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\begin{document}
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\begin{abstract} Fix $a \in \Z$, $a\notin \{0,\pm 1\}$. A simple argument shows that for each $\epsilon > 0$, and almost all (asymptotically 100\% of) primes $p$, the multiplicative order of $a$ modulo $p$ exceeds $p^{\frac12-\epsilon}$. It is an open problem to show the same result with $\frac12$ replaced by any larger constant. We show that if $a,b$ are multiplicatively independent, then for almost all primes $p$, one of $a,b,ab, a^2b, ab^2$ has order exceeding $p^{\frac{1}{2}+\frac{1}{30}}$. The same method allows one to produce, for each $\epsilon > 0$, explicit finite sets $\Aa$ with the property that for almost all primes $p$, some element of $\Aa$ has order exceeding $p^{1-\epsilon}$. Similar results hold for orders modulo general integers $n$ rather than primes $p$.
\end{abstract}

\maketitle

\section{Introduction}
Let $\ell_a(p)$ denote the multiplicative order of the integer $a$ modulo the prime number $p$. A celebrated 1927 conjecture of Artin (\textsf{Artin's primitive root conjecture})  asserts that if $a \in \Z$, with $a$ not a square and $a\ne -1$, then there are infinitely many primes $p$ for which $\ell_a(p) = p-1$. Artin's conjecture remains unresolved, but there has been significant progress. Hooley showed in 1967 that the conjecture is implied by the Riemann Hypothesis for a certain class of Dedekind zeta functions \cite{hooley67}, while Heath-Brown \cite{HB86} (building on earlier work of Gupta and Murty \cite{GM84}) showed in 1986 that Artin's conjecture holds for all prime values of $a$ with at most two exceptions.

Hooley's ideas in \cite{hooley67} have several other nice consequences for the distribution of the numbers $\ell_a(p)$. Of interest to us is the following example: Assume the same Generalized Riemann Hypothesis as in \cite{hooley67}, and fix an integer $a\notin \{0,\pm 1\}$. If $\psi(x)$ is any function of $x$ tending to infinity as $x\to\infty$, then $\ell_a(p) > p/\psi(p)$ for almost all primes $p\le x$, meaning all  $p\le x$ with at most $o(x/\log{x})$ exceptions. (See \cite[Theorem 23]{KP05} for a quantitatively precise statement along these lines.) Thus, loosely speaking, $a$ is `nearly' a primitive root mod $p$ for almost all primes $p$.

The known unconditional results in this direction are considerably weaker. The following easy observation is implicit in \cite{hooley67} (see p.\ 212 there). Fix an integer $a\notin \{0,\pm 1\}$. Then for each $y\ge 1$, the product $\prod_{n \le y} (a^n-1)$ has absolute value $\exp(O(y^2))$, and so has $O(y^2)$ distinct prime factors. But that product is divisible by all primes $p$ with $\ell_a(p) \le y$. Taking $y=x^{1/2}/\log{x}$, we deduce that all but $o(x/\log{x})$ primes $p\le x$  have $$\ell_a(p) > x^{1/2}/\log{x} > p^{1/2}/\log{p}.$$

In \cite{erdos76}, Erd\H{o}s proves by a substantially more intricate argument that $\ell_a(p) > p^{1/2}$ for almost all primes $p$. At the end of the same article, he claims that if $\epsilon(x)$ is any positive-valued function tending to $0$ as $x\to\infty$, then \begin{equation}\label{eq:epsilonimprovement} \ell_a(p) > p^{\frac{1}{2}+\epsilon(p)} \end{equation} for almost all primes $p$. The proof appeared 20 years later in a joint a paper with Ram Murty \cite{EM99}.  It seems that \eqref{eq:epsilonimprovement} still holds the record as far as a lower bound for $\ell_a(p)$ that holds almost always, and that a new idea will be required to replace  $\frac{1}{2}$ with $\frac{1}{2}+\delta$ for some $\delta\gg 1$. The purpose of this article is to observe that, by considering simultaneously multiple values of $a$, one can overcome this barrier.

In our first theorem, we produce sets of $5$ integers, at least one of which almost always has order
$$ \gtrapprox p^{\frac{8}{15}} = p^{\frac{1}{2} + \frac{1}{30}}. $$ Recall that $a_1,\dots,a_k \in \Q^{\times}$ are called \textsf{multiplicatively independent} if whenever $a_1^{e_1} \cdots a_k^{e_k}=1$ with integers $e_1,\dots,e_k$, we have $e_1=e_2=\dots=e_k=0$.

\begin{thm}\label{thm:basic} Let $a$ and $b$ be nonzero integers that are multiplicatively independent. For almost all primes $p$, at least one of $a$, $b$, $ab$, $a^2 b$, and $ab^2$ has multiplicative order exceeding
\begin{equation}\label{eq:basically815} p^{8/15}/\exp(2\sqrt{\log{p}}). \end{equation}
\end{thm}

It would be easy to prove Theorem \ref{thm:basic} with a somewhat smaller quantity in the denominator of \eqref{eq:basically815}, but this is of no importance. Fix $\delta \in (0,1)$. Then as $\epsilon\to 0$, the lower density of primes $p$ (relative to the full set of primes) for which
\begin{equation*}
\text{$p-1$ has no divisor in the interval $[p^{\delta-\epsilon}, p^{\delta+\epsilon}]$}
\end{equation*}
tends to $1$. (This is essentially Erd\H{o}s and Murty's Theorem 2 in \cite{EM99}. It also follows from Proposition \ref{prop:ford} below.) From this, we deduce immediately that Theorem \ref{thm:basic}, with any function of size $p^{o(1)}$ in the denominator of \eqref{eq:basically815}, can be bootstrapped to yield the following result.

\begin{manualtheorem}{1$'$}\label{thm:basic2} Let $\epsilon(x)$ be a positive-valued function of $x\ge 2$ that tends to $0$ as $x\to\infty$. Let $a$ and $b$ be nonzero integers that are multiplicatively independent. Then for almost all primes $p$, at least one of $a$, $b$, $ab$, $a^2 b$, and $ab^2$ has multiplicative order exceeding
\begin{equation*} p^{\frac{8}{15}+\epsilon(p)}. \end{equation*}
\end{manualtheorem}

Matthews proved in 1982 that if $a$ and $b$ are multiplicatively independent integers, then the subgroup of ${\F_p}^{\times}$ generated by $a$ and $b$ has size at least $p^{2/3}/\log{p}$ for almost all primes $p$. Of course, $\frac{2}{3} > \frac{8}{15}$. What is novel about Theorem \ref{thm:basic} (and Theorem \ref{thm:basic2}) is that we can pinpoint an \emph{explicit, finite} subset of  $\langle a,b\rangle$ one of whose elements almost always generates a subgroup of size substantially larger than $p^{1/2}$.

In fact, Matthews's result implies that if $a_1,\dots,a_k$ is any finite list of multiplicatively independent integers, then $\langle a_1,\dots,a_k\rangle$  almost always has size at least $p^{1-\frac{1}{k+1}}/\log{p}$. We prove the following.

\begin{thm}\label{thm:general} Let $a_1,\dots,a_k$ be nonzero, multiplicatively independent integers. Let $N$ be a positive integer, and let
\[ \Aa = \{a_1^{e_1} a_2^{e_2} \cdots a_k^{e_k}: \text{each }0 \le e_i < N, \text{not all $e_i=0$}\}. \]
For almost all primes $p$, there is an $a\in \Aa$ with
\[ \ell_a(p) > p^{\delta}, \quad\text{where}\quad \delta = \left(1-\frac{1}{k+1}\right)\left(1-\frac{1}{N}\right). \]
\end{thm}

\noindent We highlight two immediate consequences of Theorem \ref{thm:general}:
\begin{enumerate}\item[(a)] For each $\epsilon \in (0,1)$, there are finite sets $\Aa$ of size $\exp(O(\frac{1}{\epsilon}\log\frac{1}{\epsilon}))$ such that, for almost all primes $p$, some $a \in \Aa$ has order exceeding $p^{1-\epsilon}$.
\item[(b)] Let $\xi(x)$ be any function that tends to infinity as $x\to\infty$. For each fixed $\epsilon > 0$, and almost all primes $p$, we have $L(p) > p^{1-\epsilon}$, where $L(p)$ is the maximum order mod $p$ of any of $1,2,3,\dots,\lfloor \xi(p)\rfloor$.
\end{enumerate}

Ska\l{}ba has conjectured that almost all primes $p$ have a positive multiple of the form $2^m+2^n+1$, with positive integers $m,n$. He proved this assuming $\ell_2(p) > p^{0.8}$ for almost all primes $p$ \cite{skalba04}. (Thus the conjecture becomes a theorem if we assume GRH.) Elsholtz has shown, unconditionally, that Ska\l{b}a's  conjecture holds with $2^m + 2^n+1$ replaced by $2^{m_1} + 2^{m_2} + \dots + 2^{m_6} + 1$ \cite{elsholtz16}. Our proof of Theorem \ref{thm:general} yields another unconditional variant of Ska\l{b}a's conjecture.

\begin{cor}\label{cor:skalbavariant} Let $A = (2\cdot 3\cdot 5\cdot 7\cdot 11)^{9}$. For almost all primes $p$, there are positive integers $m,n$, such that $p$ divides $\prod_{a \mid A,\,a>1} (a^m + a^n + 1)$.
\end{cor}

Our method also has implications for orders modulo $n$ when $n$ is composite. For each integer $a$ and each positive integer $n$, the sequence $a, a^2, a^3, \dots$ is eventually periodic modulo $n$. Let $\ell_a(n)$ denote the length of the period. When $a$ and $n$ are coprime, $\ell_a(n)$ is the order of $a$ modulo $n$, while in general, $\ell_a(n)$ is the order of $a$ modulo $n'$, where $n'$ is the largest positive divisor of $n$ coprime to $a$. We show the following analogue of Theorem 1$'$.

\begin{thm}\label{thm:composite} Let $\epsilon(x)$ be a positive-valued function of $x\ge 1$ that tends to $0$ as $x\to\infty$. Let $a$ and $b$ be nonzero integers that are multiplicatively independent. Then for almost all natural numbers $n$ {\rm (}meaning all but $o(x)$ integers $n\le x$, as $x\to\infty${\rm )},
\begin{equation*} \max\{\ell_a(n), \ell_b(n), \ell_{ab}(n), \ell_{a^2 b}(n), \ell_{ab^2}(n)\} > n^{\frac{8}{15}+\epsilon(n)}. \end{equation*}
\end{thm}

The obvious analogue of Theorem \ref{thm:general} also holds; see
 the remarks following our proof of Theorem \ref{thm:composite} for a somewhat more precise result.

\subsection*{Notation} Throughout, the letters $p$ and $q$ are reserved for primes.

\section{Elements of order $\gtrapprox p^{8/15}$: Proof of Theorem \ref{thm:basic}}

For a prime $p$ and integers $a_1,\dots,a_k$ not divisible by $p$, we let $\ell_{a_1,\dots,a_k}(p)$ denote the order of the subgroup of $\F_p^{\times}$ generated by $a_1,\dots,a_k$. The following lemma is a special case of a result of Matthews \cite{matthews82} (alluded to in the introduction), and also of Lemma 1 of Murty and Srinivasan's paper \cite{MS87}.

\begin{lem}\label{lem:matthews} Let $a_1, \dots, a_k$ be nonzero, multiplicatively independent integers. There is a positive constant $C = C_{a_1,\dots,a_k}$ such that, for every $y \ge 1$, the number of primes $p$ with  $\ell_{a_1,\dots,a_k}(p) \le y$ does not exceed $C y^{1+1/k}$.
\end{lem}

Recall that a number $n$ is said to be \textsf{$z$-smooth} (or \textsf{$z$-friable}) if none of its prime factors exceed $z$. The next estimate appears as Theorem 07 on p.\ 4 of Hall and Tenenbaum's monograph \cite{HT88}.

\begin{lem}\label{lem:HT} For certain positive constants $c_1, c_2$ and all real numbers $x\ge y\ge z \ge 2$,  the number of $n\le x$ having a $z$-smooth divisor of size least $y$ is at most
\[ c_1 x \exp(-c_2 \log{y}/\log{z}). \]
\end{lem}

As our final piece of preparation, we recall an elementary fact from group theory that plays a key role in the argument.
\begin{lem}\label{lem:grouplem} Let $G$ be a finite abelian group, and let $g,h \in G$. If $q$ is a prime dividing the order of $g$ but not the order of $h$, then $q$ divides the order of $gh$.
\end{lem}
\begin{proof}[Proof (sketch)] If $g$ and $h$ are commuting elements of a group, with respective orders $m$ and $n$, it is well-known that the order of $gh$ is a multiple of $mn/\gcd(m,n)^2$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:basic}] We will show that for all large $x$, all but $o(x/\log{x})$ primes $p \le x$ are such that one of $a, b, ab, a^2 b$, or $ab^2$ has order exceeding $x^{8/15}/\exp(2\sqrt{\log x})$. For notational simplicity, set
\[ \xi = \log\log{x}. \]

At the cost of discarding $o(x/\log{x})$ `bad' primes $p\le x$, we can assume all of the following conditions hold:
\begin{enumerate}
\item the $\xi$-smooth part (largest $\xi$-smooth divisor) of $p-1$  is at most $\exp(\sqrt{\log x})$,
\item the $\xi$-rough part (the cofactor of the $\xi$-smooth part) of $p-1$  is squarefree,
\item $\ell_a(p), \ell_b(p) > x^{1/2}/\log{x}$, and $\ell_{a,b}(p) > x^{2/3}/\log{x}$,
\end{enumerate}
Indeed, Lemma \ref{lem:HT} implies that the total number of $m\le x$ with $\xi$-smooth part exceeding $\exp(\sqrt{\log x})$ is $o(x/\log{x})$. So the same upper bound certainly holds for the number of these $m$ of the form $p-1$, which handles condition (i). Keeping in mind the Brun--Titchmarsh inequality, we see that the number of primes excluded by (ii) is at most
\begin{align*} \sum_{q > \xi} \pi(x;q^2,1) &\le \sum_{\xi < q \le x^{1/4}} \pi(x;q^2,1) + \sum_{q > x^{1/4}} \frac{x}{q^2} \\
&\ll \frac{x}{\log{x}}\sum_{\xi < q \le x^{1/4}} \frac{1}{q^2} + x^{3/4} \ll \frac{x}{\xi \log{x}};   \end{align*}
this is also $o(x/\log{x})$. That (iii) excludes $o(x/\log{x})$ primes is immediate from Lemma \ref{lem:matthews}.

In the remainder of the proof, we study the prime factorization of the product $$\ell_a(p) \ell_b(p) \ell_{ab}(p) \ell_{a^2 b}(p) \ell_{ab^2}(p).$$

Let $q$ be a prime dividing $p-1$ with $q > \xi$. Then $q\parallel p-1$ (by condition (ii) above). Since the unit group mod $p$ is cyclic, it follows that for each integer $g$ coprime to $p$, the order $\ell_g(p)$ is divisible by $q$ precisely when $g$ is not a $q$th power mod $p$.

Let us use this observation to show that every prime $q>\xi$ dividing $\ell_a(p) \ell_b(p)$ divides at least four of the numbers $\ell_a(p)$, $\ell_b(p)$, $\ell_{ab}(p)$, $\ell_{ab^2}(p)$, $\ell_{a^2b}(p)$. If $q$ divides exactly one of $\ell_a(p)$ and $\ell_b(p)$, then $q$ divides all of $\ell_{ab}(p)$, $\ell_{ab^2}(p)$, $\ell_{a^2b}(p)$ by Lemma \ref{lem:grouplem}. (We use here that $q>2$, which is guaranteed since $x$ is large and $q > \xi$.) Thus the claim holds in this case. So suppose that $q$ divides both $\ell_a(p)$ and $\ell_b(p)$. In that case, $q$ must divide at least two of $\ell_{ab}(p)$, $\ell_{ab^2}(p)$, $\ell_{a^2b}(p)$. Otherwise, at least two of $ab, ab^2$, and $a^2 b$ are $q$th powers mod $p$. But then $a,b$ themselves are $q$th powers mod $p$, contradicting that $q\mid \ell_a(p)$ and $q\mid \ell_b(p)$. (We use here that $q>3$.) So the claim holds in this  case also.

