text:
\begin{document}

\title[Principally polarized squares of elliptic curves]{Principally polarized squares of elliptic curves\\ with field of moduli equal to $\QQ$}
\date{}

\author[A. G\'elin]{Alexandre G\'elin}
\address{Laboratoire de Math\'ematiques de Versailles,
         UVSQ, CNRS, Universit\'e Paris-Saclay,
         45 avenue des \'Etats-Unis,
         78035 Versailles, France.}
\email{\href{mailto:alexandre.gelin@uvsq.fr}{alexandre.gelin@uvsq.fr}}
\urladdr{\url{https://alexgelin.github.io/}}
\thanks{This work was supported in part by a public grant as part of the \emph{Investissement d'avenir} project, reference ANR-11-LABX-0056-LMH, LabEx LMH}

\author[E.\,W. Howe]{Everett W. Howe}
\address{Center for Communications Research,
         4320 Westerra Court,
         San Diego, CA 92129-1967 U.S.A.}
\email{\href{mailto:however@alumni.caltech.edu}{however@alumni.caltech.edu}}
\urladdr{\href{http://alumnus.caltech.edu/~however}{http://alumnus.caltech.edu/\lowtilde{}however}}

\author[C. Ritzenthaler]{Christophe Ritzenthaler}
\address{IRMAR,
         Universit\'e de Rennes 1,
         Campus de Beaulieu,
         263 avenue du G\'en\'eral Leclerc,
         35042 Rennes Cedex, France.}
\email{\href{mailto:christophe.ritzenthaler@univ-rennes1.fr}{christophe.ritzenthaler@univ-rennes1.fr}}
\urladdr{\url{https://perso.univ-rennes1.fr/christophe.ritzenthaler/}}

\subjclass[2010]{Primary 11G15; Secondary 14H25, 14H45}

\begin{abstract}
We give equations for $13$ genus-$2$ curves over~$\QQbar$, with models over~$\QQ$, whose unpolarized Jacobians are isomorphic to the square of an elliptic curve with complex multiplication by a maximal order. If the Generalized Riemann Hypothesis is true, there are no further examples of such curves. More generally, we prove under the Generalized Riemann Hypothesis that there exist exactly $46$ genus-$2$ curves over $\QQbar$ with field of moduli $\QQ$ whose Jacobians are isomorphic to the square of an elliptic curve with complex multiplication by a maximal order.
\end{abstract}

\maketitle

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

For $g>1$, let $\MM_g$ (resp.~$\Aa_g$) be the moduli space classifying absolutely irreducible projective smooth curves of genus~$g$ (resp.~principally polarized abelian varieties of dimension~$g$) over $\QQbar$. These spaces are quasi-projective varieties defined over~$\QQ$, linked by the Torelli map, which associates to a curve its Jacobian. To explain the modular interpretation of rational points on these spaces, we must define the terms \emph{field of definition} and \emph{field of moduli}. If $X$ is a curve or polarized abelian variety over $\QQbar$, we say that a field $F \subseteq \QQbar$ is a \emph{field of definition} of $X$ if there exists a variety~$X_0/F$ --- called a \emph{model} of $X$ over $F$ --- such that $X_0 \simeq_{\QQbar} X$. Since $\QQbar$ is a field of characteristic~$0$, by~\cite[Corollary~3.2.2, p.~54]{Koi72} we can define the \emph{field of moduli} of $X$ to be either
\begin{itemize}
\item the field fixed by the subgroup $\{\sigma \in \Gal(\QQbar/\QQ) ~|~ X \simeq X^{\sigma}\}$, or
\item the intersection of the fields of definition of $X$.
\end{itemize}
With these terms defined, we can say that the rational points on $\MM_g$ (resp.~$\Aa_g$) correspond to the isomorphism classes of curves (resp.~principally polarized abelian varieties) over $\QQbar$ that have field of moduli~$\QQ$~\cite{Bai62}.

There are a number of interesting sets of rational points on $\Aa_g$, but the complex multiplication (CM) abelian varieties --- that is, the principally polarized abelian varieties having endomorphism rings containing an order in a number field of degree~$2g$ over~$\QQ$ --- have attracted the most interest. When such a point on~$\Aa_g$ lies in the image of $\MM_g$, the corresponding curve is called a \emph{CM-curve}. For $g=2$, the set of \emph{simple} CM-abelian varieties with field of moduli $\QQ$ is known, and for those varieties that are Jacobians explicit equations have been computed for the corresponding curves~\cite{Spa94,Wam99,MU01,KS15,BS17}; for $g=3$ the similar set of possible CM maximal orders is determined in~\cite{Kil16} and conjectural equations for the curves are given in~\cite{Wen01,KW05,BILV16,LS16,KLL+17}. (And while we have avoided the case $g=1$ in the discussion above for technical reasons, it is still of course true that the CM-elliptic curves with rational $j$-invariants are known as well~\cite[Appendix~A.3]{Sil94}.)

In this article we consider genus-$2$ curves whose Jacobians are non-simple CM-abelian surfaces. Every such surface is isogenous to the square of a CM-elliptic curve, but we restrict our attention in two ways: first, we look only at surfaces that are \emph{isomorphic} (and not just isogenous) to $E^2$ for a CM-elliptic curve~$E$, and second, we only consider $E$ that have CM by a maximal order. The second restriction is not essential to our methods, and we impose it here in order to simplify some of our calculations. Note that if the elliptic curve $E$ has no CM --- \ie $\End(E)\simeq\ZZ$, then $E^2$ cannot be isomorphic to the Jacobian of a genus-$2$ curve, because $E^2$ has no indecomposable principal polarizations~\cite[Corollary~4.2, p.~159]{Lan05}.

\subsection*{Main Contributions.}
We prove under the Generalized Riemann Hypothesis that there exist exactly $46$ genus-$2$ curves over $\QQbar$ with field of moduli $\QQ$ whose Jacobians are isomorphic to the square of an elliptic curve with CM by a maximal order. We show that among these $46$ curves exactly $13$ can be defined over $\QQ$, and we give explicit equations for them. In order to accomplish this, we develop an algorithm to compute, for an imaginary quadratic maximal order $\Ocal$, canonical forms for all positive definite unimodular Hermitian forms on $\Ocal\times\Ocal$. Such Hermitian forms correspond to principal polarizations $\varphi$ on $E^2$, and our algorithm computes the automorphism group of the polarized variety $(E^2,\varphi)$ and identifies the polarizations that come from genus-$2$ curves.

\subsection*{Related work.}
Hayashida and Nishi~\cite{HN65} consider in particular when a product of two elliptic curves, with CM by the same maximal order~$\Ocal$, is the Jacobian of a curve over~$\CC$, and they find that this happens if and only if the discriminant of~$\Ocal$ is different from $-1$, $-3$, $-7$, and $-15$. Hayashida~\cite{Hay68} gives the number of indecomposable principal polarizations on $E^2$ where $E/\CC$ is an elliptic curve with CM by a maximal order. More recently, Kani~\cite{Kan14,Kan16} gives existence results on Jacobians isomorphic to the product of two elliptic curves with control on the polarization, and Schuster~\cite{Sch90} and Lange~\cite{Lan05} study generalizations to higher dimensions. Rodriguez-Villegas~\cite{Rod00} considers the same situation as Hayashida and Nishi, and in the case where $\Ocal$ has class number~$1$ and odd discriminant, he gives an algorithm (relying on quaternion algebras) for producing curves with field of moduli~$\QQ$. Note finally that Fit\'e and Guitart~\cite{FG18} determine when there exists an abelian surface $A/\QQ$ that is $\QQbar$-isogenous to $E^2$, with $E/\QQbar$ a~CM-curve.

\subsection*{Outline.}
Our article proceeds as follows. Torelli's theorem (see \cite[Appendix]{LS01}) implies that our genus-$2$ curve $C$ has field of moduli $\QQ$ if and only if its principally polarized Jacobian $(E^2,\varphi)$ has field of moduli~$\QQ$. We therefore need to find all elliptic curves $E$ with CM by a maximal order $\Ocal$ and all polarizations $\varphi$ of $E^2$ such that $(E^2,\varphi)$ is isomorphic to all of its $\Gal(\QQbar/\QQ)$-conjugates. Proposition~\ref{prop:exp2} shows that if $E^2$ is isomorphic to all of its Galois conjugates --- even just as an abelian variety without polarization --- then the class group of $\Ocal$ has exponent at most~$2$. Under the Generalized Riemann Hypothesis, this gives us an explicit finite list of possible orders (Table~\ref{tab:order}). For each of these orders~$\Ocal$, one can identify the indecomposable principal polarizations $\varphi$ on $E^2$ and describe them as certain~\mbox{$2$-by-$2$} matrices $M$ with coefficients in $\Ocal$ (Proposition~\ref{prop:matM}). Tables of such matrices were computed by Hoffmann~\cite{Hof91} and Schiemann~\cite{Sch98} and were published online,\footnote{Available at \url{https://www.math.uni-sb.de/ag/schulze/Hermitian-lattices/}.} but they only include a fraction of the discriminants that we must consider. We therefore describe an algorithm, using a method different from that of Hoffmann and Schiemann, that we use to recompute these tables of matrices (Section~\ref{sec:HowPol}). Given such a matrix~$M$, we find explicit algebraic conditions on~$M$ for the principally polarized abelian surface~$(E^2,\varphi)$ to have field of moduli~$\QQ$ (Section~\ref{sec:cond}). We check whether these conditions are satisfied for each $M$ on our list.

We conclude the article with three more results: we heuristically compute the Cardona--Quer invariants~\cite{CQ05} of the associated curves~$C$ and see that the factorization of their denominators reveals interesting patterns; we show that the field of moduli is a field of definition if and only if $C$ has a non-trivial group of automorphisms (\ie of order greater than~$2$, see Section~\ref{prop:fielddef}); and for the curves $C$ defined over~$\QQ$, we compute equations and prove that they are correct.

\subsection*{Notation.}
In the following, $E$ is an elliptic curve over~$\QQbar$ with complex multiplication by a maximal order $\Ocal$ of discriminant $\Delta$ and with fraction field~$K$, which we sometimes call the \emph{CM-field}.

\section{Condition on \texorpdfstring{$E^2$}{E2}} \label{sec:woPol}

We are interested in the field of moduli $\Mb$ of a principally polarized abelian surface $(E^2,\varphi)$. As outlined above, we first consider the abelian surface $E^2$ alone and we give a necessary condition for $\Mb$ to be contained in the CM-field $K$. If~$\Mb\subseteq K$ then in particular we have $E^2\simeq(E^\sigma)^2$ for all $\sigma \in \Gal\left(\QQbar/K\right)$. The class group~$\Cl(\Ocal)$ acts simply transitively on the set of elliptic curves with CM by~$\Ocal$~\cite[Proposition~1.2, p.~99]{Sil94}. Since $\End(E^{\sigma})=\End(E)=\Ocal$, for each \mbox{$\sigma \in \Gal(\QQbar/K)$}, there exists a unique class of ideals $I_\sigma \in \Cl (\Ocal)$ such that~\mbox{$E^\sigma \simeq E / I_\sigma$}.

Using a result of Kani~\cite[Proposition~65, p.~335]{Kan11}, we get that, for $E$, $\sigma$ and~$I_\sigma$ defined as above, \[E^2 \simeq (E / I_\sigma)^2 \quad\Longleftrightarrow\quad I_\sigma^2 = \left[\Ocal\right],\] where the last equality is in $\Cl(\Ocal)$. Note that since we only work with maximal orders, the conditions on the conductors in Kani's result are trivially satisfied. Moreover by~\cite[Theorem~4.3, p.~122]{Sil94}, since for any $I \in \Cl(\Ocal)$ there exists $\sigma \in \Gal(\QQbar/K)$ (actually even in $\Gal\left(K\big(j(E)\big)/K\right)$) such that $E/I=E^{\sigma}$, we get the following proposition.

\begin{proposition} \label{prop:exp2}
A necessary condition for $\Mb \subseteq K$ is that the class group of $\Ocal$ has exponent at most $2$.
\end{proposition}

Louboutin~\cite{Lou90} shows that under the assumption of the Generalized Riemann Hypothesis, the discriminant $\Delta$ of an imaginary quadratic field whose class group is of exponent at most $2$ satisfies $|\Delta|\le 2\cdot 10^7$. In Table~\ref{tab:order} we list the $65$ fundamental discriminants satisfying this bound that give class groups of exponent at most~$2$.

{\small
\begin{table}[ht]
\centering
\begin{tabular}{cl}
\toprule
$\# \Cl(\Ocal)$ & \multicolumn{1}{c}{Discriminants $\Delta$} \\
\midrule
  $2^0$ &    $-3$,    $-4$,    $-7$,    $-8$,   $-11$,   $-19$,   $-43$,   $-67$, $-163$          \\[1ex]
  $2^1$ &   $-15$,   $-20$,   $-24$,   $-35$,   $-40$,   $-51$,   $-52$,   $-88$,  $-91$, $-115$, \\
        &  $-123$,  $-148$,  $-187$,  $-232$,  $-235$,  $-267$,  $-403$,  $-427$  \\[1ex]
  $2^2$ &   $-84$,  $-120$,  $-132$,  $-168$,  $-195$,  $-228$,  $-280$,  $-312$, \\
        &  $-340$,  $-372$,  $-408$,  $-435$,  $-483$,  $-520$,  $-532$,  $-555$, \\
        &  $-595$,  $-627$,  $-708$,  $-715$,  $-760$,  $-795$, $-1012$, $-1435$  \\[1ex]
  $2^3$ &  $-420$,  $-660$,  $-840$, $-1092$, $-1155$, $-1320$, $-1380$,          \\
        & $-1428$, $-1540$, $-1848$, $-1995$, $-3003$, $-3315$                    \\[1ex]
  $2^4$ & $-5460$ \\
\bottomrule
\end{tabular}
\bigskip
\caption{Discriminants $\Delta$ of the imaginary quadratic maximal orders $\Ocal$ of exponent at most~$2$, conditional on the Generalized Riemann Hypothesis.} \vspace{-5.5ex}
\label{tab:order}
\end{table}}

\section{Polarized abelian surfaces} \label{sec:Pol}

\subsection{Polarizations on the square of an elliptic curve}

We now consider the principal polarizations on the product surface $A = E^2$. A principal polarization on $A$ is, in particular, an isogeny of degree $1$ from $A$ to the dual $\Adual$ of $A$, but not every isomorphism $A\to\Adual$ is a principal polarization; other properties must be satisfied as well (see~\cite[\S~4.1]{BL04}). One such polarization is the product polarization $\varphi_0 = \varphi_E \times \varphi_E$. Given any other principal polarization $\varphi$, we can consider the automorphism $M = \varphi_0^{-1}\varphi$ of $A$, which (in light of the isomorphism $A = E^2$) we view as a matrix\footnote{All matrices in this paper act on the left.} in $\GL_2(\Ocal)$. Our first result characterizes the matrices that arise in this way; the statement is not new, but we provide a proof here because it introduces some of the ideas used in the sequel. (Recall~\cite[Exercise~7, p.~134]{Hal74} that two matrices $M_1$ and~$M_2$ in~$\GL_2(\Ocal)$ are said to be \emph{congruent} if there exists a matrix $P \in \GL_2(\Ocal)$ such that~$P^* M_1 P = M_2$, where $P^*$ is the conjugate transpose of~$P$.)

\begin{proposition} \label{prop:matM}
The map $M \mapsto \varphi_0 \cdot M$ defines a bijection between the positive definite unimodular Hermitian matrices with coefficients in $\Ocal$ and the principal polarizations on~$A$. Two principal polarizations are isomorphic to one another if and only if their associated matrices are congruent to one another.
\end{proposition}

\begin{proof}
By~\cite[Theorem~5.2.4, p.~121]{BL04}, the matrices $M$ corresponding to principal polarizations are totally positive symmetric endomorphisms of norm~$1$. Here the symmetry is with respect to the Rosati involution of $\End(A)$ associated to the polarization $\varphi_0$, which is the conjugate-transpose involution under the identification $\End(A) = M_2(\Ocal)$. Thus, the matrices $M$ corresponding to principal polarizations are exactly the positive definite unimodular Hermitian matrices.

