Julian Henry

polyglot / software engineer / author

Sphere EversionIIIManifolds and Immersions

17 Aug 2026

Sphere Eversion: I · II · III · IV · V · VI

Calculus enters. Charts, tangent vectors, the Jacobian that must not drop rank, and the first honest definition of the word “eversion”.

In Vol. I we built topological spaces and continuity; in Vol. II we built paths, homotopies, and $\pi_{1}$. Topology alone cannot see a crease: a homotopy is allowed to squash a sphere flat, fold it, and unfold it. The eversion problem is only interesting once we ask for smooth maps whose derivative never degenerates. This volume adds exactly that much calculus.

The boxed questions are from the Phi-3 transcript described in Vol. I; the answers are the corrections.

Contents
  1. the two-sphere, locally flat
  2. immersions, and the matrix that must not drop rank
  3. the homotopy that is not regular

the two-sphere, locally flat

Student. What is a manifold? What is a differentiable manifold? Give an example. Then give differentiable coordinates on $S^2$ and take a derivative at $(1,0,0)$.
Teacher. A manifold is a space that is Euclidean in the small. Formally, a smooth $n$-manifold is a Hausdorff space covered by charts $\varphi_{\alpha}: U_{\alpha}\to\mathbb{R}^n$ whose transition maps $\varphi_{\beta}\circ\varphi_{\alpha}^{-1}$ are $C^\infty$. You already know one: the surface of the Earth. No single paper map covers it, but an atlas does, and the overlap rules are smooth.

The 2-sphere is the set of unit vectors,

\[S^2=\bigl\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=1\bigr\}.\]

Spherical coordinates are a pair of charts (you need at least two: the azimuth $\varphi$ is not a global coordinate). On the complement of a meridian,

\[\mathbf{r}(\theta,\varphi)=\bigl(\sin\theta\cos\varphi,\;\sin\theta\sin\varphi,\;\cos\theta\bigr),\qquad \theta\in(0,\pi),\;\varphi\in(0,2\pi).\]

A point is not a function of $(\theta,\varphi)$. The model that answered me computed $\partial_{\theta}(1,0,0)=\mathbf{0}$ and called it a day. The object that has derivatives is the chart:

\[\begin{aligned} \mathbf{r}_\theta&=(\cos\theta\cos\varphi,\;\cos\theta\sin\varphi,\;-\sin\theta),\\ \mathbf{r}_\varphi&=(-\sin\theta\sin\varphi,\;\sin\theta\cos\varphi,\;0). \end{aligned}\]

At $(1,0,0)$ one has $(\theta,\varphi)=(\pi/2,\,0)$, so

\[\mathbf{r}_\theta=(0,0,-1),\qquad \mathbf{r}_\varphi=(0,1,0),\qquad \mathbf{r}_\theta\times\mathbf{r}_\varphi=(1,0,0)=\mathbf{r}.\]

Those two vectors are a basis of the tangent plane $T_{(1,0,0)}S^2$, the $yz$-plane. Drag the sliders; the blue and orange arrows are exactly $\mathbf{r}_{\theta}$ and $\mathbf{r}_{\varphi}$.

r, rθ, rφ

The sphere is the image of $\mathbf{r}$. Blue is $\mathbf{r}_{\theta}$, orange is $\mathbf{r}_{\varphi}$, gold is their cross product $\mathbf{r}_{\theta}\times\mathbf{r}_{\varphi}=\sin\theta\,\mathbf{r}$. At the poles $\sin\theta=0$ and the chart is singular; the sphere is not. That distinction is the whole subject.

Student. Give a nonzero derivative — a function that wraps around the sphere.
Teacher. Take the equator as a curve $\gamma:[0,2\pi]\to S^2$, $$\gamma(t)=(\cos t,\;\sin t,\;0).$$ Then $\gamma'(t)=(-\sin t,\;\cos t,\;0)\neq\mathbf{0}$. At $t=0$ this is $(0,1,0)$, which is $\mathbf{r}_{\varphi}$ at $(1,0,0)$. The model wrote $\gamma(t)=(\sin(\pi t),0,\cos(\pi t))$ and claimed it was the equator with $\varphi$ fixed at $0$; that is a meridian, a semicircle from north pole to south, and at its midpoint the velocity is $(0,0,-\pi)$, not a trip around the equator. Wrapping around $S^2$ means the image of $\gamma$ is a closed loop that is not contractible in $S^2\setminus\{\text{two poles}\}$. We will need that loop.

immersions, and the matrix that must not drop rank

Student. What is an immersion? What is a Jacobian? What happens when the Jacobian determinant is zero, or switches sign?
Teacher. For $f:\mathbb{R}^n\to\mathbb{R}^m$ the Jacobian is the $m\times n$ matrix of first partials — rows are outputs, columns are inputs. The model wrote $n\times m$. An immersion $f:M\to N$ is a smooth map whose differential $df_{p}:T_{p}M\to T_{f(p)}N$ is injective at every $p$. For $f:S^2\to\mathbb{R}^3$ that is the statement that the $3\times 2$ matrix $$ J_f=\begin{pmatrix} \partial_u f_1 & \partial_v f_1 \\ \partial_u f_2 & \partial_v f_2 \\ \partial_u f_3 & \partial_v f_3 \end{pmatrix} $$ has rank $2$ everywhere, equivalently $\mathbf{f}_{u}\times\mathbf{f}_{v}\neq\mathbf{0}$. There is no $2\times 2$ determinant to watch. The only vanishing that matters is that cross product.

