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Same-sign intersection points cannot be cancelled orientedly

Statement refuted

"If two closed connected oriented complementary submanifolds $A, B$ of a closed oriented manifold meet transversely in exactly two points whose mod-two contribution is even, then an isotopy of $A$ can make $A$ disjoint from $B$ ."

Facts & Assumptions

[F1]

The local sign compares the ordered tangent spaces of the two sheets with the ambient orientation. The local oriented intersection sign

[F2]

The oriented intersection number. The oriented intersection number

[F3]

The oriented intersection number is homotopy invariant. The oriented intersection number is homotopy invariant

Counterexample

Let $T^{2} = R^{2} / Z^{2}$ , let $A = S^{1} \times {0}$ , and let $B$ be the graph of the degree-two covering map $θ \mapsto 2 θ$ of the circle, i.e. $B = {(θ, 2 θ) : θ \in S^{1}} \subset T^{2}$ . Then $B$ is a closed embedded circle, $A \cap B$ consists of exactly two points with local signs $+ 1, + 1$ , and $I (A, B) = 2 \neq 0$ . Since $I (\cdot, B)$ is invariant under homotopies of the first factor, no homotopy (hence no isotopy) of $A$ can produce a disjoint configuration, and in particular the pair cannot be removed by a Whitney move: the sign condition $ε (x) = - ε (y)$ of the Whitney trick fails for the only pair of points. Thus equal signs are an obstruction beyond parity.

Given: Countable choice for homotopy invariance, the oriented torus, and its two specified embedded circles.

1.1givenconstructalgebraF1F2

The map $θ \mapsto (θ, 2 θ)$ is an embedding because its first coordinate is the identity of $S^{1}$ . Its image meets $A = {y = 0}$ only at $θ = 0, 1 / 2$ . At both points the ordered tangent vectors are $(1, 0)$ and $(1, 2)$ , whose determinant is $2 > 0$ . Thus the intersections are transverse with local signs $+ 1, + 1$ , their mod-two count is zero, and their integer count is two.

2.1step 1.1constructF3∎

Homotopy invariance of the oriented intersection number with the fixed $B$ preserves this count under any homotopy of $A$ , hence under any isotopy. A disjoint endpoint would have the empty signed sum zero, contradicting the value two. Thus the asserted cancellation fails, beyond the parity condition, and the necessary opposite-sign hypothesis is not satisfied. Countable choice is used exactly through the published homotopy-invariance supplier.

Depends on

Used by

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Sources

Andrew Ranicki, Algebraic and Geometric Surgery (Oxford Mathematical Monographs, Oxford University Press 2002; complete electronic copy) (standard reference, not scraped)
Wolfgang Lück, A Basic Introduction to Surgery Theory (complete lecture notes, ICTP/Münster) (standard reference, not scraped)