Khovanov Homology and Periodic Knots

Khovanov homology and periodic knots

Maciej Borodzik www.mimuw.edu.pl/~mcboro joint with Wojciech Politarczyk

Institute of Mathematics, University of Warsaw

Bern, June 2017 Periodic knots

Deﬁnition A knot K ⊂ S3 is p–periodic if it admits a rotational symmetry with the symmetry axis disjoint from K .

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 2 / 20 Periodic knots

Deﬁnition A knot K ⊂ S3 is p–periodic if it admits a rotational symmetry with the symmetry axis disjoint from K .

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 2 / 20 Often one can ﬁnd factors of ∆ over integers.

Alexander polynomial obstruction

Theorem (Murasugi criterion) Suppose K ⊂ S3 is a p-periodic knot with p a prime. Let ∆ be the Alexander polynomial of K and ∆0 be the Alexander polynomial of the quotient knot K /Zp. Let l be the absolute value of linking number of K with the symmetry axis. Then ∆0|∆ and up to multiplication by a power of t we have p l−1 p−1 ∆ ≡ ∆0(1 + t + ... + t ) mod p. (1)

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 3 / 20 Alexander polynomial obstruction

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 3 / 20 m Then Zp acts on Σ (K ) – k–fold branched cover; m Then Zp acts on H1(Σ (K )). m Look at Zqm summands of H1(Σ (K )). Zp can ﬁx it, or permute different such summands. The number of ﬁxed components is controlled by |∆0(−1)|. m Restrictions for H1(Σ (K )) of periodic knots.

Naik’s criterion

3 Suppose Zp acts on S keeping K ;

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 4 / 20 m Then Zp acts on H1(Σ (K )). m Look at Zqm summands of H1(Σ (K )). Zp can ﬁx it, or permute different such summands. The number of ﬁxed components is controlled by |∆0(−1)|. m Restrictions for H1(Σ (K )) of periodic knots.

Naik’s criterion

3 Suppose Zp acts on S keeping K ; m Then Zp acts on Σ (K ) – k–fold branched cover;

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 4 / 20 m Look at Zqm summands of H1(Σ (K )). Zp can ﬁx it, or permute different such summands. The number of ﬁxed components is controlled by |∆0(−1)|. m Restrictions for H1(Σ (K )) of periodic knots.

Naik’s criterion

3 Suppose Zp acts on S keeping K ; m Then Zp acts on Σ (K ) – k–fold branched cover; m Then Zp acts on H1(Σ (K )).

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 4 / 20 Zp can ﬁx it, or permute different such summands. The number of ﬁxed components is controlled by |∆0(−1)|. m Restrictions for H1(Σ (K )) of periodic knots.

Naik’s criterion

3 Suppose Zp acts on S keeping K ; m Then Zp acts on Σ (K ) – k–fold branched cover; m Then Zp acts on H1(Σ (K )). m Look at Zqm summands of H1(Σ (K )).

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 4 / 20 The number of ﬁxed components is controlled by |∆0(−1)|. m Restrictions for H1(Σ (K )) of periodic knots.

Naik’s criterion

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 4 / 20 m Restrictions for H1(Σ (K )) of periodic knots.

Naik’s criterion

3 Suppose Zp acts on S keeping K ; m Then Zp acts on Σ (K ) – k–fold branched cover; m Then Zp acts on H1(Σ (K )). m Look at Zqm summands of H1(Σ (K )). Zp can ﬁx it, or permute different such summands. The number of ﬁxed components is controlled by |∆0(−1)|.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 4 / 20 Naik’s criterion

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 4 / 20 Zp acts on the spin-c structures. Acts on Ozsváth–Szabó d–invariants. These invariants appear with multiplicities. If K is quasi-alternating, then Σ2(K ) is an L–space and d-invariants are computable.

Jabuka–Naik criterion

Consider Σk (K );

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 5 / 20 Acts on Ozsváth–Szabó d–invariants. These invariants appear with multiplicities. If K is quasi-alternating, then Σ2(K ) is an L–space and d-invariants are computable.

