Nonnegatively Curved Alexandrov Spaces with Souls of Codimension Two

In this paper, we study a complete noncompact nonnegatively curved Alexandrov space $A$ with a soul $S$ of codimension two. We establish some structural results under additional regularity assumptions. As an application, we conclude that in this case Sharafutdinov retraction, $\pi:\ A\to S$, is a submetry.


Introduction
We begin with the classical Soul Theorem of ) on complete noncompact Riemannian manifolds of nonnegative sectional curvature: Theorem 0.1. Let M be a complete noncompact Riemannian n-manifold with sectional curvature sec(M) 0. Then M contains a compact totally geodesic submanifold S (called a soul of M) such that M is diffeomorphic to the normal bundle of S.
When sec(M) > 0, Gromoll and Meyer ([15]) earlier showed that a soul is a point, and thus M is diffeomorphic to R n . Cheeger and Gromoll proposed the following so called Soul Conjecture: If a complete noncompact nonnegatively curved Riemannian manifold has strictly positive sectional curvature around a point, then a soul is a point.
In 1994, Perelman ([26]) proved the following theorem which implies the Soul Conjecture: Theorem 0.2. Let M be a complete noncompact Riemannian n-manifold with sec(M) 0, and let S be a soul. If P : M → S is a distance nonincreasing map, then the following properties hold: (0.2.1) For any x ∈ S and any unit vector v at x normal to S, P (exp x (tv)) = x, for all t 0. In this paper, we will investigate the structure of a complete noncompact nonnegatively curved Alexandrov space which is topologically nice and whose a soul has codimension 2. A point in an Alexandrov space is called topologically nice if the iterated spaces of directions are all homeomorphic to spheres. An Alexandrov space is called topologically nice if all points on it are topologically nice. The limit space of a sequence of noncollapsed Riemannian manifolds with sectional curvature bounded from below uniformly is topologically nice.
We now begin to state the main results in this paper.
Theorem A. Let A be a complete noncompact nonnegatively curved Alexandrov space, and let π : A → S be the Sharafutdinov retraction. Suppose that A is topologically nice and that S is of codimension 2. Then π is a submetry.
We point out that the regularity assumption in Theorem A is used to classify the space of directions of points on S, which is crucial in the proof of Theorem A. Using Theorem A, one easily gets an affirmative answer to Open Problem (0.4.1) in the following 4-dimensional topological manifold case.
Corollary 0.6. Let A be a complete noncompact nonnegatively curved 4-dimensional Alexandrov space. Suppose A is a topological manifold. If A has positive curvature around a point, then a soul is a point.
We now explain the main ideas in the proof of Theorem A. We may assume that A is simply connected. (If A is not simply connected, one can pass to the universal cover, see Lemma 1.5.) For p ∈ S, and v ∈⇑ ∂Ωc p (all directions at p of minimizing geodesics from p to ∂Ω c ), where c is a fixed noncritical value of the Busemann function f (defined in Section 1.1) and ∂Ω c = f −1 (c), there is always a ray σ at p such that σ + (0) = v. We call such ray a special normal ray to S. Let F ⊆ A be the union of points on all such rays F = {x ∈ A | x ∈ σ : a ray with σ(0) = p ∈ S, σ + (0) = v ∈⇑ ∂Ωc p }.
Observe that in the special case F = A, Theorem A follows easily from Theorem 0.5 (see the proof following Lemma 1.1.).
If F = A, we set F v = ∪ {x | x ∈ flat totally geodesic strips in A spanned by σ and all minimal geodesics in S from p to all the points in S}.
case in Theorem A, then the proof of Theorem A is a little simpler), where the regularity conditions are required. Assuming Key Lemma 0.7, we can choose F = ∪ 1≤i≤l F v i , l ≤ 3 such that the distance function from F , dist F , is concave in A\F (see Lemma 2.8). For any given point x ∈ A\F , letx ∈ (S, a) ⊂ F u ⊂ F such that |xx|= |xF |. When a = 0, using the concavity of dist F , we can construct a "gradient flow" of dist F from (S, a) passing x, denoted by Ψ t a , which is distance nonincreasing (cf. [30]).
