Isotopy Extension and Embedding Theory Beyond Whitney — Examples

Fix $x\in {\mathbb{R}}^3$ and put $\phi (t):=\det d{H}_t(x)$ for $t\in I$. The function $\phi$ is continuous on $I$ by [F2], and it never vanishes because each $d{H}_t(x)$ is invertible by [F1]; since $\phi (0)=\det \mathrm{i}\mathrm{d}=1>0$, the intermediate value property [L1] gives $\phi (t)>0$ for every $t\in I$. Hence $d{H}_t(x)$ is orientation-preserving at every $(x,t)$, and in particular $H_1$ preserves the orientation of ${\mathbb{R}}^3$ at every point.

Since $H_1$ is a diffeomorphism and $H_1(B^3)=B^3$, [L2] shows that $H_1$ maps the interior of $B^3$ onto the interior and the sphere $S^2=\mathrm{\partial }{B}^3$ onto itself; thus $H_1{\mathrm{\mid }}_{S^2}:S^2\to S^2$ is a diffeomorphism. Moreover $d{H}_1$ carries the outward transverse direction of $S^2$ at each $p$ to the outward transverse direction at $H_1(p)$: the interior maps to the interior, so a tangent vector pointing into the ball maps to a vector pointing into the ball, and an outward transverse vector maps to an outward transverse vector: the inward boundary coordinate of the image vanishes at the boundary, is positive on the interior side, and has nonzero normal derivative by invertibility, so that derivative is positive. Orthogonality to the sphere need not be preserved.

The restriction $H_1{\mathrm{\mid }}_{S^2}$ preserves the outward-normal-first boundary orientation of [L3]: if $(v_1,v_2)$ is a positive basis of $T_p{S}^2$, so that $(n_p,v_1,v_2)$ is a positive basis of $T_p{\mathbb{R}}^3$ with $n_p$ outward, then step 1.1 makes $(d{H}_1(n_p),d{H}_1(v_1),d{H}_1(v_2))$ positive at $H_1(p)$, and $d{H}_1(n_p)$ is an outward vector by step 2.1, so $(d{H}_1(v_1),d{H}_1(v_2))$ is a positive basis of $T_{H_1(p)}{S}^2$. Hence $H_1{\mathrm{\mid }}_{S^2}$ is an orientation-preserving diffeomorphism of $S^2$, and [L4] gives $\mathrm{deg}(H_1{\mathrm{\mid }}_{S^2})=+1$.

Let $r:{\mathbb{R}}^3\to {\mathbb{R}}^3$ be a linear reflection, $\det r=-1$, preserving $S^2$. By [L4] the restriction $r{\mathrm{\mid }}_{S^2}$ reverses the boundary orientation and has degree $-1$, while every diffeomorphism $H_1{\mathrm{\mid }}_{S^2}$ arising as above has degree $+1$ by step 3.1; therefore $H_1{\mathrm{\mid }}_{S^2}=r{\mathrm{\mid }}_{S^2}$ is impossible, and no such ambient isotopy can restrict to the reflection.

An orthogonal $3\times 3$ matrix $Q$ of determinant one has a unit fixed axis $a$: its eigenvalues have modulus one, the nonreal ones occur in conjugate pairs, and their product together with the real eigenvalues is one, so one real eigenvalue is $+1$. On $a^{\perp }$ its restriction is a plane rotation through some angle $\theta$. Fix an orthonormal basis of that plane and let $Q_t$ fix $a$ and rotate the plane through $t\theta$; its sine and cosine entries give a smooth path with $Q_0=I$, $Q_1=Q$. Define $F_t(x)=tc+((1-t)+t\lambda )Q_t{f}_0(x)$. Each slice is the restriction of an invertible affine map because $(1-t)+t\lambda >0$, hence is an embedding. The family is smooth and satisfies $F_0=f_0$, $F_1=f_1$.

By compactness of $S^1$, [L1] extends $F$ to an ambient isotopy supported in a compact subset of $W$, with $H_1\circ f_0=f_1$. Every affine parametrization of a round circle has the form $c+\lambda (u\cos s+v\sin s)$ for an ordered orthonormal pair $u,v$; completing it by $u\times v$ gives a matrix in $SO(3)$, including when the circle parameter orientation is reversed. Thus the construction covers all such round circles. The neighbourhood must contain the whole motion, since a disconnected neighbourhood of disjoint endpoint circles cannot support a motion between its components.

