TITLE: Is it possible to build an optical system that increases the perceived surface brightness?
QUESTION [3 upvotes]: So is it possbile to build a system from lenses and mirrors that can make faint gas nebulas brighter or can be used as nightvision?
If you increase the size of the aperture of a telescope it will collect more light, but the exit pupil will be also bigger, so not all light will enter the eye.
In order to direct all light into the eye you'll need to shrink the exit pupil and you'll need a stronger eyepiece. But this will increase the magnification too and the collected light will spread on a larger image so the surface brightness will remain the same.
Is it possible to work this limitation around?

REPLY [1 votes]: It is not possible. If you could build such an optical system, you would have a perpetual motion machine of the second kind.
Suppose you would have a passive (no batteries allowed) optical system capable of creating images brighter than the source at which you aim it. You aim this device at the sun and thereby create an image brighter and hotter than the sun. That goes against the second law of thermodynamics which tells us that entropy increases and therefore heat 'evens out'.
The concept of entropy translates into a key optical concept: 'etendue'. When light propagates through an optical system, the etendue can not decrease. It basically is the second law of thermodynamics applied to optics.


TITLE: Convolution of two Schwartz functions is Schwartz
QUESTION [9 upvotes]: I am trying to show directly (i.e., not using the Fourier transform) that if $S=S(\mathbb{R}^n)$ is the class of Schwartz functions then $f,g \in S$ implies $f*g \in S$.
The definition I have is $f \in S$ iff for all $N$, $\|f\|_N:=\sup_{x \in \mathbb{R}^d, |\alpha|,|\beta| \leq N} |x^\beta\partial_x^\alpha f(x)|<\infty$. I have shown that $\|f\|_N<\infty$ is equivalent to: for all $|\alpha|,|\beta| \leq N$, there exists a constant $C_N$ such that $|\partial_x^\alpha f(x)| \leq \frac{C_N}{\prod_{i=1}^n (1+|x_i^{\beta_i}|)}$.
I also know that $\partial_x^\alpha (f*g)(x) = \int_{\mathbb{R}^n} \partial_x^\alpha f(x-t) g(t) \ dt$. Since $\partial_x^\alpha f \in S$, I believe it suffices to show that for all $f,g \in S$, all $N$, and all $|\beta| \leq N$, we have $|(f*g)(x)| \leq \frac{C_N}{\prod_{i=1}^n (1+|x_i^{\beta_i}|)}$. I can prove this in the case $\mathbb{R}^n=\mathbb{R}$ as follows: Let $C_N$ be such that for all $b \leq N$, $|f(x)|,|g(x)| \leq \frac{C_N}{1+|x|^b}$. Then
\begin{align*}
\left| \int_{\mathbb{R}} f(x-t)g(t) \ dt \right| & \leq \int_{\mathbb{R}} \left| f(x-t)g(t)\right| \ dt\\
&= \int_{|x-t| \geq |x|/2} \left| f(x-t)g(t)\right| \ dt + \int_{|x-t| < |x|/2} \left|f(x-t)g(t)\right| \ dt\\
&\leq \sup_{|x-t| \geq |x|/2} |f(x-t)|\int_{\mathbb{R}}|g(t)| \ dt + \sup_{|t| \geq |x|/2} |g(t)| \int_{\mathbb{R}} |f(x-t)| \ dt\\
&\leq \sup_{|x-t| \geq |x|/2}\frac{C_N}{1+|x-t|^b} \|g\|_{L^1} + \sup_{|t| \geq |x|/2}\frac{C_N}{1+|t|^b} \|f\|_{L^1}\\
&\leq \frac{C_N(\|g\|_{L^1}+\|f\|_{L^2})}{1+(|x|/2)^b}\\
& \leq \frac{2^NC_N(\|g\|_{L^1}+\|f\|_{L^2})}{1+|x|^b}.
\end{align*}
(Note that $|x-t|<|x|/2$ implies $|t|>|x|/2$.)
However, I am struggling with how to split up the integral in $\mathbb{R}^n$. In order to do the step  $\frac{C_N}{\prod_{i=1}^n (1+|x_i-t_i|^{\beta_i})} \leq \frac{C_N}{\prod_{i=1}^n (1+(|x_i|/2)^{\beta_i})}$ I need $|x_i-t_i| \geq |x_i|/2$ for all $i$. But in order to do the step $\frac{C_N}{\prod_{i=1}^n (1+|t_i|^{\beta_i})} \leq \frac{C_N}{\prod_{i=1}^n (1+(|x_i|/2)^{\beta_i})} $ I need $|t_i| \geq |x_i|/2$ for all $i$. However these two sets do not encompass all of $\mathbb{R}^n$ (in particular, we miss all $t$ such that $|x_i-t_i| < |x_i|/2$ for some $i$ and $|t_j| < |x_j|/2$ for some $j$). I have tried integrating over one copy of $\mathbb{R}$ at a time, but as soon as we replace $f$ and $g$ by their upper bounds in the innermost integral, they are no longer guaranteed to have finite $L^1$ norm, so I cannot continue. (Note: I can show the inequality holds for $N=1$, i.e., $x^\beta = x_i$ for some $i$. Is that enough?) Any suggestions would be greatly appreciated.

REPLY [4 votes]: One can proceed as follows:
For $N > 0$ and $x \in \mathbb{R}^{n}$
\begin{align*}
(1 + |x|)^{N}|f \star g(x)| &\leq \int_{\mathbb{R}^{n}} (1 + |x|)^{N}|f(x-t)||g(t)|\,dt\\
&\leq \int_{\mathbb{R}^{n}} (1 + |x-t|)^{N}(1 + |t|)^{N}|f(x-t)||g(t)|\,dt\\
&\leq ||f||_{N}\int_{\mathbb{R}^{n}}(1+|t|)^{N}|g(t)|\,dt\\
&\leq C_{N+2}||f||_{N}||g||_{n+N+1}\int_{\mathbb{R}^{n}}\frac{1}{(1 + |t|)^{n+1}}\,dt
\end{align*}
Here we have made use of the fact that $(1+ |x|) \leq (1+ |x-t|)(1+ |t|)$.


TITLE: $C^2$ function on closed domain has finite solutions
QUESTION [2 upvotes]: Let $\Omega $ be a bounded open set of $R^n$ and $f\in C^2(\overline\Omega)$. Define $N_f=\{x|x\in\Omega,J_f(x)=0\}$, where $J_f$ is the Jacobian of $f$.
How can I show that

For all $ p\notin f(N_f)$ with
  $\inf\limits_{x\in\partial\Omega}||f(x)-p||>0,$ the equation $f(x)=p $
  has a finite number of solutions ?

REPLY [1 votes]: Suppose we have an infinite number of solutions arranged in a sequence $\{x_n\}$. As $\Omega$ is bounded, there is a convergent subsequence $\{x_{n_i}\}$. Set $x_0=\lim\limits_{i\rightarrow\infty}x_{n_i}$. Then, $f(x_0)=p$ by continuity and $x_0\in\Omega$ , since $\inf\limits_{x\in\partial\Omega}||f(x)-p||>0$.
Because $p\notin f(N_f)$, $J_f(x_0)\ne0$, but this contradicts the inverse function theorem.


TITLE: Prove that $f(x)=\sqrt[n]{x}$ is not Lipschitz continuous
QUESTION [5 upvotes]: Show that none of the functions $f(x)=\sqrt[n]{x}$, for $n\in \Bbb N$, $n\ge2$ are Lipschitz continuous.
So i did it this way:
$|\sqrt[n]{x}-\sqrt[n]{y}|=|x-y||\frac{1}{x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}y^{\frac{1}{n}}+...+x^{\frac{1}{n}}y^{\frac{n-2}{n}}+y^{\frac{n-1}{n}}}| \le (?)L|x-y|$
and argued that for very small x and y (close to 0) the denominator is very small, therefore the whole expressions is big, and the whole fraction cannot be limited by a real number (L),  so these functions are not Lipschitz continuous.
Am I corrected? Is there any nicer, more formal way to show it?

REPLY [0 votes]: That looks fine to me. A similar way is to note that being Lipschitz would mean there's an $L$ such that
$$\bigg|{\sqrt[n]{y} - \sqrt[n]{x} \over y - x} \bigg| \leq L$$
Letting $y = 2x$ means that for all $x$ one must have
$$(\sqrt[n]{2} - 1)x^{{1 \over n} - 1} \leq L$$
Letting $x \rightarrow 0$ gives a contradiction
If you're including $x = 0$ in your domain (you didn't say) then you could take $x = 0$ in the above and let $y \rightarrow 0$ to get a similar contradiction.


TITLE: What is the number of ways to divide a rectangle into $n$ smaller rectangles line by line?
QUESTION [10 upvotes]: The original problem was to consider how many ways to make a wiring diagram out of $n$ resistors. When I thought about this I realized that if you can only connect in series and shunt. - Then this is the same as dividing an area with $n-1$ horizontal and vertical lines. When each line only divides one of the current area sections into two smaller ones.
This is also the same as the number of ways to make a set of $n$ (and only $n$) rectangles into a bigger rectangle. If the rectangles can be drawn by dividing the big rectangle, line by line, into the set of rectangles without lose endpoints of the line. - Can someone come to think of "a expression of $n$" which equals this amount, independent of the order of the rectangles or position?
(It is only the relations between the area sections that matters and not left or right, up or down. However dividing an area with a horizontal line is not the same as dividing it with a vertical line.)

REPLY [0 votes]: Since you have the big rectangle, and you want to cut it into ‘n’ number of rectangles, the number of cuts will be simply n-1. This is similar to adding a cosecutive series of numbers


text:
\begin{document}

\allowdisplaybreaks

\newcommand{\arXivNumber}{1406.4652}

\renewcommand{\PaperNumber}{009}

\FirstPageHeading

\ShortArticleName{Bipolar Lawson Tau-Surfaces and Generalized Lawson Tau-Surfaces}

\ArticleName{Bipolar Lawson Tau-Surfaces\\ and Generalized Lawson Tau-Surfaces}

\Author{Broderick {CAUSLEY}}

\AuthorNameForHeading{B.~Causley}

\Address{Department of Mathematics and Statistics, McGill University, Burnside Hall,\\ 805 Sherbrooke Street West,
Montreal, QC H3A 0B9, Canada}
\Email{\href{mailto:broderick.causley@mail.mcgill.ca}{broderick.causley@mail.mcgill.ca}}

\ArticleDates{Received November 27, 2015, in f\/inal form January 21, 2016; Published online January 25, 2016}

\Abstract{Recently Penskoi [\textit{J.~Geom.~Anal.}~\textbf{25} (2015), 2645--2666, arXiv:1308.1628]  ge\-ne\-ra\-li\-zed the well known two-parametric family of Lawson tau-surfaces $\tau_{r,m}$ minimally immersed in spheres to a three-parametric family $T_{a,b,c}$ of tori and Klein bottles minimally immersed in spheres. It was remarked that this family includes surfaces carrying all extremal metrics for the f\/irst non-trivial eigenvalue of the Laplace--Beltrami operator on the torus and on the Klein bottle: the Clif\/ford torus, the equilateral torus and surprisingly the bipolar Lawson Klein bottle $\tilde{\tau}_{3,1}$. In the present paper we show in Theo\-rem~\ref{theorem1} that this three-parametric family $T_{a,b,c}$ includes in fact all bipolar Lawson tau-surfaces $\tilde{\tau}_{r,m}$. In Theo\-rem~\ref{theorem3} we show that no metric on generalized Lawson surfaces is maximal except for~$\tilde{\tau}_{3,1}$ and the equilateral torus.}

\Keywords{bipolar surface; Lawson tau-surface; minimal surface; extremal metric}

\Classification{58E11; 58J50; 49Q05; 35P15}

\section{Introduction}

Minimal surfaces in the Euclidean space~$\mathbb{R}^3$ is one of the most classical subjects of dif\/ferential geometry. Surprisingly, interest to minimal surfaces in spheres~$\mathbb{S}^n$ appeared much later and a~serious investigation in this direction started only in the second half of the 20th century. One important paper in the early period of this investigation was the famous paper~\cite{Lawson1970} by Lawson, where several examples of immersed minimal surfaces in~$\mathbb{S}^3$ were constructed, including examples of surfaces of any genus except the projective plane. These examples include a family $\tau_{m,n}$ of minimal tori and Klein bottles.

\begin{definition}
A Lawson tau-surface $\tau_{m,n}\looparrowright\mathbb{S}^3$ is def\/ined as the image of the doubly-periodic immersion $\Psi_{m,n}\colon \mathbb{R}^2\looparrowright\mathbb{S}^3\subset\mathbb{R}^4$ given by the explicit formula
\begin{gather}\label{immersion}
\Psi_{m,n}(x,y)=(\cos(mx)\cos y, \sin(mx)\cos y, \cos(nx)\sin y,\sin(nx)\sin y).
\end{gather}
\end{definition}

For each unordered pair of positive integers $(m,n)$ with $(m,n)=1$, the surface $\tau_{m,n}$ is a~distinct compact minimal surface in $\mathbb{S}^3$. Let us assume $(m,n)=1$. If both integers~$m$ and~$n$ are odd then $\tau_{m,n}$ is a~torus (we call it Lawson torus). If one of integers~$m$ and~$n$ is even then~$\tau_{m,n}$ is a~Klein bottle (we call it a~Lawson Klein bottle). The torus~$\tau_{1,1}$ is the Clif\/ford torus.