Comparing prime factorizations, we deduce that \[ (\ell_a(p) \ell_b(p) \ell_{ab}(p) \ell_{a^2 b}(p) \ell_{ab^2}(p))^{1/4} \ge \prod_{\substack{q > \xi \\ q\mid \ell_a(p) \ell_b(p)}} q.  \]
Since the $\xi$-rough part of $p-1$ is squarefree, and $\ell_a(p), \ell_p(b)$ divide $p-1$, the right-hand product is the $\xi$-rough part of $\mathrm{lcm}[\ell_a(p),\ell_b(p)] = \ell_{a,b}(p)$, which by (i) and (iii) has size at least
\[ \ell_{a,b}(p)/\exp(\sqrt{\log x}) > x^{2/3}/\exp(2\sqrt{\log x}). \]
This, along with the preceding display, implies that the geometric mean of $\ell_a(p)$, $\ell_b(p)$, $\ell_{ab}(p)$, $\ell_{a^2 b}(p)$ and $\ell_{ab^2}(p)$ is at least $x^{8/15}/\exp(\frac{8}{5}\sqrt{\log x})$. The theorem follows.
\end{proof}

\begin{proof}[Proof of Theorem 1.2] The proof is very similar to that of Theorem \ref{thm:basic}. We keep the notation $\xi=\log\log{x}$. We impose conditions (i) and (ii) from the proof of Theorem \ref{thm:basic}, but replace (iii) with
\begin{enumerate}
\item[(iii$'$)] $\ell_{a_i}(p) > x^{1/2}/\log{x}$ for all $i$, and $\ell_{a_1,\dots,a_k}(p) > x^{k/(k+1)}/\log{x}$.
\end{enumerate}
Conditions (i), (ii), and (iii$'$)  exclude only $o(x/\log{x})$ primes $p\le x$, as $x\to\infty$.

Let $p\le x$ be one of the surviving primes, and let $q>\xi$ be a prime dividing $\ell_{a_1}(p) \cdots \ell_{a_k}(p)$. We claim that $q$ divides $\ell_{a}(p)$ for all but most $N^{k-1}-1$ elements $a \in \Aa$. To see this, let $g$ be a primitive root mod $p$, and write $\log_g(\cdot)$ for the discrete logarithm mod $p$ to the base $g$, which is well-defined (on inputs prime to $p$) as an integer modulo $p-1$. In order for $q$ not to divide $\ell_a(p)$, it must be that $q$ divides $\log_g(a)$. So if we write $a=a_1^{e_1}\cdots a_k^{e_k}$, the number of $a \in \Aa$ for which $q$ does not divide $\ell_a(p)$ is bounded above by the number of $e_1,\dots, e_k$ satisfying the congruence
\begin{equation}\label{eq:boxcongruence} e_1 \log_g(a_1) + \dots + e_k \log_g(a_k) \equiv 0 \pmod{q}, \end{equation}
where $0 \le e_i < N$ for all $i$ and not all $e_i=0$.

By assumption, $q\mid \ell_{a_i}(p)$ for some $i$, and so $q\nmid \log_{g}{(a_i)}$. Relabeling, we can suppose $i=1$. Thus, if we pick $e_j$ arbitrarily for $j=2,\dots,k$, then \eqref{eq:boxcongruence} determines $e_1$ mod $q$. We are assuming $x$ is large, and so $q > \xi > N$. It follows that for any choice of $e_2,\dots,e_k$, there is at most one $e_1 \in [0,N)$ satisfying \eqref{eq:boxcongruence}. Hence, the number of solutions to \eqref{eq:boxcongruence} satisfying our constrains on the $e_i$ is at most $N^{k-1}-1$. Here the `$-1$' comes from noticing that the choice $e_2=e_3=\dots=e_k=0$ forces $e_1=0$, while $e_1=e_2=\dots=e_k=0$ is not allowed.

Since $|\Aa| = N^k-1$, we conclude that each prime $q>\xi$ dividing $\ell_{a_1}(p) \cdots \ell_{a_k}(p)$ divides $\ell_{a}(p)$ for at least $|\Aa| - (N^{k-1}-1) = N^k-N^{k-1}$ choices of $a\in \Aa$. Now arguing as in the proof of Theorem \ref{thm:basic}, we find that as $x\to\infty$,
\begin{align*} \left(\prod_{a \in\Aa} \ell_a(p)\right)^{\frac{1}{N^k-N^{k-1}}} &\ge \prod_{\substack{q > \xi\\ q\mid \ell_{a_1}(p) \cdots \ell_{a_k}(p)}} q \ge \ell_{a_1,\dots,a_k}(p)/\exp(\sqrt{\log x})
\ge x^{1-\frac{1}{k+1} + o(1)},\end{align*}
so that
\[ \left(\prod_{a \in\Aa} \ell_a(p)\right)^{1/|\Aa|} \ge x^{\delta' + o(1)},  \]
where
\begin{equation}\label{eq:deltaprimedef} \delta':= \left(1-\frac{1}{k+1}\right) \frac{N^k-N^{k-1}}{|\Aa|}. \end{equation}
Since $\delta' > \left(1-\frac{1}{k+1}\right) \frac{N^k-N^{k-1}}{N^k} = \left(1-\frac{1}{k+1}\right) \left(1-\frac{1}{N}\right) =\delta$, the exponent on $x$ in the last display exceeds $\delta$ once $x$ is sufficiently large. The theorem follows.
\end{proof}

\section{Primes dividing $\prod_{a\mid A}(a^m+a^n+1)$: Proof of Corollary \ref{cor:skalbavariant}}

The following lemma is due to Ska\l{}ba \cite{skalba04}.

\begin{lem}\label{lem:skalba} let $p$ be a prime, and let $a$ be an integer not divisible by $p$. Suppose that $\ell_a(p) > p^{3/4}$. Then $p$ divides $a^m + a^n+1$ for some positive integers $m,n$.
\end{lem}
\begin{proof} We include the proof for completeness. Note that the hypothesis $\ell_a(p) > p^{3/4}$ implies that $p$ is odd. Let $d = (p-1)/\ell_a(p)$, so that $d < p^{1/4}$. If $d=1$, then $g$ generates $\F_p^{\times}$, and the result is easy: We choose $m$ with $a^m = -1/2$ in $\F_p$ and then take $n=m$. So suppose that $d>1$. The subgroup of ${\F_p}^{\times}$ generated by $a$ coincides with the collection of nonzero $d$th powers in $\F_p$. By Theorem 5 on p.\, 103 of \cite{IR90}, the number of solutions $(x,y)$ to $x^d+y^d=-1$ over $\F_p$ is at least $p-(d-1)^2\sqrt{p}$, and so the number of solutions with $x,y\ne 0$ is at least
\[ p-(d-1)^2\sqrt{p} - 2d > p - d^2 \sqrt{p} > 0. \qedhere\]
\end{proof}

\begin{proof}[Proof of Corollary \ref{cor:skalbavariant}] We keep the notation from the statement and proof of Theorem \ref{thm:general}. We follow the proof of that theorem with $k=5$, with $a_1=2, a_2 = 3, a_3 = 5, a_4 = 7, a_5=11$, and  with $N=10$. That argument shows that for each fixed $\epsilon > 0$, and almost all primes $p$, there is a divisor $a>1$ of $A = (2\cdot3 \cdot 5 \cdot 7\cdot 11)^{9}$ such that
\[ \ell_a(p) \ge p^{\delta'- \epsilon}. \]
Since $\delta' > \delta$, and $\delta = (1-\frac{1}{6})(1-\frac{1}{10})= \frac{3}{4}$, we may complete the proof by fixing $\epsilon$ sufficiently small and applying Lemma \ref{lem:skalba}.
\end{proof}

\begin{rmk} In contrast with Corollary \ref{cor:skalbavariant}, for every positive integer $A$, there is a positive density set of primes $p$ not dividing $\prod_{a \mid A}{(a^n+1)}$ for any integer $n$. Indeed, using quadratic reciprocity one can construct a coprime residue class such that for each  $p$ lying in this class, every prime dividing $A$ is a square mod $p$ but $-1$ is not a square mod $p$. Then the exponential congruence $a^n \equiv -1\pmod{p}$ is not solvable for any $a \mid A$.
\end{rmk}

\section{Composite integers: Proof of Theorem \ref{thm:composite}}

Our argument is modeled on Kurlberg and Pomerance's proof of Theorem 1(a) in \cite{KP05}, which asserts that for each fixed $a\notin \{0,\pm 1\}$, we have $\ell_a(n) > n^{1/2+\epsilon(n)}$ for almost all $n$.

We make crucial use of the following estimate of Kurlberg and Rudnick \cite[\S5]{KR01} (see \cite[Lemma 5]{KP05} for a shorter proof), bounding $\ell_{a}(n)$ from below in terms of the numbers $\ell_a(p)$ for primes $p$ dividing $n$. Here $\lambda(n)$ is \textsf{Carmichael's function}, i.e., the exponent of the multiplicative group mod $n$.

\begin{prop}\label{prop:KR} For each nonzero integer $a$ and each positive integer $n$,
\[ \ell_a(n) \ge \frac{\lambda(n)}{n}\prod_{p\mid n,~p\nmid a} \ell_a(p). \]
\end{prop}

\noindent In our application of Proposition \ref{prop:KR}, we will replace $\frac{\lambda(n)}{n}$ with the lower bound from the  next result, which appears as Lemma 10 of \cite{KP05}.

\begin{lem}\label{lem:lambda} For all large $x$, all but $O(x/(\log{x})^{10})$ values of $n\le x$ satisfy
\[ \lambda(n) > n \exp(-(\log\log{n})^3).\]
\end{lem}

We also use the following result of Ford concerning the number of shifted primes $p-1$ with a divisor from a given interval (see Theorems 1(v) and 6 in \cite{ford08}).

\begin{prop}\label{prop:ford} Suppose $x, y \ge 10^5$ with $y \le \sqrt{x}$ and $2y\le z \le y^2$. Write $z=y^{1+u}$. The proportion of primes not exceeding $x$ for which $p-1$ has a divisor from the interval $(y,z]$ is
\[ \ll u^{\eta} \left(\log\frac{2}{u}\right)^{-3/2},\]
where
\[ \eta := 1-\frac{1+\log\log{2}}{\log{2}} \quad (\approx 0.086). \]
\end{prop}

We now embark on the proof proper of Theorem \ref{thm:composite}.
Replacing the function $\epsilon(t)$ with $\max_{n\ge t} \epsilon(n)$, we can assume that $\epsilon(t)$ is (weakly) decreasing for all $t\ge 2$. Then replacing $\epsilon(t)$ with $\max\{\epsilon(t), 1/\log\log\log{(100 t)}\}$, we can further assume that
\begin{equation}\label{eq:epsilonlower}
 \epsilon(t) \ge 1/\log\log\log(100t)\qquad\text{for all $t\ge 1$}.
\end{equation}

Throughout this proof, we view $a$, $b$, and the function $\epsilon(t)$ as fixed. In particular, when we speak of ``large'' parameters, their required size may depend on $a$, $b$, and $\epsilon(t)$.

Now let $x$ be a large real number, and let $\xi=\log\log{x}$ as before. For primes $p\le x$, we introduce conditions (i)--(iii) defined as follows:
\begin{enumerate}
\item[(i)] the $\xi$-smooth part of $p-1$ is at most $\exp(\sqrt{\log{p}})$,
\item[(ii)] the $\xi$-rough part of $p-1$ is squarefree,
\item[(iii)] $\ell_a(p), \ell_b(p) > p^{1/2}/\log{p}$ and $\ell_{a,b}(p) > p^{2/3}/\log{p}$.
\end{enumerate}
(These are slight variants of conditions (i)--(iii) appearing in the proof of Theorem \ref{thm:basic}.) We  partition the $p\le x$ into classes $U,V,W$, where
\[ U = \{p\le x: p \mid ab\text{ or } p \le \xi \text{ or at least one of (i)--(iii) fails}\}, \]
\[ V = \{p \le x: p \notin U \text{ and } p^{2/3}/\log{p} < \ell_{a,b}(p) \le p^{2/3+5\epsilon(p)}\}, \]
\[ W = \{p \le x: p\notin U \text{ and }\ell_{a,b}(p) > p^{2/3+5\epsilon(p)}\}. \]
We write $\pi_U(t), \pi_V(t)$, and $\pi_W(t)$ for the counts of primes $p\le t$ in the sets $U,V$, and $W$, respectively.

\begin{lem}\label{lem:piUbound} For all large $x$,
\[ \pi_U(t) \ll \frac{\log{x}}{\log\log{x}} + \frac{t}{(\log{t})^{3/2}} + \frac{t}{\log{t} \cdot \log\log{x}} \]
uniformly for $2\le t \le x$.
\end{lem}
\begin{proof} Let $t_0:=\log{x}$. If $t \le t_0$, then $\pi_U(t) \le \pi(t_0) \ll \log{x}/\log\log{x}$, and the lemma  holds. So we will assume that $t > t_0$. Now let $p$ be a prime in $U$ with $p > t_0$. Since $x$ is large, $p > t_0 > |ab|$, and so $p\nmid ab$. Also, $p>t_0 > \xi$. So it must be that one of (i)—(iii) fails.

Let count how many $p \in (t_0,t]$ are such that (i) fails. Since $p > t_0$, the prime $p$ belongs to some interval $(T_j, 2T_j]$, where $T_j = 2^j t_0$ and $j$ is a nonnegative integer. Since (i) fails, the $\xi$-smooth part of $p-1$  exceeds $\exp(\sqrt{\log p})$, which in turn exceeds $\exp(\sqrt{\log T_j})$. But the number of $p\le 2T_j$ for which this occurs is, by Lemma \ref{lem:HT},
\[ \ll 2T_j \exp(-c_2 \sqrt{\log T_j}/\log \xi) \ll T_j \exp(-c_2\sqrt{\log T_j}/\log\log\log x). \]
Since $T_j \ge t_0$, we have $\log\log{T_j} \ge \log\log\log{x}$, and so the last displayed expression is
\[ \ll T_j \exp(-c_2 \sqrt{\log T_j}/\log \log T_j) \ll T_j/(\log T_j)^2. \]
Now we sum on nonnegative integers $j\ge 0$, stopping once the intervals $(T_j,2T_j]$ cover $(t_0,t]$. This gives an upper bound of $O(t/(\log{t})^2)$ on the number of failures of (i).

Condition (ii) is easier to deal with. By Brun--Titchmarsh, the number of $p \in (t_0,t]$ for which (ii) fails is at most
\begin{align*} \sum_{q > \xi} \pi(t;q^2,1) &\le \sum_{\xi < q < t^{1/4}} \pi(t;q^2,1) + \sum_{q > t^{1/4}}\frac{t}{q^2}
\\ &\ll \frac{t}{\log{t}} \sum_{q > \xi} \frac{1}{q^2} + t^{3/4} \ll \frac{t}{\log{t} \cdot \log\log{x}} + t^{3/4}.  \end{align*}

Consider finally the $p \in (t_0,t]$ where (iii) fails. Again, say that $p \in (T_j, 2T_j]$, where $T_j = 2^j t_0$. Then either $\ell_a(p) \le (2T_j)^{1/2}/\log(2T_j)$, $\ell_b(p) \le (2T_j)^{1/2}/\log(2T_j)$, or $\ell_{a,b}(p) \le (2T_j)^{2/3}/\log(2T_j)$. By Lemma \ref{lem:matthews}, the number of $p$ satisfying any of these conditions is $O(T_j/\log(2T_j)^{3/2})$. Summing on $j$ shows that the number of such $p \in (t_0,t]$ is $O(t/(\log{t})^{3/2})$.

Collecting estimates, the number of $p \in (t_0,t]$ belonging to $U$ is
\[ \ll \frac{t}{(\log{t})^{3/2}} + \frac{t}{\log{t}\cdot\log\log{x}}.  \]
Since there  $O(\log{x}/\log\log{x})$ primes not exceeding $t_0$, the lemma follows.
\end{proof}

We now turn to estimating $\pi_V(t)$. Clearly, $\pi_V(t)=0$ if $t \le \xi$.

\begin{lem}\label{lem:piVbound} For all large $x$, and uniformly for $\xi < t\le x$,
\[ \pi_V(t) \ll \epsilon(t')^{1/12} \cdot \frac{t}{\log{t}}, \]
where $t'=t/\log{t}$.
\end{lem}

\begin{proof} Suppose that $p \in V$ with $t/\log{t} < p \le t$.

