Smooth Surgery Traces and Handle Trading — Examples

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

Reading the standard decomposition of the square's boundary in [F2], the two intervals $S^0\times D^1$ are the two vertical sides and the two intervals $D^1\times S^0$ are the two horizontal sides; the four corners are $S^0\times S^0$. 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.

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 $D^1\times \{-1\}$ joins the two endpoints of one horizontal side and the glued interval $D^1\times \{+1\}$ 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 $S^1\bigsqcup S^1$.

Equivalently, the same computation reads $D^1\times S^0\cup D^1\times S^0=S^0\times \left(D^1{\cup }_{S^0}{D}^1\right)=S^0\times S^1=S^1\bigsqcup S^1,$ the two copies of $D^1\times S^0$ being the glued-in piece and the complementary arcs of the standard decomposition of $\mathrm{\partial }(D^1\times D^1)$; the gluing is the one induced by the framing, and by [F3] the smooth structure is the canonical one.

The disk chart at $y_0$ exhibits the embedding $\phi :S^p\times D^q\hookrightarrow S^p\times S^q$, $\phi (x,y)=(x,y)$ in the chart, with image in the interior; its restriction to the disk factor is the product trivialization, so $\phi$ is a framed embedded surgery sphere with underlying sphere $S^p\times \{y_0\}$, framed by the $S^q$-factor.

Use the hemisphere parameterizations given by the map of [F3]. The removed neighbourhood in the second factor is one labelled closed hemisphere and its closed complement is the other, with matching boundary coordinate $S^{q-1}$. Thus removing the open tube leaves exactly $S^p\times D^q$ with the boundary identification specified by the product framing; no complement assertion for an arbitrary disk chart is needed.

Glue $D^{p+1}\times S^{q-1}$ to the complement of step 1.2 by the product boundary identification. By [F2] this is the rounded boundary of $D^{p+1}\times D^q$. Choose a convex rounding of the product corners: its boundary is transverse to each ray from the origin, so it is $\{\rho (u)u:u\in S^m\}$ for a positive smooth $\rho$. The radial map $u\mapsto \rho (u)u$ has smooth inverse $z\mapsto z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$, proving that this boundary is diffeomorphic to $S^m$. Rounding independence gives the same diffeomorphism type for other compatible roundings. This proves the first computation, including $p=0$ and $q=1$.

For the dual reading, regard $S^m=\mathrm{\partial }(D^{p+1}\times D^q)$ with the decomposition of [F2]; the standard framed $S^p=S^p\times \{0\}$ lies in the solid piece $S^p\times D^q$ and has tubular neighbourhood $S^p\times D^q\subseteq S^m$, framed by the $D^q$-factor. The $p$-surgery on $S^m$ along this sphere removes $S^p\times \mathrm{int}{D}^q$ and glues in $D^{p+1}\times S^{q-1}$ along $S^p\times S^{q-1}$, leaving two copies of $D^{p+1}\times S^{q-1}$ glued along their common boundary; that double is the product $S^{p+1}\times S^{q-1}$ of the double of $D^{p+1}$, which is $S^{p+1}$ by [F3], with the closed factor $S^{q-1}$, the gluing being the product identification. Hence the surgery on $S^m$ along the standard framed $S^p$ produces $S^{p+1}\times S^{q-1}$.

For $p\ge 1$, the starting product is connected, so [F5] shows that the inverse of the first computation uses the belt sphere of dimension $q-1$ in $S^m$ and returns $S^p\times S^q$. For $p=0$, compute that inverse directly: in $S^m=\mathrm{\partial }(D^1\times D^q)$, remove the interior of the belt tube $D^1\times S^{q-1}$ and insert $S^0\times D^q$. The complement is another $S^0\times D^q$, with the product boundary identification; their union is $S^0\times (D^q{\cup }_{S^{q-1}}{D}^q)\cong S^0\times S^q$ by [F3]. The second computation operates on the other sphere, of dimension $p$, in $S^m$, and produces $S^{p+1}\times S^{q-1}$; its inverse returns $S^m$. Thus the two $p$-surgeries are not identified with each other's dual operations.