Many ef\/forts were later applied to f\/ind minimal surfaces in spheres. There are several constructions of paticular types of minimal surfaces in spheres.  For example, the Hsiang--Lawson approach~\cite{Hsiang-Lawson1971} constructs surfaces with additional symmetry. However it is only minimal tori that are completely described. This description is given by the theory of integrable systems using algebraic geometry. This was done in several papers by many authors starting from Hitchin's paper~\cite{Hitchin1990} dealing with a particular case of~$\mathbb{S}^3$ and f\/inishing with Burstall's paper~\cite{Burstall1995} dealing with the general case $\mathbb{S}^n$. In fact this investigation was a part of more general investigation of harmonic maps from tori into symmetric spaces. A recent paper by Carberry~\cite{Carberry2012} contains a~review of the current state of this approach.

We should remark however that {\em only a very limited number of explicitly parametrized minimal surfaces in spheres is known}. Unfortunately, most approaches to minimal surfaces in spheres give only quite implicit descriptions of minimal surfaces. This pertains to the Hsiang--Lawson construction and description of minimal tori using integrable systems mentioned above. This makes the question of f\/inding {\em explicitly parametrized} minimal surfaces in spheres interesting.

Let us mention here several known results about explicitly parametrized minimal surfaces in spheres. We mentioned already Lawson surfaces $\tau_{m,n}$ given by the explicit parametrization~\eqref{immersion}. Another example is bipolar Lawson surfaces. Starting from an immersed surface $I\colon \Sigma\looparrowright\mathbb{S}^3\subset\mathbb{R}^4$ one can construct a new surface $\tilde{\Sigma}$ called bipolar to $\Sigma$ and def\/ined by the immersion $I\wedge I^*\colon \Sigma\looparrowright\mathbb{R}^4\wedge\mathbb{R}^4\cong\mathbb{R}^6$, where $I^*$ is a Gaussian map, i.e., $I^*(p)$ is a unitary normal vector to $\Sigma$ in $\mathbb{S}^3$ at the point~$p$. It turns out that the image is in the sphere $\mathbb{S}^5\subset\mathbb{R}^6$, and if the surface~$\Sigma$ is minimal in $\mathbb{S}^3$, then its bipolar surface $\tilde{\Sigma}$ is minimal in~$\mathbb{S}^5$ (see~\cite{Lawson1970}). As was shown by Lawson, bipolar surfaces $\tilde{\tau}_{m,n}$ are in fact lying in an equatorial sphere $\mathbb{S}^4\subset\mathbb{S}^5$. Details are provided below in Section~\ref{section2}.

Another example is given by a construction by Mironov~\cite{Mironov2004,Mironov2010} of Hamiltonian-minimal Lagran\-gian embeddings in~$\mathbb{C}^N$ based on intersections of real quadrics of a special type. As was remarked by Karpukhin in~\cite{kar2}, as a by-product of this construction, one obtains minimal surfaces in spheres.

Recently Penskoi constructed a generalization of Lawson surfaces~$\tau_{m,n}$ in~\cite{penskoi1}. This gene\-ra\-li\-zation is given by an explicit formula. In complex form, it is written as
\begin{gather}
\left(\sqrt{\frac{b^2+c^2-a^2}{2(c^2-a^2)}}e^{iax}\sin y,
\sqrt{\frac{a^2+c^2-b^2}{2(c^2-b^2)}}e^{ibx}\cos y,\right.\notag\\
\left.\qquad{} \sqrt{\frac{a^2+b^2-c^2}{2(b^2-c^2)}}e^{icx}
\sqrt{1-\frac{b^2-a^2}{c^2-a^2}\sin^2y}\right)
\in\mathbb{S}^5\subset\mathbb{C}^3\cong\mathbb{R}^6.\label{Tabcexplicit}
\end{gather}
It is proven in~\cite{penskoi1} that
\begin{itemize}\itemsep=0pt
\item if either $a$, $b$, $c$ are integers and $|c|>\sqrt{a^2+b^2}$ or
\item if $a$, $b$ are nonzero integers and $|c|=\sqrt{a^2+b^2}$,
\end{itemize}
then the parametrization~\eqref{Tabcexplicit} def\/ines a minimal torus or a minimal Klein bottle in $\mathbb{S}^5$, denoted by $T_{a,b,c}$. Moreover, if $|c|=\sqrt{a^2+b^2}$, then $T_{a,b,c}$ coincides with the Lawson surface~$\tau_{a,b}$. This follows immediately by composing~\eqref{immersion} and~\eqref{Tabcexplicit}. Thus, this three-parametric family of minimal tori or Klein bottles in fact generalizes the famous two-parametric Lawson family $\tau_{m,n}$.

Penskoi's construction of generalized Lawson tori was deeply motivated by the relation between minimal surfaces in spheres and spectral geometry. We will expose this relation below, but let us also remark at this moment that it was observed in~\cite{penskoi1} that bipolar Lawson surfa\-ce~$\tilde{\tau}_{3,1}$ is isometric to $T_{1,0,2}$. The proof of this fact was indirect and was based on the uniqueness of the extremal metric for the f\/irst eigenvalue of the Laplace--Beltrami operator~$\Delta$ on the Klein bottle, proved in~\cite{sgj1}.

This observation is quite intriguing and leads us into an investigation of the following natural question: is it true that not only is the two-parametric family of Lawson tau-surfaces~$\tau_{m,n}$ a~subfamily of the three-parametric family~$T_{a,b,c}$ of generalized Lawson surfaces, but is the family of bipolar Lawson surfa\-ces~$\tilde{\tau}_{m,n}$ also a subfamily of the three-parametric family~$T_{a,b,c}$? The list of known explicitly parametrized minimal surfaces in spheres is quite short and this makes it interesting to understand relations among dif\/ferent known families of explicitly parametrized minimal surfaces in spheres.

The f\/irst result of the present paper is an af\/f\/irmative answer to this question.

\begin{theorem}\label{theorem1}\quad
\begin{enumerate} \itemsep=0pt
  \item[$1.$] If $rm\equiv 0\!\pmod{2}$ then the bipolar Lawson torus $\tilde{\tau}_{r,m}$ is isometric to
  the surface~$T_{a,b,c}$, where $a=r-m$, $b=0$, $c=r+m$.
  \item[$2.$] If $rm\equiv 1\!\pmod{4}$ then the bipolar Lawson torus $\tilde{\tau}_{r,m}$ is isometric to
  the surface~$T_{a,b,c}$, where $a=\frac{r-m}{2}$, $b=0$, $c=\frac{r+m}{2}$.
  \item[$3.$] If $rm\equiv 3\!\pmod{4}$ then the bipolar Lawson Klein bottle $\tilde{\tau}_{r,m}$ is isometric to the surfa\-ce~$T_{a,b,c}$, where $a=\frac{r-m}{2}$, $b=0$, $c=\frac{r+m}{2}$.
\end{enumerate}
\end{theorem}

Let us now explain the relation between minimal surfaces in spheres and spectral geometry in more detail.

Let $M$ be a closed surface and $g$ be a Riemannian metric on $M$. Let us consider the associated Laplace--Beltrami operator $\Delta \colon C^{\infty}(M)\rightarrow C^{\infty}(M)$,
\begin{gather*}
\Delta f=-\frac{1}{\sqrt{|g|}}\frac{\partial}{\partial x^{i}}\left(\sqrt{|g|}g^{ij}\frac{\partial f}{\partial x^{j}}\right).
\end{gather*}

The spectrum of $\Delta$ is non-negative and consists only of eigenvalues, where each eigenvalue has a f\/inite multiplicity and the associated eigenfunctions are smooth. Denote the eigenvalues of~$\Delta$ by
\begin{gather*}
0=\lambda_{0}(M,g)<\lambda_{1}(M,g)\le \lambda_{2}(M,g)\le \lambda_{3}(M,g)\le \cdots,
\end{gather*}
where eigenvalues are written with multiplicities.

Let us f\/ix the surface $M$ and consider $\Lambda_{i}(M,g)$ as a functional $g\mapsto\Lambda_{i}(M,g)$ on the space of all Riemannian metrics on~$M$. The eigenvalues possess the following rescaling property
\begin{gather*}
\forall \, t>0,\qquad \lambda_{i}(M,tg)=\frac{\lambda_{i}(M,g)}{t}.
\end{gather*}

To get scale-invariant functionals on the space of Riemannian metrics one has to normalize the eigenvalue functionals. It is most natural to normalize the functionals by multiplying by the area
\begin{gather*}
\Lambda_{i}(M,g)=\lambda_{i}(M,g)\operatorname{Area}(M,g).
\end{gather*}
The functionals $\Lambda_{i}(M,g)$ are invariant under the rescaling transformation $g\mapsto tg$.

If we consider the functional $\Lambda_{i}(M,g)$ over the space of Riemannian metrics~$g$ with a f\/ixed surface~$M$, the question about the value of supremum $\sup\Lambda_{i}(M,g)$ is interesting. It is a very dif\/f\/icult question with a limited number of known results. It follows from Yang and Yau~\cite{yy1} and Korevaar~\cite{k1} that this supremum is f\/inite.

\begin{definition}
A metric $g_{0}$ on a f\/ixed surface $M$ is called maximal for the functional $\Lambda_{i}(M,g)$ if
\begin{gather*}
\sup \Lambda_{i}(M,g)=\Lambda_{i}(M,g_{0}),
\end{gather*}
where the supremum is taken over the space of Riemannian metrics~$g$ on the f\/ixed surface~$M$.
\end{definition}

Currently, we only know maximal metrics for $\Lambda_{1}(\mathbb{S}^2,g)$, $\Lambda_{1}(\mathbb{R}P^2,g)$, $\Lambda_{1}(\mathbb{T}^2,g)$, $\Lambda_{2}(\mathbb{S}^2,g)$, and $\Lambda_{3}(\mathbb{S}^2,g)$. For more details please see the recent survey by Penskoi~\cite{penskoireview} and the recent paper by Nadirashvili and Sire~\cite{n2}.

A problem in the study of $\Lambda_{i}(M,g)$-maximal metrics is that the functional $\Lambda_{i}(M,g)$ depends continuously on metric $g$ but is not dif\/ferentiable. However, for any analytic deformation $g_{t}$, the left and right derivatives of the functional $\Lambda_{i}(M,g)$ with respect to \textit{t} exist (see Berger~\cite{b1}, Bando and Urakawa~\cite{bu1}, El Souf\/i and Ilias~\cite{si2}).

\begin{definition}[\cite{si1,si2, n1}] A Riemannian metric $g_{0}$ on a closed surface $M$ is called an extremal metric for the functional $\Lambda_{i}(M, g)$ if for any analytic deformation $g_{t}$ the following inequality holds
\begin{gather*}
\frac{d}{dt}\Lambda_{i}(M,g_{t})\bigg|_{t=0+}\cdot\frac{d}{dt}\Lambda_{i}(M,g_{t})\bigg|_{t=0-}\le 0.
\end{gather*}
\end{definition}

The mentioned metrics above are extremal since they are global maxima. However, extremal metrics are not necessarily maximal. For example, El Souf\/i and Ilias proved in \cite{si1} that the only extremal metric for $\Lambda_{1}(\mathbb{T}^2,g)$ dif\/ferent from the maximal one is the metric on the Clif\/ford torus.

