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Localized metabelian rho invariants as obstructions totorsion in the knot concordance group.

Chris Davis, Rice University

July 22, 2010

Chris Davis, Rice University () Localized metabelian rho invariants as obstructions to torsion in the knot concordance group.July 22, 2010 1 / 37

Goals

1 Introduce some ρ-invariants corresponding to localizations of theAlexander module of a knot.

2 Prove that they obstruct knots which are torsion in the algebraicconcordance group from being torsion in the concordance group.

3 Prove computable bounds on the value of these invariants.

4 Do a computation to prove the theorem about twist knots which is onthe next slide.

5 State a theorem regarding these invariants and iterated infection.


An application: Twist Knots

Theorem (D.)

No nontrivial linear combination of the twist knots,T−7,T−13,T−21 . . .T−x2−x−1 (x ≥ 2), is topologically slice. These knotsare each of algebraic order 2.

This is the mirror image of what others refer to as the −n twist knot.

n

Figure: Tn: the n-twist knot.


Previous related results

Theorem (A. Tamulis, 2000)

No nontrivial combination of the twist knots T−n with n ≥ 3 and 4n + 1prime is topologically slice. Each of these twist knots is algebraically oforder 2.

Theorem (C. Livingston, 2001)

Let pi be an enumeration of the primes congruent to 3 mod 4. Letni = p2i−1p2i − 1. No nontrivial linear combination of the knots T−ni isslice. Each of these twist knots is algebraically of order 4.

Theorem (S. Kim, 2005)

No nontrivial linear combination of the twist knots, except for the 0 twistknot (unknot), the −1 twist knot (figure eight) and the −2 twist knot(stevedore) is ribbon.


Twist knots and infection

The computation I do to prove linear independence for these twist knotscan be used to prove the following theorem about infection:

Theorem

For a sufficiently nice slice knot, R, and a sufficiently nice infecting curve,η, the set resulting from iterated infection of R along η by each ofT−7,T−21, . . . ,T−x2−x−1 is linearly independent in the concordance group.

example of niceness: any slice knot with Alexander module of the formQ[t±1]

p(t)2with p(t) prime and symmetric and η a generator of the Alexander

module.Tristram-Levine signature of the deepest knot in the infection iseverywhere zero.




Theorem




module.

Tristram-Levine signature of the deepest knot in the infection iseverywhere zero.




Theorem




module.Tristram-Levine signature of the deepest knot in the infection iseverywhere zero.


Invariants of interest: ρ0

For L an n component link with zero linking numbers, M(L) denotes zero

surgery along L.

Let φ : π1(M(L))→ π1(M(L))

π1(M(L))(1)= Zn be the

Abelianization map.

Definition

ρ0(L) := ρ (M(L), φ)

For a knot it is the integral of the Tristram-Levine Signature.(Cochran-Orr-Teichner ’04)

For links it is computable: In a nice enough case a computer can betaught how to do it. (More on this later)




surgery along L. Let φ : π1(M(L))→ π1(M(L))


Abelianization map.

Definition

ρ0(L) := ρ (M(L), φ)








Abelianization map.

Definition

ρ0(L) := ρ (M(L), φ)








Abelianization map.

Definition

ρ0(L) := ρ (M(L), φ)








Abelianization map.

Definition

ρ0(L) := ρ (M(L), φ)





For K a knot, let φ1 : π1(M(K ))→ π1(M(K ))

π1(M(K ))(2)be the quotient map.

Definition

ρ1(K ) := ρ(M(K ), φ1

)Considered by Cochran-Harvey-Leidy ’08 together with quotientscorresponding to isotropy of the Blanchfield form as a sliceobstruction.

Hard to compute. There is no known procedure.





Definition

ρ1(K ) := ρ(M(K ), φ1

)

Considered by Cochran-Harvey-Leidy ’08 together with quotientscorresponding to isotropy of the Blanchfield form as a sliceobstruction.






Definition

ρ1(K ) := ρ(M(K ), φ1







Definition

ρ1(K ) := ρ(M(K ), φ1




Invariants of interest: ρ1p

Let p(t) be a polynomial.

Rp :=

{f

g∈ Q(t) : (g , p) = 1

}is the localization of Q[t±1] at p.

Let Ap0(K ) = A0(K )⊗ Rp be the localized Alexander module of K .

Let π1(M(K ))(2)p(t) be the kernel of the composition

π1(M(K ))(1) → π1(M(K ))(1)

π1(M(K ))(2)↪→ A0(K ) →

Id⊗1Ap

0(K ).

Let φ1p : π1(M(K ))→ π1(M(K ))

π1(M(K ))(2)p

be the quotient map.

Definition

ρ1p(t)(K ) := ρ

(M(K ), φ1

p

)




Rp :=

{f

g∈ Q(t) : (g , p) = 1




π1(M(K ))(1) → π1(M(K ))(1)

π1(M(K ))(2)↪→ A0(K ) →

Id⊗1Ap

0(K ).

Let φ1p : π1(M(K ))→ π1(M(K ))

π1(M(K ))(2)p


Definition

ρ1p(t)(K ) := ρ

(M(K ), φ1

p

)




Rp :=

{f

g∈ Q(t) : (g , p) = 1




π1(M(K ))(1) → π1(M(K ))(1)

π1(M(K ))(2)↪→ A0(K ) →

Id⊗1Ap

0(K ).

