text:
\subsection{The AC Optimal Power Flow Problem}
An AC power system in steady state can be modeled as a directed graph where the set of nodes $\N = \{1,2,\ldots,n\}$ corresponds to a set of $n$ power buses, and the set of edges $ \L \in \N\times \N$ corresponds to transmission lines, \ie, $ (k,l) \in \L $ if there is a line from bus $k$ to bus $l$. The set
$\F \subseteq \L$ consists of all transmission lines with a flow constraint, $\PA \subseteq \L$ consists of all transmission lines with a phase-angle difference constraint, $ \G_k $ denotes a (possibly empty) set of generators associated with bus $k$, and $ \G = \bigcup_{k \in \N} \G_k$ is the set of all generators. The power produced by generator $g \in \G$ is $s_g = p_g + \jmath q_g $, and at each power bus $k \in \N$, we define a complex load (\ie, demand) $ S_k^d = P_k^d + \jmath Q_k^d $, a complex voltage $ v_k $, and a complex current $ i_k $. To simplify notation, we define a vector of voltages $v = (v_1,v_2,\ldots,v_n)$ and a vector of currents $ i = (i_1,i_2,\ldots,i_n) $. With this notation, the ACOPF problem can be expressed as
\begin{subequations} \label{eq:opf}
	\begin{equation}
		\mbox{minimize} \ \ \sum_{g\in\G} f_g(p_g) \
	\end{equation}
	\mbox{subject to} \\
	\begin{align}
							i_k^* v_k & = \sum_{g \in \G_k} s_g - S_k^d, & k&\in \N \label{eq:power_balance} \\
						 	P_g^{\min}	& \le p_g \le P_g^{\max}, & g&\in \G \label{eq:opf:realpower}\\
						 	Q_g^{\min}	& \le q_g \le Q_g^{\max}, & g&\in \G\label{eq:opf:reactivepower} \\
						 	V_k^{\min}	& \le \abs{v_k} \le V_k^{\max}, & k&\in \N \label{eq:opf:voltage}\\
						   \abs{S_{k,l}^{\text{fl}} (v)} & \le S_{k,l}^{\max}, & (k,l)&\in\F	\label{eq:flow_con1}\\
						   \abs{S_{l,k}^{\text{fl}} (v)} & \le S_{l,k}^{\max}, & (k,l)&\in\F \label{eq:flow_con2} \\
						 \phi_{k,l}^{\min} & \le \angle (v_k v_l^*) \le \phi_{k,l}^{\max}, & (k,l) &\in \PA \label{eq:opf:phaseanglediff} \\
						 i &= Yv
						 \label{eq:opf:ohm}
	\end{align}
\end{subequations}
with variables $i \in \cmplx^n$, $v \in \cmplx^n$, and $s \in \cmplx^{|\G|}$, and where $i=Yv$ corresponds to Ohm's law in matrix form, given the network admittance matrix $Y \in \cmplx^{n \times n}$. The cost of generation for generator $g$ is given by $f_g(p_g)$, and we will restrict our attention to convex quadratic generation cost functions, \ie,
\begin{equation}
	 f_g(p_g) = \alpha_g p_g^2 + \beta_g p_g + \gamma_g,
\end{equation}
where the parameters $ \alpha_g \ge 0 $, $ \beta_g $, and $\gamma_g$ are given.
The constraints \eqref{eq:power_balance} are power balance equations, \eqref{eq:opf:realpower} and \eqref{eq:opf:reactivepower} are generation limits, \eqref{eq:opf:voltage} are voltage magnitude limits, \eqref{eq:flow_con1} and \eqref{eq:flow_con2} are transmission line flow constraints, and \eqref{eq:opf:phaseanglediff} are phase-angle difference constraints. The flow from bus $k$ to bus $l$ is given by $ S_{k,l}^{\text{fl}} (v) = v^H T_{k,l} v + \jmath v^H \widetilde{T}_{k,l}v $ (provided that $(k,l) \in \F$ or $(l,k) \in \F$) where $ T_{k,l} \in \herm^n$ and $ \widetilde{T}_{k,l} \in \herm^n$ are given.

\subsection{Semidefinite Relaxation}
Roughly following the steps described in \citep{AHV:14}, we start by reformulating the ACOPF problem \eqref{eq:opf}. Specifically, we perform the following steps:
\begin{enumerate}
	\item Eliminate $i = Yv$ and substitute $P_g^{\min} + p_g^{\lb}$ for $p_g$, $Q_g^{\min} + q_g^{\lb}$ for $q_g$, and $X$ for $vv^H$.
	\item Drop constant terms in the objective:
	\begin{align*}
		f(p_g) &= \alpha_g(P_g^{\min} + p_g^{\lb})^2 + \beta_g (P_g^{\min} + p_g^{\lb}) + \gamma_g\\
		 &= \alpha_g (p_g^{\lb})^2 + \tilde \beta_g p_g^{\lb} + \text{const.}
	\end{align*}
	where $\tilde \beta_g = (\beta_g + 2\alpha_gP_g^{\min})$.
	\item Introduce an auxiliary variable $t_g$ for each $g \in \G^{\text{quad}} = \{g\in\G \,|\, \alpha_g > 0 \}$ and include epigraph constraint
	\[ \alpha_g(p_g^{\lb})^2 \leq t_g \ \Leftrightarrow \begin{bmatrix} 1/2 + t_g\\1/2-t_g \\ \sqrt{2\alpha_g} p_g^{\lb} \end{bmatrix} \in \qcone^3. \]
	\item Introduce slack variables to obtain a standard-form formulation.
\end{enumerate}
These steps yield the equivalent problem
\begin{subequations} \label{eq:opf-reform} \allowdisplaybreaks
	\begin{equation}
		\mbox{minimize} \ \ \sum_{g\in\G} \tilde \beta_g  p_g^{\lb} + \sum_{g\in\G^{\text{quad}}} t_g
	\end{equation}
	\mbox{subject to} \\
	\begin{align}
		\Tr(Y_k X) & = \sum_{g \in \G_k} (P_g^{\min} + p_g^{\lb}) - P_k^d, &  k &\in \N \label{eq:opf-reform-ls} \\
		\Tr(\widetilde Y_k X) & = \sum_{g \in \G_k} (Q_g^{\min} + q_g^{\lb}) - Q_k^d, & k &\in \N \\
	 	p_g^{\lb} + p_g^{\ub} &=  P_g^{\max}-P_g^{\min}, & g &\in \G \\
		q_g^{\lb} + q_g^{\ub} &=  Q_g^{\max}-Q_g^{\min}, & g &\in \G \\
	 	 X_{kk} - \nu_k^{\lb}	&= (V_k^{\min})^2,  & k &\in \N \\
		 X_{kk} + \nu_k^{\ub} &= (V_k^{\max})^2,  & k &\in \N \\
		 z_{k,l} &= \begin{bmatrix} S_{k,l}^{\max} \\ \Tr(T_{k,l} X) \\ \Tr(\widetilde{T}_{k,l} X)\end{bmatrix}, & (k,l)&\in\F	\\
		 z_{l,k} &= \begin{bmatrix}	S_{l,k}^{\max} \\ \Tr(T_{l,k} X) \\ \Tr(\widetilde{T}_{l,k} X)\end{bmatrix}, & (k,l)&\in\F\\
  	w_g &= \begin{bmatrix}1/2 + t_g \\ 1/2- t_g \\ \sqrt{2\alpha_g} p_g^{\lb}\end{bmatrix}, & g &\in \G^{\text{quad}} \\
		\Im(X_{kl}) &=	\tan(\phi_{k,l}^{\min}) \Re(X_{kl}) + y_{k,l}^{\lb}, & (k,l) &\in \PA  \\
	 	\Im(X_{kl}) &=  \tan(\phi_{k,l}^{\max}) \Re(X_{kl}) - y_{k,l}^{\ub}, & (k,l) &\in \PA \label{eq:opf-reform-le} \\
	 p_g^{\lb}, p_g^{\ub} & \ge 0, & g&\in\G \\
	 q_g^{\lb}, q_g^{\ub} & \ge 0, & g&\in\G \\
	\nu_k^{\lb}, \nu_k^{\ub} & \ge 0, & k&\in\N \\
	y_{k,l}^{\lb}, y_{k,l}^{\ub} &\geq 0, & (k,l) &\in \PA\\
	z_{k,l},z_{l,k} &\in \qcone^3, & (k,l) &\in \F \\
  w_g & \in \qcone^3, & g &\in \G^{\text{quad}} \\
	X & = vv^H \label{eq:opf-reform-rank1}
	\end{align}
\end{subequations}
with variables $p^{\lb},p^{\ub},q^{\lb},q^{\ub} \in \reals^{|\G|}$, $t\in \R^{|\G^{\text{quad}}|}$, $\nu^{\lb}, \nu^{\ub} \in \reals^{|\N|}$,
$y^{\lb},y^{\ub} \in \reals^{|\PA|}$, $z_{k,l},z_{l,k} \in \qcone^3$ for $(k,l) \in \F$, $w_g \in \qcone^3$ for $g \in \G^{\text{quad}}$, $X \in \herm^n$, and $v \in \cmplx^n$. Notice that the constraints \eqref{eq:opf-reform-ls}-\eqref{eq:opf-reform-le} are all linear. We refer the reader to \citep{AHV:14} for a definition of the data matrices $Y_k, \widetilde Y_k, T_{k,l}$, and $\widetilde T_{k,l}$.

The only non-convex constraint in \eqref{eq:opf-reform} is the rank-1 condition \eqref{eq:opf-reform-rank1}. An SDR of \eqref{eq:opf-reform} is readily obtained by replacing \eqref{eq:opf-reform-rank1} by the positive semidefiniteness constraint $X \succeq 0$. The resulting SDR is a so-called cone linear program (CLP) that can be expressed as
\begin{align} \label{eq:cone-lp}
	\begin{array}{ll}
	\mbox{minimize}   & c^Tx \\
	\mbox{subject to} & Ax = b\\
                  	& x \in \K
	\end{array}
\end{align}
where $x$ is the vector of variables and the cone $\cone$ is a Cartesian product of three types of cones, \ie,
\[ \mathcal{K} = \reals_+^{n_l} \times \underbrace{\qcone^3 \times \cdots \times \qcone^3}_{n_q} \times \cone_{\mathrm{h}}^n.\]
Thus, the number of variables is $N = n_l + 3n_q +n^2$ where $n_l = 4|\G| + |\G^{\text{quad}}| + 2|\N| + 2|\PA|$ and $n_q = 2|\F|+|\G^{\text{quad}}|$, and the number of equality constraints is
$M = 4 |\N| + 2|\G| + 2 |\PA| + 3n_q$.

\subsection{Conversion}
The computational cost of solving \eqref{eq:cone-lp} with a general-purpose interior-point method becomes prohibitively large when $n$ is large: the cost of an interior-point iteration is at least $O(n^3)$.
Fortunately, the problem \eqref{eq:cone-lp} is generally very sparse in practice, and hence the conversion method of \citet{FKMN:01} may be used to rewrite \eqref{eq:cone-lp} as an equivalent CLP
\begin{align} \label{eq:cone-lp-convert}
	\begin{array}{ll}
	\mbox{minimize}   & \tilde c^T \tilde x \\
	\mbox{subject to} & \tilde A \tilde x = b\\
										& E \tilde x = 0 \\
                  	& \tilde x \in \widetilde \K
	\end{array}
\end{align}
with
\[ \widetilde \K = \reals_+^{n_l} \times \underbrace{\qcone^3 \times \cdots \times \qcone^3}_{n_q} \times \cone_{\mathrm{h}}^{r_1} \times \cdots \times \cone_{\mathrm{h}}^{r_m}. \]
The conversion essentially decomposes the cone $\cone_{\mathrm{h}}^n$ into a Cartesian product of a number of lower-dimensional cones $\cone_{\mathrm{h}}^{r_1}\times \cdots \times \cone_{\mathrm{h}}^{r_m}$ at the expense of a set of coupling constraints $E \tilde x = 0$. This reformulation of the problem can have a dramatic effect on the computational cost of solving the SDR of the ACOPF problem, and it effectively mitigates the $O(n^3)$ bottleneck that arises with the formulation \eqref{eq:cone-lp}. Moreover, the conversion technique often induces sparsity in the system of equations that define the search direction at each interior-point iteration, reducing the cost per iteration further if the solver can exploit this type of sparsity. The conversion technique was first applied to SDRs of the ACOPF problem by \citet{Jab:12}.

\subsection{Implementation} \label{implementation}
Before turning to our numerical experiments, we briefly outline our implementation \citep{opfsdr}. The code is written in Python and performs the following steps:
\begin{enumerate}
	\item Read case file and build the CLP \eqref{eq:cone-lp}.
	\item Apply conversion method: convert \eqref{eq:cone-lp} to \eqref{eq:cone-lp-convert}.
	\item Apply Hermitian-to-symmetric transformation: map $\cone_{\mathrm{h}}^{r_i}$ to $\cone_{\mathrm{s}}^{2r_i}$ for $i=1,\ldots,m$.
	\item Scale the problem data to improve conditioning.
\end{enumerate}
As part of the first step, we allow some preprocessing of the data: (i) slack variables $p_g$ (or $q_g$) for which $P_g^{\min} = P_g^{\max}$ (or $Q_g^{\min} = Q_g^{\max}$) may be eliminated, (ii) numerical proxies for infinity which are used to indicate the absence of limits (\eg, on generation) may be truncated, and (iii) a minimum resistance of transmission lines may be enforced. The Hermitian-to-symmetric transformation is a well known trick  that is only necessary because the solvers used in our experiments cannot directly handle cones of Hermitian positive semidefinite matrices; see \eg\ \citep{BoV:04}. The scaling of the problem data in step 4 yields an equivalent problem, and we found that for some solvers, this can reduce the computational time by roughly a factor of two; we briefly return to the topic of scaling in Section \ref{sec:discussion}.


TITLE: Prove the sequence $\sum_{k=0}^n \frac{1}{(n+k)^2}$ is convergent
QUESTION [2 upvotes]: I'm having a hard time trying to prove that the sequence $\{a_n\}$ whose general term $a_n$ is
$$\sum_{k=0}^n \frac{1}{(n+k)^2}$$
is convergent. I'm trying to prove it by definition, that is to say, by finding a lower/upper bound and by proving that it is decreasing/increasing using induction.
By subtracting $a_n$ from $a_{n+1}$ we obtain the following for $n=m$ (if I'm not mistaken):
$$
\frac{1}{(2m+1)^2} + \frac{1}{(2m+2)^2} - \frac{1}{m^2}
$$
Which is less than $0$ for $n=1$ from which I have assumed the sequence is decreasing and therefore trying to prove that $a_{n+1} - a_n < 0$ using induction. But I am terribly stuck! Thanks in advance for any suggestion.