Jakobson, Nadirashvili and Polterovich proved in \cite{jnp1} that the metric on the Klein bottle realized as the bipolar Lawson surface $\tilde{\tau}_{3,1}$ is extremal for $\Lambda_{1}(\mathbb{KL},g)$. Using this result El Souf\/i, Giacomini and Jazar proved in \cite{sgj1} that this metric is the unique extremal metric.

The following extremal metrics on families of tori and Klein bottles were investigated recently: Lapointe investigated metrics on bipolar Lawson surfaces $\tilde{\tau}_{r,m}\looparrowright\mathbb{S}^{4}$ in~\cite{lapointe}, these surfaces are described below in Section~\ref{section2}; Penskoi investigated extremal metrics on Lawson surfaces $\tau_{m,n}\looparrowright\mathbb{S}^{3}$ and on Otsuki tori $O_{\frac{p}{q}}\looparrowright\mathbb{S}^{3}$ in~\cite{penskoi2} and~\cite{penskoi3} respectively; Karpukhin investigated metrics on bipolar Otsuki tori $\tilde{O}_{\frac{p}{q}}\looparrowright\mathbb{S}^{4}$ and on a family of tori $M_{m,n}\looparrowright\mathbb{S}^{5}$ in~\cite{kar1} and~\cite{kar2} respectively; and Karpukhin proved that the metrics on $\tau_{m,n}$, $\tilde{\tau}_{r,m}$, $O_{\frac{p}{q}}$, $\tilde{O}_{\frac{p}{q}}$, and $M_{m,n}$ are not maximal except metrics on $M_{1,1}$ (the equilateral torus) and $\tilde{\tau}_{3,1}$ in~\cite{kar3}. Here $\looparrowright$ denotes an immersion.

As it was mentioned above, in~\cite{penskoi1}, Penskoi introduced the new three-parametric family~$T_{a,b,c}$ of minimal surfaces in spheres, generalizing Lawson tau-surfaces, and investigated their extremal spectral properties.

\begin{theorem}[Penskoi,~\cite{penskoi1}]\label{theorem2}  Let $F_{a,b,c}\colon \mathbb{R}^2\rightarrow\mathbb{S}^{5}\subset\mathbb{R}^{6}$ be a~three-parametric doubly-periodic immersion of the plane to the $5$-dimensional sphere of radius~$1$ defined by the formula~\eqref{Tabcexplicit}, where
\begin{itemize}\itemsep=0pt
\item[$a)$] either $a$, $b$, $c$ are integers and $|c|>\sqrt{a^2+b^2}$,
\item[$b)$] or $a$, $b$ are nonzero integers and $|c|=\sqrt{a^2+b^2}$.
\end{itemize}

 Then the following statements hold:
\begin{enumerate}\itemsep=0pt
\item[$1)$] The image $T_{a,b,c}=F_{a,b,c}(\mathbb{R}^2)$ is a minimal compact surface in the $5$-dimensional sphere~$\mathbb{S}^5$.

\item[$2)$]  The case $b)$ corresponds to Lawson tau-surfaces $\tau_{a,b}\cong T_{a,b,\sqrt{a^2+b^2}}$. Distinct Lawson tau-surfaces correspond to unordered pairs $a,b\geqslant1$ such that $(a,b)=1$. The surface $T_{a,b,\sqrt{a^2+b^2}}$ is a Lawson torus $\tau_{a,b}$ if $a$ and $b$ are odd and $T_{a,b,\sqrt{a^2+b^2}}$ is a Lawson Klein bottle $\tau_{a,b}$ if either $a$ or $b$ is even, where we assume $(a,b)=1$.

\item[$3)$]  In the case $b)$ the metric induced on $\tau_{a,b}\cong T_{a,b,\sqrt{a^2+b^2}}$ is extremal for the functionals $\Lambda_j(\mathbb{T}^2,g)$ if $\tau_{a,b}$ is a Lawson torus or $\Lambda_j(\mathbb{KL},g)$ if $\tau_{a,b}$ is a Lawson Klein bottle, where $j=2\left[\frac{\sqrt{a^2+b^2}}{2}\right]+a+b-1$ and $[\cdot]$ denotes the integer part. The corresponding value of the functional is $\Lambda_j(\tau_{a,b})=8\pi aE\left(\frac{\sqrt{a^2-b^2}}{a}\right)$.

\item[$4)$]  In the case $a)$ for an integer $k\geqslant1$ one has $T_{a,b,c}=T_{ka,kb,kc}$. Moreover, $T_{-a,b,c}$, $T_{a,-b,c}$ $T_{a,b,-c}$ and $T_{b,a,c}$ are isometric to $T_{a,b,c}$. Hence, it is sufficient to consider non-negative integer $a$, $b$, $c$ satisfying conditions $a)$ such that $(a,b,c)=1$ and assume that $(a,b,c)$ and $(b,a,c)$ are equivalent.
\end{enumerate}
\end{theorem}

The second result of the present paper concerns the maximality of metrics induced on $T_{a,b,c}$.

\begin{theorem}\label{theorem3}
All metrics induced on $T_{a,b,c}$ are not maximal except for the metric of $\tilde{\tau}_{3,1}$ and the metric of the equilateral torus.
\end{theorem}

The paper is organized in the following way. In Section~\ref{section2} we recall a construction of bipolar Lawson tau-surfaces following Lapointe's paper~\cite{lapointe}. In Section~\ref{section3} we provide a~proof of Theorem~\ref{theorem1}. Finally, nonmaximality of metrics on $T_{a,b,c}$ is shown in Section~\ref{section4}.

\section{Construction of bipolar Lawson surfaces}
\label{section2}

Let us now recall the construction of bipolar Lawson surface $\tilde{\tau}_{r,m}$ following Lapointe's paper~\cite{lapointe}. The Lawson surface $\tau_{r,m}$, with $r>m>0$ and $(r,m)=1$, is minimally immersed into $\mathbb{S}^{3}$ by $I\colon \mathbb{R}^2\rightarrow\mathbb{R}^{4}$, where
\begin{gather*}
I(u,v)=(\cos ru\cos v, \sin ru\cos v, \cos mu\sin v, \sin mu\sin v).
\end{gather*}

The bipolar minimal surface $\tilde{\tau}_{r,m}$ of $\tau_{r,m}$ is the image of an exterior product of $I$ and $I^{*}$, where $I^{*}$ is a unit vector normal to~$\tau_{r,m}$ and tangent to~$\mathbb{S}^{3}$,
\begin{gather*}
I^{*}(u,v)=\frac{(m \sin ru\sin v, -m \cos ru\sin v, -r \sin mu\cos v, r \cos mu\cos v)}{\sqrt{r^2\cos^2 v+m^2\sin^2 v}}.
\end{gather*}

The explicit formula for $\tilde{I}=I\wedge I^{*}\colon \mathbb{R}^2\rightarrow\mathbb{S}^{5}\subset\mathbb{R}^{6}$ is then
\begin{gather*}
\tilde{I}=\frac{1}{\sqrt{r^2\cos^2 v+m^2\sin^2 v}}\begin{pmatrix} -m\sin v\cos v\\ r\sin v\cos v\\ -r\cos^2 v\sin mu\cos ru-m\sin^2 v\sin ru\cos mu\\ r\cos^2 v\cos mu\sin ru+m\sin^2 v\cos ru\sin mu\\ -r\cos^2 v\sin mu\sin ru+m\sin^2 v\cos ru\cos mu\\ r\cos^2 v\cos mu\cos ru-m\sin^2 v\sin ru\sin mu \end{pmatrix}.
\end{gather*}

It is known that $\tilde{\tau}_{r,m}$ actually lies in $\mathbb{S}^{4}$, seen as an equator of~$\mathbb{S}^{5}$.

In~\cite{lapointe} Lapointe proved that for the bipolar surface $\tilde{\tau}_{r,m}$ of a Lawson torus or Klein bottle~$\tau_{r,m}$:
\begin{enumerate}\itemsep=0pt
\item[(1)] if $rm\equiv 0\!\pmod{2}$, $\tilde{\tau}_{r,m}$ is a torus with an extremal metric for $\Lambda_{4r-2}$,

\item[(2)] if $rm\equiv 1\!\pmod{4}$, $\tilde{\tau}_{r,m}$ is a torus with an extremal metric for $\Lambda_{2r-2}$,

\item[(3)] if $rm\equiv 3\!\pmod{4}$, $\tilde{\tau}_{r,m}$ is a Klein bottle with an extremal metric for $\Lambda_{r-2}$.
\end{enumerate}

The value of functional $\Lambda_{i}(\tilde{\tau}_{r,m})$ can be calculated as follows \cite{lapointe}:
\begin{enumerate}\itemsep=0pt
\item[(1)] if $rm\equiv 0\!\pmod{2}$, $\Lambda_{4r-2}(\tilde{\tau}_{r,m})=16\pi rE\left(\frac{\sqrt{r^2-m^2}}{r}\right)$,

\item[(2)] if $rm\equiv 1\!\pmod{4}$, $\Lambda_{2r-2}(\tilde{\tau}_{r,m})=8\pi rE\left(\frac{\sqrt{r^2-m^2}}{r}\right)$,

\item[(3)] if $rm\equiv 3\!\pmod{4}$, $\Lambda_{r-2}(\tilde{\tau}_{r,m})=4\pi rE\left(\frac{\sqrt{r^2-m^2}}{r}\right)$.
\end{enumerate}

\section{Proof of Theorem~\ref{theorem1}}
\label{section3}

Let us prove that the surface $T_{a,b,c}$ where $a=r-m, b=0$, and $c=r+m$ is isometric to the bipolar Lawson surface $\tilde{\tau}_{r,m}$ when $rm\equiv 0\!\pmod{2}$.

The induced metric $g$ on $T_{a,b,c}$ and $\tilde{g}$ on $\tilde{\tau}_{r,m}$ are given by the formulas~\cite{lapointe,penskoi1}
\begin{gather*}
g=\frac{1}{2}\big(c^2+\big(b^2-a^2\big)\cos 2y\big)dx^2+ \frac{c^2+\big(b^2-a^2\big)\cos 2y}{2c^2-a^2-b^2+\big(b^2-a^2\big)\cos 2y}dy^2,
\\
\tilde{g}=\frac{\big(r^2-\big(r^2-m^2\big)\sin^2 v\big)^2+r^2 m^2}{r^2-\big(r^2-m^2\big)\sin^2 v}\left(du^2+\frac{dv^2}{r^2-(r^2-m^2)\sin^2 v}\right).
\end{gather*}

The task is to f\/ind the change of variable between metrics $g$ and $\tilde{g}$. Let $a=r-m$, $b=0$, $c=r+m$ and apply the change of variable $\sin y=\operatorname{sn}(z,k)$, where $k=\frac{a}{\sqrt{a^2-c^2}}$ and $\operatorname{sn}(z,k)$ is a~Jacobi elliptic function~\cite{ellipticbook}. This implies
\begin{gather*}
g=\frac{1}{2}\big(c^2-a^2+2a^2 \operatorname{sn}^2(z,k)\big)\left(dx^2+ \frac{dz^2}{c^2-a^2}\right).
\end{gather*}
Use the following transformation
\begin{gather}
\label{transform1}
H_{1}(x,z)=\left(u,\frac{2\sqrt{c^2-a^2}}{a+c}w+K(k)\right).
\end{gather}
This implies that
\begin{gather*}
g=\left(2rm+(r-m)^2\frac{1-\operatorname{sn}^2\big(2\sqrt{\frac{m}{r}}w,k\big)}{1-k^2\operatorname{sn}^2\big(2\sqrt{\frac{m}{r}}w,k\big)}\right)
\left(du^2+\frac{dw^2}{r^2}\right)
\\
\hphantom{g=} =\frac{\big(r^2-\big(r^2-m^2\big)\operatorname{sn}^2\big(w,\frac{\sqrt{r^2-m^2}}{r}\big)\big)^2+(rm)^2}{r^2-\big(r^2-m^2\big)
\operatorname{sn}^2\big(w,\frac{\sqrt{r^2-m^2}}{r}\big)} \left(du^2+\frac{dw^2}{r^2}\right),
\end{gather*}
which for $k'=\sqrt{1-k^2}$, used the identities~(13.17), (13.22), (13.23) from~\cite{ellipticbook}:
\begin{alignat*}{3}
& \operatorname{cn}^2(t,k)=1-\operatorname{sn}^2(t,k) ,\qquad &&
\operatorname{sn}\left(k'u,\frac{ik}{k'}\right)=k'\frac{\operatorname{sn}(u,k)}{\operatorname{dn}(u,k)},&\\
& \operatorname{dn}^2(t,k)=1-k^2\operatorname{sn}^2(t,k) ,\qquad && \operatorname{sn}\left((1+k')u,\frac{1-k'}{1+k'}\right)=(1+k')\frac{\operatorname{sn}(u,k) \operatorname{cn}(u,k)}{\operatorname{dn}(u,k)} .&
\end{alignat*}

