Sphere Eversion: I · II · III · IV · V · VI
The existence proof. Record the whole derivative, find the space it lives in, and show that space has no room for an obstruction.
Vol. IV ended with a question. For plane curves the obstruction was the direction of the whole derivative $\gamma’$, counted in $\pi_{1}(S^1)$. The Gauss map of a surface records only its normal, and its degree is always $1$. So record the whole derivative of the surface. This volume builds the three tools that make that computable: fibrations, the rotation group, and the long exact sequence. Then it runs Smale’s argument.
The whole derivative
The pair $(\mathbf{f}_{u},\mathbf{f}_{v})$ lies in $V_{3,2}$. One dimension down, the velocity $\gamma’$ of an immersed plane curve lies in $V_{2,1}=\mathbb{R}^2\setminus\lbrace 0\rbrace$, which Vol. II’s homotopy $(1-s)v+s\,v/\lVert v\rVert$ squeezes onto the unit circle $S^1$, and the class of $\gamma’$ in $\pi_{1}(V_{2,1})=\mathbb{Z}$ is precisely the turning number. Whitney–Graustein is the one-dimensional case of the theorem this volume proves.
There is a catch. A frame needs coordinates $(u,v)$, and the sphere has no global coordinates with independent partials: a nowhere-zero field $\partial_{u}$ on all of $S^2$ would be a hairy ball combed flat, and the hairy ball theorem of Poincaré and Brouwer forbids it: $S^2$ admits no continuous tangent vector field (a choice, at each point $p$, of a vector in the tangent plane at $p$) that is nowhere zero. So we cannot simply map the whole sphere to $V_{3,2}$. Smale’s invariant works around this by using a disk, where coordinates do exist. We will get there; first, the tools.
Fibre bundles and fibrations
Locally a product, globally perhaps twisted. The cylinder $S^1\times[0,1]$ and the Möbius band are both bundles over the circle with fibre an interval; only the first is a product. Three examples:
- The covering $\mathbb{R}\to S^1$ from Vol. II. Fibre $\mathbb{Z}$, discrete.
- The quaternion double cover $S^3\to SO(3)$, below. Fibre $\lbrace\pm1\rbrace$, discrete.
- $SO(3)\to S^2$, $R\mapsto R\,e_{3}$ (where does the rotation send the north pole). Fibre: the rotations fixing $e_{3}$, which are the rotations of the $xy$-plane, a circle’s worth called $SO(2)$. ($SO(3)$, the rotation group, is defined in the next section.)
Bundles have the homotopy lifting property (Vol. II proved it for $\mathbb{R}\to S^1$ by chopping the square into small pieces; the general proof uses the same idea). A map with that property is called a fibration, and every fibration comes with a machine for computing homotopy groups:
The unlabeled arrows are the obvious ones: $F\to E$ is inclusion and $E\to B$ is the bundle map, and since a map sends spheres to spheres it induces a map of homotopy groups. This theorem is not proved here; every case of it that the series uses is proved directly below. Exact means: at each spot, the things killed by the outgoing arrow are exactly the things hit by the incoming one. The map $\pi_{n}(B)\to\pi_{n-1}(F)$ is the interesting one: view an $n$-sphere in $B$ as an $n$-disk with its boundary crushed to the base point, lift the disk to $E$ (homotopy lifting), and look at where its boundary lands: an $(n-1)$-sphere in the fibre over the base point.
The only consequence we need: in a stretch $0\to X\to Y\to 0$ of the sequence, the arrow $X\to Y$ is an isomorphism (at the $\pi_{0}$ end a one-point set $\ast$ plays the role of $0$). Try it on Vol. II’s covering. With $E=\mathbb{R}$ (contractible, all $\pi_{n}=0$) and $F=\mathbb{Z}$:
\[0=\pi_1(\mathbb{R})\to\pi_1(S^1)\to\pi_0(\mathbb{Z})\to\pi_0(\mathbb{R})=\ast,\]and $\pi_{1}(S^1)\cong\pi_{0}(\mathbb{Z})=\mathbb{Z}$. That is the lifting argument of Vol. II, compressed into one line.
