathcal H^2$ by $H_u(f)=P_-(uf)$, $f\in\mathcal H^2$, where $P_-=Id-P_+$.

A well-known theorem of Hartman \cite[Corollary 8.5]{Peller} says that $H_u$ is compact if and only if $u\in H^\infty+C(\dot\R)$, where $H^\infty$ is the space of analytic and bounded functions of $\mathbb C_+$ and
$$
C(\dot\R)=\{f \in C(\R): \displaystyle\lim_{x \rightarrow \pm \infty} f(x) \mbox { exist  and are equal}\}.
$$
It turns out that the space $H^\infty+C(\dot\R)$ is indeed a closed  subalgebra of $L^\infty(\R)$ (which is not obvious at the first sight but can be deduced from Hartman's theorem). We also recall a part of the Devinatz-Widom criterion for the invertibility of the Toeplitz operator. For the complete result, we refer the reader to (\cite{nikolski2009functions}, page 250). Let us recall that if $b\in L^1_\Pi$, that is
$$
\int_{-\infty}^{\infty}\frac{|b(t)|}{1+t^2}\,dt<\infty,
$$
the Hilbert transform of $b$ is defined as the singular integral
$$
\tilde b(x)=\lim_{\varepsilon\to 0}\frac{1}{\pi}\int_{|x-t|>\varepsilon}\left(\frac{1}{x-t}+\frac{t}{1+t^2}\right)b(t)\,dt.
$$
\begin{thm}[Devinatz-Widom]
Let $u\in L^\infty(\mathbb R)$, $|u|=1$ a.e. on $\mathbb R$. The following are equivalent.
\begin{enumerate}
\item The Toeplitz operator $T_u$ is invertible;
\item there exist real valued bounded functions $a,b$ on $\mathbb R$ where $\|a\|_{\infty}<\pi/2$ and a real constant $c$ such that $u = e^{i(c+a+\tilde b)}$.
\item There is an outer function $h \in H^\infty$ such that $\|u-h\|_\infty < 1$
\end{enumerate}
\end{thm}

Note that in \cite{nikolski2009functions}, the result is expressed in the context of the unit disc $\mathbb D$. To transfer the result to the upper-half plane, it is sufficient to use the conformal map (Cayley transform) $\phi(z)=(z-i)/(z+i)$ from $\mathbb C_+$ onto $\mathbb D$, the map $\mathcal M(f)(t)=\frac{\sqrt 2}{t+i}(f\circ \phi)(t)$ for $f\in H^2(\mathbb D)$, which is a unitary map from the Hardy space of the unit onto the Hardy space of the upper-half plane, and note that $T_u=\mathcal M T_{u\circ\phi^{-1}}\mathcal M^{-1}$.

The following lemma gives a sufficient condition for invertibility.
\begin{lem}\label{invertible}
If $u=e^{i(\alpha + \tilde\beta)}$, for some real-valued functions $\alpha,\beta\in C(\dot\R)$, then $T_{u}$ is invertible.
\end{lem}
\begin{proof}
Since $\alpha \in C(\dot\R)$, at $\pm \infty$, it admits the same limit, say $\ell$. Then, we can find a real-valued function $v\in C_c^1(\R)$, the space of compactly supported $C^1$ functions on $\R$, such that
$$
\|\alpha-\ell-v\|_{\infty}<\pi/2.
$$
Note that since $v\in C^1_c(\R)$, $\tilde v$ is a real-valued continuous function which tends to zero at infinity and we have $\tilde{\tilde v}=-v$ (see \cite[Theorem 14.1 \& Corollary 14.9]{mashreghi}).
Then,
$$
\alpha + \tilde\beta = \alpha-\ell-v+\widetilde{\beta-\tilde{v}}+\ell.
$$
Hence
$$
u=e^{i\ell} e^{i(\alpha-\ell-v+\widetilde{\beta-\tilde{v}})}.
$$
We can apply the criterion of Devinatz-Widom to conclude that $T_u$ is invertible.
\end{proof}

When we try to characterize the property for a Toeplitz operator to be unitary modulo the compacts, the class $QC$ of \emph{quasicontinuous functions} naturally appears:
$$
QC=(H^\infty+C(\dot\R))\cap \overline{(H^\infty+C(\dot\R))}.
$$

\begin{lem}[Douglas]\label{unimodcom}
Let $u \in L^\infty$, $|u|=1$ a.e. on $\mathbb R$. Then, $T_u$ is unitary modulo the compacts iff $u \in QC$ and $T_u$ is invertible.
\end{lem}
\begin{proof}
First, let us assume that $u\in QC$ and $T_u$ is invertible. Using the polar decomposition of bounded invertible operators, we can write $$ T_u = U |T_u|,$$ where $U$ is a unitary operator and $|T_u|=(T^*_u T_u)^{\frac{1}{2}}$. Furthermore, we notice the following equality
$$ I - (T^*_u T_u)^{\frac{1}{2}} = (I + (T^*_u T_u)^{\frac{1}{2}})^{-1} (I-T^*_uT_u), $$
using the positivity of $(T^*_u T_u)^{\frac{1}{2}}$ to ensure that the operator $I + (T^*_u T_u)^{\frac{1}{2}}$ is invertible. We recall that $I-T^*_uT_u=H^*_uH_u$ and using Hartman's theorem, we get that $I-(T^*_uT_u)^{\frac{1}{2}}$ is compact. Hence, $|T_u|$ is identity modulo the compacts, ensuring that $T_u=U|T_u|$ is unitary modulo the compacts.

Conversely, let $T_u=U+K$, where $U$ is unitary and $K$ is compact. Then, $T_u^* = U^* + K^*$ and
\begin{eqnarray*}
T_uT^*_u &=& (U+K)(U^*+K^*)\\
&=& I+K_1,
\end{eqnarray*}
where $K_1=UK^* + KU^* + KK^*$ is compact. Hence, $I-T_uT_u^*$ is compact. Similarly, we can show that $I-T_u^*T_u$ is compact. Thus, $H_u^*H_u= I-T_u^*T_u$ and $H^*_{\overline u}H_{\overline u}= I- T_uT^*_u$ are both compact. But we recall the fact that an operator $T$ is compact iff $T^*T$ is compact. This means that both $u$ and $\overline u$ belong to the space $H^\infty + C(\dot\R)$. That is to say, $u\in QC$. It remains to prove that $T_u$ is invertible. The representation $T_u = U+K$ allows us to reason that since $U$ is Fredholm with index $0$ and $K$ is compact, then $T_u$ is also Fredholm with index $0$. Similarly, $T^*_u$ is also Fredholm with index $0$. We now use Coburn's lemma to claim that either $\ker T_u=\{0\}$ or $\ker T^*_u=\{0\}$. Thus, either and hence both $T_u$ and $T^*_u$ are invertible.
\end{proof}

We revisit the following theorem of Sarason's \cite{sarason1973algebras} that completely characterizes the unimodular functions that belong to the space QC. We note that the Cayley transform $\phi(z)=(z-i)/(z+i)$ is also a Blaschke factor vanishing at $i$.
\begin{lem}[Sarason]\label{sarason}
Let $u\in L^\infty(\R)$ be a unimodular function. Then, $u \in QC$ if and only if $u=\phi^ne^{i(a+\tilde b)}$, for some $n\in \mathbb Z$ and two real valued functions $a,b \in C(\dot\R)$.
\end{lem}
\begin{proof}
Let us assume that $u$ is of the form $\phi^ne^{i(a+\tilde b)}$. We can rewrite $u$ as follows. $$u=\phi^ne^{(b+i\tilde b)+(ia-b)}, $$
where $\phi^ne^{ia-b} \in C(\dot\R)$, $e^{b+i\tilde b} \in H^\infty$. Since $ H^\infty + C(\dot\R)$ is an algebra, we have that $u \in H^\infty + C(\dot\R)$. In a similar manner, we can show that $\overline u \in  H^\infty + C(\dot\R)$. Therefore, $u \in QC$.

Conversely, assume that $u \in QC$. We can easily check that the following relationships are true.
\begin{eqnarray*}
I-T_uT_{\overline u} &=& H^*_{\bar u}H_{\bar u}\\
I-T_{\overline u}T_u &=& H^*_{ u}H_{u}.
\end{eqnarray*}
Since $u,\bar u\in H^\infty + C(\dot\R)$, by Hartman's theorem $H_u$ and $H_{\bar u}$ are compact. This means that $I-T_uT_{\bar u}$ and $I-T_{\bar u}T_u$ are both compact. Thus, $T_u$ is Fredholm. Let $n=-$ind${T_u}$. First let us assume that $n\geq 0$. Let us define a new function $$v:=\overline{\phi^n} u. $$ Then, $T_v=T_{\overline{\phi^n}}T_u$ is also Fredholm. Moreover, $$\mbox{ind}(T_{v})=\mbox{ind}(T_{\overline{\phi^n}}) +  \mbox{ind}(T_u)=n-n=0.$$
By Coburn's result, either $\ker T_v=\{0\}$ or $\ker T_{\bar v}=\{0\}$. Since the index is zero, we finally deduce that both kernels are trivial and  $T_v$ will be invertible. Thus, by the Devinatz-Widom theorem, there is an outer function $h\in H^\infty$ such that $$ \|v-h\|_{\infty}=\|1-\bar v h\|_{\infty}<1. $$ Consequently, $\bar v h$ has a logarithm in the Banach algebra $H^\infty + C(\dot\R)$ (see Lemma 2.13, page 34 in \cite{douglas1972banach}). Therefore, there is a $g\in H^\infty$ and $f\in C(\dot\R)$ such that
$$\frac{1}{\bar v h}=\frac{v}{h}=e^{f+g}. $$
Hence, $$v=he^{f+g}=e^{ic+\log|h|+i\widetilde{log|h|}+f+g},$$
for some $c\in \R.$ Since $|v|=1$ a.e. on $\mathbb R$, it follows that $\log|h|+\Re f+\Re g=0.$ Let us denote $$b:=\log|h| + \Re(g) = -\Re(f) \in C(\dot\R). $$
Since, $g \in H^\infty$, we have
$$\tilde b=\widetilde{\log|h|}+\Im(g)-\Im(g(i)). $$
Denoting $a:=\Im(f) + c + \Im(g(i)) \in C(\dot\R)$. Finally, we have that $$u=\phi^n v = \phi^n e^{i(a+\tilde b)}.
$$ In the case $n<0$, we let $v=\phi^n u$ and follow the same argument as before to arrive at the same conclusion.
\end{proof}

Finally, we can prove the following result.
\begin{thm}\label{umodcom2}
Let $u\in L^\infty(\R)$, $|u|=1$ a.e. on $\mathbb R$. The following are equivalent.
\begin{enumerate}
\item $T_u$ is unitary modulo the compacts.
\item There exist real valued functions $a,b \in C(\dot\R)$ such that $u=e^{i(a+\tilde b)}$.
\end{enumerate}
\end{thm}
\begin{proof}
First let us assume that there exist real valued functions $a,b \in C(\dot\R)$ such that $u=e^{i(a+\tilde b)}$. Using Lemma \ref{invertible}, we can conclude that $T_u$ is invertible. Furthermore, using Lemmas \ref{sarason} and \ref{unimodcom}, we can conclude that $T_u$ is indeed unitary modulo the compacts.