Consider the composition i•π•Ψ t a : (S, a) → (S, a), (where i : S → (S, a) is the natural isometry,) which is distance nonincreasing and a deformation, thus onto since t = 0 is onto. A standard argument shows that π| Ψ |xx| a ((S,a)) is an isometry. We denote Ψ |xx| a ((S, a)) by S x . When a = 0, we use a limit argument (see 2.4) to get a S x ∋ x, such that π| Sx is an isometry.
With the above preparations, we are ready to explain that the Sharafutdinov retraction π : A → S is a submetry. First π is distance nonincreasing (Theorem 0.3). For anyȳ ∈ S, it suffices to find y ∈ A such that |xy|= |π(x)ȳ| and π(y) =ȳ. Now it is clear that y = S x ∩ π −1 (ȳ) satisfies the desired condition.
Our argument can be viewed as a generalization of [37], where noncompact nonnegatively curved Alexandrov spaces with souls of codimension 1 are classified.
We organize the rest of the paper as follows: In Section 1, we will collect some basic notions and properties which will be used throughout the paper.
In Section 2, we will prove Theorem A by assuming Proposition 2.1. In Section 3, some applications are proved there.
In Section 4, we will prove some structural results for spaces of directions of points on S and verify Proposition 2.1 at the end.

Preliminaries
We start this section with fixing some notations: dist x (y) = |xy|: the distance between points x, y ∈ A Alex n (κ): the class of complete n-dimensional Alexandrov spaces with curvature ≥ κ ∂A: the boundary of A, A ∈ Alex m (κ) S n (κ): the n-space form of curvature κ B(p, r) = {x | |xp|≤ r} S(p, r) = {x | |xp|= r} F r(C): the union of points whose any neighborhood contains points in C and in the complement of C [pq]: a minimal geodesic from p to q, p, q ∈ A ↑ q p : a direction at p of a minimizing geodesic from p to q ⇑ q p : the set of all directions at p of minimizing geodesics from p to q (↑ x y , ↑ z y ): the angle between ↑ x y and ↑ z ỹ k (x, y, z): the corresponding comparison angle on space form S k Flat totally geodesic strip P in X, X ∈ Alex: P is the image of an isometric embedding from {(x, y) ∈ R 2 | 0 ≤ x ≤ 1, y ≥ 0} with the standard flat metric to X For basic notions related to Alexandrov spaces, we refer to [2], [3], [29], [31] and [35].
In the following, we shall briefly recall the construction of souls using Busemann function and the construction of the Sharafutdinov retractions. Then we shall establish some properties which will be used in our proof or which may not be found in literature.

Souls and Sharafutdinov retractions.
Throughout this paper, we say a subset C convex, if for any p, q in C, C contains at least one minimal geodesic joining p, q.
Let A ∈ Alex n (0) be noncompact, and let p ∈ A. The Busemann function at p is defined by and f is a proper concave function with definite maximum a 0 = max x∈A {f (x)}.
Then C 0 = f −1 (a 0 ) satisfies that for any two points, all minimal geodesics joining them are contained in C 0 (and thus C 0 is convex).
. Repeating the above process for C 1 , and after a finite number of steps we obtain C k = S, a convex subset without boundary.
Next we will recall the construction of a distance nonincreasing deformation retraction from A to S, the so-called Sharafutdinov retraction.
The goal of this subsection is to give the following unbounded version of Theorem 0.5, which is known to experts ( [42]). Since we can not find a complete proof in literature, for the convenience of readers, we include a proof here.
A useful alternative expression of f is: for any c < a 0 = max Lemma 1.1. Let A ∈ Alex n (0) be noncompact, and let f be a Busemann function. Then the following properties hold: can be extended to a ray γ, with γ(0) = p and |pγ(t)| = p∂Ω f (γ(t)) , for any t ≥ 0. Proof of Theorem A for the case that F = A. For any x ∈ A, we have that x ∈ γ: a special normal ray fromx. Hence π(x) =x (see Lemma (1.1.1). For anȳ y ∈ S, by Lemma (1.1.2), there is a flat totally geodesic strip determined by {γ, [xȳ]}; in which we can find y ∈ π −1 (ȳ), such that |xȳ|= |xy|.