REPLY [0 votes]: $\displaystyle\sum_{k = 0}^{n}{1 \over \left(n + k\right)^{2}} =
{1 \over n}\left[{1 \over n}\sum_{k = 0}^{n}{1 \over
\left(1 + k/n\right)^{2}}\right]
\,\,\,\stackrel{\color{red}{\mathrm{as}\ n\ \to\ \infty}}{\sim}\,\,\,
{1 \over n}\int_{0}^{1}{\mathrm{d}x  \over \left(1 + x\right)^{2}} =
{1 \over 2n}$


TITLE: How to find the original function, if the tangent is given
QUESTION [0 upvotes]: So the question goes find $f(x)$ if

$\frac{df}{dx} = 4x - 3$ and the line $y = 5x - 7$ is tangent to $f(x)$.

To find the original function one would integrate the derivative to get $f(x) = 2x^2 - 3x + c$
How do I find $c$ ? And how is the tangent function a part of the question?
Using desmos with trial and error I found c to be 1, but I am unsure how to reach this conclusion mathematically.
Thanks.

REPLY [1 votes]: Slope of straight line is $5$. Set it equal to $4 x -3$. That gives you $x=2$. Corresponding $y$ value is $10-7=3.$ Substitute point coordinates  $(2,3)$ into $ y= 2x^2-3x+c $ and that gives you $c=1$.


TITLE: For a random pile of cylindrical batteries, what is the probability of a short circuit?
QUESTION [0 upvotes]: The question is not really an engineering issue, even if it seems to be on the first sight. I'm not trying to solve a practical problem, I try to understand whether this can even handled as a well defined geometrical problem.
Imagine a large number of AA battery cells in a non conducting container.
They have been put into the container by dropping each of them in a random orientation at a random location. The batteries are cylindrical and have a cylindrical extension as "-" contact at one end. The other end is the "+" contact, except the edge.
I try to get at least an intuition of how probable it is that there is a short circuit in the the volume of batteries.
I think it could be simulated, but is it possible to calculate the probability otherwise?
A short circuit is possible with batteries touching "+" and "-" contacts in a flat circle.
I'm thinking about the geometric part of the question, so there is a further simplification needed: The friction. No friction at all would create a parallel packing, I think even an optimal packing. Perfect stickiness is also not a good simplification, because it would involve rolling of batteries or an extremely loose packing. Choose any friction simplification that possibly works. But at this point, it seems to be a question that depends more on physics than math.
Use any simplification that helps*.
A calculation is surely not practically possible for even a hand full of batteries, I'm not trying to solve a practical problem. But is it even possible to handle is as a well defined geometrical problem?

* Note that I do not even specify the length and radius of the battery cells. It would implicitly make it an engineering question, I think.

REPLY [0 votes]: Okay, so I took out some batteries from my drawer and I found out that I need a minimum of 9 AA batteries to form a loop (because of the nubby top part and ring of plastic around the bottom).
Assuming a 2D plane and 5 degrees of error.
5 degrees / 360 degrees = 0.014.
The first battery can be in any orientation.
The second battery has to be in one or two orientations.
The third battery on can only be in one orientation.
So 1 * 0.028 * 0.014^7 = 3 * 10^-15 which is a chance of one in 333 trillion.
Of course there are problems to this solution. It's possible to form a loop with more than 9 batteries, but the chance of that happening is magnitudes lower.
The five degrees of error is completely arbitrary, but a small degree of error makes sense because the minimum angle between two batteries is 140 degrees. If the angle was any bigger, then the loop won't close with 9 batteries.
Also, I only considered the rotation of the battery and not its position. If the position was accounted for, the chances of forming a loop would probably be lower.
Finally, I limited my solution to a 2D plane. In a 3D space, the chances of forming a loop would be much lower because there are simply more angles.


TITLE: Bisection Method - True error versus Approximate error
QUESTION [0 upvotes]: From the book "Numerical Methods for Engineers", by Steven C. Chapra, they state the true error is always less than the approximate error, and therefore, it is safe to use the Scarborough Error formula for asserting at least $n$ significant figures:
$$\epsilon_t\leq \epsilon_a \lt (0,5*10^{2-n}) \%$$
I can see that always holds for the absolute true error and the absolute approximate error, since the following also holds:

Let $X_R$ be the real root between the upper limit $x_u$ and lower limit $x_l$
Let $\Delta X$ be the distance between the limits $\Delta X = x_u-x_l$
Let $X_r^{new}$ the new calculated approximate root $(x_u+x_l)/2$
Let $X_r^{old}$ be the old approximate root. In this context, $x_l$ or $x_u$ depending on the last iteration.

Since the real root won't overflow the limits, it is clear to see that:
$$|X_r^{new}-X_r^{old}| = \frac{\Delta X}{2}$$
$$|X_R-X_r^{new}| \leq \frac{\Delta X}{2} = |X_r^{new}-X_r^{old}|$$
$$\therefore |E_t| = |X_R-X_r^{new}| \leq |E_a| = \frac{\Delta X}{2}$$
So the statement is true for the absolute error. What I haven't been able to proof (and I am having my doubts that is always true), is the same statement for the relative errors:
$$\epsilon_a = |\frac{X_r^{new}-X_r^{old}}{X_r^{new}}|\times 100\%$$
$$\epsilon_t = |\frac{X_R - X_r^{new}}{X_R}|\times 100\%$$
Even more, I've been having confusions on what happens when the real root is very close to zero and the estimated root is still far away from it. Wouldn't it make it possible for the real true relative error to be higher than the approximate relative one?
Final questions are, how can I proof that statement is true for the relative errors, and how can I be sure it is safe to use the first inequality given as a stop condition for the algorithm if the statement doesn't always hold?
Thanks in advance.

REPLY [1 votes]: You are correct that relative error may actually be much higher than expected (infinite even if the actual root is zero). Because of this, it is impossible to assert the claimed inequalities.
If the root is non-zero and bound bounds have the same sign, then an upper bound on the relative error may be found by simply dividing by the estimate closer to zero.
To compensate for both cases (small and large answers), usually a combination of absolute and relative errors are used. This prevents the algorithm from failing to terminate if the root is zero (relative error failure) or very large (absolute error failure). Commonly one uses a stopping criterion in the form of
$$|x_u-x_l|\le\epsilon_\mathrm{abs}+\epsilon_\mathrm{rel}\cdot x_r$$
where $\epsilon_\mathrm{abs}$ and $\epsilon_\mathrm{rel}$ are the desired absolute and relative errors.


TITLE: Can we find a regular ($C^k$) parametrization for this surface?
QUESTION [2 upvotes]: I have here a surface whose curvature properties I want to study, represented in cylindrical coordinates:
$$f(r,\theta) = r^2\cos4\theta$$
The problem, however, is that the parametrization is not differentiable (indeed, not defined) at exactly the point I want to look at - the origin! I have a plot here made by Mathematica:

Trying to use the usual conversion to rectangular coordinates doesn't work either - at best, I can get $f(x,y) = (x^2 + y^2)\cos\cos^{-1}\left(\frac{x}{\sqrt{x^2+y^2}}\right)$, which is also not differentiable at the origin. Now, I'm not sure the surface itself is actually $C^\infty$, but I'm pretty sure it's at least $C^2$. How can we find a parametrization that's differentiable at the origin?

REPLY [2 votes]: Note that by setting $z = x + iy = re^{i\theta}$, your expression is equal to
$$ f(x,y) = \operatorname{Re} \left( \frac{z^4}{|z|^2} \right) = \frac{\operatorname{Re} \left( (x + iy)^4 \right)}{x^2 + y^2} = \frac{x^4 - 6x^2y^2 + y^4}{x^2+y^2} = \frac{8x^4}{x^2+y^2} + y^2 - 7x^2. $$
Now, a priori $f$ is not defined at $(0,0)$. However, the polar expression (and your plot) of $f$ makes it clear that $\lim_{(x,y) \to (0,0)} f(x,y) = 0$ so you can define $f(0,0) = 0$ and then $f$ becomes continuous. How about higher regularity? It is somewhat easier in my opinion to continue to work in polar coordinates so let me state two useful observations:

If you are given a function $f(x,y)$ defined in a neighborhood $U$ of $(x_0,y_0)$ and (continuously) differentiable in the punctured neighborhood $U \setminus \{(x_0,y_0)\}$ and the limit $\lim_{(x,y) \to (x_0,y_0)} (\nabla f)(x,y)$ exists, then $f$ is (continuously) differentiable on $U$ and $(\nabla f)(x_0,y_0) = \lim_{(x,y) \to (x_0,y_0)} (\nabla f)(x,y)$. This follows from the mean value theorem.
Assume you are given a function $h(x,y)$ with $h(0,0) = 0$ that can be represented in polar coordinates as $h(r,\theta) = r^k p(\cos(\theta),\sin(\theta))$ where $k > 1$ and $p(u,v)$ is a continuously differentiable function defined on $(-1-\varepsilon,1+\varepsilon)^2$. Then $h$ is continuous and using polar coordinates, we can compute

$$ (\nabla h)(r,\theta) = \frac{\partial f}{\partial r} \hat{e}_r + \frac{1}{r} \frac{\partial f}{\partial \theta} \hat{e}_{\theta} = kr^{k-1}p(\cos(\theta),\sin(\theta))\hat{e}_r + r^{k-1} \left( \frac{\partial p}{\partial v}(\cos(\theta),\sin(\theta))\cos(\theta) - \frac{\partial p}{\partial u}(\cos(\theta),\sin(\theta))\sin(\theta) \right)\hat{e}_{\theta} = r^{k-1} \left( kp(\cos(\theta)\hat{e}_x + \sin(\theta) \hat{e}_y) + \left( \frac{\partial p}{\partial v}\cos(\theta) - \frac{\partial p}{\partial u}\sin(\theta) \right)(-\sin(\theta) \hat{e}_x + \cos(\theta) \hat{e}_y) \right) =
r^{k-1} \left( \left( kp\cos(\theta) + \frac{\partial p}{\partial u}\sin^2(\theta) - \frac{\partial p}{\partial v}\sin(\theta)\cos(\theta) \right) \hat{e}_x + \\ \left( kp\sin(\theta) +
\frac{\partial p}{\partial v}\cos^2(\theta) - \frac{\partial p}{\partial u}\sin(\theta)\cos(\theta) \right)\hat{e}_y \right). $$
Since $k > 1$ and the expressions involving $\theta$ are bounded, we have $\lim_{(x,y) \to (0,0)} (\nabla h)(x,y) = (0,0)$ and the first observation implies that $h$ is continuously differentiable and $(\nabla h)(0,0) = (0,0)$.
In your case, applying it to $h = f$ we see that $f$ is continously differentiable at the origin and $(\nabla f)(0,0) = 0$. However, the second derivatives of $f$ are not continuous as we have
$$ \frac{\partial^2 f}{\partial x^2} = \frac{2(x^6 + 3x^4y^2 + 27x^2y^4 - 7y^6)}{(x^2+y^2)^3}. $$
The second derivative $\frac{\partial ^2 f}{\partial x^2}$ is homogeneous of degree zero. In polar coordinates, it has the form $\frac{\partial ^2 f}{\partial x^2}(r,\theta) = q(\theta)$ and so it depends only on the angle $\theta$. Since it is non-constant, it cannot be extended continuously to the origin.


TITLE: Behavior of differential equation as argument goes to zero
QUESTION [3 upvotes]: I'm trying to solve a coupled set of ODEs, but before attempting the full numerical solution, I would like to get an idea of what the solution looks like around the origin.
The equation at hand is:
$$ y''_l - (f'+g')y'_l  + \biggr[ \frac{2-l^2-l}{x^2}e^{2f}  - \frac{2}{x}(f'+g') - \frac{2}{x^2} \biggr]y_l = \frac{4}{x}(f'+g')z_l$$
$y,f,g,z$ are all functions of $x$, which has the domain ($0, X_0$). If I specifically take the $l=2$ case I of course have
$$ y''_2 - (f'+g')y'_2  + \biggr[ \frac{-4}{x^2}e^{2f}  - \frac{2}{x}(f'+g') - \frac{2}{x^2} \biggr]y_2 = \frac{4}{x}(f'+g')z_2$$
To avoid issues with singularities, I multiply both sides of the equation by $x^2$, to get, lets call it EQ1.
$$ x^2 y''_2 - x^2(f'+g')y'_2  + \biggr[ -4e^{2f}  - x(f'+g') - 2 \biggr]y_2 = 4 x(f'+g')z_2$$
Now if by some magic I know by that as $x \rightarrow 0$, $y_l = x^{l+1}$ so that $y_2 = x^3$. How would I determine what the functions $z_2$ is around the origin?
Actually I already know the answer: $z_2$ should also go as $x^3$, but I have not been able to show it.
Any help is much appreciated.
My attempt at a solution:
 I have tried to keep things general so I expand $f[x] = 1 + f_1 x + f_2 x^2 +f_3 x^3$, and similarly $g[x] = 1 + g_1 x + g_2 x^2 +g_3 x^3$ where $f_1,f_2,f_3, g_1,g_2,g_3$ are constants. I have kept up to third order because I want to substitute $y_2 = x^3$, so I figured I should take the other functions to that order as well.
As for the $e^{2f}$ term, I use a truncated Taylor series for the exponential, $1+\frac{x^2}{2!} + \frac{x^3}{3!}$, into which I substitue my expansion of $f[x]$. After expanding everything out and eliminating terms of higher order than $x^3$ from the right hand side of EQ1, I get something like $constant*x^3$, while the left hand side is third order polynomial times $z_2$.
I really just don't know how to proceed. Should I have left higher order terms in the RHS, so that I could divide by a third order polynomial and still end up with $z_2 \propto x^3$. I don't know how this would work because on the RHS I had terms has high as $x^{12}$.

REPLY [1 votes]: My solution is as follows. I essentially tried to reproduce 5PM's work in Maple with some mathematica code.
First I create the generic expansions of the functions $f$ and $g$ using
f = Sum{ff[k]*x^k, {k,0,5}]
g = Sum{gf[k]*x^k, {k,0,5}]

I had to create the variables "ff", and "gg" so as to avoid recursions limit errors in mathematica. Then I substituted these expansions into equation 1 above and then then divided the right hand side so get $z_2$ by itself. Then I used
Series[z2[r], {r,0,5}]

For which I got the following output. (I have cleaned it up a bit, having only kept terms of order three, and replacing the "ff" and "gg" with just "f" and "g".)
$
z_2= -\frac {(e^{2 f_0} - 1 ) x^2} {f_1 + g_1} + \frac {1} {4} x^3  \biggr \{ \frac {2 (4e^{2 f_0} - 4)(f_2 + g_2))} {(f_1 + g_1)^2} - \frac {8 e^{2 f_0} f_1 + 5 f_1 + 5 g_1} {f_1 + g_1}\biggr \}$
I esentially get the same result as 5PM. From this analysis it seems that $z_2$ goes as $x^2$.
The reason this disagrees with the paper I was reading, is that I forgot that physically speaking $f_0$ is very close to zero, and so the x^2 term does go to zero, and so the physical answer is
$
z_2=  \frac {1} {4} x^3  \biggr \{\frac {8 e^{2 f_0} f_1 + 5 f_1 + 5 g_1} {f_1 + g_1}\biggr \}$ .