Steps 1.1 and 2.1 exhibit the required compactly supported ambient isotopy carrying $f_0$ to $f_1$, verifying the hypothesis check of The isotopy extension theorem in this example.

By [L1] let $H$ be an ambient diffeomorphism with $H\circ f_0=f_1$. Its differential restricts to a smooth bundle isomorphism $dH:TN{\mathrm{\mid }}_{f_0(M)}\to TN{\mathrm{\mid }}_{f_1(M)}$ covering $f_1\circ f_0^{-1}$.

On the level of $M$ the map $dH$ induces a bundle map ${\nu }_0\to {\nu }_1$ over the identity of $M$: by the chain rule, $dH$ carries the summand $d{f}_0(TM)\subseteq TN{\mathrm{\mid }}_{f_0(M)}$ isomorphically onto $d{f}_1(TM)\subseteq TN{\mathrm{\mid }}_{f_1(M)}$, so it descends to an isomorphism of the fibrewise quotients ${\nu }_0=f_0^{\ast }TN\mathrm{/}TM\to {\nu }_1=f_1^{\ast }TN\mathrm{/}TM$ over ${\mathrm{i}\mathrm{d}}_M$. A bundle map that is a linear isomorphism on each fibre is a bundle isomorphism, so ${\nu }_0\cong {\nu }_1$; consequently the characteristic classes natural under this bundle isomorphism agree. For the characteristic-class construction assume additionally AC. Orientation-dependent classes agree when orientations are transported by it; unrelated choices of orientations are not being compared.

The complement statement is the second conclusion of [L1]: $H$ restricts to a diffeomorphism $N\setminus f_0(M)\to N\setminus f_1(M)$ with smooth inverse. For the standard sphere and its reflection, the radial vectors at their image points give nowhere-zero smooth frames of the normal line bundles, so both are trivial. The reflected-sphere counterexample on this page proves they are not isotopic, establishing the parenthetical failure of the converse.

The normal bundles are isomorphic and the complements diffeomorphic, which is what the example claims.

Both $\iota$ and $r\circ \iota$ are smooth embeddings of $S^2$ into ${\mathbb{R}}^3$: $\iota$ is the inclusion of an embedded submanifold, and $r$ is a linear isomorphism, hence a diffeomorphism of ${\mathbb{R}}^3$ whose composite with $\iota$ is again an embedding.

The two embeddings are regularly homotopic: by [L1], applied with $m=2$ and $n=3$, all immersions $S^2\to {\mathbb{R}}^3$ are regularly homotopic because ${\pi }_2(SO(3))=0$, and $\iota$ and $r\circ \iota$ are such immersions; the reflection is, up to an orientation-preserving rotation of the target, the antipodal reparametrisation of the standard inclusion, and the direct formal-data computation identifies the obstruction as a class in ${\pi }_2(SO(3))=0$, so the two formal data are homotopic and the formal-data criterion gives the regular homotopy. Hence clause 1 of the counterexample holds.

The two embeddings are not isotopic. Suppose an isotopy of embeddings from $\iota$ to $r\circ \iota$ existed. Since $S^2$ is compact and ${\mathbb{R}}^3$ is boundaryless, [L2] produces an ambient isotopy $H$ of ${\mathbb{R}}^3$ with $H_0=\mathrm{i}\mathrm{d}$ and $H_1\circ \iota =r\circ \iota$, that is $H_1{\mathrm{\mid }}_{S^2}=r{\mathrm{\mid }}_{S^2}$.

The sphere complement has precisely the two connected components $U=\{\mathrm{\mid }x\mathrm{\mid }<1\}$ and $V=\{\mathrm{\mid }x\mathrm{\mid }>1\}$: $U$ is convex, and in $V$ radial paths to a common large sphere followed by great-circle arcs on that sphere (for antipodal endpoints choose a perpendicular unit vector by normalizing the first nonzero coordinate-vector projection) connect any two points. Since $H_1(S^2)=S^2$, the homeomorphism permutes these components. It cannot send $U$ to $V$, since $H_1(\bar{U})$ is compact and therefore bounded, whereas $V$ is unbounded. Consequently $H_1(U)=U$ and $H_1(B^3)=B^3$.

