Definition. x ∈ ℝ^{m} is called a limit point of D if there exists a path γ: [0, 1] → ℝ^{m} and γ̃: [0, 1) → M̃ such that D ∘ γ̃ = γ on [0, 1), γ(1) = x and γ̃(1) cannot be defined continuously. The set of limit points of D denoted by L is called a limit set.

Note that the restriction of D on M̃− 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 M̃ by D. In particular, if L = ∅ then D is a covering map onto Ω(= D(M̃)). If Ω ≠ ℝ^{m} then some point of ∂Ω lies in L. So we can see that Ω = ℝ^{m} and D is a diffeomorphism. Hence if L = ∅ then M is a complete affine manifold.

For U ⊂ ℝ^{m}, Ũ ⊂ M̃ is called a copy of U if D|Ũ: Ũ → U is a homeomorphism with respect to the relative topology.

Proposition 1. If D|Ũ_{1}, D|Ũ_{2} are homeomorphisms and Ũ_{1}∩Ũ_{2} ≠ ∅ with D(Ũ_{1})∩ D(Ũ_{2}) path connected then D|_{Ũ_{1}∪Ũ_{2}} is a homeomorphism, that is, Ũ_{1} ∪ Ũ_{2} is a copy of D(Ũ_{1} ∪ Ũ_{2}).

Proof. Suppose that for x_{1} ∈ Ũ_{1}, x_{2} ∈ Ũ_{2}, D(x_{1}) = D(x_{2}). Then D(x_{1}) lies in D(Ũ_{1}) ∩ D(Ũ_{2}). Choose a point y in Ũ_{1} ∩ Ũ_{2} with D(y) ≠ D(x_{1}). Make a path γ : [0, 1] → D(Ũ_{1}) ∩ D(Ũ_{2}) with γ(0) = D(y) and γ(1) = D(x_{1}). Then there exist paths γ_{1}, γ_{2} in Ũ_{1}, Ũ_{2} respectively such that D(γ_{1}(t)) = D(γ_{2}(t)) = γ(t) for all t ∈ [0, 1], γ_{1}(0) = γ_{2}(0) = y, γ_{1}(1) = x_{1} and γ_{2}(1) = x_{2}. If x_{1} ≠ x_{2} then x_{1} ∈ Ũ_{1}\Ũ_{1} ∩ Ũ_{2} and x_{2} ∈ Ũ_{2}\Ũ_{1} ∩ Ũ_{2}. γ_{1} and γ_{2} must be same in Ũ_{1} ∩ Ũ_{2} because they are the pre images of γ. So inf {t : γ_{1}(t) ∉ Ũ_{1} ∩ Ũ_{2}} equals to inf{t : γ_{2}(t) ∉ Ũ_{1} ∩ Ũ_{2}} and let it be t_{0}.Because D is a local homeomorphism, for some ϵ > 0, γ_{1}(t) = γ_{2}(t) for all t ∈ [0, t_{0} + ϵ) and lies in Ũ_{1} ∩ Ũ_{2} which is a contradiction. □

For x ∈ ℝ^{k} × ℝ^{n}, we will adopt the notation x = (x^{k}, x^{n}) for the coordinate of x.

Lemma 1. For each point x̃ ∈ M̃, there exists a copy of ℝ^{k} × D(x̃)^{n} containing x̃.

Proof. Suppose not. Then there exists a path γ̃: [0, 1) → M̃, γ: [0, 1] → ℝ^{m} such that D ∘ γ̃ = γ on [0, 1), γ̃(1) cannot be defined continuously and γ is a line segment in ℝ^{k} × D(x̃)^{n}. Note that γ̃ lies in a maximal homeomorphic copy of a disc on ℝ^{k} × D(x̃)^{n} centered at D(x̃). Let F be a fundamental domain in M̃ with \overline{F} compact. Then γ̃ passes through infinitely many g_{i}(\overline{F}) where g_{i} is an element of Π, the deck transformation group. For if not, γ̃ lies in a compact set and we can choose γ̃(1) continuously because D is a local homeomorphism. We can assume that g_{i} is distinct for all i. Let x_{i} ∈ γ̃([0, 1]) ∩ g_{i}(\overline{F}) which is distinct from each other. Then g_{i}^{−1}(x_{i}) ∈ \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