Conversely, assume that $T_u$ is unitary modulo the compacts. It follows from Lemma \ref{unimodcom} that $u\in QC$ and $T_u$ is invertible. Next, using Lemma \ref{sarason}, we see that $u=\phi^ne^{i(a+\tilde b)},$ for some real valued functions $a,b \in C(\dot\R)$ and some $n\in \mathbb Z$. It remains to prove that $n=0$. Let us denote by $u_1 = e^{i(a+ \tilde b)}$. By Lemma \ref{invertible}, the Toeplitz operator $T_{u_1}$ is invertible. If $n>0$, then
$$T_u = T_{u_1 \phi^n} = T_{u_1}T_{\phi^n}, $$
whence $T_{\overline{\phi^n}}=T^*_{\phi^n} = T^*_u (T^{-1}_{u_1})^*$ should also be invertible. But this is absurd because $\ker T_{\overline{\phi^n}}= K_{\phi^n}$. If $n<0$, then $T_u= T_{\phi^n}T_{u_1}$, that is to say
$$ T_{\phi^n}=T_u (T_{u_1})^{-1}$$ is invertible, which is also absurd because $\ker T_{\phi^n}=K_{\phi^{-n}}.$
\end{proof}

\section{Riesz bases}
In this section we revisit a recent work of Mitkovski \cite{mitkovski2017basis} on systems of exponential systems which form a Riesz basis. We provide a generalization of his result, while maintaining the techniques of the proofs.

\begin{thm}\label{Thm:analogue-mishko}
Let $(\lambda_n)_{n\geq 1}$ be a real sequence satisfying \eqref{eq:lambda_tecn1} and \eqref{eq:lambda_tecn2} and let $I$ be the meromorphic inner function defined by \eqref{eq:definition-I}. Let $\Theta$ be a meromorphic inner function such that $\Theta'\in L^\infty(\mathbb R)$ and $\inf_{n\geq 1}|\Theta'(\lambda_n)|>0$. Then,
\begin{enumerate}
\item $(k_{\lambda_n}^\Theta)_{n\geq 1}$ is complete in $\mathcal K_\Theta$ if and only if $T_{\Theta \bar I}:\HH^2\longrightarrow \HH^2$ has a dense range.
\item $(k_{\lambda_n}^\Theta)_{n\geq 1}$ is a Riesz sequence in $\mathcal K_\Theta$ if and only if $T_{\Theta \bar I}:\HH^2\longrightarrow \HH^2$ is injective with closed range.
\item $(k_{\lambda_n}^\Theta)_{n\geq 1}$ is a Riesz basis for $\mathcal K_\Theta$ if and only if $T_{\Theta \bar I}:\HH^2\longrightarrow \HH^2$ is invertible.
\end{enumerate}
\end{thm}

\begin{proof}
Let us first show that the upper inequality in Riesz basis property \eqref{eq:defn-RS} is satisfied for $(k_{\lambda_n}^\Theta)_n$ under the hypothesis of the theorem. Indeed, according to Lemma~\ref{Lem:identite-clee}, we have
$$
(1-I)\sum_{n\geq 1}a_n k_{\lambda_n}^\Theta=\sum_{n\geq 1}a_n\eta_n k_{\lambda_n}^I -\Theta \sum_{n\geq 1}a_n \eta_n \overline{\Theta(\lambda_n)} k_{\lambda_n}^I,
$$
where $\eta_n=|I'(\lambda_n)|^{1/2} |\Theta'(\lambda_n)|^{-1/2}\asymp 1$ by hypothesis and \eqref{eq:bounded-below-above}.
Using Lemma~\ref{Lem:bounded-below-T-1-I},
\begin{eqnarray*}
\left\|\sum_{n\geq 1}a_n k_{\lambda_n}^\Theta\right\|_2&\lesssim& \left\|(1-I)\sum_{n\geq 1}a_n k_{\lambda_n}^\Theta\right\|_2\\
&\leq & \left\|\sum_{n\geq 1}a_n \eta_n k_{\lambda_n}^I\right\|_2+\left\|\sum_{n\geq 1}a_n \eta_n \overline{\Theta(\lambda_n)}k_{\lambda_n}^I\right\|_2.
\end{eqnarray*}
Then, using the fact that $(k_{\lambda_n}^I)_{n\geq 1}$ is an orthonormal basis for $\mathcal K_I$ and $|\Theta(\lambda_n)|=1$, $n\geq 1$, we get
\begin{equation}\label{eq:upper-inequality-rb}
\left\|\sum_{n\geq 1}a_n k_{\lambda_n}^\Theta\right\|_2  \lesssim \left(\sum_{n\geq 1}|a_n|^2\right)^{1/2}.
\end{equation}
\\

Let us  now prove (1). First, assume that $T_{\Theta\bar I}$ has dense range and assume that there exists $f\in \mathcal K_\Theta$, $f(\lambda_n)=0$, $n\geq 1$, and $f\not\equiv 0$. According to Lemma~\ref{Lem:bounded-below-T-1-I}, the function $g:=f/(1-I)$ belongs to $\mathcal K_\Theta$ and, since $\mathcal K_\Theta=\HH^2\cap\Theta\overline{\HH^2}$, $\bar\Theta Ig=-\bar\Theta (1-I)g+\bar\Theta g=-\bar\Theta f+\bar\Theta g\in\overline{\HH^2}$. In particular,
$T_{I\bar\Theta}g=0$ and $T_{I\bar\Theta}$ is not injective which contradicts the fact that $T_{\Theta\bar I}=T_{I\bar\Theta}^*$ has dense range. Conversely, suppose that $T_{\Theta\bar I}$ does not have a dense range. Then, there is a function $g\in\ker(T_{I\bar\Theta})$, $g\not\equiv 0$. Hence $Ig\bar\Theta\in\overline{\HH^2}$, which gives that both $g$ and $Ig$ belong to $\mathcal K_\Theta$. So $f:=(1-I)g$ is also in $\mathcal K_\Theta$, $f\not\equiv 0$ and
$$
f(\lambda_n)=(1-I(\lambda_n))g(\lambda_n)=0,\qquad n\geq 1,
$$
which proves that $(k_{\lambda_n}^\Theta)_{n\geq 1}$ is not complete in $\mathcal K_\Theta$.
\\

Let us now prove (2). Assume that $(k_{\lambda_n}^\Theta)_{n\geq 1}$ is not a Riesz sequence in $\mathcal K_\Theta$. According to \eqref{eq:upper-inequality-rb}, it means that the lower inequality in \eqref{eq:defn-RS} is not satisfied. Then, for any $\varepsilon>0$, there exists $\{b_n\}_{n\geq 1}\in\ell^2$ such that
$$
\left\|\sum_{n\geq 1}b_n k_{\lambda_n}^\Theta\right\|^2<\varepsilon \sum_{n\geq 1}|b_n|^2.
$$
Take $a_n=b_n/\|b_n\|_{\ell^2}$. Then $\sum_{n\geq 1}|a_n|^2=1$ and
$$
\left\|\sum_{n\geq 1}a_n k_{\lambda_n}^\Theta\right\|^2<\varepsilon.
$$
In particular, we can construct a sequence $(a_n^\ell)_{n\geq 1}$ of elements of $\ell^2$ such that for every $\ell\geq 1$, $\sum_{n\geq 1}|a_n^\ell|^2=1$ and
$$
\left\|\sum_{n\geq 1}a_n^\ell k_{\lambda_n}^\Theta\right\|_2\to 0,\quad \mbox{as }\ell\to\infty.
$$
Let $g_\ell:=\sum_{n\geq 1}a_n^\ell k_{\lambda_n}^\Theta$. Then $g_\ell\in \mathcal K_\Theta$ and $\|g_\ell\|_2\to 0$ as $\ell\to\infty$. According to Lemma~\ref{Lem:identite-clee}, we have
$$
(1-I)g_\ell=\sum_{n\geq 1}a_n^\ell \eta_n k_{\lambda_n}^I -\Theta\sum_{n\geq 1}a_n^\ell \eta_n \overline{\Theta(\lambda_n)}k_{\lambda_n}^I.
$$
Multiply by $\bar I$ and take the Riesz projection gives
$$
P_+(\bar I(1-I)g_\ell)=\sum_{n\geq 1}a_n^\ell \eta_n P_+(\bar I k_{\lambda_n}^I)-P_+(\Theta\bar I\sum_{n\geq 1}a_n^\ell \eta_n \overline{\Theta(\lambda_n)}k_{\lambda_n}^I).
$$
But $k_{\lambda_n}^I\in \mathcal K_I=\HH^2\cap I\overline{\HH^2}$, hence $P_+(\bar I k_{\lambda_n}^I)=0$ and, then
$$
P_+(\bar I(1-I)g_\ell)=-T_{\Theta \bar I}(\sum_{n\geq 1}a_n^\ell \eta_n \overline{\Theta(\lambda_n)}k_{\lambda_n}^I).
$$
On one hand, we have
$$
\|P_+(\bar I(1-I)g_\ell)\|_2\to 0,\quad \mbox{as }\ell\to\infty.
$$
And on the other hand, since $\Theta'\in L^\infty(\mathbb R)$, we also have
\begin{eqnarray*}
\left\|\sum_{n\geq 1}a_n^\ell \eta_n \overline{\Theta(\lambda_n)}k_{\lambda_n}^I \right\|_2^2&=&\sum_{n\geq 1}|a_n^\ell|^2 \eta_n^2 |\Theta(\lambda_n)|^2 \\
&=&\sum_{n\geq 1}|a_n^\ell|^2 \eta_n^2 \\
&\gtrsim & \sum_{n\geq 1}|a_n^\ell|^2=1,
\end{eqnarray*}
which proves that $T_{\Theta\bar I}$ is not bounded below.  Conversely, assume that $T_{\Theta\bar I}$ is not bounded below. Hence there exists a sequence $(f_\ell)_{\ell\geq 1}$, $f_\ell\in\HH^2$, $\|f_\ell\|_2=1$, such that $\|T_{\Theta\bar I}f_\ell\|_2\to 0$, as $\ell\to \infty$. Let $h_\ell=T_{\Theta \bar I}f_\ell$. Then, $\|h_\ell\|_2\to 0$ as $\ell\to\infty$, and $g_\ell:=\Theta \bar I f_\ell-h_\ell\in \overline{\HH^2}$ with $\|g_\ell\|_2\asymp 1$ (because $\|f_\ell\|_2=1$ and $\|h_\ell\|_2\to 0$). Note that
$$
Ig_\ell=\Theta f_\ell-Ih_\ell\in \HH^2\cap I\overline{\HH^2}=\mathcal K_I.
$$
Since $(k_{\lambda_n}^I)_{n\geq 1}$ is an orthonormal basis for $\mathcal K_I$, there exists $(g_n^\ell)_{n\geq 1}\in\ell^2$ such that
$$
Ig_\ell=\sum_{n\geq 1}g_n^\ell k_{\lambda_n}^I.
$$
Note also that
$$
\sum_{n\geq 1}|g_n^\ell|^2=\|Ig_\ell\|_2^2=\|g_\ell\|_2^2\asymp 1.
$$
Then
$$
Ih_\ell=\Theta f_\ell-I g_\ell=\Theta f_\ell-\sum_{n\geq 1}g_n^\ell k_{\lambda_n}^I,
$$
and using one more time Lemma~\ref{Lem:identite-clee}, we get
\begin{equation}\label{eq22334:preuve-thm-riesz-basis}
Ih_\ell=\Theta f_\ell-\left((1-I)\sum_{n\geq 1}\frac{g_n^\ell}{\eta_n}k_{\lambda_n}^\Theta+\Theta\sum_{n\geq 1}g_n^\ell \overline{\Theta(\lambda_n)} k_{\lambda_n}^I\right).
\end{equation}
Thus
$$
\Theta f_\ell-\Theta\sum_{n\geq 1}g_n^\ell \overline{\Theta(\lambda_n)} k_{\lambda_n}^I-\sum_{n\geq 1}\frac{g_n^\ell}{\eta_n}k_{\lambda_n}^\Theta=Ih_\ell-I\sum_{n\geq 1}\frac{g_n^\ell}{\eta_n}k_{\lambda_n}^\Theta.
$$
The difference of the first two terms on the left side is in $\Theta\HH^2$, while the third term is in $\mathcal K_\Theta$. Using orthogonality and triangle's inequality, we obtain
\begin{eqnarray*}
\left\| \Theta f_\ell-\Theta\sum_{n\geq 1}g_n^\ell \overline{\Theta(\lambda_n)} k_{\lambda_n}^I\right\|_2^2+\left\| \sum_{n\geq 1}\frac{g_n^\ell}{\eta_n}k_{\lambda_n}^\Theta\right\|_2^2 &\leq & \left(\|Ih_\ell\|_2+
\left\| I \sum_{n\geq 1}\frac{g_n^\ell}{\eta_n}k_{\lambda_n}^\Theta\right\|_2\right)^2 \\
&=&\left(\|h_\ell\|_2+ \left\| \sum_{n\geq 1}\frac{g_n^\ell}{\eta_n}k_{\lambda_n}^\Theta\right\|_2\right)^2.
\end{eqnarray*}
Since by \eqref {eq:upper-inequality-rb}, we have
$$
\left\| \sum_{n\geq 1}\frac{g_n^\ell}{\eta_n}k_{\lambda_n}^\Theta\right\|_2\lesssim \left(\sum_{n\geq 1}\frac{|g_n^\ell|^2}{\eta_n^2}\right)^{1/2}\lesssim 1,
$$
and since $\|h_\ell\|_2\to 0$ as $\ell\to\infty$, we deduce
$$
\left\|\Theta f_\ell-\Theta \sum_{n\geq 1}g_n^\ell \overline{\Theta(\lambda_n)}k_{\lambda_n}^I \right\|_2\to 0,\qquad \mbox{as }\ell \to\infty.
$$
Using this and \eqref{eq22334:preuve-thm-riesz-basis}, we obtain
$$
\left\|(1-I)\sum_{n\geq 1}\frac{g_n^\ell}{\eta_n}k_{\lambda_n}^\Theta\right\|_2\to 0,\quad \mbox{as }\ell\to\infty.
$$
Lemma~\ref{Lem:bounded-below-T-1-I} implies now that
$$
\left\|\sum_{n\geq 1}\frac{g_n^\ell}{\eta_n}k_{\lambda_n}^\Theta\right\|_2\to 0,\quad \mbox{as }\ell\to\infty,
$$
which together with
$$
\sum_{n\geq 1}\frac{|g_n^\ell|^2}{\eta_n^2}\asymp \sum_{n\geq 1}|g_n^\ell|^2\asymp 1,
$$
implies that $(k_{\lambda_n}^\Theta)_{n\geq 1}$ cannot be a Riesz sequence in $\mathcal K_\Theta$.

