The Whitney Trick and Surgery Below the Middle Dimension — Examples

Form $X$ by removing a small $6$-ball disjoint from $A=S^3\times \{y_0\}$ and $C_1\cup C_2$ in the product, removing a ball from $S^1\times S^5$, and identifying their boundary $5$-spheres by an orientation-reversing diffeomorphism. This explicitly defines the smooth oriented connected sum. Removing either ball does not change the fundamental group: apply van Kampen to the punctured manifold and the ball, with collar overlap homotopy equivalent to the simply connected $S^5$. The same theorem across the neck gives ${\pi }_1(X)=1\ast \mathbb{Z}=\mathbb{Z}$. Here $S^3,S^5$ are simply connected, and projection of $S^1\times S^5$ onto $S^1$ gives its fundamental group: a based loop is a pair of coordinate loops; the second contracts because $S^5$ is simply connected, while the first lifts to $\mathbb{R}$ and its integer endpoint displacement classifies based homotopy. Choose its generator orientation below.

Choose small $3$-balls $D_i\subset C_i$ away from $A$ and paths in $C_i$ from their centres $u_i$ to $p$ or $q$, respectively. Fix an embedded arc $\alpha$ in $A$ from $p$ to $q$. A reference arc from $u_2$ to $u_1$ in the product, otherwise missing $A,C_1,C_2$, can be chosen in product coordinates; the loop obtained by adjoining the fixed sheet paths and $\alpha$ is null-homotopic since the product is simply connected. Replace a short segment of this reference arc by a detour through the connected-sum neck, around one generator of the $S^1$ factor, and back through the neck. Two parallel lanes make the outward and return portions disjoint. More formally, relative endpoint smoothing followed by the compact-arc embedding supplier gives an embedded representative of this path class; make its interior transverse to each of the three $3$-dimensional sheets, keeping short fixed endpoint collars normal to $C_i$. Since $1+3-6=-2$, its interior misses every sheet. Finitely many compactly supported perturbations suffice and preserve its relative path class and embeddedness. Denote the resulting embedded core arc by $\eta :u_2\to u_1$. By the construction, closing $\eta$ using the fixed sheet paths and $\alpha$ gives the generator $t$, not a null loop.

A sufficiently thin tubular neighbourhood of $\eta$ is $I\times D^5$: its normal bundle is trivial by projecting onto it in a Euclidean ambient embedding and transporting an initial basis by the skew matrix ODE $U^{\mathrm{\prime }}=[P^{\mathrm{\prime }},P]U$ along the interval. Choose a rank-$3$ subbundle in that normal bundle agreeing with the tangent $3$-planes of $C_i$ at its endpoints. Such a choice exists because the space of $3$-planes in ${\mathbb{R}}^5$ is path-connected; endpoint frames can be joined and interpolated on the interval. The resulting $I\times D^3$ has end balls $D_i$ after shrinking and straightening in endpoint charts. Remove their interiors from $C_1\cup C_2$ and insert the lateral cylinder $I\times S^2$, rounding its corners. Use the gluing that extends the specified orientations $C_1,-C_2$; an endpoint reflection realizes the required orientation convention. The tube and all rounding lie away from $A$ and the rest of the sheets. Each punctured $C_i$ is a $3$-ball, and two such balls joined by $S^2\times I$ form $S^3$. Thus the result is an embedded oriented sphere $B$; it agrees with $C_1$ near $p$ and with $-C_2$ near $q$. Their product tangent spaces are complementary to $TA$, so the only intersections are $p,q$ with signs $+1,-1$.

Take an embedded arc $\beta$ in $B$ from $q$ to $p$ running through the tube. Its part in the tube is homotopic relative endpoints to its core $\eta$ inside the tubular neighbourhood; its end parts are the fixed sheet paths up to homotopy in the punctured spheres. Consequently $[\alpha \ast \beta ]=t$ by step 2.1. Any other paths with the same endpoints in $A$ and $B$ are homotopic relative endpoints to these, since both sheets are $S^3$. In particular every admissible arc system gives the same nontrivial class. Normalize the label at $p$ to $1$; the label comparison lemma then gives the other label $t$, up to replacing the generator by its inverse under the opposite convention. The indices $1,-t$ are not negatives of each other, although their augmentation is zero. A disk filling a Whitney circle would contract $t$, impossible. Hence no Whitney disk or Whitney move exists for this pair.