Let $\varphi_1$ and $\varphi_2$ be two principal polarizations on $A$, corresponding to matrices~$M_1$ and $M_2$. The polarizations $\varphi_1$ and $\varphi_2$ are isomorphic to one another if and only if there exists an automorphism $\alpha\colon A\to A$ such that $\alphadual\varphi_1\alpha = \varphi_2$, where $\alphadual\colon\Adual\to\Adual$ is the dual of $\alpha$. This last condition is equivalent to
$(\varphi_0^{-1}\alphadual\varphi_0) (\varphi_0^{-1}\varphi_1\alpha) = \varphi_0^{-1}\varphi_2$. Now, $\varphi_0^{-1} \alphadual\varphi_0$ is nothing other than the Rosati involute of $\alpha$, so if we write $\alpha$ as a matrix $P \in \GL_2(\Ocal)$, the condition that determines whether $\varphi_1$ and $\varphi_2$ are isomorphic is simply $P^* M_1 P = M_2$.
\end{proof}

The principal polarizations on $A$ come in two essentially different types.

\begin{definition}
A polarization $\varphi$ on an abelian variety $A$ over a field $k$ is said to be \emph{geometrically decomposable} if there exist two abelian varieties $A_1$ and $A_2$ over $\bar{k}$ of positive dimension, together with polarizations $\varphi_1$ and $\varphi_2$, such that $(A,\varphi)$ and $(A_1 \times A_2, \varphi_1 \times \varphi_2)$ are isomorphic over $\bar{k}$. A polarization that is not geometrically decomposable is \emph{geometrically indecomposable}. For brevity's sake, in this paper we drop the adjective \emph{geometrically} and simply use the terms \emph{decomposable} and \emph{indecomposable} for these concepts.
\end{definition}

Results in~\cite{Wei57,Hoy63,OU73} show that a principally polarized abelian surface is the Jacobian of a curve if and only if the polarization is indecomposable. In the remainder of this section we show how we can easily compute representatives for the congruence classes of matrices representing the decomposable polarizations on~$E^2$; we focus on the indecomposable polarizations in later sections.

\begin{proposition} \label{prop:decomposable}
If $\varphi$ is a decomposable polarization on $E^2$, then there exist elliptic curves $F$ and $F'$ that have CM by $\Ocal$ such that $\varphi$ is the pullback to $E^2$ of the product polarization on $F\times F'$ via some isomorphism $E^2 \simeq F\times F'$. The pair~$(F,F')$ giving rise to a given decomposable polarization is unique up to interchanging $F$ and~$F'$ and up to isomorphism for each elliptic curve. Moreover, for every~$F$ with CM by $\Ocal$ there exists an $F'$ with CM by $\Ocal$ such that $E^2 \simeq F\times F'$.
\end{proposition}

\begin{proof}
First we note that by definition, if $\varphi$ is a decomposable polarization on $E^2$ there must exist elliptic curves $F$ and $F'$, isogenous to $E$, such that $\varphi$ is the pullback of the product polarization on $F\times F'$ under some isomorphism $E^2\simeq F\times F'$. Now, the center of $\End(E^2)$ is $\End(E) = \Ocal$, while the center of $\End(F\times F')$ is~$\End(F)\cap\End(F')$; since $\Ocal$ is a maximal order, $F$ and $F'$ both have CM by~$\Ocal$.

If $(\alpha,\beta)\colon G\to F\times F'$ is an embedding of an elliptic curve $G$ into $F\times F'$, then the pullback of the product polarization to $G$ is the morphism
\[\begin{bmatrix} \hat\alpha & \hat\beta\end{bmatrix} \begin{bmatrix} 1& 0 \\ 0 & 1\end{bmatrix} \begin{bmatrix} \alpha \\ \beta\end{bmatrix}
 = \hat\alpha \alpha + \hat\beta \beta = \deg(\alpha) + \deg(\beta);\]
that is, the pullback is the multiplication-by-$d$ map, with ${d = \deg(\alpha) + \deg(\beta)}$. It follows that if $\varphi$ is the pullback to $E^2$ of the product polarization on $F\times F'$ via some isomorphism $E^2 \simeq F\times F'$, then the set of elliptic curves~$G$ for which there exists an embedding $\epsilon\colon G\to E^2$ such that $\epsilon^*\varphi$ is a principal polarization is simply~$\{F,F'\}$. Thus, for a given decomposable principal polarization, the pair $(F,F')$ is unique up to order and isomorphism.

As we noted at the beginning of Section~\ref{sec:woPol}, the set of elliptic curves with CM by~$\Ocal$ is a principal homogenous space for the class group of $\Ocal$. Given an $F$ with CM by $\Ocal$, let~$I\in\Cl(\Ocal)$ be the ideal class that takes $E$ to $F$. If $F'$ is an elliptic curve with CM by $\Ocal$, say corresponding to an ideal class $I'\in\Cl(\Ocal)$, then $E^2 \simeq F\times F'$ if and only if $I'$ is the inverse of $I$ (see~\cite[Proposition~65, p.~335]{Kan11}). This proves the final statement of the proposition.
\end{proof}

\begin{corollary} \label{cor:decomposable}
Let $h$ denote the class number of $\Ocal$, and let $t$ denote the size of the~$2$-torsion subgroup of the class group. The number of decomposable polarizations on $E^2$ is equal to $(h+t)/2$.
\end{corollary}

\begin{proof}
The proof of Proposition~\ref{prop:decomposable} shows that the unordered pairs $(F,F')$ with $E^2 \simeq F\times F'$ correspond to unordered pairs $(I,I^{-1})$, where $I\in \Cl(\Ocal)$. The number of such pairs is $(h+t)/2$.
\end{proof}

Let $F$ be an elliptic curve with CM by $\Ocal$ and let $I$ be the ideal class that takes~$E$ to $F$. Let $\afrak$ be an ideal of $\Ocal$ representing~$I$, such that $\afrak$ is not divisible by any nontrivial ideal of $\ZZ$. We may write $\afrak = (n,\alpha)$, where $n=\Norm(\afrak)\in\ZZ$ and where~$\alpha\in\afrak$ is chosen so that the ideal $\alpha\afrak^{-1}$ is coprime to $n\Ocal$; then there exist~$x,y\in\ZZ$ such that $xn^2 - y\Norm(\alpha) = n$. Let~$F'$ be the elliptic curve such that $E^2 \simeq F\times F'$. We prove the following corollary in~Section~\ref{sec:cond}.

\begin{corollary} \label{cor:explicit-decomposable}
In the notation of the paragraph above, the isomorphism class of the decomposable polarization on $E^2$ obtained from pulling back the product polarization on $F\times F'$ is represented by the congruence class of the matrix \[\begin{pmatrix} n + \frac{\Norm(\alpha)}{n} & (x + y)\alpha \\ (x + y)\bar{\alpha} & x^2 n + y^2\frac{\Norm(\alpha)}{n}\end{pmatrix}.\]
\end{corollary}

\subsection{How to find the polarizations?} \label{sec:HowPol}

In Section~\ref{sec:woPol}, we identified 65 orders~$\Ocal$ for which we need to compute the set of indecomposable principal polarizations, or equivalently, representatives of the congruence classes of indecomposable positive definite unimodular Hermitian matrices with coefficients in $\Ocal$. In this section we describe how we computed these representatives.

Fix an embedding $\epsilon_0$ of $K$ into the complex numbers. For any $\alpha\in\Ocal$, we write~$\alpha>0$ if either the trace of $\alpha$ is positive, or the trace of $\alpha$ is $0$ and $\epsilon_0(\alpha)$ has positive imaginary part. Then for $\alpha,\beta\in\Ocal$ we write $\alpha>\beta$ if $\alpha-\beta > 0$. Clearly this gives us a total ordering on $\Ocal$.

Let $\Hcal$ denote the set of positive definite unimodular Hermitian matrices with coefficients in $\Ocal$. Let $\chi\colon\Hcal\to\NN\times\NN\times\Ocal$ be the map that sends a matrix
$M = \left(\begin{smallmatrix} a&b\\ \bar{b}&d \end{smallmatrix}\right)$ to the triple $(a,d,b)$. We define a total ordering on $\Hcal$ by saying that $M_1 < M_2$ if~$\chi(M_1) < \chi(M_2)$ in the lexicographic ordering on $\NN\times\NN\times\Ocal$.

Given any $M\in\Hcal$, we say that $M$ is \emph{reduced} if $M\le M'$ for all $M'$ congruent to~$M$. Clearly every $M\in\Hcal$ is congruent to a unique reduced matrix. The following algorithm produces the reduced matrix that is congruent to a given~$M$.

\begin{algorithm} \label{alg:reduced-form}
\begin{algtop}
\algin A positive definite unimodular Hermitian matrix $M$ with coefficients
       in $\Ocal$, specified by $a,d\in\ZZ$ and $b\in\Ocal$ such that
       $M = \left(\begin{smallmatrix}a&b\\\bar{b}&d\end{smallmatrix}\right)$.
\algout The reduced matrix congruent to $M$.
\end{algtop}
\begin{alglist}
\item Set $a' = 1$.
\item \label{reduced-short1}
      Compute the set $A'$ of vectors
      $\xx = (x_1,x_2) \in\Ocal^2$ such that $\xx^* M \xx = a'$
      and such that $x_1$ and $x_2$ generate the unit ideal of $\Ocal$.
      If $A'=\emptyset$, increment $a'$ and repeat.
\item Set $d'= a'$.
\item \label{reduced-short2}
      Compute the set $D'$ of vectors
      $\yy = (y_1,y_2) \in\Ocal^2$ such that $\yy^* M \yy = d'$
      and such that $y_1$ and $y_2$ generate the unit ideal of $\Ocal$.
      If $D'=\emptyset$, increment $d'$ and repeat.
\item Initialize $\Mcal$ to be the empty set.
\item For each $\xx \in A'$ and $\yy \in D'$ such that $\xx$ and $\yy$
      generate $\Ocal^2$ as an $\Ocal$-module,
      let~$M'$ be the matrix representing the Hermitian form $M$
      written on the basis $\xx$, $\yy$ of $\Ocal^2$,
      and add $M'$ to the set $\Mcal$.
\item If $\Mcal$ is empty, increment $d'$ and return to
      Step~\eqref{reduced-short2}.
\item Find the smallest element $M'$ of $\Mcal$ under the ordering
      of $\Hcal$ defined above.
\item Output $M'$.
\end{alglist}
\end{algorithm}

\begin{remark} \label{rem:vect}
In Steps~\eqref{reduced-short1} and~\eqref{reduced-short2} of Algorithm~\ref{alg:reduced-form}, we need to find vectors in $\Ocal^2$ of a given length under the quadratic form specified by $M$. We note that this is a finite computation: if $\xx = (x_1, x_2)$ satisfies $\xx^* M \xx = n$, with $M = \left(\begin{smallmatrix}a&b\\\bar{b}&d\end{smallmatrix}\right)$, then \[\Norm(ax_1 + bx_2) + \Norm(x_2) = an.\]
Thus, to solve $\xx^* M \xx = n$, we can simply enumerate all pairs $(u,v)\in\Ocal^2$ with $\Norm(u)+\Norm(v) = an$, and keep those pairs for which $u-bv$ is divisible by~$a$.

Note that solving $\xx^* M \xx = n$ can be done more quickly when the value of $a$ is small. Thus, in Algorithm~\ref{alg:reduced-form}, once one finds a short vector $\xx = (x_1,x_2)$ with~$x_1$ and $x_2$ coprime, it is worthwhile to compute \emph{any} vector $\yy$ such that $\xx$ and $\yy$ generate~$\Ocal$, and to replace $M$ with the congruent form obtained by rewriting $M$ on the basis $\xx,\yy$.
\end{remark}

\begin{theorem} \label{thm:reduced-form}
Algorithm~\textup{\ref{alg:reduced-form}} terminates with the correct result.
\end{theorem}

\begin{proof}
Let $M' = \big(\begin{smallmatrix}a'&b'\\\bar{b}'&d'\end{smallmatrix}\big)$ be the reduced matrix congruent to $M$. If $P = \left(\begin{smallmatrix}x_1&y_1\\x_2&y_2\end{smallmatrix}\right)$ is an element of $\GL_2(\Ocal)$ such that $P^* M P = M'$, and if we set $\xx = (x_1,x_2)$ and~$\yy = (y_1,y_2)$, then $a' = \xx^* M \xx$ and $d' = \yy^* M \yy$. By the very definition of the ordering on $\Hcal$, then, we want to find vectors $\xx$ and $\yy$, each with coordinates that are coprime to one another, such that $\xx^*M\xx$ is as small as possible and $\yy^*M\yy$ is as small as possible, given that $\xx$ and $\yy$ generate $\Ocal^2$ as an $\Ocal$-module. This is what the algorithm does. Finally, among all possible such pairs $(\xx,\yy)$, we simply need to choose the one that gives the smallest matrix.
\end{proof}

Hayashida~\cite{Hay68} gives a formula for the number of isomorphism classes of indecomposable principal polarizations on $E^2$ in the case where $E$ has CM by a maximal order.\footnote{There is a typographical error in Hayashida's paper. In the second line of page 43, the term $(1/4)(1 - (-1))^{(m^2-1)/8}$ should be $(1/4)(1 - (-1)^{(m^2-1)/8})h$. Note that the correction involves both moving a parenthesis and adding an instance of the variable~$h$.} Hayashida's proof does not immediately lead to a constructive method of finding polarizations representing the isomorphism classes, but simply knowing the number of isomorphism classes is the key to a straightforward algorithm for producing such representatives.

\begin{algorithm} \label{alg:all-polarizations}
\begin{algtop}
\algin A fundamental discriminant $\Delta<0$.
\algout A list of reduced matrices representing the distinct congruence classes of positive definite unimodular Hermitian matrices with entries in the order $\Ocal$ of discriminant $\Delta$, separated into the decomposable and indecomposable classes.
\end{algtop}
\begin{alglist}
\item Compute the number $N$ of indecomposable polarizations on $E^2$ using
      Hayashida's formula.
\item Compute the set $\Dcal$ of reduced matrices representing decomposable
      polarizations, using Corollary~\textup{\ref{cor:explicit-decomposable}}
      and Algorithm~\textup{\ref{alg:reduced-form}}.
\item Initialize $\Ical$ to be the empty set and set $P = 0$.
\item \label{all-product}
      Increment $P$, and
      compute the set $S$ of elements of $\Ocal$ of norm $P-1$.
\item For every divisor $a$ of $P$ with $a\le P/a$, and for every $b \in S$\textup{:}
      \begin{algsublist}
      \item Compute the reduced form $M$ of the matrix
            $\left(\begin{smallmatrix}a&b\\\bar{b}&P/a\end{smallmatrix}\right)$.
      \item If $M$ is not contained in $\Dcal\cup \Ical$,
            then add $M$ to the set $\Ical$.
      \end{algsublist}
\item If $\#\Ical < N$, then return to Step~\eqref{all-product}.
\item Return $\Dcal$ and $\Ical$.
\end{alglist}
\end{algorithm}

Of course, for our goal of producing genus-$2$ curves over $\QQ$ with Jacobians isomorphic to $E^2$, we only need the indecomposable polarizations.