The part (3) follows immediately from (1) and (2).

\end{proof}

\section{Asymptotically orthonormal bases.}

In that section, we give a result on sequences of reproducing kernels which form an AOB. We need to recall the notion of angle between two closed subspaces $\mathcal F,\mathcal G$ of an Hilbert space $\mathcal H$. The angle $\langle \mathcal F,\mathcal G \rangle$ between $\mathcal F$ and $\mathcal G$ is defined by the condition:
$$
\langle \mathcal F,\mathcal G\rangle\in [0,\pi/2]\quad \cos\langle \mathcal F,\mathcal G\rangle=\sup_{f\in\mathcal F,g\in\mathcal G}\frac{|\langle f,g \rangle|}{\|f\|\|g\|}.
$$
It is easy to see that $\cos\langle \mathcal F,\mathcal G\rangle=\|P_{\mathcal F} P_{\mathcal G}\|$, where $P_{\mathcal F}$ and $P_{\mathcal G}$ are the corresponding orthogonal projections.
\begin{thm}
Let $(\lambda_n)_{n\geq 1}$ be a real sequence satisfying \eqref{eq:lambda_tecn1} and \eqref{eq:lambda_tecn2} and let $I$ be the meromorphic inner function defined by \eqref{eq:definition-I}. Let $\Theta$ be a meromorphic inner function such that $\Theta'\in L^\infty(\mathbb R)$ and $\eta_n=|I'(\lambda_n)|^{1/2} |\Theta'(\lambda_n)|^{-1/2}\to 1$ as $n\to\infty$. Then, the following assertions are equivalent:
\begin{enumerate}
\item there exists two real-valued functions $\alpha,\beta\in C(\dot\R)$ satisfying
$$
\arg\Theta-\arg I=\alpha+\widetilde\beta;
$$
\item the operator $T_{\Theta \bar I}$ is of the form unitary plus compact.
\item \begin{enumerate}
\item[(i)] the sequence $(k_{\lambda_n}^\Theta)_{n\geq 1}$ is a complete AOB for $\mathcal K_\Theta$;
\item [(ii)]$
\cos\langle \overline{\mbox{span}}(k_{\lambda_n}^\Theta:n\geq N), I\HH^2\rangle\to 0\quad\mbox{as }N\to\infty.
$
\end{enumerate}
\end{enumerate}
\end{thm}

\begin{proof}
The equivalence between (1) and (2) is proved in Theorem~\ref{umodcom2}. \\

Let us now prove $(2)\Longrightarrow (3)$. So, we assume that $T_{\Theta \bar I}$ is of the form unitary plus compact. We have already seen that necessarily $T_{\Theta \bar I}$ is invertible (see Lemma ~\ref{unimodcom}) and by Theorem~\ref{Thm:analogue-mishko}, we get that the sequence $(k_{\lambda_n}^\Theta)_{n\geq 1}$ is a Riesz basis for $\mathcal K_\Theta$. Observe now that
$$
(T_{\bar I}|\mathcal K_\Theta)^*=P_\Theta I|\mathcal K_\Theta=\Theta P_- \bar\Theta I |\mathcal K_\Theta=\Theta H_{I\bar\Theta}|\mathcal K_\Theta.
$$
But Lemma~\ref{unimodcom} implies $I\bar\Theta\in QC$ and by Hartman's theorem, we deduce that $H_{I\bar\Theta}$ is compact. Hence $T_{\bar I}|\mathcal K_\Theta$ is compact. Using Lemma~\ref{Lem:identite-clee}, we have
$$
(1-I)k_{\lambda_n}^\Theta=\eta_n k_{\lambda_n}^I -\Theta \overline{\Theta(\lambda_n)} \eta_n k_{\lambda_n}^I,
$$
where $\eta_n\to 1$ as $n\to\infty$. If we apply the operator $T_{\bar I}$, using the fact that $\ker T_{\bar I}=\mathcal K_I$, we obtain
$$
(Id-T_{\bar I})k_{\lambda_n}^\Theta=T_{\Theta\bar I}(\eta_n \overline{\Theta(\lambda_n)}k_{\lambda_n}^I).
$$
Since $(\eta_n \overline{\Theta(\lambda_n)}k_{\lambda_n}^I)_{n\geq 1}$ is an orthogonal basis for $\mathcal K_I$, since $(k_{\lambda_n}^\Theta)_{n\geq 1}$ is a Riesz basis of $\mathcal K_\Theta$ and $T_{\Theta\bar I}$ is invertible, the last equation implies that the operator $(Id-T_{\bar I})|\mathcal K_\Theta$ is also invertible. Therefore, we can write
$$
k_{\lambda_n}^\Theta=((Id-T_{\bar I})|\mathcal K_\Theta)^{-1}T_{\Theta\bar I}(\eta_n \overline{\Theta(\lambda_n)}k_{\lambda_n}^I).
$$
Now observe that the operator $((Id-T_{\bar I})|\mathcal K_\Theta)^{-1}T_{\Theta\bar I}$ is of the form unitary plus compact and the sequence $((\eta_n \overline{\Theta(\lambda_n)}k_{\lambda_n}^I)_{n\geq 1}$ is a complete AOB for $\mathcal K_I$. It implies that the sequence $(k_{\lambda_n}^\Theta)_{n\geq 1}$ is a complete AOB for $\mathcal K_\Theta$. To prove (ii), note that
\begin{eqnarray*}
\cos\langle \overline{\mbox{span}}(k_{\lambda_n}^\Theta:n\geq N), I\HH^2\rangle&=&\left\|P_{I\HH^2}|\overline{\mbox{span}}(k_{\lambda_n}^\Theta:n\geq N)| \right\|\\
&=&\left\|IP_+\bar I |\overline{\mbox{span}}(k_{\lambda_n}^\Theta:n\geq N)\right\|\\
&=&\left\|T_{\bar I} |\overline{\mbox{span}}(k_{\lambda_n}^\Theta:n\geq N)\right\|.
\end{eqnarray*}
Since $(k_{\lambda_n}^\Theta)_{n\geq 1}$ is a complete AOB for $\mathcal K_\Theta$ and $T_{\bar I}|\mathcal K_\Theta$ is compact, we can use \cite[Lemma 10.31]{fricain2016theory} to conclude that
$$
\left\|T_{\bar I} |\overline{\mbox{span}}(k_{\lambda_n}^\Theta:n\geq N)\right\|\to 0,\quad \mbox{as }N\to\infty.
$$

It remains to prove $(3)\Longrightarrow (2)$. Let us denote by $P_N^\Theta$ the projection from $\mathcal K_\Theta$ onto $\overline{\mbox{span}}(k_{\lambda_n}^\Theta:n\geq N)$. We have
\begin{eqnarray*}
\left\|P_{I\HH^2}P_N^\Theta\left(\sum_{n\geq 1}a_n k_{\lambda_n}^\Theta\right)\right\|_2^2&=&\left\|P_{I\HH^2}\left(\sum_{n\geq N}a_n k_{\lambda_n}^\Theta\right)\right\|_2^2\\
&\leq & \varepsilon_N^2 \left\|\sum_{n\geq N}a_n k_{\lambda_n}^\Theta\right\|_2^2,
\end{eqnarray*}
where $\varepsilon_N=\cos\langle \overline{\mbox{span}}(k_{\lambda_n}^\Theta:n\geq N), I\HH^2\rangle\to 0$ as $N\to\infty$. Moreover, since $(k_{\lambda_n}^\Theta)_{n\geq 1}$ is a complete AOB for $\mathcal K_\Theta$, then
\begin{eqnarray*}
\left\|P_{I\HH^2}P_N^\Theta\left(\sum_{n\geq 1}a_n k_{\lambda_n}^\Theta\right)\right\|_2^2&\lesssim& \varepsilon_N^2\sum_{n\geq N}|a_n|^2\\
&\leq & \varepsilon_N^2\sum_{n\geq 1}|a_n|^2\\
&\lesssim&\varepsilon_N^2 \left\|\sum_{n\geq 1}a_n k_{\lambda_n}^\Theta \right\|_2^2.
\end{eqnarray*}
Therefore, we get $\|P_{I\HH^2}P_N^\Theta\|\lesssim\varepsilon_N\to 0$, $N\to\infty$. Since
$$
P_{I\HH^2}|\mathcal K_\Theta=P_{I\HH^2}P_N^\Theta+P_{I\HH^2}(Id-P_N^\Theta),
$$
and $Id-P_N^\Theta$ is a finite rank operator, we get that $P_{I\HH^2}|\mathcal K_\Theta$ can be approximated in norm by a sequence of finite rank operators and hence $P_{I\HH^2}|\mathcal K_\Theta$ is compact. We deduce that $T_{\bar I}|K_\Theta=\overline{I}P_{I\mathcal H^2}|\mathcal K_\Theta$ is also compact. Since $(k_{\lambda_n}^\Theta)_n$ is an AOB for $K_\Theta$ and $(\eta_n \overline{\Theta(\lambda_n)}k_{\lambda_n}^I)_n$ is an AOB for $\mathcal K_I$, there is an isomorphism $U:\mathcal K_I\longrightarrow K_\Theta$ of the form unitary plus compact such that $U(\eta_n\overline{\Theta(\lambda_n)}k_{\lambda_n}^I)=k_{\lambda_n}^\Theta$, $n\geq 1$. Using one more time the relation $$
(Id-T_{\bar I})k_{\lambda_n}^\Theta=T_{\Theta\bar I}(\eta_n \overline{\Theta(\lambda_n)}k_{\lambda_n}^I),
$$
we then deduce that the operator $T_{\Theta \bar I}:\mathcal K_I\longrightarrow K_\Theta$ coincide with $(Id-T_{\bar I})U$. In particular, it is of the form unitary plus compact. Note now that the operator $T_{\Theta\bar I}:I\mathcal H^2\longrightarrow \Theta\mathcal H^2$ is unitary. Hence, we get that the operator $T_{\Theta\bar I}:\mathcal H^2\longrightarrow \mathcal H^2$ is unitary plus compact.

