Intersection form ($Q_M$)

For a closed manifold $M$ with an orientation and dimension $4k$, its integral intersection form is the symmetric bilinear form onobtained by pairing complementary middle-dimensional homology classes. Under Poincare duality, it is the cup product pairingon degree-$2k$ integral cohomology modulo its torsion subgroup. The even degree makes the pairing symmetric; integral duality makes it a unimodular intersection pairing. When representatives are transverse oriented submanifolds, the same value is the signed count of their intersection points.

The intersection form of a 4-manifold is the case $k=1$. On the Complex projective plane its matrix is $(1)$, whereas on $S^2\times S^2$ the factor spheres give $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$.

For a compact manifold with boundary, the perfect intersection pairing pairs absolute and relative homology, $H_{2k}(M)$ with $H_{2k}(M,\partial M)$; the induced absolute self-pairing can be degenerate. For instance the core sphere in $S^2\times D^2$ generates $H_2$ but has self-intersection zero.

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.

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.

Here $H$ is the hyperbolic form with matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. The degree-two cohomology consists of the two factor orientation classes and $4gh$ products of degree-one classes. The factor classes pair once and have zero squares. In symplectic bases $a_i,b_i$ and $c_j,d_j$, $(a_i{c}_j)(b_k{d}_l)=-{\delta }_{ik}{\delta }_{jl}{\omega }_1{\omega }_2$ and $(a_i{d}_j)(b_k{c}_l)={\delta }_{ik}{\delta }_{jl}{\omega }_1{\omega }_2$. Each pair $(i,j)$ contributes two hyperbolic forms; the fiber classes contribute one more. The sign in the first formula comes from interchanging two odd-degree factors.

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.

When compact oriented four-manifolds are glued along an entire closed boundary three-manifold, the signature of the result is the sum of their intersection form signatures. It applies to $X_0{\cup }_{\partial X_0}(-X_1)$ with the difference of the two signatures. Gluing along only part of a boundary can require a correction; that is a different situation.

An intersection pairing counts signed intersections of transverse representatives of complementary homology classes, using one relative representative when the manifold has boundary. With ${\mathbb{F}}_2$ coefficients it is defined without orientation choices. Poincare-Lefschetz duality makes this a perfect pairing for compact manifolds.

The intersection matrix records an intersection pairing on chosen classes. Its nonsingularity proves linear independence: a relation between the classes, paired with each class in turn, gives a vector in the matrix's kernel.

In the product $(S^{2d}{)}^{2\ell }$, for $d,\ell \ge 1$, vary the factors indexed by an $\ell$-element subset $I$ and fix points in the other factors. Orient the resulting submanifold $S_I$ in increasing factor order. These homology classes form an integral middle-dimensional basis by the Künneth theorem. Complementary subsets give a transverse intersection at one point, with positive sign since all factor dimensions are even. For noncomplementary subsets some factor is fixed in both representatives; moving that point makes them disjoint. In particular every basis class has zero self-intersection number. Complementary pairs give hyperbolic blocks in the intersection form.

For a compact three-manifold $W$, let $L=\ker (H_1(\partial W;{\mathbb{F}}_2)\to H_1(W;{\mathbb{F}}_2))$. The boundary intersection pairing satisfies $L^{\perp }=L$: the long exact sequence in relative homology identifies $L$ with boundaries of relative surfaces, and the boundary-interior adjoint identity identifies its orthogonal complement with the same kernel. Hence $2\dim L=\dim H_1(\partial W;{\mathbb{F}}_2)$. This also implies that a closed surface with odd first mod-two Betti number, such as ${\mathbb{RP}}^2$, cannot be the entire boundary of a compact three-manifold.

For a map $f:M\to N$ of connected closed oriented four-manifolds, naturality of the cup product givesWhen $\mathrm{deg}f\ne 0$, nondegeneracy makes $f^{\ast }:H^2(N;\mathbb{Q})\to H^2(M;\mathbb{Q})$ injective. Consequently the target intersection form, multiplied by the mapping degree, must embed into the source form. For equal second Betti numbers, a matrix $P$ for pullback satisfies $P^{\mathsf{T}}{Q}_{M}P=(\mathrm{deg}f)Q_N$; taking determinants often rules out nonzero degrees. In particular, a two-dimensional indefinite source form cannot realize a nonzero multiple of a two-dimensional definite target form.

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.

On a closed oriented Riemannian manifold of dimension four, the Hodge star operator preserves harmonic two-forms and splits them into orthogonal spaces ${\mathcal{H}}^{+}$ and ${\mathcal{H}}^{-}$ with eigenvalues $+1$ and $-1$. The real de Rham intersection form in dimension four equals the positive $L^2$ inner product on ${\mathcal{H}}^{+}$ and its negative on ${\mathcal{H}}^{-}$. It is nondegenerate, since $Q(h,\ast h)=\Vert h{\Vert }_2^2$. Its signature pair is $(\dim {\mathcal{H}}^{+},\dim {\mathcal{H}}^{-})$.