Regular Homotopy and Sphere Eversion — Examples

For $k\mathrm{\ne }0$, ${\gamma }_k$ has velocity $k(-\sin k\theta ,\cos k\theta )$ of norm $\mathrm{\mid }k\mathrm{\mid }>0$. Its normalized velocity is $\mathrm{sgn}(k)i{e}^{ik\theta }$, a constant rotation of the degree-$k$ power map, so $\mathrm{rot}({\gamma }_k)=k$. For negative $k$ the extra factor is $-1$; it preserves degree.

The velocity of $\delta$ is $p(\sin \theta )$ with $p(u)=(-u,2-4u^2)$. This never vanishes, and $p((1-t)\sin \theta )$ contracts it to $(0,2)$ through nonzero loops, giving rotation number zero. If $\delta (\theta )=\delta (\varphi )$, equality of cosines gives $\varphi =\theta$ or $\varphi =-\theta$ modulo $2\pi$. In the second case equality of $\sin 2\theta$ and $-\sin 2\theta$ requires $\sin 2\theta =0$; the only distinct pair is $\pi \mathrm{/}2,3\pi \mathrm{/}2$, both mapping to the origin. Thus this is its unique double point.

For $k<0$, ${\gamma }_k(\theta )={\gamma }_{\mathrm{\mid }k\mathrm{\mid }}(-\theta )$ is the $\mathrm{\mid }k\mathrm{\mid }$-fold circle with reversed domain orientation, consistent with its rotation number $k$ in step 1.1.

By [F3], under its countable-choice hypothesis, ${\gamma }_k$ and ${\gamma }_l$ for nonzero $k,l$ are regularly homotopic exactly when $k=l$, and no ${\gamma }_k$ is regularly homotopic to $\delta$. Steps 1.1 and 1.2 realise every integer with exactly one member of the displayed family. In particular rotation number zero does not force injectivity.

$\delta (\theta )=(\cos \theta ,\sin 2\theta )$ has velocity $(-\sin \theta ,2\cos 2\theta )$, whose squared norm ${\sin }^2\theta +4{\cos }^{2}2\theta$ never vanishes: if $\sin \theta =0$ then $\cos 2\theta =1$. It is therefore an immersion. If $\delta (\theta )=\delta (\varphi )$, equality of cosines gives $\varphi =\theta$ or $\varphi =-\theta$ modulo $2\pi$. In the latter case $\sin 2\theta =0$; the only distinct pair is $\pi \mathrm{/}2,3\pi \mathrm{/}2$, both mapping to $(0,0)$. Thus the origin is its unique double point. The round circle is an injective immersion with unit-speed velocity.

The velocity of $\delta$ is $p(\sin \theta )$ with $p(u)=(-u,2-4u^2)$. The map $p$ never vanishes, and $p((1-t)\sin \theta )$ contracts the velocity loop to $(0,2)$ through nonzero vectors. Hence $\mathrm{rot}(\delta )=0$. The unit tangent of ${\gamma }_1$ is $(-\sin \theta ,\cos \theta )$, a rotation of the identity circle map and of degree $1$.

Any smooth homotopy $H:S^1\times [0,1]\to {\mathbb{R}}^2$ through immersions has continuous nonzero velocity ${\mathrm{\partial }}_{\theta }H(\theta ,t)$, so its normalized velocity $\tau (\theta ,t)$ is a continuous homotopy of circle maps. In complex notation ${\beta }_t(u)=\tau (2\pi u,t)\mathrm{/}\tau (0,t)$, $0\le u\le 1$, is a based circle loop, and $(u,t)\mapsto {\beta }_t(u)$ is a homotopy fixing both endpoints at $1$. By [F1] its degree is $\mathrm{rot}(H_t)$, since the target rotation used to base it does not change degree. By [F3] these degrees agree at the two ends. As step 1.2 gives $0\mathrm{\ne }1$, no such regular homotopy joins $\delta$ to ${\gamma }_1$, refuting the false claim without a choice hypothesis or the sufficiency direction of classification.