\begin{theorem} \label{thm:all-polarizations}
Algorithm~\textup{\ref{alg:all-polarizations}} terminates with the correct result.
\end{theorem}

\begin{proof}
The algorithm is very straightforward. Every isomorphism class of principal polarization appears somewhere on the countable list that we are considering, and we simply enumerate the polarizations and compute their reduced forms until we have found the right number of isomorphism classes.
\end{proof}

\begin{remark}
In our applications, when the class group of $\Ocal$ has exponent at most~$2$, we can speed up our algorithm as follows: once we have a principal polarization $M$ on $E^2$, we can view the same matrix as giving a polarization on $F^2$ for any elliptic curve $F$ with CM by $\Ocal$. Since the class group has exponent at most~$2$, there exists an isomorphism $E^2 \to F^2$, and pulling $M$ back to $E^2$ via such an isomorphism gives a new positive definite unimodular Hermitian matrix~$M'$. Each time we find a new reduced polarization~$M$, we compute the reduced forms of the polarizations~$M'$ associated to all the curves $F$ isogenous to~$E$, and add these reduced forms to the set $\Dcal$ if they are new.
\end{remark}

If $\varphi$ is a principal polarization on $E^2$ and $M$ is the corresponding Hermitian matrix, then the automorphism group of the polarized abelian variety $(E^2,\varphi)$, denoted by $\Aut(E^2,\varphi)$, is isomorphic to the group $\{ P \in \GL_2(\Ocal) ~|~ P^* M P=M\}$. Note that if $\varphi$ is indecomposable, so that $(E^2,\varphi)$ is the polarized Jacobian of a curve~$C$, then Torelli's theorem~\cite[Appendix]{LS01} shows that this group is also isomorphic to $\Aut(C)$. In any case, computing $\Aut(E^2,\varphi)$ is straightforward:

\begin{algorithm} \label{alg:automorphisms}
\begin{algtop}
\algin A positive definite unimodular Hermitian matrix
       $M = \left(\begin{smallmatrix}a&b\\\bar{b}&d\end{smallmatrix}\right)$
       with entries in an imaginary quadratic maximal order $\Ocal$.
\algout A list of all matrices $P \in \GL_2(\Ocal)$ such that $P^* M P=M$.
\end{algtop}
\begin{alglist}
\item Compute the set $A$ of vectors
      $\xx = (x_1,x_2) \in\Ocal^2$ such that $\xx^* M \xx = a$
      and such that $x_1$ and $x_2$ generate the unit ideal of $\Ocal$.
\item Compute the set $D$ of vectors
      $\yy = (y_1,y_2) \in\Ocal^2$ such that $\yy^* M \yy = d$
      and such that $y_1$ and $y_2$ generate the unit ideal of $\Ocal$.
\item Initialize $\Acal$ to be the empty set.
\item For each $\xx \in A$ and $\yy \in D$ such that $\xx$ and $\yy$
      generate $\Ocal^2$ as an $\Ocal$-module\textup{:}
      \begin{algsublist}
      \item Compute $b' = \xx^* M \yy$.
      \item If $b' = b$ then add the matrix
      $\left(\begin{smallmatrix}x_1&y_1\\x_2&y_2\end{smallmatrix}\right)$
      to the set $\Acal$.
      \end{algsublist}
\item Output $\Acal$.
\end{alglist}
\end{algorithm}

\noindent(See Remark~\ref{rem:vect} for an explanation of how to implement the two first steps.)

\begin{theorem} \label{thm:automorphisms}
Algorithm~\textup{\ref{alg:automorphisms}} terminates with the correct result.
\end{theorem}

\begin{proof}
If $P=\left(\begin{smallmatrix} x_1&y_1\\ x_2&y_2\end{smallmatrix}\right)\in\GL_2(\Ocal)$ satisfies $P^* M P=M$, then $\xx = (x_1,x_2)$ and $\yy = (y_1,y_2)$ are vectors in $\Ocal^2$ such that $\xx^* M \xx = a$ and $\yy^* M \yy = d$ and $\xx^* M \yy = b$. The algorithm simply enumerates all $\xx$ and $\yy$ that meet the first two conditions, and checks to see whether they meet the third.
\end{proof}

\subsection{Conditions on the polarization} \label{sec:cond}

Throughout this section, $E$ is an elliptic curve with CM by a maximal order $\Ocal$ of an imaginary quadratic field $K$ whose class group has exponent at most~$2$. Also $\varphi$ is a principal polarization on $E^2$ corresponding (as in Proposition~\ref{prop:matM}) to a positive definite unimodular Hermitian matrix $M$ with entries in $\Ocal$ and $\Mb$ is the field of moduli of the polarized abelian variety $(E^2,\varphi)$. We resume our analysis of the condition that $\Mb = \QQ$.

\begin{proposition} \label{prop:polarization}
Let $\afrak_1$, $\dotsc$, $\afrak_h$ be ideals of $\Ocal$ representing all of the elements of the class group of $\Ocal$, and for each $i$ let $n_i\in \ZZ_{>0}$ generate $\Norm(\afrak_i)$. Then $\Mb = \QQ$ if and only if for every $i$ there exists a matrix $P_i \in \GL_2(K)$, with entries in $\afrak_i$, such that $n_i M = P_i^* M P_i$.
\end{proposition}

\begin{proof}
Lemma~\ref{lemma:KvsQ} below shows that $\Mb = \QQ$ if and only if $\Mb\subseteq K$, and this is the case if and only if for every $\sigma\in\Gal(\QQbar/K)$ there exists an isomorphism $\alpha_\sigma\colon (E^2,\varphi) \to \big((E^\sigma)^2,\varphi^{\sigma}\big)$. To understand this condition, we use the classical theory of complex multiplication of abelian varieties; the book of Shimura and Taniyama~\cite{Shi61} is one possible reference, especially Chapter~II.

Under the embedding $\epsilon_0\colon K\to \CC$ we chose earlier, the isomorphism classes of elliptic curves over $\QQbar\subset\CC$ with CM by $\Ocal$ correspond to the lattices $\epsilon_0(\afrak)$ up to scaling, for fractional ideals $\afrak$ of $\Ocal$. Since the class group of the order $\Ocal$ is $2$-torsion, we have $E^2 \simeq F^2$ for every $E$ and $F$ with CM by $\Ocal$, so we may as well choose our~$E$ so that it corresponds to the trivial ideal~$\Ocal$.

Let $\Delta$ be the discriminant of $\Ocal$ and let $\delta\in\Ocal$ be a square root of $\Delta$, chosen so that $\epsilon_0(\delta)$ is positive imaginary. Note that the trace dual $\afrak^\dagger$ of an arbitrary fractional $\Ocal$-ideal $\afrak$ is $(1/\delta)\afrak^{-1}$. If $F$ is the elliptic curve corresponding to $\afrak$, then the dual of $F$ is the elliptic curve corresponding to the complex conjugate of $\afrak^\dagger$, and the canonical principal polarization of $F$ is the isomorphism $\afrak\to (1/\delta)\bar{\afrak}^{-1}$ given by $x\mapsto x/(n\delta)$, where $n\in\QQ$ is the positive generator of $\Norm(\afrak)$. (See~\cite[\S~6.3]{Shi61} for more details.)

Let $\varphi_0$ be the product polarization on $E^2$. For $\alpha_\sigma\colon E^2\to (E^\sigma)^2$ to give an isomorphism between $(E^2,\varphi)$ and $\big((E^\sigma)^2,\varphi^\sigma\big)$, the following diagram must be commutative:
\[\xymatrix{ E^2 \ar[rr]^{M} \ar[d]_{\alpha_\sigma} && E^2\ar[rr]^{\varphi_0} && \widehat{E}^2\\ (E^\sigma)^2 \ar[rr]^{M} && (E^\sigma)^2 \ar[rr]^{\varphi_0^\sigma} && (\widehat{E}^\sigma)^2\ar[u]_{ \widehat{\alpha}_{\sigma}}.}\]
To express this diagram in terms of lattices, we let $\afrak$ be an ideal corresponding to~$E^\sigma$, we let $n = \Norm(\afrak)$, and we let $P_\sigma$ be the matrix in $\GL_2(K)$ corresponding to~$\alpha_\sigma$. Then the preceding diagram becomes
\[\xymatrix{ \Ocal\times\Ocal \ar[rr]^{M} \ar[d]_{P_\sigma} && \Ocal\times\Ocal \ar[rr]^{1/\delta\qquad} && (1/\delta)(\Ocal\times\Ocal)\\
\afrak\times\afrak \ar[rr]^{M} && \afrak\times\afrak \ar[rr]^{1/(n\delta)\qquad} && (1/\delta)(\bar{\afrak}^{-1}\times\bar{\afrak}^{-1}) \ar[u]_{ P^*_{\sigma}}.}\]
Thus, there exists an isomorphism $(E^2,\varphi)\to \big((E^\sigma)^2,\varphi^\sigma\big)$ of polarized varieties if and only if there exists a matrix $P$, with entries in $\afrak$, such that $n M = P^* M P$. Since the Galois group of $\QQbar/K$ acts transitively on the set of elliptic curves with~CM by~$\Ocal$, the field of moduli of $(E^2,\varphi)$ is contained in $K$ if and only if we can find such a matrix $P$ for each of the ideals $\afrak_1$, $\dotsc$, $\afrak_h$.
\end{proof}

\begin{lemma} \label{lemma:KvsQ}
Let $E$, $\varphi$, and $\Mb$ be as mentioned at the beginning of this section. Then $\Mb = \QQ$ if and only if $\Mb\subseteq K$.
\end{lemma}

\begin{proof}
Let us assume that $\Mb\subseteq K$; we must show that $\Mb = \QQ$. Since $\Ocal$ has a class group of exponent at most~$2$, \cite[Exercise~5.8, p.~124]{Shi71} implies that $\QQ\big(j(E)\big)$ is totally real. Let $\iota$ be any complex conjugation in~$\Gal(\QQbar/\QQ)$, so that $\iota$ acts trivially on $\QQ(j(E))$ and nontrivially on~$K$. Given any $\sigma\in \Gal(\QQbar/\QQ)$, we want to show that $(E^2,\varphi)\simeq((E^\sigma)^2,\varphi^\sigma)$.

If $\sigma$ acts trivially on~$K$, then such an isomorphism exists, because $\Mb\subseteq K$. Otherwise, $\sigma\iota$ acts trivially on $K$, and we have $(E^2,\varphi)\simeq ((E^{\sigma\iota})^2,\varphi^{\sigma\iota})$, and therefore $((E^{\iota})^2,\varphi^{\iota})\simeq ((E^\sigma)^2,\varphi^\sigma)$. So it is enough for us to show that \mbox{$(E^2,\varphi)\simeq((E^{\iota})^2,\varphi^{\iota})$}. If we choose our model of $E$ to be defined over $\QQ(j(E))$, then $E^\iota = E$, and we simply need to show that there exists an element $P$ of $\GL_2(\Ocal)$ such that~$\bar{M} = P^*MP$. If $M = \left(\begin{smallmatrix} a& b \\\bar{b} & d\end{smallmatrix}\right)$, we can simply take $P= \big(\begin{smallmatrix} b & d \\ -a & -\bar{b} \end{smallmatrix}\big)$.
\end{proof}

At this point, we have reviewed enough CM theory to prove Corollary~\ref{cor:explicit-decomposable}.

\begin{proof}[Proof of Corollary~\textup{\ref{cor:explicit-decomposable}}]
We are given an ideal $\afrak = (n,\alpha)$ of $\Ocal$, where $n\in\ZZ$ is the norm of $\afrak$ and where $\alpha\in\Ocal$, and we have $x,y\in\ZZ$ such that $xn^2 - y\Norm(\alpha) = n$. The complex conjugate $\bar{\afrak}$ of $\afrak$ represents the inverse of the class of $\afrak$ in $\Cl(\Ocal)$, and the matrix $P = \left(\begin{smallmatrix}n & y\alpha \\ \bar{\alpha} & xn\\\end{smallmatrix}\right)$ takes the lattice $\Ocal\times\Ocal\subset K^2$ onto the lattice $\afrak\times\bar{\afrak}$. The dual lattice for $\afrak\times\bar{\afrak}$ is $(n\delta)^{-1} \cdot (\afrak\times\bar{\afrak})$ (where $\delta$ is the positive imaginary square root of~$\Delta$ as in the proof of Proposition~\ref{prop:polarization}) and the product polarization from~$\afrak\times\bar{\afrak}$~to~its~dual is simply multiplication by $1/(n\delta)$. Pulling this polarization back to $\Ocal\times\Ocal$ via~$P$~gives us the polarization $(n\delta)^{-1} P^* P$. Since the product polarization on $\Ocal\times\Ocal$ is $1/\delta$, the pullback polarization is represented by the endomorphism $(1/n)P^*P$ of $\Ocal\times\Ocal$, and we compute that $(1/n) P^* P$ is the matrix given in the statement of the corollary.
\end{proof}

We close this section by indicating how we can check the criterion given in Proposition~\ref{prop:polarization}: namely, given the polarization matrix $M$ and an ideal $\afrak$ with $\Norm(\afrak) = n\ZZ$, how can we determine whether there exists a matrix $P\in M_2(\afrak)$ that satisfies $n M = P^* M P$?

Suppose there exists such a matrix $P$. If $M=\left(\begin{smallmatrix} a & b \\ \bar{b} & d \end{smallmatrix}\right)$ let us take $L = \left(\begin{smallmatrix} a & b \\ 0 & 1 \end{smallmatrix}\right)$, so that $L^*L = aM$. Let $Q = LPL^{-1}$. Then the condition $n M = P^* M P$ becomes the condition $n\Id = Q^*Q$. This equality can only hold if $Q$ is of the form
\[Q = \begin{pmatrix} x & y \\ z & t \end{pmatrix} \in GL_2(K)\]
where $x,y,z,t \in K$ satisfy $\Norm(x) + \Norm(z) = \Norm(y) + \Norm(t) = n$ and $\bar{x}y + \bar{z}t = 0$. Since we have
\[P = L^{-1} Q L = \begin{pmatrix} x - bz & \frac{bx + y - b^2 z - bt}{a}\\ az & bz + t \end{pmatrix} \in M_2(\afrak),\]
we see that we must have $x=X/a$, $y=Y/a$, $z=Z/a$, and $t=T/a$ with $X,Y,Z,T \in \afrak$.

Therefore, to check whether a matrix $P$ with the desired properties exists, it suffices to compute and store all solutions $\left(X,Z\right) \in \afrak\times\afrak$ to the norm equation $\Norm(X) + \Norm(Z) = a^2n$ (which can be done efficiently). Then, for every two solutions $(X,Z)$ and $(Y,T)$ satisfying $\bar{X} Y + \bar{Z} T =0$, we can check whether the corresponding matrix $P$ lies in $M_2(\afrak)$. If we obtain such a $P$ for each of the ideals~$\afrak_i$ from Proposition~\ref{prop:polarization}, then the field of moduli for $(E^2,\varphi)$ is~$\QQ$. In fact, we need only find a $P$ for each $\afrak_i$ in a set that generates the class group of~$\Ocal$.

\subsection{Results}

We have implemented the algorithms described in the previous sections. We were able to test all polarizations on the $65$ possible orders identified in Section~\ref{sec:woPol}. The results are presented in Table~\ref{tab:res}.