Jabuka–Naik criterion

Consider Σk (K );

Zp acts on the spin-c structures.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 5 / 20 These invariants appear with multiplicities. If K is quasi-alternating, then Σ2(K ) is an L–space and d-invariants are computable.

Jabuka–Naik criterion

Consider Σk (K );

Zp acts on the spin-c structures. Acts on Ozsváth–Szabó d–invariants.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 5 / 20 If K is quasi-alternating, then Σ2(K ) is an L–space and d-invariants are computable.

Jabuka–Naik criterion

Consider Σk (K );

Zp acts on the spin-c structures. Acts on Ozsváth–Szabó d–invariants. These invariants appear with multiplicities.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 5 / 20 Jabuka–Naik criterion

Consider Σk (K );

Zp acts on the spin-c structures. Acts on Ozsváth–Szabó d–invariants. These invariants appear with multiplicities. If K is quasi-alternating, then Σ2(K ) is an L–space and d-invariants are computable.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 5 / 20 Computable, if we know a representation. It is known when knot group admits a representation into a dihedral group. Other representations are sometimes harder to ﬁnd. None of the above criteria can be used for ∆ = 1 knots. The TAP criterion is possible, but requires ﬁnding non-trivial representations.

Twisted Alexander Polynomial criterion

Hillman, Livingston and Naik generalize the Murasugi’s periodicity criterion for twisted Alexander polynomials.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 6 / 20 It is known when knot group admits a representation into a dihedral group. Other representations are sometimes harder to ﬁnd. None of the above criteria can be used for ∆ = 1 knots. The TAP criterion is possible, but requires ﬁnding non-trivial representations.

Twisted Alexander Polynomial criterion

Hillman, Livingston and Naik generalize the Murasugi’s periodicity criterion for twisted Alexander polynomials. Computable, if we know a representation.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 6 / 20 Other representations are sometimes harder to ﬁnd. None of the above criteria can be used for ∆ = 1 knots. The TAP criterion is possible, but requires ﬁnding non-trivial representations.

Twisted Alexander Polynomial criterion

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 6 / 20 Twisted Alexander Polynomial criterion

Hillman, Livingston and Naik generalize the Murasugi’s periodicity criterion for twisted Alexander polynomials. Computable, if we know a representation. It is known when knot group admits a representation into a dihedral group. Other representations are sometimes harder to ﬁnd. None of the above criteria can be used for ∆ = 1 knots. The TAP criterion is possible, but requires ﬁnding non-trivial representations.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 6 / 20 Effective for p > 3. Theorem (HOMFLYPT criterion)

±1 ±1 −1 a+a−1 Let R be a unital subring in Z[a , z ] generated by a, a , z and p z. For a prime number p let Ip be the ideal in R generated by p and z . If a knot K is p-periodic and P(a, z) is its HOMFLYPT polynomial, then

−1 P(a, z) ≡ P(a , z) mod Ip. Przytycki shows an effective way of applying Theorem 4.

Jones and HOMFLYPT

Theorem −1 p −p If K is p–periodic, then JK (q) − JK (q) ≡ 0 mod (q − q , p).

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 7 / 20 Theorem (HOMFLYPT criterion)

−1 P(a, z) ≡ P(a , z) mod Ip. Przytycki shows an effective way of applying Theorem 4.

Jones and HOMFLYPT

Theorem −1 p −p If K is p–periodic, then JK (q) − JK (q) ≡ 0 mod (q − q , p). Effective for p > 3.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 7 / 20 Przytycki shows an effective way of applying Theorem 4.