The obstruction also distinguishes maps with the same image: ${\gamma }_2(\theta )=(\cos 2\theta ,\sin 2\theta )$ is an immersion with the round circle as its image and no transverse self-intersection, yet its unit tangent $(-\sin 2\theta ,\cos 2\theta )$ has degree $2$. The invariance argument of step 2.1 therefore separates it from ${\gamma }_1$ as well. Thus neither the image nor its number of transverse self-intersections determines the regular homotopy class; rotation number separates the figure-eight from the round circle.

Use [F1] to move the two sections to a common evaluated value and transport their characteristic-disk models to constant-boundary maps. The difference of their classes in ${\pi }_2(V_2({\mathbb{R}}^3))$ has a based representative $F:S^2\to V_2({\mathbb{R}}^3)$. Its class vanishes exactly when the original sections are homotopic. This construction does not identify the two raw hemisphere restrictions without their boundary transport.

Completing an orthonormal pair to $(u,v,u\times v)$ identifies $V_2({\mathbb{R}}^3)$ with $\mathrm{S}\mathrm{O}(3)$ by the cited vanishing lemma. A constant left translation makes the based representative take value $I$ at its basepoint. Denote the translated map again by $F$.

$F$ lifts along $\rho$: $S^2$ is path connected and simply connected by [F6], so the lifting criterion [F3] applies with $Y=S^2$, $B=\mathrm{S}\mathrm{O}(3)$ and the covering $p=\rho$ of [F2]: the subgroup of ${\pi }_1(S^2)$ is trivial, hence contained in ${\rho }_{\ast }{\pi }_1(S^3)$, and a lift $\tilde{F}:S^2\to S^3$ with the prescribed basepoint exists.

$\tilde{F}$ is nullhomotopic: every based map $S^2\to S^3$ is nullhomotopic through based maps by [F5], so the class of $\tilde{F}$ in ${\pi }_2(S^3)$ is the distinguished element.

Project a based nullhomotopy $\tilde{H}$ of $\tilde{F}$ through $\rho$. The composite $\rho \circ \tilde{H}$ contracts $F$ to $I$ and fixes the basepoint. Thus the difference class vanishes and [F1] gives a homotopy of the two formal sections.

The two possible lifts differ by the deck transformation $q\mapsto -q$. Both are based-nullhomotopic at their respective basepoints because every based map $S^2\to S^3$ is nullhomotopic. Different disk frames and reference transports may change the representative difference map, but the evaluation lemma preserves its vanishing criterion. The calculation proves the formal obstruction is zero; it does not construct a regular homotopy of immersions.

For every $t<1$ the slice $H_t(\theta )=((1-t)\cos \theta +t,(1-t)\sin \theta )$ has derivative $H_t^{\mathrm{\prime }}(\theta )=(-(1-t)\sin \theta ,(1-t)\cos \theta )$ of norm $1-t>0$, hence is an immersion with image the circle of radius $1-t$ centred at $(t,0)$; at $t=0$ this is the standard unit circle.

At $t=1$ the slice is the constant map $H_1(\theta )=(1,0)$, whose derivative vanishes identically: the differential has rank $0<1=\dim S^1$ at every point. So the family contains a rank drop at the final time and $H_1$ is not an immersion.

Therefore $H$ is not a regular homotopy in the sense of [F1], since a regular homotopy requires every slice to be immersive, and it cannot certify any regular-homotopy claim: in particular this shrinking family does not show that the circle is regularly homotopic to a point, because the point is not an immersion and the rotation number (here $1$ at the initial slice) would have to remain constant while no rotation number is defined for the final slice. The failed conclusion is exactly the rank-drop phenomenon separated in [F3]: the final slice fails the injective-differential condition.