{\small
\begin{table}[ht]
\centering
\begin{tabular}{r@{\hbox to 0.5em{}}rrcrr@{\hbox to 0.5em{}}rrcrrrrc}
\toprule
$h$ & $\Delta$ & $\#\varphi$ & $\#C$\hbox to 0.5em{} &&
$h$ & $\Delta$ & $\#\varphi$ & $\#C$\hbox to 0.5em{} &&
$h$ & $\Delta$ & $\#\varphi$ & $\#C$\hbox to 0.5em{} \\
\cmidrule(l{0.25em}r{0.75em}){1-4}\cmidrule(l{0.25em}r{0.75em}){6-9}\cmidrule(l{0.75em}r{0.75em}){11-14}
$1$ &    $-3$ &   $0$ & $0$ && $4$ &   $-84$ &   $2$ & $0$ && $8$  &  $-420$ &  $10$ & $0$ \\
    &    $-4$ &   $0$ & $0$ &&     &  $-120$ &   $5$ & $3$ &&      &  $-660$ &  $16$ & $0$ \\
    &    $-7$ &   $0$ & $0$ &&     &  $-132$ &   $3$ & $1$ &&      &  $-840$ &  $22$ & $0$ \\
    &    $-8$ &   $1$ & $1$ &&     &  $-168$ &   $4$ & $0$ &&      & $-1092$ &  $22$ & $0$ \\
    &   $-11$ &   $1$ & $1$ &&     &  $-195$ &   $8$ & $0$ &&      & $-1155$ &  $32$ & $0$ \\
    &   $-19$ &   $1$ & $1$ &&     &  $-228$ &   $5$ & $1$ &&      & $-1320$ &  $36$ & $0$ \\
    &   $-43$ &   $2$ & $2$ &&     &  $-280$ &  $14$ & $0$ &&      & $-1380$ &  $34$ & $0$ \\
    &   $-67$ &   $3$ & $3$ &&     &  $-312$ &  $11$ & $1$ &&      & $-1428$ &  $28$ & $0$ \\
    &  $-163$ &   $7$ & $7$ &&     &  $-340$ &  $14$ & $0$ &&      & $-1540$ &  $46$ & $0$ \\
    &         &       &     &&     &  $-372$ &   $8$ & $0$ &&      & $-1848$ &  $46$ & $0$ \\
$2$ &   $-15$ &   $0$ & $0$ &&     &  $-408$ &  $14$ & $0$ &&      & $-1995$ &  $56$ & $0$ \\
    &   $-20$ &   $1$ & $1$ &&     &  $-435$ &  $16$ & $0$ &&      & $-3003$ &  $72$ & $0$ \\
    &   $-24$ &   $1$ & $1$ &&     &  $-483$ &  $12$ & $0$ &&      & $-3315$ & $128$ & $0$ \\
    &   $-35$ &   $2$ & $0$ &&     &  $-520$ &  $25$ & $3$ &&      &         &       &     \\
    &   $-40$ &   $2$ & $2$ &&     &  $-532$ &  $14$ & $0$ && $16$ & $-5460$ & $128$ & $0$ \\
    &   $-51$ &   $2$ & $0$ &&     &  $-555$ &  $20$ & $0$ &&      &         &       &     \\
    &   $-52$ &   $2$ & $2$ &&     &  $-595$ &  $28$ & $2$ &&      &         &       &     \\
    &   $-88$ &   $4$ & $2$ &&     &  $-627$ &  $16$ & $0$ &&      &         &       &     \\
    &   $-91$ &   $4$ & $0$ &&     &  $-708$ &  $15$ & $1$ &&      &         &       &     \\
    &  $-115$ &   $6$ & $0$ &&     &  $-715$ &  $36$ & $0$ &&      &         &       &     \\
    &  $-123$ &   $4$ & $0$ &&     &  $-760$ &  $41$ & $1$ &&      &         &       &     \\
    &  $-148$ &   $5$ & $3$ &&     &  $-795$ &  $28$ & $2$ &&      &         &       &     \\
    &  $-187$ &   $8$ & $0$ &&     & $-1012$ &  $28$ & $0$ &&      &         &       &     \\
    &  $-232$ &   $9$ & $5$ &&     & $-1435$ &  $64$ & $0$ &&      &         &       &     \\
    &  $-235$ &  $12$ & $0$ &&     &         &       &     &&      &         &       &     \\
    &  $-267$ &   $8$ & $0$ &&     &         &       &     &&      &         &       &     \\
    &  $-403$ &  $18$ & $0$ &&     &         &       &     &&      &         &       &     \\
    &  $-427$ &  $16$ & $0$ &&     &         &       &     &&      &         &       &     \\
\bottomrule
\end{tabular}
\bigskip
\caption{The number of indecomposable principal polarizations~ $\varphi$ and the number of isomorphism classes of curves $C$ with field of moduli $\QQ$ for each discriminant $\Delta$, grouped by class number~$h$.}
\label{tab:res}
\end{table}}

There exist $1226$ indecomposable polarizations, in total. Our algorithms, implemented in \verb?Magma? on a laptop with a 2.50 GHz Intel Core i7-4710MQ processor, took less than $21$ minutes to compute all of the polarizations; about $10$ minutes of that time was spent on the largest discriminant. The computation required about~$2.8$~GB of memory.

Once we computed the polarizations, it took about $26$ minutes (on the same laptop) to check the conditions of Proposition~\ref{prop:polarization}. For this calculation, the largest discriminant represented more than two-thirds of the computation time.

In the end, we obtained exactly $46$ polarizations $\varphi$ such that the principally polarized abelian surface $(E^2,\varphi)$ is isomorphic to the Jacobian of a curve $C$ with field of moduli $\QQ$.
These $46$ curves are obtained only from orders whose class groups have order $1$, $2$, or $4$.

\section{Computation of invariants and final remarks} \label{sec:FinRes}

\subsection{Invariants of the genus-\texorpdfstring{$2$}{2} curves \texorpdfstring{$C$}{C}}

A genus-$2$ curve $C$ has field of moduli~$\QQ$ if and only if all of its absolute invariants are defined over $\QQ$ (see for example~\cite[\S~3]{LRS12}). This is in particular true for the triplet $(g_1,g_2,g_3)$ of invariants defined by Cardona and Quer in~\cite{CQ05}, which characterizes a genus-$2$ curve up to~$\QQbar$-isomorphism and enables one to find an equation $y^2=f(x)$ for the curve. We quickly review here a strategy for obtaining the Cardona--Quer invariants for the~$46$ curves whose invariants are $\QQ$-rational.

The first quantity we are able to derive is a Riemann matrix $\tau$, using the same method as~\cite[\S~3.3]{Rit10}. Starting with the positive definite unimodular Hermitian matrix $M$ corresponding to the polarization $\varphi = \varphi_0 \cdot M$, we obtain the Riemann matrix $\tau$ associated to $\varphi$ and the CM-elliptic curve $E \simeq \CC/(\ZZ+\ZZ \omega)$ where~\mbox{$\omega=(1+\sqrt{\Delta})/2$} if $\Delta$ is odd and $\omega=\sqrt{\Delta}$ otherwise.

This matrix we get is defined up to the action of the symplectic group $\Sp_4(\ZZ)$. One then works out a matrix $\tau_0$ in the orbit of $\tau$ for which the computation of the theta constants $(\theta_i )_{0 \leq i \leq 9}$ at $\tau_0$ is fast (see~\cite{Lab16} for instance).

A complex model of a curve $C : y^2 = x(x-1)(x-\lambda_1)(x-\lambda_2)(x-\lambda_3)$ with Riemann matrix $\tau_0$ can then be classically approximated using Rosenhain's formulas~\cite[p.~417]{Ros51}
\[\lambda_1 = \frac{\theta_0^2\theta_2^2}{\theta_1^2\theta_3^2}, \qquad \lambda_2 = \frac{\theta_2^2\theta_7^2}{\theta_3^2\theta_9^2}, \qquad \text{and} \qquad \lambda_3 = \frac{\theta_0^2\theta_7^2}{\theta_1^2\theta_9^2}.\]

By computing the theta constants to higher and higher precision, we are able to get a sufficiently good approximation of the Cardona--Quer invariants to recognize them as rationals. The numbers we get are \emph{a priori} only heuristic as there is no bound known for the denominators of these rationals; however, we can sometimes prove that these heuristic values are correct, as follows.

Given a set of Cardona--Quer invariants that we suspect are equal to the invariants of a curve whose Jacobian is isomorphic to $E^2$ for an $E$ with complex multiplication, we can easily produce a curve $C$ having those invariants. Then we can use the techniques of~\cite{CMSV17} to provably compute the endomorphism ring of the Jacobian of $C$. If this endomorphism ring is isomorphic to the ring $M_2(\End E)$, then we have provably found a curve of the type we are looking for.

We computed heuristic values for the Cardona--Quer invariants of our $46$ principally polarized abelian surfaces, and the list of these invariants is available on authors' web pages, together with all the programs to compute them. We are grateful to J.~Sijsling for computing the endomorphism rings for the Jacobians of~$13$ of our $46$ curves; he is currently developing a faster and more robust algorithm which should be able to handle the remaining cases. For each of these $13$ curves, the endomorphism ring was $M_2(\End E)$, so the heuristic values of the Cardona--Quer invariants of these curves are provably correct.

We observe for that the $13$ provably-correct sets of invariants, all the denominators are smooth integers. It would be very interesting, in the same spirit as~\cite{GL07,LV15} for the CM genus-$2$ case, to find formulas to explain the prime powers dividing these denominators. An example of such a closed formula appears in the introduction of~\cite{Rod00} without any details. The denominators of the $33$ sets of invariants that we have not proven to be correct also are smooth, which provides some further heuristic evidence that the values are correct.

We present in Table~\ref{tab:curves} the invariants for a few of the curves we could provably compute.

{\small
\begin{table}[ht]
\centering
\begin{tabular}{r@{\qquad}cc}
\toprule
 $\Delta$ & $M$ & Cardona--Quer invariants $[g_1,g_2,g_3]$ \\
 \midrule
$ -8$ & $\begin{pmatrix} 2 & \omega + 1 \\ -\omega + 1 & 2 \end{pmatrix}$ & $\left[2^4\cdot 5^5,2\cdot 3\cdot 5^4,-5^3\right]$ \\[3ex]
$-11$ & $\begin{pmatrix} 2 & \omega     \\ -\omega + 1 & 2 \end{pmatrix}$ & $\displaystyle\left[\frac{19^5}{2^2},\frac{3^2\cdot 11\cdot 19^3}{2^5},-\frac{19^2\cdot 47}{2^6}\right]$ \\[3ex]
$-19$ & $\begin{pmatrix} 2 & \omega     \\ -\omega + 1 & 3 \end{pmatrix}$ & $\displaystyle\left[\frac{5^5\cdot 29^5}{2^2\cdot 3^7},\frac{5^3\cdot 7\cdot 29^3\cdot 31\cdot 73}{2^5\cdot 3^8},-\frac{5^2\cdot 17\cdot 29^2\cdot 2719}{2^6\cdot 3^{10}}\right]$ \\[3ex]
$-20$ & $\begin{pmatrix} 2 & \omega     \\ -\omega     & 3 \end{pmatrix}$ & $\displaystyle\left[\frac{5^5\cdot 7^5}{2^2},\frac{5^5\cdot 7^3\cdot 11}{2^5},-\frac{3\cdot 5^3\cdot 7^2}{2^6}\right]$ \\[3ex]
$-24$ & $\begin{pmatrix} 2 & \omega + 1 \\ -\omega + 1 & 4 \end{pmatrix}$ & $\displaystyle\left[\frac{2^4\cdot 23^5}{3},\frac{2\cdot 23^3\cdot 421}{3^2},-\frac{23^2\cdot 37}{3^4}\right]$ \\[3ex]
$-40$ & $\begin{pmatrix} 2 & \omega + 1 \\ -\omega + 1 & 6 \end{pmatrix}$ & $\displaystyle\left[\frac{2^4\cdot 5^5\cdot 43^5}{3^7},\frac{2\cdot 5^4\cdot 43^3\cdot 6977}{3^8},-\frac{5^4\cdot 13\cdot 43^2}{3^{10}}\right]$ \\[3ex]
$-52$ & $\begin{pmatrix} 2 & \omega     \\ -\omega     & 7 \end{pmatrix}$ & $\displaystyle\left[\frac{5^5\cdot 173^5}{2^2\cdot 3^7},\frac{5^4\cdot 173^3\cdot 112061}{2^5\cdot 3^ 8},-\frac{5^3\cdot 7\cdot 37\cdot 173^2}{2^6\cdot 3^{10}}\right]$ \\[3ex]
\bottomrule
\end{tabular}
\bigskip
\caption{Cardona--Quer invariants for seven of the $46$ genus-$2$ curves with field of moduli $\QQ$ whose Jacobians are isomorphic to~$E^2$, where $E$ has CM by a maximal order~$\Ocal$. The discriminant of $\Ocal$ is $\Delta$, the corresponding principal polarization on $E^2$ is~$\varphi_0\cdot M$, and $\omega$ denotes either $\sqrt{\Delta}/2$ or $(1 + \sqrt{\Delta})/2$, depending on whether $\Delta$ is even or odd.}
\label{tab:curves}
\end{table}}

\subsection{When is \texorpdfstring{$\QQ$}{Q} also a field of definition for \texorpdfstring{$C$}{C}?}

To conclude let us consider any of the $46$ previous pairs $(A,\varphi)$. We know that there exists a genus-$2$ curve $C/\QQbar$ with field of moduli $\QQ$ such that $(\Jac(C),j) \simeq_{\QQbar} (A,\varphi)$, where $j$ is the canonical polarization on $\Jac(C)$. If the order of $\Aut(A,\varphi) \simeq \Aut(C)$ is larger than~$2$, then it is known~\cite{CQ05} that the field of moduli of $C$ is a field of definition and that there exists a genus-$2$ curve $C_0\colon y^2=f(x)$ with $f \in \QQ[x]$ such that \mbox{$(\Jac(C_0),j_0) \simeq_{\QQbar} (A,\varphi)$}. In particular $\QQ$ is also a field of definition for $(A,\varphi)$.

\begin{proposition}[{Compare to~\cite[\S~4]{Rod00}}] \label{prop:fielddef}
The field $\QQ$ is a field of definition of~$C$ --- and therefore of $(A,\varphi)$~--- if and only if
the order of \mbox{$\Aut(A,\varphi)\simeq \Aut(C)$} is larger than~$2$.
\end{proposition}

\begin{proof}
It remains to prove that when $\Aut(A,\varphi) = \{\pm 1 \}$, there is no model of~$(A,\varphi)$ over $\QQ$. Actually we show there is even no model $(B,\mu)$ over $\RR$. Indeed, an isomorphism $\psi : (A,\varphi)/\CC \to (B,\mu)/\RR$, defined over~$\CC$, would induce an isomorphism \[\alpha_{\iota} = (\psi^{-1})^{\iota} \circ \psi : (A,\varphi) \to (A,\varphi)^{\iota},\] for the complex conjugation $\iota$, such that ${\alpha_{\iota}}^{\iota} \circ \alpha_{\iota} = \left((\psi^{-1}) \circ \psi^{\iota}\right) \circ \left((\psi^{-1})^{\iota} \circ \psi\right) = \Id$.