Let φ1p : π1(M(K ))→ π1(M(K ))

π1(M(K ))(2)p


Definition

ρ1p(t)(K ) := ρ

(M(K ), φ1

p

)




Rp :=

{f

g∈ Q(t) : (g , p) = 1




π1(M(K ))(1) → π1(M(K ))(1)

π1(M(K ))(2)↪→ A0(K ) →

Id⊗1Ap

0(K ).

Let φ1p : π1(M(K ))→ π1(M(K ))

π1(M(K ))(2)p


Definition

ρ1p(t)(K ) := ρ

(M(K ), φ1

p

)




Rp :=

{f

g∈ Q(t) : (g , p) = 1




π1(M(K ))(1) → π1(M(K ))(1)

π1(M(K ))(2)↪→ A0(K ) →

Id⊗1Ap

0(K ).

Let φ1p : π1(M(K ))→ π1(M(K ))

π1(M(K ))(2)p


Definition

ρ1p(t)(K ) := ρ

(M(K ), φ1

p

)




Rp :=

{f

g∈ Q(t) : (g , p) = 1




π1(M(K ))(1) → π1(M(K ))(1)

π1(M(K ))(2)↪→ A0(K ) →

Id⊗1Ap

0(K ).

Let φ1p : π1(M(K ))→ π1(M(K ))

π1(M(K ))(2)p


Definition

ρ1p(t)(K ) := ρ

(M(K ), φ1

p

)




Rp :=

{f

g∈ Q(t) : (g , p) = 1




π1(M(K ))(1) → π1(M(K ))(1)

π1(M(K ))(2)↪→ A0(K ) →

Id⊗1Ap

0(K ).

Let φ1p : π1(M(K ))→ π1(M(K ))

π1(M(K ))(2)p


Definition

ρ1p(t)(K ) := ρ

(M(K ), φ1

p

)Chris Davis, Rice University () Localized metabelian rho invariants as obstructions to torsion in the knot concordance group.July 22, 2010 8 / 37

A pair of examples. 1: localize away from the Alexanderpolynomial

Let ∆ be the Alexander polynomial of K .

Suppose (p,∆) = 1. Then1

∆∈ Rp. ∆ annihilates A0, so A0 ⊗ Rp = Ap

0

vanishes.π1(M(K ))

(2)p(t) is the kernel of the composition

π1(M(K ))(1) → π1(M(K ))(1)

π1(M(K ))(2)↪→ A0(K )→Ap

0(K ) = 0.

π1(M(K ))(2)p(t) = π1(M(K ))(1), so ρ1

p(K ) = ρ0(K )


A pair of examples. 2: localize at the Alexanderpolynomial

Suppose p = ∆. Then A0(K ) →Id⊗1

A0(K )⊗ Rp is injective.

π1(M(K ))(2)p(t) is the kernel of the composition

π1(M(K ))(1) → π1(M(K ))(1)

π1(M(K ))(2)↪→ A0(K ) ↪→

Id⊗1Ap

0(K ).

π1(M(K ))(2)p(t) = π1(M(K ))(2), so ρ1

p(K ) = ρ1(K )


Interactions between ρ1p and the Alexander polynomial

Proposition

Let ∆ be the Alexander polynomial of a knot, K.

For any p which is relatively prime to ∆, ρ1p(K ) = ρ0(K ).

ρ1∆(K ) = ρ1(K ).


ρ1p under infection and connected sums

Proposition (D.)

ρ1p(J#K ) = ρ1

p(J) + ρ1p(K ).

Proposition (D.)

For η an unknot representing an element of the localized Alexander

module of J, ρ1p(Jη(K )) =

{ρ1p(J) if η = 0 in Ap

0(J)ρ1p(J) + ρ0(K ) if η 6= 0 in Ap

0(J)

From the second proposition I get the following discouraging example:


ρ1p under infection and connected sums

Proposition (D.)

ρ1p(J#K ) = ρ1

p(J) + ρ1p(K ).

Proposition (D.)

For η an unknot representing an element of the localized Alexander

module of J, ρ1p(Jη(K )) =

{ρ1p(J) if η = 0 in Ap

0(J)ρ1p(J) + ρ0(K ) if η 6= 0 in Ap

0(J)

From the second proposition I get the following discouraging example:


A pair of slice knots with differing ρ1 invariants

Consider the pair of slice knots below. ρ1(Rη(A)) = ρ1(R) + ρ0(A). Inparticular ρ1(R) and ρ1(Rη(A)) cannot both be zero. ρ1 is not welldefined on the concordance group.

Despite this fact, I will provide in the next few slides a setting in whichthese invariants provide useful concordance information.

η A

R Rη(A)

Figure: A pair of slice knots with differing ρ1.


A pair of slice knots with differing ρ1 invariants

Consider the pair of slice knots below. ρ1(Rη(A)) = ρ1(R) + ρ0(A). Inparticular ρ1(R) and ρ1(Rη(A)) cannot both be zero. ρ1 is not welldefined on the concordance group.Despite this fact, I will provide in the next few slides a setting in whichthese invariants provide useful concordance information.

η A

R Rη(A)

Figure: A pair of slice knots with differing ρ1.