The family $H$ is smooth in $(\theta ,t)$, but its final slice is constant. Any perturbation that keeps this final slice still has zero final derivative, so it still fails to be a regular homotopy, independently of its size. Self-intersections are compatible with immersive slices; a rank drop is not.

Put $d(w)=D_6(w)$. Its roots satisfy $w^3=u_{\pm }=(-\sqrt{5}\pm 3)\mathrm{/}2$. Exactly the three cube roots of $u_{+}$ lie in the open unit disk, and all are simple since $d^{\mathrm{\prime }}(w)=3w^2(2w^3+\sqrt{5})=9w^2$ there. Put $A=w\mathrm{/}d$, $B=w^5\mathrm{/}d$ and $C=(1+w^6)\mathrm{/}d$. If $g_1=g_2=0$ then $\mathrm{Im}(A-B)=0$ and $\mathrm{Re}(A+B)=0$, hence $B=-\bar{A}$. For $w\mathrm{\ne }0$ their absolute values force $\mathrm{\mid }w\mathrm{\mid }=1$. On that circle, writing $w=e^{i\theta }$ gives $C=2\cos 3\theta \mathrm{/}(\sqrt{5}+2i\sin 3\theta )$, so $\mathrm{\mid }\mathrm{Im}C\mathrm{\mid }=\mathrm{\mid }4\cos 3\theta \sin 3\theta \mathrm{\mid }\mathrm{/}(5+4{\sin }^{2}3\theta )\le 2\mathrm{/}5<1\mathrm{/}2$. Thus $g_3=\mathrm{Im}C-1\mathrm{/}2\mathrm{\ne }0$. At $w=0$ it is $-1\mathrm{/}2$. Consequently $g$ never vanishes at a finite non-pole.

Write $g=\mathrm{Re}h$, where $h=(\frac{3}{2}i(A-B),-\frac{3}{2}(A+B),-iC-\frac{1}{2})$ is holomorphic off the poles. Differentiation gives $h^{\mathrm{\prime }}=d^{-2}(\frac{3}{2}i{N}_1,-\frac{3}{2}{N}_2,-i{N}_3)$, with $N_1=w^{10}-2\sqrt{5}{w}^7-5w^6+5w^4-2\sqrt{5}{w}^3-1$, $N_2=-w^{10}+2\sqrt{5}{w}^7-5w^6-5w^4-2\sqrt{5}{w}^3-1$, and $N_3=3w^2(\sqrt{5}{w}^6-4w^3-\sqrt{5})$. The identities $N_2-N_1=-2w^4(w^3-\sqrt{5}{)}^2$ and $N_2+N_1=-2(\sqrt{5}{w}^3+1{)}^2$ give $\frac{9}{4}(N_2^2-N_1^2)=N_3^2$, hence $h^{\mathrm{\prime }}\cdot h^{\mathrm{\prime }}=0$ for the complex bilinear dot product. The first two numerators never vanish simultaneously: at $w=0$ both equal $-1$, and otherwise their sum and difference would require both $w^3=\sqrt{5}$ and $w^3=-1\mathrm{/}\sqrt{5}$. Thus $h^{\mathrm{\prime }}\mathrm{\ne }0$. Its real and imaginary parts are orthogonal and have the same positive norm, so the real derivatives $g_x=\mathrm{Re}{h}^{\mathrm{\prime }}$ and $g_y=-\mathrm{Im}{h}^{\mathrm{\prime }}$ are independent. Inversion $J(v)=v\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}^2$ has derivative $\mathrm{\mid }v{\mathrm{\mid }}^{-2}(I-2v{v}^{\ast }\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}^2)$, an invertible scaled reflection. By step 1.1, $P=J\circ g$ therefore has injective differential away from the poles.