Since we have seen that $E^{\iota} = E$, the isomorphism $\alpha_{\iota}$ can be represented as a matrix $P \in \GL_2(\Ocal)$ such that $\bar{P} P = \Id$. Moreover the commutativity of the diagram
\[\xymatrix{E^2 \ar[rr]^{\varphi} \ar[d]_{\alpha_{\iota}} && \widehat{E}^2 \\E^2 \ar[rr]^{\varphi^\iota} && \widehat{E}^2\ar[u]_{ \widehat{\alpha}_{\iota}}.\\}\]
translates into the equality $P^* \bar{M} P = M$. If we denote $M=\left(\begin{smallmatrix} a & b \\ \bar{b} & d \end{smallmatrix}\right)$, then it is easy to see that the matrix $P_0= \big(\begin{smallmatrix} \bar{b} & d \\ -a & -b \end{smallmatrix}\big)$ satisfies the last equality. Any other $P=P_0 R$ differs from $P_0$ by an automorphism $R$ of $(A,\varphi)$ since \mbox{$R^* P^* \bar{M} P R= R^* M R = M$}.
Because the automorphism group of $(A,\varphi)$ is $\{\pm 1\}$, this means that the only possible $P$ are $\pm P_0$. It is easy to check that $P_0\bar{P_0} = (-P_0)(-\bar{P_0}) = -\Id$, so the condition~$\bar{P} P = \Id$ cannot be satisfied.
\end{proof}

\subsection{Provably correct equations for the curves defined over \texorpdfstring{$\QQ$}{Q}}

Using Proposition~\ref{prop:fielddef} we found that exactly $13$ of our curves can be defined over~$\QQ$, and these~$13$ are precisely the curves for which we could provably compute the invariants. This is no coincidence, as having an equation over $\QQ$ definitely simplifies the computation. We present these curves in~Table~\ref{tab:resQ}.

{\small
\begin{table}[ht]
\centering
\begin{tabular}{cc@{\ }c@{\ }l}
\toprule
     $\Delta$ & $M$ &$d$ &\quad Equation for $C$\\
\midrule
  $-8$ & $\begin{pmatrix} 2 & \omega + 1 \\ -\omega + 1 &  2 \end{pmatrix}$ & $1$            & \begin{tabular}{l}{$y^2 = x^5 + x$}\end{tabular}\\[3ex]
 $-11$ & $\begin{pmatrix} 2 & \omega     \\ -\omega + 1 &  2 \end{pmatrix}$ & $(-11)^{1/3}$  & \begin{tabular}{l}{$y^2 = 2 x^6 + 11 x^3 - 2\cdot11$}\end{tabular}\\[3ex]
 $-19$ & $\begin{pmatrix} 2 & \omega     \\ -\omega + 1 &  3 \end{pmatrix}$ & $-19$          & \begin{tabular}{l}{$y^2 = x^6 + 1026 x^5 + 627 x^4 + 38988 x^3$}\\
                                                                                {$\phantom{y^2 = x^6} - 627\cdot19 x^2 + 1026\cdot19^2 x - 19^3$}\end{tabular}\\[3ex]
 $-43$ & $\begin{pmatrix} 2 & \omega     \\ -\omega + 1 &  6 \end{pmatrix}$ & $-43$          & \begin{tabular}{l}{$y^2 = x^6 + 48762 x^5 + 1419 x^4 + 4193532 x^3$}\\
                                                                                {$\phantom{y^2 = x^6} - 1419\cdot43 x^2 + 48762\cdot43^2 x - 43^3$}\end{tabular}\\[3ex]
 $-67$ & $\begin{pmatrix} 2 & \omega     \\ -\omega + 1 &  9 \end{pmatrix}$ & $-67$          & \begin{tabular}{l}{$y^2 = x^6 + 785106 x^5 + 2211 x^4 + 105204204 x^3$}\\
                                                                                {$\phantom{y^2 = x^6} - 2211\cdot67 x^2 + 785106\cdot67^2 x - 67^3$}\end{tabular}\\[3ex]
$-163$ & $\begin{pmatrix} 2 & \omega     \\ -\omega + 1 & 21 \end{pmatrix}$ & $-163$         & \begin{tabular}{l}{$y^2 = x^6 + 1635420402 x^5 + 5379 x^4$}\\
                                                                                {$\phantom{y^2 = x^6} + 533147051052 x^3 - 5379\cdot163 x^2$}\\
                                                                                {$\phantom{y^2 = x^6} + 1635420402\cdot163^2 x - 163^3$}\end{tabular}\\[3ex]
 $-20$ & $\begin{pmatrix} 2 & \omega     \\ -\omega     &  3 \end{pmatrix}$ & $\sqrt{5}$     & \begin{tabular}{l}{$y^2 = x^5 + 5 x^3 + 5 x$}\end{tabular}\\[3ex]
 $-24$ & $\begin{pmatrix} 2 & \omega + 1 \\ -\omega + 1 &  4 \end{pmatrix}$ & $\sqrt{2}$     & \begin{tabular}{l}{$y^2 = 3 x^5 + 8 x^3 + 3\cdot2 x$}\end{tabular}\\[3ex]
 $-40$ & $\begin{pmatrix} 2 & \omega + 1 \\ -\omega + 1 &  6 \end{pmatrix}$ & $\sqrt{5}$     & \begin{tabular}{l}{$y^2 = 9 x^5 + 40 x^3 + 9\cdot5 x$}\end{tabular}\\[3ex]
 $-52$ & $\begin{pmatrix} 2 & \omega     \\ -\omega     &  7 \end{pmatrix}$ & $\sqrt{13}$    & \begin{tabular}{l}{$y^2 = 9 x^5 + 65 x^3 + 9\cdot13 x$}\end{tabular}\\[3ex]
 $-88$ & $\begin{pmatrix} 2 & \omega + 1 \\ -\omega + 1 & 12 \end{pmatrix}$ & $\sqrt{2}$     & \begin{tabular}{l}{$y^2 = 99 x^5 + 280 x^3 + 99\cdot2 x$}\end{tabular}\\[3ex]
$-148$ & $\begin{pmatrix} 2 & \omega     \\ -\omega     & 19 \end{pmatrix}$ & $\sqrt{37}$    & \begin{tabular}{l}{$y^2 = 441 x^5 + 5365 x^3 + 441\cdot37 x$}\end{tabular}\\[3ex]
$-232$ & $\begin{pmatrix} 2 & \omega + 1 \\ -\omega + 1 & 30 \end{pmatrix}$ & $\sqrt{29}$    & \begin{tabular}{l}{$y^2 = 9801 x^5 + 105560 x^3 + 9801\cdot29 x$}\end{tabular}\\
\bottomrule
\end{tabular}
\bigskip
\caption{Genus-$2$ curves defined over $\QQ$ with Jacobian isomorphic over $\QQbar$ to $E^2$, where $E$ has CM by a maximal order $\Ocal$. The discriminant of $\Ocal$ is $\Delta$, the corresponding principal polarization on~$E^2$ is $\varphi_0\cdot M$, and $\omega$ denotes either $\sqrt{\Delta}/2$ or $(1 + \sqrt{\Delta})/2$, depending on whether $\Delta$ is even or odd. This list is complete if the Generalized Riemann Hypothesis holds. Each curve is a double cover of its corresponding $E$ (as can be seen by the fact that the upper-left entry of each polarization matrix is~$2$), and the associated involution of $C$ is given by $(x,y)\mapsto (d/x, d^{3/2} y/x^3)$ for the value of $d$ given in the third column.}
\label{tab:resQ}
\end{table}}

\bibliographystyle{alphadoi}
\bibliography{GelinEtAl}
\end{document}

config: arxiv


TITLE: Convergence of maximum of dependent RVs
QUESTION [1 upvotes]: I've been stumped on the following for a while and I am.hoping you could be of some help. We have some (not necessarily independent) real valued random variables $Y_i$ such that $\frac{Y_n}{\sqrt{n}}\to 0$ and $\frac{Y_0}{\sqrt{n}}\to 0$. I was wondering if we could conclude that $\frac{\max\limits_{k=0,...,n} |Y_k|}{\sqrt{n}}\to 0$. I suspect it does hold (it is not too difficult to show if we have sequences of real numbers), but I am having trouble to show it in this case.
Could anyone help?
Thanks!

REPLY [1 votes]: If you deal with almost sure convergence, then it follows from the case of real numbers.
If you deal with convergence in probability, then this may not be true. Define $Y_n=\sqrt n\mathbf 1_{A_n}$, where $p_n:=\mathbb P\left(A_n\right)\to 0$ as $n$ goes to infinity. Then $Y_n/\sqrt n\to 0$ in probability but if $\left(\mathbb P\left(\bigcup_{j=n+1}^{2n}A_j\right)\right)_{n\geqslant 1}$ does not converge to $0$, then the sequence $\left(\max_{1\leqslant j\leqslant 2n} \left|Y_j\right| /(2\sqrt n) \right)_{n\geqslant 1}$ does not converge to $0$ in probability. To get an explicit example, consider independent sets $(A_n)_{n\geqslant 1}$ such that $p_j=1/2$ and use Bonferroni's inequality to get
$$ \mathbb P\left(\bigcup_{j=n+1}^{2n}A_j\right ) \geqslant
\sum_{j=n+1}^{2n}p_j-\sum_{n+1\leqslant i\lt j\leqslant 2n}p_ip_j \geqslant \sum_{j=n+1}^{2n}p_j-\frac 12\left(\sum_{j=n+1}^{2n}p_j\right)^2.$$


TITLE: Is it possible to compute the variance without computing the mean first?
QUESTION [1 upvotes]: I have a list of values of a random variable $x \in \mathbb R$. Is it possible to find the varience $\overline{(x - \overline x)^2}$ without computing the mean $\overline x$ first? That is to process the list only once.

REPLY [3 votes]: How about sum of squared pairwise differences ? Indeed, you can check by direct computation that
$$
2v_X = \frac{1}{n(n-1)}\sum_{1 \le i < j \le n}(x_i  - x_j)^2.
$$
Disclaimer: This is not meant to be efficient, just another way to represent the variance, without first computing the mean.


TITLE: Homework Help - Finding a Vector when given two points, and then finding a unit vector in the same direction
QUESTION [5 upvotes]: I've attempted to solve the problem, but I got $\langle \frac{1}{\sqrt{\frac{29}{4}}}, \frac{5}{\sqrt{\frac{29}{4}}}\rangle$, which is incorrect. There is not a similar problem in my textbook that I can reference.
I know that to find a unit vector, we first find the length/magnitude of the given vector, and multiply $$1/\sqrt{magnitude}$$ by the original vector.
$$L = \sqrt{x^2 + y^2}.$$
Can anyone give me any ideas on how to solve this problem?
Find the unit vector that has the same direction as the vector from the point A = (-1,2) to the point B = (3,3).
Thank you in advance.

REPLY [4 votes]: The vector that goes from $A$ to $B$ is the vector $B-A$: to see this, notice that if you add vectors using the parallelogram rule, then adding the vector $V$ you are looking for to $A$ should give you $B$, so $A+V = B$, giving you $V=B-A$.
So the vector you are looking for is $V = B-A = (3,3) - (-1,2) = (4,1)$.
Now that you know the vector, finding the unit vector in the same direction is done as you indicate: find the magnitude of $V$, divide by the magnitude.
(Looks like you took $A+B$ instead of $B-A$)

REPLY [2 votes]: When you took the vector from A to B it looks like you added instead of subtracting.  It should be (4,1).


TITLE: Do these axioms fully describe the integers?
QUESTION [3 upvotes]: Here, I use Peano-like axioms to describe the set of integers $Z$. They are based on two successor functions, each starting with a common point of $0$, and a principle of induction for the integers.
Let $Z$, $Pos$, $Neg$, $s$, $s'$ and $0$ be such that:
$Pos\subset Z$
$Neg\subset Z$
$Z=Pos\cup Neg$ (edit)
$\forall x (x\in Pos \wedge x\in Neg \leftrightarrow x=0)$
$s:Pos\rightarrow Pos$
$s$ is injective
$s':Neg\rightarrow Neg$
$s'$ is injective
$\forall x\in Pos (s(x)\neq 0)$
$\forall x\in Neg (s'(x)\neq 0)$
$\forall m ((0\in m\wedge \forall x\in Pos (x\in m\rightarrow s(x)\in m) \wedge \forall x\in Neg (x\in m\rightarrow s'(x)\in m))\rightarrow \forall x\in Z (x\in m)) $
Note that, contrary to the usual convention, I have had to include $0$ in both sets $Pos$ and $Neg$.
Lemma: $0\in Z, Pos, Neg$
See my follow-up below

REPLY [1 votes]: Follow-up
After much tinkering, I have settled on the following Peano-like axioms for the integers:
Let $Z, L, R, 0, s$ be such that:
$R\subset Z$, the non-negative integers (right)
$L\subset Z$, the non-positive integers (left)
$Z=R\cup L$
$\forall x (x\in R \wedge x\in L \leftrightarrow x=0)$
$s: Z\rightarrow Z$, a bijection
$\forall x (x\in R \rightarrow s(x)\in R)$
$\forall x (x\in L \rightarrow s^{-1}(x)\in L)$
$\forall x (x\in R \rightarrow s(x)\neq 0)$
$\forall x (x\in L \rightarrow s^{-1}(x)\neq 0)$
$\forall P ((P\subset Z \wedge 0\in P\wedge \forall x (x\in P\rightarrow s(x)\in P) \wedge \forall x (x\in P\rightarrow s^{-1}(x)\in P) ) \rightarrow Z\subset P) $


TITLE: Sum of Jointly Distributed Normal Variables is Normal
QUESTION [0 upvotes]: I am wondering how to prove that if $X,Y$ are normal with mean $0$ and variance $\sigma_X^2, \sigma_Y^2$ respectively, and I know $\mathbb{E}[XY]$, how do I prove that $X - Y$ is also normally distributed using characteristic functions.
I know that
$$\Phi_{X-Y}(\theta) = \mathbb{E}[e^{i\theta(X-Y)}] = \mathbb{E}[e^{i\theta X} e^{-i \theta Y}]$$
but since $X,Y$ are not independent I can't split this into the product of the characteristic function of $X$ and of $Y$. Thank you

REPLY [1 votes]: You won't be able to prove it because if you don't assume anything on the joint distribution of $(X,Y)$ (e.g. independence) then $X-Y$ may not be gaussian.
The well-known example is $Y=\varepsilon X$ for a random variable $\varepsilon$ independent of $X$, equal to $-1$ or $1$ both with probability $1/2$. Then $X-Y=(1-\varepsilon)X$ is equal to $0$ with probability $1/2$ so it cannot be gaussian.


text:
\begin{document}
\title{A remark on the Wiener-Ikehara Tauberian Theorem}
\author{Ryo Kato}
\date{}

\maketitle

\textbf{Abstract.} In this paper we point out that the proof of
Kable's extension of the Wiener-Ikehara Tauberian theorem can
be applied to the case where the Dirichlet series has a pole
of order "$l/m$" without much modification
(Kable proved the case $l = 1$).

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

We use the notation $\N,\Z,\R,\C$ for the sets of positive integers,
all integers, real numbers, complex numbers respectively.
Let  $\R_{> 0}$ denote the set of positive real numbers.
For $x \in \R$ we put
$\ceil{x} = \min\{ n \in \mathbb{Z} \ | \ n \ge x \}$.

Let $\{a_{n}\}$ be a sequence of non-negative real numbers,
$d$ a positive real number, and $m, l \in \mathbb{N}$. Suppose that the Dirichlet series

\begin{equation*}
L(s) = \sum_{n=1}^{\infty} \frac{a_{n}}{n^s}
\end{equation*}

converges absolutely for $\re(s) > d$. Also suppose that $L^m$
has a meromorphic continuation to an open set containing the
closed half-plane $\re(s) \ge d$ and holomorphic except for
a pole of order $l$ at $s = d$. In this paper, if $L^m$ has a
pole of order $l$, then we say that $L$ has a pole of order $l/m$.
For functions $f$ and $g : \R \to \R $, we denote
$f(x) \sim g(x)$ if $\lim_{x \to +\infty} {f(x)}/{g(x)} = 1$.

Our purpose is to determine the asymptotic behavior of
$\sum_{n \le X} a_{n}$ as $X \to \infty$ by properties of $L(s)$.

