The Concordance Genus of Knots

Transcription

1 The Concordance Genus of Knots Finding the least genus of a knot within a concordance class Chuck Livingston Brandeis University June, 2008

2 Basics. Concordance genus. Given K S 3, what is the minimal genus of a knot concordant to K? g 3 (K) = 3 genus. g 4 (K) = 4 genus. g c (K) = concordance genus, the minimal genus of a knot concordant to K. Theorem. g 3 (K) g c (K) g 4 (K). If K is slice, g c (K) = g 4 (K) = 0. Proof.

3 History. Does g c (K) = g 4 (K)? (Gordon problem list, Answer: No; Casson, 1979.) For every N, there exists a knot with g c (K) g 4 (K) N. (Nakanishi, 1981.) Find concordance relations among pairs of low crossing number knots. (eg. Conway found many concordance relations among 10 crossing knots, 1973.) Computations of g c for low-crossing number knots (c 10) with three exceptions: 8 18,9 40, (2001) Determination of g c for 8 18 and (2008) 10 82? Continued identification of concordance relations among low crossing number knots.

4 Summary. Examples K = 6 2 : g 4 (K) = 1, g c (K) = 2. (Application of Alexander polynomials, Fox-Milnor theorem.) K = 6 2 # 6 2 : g 4 (K) = 2, g c (K) = 4. (Application of Fox-Milnor Theorem and Murasugi-Levine-Tristram signatures.) K = 9 40,8 18 : g 4 (K) = 1, g c (K) = 3. (Levine s rational algebraic concordance group of isometric structures.) K = : g 4 (K) = 2, g 3 (K) = 4, g c (K) =?? (All integral algebraic concordance invariants consistent with g c (K) = 2.) K = and other algebraically trivial cases. (Application of Details Casson-Gordon invariants.) Levine s invariants of knot cobordism are efficiently computable.

5 Example 1: K = 6 2. g 4 (K) = 1, g c (K) = 2. K (t) = 1 3t + 3t 2 3t 3 + t 4, (irreducible), g 3 (K) = 2: Deg( K (t)) 2g 3 (K). g 4 (K) 1: since K has unknotting number 1. g 4 (K) = 1: Fox-Milnor Theorem: If K is slice, K (t) = t k f (t)f (t 1 ) for some f and k: K (t) is a norm. If K is concordant to J, then K # J is slice and so K (t) J (t) = t k f (t)f (t 1 ) for some f. Thus, for any such J, deg J (t) 4, implying g 3 (J) 2. Hence, g c (K) = 2.

6 Generalization: g c (K) g 4 (K) arbitrarily large. Theorem For every N, there exists a K with unknot(k) = 1 and g c (K) > N. (Nakanishi, 1981.) n - n - 1 (3) (- 4 ) K (t) = 1 + t + - t t 2n mod 2 Theorem (Nakanishi) For every Alexander polynomial (t) there is a unknotting number one knot K with K (t) = (t).

7 Example 2: K = 6 2 # 6 2. g 4 (K) = 2, g c (K) = 4. σ(k) = 4, where σ(k) is the Murasugi signature, sign(v K + V t K ). g 4 (K) = 2, since g 4 (K) σ(k)/2. K (t) = (1 3t + 3t 2 3t 3 + t 4 ) 2, so Fox-Milnor no longer applies. (There is a slice knot with this polynomial.) Since σ(k) 0, the Levine-Tristram signature function, σ K (ω) = signature((1 ω)v K + (1 ω)v t K ), ω S1, jumps at a root of K (t). (Note: σ K (1) = 0, σ K ( 1) = 4.) If K is concordant to J, then σ J (ω) must jump at the same root: (1 3t + 3t 2 3t 3 + t 4 ) divides J (t). (1 3t + 3t 2 3t 3 + t 4 ) 2 divides J (t). (Fox-Milnor) g3 (J) 4.

8 Example 3: K = g c (K) = 3. The Alexander polynomial and signatures are sufficient to determine g c for all 10 crossing prime knots, except 8 18,9 40 and g 3 (K) = 3, g 4 (K) = 1, σ(k) = 2, K (t) = (t 2 t + 1)(t 2 3t + 1) 2, Signatures do not help, since t 2 3t + 1 does not have roots on the unit circle. K 3 1 # 4 1 # ? Need to work with finer invariants arising from Levine s Invariants of Knot Cobordism.

