On the structure of self-affine Jordan arcs in ℝ²

Theorem 1

Let γ ⊂ R 2 be a locally self-affine Jordan arc, which does not satisfy weak separation condition. Then, γ is a segment of a parabola or a straight line.

Theorem 2

Let S = { S 1 , … , S m } be a system of affine contraction maps in R 2 , whose attractor is a Jordan arc γ , which is not a segment of a parabola or a straight line. There is a finite affine multizipper Z = { S i j k } such that γ is the attractor of Z . All the maps from the multizipper Z are elements of the system S .

Definition 3

A system S = { S 1 , … , S m } of contraction mappings of R d to itself is called a zipper with vertices { z 0 , … , z m } and signature ε = ( ε 1 , … , ε m ) , ε i ∈ { 0 , 1 } if, for i = 1 … m , S i ( z 0 ) = z i − 1 + ε i and S i ( z m ) = z i − ε i .

Definition 4

Let { X u , u ∈ V } be a system of spaces, all isomorphic to R d . For each X u , let a finite array of points be given as { x 0 ( u ) , … , x m u ( u ) } . Suppose for each u ∈ V and 0 ≤ k ≤ m u , we have some v ( u , k ) ∈ V and ε ( u , k ) ∈ { 0 , 1 } and a map S k ( u ) : X v → X u such that

S k ( u ) ( x 0 ( v ) ) = x k − 1 ( u ) or x k ( u ) and S k ( u ) ( x m v ( v ) ) = x k ( u ) or x k − 1 ( u ) , depending on the signature ε ( u , r ) .

The graph-directed IFS defined by the maps S k ( u ) is called a multizipper Z .

Theorem 5

Let Z 0 = { S k ( u ) } be a multizipper with node points x k ( u ) and with a signature ε = { ( v ( u , k ) , ε ( u , k ) ) , u ∈ V , k = 1 , … , m u } . If for any u ∈ V and any i , j ∈ { 1 , 2 , … , m u } , the set K ( u , i ) ∩ K ( u , j ) = ∅ if ∣ i − j ∣ > 1 and is a singleton if ∣ i − j ∣ = 1 , then any linear parametrization { f u : I u → K u } is a homeomorphism and each K u is a Jordan arc with endpoints x 0 ( u ) , x m ( u ) .

Lemma 6

There is a homeomorphism Ψ of the set U to a neighborhood V of zero map in A ( R 2 ) such that for any f ∈ U , the map g = Ψ ( f ) satisfies the following condition: for any x ∈ R 2 , the solution y ( t ) of the Cauchy problem { y ˙ = g ( y ) , y ( 0 ) = x } is equal to f ( x ) at t = 1 .

Lemma 7

If g = f i − 1 f j ∈ ℱ ( γ ) and g ( γ ˙ ) ∩ ( γ ˙ ) ≠ ∅ , and g preserves the orientation on γ , then g has no fixed points on γ .

Proof

Indeed, let a , b be the endpoints of γ = f i − 1 f j . Let x ∈ γ ˙ be a fixed point of g . Then, say, f i ( a ) < f j ( a ) < f i ( x ) = f j ( x ) < f i ( b ) < f j ( b ) . So g ( γ ( a , x ) ) ⊂ γ ( a , x ) and g − 1 ( γ ( x , b ) ) ⊂ γ ( x , b ) . Therefore, lim g n ( a ) = x and lim g − n ( b ) = x , so x is a saddle point for g . Therefore, γ ( a , x ) and γ ( x , b ) should be non-collinear line segments, which is impossible.□

Theorem 8

Let γ = γ ( a 0 , a 1 ) be a locally self-affine Jordan arc with endpoints a 0 , a 1 in R 2 such that there is a sequence of affine shifts f k of γ converging to Id . Then γ is a parabolic or a straight line segment.

Corollary 9

Let γ be a locally self-affine Jordan arc in R 2 . Let γ ′ ⊂ γ ˙ be its subarc with endpoints x 0 , x 1 . If there is a sequence of affine maps g n such that γ ′ ⊂ g n ( γ ) ⊂ γ , then γ is a segment of a straight line or a parabola.