Blanchfield forms and isotropy

Q[t±1], and thus Rp when p is symmetric, have an involution given byq(t) = q(t−1)

With respect to this involution, the localized Alexander module has ahermitian form

Blp : Ap0(K )× Ap

0(K )→ Q(t)

Rp.

given by extending the Blanchfield form on A0(K )

Definition

P ⊆ Ap0(K ) is called isotropic if Blp(a, b) = 0 for all a, b ∈ P

Definition

K is called p-anisotropic if there is no nontrivial isotropic submodule ofAp

0(K ).



Q[t±1], and thus Rp when p is symmetric, have an involution given byq(t) = q(t−1)With respect to this involution, the localized Alexander module has ahermitian form

Blp : Ap0(K )× Ap

0(K )→ Q(t)

Rp.


Definition


Definition


0(K ).




Blp : Ap0(K )× Ap

0(K )→ Q(t)

Rp.


Definition


Definition


0(K ).




Blp : Ap0(K )× Ap

0(K )→ Q(t)

Rp.


Definition


Definition


0(K ).


Exploration of anisotropy.

If p is a symmetric prime which divides divides ∆ with multiplicity 1, then

Ap0(K ) is cyclic: Ap

0(K ) =Rp

(p). Let α generate Ap

0(K ).

The Blanchfield form is given by Blp(qα, rα) =qra

pwith (a, p) = 1. If

Blp(qα, rα) = 0 then p divides q or r . By symmetry of p, p|q if and onlyif p|qThus Blp(a, b) = 0 if and only in a or b is zero in Ap

0(K ). In particularthere are no nontrivial isotropic submodules.This proves the following in the case that p is prime.

Proposition

If the Alexander polynomial of a knot is squarefree, then the knot isp-anisotropic for every symmetric polynomial, p.





0(K ) =Rp


0(K ).



Blp(qα, rα) = 0 then p divides q or r . By symmetry of p, p|q if and onlyif p|q

Thus Blp(a, b) = 0 if and only in a or b is zero in Ap0(K ). In particular

there are no nontrivial isotropic submodules.This proves the following in the case that p is prime.

Proposition






0(K ) =Rp


0(K ).




0(K ). In particularthere are no nontrivial isotropic submodules.

This proves the following in the case that p is prime.

Proposition






0(K ) =Rp


0(K ).




0(K ). In particularthere are no nontrivial isotropic submodules.This proves the following in the case that p is prime.

Proposition



First big theorem: ρ1p as an obstruction to linear

dependence in CTheorem (D.)

Given a symmetric polynomial p(t) and (not necessarily distinct)p-anisotropic knots K1, . . .Kn, if K1# . . .#Kn is slice, thenn∑

i=1

ρ1p(t)(Ki ) = 0

Corollary (D.)

If K is of finite concordance order and is p-anisotropic, then ρ1p(K ) = 0

Corollary (D.)

If K1, . . . ,Kn are of finite algebraic order and have coprime, squarefreeAlexander polynomials, then if ρ1(Ki ) is nonzero for each i , {K1, . . . ,Kn}is linearly independent in C.





i=1

ρ1p(t)(Ki ) = 0

Corollary (D.)


Corollary (D.)






i=1

ρ1p(t)(Ki ) = 0

Corollary (D.)


Corollary (D.)



Proof of second corollary

Let pj the the Alexander polynomial of Kj .

If #aiKi is slice, then

0 =n∑

i=1

aiρ1pj

(Ki ) for each j . Since these knots have coprime Alexander

polynomials, ρ1pj

(Ki ) = ρ0(Ki ) if j 6= i and ρ1pj

(Kj) = ρ1(Kj) 6= 0. Since

each of the knots are of finite algebraic order, ρ0(Ki ) = 0. Thus,0 = ajρ

1(Kj). Since ρ1(Kj) 6= 0, aj = 0.No nontrivial combinations of the knots Ki is slice and the set is linearlyindependent



Let pj the the Alexander polynomial of Kj . If #aiKi is slice, then

0 =n∑

i=1

aiρ1pj

(Ki ) for each j .

Since these knots have coprime Alexander

polynomials, ρ1pj








0 =n∑

i=1

aiρ1pj


polynomials, ρ1pj


(Kj) = ρ1(Kj) 6= 0.

Since






0 =n∑

i=1

aiρ1pj


polynomials, ρ1pj



each of the knots are of finite algebraic order, ρ0(Ki ) = 0.

Thus,0 = ajρ





0 =n∑

i=1

aiρ1pj


polynomials, ρ1pj




1(Kj). Since ρ1(Kj) 6= 0, aj = 0.

No nontrivial combinations of the knots Ki is slice and the set is linearlyindependent




0 =n∑

i=1

aiρ1pj


polynomials, ρ1pj






Comparison of second corollary to polynomial splittingtheorems

The second corollary can be compared to the polynomial splittingtheorems for Casson-Gordon invariants (S. Kim, 2003) and for ρ-invariantscorresponding to isotropy in the Alexander module (S. Kim-T. Kim, 2007).The idea of the comparison: Each theorem says that if some knots withcoprime Alexander polynomials have non-vanishing obstructions then theirconnected sum has non-vanishing obstructions.


Example: a linearly independent family of knots in C viainfection.

Consider a family of knots of finitetopological order, J1, J2, . . . withsquare-free, coprime Alexanderpolynomials.

By first corollary, ρ1(Jn) = 0

Let Kn be given from Jn by infection alonga curve which is nonzero in A0(Jn) using aknot A with ρ0(A) 6= 0.