If $p^{2/3}/\log{p} < \ell_{a,b}(p) \le p^{2/3}$, then $d:=\frac{p-1}{\ell_{a,b}(p)}$ is a divisor of $p-1$ with
\[ t^{1/3}/\log{t} < \frac{1}{2}p^{1/3} \le d < p^{1/3}\log{p} \le t^{1/3}\log{t}. \]
By Proposition \ref{prop:ford}, the number of primes $p \le t$ for which $p-1$ has a divisor in the interval $(t^{1/3}/\log{t},t^{1/3}\log{t}]$ is $O(\pi(t) (\log{t})^{-\eta})$.

On the other hand, if
$p^{2/3} < \ell_{a,b}(p) \le p^{2/3+5\epsilon(p)}$, then $d:=\frac{p-1}{\ell_{a,b}(p)}$ satisfies
\[ \frac{1}{2} t'^{1/3 - 5\epsilon(t')} < \frac{1}{2} p^{1/3-5\epsilon(p)} \le d < p^{1/3} \le t^{1/3}, \]
where $t':=t/\log{t}$. We now apply Proposition \ref{prop:ford} with $y = \frac{1}{2} t'^{1/3 - 5\epsilon(t')}$ and $z=t^{1/3}$. A short calculation, keeping in mind \eqref{eq:epsilonlower}, shows that $z=y^{1+u}$ with $u \le 16\epsilon(t')$. Hence, the number of $p\le t$ for which $p-1$ has a divisor in $(y,z]$ is $O(\pi(t) \epsilon(t')^{\eta})$.

We conclude that the number of $p \le t$ belonging to $V$ is
\[ \ll \pi(t/\log{t}) + \pi(t) (\log{t})^{-\eta} + \pi(t) \epsilon(t')^{\eta}. \]
By \eqref{eq:epsilonlower}, the final summand dominates. The lemma follows upon noting that $\eta > \frac{1}{12}$.
\end{proof}

For each natural number $n\le x$, we let $n_U, n_V, n_W$ be the largest divisors of $n$ supported on the primes in $U, V, W$, respectively. Thus, $n=n_U n_V n_W$.

At the cost of excluding $o(x)$ values of $n\le x$, we can assume that $n/n_U$ is squarefree. Indeed, since $U$ contains all primes up to $\xi$, if this condition fails then $n$ is divisible by $p^2$ for some $p > \xi$, and the number of such $n$ is $O(x\sum_{p>\xi} p^{-2}) = O(x/\xi)$, which is $o(x)$.

Next, we use a first-moment argument to show we can assume, with $o(x)$ exceptions, \[ n_U \le \exp(\log{x}/(\log\log{x})^{1/2}).\]
With $\Lambda(d)$ the von Mangoldt function,
\[ \sum_{n \le x} \log n_U = \sum_{n \le x} \sum_{d \mid n_U} \Lambda(d) = \sum_{d=p^k,~p \in U} \log{p} \sum_{\substack{n \le x \\ d \mid n}}1 \le x \sum_{p \in U} \frac{\log{p}}{p} + O(x). \]
Partial summation, together with the estimate of Lemma \ref{lem:piUbound} for $\pi_U(t)$, shows that
\[ \sum_{p \in U} \frac{\log{p}}{p} \ll \frac{\log{x}}{\log\log{x}}, \]
and thus $\sum_{n\le x} \log n_U \ll x \log{x}/\log\log{x}$. Consequently, the number of $n\le x$ for which $\log n_U > \log{x}/(\log\log{x})^{1/2}$ is $O(x/\sqrt{\log\log{x}})$, which is $o(x)$.

Lemma \ref{lem:piVbound} implies, via partial summation, that $\sum_{p \in V} \frac{\log{p}}{p} = o(\log{x})$, as $x\to\infty$. So by an argument analogous to that in the last paragraph, we have
\[ n_V \le x^{1/10} \]
for all but $o(x)$ values of $n\le x$.

We will also assume that
\[ n \ge x^{1/2}, \]
that
\[ \lambda(n) \ge n \exp(-(\log\log{x})^3), \]
and that
\[ \omega(n) \le 2\log\log{x}.\]
That the first condition excludes only $o(x)$ values of $n\le x$ is clear. That the same holds for the second condition follows from Lemma \ref{lem:lambda}. That the third condition admits only $o(x)$ exceptions is a consequence of as well-known theorem of Hardy and Ramanujan.

Since $n\ge x^{1/2}$, while $n_U n_V \le x^{1/5}$ (say) once $x$ is large, our assumptions imply that
\[ n_W = \frac{n}{n_U n_V} \ge x^{3/10} \]

Suppose that $x$ is large, and that the natural number $n\le x$ is not in any of the exceptional sets indicated so far.  By Proposition \ref{prop:KR},
\[ \ell_a(n)\ell_b(n)\ell_{ab}(n) \ell_{a^2 b}(n) \ell_{ab^2}(n) \ge \exp(-5(\log\log{x})^3) \prod_{p \mid n_V n_W} \ell_a(p)\ell_b(p)\ell_{ab}(p) \ell_{a^2 b}(p) \ell_{ab^2}(p). \]
The argument used in the proof of Theorem \ref{thm:basic}, now using our modified conditions (i)--(iii), shows that for each $p$ dividing $n_V$,
\[ (\ell_a(p)\ell_b(p)\ell_{ab}(p) \ell_{a^2 b}(p) \ell_{ab^2}(p))^{1/4} \ge \ell_{a,b}(p)/\exp(\sqrt{\log{p}}), \]
and thus (for large $x$)
\[ \ell_a(p)\ell_b(p)\ell_{ab}(p) \ell_{a^2 b}(p) \ell_{ab^2}(p) \ge p^{8/3}/\exp(5\sqrt{\log{p}}).  \]
Using the better lower bound for $\ell_{a,b}(p)$ available for $p$ in $W$, the same argument shows that for each $p$ dividing $n_W$,
\[ \ell_a(p)\ell_b(p)\ell_{ab}(p) \ell_{a^2 b}(p) \ell_{ab^2}(p) \ge p^{8/3 + 20\epsilon(p)}/\exp(4\sqrt{\log p}).\]
Substituting back above the results of the last two displays, and using that $\epsilon(p) \ge \epsilon(n)$ for each $p$ dividing $n$, we find that
\begin{multline*} \ell_a(n)\ell_b(n)\ell_{ab}(n) \ell_{a^2 b}(n) \ell_{ab^2}(n) \ge \exp(-5(\log\log{x})^3) \cdot \exp\bigg(-5\sum_{p\mid n_V n_W} \sqrt{\log{p}}\bigg) \\
\times \prod_{p \mid n_V n_W} p^{8/3}  \prod_{p \mid n_W} p^{20\epsilon(n)}.  \end{multline*}
Recall that $n_V n_W= n/n_U$ is squarefree. Hence,
\begin{align*} \prod_{p \mid n_V n_W} p^{8/3} \prod_{p \mid n_W} p^{20\epsilon(n)} &= (n_V n_W)^{8/3} n_W^{20 \epsilon(n)} \\
&= (n^{8/3}/n_U^{8/3}) \cdot {n_W}^{20\epsilon(n)}\\
&\ge n^{8/3} \exp(-3\log{x}/(\log\log{x})^{1/2}) \cdot x^{6\epsilon(n)} \\
&\ge n^{8/3 + 6\epsilon(n)} \exp(-3\log{x}/(\log\log{x})^{1/2}).
 \end{align*}
Moreover,
\[ \exp\bigg(-5\sum_{p \mid n_V n_W} \sqrt{\log{p}}\bigg) \ge \exp(-5\omega(n)\sqrt{\log{x}}) \ge \exp(-10\sqrt{\log x}\log\log{x}). \]

Putting together our estimates, we find that (for large $x$)
\[ \ell_a(n)\ell_b(n)\ell_{ab}(n) \ell_{a^2 b}(n) \ell_{ab^2}(n) \ge n^{8/3+5\epsilon(n)}\cdot (n^{\epsilon(n)}  \exp(-4\log{x}/(\log\log{x})^{1/2})). \]
Using that $n \ge x^{1/2}$ along with the lower bound \eqref{eq:epsilonlower}, we see that the parenthesized right-hand factor is larger than $1$ (for large $x$). So taking fifth roots, the geometric mean of $\ell_a(n)$, $\ell_b(n)$, $\ell_{ab}(n)$,  $\ell_{a^2 b}(n)$, and $\ell_{ab^2}(n)$ exceeds $n^{8/15 + \epsilon(n)}$.

\begin{rmk} An analogous argument will establish the following analogue of Theorem \ref{thm:general}. Let $\epsilon(x)$ be a positive-valued function of $x\ge 1$ that tends to $0$ as $x\to\infty$. Let $a_1,\dots,a_k$ be nonzero integers that are multiplicatively independent. Let $N$ be a positive integer, and let
\[ \Aa = \{a_1^{e_1} a_2^{e_2} \cdots a_k^{e_k}: \text{each }0 \le e_i < N, \text{not all $e_i=0$}\}. \]
For almost all $n$, there is an $a\in \Aa$ with
\[ \ell_a(n) > n^{\delta' + \epsilon(n)}, \]
where $\delta'$ is defined as in \eqref{eq:deltaprimedef}.
\end{rmk}

\bibliographystyle{amsplain}
\bibliography{nearprimitive}

\end{document}


TITLE: Frobenius method for linear second order differential equation
QUESTION [2 upvotes]: I am trying to solve $x^{2}y''+xy'-9y=0$ using Frobenius' method. Plugging in the derivatives of $y=x^{s}\sum ^{\infty }_{i=0}c_{i}x^{i}$ to the equation, I get, for the lowest power $x^s$, the indicial equation $s(s-1)c_{0}+sc_{0}-9c_{0}$ which yields $s=\pm3$.
I can also solve the equation given by the coefficients of $x^{s+1}$ and $x^{s+2}$ as giving $c_{1}=0$ and $c_{2}=0$.
But when I do the general term $x^{s+n}$ I get $(nc_{n})(2s+1)+n(n-1)c_{n}=0$, which does not specify $c_{n}$.
Without a recurrence relation, I do not know what is happening, nor how to proceed with the solution to the equation.
Why does a recurrence relation not manifest from the general term?

REPLY [0 votes]: If what you calculated is ok for $s$ which I call $r$, then a solution for $r=3$ is the following. Do the same for $r=-3$.


TITLE: Matrix calculation from an equation
QUESTION [0 upvotes]: I am trying to find matrix y from an equation
For example, I know (python)
>>> x = matrix([[5,1],[2,4]])
>>> y = matrix([[1,5],[3,3]])
>>> print x
[[5 1]
 [2 4]]
>>> print y
[[1 5]
 [3 3]]
>>> print x*y % 7
[[1 0]
 [0 1]]

Assuming, I don't know what y is. But I know xy = I mod7
How do I calculate y?

REPLY [0 votes]: Let
$$
y=\begin{pmatrix} a & b\\ c& d\end{pmatrix}.
$$
Then $XY$ sets up some equations:
\begin{align}
5a+c &\equiv 1\,\, (\operatorname{mod} 7)\\
5b+d &\equiv 0\,\, (\operatorname{mod} 7)\\
2a+4c &\equiv 0\,\, (\operatorname{mod} 7)\\
2b+4d &\equiv 1\,\, (\operatorname{mod} 7).
\end{align}
You can solve this like you would solve any system of equations, only you have to remember to take every thing $\operatorname{mod} 7$.  For example, take the second equation and multiply it by $4$ to get:
$$
20b+4d \equiv 0\,\, (\operatorname{mod} 7).
$$
But, $20\equiv 6\,\,(\operatorname{mod} 7)$. So, this becomes
$$
6b+4d \equiv 0\,\, (\operatorname{mod} 7).
$$
Then subtract fourth equation from this to get:
$$
4b \equiv -1\,\, (\operatorname{mod} 7).
$$
But, $-1\equiv 6\,\,(\operatorname{mod} 7)$.  From above $b\equiv 5\,\,(\operatorname{mod} 7)$.  Now substitute $b$ into those equations and just keep going.
I'm not sure what you are comfortable with, but the integers mod $7$ are a field. So, all this works.

REPLY [0 votes]: First you have to know how to do arithmetic $\pmod 7$.
For example:
adding: $4 + 5 \equiv 2 \pmod 7$
multiplying : $4 \cdot 5 \equiv 6 \pmod 7$
subtracting: $4 - 5 \equiv 6\pmod 7$
dividing: $4 / 5 \equiv 5 \pmod 7$
Most of those should be obvious, you might have to do some studying to learn how to do modular division (sometimes called modular inverse).
Then all you have to do is solve the matrix like you normally would, except substitute the modular arithmetic for the regular arithmetic that you are used to.  You just use basic Gaussian elimination.  When you multiply a row by a constant, multiply each element $\pmod 7$.  When you add 2 rows, add them using modular arithmetic.
It is not a coincidence that all matrix operations work if you use modular arithmetic instead of regular arithmetic.  They are actually fundamentally defined that way.


TITLE: For any $n \in \mathbb{Z^+}$ Not extis $P(x) \in \mathbb{R}[x]$ with coefficients in $B$ and all roots of $P(x)$ in $A$
QUESTION [0 upvotes]: Problem: Let $A=\{a_1,a_2,..,a_m\}$ and $B=\{b_1,b_2,...,b_p\}$ where $a_1,a_2,...,a_m,b_1,b_2,...,b_p \in \mathbb{R}$ Prove that ,
  the following statements is bad :
for any $n \in \mathbb{Z^+}$ extis $P(x) \in \mathbb{R}[x]$ with
  coefficients in $B$ and all roots of $P(x)$ in $A$

*Solution
Assume for any $n$ exits $t_1,t_2,...,t_m$ are nonegative integer numbers has sum is $n$ such that:
$P_n(x)=(x-a_1)^{t_1}(x-a_2)^{t_2}...(x-a_m)^{t_m}$ has coefficients in $B$.
Finding coefficient of $x^{n-1}$ and $x^{n-2}$ we have:
$S=-\sum_{i=1}t_ia_i^m$ and $T=\sum_{i=1}^m C_{t_1}^2a_i^2+\sum_{i<j}^mt_it_j$ and $S,T \in B$.
We have : $T=\sum_{i=1}^m C_{t_1}^2a_i^2+\sum_{i<j}^mt_it_j=\frac{1}{2}S^2-\frac{1}{2}\sum_{i=1}^m t_ia_i^2$
Let $m=\min \{a_i^2:a_i \in A\}$, then
$T \le \frac{S^2}{2}-\frac{\sum_{i=1}^mt_i}{2}m^2=\frac{S^2}{2}-\frac{n}{2}m^2$
Let $n \to +\infty$ we have a contraction.
I don' understand this solution because when $n \to +\infty$ then $S$ changes ,too

REPLY [1 votes]: Recall that for all $n$ you have $T,S \in B$. If $a=\min B$ and $b=\max B$, then you have
$$a \leq T \leq \frac{S^2}{2} - \frac{n}{2}m^2 \leq \frac{b^2}{2} - \frac{n}{2}m^2 \longrightarrow -\infty$$
so you have a contradiction.


TITLE: Two morphisms $f, g : M \to L$ are equal as long as they are equal under the limit $L$.
QUESTION [0 upvotes]: If $(L, (p_D)_{D \in \mathcal{D}})$ is a limit of the functor $F : \mathcal{D} \to \mathcal{C}$, two morphisms $f, g : M \to L$ are equal as long as for every object $D \in \mathcal{D}$, $p_D \circ f = p_D \circ g$.    Proof.  $f$ and $g$ are two factorizations of the cone $(M, (p_D \circ f)_{D \in \mathcal{D}})$ through the limit.

I am having trouble getting this proof.  $(M, \dots)$ is a cone so for all $d : D \to D'$ in $\mathcal{D}$, $Fd \circ (p_D \circ f) = p_D \circ f$ and if $(M, (p_D \circ g)_{D \in \mathcal{D}})$ is another cone on $F$ then there is unqiue $u : M \to M$ such that $p_D \circ f \circ u = p_D \circ g = p_D \circ f$.  I'm not seeing how to remove $p_D$ from the left of $f, g$.

