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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Zero-surgery on the circle

Example

Assume ACω, as in the surgery definition. Take M=S1, p=0, q=1, and the standard framed embedding φ:S0×D1↪S1 that is the inclusion of the vertical sides in the standard decomposition (with corner charts rounded compatibly) ∂(D1×D1)=(S0×D1)∪(D1×S0) of the square's boundary, whose image is the union of two disjoint closed intervals. The framing is part of this data, and it is the one used in the verification below. The range 0≤p≤m−1 holds with m=1. The 0-surgery removes the interiors of the two intervals and glues D1×S0, that is, two intervals, along S0×S0, that is, four points. The result is S1⊔S1.

Verification

Given: M=S1 with the standard framed embedding φ:S0×D1↪S1 of the two closed intervals.

[F1] p-surgery on a smooth m-manifold: the p-surgery is Mφ=(M∖φ(Sp×int⁡Dq))∪φ∣Sp×Sq−1(Dp+1×Sq−1), with p=0, q=1 this removes the interiors of the two intervals and glues two intervals along four points; the construction takes place in the interior of M.

[F2] The outgoing boundary of a handle attachment trades the disk factors: for k=1, n=2 the boundary of the square is ∂(D1×D1)=(S0×D1)∪(D1×S0), two pairs of opposite sides meeting in the four corners S0×S0, and the trade lemma identifies the complement of the open attaching region with the complementary pair of sides.

[F3] The surgery gluing has a canonical smooth structure up to diffeomorphism: the gluing along the common boundary gives a smooth 1-manifold, and its diffeomorphism type is the one fixed by the identification on the overlap.

1.1F1given

The normal bundle of a point in a 1-manifold is trivial, so a framing of the 0-sphere S0×{0} is exactly the product structure exhibited by φ; the image of φ is the union of two disjoint closed intervals of S1, whose complement after removal of their interiors is a union of two disjoint closed arcs.

2.1F1F2step 1.1

Reading the standard decomposition of the square's boundary in [F2], the two intervals S0×D1 are the two vertical sides and the two intervals D1×S0 are the two horizontal sides; the four corners are S0×S0. The 0-surgery removes the open vertical sides and glues in the two horizontal sides, identifying their endpoints with the four corners by the framing, so the result is exactly the boundary of the square with the vertical sides replaced by the horizontal sides.

3.1F2step 2.1algebra

The complement of the open vertical sides in the square's boundary is the union of the two horizontal sides, each a closed arc; the glued interval D1×{−1} joins the two endpoints of one horizontal side and the glued interval D1×{+1} joins the two endpoints of the other, so each horizontal side is closed up by one glued interval into a circle. There are no other points, and the two circles are disjoint because the four corners are distributed two to each. Hence the surgered manifold is S1⊔S1.

4.1F1F2F3step 3.1∎

Equivalently, the same computation reads D1×S0∪D1×S0=S0×(D1∪S0D1)=S0×S1=S1⊔S1, the two copies of D1×S0 being the glued-in piece and the complementary arcs of the standard decomposition of ∂(D1×D1); the gluing is the one induced by the framing, and by [F3] the smooth structure is the canonical one.

The example exercises the endpoint p=0 of the definition and the case p=m−1 (q=1) of the trace construction: the trace is the cylinder S1×[0,1] with a single 1-handle attached, in accordance with the index shift of the trace definition, and its outgoing face is the two-circle manifold just computed.

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