Rotations, quaternions, and the belt
$V_{3,2}$ is the rotation group, up to homotopy. The rotation group $SO(3)$ is the set of $3\times 3$ real matrices $R$ with $R^{\mathsf T}R=I$ and $\det R=1$: the linear maps that preserve lengths and orientation. Its columns form an orthonormal basis $(v_{1},v_{2},v_{3})$ with $v_{3}=v_{1}\times v_{2}$. Gram–Schmidt turns a pair of independent vectors into an orthonormal pair, continuously, and the straight-line path from each pair to its Gram–Schmidt output stays independent. So $V_{3,2}$ deformation-retracts onto orthonormal pairs. An orthonormal pair $(v_{1},v_{2})$ extends uniquely to a rotation matrix with columns $(v_{1},\,v_{2},\,v_{1}\times v_{2})$. So
\[V_{3,2}\simeq SO(3).\]$SO(3)$ is a sphere with antipodes glued. Quaternions are expressions $a+bi+cj+dk$ with real coefficients, multiplied like polynomials subject to $i^2=j^2=k^2=ijk=-1$. The length of a quaternion is the length of $(a,b,c,d)$ in $\mathbb{R}^4$, and lengths multiply. A unit quaternion $q=\cos\tfrac\theta2+\sin\tfrac\theta2\,(a\,i+b\,j+c\,k)$ acts on vectors $v\in\mathbb{R}^3$, written as pure quaternions $xi+yj+zk$, by $v\mapsto qvq^{-1}$, where for a unit quaternion $q^{-1}$ is $q$ with the signs of its $i$, $j$, $k$ parts flipped. That map preserves length and fixes the axis $(a,b,c)$, and a direct computation shows it is the rotation by angle $\theta$ about that axis. Unit quaternions form $S^3\subset\mathbb{R}^4$. Every rotation turns space by some angle about some axis (a $3\times 3$ rotation matrix always has eigenvalue $1$), so every rotation arises, from exactly two quaternions, $q$ and $-q$. So the map $S^3\to SO(3)$ is a two-sheeted covering, and
\[SO(3)\cong S^3/\lbrace\pm1\rbrace=\mathbb{RP}^3.\]Here $S^3/\lbrace\pm1\rbrace$ is the sphere with each point $q$ glued to its antipode $-q$, called real projective space $\mathbb{RP}^3$: a point of it is a pair $\lbrace q,-q\rbrace$, or equivalently a line through the origin of $\mathbb{R}^4$. The map from pairs to rotations is a continuous bijection from a compact space, so its inverse is continuous too, by the fact from Vol. III, and $\cong$ here is a homeomorphism.
The long exact sequence of this covering, with fibre $\lbrace\pm1\rbrace$ and $\pi_{1}(S^3)=0$, gives $\pi_{1}(SO(3))\cong\mathbb{Z}/2$, the group with two elements in which the non-identity element times itself is the identity. Directly: a loop of rotations lifts to a path in $S^3$ that either closes up or ends at $-1$ times its start. Homotopies lift, and the far end moves within the discrete fibre, so it cannot move: which case occurs is an invariant. A loop whose lift closes up is trivial, because the closed lift shrinks in $S^3$ ($\pi_{1}(S^3)=0$) and the shrinking projects down. So there are exactly two classes. Rotating by $2\pi$ about a fixed axis is the nontrivial loop (its lift $\cos\tfrac\theta2+\sin\tfrac\theta2\,k$ runs from $1$ to $-1$ as $\theta$ goes from $0$ to $2\pi$); rotating by $4\pi$ is the trivial one. You can feel this with a belt, or a plate held flat on your palm (Dirac’s belt trick, the plate trick):
belt, back
The belt is a path of rotations: the frame at distance $u$ along it is $p_{\lambda}(u)\in SO(3)$, drawn as the direction the belt faces. The wall end is fixed at the identity. At $\lambda=0$ the belt carries a full $4\pi$ (or $2\pi$) twist. Slide $\lambda$: for $4\pi$ the explicit homotopy to the constant loop $p_{\lambda}(u)=q_{\lambda}(0)^{-1}q_{\lambda}(u)$ untwists it completely with the plate held still. For $2\pi$ the same trick forces the plate to turn, by up to $180^\circ$. No trick can avoid that.
