Knots and surfaces. Michael Eisermann. Universität Stuttgart 3. Juni 2014

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1 1/19 Knots and surfaces Michael Eisermann Universität Stuttgart 3. Juni 2014 Kolloquium über Reine Mathematik Fachbereich Mathematik, Universität Hamburg

2 Low-dimensional manifolds One- and two-dimensional manifolds (c.c.o.): S1 and [0, 1], Sg+ = + Sg,n =..., H1 = Z2g+n 1 1 g 2 n A link is an embedding {1,..., n} S1, R3. Surfaces embedded in 4 space, e.g. f : C, C2 with f (z) = (z 2, z 3 ). Here K = f (C) S3 is the trefoil knot. Is the singularity removable? 2/19

Links and surfaces in 3 space (d) ribbon singularity (e) 3 1 3 1 (f) 6 1 3/19 Every link {1,.

3 Links and surfaces in 3 space (d) ribbon singularity (e) (f) 6 1 3/19 Every link {1,..., n} S 1 R 3 bounds an embedded surface S R 3 : SeifertView, Jarke van Wijk, TUE (a) trivial knot, (b) trefoil knot, 3 1 (c) figure eight, 4 1 A ribbon surface S R 3 is immersed having only ribbon singularities.

4 Links and surfaces in 4 space abstract surface maximum surface embedded in R 4 + maximum isotopy saddle point h = 0 R 3 0 abstract surface surface embedded in R 4 + A link L R 3 bounds a ribbon surface S R 3 if and only if it bounds a smoothly embedded surface S R 4 + without local minima. 4/19

5 5/19 The slice vs ribbon problem (Fox 1962) Seifert genus (dimension 3): g 3 (K) := min{ g(s) S R 3 embedded surface, S = K } g 3 (K) = 0 K is trivial Ribbon genus (dimension 3 1 /2): g r (K) := min{ g(s) S R 3 ribbon surface, S = K } g r (K) = 0 K is ribbon Smooth genus (dimension 4): g 4 (K) := min{ g(s) S R 4 + smoothly embedded, S = K } g 4 (K) = 0 K is slice Question (Slice vs ribbon problem, Fox 1962) Obviously g 3 (K) g r (K) g 4 (K). >! =? Does slice imply ribbon?

6 Geometric constructions (a) Kinoshita Terasaka 1957 (b) Conway 1969 Is the Conway knot slice? or ribbon? or a symmetric union? 6/19

7 Geometric constructions 7/19 Is the knot K = 12a631 slice? ribbon? a symmetric union? Yes!

8 Geometric constructions 8/19 We have three closely related geometric properties / constructions: {symmetric unions} =? {ribbon links} =? {slice links} Empirical evidence for small prime knots: crossings total cand. slice ribbon symmetric union (Lamm 2000) (Lamm, Seeliger) (Seeliger 2013) Two-bridge slice knots: 3 families (Casson Gordon 1975, Lisca 2007). Each of these knots is a symmetric union, hence ribbon. (Lamm 2005)

9 9/19 Geometric constructions Gompf Scharlemann Thompson (2010): This link is slice! Is it ribbon? Is it a symmetric union?

10 10/19 Algebraic obstructions Classical algebraic topology: Alexander polynomial (K) Seifert surface and Seifert form Fundamental group of R 3 K Infinite cyclic covering Heegaard Floer homology Quantum invariants: Jones polynomial V (L) and its generalizations combinatorial definition and calculation representation of tangle category Kauffman bracket, state sums Khovanov homology

11 Algebraic obstructions Theorem (Seifert 1934, Pontryagin 1930) Every link L R 3 bounds a c.c.o. surface S R 3. S S Any two such surfaces are equivalent by embedded surgery in R 3. To each surface S R 3 we associate its Seifert form: θ S : H 1 (S) H 1 (S) Z, θ S (a, b) = lk(a, b ) Surgery S S corresponds to stabilisation: θ S 0 θ S + θ 0 θ S θ S = 0, θ S + θs = We can thus extract invariants of the link L = S. 11/19