Proof

By compactness, we can choose a subsequence, denoted the same way, such that the arcs g n ( γ ) converge to some arc γ ″ with endpoints x 0 ′ , x 1 ′ and γ ′ ⊂ γ ″ ⊂ γ . If there is a sequence of n such that g n − 1 g n + 1 are affine shifts of γ , then γ is a segment of parabola by Theorem 8. Otherwise (by passing to a subsequence), we can assume that for any n , g n + 1 ( γ ) ⊂ g n ( γ ) (or otherwise) and the fixed points y n of g n − 1 g n + 1 belong to a sufficiently small subarc δ ⊂ γ so that S k ( γ ) ∩ δ = ∅ for k = 1 or m . Then, the maps S k − 1 g n − 1 g n + 1 S k form a sequence of affine shifts of γ , converging to identity, which proves the corollary.□

Definition 10

We say that α ∈ R / π Z defines a transversal direction to the arc γ , if for any line l , which intersects real axis in the angle α , # ( γ ∩ l ) ≤ 1 .

The set T ( γ ) of all transversal directions α to the arc γ is called the range of transversal directions to γ or T-range of γ .

We say that γ has empty T-range for all subarcs if, for any non-degenerate subarc γ ′ ⊂ γ , its T-range is empty.

Lemma 11

If γ is a locally self-affine Jordan arc in the plane, then either T ( γ ) ≠ ∅ or γ has an empty T-range for all its subarcs.

Lemma 12

Let γ be a Jordan arc that lies completely in the upper half-plane Im z > 0 except for its end points z 0 < z 1 , which lie on the real axis. Then, for any positive r < z 1 − z 0 , there are z 0 ′ , z 1 ′ ∈ γ such that z 1 ′ − z 0 ′ = r .

Proof

Without loss of generality, we assume for (d1) and (d2) that z 0 < z 1 .

Take r < z 1 − z 0 . The points z 0 + r and z 1 + r do not belong to γ ; therefore, there is δ > 0 such that the neighborhoods V δ ( z 0 + r ) and V δ ( z 1 + r ) are disjoint from γ . If 0 < ε < δ , then z 0 + r + i ε lies in the domain D bounded by the arc γ and a segment [ z 0 z 1 ] , while the point z 1 + r + i ε lies in the complement to the closed domain D ¯ .

Therefore, the image of γ under the translation by r + i ε intersects the arc γ . Reducing ε to 0 , we obtain that ( γ + r ) ∩ γ ≠ ∅ . This intersection does not contain points with real coordinates, and therefore, it lies in γ \ { z 0 , z 1 } . Take z 1 ′ ∈ γ ∩ ( γ + r ) . The point z 0 ′ = z 1 ′ − r is also in γ , which proves the lemma.□

Corollary 13

Let γ be a Jordan arc in C with endpoints z 0 , z 1 such that γ ∩ [ z 0 , z 1 ] = { z 0 , z 1 } .

Then,

1. for any positive r < ∣ z 1 − z 0 ∣ , there are z 0 ′ , z 1 ′ ∈ γ such that ∣ z 1 ′ − z 0 ′ ∣ = r and Arg ( z 1 ′ − z 0 ′ ) = Arg ( z 1 − z 0 ) ;

2. there are h , ε > 0 such that for any r < h and any ∣ θ ∣ < ε , there are z 0 ′ , z 1 ′ ∈ γ for which z 1 ′ − z 0 ′ = r e i θ ( z 1 − z 0 ) .

Proof

Consider the intersection of γ and the line l passing through the points z 0 z 1 . There is at least one pair of points z 0 ′ , z 1 ′ ∈ l such that [ z 0 , z 1 ] ∈ [ z 0 ′ , z 1 ′ ] and the subarc γ ′ ⊂ γ with endpoints z 0 ′ , z 1 ′ has no other common points with l . Applying Lemma 12 to the arc γ ′ and the half-plane bounded by l and containing γ ′ , we come to (d1).

Suppose for convenience that z 0 , z 1 ∈ R and z 0 < z 1 . Take some segment [ u 0 , u 1 ] ⊂ ( z 0 , z 1 ) and put h = u 1 − u 0 . By the compactness of [ u 0 , u 1 ] , there is λ > 0 such that a rectangle P = [ u 0 , u 1 ] × [ 0 , λ ] does not intersect γ . For any ray l + starting at the point z 0 (resp. l − starting at z 1 ), which intersects both vertical sides of the rectangle P , the set l ∩ P is contained in some interval ( ξ 0 ( l ) , ξ 1 ( l ) ) for which ξ i ( l ) ∈ γ 1 , ( ξ 0 ( l ) , ξ 1 ( l ) ) ∩ γ 1 = ∅ , and ‖ ξ 0 ( l ) − ξ 1 ( l ) ‖ ≥ h . Then, by (d1), for any r < h , there are z 0 ′ , z 1 ′ ∈ γ such that the interval [ ( z 0 ′ , z 1 ′ ) ] is parallel to l and ∣ z 1 ′ − z 0 ′ ∣ = r . Take ε > 0 such that the ray l + , which forms an angle ε with ( z 0 , z 1 ) , and the ray l − ∋ z 1 , which forms an angle − ε with ( z 1 , z 0 ) , intersect both vertical sides of the rectangle P . Then, h and ε fit the statement (d2) of the lemma.□

Corollary 14

Let γ ⊂ R 2 be a Jordan arc R 2 that does not contain a line segment. Then, T ( γ ) is a closed subarc in R / π Z .

