Satellite operators and concordance filtrations — Examples and Counterexamples

The group $\mathrm{\Gamma }$ is torsion-free abelian, hence PTFA with its one-step series. The ring $\mathrm{\Lambda }$ is a commutative domain: after multiplying nonzero Laurent polynomials by monomials, lexicographically leading nonzero terms of ordinary polynomials have nonzero product over $\mathbb{Z}$. Its fraction field is $K$. Give each circle its one-vertex, one-edge cell structure, use their eight product cells on $T^3$, and lift them to the oriented coordinate cells of ${\mathbb{R}}^3$. Opposite faces differ by $x,y,1$, respectively. Thus [F1] identifies the coefficient complex with the exterior algebra on $e_1,e_2,e_3$, with differential the graded derivation $d(e_1)=a$, $d(e_2)=b$, $d(e_3)=0$. In particular $d_1(u,v,w)=au+bv$ and $d_2(p,q,r)=(-bp,ap,aq+br)$ in the ordered bases $(e_1,e_2,e_3)$ and $(e_1\wedge e_2,e_1\wedge e_3,e_2\wedge e_3)$.

Evaluation at $x=1$ identifies $\mathrm{\Lambda }\mathrm{/}(a)$ with the domain $\mathbb{Z}[y^{\pm 1}]$; its kernel is exactly $(a)$ by multiplying by a power of $x$ and applying the ordinary identity $x^j-1=(x-1)(1+\cdots +x^{j-1})$. If $au+bv=0$, reduction modulo $a$ gives $b\bar{v}=0$, whence $v=at$ and $u=-bt$. Therefore every degree-one cycle has a unique expression $(-bt,at,w)$. The image of $d_2$ consists precisely of those expressions with arbitrary $t=p$ and $w\in I$. Consequently $H_1(M;\mathrm{\Lambda })\cong \mathrm{\Lambda }\mathrm{/}I\cong \mathbb{Z}$, generated by the third circle. It is nonzero, and all of it is $\mathrm{\Lambda }$-torsion because $a\mathrm{\ne }0$ annihilates it.

Over $K$, multiplication by $a$ is invertible. The operator $h(\omega )=a^{-1}{e}_1\wedge \omega$ satisfies $dh+hd=1$ by the displayed graded-derivation differential. Hence the entire coefficient chain complex over $K$ is contractible and in particular $H_1(M;K)=0$. This verifies the vanishing assumption of the refuted statement without any finiteness or rank inference.

Any $\mathrm{\Lambda }$-linear map $\mathrm{\Lambda }\mathrm{/}I\to K\mathrm{/}\mathrm{\Lambda }$ is determined by the image $q+\mathrm{\Lambda }$ of $1+I$, where $aq,bq\in \mathrm{\Lambda }$. Set $r=aq$ and $s=bq$. Then $br=as$. Reducing this identity modulo $a$ and using the domain $\mathrm{\Lambda }\mathrm{/}(a)$ from step 2.1 gives $r=at$ for some $t\in \mathrm{\Lambda }$. Since $a\mathrm{\ne }0$ in the field $K$, the identity $aq=at$ forces $q=t\in \mathrm{\Lambda }$. Thus the image of $1+I$ is zero, and ${\mathrm{Hom}}_{\mathrm{\Lambda }}(\mathrm{\Lambda }\mathrm{/}I,K\mathrm{/}\mathrm{\Lambda })=0$. The conjugate dual is also zero; the involution $x\mapsto x^{-1},y\mapsto y^{-1}$ preserves $I$ because $x^{-1}-1=-x^{-1}a$ and $y^{-1}-1=-y^{-1}b$.

By steps 2.1 and 3.1, the adjoint of any sesquilinear pairing on $T{H}_1(M;\mathrm{\Lambda })$ with target $K\mathrm{/}\mathrm{\Lambda }$ has nonzero source and zero target, and therefore is not an isomorphism. This includes the coefficient Bockstein–duality pairing whenever expressed in those conventions. Together with step 2.2 this refutes the asserted implication. Every object and calculation is explicit and finite; no choice assumption is needed.

The identity core meets a meridional disk once positively. A (p,1)-cable represents p times the core class, and the local knot changes the core by a knot tied into a ball, not its homology class. Thus the listed winding numbers are 1,p,1.

The (p,1) torus knot is the unknot, so its closure bounds the standard disk in B⁴ and is Z[1/|p|]-slice. The identity map is K↦K.

For the local-knot pattern the closure complement is the exterior of J with the axis meridian identifying to a meridian of J. Meridional Dehn filling gives S³; van Kampen therefore says the normal closure of this meridian is the whole knot group. J can be nontrivial, so knotted closure does not preclude strong winding.

The pattern P_J has winding number 1 because tying a local knot does not change its homology class in the solid torus. Its closure is J, which is not the unknot by the choice of J.

The axis meridian is a meridian of J in this local-knot construction. Meridional filling of the knot exterior yields S³, whose fundamental group is trivial; van Kampen implies that the normal closure of the meridian is the entire group. Hence P_J is strong winding +1 although P_J(U) is not the unknot.

This disproves the forward implication from strong winding to an unknotted closure, while leaving the reverse implication untouched. The counterexample makes no claim that ordinary winding ±1 always implies strong winding ±1.