12 12/19 Algebraic obstructions We can thus extract invariants of the link L = S: sign(l) := sign(θ S + θ S ) null(l) := null(θ S + θ S ) det(l) := det(θ S + θ S ) det(l) := det(θ S + θ S ) ( i) 2g+n 1 Z N N Z[i] (L) := det(q 1 θ S q +1 θ S ) Z[q ± ] Example: For the trefoil knot K = T 3,2 we find sign(k) = 2, null(k) = 0, det(k) = 3, (K) = q q 2 For every knot K R 3 and every link L R 3 we have: width (K) 4 g 3 (K) width (L) 2 [ 2 g 3 (L) + n 1 ] sign(k) 2 g 3 (K) sign(l) 2 g 3 (L) + n 1 det(k) 1 mod 4 null(l) n 1

13 13/19 Algebraic obstructions Theorem (Fox Milnor 1966) If L = L 1 L n R 3 is an n component slice link, then sign(l) = 0 and null(l) = n 1. If K R 3 is a slice knot, then its Alexander polynomial satisfies (K) = P (q) P (q 1 ) for some P Z[q]. In particular, det(k) = (K) q i is a square, so det(k) 1 modulo 8. Application to singularities like f : C C 2 with f(z) = (z 2, z 3 ). Here K = f(c) S 3 is the trefoil knot, which is not slice.

14 Jones polynomial and Kauffman bracket 14/19 Theorem (Alexander 1928, Conway 1969) There exists a unique invariant : L Z[q ± ] with ( ) = 1 and ( ) ( ) ( ) = (q +1 q 1 ). Theorem (Jones 1984) There exists a unique invariant V : L Z[q ± ] with V ( ) = 1 and ( ) ( ) ( ) q 2 V q +2 V = (q 1 q +1 ) V. Simplest construction via the Kauffman bracket D : = A + A 1 normalized by = 1 and n = ( A +2 A 2 ) n 1. We obtain V (L) (q A 2 ) = D ( A 3 ) writhe(d).

15 The Jones polynomial of ribbon links S = S = Theorem (E 2009) Suppose L R 3 bounds a ribbon surface S R 3 of positive Euler characteristic n > 0. Then the Jones polynomial V (L) is divisible by the Jones polynomial V ( n ) = (q 1 + q +1 ) n 1 of the trivial link. Examples: V (L) = (q +1 + q +5 ) 2 (q 1 + q 5 ) 2. Thus L bounds ribbon surfaces with χ 1 but not with χ 2. V (L ) = (q 1 + q +1 ) (q 6 q 4 + 2q 2 + 2q 2 q 4 + q 6). Thus L bounds ribbon surfaces with χ 2 but not χ 3. 15/19

16 16/19 The Jones polynomial of ribbon links Proof: We show V ( n ) V (L) by induction on r = ribbon singularities. For r = 0 the surface S is embedded, whence L = L 0 n. For r 1 we apply the Kauffman bracket to ribbon diagrams: = A +4 + A 4 + A +2 + A +2 + A 2 + A 2 + A +2 + A +2 + A 2 + A 2.

17 The Jones polynomial of ribbon links Theorem (E 2009) Let L = L 1 L n be a ribbon link with n components. Then V (L) is divisible by V ( n ) = (q + + q ) n 1 and we have det V (L) := [V (L)/V ( n )] q i det(l 1 ) det(l n ) mod 32. Questions Does this yield a useful obstruction for ribbon links? slice links? Why 32? Possible generalizations? Khovanov? Kauffman? Homflypt? Does the Seifert nullity null(l) = n imply divisibility P ( n ) P (L)? Theorem Let L = L 1 L n be a boundary link with n components. Then the Homflypt polynomial P (L) is divisible by P ( n ). 17/19

18 The congruence modulo 32 is optimal Here det V (L ) = 15 whereas det(k 1 ) = det(k 2 ) = 1. 18/19 Examples: L = L = The link L = 10n36 is ribbon. We find V (L) = (q + +q )( q +8 +2q +6 3q +4 +4q q 2 3q 4 +2q 6 q 8 ). Here det V (L) = 23 whereas det(k 1 ) = 1 and det(k 2 ) = 9. The link L = 10n57 is not ribbon. We find V (L ) = (q + + q )(q +6 2q q q 2 2q 4 + 2q 6 q 8 )

19 19/19 Thank you for your attention!

Surface invariants of finite type

Surface invariants of finite type Michael Eisermann Institut Fourier, UJF Grenoble 17 September 2008 i f INSTITUT FOURIER Michael Eisermann www-fourier.ujf-grenoble.fr/ eiserm Summary 1 Definitions and