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\document
\def\~{\widetilde}
\def\gam{\gamma}
\def\vec{\overrightarrow}
\def\biarrow{\overleftrightarrow}
\proclaim {Definition}
$x \in$ $\Bbb R^m$ is called a limit point of $D$ if there exists a path
$\gam \colon [0,1] \rightarrow \Bbb R^m$ and $\~{\gam} \colon [0,1) \rightarrow
\~M$ such that $D \circ \~{\gam} = \gam$ on $[0,1)$, $\gam(1) = x$ and $\~{\gam}(1)$ cannot
be
defined continuously. The set of limit points of $D$ denoted by $L$ is called a limit set.
\endproclaim
Note that the restriction of D on $\~M - D^{-1} (L)$ is a covering map onto its image
because $D$ is a local diffeomorphism and every path in $R^m$ can be lifted
to $\~M$ by $D$.
In particular, if $L = \emptyset$ then $D$ is a covering map onto
$\Omega(=D(\~M))$.
If $\Omega \not= \Bbb R^m$ then some point of $\partial \Omega$ lies in $L$.
So we can see that $\Omega = \Bbb R^m$ and $D$ is a diffeomorphism. Hence if
$L = \emptyset$ then $M$ is a complete affine manifold.
For $U \subset \Bbb R^m$, $\~U \subset \~M$ is called a copy of $U$ if $D|_{\~U} \colon
\~U \rightarrow U$ is a homeomorphism with respect to the relative topology.
\proclaim {Proposition 1}
If $D|_{\~U_1}$, $D|_{\~U_2}$
are homeomorphisms and $\~U_1 \cap \~U_2 \not= \emptyset$ with $D(\~U_1)
\cap D(\~U_2)$ path connected then $D|_{\~U_1 \cup \~U_2}$ is a
homeomorphism, that is, $\~U_1 \cup \~U_2$ is a copy of $D(\~U_1 \cup
\~U_2)$.
\endproclaim
\demo{Proof}
Suppose that for $x_1 \in \~U_1 , x_2 \in \~U_2$, $D(x_1) = D(x_2)$. Then $D(x_1)$ lies
in $D(\~U_1) \cap D(\~U_2)$. Choose a point $y$ in $\~U_1 \cap \~U_2$ with $D(y) \not=
D(x_1)$. Make a path $\gamma:[0,1] \rightarrow
D(\~U_1) \cap D(\~U_2)$ with $\gamma(0)= D(y)$ and $\gamma(1) = D(x_1)$. Then there
exist paths $\gamma_1$, $\gamma_2$ in $\~U_1$, $\~U_2$ respectively such
that $D(\gamma_1 (t))
= D(\gamma_2 (t)) = \gamma(t)$ for all $t \in [0,1]$, $\gamma_1(0) = \gamma_2(0) = y$, $\gamma_1(1) = x_1$ and
$\gamma_2(1) = x_2$. If $x_1 \not= x_2$ then $x_1 \in \~U_1 \backslash \~U_1 \cap \~U_2$ and $x_2 \in \~U_2 \backslash
\~U_1 \cap \~U_2$.
$\gamma_1$ and $\gamma_2$ must be same in $\~U_1 \cap \~U_2$ because they are the pre images of $\gam$. So inf $\{t : \gamma_1(t)
\notin \~U_1 \cap \~U_2 \}$ equals to inf$ \{t : \gamma_2(t) \notin \~U_1 \cap
\~U_2 \}$ and let it be $t_0$.Because $D$ is a local homeomorphism, for some $\epsilon > 0$, $\gam_1 (t) = \gam_2 (t)$ for all $t \in [0,t_0 + \epsilon)$ and lies in $\~U_1 \cap \~U_2$ which is a contradiction.
$\square$
\enddemo
For $x \in \Bbb R^k \times \Bbb R^n$, we will adopt the notation $x = (x^k , x^n)$ for
the coordinate of $x$.
\proclaim {Lemma 1}
For each point $\~x \in \~M$, there exists a copy of $\Bbb R^k \times D(\~x)^n$
containing $\~x$.
