= Solution
An index-$\lambda$ <handle> in dimension $d$ is $D^\lambda\times D^{d-\lambda}$. A <handle attachment> glues its attaching region $S^{\lambda-1}\times D^{d-\lambda}$ to a corresponding neighborhood in the boundary of the preceding <manifold with boundary>. Besides the embedding of the <attaching sphere>, one specifies a <framing of an embedded sphere>, namely the normal-disk directions used in this gluing. Corners are then smoothed.
In dimension four, a <two-handle> is $D^2\times D^2$ and attaches along $S^1\times D^2$. Its attaching circle is a <knot> in the boundary of the <zero-handle> $D^4$. Identify that boundary with $S^3$ and remove a point away from the circles to draw them in $\mathbb R^3$. Disjoint attaching circles give a <framed link>. Each integer records the twist of its normal framing relative to the zero <Seifert framing>; the integer is also the <linking number> of the circle and its framed push-off. The resulting boundary change is <Dehn surgery on a framed link>. This is the two-handle part of a <Kirby diagram>.
For $S^2\times S^2$, use the following explicit <handle decomposition>. Write each factor as two disks, $S^2=D_-^2\cup D_+^2$, glued along their boundary circles. The four product pieces give
$$
h^0=D_-^2\times D_-^2,\qquad h^2_1=D_+^2\times D_-^2,\qquad h^2_2=D_-^2\times D_+^2,\qquad h^4=D_+^2\times D_+^2.
$$
The first is a four-dimensional <zero-handle> after rounding its corners. Its boundary is the union of the solid tori $S^1\times D^2$ and $D^2\times S^1$. The two middle product pieces attach along these tori. Their attaching circles $S^1\times\{0\}$ and $\{0\}\times S^1$ form the standard <Hopf link>, with <linking number> one after choosing orientations.
Both framings are zero. For example, the normal disk of $S^1\times\{0\}$ is the second disk factor. Its constant normal push-off has linking number zero with the original circle: the circle bounds $D^2\times\{0\}$ in the four-ball, and a parallel push-off bounds a disjoint translated disk. Thus the framed self-intersection is zero; the same argument applies with the factors interchanged. These product framings are the zero <Seifert framings>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-27-hopf.png]
{title=Zero-framed Hopf link for the two two-handles of S2 times S2; the upper crossing has blue over orange and the lower crossing has orange over blue}
After these two <two-handles>, the remaining product piece $D_+^2\times D_+^2$ is a four-ball. Its boundary is exactly the boundary of the assembled first three pieces, so it attaches as a <four-handle>. Equivalently, <zero surgery on the Hopf link> gives $S^3$, which this handle caps. We have therefore constructed the actual product, proving
$$
\boxed{S^2\times S^2=\text{one zero-handle}+\text{two zero-framed Hopf-linked two-handles}+\text{one four-handle}.}
$$
There are no one- or three-handles. As a check, the two factor spheres have self-intersection zero and meet once, giving the <intersection form> $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$. The explicit product construction, rather than the intersection form alone, identifies the smooth four-manifold. This is the <zero-framed Hopf-link handle decomposition of the sphere product>.
Back to article page
