Smooth 4-dimensional Schoenflies problem

Smooth 4-dimensional Schoenflies problem

Definitions and mathematical statement

A smooth embedding $e:S^3\hookrightarrow S^4$ has two complementary components; write $W_+$ and $W_-$ for their closures, regarded as smooth manifolds with boundary using collars along $e(S^3)$. The selected Schoenflies statement asks whether \[ W_+\cong_{\mathrm{diff}}B^4\quad\text{and}\quad W_-\cong_{\mathrm{diff}}B^4 \] for every such embedding. Here $B^4=\{x\in{\mathbb{R}}^4:\|x\|\le1\}$ has its standard smooth structure. The equivalence to smooth unknottedness is the one in the cited statement. Replacing diffeomorphism by homeomorphism, or assuming an extension of the boundary embedding in advance, changes the problem.

Formulation and concept references: [S1].

Short English statement

Does every smoothly embedded three-sphere in the standard four-sphere bound a standard smooth four-ball on each side?

Research attempt

Current frontier.

The distinction between an abstract contractible manifold, a homotopy ball embedded in the standard sphere, and a standard smooth ball is essential here. Lambert-Cole proves that a smoothly embedded homotopy 4-ball can be re-embedded in ${\mathbb{C}}^2$ as a union of three pseudoconvex domains. His Theorem 1.1 recalls that a homotopy 4-ball which is itself pseudoconvex is standard, whereas Theorem 1.2 supplies a union, not that stronger conclusion [S1]. The source does not assert the full Schoenflies theorem. I did not locate a verified general resolution in the additional current search. The attempt below separates a geometric handle-cancellation condition from its much weaker homological shadow; the handle conventions are those of [E1], Chapters 4--5.

Attack route.

A minimal-counterexample approach might try to cancel the handles of one complementary closure. Its missing input is geometric removal of extra intersections, not merely a unimodular boundary matrix. An analytic approach might turn the three-domain re-embedding into a single pseudoconvex domain; its missing input is a globally appropriate plurisubharmonic defining function. I investigate both, including a concrete test showing why a natural union construction fails.

Attempt: what the ambient sphere already forces.

Write $W=W_+$ and $W'=W_-$. Collars allow van Kampen to be applied to $S^4=W\cup_{S^3}W'$. Since the common boundary is simply connected, \begin{equation} 1=\pi_1(S^4)=\pi_1(W)*\pi_1(W'), \end{equation} so each factor is trivial. Mayer--Vietoris gives $H_i(W;{\mathbb{Z}})= H_i(W';{\mathbb{Z}})=0$ for $i=1,2$. In the remaining segment, the connecting map $H_4(S^4;{\mathbb{Z}})\to H_3(S^3;{\mathbb{Z}})$ sends the fundamental class to the oriented separating boundary and is an isomorphism. Thus $H_3$ also vanishes on each side. The absolute fourth homology vanishes for a connected compact manifold with nonempty boundary. Each closure is a simply connected acyclic finite CW complex, and hence contractible: if a higher homotopy group were the first nonzero one, the Hurewicz theorem would contradict acyclicity; Whitehead's theorem then applies.

This calculation establishes the hypotheses for studying a homotopy ball, but does not establish its smooth type. In particular, it supplies no embedded disks for cancelling handles.

Attempt: a geometric cancellation certificate.

Temporarily make the additional hypothesis that $W$ has a handle presentation consisting of one 0-handle, $n$ 1-handles, and $n$ 2-handles, with no 3- or 4-handles. The equality of the two numbers follows from $\chi(W)=1$ if such a presentation exists. The cellular boundary matrix \begin{equation} D:{\mathbb{Z}}^n\longrightarrow{\mathbb{Z}}^n \end{equation} is unimodular, since $H_1(W)=H_2(W)=0$. Handle slides and orientation changes realize the elementary integral row and column operations, so one can arrange $D=I_n$. This is a statement about algebraic intersection numbers, not yet about actual attaching curves [E1].

In the boundary of the 0- and 1-handles, let $B_j$ be the belt 2-sphere of the $j$th 1-handle and $K_i$ the attaching circle of the $i$th 2-handle. Put them in transverse position. After the normalization, \begin{equation} K_i\mathbin{\cdot} B_j=\delta_{ij},\qquad s=\sum_{i,j}\#(K_i\cap B_j)-n\in 2{\mathbb{Z}}_{\geq0}. \end{equation} Call a cancellation bigon clean when it is an embedded disk in this three-dimensional boundary whose two boundary arcs lie on one $K_i$ and one $B_j$, whose endpoints are an opposite-sign intersection pair, and whose interior and relevant boundary interiors avoid all other attaching circles and belt spheres. Require the usual local bigon position, so pushing the arc of $K_i$ across the disk removes exactly those two intersections. The framing is carried along by this isotopy.