TITLE: How to prove that if the expectation value of $A$ in any state is real, then $A$ is Hermitian?
QUESTION [4 upvotes]: If the expectation value of operator $A$ in any state is real, then $A$ is Hermitian.

there is an uncompleted proof:
$$ \int(c_1\psi_1+c_2\psi_2)^* A (c_1\psi_1+c_2\psi_2)dx$$
$$=|c_1|^2\int\psi_1^* A \psi_1dx + |c_2|^2\int\psi_2^* A \psi_2dx
+c_1^*c_2\int\psi_1^* A \psi_2dx +c_2^*c_1\int\psi_2^* A \psi_1dx
$$
My teacher said that because
$$c_1^*c_2\int\psi_1^* A \psi_2dx +c_2^*c_1\int\psi_2^* A \psi_1dx$$
is real, then we can conclude that
$$\int\psi_1^* A \psi_2dx = (\int\psi_2^* A \psi_1dx)^*$$
but I don't understand this last step, can someone explain it, or give another proof?

REPLY [3 votes]: Here is another version of the same proof.
If $\langle \psi | A \psi \rangle \in \mathbb R$ for all $\psi \in \cal H$, then
$\langle \psi | A \psi \rangle^* =  \langle \psi | A \psi \rangle$ for all $\psi \in \cal H$. Since $\langle \psi | \phi \rangle^* = \langle \phi| \psi\rangle$  we have that $\langle \psi | A \psi \rangle = \langle A\psi |  \psi \rangle$ that is $^1$ $\langle \psi | A \psi \rangle = \langle \psi | A^\dagger \psi \rangle$, namely:
$$\langle \psi | (A-A^\dagger) \psi \rangle =0 \quad \forall \psi \in {\cal H}\:.\tag{1}$$
Now consider $\psi = \phi + \chi$, where $\phi,\chi \in {\cal H}$ are arbitrary, obtaining:
$$\langle\phi + \chi| (A-A^\dagger) (\phi + \chi) \rangle =0 \tag{2}\:.$$
using the fact that $\langle \phi | (A-A^\dagger) \phi \rangle = \langle \chi | (A-A^\dagger) \chi \rangle 0$ in view of (1), and expanding (2) taking the real bilinerity of the scalar product into account we find
$$\langle \phi | (A-A^\dagger) \chi \rangle + \langle \chi | (A-A^\dagger) \phi \rangle=0\:.\tag{3}$$
The procedure can be implemented another time, starting again from (1) but now using $\psi = \phi + i\chi$. Exploiting the fact that the scalar product is anti linear in the left argument and linear in the right one, this time we end up with
$$i\langle \phi | (A-A^\dagger) \chi \rangle -i \langle \chi | (A-A^\dagger) \phi \rangle=0\:,$$
that is
$$\langle \phi | (A-A^\dagger) \chi \rangle - \langle \chi | (A-A^\dagger) \phi \rangle=0\:,$$
Together with (3) it entails in particular that
$$\langle \chi | (A-A^\dagger) \phi \rangle =0 \quad \forall \phi, \chi \in {\cal H}\:.$$
We are free to choose $\chi =  (A-A^\dagger) \phi$ obtaining
$$||   (A-A^\dagger) \phi||^2 = \langle (A-A^\dagger) \phi | (A-A^\dagger) \phi \rangle =0 \quad \forall \phi \in {\cal H}\:,$$
which immediately implies
$$ (A-A^\dagger) \phi =0 \quad \forall \phi \in {\cal H}\:.$$
Summing up, we have obtained
$$A^\dagger \phi = A\phi\quad  \forall \phi \in {\cal H}$$
that is the thesis. If the domain of $A$ is not the whole Hilbert space, the above reasoning holds however in that domain as it is a linear subspace by hypotheses.

(1) I just used the defintion of adjoint, $A^\dagger$, of $A$ that implies both
$\langle \psi | A \phi \rangle =  \langle A^\dagger \psi |  \phi \rangle$ and
$\langle A\phi |  \psi \rangle =  \langle  \phi |  A^\dagger\psi \rangle$.


TITLE: Explicit equations for $Y(N)$ for small $N$
QUESTION [5 upvotes]: Consider the congruence subgroup
$$\Gamma(N) = \left\{\left(\begin{array}{cc}
                            a & b \\
                             c & d\end{array} \right) \in SL_2(\mathbb{Z})\ ;\ \left(\begin{array}{cc}
                            a & b \\
                             c & d\end{array} \right) \equiv  \left(\begin{array}{cc}
                            1 & 0 \\
                             0 & 1\end{array} \right)\mod N\right\}$$
acting on the upper half-plane $\mathfrak{h} = \{\tau \in \mathbb{C} \ ;\  \Im(\tau)>0\}$ by fractional linear transformations:
$$
\left(\begin{array}{cc}
                            a & b \\
                             c & d\end{array} \right) \cdot \tau = \frac{a\tau + b}{c\tau + d}
$$
It is known that the quotient $Y(N) = \Gamma(N)\backslash\mathfrak{h}$ is an algebraic curve for every $N$. For instance,
$$Y(2) = \mathbb{P}^1 \setminus \{0,1,\infty\}$$
which can be proved by considering modular function $\lambda : \mathfrak{h} \to \mathbb{C}$. It is also well known that the compactification $X(7)$ of $Y(7)$ is isomorphic to Klein's quartic in $\mathbb{P}^2$.
What about the other small values of $N$? How do I compute $Y(N)$ for $N=3,4,5,6$? I know I can look at the genus of the compactification $X(N)$, but is there a way of getting the precise form of $Y(N)$ as in the example for $N=2$?
Bonus: What about the universal elliptic curve over $Y(N)$, is it also computable in these small cases?

REPLY [1 votes]: Let me try to give a complete answer.
First, for $ N = 3,4,5 $ the genus of the compactification $X(N)$ is $0$, therefore it is isomorphic to $\mathbb{P}^1$. Since $X(N) = Y(N) \cup C(N)$, where $C(N)$ is the finite set of cusps of $\Gamma(N)$, namely $C(N) = \Gamma(N) \backslash \mathbb{P}^1(\mathbb{Q})$, we see that $Y(N) = \mathbb{P}^1 \setminus C(N)$, where $C(N)$ is a finite set of points. Specifically, computing the number of cusps, we see that $|C(2)| = 3, |C(3)| = 4, |C(4)| = 6$ and $|C(5)| = 12$.
Moreover, we know that the isomorphism $X(1) \rightarrow \mathbb{P}^1$ is given by the $j$-map. We can compute the natural covering map $X(N) \rightarrow X(1)$, and by composition obtain the natural corresponding isomorphism. (This is the modular function $\lambda : \mathfrak{h} \rightarrow \mathbb{C}$ you have described above). This map also allows you to represent the universal elliptic curve, as it describes $j$ as a rational function in $\lambda$, and one may simply substitute it in the description of $E_j$.
This was done for several cases, using towers of prime-power level subgroups, basically using two methods.
In the paper https://projecteuclid.org/euclid.ant/1513090725 [Sutherland, Andrew; Zywina, David. Modular curves of prime-power level with infinitely many rational points. Algebra Number Theory 11 (2017), no. 5, 1199--1229. doi:10.2140/ant.2017.11.1199] the authors used Siegel functions to describe the $j$-map for $X(2), X(3), X(4), X(5)$. The tables in the end of the paper show the covering maps to intermediate covers, but composing them all together one obtains the result.
In the paper https://link.springer.com/article/10.1007/s40993-015-0013-7 [Rouse, J., Zureick-Brown, D. Elliptic curves over ℚ
and 2-adic images of Galois. Res. Number Theory 1, 12 (2015). https://doi.org/10.1007/s40993-015-0013-7] the authors used Eisenstein series to describe the $j$-map for $X(2)$ and $X(4)$, again by maps to intermediate covers.
Finally, for $N = 6$, the genus of $X(6)$ is $1$, and since it has rational points, we know it is an elliptic curve. As $\Gamma(6)$ has $12$ cusps, $Y(6)$ is this elliptic curve without $12$ points.
Generalizing and adapting the methods of Rouse and Zureick-Brown, I calculated that $X(6)$ is the Elliptic Curve $y^2 = x^3 + 1$, and the cusps are mapped to the $12$ points
$$ S = \left \{ \infty, (-1,0), (2, \pm 3), (0,\pm 1), (\zeta_6^{\pm1},0
), \left(-2 + \sqrt{-3}, \pm (3 + \sqrt{-3}) \right), \left(-2 - \sqrt{-3}, \pm (-3 + \sqrt{-3}) \right)
\right \}$$ so that
$$
Y(6) = E \setminus S
$$
I also ran it for the other curves above to see where the cusps map to. One obtains the following models.
$$
Y(3) = \mathbb{P}^1 \setminus \left \{ 0, \frac{1}{3}, \frac{1}{2} \left( 1 \pm \frac{\sqrt{-3}}{3} \right) \right \}
$$
$$
Y(4) = \mathbb{P}^1 \setminus \left \{ 0, \pm 1, \infty, \pm i \right \}
$$
$$
Y(5) = \mathbb{P}^1 \setminus \left \{ 0, \infty, \phi \zeta_5^j, -\phi^{-1} \zeta_5^j \right \}
$$
where $\phi = \frac{1 + \sqrt{5}}{2}$, and $\zeta_5$ is a $5$-th root of unity.
As for the universal elliptic curve, it's slightly tedious to explicitly write it down here, but it is simply substituting the $j$-map in each of the cases.


TITLE: Compute $\lim_{n\to\infty}\int_\limits{0}^{\infty}\frac{dx}{x^{n}+1}$
QUESTION [4 upvotes]: Compute the following limit: $$\lim_{n\to\infty}\int_\limits{0}^{\infty}\frac{dx}{x^{n}+1}$$.

So I thought to use the result:

a) If $\{F_n\}$ converges uniformly on $S=[a,b]$ to $F$ and $F_n$ is integrable $\forall n$. Then $\int_\limits{a}^{b}F(x)dx=\lim\limits_{n\to\infty}\int_\limits{a}^{b}F_n(x)dx$.

$$
\lim_{n\to\infty}\int_\limits{0}^{\infty}\frac{dx}{x^{n}+1}=\lim_{n\to\infty}\int_\limits{0}^{1-\delta}\frac{dx}{x^{n}+1}+\lim_{n\to\infty}\int_\limits{1-\delta}^{1+\delta}\frac{dx}{x^{n}+1}+\lim_{n\to\infty}\int_\limits{1+\delta}^{\infty}\frac{dx}{x^{n}+1}
$$
I can easily pass the limit under the integral in the first and third integral which gives me the results  of $1-\delta$ and $0$ respectively.
However my problems lie in the second integral: $\lim\limits_{n\to\infty}\int_\limits{1-\delta}^{1+\delta}\frac{dx}{x^n+1}$
I changed the variable $y=1-x$ which led me to $\lim\limits_{n\to\infty}\int_\limits{-\delta}^{\delta}\frac{dy}{(1-y)^{n}+1}=?$
Question:
How do I compute  $\lim\limits_{n\to\infty}\int_\limits{-\delta}^{\delta}\frac{dy}{(1-y)^{n}+1}=$? As the delta is arbitrary I do not know how $1-y$ is going to behave.

REPLY [1 votes]: Note that
$$\left|\int_\limits{0}^{\infty}\frac{dx}{x^{n}+1}-1 \right|\leq \left| \int_\limits{0}^{1-\delta}\frac{dx}{x^{n}+1} -(1-\delta)\right|+ \left| \int_\limits{1-\delta}^{1+\delta}\frac{dx}{x^{n}+1}-2 \delta \right|+\left| \int_\limits{1+\delta}^{\infty}\frac{dx}{x^{n}+1}\right|+3 \delta$$
and
$$0 \leq \int_\limits{1-\delta}^{1+\delta}\frac{dx}{x^{n}+1} \leq \int_\limits{1-\delta}^{1+\delta}dx$$


TITLE: Find those values 'a' which belongs to the Convex Hull
QUESTION [0 upvotes]: Find those values of 'a' for which (1,a,1) belongs to the convex hull of
$$\{(0,0,0), (1,1,2),(2,4,-6), (1,3,8)\}$$
Give me hints as much as you can, I would like to understand the mindset rather than the mechanical way of solving it. Thank you in advance!

REPLY [0 votes]: It's fairly easy to see that the plane $x=1$ cuts the tetrahedral solid generated by $\{(0,0,0),(1,1,2),(2,4,-6),(1,3,8)\}$ in a triangular section with vertices $(1,1,2),(1,2,-3),(1,3,8)$.
The plane $z=1$ then cuts this triangle along a line segment with an endpoint between the first and second of these vertices:
$$ \frac{4}{5} \cdot (1,1,2) + \frac{1}{5} \cdot (1,2,-3) = (1,\frac{6}{5},1) $$
and another endpoint between the second and third vertices:
$$ \frac{7}{11} \cdot (1,2,-3) + \frac{4}{11} \cdot (1,3,8) = (1,\frac{26}{11},1) $$
Thus "by inspection" the feasible interval for $a$ is:
$$  \frac{6}{5} \le a \le \frac{26}{11} $$
We could alternatively formulate the problem as a pair of linear programs, seeking the minimum (resp. maximum) value of $a$ such that:
$$ (1,a,1) = b\cdot (1,1,2) + c\cdot (2,4,-6) + d\cdot (1,3,8) $$
$$ b + c + d \le 1 $$
subject to $b,c,d \ge 0$ nonnegativity constraints.  The barycentric coordinates of $(1,a,1)$ with respect to vertices $(0,0,0),(1,1,2),(2,4,-6),(1,3,8)$ are $(1-b-c-d),b,c,d$ respectively.


TITLE: Injective module
QUESTION [2 upvotes]: Let $R=M_n(D)$ the Matrix ring over a division ring and consider $R$ as a left module over itself. Is $R$ an injective module?
I know that $R$ is free, hence it is projective.
Is $R$ injective? I were thinking about using Baer's criterion, but don't get it to work. Any hints on that?

REPLY [1 votes]: Suppose $R$ is not injective as a left module over itself. Then the injective envelope is a proper essential extension: $R\subset E(R)$.
But $R$ is semisimple, so all $R$ modules, including $E(R)$ are semisimple, so $R$ is a direct summand of $E(R)$. But if $R\oplus C=E(R)$, $R$ isn't an essential submodule unless $C=\{0\}$, a contradiction since $E(R)$ doesn't properly extend $R$ in that case.


TITLE: Counterexample for statements regarding to Raabe and convergence
QUESTION [0 upvotes]: Is there seq $\{a_n\}$ which absolutely converges and satisfies following condition:

For any given $L>1$, there exists $n_{L}\in \mathbb{N}$ such that $1-1/n \geq |a_{n+1}/a_{n}| \geq 1-L/n$ for $n\geq n_{L}$.

