2-string free tangles and incompressible surfaces

Transcription

1 Prev Next Last Go Back ℵ The Fifth East Asian School of Knots and Related Topics 2-string free tangles and incompressible surfaces Page 1 / 19 Li, Yannan Dalian Jiaotong University

2 Prev Next Last Go Back 1. Background Page 2 / Basic concepts 3. Construction of the surfaces

3 Prev Next Last Go Back 1 Background (1) W. Joca: incompressible surfaces of arbitrarily high genera in S 1 -bundles of closed surfaces; (2) H. Lyon: closed incompressible surfaces of arbitrarily high odd genera in the complements of fibered knots; Page 3 / 19 (3) U. Oertel: closed incompressible surfaces of arbitrarily high genera in complements of star links(or knots).

4 Prev Next Last Go Back (4) T. Kobayashi, R. Qiu, R. Rieck, and S. Wang: for any 3-manifold M there is a link of three components such that the complement of the link contains closed incompressible surfaces of arbitrarily odd genera; Page 4 / 19 (5) J. Ma and R. Qiu: for any 3-manifold M, there is a link of four components such that the complement of the link contains closed incompressible surfaces of all positive genera.

5 Prev Next Last Go Back Theorem 1. Suppose k 1 is a knot which admits a 2- string essential free tangle decomposition, and k 2 is a nontrivial knot. Let k = k 1 #k 2, then E(k) contains a closed incompressible surface of genus n for each positive integer n. Page 5 / 19 Remark Note that any closed incompressible surfaces survive in the exterior of every satellite knot of k, and that there are infinitely many satellite knots of k which are prime. Hence Theorem 1 implies that there are infinitely many prime knots, each of which contains a closed incompressible surface of genus n for each positive integer n.

6 Prev Next Last Go Back 2 Basic concepts A closed surface F properly embedded in the 3- manifold M(or is a component of M) is compressible in M if (1) F is a 2-sphere and F bounds a 3-cell in M, or Page 6 / 19 (2) there exists a disk D M such that D F = D and D is essential in F. Otherwise, F is incompressible in M.

7 Prev Next Last Go Back A surface F in a 3-manifold M is -compressible if either (1) F is a disk and F is parallel to a disk in M or Page 7 / 19 (2) F is not a disk and there exists a disk D M such that D = a b, where a = D F is an essential arc in F and b M. Otherwise, F is -incompressible in M.

8 Prev Next Last Go Back Page 8 / 19 The pair (B, t 1 t 2 t n ) is called an n-string tangle, where B is a 3-ball and t 1, t 2,, t n are n arcs properly embedded in B. A tangle (B, t 1 t 2 t n ) is essential if cl( B N(t 1 t 2 t n )) is incompressible in cl(b N(t 1 t 2 t n )). A tangle (B, t 1 t 2 t n ) is trivial if it is homeomorphic to (D 2 I, {x 1, x 2,, x n } I). A tangle (B, t 1 t 2 t n ) is free if cl(b N(t 1 t 2 t n )) is homeomorphic to a genus n handlebody.

9 Prev Next Last Go Back A knot k has an n-string tangle decomposition if (S 3, k) is decomposed into two n-string tangles (B 1, t 1 1 t2 1 t n 1 ) (B 2, t 1 2 t2 2 tn 2 ). The decomposition is free if both tangles are free. Page 9 / 19 The decomposition is essential if both tangles are essential.

10 Prev Next Last Go Back 3 Construction of the surfaces Theorem 1. Suppose k 1 is a knot which admits a 2- string essential free tangle decomposition, and k 2 is a nontrivial knot. Let k = k 1 #k 2, then E(k) contains a closed incompressible surface of genus n for each positive integer n. Page 10 / 19 E(k) = E(k 1 ) A 1 =A 2 E(k 2), where A i is a subannulus of E(k i ) for i = 1, 2.

11 Prev Next Last Go Back Step 1. Constructing P in E(k 1 ), where P is a twice punctured torus with P are meridians of N(k 1 ). (S 3, k 1 ) = (B 1, t 1 1 t2 1 ) (B 2, t 1 2 t2 2 ). Page 11 / 19 Figure 1.

12 Prev Next Last Go Back Suppose N(t 1 1 ) = t1 1 D : regular neighborhood of t1 1 in B 1. N (t 1 1 ) = t1 1 D : fatter regular neighborhood of t 1 1. Page 12 / 19 That is N(t 1 1 ) int N (t 1 1 ). Let P = (Q t 1 1 D ) (t 1 1 D ), where Q = cl( B 1 N(t 1 1 t2 1 )). Lemma 1 P is incompressible and -incompressible in E(k 1 ).

13 Prev Next Last Go Back Step 2. Getting a properly embedded annulus A in E(k 2 ) such that A is parallel to E(k 2 ) A 2. Then A is not parallel to A 2. Page 13 / 19 A is incompressible in E(k 2 ).

14 Prev Next Last Go Back Step 3. The construction of S n. Page 14 / 19 Let P 1, P 2,, P n be n copies of P in E(k 1 ) such that they appear consecutively in E(k 1 ). Their boundaries P i inta 1 appeared as in Fig. 2.

15 Prev Next Last Go Back Let A 1, A 2,, A n be n copies of A in E(k 2 ) such that they appear consecutively in E(k 2 ). Page 15 / 19 Their boundaries A j inta 2 appeared as in Fig. 2.

16 Prev Next Last Go Back E(k) = E(k 1 ) A 1 =A 2 E(k 2) Page 16 / 19 Figure 2.

17 Prev Next Last Go Back A properly isotopy of P i and A j : 1 P 1 = 1 A 2, 1 P 2 = 1 A 3,, 1 P n 1 = 1 A n, 2 P 1 = 1 A 1, 2 P 2 = 2 A 1, 2 P n = 2 A n 1, 1 P n = 2 A n. Page 17 / 19 Figure 3.

18 Prev Next Last Go Back S n = 1 i,j n P i A j. Conclusion: (1) S n is connected and the genus of S n is n + 1; Page 18 / 19 (2) S n is incompressilbe in E(k).

19 Prev Next Last Go Back Thanksœ Page 19 / 19