Homeomorphic Arrangements of Smooth Manifolds

Symmetry between mathematical constructions is a very desired phenomena in mathematics in general, and in algebraic geometry in particular. For line arrangements, symmetry between topological characterizations and the combinatorics of the arrangement has often been studied, and the first counterexample where symmetry breaks is in dimension 13. In the first part of this paper, we shall prove that two arrangements of smooth compact manifolds of any dimension that are connected through smooth functions are homeomorphic. In the second part, we prove this in the affine case in dimension 4.


Introduction
An arrangement is a finite collection of affine subspaces (of possibly varying dimensions) in R l or C l , or a collection of linear subspaces of a projective space CP l−1 or RP l−1 . The topology of the complement of the union of the planes is of considerable interest. In 1989, Randell [1] proved a deep theorem, which shows that two arrangements have the same topology if they can be transferred from one to the other using a smooth one parameter family of arrangements. This resulted in the invention of a new invariant-the moduli space. Randell's study was very fruitful and resulted in many important theorems concerning symmetry rope. For instance, in [2] it was implemented for lines which in this case are a linear subspace of C 2 , for example, the solution in C 2 to the equation y = (5 + i)x + 2 + 4i. In [3,4], it was proven that the combinatorial structure determines the fundamental group of the complement for a six line arrangement. Using Van Kampen theorems [5] and the Moishezon-Teicher algorithm [6], it was extended to seven and eight lines [7,8]. Later, it was generalized to nine lines in [9], ten lines in [10] and eleven lines in [11].
In this paper, we are going to improve Randell's theorem and we move to diffeomorphisms of smooth manifolds in general, a contemporary topic (see, e.g., [12,13]). Whereas Randell talks about changing hyperplane arrangements through smooth families of hyperplanes, we are going to show that we can also transfer from one arrangement of smooth manifolds to another using smooth families of hyperplane arrangements. As a consequence, we show that we can transform arrangements of lines to other arrangements of lines through symmetry of arrangements to polynomials, families of polynomials of any degree. Since lines in the complex planes are homeomorphic to two dimensional sets in R 4 , our theorems will use that idea for another improvement of the theorem.

Definitions and Notations
The following theorems and definitions are well known.
Theorem 1 (Topological invariance of dimension [14]). A non-empty n-dimensional topological manifold cannot be homeomorphic to an m-dimensional manifold unless m = n. Definition 1 ([14]). For any smooth manifold M, we define an open submanifold of M to be any open subset with the subspace topology and with the smooth charts obtained by restricting those of M.
Theorem 2 (Open sub-manifold [14]). Suppose M is a smooth manifold. The embedded submanifolds of codimension 0 in M are exactly the open submanifolds. Proposition 1 ([14]). Suppose M and N are smooth manifolds with or without boundary, and F : M → N is an injective smooth immersion. If any of the following holds, then F is a smooth embedding: F is an open or closed map; M has empty boundary and dim M = dim N.
Theorem 3 (Proper continuous maps are closed [14]). Suppose X is a topological space and Y is a locally compact Hausdorff space. Then every proper continuous map F : X → Y is closed.

Definition 2.
A stratification of a manifold is a partitioning of the manifold into a finite collection of submanifolds {U} (called the strata) so that the following frontier condition is satisfied: Whenever U and V are strata with V ∩ cl(U) = ∅, then V ⊂ cl(U).

Definition 3 ([14]).
A stratification is called a Whitney stratification if it satisfies Whitney's condition (b): For all strata U, V, with V ∩ cl(U) = ∅, and for all x ∈ V, whenever x i and y j are sequences in V and U, respectively, with x i = y i so that x i converges to x and y i converges to x, so that the secants x i y i converge to l ∈ RP n−1 and so that T y i U converges to τ in the Grassmannian of dimension U planes in R n , then l ⊂ τ.

Theorem 4 ([1]
). Any stratification in which the closure of every stratum is a smooth submanifold is a Whitney stratification.
Theorem 5 (Thom's first isotopy theorem [15]). Let f : M → R be a proper, smooth map which is a submersion on each stratum of a Whitney stratification of M. Then there is a stratum-preserving homeomorphism h : M → R × ( f −1 (0) ∩ M) which is smooth on each stratum and commutes with the projection to R. In particular, the fibers of f are homeomorphic by a stratum-preserving homeomorphism.