In the proof of Lemma 1.1 we will use the following lemma. . Let Σ ∈ Alex n (1), and let C ⊂ Σ be a locally convex closed subset without boundary with positive dimension. If v ∈ Σ such that |vC|≥ π 2 , then |vξ|= π 2 , for any ξ ∈ C. We emphasize that Lemma 1.2 will be frequently used throughout the paper. Let X be an Alexandrov space. For p ∈ X, let T p X (or T p ) denote the tangent cone of X at p, and let Σ p X (or Σ p ) denote the space of directions of X at p.
] is a minimal geodesic with the desired property. Iterating this process, one can get the desired ray γ(t).
Note that flat totally geodesic strip in Lemma (1.1.2) may not be unique, see example [42, 14.8].
Remark 1.3. Inspecting the proof of Lemma 1.2, one can see that when ∂C = ∅ the following holds: Let x ∈ C be a point such that |vC|= |vx|. If x ∈ ∂C, then |vξ|= π 2 , for any ξ ∈ C. We will use this observation in Section 4.

A reduction.
The goal here is to reduce the proof of Theorem A to the simply connected case.
Let A ∈ Alex n (κ). Recall that p ∈ A is topologically regular, if Σ p A is homeomorphic to a sphere. A topologically regular point has a neighborhood homeomorphic to a Euclidean ball. A is called topologically regular, if all points are topologically regular, and thus A is a topological manifold.
A point p ∈ A is called topologically nice, if the iterated spaces of directions, i.e., Σ p A, Σ v 1 Σ p A, ... are all homeomorphic to spheres. A is called topologically nice, if all points are topologically nice. Topologically nice implies topologically regular, but the converse may not be true.
Proof. LetS be the universal cover of S. Denote the covering map by φ. Let φ * (A) = {(p, e) ∈S × A | φ(p) = π(e)} ⊂S × A, with the induced topology. Then by a standard argument, we obtain that φ * (A) is the universal cover of A, andφ : φ * (A) → A, defined byφ((p, e)) = e, is the covering map.
First we assume that C 0 = S.
Ω c , by the construction ofÃ, there exists a curve inΩ c with length ≤ |xy|, a contradiction. Thus we get the sublemma.
Letf = dist ∂Ωc . By the property of covering space,f ((p, e)) = |(p, e), ∂Ω c |= |e, ∂Ω c |= f (e)−c. It follows that ∂Ω c are level sets off . And by the local isometry, IfS is compact, by the assumption of the lemma, we can see thatπ is a submetry (sinceÃ is topologically nice). Hence π is also a submetry.
IfS is not compact, by the splitting theorem [23], there is an isometric splitting S = R k × S 0 , where S 0 is simply connected and compact, exactly as the proof of Riemannian case. It follows thatÃ , as can be seen in the proof of [32,Theorem D]. Thus by the assumption of the lemma, we know that π 0 : A 0 → S 0 is a submetry. It follows thatπ is a submetry.
Finally, we will verify the claim: for any x 0 ∈ A 0 and y = ( By the definition of gradient the claim follows. If C 0 = S, consider dist ∂Ωa 0 instead of f , we can get the same conclusion. Remark 1.7. If A is only topologically regular, then A 0 may not be a topological manifold, evenÃ = R k × A 0 is a topological manifold ( [21]).

Proof of Theorem A
In our proof of Theorem A, the following structural results on spaces of directions of points on soul plays a curial role. Proposition 2.1. Let the assumptions be as in Theorem A. For p ∈ S, let Σ p 0 = Σ p S, and let Σ p 1 is convex 1 and isometric to one of the following: Because the proof of Proposition 2.1 is technical and long, we will postpone the proof to the next section. Below we shall prove Theorem A by assuming Proposition 2.1.

Proof of Key Lemma 0.7.
Recall that F v = ∪ {x | x ∈ flat totally geodesic strips in A spanned by σ and all minimal geodesics in S from p to all the points in S}.