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open neighborhood U of x_{0} which is mapped homeomorphically to its image by D. Within F′∩U, we can choose a line copy l_{i} containing g_{i}^{−1}(x_{i}) with a small constant length δ in the pull back metric of Euclidean metric on ℝ^{m} such that D(g_{i}(l_{i})) has a same direction as γ. Then g_{i}(l_{i}) is distinct and lies in γ̃([0, 1)). Because the holonomy group acts isometrically in ℝ^{k}-direction, the length of D(g_{i}(l_{i})) is δ. This means that the length of γ([0, 1]) is infinite which is a contradiction. □

For x̃_{0} ∈ M̃, Choose a neighborhood B̃ of x̃_{0} such that B̃ is mapped to B = {x ∈ ℝ^{m} | |x − D(x̃_{0})| < r} homeomorphically by D. For each x̃ ∈ M̃, let Fx̃ be a copy of ℝ^{k} × D(x̃)^{n}. Then Lemma 1 implies that C̃ = ⋃ ; x̃ Fx̃ is a cylindrical copy

with center x̃ which is homeomorphic to C = {(x^{k}, x^{n}) ∈ ℝ^{m} | |x^{n} − D(x̃)^{n}| < r}. If L ≠ ∅ then there exists a maximal cylindrical copy with center x̃. Note that if x is a limit point then ℝ^{k} × x^{n} lies in L.

Proposition 2. If L ≠ ∅, D is a covering map onto ℝ^{m}\L and L is an affine subspace of ℝ^{m}.

Proof. We follow the proof of Matsumoto. For each point x̃ ∈ M̃, let r(x̃) be the radius of maximal cylindrical copy with center x̃. Define a metric on M̃ by

g_{F} = D^{∗}g_{R^{k}} + \frac{D^{∗}g_{R^{n}}}{r(x̃)^{2}}

where g_{R^{l}} is an Euclidean metric on R^{l}. Then the deck transformation group Π 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(x̃) = e = (0, 0, ⋯, 1), r(x̃) = 1 and o, the origin, is a limit point. Let

C = {(x^{k}, x^{n}) ∈ ℝ^{m} | | x^{n} − e^{n} |< 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}} + \frac{g_{R^{n}}}{| x^{n} |^{2}} .

Let

P = {(x^{k}, x^{n}) ∈ ℝ^{m} | x^{n} = (0, 0, ⋯, t), 0 ≤ t ≤ 2}.

We abuse the notation d_{F} for the metric on C which is homeomorphic to C_{0}. Then for each x ∈ C, d_{F} (x, P) ≥ d_{G}(x, P) because | x^{n} |≥ r(x). Let x_{i} ∈ M̃ be a sequence such that D(x_{i}) = (0, 0, ⋯, d_{i}), d_{i} ↘ 0 as with the proof of Lemma 1. Let K be a compact subset such that Π · K = M̃. Then there exist γ_{i} ∈ Π such that γ_{i}^{−1}(x_{i}) ∈ K. Because K is compact, we can assume that γ_{i}^{−1}(x_{i}) converges to some point x_{0} ∈ K in g_{F} metric. Let g_{i} be an element of the holonomy group G corresponding to γ_{i} by D and d = D(x_{0}). Then d_{F} (γ_{i}(x_{0}), x_{i}) → 0 and hence d_{G}(g_{i}(d), P) → 0 and g_{i}(d) → 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 γ_{i}(x_{0}) lies in C_{0}. Since for x ∈ C, r(x) ≤ |x^{n}|,

∥g_{i}∥ = \frac{r(g_{i}(d))}{r(d)} ≤ \frac{|g_{i}(d)^{n}|}{r(d)} → 0

where ∥g_{i}∥ is the norm of the similarity part of g_{i}. So for i ≫ 1, j ≫ 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

( \begin{matrix}I_{k} & 0 \\ 0 & λ_{ij}I_{n} \end{matrix} ) where λ_{ij} → ∞.