For every $x$, the function $t\mapsto \det d{H}_t(x)$ is continuous, never zero, and equals one at zero, so [L4] makes it positive for all $t$. Thus $H_1$ preserves the ambient orientation. Because it maps the ball's interior onto itself, its differential takes an outward transverse vector to an outward transverse vector at the sphere: in a boundary chart the inward normal coordinate has positive inward derivative, by invertibility and preservation of the interior. The outward-normal-first rule in [L4] therefore makes $H_1{\mathrm{\mid }}_{S^2}$ orientation preserving. By [L3] its degree is $+1$, contradicting the reflection's degree $-1$. This proves the orientation argument locally, without a B-page prerequisite.

Therefore $\iota$ and $r\circ \iota$ are regularly homotopic but not isotopic, so the statement refuted is false: regular homotopy of embeddings is strictly coarser than isotopy of embeddings for $S^2$ in ${\mathbb{R}}^3$. The historically first instance, a knotted circle versus the round circle in ${\mathbb{R}}^3$, is recorded as a boundary rather than proved here, because its non-isotopy invariant ${\pi }_1({\mathbb{R}}^3\setminus L)$ belongs to the low-dimensional knot track and not to this run's closure.

Clause 1: with $m=2$ and $n=4$ the expected dimension is $2m-n=0$, so [F1] makes ${\mathrm{\Delta }}_2(f)$ a closed embedded $0$-dimensional submanifold of $M\times M\setminus {\mathrm{\Delta }}_M$; by [F5] each of its points is isolated.

A neighbourhood of the diagonal free of double points: by [F3] every $x\in M$ belongs to an open set $U$ on which $f$ is injective. Thus the family of all $U\times U$ with this property is an ambient-open cover of ${\mathrm{\Delta }}_M$. By [F4] finitely many $U_1\times U_1,\dots ,U_k\times U_k$ cover the compact diagonal; their union $N$ is an open neighbourhood of ${\mathrm{\Delta }}_M$. If $(y,z)\in N\cap {\mathrm{\Delta }}_2(f)$, then $(y,z)\in U_i\times U_i$ for some $i$ and $f(y)=f(z)$ with $y\mathrm{\ne }z$, contradicting injectivity on $U_i$. Hence $N\cap {\mathrm{\Delta }}_2(f)=\mathrm{\varnothing }$. No family of neighbourhoods indexed by all source points was selected.

Clause 2: with $m=2$ and $n=5$ one has $n>2m$, so [F2] makes $f$ injective; equivalently the expected dimension $2m-n=-1$ is negative and [F1] gives ${\mathrm{\Delta }}_2(f)=\mathrm{\varnothing }$ directly. Since $M$ is closed, every compact subset of ${\mathbb{R}}^5$ has compact preimage under the continuous map $f$ (the preimage is closed in the compact space $M$ and hence compact by [F4]), so $f$ is proper; being a proper injective immersion, $f$ is a smooth embedding by [F3].

Finiteness of the double point set: by step 1.2 one has ${\mathrm{\Delta }}_2(f)\subseteq M\times M\setminus N$; the set $M\times M\setminus N$ is closed in $M\times M\setminus {\mathrm{\Delta }}_M$ because $N$ is open and contains ${\mathrm{\Delta }}_M$, and ${\mathrm{\Delta }}_2(f)$ is closed in $M\times M\setminus {\mathrm{\Delta }}_M$ by step 1.1, so ${\mathrm{\Delta }}_2(f)$ is closed in $M\times M\setminus N$, while $M\times M\setminus N$ is compact by [F4] as a closed subset of the compact space $M\times M$; hence ${\mathrm{\Delta }}_2(f)$ is compact. By step 1.1 it is a discrete subspace, and a compact discrete subspace is finite, so $\mathrm{\Sigma }(f)=f({\mathrm{pr}}_1{\mathrm{\Delta }}_2(f))$ is finite as the image of a finite set. The double points are isolated by step 1.1 and transverse because $f$ is self-transverse by hypothesis and the branch tangents span the target tangent space at each double point.

Both clauses hold: clause 1 is steps 1.1, 1.2 and 2.1, and clause 2 is step 1.3. Hence the critical dimension for surfaces is four, where the double point locus is a finite set, while in five-space self-transversality already forces an embedding.