REPLY [1 votes]: For any cone, $(M,(q_D)_{D\in\mathcal{D}})$ there is a unique arrow $k : M\to L$ where $(L,(p_D)_{D\in\mathcal{D}})$ is the limiting cone such that $q_D=p_D\circ k$ for all $D\in\mathcal{D}$. This is the universal property of the limit. If $q_D=p_D\circ f$ for $f: M \to L$, then $k = f$. Of course, by exactly the same logic, if $q_D=p_D\circ g$ for $g:M\to L$ then $k = g$. Since $p_D\circ f = p_D\circ g$ for all $D\in\mathcal{D}$, $(M,(p_D\circ f)_{D\in\mathcal{D}})$ and $(M,(p_D\circ g)_{D\in\mathcal{D}})$ are the same cone and induce the same unique arrow $M \to L$.
I don't really care for talking about (co)cones though. We can view a cone, $(M,(q_D)_{D\in\mathcal{D}})$ over the diagram $F$ as a natural transformation $q:\Delta M\to F$. The universal property of the limiting cone says given a natural transformation $\Delta M\to F$ there exists a unique arrow $M\to L$, that is there is a bijection $\varphi:\mathsf{Nat}(\Delta M,F)\cong\mathsf{Hom}(M,L)$. The constraint that $q_D=p_D\circ k$ where $k\in\mathsf{Hom}(M,L)$ is handled by requiring this bijection to be natural in $M$. In this case, the cones $f$ and $g$ induce are exactly $\varphi^{-1}(f)$ and $\varphi^{-1}(g)$, and the equality assumption is just that $\varphi^{-1}(f)=\varphi^{-1}(g)$, from which it clearly follows that $f=g$ by applying $\varphi$ to both sides.


text:
\begin{document}

\title[The Gauss map of minimal surfaces in $\s^2 \times \R$]{The Gauss map of minimal surfaces in $\s^2 \times \R$}

\author[Iury Domingos]{Iury Domingos$^{1,2}$}

\address{$^1$ Universit\'e de Lorraine \\ CNRS, IECL\\
F-54000 Nancy, France.}

\address{$^2$ Universidade Federal de Alagoas \\ Instituto de Matem\'{a}tica  \\ Campus A. C. Sim\~{o}es, BR 104 - Norte, Km 97, 57072-970.
Macei\'o - AL, Brazil.}

\email{domingos1@univ-lorraine.fr}

\thanks{
This work was partially conducted during a scholarship supported by the International Cooperation Program CAPES/COFECUB Foundation at the University of Lorraine; financed by CAPES -- Brazilian Federal Agency for Support and Evaluation of Graduate Education within the Ministry of Education of Brazil.}

\keywords{Minimal surface, Gauss map, conformal immersions, homogenous 3-manifolds}
\subjclass[2010]{Primary 53C42; Secondary 53A10, 53C30}

\begin{abstract}
In this work, we consider the model of $\S2R$ isometric to $\R^3\setminus \{0\}$,\break endowed with a metric conformally equivalent to the Euclidean metric of $\R^3$,
and we define a Gauss map for surfaces in this model likewise in the $3-$Euclidean space. We show as a main result that any two minimal conformal immersions in $\S2R$ with the same non-constant Gauss map differ by only two types of ambient isometries: either $f=(\id,T)$, where $T$ is a translation on $\R$, or $f=(\mathcal{A},T)$, where $\mathcal{A}$ denotes the antipodal map on $\s^2$. Moreover, if the Gauss map is singular, we show that it is necessarily constant, and then only vertical cylinders over geodesics of $\s^2$ in $\S2R$ appear with this assumption. We also study some particular cases, among them we prove that there is no minimal conformal immersion in $\S2R$ which the Gauss map is a non-constant anti-holomorphic map.
\end{abstract}

\maketitle

\section{Introduction}

A fundamental contribution to the theory of minimal surfaces in $\R^3$ is due to K. Weierstrass: Given $g:\Omega\subset\C\to\bar{\C}$ a meromorphic function and $\eta$ a holomorphic $1-$form defined on an open set $\Omega\subset\C$, such that whenever that $g$ has a pole of order $m$ at $z\in\Omega$ implies that $\eta$ has a zero of order $2m$ at $z\in\Omega$, the Weierstrass Representation states that
\[X(z)=X(z_0)+\re{\int_{z_0}^{z}\Big(\frac{1}{2}(1-g^2)\eta,\frac{i}{2}(1+g^2)\eta,g\eta\Big)}\]
define a conformal minimal surface in $\R^3$ with Gauss map $g$. In light of this, the relationship between the geometry of minimal surfaces and its Gauss map $g$ has been studied in the last century, and some Weierstrass-type Representations has been established for minimal surfaces in others $3-$dimensional ambient Riemannian manifolds.

In general, given an isometric (or conformal) immersion into some $3-$dimensional ambient Riemannian manifold, the first problem is to define a Gauss map of a such surface into the unit $2-$sphere $\s^2$, preserving some proprieties such as the Gauss map defined in $\R^3$, whenever possible. The second problem is how ``good'' is this definition, that is, if the Gauss map defined carries important informations about the surface.
However, the harmonicity condition of this map is not satisfied generally for an arbitrary ambient manifold.

For simply connected $3-$homogeneous manifolds with a $4-$dimensional isometry group, i.e., a Riemannian fibration of bundle curvature $\tau$ over a $2-$space form $\mathbb{M}^2_\kappa $ with sectional curvature $\kappa$, satisfying $\kappa-4\tau^2\neq 0$, the topic of Gauss map theory is very active and several results were established in the last decade. For instance, by B. Daniel \cite{Daniel11} on the Heisenberg Group $\mathrm{Nil}_3$, by B. Daniel, I. Fernandéz and P. Mira \cite{DanielFernandezMira15} on $\psl2$, by I. Fernandéz and P. Mira \cite{FernandezMira07} on $\h^2\times\R$ and by M. Leite and J. Ripoll \cite{LeiteRipoll11} on $\S2R$ and $\h^2\times\R$.

W. Meeks and J. Pérez studied surfaces with constant mean curvature in $3-$
dimensional simply connected metric Lie groups in the survey \cite{MeeksPerez12}. Due to the Lie group structure, they presented a \emph{left invariant Gauss map} for an oriented surface in metric Lie groups. This map assumes values in the unit
$2-$sphere of the metric Lie algebra and it can be generalised to higher dimensions. In \cite{MeeksMiraPerezRos13}, W. Meeks, P. Mira, J. Pérez and A. Ros found a $2-$order equation satisfied by the \emph{left invariant Gauss Map}. They established, from this equation, a Weierstrass-type representation for surfaces with constant mean curvature in a $3-$dimensional metric Lie groups. However, since $\s^2\times\R$ is the only simply connected $3-$homogeneous Riemannian manifold that does not carry a structure of $3-$dimensional simply connected Lie group, with a left invariant metric, this ambient manifold was excluded from this list.

Motivated by the exceptional case of $\S2R$ not discussed in \cite{MeeksMiraPerezRos13,MeeksPerez12}, we define a new Gauss map for surfaces in $\S2R$. We use the model of $\S2R$ isometric to $\R^3\setminus \{0\}$, endowed with a metric conformally equivalent to the Euclidean metric of $\R^3$. Therefore, given a surface in $\S2R$ we define the Gauss map likewise in $\R^3$, identifying each unit $2-$sphere on each tangent plane to this surfaces with the unit $2-$sphere $\s^2\subset\R^3$.

The goal of this paper is to study minimal surfaces in $\S2R$ from this Gauss map.
More  specifically,  we  are  interested  to  know  when  a  such  minimal  immersion  is determined by its conformal structure and its Gauss map. As a main result, we prove that:

\begin{th-A}
Let $X:\Sigma\to\S2R$ be a minimal conformal immersion and $g$ be its non-constant Gauss map. If $\hat{X}:\Sigma\to\S2R$ is another minimal conformal immersion with the same Gauss map of $X$, then $\hat{X}=f\circ X$, with $f\in\iso{\S2R}$ given either by $f=(\id,T)$ or $f=(\mathcal{A},T)$, where $\mathcal{A}$ denotes the antipodal map on $\s^2$ and $T$ is a translation on $\R$.
\end{th-A}

Unlike what happens in $\R^3$, any deformation of a minimal conformal immersion preserving this Gauss map is in fact rigid, i.e., it is given by an isometry of $\S2R$. However, we show that the zeroes of the curvature $K$ of the metric induced by $X$ are the singular points of $g$, hence either $K=0$ or its zeroes are isolated, likewise what happens in $\R^3$.

Under hypothesis of anti-holomorphic non-constant Gauss map $g$, we prove that this condition is an obstruction to the existence of a minimal conformal immersion into $\S2R$. For instance:

\begin{prop-antiholo}
There is no minimal conformal immersion $X:\Sigma\to\S2R$ which its Gauss map $g$ is an anti-holomorphic non-constant map.
\end{prop-antiholo}

In the case of minimal conformal immersion into $\S2R$ with a singular Gauss map, we show that only vertical cylinders over geodesics of $\s^2$ in $\S2R$ appear, and then the Gauss map is necessarily constant.

\begin{prop-cylinder}
Let $X:\Sigma\to\S2R$ be a minimal conformal immersion and $g$ be its Gauss map. If $g$ is a singular map then $X(\Sigma)$ is part of a vertical cylinder over a geodesic of $\s^2$ in $\s^2\times\R$. In particular, $g$ is constant.
\end{prop-cylinder}

It is worthwhile to point out our definition of Gauss map does not coincide with the \emph{Twisted normal map} defined by M. Leite and J. Ripoll in \cite{LeiteRipoll11}. In fact, the \emph{Twisted normal map} assumes values into the unit $3-$sphere $\s^3$ and the harmonicity of this map is equivalent to affirm that the surface has constant mean curvature in $\s^2\times\R$. In our case, only totally geodesic $2-$spheres in $\s^2\times\R$ have non-constant harmonic Gauss map.

\begin{cor-g-harmonic}
Let $X:\Sigma\to\S2R$ be a minimal conformal immersion and $g$ be its Gauss map. If $g$ is a non-constant harmonic map then $X(\Sigma)$ is part of a totally geodesic $2-$sphere in $\s^2\times\R$.
\end{cor-g-harmonic}

\section{The model of \texorpdfstring{$\S2R$}{S2xR}}
Let $\s^2$ be the unit $2-$sphere in $\R^3$ endowed with the induced metric, and consider the Riemannian product manifold $\S2R$ endowed with the product metric (referred to as the standard model of $\S2R$). We denote by $\Pi_1:\S2R\to\s^2$ and $\Pi_2:\S2R\to\R$ the projections into the factors $\s^2$ and $\R$, respectively, given by $\Pi_1(y,t)=y$ and $\Pi_2(y,t)=t$.

We consider $\R^3\setminus\{0\}$ endowed with the metric
\[\dif \mu^2=\frac{1}{x_1^2+x_2^2+x_3^2}(\dif x_1^2+\dif x_2^2+\dif x_3^2)\]
and the smooth map $\phi:\S2R\to\R^3\setminus\{0\}$, defined by
$\phi(y,t)=e^ty$, with inverse map given by $\phi^{-1}(x)=(x/|x|,\log|x|)$, for $x=(x_1,x_2,x_3)\in\R^3\setminus\{0\}$. In addition, we define $\pi_j=\Pi_j\circ\phi^{-1}$ for $j=1,2$, i.e.,
\[x  \mapsto \pi_1(x)=\frac{x}{|x|} \text{ \ and \ } x\mapsto \pi_2(x)=\log|x|.\]
A straightforward computation shows that $\phi$ is a global isometry between $\S2R$ and $\R^3\setminus\{0\}$, therefore, we may consider $\S2R$ as $(\R^3\setminus\{0\},\dif \mu^2)$.

Let $\{\partial_{x_j}\}_{j=1,2,3}$ be the canonical orthonormal frame of $\R^3$ and we set $E_j(x)=|x|\partial_{x_j}$, for $x\in\S2R$ and $j=1,2,3$. The frame $\{E_j(x)\}_{j=1,2,3}$ is an orthonormal frame of $\S2R$. If $\bar{\nabla}$ denotes the Riemannian connection of $\S2R$,  by a standard computation we get
\[\everymath={\displaystyle}\begin{array}{ccc}
  \bar{\nabla}_{E_1}E_1=\frac{1}{|x|}(x_2E_2+x_3E_3), & \bar{\nabla}_{E_2}E_1=-\frac{x_1}{|x|}E_2, & \bar{\nabla}_{E_3}E_1=-\frac{x_1}{|x|}E_3,
\end{array}\]

\[\everymath={\displaystyle}\begin{array}{ccc}
  \bar{\nabla}_{E_1}E_2=-\frac{x_2}{|x|}E_1, & \bar{\nabla}_{E_2}E_2=\frac{1}{|x|}(x_1E_1+x_3E_3), & \bar{\nabla}_{E_3}E_2=-\frac{x_2}{|x|}E_3,
\end{array}\]

\[\everymath={\displaystyle}\begin{array}{ccc}
  \bar{\nabla}_{E_1}E_3=-\frac{x_3}{|x|}E_1, & \bar{\nabla}_{E_2}E_3=-\frac{x_3}{|x|}E_2, & \bar{\nabla}_{E_3}E_3=\frac{1}{|x|}(x_1E_1+x_2E_2).
\end{array}\]

Given a vector $V=v_1\partial_{x_1}+v_2\partial_{x_2}+v_3\partial_{x_3}$ tangent at the point $x=(x_1,x_2,x_3)$ of $\S2R$, we will use brackets to express the coordinates of $V$ in the frame $\{E_j(x)\}_{j=1,2,3}$, that is,
\[v_1\partial_{x_1}+v_2\partial_{x_2}+v_3\partial_{x_3}=\frac{1}{|x|}
                                          \begin{bmatrix}
                                            v_1 \\
                                            v_2 \\
                                            v_3 \\
                                          \end{bmatrix}.\]

The isometry group of $\S2R$ is a $4-$dimensional group isomorphic to $\iso{\s^2}\times\iso{\R}$. In particular, it has $4$ connected components: a given isometry can either preserves or reverses the orientation in each factor $\s^2$ and $\R$. Consider $f=(M,T)\in\iso{\s^2\times\R}$ an ambient isometry. If $f$ preserves the orientation of $\R$, i.e., when $T$ is given by $T(t)=s+t$, then $\phi^{-1}\circ f\circ\phi$ is the correspondent isometry of $f$ in our model of $\s^2\times\R$, given by
\[x\in\R^3\setminus\{0\}\mapsto e^{s}M(x),\]
with $s\in\R$ and $M\in O(3)$, where $O(3)$ denotes the $3-$dimensional Orthogonal Group. If $f$ reverses the orientation of $\R$, i.e., when $T$ is given by $T(t)=s-t$, then the correspondent isometry of $f$ in our model of $\s^2\times\R$ is given by
\[x\in\R^3\setminus\{0\}\mapsto \frac{e^{s}}{|x|^2} M(x),\]
with $s\in\R$ and $M\in O(3)$. In order to simplify our notation, given an isometry $f\in\iso{\s^2\times\R}$, we identify $f$ and its correspondent $\phi^{-1}\circ f\circ\phi$ in our model of $\S2R$.

We say that an isometry $f\in\iso{\S2R}$ is a vertical translation of $\S2R$ when it is the identity on the factor $\s^2$ and it preserves the orientation of $\R$, that is, $f=(\id,T)$ where $T(t)=s+t$ is a translation on $\R$, with $s\in\R$ fixed.

\section{The Gauss map}

In this section, we identify
\[\S2R \text{ \ with \ }(\R^3\setminus\{0\},\dif \mu^2).\]

In our model of $\S2R$, we consider $\Pi:\S2R\to\C$ the projection into $\R^2$ (identified with $\C$) and $\pi:\S2R\to\R$ the projection into $\R$, respectively, given by  $\Pi(x_1,x_2,x_3)=x_1+ix_2$ and $\pi(x_1,x_2,x_3)=x_3$.

Let $\Sigma$ be an oriented Riemmanian surface and $z=u+iv$ be a conformal coordinate on $\Sigma$ and consider $X:\Sigma\to\S2R$ a conformal immersion. We define the projections of $X$ on $\C$ and $\R$ by $F=\Pi\circ X: \Sigma\to\C$ and $h=\pi\circ X:\Sigma\to\R$, and then the conformal immersion $X$ is written as $X=(F,h)$.
We denote by $N:\Sigma\to \textrm{U}(\S2R)$ the unit normal field to $\Sigma$, where $\textrm{U}(\S2R)$ is the unit tangent bundle to $\S2R$.