Jones and HOMFLYPT

Theorem −1 p −p If K is p–periodic, then JK (q) − JK (q) ≡ 0 mod (q − q , p). Effective for p > 3. Theorem (HOMFLYPT criterion)

−1 P(a, z) ≡ P(a , z) mod Ip.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 7 / 20 Jones and HOMFLYPT

Theorem −1 p −p If K is p–periodic, then JK (q) − JK (q) ≡ 0 mod (q − q , p). Effective for p > 3. Theorem (HOMFLYPT criterion)

−1 P(a, z) ≡ P(a , z) mod Ip. Przytycki shows an effective way of applying Theorem 4.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 7 / 20 Take a ring R. Then CKh(D; R) has a structure of Λ = R[Zp]–module.

Deﬁnition (Politarczyk) For any Λ–module M deﬁne the equivariant Khovanov homology as

EKh(L; M) = ExtΛ(M, CKh(D; R)).

Does not depend on the choice of the diagram. In a similar way one can show the existence of equivariant Lee theory. Most important example: M = Λ.

Equivariant Khovanov homology

For a periodic diagram D, the group Zp permutes the cube of resolution.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 8 / 20 Deﬁnition (Politarczyk) For any Λ–module M deﬁne the equivariant Khovanov homology as

EKh(L; M) = ExtΛ(M, CKh(D; R)).

Does not depend on the choice of the diagram. In a similar way one can show the existence of equivariant Lee theory. Most important example: M = Λ.

Equivariant Khovanov homology

For a periodic diagram D, the group Zp permutes the cube of resolution. Take a ring R. Then CKh(D; R) has a structure of Λ = R[Zp]–module.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 8 / 20 Does not depend on the choice of the diagram. In a similar way one can show the existence of equivariant Lee theory. Most important example: M = Λ.

Equivariant Khovanov homology

For a periodic diagram D, the group Zp permutes the cube of resolution. Take a ring R. Then CKh(D; R) has a structure of Λ = R[Zp]–module.

Deﬁnition (Politarczyk) For any Λ–module M deﬁne the equivariant Khovanov homology as

EKh(L; M) = ExtΛ(M, CKh(D; R)).

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 8 / 20 In a similar way one can show the existence of equivariant Lee theory. Most important example: M = Λ.

Equivariant Khovanov homology

For a periodic diagram D, the group Zp permutes the cube of resolution. Take a ring R. Then CKh(D; R) has a structure of Λ = R[Zp]–module.

Deﬁnition (Politarczyk) For any Λ–module M deﬁne the equivariant Khovanov homology as

EKh(L; M) = ExtΛ(M, CKh(D; R)).

Does not depend on the choice of the diagram.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 8 / 20 Most important example: M = Λ.

Equivariant Khovanov homology

For a periodic diagram D, the group Zp permutes the cube of resolution. Take a ring R. Then CKh(D; R) has a structure of Λ = R[Zp]–module.

Deﬁnition (Politarczyk) For any Λ–module M deﬁne the equivariant Khovanov homology as

EKh(L; M) = ExtΛ(M, CKh(D; R)).

Does not depend on the choice of the diagram. In a similar way one can show the existence of equivariant Lee theory.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 8 / 20 Equivariant Khovanov homology

For a periodic diagram D, the group Zp permutes the cube of resolution. Take a ring R. Then CKh(D; R) has a structure of Λ = R[Zp]–module.

Deﬁnition (Politarczyk) For any Λ–module M deﬁne the equivariant Khovanov homology as

EKh(L; M) = ExtΛ(M, CKh(D; R)).

Does not depend on the choice of the diagram. In a similar way one can show the existence of equivariant Lee theory. Most important example: M = Λ.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 8 / 20 i If R = Zm and p is invertible in R, then ExtΛ = 0 for i > 0 and EKh(L; Λ) = Kh(L; R). On the other hand we have a Schur decomposition of HomΛ(Λ; CKh(D)).

Equivariant Khovanov. Properties.

We can deﬁne EKhd (L) = EKh(L; Z[ξd ]) for any d|p. This is the third gradation, coming from representations of Zp.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 9 / 20 On the other hand we have a Schur decomposition of HomΛ(Λ; CKh(D)).

Equivariant Khovanov. Properties.