Is is possible :$c_n>0, b_n \neq 0$, $lim_{n \rightarrow \infty} b_n/c_n \neq 1$ and $\sum_{n=1}^{\infty} (b_n-c_n)$ converges but $\sum_{n=1}^{\infty} c_n$ diverges.
(Here, $lim_{n \rightarrow \infty} b_n/c_n$ converges)

REPLY [1 votes]: Let $a_n=\dfrac{1}{n(\log n)^2}$. Then $\sum a_n<\infty$ and
$$
1-\frac{a_{n+1}}{a_n}=\frac{1}{n}+\frac{2}{n\log n}+O\Bigl(\frac{1}{n^2}\Bigr).
$$

2.
$$
b_n=\frac{2+(-1)^n}{n},\quad c_n=\frac{2}{n}.
$$
In this example $b_n/c_n$ does not converge.
2'. Let $d_n=b_n-c_n$. The we want to find $c_n>0$ and $d_n$ such that $\sum c_n=+\infty$, $c_n+d_n\ne0$, $\sum d_n$ converges and $\lim_{n\to\infty}d_n/c_n=\ell$ exists and $\ell\ne0$. If $\ell>0$ then $d_n>0$ for all $n$ large enough, and by the comparison test $\sum c_n$ and $\sum d_n$ have the same character. Similarly if $\ell<0$. Thus there is no such counterexample. This works also if $\ell=\pm\infty$.


text:
\begin{document}

\date{\today}

\maketitle

\pagebreak

\centerline{\it Dedicated to Professor K. Kodaira}

\

\section{Introduction}

\

\

Let $({\mathcal W}, \omega)$ be a smooth projective algebraic
variety of dimension $2n$ over $\C$ together with a holomorphic
$(2,0)$-form of maximal rank $2n$. A subvariety $X\subset {\mathcal W}$
is called {\em weakly lagrangian} if $\dim X \le n$ and if
the restriction of $\omega$ to $X$ is trivial (notice that
$X$ can be singular).
An $n$-dimensional subvariety
$X\subset \W$ with this property is called {\em lagrangian}.
For example, any curve $C$ contained in
a K3 or abelian surface $S$ is lagrangian.
Further examples of lagrangian subvarieties are
obtained by taking a curve $C\subset S$ and by
considering the corresponding symmetric products.
Alternatively, one could look at a product of
different curves (of genus $> 1$) inside a product
of abelian varieties.
We will say that a variety
$X\subset \W$ is {\em fibered} if it admits a dominant map
onto a curve of genus $ > 1$.
In this note we construct examples of nonfibered
lagrangian surfaces in abelian varieties.
Our motivation comes from the following

\begin{prob}
Find examples of projective surfaces
with a nontrivial fundamental group. In particular,
find examples where the fundamental group
has a nontrivial nilpotent tower.
\end{prob}

If $X$ is fibered over a curve
$C$ of genus $>1$ then the fundamental group
$\pi_1(X)$ surjects onto a subgroup of
finite index in $\pi_1(C)$ and consequently
both $\pi_1(X)$ and its nilpotent tower are big.
Therefore, we are interested in examples of surfaces where the
nontriviality of $\pi_1(X)$ is not induced from curves.
Consider the map to the Albanese variety $\alb\,:\, X\ra \Alb(X)$.
One is interested in situations where the natural map
$$
H^2(\Alb(X),\C)\ra H^2(X,\C)
$$
has a nontrivial kernel -- the triviality of the  kernel implies
the triviality of the nilpotent tower (tensor $\Q$).
Such examples were given by
Campana (\cite{campana}, Cor. 1.2) and
Sommese-Van de Ven (\cite{sommese-vandeven}).
However, there the kernel was found in the map
$$
\Pic(\Alb(X))\ra \Pic(X).
$$
In their construction the fundamental group of the
variety is a central extension of an abelian group
(and the lower central series has only two steps).
The lagrangian property of $\alb(X)\subset \Alb(X)$ implies
that there is a nontrivial kernel in $H^{2,0}$ (rather than
on the level of the Picard groups).
We produce an infinite series of surfaces $X$
of different topological types which
are contained in abelian varieties and
are lagrangian (or weakly lagrangian)
with respect to a nondegenerate $(2,0)$-form.
The weakly lagrangian property for a subvariety $X$
of an abelian variety $A$ is related to the nontriviality
of the fundamental group $\pi_1(X)$ as follows:
the number of linearly independent generators
of the second quotient of the central series
of $\pi_1(X)$ is bounded from below
by the dimension of the rational envelope of
the space of those $(2,0)$-forms on $A$ which restrict
trivially to $X$.

Our construction uses dominant maps
between K3 surfaces. Let $S$ be a K3 surface
and $g_1,g_2$ dominant rational maps of $S$ to
Kummer K3 surfaces $S_1,S_2$ (blowups of quotients of
abelian surfaces $A_1,A_2$ by standard involutions).
Then the (birational) preimage $X$ of
$(g_1,g_2)(S)\subset S_1\times S_2$
in $A_1\times A_2$ is a lagrangian
surface (in general, singular). For special choices of
$g_1,g_2$ we can compute some basic invariants of $X$
and, in particular, show that $X$ is nonfibered.
For example, let $A$ be an abelian surface which is not
isogenous to a product of elliptic curves and $S$
the associated Kummer surface.  Assume that
$g_1\,:\,S\ra S$ is an isomorphism
and $g_2\,:\,S\ra S$ is not an isomorphism.
Then $X\subset A\times A$ is not fibered, $\Alb(X)$ is isogenous
to $A\times A$ and $X$ is lagrangian with
respect to exactly one 2-form on $A\times A$.
We analyze other constructions, with $S$ an elliptic
Kummer surface and $g_j$ induced from the group law.
We don't determine the actual structure of $\pi_1(X)$
(it presumably depends on $X$), but it seems quite
plausible that for some $X$ from our list
$\pi_1(X)$ has a rather nontrivial nilpotent tower.

\

{\bf Acknowledgments.} The first author was partially
supported by the NSF. The second author was partially supported
by the NSA.  The paper was motivated by a
question raised by F.~Catanese. We would like
to thank him for useful discussions in the
early stages of this work. We thank the referee for comments
which helped to improve the exposition.

\

\section{Preliminaries}
\label{sect:2}

Let $V_{\Q}$ be a finite dimensional $\Q$-vector space and
$V_{\C}$ its complexification. Let $w\in V_{\C}$ be
a vector. We denote by
$L(w)\subset V_{\Q}$ its rational envelope, i.e.,
the smallest linear subspace such that $w\in L(w)_{\C}$.
More generally, if $W\subset V_{\C}$ is any
set, we will denote by $L(W)\subset V_{\Q}$ the smallest
linear subspace such that $L(w)\subset L(W)$ for all $w\in W$.
An element $w\in V_{\C}$ will be
called {\em k-generic} if the dimension of $L(w)$
is $\dim V_{\Q} - k$.

\begin{rem}
The set of rationally defined subspaces in $V_{\C}$ is
countable. Therefore, for any linear subspace $W\subset V_{\C}$
we have $L(W)=L(w)$ for all  $w\in W$ which
are not contained in a countable number of linear subspaces.
\end{rem}

\begin{prop}
Let $(A,\omega)$ be an abelian variety of dimension $2n$
together with a nondegenerate holomorphic $(2,0)$-form $\omega$.
Assume that $\omega$ defines a $k$-generic
element in $H^{2,0}(A,\C)$ with $k  < 3 $. Then there are no
weakly lagrangian
surfaces in $A$.
\end{prop}

{\em Proof.}
Assume that we have a weakly lagrangian surface
$i\,:\, X\hookrightarrow A$. Consider the induced
$\Q$-rational homomorphism
$$
i^* \,:\, H^2(A,\Q)\ra H^2(X,\Q).
$$
The class of $\omega$ is contained in
the ($\Q$-rationally defined)
kernel of $i^*$. Since $\omega$ is $k$-generic in $H^{2,0}(A,\C)$
for $k < 3$ the image of $H^2(A,\C) $
in $H^2(X,\C)$ has rank at most
$2$. If we had  a
holomorphic form $w'\in H^{2, 0}(A,\C) $
restricting nontrivially to $X$ then the image of $H^2(A,\C) $ in
$H^2(X,\C) $ would contain at least three linearly independent
forms $ w', \bar w'$ and the image of
a polarization from $\Pic(A)$.
Thus the triviality of $\omega$ on $X$ implies that
{\em all} $(2,0)$-forms restrict trivially to $X$.
Consequently, the dimension of $X$ is $\le 1$, contradiction.
{}\hfill $\square$

\

We end this section with a simple description
of maps of fibered surfaces onto curves (see also \cite{catanese}).

\begin{prop}\label{lemm:factoring}
For every smooth projective
algebraic surface $V$ there exists
a universal algebraic variety ${\cal U}(V)$ such that
every dominant rational map $V\ra C$ onto a smooth
curve of genus $\ge 2$ factors through ${\cal U}(V)$.
\end{prop}

{\em Proof.}
First observe that the set of dominant rational maps
of $V$ onto
smooth curves with generically irreducible fibers is
countable. Indeed, the class $f^*(L)$ (where $L$ is
some polarization on the image-curve) defines $f$.
If two such classes $[f]$
and $[f']$ differ by an element in $\Pic^0(V)$,
then the maps must be the same. Otherwise,
some fiber of $f$ would surject onto the image of $f'$
and therefore, the degrees of the two classes
$[f]$  and $[f']$ on this
fiber would be  different.

Further, dominant maps onto smooth curves of genus $\ge 2$
define a linear subspace $W\subset H^0(V,\Omega^1)$ by
the property that wedge-products of linearly independent
forms $w,w'\in W$ are trivial.
The set of maximal subspaces $W$ in $H^0(V,\Omega^1)$
with this property is an algebraic variety.
Remark that for two forms $\omega, \omega'$
with $\omega\wedge \omega'=0$  their ratio $\omega/\omega'$ is
a nonconstant rational function, which is constant on the fibers
of the foliations on $V$ defined by $\omega$, resp. $\omega'$.
Thus, any such subspace $W$ defines a foliation ${\cal F}_W$
on $V$ with compact nonintersecting fibers (locally the
form defines
a holomorphic map, its fibers are the fibers of ${\cal F}_W$).
Therefore, ${\cal F}_W$ defines a dominant rational map
onto a curve $C_W$ (which lifts to a map of $V$ onto the
normalization of $C_W$).
Different $W$ define different foliations
and different morphisms
(the sum of two spaces $W$ and $W'$ with
the same foliations ${\cal F}_W={\cal F}_{W'}$
has the same property, contradicting the maximality of
$W$ and $W'$). Since the
set of such spaces $W$ is on the one hand
algebraic (a finite union of subvarieties of a Grassmannian)
and on the other hand countable, it must be finite. \hfill $\square$

\begin{coro}
\label{coro:factoring}
For every smooth projective
algebraic surface $V$ there exists
a universal algebraic variety ${\cal U}(V)$ such that
every dominant rational map $V\ra V'$, where $V'$ is a
product of smooth curves of genus $\ge 2$,
factors through ${\cal U}(V)$.
\end{coro}

{\em Proof.} We have shown in Lemma~\ref{lemm:factoring}
that there is a finite
number of dominant rational maps onto curves of genus $\ge 2$.
The product of these curves and maps is ${\cal U}(V)$.
Universality follows. \hfill $\square$

\section{Construction}
\label{sect:construction}

Let $A$ be an abelian surface and $\tau$ the involution
$\tau(a) = -a$ (for $a\in A$). This involution
acts on $A$ with 16 fixed points.
We denote by $\tilde{S}$ the (singular) quotient $A/\tau$.
Let $A^*$ be the blowup of $A$ in the 16 points.
The involution $\tau$ extends to a fixed point free
action on $A^*$ and the quotient $A^*/\tau$ is a K3 (Kummer) surface $S$,
a blowup of $\tilde{S}$.
It contains 16 exceptional curves and we
will denote by $\Ex=\cup_{k=1}^{16} \Ex^k$ their union.
We shall denote by  $\delta\,:\, A\ra \tilde{S}$  the
double cover.
Every K3 surface has a unique (up to constants)
nondegenerate holomorphic $(2,0)$-form.

\

We start with two {\em simple} abelian surfaces $A_1$ and
$A_2$ (in particular,
$A_j$ don't contain elliptic curves - this assumption simplifies
the discussion in Section~\ref{sect:details}).
We consider the corresponding Kummer surfaces $S_j$ and
we assume that there exists another K3 surface
$S$ together with two dominant rational maps $g_j\,:\, S\ra S_j$.
We denote by $\tilde{g}_j$ the induced maps $S\ra \tilde{S}_j$.
Now consider the map
$$
\tilde{\varphi}
=(\tilde{g}_1,\tilde{g}_2)\,:\, S\ra \tilde{S}_1\times \tilde{S}_2.
$$
We have two double covers
$$
(\delta_1,1)\,:\,
A_1\times \tilde{S}_2 \ra \tilde{S}_1\times \tilde{S}_2
$$
and
$$
(1,\delta_2)\,:\, \tilde{S}_1\times A_2\ra \tilde{S}_1\times \tilde{S}_2.
$$
Denote by $\tilde{Y}_1=(\delta_1,1)^{-1}(\tilde{\varphi}(S))$
and by $\tilde{Y}_2=(1,\delta_2)^{-1}(\tilde{\varphi}(S))$.
Let
$$
\tilde{X}=\tilde{X}_{\tilde{\varphi}}:=(\delta_1,\delta_2)^{-1}(
\tilde{\varphi}(S))
$$
be the  preimage of
$S$ in $A_1\times A_2$.
We see that the
$(2,2)$-covering $\tilde{X}\ra \tilde{\varphi}(S)$
is
$$
\tilde{X}=\tilde{Y}_1\times_{\tilde{\varphi}(S)}
\tilde{Y}_2\subset A_1\times A_2.
$$
The surface $\tilde{X}$ is, in general, singular.

\begin{lemm}\label{lemm:largange}
There exists a nondegenerate holomorphic 2-form  $\omega'$
on $A_1\times A_2$ such that
$\tilde{X}$ is lagrangian with respect to $\omega'$.
\end{lemm}

{\em Proof.}
The singular surface $\tilde{S}_j$ carries a
nondegenerate holomorphic 2-form, which we again denote by
$\omega_j$.
Evidently, $\tilde{g}_j^*\omega_j=\lambda_j\omega$ for
some nonzero numbers $\lambda_j$.
Consider the form
$$
\omega'= \lambda_2 \omega_1 -
\lambda_1 \omega_2
$$
on $\tilde{S}_1\times \tilde{S}_2$. The form $\omega'$ is
identically zero on  the image $\tilde{\varphi}(S)$
(since it is trivial on the open part of $S$ where $\tilde{\varphi}$
is smooth).
Since both forms $\omega_j$ lift to nondegenerate forms on the
abelian surfaces $A_j$ the form $\omega'$ lifts
to a nondegenerate form on $A_1\times A_2$.
The restriction of the lift of $\omega'$ to $\tilde{X}$ is
identically zero on the smooth points of $\tilde{X}$.
\hfill $\square$

\begin{exam} If $A_1=A_2=A$ and the maps
$\tilde{g}_j={\rm id}$ then the resulting surface $\tilde{X}$ is
the abelian surface $A$, embedded diagonally into $A\times A$.
\end{exam}

\begin{exam}\label{exam:unique}
Let $S$ be a K3 surface, which is simultaneously Kummer and
a double cover of $\P^2$. Denote
by $\theta$ the covering involution on $S$.
Put $S_1=S_2=S$, $\tilde{g}_1={\rm id}$ and
$\tilde{g}_2=\theta$.
Then the corresponding surface $\tilde{X}$ is lagrangian. A special case
of this construction is obtained as follows:
Let $C$ be a curve of genus $2$
and $\sigma$ a hyperelliptic involution on $C$.
Consider $C\times C$, together with the involutions:
$\sigma_{12}$ interchanging the factors and
$$
\sigma_a\,:\,
(c_1,c_2)\ra (\sigma(c_2),\sigma(c_1)).
$$
Denote by  $Y_1=C\times C/\sigma_{12}$ and $Y_2=C\times C/\sigma_a$.
We see that both $Y_1$ and $Y_2$ are isomorphic to a symmetric square
of $C$, which is birational to the same abelian surface $A$.
Denote by $\tilde{S}$ the quotient of $A$ by the standard involution $\tau$.
Observe that $S$ is realized as a double cover of
$\P^2=\P^1\times \P^1/\sigma_{12}$. Then
$$
(\tau,\tau)^{-1}((1,\theta)(S))=C\times C\subset A\times A.
$$
\end{exam}

In the following sections we will show that
we can arrange a situation where $\tilde{X}$ is not contained in
any abelian subvariety of $A_1\times A_2$ and
where it does not admit any dominant
morphisms onto a curve of genus $\ge 2$.