ρ1(Kn) = ρ1(Jn) + ρ0(A) = ρ0(A) 6= 0

by the second corollary, {K1,K2, . . . } islinearly independent.

n −n

A

Jn

Kn






ρ1(Kn) = ρ1(Jn) + ρ0(A) = ρ0(A) 6= 0


n −n

A

Jn

Kn






ρ1(Kn) = ρ1(Jn) + ρ0(A) = ρ0(A) 6= 0


n −n

A

Jn

Kn






ρ1(Kn) = ρ1(Jn) + ρ0(A) = ρ0(A) 6= 0


n −n

A

Jn

Kn






ρ1(Kn) = ρ1(Jn) + ρ0(A) = ρ0(A) 6= 0


n −n

A

Jn

Kn


Proof of first big theorem

I start by constructing a 4-manifold whose signature defect is the sum ofρ-invariants in question. Take M(K1)× [0, 1] t . . .M(Kn)× [0, 1] andconnect it by gluing together neighborhoods of meridians. Glue a slice diskcompliment for K1# . . .#Kn to the M(K1# . . .#Kn)-component of theboundary.It is the anisotropic assumption that lets us cap the cobordism safely.

M(K1) M(K2) M(K3)

M(K1#K2#K3)

W0

W V coral reef upside-down

= jellyfish



I start by constructing a 4-manifold whose signature defect is the sum ofρ-invariants in question.

Take M(K1)× [0, 1] t . . .M(Kn)× [0, 1] andconnect it by gluing together neighborhoods of meridians. Glue a slice diskcompliment for K1# . . .#Kn to the M(K1# . . .#Kn)-component of theboundary.It is the anisotropic assumption that lets us cap the cobordism safely.

M(K1) M(K2) M(K3)

M(K1#K2#K3)

W0


= jellyfish



I start by constructing a 4-manifold whose signature defect is the sum ofρ-invariants in question. Take M(K1)× [0, 1] t . . .M(Kn)× [0, 1]

andconnect it by gluing together neighborhoods of meridians. Glue a slice diskcompliment for K1# . . .#Kn to the M(K1# . . .#Kn)-component of theboundary.It is the anisotropic assumption that lets us cap the cobordism safely.

M(K1) M(K2) M(K3)

M(K1#K2#K3)

W0


= jellyfish



I start by constructing a 4-manifold whose signature defect is the sum ofρ-invariants in question. Take M(K1)× [0, 1] t . . .M(Kn)× [0, 1] andconnect it by gluing together neighborhoods of meridians.

Glue a slice diskcompliment for K1# . . .#Kn to the M(K1# . . .#Kn)-component of theboundary.It is the anisotropic assumption that lets us cap the cobordism safely.

M(K1) M(K2) M(K3)

M(K1#K2#K3)

W0


= jellyfish



I start by constructing a 4-manifold whose signature defect is the sum ofρ-invariants in question. Take M(K1)× [0, 1] t . . .M(Kn)× [0, 1] andconnect it by gluing together neighborhoods of meridians. Glue a slice diskcompliment for K1# . . .#Kn to the M(K1# . . .#Kn)-component of theboundary.

It is the anisotropic assumption that lets us cap the cobordism safely.

M(K1) M(K2) M(K3)

M(K1#K2#K3)

W0

W V

coral reef upside-down

= jellyfish




M(K1) M(K2) M(K3)

M(K1#K2#K3)

W0

W V

coral reef upside-down

= jellyfish




M(K1) M(K2) M(K3)

M(K1#K2#K3)

W0


= jellyfish




M(K1) M(K2) M(K3)

M(K1#K2#K3)

W0

W V coral reef upside-down = jellyfish



W0

W

V

A coefficient system on W must be found.

π1(W ) ?

π1(M(Ki ))π1(M(Ki ))

π1(M(Ki ))(2)p

//

OO

//

?�

OO



W0

W

V

Every inclusion on first homology is anisomorphism. (So Ap

0(W ) = H1(W ; Rp) makessense.)

π1(V ) H1(V ) = Z

π1(M(#Ki )) H1(M(#K1)) = Z

π1(W0) H1(W0) = Z

π1(M(Ki )) H1(Mi ) = Z

//

OO

��

//

��∼=

OO ∼=

//

OO

//

OO ∼=



W0

W

V

Claim: ker(Ap0(Ki )→ Ap

0(W )) is isotropicand thus, zero.

P = ker(Ap

0(#Ki )→ Ap0(V )

)is isotropic.

Ap0(V )

⊕Ap0(Kj)

P

Ap0(#Ki ) ⊕Ap

0(Kj)

Ap0(W0) ⊕Ap

0(Kj)

Ap0(Ki )

//∼=

��

∼=

OO

//∼=

OO

Id

��Id

//∼=

OO ::ttttttt [0,...,0,1,0... ]



W0

W

V


0(W )) is isotropicand thus, zero.P = ker

(Ap

0(#Ki )→ Ap0(V )

)is isotropic.

Ap0(V )

⊕Ap0(Kj)

P

Ap0(#Ki ) ⊕Ap

0(Kj)

Ap0(W0) ⊕Ap

0(Kj)

Ap0(Ki )

//∼=

��

∼=

OO

//∼=

OO

Id

��Id

//∼=

OO ::ttttttt [0,...,0,1,0... ]



W0

W

V


0(W )) is isotropicand thus, zero.P = ker

(Ap

0(#Ki )→ Ap0(V )

)is isotropic.