We can deﬁne EKhd (L) = EKh(L; Z[ξd ]) for any d|p. This is the third gradation, coming from representations of Zp. i If R = Zm and p is invertible in R, then ExtΛ = 0 for i > 0 and EKh(L; Λ) = Kh(L; R).

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 9 / 20 Equivariant Khovanov. Properties.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 9 / 20 Equivariant Lee homology for knots is easy. The equivariant Khovanov polynomial and the equivariant Lee polynomial differ by a speciﬁc polynomial. And EKh(L; Λ) splits as a sum over different representations of Λ.

Equivariant Lee spectral sequence

There exists an equivariant Lee spectral sequence (if R = Zq, q > 2, or R = Z or R = Q).

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 10 / 20 The equivariant Khovanov polynomial and the equivariant Lee polynomial differ by a speciﬁc polynomial. And EKh(L; Λ) splits as a sum over different representations of Λ.

Equivariant Lee spectral sequence

There exists an equivariant Lee spectral sequence (if R = Zq, q > 2, or R = Z or R = Q). Equivariant Lee homology for knots is easy.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 10 / 20 And EKh(L; Λ) splits as a sum over different representations of Λ.

Equivariant Lee spectral sequence

There exists an equivariant Lee spectral sequence (if R = Zq, q > 2, or R = Z or R = Q). Equivariant Lee homology for knots is easy. The equivariant Khovanov polynomial and the equivariant Lee polynomial differ by a speciﬁc polynomial.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 10 / 20 Equivariant Lee spectral sequence

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 10 / 20 Main criterion

Theorem (—,Politarczyk, 2017) Let K be a pn-periodic, where p is an odd prime. Suppose that F = Q or Fr , for a prime r such that r 6= p, and r has maximal order in the n multiplicative group mod p . Set c = 1 if F = F2 and c = 2 otherwise. Then the Khovanov polynomial KhP(K ; F) decomposes as

n X j j−1 KhP(K ; F) = P0 + (p − p )Pj , (2) j=1

where ±1 ±1 P0, P1,..., Pn ∈ Z[q , t ], are Laurent polynomials such that

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 11 / 20 2 P∞ 2cj Pk = j=1(1 + tq )Skj (t, q) and the polynomials Skj have non-negative coefﬁcients for 1 ≤ k ≤ n,

3 c If the width of Kh(K ; F) is equal to w, then Skj = 0 for j > 2 w. −1 −1 4 Pk (−1, q) − Pk+1(−1, q) ≡ Pk (−1, q ) − Pk+1(−1, q ) − − (mod qpn k − q−pn k ); Without (4) the condition is trivial!

Main criterion

1 s(K ,F) −1 P∞ 2cj P0 = q (q + q ) + j=1(1 + tq )S0j (t, q), and the polynomials S0j have non-negative coefﬁcients;

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 12 / 20 3 c If the width of Kh(K ; F) is equal to w, then Skj = 0 for j > 2 w. −1 −1 4 Pk (−1, q) − Pk+1(−1, q) ≡ Pk (−1, q ) − Pk+1(−1, q ) − − (mod qpn k − q−pn k ); Without (4) the condition is trivial!

Main criterion

1 s(K ,F) −1 P∞ 2cj P0 = q (q + q ) + j=1(1 + tq )S0j (t, q), and the polynomials S0j have non-negative coefﬁcients;

2 P∞ 2cj Pk = j=1(1 + tq )Skj (t, q) and the polynomials Skj have non-negative coefﬁcients for 1 ≤ k ≤ n,

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 12 / 20 −1 −1 4 Pk (−1, q) − Pk+1(−1, q) ≡ Pk (−1, q ) − Pk+1(−1, q ) − − (mod qpn k − q−pn k ); Without (4) the condition is trivial!