\end{proof}


TITLE: How does Young's inequality reduce to the arithmetic-geometric mean inequality?
QUESTION [1 upvotes]: I'm going through Carothers Real Analysis to study for my qualifying exams, and came across this unjustified statement, that Youngs inequality (for $p > 1$ and q such that $\frac{1}{p} + \frac{1}{q} = 1$ we have that for $a, b \geq 0$, $ab \leq \frac{a^p}{p} + \frac{a^q}{q}$), when $p=q=2$, reduces to the arithmetic geometric mean inequality $\sqrt{ab} \leq \frac{a+b}{2}$. I tried manipulating youngs inequality but cant seem to derive the arithmetic geometric mean inequality from it. How is it done?

REPLY [3 votes]: Substitute $\sqrt a$, $\sqrt b$ for $a,b$, respectively.


TITLE: Substructures and universal sentences.
QUESTION [0 upvotes]: On page 35 in Ziegler, Tent: Model Theory it is stated:

Let $T$ be an $L$-theory. It follows from 3.1.2. that the models of
$T_{\forall}$ are the substructures of models of $T$.

with

Lemma 3.1.2. Let $T$ be a theory, $\mathfrak{A}$ a structure and
$\Delta$ a set of formulas closed under existential quantification,
conjunction and substitution of variables. Then the following are
equivalent:
a) All sentences $\varphi\in\Delta$ which are true in
$\mathfrak{A}$ are consistent with $T$.
b) There is a model
$\mathfrak{B}\vDash T$ and a map $f:\mathfrak{A}\to_\Delta \mathfrak{B}$.

I don't see why it follows directly from 3.1.2.
I try to argue by this, but I am not sure if it makes much sense:
Assume that $T$ is consistent. As $T_\forall=\{\psi_\forall|\psi_\forall\text{ universal and } T\vdash\psi_\forall\}$ we have that $T_\forall\subseteq\text{Ded}(T)$ (deductive closure) and therefore $T_\forall\cup\text{Ded}(T)$ and also $T_{\forall}\cup T$ are consistent. Now, pick a model $\mathfrak{A}_\forall\vDash T_\forall$. By the equivalence of 3.1.2. we have that there is a model $\mathfrak{B}\vDash T$ with a map $f:\mathfrak{A}_{\forall}\to_{T_\forall}\mathfrak{B}$. I see no reason why I can pick $f$ to be the inclusion map.
Conversely, I don't see why if I have $\mathfrak{B}\vDash T$ and the inclusion $\iota:\mathfrak{A}\to_{T_\forall}\mathfrak{B}$, that then $\mathfrak{A}\vDash T_\forall$.

REPLY [1 votes]: I think the authors' intention is to apply 3.1.2 with $\Delta$ being the set of existential formulas.  Then the proof of the alleged consequence would go like this. Suppose $\mathfrak A\models T_\forall$. Then I claim clause (a) of 3.1.2 is true. Indeed, if $\phi$ is an existential sentence true in $\mathfrak A$, then $\neg\phi$ is (up to logical equivalence) a universal sentence that is false in $\mathfrak A$ and therefore is not in $T_\forall$. So $\phi$ is consistent with $T$. By 3.1.2, we infer that clause (b) is also true, so we have $\mathfrak B\models T$ and $f:\mathfrak A\to_\Delta\mathfrak B$.
Now we need another (standard but easy to get wrong) maneuver to get $f$ to be an inclusion. As it stands, $f$ is an isomorphism from $\mathfrak A$ to a substructure $\mathfrak A'$ of $\mathfrak B$ (because all atomic formulas and their negations are in $\Delta$).  The idea is to modify $\mathfrak B$ by replacing this copy $\mathfrak A'$ of $\mathfrak A$ with $\mathfrak A$ itself. That way you get an isomorphic copy of $\mathfrak B$ (hence still a model of $T$) that contains literally $\mathfrak A$ rather than a copy $\mathfrak A'$ of it.  The problem (the reason it's easy to get wrong) is that, if we're unlucky, $\mathfrak B$ might already contain some elements of $\mathfrak A$, and not in the place where we want them; then if we just replace $\mathfrak A'$ with $\mathfrak A$ we'll end up with some elements of $\mathfrak A$ trying to be in two different places in $\mathfrak B$.  Fortunately, the cure for this problem is easy: First replace $\mathfrak B$ by an isomorphic copy that is disjoint from $\mathfrak A$, and then proceed as above.
Finally, to show the converse, that all substructures of models of $T$ are models of $T_\forall$, prove, by induction on universal formulas $\theta$, that, if $\mathfrak A$ is a substructure of $\mathfrak B$ and if some assignment of values for variables in $\mathfrak A$ satisfies $\theta$ in $\mathfrak B$, then it also satisfies $\theta$ in $\mathfrak A$.  The only nontrivial step in the induction is the universal quantifier step, and there you just use that all elements of $\mathfrak A$ (that might serve as counterexamples) are also in $\mathfrak B$ (and therefore cannot be counterexamples).


TITLE: How to solve the equation $n^2 \equiv 0 \pmod{584}$?
QUESTION [1 upvotes]: Well, I've confused when trying to solve this equation can anybody help me :
$n^2 \equiv 0 \pmod{584}$
I tried to factorize the $584$ i got $584=2^3\times73$.
so $n^2$ has to be divisible by $2^3$ and $73$ in this same time.
here i get stuck.

REPLY [0 votes]: ${\bf Hint}\ \ {\rm If}\ \ p\ \ {\rm is\ prime,}\ \ n = p^{\large k} m,\,\ p\nmid m\ \ {\rm   then}\ \  p^{\large 2j-1}\!\mid n^{\large 2}\!\iff\! \color{#0a0}{p^{\large 2j}\mid n^{\large 2}}\!\iff \color{#c00}{p^{\large j}\mid n},\ $ since
$$\begin{eqnarray}\,\ p^{\large 2j-1}\!\mid n^{\large 2}\! = p^{\large 2k} m^{\large 2}  \!\iff\! 2j\!-\!1\le 2k &\iff& \ \color{#0a0}{2j\le 2k} &\iff& \ \color{#c00}{j\le k}\\
 &\overset{\phantom{I^I}}\iff& \color{#0a0}{p^{\large 2j}\mid \smash[b]{\underbrace{p^{\large 2k}m^2}_{\Large n^2}}}\!\! &\iff& \color{#c00}{p^{\large j}\mid \smash[b]{\underbrace{p^{\large k}m}_{\Large n}}} \\ \\ \end{eqnarray}$$
by the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations)
$\!\begin{eqnarray}{\rm Therefore}\ \ \ 2^3\cdot 73\mid n^2 &\iff& 2^3\mid n^2,\,\ 73\mid n^2 \ &&\text{by lcm = product for coprimes}\\
&\iff& \color{#c00}{2^2\mid n,\ \ \ 73\mid n}&&\text{by above Lemma}\\
&\iff& 2^2\cdot 73\mid n &&\text{by lcm = product for coprimes}
\end{eqnarray}$


TITLE: Are there combinatorial games of finite order different from $1$ or $2$?
QUESTION [4 upvotes]: Are there any combinatorial games whose order (in the usual addition of combinatorial games) is finite but neither $1$ nor $2$?
Finding examples of games of order $2$ is easy (for example any impartial game), but I have not been able to think up an example with finite order where the order did not come from some sort of symmetry (for example even though Domineering is not impartial, it is easy to see that any square board will give a game of order $1$ or $2$), and such a symmetry only gives $1$ or $2$ as the possible orders.

REPLY [3 votes]: Chapter III section 3 of Siegel's "Combinatorial Game Theory" contains a straightforward construction to make games with orders that are arbitrarily high powers of $2$.
If $G$ has finite order, and $n$ is three times the birthday of $G$, then $h(G)=\{n+G|-n\}$ satisfies $h(G)+h(G)=G$. This isn't too hard to prove if you are used to standard CGT proof techniques: assume $G$ is in canonical form and look at the game $h(G)+h(G)-G$ and a few cases. $n$ is made large enough that when you're looking at a subposition of $G+G+G$ you don't have to worry about which subposition it is.
This construction turns $0$ into $*=\{0|0\}$, $*$ into $\{3*|-3\}$ (which has order 4), which becomes $\{18|12*\|-15\}$ (order $8$)$\ldots$
What about odd orders? That's another question.


TITLE: Creating a 3D surface from 2D graphs
QUESTION [1 upvotes]: So I have two sets of equations:
$\mathcal{A}$ = \begin{equation}
    \{ f(y_{0},x), \, f(y_{1},x) , \;... \;, f(y_{n},x) \}
    \end{equation}
$\mathcal{B}$ = \begin{equation}
    \{ g(y,x_{0}), \, g(y,x_{1}) , \;... \;, g(y,x_{n}) \}
    \end{equation}
And I created a surface out of these equations in a plot. I know I can find an equation for this surface by fitting the plot data to known surfaces, but that is not what I want.
I want to be able to directly derive an equation for the surface from just these equations. Is this possible? How can I do this?

REPLY [1 votes]: You are not given "equations", but $2n+2$ curves in three-space, namely the curves
$$\gamma_j:\quad x\mapsto\bigl(x,y_j,f(x,y_j)\bigr)\qquad(0\leq j\leq n)$$
in the vertical planes $y=y_j$, and the curves
$$\gamma_k'\quad y\mapsto\bigl(x_,y,g(x_k,y)\bigr)\qquad(0\leq k\leq n)$$
in the vertical planes $x=x_k$. These $2n+2$ curves form a wire mesh over the $(x,y)$-plane which from above looks like an orthogonal grid. Now you are in search of a function $$\phi: \ (x,y)\mapsto z=\phi(x,y)\ ,$$ such that the graph of $\phi$ (a surface over the $(x,y)$-plane) contains all these wires $\gamma_j$ and $\gamma_k'$.
In the first place such a $\phi$ is not uniquely determined at all. There are various principles to choose an "optimal" $\phi$ corresponding to the given data. Which of these principles suits best in your situation has to be judged by an "expert". In any case the problem formulated here is a problem of utmost interest in practical applications (e.g., sculpting of car fenders), and there is a large variety of numerical techniques (and software) available to deal with it.


TITLE: A subset $D$ of $\mathbb R^2$ is called open if none of its boundary points are in the subset, that is, if $\delta D \cap D = \varnothing$.
QUESTION [1 upvotes]: A subset $D$ of $\mathbb R^2$ is called open if none of its boundary points are in the subset, that is, if $\delta D \cap D = \varnothing$.
This is in the context of multivariable maxima and minima.
My question is, what does $\delta D \cap D$ of $\delta D \cap D = \varnothing$ mean? I was under the impression that $\cap$ stood for 'intersect', which basically means AND. However, I'm having trouble understanding what it means in this case, as $\delta D \cap D$.
I'd appreciate any clarification. Thank you.

REPLY [1 votes]: "$\delta D\cap D$" means "all the points contained in both $\delta D$ and $D$. "$\delta D$" means the "boundary of $D$". So if $D$ does not contain any of the points found in the boundary $\delta D$, we say it is open.

