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Plumbing identifies a base disk times a fiber disk in one oriented rank-two disk bundle with such a neighborhood in another, exchanging base and fiber coordinates. Their zero sections then intersect transversely once. For disk bundles over $S^2$ arranged along a chain with Euler numbers $-a_i$, the intersection form has diagonal entries $-a_i$, adjacent entries one and other entries zero. The same manifold has a Kirby diagram consisting of adjacent Hopf-linked unknots with framings $-a_i$. Its boundary is described by the corresponding negative continued fraction, with the orientation fixed by the surgery convention.

For a closed oriented smooth four-manifold, $b_2^{+}$ is the dimension of a maximal positive-definite subspace of its real intersection form. A symplectic form forces $b_2^{+}\ge 1$, because its cohomology class has positive square. Reversing orientation exchanges the positive and negative indices.

The fundamental class of a complex projective line generates $H_2({\mathbb{CP}}^2;\mathbb{Z})$. Two distinct projective lines are homologous and meet at one transverse point. In complex affine coordinates their two complex tangent lines concatenate to the complex orientation of the plane, so the signed intersection is $+1$. The integral intersection form therefore has matrix $(1)$, giving a unimodular intersection pairing. This computes a self-intersection through distinct representatives of the same class.

For a simply connected closed four-manifold with second Betti number one, Poincare duality and the universal coefficient theorem for cohomology make $H_2$ torsion-free of rank one. The integral intersection form is unimodular, so its one-by-one matrix is $[1]$ or $[-1]$. Hence the square of a generator $u\in H^2$ generates $H^4$, giving the displayed cohomology ring. This is a ring isomorphism; it need not preserve a preassigned orientation class.

On a compact oriented smooth manifold $N$ of dimension four without boundary, set $Q([\alpha ],[\beta ])={\int }_N\alpha \wedge \beta$ for closed real two-forms. The Stokes theorem makes the expression independent of representatives, and the even degrees make it symmetric. It is the real de Rham cohomology version of the intersection form. The Hodge decomposition theorem and Hodge star operator express it as $Q(h,k)=(h,\ast k{)}_{L^2}$ on harmonic representatives.

A smooth four-manifold is a four-dimensional smooth manifold. Its intersection form and gauge-theoretic invariants impose strong constraints on its possible smooth and symplectic structures.

The surgery linking matrix has the framing coefficients on its diagonal and pairwise linking numbers off the diagonal. It represents the intersection form on the second homology of the surgery trace, presents the first homology of its boundary, and has kernel isomorphic to the boundary's second homology.

An integral intersection form on a finite-rank free abelian group $L$ is unimodular if its adjoint $x\mapsto \lambda (x,-)$ is an isomorphism to the integral dual. In an integral basis, this is equivalent to the pairing matrix having determinant $1$ or $-1$, not merely nonzero determinant. For example, projective lines generate a unimodular intersection form on the Complex projective plane, with matrix $(1)$ in its complex orientation.

For $k\ge 1$, the rational middle intersection pairing induces an isomorphism ${\Omega }_{4k}^{\mathrm{Witt}}\cong W(\mathbb{Q})$. The restriction $k>0$ matters: the degree-zero group is $\mathbb{Z}$. In positive dimension congruent to zero modulo four, a space is Witt null-bordant precisely when its middle intersection form is metabolic.

Write each sphere as two disks, $S^2=D_{-}^2\cup D_{+}^2$. In their product, $D_{-}^2\times D_{-}^2$ is a zero-handle; $D_{+}^2\times D_{-}^2$ and $D_{-}^2\times D_{+}^2$ are two-handles. Their attaching circles are the core circles of the complementary solid tori in $\partial (D^2\times D^2)$, hence a Hopf link. Constant normal directions from the disk factors give both components the zero Seifert framing. The remaining $D_{+}^2\times D_{+}^2$ is a four-handle. This product construction establishes the actual diffeomorphism type, rather than relying on its intersection form alone. The resulting intersection matrix is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$.