Topological and Geometric Aspects of Manifolds
With the previous theorems in mind we prove the following. Lemma 1. Let A, B and C 1 , . . . , C n be manifolds such that B is a closed submanifold of A and C 1 , . . . , C n are closed submanifolds of B. Assume that for all Proof. First we prove that for any 1 ≤ i ≤ n, C i is nowhere dense. Indeed, since C i is closed it is sufficient to show that it has an empty interior. Assume to the contrary that S is an open set of B contained in C i , then from the definition of subspace topology it is an open set also in C i by Theorem 2. We get that S is a manifold of dimension equal to the dimension of B and also a manifold with a dimension equal to C i . The contradiction to Theorem 1 proves the statement. Since a finite union is nowhere dense so is n i=1 C i and since ) then h is continuous and proper.
Proof. Let p be the projection of C × D to D and let S be some set. Then, an element open. Next we will prove that h is proper: let S be a compact set. In particular it is closed p is continuous, P(S) is compact and since h is proper g −1 p(S) is also compact. Since A and g −1 p(S) are compact so is A × g −1 p(S). To conclude, we get that h −1 (S) is closed and a subset of compact subset and, therefore, compact.

Proposition 3.
Let Q and M be smooth manifolds and f : Q × R → M a smooth function. For any t ∈ R, let f t : Q → M be the function sending q to f (q, t). Let F : if f is smooth and for any t ∈ R f t is an immersion then F is an immersion; 4.
if f is smooth and dim(Q) = 0 then F is an immersion; 5.
if Q is compact,f is smooth, for any t ∈ R f t is injective and if dim(Q) > 0 it is also an immersion then F is an embedding with a closed image; 6.
if F is an embedding and π is the projection of M × R to the second factor then π| Im(F) is a submersion.
. Therefore, t 1 = t 2 and we denote it as t, so we get that f (q 1 , t) = f (q 2 , t). By the definition of f t this implies that which is equal to Let us assume that M is of order k. Then . . , f k which are all smooth. So we can write F as k + 1 components f 1 , . . . , f k , p where p is the projection from Q × R to R. We can see that all the components are smooth and, therefore, F is smooth. 3.
Let p be a point as in the previous paragraph and that . , x m , t when we assume that the order of Q is m. Let 1 ≤ i ≤ k and 1 ≤ j ≤ m, then the partial derivative of f i in the coordinate x j is equal to We can see that the partial derivative of f i in the coordinate x j is equal to the partial derivative of f t i in the coordinate x j . The x j partial derivative of t is 0 and the t-th derivative of t is 1. To conclude, the Jacobian of F in the point p is Since dim(Q) = 0, it is a union of points with the discrete topology and every point has a homeomorphism to R 0 which is precisely one point by a function that sends one point to the other. So if we take one point a × R and the homeomorphism to R as the natural one, then we have that one function f (t) = z where z is the variable of R in M × R which is obviously an immersion. 5.
By paragraphs (1)-(4) F is injective immersion. Since the identity is obviously proper, then by Lemma 3 F is proper. So by Proposition 1 it is a smooth embedding. By Theorem 3 it is closed; in particular, it has a closed image. 6.
It is sufficient to prove that F • π is a submersion but F • π is a projection of the second factor of Q × R which is known to be a submersion.

Main Theorem
In order to prove the main Theorem 4, we need the following lemma. h since by our assumption every h that is strictly contained in h 1 has a smaller dimension by Lemma 1 Cl(w) = h 1 . Similarly, since for every h ∈ H dim(h) < dim(M) and M = M \ ∪H. Then by Lemma 1 Cl(M) = M. Assume now that we have two elements w 1 , w 2 such that Cl(w 1 ) ∩ w 2 = ∅. Then, by our last statement, Cl(w 1 ) = h 1 for some h 1 ∈ H and h 2 , and, therefore, h 1 ∩ h 2 ∩ w 2 = ∅. Since w 2 ⊆ h 2 we get that h 2 ∩ w 2 = w 2 and, therefore, h 1 ∩ w 2 = ∅. This contradiction forces us to say that h 1 ∩ h 2 = h 2 which implies that h 2 ⊆ h 1 and, therefore, w 2 ⊆ h 1 . Since Cl(M) = M, it is obvious that every element in W is a subset of cl(M). To conclude, we get that W is a stratification. Since for every w ∈ W there is h ∈ H such that Cl(w) = h and every h ∈ H is a smooth submanifold and the closure of M = M, which is also a smooth manifold, by Theorem 4 W is also a Whitney stratification. Since submersion is a local property and s|h and are submersions, this is also true for their open subspaces, namely the elements in W.

