Middle-dimensional surgery can change an intersection form

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.