REPLY [1 votes]: You have it right. $\delta D$ is the boundary of $D$, which consists of points $x$ so that any ball centered at $x$ contains both points of $D$ and points outside $D$. To say that $D$ is open is to say that it contains none of its boundary points. (So the set of points $x\in\Bbb R^2$ with $\|x\|<1$ is open, but the set of points $x$ with $\|x\|\le 1$ is not.)


TITLE: How to prove that prime $q \neq p$ isn't ramified in $\mathbb{Z}_p$-extension of $\mathbb{Q}$?
QUESTION [2 upvotes]: I want to show that there is only one $\mathbb{Z}_p$-extension of the field of rational numbers - $\mathbb{Q}$. The cyclotomic one.
For that I need to show that if the field $\mathbb{K}$ is $\mathbb{Z}_p$-extension of $\mathbb{Q}$, then extension $\mathbb{K}/\mathbb{Q}$ is unramified in any prime number $q \neq p$.
I have found two different proof of this fact, but I fail to understand the ending of both of them. One is in the book "L. C. Washington - Introduction to Cyclotomic Fields", on page 265 - Proposition 13.2, and the other one is in the article "On $\Gamma$-extensions of algebraic number fields. Bull. Amer. Math. Soc., 65 (1959), 183-226." of Kenkichi Iwasawa, Lemma 7.1.

REPLY [1 votes]: For an abelian extension $K/\Bbb{Q}$ then (Kronecker Weber) $Gal(K/\Bbb{Q})$ is a quotient of $Gal(\Bbb{Q}(\zeta_{\infty})/\Bbb{Q}) = \hat{\Bbb{Z}}^\times$.

We assume $Gal(K/\Bbb{Q}) \cong \Bbb{Z}_p$ as profinite topological groups

$Gal(K/\Bbb{Q})$ is torsion free thus the kernel of $Gal(\Bbb{Q}(\zeta_\infty) / \Bbb{Q}) \to Gal(K/\Bbb{Q})$ contains the profinite completion of $Gal(\Bbb{Q}(\zeta_\infty) / \Bbb{Q})_{tors} = \hat{\Bbb{Z}}^\times_{tors} = \prod_q'(\Bbb{Z}_q^\times)_{tors}$ so that $Gal(K/\Bbb{Q})$ is a quotient of $$ \hat{\Bbb{Z}}^\times/\overline{\hat{\Bbb{Z}}^\times_{tors} } =  \prod_q \Bbb{Z}_q^\times/(\Bbb{Z}_q^\times)_{tors}=(1+4\Bbb{Z}_2)\prod_{q\ne 2}(1+q\Bbb{Z}_q)$$
(think to the RHS not as a subgroup but as a quotient of $\hat{\Bbb{Z}}^\times$)
$Gal(K/\Bbb{Q}) \cong \Bbb{Z}_p$ tells us $K = \bigcup_n K_n$ where $Gal(K_n/\Bbb{Q}) \cong \Bbb{Z}/p^n\Bbb{Z}$,
Since $Gal(K_n/\Bbb{Q})$ is of order $p^n$ we obtain that $K_n$ is fixed by $$(\hat{\Bbb{Z}}^\times/\overline{\hat{\Bbb{Z}}^\times_{tors} } )^{p^n}=(1+4\Bbb{Z}_2)^{p^n}\prod_{q\ne 2}(1+q\Bbb{Z}_q)^{p^n}= (1+p^{n+1}\Bbb{Z}_p) (1+4\Bbb{Z}_2) \prod_{q\ne 2,p}(1+q\Bbb{Z}_q)$$ and hence $K= \bigcup_n K_n$ is fixed by $$\bigcap_n (\hat{\Bbb{Z}}^\times/\overline{\hat{\Bbb{Z}}^\times_{tors} } )^{p^n}=(1+4\Bbb{Z}_2) \prod_{q\ne 2,p}(1+q\Bbb{Z}_q)=Gal(\Bbb{Q}(\zeta_\infty)/\Bbb{Q}(\zeta_{p^\infty})^{\langle \omega_{p-1}\rangle})$$ Whence $ K $ is a subfield of $ \Bbb{Q}(\zeta_{p^\infty})^{\langle \omega_{p-1}\rangle}$ and since the later is a $\Bbb{Z}_p$-extension, its subfields are finite extensions of $\Bbb{Q}$ and hence $$K=\Bbb{Q}(\zeta_{p^\infty})^{\langle \omega_{p-1}\rangle}$$


TITLE: Eigenvalues of matrix sums
QUESTION [78 upvotes]: Is there a relationship between the eigenvalues of individual matrices and the eigenvalues of their sum? What about the special case when the matrices are Hermitian and positive definite?
I am investigating this with regard to finding the normalized graph cut under general convex constraints. Any pointers will be very helpful.

REPLY [4 votes]: Here is a trivial case with a simple solution. Applicable in Quantum Mechanics, for one.
Given two matrices of the form $A \otimes Id$, $Id \otimes B$, the eigenvalues of their sum are all combinations $a_i+b_j$, where $A\vec{a}_i=a_i\vec{a}_i$ and $B\vec{b}_i=b_i\vec{b}_i$. The eigenvectors are all tensor products of the individual eigenvectors of $A$ and $B$.
This is a one-liner to show:
$$(A \otimes Id + Id \otimes B) \,\vec{a}_i\otimes \vec{b}_j=(a_i+b_j)\,\vec{a}_i\otimes \vec{b}_j$$


TITLE: Debunking thought experiment purporting to show that gravity is not classical
QUESTION [2 upvotes]: As far as I know, there are no known observations that require a quantum theory of gravity. Now, that's not to say we don't need a quantum theory of gravity. In fact, I get the sense that we do, at least in part, because GR couples energy density to spacetime curvature, so we expect spacetime to be influenced by quantum fields, which begs for a quantum description of gravity. However, this doesn't imply that gravity is quantized. Gravity could, as far as I know, be classical, which I mean in the information-theoretic sense of the word: gravitational degrees of freedom cannot be entangled.
I am aware of Eppley-Hannah's paper that argues that gravity cannot be classical. Roughly, they show that if it were, then one could violate the uncertainty principle using a gravitational wave measurement device (which classically could have arbitrarily low momentum and wavelength). However, there are many rebuttals out there, including Mattingly's gr-qc/0601127. I came up with an alternative thought experiment, and I am interested to know what is wrong with it.
Suppose gravity is classical. Consider a box with a massive particle inside, in a state that makes it equally probable for it to be on the left or right side of the box. Now, partition the two sides with an impenetrable barrier. Then the wavefunction is
$$
\psi(x) = \frac{1}{\sqrt{2}}\big(\psi_L(x) + \psi_R(x)\big),
$$
where $\psi_L(x)$ and $\psi_R(x)$ are the same, except they only have support on their respective side of the box. Now, with your friend, Alice, separate the two sides of the box as far as you can before considerable decoherence ruins the superposition. I see two possibilities, based on whether gravity contributes to decoherence or not.
If gravity does contribute to decoherence, then the wavefunction must "collapse" before the boxes were separated (since the collapse was induced either by the particle's self-gravity or the acceleration imparted on the box when the boxes were first separated). But this conflicts with experiment, because we can create nonlocal superpositions that were initially local (e.g., in tests of entanglement).
If gravity does not contribute to decoherence, then, because gravity is classical, spacetime cannot warp under both boxes (and exist in a superposition like the particle that induces the warping). It can only warp under one box, for otherwise if Alice looks and does not find the electron, then spacetime would have to either suddenly unwarp to accommodate Alice's observation (which is nonclassical and would presumably send gravitational waves in all directions) or spacetime would remain warped under both boxes, despite there not being a source of energy to cause the warping around Alice's box. The only conclusion I see is that spacetime is warped under the box with the electron "in" it, which means that gravity "collapsed" the wavefunction early on --- a contradiction.
I'd like to know what is fundamentally flawed here. This argument seems too easy and obvious to rule out classical gravity. I suspect it's my assumption about how the energy density of the particle is allocated across spacetime pre-measurement. But I could see this as a non-issue, for is it not true that, however the energy density is allocated across spacetime pre-measurement, after Alice looks into her box to find it's (not) there, the spacetime must be warped under her (your) box, so my second argument above still applies?

REPLY [2 votes]: I'd like to know what is fundamentally flawed here…

In my opinion the main flaw is that what is described is not a thought experiment, but only a start of one strung together with some “moral” reasoning on how the gravitational field have to behave to accommodate situation described.
For an experiment (even thought ones) there has to be some sort of measurements, e.g. of spacetime “warping/unwarping” and/or detection of gravitational waves (if present). If these measurements once performed would lead to violation of some well established laws of Nature, such as uncertainty principle or relativistic causality, if the gravity remains classical, only then we can conclude that this thought experiment could be considered an argument in favor of quantization of gravity.


TITLE: Resolution of ADE- singularities: The second blow-up for the A_2 singularity
QUESTION [0 upvotes]: I try to resolve the surface singularity $X\subset \mathbb{A}^3$ of type $A_2$. It is defined by $x^2+y^2+z^3=0$. I expect to need at least two blow-ups. But after the first blow-up of $X$ the strict transform $\tilde{X}$ seems to be already non-singular: As usual I check
$$\tilde{X} \cap (\mathbb{A}^3\times \mathbb{P}^2)$$
with respect to each of three affine standard sets of $\mathbb{P}^2$.
What is wrong? Do I have to blow-up for the second blow-up an infinitesimal near point, i.e. a point on the exceptional divisor of the first blow-up?

REPLY [0 votes]: Indeed: The second blow-up has to blow-up a singular point of multiplicity $=2$ of the exceptional divisor $E_1$ of the first blow-up. Hence one gets a second exceptional divisor $E_2$ with self-intersection -2, and both exceptional divisors have intersection number $=1$.The dual graph of the two exceptional divisors is the Coxeter graph $A_2$ with annotation -2.


TITLE: How to find general solution for this matrix
QUESTION [0 upvotes]: I'm trying to find the general solution to this matrix
\begin{bmatrix}1&-2&1&3&0\\2&-4&4&6&4\\ -2&4&-1&-6&2\\1&-2&-3&3&-8\end{bmatrix}
Ax=$\begin{bmatrix}1&6&0&-7&\end{bmatrix}^T$
I think I'm supposed to get it in x=x*+z format, I'm still not sure if this is the correct way to do it.
But I ended up getting this matrix in row echelon form.
-2(r1)+(r2)
2(r1)+(r3)
-(r1)+(r4)
\begin{bmatrix}1&-2&1&3&0\\0&0&2&0&4\\ 0&0&1&0&2\\0&0&-4&0&-8\end{bmatrix}
And then
2(r2)+(r4)
-1/2(r2)+(r3)
\begin{bmatrix}1&-2&1&3&0\\0&0&2&0&4\\ 0&0&0&0&0\\0&0&0&0&0\end{bmatrix}
then
1/2(r2)
\begin{bmatrix}1&-2&1&3&0\\0&0&1&0&1\\ 0&0&0&0&0\\0&0&0&0&0\end{bmatrix}
and lastly  -r2 + r2
\begin{bmatrix}1&-2&0&3&-2\\0&0&1&0&1\\ 0&0&0&0&0\\0&0&0&0&0\end{bmatrix}
after doing some algebra i ended up getting
x1 = 2(x2)-3(x4)-2(x5)
x3 = -2(x5)
and set x5 = x4 = x2 = 1
and got z =  $\begin{bmatrix}1&1&-2&1&1\end{bmatrix}^T$
But I'm when i try to solve for Ax = $\begin{bmatrix}1&6&0&-7\end{bmatrix}^T$
The last two rows are full of zeros
so I can't have 0=-7
How would i solve this?