Now apply change of variable $\sin v=\operatorname{sn}(w,\tilde{k})$, where $\tilde{k}=\frac{\sqrt{r^2-m^2}}{r}$, implying
\begin{gather*}
g=\frac{\big(r^2-\big(r^2-m^2\big)\sin^2 v\big)^2+r^2 m^2}{r^2-\big(r^2-m^2\big)\sin^2 v}\left(du^2+\frac{dv^2}{r^2-\big(r^2-m^2\big)\sin^2 v}\right)=\tilde{g}.
\end{gather*}

When $rm\equiv 0\!\pmod{2}$, $a=r-m$ and $c=r+m$ are both odd since $(r,m)=1$. We have that $T_{a,b,c}$ is a torus and $\tilde{F}_{a,b,c}\colon \mathbb{R}^2/\mathcal{L}\rightarrow T_{a,b,c}$, where $\mathcal{L}=\{(2\pi n,2\pi m)\,|\,n,m\in \mathbb{Z}\}$, is a one-to-one map~\cite{penskoi1}. Apply change of variable $\sin y=\operatorname{sn}(z,k)$, and we alternatively have $\hat{F}_{a,b,c}\colon \mathbb{R}^2/\mathcal{\hat{L}}\rightarrow T_{a,b,c}$, where $\mathcal{\hat{L}}$ $=\{(2n\pi,4mK(k))\,|\,n,m\in \mathbb{Z}\}$. There is now a one-to-one correspondence between the rectangle~$[0,2\pi)\times [K(k),5K(k))$ and $T_{a,b,c}$. Our linear transformation~\eqref{transform1} maps this rectangular domain as follows {\samepage
\begin{gather*}
 H_{1}\left([0,2\pi)\times [K(k),5K(k))\right)=[0,2\pi)\times \left[0,\frac{2(a+c)K(k)}{\sqrt{c^2-a^2}}\right)=[0,2\pi)\times \big[0,2K\big(\tilde{k}\big)\big),
\end{gather*}
which used the identities $K(k)=\frac{1}{k'}K(\frac{i k}{k'})$, $K(\tilde{k})=\frac{2}{1+\tilde{k}'}K\big(\frac{1-\tilde{k}'}{1+\tilde{k}'}\big)$. }

Let us remark that when $rm\equiv 0\!\pmod{2}$ and after change of variable $\sin v=\operatorname{sn}(w,\tilde{k})$, the bipolar Lawson torus $\tilde{\tau}_{r,m}$ has a one-to-one correspondence with $[0,2\pi)\times [0,2K(\tilde{k}))$.

Thus we obtained the required isometry.

We recall now the explicit def\/inition of $S(a,b,c)$ \cite{penskoi1}:
\begin{gather}
\label{Sabc}
S(a,b,c)=\frac{4\pi}{\sqrt{c^2-a^2}}\left(2\big(c^2-a^2\big)E\left(\sqrt{\frac{b^2-a^2}{c^2-a^2}}\right)-
\big(c^2-a^2-b^2\big)K\left(\sqrt{\frac{b^2-a^2}{c^2-a^2}}\right)\right),
\end{gather}
where $K(k)$, $E(k)$ are the complete elliptic integrals of the f\/irst, second kind~\cite{ellipticbook}.

Using the values $\Lambda_{2(a+c)-2}(T_{a,b,c})$ and $\Lambda_{4r-2}(\tilde{\tau}_{r,m})$ from~\cite{penskoi1} and~\cite{lapointe}, which are twice the areas of these surfaces, we can check that the areas of corresponding surfaces are the same. Note that $S(a,b,c)=S(b,a,c)$ since $T_{b,a,c}\cong T_{a,b,c}$,
\begin{gather*}
\Lambda_{2(a+c)-2}(T_{a,0,c})=2S(0,a,c)=\frac{8\pi}{c}\left[2c^2 E\left(\frac{a}{c}\right)-\big(c^2-a^2\big)K\left(\frac{a}{c}\right)\right] \\
\hphantom{\Lambda_{2(a+c)-2}(T_{a,0,c})}{}
=8\pi(a+c) E\left(\frac{2\sqrt{ac}}{a+c}\right)=\Lambda_{4r-2}(\tilde{\tau}_{r,m}),
\end{gather*}
which uses the identity $E\big(\frac{2\sqrt{k}}{1+k}\big)=\frac{2E(k)-(1-k^2)K(k)}{1+k}$.

We now consider the cases when $rm\equiv 1\!\pmod{4}$ and $rm\equiv 3\!\pmod{4}$.
For the case of $rm\equiv 1\!\pmod{4}$, we have that $\tilde{\tau}_{r,m}$ is a bipolar Lawson torus, and it is isometric to the surfa\-ce~$T_{a,b,c}$, where $a=\frac{r-m}{2}$, $b=0$, and $c=\frac{r+m}{2}$.

Remark that the change of variable on $y$ and $v$ are the same as in the case $rm\equiv 0\!\pmod{2}$. The method of proof is the same using the transformation
\begin{equation}
\label{transform2}
H_{2}(x,z)=\left(2u,\frac{2\sqrt{c^2-a^2}}{a+c}w+K(k)\right).
\end{equation}

When $rm\equiv 1\!\pmod{4}$, $a=\frac{r-m}{2}$ is even and $c=\frac{r+m}{2}$ is odd. We have that $T_{a,b,c}$ is again a torus and $\tilde{F}_{a,b,c}\colon \mathbb{R}^2/\mathcal{L}\rightarrow T_{a,b,c}$ is a one-to-one map~\cite{penskoi1}. Let us again use the rectangle $[0,2\pi)\times [K(k),5K(k))$ and $\hat{F}_{a,b,c}\colon \mathbb{R}^2/\mathcal{\hat{L}}\rightarrow T_{a,b,c}$, where $\mathcal{\hat{L}}$ $=\{(2n\pi,4mK(k))\,|\,n,m\in \mathbb{Z}\}$. Now, our linear transformation~\eqref{transform2} maps this rectangular domain as follows
\begin{gather*}
H_{2}\big([0,2\pi)\times [K(k),5K(k))\big)=[0,\pi)\times \left[0,\frac{2(a+c)K(k)}{\sqrt{a^2-c^2}}\right)=[0,\pi)\times \big[0,2K\big(\tilde{k}\big)\big) .
\end{gather*}

When $rm\equiv 1\!\pmod{4}$ and after change of variable $\sin v=\operatorname{sn}(w,\tilde{k})$, the bipolar Lawson torus $\tilde{\tau}_{r,m}$ has a~one-to-one correspondence with $[0,\pi)\times [0,2K(\tilde{k}))$.

Thus we obtained the required isometry.

Now, for the case of $rm\equiv 3\!\pmod{4}$, we have that $\tilde{\tau}_{r,m}$ is a bipolar Lawson Klein bottle, and it is isometric to $T_{a,b,c}$, where $a=\frac{r-m}{2}$, $b=0$, and $c=\frac{r+m}{2}$.

Remark that we need the same change of variable on $y$ and $v$ and transformation~\eqref{transform2} as the case $rm\equiv 1\!\pmod{4}$.

When $rm\equiv 3\!\pmod{4}$, $a=\frac{r-m}{2}$ is odd and $c=\frac{r+m}{2}$ is even. We have that $T_{a,b,c}$ is a Klein bottle and $\tilde{F}_{a,b,c}\colon \mathbb{R}^2/\mathcal{L}\rightarrow T_{a,b,c}$ is a double covering~\cite{penskoi1}. With change of variable $\sin y=\operatorname{sn}(z,k)$ we now have as before a one-to-one correspondence between the rectangle $[0,\pi)\times [K(k),5K(k))$ and $T_{a,b,c}$. Our linear transformation~\eqref{transform2} maps this rectangular domain as follows
\begin{gather*}
H_{2}\big([0,\pi)\times [K(k),5K(k))\big)=\left[0,\frac{\pi}{2}\right)\times \left[0,\frac{2(a+c)K(k)}{\sqrt{a^2-c^2}}\right)
=\left[0,\frac{\pi}{2}\right)\times \big[0,2K\big(\tilde{k}\big)\big).
\end{gather*}

When $rm\equiv 3\!\pmod{4}$ and after change of variable $\sin v=\operatorname{sn}(w,\tilde{k})$, the bipolar Lawson Klein bottle $\tilde{\tau}_{r,m}$ has a one-to-one correspondence with $\big[0,\frac{\pi}{2}\big)\times [0,2K(\tilde{k}))$.

This completes the proof of Theorem~\ref{theorem1}.

\section{Proof of Theorem~\ref{theorem3}}
\label{section4}

In~\cite{kar3}, Karpukhin studied nonmaximality of known extremal metrics including metrics on Otsuki tori $O_{\frac{p}{q}}$, Lawson tori and Klein bottles~$\tau_{r,m}$, generalized tori~$M_{r,m}$, bipolar Lawson surfaces~$\tilde{\tau}_{r,m}$, and bipolar Otsuki tori~$\tilde{O}_{\frac{p}{q}}$. Although the family $T_{a,b,c}$ of tori and Klein bottles include the families $\tau_{r,m}$, $M_{r,m}$ and now $\tilde{\tau}_{r,m}$, there has been no general study of nonmaximality of metrics on~$T_{a,b,c}$. We prove that there are no maximal metrics among $T_{a,b,c}$ apart from the equilateral torus $T_{1,1,2}\cong M_{1,1}$ and Klein bottle $T_{1,0,2}\cong\tilde{\tau}_{3,1}$.

\begin{proposition}[Karpukhin, \cite{kar3}]\label{proposition1} The following inequalities hold
\begin{gather*}
\sup \Lambda_{n}\big(\mathbb{T}^2,g\big)\geq 8\pi\left(n-1+\frac{\pi}{\sqrt{3}}\right), \\
\sup \Lambda_{n}(\mathbb{KL},g)\geq 8\pi(n-1)+12\pi E\left(\frac{2\sqrt{2}}{3}\right),
\end{gather*}
where $E(k)$ is the complete elliptic integral of the second kind~{\rm \cite{ellipticbook}}.
\end{proposition}

This result can also be found by Colbois and El Souf\/i in~\cite[Theorem~B:4]{colelso}. Recent computational evidence~\cite{compu} further suggests that this construction is in fact maximal.  For our purposes, we use Proposition~\ref{proposition1} to show nonmaximality on~$T_{a,b,c}$.

\begin{proposition}\label{proposition2}
Let $a b c \ne 0$, and $|c|>\sqrt{a^2+b^2}$. If $a$ and $b$ have different parity and~$c$ is even, then the following inequality holds
\begin{gather*}
\Lambda_{a+b+c-3}(T_{a,b,c})<8\pi(a+b+c-4)+12\pi E\left(\frac{2\sqrt{2}}{3}\right).
\end{gather*}
If $a$ and $b$ are odd and $c$ is even, then the following inequality holds
\begin{gather*}
\Lambda_{a+b+c-3}(T_{a,b,c})<8\pi\left(a+b+c-4+\frac{\pi}{\sqrt{3}}\right).
\end{gather*}
Otherwise, the following inequality holds
\begin{gather*}
\Lambda_{2(a+b+c)-3}(T_{a,b,c})<8\pi\left(2(a+b+c)-4+\frac{\pi}{\sqrt{3}}\right).
\end{gather*}
\end{proposition}

