Zero traces and the Conway nonsliceness theorem — Examples and Counterexamples

The standard spanning disk for $U$ in $B^4$ is a smooth proper slice disk. By [F4], its zero-trace embeds smoothly in $S^4$.

Write the standard $0$-handle as the rounded product $D^2\times D^2$, with $U=\mathrm{\partial }{D}^2\times \{0\}$. Its product attaching region is $\mathrm{\partial }{D}^2\times D^2$. A section pushoff $\mathrm{\partial }{D}^2\times \{z\}$ for $z\in \mathrm{\partial }{D}^2$ bounds $D^2\times \{z\}$ in the other boundary face, so the product framing is the zero framing of [F1]. Attaching the second product $D^2\times D^2$ by this framing glues the two first-factor disks along their boundaries without twisting the second factor. After rounding, the result is $(D^2{\cup }_{S^1}{D}^2)\times D^2\cong S^2\times D^2$.

Each of $B$ and $G$ bounds its chosen normal disk meeting the core $R$ once, so both indicated sublinks are a knot and its meridian. The disks are disjoint, so $B$ and $G$ have zero linking number. Their framings are zero by the stated preferred-longitude pushoffs. Thus $(B,G,R)$ satisfies [F2].

In the round model $S^4\subset {\mathbb{R}}^5$, take the standard equatorial sphere $S^2=\{(x_1,x_2,x_3,0,0):x_1^2+x_2^2+x_3^2=1\}$. The constant vectors $e_4,e_5$ are tangent to $S^4$ at every point of this sphere and orthogonal to its tangent plane, so they form its normal frame. Explicitly, for small $v=(v_1,v_2)\in D^2$, the map $(x,v)\mapsto (\sqrt{1-\parallel v{\parallel }^2}x,v_1,v_2)$ is a product tubular embedding. Its image is an unknotted-sphere neighborhood diffeomorphic to $S^2\times D^2$, hence gives the stated embedding of $X_0(U)$.

Put $R$ in its standard planar position, with spanning disk $D_R$, and carry the two small normal meridians along it. Along an arc of $R$ between their base points, the normal circles sweep a product annulus. Replace its $B$ boundary by a close zero-framed parallel $B_f$ toward $G$, so the annulus misses the retained $B$. The annulus meets $D_R$ in its inward longitudinal arc. Take a narrow strip about that arc joining its two boundary circles. Slide $G$ over the oppositely oriented $B_f$ along this strip, with the product framing, as permitted by [F5]. The new circle bounds the disk obtained by cutting the annulus open along the strip; push its boundary slightly to round the band seams. This disk misses $D_R\cup B\cup R$, and its disk framing is the slide framing: the two old zero framings are the product annulus framings at its ends, and the product strip adds no twist. Thus it and its framed boundary can be kept outside a sufficiently thin cancellation neighborhood. Now [F3] cancels $(R,B)$ using the single geometric intersection, leaving this disk-bounding zero-framed unknot on the $4$-ball. By [F1] the remaining attachment is $X_0(U)$. Interchanging $B,G$ gives the other cancellation. This explicitly transports the remaining attaching data; the original normal disk meeting $R$ is not used as a disk disjoint from the cancellation region.

The oriented knots $C,K_P$ of [F4] have diffeomorphic zero-traces. Suppose they were smoothly concordant. By [F1] the concordance is a properly embedded product-collared annulus, hence a genus-zero cobordism.

Apply [F3] with $g=0$. It gives $\mathrm{\mid }s(C)-s(K_P)\mathrm{\mid }\le 0$, so the two invariants must agree.

But [F2] and [F4] give $s(C)=0$, $s(K_P)=2$. Their difference has absolute value two, contradicting step 2.1. Thus these actual equal-zero-trace knots are not smoothly concordant, refuting the unchanged universal Statement.