Each ${\phi }_k$ is a smooth embedding with inverse $(x,y)\mapsto (x,x^{-k}y)$ on $U$, and all have core $S^1\times \{0\}$. Writing the replacement torus as $T=D_a^2\times S_b^1$, its boundary gluing to $V$ is $g_k(a,b)=(a,a^{k}b)$. This follows directly from [F1], using $a$ as the attaching-sphere coordinate and $b$ as the normal-circle coordinate.

For $k=0$, $g_0$ is the product identification. Thus $V{\cup }_{g_0}T=(D^2{\cup }_{S^1}{D}^2)\times S^1\cong S^2\times S^1$. To check smoothness of the disk double identification, map polar disk coordinates $(r,u)$ in the two copies to $(\sin (\pi r\mathrm{/}2)u,\pm \cos (\pi r\mathrm{/}2))$ on $S^2$; it is smooth and invertible at the centres and in the signed collar coordinate at the seam.

For $k=1$, change coordinates by diffeomorphisms of the solid tori themselves: $L:V\to V$, $L(x,y)=(x{y}^{-1},y)$, and $R:T\to T$, $R(a,b)=(a{b}^{-1},b)$. The transformed boundary gluing is $L\circ g_1\circ R(a,b)=(b^{-1},a)$, as direct multiplication shows. This exchanges the two boundary circle factors with one reversal. In the boundary decomposition of [F2], identify the first solid torus $S^1\times D^2$ with $T=D^2\times S^1$ by $(a,b)\mapsto (b^{-1},a)$; it is a diffeomorphism, so the transformed gluing produces the boundary of $D^2\times D^2$. A convex corner rounding is radially transverse to all rays from the origin; write its boundary as $\rho (u)u$ for smooth positive $\rho$ on $S^3$. The radial map and its inverse $z\mapsto z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$ prove that boundary is diffeomorphic to $S^3$. Hence one-twist surgery gives $S^3$.

By [F3] and [F4], the manifolds computed in steps 2.1 and 2.2 cannot be diffeomorphic. These two valid framings of the same core knot therefore give different diffeomorphism types.

The diagonal $\mathrm{\Delta }\subseteq S^2\times S^2$ is a closed embedded $2$-sphere with $2\dim \mathrm{\Delta }=4=\dim (S^2\times S^2)$, so it has half the ambient dimension and both $\mathrm{\Delta }\cdot \mathrm{\Delta }$ and the normal-bundle statements apply to it.

By [F1] the normal bundle of $\mathrm{\Delta }$ is canonically isomorphic to $T{S}^2$. By [F6] the tangent bundle $T{S}^2$ has no nowhere-zero global section and is therefore not trivial, so ${\nu }_{\mathrm{\Delta }}$ is a nontrivial rank-two bundle over $\mathrm{\Delta }\cong S^2$.

By [F5] the existence of a framing of ${\nu }_{\mathrm{\Delta }}$, equivalently of an extension of the inclusion $\mathrm{\Delta }\hookrightarrow S^2\times S^2$ to an embedding $S^2\times D^2\hookrightarrow S^2\times S^2$, is equivalent to triviality of ${\nu }_{\mathrm{\Delta }}$. Since ${\nu }_{\mathrm{\Delta }}$ is nontrivial by step 2.1, no such extension exists: $\mathrm{\Delta }$ is not the underlying sphere of any framed embedded surgery sphere, so it is not a valid surgery datum for the construction of this page.

The same obstruction has a geometric form. By [F2] applied to the manifold $S^2$ of $S^2\times S^2$, the self-intersection of the diagonal is $\mathrm{\Delta }\cdot \mathrm{\Delta }=⟨e(T{S}^2),[S^2]⟩$, the Euler number of the tangent bundle of the $2$-sphere, and [F1] with [F3] gives the same value as $⟨e({\nu }_{\mathrm{\Delta }}),[\mathrm{\Delta }]⟩$. Had ${\nu }_{\mathrm{\Delta }}$ been trivial, [F3] combined with [F4] would have forced $\mathrm{\Delta }\cdot \mathrm{\Delta }=0$, so the nontriviality of ${\nu }_{\mathrm{\Delta }}$ detected in step 2.1 is exactly the obstruction that the self-intersection form measures in the middle dimension.