Certificate lemma. If there is a finite sequence of such clean bigon isotopies, each reducing $s$ by two and ending with $s=0$, then $W$ is diffeomorphic to $B^4$.

Proof. At the end, every $K_i$ meets $B_i$ exactly once and is disjoint from every other $B_j$, because the signed intersection matrix remains $I_n$ and there are exactly $n$ geometric intersections in total. The geometric 1-/2-handle cancellation theorem now cancels the pairs. The other attaching circles miss the belt sphere of a pair being cancelled, so cancellation introduces no new intersection requirement for them. Repeating leaves the original 0-handle. For $n=0$ the conclusion is immediate. This is an application of the geometric cancellation theorem, not its replacement by homology [E1], Chapter 5. $\square$

Consequently the following would suffice for the target: for each closure arising from an embedding in $S^4$, produce the assumed handle presentation and, after suitable slides, such a clean-bigon certificate. Neither existence assertion is proved here. Even the ability to eliminate all 3-handles must not be silently assumed.

Why the algebraic argument stops.

The condition $D=I_n$ gives an opposite-sign pair whenever $s>0$; it does not give a clean disk connecting that pair. A putative disk may meet other attaching curves, run through belt spheres, or fail to exist as an embedded disk in the required position. Normal generation of the fundamental-group relators does not repair this geometric defect. A Whitney-disk argument in dimension four must supply framed embedded disks and remove their intersections; the high-dimensional Whitney trick is not available just from simple connectivity. Thus the statement "$s>0$ implies a clean bigon after permitted moves" is precisely an unproved geometric lemma, not an induction step already justified by unimodularity. The certificate lemma proves a checkable sufficient condition, but has not shown that every actual Schoenflies complement satisfies it.

Attempt: stress-testing the pseudoconvex-union route.

A tempting way to use [S1], Theorem 1.2, is to take defining functions for the pieces and combine them. The following local test rules out the most immediate version. On ${\mathbb{C}}^2$, put \begin{equation} u(z,w)=|z|^2+|w|^2+\Re z-\tfrac14,\qquad v(z,w)=|z|^2+|w|^2-\Re z-\tfrac14. \end{equation} Both are strictly plurisubharmonic. Their negative sublevel sets are overlapping balls, and their union is the negative sublevel set of \begin{equation} m(z,w)=\min(u,v)=|z|^2+|w|^2-|\Re z|-\tfrac14. \end{equation} On the complex line $w=0$, the average of $m$ around the circle $|z|=r$ is \begin{equation} r^2-\frac{2r}{\pi}-\frac14<-\frac14=m(0,0) \qquad(0<r<2/\pi). \end{equation} It violates the submean inequality, so $m$ is not plurisubharmonic. This example concerns the proposed defining function; it is not a claim that this particular union has no alternative good description. A regularized maximum preserves plurisubharmonicity but naturally models an intersection of negative sublevels, not their union. No locally uniform plurisubharmonic approximation to this $m$ can fix the issue, since such a limit would again satisfy the submean inequality.

Thus the three-domain theorem does not by itself produce a single pseudoconvex homotopy ball. An additional global construction, possibly changing the embedding and the domains, would be required before the standardness theorem in [S1], Theorem 1.1, could be applied.

Stress test.

Both closures must be addressed, not merely one selected diagram. A contractible manifold with a different boundary is not a counterexample to the printed assertion; nor is an exotic object which has not been embedded in the standard $S^4$. The handle proof uses actual geometric cancellation only after stating its full extra hypotheses. The analytic test does not disprove the existence of a better re-embedding. Neither route imports the smooth four-dimensional Poincaré conjecture or a smooth $h$-cobordism theorem as an available fact. Any universal theorem supplying the two missing handle properties would already resolve the Schoenflies target and must be regarded as the deep step, not routine handle bookkeeping.

Outcome.

Outcome: Exploratory approach with unresolved gap.

The attempt proves contractibility directly from the ambient embedding, formulates and proves a geometric cancellation-certificate lemma, and exhibits a precise failure of the minimum-of-defining-functions approach to the analytic route. These are deductions and diagnostics, with no claim of novelty for the underlying handle facts. What remains is a universal construction of the clean geometric certificate, or a global pseudoconvex re-embedding of each closure. No nonstandard complementary ball or complete proof candidate has been obtained.

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Sources

• S1 Peter Lambert-Cole, "Stein trisections and homotopy 4-balls," arXiv:2104.02003 [math.GT], April 2021. https://arxiv.org/abs/2104.02003. Catalogue formulation source.
• E1 Robert E. Gompf and András I. Stipsicz, 4-Manifolds and Kirby Calculus, Graduate Studies in Mathematics 20, American Mathematical Society, 1999. Chapters 4--5: handle presentations, slides, and geometric cancellation. https://doi.org/10.1090/gsm/020. Reference for the standard handle-calculus facts, not a theorem that the clean disks postulated in the notebook exist.