In the case where $L$ has a simple pole at $s = d$ (i.e., $m = l = 1$),
Wiener and Ikehara proved that

\begin{equation*}
 \sum_{n \le X} a_{n} \sim \frac{A}{d}X^d
\end{equation*}

as $X \to \infty$, where $A$ is residue of $L$ at $s = d$.
This result was proved in 1932(\cite{wiener}) and is called the Wiener-Ikehara theorem.

In the general case,
an extension was given by Delange (\cite[p.235, TH\'EOR\`EM III]{delange})
in 1954. Delange considered the case where the order of the
pole is a positive real number in some sense. However,
to apply his theorem to $L$ satisfying above conditions,
an extra condition about zeros of $L^m$ on $\re(s) = d$ is required.
Kable has given an extension for the case where the order of the pole is
$1/m$  without the condition.
In \cite{kable}, he used the notion of functions of bounded variation.

The result of Delange and Kable is as follows.
Let  $\alpha$ = $1/m$ or $l$.
If $\alpha = 1/m$, then suppose that the residue of $L^m$ at $s = d$ is $A^m$, where $A > 0$.
If $\alpha = l$, then let $A = \lim_{s \to d} L(s)(s -d)^l \ (A>0)$.
Then

\begin{equation}
\sum_{n \le X} a_{n} \sim \frac{AX^d}{d\Gamma(\alpha)(\log(X))^{1-\alpha}} \label{1}
\end{equation}

as $X \to \infty$.
We shall point out in this paper that the proof of Kable's result
works for the case, without much modification,
where the order of the pole is $l/m$, and
that (\ref{1}) holds with $\alpha=l/m$.

The organization of this paper is as follows. In Section \ref{sec:preliminaries},
we define symbols and functions used in this paper. Then we slightly
extend \cite[p.140, T{\footnotesize HEOREM} 1]{kable}.
In Section \ref{sec:main-theorem}, we apply the result in Section \ref{sec:preliminaries}
to obtain the same result as \eqref{1} for the case $\alpha = l/m$.
In Section \ref{sec:an-example}, we give an example
of a Dirichlet series which has a pole of order $2/3$ and apply
the main theorem (Theorem \ref{thm:main-theorem}).

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

Let $\lambda$ be $(2 \pi)^{-\frac{1}{2}}$ times the Lebesgue measure
on $\R$ and $\mathcal{S}(\R)$ the space of Schwartz functions on $\R$.
For $\Phi \in  \mathcal{S}(\R)$ define the Fourier transform by

\begin{equation*}
\mathcal{F}(\Phi)(t) = \int_{\R} \Phi(u) e^{-iut} d\lambda(u).
\end{equation*}

We define the inverse Fourier transform by

\begin{equation*}
\mathcal{F}^{-1}(\Phi)(t) = \int_{\R} \Phi(u) e^{iut} d\lambda(u).
\end{equation*}

Then

\begin{equation*}
\mathcal{F}^{-1}(\mathcal{F}(\Phi))(t) = \Phi(t) \quad i.e.,\ \
\Phi(t) = \int_{\R} \mathcal{F}(\Phi)(u) e^{iut} d\lambda(u).
\end{equation*}

For $\Phi, \Psi \in \mathcal{S}(\R)$, we define the convolution by

\begin{equation*}
(\Phi * \Psi)(t) = \int_{\R} \Phi(t - u) \Psi(u) d \lambda (u).
\end{equation*}

It is well known (see \cite[p.183, 7.2 Theorem]{fourier}) that

\begin{equation*}
\mathcal{F}(\Phi * \Psi) = \mathcal{F}(\Phi) \mathcal{F}(\Psi).
\end{equation*}

For $s \in \C -(-\infty , 0]$ we choose the branch of $\log(s)$
so that $-\pi < arg(s) < \pi$. Let $\alpha \in \R$ and $x \in \C$.
For $s \in \C - \{ s \in \C \ | \ s - x \in (\infty, 0] \}$
we define $(s-x)^{\alpha} = \exp(\alpha \log (s-x))$.
Then $(s-x)^\alpha$ is holomorphic on
$\C - \{ s \in \C \ | \ s - x \in (\infty, 0] \}$
and has positive real values for $s \in \{s \in \C \ | \ s -x \in \R_{> 0}\}.$

Let $\mathcal{P}([a, b])$ denote the set of all partitions of $[a, b]$
(i.e., sequences $x_0=a<x_1<\cdots<x_n=b$).
For a function  $f :[a, b] \to \R$ we define
$V_{a}^{b} f \in \R \cup \{\infty\}$ by

\begin{equation*}
V_{a}^{b} f = \sup \left\{ \sum_{i = 1}^{n}
\left| f(x_{i}) - f(x_{i-1}) \right| \bigg|
P = \{x_{i} \ | \ i = 0,1,\cdots,n\} \\
\in \mathcal{P}([a, b])
\right\}.
\end{equation*}

We say that a function $f :[a, b] \to \R$ is of bounded variation
if $V_{a}^{b} f < + \infty$.
Also we say that a function $f :[a, b] \to \C$ is of bounded variation
if the real part and the imaginary part of $f$ are of bounded variation.
If a function $f :\R \to \mathbb{C}$  is of bounded
 variation on any closed-interval $[a , b]$, then
we say that $f$ is locally of bounded variation.

The following theorem is an extension of the Wiener-Ikehara
theorem and plays a key role to prove the main theorem in Section 3.
This theorem slightly extends \cite[p.140, T{\footnotesize HEOREM} 1]{kable}.

\begin{thm}
\label{thm:preliminary-theorem}
Let $f : \mathbb{R} \to [0,\infty)$ be a non-decreasing function such that
the Laplace transform

\begin{equation*}
F(s) = \int_{0}^{\infty} f(u) e^{-su} du
\end{equation*}

converges absolutely for $\re(s) > 1$.
Let $\alpha, \alpha_{i} \in \mathbb{R}_{ > 0} ( i = 1, 2, \cdots, r ),
A \in \mathbb{R}_{> 0}, A_{i} \in \mathbb{C} $ and
$\alpha > \alpha_{i}$ for all $i$. We define

\begin{equation*}
G(s) = F(s) - \frac{A}{(s - 1)^{\alpha}} - \sum_{i=1}^{r} \frac{A_{i}}{(s - 1)^{\alpha_{i}}}
\end{equation*}

for $\re(s) > 1$.
Suppose that the function G extends continuously to the set
$\{s \in \mathbb{C}$ $|$ $\re(s) \ge 1\}$.
If $\alpha < 1$ then assume, in addition,
that once so extended, the function $ t \mapsto G(1 + it)$ is locally of bounded variation.
Then

\begin{equation*}
\lim_{u \to \infty} u^{1 - \alpha} e^{-u} f(u) = \frac{A}{\varGamma(\alpha)}.
\end{equation*}

\end{thm}
\begin{proof}

Define a function $h : \mathbb{R} \to [ 0,\infty)$
by $h(u) = u^{1 - \alpha} e^{-u} f(u)$ and let
$C_{i} = A_{i} / \varGamma(\alpha_{i}), \ C = A / \varGamma(\alpha)$.
By \cite[p.140, L{\footnotesize EMMA} 2]{kable}

\begin{equation*}
\frac{1}{(s - 1)^{\alpha}} = \frac{1}{\varGamma(\alpha)}
\int_{0}^{\infty} e^{-(s-1) u} u^{\alpha - 1} du
\end{equation*}

for $\re(s) > 1$.  By definition

\begin{equation*}
F(s) = \int_{0}^{\infty} h(u) e^{-(s-1) u} u^{\alpha - 1} du.
\end{equation*}

It follows that

\begin{equation*}
G(s) = \int_{0}^{\infty} \left(h(u) - C - \sum_{i=1}^{r}
\frac{C_{i}}{u^{\alpha - \alpha_{i}}} \right) e^{-(s-1)u} u^{\alpha - 1} du
\end{equation*}

for $\re(s) > 1$.
Let $\Psi$ be an even Schwartz function with
compact support and $\Phi = \mathcal{F}(\Psi)^2$
.
Since $\mathcal{F}(\Phi) = \Psi*\Psi$,
$\mathcal{F}(\Phi)$ is even and has compact support.

Suppose that $\sigma > 1$ and $x > 0$. Since

\begin{equation*}
F(\sigma) \ge \int_{x}^{\infty} f(u) e^{-\sigma u} du  \ \ , \ \ f(u) \ge 0
\end{equation*}

and $f(u)$ is monotone increasing, we obtain $f(x) \leq \sigma e^{\sigma x}F(\sigma)$. Hence,
\begin{equation}
h(x) \leq \sigma F(\sigma) e^{(\sigma - 1) x} x^{1 - \alpha} \label{h,h}
\end{equation}
for  any $x > 0$.

Let $\varepsilon > 0$ be a small number.
We choose $\sigma$ so that $0 < \sigma - 1 < \varepsilon$.
For any $v \in \mathbb{R}_{ > 0}$, by using \eqref{h,h}, we obtain

\begin{equation}
\int_{0}^{\infty} h(u) \Phi(v -u) e^{-\varepsilon u} u^{\alpha - 1} du
\leq \sigma F(\sigma) \int_{0}^{\infty} \Phi(v -u) e^{-\big( \varepsilon - (\sigma - 1)\big) u} du.\label{h,ab}
\end{equation}

Since $\Phi$ is bounded and  non-negative, the integral
on the right-hand side of \eqref{h,ab} converges.
Since $h(u)$ is also non-negative,
the integral on the left-hand side converges absolutely. Moreover the integral

\begin{equation*}
\quad \int_{0}^{\infty} \Phi(v - u) e^{-\varepsilon u} u^{\alpha_{i} - 1} du
\end{equation*}

 also converges absolutely.
Therefore, by Fubini's theorem, the following equation holds.

\begin{equation}
\begin{split}
&\int_{\mathbb{R}} \left( h(u) - C - \sum_{i=1}^{r}
\frac{C_{i}}{u^{\alpha - \alpha_{i}}}\right) \Phi(v - u)
e^{- \varepsilon u} u^{\alpha -1} du \\
 &=\int_{0}^{\infty} \left(h(u) - C - \sum_{i=1}^{r}
\frac{C_{i}}{u^{\alpha - \alpha_{i}}}\right)
e^{-\varepsilon u} u^{\alpha -1}
\int_{\mathbb{R}} \mathcal{F}(\Phi)(t) e^{i(v - u)t} d\lambda(t) du \\
 &=  \int_{\mathbb{R}} \mathcal{F}(\Phi)(t) e^{ivt} \int_{0}^{\infty}
\left( h(u) - C - \sum_{i=1}^{r}\frac{C_{i}}{u^{\alpha - \alpha_{i}}}
 \right)
e^{-(it + \varepsilon) u} u^{\alpha -1} du d\lambda(t) \\
 &= \int_{\mathbb{R}} \mathcal{F}(\Phi)(t) e^{ivt}
G(1 + \varepsilon + it) d\lambda(t) . \label{FH}
\end{split}
\end{equation}

By assumption, $G(s)$ is continuous on $\re (s) \ge 1$
and the support of $\mathcal{F}(\Phi)$ is compact.
Thus, $G(1 + \varepsilon + it)$ converges to  $G(1 + it)$
uniformly on the support of $\mathcal{F}(\Phi)(t)$ as $\varepsilon \to
 +0$.
Therefore,

\begin{equation}
\lim_{\varepsilon \to +0} \int_{\mathbb{R}} \mathcal{F}(\Phi)(t)
e^{ivt} G(1 + \varepsilon + it) d\lambda(t)
= \int_{\mathbb{R}} \mathcal{F}(\Phi)(t) e^{ivt} G(1 + it) d\lambda(t). \label{F}
\end{equation}

As far as the left-hand side of \eqref{FH} is concerned,
by using the monotone convergence theorem to each term,
we have

\begin{equation}
\begin{split}
\lim_{\varepsilon \to +0} \int_{0}^{\infty}
\Bigg( h(u) - &C - \sum_{i=1}^{r} \frac{C_{i}}{u^{\alpha - \alpha_{i}}}\Bigg)
\Phi(v - u) e^{- \varepsilon u} u^{\alpha -1} du \\
&= \int_{\mathbb{R}} \left( h(u) - C -
\sum_{i=1}^{r} \frac{C_{i}}{u^{\alpha - \alpha_{i}}} \right)
\Phi(v - u) u^{\alpha -1} du . \label{H}
\end{split}
\end{equation}

We may choose a closed-interval $[a,b]$ containing the support of $\mathcal{F}(\Phi)$.
By \cite[p.140 L{\footnotesize EMMA} 3]{kable}, if $\alpha < 1$

\begin{equation}
\begin{split}
\lim_{v \to \infty}
& \ v^{1 - \alpha} \int_{\mathbb{R}} \mathcal{F}(\Phi)(t)
e^{ivt} G(1 + it) d\lambda(t) \\
= & \lim_{v \to \infty} \ v^{1 - \alpha}
\int_{a}^{b} \mathcal{F}(\Phi)(t) e^{ivt} G(1 + it) d\lambda(t) \\
= & \ 0 \label{vF}.
\end{split}
\end{equation}

If $\alpha \ge 1$, then we have the same
equation by \cite[p.185, 7.5 Theorem]{fourier}.
Since the right-hand sides of \eqref{F} and
\eqref{H} are equal, by using \eqref{vF}, we obtain
\begin{equation}
\lim_{v \to \infty} v^{1 - \alpha} \int_{0}^{\infty} \left(h(u)
- C - \sum_{i=1}^{r} \frac{C_{i}}{u^{\alpha - \alpha_{i}}}\right)
\Phi(v - u) u^{\alpha -1} du = 0 \label{A}
\end{equation}
for any $\alpha > 0$.

By \cite[p.139, Lemma 1]{kable} we have

\begin{equation}
\lim_{v \to \infty} v^{1 - \alpha} \int_{0}^{\infty}
\Phi(v - u) u^{\alpha - 1} du =  \int_{\mathbb{R}} \Phi(u) du \label{k.phi}
\end{equation}

for any $\alpha > 0$. So,

\begin{equation}
\begin{split}
\lim_{v \to \infty} v^{1 - \alpha}& \int_{0}^{\infty}
\frac{C_{i}}{u^{\alpha - \alpha_{i}}} \Phi(v - u) u^{\alpha -1} du \\
&= \lim_{v \to \infty} \frac{1}{v^{\alpha - \alpha_{i}}}
v^{1 - \alpha_{i}}\int_{0}^{\infty} C_{i}\Phi(v - u) u^{\alpha_{i} -1} du  \\
&= \ 0 	\label{C}
\end{split}
\end{equation}
for $i = 1, 2, \cdots, r$. Therefore, by \eqref{A}, \eqref{k.phi}, and \eqref{C}, we have
\begin{equation}
\begin{split}
\lim_{v \to \infty} v^{1 - \alpha}\!\!\int_{0}^{\infty} h(u) \Phi(v - u) u ^{\alpha -1} du
&=\lim_{v \to \infty} C \ v^{1 - \alpha} \!\!\int_{0}^{\infty}\! \Phi(v - u) u^{\alpha - 1} du \notag \\
&= C \int_{\mathbb{R}} \Phi(u) du .
\end{split}
\end{equation}

The rest of the argument is similar to that
in \cite[pp.142-143]{kable}, and we can conclude that

\begin{equation*}
\lim_{u \to \infty} h(u) = C.
\end{equation*}

Therefore, Theorem \ref{thm:preliminary-theorem} is proved.
\end{proof}

\section{Main Theorem}
\label{sec:main-theorem}

In this section, we apply Theorem \ref{thm:preliminary-theorem} to Dirichlet series
which are obtained by sequences of non-negative real numbers and satisfy some conditions.
The following theorem is the main application of Theorem \ref{thm:preliminary-theorem}.

\begin{thm}
\label{thm:main-theorem}
Let $\{a_{n}\}$ be a sequence of non-negative real numbers,
$d \in \mathbb{R}_{> 0}$ and $m$ a positive integer.
Suppose that the Dirichlet series

\begin{equation*}
L(s) = \sum_{n=1}^{\infty} \frac{a_{n}}{n^s}
\end{equation*}

converges absolutely for $\re(s) > d$.
Also suppose that $L^m$ has a meromorphic continuation
to an open set containing the closed half-plane
$\re(s) \ge d$ and holomorphic except for a pole of order $l$ at $s = d$ with
$\lim_{s \to d} L(s)^m (s - d)^l = A^m$, where $A > 0$. Then we have

\begin{equation*}
\sum_{n \le X} a_{n} \sim \frac{AX^d}{d \varGamma (l / m) (\log(X))^{1 - \frac{l}{m}}}
\end{equation*}

as $X \to \infty$.
\end{thm}
\begin{proof}
For a subset $S \subset \R $, let $\phi_{S}$ be the characteristic function of $S$.
Define a function $f : \mathbb{R} \to [0 ,\infty)$ by

\begin{equation*}
f(u) = \sum_{n=1}^{\infty} a_{n} \phi_{[d \log(n),\infty)}(u).
\end{equation*}

By direct computation, we have

\begin{equation*}
F(s) = \int_{0}^{\infty} f(u) e^{-su} du = \frac{1}{s} L(ds)
\end{equation*}

for $\re(s) > 1$.