At each interior pole $w_j$, write $u=w-w_j$ and $h=c\mathrm{/}u+h_0(u)$. The residue vector $c$ is nonzero because the second numerator $w_j(1+w_j^4)$ is nonzero ($0<\mathrm{\mid }{w}_j\mathrm{\mid }<1$). The leading term in $h^{\mathrm{\prime }}\cdot h^{\mathrm{\prime }}=0$ gives $c\cdot c=0$. Thus $\mathrm{Re}c$ and $\mathrm{Im}c$ are independent, orthogonal and have common squared norm $\lambda >0$. Set $N(u)=\mathrm{Re}(c\bar{u})$ and $H(u)=\mathrm{Re}{h}_0(u)$; then $\mathrm{\mid }N(u){\mathrm{\mid }}^2=\lambda \mathrm{\mid }u{\mathrm{\mid }}^2$ and $g=N(u)\mathrm{/}\mathrm{\mid }u{\mathrm{\mid }}^2+H(u)$. It follows that $P=(N(u)+\mathrm{\mid }u{\mathrm{\mid }}^{2}H(u))\mathrm{/}(\lambda +2⟨N(u),H(u)⟩+\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{\mid }H(u){\mathrm{\mid }}^2)$ extends real-analytically to $u=0$, with value $0$ and differential $N\mathrm{/}\lambda$, which is injective. This verifies all three ends, including the two nonreal poles.

For $t=-1\mathrm{/}\bar{w}$, direct substitution gives $d(t)=-\overline{d(w)}\mathrm{/}{\bar{w}}^6$, $(t-t^5)\mathrm{/}d(t)=-\overline{(w-w^5)\mathrm{/}d(w)}$, $(t+t^5)\mathrm{/}d(t)=\overline{(w+w^5)\mathrm{/}d(w)}$, and $(1+t^6)\mathrm{/}d(t)=-\overline{(1+w^6)\mathrm{/}d(w)}$. Taking the indicated real and imaginary parts proves $g(t)=g(w)$ and $P(t)=P(w)$; on $\mathrm{\mid }w\mathrm{\mid }=1$ this is $P(-w)=P(w)$. Near the boundary these identities hold on a two-sided annulus, not merely on the circle. At infinity use the coordinate $t=-1\mathrm{/}\bar{w}$ near $0$; the same identity gives a smooth immersive extension there. Thus the extended map on the Riemann sphere is invariant under its free antipodal involution and descends through the local quotient charts to a smooth immersion $\beta :{\mathbb{R}\mathbb{P}}^2\to {\mathbb{R}}^3$. The disk model in [F2] is a fundamental domain for this involution.

The three interior poles are distinct points of the projective-plane domain: their antipodes lie outside the disk. All have image $0$ by step 3.1, so $\beta$ has a triple point and is not injective. This is the Bryant–Kusner Boy surface; Karcher's source, PDF p.2, describes the three antipodal pairs of planar ends and their common image after inversion. The formulas above verify its immersedness directly.

For any immersion $f:{\mathbb{R}\mathbb{P}}^2\to {\mathbb{R}}^3$, a global nonzero normal field would orient each tangent plane by the sign of $\det (d{f}_x{v}_1,d{f}_x{v}_2,n(x))$, continuously and consistently. This contradicts [F2], so its normal line is nontrivial by [F4]. For the characteristic-class description assume AC as in [F5]. The ambient volume form gives $\det (T{\mathbb{R}\mathbb{P}}^2)\otimes {\nu }_f\cong {\epsilon }^1$, so [F5] yields $w_1({\nu }_f)=w_1(T{\mathbb{R}\mathbb{P}}^2)\mathrm{\ne }0$. To identify this with the tautological class, represent a point by a unit $x\in S^2$ and tangent vectors by $v,w\in x^{\perp }$; the map $v\wedge w\mapsto \det (x,v,w)x$ identifies the determinant tangent line with the tautological line. It is unchanged under $(x,v,w)\mapsto (-x,-v,-w)$ and is a fibrewise isomorphism. Thus their $w_1$ classes agree by [F5], giving precisely the tautological degree-one class.