Let a_{i} = (g_{i}(d)^{k}, 0). Then the line \overleftrightarrow{a_{j}g_{j}(d)} is almost same as \overleftrightarrow{oe} because the angle between two line is very small and g_{j}(d) converges to o. And g_{i}g_{j}^{−1}(\overleftrightarrow{a_{j}g_{j}(d)}) is almost same as \overleftrightarrow{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 \overleftrightarrow{oe}. The tangent space of g_{i}g_{j}^{−1}(∂C) at g_{i}g_{j}^{−1}(a_{j}) is almost parallel to the tangent space of ∂C at o. But for arbitrary indices i and j, γ_{i}γ_{j}^{−1}C_{0} ∩ C_{0} is nonempty ( γ_{i}(x_{0}) lies in this set) and γ_{i}γ_{j}^{−1}(C_{0}) ∪ C_{0} is a copy of g_{i}g_{j}^{−1}(C) ∪ C by Proposition 1 so o ∉ g_{i}g_{j}^{−1}(C) and g_{i}g_{j}^{−1}(a_{j}) ∉ 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 γ_{i}γ_{j}^{−1}(C_{0}) ∪ C_{0} is a copy of g_{i}g_{j}^{−1}(C) ∪ C, we can deduce the claim.

We see that this half space copy containing C_{0} is uniquely determined and denote this by H̃(x̃). For x̃ ∈ M̃, let H(x̃) = D(H̃(x̃)) and L(x̃) = ∂H(x̃)\D(Fr H̃(x̃)) where Fr H̃(x̃) is a frontier of H̃(x̃). Denote p(x̃) by the point of ∂H(x̃) corresponding to o of step1. Note that p(x̃) lies in L(x̃) and ⋃ ; x̃ L(x̃) ⊂ L.

Step2: We will show that for y ∈ \overline{H(x̃)}\L(x̃) and D(ỹ) = y, ∂H(ỹ) passes through p(x̃). If we consider p(x̃) as o, from Step1, we see that g_{j}g_{i}^{−1} is almost same as

( \begin{matrix}I_{k} & 0 \\ 0 & μ_{ij}I_{m} \end{matrix} ) ( \begin{matrix}0 \\ 0 \end{matrix} )

where μ_{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}(∂H(ỹ)) is almost parallel to ∂H(ỹ). If p(x̃) ∉ ∂H(ỹ) then for large i, j, g_{j}g_{i}^{−1}(∂H(ỹ)) is much closer to p(x̃) than ∂H(ỹ). Hence g_{j}g_{i}^{−1}(H(ỹ)) ∪ H(ỹ) becomes a copy containing p(ỹ) which is a contradiction.

Step3: L(x̃) is an affine subspace. Choose a, b ∈ L(x̃). Then a = p(c̃) for some c ∈ H(x̃) such that H(x̃) = H(c̃). Then L(x̃) = L(c̃) by definition of L(x̃). Likewise for b. If y lies on the line \overleftrightarrow{ab} but y ∉ L(x̃) then ∂H(ỹ) passes through a and b by Step2. This says y lies in ∂H(ỹ) which is a contradiction.

Step4: We claim that for ỹ ∈ H̃(x̃), L(ỹ) = L(x̃). By Step2, ∂H(ỹ) passes through p(x̃) and p(x̃) ∈ L(ỹ). If some point a ∈ L(ỹ) lies in ℝ^{m}\\overline{H(x̃)} then H(x̃) contains a point in the line \overleftrightarrow{ap(x̃)} which lies in L(ỹ) by Step3. Since H(x̃)∩L(ỹ) = ∅, this cannot happen. So L(ỹ) ⊂ ∂H(x̃) and hence L(ỹ) ⊂ L(x̃). By Step2, x̃ lies in H̃(ỹ). Hence by the same argument, L(x̃) ⊂ L(ỹ). Thus L(x̃) = L(ỹ).