, Key Lemma 0.7 holds. To prove the independency, we will first show that φ [pq] is an isometry. (We point out that the method we used in the proof of Lemma 2.2 below was previously used in [29].) Lemma 2.2. Let the assumptions be as in Theorem A. For every x ∈ S and every y ∈ S, φ [xy] : ⇑ ∂Ωc x →⇑ ∂Ωc y , is an isometry, for any minimal geodesic [xy].
Proof. For u, v ∈⇑ ∂Ωc x , and ε > 0, letv ε = |y, exp x (εv)|↑ Since by Proposition (2.1.3) and by the property of flat totally geodesic strip, Then |ūv| Ty ≥ |uv| Tx , and |ū|= |u|, |v|= |v|. Thus we get that |φ [xy] Similarly, the opposite inequality holds. Hence Remark 2.3. It seems that Lemma 2.2 can be strengthened to that the isometric class of Σ p 1 is independent of p. Proof of Key Lemma 0.7. Define a map, ψ : In view of the simply connectedness of S (because A is simply connected), first we will show that F v is a product locally, it suffices to show that locally φ [pq] (v) is independent of q ∈ S (all p). Precisely, for x ∈ S, there exists ε > 0 (ε depends on x), such that for any y, z ∈ B(x, ε), Thus each point in S has just one special normal ray to S, clearly g = id.
For other cases we will argue by contradiction. Suppose that for a sequence and by the property of flat totally geodesic strips, we have that |exp If Σ x 1 = S 1 (r) with r ≤ 1, by Lemma 2.4 below, we have that every g i is the restriction of an isometry,ḡ i : S 1 → S 1 , which is a rotation or a reflection. By passing to a subsequence, we can suppose that everyḡ i is a rotation or everyḡ i is a reflection.
(a): Everyḡ i is a rotation.
For v ∈⇑ ∂Ωc x , |exp x (tv) exp x (g i (tv))|→ 0, for any t ≥ 0, as ε i → 0, which can be seen in the above case. Hence |vg i (v)|→ 0, i.e. |vḡ i (v)|→ 0, as ε i → 0. Then by the closeness of ⇑ ∂Ωc x , we can get that ⇑ ∂Ωc x = S 1 . This is a contradiction, since F = A and Lemma 2.2 imply that ⇑ ∂Ωc q = S 1 , for any q ∈ S. (b): Everyḡ i is a reflection. By passing to a subsequence, we can assume thatḡ i → h, which is also a reflection. Observe that there is If Σ x 1 = [ab], likewise by Lemma 2.4, each g i just can be the restriction of the reflection of [ab]. Similarly as above, we can get the conclusion. Then Finally, we will show that F v is a product globally. For any q, r ∈ S, and for three fixed geodesics Let {s 0 = 0 < s 1 < · · · < s n = 1} and {t 0 = 0 < t 1 < · · · < t n = 1} be two partitions of There are corresponding g i . We can see that g i (w), for any w ∈⇑ ∂Ωc x , are the same for any i. It follows that g 1 (w) = w, i.e., If there exist v, w ∈ N such that v, w are not antipodal, then for any u ∈ M, u is uniquely determined by |uv|, |uw|. Thus g is uniquely determined by g(v), g(w). Hence g can be extended toḡ, byḡ(u) = x, where x is the unique point such that |xg(v)|= |uv| and |xg(w)|= |uw|.
If not, then N = {v, w} with v, w antipodal. Clearly g is extendable.
If x, y ∈ (S, a), with a = 0, then the same as above we have that (↑ y x , ↑x x ) = (y, x,x) = π 2 . Therefore there exists a flat triangle bounded by y, x,x for the given [xy], with [yx] • F . By the above case, we get a contradiction.
If x ∈ (S, a), y ∈ (S, b), with a = 0, b = 0 and a = b, without loss of generality, we can assume that a < b. Let s = [yȳ] ∩ (S, a). Similarly as the above two cases, we can also get a contradiction.
As seen following Lemma 1.2, the remaining case in the proof of Theorem A is that F = A and π 1 (A) = 0, which implies that ⇑ ∂Ωc p = S 1 . 2.2. The concavity of dist F .
As seen in the introduction, F is the union of several F v 's. We point it out that the selection of these F v 's is crucial for the desired concavity of dist F ; see following for details.