Proposition 4.
Let M be a compact smooth manifold, J a finite set. For every j ∈ J, Q j is a compact smooth manifold and g j : Q j × R → M is a smooth function. For every j ∈ J and t ∈ R, let g t j : Q j → M be the function that sends q to g j (q, t). f : J × J → P(J) is a function such that: 1.
for any (i, j) ∈ J × J and for all t ∈ R Im(g t i ) ∩ Im(g t j ) = k∈ f ((i,j)) Im(g t k ); 2. for all j ∈ J and t ∈ R g t j is injective; 3.
for all j ∈ J and t ∈ Proof. We define for any j ∈ J, G j : Q j × R → M × R by G j (x, t) := (g t j (x), t). Since g j is smooth, Q j is compact. For all t ∈ R, g t j is injective and if dim(Q j ) > 0 then it is also an immersion. Then, by Proposition 3 G j is an embedding with a closed image, and for π equal to the projection of M × R to the second factor π| Im(G j ) is a submersion. Now we are going to show that for any (i, j) ∈ J × J, (1) . We would like to show that dim(Im(G i )) < dim(Im(G i )). We know that (x, t) is in Im(G i ) if and only if x ∈ g t j and the same is true for j. So if (x, t) ∈ Im(G j ) \ Im(G i ), then x ∈ Im(g t i ) \ Im(g t j ) and for this specific t if x ∈ g t i then (x, t) ∈ Im(G i ) which implies that (x, t) ∈ Im(G i ) and, therefore, x ∈ g t t . Combining these two facts, we get that Im(g t i ) Im(g t j ) which implies by our condition that dim(Q j ) < dim(Q l ) which imply that dim(Q j × R) < dim(Q l × R) which means that dim(Im(G j )) < dim(Im(G l )), as needed. Let H := Im(G j ) | j ∈ J . Then H is a finite subset of closed smooth manifolds. For all h 1 , h 2 , there exist H 1 such that h 1 ∩ h 2 = ∪H, for all h 1 , h 2 ∈ H such that h 1 h 2 dim(h 1 ) < dim(h 2 ) and for all h ∈ H dim(h) < dim(N).
We denote W = {h \ Then by Lemma 2 W is a Whitney stratification and for all w ∈ W, π| w is a submersion, and since M is compact. π is proper on M × R. Therefore, by Theorem 5, there is a homeomorphism from π −1 (0) to π −1 (1) which is a stratum-preserving homeomorphism, so we can see that π −1 (0) = M. If h ∈ H then there exist k ∈ P(I) \ ∅ such that h = H(k) In the same way, π −1 (0) ∩ N = M \ i∈I h i 0 and, therefore, they are homeomorphic.

Adaptation of the Main Theorem to Curves
This section is a corollary of the main theorem in the case of curves. We are going to use the following definition for simplification. Definition 4. Let F : X × R → Y be a function. We define for every t ∈ R, F t : X → Y by F t (x) = F(x, t). Then we say that F satisfies condition 1 if it satisfies the following conditions: 1.
for every t ∈ R, F t is injective; 3.
if dim(X) > 0 then for every t ∈ R, F t is injective.
, such that H 1 , H 2 are polynomials of the variables x, y over the smooth function with one variable, k = max{deg(H 1 ), deg(H 2 ))} and (p, t) sends to p × p. Then F satisfies condition 1.