Since M̃ is connected, L(x̃) is independent of x̃ ∈ M̃. Denote this space by L. Then it follows easily that D becomes a covering map onto R^{m}\L and L is a limit set of D. □

Lemma 2. L is an affine subspace of dimension k.

Proof. Suppose that L is an affine subspace of dimension k + l, l > 0. We can assume that L = ℝ^{k} × (ℝ^{l} × (0, ⋯, 0)). Define a metric g_{F} on ℝ^{m}\L by

g_{F} = g_{R^{k}} + \frac{g_{R^{n}}}{d(x)}

where d(x) = x_{k+l+1}^{2}+⋯+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 ℝ^{m}\L defined by X(x) = (0, (0, x_{k+l+1}, ⋯, x_{k+n})). Then X is also Ginvariant vector field. Hence it defines a vector field on M. The volume form ω with respect to g_{F} is h(x)dx^{1} ∧ ⋯ ∧ dx^{m} where h(x) = d(x)^{− \frac{n}{2}} . We calculate div(X) from the equation L_{X}ω = div(X)ω. Note that L_{X}ω = dι_{X}ω + ι_{X}dω = dι_{X}ω.

ι_{X}ω = ∑_{i}(−1)^{i−1}h(x)dx^{1} ∧ ⋯ ∧ ι_{X}(dx^{i}) ∧ ⋯ ∧ dx^{m}. We see that

ι_{X}(dx^{i}) = \{ \begin{matrix}0, i ≤ k + l \\ x^{i}, i > k + l. \end{matrix}

dι_{X}ω = ∑_{i>k+l}(−1)^{i−1}d(h(x)x^{i})dx^{1} ∧ ⋯ ∧ d̂x^{i} ∧ ⋯ ∧ dx^{m}

= ∑_{i>k+l}(x^{i} \frac{∂h(x)}{∂x^{i}} + h(x))dx^{1} ∧ ⋯ ∧ dx^{n} = ∑_{i>k+l}(\frac{x^{i}}{h(x)} \frac{∂h(x)}{∂x^{i}} + 1)ω

= (−n + (n − l))ω = −lω.

Since M is a closed manifold, ∫_{M} div(X)ω = 0. But we compute that div(X) is nonzero constant which is a contradiction. □

We may assume that D is a covering map onto

ℝ^{m} − ℝ^{k} = {(x^{k}, x^{n}) ∈ ℝ^{n} × ℝ^{k} | x^{n} ≠ 0}.

Let

G = E(k) × (O(n) × A)

where A = {e^{t}I_{n} : t ∈ R}.

Proposition. Suppose π is a discrete subgroup of G acting on E^{k} × (E^{n} − 0) as a covering transformation, with compact quotient. Then π is virtually Z^{k} × Z, where Z^{k} acts on E^{k} as translations and Z acts on E^{n}−0 as dilations; both factors act diagonally.

If n ≠ 2, then the developing image D(M̃) 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 D̃ where D̃ maps M̃ to the universal covering space of ℝ^{k+2} − ℝ^{k} which can be identified with ℝ^{k} × ℍ^{2} where ℍ^{2} is an upper half subspace of ℝ^{2}.And this covering map is given by

\begin{matrix}p : ℝ^{k} × ℍ^{2} & → ℝ^{k+2} − ℝ^{k} \\ (x, (θ, r)) & ↦ (x, (r cos θ, r sin θ)) \end{matrix}

Then the element of E(k) × (SO(2) × A), ( α, λ ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) ) is lifted to

( α, ( ( \begin{matrix}1 & 0 \\ 0 & λ \end{matrix} ) , ( \begin{matrix}θ \\ 0 \end{matrix} ) ) )

which is an element of E(k + 1) × (SO(1) × A). Thus M is given a new geometric structure on which we can apply above proposition.

Conclusion. If M is an affine manifold with a partial dilation structure then M is finitely covered by T^{k} × S^{1} × S^{n−1}.