Proof

Suppose α ∈ C T ( γ ) . Then, there is a line l that intersects γ in at least two points. The complement γ \ l is a disjoint union of subarcs with endpoints on l but no other intersection points with l . Therefore, by the statement (d2) of Corollary 13, there is a neighborhood ( α − ε , α + ε ) ⊂ C T ( γ ) . This shows that C T ( γ ) is open in R / π Z .

Let α , β ∈ T ( γ ) . Take a point x ∈ γ and let l α , l β be the lines passing through x in directions α and β . These lines divide the plane into four angles. The set γ \ { x } consists of two components, which are contained in two opposite angles of these four angles. As x travels along γ , these four angles remain the same; therefore, all values between α mod π and β mod π that correspond to the remaining two opposite angles belong to T ( γ ) . Since C T ( γ ) is open, T ( γ ) is a closed subarc in R / π Z .□

Lemma 15

Let γ ⊂ R 2 be a Jordan arc that has empty T-range for all subarcs.

1. For any line l ⊂ R 2 , l ∩ γ is nowhere dense in l and in γ .

2. For any line l ⊂ R 2 and for any n ∈ N , the set of those lines l ′ parallel to l for which # ( l ′ ∩ γ ) < n is nowhere dense.

Proof

The first statement is obvious, because the set γ ∩ l is closed and does not contain any straight line interval. We prove the second statement by induction in n .

Let l be a line and let a be a vector orthogonal to l such that for any t ∈ ( 0 , 1 ) , γ ∩ ( l + t a ) ≠ ∅ . We show that the set E 1 of those t ∈ ( 0 , 1 ) for which # ( γ ∩ ( l + t a ) ) = 1 is nowhere dense in ( 0 , 1 ) . Suppose that the set E 1 is dense in some interval ( t 1 , t 2 ) ⊂ ( 0 , 1 ) . This interval defines an open strip S , bounded by the lines l + t 1 a , l + t 2 a . Consider the intersection γ ∩ S = γ ′ . Note that if for some t ∈ ( t 1 , t 2 ) , the set γ ∩ ( l + t a ) is disconnected, then it is disconnected for any t ′ in one of the intervals ( t − ε , t ] , [ t , t + ε ) for some ε > 0 . Therefore, γ ′ is a Jordan arc, and for any t ∈ ( t 1 , t 2 ) , the intersection γ ′ ∩ ( l + t a ) is either a point or a line segment contained in l + t a . The second case is impossible, because each line segment of that kind is a subarc in γ whose T-range is non-empty. If for each t , # ( γ ′ ∩ ( l + t a ) ) = 1 , then T ( γ ′ ) is non-empty.

Therefore, for any Jordan arc γ , which has empty T-range for all subarcs, and for any line l and its orthogonal vector a , the set { t : # ( γ ∩ ( l + t a ) = 1 ) } is nowhere dense in R .

Now suppose that for any subarc γ ′ ⊂ γ , the set { t : 0 < # ( γ ′ ∩ ( l + t a ) ) < n } is nowhere dense in ( 0 , 1 ) , whereas the set { t : 0 < # ( γ ∩ ( l + t a ) ) = n } is dense in some interval ( t 1 , t 2 ) ⊂ ( 0 , 1 ) . Let S be the open strip bounded by the lines l + t 1 a , l + t 2 a . Since γ intersects both these lines, one of the components of γ ∩ S is a subarc γ 0 with endpoints on different sides of S . Choose an endpoint x of this component so that both components γ 1 and γ 2 of γ \ { x } have nonempty intersection with S . The arc γ 0 is a subarc of one of these components, say, γ 1 . Suppose ( t 1 ′ , t 2 ′ ) ⊂ ( t 1 , t 2 ) is such that for any t ∈ ( t 1 ′ , t 2 ′ ) , γ 2 ∩ ( l + t a ) ≠ ∅ . Being contained in the intersection of the sets { t : 0 < # ( γ 1 ∩ ( l + t a ) ) < n } and { t : 0 < # ( γ 2 ∩ ( l + t a ) ) < n } , which are both nowhere dense in ( t 1 ′ , t 2 ′ ) , the set { t ∈ ( t 1 ′ , t 2 ′ ) : # ( γ ∩ ( l + t a ) ) = n } is nowhere dense in ( t 1 ′ , t 2 ′ ) too.□