Since $\{E_j(x)\}_{j=1,2,3}$ is a global orthonormal frame of  $\S2R$, then for each $x\in\S2R$, we identify the unit $2-$sphere of $T_{x}(\S2R)$  with the unit $2-$sphere $\s^2\subset \R^3$, i.e., $v_1E_1+v_2E_2+v_3E_3 \in\s^2\subset T_x(\S2R)$
is identified with $v_1\partial_{x_1}+v_2\partial_{x_2}+v_3\partial_{x_3}\in\s^2\subset\R^3$.
Therefore, we may consider the unit normal vector $N$ into the $2-$sphere $\s^2\subset\R^3$ up to this identification.

\begin{definition}The Gauss map of $X$ is the map $g=\varphi\circ N:\Sigma\to\bar{\C}$, where $\varphi: \s^2\to\bar{\C}$ is the stereographic projection with respect to the southern pole, that is, if $N=N_1E_1+N_2E_2+N_3E_3$ then
\[g=\frac{N_1+iN_2}{1+N_3}:\Sigma\to\bar{\C}\]
satisfying
\[N=\frac{1}{1+|g|^2}\begin{bmatrix}
                       g+\bar{g}\\
                       i(\bar{g}-g)\\
                       1-|g|^2
                     \end{bmatrix}.\]
\end{definition}

\begin{remark}
For the other choice of $N$, we replace $g$ by $\widetilde{g}=-1/\bar{g}$.
\end{remark}

\begin{remark}\label{isometries}
Let $f=(M,T)\in\iso{\S2R}$, where $T$ is a translation by $s$ on $\R$, and consider the conformal immersion $\hat{X}=f\circ X$. Then $\hat{N}=(\det M)M\circ N$. In fact, since $f:\S2R\to\S2R$, given by $f(x)=e^sM(x)$, is a linear map then, for $V\in \mathrm{T}_x(\S2R)$, we have $\dif f_x (V) = e^s M(V)$ and the unit normal $\hat{N}$ is given by
\[\hat{N} = \frac{\dif f_{X(z)}(X_u)\times\dif f_{X(z)}(X_v)}{\|\dif f_{X(z)}(X_u)\times\dif f_{X(z)}(X_v)\|}. \]
However,
\[\dif f_{X(z)}(X_u)\times\dif f_{X(z)}(X_v) = e^{2s}(\det M)M(X_u\times X_v)\]
and
\[\|\dif f_{X(z)}(X_u)\times\dif f_{X(z)}(X_v)\| = e^{2s}\|X_u\times X_v\|,\]
once $M\in O(3)$, thus $\hat{N}=(\det M)M\circ N$. Moreover, since $(\det M)M\in SO(3)$, there is $R_M\in SU(2)$ such that $(\det{M})M(z_1)=z_2$ implies $R_M(\varphi(z_1))=\varphi(z_2)$, for $z_1,z_2\in \s^2$,  where $SU(2)$ denotes the Special Unitary Group of degree $2$. Consequently, we get that $\hat{g}=R_M\circ g$. In particular, when $M$ is the antipodal map $\mathcal{A}$, then $\hat{N}=N$ and hence $\hat{g}=g$.
\end{remark}

In the following computations, we adapt the method used by Daniel in \cite{Daniel11} to study the Gauss map of surfaces in the Heisenberg group $\mathrm{Nil}_3$. For sake of clarity, we maintain the Daniel's notations.

We consider the complex function $\eta = 2h_z$. For a conformal immersion $X$, we have
\[X_z=\frac{1}{2|X|}\begin{bmatrix}
    (F+\bar{F})_z \\
    i(\bar{F}-F)_z \\
    \eta
  \end{bmatrix}
  \text{ \ and \ }
  X_{\bar{z}}=\frac{1}{2|X|}\begin{bmatrix}
    (F+\bar{F})_{\bar{z}} \\
    i(\bar{F}-F)_{\bar{z}} \\
    \bar{\eta}
  \end{bmatrix},\]
where $|X|=\sqrt{|F|^2+h^2}$. We note that $X$ is a conformal immersion if, and only if, $|X_z|=0$, that is,
\begin{equation}\label{eq-conformality}
F_z\bar{F}_z=-\frac{\eta^2}{4}.
\end{equation}

We compute
\[X_z\times X_{\bar{z}}=\frac{i}{2|X|^2}\begin{bmatrix}
                                          \re(\eta F_{\bar{z}}-\bar{\eta}F_z) \\
                                          \im(\eta F_{\bar{z}}-\bar{\eta}F_z) \\
                                          |F_z|^2-|F_{\bar{z}}|^2
                                        \end{bmatrix}\]
and we get
\[X_u\times X_v=-2iX_z\times X_{\bar{z}}=\frac{1}{|X|^2}\begin{bmatrix}
                                          \re(\eta F_{\bar{z}}-\bar{\eta}F_z) \\
                                          \im(\eta F_{\bar{z}}-\bar{\eta}F_z) \\
                                          |F_z|^2-|F_{\bar{z}}|^2
                                        \end{bmatrix}.\]

We note that $g=0$ if, and only if, $N$ is the northern pole, equivalently
\begin{align*}
  \eta F_{\bar{z}}-\bar{\eta}F_z &= 0, \\
  |F_z|^2-|F_{\bar{z}}|^2 &> 0.
\end{align*}
Then by equation \eqref{eq-conformality} we get $\bar{\eta}(|F_z|^2+|\eta|^2/4)=0$ that implies $\eta =0$, since $|F_z|^2>0$. Again, by \eqref{eq-conformality} we get $F_z F_{\bar{z}}=0$, that is, $F_{\bar{z}}=0$. Analogously, $g=\infty$ if, and only if, $N$ is the southern pole, equivalently, $\eta =0$ and $F_z=0$. Therefore $g=0$ or $\infty$ if, and only if, $\eta = 0$. Moreover, $g=0$ if, and only if, $F_{\bar{z}}=0$ and $g=\infty$ if, and only if, $F_z=0$.

We compute
\[\|X_u\times X_v\|=2\langle X_z,X_{\bar{z}}\rangle=\frac{1}{|X|^2}\Big(|F_z|^2+|F_{\bar{z}}|^2+\frac{1}{2}|\eta|^2\Big).\]

Then, when $g\neq \infty$, we get
\[\frac{2g}{1+|g|^2}=\frac{\eta F_{\bar{z}}-\bar{\eta}F_z}{|F_z|^2+|F_{\bar{z}}|^2+\frac{1}{2}|\eta|^2}\text{ \ and \ }
\frac{1-|g|^2}{1+|g|^2}=\frac{|F_z|^2-|F_{\bar{z}}|^2}{|F_z|^2+|F_{\bar{z}}|^2+\frac{1}{2}|\eta|^2},\]
therefore
\[g=\frac{\eta F_{\bar{z}}-\bar{\eta}F_z}{2|F_z|^2+\frac{1}{2}|\eta|^2}.\]

Using \eqref{eq-conformality} in the expression above, we get
\begin{equation}\label{eq-F}
\bar{g}F_z=-\frac{\eta}{2}\text{ \ and \ }F_{\bar{z}} = \frac{g\bar{\eta}}{2},
\end{equation}
which implies that the functions $\eta/\bar{g}$ and $g\bar{\eta}$ can be extended smoothly to the points where $g=0$ or $\infty$. Furthermore, at the points where $g\neq0$ or $\infty$, the metric induced by $X$ is given by
\begin{align}\label{eq-inducedmetric1}
\dif s^2 &= \frac{(1+|g|^2)^2|\eta|^2}{4|g|^2|X|^2}|\dif z|^2.
\end{align}

\section{Minimal immersions into \texorpdfstring{$\S2R$}{S2xR}}

We devote this section to study minimal surfaces in $\S2R$ from the Gauss map defined in the previous section. We recall that, unless otherwise stated, we identify
\[\S2R \text{ \ with \ }(\R^3\setminus\{0\},\dif \mu^2).\]

From now on, we suppose that $X:\Sigma\to\S2R$ is a minimal conformal immersion. As a first result, we translate the minimality condition of $X$ in terms of its Gauss map $g$ and its complex function $\eta$.

\begin{lemma}\label{lemma-g-eta}
Let $X:\Sigma\to\S2R$ be a minimal conformal immersion. Then the Gauss map $g$ and the complex function $\eta$ satisfy
\begin{equation}\label{eta}
\frac{1-|g|^2}{1+|g|^2}(\log \bar{g})_{\bar{z}}=(\log \eta)_{\bar{z}}
\end{equation}
when $g\neq \infty$.
\end{lemma}

\begin{proof}
We consider $U\subset\Sigma$ an open set which $g\neq 0$. Firstly, note that by equations \eqref{eq-F}, we can write
\[X_z=\frac{\eta}{4|X|}\begin{bmatrix}\everymath={\displaystyle}
    \bar{g}-\bar{g}^{-1} \\
    i(\bar{g}+\bar{g}^{-1}) \\
    2
  \end{bmatrix}
  \text{ \ and \ }
  X_{\bar{z}}=\frac{\bar{\eta}}{4|X|}\begin{bmatrix}\everymath={\displaystyle}
    g-g^{-1} \\
    i(g+g^{-1}) \\
    2
  \end{bmatrix}.\]
Therefore, setting
\[A_1 = \frac{\eta}{4|X|}\Big(\bar{g}-\frac{1}{\bar{g}}\Big), \ A_2 = \frac{\eta}{4|X|}\Big(\bar{g}+\frac{1}{\bar{g}}\Big)
 \ \text{ and } \ A_3=\frac{\eta}{2|X|},\]
we have that $\displaystyle X_z = \sum_{j=1}^{3} A_jE_j$ and $\displaystyle X_{\bar{z}} = \sum_{j=1}^{3}\bar{A}_j E_j$.

Given $X:\Sigma\to\S2R$ a conformal immersion, a necessary and sufficient condition for $X$ to be minimal is
\[\bar{\nabla}_{X_{\bar{z}}}X_z=0,\]
that is,
\begin{equation}\label{minimality-condition}
\sum_{k=1}^{3}(A_k)_{\bar{z}}E_k+\sum_{\ell,j=1}^{3} \bar{A}_{\ell} A_j \bar{\nabla}_{E_{\ell}}E_j = 0.
\end{equation}

A straightforward computation gives:
\begin{multline*}
(A_1)_{\bar{z}} = \frac{1}{16|X|^3|g|^2\bar{g}}\Big\{4|X|^2|g|^2(\bar{g}^2-1)\eta_{\bar{z}}+4|X|^2(\bar{g}^2+1)g\eta\bar{g}_{\bar{z}} \\
  -|\eta|^2(\bar{g}^2-1)(|g|^2g\bar{F}-\bar{g}F+2h|g|^2)\Big\},
\end{multline*}
\begin{multline*}
(A_2)_{\bar{z}} = \frac{i}{16|X|^3|g|^2\bar{g}}\Big\{4|X|^2|g|^2(\bar{g}^2+1)\eta_{\bar{z}}+4|X|^2(\bar{g}^2-1)g\eta\bar{g}_{\bar{z}} \\
  -|\eta|^2(\bar{g}^2+1)(|g|^2g\bar{F}-\bar{g}F+2h|g|^2)\Big\}
\end{multline*}
and
\begin{equation*}
(A_3)_{\bar{z}} = \frac{1}{8|X|^3g}\Big\{4|X|^2g\eta_{\bar{z}}-|\eta|^2(g^2\bar{F}-F+2hg)\Big\}.
\end{equation*}
Thus
\begin{multline*}
\sum_{j,k} \bar{A}_j A_k \bar{\nabla}_{E_j}E_k =\\ =\frac{|\eta|^2}{16|X|^3|g|^2}\Big\{(|g|^4+g^2)\bar{F}+(\bar{g}^2+1)F-2h\bar{g}(g^2-1)+4|g|^2\re{F}\Big\}E_1 \\
+\frac{|\eta|^2}{16|X|^3|g|^2}\Big\{i(|g|^4-g^2)\bar{F}+i(\bar{g}^2-1)F+2ih\bar{g}(g^2+1)+4|g|^2\im{F}\Big\}E_2\\
+\frac{|\eta|^2}{8|X|^3|g|^2}\Big\{(1+|g|^4)h-|g|^2\bar{g}F+g\bar{F}\Big\}E_3.
\end{multline*}

The vertical part of \eqref{minimality-condition} is equal to zero if, and only if,
\[4|X|^2|g|^2\eta_{\bar{z}}-|\eta|^2(|g|^2g\bar{F}-\bar{g}F+2|g|^2h)+|\eta|^2((1+|g|^4)h-|g|^2\bar{g}F+g\bar{F})=0,\]
i.e.,
\begin{equation*}
  4|X|^2|g|^2\bar{\eta}_{z}=-|\eta|^2(1-|g|^2)\Omega,
\end{equation*}
where $\Omega = F\bar{g}+\bar{F}g+h(1-|g|^2)\in \R$.

The horizontal part of \eqref{minimality-condition} is equal to zero if, and only if,
\begin{multline*}
-4|X|^2|g|^2\eta_{\bar{z}}+4|X|^2\eta g\bar{g}_{\bar{z}}+|\eta|^2(|g|^2g\bar{F}-\bar{g}F+2h|g|^2) \\
  +|\eta|^2(|g|^2g\bar{F}+\bar{g}F-2|g|^4h+2|g|^2\bar{g}F)=0,
\end{multline*}
i.e.,
\[4|X|^2\eta g\bar{g}_{\bar{z}}-4|X|^2|g|^2\eta_{\bar{z}}+2|\eta|^2|g|^2\Omega=0.\]
Using \eqref{eqminimality-1} in the equation above we get
\begin{equation*}
  4|X|^2g\bar{g}_{\bar{z}} = -\bar{\eta}(1+|g|^2)\Omega.
\end{equation*}

By continuity, equations \eqref{eqminimality-1} and \eqref{eqminimality-2} also hold on a neighbourhood of a point which $g=0$. Therefore,
\begin{align}
4|X|^2g\bar{g}_{\bar{z}}
&= -\bar{\eta}(1+|g|^2)\Omega \label{eqminimality-2},\\
4|X|^2|g|^2\bar{\eta}_{z} &= -|\eta|^2(1-|g|^2)\Omega\label{eqminimality-1}
\end{align}
are necessary and sufficient conditions for $X$ to be minimal, when $g\neq \infty$.

Suppose that $X$ is a minimal conformal immersion. At a point which $g\neq 0$, we have that $\eta \neq 0$, then by \eqref{eqminimality-2} and \eqref{eqminimality-1} we have
\[-\frac{\bar{\eta}\Omega}{4|X|^2g} = \frac{\bar{g}_{\bar{z}}}{1+|g|^2}\text{ \ and \ }-\frac{\bar{\eta}(1-|g|^2)\Omega}{4|X|^2g}=\frac{\bar{g}\bar{\eta}_z}{\eta}.\]
Multiplying the first equation by $(1-|g|^2)$ and substituting it in the second one, we get our assertion. Again by continuity, \eqref{lemma-g-eta} holds at a point where $g=0$.
\end{proof}

\begin{example}[Horizontal surfaces and vertical cylinders]
The simplest examples of minimal surfaces in $\s^2\times\R$ are the horizontal surfaces and the vertical cylinders over geodesics of $\s^2$.

The totally geodesic $2-$spheres $\s^2\times\{a\}$ in the standard model of $\s^2\times\R$, for $a\in\R$, are $2-$spheres of radius $e^a$ centered at the origin of $\R^3$ in our model of $\S2R$.

The vertical cylinders $\gamma\times\R$ over geodesics of $\s^2$ in the standard model of $\s^2\times\R$ are planes passing through the origin of $\R^3$ in our model of $\S2R$.
\end{example}

\begin{proposition}\label{cylinders2}
Let $X:\Sigma\to\S2R$ be a minimal conformal immersion and $g$ be its Gauss map. Then $g$ is constant if, and only if, $X(\Sigma)$ is part of a vertical cylinder over a geodesic of $\s^2$ in  $\s^2\times\R$.
\end{proposition}
\begin{proof}
Suppose that $g$ is constant. Then the normal vector $N$ is also constant on $\Sigma$, so $X(\Sigma)$ is contained in a plane $P$ of $\R^3$. If $P$ does not pass through the origin of $\R^3$, there is a $2-$sphere centered at the origin of $\R^3$ which it is tangent to $P$ at a point $z_0$. However, by the Tangency Principle, $P$ must coincide with this $2-$sphere on a neighbourhood of $z_0$, that is absurd. Therefore, $P$ passes through the origin of $\R^3$ and then $X(\Sigma)$ is part of a vertical cylinder over a geodesic of $\s^2$ in $\s^2\times\R$.

