Vanishing stable characteristic classes do not make two embeddings isotopic

For the standard embedding $i(p)=p$ the radial field $n_i(p)=p$ spans its orthogonal normal line and is smooth and nowhere zero. For the reflected embedding $r$, the field $n_r(p)=r(p)$ is normal because reflection preserves inner products: $⟨r(p),d{r}_p(v)⟩=⟨p,v⟩=0$. It is again a smooth nowhere-zero frame. Thus both normal lines are trivial. The embedding normal-bundle identity identifies their characteristic classes with the stable normal classes of $S^2$, and the trivial line has total Stiefel--Whitney and Pontryagin class $1$ and Euler class $0$.

The two embeddings have identical stable characteristic classes. Both are embeddings of the same closed oriented surface $S^2$ into ${\mathbb{R}}^3$ with trivial normal line bundle by [F1], so their normal bundles are isomorphic (both trivial). The normal classes $\bar{w}(S^2)=1$ and $\bar{p}(S^2)=1$ of [F1] are classes of the manifold $S^2$ and therefore the same for both embeddings, and the codimension-one Euler class of the trivial normal line is zero. Consequently every characteristic-class test considered on this page — the mod-two normal classes, the rational normal Pontryagin classes and the oriented Euler class — evaluates trivially for both $i$ and $r$, so no such test distinguishes them.

Suppose, for contradiction, that $i$ and $r$ were isotopic: there is a smooth isotopy of embeddings $F:S^2\times I\to {\mathbb{R}}^3$ with $F_0=i$ and $F_1=r$. Replacing $F$ by a reparametrisation in $t$ that is constant near the ends, the isotopy is constant near the ends without changing its endpoints; by [F2] and [F5] it extends to an ambient isotopy $H:{\mathbb{R}}^3\times I\to {\mathbb{R}}^3$ with $H_0={\mathrm{id}}_{{\mathbb{R}}^3}$, every $H_t$ a diffeomorphism, and $H_t\circ i=F_t$ for every $t$. In particular $H_1\circ i=r$.

The time-one map $H_1$ is an orientation-preserving diffeomorphism of ${\mathbb{R}}^3$ by [F3]. Since $H_1\circ i=r$ and $i,r$ have image $S^2$, it satisfies $H_1(S^2)=S^2$. As a diffeomorphism it maps the two connected components of ${\mathbb{R}}^3\setminus S^2$ onto the two components, and $H_1(\mathrm{int}{B}^3)$ is a component whose closure $H_1(B^3)$ is compact because $B^3$ is compact and $H_1$ is continuous: hence $H_1(\mathrm{int}{B}^3)$ is the bounded component $\mathrm{int}{B}^3$ and $H_1(B^3)=B^3$.

Since $H_1$ is an orientation-preserving diffeomorphism of ${\mathbb{R}}^3$ mapping $B^3$ onto itself, it maps the boundary $S^2$ to itself and its differential carries outward-pointing boundary vectors to outward-pointing boundary vectors; therefore the restriction $H_1{\mathrm{\mid }}_{S^2}:S^2\to S^2$ preserves the outward-normal-first boundary orientation. By [F3] and [F4], an orientation-preserving diffeomorphism of the connected closed oriented surface $S^2$ has degree $+1$.

On the other hand $H_1\circ i=r$ means $H_1{\mathrm{\mid }}_{S^2}=r{\mathrm{\mid }}_{S^2}$ as maps $S^2\to S^2$, and by [F4] the reflection has degree $-1$. This contradicts the degree $+1$ computed in step 3.1, so no isotopy from $i$ to $r$ exists. Hence two embeddings with identical (indeed trivial) normal bundles and identical stable characteristic classes need not be isotopic, and these characteristic classes do not determine isotopy. AC is inherited from the characteristic-class suppliers and supplies the countable choice used by isotopy extension.