The same sequence one step higher is the one we need. $\pi_{2}$ of the fibre $\lbrace\pm1\rbrace$ is $0$ and $\pi_{1}$ of it is $0$ too (a map from a sphere into a discrete set is constant), so
\[0\to\pi_2(S^3)\xrightarrow{\ \cong\ }\pi_2(SO(3))\to 0.\]The same conclusion comes straight from the lifting of Vol. II, with no exact sequence. Take a map $g:S^2\to SO(3)$ and lift it to $S^3$ one point at a time: join $x$ to the base point by a path in $S^2$, lift the image path, and let $\tilde g(x)$ be where the lift ends. Another path gives the same endpoint. Two paths with the same ends differ by a loop, the loop shrinks in $S^2$ because $\pi_{1}(S^2)=0$, and the shrinking lifts; its far end moves within the discrete fibre over $g(x)$, so it cannot move at all. So $g$ is $\tilde g:S^2\to S^3$ followed by the covering. Since $\pi_{2}(S^3)=0$, $\tilde g$ shrinks to a point in $S^3$, and projecting the shrinking down shrinks $g$. Every map $S^2\to SO(3)$ is homotopic to a constant.
Why the obstruction vanishes
First the invariant, stated so that it avoids the hairy-ball catch. Take two immersions $f,g:S^2\to\mathbb{R}^3$. After a regular homotopy of one of them, they agree on a small disk $D_{-}$ around the south pole. (Translate $g$ so the two south poles coincide. Its frame there can be moved to $f$’s along a path in $V_{3,2}$, which is connected because $SO(3)$ is: shrinking the angle joins any rotation to the identity. Once the two maps share a point and a frame, both are nearly linear on a tiny disk, and blending them there keeps rank $2$ by the openness of Vol. I.) The rest of the sphere, $D_{+}=S^2\setminus D_{-}$, is a disk, and on a disk there are coordinates $(u,v)$. Each immersion gives a frame field on it, $(\mathbf{f}_{u},\mathbf{f}_{v})$ and $(\mathbf{g}_{u},\mathbf{g}_{v})$: two maps $D_{+}\to V_{3,2}$ that agree on the boundary circle. Glue two disks along their common boundary and you get a sphere. The glued map $S^2\to V_{3,2}$ has a homotopy class
\[\Omega(f,g)\in\pi_2(V_{3,2}).\]The computation, written so each symbol is a space you can name:
- $V_{3,2}\simeq SO(3)$, by Gram–Schmidt and the cross product.
- $SO(3)\cong\mathbb{RP}^3$, by the quaternion double cover. This is the belt.
- The covering $S^3\to\mathbb{RP}^3$ has discrete fibre $\lbrace\pm 1\rbrace$, so maps of $S^2$ lift through it, and $\pi_{2}(\mathbb{RP}^3)\cong\pi_{2}(S^3)$.
- $\pi_{k}(S^n)=0$ for $k<n$ (Vol. II). In particular $\pi_{2}(S^3)=0$: a 2-sphere in a 3-sphere has room to shrink.
Therefore
\[\pi_2(V_{3,2})\cong\pi_2(SO(3))\cong\pi_2(\mathbb{RP}^3)\cong\pi_2(S^3)=0.\]So $\Omega(f,g)=0$ for every pair. In particular $\iota(p)=p$ and $\alpha(p)=-p$ lie in the same path component of $\operatorname{Imm}(S^2,\mathbb{R}^3)$. Step through it:
What the theorem says, object by object. The computation is the easy half, and it is proved above. The hard half is Smale’s: that $\Omega(f,g)=0$ really does produce a regular homotopy. Its proof is serious analysis and is not reproduced here, but its statement uses only objects we can now name.
- Immersions of a disk. $\operatorname{Imm}(D^2,\mathbb{R}^3)$ is the space of immersions of the closed disk $D^2$ (the disk together with its boundary circle), with the $C^1$ distance of Vol. I. It is homotopy equivalent to $V_{3,2}$: shrinking the disk toward its centre deforms any immersed disk into a smaller and smaller one, which in the limit is nothing but its frame at the centre.
- Boundary data. Restrict an immersion of the disk to its boundary circle, but keep the first derivatives there: the curve the rim traces, and at each of its points the frame $(\mathbf{f}_{u},\mathbf{f}_{v})$, which says how the surface leaves the rim. Call this restriction map $\rho$.
- A fibration. Smale’s main technical theorem is that $\rho$ has the homotopy lifting property. In words: however the boundary data moves, an immersed disk with that boundary can be moved along with it, starting from any disk you like.