REPLY [0 votes]: You have to work with the augmented matrix, and do the same operations on the rows of this matrix (including the supplementary column). Here is how it goes:
\begin{align}
&\begin{bmatrix}\begin{array}{@{}rrrrr|r@{}}
1&-2&1&3&0&1\\2&-4&4&6&4&6\\-2&4&-1&-6&2&0\\1&-2&-3&3&-8&-7
\end{array}\end{bmatrix}\rightsquigarrow
\begin{bmatrix}\begin{array}{@{}rrrrr|r@{}}
1&-2&1&3&0&1\\0&0&2&0&4&4\\0&0&1&0&2&2\\0&0&-4&0&-8&-8
\end{array}\end{bmatrix}\rightsquigarrow\\
&\begin{bmatrix}\begin{array}{@{}rrrrr|r@{}}
1&-2&1&3&0&1\\0&0&1&0&2&2\\0&0&0&0&0&0\\0&0&0&0&0&0
\end{array}\end{bmatrix}\rightsquigarrow\begin{bmatrix}\begin{array}{@{}rrrrr|r@{}}
1&-2&0&3&-2&-1\\0&0&1&0&2&2\\0&0&0&0&0&0\\0&0&0&0&0&0
\end{array}\end{bmatrix}
\end{align}
so the solutions are given by the equations
\begin{cases}
x_1=2x_2-3x_4-2x_5-1,\\
x_3=-2x_5+2.
\end{cases}


TITLE: Solve a simple differential equation
QUESTION [1 upvotes]: This is an equation I met while solving a probability theory problem:
$ dy = 2(m/x-y) \cdot dx/x $
m is a constant
Also it is known that $y(m) =1$.
I have the function which satisfies the equation and I verified that it works. But I haven't solved a single differential equation in many years so... I just got curious how one can derive this solution.
Is this equation of some particular type which is solvable through a well-known procedure? Any reference?
Sorry... I am typing on my phone... I will try to improve the equation outlook now.

REPLY [0 votes]: $$dy = 2(m/x-y) \cdot dx/x$$
$$xdy -2(\frac mx-y)dx=0$$
$$xdxy -(2m-xy)dx=0$$
I's separable
$$\int \frac {dxy}{(xy-2m)}=-\int \frac {dx}x$$
$$\ln(xy-2m)=-\ln(x)+K$$
$$(xy-2m)=\frac Kx$$
$$\implies y(x)=\frac {K+2mx}{x^2}$$

As @haqnatural has pointed out
$$xdy -2(m/x-y) dx=0$$
the differential is exact with integrating factor $\mu(x)=x$
$$x^2dy -2(m-xy) dx=0$$
$$\implies \partial_x(x^2)=\partial_y(2xy-2m)$$


TITLE: If $f:M\to M,m\mapsto mr$ is injective, then show that $\text{Hom}_R(M,E)\stackrel{r}{\longrightarrow}\text{Hom}_R(M,E)$ is surjective
QUESTION [1 upvotes]: Let $R$ be a commutative ring, $M$ an $R$-module and $r\in R$.  If $f:M\to M$ defined by $f(m)=mr$  is an  injective $R$-module endomorphism, then show that the mapping $\text{Hom}_R(M,E)\stackrel{r}{\longrightarrow}\text{Hom}_R(M,E)$ is surjective, where $E$ is the injective cogenerator of $R$.
I have tried to search for injective generators and what I know is that
(1) an injective $R$-module $E$ is called an {\it injective cogenerator} of $R$ if, for every $R$-module $M$ and for every non-zero $m\in M$, there is a homomorphism $\phi:M\to E$ such that $\phi(m)\neq0$.
(2) since $R$ is commutative, $\text{Hom}_R(M,E)$ is also an $R$-module whose elements are maps $\phi\in \text{End}_R(M)$.
However, I cannot figure out the surjectivity of that multiplication map in $\text{Hom}_R(M,E)\stackrel{r}{\longrightarrow}\text{Hom}_R(M,E)$ comes about.

REPLY [0 votes]: Exercise 2.19 (Rotman): the functors $\operatorname{Hom}_R(-, -)$ preserve multiplication. Explicitly, if $\mu_r : M \to M$ is the multiplication map $\mu_r(m) = r \cdot m$ from a module $M$ over a commutative ring $R$ to itself, then for any $R$-module $N,$ the induced map $\operatorname{Hom}(\mu_r) : \operatorname{Hom}_R(M, N) \to \operatorname{Hom}_R(M, N)$ is the multiplication map $\varphi \mapsto r \cdot \varphi.$
Considering that $E$ is an injective cogenerator of $R,$ it is by definition an injective $R$-module, and this implies that $\operatorname{Hom}_R(-, E)$ is an exact functor, i.e., $\operatorname{Hom}_R(-, E)$ is right exact.
Consequently, the exact sequence $$0 \to M \xrightarrow{r \cdot} M \to \frac{M}{rM} \to 0$$ gives rise to the exact sequence $$0 \to \operatorname{Hom}_R \biggl(\frac{M}{rM}, E \biggr) \to \operatorname{Hom}_R(M, E) \xrightarrow{r \cdot} \operatorname{Hom}_R(M, E) \to 0,$$ as $\operatorname{Hom}_R(-, E)$ is contravariant. But this says that $\operatorname{Hom}_R(M, E) \xrightarrow{r \cdot} \operatorname{Hom}_R(M, E)$ is surjective.


TITLE: Detection of day or night?
QUESTION [1 upvotes]: I was wondering, are there "particles" that can be detected, when one's side of earth is shined on by the sun?
I understand that there are cosmic radiations present, which probably will drown the little effect the sun has, but is there any way to detect something specific to the sun?
I am also assuming the following:
These "particles" (if in existence) are strong enough to penetrate the earth's outer layers (thus picked up by a sensor of sort no matter if it's under direct sun light or not), but not strong enough to be picked up on the other side of earth.
Please also note that I have very little knowledge in this area!

REPLY [3 votes]: The sun blasts us continuously with charged particles. When these strike the upper reaches of our atmosphere, they scatter off of and ionize some of the air molecules there. This renders those particular layers of the atmosphere somewhat electrically conductive- and in that state, radio waves beamed from the ground will get refracted off those layers and go back down towards the earth instead of shooting straight out into space.
Amateur radio hobbyists use the shifting daily/nightly balance of ionization in the different layers of the atmosphere to bounce their signals all the way around the world, and to do so they monitor the daily relative activity level of the sun's photosphere (from which the charged particles get ejected).
So this furnishes a simple and direct measure of the particle output of the sun; note that those ionized layers undergo recombination after the sun goes down which shuts off these "skip channels" shortly after dark.


TITLE: How to solve $T(n)=T((4/5)n+1)+n^2$.
QUESTION [0 upvotes]: I have a problem with this recurrence relation. I have to find the Theta notation. I tried to solve it with the iterative method. The generic formula I calculated is:
\begin{equation}
T(n)=T[(4/5)^k n]+[(4/5)^{k+1} n]^2+[(4/5)^{k+2} n]^2+\cdots+n^2
\end{equation}
$n=(4/5)^k\Rightarrow k=\log_{4/5}(n)$.
At this point I tried to calculate the geometric series, forgetting the square elevation. I don't know how to continue.
Thanks for your attention.

REPLY [2 votes]: Hint.
Asymptotically $T(n)\approx T(an)-n^2$ with $a= \frac 45$.
Now making $n=a^m$ and substituting we arrive to the equivalent recurrence
$$
R(m) = R(m-1) - \frac{a^{2m}}{a^2}
$$
with solution
$$
R(m) = c_0-\frac{a^{2m}-1}{a^2-1}
$$
now we can go backwards making $m = \log_a n$, hence $T(n) = \mathcal{O}(n^2)$
Follows a MATHEMATICA script simulating the recurrence (red) as well as the asymptotic approximation (blue)
Clear[Rec, f]
a = 4/5;
f[n_] := (n^2 - 1)/(1 - a^2)
nmin = 5;
nmax = 2000;

Rec[n_] := Module[{}, If[n < nmin, Return[0]];
  If[n == nmin, Return[0]];
  Return[Rec[Floor[a n + 1]] + n^2]
  ]

tab = Table[{n, Rec[n]}, {n, nmin, nmax}];
gr1 = ListPlot[tab, PlotRange -> All, Filling -> Axis, PlotStyle -> {Thick, Red}];
gr2 = Plot[f[x], {x, nmin, nmax}, PlotStyle -> {Blue, Thick}];
Show[gr1, gr2, PlotRange -> All]


TITLE: No solution to Laplace equation boundary value problem
QUESTION [0 upvotes]: I am working on solving the following boundary value problem over an annulus with inner radius 1 and outer radius 2.
$$\frac{\partial^2 u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}=0, 1<r<2, -\pi \le \theta \le \pi$$
$$u(1,\theta)=\sin{4\theta}-\cos{\theta}, -\pi \le \theta \le \pi$$
$$u(2,\theta)=\sin{\theta}, -\pi \le \theta \le \pi$$
The separation of variables $u(r,\theta)=R(r)T(\theta)$ yields the following equations for some constant $\lambda$.
$$\frac{r^2R''(r)+rR'(r)}{R(r)}=-\frac{T''(\theta)}{T(\theta)}=\lambda$$
The continuity equations $T(-\pi)=T(\pi)$ and $T'(-\pi)=T'(\pi)$ shows that $\lambda$ must be non-negative for a nontrivial solution. For $\lambda=0$, $T(\theta)=A$, and for $\lambda>0$:
$$T(\theta)=A_n\cos{\sqrt{\lambda}\theta}+B_n\sin{\sqrt{\lambda}\theta}$$
From this, assuming a nontrivial solution implies that $\sqrt{\lambda}=n$ for some positive integer $n$.
Using $\lambda=n^2$, the equation for $R(r)$ is the Cauchy-Euler equation:
$$r^2R''(r)+rR'(r)-n^2R(r)=0$$
With $\lambda=0$ leading to the solution $R_0(r)=C+D\ln{r}$, and $\lambda>0$ leading to the solution $R_n(r)=C_nr^n+D_nr^{-n}$.
Thus, the general solution for $u(r,\theta)$ is given as follows:
$$u(r,\theta)=C+D\ln{r}+\sum_{n=1}^{\infty}(C_nr^n+D_nr^{-n})(A_n\cos{n\theta}+B_n\sin{n\theta})$$
Since the set of functions $\{1, \cos{\theta}, \cos{2\theta},..., \sin{\theta}, \sin{2\theta}, ...\}$ is linearly independent, the $n\notin\{1,4\}$ terms are zero in the $r=1$ boundary condition, producing:
$$\sin{4\theta}-\cos{\theta}=(C_4+D_4)(A_4\cos{4\theta}+B_4\sin{4\theta})+(C_1+D_1)(A_1\cos{\theta}+B_1\sin{\theta})$$
And then coefficient matching yields the system:
$$0=(C_4+D_4)A_4$$
$$1=(C_4+D_4)B_4$$
$$-1=(C_1+D_1)A_1$$
$$0=(C_1+D_1)B_1$$
Thus, $A_4=0$ and $B_1=0$. Likewise, the $r=2$ boundary condition produces:
$$\sin{\theta}=(C_1 2^1+D_1 2^{-1})(A_1\cos{\theta}+B_1\sin{\theta})$$
Which yields the equation $1=(C_1 2^1+D_1 2^{-1})B_1$ in direct contradiction of $B_1=0$ found from the previous boundary condition. Since the trivial solution does not satisfy the boundary conditions, it seems like there is no solution to this problem.

Edit: It turns out that the listen solution (given in the answer below) does satisfy the boundary condition at $r=2$, my mistake. However, the issue I have with this a solution of the form
$$(\alpha_1r+\gamma_1r^{-1})\cos{\theta}+(\beta_1r+\delta_1r^{-1})\sin{\theta}$$
is not necessarily factorable into the form $(C_1r+D_1r^{-1})(A_1\cos{\theta}+B_1\sin{\theta})$ and, in fact, with the stated solution, it is not. This is where my confusion lies.