Main criterion

1 s(K ,F) −1 P∞ 2cj P0 = q (q + q ) + j=1(1 + tq )S0j (t, q), and the polynomials S0j have non-negative coefﬁcients;

2 P∞ 2cj Pk = j=1(1 + tq )Skj (t, q) and the polynomials Skj have non-negative coefﬁcients for 1 ≤ k ≤ n,

3 c If the width of Kh(K ; F) is equal to w, then Skj = 0 for j > 2 w.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 12 / 20 Without (4) the condition is trivial!

Main criterion

1 s(K ,F) −1 P∞ 2cj P0 = q (q + q ) + j=1(1 + tq )S0j (t, q), and the polynomials S0j have non-negative coefﬁcients;

2 P∞ 2cj Pk = j=1(1 + tq )Skj (t, q) and the polynomials Skj have non-negative coefﬁcients for 1 ≤ k ≤ n,

3 c If the width of Kh(K ; F) is equal to w, then Skj = 0 for j > 2 w. −1 −1 4 Pk (−1, q) − Pk+1(−1, q) ≡ Pk (−1, q ) − Pk+1(−1, q ) − − (mod qpn k − q−pn k );

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 12 / 20 Main criterion

1 s(K ,F) −1 P∞ 2cj P0 = q (q + q ) + j=1(1 + tq )S0j (t, q), and the polynomials S0j have non-negative coefﬁcients;

2 P∞ 2cj Pk = j=1(1 + tq )Skj (t, q) and the polynomials Skj have non-negative coefﬁcients for 1 ≤ k ≤ n,

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 12 / 20 Example

Consider knot 15n135221.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 13 / 20 Example

Consider knot 15n135221.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 13 / 20 No Naik criterion either. No Jabuka–Naik criterion. Passes MPT criterion for period 5. Let’s check Khovanov. We do not need to calculate equivariant Khovanov.

We work over Z3.

∆(15n1335221) = 1. No Alexander criterion.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 14 / 20 No Jabuka–Naik criterion. Passes MPT criterion for period 5. Let’s check Khovanov. We do not need to calculate equivariant Khovanov.

We work over Z3.

∆(15n1335221) = 1. No Alexander criterion. No Naik criterion either.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 14 / 20 Passes MPT criterion for period 5. Let’s check Khovanov. We do not need to calculate equivariant Khovanov.

We work over Z3.

∆(15n1335221) = 1. No Alexander criterion. No Naik criterion either. No Jabuka–Naik criterion.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 14 / 20 Let’s check Khovanov. We do not need to calculate equivariant Khovanov.

We work over Z3.

∆(15n1335221) = 1. No Alexander criterion. No Naik criterion either. No Jabuka–Naik criterion. Passes MPT criterion for period 5.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 14 / 20 We do not need to calculate equivariant Khovanov.

We work over Z3.

∆(15n1335221) = 1. No Alexander criterion. No Naik criterion either. No Jabuka–Naik criterion. Passes MPT criterion for period 5. Let’s check Khovanov.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 14 / 20 We work over Z3.

∆(15n1335221) = 1. No Alexander criterion. No Naik criterion either. No Jabuka–Naik criterion. Passes MPT criterion for period 5. Let’s check Khovanov. We do not need to calculate equivariant Khovanov.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 14 / 20 ∆(15n1335221) = 1. No Alexander criterion. No Naik criterion either. No Jabuka–Naik criterion. Passes MPT criterion for period 5. Let’s check Khovanov. We do not need to calculate equivariant Khovanov.

We work over Z3.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 14 / 20 Khovanov homology

13 11 3 9 5 7 7 3 5 3 8 5 3 5 9 7 1 8 10 8 −1 8 11 8 −3 8 10 7 −5 7 9 5 −7 5 8 3 −9 3 7 −11 5 −13 3 −15 t −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 15 / 20 Khovanov Polynomial

KhP = q + q−1 + (1 + tq4)(t−7q−15 + 3t−6q−13 + t−5q−11+ + 3t−4q−9 + t−3q−9 + 3t−2q−7 + t−1q−5 + 3t−1q−3 + q−3 + q−1 + 3tq+ + t2q3 + 3t3q3 + t4q5 + 3t5q7 + t6q9 + 4(t−5q−11 + t−4q−9 + 2t−3q−7 + 2t−2q−5+ + t−1q−5 + t−1q−3 + 2tq−1 + q−3 + q−1 + 2t2q + t3q3 + t4q5)).