Let $L(s)^m (s - d)^l = Q(s)$.
Since $L(s)^m$ has a pole of order $l$ at $s = d$,
$Q(s)$ is holomorphic around $s = d$ and $Q(d) = A^{m} ( \neq 0)$.
So there exists a holomorphic function $P(s)$
defined on an open disc $D$  with center at $s = d$
such that $P(s)^{m} = Q(s)$. Since $L(s)(s - d)^{\frac{l}{m}}$
is holomorphic on $\re(s) > d$,
there exists an $m$-th root of unity $\zeta$ such that
$L(s)(s - d)^{\frac{l}{m}} = \zeta\cdot P(s)$ on
$\{s \in \C \ | \ \re(s) > d \} \cap D$.
Therefore, $L(s)(s - d)^{\frac{l}{m}}$ can be
extended to a holomorphic function around $s = d$,
and $\lim_{s \to d} L(s)(s -d)^{\frac{l}{m}} = A \ (A \in
\mathbb{R}_{>0})$
since $L(s)(s - d)^{\frac{l}{m}}$
has positive real values for $s \in \mathbb{R}_{> d}$.
Hence, $\frac{1}{s}L(ds) (d s - d)^{\frac{l}{m}}$ is holomorphic around $s = 1$.
Since

\begin{equation*}
F(s)(s - 1)^{\frac{l}{m}} = \frac{1}{s} L(ds) (s - 1)^{\frac{l}{m}}
= \frac{1}{sd^{\frac{l}{m}}} L(ds) (ds - d)^{\frac{l}{m}},
\end{equation*}

$F(s)(s - 1)^{\frac{l}{m}}$ is holomorphic around $s = 1$.

Therefore, there exists a holomorphic function $B(s)$
defined on an open disc $D'$ with center at $s = 1$ such that

\begin{equation*}
F(s)(s - 1)^{\frac{l}{m}} = \sum_{i=0}^{r} A_{i}(s -1)^i + (s -1)^{r + 1} B(s),
\end{equation*}

where $r = \ceil{ l /m } -1$ and $A_{i} \in \C$ for $i = 0, 1, \cdots, r$.
Then

\begin{equation*}
\lim_{s \to 1} F(s)(s - 1)^{\frac{l}{m}} = \frac{A}{d^{\frac{l}{m}}}
\ \ \ i.e., \ \ A_{0} = \frac{A}{d^{\frac{l}{m}}}.
\end{equation*}

Thus, we have

\begin{equation*}
F(s) = \frac{A_{0}}{(s - 1)^{ \frac{l}{m}}} + \sum_{i=1}^{r}
\frac{A_{i}}{(s - 1)^{ \frac{l}{m} - i}} + (s -1)^{r + 1 - \frac{l}{m}} B(s)
\end{equation*}

for $s \in D' \cap \{ s \in \mathbb{C} \ | \ \re(s) > 1 \}$.

Now, we define
\begin{equation*}
\begin{split}
G(s) &= F(s) - \frac{A_{0}}{(s - 1)^{\frac{l}{m}}}
- \sum_{i=1}^{r} \frac{A_{i}}{(s - 1)^{\frac{l}{m} - i}}\\
&= \frac{1}{s} L(ds) - \frac{A_{0}}{(s - 1)^{\frac{l}{m}}}
- \sum_{i=1}^{r} \frac{A_{i}}{(s - 1)^{\frac{l}{m} - i}}
\end{split}
\end{equation*}
for $\re(s) > 1$.

We shall prove the following (a), (b).
\begin{flushleft}
\begin{itemize}
\item[(a)]
The function $G(s)$ extends continuously to
the set $\{ s \in \mathbb{C} \ | \ \re(s) \ge 1 \}$.
\item[(b)]
The function $x \mapsto G(1 + ix)$ $(x \in \R)$ is locally of bounded variation.
\end{itemize}
\end{flushleft}
Then $f(s)$, $F(s)$, and $G(s)$ satisfy the
condition of Theorem \ref{thm:preliminary-theorem}.
It is enough to prove (a), (b) locally.

We first consider the neighborhood of $s = 1$.
By definition

\begin{equation*}
G(s) = (s - 1)^{r + 1 - \frac{l}{m}} B(s)
\end{equation*}

for $s \in D' \cap \{ s \in \mathbb{C} \ | \ \re(s) > d \}$.

By Theorems 10.2 of \cite[p.192]{real},
the function $x \mapsto B(1 + ix)$ is
of bounded variation on any closed-interval $[a,b]$,
where $\{1 + ix \in \mathbb{C} \ | \ x \in [a,b] \}  \subset D'$.
For any $\alpha > 0$ we can obviously
extend $(s - 1)^{\alpha}$ to a continuous function on
$\re(s) \ge 1$.  By a similar argument as in \cite[p.144]{kable},
the function $x \mapsto \bigl((1 + ix) - 1\bigr)^{\alpha}$ $(x \in \R)$
is locally of bounded variation.

Therefore, if $r + 1 - \frac{l}{m} > 0$, then $G(s)$ extends
continuously to the set $D' \cap \{ s \in \C \ | \ \re(s) \ge 1 \}$
and the function $x \mapsto G(1 + ix)$ ($x \in \mathbb{R}$) is of
bounded variation on any closed-interval $[a,b]$, where
$\{1 + ix \in \mathbb{C} \ | \ x \in [a,b] \}  \subset D'$.
If $r + 1 - \frac{l}{m} = 0$, then $G(s)$ has the same properties
since $G(s) = B(s)$ on $D' \cap \{ s \in \C \ | \ \re(s) > 1 \} $.
Thus, (a), (b) are proved around $s = 1$.

Next we consider the neighborhood of $s = 1 + ix_{0}$,
where $x_{0} \in \R - \{0\}$. By the definition of $G(s)$,
it is enough to prove (a), (b) for $L(ds)$ instead of $G(s)$.

Since $L^{m}$ is holomorphic around $s = d + ix_{0}$,
there exists a holomorphic function $R(s)$
defined for $s$ near $d + ix_{0}$ such that

\begin{equation*}
L(s)^m = \bigl(s - (d + ix_{0}) \bigr)^k R(s),  \ \
R(d + ix_{0}) \neq 0, \  \ and \ \ \  k \ge 0.
\end{equation*}

Since $R(d + ix_{0}) \neq 0$, there exists a
holomorphic function $T(s)$ defined on an open disc
$D''$ with center at $s = d + ix_{0}$ such that $T(s)^{m} = R(s)$.
Then, there exists an $m$-th root of unity  $\zeta$ such that

\begin{equation*}
L(s) = \zeta \cdot \bigl(s - (d + ix_{0}) \bigr)^{\frac{k}{m}} T(s)
\end{equation*}

on $D'' \cap \{ s \in \mathbb{C} \ | \ \re(s) > d\}$.

Now, for any $x_{0} \in \R - \{0\}$ we  can extend
$\bigl(s - (d + ix_{0}) \bigr)^{\frac{k}{m}}$
to a continuous function on $\re(s) \ge d$ and
the function $x \mapsto \bigl(d + ix - (d + ix_{0})\bigr)^{\frac{k}{m}}$
is locally of bounded variation also as in \cite[p.144]{kable}.
Thus, $L(s)$ extends continuously to the set
$D'' \cap \{ s \in \mathbb{C} \ | \ \re(s) \ge d\}$
and the function $x \mapsto L(d + ix)$ is of
bounded variation around $x_{0}$.  It follows
that we can extend  $L(ds)$ to a continuous function
on $\re(s) \ge 1$ except $s = 1$
and the function $x \mapsto L(d + idx)$ is of
bounded variation on any closed-interval $[a,b]$
not containing $x = 0$. Thus, we have proved (a), (b).

Therefore, we can apply Theorem \ref{thm:preliminary-theorem}
to $f(s)$, $F(s)$, $G(s)$, and  $\alpha = l / m$. So we have

\begin{equation}
\lim_{u \to \infty} u^{1 - \frac{l}{m}} e^{-u} f(u)
= \frac{A_{0}}{\varGamma(l / m)}
= \frac{A}{d^{\frac{l}{m}}\varGamma(l / m)}. \label{last}
\end{equation}

Since $f(d \log X ) = \sum_{n \le X} a_{n}$, by substituting
$d\log X$ for $u$ in \eqref{last}, we have

\begin{eqnarray*}
\lim_{X \to \infty} (d \log X)^{1 - \frac{l}{m}} e^{-(d \log X)} f(d\log X)
= \frac{A}{d^{\frac{l}{m}}\varGamma(l / m)}, \\
\lim_{X \to \infty} d^{1 - \frac{l}{m}} (\log X)^{1 - \frac{l}{m}} X^{-d}
\sum_{n \le X} a_{n} = \frac{A}{d^{\frac{l}{m}}\varGamma(l / m)}.
\end{eqnarray*}

Therefore,

\begin{equation*}
\sum_{n \le X} a_{n} \sim \frac{AX^d}{d \varGamma (l / m) (\log(X))^{1 - \frac{l}{m}}}.
\end{equation*}

\end{proof}

\section{An Example}
\label{sec:an-example}

We give an example of a Dirichlet series having a pole of
order "$2 / 3$" and apply Theorem \ref{thm:main-theorem}.

If $n = {p_{1}}^{e_{1}}{p_{2}}^{e_{2}}\cdots{p_{r}}^{e_{r}}$
is a prime decomposition of a positive integer $n$
where $p_1,\ldots,p_r$ are distinct primes, then
we define $f(n) = \sum_{n = 1}^{r} e_{i}$.
We define a Dirichlet series $L$ by

\begin{equation*}
L(s) = \prod_{p : prime} \left(1 - \frac{2}{3} p^{-s} \right)^{-1}
= \sum_{n = 1}^{\infty} \left(\frac{2}{3} \right)^{f(n)} n^{-s}.
\end{equation*}

Then the function $L(s)$ is holomorphic on $\re(s) > 1$
since $\left(2/3 \right)^{f(n)} \le 1$.
We claim that $L(s)^3$ has a meromorphic continuation
to an open set containing $\re(s) \ge 1$ and
holomorphic except for a pole of order $2$ at $s = 1$.

By computation

\begin{equation*}
L(s)^{3} = \prod_{p : prime} \left(1 - 2p^{-s}
+ \frac{4}{3}p^{-2s} - \frac{8}{27}p^{-3s} \right)^{-1},
\end{equation*}

and
\begin{equation*}
\begin{split}
 \frac{L(s)^{3}}{\zeta(s)^2} &= \prod_{p : prime} \frac{(1 - p^{-s})^{2}}
{ \left(1 - 2p^{-s} + \frac{4}{3}p^{-2s} - \frac{8}{27} p^{-3s}\right)}\\
&= \prod_{p : prime} \frac{(1 - p^{-s})^2(1 + p^{-s})^{2}}{ \left(1 -
2p^{-s} + \frac{4}{3}p^{-2s} - \frac{8}{27}p^{-3s}\right)(1 + p^{-s})^2}\\
&= \prod_{p : prime} \frac{(1 - p^{-2s})^{2}}{ \left(1 - \frac{5}{3}p^{-2s}
+ \frac{10}{27}p^{-3s} + \frac{20}{27}p^{-4s} - \frac{8}{27}p^{-5s} \right) }\\
&= \zeta(2s)^{-2} \prod_{p : prime}\left(1 - \frac{5}{3}p^{-2s}
+ \frac{10}{27}p^{-3s} + \frac{20}{27}p^{-4s} - \frac{8}{27}p^{-5s} \right)^{-1}.
\end{split}
\end{equation*}
Let

\begin{equation*}
F(s) = \prod_{p : prime} \left(1 - \frac{5}{3}p^{-2s} + \frac{10}{27}p^{-3s}
+ \frac{20}{27}p^{-4s} - \frac{8}{27}p^{-5s} \right).
\end{equation*}

If $\re(s) \ge 0$, then we have

\begin{equation*}
\left|- \frac{5}{3}p^{-2s} + \frac{10}{27}p^{-3s} + \frac{20}{27}p^{-4s}
- \frac{8}{27}p^{-5s} \right| \le \frac{83}{27}|p^{-2s}|.
\end{equation*}

Thus, if $\re(s) > 1/2$, then

\begin{equation*}
\begin{split}
\sum_{p : prime } \left| -\frac{5}{3}p^{-2s} + \frac{10}{27}p^{-3s}
+ \frac{20}{27}p^{-4s} - \frac{8}{27}p^{-5s}\right|
&\le \sum_{p : prime } \frac{83}{27}p^{-2\re(s)}\\
&< +\infty.
\end{split}
\end{equation*}

Therefore,

\begin{equation*}
\sum_{p : prime } \left|- \frac{5}{3}p^{-2s} + \frac{10}{27}p^{-3s}
+ \frac{20}{27}p^{-4s} - \frac{8}{27}p^{-5s}\right|
\end{equation*}

converges absolutely and uniformly on any compact
subset of the open half-plane $\re(s) > 1 / 2$.
It follows that

\begin{equation*}
\prod_{p : prime }  \left(1 - \frac{5}{3}p^{-2s} + \frac{10}{27}p^{-3s}
+ \frac{20}{27}p^{-4s} - \frac{8}{27}p^{-5s}\right)
\end{equation*}

converges uniformly on any compact subset of the open
half-plane $\re(s) > 1 / 2$ and is holomorphic on
$\re(s) > 1/2$ (see \cite[p.300, 15.6 Theorem ]{complex}).
Thus, $F(s)$ is holomorphic on $\re(s) > 1/2$.

For any prime number $p$, since $(1 -\frac{2}{3}p^{-s})$,
$(1 + p^{-s}) \neq 0$ on $\re(s) > 1/2$,

\begin{equation*}
\left(1 - \frac{5}{3}p^{-2s} + \frac{10}{27}p^{-3s}
+ \frac{20}{27}p^{-4s} - \frac{8}{27}p^{-5s} \right)
= \left(1 + p^{-s}\right)^{2}\left(1 -\frac{2}{3}p^{-s}\right)^{3} \neq 0
\end{equation*}

on $\re(s) > 1/2$.
Hence $F(s) \neq 0$ on $\re(s) > 1/2$ i.e., $F(s)^{-1}$ is
holomorphic on $\re(s) > 1/2$ (see \cite[p.300, 15.6 Theorem]{complex}).