\section{Details and proofs}
\label{sect:details}

\subsection{Desingularization}
\label{sect:desingularization}

To analyze the surface $\tilde{X}$
constructed in Section~\ref{sect:construction}
we will need its sufficiently explicit partial
desingularization.  It will be constructed in two steps.

\

Let $\Sigma $ be the minimal finite set of points in $S$ such
that both maps $g_1,g_2$ are well defined on
the complement $S^0=S\setminus \Sigma$.
On $S^0$ the maps $g_1$ and $g_2$ are local isomorphisms.
Consider the divisors
$\Ex_j\subset S_j$.
The preimages $D_j^0$ of these divisors
(under the maps $g_j$) in
$S^0$ are smooth and the components of $D^0_j$
don't intersect in $S^0$. However, the divisors $D_1^0$ and $D^0_2$ do
intersect in $S^0$, and we denote by $p_1,...,p_l\in S^0$
their intersection points. Let $D_j$ be the closure
of $D^0_j$ in $S$.
We fix a blowup  $\bar{S}$ of $S$
with  centers supported in $\Sigma$
such that the preimage of the intersection
$D_1\cap D_2$ in the neighborhood
of every point in $\Sigma$ is a normal crossing divisor.
Of course, $\bar{S}$ has a map to $\tilde{S}_j$ and we denote by
$\bar{Y}_j$ the fibered product $\bar{S}\times_{\tilde{S}_j}A_j$.
The surface $\bar{Y}_j$ is a double cover
of $\bar{S}$ and it has at most $A1$-singularities (double points), since
$D_j^0$ is smooth in $S^0$. Consider the fibered product
$\bar{X}= \bar{Y}_1\times_{\bar{S}}\bar{Y}_2$. All singular
points of $\bar{X}$ which are not of type $A_1$ lie
over the intersection points of $D_1^0$ and $D_2^0$ in $S^0$.
The surface $\bar{X}$ has a natural action of $\Z/2+\Z/2$. It admits
equivariant surjective maps $\bar{\delta}_j\,:\, \bar{X}\ra\bar{Y}_j$,
where $\bar{Y}_j$ is the quotient of $\bar{X}$ by
the involution $\bar{\tau}_j$. We denote by $\bar{Y}_{12}$ the
quotient of $\bar{X}$ by
$\bar{\tau}_{12}=\bar{{\tau}}_1\bar{{\tau}}_2$,
it is still singular.

\

Let $\hat{S}$ be the minimal
blowup of $\bar{S}$ with support
in the points $p_1,...,p_l\in S^0\subset \bar{S}$
such that proper transforms of the irreducible
components $D_1^0$ and $D_2^0$ are disjoint in the preimage of $S^0$ in
$\hat{S}$. Now we define $Y_j$ as
the induced (from the open part) double covers of $\hat{S}$.
Their ramification is contained in
the full transforms of the divisors
$D_j$. Define $X$ as the fibered product
$Y_1\times_{\hat{S}}Y_2$. We have
the induced involutions
(again denoted by  $\tau_j$, $\tau_{12}$) on $X$ and
we define
$Y_{12} $ as the quotient of $X$ under $\tau_{12}$
(it admits a map onto $\hat{S}$).
By construction, the
surfaces $X$, $Y_j$ and $Y_{12}$ all have
at most $A1$-singularities.
One has surjective regular maps $X \ra \tilde{X},
Y_j\ra\tilde{Y}_j$ etc.

\

\centerline{
\xymatrix{
          &  \tilde{X}
\ar[dl]_{\tilde{\tau}_1}\ar[d]^{\tilde{\tau}_{12}}
\ar[dr]^{\tilde{\tau}_2}
&    &  {}\ar@/_/[l]  &  \bar{X}
\ar[dl]_{\bar{\tau}_1}\ar[d]^{\bar{\tau}_{12}}\ar[dr]^{\bar{\tau}_2}
&    & {}\ar@/_/[l]   & X
\ar[dl]_{{\tau}_1}\ar[d]^{{\tau_{12}}}\ar[dr]^{{\tau}_2}
&
\\
  \tilde{Y}_1  \ar[dr]     &  \tilde{Y}_{12}  \ar[d] &
\tilde{Y}_2\ar[dl] &
   \bar{Y}_1 \ar[dr] &  \bar{Y}_{12}    \ar[d] &   \bar{Y}_2\ar[dl] &
       {Y}_1 \ar[dr] &  {Y}_{12}        \ar[d] &   {Y}_2   \ar[dl]
\\
         &    \tilde{S}    &  &  & \bar{S} &  &  & \hat{S}
}
}

\

\begin{lemm}\label{lemm:compon}
For all $k=1,...,l$ every irreducible component
of the preimage of $p_k$ in $Y_{12}$
is a rational curve.
\end{lemm}

{\em Proof.}
Notice that $D_j^0$ are smooth in $S^0$ and that their
components don't intersect in $S^0$. Every point $p_k$
is a point of intersection of an irreducible component of $D_1^0$
with an irreducible component of $D_2^0$.
In the neighborhood of $p_k$ these two divisors
have a canonical form: $D_1$ is given by $x=0$ and $D_2$ is given
by $y=x^n$, where $n$ is the order of tangency.
There is a standard
chain of blowups separating the proper transforms of $D_1$ and $D_2$
over $p_k$ (which intersect one of the ends of the chain in two
distinct points). The induced double cover on every component
of the preimage of $p_k$ is ramified in at most three points (hence
in fact, two) and is therefore a rational curve.
\hfill $\square$

\begin{coro}\label{coro:factors}
The map $\alb\,:\, Y_{12}\ra \Alb(Y_{12})$ factors
through $\bar{Y}_{12}$.
\end{coro}

{\em Proof.} Indeed, the natural map
$Y_{12}\ra \bar{Y}_{12}$ contracts
connected graphs of rational curves to distinct points in
$\bar{Y}_{12}$.
Since these connected graphs of rational curves
map into points in $\Alb(Y_{12})$  our claim follows.
\hfill $\square$

\

\subsection{Elliptic fibrations}
\label{sect:fibr}

Let $\E\ra \P^1$ be a Jacobian elliptic fibration and $M_1, M_2$
two irreducible
horizontal divisors on $\E$. Let $M_{12}\subset \E$
be the divisor
of pairwise differences: $M_{12}\cap \E_b$ is the set of all points
of the form $p_1-p_2$, where $p_j\in  M_j\cap \E_b$.
The divisor $M_{12}$ may
have several irreducible components.
We shall say that the divisor $M_{12}$ is torsion
if every irreducible component of $M_{12}\subset \E$
consists of torsion points.

\begin {lemm}
\label{lemm:restriction}
Consider the
restriction map to the generic fiber
$$
\eta^*\,:\, \Pic(\E)\ra \Pic(\E_{\eta}).
$$
If the divisor $M_{12}$ is torsion
then there exists a positive integer
$N$ such that the class
$\eta^* ([M_{12}])\in \Pic^{(0)}(\E_{\eta})$
is annihilated by $N$.
\end{lemm}

{\em Proof.}
For every irreducible component of $M_{12}$ there exists
a positive integer $N'$ such that all points $p$ in this component
are annihilated by $N'$.
Since the divisor $M_{12}$ has only a finite
number of irreducible components we can find an $N$
annihilating all points in $M_{12}$.
Consider the class $N[M_{12}]$. The
corresponding divisor is trivial upon restriction to
the generic fiber $\E_{\eta}$. \hfill $\square$

\begin{coro} \label{coro:eta}
The kernel of the map
$\eta^*\,:\,\Pic(\E)\ra \Pic(\E_{\eta})$
is isomorphic to the subgroup of $\Pic(\E)$ generated
by the components of the singular fibers.
\end{coro}

\begin{lemm}\label{lemm:rank}
Let $\E\ra \P^1$ be a Jacobian elliptic fibration with
singular fibers of simple multiplicative type (irreducible nodal curves).
Let $M_1,M_2,M_3\subset S$ be three different irreducible
horizontal divisors which are linearly independent
in $\Pic(\E)$. Then at most one of the divisors
$M_{ij},$ ($i,j = 1,2,3,$ and $i\neq j$) is torsion.
\end{lemm}

{\em Proof.}
Indeed if two of the above divisors, for example $M_{12}$ and $M_{13}$,
are torsion then there is a positive integer
$N$ which annihilates both
$\eta^*([M_{12}])$ and $ \eta^*([M_{13}])$.
Hence the kernel of $\eta^*$
has rank $\ge 2$. By assumption, the singular fibers
of $\E$ generate a subgroup of rank 1. Contradiction.
\hfill $\square$

\

Let $\E\ra \P^1$ be a nonisotrivial elliptic fibration and
$r$ the order of $\E$ in the Tate-Shafarevich
group of the corresponding Jacobian elliptic fibration
${\cal J}(\E)$.
Recall that for each integer $r'$
we have a principal homogeneous fibration
${\cal J}^{(r')}(\E)$ of (relative)
zero cycles of degree $r'$, together
with natural maps
$$
{\cal J}^{(r')}(\E)\times {\cal J}^{(r'')}(\E)\ra {\cal J}^{(r'+r'')}(\E)
$$
and an identification of ${\cal J}^{(0)}(\E)$ and
${\cal J}^{(r)}(\E)$, depending on the choice of a global section
of the Jacobian fibration ${\cal J}(\E)={\cal J}^{(0)}(\E)$.
After fixing the identification, we get
for every integer $m\equiv 1 \mod r$
a rational map
$$
\phi_{m}\,:\, \E\ra \E
$$
of degree $m^2$, well defined up to the action of $H^0(\P^1,\E)$.
This map is regular \'etale on the open (grouplike) part
$\E^0$ (complement to the singular points of the
singular fibers), but highly nonregular on $\E$.

\

\begin{lemm}\label{lemm:irreducible}
Let $\E\ra \P^1$ be a nonisotrivial elliptic fibration
and $M$ a horizontal irreducible divisor.
Then for all but finitely many primes (congruent to
1 mod $r$) the preimage
$\phi_p^{-1}(M)$ is also irreducible.
\end{lemm}

{\em Proof.}
After base change, we can assume that $M$ is a nonzero section.
The global monodromy group of $\E_M$ over $M$ is a
subgroup of a finite index in $\SL(2,\Z)$.
Let ${\mathbb F}$ be the fundamental
group of the complement $M \setminus {\rm Sing}$ where ${\rm Sing}$
is a subset of $M$ corresponding to singular fibers of $\E\ra M$.
Every section $M'\in \E_M$
defines a cocycle $s_{M'}\in H^1({\mathbb F}, (\Q/\Z)^2)$
(corresponding to the principal $(\Q/\Z)^2$-fibration,
whose fiber over $b\in M\setminus {\rm Sing}$
is the set of points differing from $M'\cap \E_b$ by
torsion).
This cocycle gives an affine action of ${\mathbb F}$ on $(\Q/\Z)^2$
and the orbits of the action of ${\mathbb F}$
correspond to the irreducible components of the preimages of $M'$ under
the maps $\phi_{m,M} : \E_M\to \E_M$ for all $m$.

If the section
$M$ is not divisible in the group of sections $H^0(M,\E_M)$
by a prime $p$ and the monodromy map
$ {\mathbb F} \to \SL(2,\Z/p)$ is surjective then
the preimage of $M$ under
the map $\phi_{p,M} : \E_M \ra \E_M$ is irreducible.
Indeed in this case either the action is affine and its orbit is
$(\Z/p)^2$ or the action is linear.
In the first case the preimage of $M$ is irreducible.
In the second case there is a section $M'$ with
$ pM' = M$ and  hence $M$ is divisible as a section
(see also \cite{bogomolov-tschinkel}). \hfill $\square$

\

\begin{lemm}\label{lemm:n1n2}
Let $\E\ra C$ be a nonisotrivial
Jacobian elliptic fibration over an affine
connected and smooth curve $C$.
Let $M_1, M_2$ be two irreducible multisections on $\E$ with the
property that there exists a smooth fiber $\E_b$  and a pair of
points $p_j\in \E_b\cap M_j$ such that $p_1-p_2$ is nontorsion in
$\E_b$. Then there exists a positive integer
$N$ (depending on $\E,M_1,M_2$) such that
for all positive integers $n_1,n_2$ with $n_1+n_2>N$ the preimage
$\phi_{n_1}^{-1}(M_1)$ intersects $\phi_{n_2}^{-1}(M_2)$.
\end{lemm}

{\em Proof.}
The proof runs in the analytic category.
It uses the following universal construction.