Conversely, since vertical cylinders over geodesics of $\s^2$ in $\s^2\times\R$ are planes in our model of $\s^2\times\R$, then the Gauss map $g$ must be constant.
\end{proof}

Given $X:\Sigma\to\S2R$ a minimal conformal immersion, we consider the smooth maps $p:\Sigma\to\bar{\C}$ and $r:\Sigma\to\R$ defined by $p(z)=\varphi\circ\pi_1\circ X(z)$ and $r(z)=\pi_2\circ X(z)$, respectively, that is,
\[z  \mapsto p(z)=\frac{F(z)}{|X(z)|+h(z)} \text{ \ and \ } z\mapsto r(z)=\log|X(z)|.\]

We observe that $p=0$ or $\infty$ if, and only if, $F=0$. Then $p$
cannot be identically equal to $0$ or $\infty$ on an open set, since $X$ is an immersion. Moreover, $p$ is a harmonic map into $(\bar{\C},4/(1+|w|^2)^2|\dif w|^2)$ and $r$ is a harmonic function on $\R$. Indeed, if $X$ is minimal then $\pi_j\circ X$ is a harmonic map, for $j=1,2$. Since the stereographic projection $\varphi$ is an isometry between $\s^2$ and $\bar{\C}$ endowed with the metric $4/(1+|w|^2)^2|\dif w|^2$, we have that
$\varphi\circ\pi_1\circ X=p$ and $\pi_2\circ X = r$ are harmonic maps.

\begin{proposition} Let $X:\Sigma\to\S2R$ be a minimal conformal immersion. Let $g:\Sigma\to\bar{\C}$  be the Gauss map of $X$ and suppose that $g\neq \infty$. Then the following equations hold:
  \begin{align}
     & (1+|p|^2)p_{z\bar{z}}-2\bar{p}p_zp_{\bar{z}}=0, \label{harmonicity-p}\\
     & (\bar{g}-\bar{p})^2p_z+(1+\bar{g}p)^2\bar{p}_z=0,\label{equation-p-1} \\
     & (1+|p|^2)^2g_z = (1+g\bar{p})^2p_z+(g-p)^2\bar{p}_z.\label{equation-p-2}
  \end{align}
These equations hold when $p\neq \infty$.
\end{proposition}
\begin{proof}
Indeed, equation \eqref{harmonicity-p} is the harmonicity condition for the map $p$.

We restrict ourselves on an open set of $\Sigma$ which $p\neq 0$. Away from the points where $g=0$ and $\infty$, by definition of $p$ and \eqref{eq-F}, we compute
\begin{align}
  p_z &=-\frac{\eta(|X|+h+\bar{g}F)^2}{4|X|(|X|+h)^2\bar{g}},\label{eq-pz} \\
  \bar{p}_z
  &=\frac{\eta(\bar{g}|X|+h\bar{g}-\bar{F})^2}{4|X|(|X|+h)^2\bar{g}}.\label{eq-barpz}
\end{align}

On the other hand, we have $|X|+h=F/p\in\R$,
\[\displaystyle(|X|+h+\bar{g}F)^2=\frac{F^2}{p^2}(1+\bar{g}p)^2
\text{ \ and \ }
\displaystyle(|X|+h+\bar{g}F)^2=\frac{F^2}{p^2}(1+\bar{g}p)^2.\]
Then, by equations \eqref{eq-pz} and \eqref{eq-pz2}, we get
\begin{align}
  4|X|\bar{g}p_z &= -\eta(1+\bar{g}p)^2, \label{eq-pz2}\\
  4|X|\bar{g}\bar{p}_z &= \eta(\bar{g}-\bar{p})^2. \label{eq-pbarz2}
\end{align}
Therefore, $(\bar{g}-\bar{p})^2\eqref{eq-pz2}+(1+\bar{g}p)^2\eqref{eq-pbarz2}=0$ that implies \eqref{equation-p-1}. By continuity, \eqref{equation-p-1} holds at a point where $g=0$. Analogously, equation \eqref{equation-p-1} holds on a neighbourhood of a point which $p=0$.

If $p=g$ or $p=-1/\bar{g}$ on an open set of $\Sigma$, then equation \eqref{equation-p-2} is trivial. Suppose that $g\neq 0$ on an open set of $\Sigma$. Then, away from the points where $p=g$ and $p=-1/\bar{g}$, we observe that $p_z$ is well defined by \eqref{eq-pz2} and it does not vanish. Differentiating equation \eqref{equation-p-1} with respect to $\bar{z}$ and multiplying by $(1+|p|^2)(1+\bar{g}p)$, we have
\begin{multline*}
2(1+|p|^2)(\bar{g}-\bar{p})(1+\bar{g}p)(\bar{g}_{\bar{z}}-\bar{p}_{\bar{z}})p_z +(\bar{g}-\bar{p})^2(1+\bar{g}p)\big[(1+|p|^2)p_{z\bar{z}}\big]\\
+2(1+|p|^2)(\bar{g}_{\bar{z}}p+\bar{g}p_{\bar{z}})\big[(1+\bar{g}p)^2\bar{p}_z\big]+(1+\bar{g}p)^3\big[(1+|p|^2)\overline{p_{z\bar{z}}}\big]=0.
\end{multline*}
From the equation above, using \eqref{harmonicity-p} and \eqref{equation-p-1} into to the brackets terms, we get
\[2(\bar{g}-\bar{p})p_z\Big\{(1+|p|^2)^2\bar{g}_{\bar{z}}- (1+\bar{g}p)^2\bar{p}_{\bar{z}}-(\bar{g}-\bar{p})^2p_{\bar{z}}\Big\}=0,\]
which implies \eqref{equation-p-2}. By continuity, \eqref{equation-p-2} holds at a point where $p=g$ or $p=-1/\bar{g}$. Analogously, equation \eqref{equation-p-2} holds on a neighbourhood of a point which $g=0$.
\end{proof}

\begin{remark}\label{2-spheres}
We notice that if $p=g$ (respec., $p=-1/\bar{g}=\widetilde{g}$) on $\Sigma$, by the definitions of $p$ and $g$, we have that $X/|X|=N$ (respec., $X/|X|=-N$), i.e., $X(\Sigma)$ is part of a totally geodesic $2-$sphere in $\S2R$. Moreover, in both cases, the Gauss map is holomorphic.
\end{remark}

In the next result, we derive an important second order equation satisfied by the Gauss map $g$ and the map $p$.

\begin{proposition}\label{prop-gaussmap}
The Gauss map $g$ and the map $p$ satisfy
  \begin{multline}\label{equation-gaussmap}
\big(|g-p|^2-|1+\bar{g}p|^2\big)(1+|g|^2)g_{z\bar{z}}+2(g-p)(1+g\bar{p})|g_z|^2\\
 +2\big(\bar{g}|1+\bar{g}p|^2-(\bar{g}-\bar{p})(1+|g|^2)\big)g_zg_{\bar{z}}=0
  \end{multline}
when $g\neq \infty$.
\end{proposition}

\begin{proof}
Firstly, note that if $p=g$ or $p=-1/\bar{g}$ on an open set of $\Sigma$, then \eqref{equation-gaussmap} is equivalent to \eqref{harmonicity-p}, and if $g_z=0$ or $g=0$ on an open set of $\Sigma$, then \eqref{equation-gaussmap} is trivial.

Suppose that $g\neq 0$ on an open set of $\Sigma$. Then, away from the points where $p=g$ and $p=-1/\bar{g}$, $p_z$ and $\bar{p}_z$ are well defined by \eqref{eq-pz2} and \eqref{eq-pbarz2} and they do not vanish. Differentiating equation \eqref{equation-p-2} and using \eqref{harmonicity-p}, we get
\begin{multline*}
(1+|p|^2)^2g_{z\bar{z}}+\frac{2\bar{p}p_{\bar{z}}}{1+|p|^2}\Big((1+|p|^2)^2g_z-(1+g\bar{p})^2p_z\Big) +2(g-p)p_{\bar{z}}\bar{p}_z\\
+\frac{2p\bar{p}_{\bar{z}}}{1+|p|^2}\Big((1+|p|^2)^2g_z-(g-p)^2\bar{p}_z\Big)-2g(1+g\bar{p})p_z\bar{p}_{\bar{z}} \\
-2\bar{p}(1+g\bar{p})p_zg_{\bar{z}}-2(g-p)\bar{p}_zg_{\bar{z}}=0.
\end{multline*}
By equation \eqref{equation-p-2}, we have
\begin{multline}\label{eq-prop3}
(1+|p|^2)^2g_{z\bar{z}}+\frac{2(g-p)(1+g\bar{p})}{1+|p|^2}(|\bar{p}_z|^2-|p_z|^2)\\
-2\bar{p}(1+g\bar{p})p_zg_{\bar{z}}-2(g-p)\bar{p}_zg_{\bar{z}}=0.
\end{multline}

On the other hand, by equation \eqref{equation-p-1}, we have
\begin{equation}\label{eq-pz-and-barpz}
p_z = -\frac{(1+\bar{g}p)^2}{(\bar{g}-\bar{p})^2}\bar{p}_z,
\end{equation}
and then \eqref{equation-p-2} implies
\begin{equation}\label{eq-g-z}
  (1+|p|^2)^2g_z = \frac{1}{(\bar{g}-\bar{p})^2}\big(|g-p|^4-|1+\bar{g}p|^4\big)\bar{p}_z.
\end{equation}

At a point where $g_z\neq 0$, we get $|g-p|^4-|1+\bar{g}p|^4\neq0$. We also note that $|g-p|^4-|1+\bar{g}p|^4=(1+|g|^2)(1+|p|^2)(|g-p|^2-|1+\bar{g}p|^2)$ and then we get
\begin{align}
  p_z &= - \frac{(1+|p|^2)(1+\bar{g}p)^2}{(1+|g|^2)(|g-p|^2-|1+\bar{g}p|^2)}g_z,\label{eq-pz-expression} \\
  \bar{p}_z &= \frac{(1+|p|^2)(\bar{g}-\bar{p})^2}{(1+|g|^2)(|g-p|^2-|1+\bar{g}p|^2)}g_z.\label{eq-barpz-expression}
\end{align}

Then, by \eqref{eq-prop3}, we have
\begin{multline}\label{eq-prop4}
(|g-p|^2-|1+\bar{g}p|^2)(1+|p|^2)(1+|g|^2)g_{z\bar{z}}+2(1+|p|^2)(g-p)(1+g\bar{p})|g_z|^2\\
  +2\Big(\bar{p}(1+\bar{g}p)|1+\bar{g}p|^2-(\bar{g}-\bar{p})|g-p|^2\Big)g_zg_{\bar{z}}=0,
\end{multline}
and since
\[\bar{p}(1+\bar{g}p)|1+\bar{g}p|^2-(\bar{g}-\bar{p})|g-p|^2 = (1+|p|^2)\Big(\bar{g}|1+\bar{g}p|^2-(\bar{g}-\bar{p})(1+|g|^2)\Big),\]
we get \eqref{equation-gaussmap} dividing \eqref{eq-prop4} by $(1+|p|^2)$. By continuity, \eqref{equation-p-2} holds on a neighbourhood of a point which $g_z=0$. If $p=g$ or $p=-1/\bar{g}$ at some point, again by continuity \eqref{equation-p-2} holds. Analogously, equation \eqref{equation-p-2} holds on a neighbourhood of a point which $g=0$.
\end{proof}

\begin{proposition}\label{anti-holomorphic}
There is no minimal conformal immersion $X:\Sigma\to\S2R$ which its Gauss map $g$ is an anti-holomorphic non-constant map.
\end{proposition}

\begin{proof}
By contradiction, suppose that exists a such immersion which $g_z=0$ on $\Sigma$. Away from the points where $p=g$ and $p=-1/\bar{g}$, by equation \eqref{eq-g-z}, we get that either $\bar{p}_z=0$ or $|g-p|^2-|1+\bar{g}p|^2=0$.

If $\bar{p}_z=0$ on $\Sigma$, then, by equations \eqref{equation-p-1} and \eqref{equation-p-2}, we get $p_z=0$. Thus $p$ is a constant function on $\Sigma$ and, by definition of $p$, we have $\varphi^{-1}(\text{const})=X/|X|$; however it is not possible since $X$ is an immersion.

On the other hand, by the definition of $p$, $|g-p|^2-|1+\bar{g}p|^2=0$ on $\Sigma$ if, and only if,
\begin{equation}\label{eq-omega}
F\bar{g}+\bar{F}g+h(1-|g|^2)=0.
\end{equation}
Differentiating equation \eqref{eq-omega} and using \eqref{eq-F}, we get $\bar{g}_z(F-hg)=0$, that is, $F-hg=0$ on $\Sigma$. Finally, differentiating $F-hg=0$ and using $\eqref{eq-F}$, we get $-\eta(1+|g|^2)/\bar{g}=0$ which holds if, and only if, $\eta=0$, i.e., if $g=0$ or $\infty$ on $\Sigma$.
\end{proof}

\begin{remark}\label{quociente} Suppose that $g$ is a non-constant map. By equations \eqref{equation-p-1} and \eqref{equation-p-2}, we get
\[(1+|p|^2)^3|g_z|^2 = (1+|g|^2)(|g-p|^2-|1+\bar{g}p|^2)(|\bar{p}_z|^2-|p_z|^2). \]
On the other hand, since \eqref{eq-pz2} and \eqref{eq-pbarz2}, we have
\[16|X|^2|g|^2(|\bar{p}_z|^2-|p_z|^2)=|\eta|^2(1+|g|^2)(1+|p|^2)(|g-p|^2-|1+\bar{g}p|^2),\]
then
\[16|X|^2|g|^2(1+|p|^2)^2|g_z|^2=|\eta|^2(1+|g|^2)^2(|g-p|^2-|1+\bar{g}p|^2)^2.\]
Since $g$ is not constant,  $\eta$ cannot vanish on an open set of $\Sigma$ and, by Proposition \ref{anti-holomorphic}, the same happens for $g_z$. At a point where $\eta\neq 0$ and $g_z \neq 0$, we have
\[\frac{(|g-p|^2-|1+\bar{g}p|^2)^2}{|g_z|^2}=\frac{16|X|^2|g|^2(1+|p|^2)^2}{|\eta|^2(1+|g|^2)^2}.\]
Therefore, by continuity, the left side of the equation above is a function that do not vanish on $\Sigma$, when $g$ is a non-constant map.
\end{remark}

\begin{corollary}\label{eq-inducedmetric2}
Let $X:\Sigma\to\S2R$ be a minimal conformal immersion. Then the metric induced by $X$ is given by
\[\dif s^2 =\frac{4(1+|p|^2)^2|g_z|^2}{(|g-p|^2-|1+\bar{g}p|^2)^2}|\dif z|^2\]
when $g\neq \infty.$
\end{corollary}

\begin{proof}
Let $U\subset\Sigma$ be an open set which $p\neq g$. At a point where $g\neq 0$, by equations \eqref{eq-inducedmetric1} and  \eqref{eq-pbarz2}, we have that the metric induced by $X$, in the minimal case, is given by
\[\dif s^2 = \frac{4(1+|g|^2)^2|\bar{p}_z|^2}{|g-p|^4}|\dif z|^2.\]
However, using equation \eqref{eq-barpz-expression}, we get
\begin{equation*}
\dif s^2 =\frac{4(1+|p|^2)^2|g_z|^2}{(|g-p|^2-|1+\bar{g}p|^2)^2}|\dif z|^2.
\end{equation*}
By continuity, this expression holds at a point where $g=0$. Analogously, equation \eqref{equation-p-2} holds on a neighbourhood of a point which $g=p$.
\end{proof}

\begin{proposition}
Let $X:\Sigma\to\S2R$ be a minimal conformal immersion with a holomorphic non-constant Gauss map $g$. Then $X(\Sigma)$ is part of a totally geodesic $2-$sphere in  $\s^2\times\R$.
\end{proposition}
\begin{proof}
If $g$ is a holomorphic non-constant function, by Proposition \ref{prop-gaussmap}, we have $(g-p)(1+g\bar{p})|g_z|^2=0$, that is, either $p=g$ or $p=-1/\bar{g}$. Therefore, $X(\Sigma)$ is part of a totally geodesic $2-$sphere in $\s^2\times\R$.
\end{proof}

\begin{lemma}\label{curvatura}
The curvature $K$ of the metric induced by $X$ is
\[K=\frac{(|g-p|^2-|1+\bar{g}p|^2)^2}{(1+|g|^2)^2(1+|p|^2)^2|g_z|^2}(|g_z|^2-|\bar{g}_z|^2).\]
This formula holds when $g\neq \infty$. In particular, $K$ vanishes at singular points of $g$.
\end{lemma}
\begin{proof}
Since $X$ is a conformal immersion, we can compute the curvature $K$ of the metric induced by $X$  through $2K=-\Delta (\log \lambda)$, i.e., $K=-(2/\lambda) (\log\lambda)_{z\bar{z}}$, where $\lambda$ is given by $\dif s^2 = \lambda |\dif z|^2$. In order to simplify our computations, we will use the expression \eqref{eq-inducedmetric1} instead of expression obtained in Corollary \ref{eq-inducedmetric2}. Then, $\lambda$ is given by
\[\lambda=\frac{(1+|g|^2)^2|\eta|^2}{4|g|^2|X|^2}.\]