Before proving Proposition~\ref{proposition2}, let us f\/ind an upper bound for $S(a,b,c)$.

\begin{proposition}\label{proposition3}
Let $|c|>\sqrt{a^2+b^2}$. The following inequality holds
\begin{gather*}
S(a,b,c)<2\pi^2(a+b+c).
\end{gather*}
\end{proposition}

\begin{proof} Using the def\/inition \eqref{Sabc} of $S(a,b,c)$, we show the proof when $a$, $b$, $c$ are each nonzero. The instance when $a,b,c$ contains at least one zero will follow as a special case. Here, we now use $a,b,c>0$ without loss of generality. Since $c>\sqrt{a^2+b^2}$ we have that our interval on~$k$ is~$[0,1)$ for both $K(k)$ and $E(k)$ in the inequality above. It is well known \cite{ellipticbook} that $K(k)$ is bounded below by~$\frac{\pi}{2}$, and $E(k)$ is bounded above by~$\frac{\pi}{2}$ on the interval~$[0,1]$. This implies that
\begin{gather*}
S(a,b,c)\le\frac{4\pi}{\sqrt{c^2-a^2}}\left(2\big(c^2-a^2\big)\frac{\pi}{2}-\big(c^2-a^2-b^2\big)\frac{\pi}{2}\right).
\end{gather*}
We now simplify this expression as follows
\begin{gather*}
S(a,b,c)\le 2\pi^2\sqrt{c^2-a^2}+\frac{2b^2\pi^2}{\sqrt{c^2-a^2}}.
\end{gather*}
We have that $\sqrt{c^2-a^2}\le\sqrt{c^2+a^2}\le\sqrt{c^2+2c a+a^2}=\sqrt{(c+a)^2}=(c+a)$. Hence, this implies that
\begin{gather*}
S(a,b,c)\le 2\pi^2(c+a)+\frac{2b^2\pi^2}{\sqrt{c^2-a^2}}.
\end{gather*}
The condition $c>\sqrt{a^2+b^2}$ implies that $\frac{b}{\sqrt{c^2-a^2}}<1$. This now implies
\begin{gather*}
S(a,b,c)<2\pi^2(c+a)+2\pi^2 b.
\end{gather*}
This concludes the proof of Proposition~\ref{proposition3}.
\end{proof}

We recall that Lawson tau-surfaces $\tau_{r,m}$ correspond to $T_{a,b,\sqrt{a^2+b^2}}$. Since $c=\sqrt{a^2+b^2}$, we have the same upper bound for $S(a,b,c)$ with the exception that the inequality is not strict.

\begin{proof}[Proof of Proposition~\ref{proposition2}]
In order to prove the f\/irst inequality we must show
\begin{gather*}
0<8\pi(a+b+c-4)+12\pi E\left(\frac{2\sqrt{2}}{3}\right)-S(a,b,c).
\end{gather*}
Using Proposition~\ref{proposition3}, this implies that we must show
\begin{gather*}
0<8\pi(a+b+c-4)+12\pi E\left(\frac{2\sqrt{2}}{3}\right)-2\pi^2(a+b+c).
\end{gather*}
This expression can be rewritten as
\begin{gather*}
\frac{16-6 E\left(\frac{2\sqrt{2}}{3}\right)}{4-\pi}<(a+b+c).
\end{gather*}
This inequality holds for $a+b+c\geq 11$. For the exceptional cases $(a,b,c)\in\{(1,2,4),(1,2,6)$, $(2,3,4)\}$, one can verify the f\/irst inequality explicitly using the tables of elliptic integrals in the book~\cite{integraltables}.

Similarily, the second inequality holds for $a+b+c\geq 11$. Remark that when $(a,b,c)=(1,1,2)$, we have the well known equilateral torus $T_{1,1,2}\cong M_{1,1}$. For the exceptional cases $(a,b,c)\in\{(1,1,4),(1,1,6),(1,1,8),(1,3,4),(1,3,6)$, $(3,3,4)\}$, one can again verify the second inequality explicitly.

The third inequality holds for $a+b+c\geq 6$. Only one exceptional case of $(a,b,c)=(1,1,3)$ should be verif\/ied for the third inequality explicitly.
\end{proof}

We now observe when exactly one of $a$ or $b$ is zero, while $|c|>\sqrt{a^2+b^2}$. Since $T_{a,b,c}\cong T_{b,a,c}$ when we assume that $S(a,b,c)=S(b,a,c)$, this corresponds to surfaces $T_{a,0,c}$. These surfaces have now been shown isomorphic to the family of bipolar Lawson surfaces $\tilde{\tau}_{r,m}$. Fortunately, these surfaces were investigated by Karpukhin in~\cite{kar3} and nonmaximality was shown for all extremal metrics except the well known bipolar Lawson Klein bottle $T_{1,0,2}\cong \tilde{\tau}_{3,1}$.

In~\cite{penskoi1}, the case of $T_{0,0,1}$ was already investigated to be the Clif\/ford torus but with a metric multiplied by $\frac{1}{2}$. Remark that the condition of $(a,b,c)=1$ prevents any other~$T_{0,0,n}$.

Finally, when $c=\sqrt{a^2+b^2}$, we have seen in Theorem~\ref{theorem2} that this corresponds to Lawson tau-surfaces $\tau_{r,m}\cong T_{a,b,\sqrt{a^2+b^2}}$. These surfaces were also investigated by Karpukhin in~\cite{kar3}, and it was shown that none of the extremal metrics are maximal.

This completes the proof of Theorem~\ref{theorem3}.

\subsection*{Acknowledgements}

This result was obtained during studies of the author at the National Research University~-- Higher School of Economics, Moscow and the author is very grateful for its hospitality.  The author also thanks A.V.~Penskoi for the statement of this problem, many useful discussions and invaluable help in preparing this manuscript.
The research of the author was partially supported by an NSERC Postgraduate Fellowship and by AG Laboratory NRU-HSE, Russian Federation government grant, ag.~11.G34.31.0023.

\pdfbookmark[1]{References}{ref}

config: arxiv


TITLE: functional calculus and spectral measure
QUESTION [0 upvotes]: Let $T$ be a normal operator and $f$ be a bounded borel function on ${\sigma}(T)$. If $E_{T}$ and $E_{f(T)}$ are the spectral decompositions of $T$ and $f(T)$ respectively, prove that for any borel set $w$ we have that $E_{f(T)}(w)=E_{T}(f^{-1}(w))$ for any borel subset $w$ of ${\sigma}(T)$

REPLY [0 votes]: The definition of $f(T)$ is
$$
         f(T) = \int_{\mathbb{R}} f(\lambda)dE(\lambda),
$$
for any bounded Borel function $f$ on $\sigma(T)$. For convenience, extend $f$ to be $0$ on $\mathbb{C}\setminus\sigma(T)$, and suppose $f$ is strictly bounded by $M$. Then $\Re f$ and $\Im f$ are strictly bounded by $M$. Let $N$ be a positive integer, and divide the rectangle $\{ \lambda \in\mathbb{C} : -M \le \Re f < M,\;-M \le \Im f < M\}$ into $N^{2}$ equal squares $S_{j,k}$ (closed on the left and bottom, open on the right and top) with centers $c_{j,k}$ for $1 \le j,k \le N$. Then
$$
              f(\lambda) - \sum_{j,k}c_{j,k}\chi_{f^{-1}(S_{j,k})}(\lambda)
$$
is uniformly bounded by $M/N$ on $\mathbb{C}$. Therefore, integrating with respect to $E$ gives
$$
                       \left\|f(T)-\sum_{j,k}c_{j,k}E(f^{-1}(S_{j,k}))\right\| \le M/N.
$$
Define $E_{f}(S)=E(f^{-1}(S))$. Then $E_{f}$ is a spectral measure, and the above implies
$$
                f(T)=\lim_{N}\sum_{j,k}c_{j,k}E_{f}(S_{j,k}).
$$
Notice also that
$$
   \left\|\sum_{j,k}c_{j,k}E_{f}(S_{j,k})-\int \lambda dE_{f}(\lambda)\right\| \le M/N
$$
Therefore,
$$
                  f(T)= \int f(\lambda)dE(\lambda)=\int \lambda dE_{f}(\lambda).
$$
By uniqueness of spectral measures $E_{f}$ is the spectral measure for $f(T)$. That is, $E_{T}(f^{-1}(S))=E_{f(T)}(S)$.


TITLE: Ultracoproducts and Cartesian products
QUESTION [8 upvotes]: Let $X$ be a metrizable compact topological space, let $\mathcal U$ be an ultrafilter, and denote by $X^{\mathcal U}$ the ultracopower of $X$ with respect to $\mathcal U$.
As a C$^*$-algebraist, I prefer to define $X^{\mathcal U}$ as the compact Hausdorff space satisfying
\begin{align*}
C(X^{\mathcal U},\mathbb C) &\cong \prod_{\mathcal U} C(X,\mathbb C) \\
&:= \ell^\infty(\mathbb N, C(X, \mathbb C))/\{(f_n)_n \mid \lim_{n\to\mathcal U} \|f_n\|_{\infty} = 0\}.
\end{align*}
It can alternatively be described, directly, as ultrafilters in $\mathbb N \times X$ which respect $\mathcal U$ (see Bankston's "Reduced
coproducts of compact Hausdorff spaces" for a more precise version of this).
My question is: if $Y$ is another metrizable compact topological space, is it the case that $(X \times Y)^{\mathcal U} \cong X^{\mathcal U} \times Y^{\mathcal U}$?
Equivalently, in terms of C$^*$-algebras, is it the case that
$$ \prod_{\mathcal U} C(X \times Y, \mathbb C) \cong \left(\prod_{\mathcal U}C(X, \mathbb C)\right) \otimes \left(\prod_{\mathcal U} C(Y, \mathbb C)\right)? $$
My intuition, looking at the C$^*$-algebra version, is that this isn't true, although I've found it difficult to prove this.
(Note: we could do an ultracoproduct with respect to a sequence of spaces instead, but this is notationally messier and, well, let's first see the answer to the question I stated before going there.)

REPLY [1 votes]: Here's another way to see that the answer is no for infinite $X$ and $Y$, from a more model-theoretic point of view.  As an ultrapower over a countably incomplete ultrafilter, $C(X\times Y)^{\mathcal{U}}$ is countably saturated.  It follows from this (actually, from the weaker property of countable degree-$1$ saturation) that $C(X \times Y)^{\mathcal{U}}$ is $SAW^*$ (see Farah and Hart, "Countable saturation of corona algebras"), and hence cannot be the tensor product of two infinite-dimensional algebras (see Saeed Ghasemi, "$SAW^*$-algebras are essentially non-factorizable).


TITLE: Find the antiderivatives
QUESTION [4 upvotes]: Find the antiderivatives:
$\int\!\left(2x^2 +3\right)^{1/3} x\,dx $
I have hit this in my book and the way I do it I get
$3/4\left(2x^2 + 3\right)^{4/3} x^2 +c $
But my book tells me it should be $3/16(2x^2 + 3)^{4/3}$
Anyone know how they get to that?

REPLY [2 votes]: Make the substitution $u=x^2+3$. Then $du=4x\,dx$ so $x\,dx=\frac{1}{4}\,du$.
Thus
$$\int (2x^2+3)^{1/3} x\,dx=\int \frac{1}{4}u^{1/3}\,du.$$
Remark: The derivative of $2x^2+3$ is "almost" part of your expression. sure, we have an $x$ instead of $4x$, but that's no problem, since $x=\frac{1}{4}(4x)$. In this kind of situation, substitution is often useful. You will meet variants often, like $\int xe^{x^2}\,dx$, or $\int (1+\sin x)^8\cos x\,dx$.

REPLY [1 votes]: Use the substitution $u=2x^2+3, du=4x$.    It is unfortunately not true that $\int f(x) g(x)dx=\int f(x)dx \int g(x)dx$; you cannot integrate the $x$ part separately.