In summary, $\mathrm{\Delta }$ is an embedded $2$-sphere in the closed smooth $4$-manifold $S^2\times S^2$ whose normal bundle is nontrivial; embeddedness alone does not make it valid framed surgery data, and the surgery step of this page cannot be applied along it. This refutes the claim that every embedded sphere is eligible surgery data.

Removing the standard product tube $S^2\times \mathrm{int}{D}^2$ from $S^2\times S^2$ leaves $S^2\times D^2$, since the complement of the open hemisphere in the second sphere is its opposite closed hemisphere. Product-framed surgery glues in $D^3\times S^1$ by the identity on $S^2\times S^1$. The resulting union is $\mathrm{\partial }(D^3\times D^2)$ by the disk-factor boundary decomposition. Choose a convex rounding transverse to rays from the origin; its boundary is $\rho (u)u$ for a smooth positive function on $S^4$. Radial projection has smooth inverse $u\mapsto \rho (u)u$, identifying the boundary with $S^4$ and proving the surgery identification directly.

The datum is middle-dimensional: $p=q=2$, so $2p+2=6>4=m$ and the below-middle hypothesis $p\le q-2$ of the killing lemma fails. The product example [F1] identifies the surgered manifold: the $2$-surgery on $S^2\times S^2$ along the standard framed $S^2\times \{y_0\}$ is $S^4$, compatible with the degree bounds of [F5] since $p=2$ and $q=2$.

Sphere homology is free, concentrated in degrees $0,2$ for $S^2$. The integral Künneth sequence therefore has vanishing Tor terms; in degree two its tensor terms are $H_2(S^2)\otimes H_0(S^2)$ and $H_0(S^2)\otimes H_2(S^2)$, each $\mathbb{Z}$, and its cross-product map sends the two generators to the factor sphere classes. Thus the degree-two homology of the source is free of rank two on the two factor sphere classes: by [F2] applied to $m=n=2$, the classes of $A=S^2\times \{y_0\}$ and $B=\{x_0\}\times S^2$ form a basis of $H_2(S^2\times S^2;\mathbb{Z})\cong {\mathbb{Z}}^2$. The target has $H_2(S^4;\mathbb{Z})=0$ by [F3].

By [F6], intersection gives a bilinear form on the two factor classes. Pushing $A=S^2\times \{y_0\}$ to $S^2\times \{y_1\}$ along a short path with $y_1\mathrm{\ne }{y}_0$ makes it disjoint from $A$, so $A\cdot A=0$ by [F4]; the analogous push-off of $B$ gives $B\cdot B=0$. At the sole intersection $(x_0,y_0)$, the ordered tangent spaces of $A$ and $B$ are precisely the two positively oriented factors of $TM$, so $A\cdot B=+1$. Exchanging the two two-dimensional blocks has sign $(-1{)}^{2\cdot 2}=+1$, giving $B\cdot A=+1$. Thus the matrix in the factor basis is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, of determinant $-1$, the hyperbolic form.

Hence the middle-dimensional intersection pairing of $M$ is a nondegenerate pairing on a rank-two free group, while the corresponding pairing of $M_{\phi }=S^4$ is a pairing on the zero group: by [F3] the middle homology vanishes, and [F6] transports this to middle cohomology, so the form of the surgered manifold is the zero form. The two pairings therefore cannot be identified by any isomorphism of the underlying groups, and the middle-dimensional intersection form is not preserved by the surgery.

Consequently framed sphere surgery does not preserve the middle-dimensional intersection form: at the middle dimension $p=q$ the form itself can change, a case that the below-middle results of this page, proved under $p\le q-2$, do not cover, and there the analysis requires the quadratic refinement and the surgery obstruction recorded in the preceding remark.