We restrict ourselves to a domain on which $g\neq 0$. We observe that, since
\[(\log \lambda)_{z\bar{z}} = 2(\log (1+|g|^2))_{z\bar{z}} + (\log\eta)_{z\bar{z}} +
(\log\bar{\eta})_{z\bar{z}} -
2r_{z\bar{z}} -
(\log g)_{z\bar{z}}-(\log\bar{g})_{z\bar{z}}\]
and $r_{z\bar{z}}=0$, once $X$ is a minimal immersion, by Lemma \ref{lemma-g-eta}, we get
\begin{multline*}
(\log \lambda)_{z\bar{z}} = \frac{2}{(1+|g|^2)^2}\Big((g_{z\bar{z}}\bar{g}+|g_z|^2+|\bar{g}_z|^2+g\bar{g}_{z\bar{z}})(1+|g|^2)-(g_z\bar{g}+g\bar{g}_z)(g_{\bar{z}}\bar{g}+g\bar{g}_{\bar{z}})\Big) \\
-\frac{2\bar{g}_{\bar{z}}}{(1+|g|^2)^2\bar{g}}(g_z\bar{g}+g\bar{g}_z)
-\frac{2g_z}{(1+|g|^2)^2g}(g_z\bar{g}+g\bar{g}_z)\\
-\frac{2g}{(1+|g|^2)\bar{g}}(\bar{g}_{z\bar{z}}\bar{g}-\bar{g}_z\bar{g}_{\bar{z}})-\frac{2\bar{g}}{(1+|g|^2)g}(g_{z\bar{z}}g-g_zg_{\bar{z}}).
\end{multline*}
Follows from a direct computation that
\[(\log \lambda)_{z\bar{z}} = \frac{2}{(1+|g|^2)^2}(|\bar{g}_z|^2-|g_z|^2).\]
Therefore, the curvature $K$ is given by
\[K = \frac{16|g|^2|X|^2}{(1+|g|^2)^4|\eta|^2}(|g_z|^2-|\bar{g}_z|^2),\]
and by continuity this formula holds at a point where $g=0$.

Let $U\subset\Sigma$ be an open set on which $p\neq g$. By equation \eqref{eq-pbarz2}, we get
\[K = \frac{|g-p|^4}{(1+|g|^2)^4|\bar{p}_z|^2}(|g_z|^2-|\bar{g}_z|^2).\]
However, by equation \eqref{eq-barpz-expression}, we get
\[K=\frac{(|g-p|^2-|1+\bar{g}p|^2)^2}{(1+|g|^2)^2(1+|p|^2)^2|g_z|^2}(|g_z|^2-|\bar{g}_z|^2),\]
and by continuity, this formula holds at a point where $p=g$. Moreover, since $\det \dif g = (|g_z|^2-|\bar{g}_z|^2)$, by Remark \ref{quociente}, we have that $K$ vanishes at singular points of $g$.
\end{proof}

\begin{proposition}\label{cylinders}
Let $X:\Sigma\to\S2R$ be a minimal conformal immersion and $g$ be its Gauss map. If $g$ is a singular map then $X(\Sigma)$ is part of a vertical cylinder over a geodesic of $\s^2$ in $\s^2\times\R$. In particular, $g$ is constant.
\end{proposition}

\begin{proof}
Since $\det g= |g_z|^2-|\bar{g}_z|^2$, if the Gauss map $g$ is singular, then  $|g_z|=|\bar{g}_z|$ on $\Sigma$, and by Lemma \ref{curvatura}, we get that $K=0$. Therefore, by \cite[Torralbo-Urbano]{TorralboUrbano15}, $X(\Sigma)$ is part of a vertical cylinder over a geodesic of $\s^2$ in $\s^2\times\R$. Hence, by Proposition \ref{cylinders2}, $g$ is constant.
\end{proof}

\begin{corollary}\label{g-harmonic}
Let $X:\Sigma\to\S2R$ be a minimal conformal immersion and $g$ be its Gauss map. If $g$ is a non-constant harmonic map then $X(\Sigma)$ is part of a totally geodesic $2-$sphere in  $\s^2\times\R$.
\end{corollary}
\begin{proof}
If $g$ is a non-constant harmonic map, then $g$ satisfies
\[(1+|g|^2)g_{z\bar{z}}-2\bar{g}g_z g_{\bar{z}}=0,\]
and, by Proposition \ref{prop-gaussmap}, we get that
\[|g-p|^2|1+\bar{g}p|^2(|g_z|^2-|g_{\bar{z}}|^2)=0.\]
Since $g$ is non-constant, then $(|g_z|^2-|g_{\bar{z}}|^2)\neq 0$, otherwise, by Proposition \ref{cylinders}, $g$ must be constant. Therefore, $g=p$ or $g=-1/\bar{p}$ and then $X(\Sigma)$ is part of a totally geodesic $2-$sphere in $\s^2\times\R$ (Remark \ref{2-spheres}).
\end{proof}

\begin{lemma}\label{lemma-r} Let $X:\Sigma\to\S2R$ be a minimal conformal immersion. Then the Gauss map $g$ and the maps $p$ and $r$ satisfy
\begin{equation}\label{eq-r}
r_{z} = \frac{2(\bar{g}-\bar{p})(1+g\bar{p})}{(1+|g|^2)(|g-p|^2-|1+\bar{g}p|^2)}g_z
\end{equation}
when $g\neq \infty$.
\end{lemma}
\begin{proof}
Let $U\subset\Sigma$ be an open set on which $p\neq g$. Then, by equation \eqref{eq-pz-and-barpz}, $\bar{p}_z$ does not vanish on $U$, otherwise $p$ is a constant function. By Lemma \ref{lemma-g-eta} and equation \eqref{eq-pbarz2}, we get
\[\frac{1-|g|^2}{1+|g|^2}(\log \bar{g})_{\bar{z}} = \Bigg(\log \frac{4|X|\bar{g}\bar{p}_z}{(\bar{g}-\bar{p})^2} \Bigg)_{\bar{z}}.\]

Away from the points where $g=0$ and $\infty$, we have
\[-\frac{2|g|^2}{1+|g|^2}\frac{\bar{g}_{\bar{z}}}{\bar{g}}
=r_{\bar{z}}+\frac{\bar{p}_{z\bar{z}}}{\bar{p}_z}-2\frac{\bar{g}_{\bar{z}}-\bar{p}_{\bar{z}}}{\bar{g}-\bar{p}}
\]
and, by equation \eqref{harmonicity-p}, we obtain
\[r_{\bar{z}}=\frac{2(1+g\bar{p})}{(\bar{g}-\bar{p})(1+|g|^2)}\bar{g}_{\bar{z}}-\frac{2(1+\bar{g}p)}{(\bar{g}-\bar{p})(1+|p|^2)}\bar{p}_{\bar{z}}.
\]
By continuity, this holds on a neighbourhood of a point which $g=0$. Then, using equation \eqref{eq-pz-expression}, we get the conjugate of our assertion and, by continuity, it also holds at a point where $p=g$.
\end{proof}

In the following we establish our main result. As was seen before in the Proposition \ref{cylinders},
vertical cylinders over geodesics of $\s^2$ in $\S2R$ are the only surfaces which the Gauss map is singular. Moreover, any two of them differ by an isometry of $\S2R$. The next result prove that the same rigidity happens for any two minimal conformal immersions into $\S2R$ with the same non-constant Gauss map.

\begin{theorem}\label{th-A}
Let $X:\Sigma\to\S2R$ be a minimal conformal immersion and $g$ be its non-constant Gauss map. If $\hat{X}:\Sigma\to\S2R$ is another minimal conformal immersion with the same Gauss map of $X$, then $\hat{X}=f\circ X$, with $f\in\iso{\S2R}$ given either by $f=(\id,T)$ or $f=(\mathcal{A},T)$, where $\mathcal{A}$ denotes the antipodal map on $\s^2$ and $T$ is a translation on $\R$.
\end{theorem}

\begin{proof}
Firstly, we can assume that $g$ is a regular map. Otherwise, by Proposition \ref{cylinders}, we get that $X(\Sigma)$ is a vertical cylinder over a geodesic of $\s^2$ in $\S2R$, and then $g$ is constant.

Suppose that $g$ is a regular map on $\Sigma$. Then $|g|$ cannot be constant on an open set $U\subset\Sigma$, otherwise $g_z\bar{g}=-g\bar{g}_z$ and follows that $g$ is a singular map. So, we restrict ourselves to this open set $U\subset\Sigma$ on which $|g|\neq 1$.

By Proposition \ref{prop-gaussmap}, for the minimal conformal immersion $X:\Sigma\to\S2R$, we get
\begin{equation}\label{eq-gaussmap-2}
  A(1-|p|^2)+Bp+C\bar{p}=0,
\end{equation}
where the coefficients $A, B$ and $C$ depend on $g,\bar{g}$ and theirs derivatives, given by
\begin{align*}
  A &= -(1-|g|^4)g_{z\bar{z}}+2g|g_z|^2-2|g|^2\bar{g}g_zg_{\bar{z}}, \\
  B &= -2\bar{g}(1+|g|^2)g_{z\bar{z}}-2|g_z|^2+2\bar{g}^2g_zg_{\bar{z}},  \\
  C &= -2g(1+|g|^2)g_{z\bar{z}}+2g^2|g_z|^2+2(1+2|g|^2)g_zg_{\bar{z}}.
\end{align*}

If $A=0$ at some $z_0\in U$, at this point we have $(1-|g|^2)Bp+(1-|g|^2)C\bar{p}=0$ and then
\[2(1+|g|^2)g_z\Big\{(\bar{g}^2g_{\bar{z}}-\bar{g}_{\bar{z}})p+(g_{\bar{z}}-g^2\bar{g}_{\bar{z}})\bar{p}\Big\}=0.\]
When $g_z(z_0)\neq 0$, the equation above has unique solution $p(z_0)=0$ if, and only if, $|\bar{g}^2g_{\bar{z}}-\bar{g}_{\bar{z}}|^2-|g_{\bar{z}}-g^2\bar{g}_{\bar{z}}|^2\neq 0,$ that is, $(1-|g|^4)(|g_z|^2-|\bar{g}_z|^2)\neq 0$ which holds in $U$. Moreover, $A$ cannot vanish identically on an open set of $\Sigma$, since this would imply that $X$ is not an immersion.

If $A\neq 0$  at $z_0\in U$, then we consider the system
\begin{align}
  A(1-|p|^2)+Bp+C\bar{p}&= 0, \label{eq-1}\\
  \bar{A}(1-|p|^2)+\bar{C}p+\bar{B}\bar{p}&=0.\label{eq-2}
\end{align}
Since $\bar{A}\eqref{eq-1}+A\eqref{eq-2}$ implies
\begin{equation}\label{eq-6}
(\bar{A}B-A\bar{C})p+(\bar{A}C-A\bar{B})\bar{p}=0
\end{equation}
then, multiplying \eqref{eq-1} by $(\bar{A}C-A\bar{B})$ and dividing by $A$, we get
\begin{equation}\label{eq-3}
(\bar{A}B-A\bar{C})p^2+(|C|^2-|B|^2)p+(\bar{A}C-A\bar{B})=0,
\end{equation}
i.e., $\alpha p^2+\beta p-\bar{\alpha}=0$, with $\alpha=\bar{A}B-A\bar{C}$ and $\beta =|C|^2-|B|^2 $. Considering that the discriminant of the equation above is $\beta^2+4|\alpha|^2$, then equation \eqref{eq-3} has two distinct solutions for $p(z_0)$ if, and only if, $\alpha\neq 0$, that is, $\bar{A}B-A\bar{C}\neq 0$ (once  $\bar{A}B=A\bar{C}$ implies $|B|^2=|C|^2$).

On the other hand, note that
\begin{align}
2\bar{g}A-(1-|g|^2)B &= 2(1+|g|^2)g_zD,\label{eq-4}\\
2gA-(1-|g|^2)C &= 2(1+|g|^2)g_z E,\label{eq-5}
\end{align}
where $D=\bar{g}_{\bar{z}}-\bar{g}^2g_{\bar{z}}$ and $E=g^2\bar{g}_{\bar{z}}-g_{\bar{z}}$. Then $\bar{A}\eqref{eq-4}-A\overline{\eqref{eq-5}}$ implies that
\[(1-|g|^2)(A\bar{C}-\bar{A}B)=2(1+|g|^2)(g_z\bar{A}D-\bar{g}_{\bar{z}}A\bar{E}).\]
When $g_z(z_0)\neq 0$, then $\alpha=0$ implies $|D|^2=|E|^2$. By definition of $D$ and $E$, $|D|^2=|E|^2$ if, and only if, $(1-|g|^4)(|g_z|^2-|\bar{g}_z|^2)= 0$ which does not happen on $U$. Therefore, equation \eqref{eq-3} has two distinct solutions for each $p(z_0)$ on $U$, when $g$ is a regular map.

Let $\hat{X}:\Sigma\to\S2R$ be a minimal conformal immersion with the same regular Gauss map $g$ of $X$. In the following, we use the symbol  $\hat{ \ }$ to refer to objects related to $\hat{X}$.

If $p=\hat{p}$ then, by definition, we have that $X/|X|=\hat{X}/|\hat{X}|$, i.e., $\hat{X}=q(z)X$, where $q:\Sigma\to\R^+$ is a smooth function, given by $q(z)=|\hat{X}(z)|/|X(z)|$. However, we observe that the function $q$ is constant.

In fact, by Lemma \ref{lemma-r}, we get that $(\log q)_{\bar{z}}=0$, since $X$ and $\widehat{X}$ have the same Gauss map. Then $q$ is a real holomorphic function, that is, $q(z)=q_0$ is constant and therefore $\hat{X}=f\circ X$, where $f=(\id,T)\in\iso{\S2R}$ is a vertical translation of $\S2R$ with $T(t)=t+q_0$.