TITLE: Concerning a Cyclic Galois Group
QUESTION [0 upvotes]: Why is it that: $\forall K \supseteq \mathbb{Q}(\mathbb{i}), G=Gal(K / \mathbb{Q}) = \langle \sigma \rangle \implies \sigma (\mathbb{i})  = - \mathbb{i}$? (Note: I am guessing that $\sigma ≠ \mathbb{1}$ here.) Moreover, why must there not be a field extension $K \supseteq \mathbb{Q}(\mathbb{i}): K/ \mathbb{Q}$ is Galois with Galois group $G$ is cyclic with order $|G|=4$? I see the connection but why could there not be another one? Perhaps because an automorphism $\tau$ that results in complex conjugation is a power of $\sigma$? That is all that I can come up with but even that does not immediately relate as far as I can tell.
I am going over notes in class and the teacher said that these things were obvious, but not to me...
Note: this formulation may not be perfectly correct because (1) it was sort of on the fly during class (rather than carefully written out), and (2) I was taking notes, so I may have written something down incorrectly.

REPLY [1 votes]: Let us assume $\;\Bbb Q(i)\subset \Bbb K\;$ is such that $\;\Bbb K/\Bbb Q\;$ is a Galois cyclic extension, and thus $\;Gal(\Bbb K/\Bbb Q)=\langle\sigma\rangle\;$.
Now, since $\;\sigma (q)=q\;\;\;\forall\,q\in\Bbb Q\;$ ,we have that
$$\sigma(i)=i\iff \sigma=\text{Id.}_{\Bbb Q(i)}$$
But this can't be by the Fundamental Theorem of Galois Extensions, since the fixed field of $\;\sigma\;$ is $\;\Bbb Q\;$ ...


TITLE: Proving connectedness of the $n$-sphere
QUESTION [5 upvotes]: My homework assignment contains the following question:

Prove that $S^n = \{x \mid x \in \mathbb R^{n+1}, d(x,0)=1\}$ is connected.

Can you give me a hint please?
What I can do is nothing..

REPLY [2 votes]: By taking $S^n$ to be the union of

$$A_1:=\{x=(x_1,...,x_{n+1}) \mid  ||x||=1, ~ x_{n+1} \geq 0\},$$
$$A_2:=\{x=(x_1,...,x_{n+1}) \mid ||x||=1, ~x_{n+1} \leq 0\},$$
we have that $S^n$ is the union of two connected sets with one point in common (in fact, a lot of points in common), hence $S^n$ is connected. To check that $A_1$ and $A_2$ are connected, just notice that both are the image of a continuous function: $A_1=f(D^n)$, where $f: D^n \rightarrow S^n$ takes $(x_1,...,x_n)$ to $(x_1,...,x_n, \sqrt{1-x_1^2-...-x_n^2})$ and $A_2=g(D^n)$, where $g: D^n \rightarrow S^n$ takes $(x_1,...,x_n)$ to $(x_1,...,x_n, -\sqrt{1-x_1^2-...-x_n^2})$.

$S^n \cong D^n/ \sim$. Since $D^n$ is connected, so is $S^n$.
$S^n$ is the one-point compactification of a non-compact connected space (namely, $\mathbb{R}^n$). The result now follows due to the fact that the closure of a connected set is connected.
$S^n$ is (again) the union of two connected sets with one point in common (and yet again, a lot of points in common): $S^n$ without the north pole and $S^n$ without the south pole, since those two sets are homeomorphic to $\mathbb{R}^n$ by the stereographic projection.
$S^n$ is pathwise connected. This can be seen since any two antipodal points $x,y$ can be joined by the path $\displaystyle \frac{(1-t)x+ty}{||1-t)x+ty||}$. If $x,y$ are antipodal points, make a path from $x$ to another point $z \neq y$ as before, then join $z$ to $y$ and we have a path from $x$ to $y$.


TITLE: What is positive-0 squared minus positive-0?
QUESTION [1 upvotes]: I've got a basic limit problem that I think I'm solving the right way, but I've run into something that looks confusing enough to make me wonder if I'm doing it right.
$$
\lim_{y\to0} \frac{1}{y^2-y} + \frac{1}{y}
$$
I tried approaching this by looking at the Left Hand Limit and Right Hand Limit. Though when I plug in $0^+$, I get
$$
\lim_{y\to0} \frac{1}{0^+-0^+} + \frac{1}{0^+}
$$
What is $0^+-0^+$? Is it still positive? or is it just 0, or in this case, would it be negative? I'm leaning toward negative because for all $0\lt y\lt 1$, $y^2$ will be less than $y$, so then wouldn't I have an infinitely small negative number as a result?
And then, if that turns out to be true, it would appear that this problem is unsolvable by that method then, seeing as I'd have $+\infty + -\infty$, which is undefined.

REPLY [3 votes]: Clearly $y^2-y\to 0$ as $y\to 0$, as does $y$, so both fractions are blowing up, and their difference is an indeterminate form. Combine them over a common denominator; if you do the algebra correctly, you’ll be able to simplify the resulting fraction to one whose limit as $y\to 0$ is easy to compute.

REPLY [1 votes]: Dividing by $0$ is nonsensical. And there is only one $0$.
Bring your expression to a common denominator and simplify.
Remark: You could get useful information from your calculator. Calculate
$\dfrac{1}{y^2-y}+\dfrac{1}{y}$ for some values of $y$ not far from $0$, but of course not equal to $0$, like $y=0.001$ or $y=-0.002$.


TITLE: Explicit triangulation of a genus 3 surface $X^4+Y^4+Z^4=0$
QUESTION [4 upvotes]: I tried to describe the Riemann surface $X^4+Y^4+Z^4=0$ as a surface in real 3-dimensional space (without self-intersection) with triangulation.
I picked $12$ points on the surface:
$P_{a} = [0:1:\zeta^a], Q_{a}= [\zeta^{a}:0:1], R_{a} = [1:\zeta^a:0]$,
where $a \in \{1,3,5,7\}, \zeta = (1+i)/\sqrt2$.
I join $P_a$ $Q_b$ by the parametlized curve:
$[X:Y:Z] = [ζ^b  t: ζ^{-a} \sqrt[4] {1-t^4}:1]$.
Similarly I join $Q_aR_B$ and $R_aP_b$.
Then I found following triangles are contractible:
$\triangle P_1Q_1R_5,
\triangle P_1Q_1R_7,
\triangle P_1Q_3R_3,
\triangle P_1Q_3R_5,
\triangle P_1Q_5R_1,
\triangle P_1Q_5R_3,
\triangle P_1Q_7R_1,
\triangle P_1Q_7R_7$
$\triangle P_3Q_1R_3,
\triangle P_3Q_1R_5,
\triangle P_3Q_3R_1,
\triangle P_3Q_3R_3,
\triangle P_3Q_5R_1,
\triangle P_3Q_5R_7,
\triangle P_3Q_7R_5,
\triangle P_3Q_7R_7$
$\triangle P_5Q_1R_1,
\triangle P_5Q_1R_3,
\triangle P_5Q_3R_1,
\triangle P_5Q_3R_7,
\triangle P_5Q_5R_5,
\triangle P_5Q_5R_7,
\triangle P_5Q_7R_3,
\triangle P_5Q_7R_5$
$\triangle P_7Q_1R_1,
\triangle P_7Q_1R_7,
\triangle P_7Q_3R_5,
\triangle P_7Q_3R_7,
\triangle P_7Q_5R_3,
\triangle P_7Q_5R_5,
\triangle P_7Q_7R_1,
\triangle P_7Q_7R_3$
(I found this by integrating $3$ holomorphic differentials which I previously described here)
Thus I have $12$ points and $48$ edges, and $32$ triangles on the surface. (The Euler characteristic shows that this is a genus 3 surface.)
So I tried to choose $12$ points on a sphere with $3$ handles, and draw $48$ edges ($8$ edges for each point) to triangulate, satisfying the relations above. But it was really difficult and I couldn't succeed. Is this actually possible (and can you draw it)?
What I tried else:
I considered 4 cones with:
Apex:$P_1$, 8-edged bottom $Q_1R_5Q_3R_3Q_5R_1Q_7R_7Q_1$
Apex:$P_3$, 8-edged bottom $Q_1R_3Q_3R_1Q_5R_7Q_7R_5Q_1$
Apex:P_5, 8-edged bottom Q_1R_1Q_7R_5Q_5R_7Q_3R_7Q_1
Apex:$P_5$, 8-edged bottom $Q_1R_1Q_3R_7Q_5R_5Q_7R_3Q_1$
Apex:P_7, 8-edged bottom Q_1R_1Q_3R_5Q_5R_3Q_7R_3Q_1
Apex:$P_7$, 8-edged bottom $Q_1R_1Q_7R_3Q_5R_5Q_3R_7Q_1$
Then patch together with the blue edges or pink edges respectively.
But it seems to me that it is not possible in 3-dimensional space without self-intersection (but I am not sure).
EDIT
I tried again with more symmetric shape and succeeded so I have posted it as an answer.

REPLY [0 votes]: I noticed that instead of considering the cone with 8 edges, I can just use the area formed by the edges, because they are homeomorphism.
I used more symmetric 4 shapes each has $8$ points and $8$ edges with the relation described in the question. The shapes are shown in the attached pictures.
I painted the inside (bottom side) by glay and the outside (top side) by light blue.
The order to patch them is:
$P_1\to P_7\to P_5\to P_3$  in the upper area containing $Q_1,R_1,Q_5,R_5$,
$P_5\to P_3\to P_1\to P_7$ in the lower area containing $Q_3,R_3,Q_7,R_7$,
in the direction from top to bottom.
At first I couldn't image that it can be realized in 3-dimensional space so I made a model by paper. I observed it is actually a surface with genus 3.


TITLE: An example of covariant functor.
QUESTION [1 upvotes]: Let $F$ be the following covariant functor from the category of sets to the category of left module over a ring $R$ with identity. For each set $X$, $F\left(X\right)$ is the free $R$-module on $X$. If $f : X\rightarrow X'$ is a function, let $F\left(f\right) : F\left(X\right)\rightarrow F\left(X'\right)$ be the unique module homomorphism $\overline f : F\left(X\right)\rightarrow F\left(X'\right)$ such that $\overline fi=f$, where $i$ is the inclusion map $X\rightarrow F\left(X\right)$.
This is an example in the Hungerford. I think that the given functor $F$ is not well-defined. For an object $X$ in the category of sets there are many free $R$-module on $X$. So it seems to consider a category of isomorphic classes of left module over a ring $R$ with identity and a covariant functor from the category of sets to this category. Is it right?

REPLY [6 votes]: No, he has in mind a specific construction of the free module on a set, so that fixes the choice.
For example, you could take the set of all functions $X\to R$ with finite support with its obvious structure of $R$-module as $F(X)$, or you could take some other fixed construction.


TITLE: What is the $2012th$ number in this pattern?
QUESTION [3 upvotes]: This is question 30 from Australian Maths 2012
$(0,1,2,1,2,3,2,3,4,1,2,3,2,3,4,3,4,5,2,3,4,...)$
What is the $2012th $ number in this list?
What I did:
I broke up the first few numbers into smaller parts to identify the pattern
$(0,1,2,1,2,3,2,3,4)$
$(1,2,3,2,3,4,3,4,5)$
From this,I can see that every 9 numbers,the pattern restarts.
So, the $2012th$ number should be in the $\frac {2012}{9} th $ group which is the $223 \frac {5}{9} th$ group.
So,the first number in the $223 th $ group is $(223-1)$.
And to further break it up to make it easier to find the $2012th$ number.
From,$(0,1,2,1,2,3,2,3,4)$
$(0,1,2)$-$x$
$(1,2,3)$-$y$
$(2,3,4)$-$z$
The last number number in $x$ is the $2nd$ number in $y$,which is also the first number in $z$.
So,$222+2 =224$ is my answer.
However,checking the answer key I got,the answer is $9$.
Where did I go wrong?

REPLY [0 votes]: If you graph this sequence of numbers, you will find an interesting curve:
(0,1,2), (1,2,3), (2,3,4), -> first cycle
(1,2,3), (2,3,4), (3,4,5), -> second cycle
(2,3,4), ...
Use Excel or squared paper.


