LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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A disconnected link projection represents a split link

Statement

If the underlying projection of a nonempty regular link diagram on $S^{2}$ is disconnected, then the represented link is split.

Statement references

Calegari, Introduction to Knot Theory, §5.2 and Exercise 5.14

Menasco, Closed incompressible surfaces in alternating knot and link complements (1984), Theorem 1(a)

Facts & Assumptions

Given: A regular link diagram whose projection graph is a nonempty finite disconnected subset $P \subset S^{2}$ .

[F1]

A regular diagram is the projection of a finite disjoint union of embedded circles, and its projection has finitely many transverse double points (Oriented links in the three-sphere and ambient isotopy, Regular oriented link diagrams).

[F2]

A link is split if a sphere disjoint from it has link points on both sides (Split and nonsplit links).

[F3]

A nonempty disconnected finite projection graph admits a Jordan curve in its complement with diagram points on both sides (Disconnected projection components admit a separating Jordan curve).

[F4]

The proof of [F3] constructs its separating curve as a smooth boundary component of a finite regular neighborhood. The choice-free smooth Jordan-disk result gives a closed disk on either side of that curve on $S^{2}$ (Choice-free smooth Jordan disk parameters and prescribed collars).

Proof

1.1givenF1F3F4choose

By [F1], the projection graph has finitely many components and the link has compact image. Choose a point of $S^{2} ∖ P$ as infinity and work in the resulting finite projection plane. Apply the regular-neighborhood construction in the proof of [F3] in this chart; its smooth separating curve $γ$ misses the projection and has diagram points on both sides. Choose the bounded closed side as the disk $Δ$ , using [F4].

2.1givenF1F2F3F4step 1.1construct∎

Regard $S^{2}$ as the projection sphere in $S^{3}$ . The product $Δ \times [- M, M]$ is a smoothly embedded 3-ball after rounding its corners, where $M$ exceeds the absolute height of every point of the compact link. Its rounded boundary consists of the side annulus over $γ$ and the two caps at heights $\pm M$ ; it is a sphere. Since $γ$ misses the projection and the caps lie above and below the link, this sphere misses the link. The two sides meet the link because $Δ$ and its complementary disk each contain projection points, so [F2] says that the link is split.

Depends on

Used by

Reduced alternating diagrams minimize crossing number Theorem

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Danny Calegari, Introduction to Knot Theory, §5.2 and Exercise 5.14, printed pp. 42–43 (standard reference, not scraped)
William W. Menasco, Closed incompressible surfaces in alternating knot and link complements, Topology 23 (1984), Theorem 1(a), pp. 37–41 (standard reference, not scraped)