For p ∈ S, by the first variation formula for the Busemann function, be the farthest points to w from both sides respectively in Σ p For the second part of the lemma: if Σ p 1 = {v} or {v 1 , v 2 }, by Lemma 2.2 one can deduce that N ′ has the desired property.
Thus we finish the proof of the lemma.
Remark 2.9. Observe that if the boundary points of each component D i are "true" boundary points, i.e., which are not interior points in the closureD i and thatD i is convex, then it follows that dist F is concave in D. In our case, we show that even if a component of D may not be convex, dist F is still concave. For example: serves an example.
In the proof of Lemma 2.8, we need the following lemma which is an analogue to the totally geodesic property in Riemannian geometry. (Lemma 2.10 below is from a helpful discussion with Shicheng Xu.) Lemma 2.10. Let X ∈ Alex n (κ), and let Y be a closed subset of X such that for x, y ∈ Y , any [xy] ⊂ Y . Then for any p ∈ Y and q ∈ Y \∂Y , we have that We don't know whether Lemma 2.10 is true for convex subset or not.
Proof of Lemma 2.10. If ∇ q dist p = 0, nothing need to prove. Hence we can assume ∇ q dist p = 0. Since ∇qdistp |∇qdistp| = {u ∈ Σ q X | |u ⇑ p q |= max v∈Σq {|v ⇑ p q |}, and by the condition of the lemma, we have that ⇑ p q ⊂ Σ q Y , it suffices to show that for V ⊂ Σ q Y , let w ∈ Σ q X be a point such that |V w|= max{|V, |} > π 2 , then we have that w ∈ Σ q Y .
Recall a standard fact in topology (cf. [5]): If X ⊂ S m is a closed (m − 1)topological manifold as a subspace, then S m − X has two connected components, each having X as its set boundary. We say that X separates S m .
Proof of Lemma 2.8. Given q ∈ D, let γ(t) ⊂ D be a minimal geodesic with γ(0) = q, and let x ∈ F be a point such that |qx|= |qF |. By a standard contradiction argument, one can get that for v =↑ From the proof of the concavity of distance function to the boundary of an Alexandrov space (cf. [30, Theorem 3.3.1], [9, Lemma 3.1]), one can deduce that if F satisfies the following two conditions: Thus it suffices to check that F satisfies the two conditions. For condition (i): if x ∈ S, by the first variation formula, we have that |↑ q x Σ x F |≥ π 2 . Specially, |↑ q x Σ x S|≥ π 2 , thus ↑ q x ∈ Σ x 1 . Hence Σ x 1 = S 1 or [ab] with |ab|= π and ↑ q x the middle point of [ab]. And by Lemma 2.7, we have that for any v ∈ Σ x , |vφ [px] (N ′ )|≤ π 2 . Hence there are two of N ′ say v 1 , v 2 (v 1 may be equal to v 2 ) such that |φ [px] ) is isometric to a triangle with three side lengths π 2 on S 2 (1). Hence there isw ∈ Σ x F such that |↑ q xw |= |↑ q x v|+|vw|= π 2 . If x∈S, Σ x F = S(Σ x S) homeo ≃ S n−2 is convex in Σ x A and separates Σ x A, and by the first variation formula, |↑ q x Σ x F |≥ π 2 . Then by Lemma 4.19 below, we have that there isw ∈ Σ x F such that |↑ q xw |= |↑ q x v|+|vw|= π 2 . For condition (ii): sincew ∈ Σ x F , without loss of generality, we can assume x →w. Let σ i be the radial curve at q i with respect to x. By [1, Chapter 15] or [27, 3.4], we know that if we can show that lim i→∞ σ i ([0, ε]) = σ([0, ε]) ⊂ F v 1 for some small ε > 0, then σ is the desired radial curve.
If x∈S, by Lemma 2.10, we can get the desired radial curve. If x ∈ S and w ∈ Σ x S, similarly by Lemma 2.10, we can get the desired radial curve in S. If 13 x ∈ S andw ∈ Σ x S, we can choose q i ∈ F v 1 − S. We claim that σ i ([0, ∞)) ⊂ F v 1 . Therefore we can get the desired radial curve.