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 16 / 20 0 −7 −15 −6 −13 −5 −11 −4 −9 −3 −9 −2 −7 S01 = t q + 3t q + t q + 3t q + t q + 3t q + t−1q−5 + 3t−1q−3 + q−3 + q−1 + 3tq + t2q3 + 3t3q3+ + t4q5 + 3t5q7 + t6q9, 0 −5 −11 −4 −9 −3 −7 −2 −5 −1 −5 S11 = t q + t q + 2t q + 2t q + t q + + t−1q−3 + 2tq−1 + q−3 + q−1 + 2t2q+ + t3q3 + t4q5.

Tentative decomposition

−1 4 0 4 0 KhP = q + q + (1 + tq )S01 + 4(1 + tq )S11, where

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 17 / 20 Tentative decomposition

−1 4 0 4 0 KhP = q + q + (1 + tq )S01 + 4(1 + tq )S11, where

0 −7 −15 −6 −13 −5 −11 −4 −9 −3 −9 −2 −7 S01 = t q + 3t q + t q + 3t q + t q + 3t q + t−1q−5 + 3t−1q−3 + q−3 + q−1 + 3tq + t2q3 + 3t3q3+ + t4q5 + 3t5q7 + t6q9, 0 −5 −11 −4 −9 −3 −7 −2 −5 −1 −5 S11 = t q + t q + 2t q + 2t q + t q + + t−1q−3 + 2tq−1 + q−3 + q−1 + 2t2q+ + t3q3 + t4q5.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 17 / 20 −1 4 0 0 Ξ(e q) = (q + q + (1 + tq )(S01 − S11)|t=−1 and set Ξ := (Ξ(e q) − Ξ(e q−1)) mod q5 − q−5. We have then

Ξ = −10q + 5q3 − 5q7 + 10q9.

0 0 Ξ is not zero, but we might get zero if we choose different S01 and S11. In general checking all possibilities requires 20736 possibilities.

Main criterion

According to the main criterion deﬁne

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 18 / 20 We have then

Ξ = −10q + 5q3 − 5q7 + 10q9.

0 0 Ξ is not zero, but we might get zero if we choose different S01 and S11. In general checking all possibilities requires 20736 possibilities.

Main criterion

According to the main criterion deﬁne −1 4 0 0 Ξ(e q) = (q + q + (1 + tq )(S01 − S11)|t=−1 and set Ξ := (Ξ(e q) − Ξ(e q−1)) mod q5 − q−5.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 18 / 20 0 0 Ξ is not zero, but we might get zero if we choose different S01 and S11. In general checking all possibilities requires 20736 possibilities.

Main criterion

According to the main criterion deﬁne −1 4 0 0 Ξ(e q) = (q + q + (1 + tq )(S01 − S11)|t=−1 and set Ξ := (Ξ(e q) − Ξ(e q−1)) mod q5 − q−5. We have then

Ξ = −10q + 5q3 − 5q7 + 10q9.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 18 / 20 In general checking all possibilities requires 20736 possibilities.

Main criterion

According to the main criterion deﬁne −1 4 0 0 Ξ(e q) = (q + q + (1 + tq )(S01 − S11)|t=−1 and set Ξ := (Ξ(e q) − Ξ(e q−1)) mod q5 − q−5. We have then

Ξ = −10q + 5q3 − 5q7 + 10q9.

0 0 Ξ is not zero, but we might get zero if we choose different S01 and S11.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 18 / 20 Main criterion

According to the main criterion deﬁne −1 4 0 0 Ξ(e q) = (q + q + (1 + tq )(S01 − S11)|t=−1 and set Ξ := (Ξ(e q) − Ξ(e q−1)) mod q5 − q−5. We have then

Ξ = −10q + 5q3 − 5q7 + 10q9.

