The intersection form on the homology of a closed oriented surface

Assume the Axiom of Choice (The Axiom of Choice), as required by the classification, Poincaré-duality, and geometric-intersection interfaces used here. Let $X$ be a nonempty compact connected topological $2$-manifold without boundary, with a specified integral orientation (Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces, R-orientation of a topological manifold) and with $H_1(X;\mathbb{Z})$ free of finite rank. For the standard surfaces on this page, the rank condition follows from Cellular homology of the one-polygon surface model; for a general compact connected orientable surface it follows from Classification of compact connected surfaces and the same cellular computation. A compact Riemann surface has the orientation determined by its complex structure (Topological classification of compact Riemann surfaces).

Write $[X]\in H_2(X;\mathbb{Z})$ for the fundamental class of the specified orientation (Fundamental class of a compact oriented manifold). Cap with this class gives the Poincaré-duality isomorphism $D_X:H^1(X;\mathbb{Z})⟶H_1(X;\mathbb{Z}),D_X(a):=a\cap [X]$ (The cap-duality map of an oriented manifold, Poincaré duality for oriented topological manifolds); write $D_X^{-1}$ for its inverse. The intersection form is $⟨\cdot ,\cdot {⟩}_X:H_1(X;\mathbb{Z})\times H_1(X;\mathbb{Z})⟶\mathbb{Z},⟨\gamma ,\delta {⟩}_X:=⟨D_X^{-1}(\gamma )⌣D_X^{-1}(\delta ),[X]⟩,$ where $⌣$ is the singular cup product (Singular cohomology ring, Singular cup product on cochains) and the outer brackets denote Kronecker evaluation (Kronecker evaluation pairing, The kronecker pairing is independent of cocycle and cycle representatives).

For genus zero, $H_1(X;\mathbb{Z})=0$ and this is the unique bilinear form on the zero group. A closed connected oriented surface has $\mathrm{rank}{H}_1=2g$, so rank one does not occur (Cellular homology of the one-polygon surface model, Classification of compact connected surfaces).

This form is well defined and biadditive because the cup product and Kronecker evaluation descend to cohomology and homology. Graded commutativity in degree one gives $⟨\gamma ,\delta {⟩}_X=-⟨\delta ,\gamma {⟩}_X$ (Singular cohomology is graded commutative); since the values lie in the torsion-free group $\mathbb{Z}$, it follows also that $⟨\gamma ,\gamma {⟩}_X=0$, so the form is alternating. For all $a,b\in H^1(X;\mathbb{Z})$ the cap-cup adjunction identity is $⟨a⌣b,[X]⟩=⟨b,D_X(a)⟩,$ where the right side is Kronecker evaluation (Poincaré duality gives a nonsingular cup pairing). Equivalently, for all $\gamma ,\delta \in H_1(X;\mathbb{Z})$, $⟨\gamma ,\delta {⟩}_X=⟨D_X^{-1}(\delta ),\gamma ⟩\mathrm{.}$

When $X$ is a closed oriented smooth surface and $A,B\subset X$ are closed oriented smooth embedded curves meeting transversely, $⟨[A],[B]{⟩}_X$ equals their algebraic intersection number $I(A,B)$, with the first-factor convention of The geometric intersection pairing on a closed oriented manifold (The geometric intersection number is the Poincare-dual cup pairing). The analogous geometric formula modulo $2$ holds for closed smooth surfaces without orientability (The geometric intersection number is the Poincare-dual cup pairing).

Reversing the orientation changes $[X]$ and both inverse-duality classes by a sign. The signs on the two cup-product factors cancel, while evaluation on $-[X]$ negates the result, so the intersection form changes sign. No unimodularity assertion is made here; it is proved with the symplectic-basis theorem below. Once the duality isomorphism is given, the formula and its algebraic identities are choice-free; AC is used only through the stated global classification, duality, and geometric-intersection interfaces.