Ap0(V )

⊕Ap0(Kj)

P

Ap0(#Ki ) ⊕Ap

0(Kj)

Ap0(W0) ⊕Ap

0(Kj)

Ap0(Ki )

//∼=

��

∼=

OO

//∼=

OO

Id

��Id

//∼=

OO ::ttttttt [0,...,0,1,0... ]



W0

W

V

π1(K )

π1(K )(2)p

= H1(M(Ki )) n Ap0(Ki )Z

π1(W )π1(W )

π1(W )(2)p


π1(M(Ki ))(2)p

//

OO

//

?�

OO



W0

W

V Thus,∑ρ1p(Ki ) = σ2(W , φ1

p)− σ(W ) = 0


Twist knotsNeed to compute ρ1(T−x2−x−1).

This is hard. The infection trick won’t work. Instead try to compute2ρ1(Tn) = ρ1(Tn#Tn).This is algebraically slice, and has a metabolizing link. The followingtheorem will allow us to get a handle on ρ1

n +1

x red strands x + 1 blue strands

n +1 n +1


Twist knotsNeed to compute ρ1(T−x2−x−1).This is hard. The infection trick won’t work.

Instead try to compute2ρ1(Tn) = ρ1(Tn#Tn).This is algebraically slice, and has a metabolizing link. The followingtheorem will allow us to get a handle on ρ1

n +1


n +1 n +1


Twist knotsNeed to compute ρ1(T−x2−x−1).This is hard. The infection trick won’t work. Instead try to compute2ρ1(Tn) = ρ1(Tn#Tn).

This is algebraically slice, and has a metabolizing link. The followingtheorem will allow us to get a handle on ρ1

n +1


n +1 n +1


Twist knotsNeed to compute ρ1(T−x2−x−1).This is hard. The infection trick won’t work. Instead try to compute2ρ1(Tn) = ρ1(Tn#Tn).This is algebraically slice, and has a metabolizing link. The followingtheorem will allow us to get a handle on ρ1

n +1


n +1 n +1


Second big theorem: computing ρ1.

Theorem (D.)

Let K1, . . . ,Kn be (not necessarily distinct) p-anisotropic knots . Supposen#i=1

Ki is algebraically slice.

Let F be a genus g seifert surface for #Ki . Let

L be a link of g curves on F representing a metabolizer for the Seifertform, such that the meridians about the bands on which the componentsof L sit form a Z-linearly independent set in Ap

0(#Ki )/P, where P is thesubmodule of Ap

0(#Ki ) generated by L. Let η(L) be the rank of theAlexander module of L.Then ∣∣∣∑(

ρ1p(Ki )

)− ρ0(L)

∣∣∣ ≤ g − 1− η(L).

ρ1p (a nonabelian ρ-invariant) is approximated by ρ0 (an abelianρ-invariant). The latter is more reasonable to expect to compute.



Theorem (D.)


Ki is algebraically slice. Let F be a genus g seifert surface for #Ki .

Let




ρ1p(Ki )

)− ρ0(L)

∣∣∣ ≤ g − 1− η(L).




Theorem (D.)


Ki is algebraically slice. Let F be a genus g seifert surface for #Ki . Let

L be a link of g curves on F representing a metabolizer for the Seifertform,

such that the meridians about the bands on which the componentsof L sit form a Z-linearly independent set in Ap



ρ1p(Ki )

)− ρ0(L)

∣∣∣ ≤ g − 1− η(L).




Theorem (D.)





0(#Ki ) generated by L.

Let η(L) be the rank of theAlexander module of L.Then ∣∣∣∑(

ρ1p(Ki )

)− ρ0(L)

∣∣∣ ≤ g − 1− η(L).




Theorem (D.)





0(#Ki ) generated by L. Let η(L) be the rank of theAlexander module of L.

Then ∣∣∣∑(ρ1p(Ki )

)− ρ0(L)

∣∣∣ ≤ g − 1− η(L).




Theorem (D.)






ρ1p(Ki )

)− ρ0(L)

∣∣∣ ≤ g − 1− η(L).




Theorem (D.)






ρ1p(Ki )

)− ρ0(L)

∣∣∣ ≤ g − 1− η(L).



More examples: Infection by a string link.Let L be a 2 component link with zero linking numbers. Jn(L)#Jn isalgebraically slice.

It has L as a metabolizer.∣∣ρ1(Jn(L)) + ρ1(Jn)− ρ0(L)∣∣ ≤ 2− 1− η(L) ≤ 1. Thus, if

∣∣ρ0(L)∣∣ > 1,

then ρ1(Jn(L)) + ρ1(Jn) 6= 0. As in the previous example, ρ1(Jn) = 0.Thus, ρ1(Jn(L)) 6= 0 and {Jn(L) : n ≥ 1} is linearly independent.

Jn(L) Jn

n −n

L

n −n


More examples: Infection by a string link.Let L be a 2 component link with zero linking numbers. Jn(L)#Jn isalgebraically slice. It has L as a metabolizer.