REPLY [2 votes]: Let's redo the algebra a little bit.  Extracting from your general solution the terms involving the functions that appear on the boundary, we posit a solution of the form
$$
u(r,\theta) = \beta_4 r^4 \sin(4\theta) + \delta_4 r^{-4} \sin(4\theta) + \alpha_1 r \cos(\theta) + \beta_1 r \sin(\theta) \\ + \gamma_1 r^{-1} \cos(\theta) +\delta_1 r^{-1} \sin(\theta).
$$
At $r=1$ the equation reads
$$
\sin(4\theta) - \cos(\theta) = (\beta_4+\delta_4) \sin(4\theta) + (\alpha_1 + \gamma_2) \cos(\theta) + (\beta_1 + \delta_1) \sin(\theta),
$$
which yields the equations
$$
\beta_4 + \delta_4 =1 \\
\alpha_1 + \gamma_1 = -1 \\
\beta_1 + \delta_1 = 0.
$$
At $r=2$ the equation reads
$$
\sin(\theta) = (2^4 \beta_4 + 2^{-4} \delta_4) \sin(4\theta) +(2\alpha_1 +2^{-1} \gamma_2) \cos(\theta) +(2\beta_2 + 2^{-1} \delta_1)\sin(\theta),
$$
which yields the equations
$$
2^4 \beta_4 + 2^{-4} \delta_4 = 0\\
2 \alpha_1 + 2^{-1} \gamma_1 =0 \\
2\beta_1 + 2^{-1} \delta_1 =1.
$$
We can solve the resulting linear system with relative ease, as it decouples into three pairs of equations with two unknowns each.  We get
$$
\beta_4 = -1/255, \delta_4 = 256/255, \alpha_1 = 1/3, \gamma_1 = -4/3, \\
\beta_1 = 2/3, \delta_1 = -2/3.
$$
Hence
$$
u(r,\theta) = \frac{1}{3} r \cos(\theta) - \frac{4}{3} r^{-1} \cos(\theta) + \frac{2}{3} r \sin(\theta) - \frac{2}{3} r^{-1} \sin(\theta) \\
- \frac{1}{255} r^4 \sin(4\theta) + \frac{256}{255} r^{-4}\sin(4\theta),
$$
and it's a simple matter to chack by hand that this does the job.
EDIT: Let's look again at the general solution.
You show above that a separated solution $u(r,\theta) = R(r) T(\theta)$ leads to solutions of the form
$$
R(r) = r^n \text{ or } R(r) = r^{-n}
$$
and
$$
T(\theta) = \cos(n\theta) \text{ or } T(\theta) = \sin(n\theta)
$$
when $n \ge 1$, and
$$
R(r) = 1 \text{ or } R(r) = \log(r) \text{ and } T(\theta) = 1
$$
when $n=0$.  From this we get four possible solutions when $n \ge 1$:
$$
u_n(r,\theta)= r^n \cos(n\theta) \\
u_n(r,\theta) = r^{-n} \cos(n\theta) \\
u_n(r,\theta) = r^n \sin(n \theta) \\
u_n(r,\theta) = r^{-n} \sin(n\theta)
$$
and two possible solutions when $n=1$:
$$
u_0(r,\theta) = 1 \text{ and } u_0(r,\theta) = \log(r).
$$
The general form of the solution is then a linear combination of these:
$$
u(r,\theta) = \omega +  \chi \log(r) \\
+ \sum_{n=1}^\infty \left[\alpha_n r^n \cos(n\theta) + \beta_n r^n \sin(n\theta) + \gamma_n r^{-n} \cos(n\theta) + \delta_n r^{-n} \sin(n\theta) \right].
$$
If we start with this then we're led naturally to my solution.  The error you're running into is due to the factored form you put the solution in.  To invert we need to solve
$$
C_n A_n = \alpha_n, C_n B_n = \beta_n, D_n A_n = \gamma_n, D_n B_n = \delta_n.
$$
I don't think we can invert this in a nice way.


TITLE: Group Law on Specific Elliptic Curve
QUESTION [3 upvotes]: Let $n$ be a fixed positive integer.
Consider the equation
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=n.$$
If $n$ is odd then there are no positive integer solutions to this equation, and if $n$ is even then the positive integer solutions can be quite large. A nice reference is this mathoverflow question.
Clearing denominators gives the equivalent equation
$$a^3+b^3+c^3+abc=(n-1)(a+b)(a+c)(b+c).$$
This is an elliptic curve (provided that you pick an identity point).

Is there a nice way to explicitly write down the group law of this elliptic curve?

Certainly it's possible to write down the group law by transforming to Weierstrass form, using the (ugly) group law formula, and transforming.
The question is whether there is a nice way to write down the group law.

REPLY [1 votes]: I worked this out a while ago, but I had forgotten about this question. I don't think that there is a nice algebraic formula for the group law, but at least you can avoid transforming to Weierstrass form.
Take $O=[-1:0:1]\in\mathbb{P}^2$. The key property of $O$ is that its tangent line is $y=6(x+z)$, which passes through $O$ with multiplicity 3 (rather than 2). Then there is a group operation on $E$ with identity element $O$, and the group operation satisfies $P+Q+R=O$ for any line $L$ that intersects $E$ at $P,Q,R$. Proving this requires some algebraic geometry.
You can compute $P+Q$ as follows:

Let $L$ be the line through $P$ and $Q$. If $P=Q$ then $L$ is the tangent line.
Let $R$ be the third point of intersection between $L$ and $E$.
Let $L^\prime$ be the line through $R$ and $O$. If $R=O$ then $L^\prime$ is the tangent line.
Then $P+Q$ is the third point of intersection between $L^\prime$ and $E$.


text:
\chapter{Wheels and the Kontsevich integral of the unknot}
\label{chap:wheels}
\section{Results}\label{sec:wheels-results}
In this chapter, we will give some explicit computations of the
Kontsevich integral to all orders for some specific knots and links.
We begin by reviewing our principal results.  The first one was
already stated in the introduction.

\setcounter{temp}{\value{theorem}}

\setcounter{theorem}{2}
\begin{theorem}[Wheels; joint with T.~Le]
The Kontsevich integral of the unknot is
\[
Z(\bigcircle) = \nu = \Omega \in \cB.
\]
\end{theorem}
\setcounter{theorem}{\value{temp}}

Recall that $\Omega$ is the ``wheels'' element from the introduction.

By changing the framing on the unknot and cabling it, we can construct
a Hopf link.  Using the results of Section~\ref{sec:conn-cable} and
the value of $Z(\bigcirc)$, we can compute the invariant of the Hopf
link from the invariant of the unknot.
There are several good formulas for the answer.

\begin{theorem}\label{thm:hopf-link}
The framed Kontsevich integral of the Hopf link can be expressed in
the following equivalent ways:
\ssp\begin{align*}
Z(\hopf{x}{y}) &=
  \begin{cases}
    \Upsilon_x\circ\Upsilon_y(\exp(\strutn{y}{x})))\cdot \Vac \\
    \Upsilon_x(\exp_{\udot}(\strutn{y}{x})\Omega_x) \cdot \Vac
  \end{cases} \\
Z(\openhopf{x}{y}) &= \exp(\strutn{y}{x})\udot\Omega_y,
\dsp\end{align*}
for some elements $\Vac \in \cA(\emptyset)$.
\end{theorem}

In the last expression, $\openhopf{x}{y}$ is the $(1,1)$ tangle whose
closure is the Hopf link, with the bead labelled by $y$ and the wire
labelled by $x$.
From this last equality in Theorem~\ref{thm:hopf-link}, we
can see exactly the map $\Phi$ from Chapter~\ref{chap:wheeling}.

\begin{corollary}
$\Phi = \Phi_0 = \chi \circ \partial_\Omega.$
\end{corollary}

\section{Useful facts}

The element $\Omega \in \cB$ is a very remarkable element.  We will
need one of its nice properties for the proof of
the Wheels theorem.  Start from the basic equality proved in the
Wheeling theorem,
\[
\Delta^x_{x_1x_2}H_0(z;x) = H_0(z;x_1) \connect_z H_0(z;x_2)
  \in \cA(\uparrow_z, \circledast_{x_1}, \circledast_{x_2})
\quad \text{where} \quad H_0(z;x) = \Omega_x \exp(\strutn{x}{z}).
\]
Now consider dropping the strand $z$, i.e., mapping all diagrams with
a $z$ vertex to 0.  (Knot-theoretically, this corresponds to dropping
the central strand in the equation ``$1+1=2$''.)  We find
\begin{equation}\label{eq:delta-omega}
\Delta \Omega = \Omega \tensor \Omega \in \cA(\circledast \circledast).
\end{equation}
Note that this equality is not true inside $\cA(\ast\ast)$.

\begin{lemma}[Pseudo-linearity of $\log\Omega$]\label{lem:pseudo-linear}
For any $D \in \cB$,
\[
\partial_D(\Omega) = \partial_D\Omega\rvert_0 \udot \Omega
	= \langle D, \Omega \rangle \Omega.
\]
\end{lemma}
\begin{proof}
\[ \partial_D(\Omega)_x = \langle D_y, \Omega_{x+y}\rangle_y
		        = \langle D_y, \Omega_x \Omega_y\rangle_y
		        = \langle D_y, \Omega_y\rangle_y \Omega_x.
\]
In the second equality, we use Equation~\ref{eq:delta-omega}.  This is
allowed, since the contraction descends to $\cA(\circledast
\circledast) \isom \cA(\circledast \uparrow)$ by
the argument of Lemma~\ref{lem:glue-welldef}.  Note that it is crucial
that $D$ is invariant for this argument.
\end{proof}

\begin{remark} Compare this lemma with standard calculus: if $D$ is
any differential operator and $f$ is a linear function, then
$De^f=(Df)(0)e^f$.  The prefix ``pseudo'' is written above because
Lemma~\ref{lem:pseudo-linear} does not hold for every $D$, but
only for $x$-invariant $D$'s.  (E.g., if $D$ were in $\cA(\ast\ast)$
rather than $\cA(\ast)$, the lemma would not be true.)
\end{remark}

Although we are interested in knots and links in $S^3$ in this thesis,
for which the appropriate space of diagrams $\cA^\bc$ from
Section~\ref{sec:weight-systems} is boundary connected, vacuum
diagrams (elements of $\cA(\emptyset)$)
appear at various points.  Notably, the wheeling map $\Upsilon$ does
not preserve the subspace of boundary connected diagrams.

Although the resulting vacuum
components can be computed explicitly,
\footnote{D.~Bar-Natan and
R.~Lawrence~\cite{Bar-NatanLawrence:RationalSurgery} have done these
computations}
they are
almost always irrelevant for us and it would just
complicate the formulas to keep track of them.  To avoid this, we will
introduce the {\em boundary-connected projection} $\pi^\bc: \cA
\rightarrow \cA^\bc$ which maps any diagram containing vacuum
components to 0 and is otherwise the identity.  Note
that $\pi^\bc$ is multiplicative.  There are similar projections, which we
also be called $\pi^\bc$, for other spaces $\cA(X)$.

