Sharpness of Rickman s Picard theorem in all dimensions

Transcription

1 Sharpness of Rickman s Picard theorem in all dimensions Pekka Pankka University of Jyväskylä XXII Rolf Nevanlinna Colloquium Helsinki August 5-9, joint work with David Drasin

2 Picard s theorem: A (non-constant) holomorphic map C S 2 omits at most 2 points.

3 Picard s theorem: A (non-constant) holomorphic map C S 2 omits at most 2 points. Liouville s theorem: For n 3, a conformal map Ω S n on a domain Ω S n is a restriction of a Möbius transformation. Corollary: For n 3, a conformal map R n S n is an embedding.

4 From conformal to quasiconformal A mapping f : R n S n is K-quasiregular if f W 1,n loc (Rn ; S n ) and Df n K det Df a.e.

5 From conformal to quasiconformal A mapping f : R n S n is K-quasiregular if f W 1,n loc (Rn ; S n ) and Df n K det Df a.e. Rickman s Picard theorem (1980): Let f : R n S n be a (non-constant) K-quasiregular mapping. Then #(S n \ f R n ) C(n, K).

6 From conformal to quasiconformal A mapping f : R n S n is K-quasiregular if f W 1,n loc (Rn ; S n ) and Df n K det Df a.e. Rickman s Picard theorem (1980): Let f : R n S n be a (non-constant) K-quasiregular mapping. Then #(S n \ f R n ) C(n, K). Remark: For n = 2, Stoilow factorization: f = (holomorphic map) (quasiconformal homeomorphism). C(2, K) = 2.

7 Sharpness of Rickman s Picard theorem Theorem (Rickman (1985)) Given a finite set P S 3, there exists a quasiregular map f : R 3 S 3 satisfying f R 3 = S 3 \ P.

8 Sharpness of Rickman s Picard theorem Theorem (Rickman (1985)) Given a finite set P S 3, there exists a quasiregular map f : R 3 S 3 satisfying f R 3 = S 3 \ P. Remarks: S n \ P simply connected for n 3. Zorich s global homeomorphism theorem f is not a local homeomorphism.

9 Sharpness of Rickman s Picard theorem Theorem Given a finite set P S n for n 3, there exists a quasiregular map f : R n S n satisfying f R n = S n \ P.

10 Sharpness of Rickman s Picard theorem Theorem Given a finite set P S n for n 3, there exists a quasiregular map f : R n S n satisfying f R n = S n \ P. Question: Is every quasiregular map R 4 S 2 S 2 onto?

11 Sharpness of Rickman s Picard theorem Theorem (Conformal version) Given a finite set P S n for n 3, there exists a quasiregular map f : R n S n satisfying f R n = S n \ P. Theorem (Metric version) Let P = {y 1,..., y p } S n, δ > 0, g a Riemannian metric on S n \ P for which each punctured Euclidean ball B n (y i, δ) S n is isometric to the Euclidean half-infinite cylinder S n 1 (δ) (0, ) in metric g. Then there exists a BLD-map R n (S n \ P, g). Figure: (S n \ P, g)

12 Why BLD? Let X and Y be metric spaces. A continuous, discrete, and open map f : X Y is BLD if there exists L 1 satisfying for all paths γ in X. Observations: 1 l(γ) l(f γ) Ll(γ) L BLD-maps between manifolds are QR. f i : X i Y i BLD for i = 1, 2 f 1 f 2 : X 1 X 2 Y 1 Y 2 is BLD. QR-maps do not admit products (in general).

13 Observations Alexander map A: R n 1 S n 1 Zorich map Z : R n R n \ {0} R n 1 R A id S n 1 R BLD (x,t) et x conf R n \ {0} 1/2-Zorich map R n 1 [0, ) A id S n 1 [0, )

14 Idea of the proof: f : R n (S n \ P, g) is a p/2-zorich map Part 0 Partition S n into n-cells E 1,..., E p containing the points y 1,..., y p (to be omitted) in their interiors. Figure: p = 3

15 Idea of the proof: f : R n (S n \ P, g) is a p/2-zorich map Part 0 Partition S n into n-cells E 1,..., E p containing the points y 1,..., y p (to be omitted) in their interiors. Part I Figure: p = 3 Partition R n into sets Ω 1,..., Ω p, with finitely many components bilipschitz to R n 1 (0, ) in their inner geometry.

16 Idea of the proof: f : R n (S n \ P, g) is a p/2-zorich map Part 0 Partition S n into n-cells E 1,..., E p containing the points y 1,..., y p (to be omitted) in their interiors. Part I Figure: p = 3 Partition R n into sets Ω 1,..., Ω p, with finitely many components bilipschitz to R n 1 (0, ) in their inner geometry. Part II Modify domains Ω 1,..., Ω p to obtain domains Ω 1,..., Ω p supporting a BLD mapping f : i Ω i i E i, which restricts to an Alexander map Ω i E i for each i.

17 Idea of the proof: f : R n (S n \ P, g) is a p/2-zorich map Part 0 Partition S n into n-cells E 1,..., E p containing the points y 1,..., y p (to be omitted) in their interiors. Part I Figure: p = 3 Partition R n into sets Ω 1,..., Ω p, with finitely many components bilipschitz to R n 1 (0, ) in their inner geometry. Part II Modify domains Ω 1,..., Ω p to obtain domains Ω 1,..., Ω p supporting a BLD mapping f : i Ω i i E i, which restricts to an Alexander map Ω i E i for each i. Part III For each i, extend f Ω i to a BLD-map Ω i (E i \ {y i }, g) using a 1/2-Zorich map.

18 Part I: Rickman partitions / caving From now p = 3! Construction of the partition (Ω 1, Ω 2, Ω 3 ) of R n has two steps. Step 1 We construct a sequence (D m ) = (D 1,m, D 2,m, D 3,m ): (1) D 1,0, D 2,0, D 3,0 are n-cubes of side length 3, (2) D m,1 D m,2 D m,3 is an n-cell (i.e. [0, 1] n ), (3) (D m,i, d Dm,i ) is an n-cell (uniformly) bilipschitz to [0, 3 m ] n, (4) dist( i j D m,i D m,j, i D m,i) < const. (5)...

19 Part I / Step 1: (D 0,1, D 0,2, D 0,3 ) (side length 3)

20 Part I / Step 1: (D 1,1, D 1,2, D 1,3 ) (side length 9)

21 Part I / Step 1: (D 2,1, D 2,2, D 2,3 ) (side length 27)

22 Part I / Step 1: (D 3,1, D 3,2, D 3,3 ) (side length 81)

23 Part I / Step 1: Idea of the induction step Figure: (D 0,1, D 0,2, D 0,3 ) Figure: (D 1,1, D 1,2, D 1,3 ) Figure: From (D 0,1, D 0,2, D 0,3 ) to (D 1,1, D 1,2, D 1,3 ).

24 Part I Step 2 Take a limit (D m,1, D m,2, D m,3 ) m (D 1, D 2, D 3 ) D 1 D 2 D 3 = [0, ) n 1 R Reflect domains D i to get domains (Ω 1, Ω 2, Ω 3 ).

25 Part II Locally modify partition (Ω 1, Ω 2, Ω 3 ) to obtain partition ( Ω 1, Ω 2, Ω 3 ) admitting a BLD-map i Ω i i E i. Step 1: Triangulate Figure: Before Figure: After

26 Step II: Rickman s sheets Figure: Before Figure: After

27 Local picture Figure: Original

28 Local picture Figure: Sheets

29 Local picture Figure: Middle region

30 Local picture Figure: Twist