∣∣ρ1(Jn(L)) + ρ1(Jn)− ρ0(L)∣∣ ≤ 2− 1− η(L) ≤ 1. Thus, if

∣∣ρ0(L)∣∣ > 1,

then ρ1(Jn(L)) + ρ1(Jn) 6= 0. As in the previous example, ρ1(Jn) = 0.Thus, ρ1(Jn(L)) 6= 0 and {Jn(L) : n ≥ 1} is linearly independent.

Jn(L) Jn

n −n

L

n −n


More examples: Infection by a string link.Let L be a 2 component link with zero linking numbers. Jn(L)#Jn isalgebraically slice. It has L as a metabolizer.∣∣ρ1(Jn(L)) + ρ1(Jn)− ρ0(L)

∣∣ ≤ 2− 1− η(L) ≤ 1.

Thus, if∣∣ρ0(L)

∣∣ > 1,then ρ1(Jn(L)) + ρ1(Jn) 6= 0. As in the previous example, ρ1(Jn) = 0.Thus, ρ1(Jn(L)) 6= 0 and {Jn(L) : n ≥ 1} is linearly independent.

Jn(L) Jn

n −n

L

n −n



∣∣ ≤ 2− 1− η(L) ≤ 1. Thus, if∣∣ρ0(L)

∣∣ > 1,then ρ1(Jn(L)) + ρ1(Jn) 6= 0.

As in the previous example, ρ1(Jn) = 0.Thus, ρ1(Jn(L)) 6= 0 and {Jn(L) : n ≥ 1} is linearly independent.

Jn(L) Jn

n −n

L

n −n



∣∣ ≤ 2− 1− η(L) ≤ 1. Thus, if∣∣ρ0(L)

∣∣ > 1,then ρ1(Jn(L)) + ρ1(Jn) 6= 0. As in the previous example, ρ1(Jn) = 0.

Thus, ρ1(Jn(L)) 6= 0 and {Jn(L) : n ≥ 1} is linearly independent.

Jn(L) Jn

n −n

L

n −n



∣∣ ≤ 2− 1− η(L) ≤ 1. Thus, if∣∣ρ0(L)

∣∣ > 1,then ρ1(Jn(L)) + ρ1(Jn) 6= 0. As in the previous example, ρ1(Jn) = 0.Thus, ρ1(Jn(L)) 6= 0 and {Jn(L) : n ≥ 1} is linearly independent.

Jn(L) Jn

n −n

L

n −n


Comparison to Casson-Gordon invariants and Metabelian ρinvariants corresponding to isotropy

(advantage) ρ1p is not defined in terms of of metabolizers/isotropic

submodules: only one computation needs to be made, instead of(potentially infinitely) many.

(disadvantage) Only applies to knots whose every prime factor isp-anisotropic. In particular this technique cannot say anything aboutprime algebraically slice knots, for which the other invariants workvery well.


The 4-manifold used to prove the second big theorem

Start with W0 as in the previous proof, consider L as a link in ∂+W0. Add2-handles to the zero framing of L. Adding a three handle to this manifoldresults in a manifold with boundary (tM(Ki )) t −M(L)

M(K1) M(K2) M(K3)

M(K1#K2#K3)M(L)



Start with W0 as in the previous proof,

consider L as a link in ∂+W0. Add2-handles to the zero framing of L. Adding a three handle to this manifoldresults in a manifold with boundary (tM(Ki )) t −M(L)

M(K1) M(K2) M(K3)

M(K1#K2#K3)

M(L)



Start with W0 as in the previous proof, consider L as a link in ∂+W0.

Add2-handles to the zero framing of L. Adding a three handle to this manifoldresults in a manifold with boundary (tM(Ki )) t −M(L)

M(K1) M(K2) M(K3)

M(K1#K2#K3)

M(L)



Start with W0 as in the previous proof, consider L as a link in ∂+W0. Add2-handles to the zero framing of L.

Adding a three handle to this manifoldresults in a manifold with boundary (tM(Ki )) t −M(L)

M(K1) M(K2) M(K3)

M(K1#K2#K3)M(L)



Start with W0 as in the previous proof, consider L as a link in ∂+W0. Add2-handles to the zero framing of L. Adding a three handle to this manifoldresults in a manifold with boundary (tM(Ki )) t −M(L)

M(K1) M(K2) M(K3)

M(K1#K2#K3)

M(L)



π1(M(L) H1(M(L))

π1(W )π1(W )

π1(W )(2)p


π1(M(Ki ))(2)p

��

//

_�

��

//

OO

//

?�

OO



Thus,∑ρ1p(Ki )− ρ0(L) = σ2(W , φ1

p)− σ(W )

σ(W ) = 0∣∣σ2(W , φ1p)∣∣ ≤ rank

(H2 (W ; Γ)

H2 (∂W ; Γ)

)= g − 1− η(L)

where Γ = Q

[π1(W )

π1(W )(2)p

]



Thus,∑ρ1p(Ki )− ρ0(L) = σ2(W , φ1

p)− σ(W )

σ(W ) = 0∣∣σ2(W , φ1p)∣∣ ≤ rank

(H2 (W ; Γ)

H2 (∂W ; Γ)

)= g − 1− η(L)

where Γ = Q

[π1(W )

π1(W )(2)p

]


Proof of main application.

For n = −x2 − x − 1, Tn#Tn is algebraically slice.

It has a metabolizerwhich I call Lx . Thus we are in the setting of the second big theorem.


n +1 n +1



For n = −x2 − x − 1, Tn#Tn is algebraically slice. It has a metabolizerwhich I call Lx .

