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@\section{Examples of higher cycles}
In this section we construct some higher cycles in $Z^2(E\times E', 1)$ where $E$ and $E'$ are elliptic curves.

\subsection{Notations}
We consider the Weierstrass family $y^2 = x^3 + ax +b$. Denote the total space by $\E$. A fiber, i.e. a single elliptic curve is denoted by $E$.  We cover $E$ by two charts. The first one is 
\[
U_0=E\setminus \{[\infty]\},
\]
which was mentioned before. The second one is 
\[
U_1=E\setminus \{(y=0)\}.
\]
The coordinates on $U_1$ are $x, z$ and the equation is
\[
z = x^3 + a x z^2 + b z^3.
\]
The gluing maps are given as follows:
\[
U_0\ra U_1: \; (x,y) \ra (x y^{-1}, y^{-1}), \;\; U_1\ra U_0: \; (x,z) \ra (x z^{-1}, z^{-1}).
\]

\subsection{The first cycle}
The first cycle will be on $E\times E$ for $b\neq 0$. To denote the coordinates on the first $E$ in $E\times E$ we will use index $1$, for the second one we use $2$. So $U_0\times U_0$ is given by two equations in $4$ variables:
\[
y_1^2 = x_1^3 + a x_1 + b,\;\; y_2^2 = x_2^3 + a x_2 + b.
\]

Consider the subvariety $W$ which is the closure in $E\times E$ of the subvariety given by the equation $x_2 y_1 + \I x_1 y_2=0$. To compute the closure first consider $U_0\times U_1$. We have equations:
\[
y_1^2 = x_1^3 + a x_1 + b,\;\; z_2 = x_2^3 + a x_2 z_2^2 + b z_2^3,\;\; x_2 z_2^{-1} y_1 + \I x_1 z_2^{-1}=0.
\]
This gives 
\[
x_1 = \I y_1 x_2,\;\; y_1^2 = -\I x_2^3 y_1^3 + \I a x_2 y_1 + b.
\]
when $(x_2, z_2) = 0$ we obtain $(x_1, y_1) = (0, \pm \sqrt{b})$. So we see that $W$ contains two more points on $U_0\times U_1$. In fact $x_2$ is a local parameter on $W$ at these points, $z_2$ has order $3$ and $x_1$ has order $1$.

Analogously $W$ contains two more points on $U_1\times U_0$, namely $(x_2, y_2) = (\pm \sqrt{b},0)$. 

It remains to look at $U_1\times U_1$. The equations are
\[
z_1 = x_1^3 + a x_1 z_1^2 + b z_1^3,\;\;z_2 = x_2^3 + a x_2 z_2^2 + b z_2^3,\;\; x_2 z_1^{-1} z_2^{-1} + \I x_1 z_1^{-1} z_2^{-1}=0.
\]
Therefore we have
\[
x_1 = \I x_2.
\]
One can check that the point $[\infty]\times[\infty]$ also belongs to $W$ and the local parameter there can be chosen as $x_1$ or $x_2$.

Let $f$ be the rational function on $W$ given as $y_1-\I y_2$. We first compute the divisor of $f$.  Since $y$ has a triple pole at $[\infty]$ and the projections $W\to E$ are unramified at $[\infty]$ we see that $f$ has triple pole at the points $[\infty]\times(0, \pm\sqrt{b})$ and $(0, \pm\sqrt{b})\times[\infty]$. To study the behaviour of $f$ at $[\infty]\times[\infty]$ we write $f$ as
\[
f = z_1^{-1} - \I z_2^{-1} = (z_2 - \I z_1) z_1^{-1} z_2^{-1}.
\]
Using the fact that $z = x^3 + a x^7 + b x^9 + \cdots$ on $U_1$ at $[\infty]$ we see that
\[
f\sim 2 b \I x_2^{3} \;\; \text{at $[\infty]\times[\infty]$,}
\]
so $f$ has a triple zero.

Finally we look for zeroes of $f$ on the set $U_0\times U_0$. We need to find all common solutions of the equations
\[
y_1^2 = x_1^3 + a x_1 + b,\;\; y_2^2 = x_2^3 + a x_2 + b,\;\;x_2 y_1 + \I x_1 y_2=0,\;\;y_1-\I y_2=0.
\]
We get
\[
y_1=\I y_2,\;\; (x_1+x_2) y_1 y_2 = 0.
\]
In the case $y_1=y_2=0$ we see that for $\lambda_1, \lambda_2, \lambda_3$~--- the distinct roots of the polynomial $x^3+ax+b$ the $9$ points $(\lambda_i,0)\times(\lambda_j,0)$ are solutions. The case $x_1+x_2=0$ does not give any solution unless $b=0$. We will not check that the $9$ zeroes are indeed simple since we already know that $f$ has total multiplicity of poles $12$ and triple zero was already found. Therefore
\begin{multline*}
\Div f = \sum_{i,j} (\lambda_i,0)\times(\lambda_j,0) + 3 [\infty]\times[\infty] - 3 [\infty]\times(0, \sqrt{b}) \\- 3 [\infty]\times(0, -\sqrt{b}) - 3 (0, \sqrt{b})\times[\infty] - 3 (0, -\sqrt{b})\times [\infty].
\end{multline*}

It is not difficult to correct $(W,f)$ and obtain a cycle. The following combination is a cycle in $Z^2(E\times E, 1)$:
\[
(W,f) - \sum_i((\lambda_i,0)\times E, y_2) - 3(E\times[\infty], y_1) + 3([\infty]\times E, x_2) + 3(E\times[\infty], x_1).
\]

\subsection{The second cycle}
The second cycle is also on $E\times E$, but it is given by a single summand. We take $W$ as the closure of the subvariety defined by equation $x_1+x_2=0$. It is clear from the equation that when $x_1$ is infinite, $x_2$ must be infinite as well and vice versa. Therefore the closure contains only one additional point $[\infty]\times[\infty]$.

The problem with this $W$ is that it is singular. Namely on $U_1\times U_1$ the equations are
\[
z_1 = x_1^3 + a x_1 z_1^2 + b z_1^3,\;\;z_2 = x_2^3 + a x_2 z_2^2 + b z_2^3,\;\; x_1 z_1^{-1} + x_2 z_2^{-1}=0.
\]
So we obtain the following equations for $W$ inside $E\times E$:
\[
x_1 z_2 = -z_1 x_2,\;\; x_1^2+x_2^2 = 2 b x_2^2 z_1^2.
\]
One can see that the function $x_1 x_2^{-1}$ belongs to the integral closure of the structure ring, but does not belong to the structure ring itself. It is easy to check that if we pass to the normalization of $W$ we obtain $2$ points over $[\infty]\times[\infty]$, namely the one with $x_1 x_2^{-1} = \I$ and the one with $x_1 x_2^{-1} = -I$.

The function $f$ remains the same,
\[
f = y_1 - \I y_2.
\]

It is clear that $f$ does not have zeroes or poles on $W\cap U_0\times U_0$. The only remaining point is $[\infty]\times[\infty]$. We use expansion $z = x^3 + a x^7 + b x^9 + \cdots$.
The second equation for $W$ gives
\[
x_1^2 + x_2^2 = 2 b x_2^2 (x_1^6 + 2 a x_1^{10} + 2 b x_1^{12}+\cdots).
\]
This implies
\[
x_2^2 = -x_1^2 - 2 b x_1^8 - 4 ab x_1^{12} - 8 b^2 x_1^{14}+\cdots.
\]
There are $2$ solutions of this equation:
\[
x_2 = \pm \I (x_1 + b x_1^7 + 2 a b x_1^{11} + \frac{7}{2} b^2 x_1^{13}+\cdots).
\]
The corresponding value of $z_2$ can be computed:
\[
z_2 = \mp \I (x_1^3 + a x_1^7 + 2 b x_1^9+\cdots).
\]

We can now compute the expansion of $f$. Along the branch $x_2\sim \I x_1$ we have
\[
f = \frac{z_2 - \I z_1}{z_1 z_2} \sim 2 x_1^{-3}.
\]
Along the branch $x_2\sim -\I x_1$ we have
\[
f = \frac{z_2 - \I z_1}{z_1 z_2} \sim b x_1^3.
\]
Therefore $\Div f = 0$ and $(W,f) \in Z^2(E\times E, 1)$.

\subsection{Equivalence of the first and the second cycles}
Denote by
\begin{align*}
D_1 &\;\; \text{the closure of the set of zeroes of} & x_2 y_1 + \I x_1 y_2,\\
D_2 &\;\; \text{the closure of the set of zeroes of} & x_1+x_2,\\
D_3 &\;\; \text{the closure of the set of zeroes of} & y_1-\I y_2,
\end{align*}
\begin{align*}
Z_1 &= [\infty]\times E,\\
Z_2 &= E\times [\infty],
\end{align*}
\begin{align*}
f_1 &= x_2 y_1 + \I x_1 y_2& (\Div f_1 = D_1 - 3 Z_1 - 3 Z_2),\\
f_2 &= x_1 + x_2& (\Div f_2 = D_2 - 2 Z_1 - 2 Z_2),\\
f_3 &= y_1-\I y_2& (\Div f_3 = D_3 - 3 Z_1 - 3 Z_2).
\end{align*}

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The first cycle will be on $E\times E$ for $b\neq 0$. To denote coordinates on the first $E$ in $E\times E$ we will use index $1$, for the second one we use $2$. So $U_0\times U_0$ is given by two equations in $4$ variables:
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