\

\noindent
{\bf Construction.} Consider the universal Jacobian elliptic curve
$u\,:\, \E_{\mathcal U}\ra {\mathcal H}$ which is obtained as a quotient
${\mathcal H}\times \C/(1,\lambda)$, where $\lambda $ is a
coordinate function in the upper halfplane ${\mathcal H}$.
This fibration is topologically trivial and has a natural trivialization
map
$$
\kappa\,:\, {\mathcal H}\times \C/(1,\lambda)\ra {\rm T}=\C/(1,i),
$$
(where $i=\sqrt{-1}$). The map $\kappa$ is not complex analytic; however,
the preimages of points $t\in {\rm T}$ are analytic sections of
the fibration
$u\,:\, \E_{\mathcal U}\ra {\mathcal H}$ (this gives a nonanalytic family
of analytic sections). Indeed, the preimage of the point $t=a+bi$
is a section of $u$ which is given in a parametric
form $(\lambda, b\lambda +a)\in {\mathcal H}\times \C$ (which descends to
$\E_{\mathcal U}$).
The torsion sections of $u$ map into points in ${\rm T}$
(since $\kappa$ is a continuous homomorphism of algebraic groups).
The construction is equivariant with respect to $\SL(2,\Z)$.
Indeed, the action of $\SL(2,\Z)$ on $\E_{\mathcal U}$ transforms
torsion sections into torsion sections, (which are dense
in the family $\kappa^{-1}({\rm T})$).
For any subgroup $\Gamma\subset \SL(2,\Z)$
of finite index which doesn't contain the center $\Z/2$
we have the induced map
$$
u_{\Gamma}\,:\, \E_{\mathcal U}/\Gamma \ra {\mathcal H}/\Gamma
$$
and a factorization map $\pi_{\Gamma}$.
We have the diagram:

\

\centerline{
\xymatrix{
\E_{\mathcal U}/\Gamma\ar[d]_{u_{\Gamma}}
& \E_{\mathcal U} \ar[l]_{\pi_{\Gamma}} \ar[d]^{u}\ar[r]^{\kappa} & {\rm T} \\
{\mathcal H}/\Gamma  &  {\mathcal H} \ar[l]_{\pi_{\Gamma}}      &
}
}

\

\noindent
For every $t\in {\rm T}$, which is
not a torsion point in ${\rm T}$
the orbit $\cup_{\gamma\in \Gamma} \gamma(t)$ is dense (in the
usual topology) in ${\rm T}$. Thus if $t$ is not torsion in ${\rm T}$
the intersection of the
set $\pi_{\Gamma}(\kappa^{-1}(\cup_{\gamma\in \Gamma} \gamma(t)))$
with every fiber $\E_b\in \E_{\mathcal U}/\Gamma$ is dense.

\

Now we return to the proof of Lemma~\ref{lemm:n1n2}.
Consider the nonisotrivial Jacobian elliptic fibration
$\E\ra C$.
We have a diagram

\

\centerline{
\xymatrix{
\E\ar[d]  & \E' \ar[l] \ar[d] & \E_U'\ar[l]\ar[d]\ar[r] &
\E_{\mathcal U}\ar[d]\ar[r]& {\rm T} \\
C         &  C' \ar[l]  & C'\times_{{\mathcal H}/\Gamma}{\mathcal H}\ar[r]
\ar[l]& {\mathcal H} &
}
}

\

Changing the base, we reduce to
the case when $M_1,M_2$ are sections of
a (topologically trivial) elliptic fibration
$\E'\ra C'$ over some
analytic curve $C'$.
Moreover, we can assume that the fibration $\E'\ra C'$
is induced from $\E_{\mathcal U}/\Gamma \ra {\mathcal H}/\Gamma$
under a dominant map $C'\ra {\mathcal H}/\Gamma$.
We can identify $M_1'$ (the pullback of $M_1$) with
the zero section of $\E'$
(changing the zero section
amounts to changing the argument of $\kappa$ by a fiberwise translation).
Then the image
$M_{2,U}$ of $M_2$
in $\E_{\mathcal U}/\Gamma$ is algebraic, and therefore
not an orbit of $\Gamma$ on
(the preimage of) some nontorsion point $t\in {\rm T}$.
This means that
$\kappa(\pi_{\Gamma}^{-1}(M_{2,U}))$
covers some open (in the usual topology)
subset $V$ of ${\rm T}$. If $n_1+n_2$ is sufficiently big, then
the translations of the set  $V$ by $n_2$ torsion points in ${\rm T}$
contain $n_1$ torsion points. \hfill $\square$

\subsection{Kummer surfaces}
\label{sect:kummer}

Let  $A$ be an abelian surface,
$\tau$ the standard involution,
$A^*$ the blowup of $A$ in the 16 fixed points of $\tau$,
$S$ the associated Kummer K3 surface and
$\Ex$ the union of the 16 exceptional $(-2)$-curves
on $S$.
We will say that  $A$ is {\em generic} if
$\Pic(A)/\Pic^0(A) = \Z$ and if the endomorphism
ring ${\rm End}(A) = \Z$.
In particular,  $A$ is not isogenous to
a product of elliptic curves.

\begin{lemm}\label{lemm:cond}
We can choose $A$ such that
\begin{enumerate}
\item $A$ is generic;
\item the elliptic fibration $S\ra \P^1$ is Jacobian;
\item all singular fibers of the elliptic fibration $S\ra \P^1$
are irreducible (consequently,
the exceptional divisor $\Ex$ is horizontal in $S\ra\P^1$).
\end{enumerate}
\end{lemm}

{\em Proof.}
The Picard lattice of a polarized Kummer surface  $S$ is
given by a sublattice (of rank at least 17) in
$3H\oplus (-E_8)\oplus(-E_8)$, where $H$ is the standard
hyperbolic lattice.
Let $h_A$ be the generator (polarization) of the Neron-Severi group
of our generic abelian surface $A$. This class is invariant
under the involution. It descends to the singular surface
$\tilde{S}$ and lifts to a class $h_S\in \Pic(S)$.
For any even positive integer $2k$ there exists
a generic $A$ such that $h_S^2 = 2k$.
For generic $A$ the Picard group of $S$ is a direct sum
$\Z h_S\oplus \Pi$, where $\Pi$ sits in the exact
sequence
$$
0\ra (-2){\rm Id}_{16}\ra \Pi\ra \Lambda^2((\Z/2)^4)\ra 0.
$$
The last projection extends naturally to a projection
$\Pic(S)\ra\Lambda^2((\Z/2)^4)$ ($h_S$ is mapped to zero).
Given a lattice vector of square zero we can find a translate
of this vector (under involutions of $\Pic(S)$
with respect to $(-2)$-classes) representing an
elliptic fibration (see \cite{nikulin}).

Remark that though the lattice
$\Pi$ is not unimodular, for any primitive
element $e\in \Pic(S)$ with a nontrivial
projection to $\Lambda^2((\Z/2)^4)$ we can find an element $x\in \Pi$
such that $(e,x)=1$.
Let $e'\in \Pic(S)$ be a class obtained from $e$ by
reflexion with respect to a $(-2)$-class. Then there still exists a
class $x'\in\Pic(S)$ with $(e',x')=1$.

Choose a class $e$
of square zero giving an elliptic fibration $f_e\,:\,S\ra \P^1$,
such that the projection of $e$ to  $\Lambda^2((\Z/2)^4)=(\Z/2)^6$
is nonzero.
The lattice $N_e:=\{n\in \Pi\,|\, (n,e)=0\}$
is negative semi-definite.
We will choose $N_e$ such that it has no elements of
square $-2$.
This implies that the singular fibers of $f_e$ are
irreducible.
Simultaneously, we will choose
the (affine) lattice $L_e:=\{l\in\Pic(S)\,|\, (l,e)=1\}$
such that $L_e$ has $(-2)$-vectors.
Since $N_e$ has no $(-2)$-classes
every class (modulo translations by $e$)
of square $(-2)$ in $L_e$
corresponds to a section of $f_e\,:\, S\ra\P^1$.

The lattice $\Pi$ contains a finite number
of $(-2)$-vectors. Thus a generic vector in this lattice
is not orthogonal to any $(-2)$-vector. Take such a
primitive vector $x$
and choose a polarization $h_S$ such that $h_S^2=-x^2$.
Now we can choose a (generic) abelian surface $A$
(with endomorphisms $\Z$)
such that the square of the generator $h_A$ of $\NS(A)$ equals $2h_S^2$.
On the corresponding Kummer K3 surface  $S$ we have
$(h_S-x)^2=0$. It follows that
$S$ has an elliptic fibration
$f\,:\, S\ra \P^1$ and  that the lattice $N_f$
is isomorphic to $N_x\oplus \Z(h_S-x)$, and therefore
has no classes of square $(-2)$ (for any $x$ as above).
Let $z$ be a vector $\Pi$ such  that $(x,z)=1$
(this is possible since $x$ is primitive).
Every vector in $L_x$ is equal to $z+n + c(h_S-x)$
where $n\in N_x$ and $c\in \Z$. Its square is
$z^2+n^2 + 2(z,n) -2c$. Since the lattice $\Pi$ is even
and $c$ an arbitrary  integer it follows that $L_x$ always contains
classes of square $-2$. \hfill $\square$

\

\begin{rem}
The same proof shows that one can construct  Jacobian
elliptic Kummer surfaces with all
singular fibers of simple multiplicative type without requiring
that the associated abelian surface is generic.
\end{rem}

We return to the general setup of Section~\ref{sect:construction}
and Section~\ref{sect:desingularization}.
Let $\E\ra \P^1$ be a Jacobian elliptic fibration, $M$
a horizontal divisor and $M_1,...,M_k$ its irreducible
components. Let $\Gamma(M)$ be the ``torsion'' graph
of $M$ -  each $M_l$ defines a vertex and
two vertices $M_{l}$ and $M_{l'}'$
are connected by an edge
if the difference divisor $M_{ll'}$ is not torsion.

\begin{lemm}\label{lemm:connected-ex}
Let $S$ be a Kummer surface as
in the statement of Lemma~\ref{lemm:cond}.
Then the torsion graph $\Gamma(\Ex)$ is connected.
\end{lemm}

{\em Proof.}
The divisor $\Ex$ consists of 16 irreducible  components,
their classes are linearly independent in $\Pic(S)$.
The rank of the subgroup of $\Pic(S)$ generated by
the components of singular fibers is 1.
Applying Lemma~\ref{lemm:rank}
we see that every component of $\Ex$ is connected
to at least 14 other components of $\Ex$.
In particular, every two vertices in $\Gamma(\Ex)$
are connected by a path of length at most 2.
\hfill $\square$

\

\begin{lemm}\label{lemm:gamma-connected}
Consider $D_1^0,D_2^0\subset S^0$, obtained
as preimages of $\Ex$ under the maps $\phi_{q_1}, \phi_{q_2}$.
For almost all pairs of distinct prime numbers
$q_1,q_2$ the intersection
graph of the irreducible components of
$D_1^0\cup D_2^0$ (in $S^0$) is connected.
\end{lemm}

{\em Proof.}
By Lemma~\ref{lemm:irreducible},
the preimages of the irredicible components of $\Ex$ under
$\phi_{q_j}$ remain irreducible for almost all pairs of primes $q_1,q_2$.
Moreover, by Lemma~\ref{lemm:n1n2}, the {\em intersection} graphs
of the (irreducible) divisors $\phi^{-1}_{q_j}(\Ex_k)$ is
connected for $q_j$ big enough.
Now take an irreducible component $\Ex_1\subset\Ex$ and a component
$\Ex_k$, which is connected to $\Ex_1$ in the
{\em torsion} graph $\Gamma(\Ex)$.
Applying Lemma~\ref{lemm:n1n2} with $n_1=q_1$ and $n_2=q_2$
(and $q_1+q_2$ big enough) we see that
$\phi_{q_1}^{-1}(\Ex_1)$ intersects $\phi_{q_2}^{-1}(\Ex_k)$ (in $S^0$).
\hfill $\square$

\begin{rem}
The difficulty was to show that $D_1^0\cup D_2^0$ is
a connected divisor in the {\em open} surface $S^0$.
The corresponding fact in the closed surface $S$ is trivial.
\end{rem}

\subsection{Surfaces in abelian fourfolds}
\label{sect:morph}

We use the existence of nontrivial maps
between elliptic Kummer surfaces (see Sections~\ref{sect:fibr}
and \ref{sect:kummer})
to construct interesting examples of surfaces
in abelian fourfolds.

\

Fix a generic abelian surface $A$ such that
the associated Kummer surface $S$ admits a
Jacobian elliptic fibration
(this is possible by Lemma~\ref{lemm:cond}).
Choose a pair $(q_1,q_2)$ of positive
integers and consider the map
$$
\tilde{\phi}_{q_1,q_2}=(\tilde{\phi}_{q_1},\tilde{\phi}_{q_2})
\,:\, S\ra \tilde{S}\times \tilde{S}.
$$
Put $\tilde{g}_1=\tilde{\phi}_{q_1}, \tilde{g}_2=\tilde{\phi}_{q_2}$ and
denote by $\tilde{X}=\tilde{X}_{q_1,q_2}$ and by $Y_1,Y_2,Y_{12}$ the surfaces
obtained through the construction in
Sections~\ref{sect:construction} and  \ref{sect:desingularization}.
The surface $\tilde{X}$
is lagrangian with respect to some
nondegenerate 2-form on $A\times A$.
We will show that for appropriate choices of
integers $q_1,q_2$ the surface $X$ (and consequently $\tilde{X}$)
is not fibered.