Finally, we will verify the claim by showing that σ i are more and more farther away from ∂F v 1 . The reason is that ( , |xσ i (t)| t ∇ σ i (t) dist x ≥ 0, the last inequality is because of the symmetry of F locally, we have that ⇑ x q are in the same half sphere as ↑ π(q) q =↑ ∂Fv 1 q in Σ q F , for any q ∈ F .
The lemma thus follows.
Since dist F is concave in D = A\F , for each x ∈ D there is a unique dist Fgradient curve from x. We call a gradient curve maximal if it is not a proper subset of another gradient curve. Note that any maximal gradient curve has empty intersection with F . We will extend maximal gradient curves to include points in F so that each point in F − (S, 0) is contained in two extended maximal gradient curve. This property plus the simply connectedness of A allow us to choose one such curve for each point in F − (S, 0), such that we can define a "flow", Ψ t a : (S, a > 0) → A, by Ψ t a ((s, a)) = γ a (t), where γ is the chosen extended maximal gradient curve at (s, a), passing any given extended maximal gradient curve at any given (s 0 , a). Our goal is to show that Ψ t a is 1-Lipschitz. To carry out the above, the key is to establish the local separation property for F − (S, 0) (see Lemma 2.11) and the local 1-Lipschitz property for Ψ t a . Before moving on, we need the following two lemmas.
Lemma 2.11. For any q ∈ (S, a) ⊂ F u , a = 0, B F (q, r) separates B(q, r), for r small enough, where B F (q, r) is a closed r-ball in F .
Proof. For r small enough, we can assume that B(q, r)∩F = B(q, r)∩F v 1 . By the local version of Perelman's stability theorem ([24, 4.7]) and Proposition (2.1.1), we can choose r sufficiently small, such that B(q, r) is homeomorphic to a r-ball on T q A, which is homeomorphic to D n , and B F (q, r) is homeomorphic to a r-ball on T q F , which is homeomorphic to D n−1 . By definition, B F (q, r) ∩ F r(B(q, r)) = (F r(B F (q, r)) in F ), which is homeomorphic to S n−2 . By considering the double of B(q, r), we get that B F (q, r) separates B(q, r).
Let q, r be as in Lemma 2.11, and letŪ q ⊂ B(q, r) be a convex closed neighborhood of q. Then B F (q, r) separatesŪ q into two components G q1 , G q2 .
Since dist F is concave in D, for dist F -gradient curves α(t), β(t), we have that |α(t)β(t)| is 1-Lipschitz if there exists a minimal geodesic joining α(t) and β(t), for any t, in the domain D. The following property will guarantee the local 1-Lipschitz property for Ψ t a .
Lemma 2.12. For x, y in the interior of the same component say G q1 \∂Ū q (denoted by G • b1 ), we have that [xy] ∩ F = ∅.
In the proof of Lemma 2.12, we need the following theorem about the relationship between the boundary of a convex subset as an Alexandrov space and the set boundary.
Proof of Lemma 2.12. Based on the local separation and the convexity of F v , it is easy to check thatḠ qi , i = 1, 2 are convex. By Theorem 2.13, we have that F ∩Ḡ qi ⊂ ∂Ḡ qi . Hence [xy] ∩ F = ∅, because any minimal geodesic between interior points of a convex set does not intersect the boundary of the convex set ( [24, 5.2]).
We are now in a position for the construction of a nonexpanding map. For x ∈ D, let q ∈ (S, a) ⊂ F u , u ∈ N ′ , be a point such that |xq|= |xF |. First suppose a = 0. We will construct extended maximal gradient curves from (S, a). For is the standard dist F -gradient flow defined in [30, 2.2]. Let P m = {0 = s 0 < s 1 < · · · < s m = l, s i − s i−1 = l m } be a partition of [0, l], with m large enough, such where Length() denotes the length of the curve. If t > ε 4 , we can repeat the procedure for [σ ε 4 (s i )σ ε 4 (s i−1 )]. Finally, we get that for any For simplicity, in the following context, we will say that γ b and γ c are in the same component with respect to U b , and we will call γ b an extended maximal gradient curve, if γ b − b is maximal.