\endproclaim
\demo{Proof}
Suppose not.
Then there exists a path $\~{\gam} \colon [0,1) \rightarrow \~M$,
$\gam \colon [0,1] \rightarrow \Bbb R^m$ such that $D \circ \~{\gam} = \gam$ on $[0,1)$,
$\~{\gam} (1)$ cannot be defined continuously and $\gam$ is a line segment in $\Bbb R^k
\times D(\~x)^n$. Note that $\~{\gam}$ lies in a maximal homeomorphic copy of a disc on $\Bbb R^k \times D(\~x)^n$ centered at $D(\~x)$.
Let $F$ be a fundamental domain in $\~M$ with $\overline{F}$ compact.
Then $\~{\gam}$ passes through infinitely many $g_i (\overline{F})$
where $g_i$ is an element of $\Pi$, the deck transformation group.
For if not, $\~{\gam}$ lies in a compact set and we can choose $\~{\gam}(1)$ continuously
because D is a local homeomorphism.
We can assume that $g_i$ is distinct for all $i$.
Let $x_i \in \~{\gam}([0,1]) \cap g_i(\overline{F})$ which is distinct from each other.
Then $g_i^{-1}(x_i) \in \overline{F}$ and we may assume that $g_i^{-1}(x_i)$
converges to some point $x_0$ in $\overline{F}$.
Choose a fundamental domain $F'$ containing $x_0$ and an open neighborhood $U$ of $x_0$ which is mapped homeomorphically to its image by $D$.
Within $F' \cap U$, we can choose a line copy $l_i$ containing $g_i^{-1}(x_i)$ with a small constant length $\delta$ in the pull back metric of Euclidean metric on $\Bbb R^m$
such that $D(g_i(l_i))$ has a same direction as $\gam$.
Then $g_i(l_i)$ is distinct and lies in $\~{\gam}([0,1))$.
Because the holonomy group acts isometrically in $\Bbb R^k$-direction, the length of $D(g_i(l_i))$ is $\delta$.
This means that the length of $\gam([0,1])$ is infinite which is a contradiction.
$\square$
\enddemo
For $\~x_0 \in \~M$, Choose a neighborhood $\~B$ of $\~x_0$ such that $\~B$ is mapped to $B = \{ x \in \Bbb R^m \ | \ |x - D(\~x_0)| < r \}$ homeomorphically by $D$. For each $\~x \in \~M$, let $F_{\~x}$ be a copy of $\Bbb R^k \times D(\~x)^n$. Then Lemma 1 implies that $\~C = \bigcup\limits_{\~x} F_{\~x}$ is a cylindrical copy with center $\~x$ which is homeomorphic to $C = \{ (x^k, x^n) \in \Bbb R^m \ | \
|x^n -D(\~x)^n | < r \}$. If $L \not= \emptyset$ then there exists a
maximal cylindrical copy with center $\~x$. Note that if $x$ is a limit point then $\Bbb R^k \times x^n$ lies in $L$.
\proclaim {Proposition 2}
If $L \not= \emptyset$, $D$ is a covering map onto $\Bbb R^m \backslash L$
and $L$ is an affine subspace of $\Bbb R^m$.
\endproclaim
\demo{Proof}
We follow the proof of Matsumoto.
For each point $\~x \in \~M$, let $r(\~x)$ be the radius of maximal
cylindrical copy with center $\~x$.
Define a metric on $\~M$ by
$$ g_F = D^* g_{R^k} + {D^* g_{R^n} \over r(\~x)^2}$$
where $g_{R^l}$ is an Euclidean metric on $R^l$.
Then the deck transformation group $\Pi$
becomes an isometry group with respect to $g_F$.
Let $C_0$ be the maximal cylindrical copy.
Step1: We will show that there exists a half space copy containing $C_0$.
We may assume that $D(\~x) = e = (0,0,\cdots,1)$, $r(\~x) = 1$ and $o$, the
origin, is a limit point.
Let $$ C = \{(x^k , x^n) \in \Bbb R^m \mid \ \mid x^n - e^n \mid < 1\}. $$
We may assume $D( C_0 ) = C$ and consider a metric $g_F$ on $C$.