9 Levine s algebraic concordance group. G Z. The previous examples showed each K is not concordant to a knot J with a smaller Seifert matrix. Algebraic concordance group G Z of integer matrices V : 1. det(v V t ) = 1; 2. V is Witt trivial (or algebraically slice) if the bilinear form defined by V vanishes on a half-rank summand. (ie. V has a half-dimensional block of zeros.) 3. V 1 V 2 if V 1 V 2 is Witt trivial; In showing g c (6 2 ) = 2 we showed that the Seifert matrix for 6 2 does not have a 2 2 representative in G Z. In showing g c (6 2 # 6 2 ) = 4 we showed that the Seifert matrix for 6 2 # 6 2 does not have a 6 6 representative if G Z.

10 Levine s group of isometric structures. G Q. The Witt group of rational isometric structures: pairs (B,T) where B is a nonsingular, symmetric rational bilinear form and T is an isometry. An isometric structure is Witt trivial if B vanishes on a half-dimensional T invariant subspace (implying B = 0 W (Q), the Witt group of symmetric nonsingular rational bilinear forms). There is an injective homomorphism G Z G Q. V (V + V t,v 1 V t ) = (B,T) V (t) = det(v ) T (t), where T (t) = char. polynomial. One studies G Q by splitting it as G Q = δ GQ δ where Gδ Q is the subgroup of isometric structures with characteristic polynomial a power of the (symmetric) irreducible polynomial δ Q[t].

11 K = 9 40, g 3 (K) = 3, K = (t 2 t + 1)(t 2 3t + 1) 2. g c (K) = 3. (B,T) = (V + V t,v 1 V t ). G Q = δ G δ Q. (B K,T K ) = (B 1,T 1 ) t2 t+1 (B 2,T 2 ) t2 3t+1. Show B 2 is nontrivial. B 2 can be diagonalized over Q: ( 1) ( 2) There is a homomorphism of Witt groups of bilinear forms: 5 : W (Q) W (Z/5) taking B 2 to [ ] This is nontrivial in W (Z/5); thus (B 2,T 2 ) t2 3t+1 is nontrivial, and the t 2 3t + 1 factor is essential, implying g c (K) = 3.

12 K = 8 18, g 3 (K) = 3, σ(k) = 0, K = (t 2 t + 1) 2 (t 2 3t + 1). (B,T) = (V + V t,v 1 V t ). G Q = δ G δ Q. (B K,T K ) = (B 1,T 1 ) t2 t+1 (B 2,T 2 ) t2 3t+1. B 1 can be diagonalized over Q: ( 2) ( 2) There is a homomorphism of Witt groups of bilinear forms: 3 : W (Q) W (Z/3) taking B 1 to [ ] This is nontrivial in W (Z/3); thus (B 1,T 1 ) t2 t+1 is nontrivial, and the t 2 t + 1 factor is essential, implying g c (K) = 3.

13 Example 4: K = g c (K) = 2, 3, 4? g 3 (K) = 4, g 4 (K) = 1, σ(k) = 2, K (t) = (t 2 t + 1) 2 (t 4 2t 3 + t 2 2t + 1) (B K,T K ) = (B 1,T 1 ) t2 t+1 (B 2,T 2 ) t4 2t 3 +t 2 2t+1. B 1 = This bilinear form, B 1, is Witt trivial. Via results of Levine and Milnor, the isometric structure (B 1,T 1 ) is Witt trivial. Thus, K does have a rank 4 (genus 2) representative in G Q.

14 Ex. 4: K = has algebraic concordance genus 2? Yes. K (t) = (t 4 2t 3 + t 2 2t + 1)(t 2 t + 1) 2 V K has a 4-d representative in G Q. How about in G Z? One can work backwards and find the genus 2 Seifert matrix representative for the algebraic concordance class of K This is not an integral Seifert matrix, but half of it is. Multiplication by 2 induces an involution of G Q. Careful application of Levine s complete set of p adic invariants shows this form is invariant under the involution. The only primes to check divide 2det(V )disc( V (t)) i.e. 2,7. Stoltzfus: Unraveling the Integral Knot Concordance Group.

