kellermt.tex

\subsection{Rationality of the cap}\label{rat-cap11}
We will now prove proposition \ref{rat-cap}.  In fact we will show
that it holds for any circle $\alpha$ contained in a torus fiber
of $e(P)$ and passing through a rational point. We would like to
thank Misha Kapovich for simplifying  our original argument. The
idea is to construct, for each component of $\partial C_x$, a
$2$-chain $A$ with that component as boundary so that $A$ is a sum
$P+ T  +\mathcal{M}(\gamma_0)$ of three simplicial $2$-chains in
$M$. We then verify that the ``parallelogram'' $P$  and the ``triangle''
$T$  have rational area and the period of $\Omega$ over the ``monodromy
chain'' $\mathcal{M}(\gamma_0)$ is zero.

In what follows we will pass from pictures in the plane involving
directed line segments, triangles and parallelograms to identities
in the space of simplicial $1$-chains $C_1(T^2)$  on $T^2$. The
principle behind this is that any $k$-dimensional subcomplex $S$
of a simplicial complex $Y$ which is the fundamental cycle of an
oriented $k$-submanifold $|S|$ (possibly with boundary) of $Y$
corresponds in a {\it unique} way to a sum of oriented $k$-simplices
in $C_k(Y)$.



In this subsection we will work with a general $3$-manifold $M$
with Sol geometry. Of course this includes all the manifolds $e'(P)$
that occur in this paper. Let $f \in \SL(2,\Z)$ be a hyperbolic
element. We will then consider the $3$-manifold $M$ obtained from
$ \R \times T^2$ (with the $2$-torus $T^2 = W/ \Z^2$) given by the
relation
\begin{equation}\label{glueing}
(s,w) \sim (s+1,f(w)).  
\end{equation}
We let $\pi: \R\times T^2 \to M$ be the resulting infinite cyclic covering. 



We now define notation we will use below. We will use Greek letters
to denote closed geodesics on $T^2$, a subscript $c$  will indicate
that the geodesic starts at the point $c$ on $T^2$. We will use the
analogous notation for geodesic arcs on $W$.  We will use $[\alpha]$
to denote the corresponding homology class of a closed geodesic $\alpha$ on $T^2$. 
If $x$ and $y$ are points on $W$ we will use $\overline{xy}$ to
denote the oriented line segment joining $x$ to $y$ and
$\overrightarrow{xy}$ to denote the corresponding (free) vector,
i.e. the equivalence class of $\overline{xy}$ modulo parallel translation. 
   
We first take care of the fact that $\alpha$ does not necessarily
pass through the origin. For convenience we will assume $\alpha$
is in the fiber over the base-point $z(x)$ corresponding to $s=0$.
Let $\alpha_0$ be the parallel translate of $\alpha$ to the origin.
Then we can find a cylinder $P$, image of an oriented parallelogram
$\widetilde{P}$ under the universal cover $W \to T^2$ with rational
vertices, such that in $Z_1(T^2,\Q)$, the group of rational $1$-cycles,
we have
\begin{equation}\label{firstrectangle}
\partial P = \alpha - \alpha_0.
\end{equation}
Since $\widetilde{P}$ has rational vertices we find $\int_{P} \Omega
= \int_{\widetilde{P}} \Omega \in \Q$.

Now we take care of the harder part of finding $A$ as above. The
key is the construction  of ``monodromy $2$-chains.''  For any closed
geodesic $\gamma_0 \subset T^2$ starting at $0$ we define the
monodromy $2$-chain  $\mathcal{M}(\gamma_0)$ to be the image of the
cylinder $\gamma_0 \times [0,1] \subset T^2 \times \R$ in $M$.
The reader will verify using \eqref{glueing} that in $Z_1(T^2,\Q)$ we have 
\begin{equation} \label{boundaryofmon}
 \partial \mathcal{M}(\gamma_0) = f^{-1}(\gamma_0) -\gamma_0.
\end{equation}
Since $f$ preserves the origin, the geodesic $f^{-1}(\gamma_0)$ is
also a closed geodesic starting at the origin. Since $f^{-1}$ is
hyperbolic we have $|\tr(f^{-1})| >2$ and hence $\det(f^{-1} -I)=
\det( I - f) = \tr(f) -2 \neq 0$. Put $N= \det(f^{-1} -I)$ and define
$[\gamma_0] \in H_1(T^2,\Z)$ by
\begin{equation}\label{invertmatrix}
f^{-1}([\gamma_0]) -[\gamma_0] = N[\alpha_0]. 
\end{equation}
Note that $[\gamma_0] = N \{(f^{-1} - I)^{-1} ([\alpha_0] \}$ is
necessarily an integer homology class. Also note that is an equation
in the first {\it homology}, it is not an equation in the group of
$1$-cycles $Z_1(T^2,\Q)$. Since any homology class contains a unique
closed geodesic starting at the origin we obtain a closed geodesic
$\gamma_0 \in [\gamma_0]$  and a corresponding  monodromy $2$-chain
$\mathcal{M}(\gamma_0)$ whence \eqref{boundaryofmon} holds in
$Z_1(T^2,\Q)$. We now solve

\begin{problem}
Find an equation in $Z_1(T^2,\Z)$ which descends to  \eqref{invertmatrix}. 
\end{problem}