0 0 Ξ is not zero, but we might get zero if we choose different S01 and S11. In general checking all possibilities requires 20736 possibilities.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 18 / 20 5 −5 i 0 Reducing modulo q − q we get Tij = (−1) Rj0 with j = j mod 10 and

9 R1 = R5 = 5(q − q ), 3 7 R3 = 10(q − q ), 3 7 9 R7 = R9 = 5(−q − q + q + q ).

Speed up

u j 0 0 0 0 Let δ = at q . The change S11 7→ S11 − δ, S01 7→ S01 + 4δ induces the change Ξ 7→ Ξ + aTij , where

i −j−4 −j j j+4 5 −5 Tij = (−1) 5(−q + q − q + q ) mod (q − q ).

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 19 / 20 Speed up

u j 0 0 0 0 Let δ = at q . The change S11 7→ S11 − δ, S01 7→ S01 + 4δ induces the change Ξ 7→ Ξ + aTij , where

i −j−4 −j j j+4 5 −5 Tij = (−1) 5(−q + q − q + q ) mod (q − q ).

5 −5 i 0 Reducing modulo q − q we get Tij = (−1) Rj0 with j = j mod 10 and

9 R1 = R5 = 5(q − q ), 3 7 R3 = 10(q − q ), 3 7 9 R7 = R9 = 5(−q − q + q + q ).

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 19 / 20 This is impossible. The knot is not periodic.

Speed up 2

0 δ is such that S11 − δ and δ have non-negative coefﬁcients. Hence the question is, whether

Ξ = a1R1 + a3R3 + a7R7

with

a1 ∈ {−1, 0, 1, 2, 3, 4, 5, 6},

a3 ∈ {−3, −2, −1, 0},

a7 ∈ {−4, −3, −2, −1, 0, 1, 2}.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 20 / 20 Speed up 2

0 δ is such that S11 − δ and δ have non-negative coefﬁcients. Hence the question is, whether

Ξ = a1R1 + a3R3 + a7R7

with

a1 ∈ {−1, 0, 1, 2, 3, 4, 5, 6},

a3 ∈ {−3, −2, −1, 0},

a7 ∈ {−4, −3, −2, −1, 0, 1, 2}.

This is impossible. The knot is not periodic.

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 20 / 20 2 Linking form on H1(Σ (K )) (generalize Naik)? Blanchﬁeld form of periodic links (A. Conway)? Twisted Blanchﬁeld forms? Is 15n166130 5–periodic?

Questions

Khovanov-Rozanski homology?

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 21 / 20 Blanchﬁeld form of periodic links (A. Conway)? Twisted Blanchﬁeld forms? Is 15n166130 5–periodic?

Questions

Khovanov-Rozanski homology? 2 Linking form on H1(Σ (K )) (generalize Naik)?

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 21 / 20 Twisted Blanchﬁeld forms? Is 15n166130 5–periodic?

Questions

Khovanov-Rozanski homology? 2 Linking form on H1(Σ (K )) (generalize Naik)? Blanchﬁeld form of periodic links (A. Conway)?

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 21 / 20 Is 15n166130 5–periodic?

Questions

Khovanov-Rozanski homology? 2 Linking form on H1(Σ (K )) (generalize Naik)? Blanchﬁeld form of periodic links (A. Conway)? Twisted Blanchﬁeld forms?

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 21 / 20 Questions

Khovanov-Rozanski homology? 2 Linking form on H1(Σ (K )) (generalize Naik)? Blanchﬁeld form of periodic links (A. Conway)? Twisted Blanchﬁeld forms? Is 15n166130 5–periodic?

Maciej Borodzik (Institute of Mathematics, UniversityKhovanov of Warsaw) homology and periodic knots Bern, June 2017 21 / 20