15 Interlude: Casson-Gordon theory. For p and q primes, let M(K,p) be the p-fold branched cover of K and let H 1 (M(K,p)) q denote the q torsion in H 1 (M(K,p)). Theorem (C-G) If K is slice, there is a square root order subgroup H H 1 (M(K,p)) q, such that for every χ : H 1 (M(K,p)) q Z q vanishing on H, τ χ (K) = 0. τ χ (K) W (Q(ζ q r)(t)) Z[ 1 q ]. Theorem (Kirk-L.) (Herald-Kirk-L.) The discriminant (determinant) of τ χ (K) is the computable order of the twisted homology group H 1 (S 3 K, {(Q(ζ q )[t,t 1 ]) p }) as a Q(ζ q )[t,t 1 ] module, for some twisted coefficient group. Disc(τ χ (K)) = K,χ (t) Q(ζ q )[t,t 1 ], the twisted polynomial. Theorem If K is slice, then for the appropriate χ, K,χ (t) = af (t)f (t 1 ) Q(ζ q r)[t,t 1 ]. r = 1 if q odd.

16 Ex. 4: 10 82, C-G theory. K (t) = (t 4 2t 3 + t 2 2t + 1)(t 2 t + 1) 2 If K is concordant to J with g 3 (J) = 3, then J (t) = (t 4 2t 3 + t 2 2t + 1)((a + 1)t a)(at (a + 1)). Considering 3 fold branched covers: H 1 (M(K,3)) 2 = Z/8 Z/8 H 1 (M(J,3)) 2 = Z/2 Z/2. Regardless of the metabolizer H H 1 (M(K# J,3)) 2, there is a nontrivial χ: H 1 (M(K# J,3)) 2 Z/2 which vanishes on H and on H 1 (M( J,3)) 2. For any nontrivial χ, K,χ (t) factors. For χ trivial on H 1 (M( J,3)), J,χ (t) = t 4 8t t 2 8t + 1. This does not factor in Q(ζ 4 )[t] but factors in Q(ζ 8 )[t]. Therefore, K # J is not ribbon and K is not ribbon concordant to a genus 3 knot. (Work in progress.)

17 Another application of C-G invariants. Theorem For all N there exists an unknotting number 1 (g 4 (K) = 1), algebraically slice knot K with g c (K) = N. Proof outline: If H 1 (M(K,2),Z/q) = (Z/q) 2N and K is concordant to a knot J of genus M < N then H 1 (M(K # J,2),Z/q) = (Z/q) 2N (Z/q) 2M, M M. For some metabolizer H K H 1 (M(K,2),Z/q) and some proper subgroup H 0 H K, the value of τ(k,χ) is constant on the cosets of H 0 in H. Construct examples by starting with simple genus N unknotting number 1 knot and repeatedly knotting the surface while preserving the unknotting number.

18 Primes and G Q Theorem (Levine): (Q,T) G Q is trivial if and only if it is trivial in G F for F = R and F = Q p for all p adic rational fields. Theorem In checking triviality for a class V K G Z, you can restrict to primes dividing 2det(V + V t )det(v )disc( T (t)). (For detecting order 4, 2 det(v + V t ) suffices, T. Morita.) If p does not divide det(v + V t ), then (V + V t ) represents a class in W(Z p ). If p does not divide det(v), then V 1 V t represents an isometry of B K over Z p. Thus, if p does not divide det(v)det(v + V t ), then (V + V t, V 1 V t ) G Zp. p : W(Q p ) W(Z/p) has kernel W(Z p ).

19 Primes and G Q Theorem In checking triviality for a class V K G Z, you can restrict to primes dividing 2det(V + V t )det(v )disc( T (t)). If p does not divide det(v )det(v + V t ) then (V + V t,v 1 V t ) G Zp. There is the Q p splitting: G Qp = δ G δ Q p. If M = (Q,T) G Zp and p does not divide disc( T (t)), then M splits in δ G δ Z p. disc( T (t)) = (α i α j ) 2, over all roots of T (t). Example: Z[t] <fg> = Z[t] f Z[t] g if Res(f,g) = (α i β j ) = 1.

20 Problems Identify smooth invariants that distinguish g c and g 4. In particular, find a knot K with K (t) = 1 and g 4 (K) g c (K). Examples showed and Is ? In general, find all concordance relations among (pairs of) low-crossing number knots. Such relations give many examples of simple knots (eg # 3 1 ) which represent torsion in G Z. Are these of infinite order in the concordance group? Ribbon vs. slice questions: