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Mod-two evenness does not by itself supply a Whitney move

Statement

Assume ${A C}_{ω}$ for the homotopy-invariance supplier. On the oriented torus $T^{2} = R^{2} / Z^{2}$ , orient $A = S^{1} \times {0}$ and $B = {(θ, 2 θ) : θ \in S^{1}}$ by increasing $θ$ . Then $A \cap B$ consists of exactly two transverse points, each of local oriented sign $+ 1$ . Thus $I_{2} (A, B) = 0$ but $I (A, B) = 2 \neq 0$ . Homotopy invariance of oriented intersection implies that no homotopy, hence no isotopy, of $A$ can make it disjoint from $B$ . In particular even parity alone does not supply a Whitney move: the only pair already fails the necessary opposite-sign condition. This example establishes that mod-two vanishing does not imply oriented cancellability; it makes no separate claim about the independence of the label and framing conditions.

Facts & Assumptions

Given: The torus $T^{2} = R^{2} / Z^{2}$ with its orientation $d θ \land d y$ , the oriented embedded circles $A = S^{1} \times {0}$ and $B = {(θ, 2 θ m o d 1) : θ \in S^{1}}$ , both oriented by increasing $θ$ , and the inclusion $i_{A} : A \to T^{2}$ .

[F1]

For compact oriented complementary-dimensional submanifolds $A, B \subseteq M$ , where one is compact and the other closed, one sets $I (A, B) : = I (i_{A}, B)$ with $i_{A}$ the inclusion, so the first factor is the submanifold $A$ (The oriented intersection number).

[F2]

For transverse oriented embedded submanifolds $A^{a}, B^{b} \subseteq M$ with $a + b = n$ one takes $f$ and $g$ to be the inclusion maps; the sign at $p \in A \cap B$ then compares $T_{p} A \oplus T_{p} B \to T_{p} M$ with $A$ first (The local oriented intersection sign).

[F3]

For compact complementary-dimensional transverse submanifolds $A, B \subseteq M$ , where one is compact and the other closed, $I_{2} (A, B) : = I_{2} (i_{A}, B)$ with $i_{A}$ the inclusion, and $I_{2}$ is the cardinality of the transverse intersection reduced modulo two (The mod 2 intersection number).

[F4]

In the common setting of compact oriented complementary submanifolds, the oriented and mod 2 intersection numbers satisfy $I (A, B) \equiv I_{2} (A, B) (m o d 2)$ (The oriented intersection number reduces to the mod 2 number).

[F5]

If a smooth family $F : [0, 1] \times X \to M$ is transverse to $Z$ , including on the boundary faces, then $I (F_{0}, Z) = I (F_{1}, Z)$ ; consequently $I (\cdot, Z)$ is well defined on homotopy classes of smooth maps, any two transverse maps in the same homotopy class give the same number, and the definition extends to all smooth maps (The oriented intersection number is homotopy invariant). That theorem assumes the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the transverse representatives it selects.

Proof

technique · direct; compute the two local signs, then apply the homotopy-invariance consequence on the oriented intersection number

1.1F2givenconstructalgebra

The parametrization $θ \mapsto (θ, 2 θ m o d 1)$ is injective on $S^{1} = R / Z$ because its first coordinate is, so $B$ is an embedded circle with tangent spanned by $(1, 2)$ , while $T_{p} A$ is spanned by $(1, 0)$ ; hence $A \cap B = {(θ, 0) : 2 θ \equiv 0} = {(0, 0), (1 / 2, 0)}$ , at both of which $T_{p} A + T_{p} B = T_{p} T^{2}$ , so the two intersections are transverse, and the isomorphism $T_{p} A \oplus T_{p} B \to T_{p} T^{2}$ with the factor $A$ first has, in the basis $(\partial_{θ}, \partial_{y})$ , the matrix with columns $(1, 0)$ and $(1, 2)$ and determinant $2 > 0$ , so both local signs equal $+ 1$ .

2.1step 1.1F1F3F4algebra

Summing the two local signs $+ 1$ over the transverse intersection as in [F1], and counting its two points modulo two as in [F3], gives $I (A, B) = 2$ and $I_{2} (A, B) = 0$ ; the two values are congruent modulo $2$ , as [F4] requires, and the example therefore has $I_{2} (A, B) = 0$ while $I (A, B) \neq 0$ .

3.1step 2.1F1F5given∎

Suppose a smooth homotopy of $i_{A}$ ended at a smooth map $F_{1}$ whose image meets $B$ in no point; then $F_{1}$ is transverse to $B$ with empty preimage, so $I (F_{1}, B) = 0$ by [F1], while the homotopy-invariance consequence [F5], applied to the two transverse maps $i_{A}$ and $F_{1}$ in the same homotopy class, gives $I (i_{A}, B) = I (F_{1}, B)$ , which with step 2.1 is the contradiction $2 = 0$ . Since an isotopy of $A$ is such a homotopy, no isotopy can make $A$ disjoint from $B$ ; in particular no Whitney cancellation of the pair — an isotopy removing the two points and creating no new ones — is available, and the necessary opposite-sign hypothesis is violated because both local signs equal $+ 1$ by step 1.1. The homotopy-invariance input [F5] assumes ${A C}_{ω}$ , inherited here through The Axiom of Countable Choice ( ${A C}_{ω}$ ), while steps 1.1-2.1 are choice-free.

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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)
John Milnor, Lectures on the h-Cobordism Theorem (notes by L. Siebenmann and J. Sondow, Princeton University Press 1965; scanned edition with searchable text layer) (standard reference, not scraped)