Lemma 16

Let γ ( t ) = ( γ x ( t ) , γ y ( t ) ) , t ∈ [ 0 , 1 ] be a Jordan arc with endpoints γ ( 0 ) = ( 0 , 0 ) , γ ( 1 ) = ( 0 , h ) whose interior γ ( ( 0 , 1 ) ) lies in the open strip S = { ( x , y ) : 0 < y < h } and has empty T-range for all subarcs. Let S + be a connected component of S \ γ whose boundary contains [ 0 , + ∞ ) .

There is a point a ∈ γ and ρ > 0 such that C ( a , ρ , 0 ) ⊂ S + .

Proof

Suppose max γ x ( t ) > 0 , and τ ∈ ( 0 , 1 ) satisfies γ x ( τ ) = max γ x ( t ) . Let a = γ ( τ ) . If ρ = min ( a y , 1 − a y ) , then C ( a , ρ , 0 ) ⊂ S + .

Suppose for any x ∈ ( 0 , 1 ) , γ x ( t ) ≤ 0 . As it follows from Lemma 15, there is a vertical line l : x = χ , χ < 0 , which intersects γ more than three times. Let S ˜ + be an unbounded component of S + \ l . The set ∂ S ˜ + \ l is an union of more than two subarcs of the arc γ . Therefore, at least one of these subarcs, say γ ′ , has both its endpoints in l . Let a ∈ γ ′ be the point at which γ x ′ reaches its maximum. Then, there is such ρ , that C ( a , ρ , 0 ) ⊂ S + .□

Lemma 17

Let γ ( t ) , t ∈ [ 0 , 1 ] be a Jordan arc that has empty T-range for all subarcs. Suppose γ ( 0 ) = z 0 , γ ( 1 ) = z 1 , and z 1 = z 0 + Re i α and γ ( ( 0 , 1 ) ) lies completely in an open half-plane, bounded by the line l containing z 0 , z 1 . Let D be a domain bounded by γ and [ z 0 , z 1 ] .

There are points z 0 ′ , z 1 ′ ∈ γ and r > 0 such that C ( z 0 ′ , r , α ) ⊂ D and C ( z 1 ′ , r , α ) ∩ D ¯ = ∅

Proof

Without loss of generality, we may assume that z 0 = 0 , z 1 = 1 , α = 0 , and the domain D lies in the upper half-space. Let h 2 = max γ y ( t ) and τ 0 = min { τ : γ y ( τ ) = h 2 } . Denote by γ 1 the subarc in γ with endpoints ( 0 , 0 ) and ( h 1 , h 2 ) = γ ( τ 0 ) . By Lemma 16, there exists a point a ∈ γ 1 such that C ( a , ρ , 0 ) ⊂ S + . There is a ball B ( ρ ′ , a ) such that B ( ρ ′ , a ) ∩ γ \ γ 1 = ∅ . The set D is a component of S + \ γ whose boundary contains γ 1 , hence C ( a , ρ ′ , 0 ) ⊂ S + ( γ 1 ) . Let now τ 1 = max { τ : γ y ( τ ) = h 2 } and h 1 ′ = γ x ( τ 1 ) . Denote by γ 2 a subarc in γ with endpoints ( 1 , 0 ) and ( h 1 ′ , h 2 ) and let S + ′ be an unbounded component of S + \ γ 2 . By Lemma 16, there is a point b ∈ γ 2 , for which there is a sector C ( b , ρ , 0 ) ⊂ S + ′ . Since S + ′ ∩ D = ∅ , the same is true for C ( b , ρ , 0 ) .□

Proposition 18

Let γ be a Jordan arc that has empty T-range for all its subarcs. For any α ∈ [ 0 , π ) , there is a subarc γ ′ ⊂ γ with endpoints z 1 , z 2 such that arg ( z 2 − z 1 ) = α and γ ′ ∪ [ z 1 z 2 ] is a closed Jordan curve bounding a domain D .

There is r > 0 and points z 0 ′ , z 1 ′ ∈ γ ′ such that C ( z 0 ′ , r , α ) ⊂ D and C ( z 1 ′ , r , α ) ∩ D ¯ = ∅ .