TITLE: $f$ is bounded and continious $\Rightarrow$ the convolution integral $\int f(\tau)g(x-\tau)\text{ d}\tau$ is bounded and continuous
QUESTION [2 upvotes]: Let $g\in L^1(\mathbb{R}^n)$ and $f:\mathbb{R}^n\to\mathbb{R}$ be bounded and continuous. Why is the convolution integral $$f*g:\mathbb{R}^n\to\mathbb{R}\;,\;\;\;\int f(\tau)g(x-\tau)\text{ d}\tau$$ also bounded and continuous?
*Proof*$\;\;\;$ It holds $$|(f*g)(x)|\leq\int |f(\tau)g(x-\tau)|\text{ d}\tau =\int |f(x-t)g(t)|\text{ dt}\le \sup_{x\in \mathbb{R}^n}|f(x)|\int |g(t)|\text{ dt}$$ Since $f$ is bounded and $g$ is integrable the supremum and the integral exists. This should be enough for at least the boundedness of $f*g$. While I would like to be sure that this is correct, I would also like to know how we show the continuity.

REPLY [1 votes]: I would think continuity is not too bad, for fix and $x$ consider take an
$|f*g(x)-f*g(y)| = |\int f(x-t)g(t)-f(y-t)g(t)\text{ dt}| \leq \int |(f(x-t)-f(y-t))||g(t)|\text{ dt}$
Consider this:
Choose an $R>0$ large enough such that $\int_{\mathbb{R}^n/B(0,R)} |g| < \frac{\epsilon}{2K}$, where $B(0,R)$ is a ball of radius $R$ centred  at the origin. Since $|f|<K$ for some $K>0$,
$|f*g(x)-f*g(y)|\leq \int_{B_R} |(f(x-t)-f(y-t))||g(t)|\text{ dt} + \int_{\mathbb{R}^n/B(0,R)} |(f(x-t)-f(y-t))||g(t)|\text{ dt} \leq \sup_{x\in B(0,R)}|(f(x-t)-f(y-t))|\int_{B(0,R)} |g| +\frac{\epsilon}{2K} 2K$
but $f$ is continuous, hence uniformly continuous on $B(0,R)$, so can choose $\delta>0$ such that $|x-y|<\delta$, then $\sup_{x\in B(0,R)}|(f(x-t)-f(y-t))|<\frac{\epsilon}{\int_{B(0,R)} |g|}$


TITLE: Prove that $|\mathcal{F}|=|\mathbb{R}|$.
QUESTION [0 upvotes]: Let $\mathcal{F} = \Big\{ \mathcal{I}_{a,b}\times\{x\} :  a,b,x\in\mathbb{R}\Big\}$, where $\mathcal{I}_{a,b}:=\{\zeta\in\mathbb{R} :  a < \zeta <b\}$. Im trying to prove that $|\mathcal{F}|=|\mathbb{R}|$ (that can be denoted by $\mathfrak{c}).$
Please, can someone help with a clue our something. Thanks.

REPLY [1 votes]: You might try to show that $|\mathcal{F}| = |\mathbb{R}^3|$. Then you need only show that $|\mathbb{R}| = |\mathbb{R}^3|$ (or, more generally, $\mathbb{R}^n$).

REPLY [1 votes]: Consider $f: \mathcal{F} \to \mathbb{R}^3$ defined by
\begin{align*}
\mathcal{I}_{a,b} \times \{x\} \mapsto (a,b,x).
\end{align*}
It should be easy to show that this is injective, so that $|\mathcal{F}|\leq |\mathbb{R}^3|$.  Also notice that for any fixed $a < b$, $\mathcal{A} = \{\mathcal{I}_{a,b} \times \{x\} \,:\, x\in \mathbb{R}\} \subseteq \mathcal{F}$.
 But there is a bijection from $\mathcal{A}$ to $\mathbb{R}$ via
\begin{align*}
\mathcal{I}_{a,b} \times \{x\} \mapsto x
\end{align*}
so $|\mathbb{R}| \leq |\mathcal{F}| \leq |\mathbb{R}^3|$.
Now it comes down to arguing that $|\mathbb{R}| = |\mathbb{R}^3|$.


TITLE: Derivation of Taylor polynomial
QUESTION [0 upvotes]: sorry for bad title.
Suppose $f:\mathbb{R} \rightarrow \mathbb R$ is differentiable twice and $f = f''$,    $ f(0) = f'(0)=0$ and I want to prove that $f=0$.
So I started with proving that for $[-1,1]$
because it's differentiable it's continuous in $[-1,1]$, and there is $M\geq |f(x)| $ for any $x$ in $[-1,1]$, suppose $f(x_1)=M$.
We can develop a Taylor polynomial around zero and say that there is a $c$ between zero and $x$ such that:
$$f(x) =f(0) + f'(0)x + \frac{f''(c)x^2}{2}= \frac{f(c)x^2}{2}$$
$$ f'(x) = (\frac{f(c)x^2}{2} )' =f(c)x$$
Now my question is:
Is it possible to derive this polynomial and treat $f(c)$ as a constant number?
Because then, since $M$ is maximum then $f'(x_1)=0$ and $f(c)x_1=0$ either way $M=0$.
Thanks for the help.

REPLY [1 votes]: Things can be saw in three ways :

1-Uniqueness of solution with given initial conditions
Knowing that for a linear homogenous differential equation there exists an unique solution given initial conditions for $f$ and $f'$.
Because $0$ is solution you're function must be 0.
2-Direct Solving
Your solution is of the form :
$$\exists(A,B), \forall x,  f(x)=A\cosh(x)+B\sinh(x)$$
So giving your conditions
$$f=0$$


TITLE: Result of a Sum Up to $\pi(n)$
QUESTION [2 upvotes]: I have been working with this sum recently:
$$\sum_{k=2}^{\pi(n)}\frac{\pi(n)-1}{p_k -1}$$In case the notation of $k=2$ is not clear, I mean this:
$$\frac{\pi(n)-1}{2} + \frac{\pi(n)-1}{4} + \frac{\pi(n)-1}{6} + ...$$
I was wondering whether it converged, or whether there was a simpler way to express it. I know that the sum of the reciprocals of the primes diverges, but I'm not sure whether this is applicable here.

REPLY [1 votes]: From $$p_k-1<p_k \Rightarrow \sum_{k=2}^{\pi(n)}\frac{\pi(n)-1}{p_k-1}>\left(\pi(n)-1\right)\sum_{k=2}^{\pi(n)}\frac{1}{p_k}\geq...$$
using this
$$...\geq\left(\pi(n)-1\right)\left(\ln{\ln{(\pi(n)+1)}}-\ln{\frac{\pi^2}{6}}-\frac{1}{p_1}\right)$$
it should diverge.


text:
\section{Introduction}

Markov chain Monte Carlo (MCMC) is a sampling algorithm that {in principle} can simulate any probability distribution. {In practice,} many MCMC algorithms are infeasible when target density evaluations are computationally intensive. In this paper, we present an algorithm that mitigates this issue, achieving two desirable properties:  (i) vastly reducing the number of target density evaluations per MCMC step, while (ii) \textit{provably} retaining a convergence rate roughly equal to that of the original expensive algorithm.

Challenges associated with computationally intractable target densities are extremely well known in the Monte Carlo literature. We now give a short survey of approaches to the problem, emphasizing the contributions of the current paper.

Replacing the intractable target density with an approximation or  ``surrogate model'' is the simplest way to significantly reduce the computational cost of MCMC. Replacing the target density with a \textit{fixed} surrogate, however, introduces a non-vanishing bias \citep{MarzoukXiu2009, Cotteretal2010, Bliznyuketal2012, LiMarzouk2014, Cui2016etal, stuart2018posterior}.

The obvious solution to the problem of non-vanishing bias is to iteratively refine the surrogate during sampling. This strategy introduces some non-trivial technical difficulties: continual surrogate refinement within MCMC results in a process that is not Markovian. Such non-Markovian processes can have surprising and terrible convergence properties even when the surrogate has very low pointwise error \citep{LatuszynskiRosenthal2014}, sometimes failing to converge at all. In our previous work \citep{Conradetal2016, Conradetal2018}, we presented an MCMC algorithm with a continually refined surrogate based on \textit{local approximations}. We showed that this algorithm avoided the worst of these convergence problems and produced asymptotically exact results under certain strong assumptions.

These strong assumptions are symptoms of a practical problem with how local approximation and MCMC interact in the ``tails'' of the target distribution. \citet{Conradetal2016, Conradetal2018} built a piecewise polynomial approximation by solving a local regression problem, using a small number of nearby exact target density evaluations. Refinements triggered by a cross-validation heuristic on the acceptance ratio added new points to the set of target density evaluations. Ideally, this approach should avoid expensive density evaluations in regions of low probability---particularly the tails of the distribution. In practice, there is an important tension: the cross-validation heuristic requires many \textit{more} tail evaluations than a good global design would suggest, but gives many \textit{fewer} tail evaluations than are sometimes required for global convergence and stability of the MCMC algorithm.

In \citet{Conradetal2016, Conradetal2018} we showed that this tension could be avoided under certain strong tail conditions, with the additional requirement of randomly triggered surrogate refinements to ensure convergence.

In more general settings, the underlying tension remains and leads to two serious issues which we address in the present paper.

First is the behavior of the cross-validation heuristic noted above:
it demands more refinements in low probability regions than the goal of ``good'' approximation of the target density (in a natural sense that we will make precise later)

would dictate. Second, and even worse, local polynomial approximations typically have very bad tail behavior, even failing to be integrable unless nearly every point in the tail has a density evaluation.

To resolve the first issue, we introduce here a new surrogate refinement strategy that relaxes the acceptable error threshold in the tails of the distribution. The refinement strategy also \emph{balances} the rate of decay of the surrogate bias with that of Monte Carlo variance, using a local error indicator to characterize the bias. To resolve the second issue, we introduce a new correction term for the Metropolis-Hastings acceptance ratio that allows the algorithm to retain good convergence properties even under very weak assumptions about tail behavior, and without requiring integrability of the surrogate.

The result is a local approximation MCMC (LA-MCMC) algorithm that tends to require far fewer density evaluations (resulting in faster runs) while delivering robust performance for a much broader class of target distributions. From a practical point of view, work such as \citet{Conradetal2016, Conradetal2018, Angelikopoulosetal2015} showed that the number of expensive target density evaluations can be reduced by orders of magnitude with minimal impact on the accuracy of target expectation estimates. Our new algorithm further reduces the number of density evaluations to nearly the rate-optimal number, without sacrificing MCMC stability.

From a theoretical point of view, we \textit{guarantee} that LA-MCMC converges quickly by providing useful bounds on the convergence of the algorithm after a \textit{finite} number of MCMC steps.  Our main result is that the mean-square error (MSE) of our LA-MCMC estimates decays at roughly the expected $1/T$ rate, where $T$ is the number of MCMC steps. We also show that LA-MCMC converges under very weak tail bounds that hold for many statistical examples. Comparatively, \citet{Conradetal2016} only established a law of large numbers that applied for special tail shapes.

Using our optimal surrogate refinement strategy, the refinement rate (i.e., the rate at which new density evaluations are demanded) decreases with $T$. This implies that the decay of the MSE as the number of density evaluations $n$ increases is not only faster than the $1/n$ rate expected for standard geometrically ergodic MCMC, but also that this convergence rate (in $n$) may accelerate as the number of MCMC steps $T$ increases.

\paragraph{Other related work.}
There have been many attempts to speed up MCMC for models that are expensive to compute. We have already discussed work based on the construction of surrogate models/approximations for the target density. Some other approaches have been rather different.

The multi-level MCMC approach of \citet{Dodwelletal2015}, and related multi-index extensions \citep{jasra2018multi}, rely on predefined hierarchies of models---e.g., corresponding to different mesh refinements of an underlying differential equation model---that induce trade-offs between computational cost and accuracy. These approaches can drastically reduce the number of expensive model evaluations and total wallclock time of an MCMC run, but require a careful understanding of numerical approximation errors to achieve optimal convergence rates. In contrast, \citet{kaipio2007statistical,chkrebtii2016bayesian} and other related work create statistical models of numerical discretization error in ODE or PDE models. Delayed-acceptance MCMC can also exploit a hierarchy of models, but instead ``screens'' MCMC proposals through the approximate model(s) and evaluates the expensive target density at least once for each accepted sample \citep{ChristenFox2005, Cuietal2011}, thus reducing cost by a constant factor.