Define another metric on $C$ by
$$ g_G = g_{R^k} + {g_{R^n} \over {\mid x^n \mid}^2}.$$
Let $$ P=\{ (x^k , x^n) \in \Bbb R^m \mid x^n = (0,0,\cdots,t), 0\leq
t\leq2\}.$$
We abuse the notation $d_F$ for the metric on $C$ which is homeomorphic to $C_0$.
Then for each $x \in C$, $d_F (x, P) \geq d_G (x, P)$
because $\mid x^n \mid \geq r(x)$.
Let $x_i \in \~M$ be a sequence such that $D(x_i) = (0,0,\cdots,d_i),
d_i \searrow 0$ as with the proof of Lemma 1.
Let $K$ be a compact subset such that $\Pi \cdot K = \~M$.
Then there exist $\gam_i \in \Pi$ such that $\gam_i^{-1} (x_i) \in K$.
Because $K$ is compact, we can assume that $\gam_i^{-1} (x_i)$ converges to
some point $x_0 \in K$ in $g_F$ metric.
Let $g_i$ be an element of the holonomy group $G$ corresponding to $\gam_i$ by $D$ and $d=D(x_0)$.
Then $d_F (\gam_i (x_0), x_i) \rightarrow 0$ and hence $d_G (g_i (d), P) \rightarrow 0$ and $g_i (d) \rightarrow o$. Because $d_G(g_i(d), P)$ is the angle between $(0,g_i(d)^n)$ and $P$ we see that $g_i(d)$ lies in $C$ and $\gam_i(x_0)$ lies in $C_0$.
Since for $x \in C$, $r(x) \leq |x^n|$,
$$\|g_i\| = {r(g_i (d)) \over r(d)}
\leq {|g_i(d)^n| \over r(d)} \rightarrow 0$$
where $\| g_i \|$ is the norm of the similarity part of $g_i$.
So for $i \gg 1$, $j \gg i$, using the compactness of $O(k)$ and $O(n)$ we may assume that the linear part of
$g_i g_j^{-1}$ is almost same as
$$ \pmatrix I_k & 0 \cr 0 & \lambda_{ij}I_n \cr \endpmatrix \ where \
\lambda_{ij} \rightarrow \infty.$$
Let $a_i = (g_i(d)^k, 0)$. Then the line $\biarrow{a_j g_j(d)}$ is almost
same as $\biarrow{oe}$ because the angle between two line is very small and
$g_j(d)$ converges to $o$.
And $g_i g_j^{-1}(\biarrow{a_j g_j(d)})$ is almost same as $\biarrow{oe}$
because it passes through $g_i(d)$ which is close to $o$ and the rotation
angle of $g_i g_j^{-1}$ is very small and hence $g_i g_j^{-1}(a_j)$ goes nearer to $\biarrow{oe}$. The tangent space of $g_i g_j^{-1}(\partial C)$ at $g_i g_j^{-1}(a_j)$ is almost parallel to the tangent space
of $\partial C$ at $o$. But for arbitrary indices i and j, $\gam_i \gam_j^{-1} C_0 \cap C_0$ is nonempty ( $\gam_i(x_0)$ lies in this set) and
$\gam_i \gam_j^{-1} (C_0) \cup C_0 $ is a copy of $g_i g_j^{-1}(C) \cup C$ by Proposition 1
so $o \not\in g_i g_j^{-1}(C)$ and $g_i g_j^{-1}(a_j) \not\in C$. Hence the two tangent space is almost same and $g_i g_j^{-1}(a_j)$ converges to $o$ (refer to [Matsumoto] for more details).
So any point of half space which has a tangent space of $C$ as a boundary is
contained in some $g_i g_j^{-1} (C)$. From that $\gam_i \gam_j^{-1} (C_0)
\cup C_0 $ is a copy of $g_i g_j^{-1}(C) \cup C$, we can deduce the claim.
We see that this half space copy containing $C_0$ is uniquely determined and
denote this by $\~H(\~x)$.