Change the middle positive crossing to a negative crossing, giving ${\sigma }_1{\sigma }_1^{-1}{\sigma }_1$. Cancel the first inverse pair by an explicit local ambient isotopy. In a braid cylinder $D^2\times I$, write its two strands as $(\pm r{e}^{i\theta (u)},u)$, where $\theta$ makes one half-turn and then its inverse and is zero near both cylinder ends. Choose a smooth cutoff $\eta$ of the squared radius, equal to one near $r^2$ and zero near the boundary value. The maps $(z,u)\mapsto (e^{-is\eta (\mathrm{\mid }z{\mathrm{\mid }}^2)\theta (u)}z,u)$ preserve radius, have inverse obtained by changing $s$ to $-s$, and are the identity near the cylinder boundary. They extend by the identity to ambient isotopies of $S^3$ and straighten these two strands at $s=1$. The remaining one-crossing closure bounds an explicit embedded disk in $S^3$: take the two disks spanning its oriented smoothing circles at separate heights and join them by the single narrow half-twisted crossing band. This surface is embedded, and two disks joined by one band connecting their components form a disk. Its boundary is exactly the one-crossing closure. Thus the knot movie from the trefoil reaches a disk-bounding knot with one crossing change and otherwise only the displayed ambient isotopy; no external Reidemeister theorem is used.

Put this movie into $S^3\times [0,\epsilon ]$ by sending a strand point at movie time $t$ to $(k_t(u),t)$. Away from the crossing-change time, each time slice is embedded and the time coordinate separates distinct slices. Near the event use spatial coordinates $(x,y,z)$ and time $t$, with the two sheets parametrized by $(u,t)\mapsto (u,0,t,t)$ and $(v,t)\mapsto (0,v,-t,t)$. They coincide only at $u=v=t=0$. Their tangent planes are spanned by $(1,0,0,0),(0,0,1,1)$ and by $(0,1,0,0),(0,0,-1,1)$, respectively; these four vectors are independent. Both branches are immersions and meet transversely at this single point. Patch this local movie to the stationary outside strands, and choose stationary time collars at both endpoints. This constructs an immersed annulus with exactly one transverse interior double point.

Use the embedded disk in the inner collar sphere constructed in step 1.1 as a cap. In a fresh inward collar write its graph as $(d(x),\epsilon +\tau (x))$, where $d:D^2\hookrightarrow S^3$ is that disk, $\tau$ vanishes on its boundary, is positive in its interior, and has positive inward derivative near its boundary. The graph is embedded because $d$ is, and its boundary is the inner movie knot. Its interior lies deeper than the movie annulus, so there are no new coincidences. Glue along their stationary boundary collars and round the corner. Annulus plus disk is a properly immersed disk whose only double point is the transverse crossing-change event of step 2.1.

The local trefoil nonsliceness lemma proves that no smooth proper embedded disk has boundary $K$: a putative slice disk would have a rationally acyclic branched double cover, but its trefoil boundary cover has first homology of order $3$, contradicting the locally proved square-order consequence of duality. Thus the immersed disk cannot be homotoped relative to its boundary to a proper embedding. This is a disk-cleaning obstruction; calling the disk a Whitney disk additionally requires sheet arcs and their boundary data, which are not part of this witness.

Use an increasing coordinate change to put the zeros at $u=-1,1$, and reflect $v$ if necessary so $f$ is negative between them and positive outside. The ratio $\lambda (u)=(u^2-1)\mathrm{/}f(u)$ extends smoothly and positively over the two zeros by their nonzero first derivatives. The map $(u,v)\mapsto (u,\lambda (u)v)$ is a plane diffeomorphism fixing the axis and taking the other arc to $v=u^2-1$. At its corners the determinant of the ordered tangent directions $(1,0),(1,2u)$ is $2u$, giving one negative and one positive intersection.