Thus we are in the setting of the second big theorem.


n +1 n +1



For n = −x2 − x − 1, Tn#Tn is algebraically slice. It has a metabolizerwhich I call Lx . Thus we are in the setting of the second big theorem.


n +1 n +1


Computation.

If∣∣ρ0(Lx)

∣∣ > 1 then ρ1(Tn) 6= 0, since∣∣2ρ1(Tn)− ρ0(Lx)

∣∣ < 2− 1− 0and the proof of linear independence will be complete, since they havedistinct, prime Alexander polynomials.

The computation of ρ0(Lx) can be reduced to planar considerations,algorithmized, and implemented on a computer. The javaimplementation may be found here.

I will do the computations by hand for an easier example, and thenstate the results for Lx . In the end, ρ0(Lx) < −1 for each x , and theproof will be complete


http://math.rice.edu/~cwd1/research/computeRho0

Computation.

If∣∣ρ0(Lx)







Computation.

If∣∣ρ0(Lx)







Simplified exampleLet K be the knot with the surgery description below. It happens tobe the left handed trefoil.

This surgery description gives M(K ) as the boundary of the4-manifold, W . The associated inclusion is an isomorphism on firsthomology.Thus, ρ0(K ) = σ2(W , φ)− σ(W ).By inspection, σ(W ) = −1.

−1

•


Simplified exampleLet K be the knot with the surgery description below. It happens tobe the left handed trefoil.This surgery description gives M(K ) as the boundary of the4-manifold, W . The associated inclusion is an isomorphism on firsthomology.

Thus, ρ0(K ) = σ2(W , φ)− σ(W ).By inspection, σ(W ) = −1.

−1

•


Simplified exampleLet K be the knot with the surgery description below. It happens tobe the left handed trefoil.This surgery description gives M(K ) as the boundary of the4-manifold, W . The associated inclusion is an isomorphism on firsthomology.Thus, ρ0(K ) = σ2(W , φ)− σ(W ).

By inspection, σ(W ) = −1.

−1

•


Simplified exampleLet K be the knot with the surgery description below. It happens tobe the left handed trefoil.This surgery description gives M(K ) as the boundary of the4-manifold, W . The associated inclusion is an isomorphism on firsthomology.Thus, ρ0(K ) = σ2(W , φ)− σ(W ).By inspection, σ(W ) = −1.

−1

•


Computing σ2(W ):

the curve, γ, lifts to the universal abelian cover of M(unknot) whereit bounds an embedded disk, D. This disk together with the core ofthe 2-handle glued to γ form the 2-sphere S which generatesH2(W ; Q[Z]).The twisted intersection form of W is given by the self intersection ofS .The self intersection of S in H2(W ; Q[Z]) is the same as theintersection (with coefficients) between D and the −1 push-off of γ inthe cover of M(unknot).counting these, 〈S ,S〉 = −t − t−1 + 1, so that [−t − t−1 + 1] is thetwisted intersection matrix of W

−1 γ

γ

γγ

t−1γ γ

tγ


Computing σ2(W ):the curve, γ, lifts to the universal abelian cover of M(unknot)

whereit bounds an embedded disk, D. This disk together with the core ofthe 2-handle glued to γ form the 2-sphere S which generatesH2(W ; Q[Z]).The twisted intersection form of W is given by the self intersection ofS .The self intersection of S in H2(W ; Q[Z]) is the same as theintersection (with coefficients) between D and the −1 push-off of γ inthe cover of M(unknot).counting these, 〈S ,S〉 = −t − t−1 + 1, so that [−t − t−1 + 1] is thetwisted intersection matrix of W

−1 γ

γ

γ

γ

t−1γ γ

tγ


Computing σ2(W ):the curve, γ, lifts to the universal abelian cover of M(unknot) whereit bounds an embedded disk, D. This disk together with the core ofthe 2-handle glued to γ form the 2-sphere S which generatesH2(W ; Q[Z]).

The twisted intersection form of W is given by the self intersection ofS .The self intersection of S in H2(W ; Q[Z]) is the same as theintersection (with coefficients) between D and the −1 push-off of γ inthe cover of M(unknot).counting these, 〈S ,S〉 = −t − t−1 + 1, so that [−t − t−1 + 1] is thetwisted intersection matrix of W

−1 γ

γ

γ

γ

t−1γ γ

tγ


Computing σ2(W ):the curve, γ, lifts to the universal abelian cover of M(unknot) whereit bounds an embedded disk, D. This disk together with the core ofthe 2-handle glued to γ form the 2-sphere S which generatesH2(W ; Q[Z]).The twisted intersection form of W is given by the self intersection ofS .

