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The geometric intersection number is the Poincare-dual cup pairing

Statement

Assume AC. Let M be a closed oriented smooth n-manifold and let Aa,Bb⊆M be closed oriented embedded submanifolds with a+b=n. Write [A]=(iA)∗[A]A, [B]=(iB)∗[B]B, and PD[A]=DM−1([A]), PD[B]=DM−1([B]). In the cohomology-first, front-evaluation cap and cup conventions, I(A,B)=⟨PD[A]⌣PD[B],[M]⟩. For nontransverse submanifolds I(A,B) is computed by a transverse map homotopic to iA; no embedded representative for that map is required. Over F2 the same formula holds without orientability. Consequently this geometric number depends only on the represented homology classes, with the factor order of The local oriented intersection sign.

Facts & Assumptions

Given: AC and M,A,B with their orientations and complementary dimensions as in the statement.

[F1]

Under Countable Choice a smooth map is homotopic to a transverse map; the resulting intersection number is well defined and homotopy invariant (The transversality homotopy theorem, The geometric intersection pairing on a closed oriented manifold).

[F2]

Cap is natural and satisfies (x⌣y)∩c=y∩(x∩c). Evaluation of a top-degree cup is therefore evaluation of its second factor on the cap by the first (Cap naturality and projection formula, Kronecker evaluation pairing).

[F3]

For the tangent-first normal Thom extension αB∈Ha(M,M∖B;R), its absolute image satisfies PD[B]=(−1)abαˉB (The normal Thom class realizes the Poincare dual of a closed submanifold).

[F4]

Excision localizes a class supported on finitely many points to disjoint disk pairs; fundamental classes restrict to their prescribed local orientation generators. A normalized Thom class restricts to the normal fibre generator (Excision for singular cohomology, Fundamental class of a compact oriented manifold, Thom class by fiberwise normalization).

[F5]

The local intersection sign compares df(TA) followed by TB with TM. Complementary transverse preimages are finite for a compact source and closed target (The local oriented intersection sign, Compact transverse complementary intersections are finite).

Proof

technique · use a transverse map on the original source, restrict the normal Thom class of $B$, and compare its local determinant with the ordered intersection sign
1.1F1F2F5givenchoose

By [F1] choose a smooth f:A→M, homotopic to iA, transverse to B. Then I(A,B)=I(f,B) and P=f−1(B) is finite by [F5]. Homotopy invariance of homology gives f∗[A]A=[A]. By [F2] and PD[A]∩[M]=[A], ⟨PD[A]⌣PD[B],[M]⟩=⟨f∗PD[B],[A]A⟩. This uses the map f on the fixed oriented source A, and does not replace A by its image.

2.1F3F4F5step 1.1algebra

Pull back αB along the map of pairs (A,A∖P)→(M,M∖B) to get β. At p∈P, the quotient derivative qp:TpA→νB,f(p) is an isomorphism. A normalized tubular chart for B and a local trivialization of its normal bundle give a normal-component map with derivative qp. The inverse function theorem makes it a local diffeomorphism. In the induced normal coordinates the fibre Thom generator pulls back to σp times the source orientation generator, where σp is the orientation-ray sign of qp: using the local diffeomorphism as a source chart proves this directly. Excision [F4] and the finite direct sum of the point-supported relative complexes then give ⟨βˉ,[A]A⟩=∑p∈Pσp. In dimension zero the same statement is multiplication of the supplied point and normal orientation units, without an inverse-function argument.

3.1F2F3F4F5step 1.1step 2.1algebra∎

Let u be a positive tangent determinant of B and v a positive normal determinant. By tangent-first normal orientation, (u,v) is positive in M. If d is a positive determinant of TpA, then (u,dfp(d)) has sign σp because quotienting its second block gives qp(d). Swapping the blocks of dimensions b,a shows that (dfp(d),u) has sign (−1)abσp. Thus [F5] gives εp=(−1)abσp, including the point-ray case. By [F3], f∗PD[B]=(−1)abβˉ. Combining with steps 1.1–2.1 yields the displayed formula. Empty P gives zero on both sides. Over F2 the same finite local evaluation applies with every orientation sign equal to one. AC is inherited through transverse representatives, Thom existence and duality.

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