- The fibre. The fibre of $\rho$ over fixed boundary data is the space of all immersed disks with that boundary: every way of filling in a fixed rim. Smale shows it is weakly homotopy equivalent to the double loop space $\Omega^2V_{3,2}$, meaning there is a map between them that induces a bijection on every $\pi_{k}$. The double loop space is the space of maps of a disk into $V_{3,2}$ that send the whole boundary circle to one fixed frame. Crushing the boundary to a point makes each such map a sphere in $V_{3,2}$, which is why its path components are the elements of $\pi_{2}(V_{3,2})$.
Put together: two fillings of the same rim are joined by a regular homotopy that holds the rim fixed exactly when they lie in the same path component of the fibre, and those path components are the elements of $\pi_{2}(V_{3,2})$. That is what $\Omega(f,g)$ measures. Once $f$ and $g$ agree on $D_{-}$, they are two fillings of the rim of $D_{+}$, and $\Omega(f,g)$ names the path component that separates them, or says that none does. Since $\pi_{2}(V_{3,2})=0$, none ever does.
Why the fibration theorem is hard, Vol. IV already shows. There, once the turning numbers matched, the proof interpolated the derivative data (the angles) and repaired the result into a closed curve by subtracting an average. That repair takes one line, because any loop of nonzero vectors with average zero is the velocity of a closed curve. A field of frames on a disk has no such freedom: a pair $(\mathbf{a},\mathbf{b})$ is $(\mathbf{f}_{u},\mathbf{f}_{v})$ for some $\mathbf{f}$ only if $\partial_{v}\mathbf{a}=\partial_{u}\mathbf{b}$, and a homotopy of frame fields has no reason to respect that. Smale’s analysis is the two-dimensional replacement for the one-line repair.
Hirsch (1959) extended this to arbitrary manifolds. A formal immersion $M\to N$ is a candidate derivative: a map $M\to N$ together with, at each point, an injective linear map from the tangent space of $M$ there to the tangent space of $N$ at the image point, varying continuously, with no requirement that it be the derivative of anything. The derivative of an immersion is one; a field of frames that fails $\partial_{v}\mathbf{a}=\partial_{u}\mathbf{b}$ is another. Hirsch’s theorem: when $\dim M<\dim N$, the space of immersions $M\to N$ is weakly homotopy equivalent to the space of formal immersions. Every candidate derivative can be deformed into a real one, and up to homotopy nothing is gained or lost. That is an early instance of what Gromov later named the h-principle: for many geometric problems, if there is no homotopy-theoretic obstruction, there is a solution. Sphere eversion is its most famous consequence.
Why $\mathbb{R}^3$ is special. For $n=4$ the same machine gives $\pi_{2}(V_{4,2})\cong\mathbb{Z}$. (An orthonormal pair $(v_{1},v_{2})$ in $\mathbb{R}^4$ is a point $v_{1}\in S^3$ together with a unit tangent vector $v_{2}$ to $S^3$ there. Unit tangent vectors to $S^3$ form $S^3\times S^2$, because $S^3$, the unit quaternions, has three independent vector fields $qi, qj, qk$ (at each $q$ these three are perpendicular to $q$ and to each other, a basis of the tangent space there); so $V_{4,2}\simeq S^3\times S^2$ and $\pi_{2}=\pi_{2}(S^2)=\mathbb{Z}$.) Immersions $S^2\to\mathbb{R}^4$ therefore have infinitely many regular homotopy classes, detected by the normal Euler number. At each point of a surface in $\mathbb{R}^4$ the normal plane is the plane perpendicular to the tangent plane, and the Euler number counts, with signs, the zeros of a generic field of normal vectors. It equals minus twice the number of double points, the points where two sheets cross, counted with sign. For $n\ge5$ the group is $0$ again, because $V_{n,2}$ is $(n-3)$-connected, meaning $\pi_{k}(V_{n,2})=0$ for every $k\le n-3$. In $\mathbb{R}^3$ connectivity does not explain the vanishing: $V_{3,2}\simeq SO(3)$ is not even simply connected, and $\pi_{2}$ vanishes because $SO(3)$ is covered by $S^3$.
Where this goes
This is an existence proof. It produces no picture. It says that the obstruction group is zero, so no obstruction can stop you, and nothing more. Bott, Smale’s advisor, at first flatly told him he was wrong; once persuaded, he still wanted to see a direct eversion. Finding explicit eversions took years more, and drawing them honestly is the last volume.