By now we can see that for each point on (S, a) there exist exactly two extended maximal gradient curves. Choose a point y ∈ (S, a), denote the two extended maximal gradient curves by γ y1 , γ y0 . For any other point say z, we will denote the two extended maximal gradient curves from z by γ z1 , γ z0 , such that γ zi are a continuation of γ yi along [yz], i.e., there is a partition of [yz], P = {y 0 = y, y 1 , · · · , y k = z}, such that γ y j ,i , γ y j+1 ,i are in the same component with respect to U y j . It doesn't depend on the choice of the partition, since for another partition P 1 , we can consider P ∪ P 1 to get the independency.
Since π 1 ((S, a)) = 0, similarly as the final part of the proof of Key Lemma 0.7, we can see that the denoting doesn't depend on the choice of [yz]. Thus we can define two "flows" Ψ t ia : (S, a) → A, by b → γ bi (t), b ∈ (S, a), and we fix one passing x, which exists, since [xq] is contained in an extended maximal gradient curve, denoted by Ψ t a : (S, a) → A. Lemma 2.14. The map Ψ t a : (S, a) → A is 1-Lipschitz. Proof. For b, c ∈ (S, a), we need to show that |bc|≥ |γ b (t)γ c (t)|. Dividing [bc] into small pieces so that the above local 1-Lipschitz property holds, one gets that |bc|≥ Length(Ψ t a ([bc])) ≥ |Ψ t a (b)Ψ t a (c)|.  l) is onto. Recall that given two 1-Lipschitz onto maps between two compact metric spaces, g : X → Y and h : Y → X, then g and h are isometries. Since Ψ t a and π are 1-Lipschitz, the desired result follows. If a = 0, consider Σ q A, from the proof of Lemma 2.8, we get that there are [pq] (u) , where a i = 0, and x i → q, y i → q, as i → ∞.
, is in the same component (defined after Lemma 2.11) as [y ′ iȳ i ] ⊂ [x iȳi ] with y ′ i ,ȳ i close enough, and let γ i (t) = Ψ t b i (ȳ i ). Then by the construction of γ i (t), one obtains that |γ i (t), [x i x](t)|≤ |ȳ i x i |, thus γ i (t) → [xq](t), for any 0 ≤ t ≤ |xq|, as i → ∞. Hence by passing to a subsequence we can suppose that Ψ Proof of Claim 2.16. It suffices to show that if z ∈ [x i y i ] ∩ F , then z ∈ S. Argue by contradiction, suppose z ∈ S. Since |x i q|≤ |x i z| and |y i q|≤ |y i z|, we have that [x i q] ∪ [qy i ] is a minimal geodesic, which is impossible, because by the choice of y i we have that ↑ x i q ⊥↑ y i q . Now we are ready for the proof of Theorem A.

Application
We will prove Corollary 0.6, and we will show that in Theorem A if F = A, then π is a bundle map. First we recall the following lemma.
(3) A is a topological manifold.
Lemma 3.1 is pointed out by Kapovitch in [21], for completeness we include a proof here.
In the proof, we shall apply the following theorem which is a sufficient condition for a bundle map: 33]). Let X, X 0 be two metric spaces and let f : X → X 0 be a continuous onto map. Suppose that for any p ∈ X 0 and ε > 0 there exits a δ(p, ε) > 0 such that for any q ∈ B(p, δ), there is a homeomorphism h : f −1 (p) → f −1 (q), with |h(x)x|≤ ε, for any x ∈ f −1 (p). If the fiber F is locally compact and separable, and the homeomorphism group of F (with some natural topology) is locally path connected, then F is a Serre fibration. If in addition, X 0 is finite dimensional ANR, then f is a locally trivial bundle map.
Since A is topologically nice and π is a submetry, if p ∈ S is a regular point (i.e., T p S is isometric to R n−2 ), then π −1 (p) is a topological manifold ( [32,Theorem D], note that the proof is local, and thus apply to non-compact cases). Since π : π −1 (p) → p is a deformation retraction, π −1 (p) homeo ≃ R 2 . And by [11, 7.3], we know that the homeomorphism group homeo(π −1 (p)) is locally path connected with the topology in Theorem 3.3. By now we are able to apply Theorem 3.3 to conclude that π is a bundle map.

