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Sphere eversion cannot be an isotopy through embeddings

Remark

Assume AC, including the countable-choice hypotheses of the eversion and divergence suppliers. No homotopy from $ι$ to $ι \circ a$ (or to $r \circ ι$ ) through embeddings of $S^{2}$ into $R^{3}$ exists, so every eversion contains slices that are not injective; self-intersections are unavoidable. The invariant is the coorientation: an embedding of $S^{2}$ into $R^{3}$ bounds a compact complementary region (indeed a ball, by Alexander's theorem, which the argument below does not need; Jordan–Brouwer separation supplies only the existence of the bounded region), and the sign of the parametrisation relative to the boundary orientation of that region is locally constant along a continuous family of embeddings: the proof below exhibits it as the sign of a continuous flux integral, so continuity of the regions themselves is never invoked. Thus $ι$ has the positively oriented parametrisation with respect to the outward normal while $ι \circ a$ and $r \circ ι$ have the opposite sign, and the two lie in different path components of the space of embeddings $Emb (S^{2}, R^{3}) \subseteq C^{\infty} (S^{2}, R^{3})$ . Equivalently, an ambient isotopy of $R^{3}$ preserves the side of the image, whereas eversion reverses the inside/outside labelling. Regular homotopy is strictly coarser than isotopy here: the two maps are in one path component of $Imm (S^{2}, R^{3})$ by the eversion theorem and in different path components of the embedding space.

Facts & Assumptions

Given: The unit sphere $S^{2}$ , the standard embedding $ι$ , the antipodal map $a$ , a reflection $r$ of $R^{3}$ with $r \circ ι = A \circ (ι \circ a)$ for a rotation $A$ , and the family of slices of a hypothetical homotopy through embeddings.

[F1]

By the eversion theorem, $ι$ and $ι \circ a$ are regularly homotopic through immersions, and every regular homotopy between them has non-injective slices; a homotopy through embeddings is a homotopy whose slices are injective immersions of the compact sphere, hence embeddings. Sphere eversion, Smooth embeddings, An injective immersion from a compact manifold is an embedding

[F2]

A regular homotopy has every slice immersive; embeddings are the injective immersions, and the orientation of the parametrisation relative to the bounded side is the coorientation sign of the remark. Regular homotopy of immersions, Immersions, submersions, and constant-rank maps, Orientable manifolds

[F3]

Jordan–Brouwer separation (AC): the image of an embedding $S^{2} ↪ R^{3}$ has exactly two complementary components, one bounded and one unbounded, with common boundary the image. Jordan–Brouwer separation, The Axiom of Choice

[F4]

The divergence theorem on a bounded $C^{1}$ Euclidean domain: the flux of the field $\frac{1}{3} x$ through the boundary equals $vol (B)$ for the outward orientation and $- vol (B)$ for the inward orientation. Divergence on a bounded C1 Euclidean domain

Proof

1.1F2F3F4

Suppose $t \mapsto H_{t}$ is a continuous path in the weak $C^{\infty}$ space of embeddings from $H_{0} = ι$ to $H_{1} = r \circ ι$ or $H_{1} = ι \circ a$ . For each $t$ the slice $H_{t}$ is an embedding of the compact sphere and bounds a bounded region $B_{t}$ by [F3]; define the flux $S (t)$ of the field $\frac{1}{3} x$ through the parametrised surface $H_{t}$ . Continuity of this path controls the values and first spatial derivatives uniformly on a finite chart cover of the compact sphere. Therefore the integrand $\frac{1}{3} H_{t} \cdot (\partial_{1} H_{t} \times \partial_{2} H_{t})$ is continuous in $(x, t)$ and hence bounded and uniformly continuous, so $t \mapsto S (t)$ is continuous; and by [F4], $S (t) = \pm vol (B_{t})$ with the sign given by the orientation of the parametrisation relative to the outward normal of $B_{t}$ , so $S (t) \neq 0$ for all $t$ and the sign of $S$ is constant.

2.1F1step 1.1∎

At the ends: for $H_{0} = ι$ in positively oriented coordinates of the unit sphere, the flux of $\frac{1}{3} x$ is the volume $\frac{4 π}{3}$ of the unit ball, so $S (0) > 0$ ; for $H_{1} = T \circ ι$ with $T = r$ or $T = - I$ , the chain rule and $(T u) \times (T v) = \det (T) T (u \times v)$ with $\det T = - 1$ give $S (1) = - S (0) < 0$ . This contradicts the constant sign of step 1.1, so no homotopy through embeddings from $ι$ to $r \circ ι$ exists, and by [F1] every regular homotopy from $ι$ to $ι \circ a$ has non-injective slices: eversion necessarily produces self-intersections. AC is inherited from Jordan–Brouwer and from the eversion assertion; it also implies the countable-choice assumption of the divergence theorem. The sign computation is elementary.

The two path components just separated are components of the space of embeddings, while the eversion theorem puts the two maps in one component of the space of immersions; this is the precise sense in which regular homotopy is coarser than isotopy for the sphere in $R^{3}$ .

Depends on

Used by

Regular homotopy allows self-intersections but never rank drop Remark

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Sources

Ralph L. Cohen, Immersions of Manifolds and Homotopy Theory (Harvard CMSA Math-Science Literature Lecture write-up, June 30 2022), §2.1 eversion paragraph (standard reference, not scraped)
Allen Hatcher, Notes on Basic 3-Manifold Topology, Theorem 1.1 (every smoothly embedded 2-sphere in R^3 bounds a ball) (standard reference, not scraped)