Apply the explicit local flow of the model definition: $g(u)=b(u)(u^2-1-\epsilon )$, with $b=1$ on $[-1,1]$, and use a compact vertical cutoff equal to one on the swept segments. The axis is taken to $v=g(u)$ at time one. For $\mathrm{\mid }u\mathrm{\mid }\le 1$, $g=u^2-1-\epsilon <u^2-1$. For $\mathrm{\mid }u\mathrm{\mid }>1$, $u^2-1>0$ and $g=b(u)(u^2-1)-b(u)\epsilon <u^2-1$, also where $b=0$. Thus the moved axis and the unchanged graph are disjoint. The vector field has compact support, so its auxiliary time maps are diffeomorphisms; the first arc is embedded throughout. Pull back by the plane normalization to obtain the asserted isotopy of the original first arc, with the second held fixed.

In complementary dimensions the sheet factors are $E={\mathbb{R}}^{a-1}$ and $H={\mathbb{R}}^{b-1}$, with sheets $\{v=0,h=0\}$ and $\{v=u^2-1,e=0\}$. Use the compact normal cutoff from the theorem; a possible intersection still forces $e=h=0$, where the preceding calculation applies. The normal factors are split sheet directions, not a simultaneous product action on both images. This verifies the exact local picture and the cancellation of the opposite-sign pair.

The map $\theta \mapsto (\theta ,2\theta )$ is an embedding because its first coordinate is the identity of $S^1$. Its image meets $A=\{y=0\}$ only at $\theta =0,1\mathrm{/}2$. At both points the ordered tangent vectors are $(1,0)$ and $(1,2)$, whose determinant is $2>0$. Thus the intersections are transverse with local signs $+1,+1$, their mod-two count is zero, and their integer count is two.

Homotopy invariance of the oriented intersection number with the fixed $B$ preserves this count under any homotopy of $A$, hence under any isotopy. A disjoint endpoint would have the empty signed sum zero, contradicting the value two. Thus the asserted cancellation fails, beyond the parity condition, and the necessary opposite-sign hypothesis is not satisfied. Countable choice is used exactly through the published homotopy-invariance supplier.

Take $A$ to be the compactification of the plane $y=0$, a standard linear $S^3$. Start with a small round $3$-sphere $B_0$ in the affine $4$-plane $e_2=e_3=0$, centred far enough in the positive $y_3$ direction to miss $A$. Near its lowest point it is a graph $y_3=h_0(e_1,y_1,y_2)>0$ over a small $3$-ball. Replace only a smaller graph cap by $y_3=h_1(e_1,y_1,y_2)$, where $h_1=e_1^2+y_1^2+y_2^2-{\rho }^2$ on a small inner ball and is positive outside it, agreeing with $h_0$ near the outer cap boundary. Such a smooth radial interpolation can be chosen positive whenever $e_1^2+y_1^2+y_2^2>{\rho }^2$, by choosing $\rho$ sufficiently small. The linear interpolation from $h_0$ to $h_1$ gives embedded graph caps throughout and fixes the outer collar; it is a local finger move of this $3$-ball. The resulting $B$ is an embedded $S^3$.

An intersection with $A$ forces $y_1=y_2=y_3=0$. On the changed cap it therefore forces $e_1^2={\rho }^2$, giving exactly $p=(\rho ,0,0,0,0,0)$ and $q=(-\rho ,0,0,0,0,0)$. Outside the changed cap $y_3>0$ wherever $y_1=y_2=0$, so there are no other intersections. At either point the tangent directions of $B$ are ${\mathrm{\partial }}_{e_1}+2e_1{\mathrm{\partial }}_{y_3},{\mathrm{\partial }}_{y_1},{\mathrm{\partial }}_{y_2}$. Together with $TA=\mathrm{span}({\mathrm{\partial }}_{e_1},{\mathrm{\partial }}_{e_2},{\mathrm{\partial }}_{e_3})$ they span the six-dimensional ambient space; their determinant differs at the two points only by the sign of $2e_1$. Thus the local signs are opposite and $I(A,B)=0$.

Here $a=b=3=m-3$, so the stable dimension inequalities are met at equality. The sphere $S^6$ is simply connected by the published sphere theorem, and the two connected sheets admit the required avoiding arcs. The high-dimensional Whitney trick therefore removes exactly this pair, producing an embedded $A^{\mathrm{\prime }}$ disjoint from $B$. The explicit cap verifies the claimed finger-move witness, rather than presupposing its intersection count.