If $p\neq \hat{p}$ then we claim that $\hat{p}=-1/\bar{p}$. Indeed, by a direct computation, we get that $-1/\bar{p}$ satisfies equation \eqref{equation-gaussmap}. Since \eqref{eq-3} has exactly two distinct solutions when $g$ is a regular map, then $\hat{p}=-1/\bar{p}$. Moreover, $\hat{p}\bar{p}=-1$ implies that $\varphi^{-1}(\hat{p})$ and $\varphi^{-1}(p)$ are antipodal points on $\s^2$, that is, $\hat{X}=q(z)\mathcal{A}\circ X$, where $\mathcal{A}$ is the antipodal map on $\s^2$ and $q(z)=|\hat{X}(z)|/|X(z)|$. However, by Lemma \ref{lemma-r}, we get $(\log q)_{\bar{z}}=0$, since $X$ and $\hat{X}$ have the same Gauss map and $p$ and $\hat{p}$ are antipodal points. Then $q$ is a real holomorphic function, that is, $q(z)=q_0$ is constant and therefore $\hat{X}=f\circ X$, where $f=(\mathcal{A},T)\in\iso{\S2R}$ with $T(t)=t+q_0$ a translation on $\R$.
\end{proof}

\begin{remark}
We notice that the minimal conformal immersion $X:\Sigma\to\S2R$ can be recovered from of the Gauss map $g$, up to isometries described on Theorem \ref{th-A},
when $g$ is a non-constant map. In fact, by definition of $p$, we have $X=|X|\varphi^{-1}(p)$. On the other hand, by Lemma \ref{eq-r}, $|X|$ can be recovered from of $g$, up to a multiplicative constant, and the map $p$ can be recovered from of $g$ as the solution of equation \eqref{eq-3}, up to change of the antipodal map on $\s^2$.
\end{remark}

\begin{remark}[Chirstoffel's problem] In 1867, E. B. Christoffel \cite{Christoffel1867} considered the following problem: Given an immersion $X:\Sigma\to \R^3$ of an oriented Riemannian surface $\Sigma$, when does $X$ possess a non-congruent immersion $\hat{X}:\Sigma\to\R^3$, such that the tangent planes at each point of $X$ and $\hat{X}$ are parallels and the metrics induced by $X$ and $\hat{X}$ are conformally equivalents? In the literature, this problem is known as Christoffel's problem (we may refer to \cite{DajczerTojeiro16}, \cite[Chapter 9]{Jensen16} and \cite{Samuel47}).

The local solution found by Chirstoffel depends of the orientation of $X$ and $\hat{X}$. Suppose that there are $X$ and $\hat{X}$ satisfying the Chirstoffel's problem. If $X$ and $\hat{X}$ induce the same orientation at each point of $\Sigma$, then $X$ is minimal and $\hat{X}$ is conformal associate of $X$. Otherwise, if $X$ and $\hat{X}$ induce opposite orientations at each point of $\Sigma$, then $X$ is isothermic and, in this case, the immersion $\hat{X}$ is called \emph{Christoffel transform} of $X$.

Since we identify $\S2R$ with $\R^3\setminus \{0\}$, endowed
with a metric conformally equivalent to the Euclidean metric of $\R^3$, a conformal immersion $X:\Sigma\to\S2R$ is a conformal immersion $X:\Sigma\to\R^3$ as well. Moreover, the definition of Gauss map in $\S2R$ coincides with the definition in $\R^3$. Then given $X$ and $\hat{X}$, the condition $g=\hat{g}$ implies that the tangent planes are parallels at each point of $\Sigma$ and $X$ and $\hat{X}$ induce the same orientation. By the Christoffel's Theorem, either $X$ and $\hat{X}$ differ by an isometry of $\R^3$ or $X$ and $\hat{X}$ are both minimal conformal immersions on $\R^3$. In the second case, we get that $g$ is anti-holomorphic and, by Proposition \ref{anti-holomorphic}, $g$ must be a constant map. Then $X(\Sigma)$ and $\hat{X}(\Sigma)$ are parts of planes as surfaces in $\R^3$ with the same constant Gauss map. Therefore, $X(\Sigma)$ and $\hat{X}(\Sigma)$ are parts of the same vertical cylinder over a geodesic of $\s^2$ in  $\s^2\times\R$.
\end{remark}

\section{Examples}

As previously shown in the last section, minimal conformal immersions with constant Gauss map are vertical cylinders over geodesic of $\s^2$ in $\S2R$. In the case where the Gauss map $g$ is a non-constant holomorphic map, we have totally geodesic $2-$spheres in $\S2R$. These ones are the simplest examples of minimal surfaces in $\S2R$.

Next, we describe briefly classical examples studied by Pedrosa and Ritoré \cite{PedrosaRitore99}, Rosenberg \cite{rosenberg2002} and Daniel \cite{danieltams09} using our model of $\S2R$.

\begin{example}[Helicoids]  Let $\beta \neq 0$. We consider the conformal immersion
\[X_{\beta}(u+iv) = \big(e^v\sin\rho(u)e^{i\beta v},e^v \cos\rho(u)\big),\]
where the function $\rho$ satisfies
\begin{equation}\label{edo-helicoid}
\rho'^2(u)=1+\beta^2\sin^2\rho(u).
\end{equation}

This conformal immersion was presented in \cite[Section 4.2]{danieltams09} and corresponds to the minimal helicoid in $\S2R$. Following Daniel's terminology, for $\beta>0$ we say that $X_{\beta}$ is a right helicoid and for $\beta<0$ we say that $X_{\beta}$ is a left helicoid.

We can assume that $\rho'(u)>0$. We compute the normal vector $N_{\beta}$ by
\[N_{\beta}(u+iv) = \frac{1}{\rho'(u)}\begin{bmatrix}
                                            \sin\beta v+\beta\sin^2\rho(u)\cos\beta v
                                             \\
                                            -\cos\beta v+\beta\sin^2\rho(u)\sin\beta v \\
                                            \beta\sin\rho(u)\cos\rho(u) \\
                                          \end{bmatrix}.\]
Then the Gauss map $g_{\beta}$ is given by
\[g_{\beta}(u+iv) = \frac{(\beta\sin^2\rho(u)-i)e^{i\beta v}}{\rho'(u)+\beta\sin\rho(u)\cos\rho(u)}.\]

Moreover, the $r_{\beta}$ is given by $r_{\beta}(u+iv)=v$ and $p_{\beta}$ is given by
\[p_{\beta}(u+iv)=\frac{\sin\rho(u)}{1+\cos\rho(u)}e^{i\beta v}.\]

We note that when $\beta\to 0$ then $g_0(u+iv) = -i$, that is, $X_0$ is a vertical cylinder over a geodesic of $\s^2$ in  $\s^2\times\R$.

In order to obtain an explicit parametrization of the minimal helicoid in $\S2R$, we consider the following change of coordinates
\[u(s)=\int_{0}^{s}\frac{1}{\sqrt{1+\beta^2\sin^2(\sigma)}}\dif\sigma,\]
and we note that $\rho\circ u(s)=s$ satisfies equation \eqref{edo-helicoid}. In terms of these coordinates, we consider the immersion $\widetilde{X}_{\beta}:\C\to\S2R$ given by
\[\widetilde{X}_{\beta}(s+iv) = \big(e^v\sin(s)e^{i\beta v},e^v \cos(s)\big),\]
and, therefore, the surface $\widetilde{X}_{\beta}(\C)$ is the analytic continuation of the minimal helicoid in $\S2R$.

\begin{figure}[!ht]
\centering
\captionsetup{justification=centering}
\caption{\bf Part of the minimal helicoid in $\S2R$, with $\beta = 4$:}
\subfigure{\includegraphics[width=70mm]{helicoid-s2xr}}\vfil
\caption{\bf Minimal helicoid in $\S2R$, with $\beta = 4$:}
\subfigure{\includegraphics[width=90mm]{helicoid-s2xr-2}}
\end{figure}
\end{example}
\newpage

\begin{example}[Unduloids] Let $\alpha\in\R\setminus[-1,1]$. We consider the conformal immersion
\[X_{\alpha}(u+iv) = \big(e^u\sin\rho(u)e^{i\alpha v},e^u \cos\rho(u)\big),\]
where the function $\rho$ satisfies
\begin{equation}\label{edo-unduloid}
\rho'^2(u)+1=\alpha^2\sin^2\rho(u).
\end{equation}

This conformal immersion was presented in \cite[Section 4.2]{danieltams09} and corresponds to the minimal unduloid in $\S2R$.

We can assume that $\rho'(u)>0$ and $\rho(u)\in (0,\pi)$. We compute the normal vector $N_{\alpha}$ by
\[N_{\alpha}(u+iv) = \frac{1}{\alpha\sin\rho(u)}\begin{bmatrix}
                                            (\rho'(u)\sin\rho(u)-\cos\rho(u))\cos\alpha v
                                             \\
                                (\rho'(u)\sin\rho(u)-\cos\rho(u))\sin\alpha v \\
                            \sin\rho(u)+\rho'(u)\cos\rho(u) \\
                                          \end{bmatrix}.\]
Then the Gauss map $g_{\alpha}$ is given by
\[g_{\alpha}(u+iv) = \frac{(\rho'(u)\sin\rho(u)-\cos\rho(u))e^{i\alpha v}}{(1+\alpha)\sin\rho(u)+\rho'(u)\cos\rho(u)}.\]

Moreover, the $r_{\alpha}$ is given by $r_{\alpha}(u+iv)=u$ and $p_{\alpha}$ is given by
\[p_{\alpha}(u+iv)=\frac{\sin\rho(u)}{1+\cos\rho(u)}e^{i\alpha v}.\]

We note that when $\alpha\to \pm 1$ then $\rho'^2(u)+\cos^2\rho(u)=0$, that is, $\rho(u)=\pi/2$. Since
\[N_{-1}=\begin{bmatrix}
           0 \\
           0 \\
           -1
         \end{bmatrix} \text{ \ and \ }
         N_{+1}=\begin{bmatrix}
           0 \\
           0 \\
           1
         \end{bmatrix},\]
we have that $g_{-1}(u+iv)=\infty$ and $g_{+1}(u+iv)=0$. Therefore, $X_{-1}$ and $X_{+1}$ are vertical cylinders over geodesics of $\s^2$ in $\s^2\times\R$.

In order to obtain an explicit parametrization of the minimal unduloid in $\S2R$, we consider the following change of coordinates
\[u(s)=\int_{0}^{s}\frac{1}{\sqrt{1+(\alpha^2-1)\sin^2(\sigma)}}\dif\sigma,\]
and we note that
\[\rho\circ u(s)
=\arcsin\Bigg(\frac{\sqrt{1+(\alpha^2-1)\sin^2(s)}}{\alpha}\Bigg)\]
satisfies equation \eqref{edo-unduloid}. In terms of these coordinates, we consider the immersion $\widetilde{X}_{\alpha}:\C\to\S2R$ given by
\[\widetilde{X}_{\alpha}(s+iv) = \frac{e^{u(s)}}{\alpha}\Big(\sqrt{1+(\alpha^2-1)\sin^2(s)} e^{i\alpha v},\sqrt{\alpha^2-1}\cos(s)\Big),\]
and, therefore, the surface $\widetilde{X}_{\alpha}(\C)$ is the analytic continuation of the minimal unduloid in $\S2R$.

\begin{figure}[!ht]
\centering
\captionsetup{justification=centering}
\caption{\bf  Generating curve of the minimal unduloid in $\S2R$, with $\alpha = 8$:}
\subfigure{\includegraphics[width=55mm]{unduloid-s2xr}}\vfil
\caption{\bf Minimal unduloid in $\S2R$, with $\alpha = 8$:}
\subfigure{\includegraphics[width=110mm]{unduloid-s2xr-2}}
\end{figure}
\end{example}
\newpage

\bibliographystyle{amsplain}
\bibliography{references}

\end{document}

config: arxiv


TITLE: Subfields of a quartic extension, $\text{Gal}(E/F) \cong V$
QUESTION [2 upvotes]: I'm trying to figure out generators of subfields given $E$ and $F$, $\text{char}(F) \neq 2$, and $E$ the splitting field of $$f(x) = x^4-2ax^2+a^2-b^2c$$ which has roots
$$\alpha_1 = \sqrt{a+b\sqrt{c}} \quad \alpha_2 = -\sqrt{a+b\sqrt{c}} \quad \alpha_3 = \sqrt{a-b\sqrt{c}} \quad \alpha_4 = -\sqrt{a-b\sqrt{c}}$$
A condition for $E$ being the splitting field I've previously found is that $a^2-b^2c \in E$ is square, since
$$\sqrt{a-b\sqrt{c}} = \frac{\sqrt{a^2-b^2c}}{\sqrt{a+b\sqrt{c}}}$$
Moreover, $\text{Gal}(E/F) \cong V$ (Klein-4 group) if $\Delta(f)$ is square in $F$, so I'm assuming that.
I've thus far figured out that $F[\sqrt{c}]$ is one subfield, corresponding to the cyclic subgroup generated by $\langle (12)(34) \rangle$, however I'm not sure how to determine the other two. I know that
$$(13)(24) \text{ fixes } \sqrt{a+b\sqrt{c}} + \sqrt{a-b\sqrt{c}}$$
$$(14)(23) \text{ fixes } \sqrt{a+b\sqrt{c}} - \sqrt{a-b\sqrt{c}}$$
But I'm unsure how to ascertain the generators of the other two subfields. Since they are quadratic extensions of $F$, I can assert that the other two generators will be of the form $\sqrt{x}$ and $\sqrt{y}$, but aside from that I'm not sure.
Would appreciate any tips.

REPLY [0 votes]: I'm a bit puzzled by your question: you start by defining $E$ to be the
splitting field, then later you give a condition for $E$ to be the splitting field. You don't define $F$: is this your base field?
Let's suppose that $F$ is the ground field. Then $F(\alpha_1)$
is the splitting field iff $a^2-b^2c$ is a square in $F$. You want
the extension to be $V_4$. The discriminant is
$$\begin{align}
&(\alpha_1-\alpha_2)^2(\alpha_1-\alpha_3)^2(\alpha_1-\alpha_4)^2(\alpha_2-\alpha_3)^2(\alpha_2-\alpha_4)^2(\alpha_3-\alpha_4)^2\\
=&16\alpha_1^2\alpha_3^2(\alpha_1-\alpha_3)^2(\alpha_1+\alpha_3)^2(-\alpha_1-\alpha_3)^2(-\alpha_1+\alpha_3)^2\\
=&16\alpha_1^2\alpha_3^2(\alpha_1^2-\alpha_3^2)^4\\
=&16(a+b\sqrt c)(a-b\sqrt c)(2b\sqrt c)^4\\
=&256(a^2-b^2c)b^4c^2
\end{align}$$
so it's a square iff $a^2-b^2c$ is a square. Let's assume this.
Write $a^2-b^2c=r^2$ where $r=\alpha_1\alpha_3$. Let
$$\beta=\alpha_1+\alpha_3=\sqrt{a+b\sqrt c}+\sqrt{a-b\sqrt c}.$$
Then
$$\beta^2=2a+2r.$$
So one of the quadratic extensions you seek is $F(\sqrt{2(a+r)})$.
Of course, the other is $F(\sqrt{2(a-r)})$.


TITLE: Prove or disprove this inequality involving integral
QUESTION [0 upvotes]: I want to prove or disprove this inequality:
For $f_1\in L_1(\mathbb{R})$ and $f_2\in L_2(\mathbb{R})$,
$$\int_{\mathbb{R}}|f_1(x)|I_{\{x:f_1^2(x)\leq f_2^2(x)\}}dx\leq\int_{\mathbb{R}} I_{\{x:|f_1 (x)|\leq \int_{\mathbb{R}}f_2^2(x)dx\}}dx$$ where $I_E$ is the indicator function of a set $E$.
My attempt is to make the $$LHS  \leq\int_{\mathbb{R}}|f_1(x)|dx=\int_0^\infty\mu\{x:|f_1|>t\}dt$$  but this seems no help cause what we need is $|f_1|$ less than a value of an integral. Could you give me some suggestions or hints to deal with this problem? Thank you!

REPLY [1 votes]: RHS is the measure of $\{x: |f_1(x)| \leq c\}$ where $c=\|f_2\|_2^{2}$. Suppose this is finite and $c \neq 0$. Note that $\mu(\{x: |f_1(x)| > c\}\leq \frac  1 c \int |f_1(x)|dx<\infty$. Adding these we see that the real line has finite measure, a  contradction. Hence, either $c=0$, which ,means $f_2=0$ a.e. or RHS is $\infty$.
The inequality is trivial when $f_2=0$ a.e.


TITLE: Weak derivative of $\operatorname{sgn}(x_1)$
QUESTION [1 upvotes]: Let $x\in \mathbb{R}^{n}, x = (x_1,\ldots,x_n)$, and $f(x) = \operatorname{sgn}(x_{1})$. Is $f$ weakly differentiable on $U = B(0,1)$, i.e. unit ball in $\mathbb{R}^{n}$, and what is the weak derivative?

REPLY [1 votes]: We see that $ sgn(x) = 2H(x) -1 $ where $ H $ is the Heaviside function with $ H(0) = \frac{1}{2}$. Then distribution derivative of $ sgn $ would be $ 2\delta_x $ which is not induced by any function. So $f$ is not weakly differentiable.