For $\~x \in \~M$, let $H(\~x) = D(\~H(\~x))$ and $L(\~x) = \partial H(\~x)
\backslash D($\rm Fr $\~H(\~x))$
where \rm Fr $\~H(\~x)$ is a frontier of $\~H(\~x)$. Denote $p(\~x)$ by the
point of $\partial H(\~x)$ corresponding to $o$ of step1.
Note that $p(\~x)$ lies in $L(\~x)$ and $\bigcup\limits_{\~x} L(\~x) \subset L$.
Step2: We will show that for $y \in \overline{H(\~x)} \backslash L(\~x)$
and $D(\~y) = y$, $\partial H(\~y)$ passes through $p(\~x)$.
If we consider $p(\~x)$ as $o$, from Step1,
we see that $g_j g_i^{-1}$ is almost same as
$$\pmatrix I_k & 0 \cr 0 & \mu_{ij}I_m \cr \endpmatrix \pmatrix 0 \cr 0 \cr \endpmatrix $$
where $\mu_{ij}$ is arbitrarily small because $g_i g_j^{-1}(a_j)$ converges to $o$ and $a_j$ converges to $o$. So $g_j g_i^{-1}(\partial H(\~y))$
is almost parallel to $\partial H(\~y)$.
If $p(\~x) \not\in \partial H(\~y)$ then for large $i$, $j$,
$g_j g_i^{-1}(\partial H(\~y))$ is much closer to $p(\~x)$
than $\partial H(\~y)$. Hence $g_j g_i^{-1}(H(\~y)) \cup H(\~y)$ becomes a copy
containing $p(\~y)$ which is a contradiction.
Step3: $L(\~x)$ is an affine subspace. Choose $a$, $b \in L(\~x)$.
Then $a = p(\~c)$ for some $c \in H(\~x)$ such that $H(\~x) = H(\~c)$. Then
$L(\~x) = L(\~c)$ by definition of $L(\~x)$. Likewise for b.
If $y$ lies on the line $\biarrow{ab}$ but $y \not\in L(\~x)$ then
$\partial H(\~y)$ passes through $a$ and $b$ by Step2. This says $y$ lies
in
$\partial H(\~y)$ which is a contradiction.
Step4: We claim that for $\~y \in \~H (\~ x)$, $L(\~y) = L(\~x)$.
By Step2, $\partial H(\~y)$ passes through $p(\~x)$ and $p(\~x) \in
L(\~y)$.
If some point $a \in L(\~y)$ lies in $\Bbb R^m \backslash \overline{H(\~x)}$
then $H(\~x)$ contains a point in the line $\biarrow{ap(\~x)}$ which lies
in $L(\~y)$ by Step3. Since $H(\~x) \cap L(\~y) = \emptyset$, this cannot
happen. So $L(\~y) \subset \partial H(\~x)$ and hence $L(\~y) \subset
L(\~x)$. By Step2, $\~x$ lies in $\~H(\~y)$.
Hence by the same argument, $L(\~x) \subset L(\~y)$. Thus $L(\~x) = L(\~y)$.
Since $\~M$ is connected, $L(\~x)$ is independent of $\~x \in \~M$.
Denote this space by $L$.
Then it follows easily that $D$ becomes a covering map onto $R^m \backslash L$
and $L$ is a limit set of $D$. $\square$
\enddemo
\proclaim{Lemma 2}
$L$ is an affine subspace of dimension $k$.
\endproclaim
\demo{Proof}
Suppose that $L$ is an affine subspace of dimension $k + l$, $l > 0$.
We can assume that $L = \Bbb R^k \times (\Bbb R^l \times (0, \cdots , 0))$.
Define a metric $g_F$ on $\Bbb R^m \backslash L$ by
$$ g_F = g_{R^k} + {g_{R^n} \over d(x)}$$
where $d(x) = x_{k+l+1}^2 + \cdots + x_{k+n}^2$ which is the square of the distance between $x$ and $L$. Then $g_F$ is an invariant by the holonomy group $G$. Thus its pullback metric is invariant by the fundamental group so it defines a metric on $M$.