It follows that $L(s)^3$ has a meromorphic continuation to
$\re(s) > 1 / 2$ and is holomorphic except
for a pole of order $2$ at $s = 1$.  Moreover,

\begin{equation*}
\begin{split}
\lim_{s \to 1} L(s)^{3}(s - 1)^2 &= \lim_{s \to 1}
\frac{L(s)^{3}}{\zeta(s)^2} \left(\zeta(s)(s - 1) \right)^2 \\
&=  \prod_{p : prime} \frac{(1 - p^{-2})^{2}}{\left(1 - \frac{5}{3}p^{-2}
+ \frac{10}{27}p^{-3} + \frac{20}{27}p^{-4} - \frac{8}{27}p^{-5}\right) }\\
&= \zeta(2)^{-2} F(1)^{-1}.
\end{split}
\end{equation*}

By applying Theorem \ref{thm:main-theorem} to $L(s)$, we have

\begin{equation*}
\sum_{n \le X} \left(\frac{2}{3} \right)^{f(n)}
\! \sim \ \ \frac{A}{\Gamma(2 / 3)} \cdot \frac{X}{\log(X)^{1/3}},
\end{equation*}

where

\begin{equation*}
A = \left( \zeta(2)^{2} \prod_{p : prime}
\left(1 - \frac{5}{3}p^{-2} + \frac{10}{27}p^{-3}
+ \frac{20}{27}p^{-4} - \frac{8}{27}p^{-5}\right) \right)^{-\frac{1}{3}}.
\end{equation*}

\bibliographystyle{plain}
\bibliography{katorefs}

\end{document}


TITLE: Show that the function $f:X\to \mathbb R$ defined by $f(x)=d(x,A)$ is continuous.
QUESTION [4 upvotes]: Let $d$ be a metric on $X$ and let $A$ be any arbitrary subset of $X$. Show that the function $f:X\to \mathbb R$ defined by $f(x)=d(x,A)$ is continuous.

Let $p\in X$. We want to show that for any open set $S$ containing $f(p)$, there exists an open neighbourhood of $p$, denoted by $O$, such that $f[O]\subset S$. If $p\in$ int $A$, then there exists an open set $O \subset A$ containing $p$. Clearly $f[O] \subset S$ since $O$ is a subset of $A$ and $f[O]= \{ 0 \}= \{f(p) \} \subset S$. If $p\in \partial A$ (boundary of A) or $p\in$ ext $A$, the situation is much harder. The reason for this being hard is that the function $d$ is not specific and is defined in terms of infimum.

REPLY [1 votes]: Lemma: $\forall x,y \in X: f(x) - f(y) \le d(x,y)$
(Intuition: this is the triangle inequality, but with the whole of the set $A$ in place of one of the points: $d(x, A) \le d(x,y) + d(y,A)$)
Proof:
$f(x) - f(y) = d(x, A) - d(y, A) = d(x, A) - \inf_{a \in A} d(y, a) = \inf_{a \in A}(d(x, A) - d(y, a))$
$ \le \inf_{a \in A}(d(x, a) - d(y, a))$ since $d(x,A) \le d(x,a)$
$ \le \inf_{a \in A}d(x, y)$ (triangle inequality, $d(x,a) \le d(x,y) + d(y,a)$)
$ = d(x,y)$
Proving the lemma.
Then $f(y) - f(x) \le d(y, x) = d(x, y)$ since $d$ is symmetric, so:
$|f(x) - f(y)| \le d(x,y)$

Now we will show that the inverse image under $f$ of an open set is open, that is to say $f$ is continuous.
Let $U \subseteq \mathbb{R}$ be an open set.
Let $x \in f^{-1}(U)$, that is $f(x) \in U$, so there exists an open ball $B$ (of radius $r$, say), around $f(x)$ that is a subset of $U$.
Let $B'$ be the open ball around $x$ with radius $r$. Then for $y \in B'$ we have $d(x, y) < r$, so $|f(x) - f(y)| \le d(x, y) < r$, hence $f(y) \in B$. So $B' \subseteq f^{-1}(B) \subseteq f^{-1}(U)$
So every element of $f^{-1}(U)$ has an open ball around it in $f^{-1}(U)$, hence $f^{-1}(U)$ is open.


TITLE: $|V|=?$ if $2\vec{V}+(\vec{V}\times(\vec{i}+2\vec{j}))=2\vec{i}+\vec{k} $
QUESTION [1 upvotes]: I am given $$2\vec{V}+(\vec{V}\times(\vec{i}+2\vec{j})=2\vec{i}+\vec{k} $$
and I need to find $$|V|=?$$ , I took dot product of equation with $\vec{V}$ but got stuck in $$2V^2  = \vec{V}.(2\vec{i}+\vec{k})$$ but there seems no way out similarly with cross product , what to do ?

REPLY [0 votes]: There is probably a quicker way but if you write $\mathbf{V}=a\,\mathbf{i}+b\,\mathbf{j}+c\,\mathbf{k}$ and compute the left-hand-side and compare is to the right-hand side you will generate equations in $a,b$ and $c$ which can be solved.


TITLE: Is there a simple and a non-simple group with same numbers of elements of each order.
QUESTION [4 upvotes]: Are there finite groups $G$ and $H$ such that:

$n:=|G|=|H|$.
$G$ is simple.
$H$ is not simple.
for every $d\mid n$, $G$ and $H$ have the same number of elements of order $d$.
?

REPLY [5 votes]: No, there is no such example. It has been shown that if 1., 2. and 4. hold, then $G$ and $H$ are isomorphic. This is not an easy result, as the proof requires classification of finite simple groups. See the slides here for references and more information.


TITLE: Is there a Poisson Summation formula for imprimitive Dirichlet characters?
QUESTION [5 upvotes]: I was wondering if there exists a Poisson Summation formula (like the one existing with primitive character) for imprimitive Dirichlet characters ?
For a primitive Dirichlet character $\chi$ we have:
$$ \sum\limits_{n=-\infty}^{\infty}\chi(n) f\bigg(\frac{n}{q}x\bigg)  =\frac{K}{x} \sum\limits_{n=-\infty}^{\infty} \overline{\chi(n)} \hat{f}\bigg(\frac{n}{x}\bigg)$$
But for imprimitive characters ?

REPLY [2 votes]: Yes, of course.
Poisson summation formula has nothing to do with characters.  If χ is any periodic function, all you need to do is to replace χ̅ by the discrete Fourier Transform of χ.


TITLE: Relative Characteristic classes
QUESTION [6 upvotes]: A pair of vector bundles over  a  base space $X$ is  a  pair $(E,F)$ where $E$ is a vector bundle over $X$ and $F$ is  a sub-bundle of $E$. Two pairs $(E_{1},F_{1})$  and  $(E_{2}, F_{2})$ are isomorphic if there is  an isomorphism from $E_{1}$ to $E_{2}$ which send $F_{1}$ onto $F_{2}$. The direct sum of two pairs has a natural definition.
Are there some type of characteristic classes with values in the cohomology of base space with the following properties?:
It is trivial $("1")$ on the trivial pairs. Isomorphism pairs have the same characteristic classes. They commute with the pull back operation. They restrict (or have  a reasonable relation)to the usual  characteristic classes on pairs $(E,0)$  and  $(\epsilon_{n}, F)$ where $\epsilon_{n}$ is the trivial $n$ dimensional bundle.

REPLY [8 votes]: You obviously have all characteristic classes of $E$, of $F$, and of $E/F$,
with some relations because $E\cong F\oplus E/F$. To see if there are more, consider the classifying space for your pairs. If you are working over $\Bbbk$, $\mathrm{rk}(F)=k$, $\mathrm{rk}(E)=\ell$, it is the colimit over $n$ of the space of partial flags of length $(k,\ell)$ in $\Bbbk^n$. For a fixed $n$, this is homotopy equivalent to
$$G_{k,\ell,n}(\Bbbk)=U(n,\Bbbk)/(U(k,\Bbbk)\times U(\ell-k,\Bbbk)\times U(n-\ell,\Bbbk))\;.$$
This space is the total space of (at least) three fibre bundles $G_{k,\ell}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{\ell,n}(\Bbbk)$ and $G_{\ell-k,n-k}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{k,n}(\Bbbk)$ and $G_{k,n+k-\ell}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{\ell-k,n}(\Bbbk)$. Hence, you can try the Leray-Serre spectral sequence to find characteristic classes not generated by those mentioned above.
In the holomorphic setting, you see Bott-Chern forms. If $E$ and $F$ are flat, you see the forms described in the appendix of Bismut-Lott.
In both settings, you can sometimes extract cohomological information.
I am not sure there are similar classes in a purely topological setting, but I would personally go for $\Bbbk=\mathbb R$ first and look at $\mathbb Z/2$-valued cohomology.
EDIT: It seems there are no additional characteristic classes that do not come from $F$ or $E/F$. For consider the fibre bundle $G_{\ell-k,n-k}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{k,n}(\Bbbk)$. For sufficiently large $n$, all classes of $G_{\ell-k,n-k}(\Bbbk)$ of small degree come from the vector bundle $E/F$, and so are pulled back from $G_{k,\ell,n}(\Bbbk)$. Hence by the Leray-Hirsch theorem, in small degrees the cohomology ring of
$G_{k,\ell,n}(\Bbbk)$ is generated by the characteristic classes of the vector bundles $F$ and $E/F$.

REPLY [6 votes]: The characteristic classes of the virtual bundle $E-F$ seem to have all the properties you asked for.
If $c$ is a characteristic class of vector bundles satisfying a sum formula
$$
c(E\oplus F) = c(E)c(F),
$$
then $c(E-F)$ can be defined via the formula
$$
c(E-F)c(F) = c(E),
$$
or if the base $X$ is compact,
$$
c(E-F) = \frac{c(E)}{c(F)}.
$$


TITLE: A series with logarithms
QUESTION [5 upvotes]: Can we express in terms of known constants the sum:
$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{\log (n+1)-\log n}{n}$$
First of all it converges , but not matter what I try or whatever technic I am about to apply it fails. In the mean time if we split it apart (let us take the partial sums) then:
$$\sum_{n=1}^{N} \frac{\log (n+1) - \log n}{n}= \sum_{n=1}^{N} \frac{\log (n+1)}{n} - \sum_{n=1}^{N} \frac{\log n}{n}$$
The last sum at the RHS does resemble a zeta function derivative taken at $1$. Of course zeta function diverges at $1$ but its PV exists, namely $\mathcal{P}(\zeta(1))=\gamma$. Maybe we have a PV for the derivative also? The other sum at the RHS is nearly the last sum at the right.
This is as much as I have noticed in this sum. Any help?
Addendum: I was trying to evaluate the integral:
$$\mathcal{J}=\int_0^1 \frac{(1-x) \log (1-x)}{x \log x} \, {\rm d}x$$
This is what I got.
\begin{align*}
\int_{0}^{1}\frac{(1-x) \log(1-x)}{x \log x} &=-\int_{0}^{1} \frac{1-x}{x \log x} \sum_{n=1}^{\infty} \frac{x^n}{n} \, {\rm d}x \\
 &= -\sum_{n=1}^{\infty}\frac{1}{n} \int_{0}^{1}\frac{x^{n-1} (1-x)}{\log x} \, {\rm d}x\\
 &=\sum_{n=1}^{\infty} \frac{1}{n} \int_0^1 \frac{x^n-x^{n-1}}{\log x} \, {\rm d}x \\
 &\overset{(*)}{=} \sum_{n=1}^{\infty} \frac{\log(n+1) -\log n}{n}  \\
 &= ?
\end{align*}
$(*)$ since it is quite easy to see that:
$$\int_{0}^{1}\frac{x^a-x^{a-1}}{\log x} \, {\rm d}x = \log (a+1) - \log a  , \; a \geq 1$$
due to DUTIES.
Maybe someone else can tackle the integral in a different manner?

REPLY [1 votes]: Just an addendum, maybe some of this will be useful to you:
$$\int_0^1 \frac{(1-x) \log (1-x)}{x \log x} \, {\rm d}x=$$
$$=1.2577468869\dots= \gamma_{1}(1,0) - \gamma=\int_0^1 \ln (1-x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx-\gamma=$$
$$=\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\sum_{k = 2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}=$$
$$= \frac{\pi^2}{4}-1-4\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}=\frac{\pi^2}{4}-1-4\int_{0}^{\pi/2} \frac{t~dt}{e^{\pi \tan t}+1} =$$
$$=\int_{1}^{\infty} \frac{\ln ([x]+1)}{x^2}dx=\int^{\infty}_0 \frac{\log x+\Gamma (0,x) + \gamma}{e^x-1}~dx$$
Among the sources of the above expressions are:
http://www.people.fas.harvard.edu/~sfinch/csolve/kz3.pdf
http://www.people.fas.harvard.edu/~sfinch/csolve/kz.pdf
https://math.dartmouth.edu/~carlp/factorial.pdf
https://books.google.com/books?id=Pl5I2ZSI6uAC&pg=PA122&lpg=PA122&dq=1.2577468869&source=bl
https://math.stackexchange.com/a/1065075/269624
https://math.stackexchange.com/a/1733543/269624


TITLE: Suppose $\frac1{\sqrt{n}}\sum_{i=1}^nY_i\overset{d}\to N(0,V).$ What is the distribution of$\frac1{\sqrt{n}}\sum_{i=1}^nG(Y_i)$?
QUESTION [1 upvotes]: Suppose
$$\frac1{\sqrt{n}}\sum_{i=1}^nY_i\overset{d}\to N(0,V).$$
Let $G(x)=\int_{-\infty}^xk(u)du$ be a kernel distribution function.
Can we obtain the asymptotic distribution of $\frac1{\sqrt{n}}\sum_{i=1}^nG(Y_i)$?
Another question is if
$$\frac1{\sqrt{n}}\sum_{i=1}^nY_i\overset{d}\to N(0,V).$$
Can we obtain that
$\frac1n\sum_{i=1}^nY_i^2\overset{p}\to V$?

REPLY [1 votes]: For the second question, the answer is no in general. Let $\left(\varepsilon_i\right)_{i\geqslant 1}$ be i.i.d. random variables which take the valued $-1$ and $1$ with probability $1/2$. Define
$$
Y_i:=\varepsilon_i+Z_i-Z_{i-1},
$$
where $\left(Z_i\right)_{i\geqslant 0}$ is an i.i.d. sequence independent of $\left(\varepsilon_i\right)_{i\geqslant 1}$. Then
$$
\frac 1{\sqrt n}\sum_{i=1}^nY_i=\frac 1{\sqrt n}\sum_{i=1}^nY_i+\frac 1{\sqrt n}\left(Z_n-Z_0\right)
$$
and since the second term goes to $0$ in probability, the central limit theorem shows that $\left(\frac 1{\sqrt n}\sum_{i=1}^nY_i\right)_{n\geqslant 1}$ converges in distribution to a centered normal law with variance $1$. Moreover,
$$
\frac 1n\sum_{i=1}^nY_i^2=\frac 1n\sum_{i=1}^n\varepsilon_i^2
+\frac 2n\sum_{i=1}^n\varepsilon_i\left(Z_i-Z_{i-1}\right)+\frac 1n\sum_{i=1}^n\left(Z_i-Z_{i-1}\right)^2.
$$
The first term is equal to $1$, the second converges to $0$ and the third to
$\mathbb E\left[\left(Z_1-Z_{0}\right)^2\right]$ which may be not zero.