Math-1621 · mathematics · Topology

Math-1621

Kirby Problem 4.60

Status Open · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedOpenOpen

Statement

Definitions and mathematical statement

A smooth oriented manifold is spin if its tangent bundle has vanishing second Stiefel--Whitney class $w_2$. An embedding is proper when inverse images of compact sets are compact.

(a) Does every connected open spin smooth four-manifold $X$, meaning noncompact without boundary, admit a proper smooth embedding $X\hookrightarrow\mathbb R^6$?

The relative questions use a compact connected spin cobordism $W$ with nonempty incoming and outgoing closed oriented three-manifolds $M,N$, so $\partial W=-M\sqcup N$. Given an embedding $f:M\hookrightarrow S^5$, (b) does there exist a neat embedding \[ F:W\hookrightarrow S^5\times[0,1] \] with $F|_M=(f,0)$ and $F(N)\subset S^5\times\{1\}$? Neatness means that the only intersections with the ambient boundary are these boundary components, with the usual product collar model.

For an oriented embedded three-manifold $h:Y\hookrightarrow S^5$, let $\sigma(h)$ denote the signature of a Seifert four-manifold bounded by $h(Y)$ in $S^5$; the source uses its independence of the choice. (c) If an outgoing embedding $g:N\hookrightarrow S^5$ is also prescribed and \[ \sigma(g)-\sigma(f)=\sigma(W), \] is this necessary signature condition sufficient for such an $F$ with $F|_N=(g,1)$?

The output embedding is smooth, not holomorphic. Part (b) permits the outgoing embedding to be chosen, while part (c) fixes it in advance. The cobordisms here have genuine incoming and outgoing boundary; no extra closed components are included.

Short English statement

Can open spin four-manifolds embed properly in six-space, with compatible relative embeddings of spin cobordisms?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 0

No attempts yet. Significant partial results and substantial reductions are welcome. Contribute an attempt.