Structure of Space of Directions
Our main efforts in this section is to prove Proposition 2.1, and thus complete the proof of Theorem A. We point out that Theorem 4.4, which classifies certain isometric class in Alex n (1), may have independent interest.
(2) If a subspace with restricted metric is isometric to an Alexandrov space, we also say that it is convex, although when the dimension is 0, there may not be a minimal geodesic in the subspace joining two given points.
(3) When we say that two metric spaces are equal, we always mean metrically, except otherwise stated. If there is no confusion, we will not mention the metric.
Let A ∈ Alex n (1) and let C ⊂ A be a closed convex subset without boundary. LetΣ p C = {v ∈ Σ p A | |vΣ p C|≥ π 2 }, and letT p C = C(Σ p C). Observe that if A and C are Riemannian manifolds, then T p A isometrically splits, i.e., T p A = T p C ×T p C or equivalently Σ p A is isometric to the join of Σ p C andΣ p C ( [3]). In Alexandrov geometry, such property doesn't hold.
Conjecture 4.5. Let A ∈ Alex n (0) be noncompact with a soul S. If p ∈ (C i − ∂C i ) is a topologically regular point, then T p X is isometric to the product T p C i × K, where C i are as in 1.1, and K is a Euclidean cone.
Using Theorem 4.4, with additional argument we can get the following classification result, which can imply Proposition 2.1.
Next we will show that g is nonexpanding. In order to do so we need the following fact: For p ∈ C 1 and q ∈ C 0 , |pq|= |pΣ 0 |. By Lemma 1.2, |↑ p q v|= π 2 , for any v ∈ Σ q Σ 0 . Hence˜ (p, q, r) = (p, q, r) = π 2 , for any r ∈ Σ 0 . By Lemma 2.6, there is a convex triangle isometric to the corresponding one on space form.
Finally, we will verify the claim. It suffices to show that if x 1 = x 2 , y 1 = y 2 , then [x 1 y 1 ] ∩ [x 2 y 2 ] = ∅. Suppose that [x 1 y 1 ] ∩ [x 2 y 2 ] = z. Hence by the above fact, [zy 1 ] ⊂ the totally geodesic triangle bounded by {x 2 , y 1 , y 2 }, which is isometric to the corresponding triangle on S 2 (1), or the geodesic will branch. Thus (x 1 , y 1 , y 2 ) < π 2 . This is a contradiction. Remark 4.8. Let Σ ∈ Alex n (1), and let Σ 0 ⊂ Σ be a convex closed subset. Sketch of the proof: we will prove Theorem 4.4 by induction on the dimension of Σ. And we will prove the inductive step according to the different situations of the dimension of Σ 0 and Σ 1 . Except two simple cases, we will first show that for each point in Σ 1 and each point in Σ 0 , there exist exactly m minimal geodesics joining them for some m > 0. Then argue by contradiction, we can get that m = 1, using this we can derive that Σ is a join.
Proof of Theorem 4.4. Denote dim(Σ 1 ) by I and dim(Σ 0 ) by J. First we will prove the following property. Sublemma 4.9. We have that Σ 1 is convex.
For the following use, we recall Alexander duality [19]: Let K ⊂ S n be a compact, locally contractible, nonempty, proper subspace. Then H i (S n − K; Z) = H n−i−1 (K; Z), for all i, where H i , H i denote the reduced homology and cohomology.
Hence in the following we can suppose that I > 0. First we will show that for any x i ∈ Σ i , i = 0, 1, there are m minimal geodesics joining x 0 and x 1 , for some m > 0.
Hence Σ 0 homeo ≃ S J /Γ 0 . Similarly, we get that for any p ∈ Σ 0 , ψ 1 :⇑ Σ 1 p → Σ 1 is an m-fold covering map, and therefore Σ 1 homeo ≃ S I /Γ 1 . By now we can see that for any x i ∈ Σ i , there are m minimal geodesics joining x 0 and x 1 .