\paragraph{Organization of the paper.} In Section~\ref{sec:numerics} we present our new algorithm, including the refinement scheme, the bias-variance tradeoff, and the tail correction. Section~\ref{sec:theory} describes our main theoretical results. Section~\ref{sec:examples} provides a range of numerical examples, beginning with simple configurations intended to illustrate specific features and variations of the algorithm, and culminating in the inference of a spatially distributed coefficient in a set of nonlinear PDEs.


TITLE: Entire function mapping a circle to a line
QUESTION [0 upvotes]: The problem says:
Let $f$ be an entire function, $S$ a circle of positive radius and $L$ a line in the complex plane such that $f(S) \subseteq L$. Show that $f$ is constant.

It would look like an application of the open mapping theorem, however I don't know if the disc enclosed by $S$ is inside of $L$ in order to apply it (this is probably false, although I can't think of an example). Also, looking a the hypothesis that wouldn't really require $f$ to be entire. There is something more to it.

REPLY [3 votes]: As Conrad said, we can assume that $f(S) \subset \Bbb R$. Then $v = \operatorname{Im}(f)$ is harmonic with $v(z) = 0$ on $S$. It follows that $v= 0$ in the interior of $S$.
Then the Cauchy-Riemann equations imply that $f$ is constant in the interior of $S$.
Finally the identity theorem shows that $f$ is constant in $\Bbb C$.
Alternatively one can consider $g(z) = e^{if(z)}$ which is entire, non-zero, and satisfies $|g(z)| = 1$ for all $z \in S$. The maximum modulus theorem implies that $g$ (and therefore $f$) is constant in the interior of $S$, which again implies that $f$ is constant everywhere.


TITLE: Norm of vector sum
QUESTION [1 upvotes]: So for real-valued vectors the statement

$$ \lVert u+v\rVert = \lVert u-v\rVert $$

holds if and only if $$ \langle u,v\rangle =0 $$
I want to state this now instead for complex-valued vectors

$$\lVert u+v\rVert =\lVert u-v\rVert \iff \langle u,v\rangle=-\langle u,v\rangle $$

Im not seeing the connection, but is the right hand side equal to the conjugate of $\langle u,v\rangle$ or?

REPLY [1 votes]: Use the polarization identities (since you're in a real vector space):

$$\langle u,v\rangle = {1\over 4}(\lVert u+v\rVert^2-\lVert u-v\rVert^2).$$

For complex:

$$\langle u,v\rangle = {1\over 4}(\lVert u+v\rVert^2-\lVert u-v\rVert^2+i\lVert u+iv\rVert^2 -i\lVert u-iv\rVert^2).$$

If all they want is a restatement then just write it out:
$$\begin{cases}\lVert u+v\rVert^2=\langle u+v,u+v\rangle=\lVert u\rVert^2+\lVert v\rVert^2+\langle u,v\rangle +\langle v,u\rangle \\
\lVert u-v\rVert^2=\langle u-v,u-v\rangle=\lVert u\rVert^2+\lVert v\rVert^2-\langle u,v\rangle -\langle v,u\rangle\end{cases}$$
Then equality means

$$\langle u,v\rangle +\overline{\langle u,v\rangle} = -(\langle u,v\rangle+\overline{\langle u,v\rangle})$$

So $\text{Re}(\langle u,v\rangle)=0$ is the condition.
You can even verify this in the basic case of $V=\Bbb C$ where the inner product is given as $\langle u,v\rangle= u\overline{v}$. Let $u=1, v=i$ then $\langle u,v\rangle =(1)(-i)$ and clearly $\lVert u+v\rVert^2 = \lVert u-v\rVert^2=2$.


TITLE: What is the difference between a loop, cycle and strongly connected components in Graph Theory?
QUESTION [24 upvotes]: Which graph is/contains a cycle? loop? strongly connected component?
I believe the following is correct in regards to graphs above:

all cycles are loops
all graphs contain cycles, but $G_3$ has two cycles: $\{(a,b), (b,c)\}$ and $\{(a,b),(b,c),(c,a)\}$?
all graphs are considered strongly connected?

But if all graphs above are considered strongly connected and all contains cycles, then what is the difference between a cycle and a strongly connected component?
Update:

A cycle is a simply closed path $v_1, ..., v_k, v_1$ with $v_1, ..., v_k$ all distinct, and $k\geq3$
A cycle is a closed path. That is, we start and end at the same vertex. In the middle, we do not travel to any vertex twice.
A graph is said to be strongly connected if every vertex is reachable from every other vertex.

For the sake of completeness, I added a graph $G_0$ with a real loop:

$G_0$: This is a directed graph with a loop since there is an edge going from vertex $a$ to itself.
$G_1$: This is a directed connected graph with a cycle, and this graph also is a strongly connected component. This is NOT a simple graph because it has a pair of vertices that have edges going in both directions of each other.
$G_2$: This is a directed simple-connected graph with a simple cycle.
$G_3$: This is a directed connected graph with multiple strongly connected components.

REPLY [17 votes]: A loop is commonly defined as an edge (or directed edge in the case of a digraph) with both ends as the same vertex.  (For example from $a$ to itself).  Although loops are cycles, not all cycles are loops.  In fact, none of the above digraphs have any loops.
Cycles are usually defined as closed walks which do not repeat edges or vertices except for the starting and ending vertex.  This definition usually allows for cycles of length one (loops) and cycles of length two (parallel edges).
Note that cycles (and walks) do not make any reference to the orientation of the edges in question.  Directed cycles (and directed walks) may only travel along the "forward" direction of the edges.  In particular, that implies that $G_3$ pictured above has a third cycle, $(\color{blue}{(a,b)},(b,c),(c,a))$ where the $\color{blue}{(a,b)}$ refers instead to the edge pointing from $b$ to $a$.
Technically, all of the graphs above except for $G_2$ are directed multigraphs since in each you have parallel edges.  Although in simple graphs (graphs with no loops or parallel edges) all cycles will have length at least $3$, a cycle in a multigraph can be of shorter length.  Usually in multigraphs, we prefer to give edges specific labels so we may refer to them without ambiguity.
As for being strongly connected, yes all of them are and your definition is correct.

Your additional question, "what is the difference between a cycle and a connected component"

The above graph contains a cycle (though not a directed cycle) yet is not strongly connected.
One can prove that if a directed multigraph is strongly connected then it contains a cycle (take a directed walk from a vertex $v$ to $u$, then a directed walk from $u$ to $v$.  Any closed walk contains a cycle).
One can also show that if you have a directed cycle, it will be a part of a strongly connected component (though it will not necessarily be the whole component, nor will the entire graph necessarily be strongly connected).


TITLE: When is an étale map proper?
QUESTION [1 upvotes]: Let $Y\to X$ be étale and let both schemes be connected, and let $X$ be locally noetherian and geometrically unibranch, or normal.
How/under what assumptions can we prove that the map is proper?
A related thing I read somewhere stated that being proper was a local property and went on to assume that $X$ was noetherian and wanting to show that $Y$ was quasi-compact, but why would that imply that the map is proper?

REPLY [1 votes]: To give a precise reference, finite is equivalent to proper + locally quasifinite (https://stacks.math.columbia.edu/tag/02LS). Etale morphisms are locally quasifinite (since they are locally given by standard etale morphisms https://stacks.math.columbia.edu/tag/02GH), so etale + proper implies etale + proper + locally quasifinite implies etale and finite.


TITLE: Are all $30n + 7$ numbers prime numbers where $n$ is a whole number?
QUESTION [1 upvotes]: So I was lying on my bed and I began to wonder about prime numbers and as I went through each prime number I found that numbers $37, 67, 97, 127, 157, \ldots$ all happen to be prime numbers (correct me if I am wrong). Is there a way to mathematically prove this?

REPLY [2 votes]: We can actually be quite precise about how often a number of the form $30n+7$ is prime, thanks to the Prime Number Theorem for Arithmetic Progressions.  In this case it shows that for large values of $x$, the number of positive integers $n \le x$ such that $30n+7$ is prime is approximately (asymptotically equal to)
$$\frac{15}{4} \frac{x}{\log x}.$$
Since that quantity goes to $\infty$ as $x\to\infty$, there are indeed infinitely many prime values of $30n+7$ (a fact that also follows from the slightly weaker Dirichlet’s theorem).  But since that same quantity is eventually smaller than $x$, $30n+7$ is not always — nor is it even usually — prime.  The fact that the approximate count doesn’t drop below $x$ until $x>42$ or so is a partial explanation for why the frequency of primes is so high for the first few values of $n$ that you believed the pattern might continue.


TITLE: Prove that the total number of subgroups of a finite cyclic group $G=\langle a \rangle$ such that $o(G)=n$ is the number of divisors of $n$.
QUESTION [0 upvotes]: Prove that the total number of subgroups of a finite cyclic group $G=\langle a \rangle$ such that $o(G)=n$ is the number of divisors of $n$.

If $G=\langle a \rangle $ is a finite cyclic group such that $o(G)=n$, then any subgroup of $G$ has the form $S=\langle g^d\rangle$ for a divisor $d\mid n$. Different $d$ give different subgroup. But how to give a concrete proof of the problem.
Please help me to solve the problem using the elementary tools specifically using Lagrange's theorem.

REPLY [1 votes]: I'm not sure how you could prove this without characterizing what the subgroups of a cyclic group look like.
Your result follows directly from the fact that a (finite) cyclic group (of order $n$) has a unique subgroup of order $k$ for each divisor $k$ of $n$. [so the number of divisors = the number of distinct subgroups.]
To prove this fact about uniqueness, typically you'll need to show:
(1) Every subgroup of a cyclic group is cyclic. [Sketch of proof: Suppose $G$ is generated by $g$. Let $H$ be a subgroup. For $H$ non-trivial, there is a unique smallest positive power $k$ of $g$ such that $g^k \in H$. Then show $H=\langle g^k \rangle$.]
(2) Given any subgroup, $H=\langle g^k \rangle$, show that $H= \langle g^k \rangle = \langle g^d \rangle$ where $d=\mathrm{gcd}(k,n)$. [Use the fact that $d=kx+ny$ for some integers $x,y$ so get this.]
(3) Note that if $k$ divides $n$, $\langle g^{n/k} \rangle$ is a subgroup of order $k$. So there is a subgroup for each divisor.
(4) Let $H$ be a subgroup of order $k$. Well, $H=\langle g^d \rangle$ for some divisor $d$ (by (1) and (2) put together). But $|H|=|\langle g^d \rangle|=n/d = k$ so $d=n/k$. This means that the order of $H$ uniquely determines $d$. Thus any two subgroups of order $k$ must be equal to the same $\langle g^d \rangle$. Thus there is only one subgroup for each divisor $k$ (namely $\langle g^{n/k} \rangle$).


TITLE: Need help understanding Chapter 2 of Joseph J. Rotman's: An Introduction to Theory of Groups
QUESTION [2 upvotes]: I am trying to self learn Group Theory and I'm starting with this book. On page 20, start of Chapter 2, the book immediately introduces a subgroup as:

 Definition. A nonempty subset of S of a group G is a  subgroup of G if $s \in G$ implies $s^{-1} \in G$ and $s,t \in G$ imply $st \in G$.

On the same page, for the proof of Theorem 2.1: Every subgroup is a group in its own right, the book starts with:

The hypothesis '$s,t \in S$ imply $st \in S$' shows that $S$...

Is the former line just a typo, and if so, is it a known typo? Or am I missing something? I always thought a subgroup was just a subset that satisfied the four group laws, am I mistaken?
I can post photos of my textbook if needed, I do not have an online copy, only a hard cover copy.

REPLY [1 votes]: Is the former line just a typo, and if so, is it a known typo? Or am I missing something?

The use of $G$ should be $S$. Yes, it is known.

I always thought a subgroup was just a subset that satisfied the four group laws, am I mistaken?

There are multiple ways for a subset of a group to be a subgroup. Consider, for example,

Theorem (One-step Subgroup Test): Suppose $S$ is a nonempty subset of a group $G$. Then TFAE:

$a,b\in S$ implies $ab^{-1}\in S$.
$S$ is a subgroup of $G$.