The self intersection of S in H2(W ; Q[Z]) is the same as theintersection (with coefficients) between D and the −1 push-off of γ inthe cover of M(unknot).counting these, 〈S ,S〉 = −t − t−1 + 1, so that [−t − t−1 + 1] is thetwisted intersection matrix of W

−1 γ

γ

γ

γ

t−1γ γ

tγ


Computing σ2(W ):the curve, γ, lifts to the universal abelian cover of M(unknot) whereit bounds an embedded disk, D. This disk together with the core ofthe 2-handle glued to γ form the 2-sphere S which generatesH2(W ; Q[Z]).The twisted intersection form of W is given by the self intersection ofS .The self intersection of S in H2(W ; Q[Z]) is the same as theintersection (with coefficients) between D and the −1 push-off of γ inthe cover of M(unknot).

counting these, 〈S ,S〉 = −t − t−1 + 1, so that [−t − t−1 + 1] is thetwisted intersection matrix of W

−1 γ

γ

γ

γ

t−1γ γ

tγ


Computing σ2(W ):the curve, γ, lifts to the universal abelian cover of M(unknot) whereit bounds an embedded disk, D. This disk together with the core ofthe 2-handle glued to γ form the 2-sphere S which generatesH2(W ; Q[Z]).The twisted intersection form of W is given by the self intersection ofS .The self intersection of S in H2(W ; Q[Z]) is the same as theintersection (with coefficients) between D and the −1 push-off of γ inthe cover of M(unknot).counting these, 〈S , S〉 = −t − t−1 + 1, so that [−t − t−1 + 1] is thetwisted intersection matrix of W

−1 γ

γ

γ

γ

t−1γ γ

tγ


Computing σ2(W ):

The transform from l2(Z) to L2(S1) (S1 ⊆ C) sending tn to zn is anisomorphism: [1− t − t−1] has the same signature as[1− z − z−1] = [1− 2 Re(z)] which is the integral of sign(1− 2 Re(z))

over S1, which is1

3.

thus, ρ0(K ) =1

3− (−1) =

4

3


Computation for the link, L2

In general this can be duplicated for any link arising as ±1 surgery alongcommutator curves on the unlink.

With this in mind, L2 can be realizedas such.

+1 +1

+1+1+1+1

•

•



In general this can be duplicated for any link arising as ±1 surgery alongcommutator curves on the unlink. With this in mind, L2 can be realizedas such.

+1 +1

+1+1+1+1

•

•




+1 +1+1+1+1+1

•

•




+1 +1+1+1+1+1

•

•



In general this can be duplicated for any link arising as ±1 surgery alongcommutator curves on the unlink. With this in mind, L2 can be realizedas such. The resulting 4-manifold has regular intersection form (byinspection)

1 0 0 00 1 0 00 0 1 00 0 0 1

,

+1 +1+1+1+1+1

•

•




1 0 0 00 1 0 00 0 1 00 0 0 1

,

and twisted intersection form (computed by taking the algorithm outlinedfor the trefoil knot and implementing it on a computer) 1 + xy 2 + y + x−1y−2 + y−1 −xy − y−2 xy + y−2

−y 2 − x−1y−1 y + y−1 −y + x−1 − y−1

y 2 + x−1y−1 −y + x − y−1 y + y−1

+1 +1+1+1+1+1

•

•




1 0 0 00 1 0 00 0 1 00 0 0 1

,


−y 2 − x−1y−1 y + y−1 −y + x−1 − y−1

y 2 + x−1y−1 −y + x − y−1 y + y−1

which have signatures 4 and ∼ .38 (numerically integrating via computer).

+1 +1+1+1+1+1

•

•




1 0 0 00 1 0 00 0 1 00 0 0 1

,


−y 2 − x−1y−1 y + y−1 −y + x−1 − y−1

y 2 + x−1y−1 −y + x − y−1 y + y−1

which have signatures 4 and ∼ .38 (numerically integrating via computer).Thus, ρ0(L2) ∼ −3.62 and 2ρ1(K−7) . −2.62 (−7 = −22 − 2− 1), and inparticular is not zero.

+1 +1+1+1+1+1

•

•


Computation for the link, Lx

Lx+1 can be realized from Lx by +1 surgery along 2x − 1 new commutatorcurves.

Proposition

If a link L′ is realized as +1 surgery along commutator curves on anotherlink L, then ρ0(L′) ≤ ρ0(L)

Thus ρ0(Lx) . −3.62 for all x ≥ 2 and ρ1(T−x2−x−1) 6= 0. The knotshave distinct, prime Alexander polynomials and so are linearly independent.




Proposition






Proposition


Thus ρ0(Lx) . −3.62 for all x ≥ 2 and ρ1(T−x2−x−1) 6= 0.

The knotshave distinct, prime Alexander polynomials and so are linearly independent.




Proposition




Anisotropic knots and infection.

Definition

ρ̃1p is some other localized ρ-invariant for knots.



Theorem

Given slice knots R1, . . .Rn, infecting curves η1, . . . ηn representingelements of A0(Rn) which sit in the sum of no pair of isotropic submodulesand p-anisotropic knots K1, . . . ,Km, ifn#i=1

R1(η1, (R2(η2, . . . (Rn(ηn(Ki )) . . . ))) is slice

thenn∑

i=1

ρ̃1p(Ki ) = 0

The following is an immediate application.



Corollary

Let R be a slice knot whose Alexander module has a unique nontrivialisotropic submodule. Let η be an arc representing an element of A0(R)which is not isotropic. Let Kn be a sequence of anisotropic knots withsquare-free, strongly coprime Alexander polynomials and nonzero ρ1

invariants (take Tn(x), for example).Then {R(η,R(η, . . .R(η(Kn)) . . . ))} is linearly independent, whereinfection is understood to take place some fixed number of times.


Localized metabelian rho invariants as ... - math.rice.edumath.rice.edu/~cwd1/research/wesleyan talk 2010/rho1v2.pdfLocalized metabelian rho invariants as ... - math.rice.edu

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