Let $X$ be a vector field on $\Bbb R^m \backslash L$ defined by $X(x) = (0,(0,x_{k+l+1}, \cdots, x_{k+n}))$. Then $X$ is also $G$-invariant vector field. Hence it defines a vector field on $M$.
The volume form $\omega$ with respect to $g_F$ is $h(x)dx^1 \wedge \cdots \wedge dx^m$ where $h(x) = d(x)^{-{n \over 2}}$.
We calculate $div (X)$ from the equation $L_X \omega = div (X) \omega$. Note that $L_X \omega = d \iota_X \omega + \iota_X d \omega = d \iota_X \omega$. $\iota_X \omega = \sum\limits_i (-1)^{i-1} h(x)dx^1 \wedge \cdots \wedge \iota_X(dx^i) \wedge \cdots \wedge dx^m$. We see that
$$\iota_X(dx^i) = \cases 0, \ i \le k+l \cr x^i, i>k+l. \cr \endcases$$
$$\eqalign{d \iota_X \omega &= \sum\limits_{i > k+l} (-1)^{i-1} d(h(x)x^i) dx^1 \wedge \cdots \wedge \hat{dx^i} \wedge \cdots \wedge dx^m \cr
&= \sum\limits_{i > k+l}(x^i {\partial h(x) \over \partial x^i} + h(x))dx^1 \wedge \cdots \wedge dx^n = \sum\limits_{i > k+l} ({x^i \over h(x)} {\partial h(x) \over \partial x^i} + 1) \omega \cr
&= (-n +(n - l)) \omega = -l \omega.} $$
Since $M$ is a closed manifold, $\int_M div(X) \omega = 0$.
But we compute that $div(X)$ is nonzero constant which is a contradiction.
$\square$
\enddemo
We may assume that $D$ is a covering map onto $$\Bbb R^m - \Bbb R^k = \{(x^k,x^n) \in \Bbb R^n \times \Bbb R^k \ | \ x^n \not= 0 \}.$$
Let
$$G = E(k) \times (O(n) \times A)$$
where $A = \{e^tI_n : t \in R\}$.
\proclaim {Proposition}
Suppose $\pi$ is a discrete subgroup of $G$ acting on $E^k \times (E^n - 0)$ as a covering transformation, with compact quotient. Then $\pi$ is virtually $Z^k \times Z$, where $Z^k$ acts on $E^k$ as translations and $Z$ acts on $E^n -0$ as dilations; both factors act diagonally.
\endproclaim
If $n \not= 2$, then the developing image $D(\~M)$ is simply connected and $D$ is a diffeomorphism. So we can apply above Proposition.
Now, consider the case when $n=2$. We can lift the covering map $D$ to $\~D$ where $\~D$ maps $\~M$ to the universal covering space of $\Bbb R^{k+2} - \Bbb R^k$ which can be identified with $\Bbb R^k \times \Bbb H^2$ where $\Bbb H^2$ is an upper half subspace of $\Bbb R^2$.And this covering map is given by $$\eqalign{ p : \Bbb R^k \times \Bbb H^2 &\rightarrow \Bbb R^{k+2} - \Bbb R^k \cr (x,(\theta,r)) &\mapsto (x,(r\cos\theta,r\sin\theta))}$$
Then the element of $E(k) \times (SO(2) \times A)$, $\pmatrix \alpha, \lambda \pmatrix \cos\theta & -\sin\theta \cr \sin\theta & \cos\theta \endpmatrix \endpmatrix$ is lifted to
$$ \pmatrix \alpha, \pmatrix \pmatrix 1 & 0 \cr 0 & \lambda \endpmatrix, \pmatrix \theta \cr 0 \endpmatrix \endpmatrix \endpmatrix$$
which is an element of $E(k+1) \times (SO(1) \times A)$. Thus $M$ is given a new geometric structure on which we can apply above proposition.
\proclaim{Conclusion}
If $M$ is an affine manifold with a partial dilation structure then $M$ is finitely covered by $T^k \times S^1 \times S^{n-1}$.
\endproclaim
\end