\subsection{Uniqueness}

\begin{prop}\label{prop:unique-form}
Let $A_1,A_2$ be simple abelian surfaces and
$\tilde{X}$ a lagrangian surface in $A_1\times A_2$.
Assume that $\tilde{X}$ is not isomorphic to an abelian surface,
that $\tilde{X}$ projects dominantly onto $A_1$ and $A_2$,
that it is stable under
the involutions $\tilde{\tau}_1,\tilde{\tau}_2$ and that
it is lagrangian with respect to at least two
nonproportional nondegenerate
forms, one of which is invariant with
respect to both involutions $\tilde{\tau}_j$.
Then $\tilde{X}$ is a finite unramified cover
of a product of two curves.
\end{prop}

{\em Proof.}
By assumption, every $(1,0)$-form on $A_1\times A_2$ restricts
nontrivially to $\tilde{X}$.
If there are two nonproportional nondegenerate  $(2,0)$-forms
which restrict trivially to $\tilde{X}$ then
there exists a
holomorphic 2-form $\omega$ of rank 2 on $A_1\times A_2$
which restricts trivially to $X$.
(Indeed, the $(2,0)$-forms of rank 2 on the abelian variety
$A_1\times A_2$ correspond to points on a quadric in the projective
space $\P^5=\Proj(H^{2,0}(A_1\times A_2,\C))$. Any line in
$\P^5$ intersects this quadric.)
Any such $\omega $ is equal to
$w'\wedge w''$, where $w',w''$ are nonproportional
$(1,0)$-forms on $A_1\times A_2$.
Thus we have a pencil of curves ${\mathcal P}_{\omega}$ on $\tilde{X}$
such that both $w'$ and $w''$
are equal to zero on the fibers of this pencil.
Therefore,
we have a family of abelian surfaces $A_{t}\subset A_1\times A_2 $
(where $t$ is a point in the surface $B_{\omega}=(A_1\times A_2)/A_0$)
such that the intersection of $\tilde{X}$ with $A_t$ is either empty or
a fiber of the pencil ${\mathcal P}_{\omega}$.
Hence, the image of $\tilde{X}$ under the projection to the base of
${\mathcal P}_{\omega}$ coincides with the image of $\tilde{X}$ in
the abelian surface $B$. The image of $\tilde{X}$ in $B$ is a
curve  $C_{\omega}$ of genus $\ge 2$.
The form $\omega $ is induced from the holomorphic
volume form on the abelian surface $B$.
Consider the action of $\tilde{\tau}_1,\tilde{\tau}_2$ on $\omega$.
Since this action transforms
$(2,0)$-forms
of rank 2 on $A_1\times A_2$ onto themselves, we have the following
possibilities: either $\omega$ is invariant under $\Z/2+\Z/2$
(modulo multiplication by a constant)  or there
is another 2-form  $\tilde{\omega}$ of rank 2,
which is trivial on $\tilde{X}$.
The first case is excluded by the assumptions
(that both projections on $A_1,A_2$ are dominant).
In the second case
we have another projection of $\tilde{X}$ onto an abelian surface,
with fibers transversal to the fibers of the first projection.
Thus we have an isogeny
$$
A_1\times A_2\ra  B_{\omega}\times B_{\tilde{\omega}},
$$
which exhibits $\tilde{X}$ as a finite
abelian covering of a product
$C_{\omega}\times C_{\tilde{\omega}}$ of two curves of genus $\ge 2$.
\hfill $\square$

\begin{coro}\label{coro:unique}
Let $\tilde{X}$ be a  surface obtained as
a $\Z/2+\Z/2$-cover of a (singular) K3-surface as
in Section~\ref{sect:construction} (in particular, we do not assume
that $S$ is elliptic).
Assume that the conditions of Proposition~\ref{prop:unique-form}
hold for $\tilde{X}$. Then $\tilde{X}$ is isomorphic to $C\times C$, where
$C$ is a curve of genus 2 and the quotient map
$$
\tilde{X}\ra S=(C\times C)/(\Z/2 + \Z/2)
$$
is described in Example~\ref{exam:unique}.
\end{coro}

{\em Proof.}
Let $\tilde{X}$ be an unramified abelian cover of the product $C_1\times C_2$
of curves of genus $\ge 2$ as in the proof of
Proposition~\ref{prop:unique-form}.
Our assumption implies that there
is a unique $(\Z/2 + \Z/2)$-equivariant $(2,0)$-form on $\tilde{X}$.
The involutions $\tau_1$ and $\tau_2$ interchange
the projections of $X$ to $C_1\times C_2$ (since
we have two projections of $X$ onto abelian surfaces $B_1,B_2$ and
$A_1,A_2$ both map surjectively onto $B_1,B_2$ and
the involution on $A_1$ interchanges the two projections).
In particular, $C_1=C_2$ and $C_1\times C_1=C\times C$.
The involution $\tau_{12}=\tau_1\tau_2$
induces an involution on the abelian cover of $C\times C$.
Therefore, $\tau_{12}=(\sigma,\sigma)$, where $\sigma $ is some involution
on $C$. Thus we have a map
$$
S=\tilde{X}/ (\Z/2 + \Z/2)\ra \Sym^2(C)/\sigma.
$$
The condition $h^{1,0}(S)=0$ implies that $\sigma $ is a hyperelliptic
involution and the condition $h^{2,0}(S)=1$ implies that $g(C)=2$.
Moreover, the map $\tilde{X}\ra C\times C$ is
in fact an isomorphism (unramified covers increase the Euler characteristic
and the number of invariant 2-forms has to increase as well).
\hfill $\square$

\begin{coro}\label{coro:prod}
We keep the notations of Section~\ref{sect:construction}.
Consider
$$
\tilde{\varphi}=(\tilde{g}_1,\tilde{g}_2)\,:\,
S\ra \tilde{S}_1\times \tilde{S}_2
$$
(and the associated maps $g_j\,:\, S\ra S_j$).
Assume that at least one of the maps ${g}_1,{g}_2$ is not
an isomorphism or (if it is) the map $g_1g_2^{-1}$ doesn't
lift to an automorphism of $C\times C$.
Then $X$ is lagrangian
with respect to exactly one (up to multiplication
by constants) nondegenerate form on $A_1\times A_2$.
\end{coro}

{\em Proof.} Indeed, the proof of Corollary~\ref{coro:unique}
shows that in the opposite case, $X$ is isomorphic to $C\times C$,
with $g(C)=2$ and the action of $\Z/2 + \Z/2$ is the product
of the hyperelliptic involution and interchanging of coordinates, modulo
automorphisms of $C\times C$. \hfill $\square$

\

\subsection{1-forms}
\label{sect:1-forms}

\begin{lemm}\label{lemm:22}
Let $\hat{S}$ be birational to a K3 surface and
$Y\ra \hat{S}$ its double cover. Assume that
$Y$ has at most $A1$-singularities (double points).
Then either $h^0(Y, \Omega^1)\le 2$ or
$\alb(Y)\subset\Alb(Y)$ is a hyperelliptic curve $C$
and the (covering) involution on $Y$ transforms into
a hyperelliptic involution on $C$.
\end{lemm}

{\em Proof.} Indeed,
since there are no 1-forms on $S$, the involution acts
by multiplication with $-1$ on $H^0(Y,\Omega^1)$.
Thus the product of any two forms is invariant
under the involution.
Therefore, the map
$$
\alpha\,:\, \wedge^2 H^0(Y,\Omega^1)\ra H^0(Y, \Omega^2)
$$
has image of dimension 1 or 0. If it is 0, then
$\alb(Y)$ has dimension 1. (Indeed, it means that
any $(2,0)$-form on $\Alb(Y)$ pulls back to a $0$-form on $Y$
and hence $\alb(Y)$ has dimension 1.)
In this case the involution acts on $C=\alb(Y)\subset \Alb(Y)$
as well as on $H^0(C,\Omega^1)$ as $(-1)$. It follows that
$C$ is hyperelliptic and that the map $Y\ra C$ transforms
the involution on $Y$ into a hyperelliptic involution on $C$.

If the dimension of the image of $\alpha$
is 1, then there is a pair of nonproportional forms
$\omega, \omega'$ such their product is nonzero.
Assume that there is another form $\omega''$, linearly
independent of $\omega,\omega'$. Then $\omega'\wedge\omega''=
f\omega\wedge\omega'$, where $f$ is a nonconstant function,
contradiction. \hfill $\square$

\begin{coro}
Let $Y_j$ be the double covers of $\hat{S}$ as above. Then
$h^0(Y_j,\Omega^1)=2$ (for $j=1,2$). Moreover,
$Y_j$ admits no dominant rational maps onto
curves of genus $> 0$.
\end{coro}

{\em Proof.}
Indeed, both surfaces $Y_j$
admit a dominant map onto an abelian surface $A$, which has
two linearly independent
1-forms. Now we apply Lemma~\ref{lemm:22}. This proves the
first statement.
By assumption, the abelian surface $A$ contains
no elliptic curves. By the previous lemma, $\Alb(Y_j)$ is
isogenous to $A$. The second statement follows. \hfill $\square$.

\

A priori, we don't know that $Y_{12}$ does not admit
dominant maps onto curves of genus $>0$.
We have to consider the following (mutually exclusive) possibilities:

$h^0(Y_{12}, \Omega^1)=0$;

$h^0(Y_{12}, \Omega^1)=2$ and the wedge product of the two
forms is nontrivial;

There is a projection $Y_{12}\ra C$, where $C$ is
a curve of genus $\ge 1$.

\

We will show that the surfaces $X_{q_1,q_2}$ constructed in
Section~\ref{sect:morph} are not of the last two types
for almost all pairs of primes $q_1,q_2$. The surfaces of
the first type are nonfibered; and there are
simple examples such surfaces:

\

\begin{lemm}\label{lemm:y12}
Assume that the map $g_1\,:\, S\ra S_1$
is an isomorphism.
Then $h^0(Y_{12}, \Omega^1) = 0$.
\end{lemm}

{\em Proof.}
The covering $Y_1$ is (birational to) the abelian
surface $A_1$. The variety $X$ is (birational to)
a fiber product of $A_1$ and some surface
$Y_2$ over $\hat{S}$.
Consider the product
$$
H^0(Y_2,\Omega^1)\times H^0(Y_{12},\Omega^1)\ra H^0(X,\Omega^2).
$$
The image is at most 1-dimensional, since it consists of
$\tau_1$-invariant forms and since there is exactly one such
form which is induced from $Y_1=A_1$.
The space $H^0(Y_2,\Omega^1)$ has dimension 2 and hence
for any nontrivial form $\omega_{12}$ in $H^0(Y_{12},\Omega^1)$
its product with some form $\omega_2\in H^0(Y_2,\Omega^1)$
is zero. This means that $\omega_2$ is trivial on
the fibers of some pencil of curves on $X$. This pencil covers
(birationally) $A_2$. The image of a generic
fiber of the pencil must be an elliptic curve in $A_2$
(being the zero set of a 1-form on $A_2$).
This leads to a contradiction,
since $A_2$ doesn't contain elliptic curves by assumption.
Thus there are no holomorphic 1-forms on $Y_{12}$.
\hfill $\square$

\begin{coro}\label{coro:g1-iso}
Assume that $g_1$ is an isomorphism and
that either $g_2$ is not an isomorphism or (if it is)
$g_1g_2^{-1}\,:\, S\ra S$ is not an involution on $S$.
Then the associated surface $X$ is not fibered
and $\tilde{X}$ is lagrangian with
respect to exactly one 2-form on $A\times A$.
\end{coro}

{\em Proof.} By Corollary~\ref{coro:unique}, $\tilde{X}$ is lagrangian
with respect to exactly one nondegenerate 2-form, unless
$\tilde{X}=C\times C$, where $C$ is a curve of genus 2.
Thus it suffices to prove that $X$ is not fibered.
Notice that $h^0(X,\Omega^1)=4$.
By Lemma~\ref{lemm:y12}, the Albanese $\Alb(X)$ is isogenous
to $A_1\times A_2$. The image  $\alb(X)$ in the Albanese is also lagrangian
with respect to exactly one nondegenerate 2-form. If $X$ were fibered
over a curve of genus $>1$ there would be another (a degenerate) 2-form on
$\Alb(X)$ which would be trivial on $\alb(X)$.
This is a contradiction. Notice also that $X$ doesn't admit
any dominant maps onto elliptic curves since $\Alb(X)$ doesn't map onto
elliptic curves. \hfill $\square$

\

In the following  sections we
will give further (more complicated)
examples of nonfibered lagrangian surfaces in
$A_1\times A_2$.

\

\subsection{Weakly lagrangian structures}

Let $\tilde{X}\subset A_1\times A_2$
be as in Section~\ref{sect:construction} and $X$ its
(partial) desingularization as in Section~\ref{sect:desingularization}.
Consider the set $L^X$ of (possibly degenerate) holomorphic
$(2,0)$-forms
on $\Alb(X)$ which restrict trivially to $\alb(X)$.
It contains the set
of weakly lagrangian
structures on $\alb(X)$.
Since the condition is linear, $L^X$ is a linear subspace
of $H^{2,0}(\Alb(X),\C)$,
The $\Z/2 + \Z/2$-action on $X$ lifts to an
action on $\Alb(X)$, leaving the space $L^X$ invariant.

From now on we assume that $\tilde{X}$ is not isomorphic to
$C\times C$,
where $C$ is a curve of genus $2$ (since in this case
$\dim L^X=2$, see Proposition~\ref{prop:unique-form}).
Furthermore, we will assume that $h^0(Y_{12},\Omega^1) \neq 0$
(if $\tilde{X}\neq C\times C$
and $h^0(Y_{12},\Omega^1)= 0$ then $\dim L^X = 1$).

We have to consider the following cases:
\begin{enumerate}
\item  $h^0(Y_{12},\Omega^1) \neq 0$ and the wedge product
is degenerate;
\item
$h^0(Y_{12},\Omega^1) = 2$
and the wedge product
$$
\wedge\,:\, H^0(Y_{12},\Omega^1)\times H^0(Y_{12},\Omega^1)\ra
H^0(Y_{12},\Omega^2)
$$
is nondegenerate.
\end{enumerate}

\begin{rem}
Notice however, that
examples of surfaces $X$ such that
$$
h^0(Y_{12},\Omega^1) \neq 0
$$
are somewhat pathological. In particular, almost all surfaces
$X_{q_1,q_2}$ are not
of this type.
\end{rem}

\

The space $L^X$ is invariant under $\Z/2 + \Z/2$ and hence
decomposes into the direct sum
$$
L^X = L^X_0 \oplus L^X_1 \oplus L^X_2 \oplus L^X_{12},
$$
where $L^X_0$ stands
for a subspace of $\Z/2 + \Z/2$-invariant forms in $L^X$, the space
$L^X_j$ is the space of $\tau_j$-invariant forms and
$L^X_{12}$ is the space of $\tau_{12}$ invariant forms.

This decomposition arises from the decomposition of
the space $H^{2,0}$ of $(2,0)$-forms on $\Alb(X)$
under the $\Z/2 +\Z/2$-action.
We will denote these spaces
using the
same indices.
We know that
$$
\begin{array}{ccc}
H^{1,0}(\Alb(X))_0    & = & 0,\\
H^{1,0}(\Alb(X))_1    & = & H^{1,0}(A_2), \\
H^{1,0}(\Alb(X))_2    & = & H^{1,0}(A_1), \\
H^{1,0}(\Alb(X))_{12} & = & H^{1,0}(Y_{12}).
\end{array}
$$
Thus by taking the exterior product we obtain
that
$$
H^{2,0}_0 =
\Lambda^2 H^{1,0}(\Alb(X))_1 +  \Lambda^2 H^{1,0}(\Alb(X))_2
+  \Lambda^2 H^{1,0}(\Alb(X))_{12}.
$$
We also have the decompositions
$$
\begin{array}{ccc}
H^{2,0}(\Alb(X))_1    & = & H^{1,0}(\Alb(X))_{12}\times H^{1,0}(A_2),\\
H^{2,0}(\Alb(X))_2     & = & H^{1,0}(\Alb(X))_{12}\times H^{1,0}(A_2), \\
H^{2,0}(\Alb(X))_{12}  & = &  H^{1,0}(A_1)\times H^{1,0}(A_2).
\end{array}
$$

\begin{prop}
\label{prop:cases}
If  $h^0(Y_{12},\Omega^1) \neq 0$ and the wedge product
is degenerate then
$$
\dim L^X = \dim \Lambda^2 H^0(Y_{12}, \Omega^1) + 1.
$$
\end{prop}

{\em Proof.} We subdivide the proof into a sequence of lemmas.

\begin{lemm}
Under the conditions in Proposition~\ref{prop:cases} we have
$$
L^X_1 = L^X_2=L^X_{12} = 0.
$$
\end{lemm}

{\em Proof.}
Since the space $L^X_{12}$ is induced from $A_1\times A_2$
it has dimension 0 (by Proposition~\ref{prop:unique-form} there is
a unique lagrangian structure on $\tilde{X}\subset A_1\times A_2$).
We have
$$
L^X_1\subset H^{2,0}(\Alb(X))_1 = H^{1,0}(\Alb(X))_{12}\times
H^{1,0}(A_2).
$$
Since  $\dim H^{1,0}(A_2) = 2$ for any element
$l\in L^X_1 $  there are two forms $w_1, w_2 \in H^{1,0}(A_2)$
and two forms $u_1,u_2 \in H^{1,0}(\Alb(X))_{12}$ with
the property
$$
w_1\wedge u_1 + w_2\wedge u_2 = l.
$$
We know that the forms $u_j$ are induced from a
map $f : X\to C$, where $C$ is some hyperelliptic curve.