If we compose Lemma~\ref{lem:pseudo-linear} with $\pi^\bc$,
we find
\begin{equation}\label{eq:pseudo-linear-2}
\pi^\bc \partial_D \Omega = \begin{cases}
    \Omega	& \text{$D$ is the empty diagram} \\
    0	& \text{otherwise}
  \end{cases}.
\end{equation}

\section{Coiling the unknot}\label{sec:coiling}

The basic equation we will use to identify $\nu$ is
``$n\cdot 0 = 0$'' from the introduction: the $n$-fold connected cable
of the unknot is the unknot with a new framing.  From
Section~\ref{sec:conn-cable}, this implies that
\begin{equation}\label{eq:n00}
\psi^{(n)}(\nu \connect \exp_{\connect}(\frac{1}{2n}\strutn{}{}))
	 = \nu \connect \exp_{\connect}(\frac{n}{2}\strutn{}{}).
\end{equation}
This equation is true for all $n \in \ZZ$, $n > 0$.  In each degree,
each side is a Laurent polynomial in $n$ of bounded degree; therefore,
the two sides are equal as Laurent polynomials.  The RHS is a
polynomial in $n$, so both sides are polynomials (i.e., have no
negative powers of $n$.)  Let us evaluate both sides at $n=0$.  On the
RHS, we get just $\nu$.  For the LHS, recall how $\psi^{(n)}$ acts in
the space $\cB$: it multiplies a diagram with $k$ legs by $n^k$.  Each
strut appearing in the product on the LHS contributes a factor of
$1/n$; each leg in the symmetrized result gives a factor of $n$.  To
bound how many legs can appear in a symmetrized product, we use the following
lemma.

\begin{lemma}\label{lem:wheels-lemma} For any elements $x_1, \dots,
x_k \in \cA(\uparrow)$ with at least one leg on the interval $\uparrow$,
$\chi^{-1}(x_1 \connect \cdots \connect x_k) \in \cB$ has at least $k$
legs.
\end{lemma}

\begin{proof} First note that any vacuum diagrams that appear in the
$x_i$'s pass through unchanged to the result; let us assume that there
are none, so that we can use the vacuum projection $\pi^\bc$ without
changing the result.  By Theorem~\ref{thm:wheeling},
\[
\pi^\bc \chi^{-1}(x_1 \connect \dots \connect x_n)
	= \pi^\bc \partial_\Omega(\Upsilon^{-1}(x_1)\udot\dots\udot\Upsilon^{-1}(x_k)).
\]
Let $y_i = \pi^\bc \Upsilon^{-1}(x_i)$.  Each $y_i$ has at least one
leg, since if
the $\partial_\Omega^{-1}$ of $\Upsilon^{-1}$ eats all the legs of
$\chi^{-1}x_i$, it also creates a vacuum diagram which is killed by
$\pi^\bc$.  Then
\[
\pi^\bc \partial_\Omega(y_1 \dots y_k) =
    \pi^\bc \langle \Omega_a, \Delta_{ab}(y_1 \dots y_k)\rangle_a.
\]
Let $\Delta_{ab} y_i = (y_i)_a + z_i$; diagrams in $z_1$ have at least
one $b$ leg.  We see that
\begin{align*}
\pi^\bc\langle\Omega_a, (y_1)_a \Delta_{ab}(y_2 \dots y_n)\rangle_a
 &= \pi^\bc\langle(\partial_{y_1}\Omega)_a, \Delta_{ab}(y_2\dots y_k)\rangle_a
	&&\text{by Lemma~\ref{lem:mult-diffop-dual}} \\
 &= 0.   &&\text{by Equation~\ref{eq:pseudo-linear-2}}
\end{align*}
Therefore
\begin{align*}
\pi^\bc \partial_\Omega(y_1 \dots y_k)
   &= \pi^\bc(\langle \Omega_a, (y_1)_a \Delta_{ab}(y_2 \dots y_k)\rangle_a
    + \langle \Omega_a, z_1 \Delta_{ab}(y_2 \dots y_k)\rangle_a) \\
   &= \pi^\bc\langle \Omega_a, z_1 \Delta_{ab}(y_2 \dots y_k)\rangle_a \\
   & = \cdots \\
   & = \pi^\bc\langle \Omega_a, z_1 \dots z_k \rangle_a.
\end{align*}
Each $z_i$ has at least one leg labelled $b$, so the product has at
least $k$ legs labelled $b$ which are the legs in the result.
\end{proof}

Consider expanding the exponential $\exp_{\connect}(\strutn{}{}/2n)$
in the LHS of Equation~\ref{eq:n00}.  In the term with $k$ struts,
there is a factor of $1/n^k$ from the factors in front of the struts.  On
the other hand, by Lemma~\ref{lem:wheels-lemma}, the
product has at least $k$ legs, or $k+1$ if there is a non-trivial
contribution from $\nu$.
Since the overall power of $n$ is $n^{k - \text{\# legs}}$, when we
evaluate at $n=0$ the term $\nu$ does not contribute at all.
\[
\psi^{(n)}(\nu \connect \exp_{\connect}(\frac{1}{2n}\strutn{}{}))|_{n=0}
    = \psi^{(n)}(\exp_{\connect}(\frac{1}{2n}\strutn{}{}))|_{n=0}.
\]
Now we want to pick out the term from $(\strutn{}{})^{\connect k}$ with
exactly $k$ legs.  We can do this computation explicitly using the
wheeling map $\Upsilon$. Alternatively, the
result (which is $\nu$) must be a diagram of degree $k$ and with $k$
legs, hence of
mapping degree 0: $\nu = \nu_0$.  $\Omega \cap \strutv{}{}$ was shown in
Chapter~\ref{chap:wheeling} the be the part of $Z(\OpenHopf)$ of
mapping degree 0.  Dropping the central strand from $\OpenHopf$ leaves
an unknot, so $\Omega =\nu_0 = \nu$.
This completes the proof of Theorem~\ref{thm:wheels}.\qed

\begin{exercise} Do the computation suggested above.  Show that
\[
\exp_{\connect}(\frac{1}{2}\strutn{}{}) =
	\Omega \udot \exp_{\udot} (\frac{1}{2}\strutn{}{}).
\]
\end{exercise}
\begin{hint} Use Lemma~\ref{lem:inspired-guess}.
\end{hint}

\section {From the unknot to the Hopf link}\label{sec:Hopf}

We will now apply Theorems~\ref{thm:wheeling}
and~\ref{thm:wheels} to compute the invariant of the Hopf link.
A little attention is required in order to perform the operations in the
correct order.  We start by computing the Kontsevich integral of the
$+1$ framed unknot.
\begin{align*}
Z(\bigcircle^{+1}) &= \Omega \connect \exp_{\connect}(\frac 12\isolatedchord)\\
	&= \partial_\Omega\left(\partial_\Omega^{-1}(\Omega) \udot
		\exp_{\udot}(\partial_\Omega^{-1}(\strutn{}{}))\right)
		&& \text{by Theorem~\ref{thm:wheeling}} \\
	&= \pi^\bc\partial_\Omega\left(\Omega \udot \exp(\strutn{}{})\right).
		&& \text{by Equation~\ref{eq:pseudo-linear-2}}
\end{align*}

To pass to the Hopf link, we double $Z(\bigcircle^{+1})$.  The
following lemma tells us how $\partial_\Omega$ intereacts with
doubling.  We use $\hat D$ as an
alternate notation for $\partial_D$ so that we can use a
subscript to indicate which variable the differential operator acts on.
\begin{lemma}\label{lem:double-diffop} For $C, D \in \cB$ with $C$ strutless,
\[
\Delta_{xy} \hat C(D) = \hat C_x(\Delta_{xy} D) = \hat C_y(\Delta_{xy} D).
\]
\qed
\end{lemma}
If we want to apply $\partial_\Omega^{-1}$ to both components of the
Hopf link, we can compute $\partial_\Omega^{-2}(Z(\bigcircle^{+1}))$. We
make an inspired guess.

\begin{lemma}\label{lem:inspired-guess}
 $\pi^\bc\partial_\Omega(\exp\frac{1}{2}\strutn{}{}) =
	\Omega\udot\exp(\frac{1}{2}\strutn{}{})$.
\end{lemma}
\begin{proof}
\begin{align*}
\pi^\bc\partial_\Omega(\exp(\frac12\strutn{}{}))
   &= \pi^\bc \langle\Omega_y, \exp(\frac{1}{2} \strutn{x+y}{x+y})\rangle_y

\\
   &= \pi^\bc\langle\Omega_y,
	\exp(\frac12\strutn xx)\exp(\strutn yx)\exp(\frac12\strutn yy)\rangle_y \\
   &= \pi^\bc\langle \partial_{\exp(\frac12\strutn{}{})}(\Omega)_y,
	\exp(\strutn xy)\rangle_y\udot\exp(\frac12\strutn xx)
			&& \text{by Lemma~\ref{lem:mult-diffop-dual}} \\
   &= \pi^\bc\langle\Omega_y, \exp(\strutn yx)\rangle_y\udot\exp(\frac12\strutn xx)
			&& \text{by Equation~\ref{eq:pseudo-linear-2}} \\
   &= \Omega\udot\exp(\frac12\strutn{}{}).
\end{align*}
\end{proof}

As a corollary, we see that
\begin{equation} \label{eq:inspired-guess}
\pi^\bc\partial_\Omega^{-2}(Z(\bigcircle^{+1})) = \exp(\frac12\strutn{}{}).
\end{equation}
We now compute.
\begin{align*}
\pi^\bc \Delta_{xy}(\hat\Omega^{-2}Z(\bigcircle^{+2}))
   &= \pi^\bc \hat\Omega_x^{-1}\hat\Omega_y^{-1} Z(\,{}_x^{+1}\!\Hopf_y^{+1})
	&&\text{by Lemma~\ref{lem:double-diffop} and Proposition~\ref{prop:disconn-cable}} \\
   &= \pi^\bc\Delta_{xy}(\exp(\frac12\strutn{}{}))
	&&\text{by Equation~\ref{eq:inspired-guess}} \\
   &= \exp(\strutn{x}{y})\exp(\frac12\strutn xx)\exp(\frac12\strutn yy).
\end{align*}
Apply $\Upsilon_x \circ \Upsilon_y$ to both sides.  We see that
\begin{align*}
Z(\,{}_x^{+1}\!\Hopf_y^{+1}) &= \pi^\bc Z(\,{}_x^{+1}\!\Hopf_y^{+1}) \\
	&= \pi^\bc\Upsilon_x \circ \Upsilon_y(\exp(\strutn{x}{y})\udot\exp(\frac12\strutn xx)
			\udot\exp(\frac12\strutn yy)) \\
	&= \pi^\bc \Upsilon_x \circ \Upsilon_y(\exp(\strutn{x}{y}))
		\connect \exp_{\connect}(\frac12\isolatedchord_x)
		\connect \exp_{\connect}(\frac12\isolatedchord_y) \\
\intertext{so}
Z(\hopf{x}{y}) &= \pi^\bc \Upsilon_x \circ \Upsilon_y(\exp(\strutn{x}{y})).
\end{align*}
This is the first equality of Theorem~\ref{thm:hopf-link}.
For the second equality,
\[
\hat\Omega_y(\exp(\strutn xy)) = \Omega_x \udot \exp(\strutn xy).
\]
so
\[
Z(\hopf{x}{y}) = \pi^\bc \Upsilon_x(\exp(\strutn{y}{x})\Omega_x).
\]
For the last equality of the theorem, multipicativity of
$\Upsilon$ implies that
\begin{align*}
\pi^\bc\Upsilon_x(\exp(\strutn yx)\Omega_x)
  &= \pi^\bc(\Upsilon_x(\exp(\strutn yx)) \connect \Upsilon_x(\Omega_x)) \\
  &= \pi^\bc(\Upsilon_x(\exp(\strutn yx))) \connect \chi(\Omega_x)
  &= \chi(\exp(\strutn yx) \udot \Omega_y) \connect \chi(\Omega_x).
\end{align*}
But by Equation~\ref{eq:11tangle},
\[
Z(\openhopf{x}{y}) = Z(\hopf{x}{y}) \connect \Omega_x^{-1}
	= \exp(\strutn yx) \udot \Omega_y.
\]
This completes the proof of Theorem~\ref{thm:hopf-link}.\qed

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