Proof. Let
such that for all i, j and k = 1, 2, h k ij is a smooth function. First, we prove that F is a smooth function. Let a ∈ S 2 \ p, then we choose for the domain the chart ϕ 1 × Id and for the image ϕ × ϕ, so we need to prove that is smooth. Indeed we get that It is easy to see that "g" is smooth. For the points (p, t), we take the domain chart to be ψ 2 × Id and the image chart to be ψ 2 × ψ 2 . So we need to prove that ( Thus, ψ 2 is injective and ψ 2 (p 1 ) = (0, 0). So if (x, t) ∈ R 2 \ (0, 0) × R and we denote z := x 2 + y 2 , we get that ((ψ 1 ) − 1 • (ϕ 1 ) × Id)(x, y, t) = ( x z , y z , t). If we apply g on the result we get: Im(κ i (0)) .
Proof. Let φ : R 4 → R 4 be the homeomorphism sending (x, y, z, w) to (x, y, z, w + ( To a i we attach the manifold S 2 and the function G i : S 2 × R → S 2 × S 2 which we defined piecewise on S 2 \ p. It will be defined as (ϕ × Id) • (ϕ × Id) • g i • φ • (ϕ × ϕ) −1 and p × R will be sent to p × p. By Theorem 6 they satisfy condition (1). For b i we attach the manifold S 2 and the functions β i : S 2 × R → S 2 × S 2 are defined as follows: β 1 (a, t) = (a, p) and β 2 (a, t) = (p, a). It is easy to see that these functions satisfy condition (1). For c and d i we attach a manifold with a single point which we denote by Pt. We define the following function: for c we define the constant function P which sends every element to p × p and for d i we define the function as follows: To complete the definition, we add thatf (X, X) := X and iff (Y, X) is defined theñ f (X, Y) =f (Y, X). Now we would like to prove that for any (i, j) ∈ J × J and for any t ∈ R if we denote the function attached to k as r k we get that r t i ∩ r t j = k∈ f ((i,j)) r t k .

1.
For every k and t ∈ R the image of G t k is equal to G k (S 2 × t), the image of p × t is p × p and the image of Since ϕ is surjective on R 2 , this is equal to φ × ID(g k • (ϕ × ϕ) −1 (R 2 × t)). So the intersection of the images of G t i and G t j is equal to a p × p union with .

2.
As in paragraph (1) Im Since p × p is also in b j and the image of ϕ × ϕ is S 2 \ p × S 2 \ p, the intersection is precisely p × p.

3.
As in paragraph (1) Im First, if there exists j we already proved that (G i (t)) ∩ (G j (t)) = κ r (t) and κ r (t) contain only one point, then G i (t) ∩ κ r (t) = κ r (t). Next if j do not exist then as in ). By our assumption, this is equal to the empty set. Applying (ϕ × ϕ) −1 • (φ × Id) will give us the desired conclusion. 5.
Trivial. 7. Since Same proof as paragraph (7). 9. Since is also empty. It is easy to see that for every j ∈ J the dimension of the manifold we attach to it is smaller than the dimension of S 2 × S 2 . Now we will prove that for all j, k ∈ J if there exist t ∈ R such that Im(r k t ) Im(r t j ), then dim(Q k ) < dim(Q j ) such that r k is the function attached to k and G k is the manifold attached to k. We note that if Im(r k t ) Im(r t j ) then Im(r k t ) ∩ Im(r t j ) = Im(r k t ). Since we know that for any (i, j) ∈ J × J and for any t ∈ R, if we denote the function attached to k as r k we get that r t i ∩ r t j = k∈ f ((i,j)) r t k . If we have i, j that contradicts our statement there must be k where k ∈ f ((i, j)) where dim(Q i ) = dim(Q j ) = dim(Q k ). It is easy to verify that this is not the case. So by Theorem 4, Im(D 1 i ) ∪ (p × S 2 ) ∪ (S 2 × p) ∪ (p × p)). We know that A \ (B ∪ C) = (A \ C) \ (B \ C). Then it turns out that We can see that S 2 × S 2 \ (p × S 2 ∪ S 2 × p) = S 2 \ p × S 2 × p. We know from (1) that Im(G t j ) = p × p ∪ φ(g k • (ϕ × ϕ) −1 (R 2 × t)), so if we subtract p × p we get Im(G t j ) \ (p × p) = (g k • φ • (ϕ × ϕ) −1 (R 2 × t)). We can see that the image is inside ((S 2 \ p) × (S 2 \ p) ∪ p × p) and, therefore, ϕ × ϕ(Im(G t j ) \ ((S 2 \ p) × (S 2 \ p) ∪ p × p)) = Im(g k • φ).