If $w_1, w_2$ are linearly dependent then
$w_1$ is trivial on the  images of the fibers of the projection
$f : X\ra C$. Since $X$ surjects upon $A_1$ and
$A_2$ this means that $w_1$ is trivial on a
family of curves in $A_2$. This can happen only
if $A_2$ is isogenous to a product of elliptic curves.
This contradicts the assumption that $A_2$ is simple.

Consider the case when $w_1,w_2$ are linearly
independent.
Since the wedge product on $H^{0,1}(\Alb(X))_{12}$ is trivial
we can write $u_2 = fu_1$ for some rational function $f$ on $X$ and
hence $ w_1 + fw_2\wedge u_1 = 0$. The function $f$
is constant on the fibers of the family $ X \ra C$ and the fibers
map into $A_1$. Since the generic fiber is not
an elliptic curve the forms
$w_1,w_2\in H^{1,0}(A_2)$ are not linearly dependent
on the fibers which
contradicts the assumption.

Similar argument yields $L^X_2 = 0$.
\hfill $\square$

\begin{coro}
Under the conditions of Proposition~\ref{prop:cases} we have
$$
L^X = L^X_0= \Lambda^2 H^{1,0}(\Alb(X))_{12} + L_A,
$$
where
$\dim L_A = 1$. The subspace $L_A = 1$ is induced from $A_1\times A_2$.
A generic element
in $L^X$ is nondegenerate if $\dim \Alb(X)$ is even and
is degenerate of corank $1$ if $\dim \Alb(X)$ is odd.
\end{coro}

{\em Proof.}
If
$$
v = v_1 + v_2,\,\,  v_1\in L_A, \,\,v_2 \in
\Lambda^2 H^{1,0}(\Alb(X))_{12}
$$
then
$\rk\, v = \rk\, v_1 + \rk\, v_2$ (since the spaces
$ L_A, L^X_0$ belong to the second exterior powers
of complementary linear subspaces in $H^{1,0}(\Alb(X))$).

The nonzero element of $L_A$ is induced from a nondegenerate
form on $A_1\times A_2$ and has rank $4$.
We are in the case when $Y_{12}$ (and consequently $X$)
admits a dominant map onto a curve $C$ of genus $\ge 1$.
If $g(C)$ is even the generic
element of $L^X$ is a nondegenerate form on $\Alb(X)$.
The dimension of $\Alb(X)$ is even iff $g(C)$ is even.
Similarly when $g(C)$ is odd.
\hfill $\square$

This finishes the proof of Proposition~\ref{prop:cases}.

\begin{lemm}\label{lemm:factor}
Under the conditions of Proposition~\ref{prop:cases}
any dominant map $ h\,:\, X\ra C' $ of $X$ onto
a curve with $g(C')>0$
is a composition
$$
h = sf,\, f\, :\, X\ra C,\,\,  s\,: \, C\ra C'.
$$
\end{lemm}

{\em Proof.}
Repeating the previous argument
we see that
the rank of a nonzero form in $L_A$ is $2$ and hence $\rk\, v \geq 2$
for
$v = v_1 + v_2,$
$v_1\in L_A, $ $ v_2 \in \Lambda^2 H^{1,0}(\Alb(X))_{12} $
with $v_1\neq 0$. Thus
any form of rank $1$ in $L^X$ belongs to
$\Lambda^2 H^{1,0}(\Alb(X))_{12}$.
Since any map $h \,:\, X\ra C',\,\, g(C') \geq 2$
gives a nontrivial
form of rank $1$ we obtain the result in this case.

If $g(C') = 1$ then there is a map $\Alb(X) \to C'$.
The variety $\Alb(X)$ is isogenous to the product
$ \Jac(C)\times A_1\times A_2$ and any map from
$A_1\times A_2$ to a curve of genus $1$ is
trivial. Hence the map $ \Jac(C)\times A_1\times A_2 \to C'$
is induced from the projection onto $\Jac(C)$.
It means that the holomorphic $(1,0)$-form lifted from $C'$
on $X$ is induced from $f : X \to C$.
This yields the result for $g(C') = 1$.
\hfill $\square$

\begin{coro}
\label{coro:factoringc}
Under the conditions of Proposition~\ref{prop:cases}
every map $X\ra C'$ factors through the map
$X\ra Y_{12}\ra C\ra C'$.
\end{coro}

\

\noindent
We consider the case when $h^0(Y_{12},\Omega^1) = 2$
and the wedge product on 1-forms is nondegenerate on $Y_{12}$.
It follows that the Albanese variety $\Alb(X)$ is isogenous
to a product of three abelian surfaces
to $A_1\times A_2\times A_{12}$. Indeed, for all 3 quotients $Y_1,Y_2,Y_{12}$
of $X$ by the nontrivial involutions the corresponding Albanese varieties
have dimension 2. These are very strong conditions; though we are not
sure that such examples exist, we
would like to analyze this potential possibility:

\

In this case $\bar{Y}_{12}$ admits a
$\Z/2$-equivariant map $\bar{Y}_{12}\ra A_{12}$,
which descends to a map $\bar{S}\ra \tilde{S}_{12}$ (here
$\tilde{S}_{12}$ is the K3 surface obtained as the quotient $A_{12}/\Z/2$).
The ramification over $S^0$ is the union of $D_1^0\cup D_2^0$.
It follows that the preimage $D_{12}^0$ of $\Ex_{12}$ in $S^0$
(under the map $S^0\ra S_{12}$) is equal to $D_1^0\cup D_2^0$.
It follows that $D_{12}^0$ is a smooth divisor and hence
$D_1^0$ and $D_2^0$ have no intersection in $S^0$.

\begin{coro}
\label{coro:disconnect}
The intersection graph of the irreducible components of $D_{12}^0$
{\em in $S^0$} is totally disconnected.
\end{coro}

\subsection{Nonfibered lagrangian surfaces}

In this section we show that for infinitely many pairs
of integers  $q_1,q_2$
the surface $X=X_{q_1,q_2}$ is not fibered
(in particular, the lagrangian surface
$\tilde{X}_{q_1,q_2}\subset A_1\times A_2$ is also not fibered).

\begin{lemm}\label{lemm:12}
Assume that $Y_{12}$ admits a dominant
$\Z/2$-equivariant (with respect to the covering $Y_{12}\ra \hat{S}$)
map $\rho$ onto a hyperelliptic curve $C$ of genus $>0$.
Then $\rho$ descends to a dominant
map $\bar{\rho}\,:\, \bar{Y}_{12}\ra C$
which is also equivariant with respect to the covering involution
$\bar{Y}_{12}\ra \bar{S}$.
\end{lemm}

{\em Proof.} Indeed, the map $\rho$ is a composition of a map
$Y_{12}\ra \Alb(Y_{12})$ and $\Alb(Y_{12})\ra \Jac(C)$.
Now we apply Corollary~\ref{coro:factors}.\hfill $\square$

\

After factorization of $\bar{Y}_{12}$ by the involution $\bar{\tau}$
we obtain a map
$\bar{\rho}_{\sigma}\,:\, \bar{S}\ra \P^1=C/\sigma$,
(where $\sigma$ is the
hyperelliptic involution on $C$). Notice that the fibers
of $\bar{\rho}_{\sigma}$ are connected (since the
same property holds for $\rho$). Denote by $R=\{p_1,...,p_{2g+2}\}\subset C$
the ramification divisor of $C\ra \P^1$.

\

\centerline{
\xymatrix{
 \bar{Y}_{12}\ar[d]_{\bar{\tau}}\ar[r]^{\bar{\rho}} & C\ar[d]^{\sigma}   \\
 \bar{S}\ar[r]_{\bar{\rho}_{\sigma}}   & \P^1
}
}

\

\begin{lemm}
The irreducible components
of $D_1^0\cup D_2^0$ are contained in the fibers of $\bar{\rho}_{\sigma}$.
Moreover, they lie over $R\subset \P^1$.
\end{lemm}

{\em Proof.}
Since the map $\bar{\rho}$ commutes with the covering involution on
$\bar{Y}_{12}$
the image of the ramification divisor in $S^0\subset \bar{S}$
under the map $\bar{\rho}_{\sigma}$
is contained in $R$.
The divisors $D_j^0$ are exactly the ramification divisors of
the double cover on $S^0$.
\hfill $\square$

\begin{coro}\label{coro:connected}
Let $I_1^0\subset D_1^0$ and $I_2^0\subset D_2^0$ be
two irreducible components of the divisors $D_1^0, D_2^0$.
Assume that
the intersection
$I_1^0\cap I_2^0\cap S^0\neq \emptyset$.
Then $I_1^0$ and $I_2^0$ are mapped to the same point
in $R\subset \P^1$.
\end{coro}

Now we describe the structure of those fibers of $\bar{\rho}_{\sigma}$
which lie over a point $p\in R$.

\begin{lemm}\label{lemm:fibers}
All the components of a
fiber in $S^0$ over a point $p\in R$, apart from components in $D_j^0$
have even multiplicities.
\end{lemm}

{\em Proof.} Indeed, the double cover $\bar{Y}_{12}\ra \bar{S}$ is
induced from the double cover $C\ra \P^1$. Therefore, the ramification
divisor of the former is the preimage of $R$.
Since we know that the ramification divisor in $S^0$ is $D_1^0\cup D_2^0$
all the other components of the preimage of $R$ have even multiplicities.
\hfill $\square$

\begin{lemm}\label{lemm:y12-dominates}
Assume that the graph
of components of $D_1^0\cup D_2^0$ is connected
in $S^0$. Then there are no dominant maps
of $Y_{12}$ onto curves of genus $>0$.
Moreover, $\Alb(Y_{12})=0$.
\end{lemm}

{\em Proof.}
Assume first that $\dim\Alb(Y_{12})=2$ and
that the wedge product of 1-forms on $Y_{12}$
is nondegenerate.
Then, by the Corollary~\ref{coro:disconnect} - this
contradicts the assumptions. Thus we have to consider
the remaining case when
$Y_{12}$ does admit a dominant
$\Z/2$-equivariant (with respect to the covering)
map $\rho$ onto a hyperelliptic curve $C$ of genus $>0$.
Then $S^0$ has a regular map onto $\P^1$ such that
components of $D_1$ and $D_2$ are mapped onto
ramification points $p_1,...,p_{2g+2}\subset R$.
By Corollary~\ref{coro:connected},
all connected (in $S^0$) components of the
union of the divisors $D_1^0$
and $D_2^0$ are mapped to one point, which we can assume to be $p_1$.
Consider the singular symmetric tensor
on $C$, which is equal to 0 at $\P^1$ and has the
singularity $dz^{2n}/z^n$
at $p_2,...,p_{2g+2}$.
Since the preimages of points $p_2,...,p_{2g+2}$ consist of components
of multiplicity two this tensor lifts
into a  nonsingular symmetric tensor on $S^0$.
Since the complement to $S^0$ in the K3 surface $S$
consists of finitely many points we can extend this tensor to
a holomorphic symmetric tensor on  $S$. However, there
are no such tensors on a K3 surface, contradiction.
\hfill $\square$

\subsection{Maps to curves}
\label{sect:maps}

\begin{thm}\label{thm:main}
For almost all pairs of distinct prime numbers
$(q_1,q_2)$ the surface $X=X_{q_1,q_2}$
is not fibered.
\end{thm}

{\em Proof.}
By Lemma~\ref{lemm:gamma-connected},
we can insure (for almost all pairs
of distinct primes $q_1,q_2$) that the intersection
graph $D_1^0\cup D_2^0$ is connected in $S^0$.
By Lemma~\ref{lemm:y12-dominates}, we have $\Alb(Y_{12})=0$.
This implies that $X$ has a dominant map onto a curve only
if $X$ is isomorphic to $C\times C$, where $C$ is a curve of genus 2.
Contradiction.
\hfill $\square$

\

\subsection{Fundamental groups}

\begin{prop}
Let $\tilde{X}\subset A_1\times A_2$ be
a surface constructed in Section~\ref{sect:construction}.
Assume that $\tilde{X}$ is not isomorphic to $C\times C$,
where $C$ is a curve of genus $\ge 2$.
Then $\pi_1(\tilde{X})$ admits a surjective
homomorphism onto a group $G$, where $G$ is a nontrivial central
extension of $\Z^8$ by $\Z^5$.
If $\Alb(\tilde{X})$ is isogenous to $A_1\times A_2$ then
$$
G= \pi_1(\tilde{X})/
[[\pi_1(\tilde{X}),\pi_1(\tilde{X})] \pi_1(\tilde{X})]
$$
(modulo torsion).
\end{prop}

{\em Proof.} For generic $A_j$ the $\Q$-linear envelope of
$\omega_j\in H^{2,0}(A_j)$ has dimension 5. Same holds
for $\lambda_2\omega_1 -\lambda_1\omega_2$
on $A_1\times A_2$ (for $\lambda_j\in\Q$). Thus the kernel is $\Z^5$.
The image is $\Z^8=H^1(\tilde{X})/{\rm torsion}$.
It remains to recall a general fact from group cohomology:
Consider the exact sequences

\centerline{
\xymatrix{ 1\ar[r] &  K_1\ar[r] & H^2(G/[G,G],\Z)\ar[r] & H^2(G,\Z)\\
           1\ar[r] &  K_2\ar[r] & H^2(G/[G,G],\Z)\ar[r] & H^2(G/[[G,G],G],\Z)
}
}

Then $K_1$ and $K_2$ are isomorphic and, by duality,
$K_2$ is isomorphic to $[G,G]/[[G,G],G]$ (tensor $\Q$).

\begin{rem}
There are two other potential cases:
when $Y_{12}$ admits a surjective map onto a curve $C$ of genus $g(C)=g$
and when $\Alb(\tilde{X})$ is isogenous to a product of 3 abelian surfaces.
In the first case we have an exact sequence
$$
0\ra \Z^{1+i +g(2g-1)}\ra G\ra \Z^{8+2g}\ra 0.
$$
In the second case
$$
0\ra \Z^{4+2i}\ra G\ra \Z^{12}\ra 0.
$$
Here in both cases $0\le i\le 3$.
We believe that these cases do occur if $A_1,A_2$ have
Picard groups of higher rank or admit nontrivial endomorphisms.
Then the rational envelope of $(2,0)$-forms has smaller
dimension.
\end{rem}

config: arxiv