A sign change of $\det J$ for a map $\mathbb{R}^n\to\mathbb{R}^n$ means the map reverses orientation. It does not mean a saddle: $f(x,y)=(e^x\cos y,\,e^x\sin y)$ has $\det J=e^{2x}>0$ everywhere (it is a local diffeomorphism, the complex exponential). At $(0,\pi/2)$ one has $\det J=1$, not $-1$, and there is no critical point. Vanishing of $\det J$ does mean the inverse-function theorem fails: the map is not a local diffeomorphism there.

Self-intersection is allowed. A figure-eight $\gamma(t)=(\sin 2t,\,\sin t)$ is an immersion — $\gamma’$ never vanishes — and it crosses itself. An embedding is an injective immersion (proper, on noncompact manifolds). The standard sphere $\iota(p)=p$ is an embedding. Mid-eversion surfaces are immersions and not embeddings.

Step through three curves. Watch the arrow, not the picture.

γγ′(t)

The gold arrow is the velocity $\gamma'(t)$, drawn at true relative length. An immersion is a curve whose arrow never shrinks to a point. Self-crossings are allowed; stopping is not.

the homotopy that is not regular

Student. Any two $C^2$ immersions of $S^2$ in $E^3$ are regularly homotopic. Here is a proof: $H(x,t)=(1-t)f_{0}(x)+t f_{1}(x)$. Also, what is $S^2\times[0,1]$?
Teacher. $S^2\times[0,1]$ is a thickened sphere: a spherical shell, not the ball. Each point of $S^2$ grows an interval. A homotopy of maps $S^2\to\mathbb{R}^3$ is a single map $$F:S^2\times[0,1]\to\mathbb{R}^3.$$ It is a regular homotopy when each slice $F(\,\cdot\,,t)$ is an immersion. Straight-line interpolation between immersions is a homotopy of smooth maps and almost never a regular homotopy. It routinely drops rank.
Definition (regular homotopy). A homotopy $F:S^2\times[0,1]\to\mathbb{R}^3$ such that every slice $f_{t}=F(\cdot,t)$ is an immersion and the derivative $df_{t}$ varies continuously with $t$. Equivalently: a path in the space $\operatorname{Imm}(S^2,\mathbb{R}^3)$ with the $C^1$ topology. That is Vol. II's homotopy, with one extra rule: no slice may crease.

The sock-push is the interpolation from the identity to reflection through the equator,

\[F_s(\theta,\varphi)=\bigl(\sin\theta\cos\varphi,\;\sin\theta\sin\varphi,\;(1-2s)\cos\theta\bigr).\]

A short computation:

\[\mathbf{F}_\theta\times\mathbf{F}_\varphi=\bigl((1-2s)\sin^2\theta\cos\varphi,\;(1-2s)\sin^2\theta\sin\varphi,\;\sin\theta\cos\theta\bigr),\] \[\bigl\|\mathbf{F}_\theta\times\mathbf{F}_\varphi\bigr\|=\lvert\sin\theta\rvert\sqrt{(1-2s)^2\sin^2\theta+\cos^2\theta}.\]

At $s=\tfrac12$ this is $\lvert\sin\theta\cos\theta\rvert$, which is zero all along the equator. The image is a disk covered twice, with a fold on the boundary. That is a crease. Colour in the canvas is $\lVert\mathbf{F}_{\theta}\times\mathbf{F}_{\varphi}\rVert$: blue is a healthy tangent plane, orange is rank drop. (The poles are orange at every $s$ for the boring reason that the chart is singular there; watch the equator.)

min ‖Fθ × Fφ‖
rank 2
rank drop

$F_{s}(\theta,\varphi)=(\sin\theta\cos\varphi,\,\sin\theta\sin\varphi,\,(1-2s)\cos\theta)$. This is not an eversion. At $s=1$ you have reflected the sphere through the $xy$-plane, and at $s=1/2$ you have left $\mathrm{Imm}(S^2,\mathbb{R}^3)$.

An eversion is a regular homotopy from the inclusion $\iota(p)=p$ to the antipodal embedding $\alpha(p)=-p$. The map $\alpha$ reverses orientation of $\mathbb{R}^3$ ($\det D\alpha=(-1)^3=-1$), and the outward normal of the image sphere at $-p$ is $-p$, while the pushed tangent frame produces the opposite normal. Inside has become outside. $F_{s}$ ends at the reflection $\rho(x,y,z)=(x,y,-z)$, which is $\alpha$ followed by a half-turn about the $z$-axis. Rotating an immersion keeps it an immersion, so reaching $\rho$ regularly would be as good as reaching $\alpha$. $F_{s}$ gets there, but it cheats: at $s=\tfrac12$ it leaves $\operatorname{Imm}(S^2,\mathbb{R}^3)$.

So the question is sharp now. $\iota$ and $\alpha$ are both immersions (both embeddings, in fact). Is there a path between them in $\operatorname{Imm}(S^2,\mathbb{R}^3)$? The crease homotopy is not one. Before attacking the sphere we drop a dimension and ask the same question of circles in the plane, where the answer is no and the reason is an integer.

Next: Vol. IV: Curves in the Plane

Sphere Eversion: Vol. I · Vol. II · Vol. III · Vol. IV · Vol. V · Vol. VI