I know that one can turn a sphere $S^2$ in $\mathbb{R}^3$ inside-out having at each time an immersion of $S^2$ into $\mathbb{R}^3$. It is called Smale paradox. There is beautiful animation about that.

Do you know if the same thing is possible for higher dimensional spheres? I mean, for a sphere $S^k$ in $\mathbb{R}^{k+1}$ for $k\geq 3$.

Thank you!


Hirsch extended Smale's theorem, his extension is called the Smale Hirsch Theorem. It applies to your question, and says that the map that sends an immersion $S^n \to \mathbb R^{n+1}$ to its derivative $TS^n \to T\mathbb R^{n+1}$. is a homotopy-equivalence. Specifically, you consider the domain of this map to be the space of immersions of the sphere in Euclidean space, and the range to be the space of bundle monomorphisms from the tangent bundle of the sphere to the tangent bundle of Euclidean space.

Using the "cross product" construction you can convert between bundle monomorphisms $TS^n \to T\mathbb R^{n+1}$ and maps $S^n \to SO_{n+1}$. So the question of whether or not a sphere can be turned inside-out converts to whether or not these two maps $S^n \to SO_{n+1}$ (for the standard sphere and mirror reflection of the standard sphere) are homotopic. So when $n=2$ we're done, since $\pi_2 SO_3 = 0$, but these maps can be non-trivial in higher dimensions.

First off, what are these elements of $\pi_n SO_{n+1}$? They come from the standard immersion via this cross-product construction I mentioned. For the standard immersion of the sphere, the element of $\pi_n SO_{n+1}$ constructed is the trivial element. How about for the mirror-reflected immersion? If you think about it a little, you see it is a composite of two mirror reflections, first the mirror reflection across a fixed vector and 2nd the mirror reflection across the plane orthogonal to the point in $S^n$. This map $S^n \to SO_{n+1}$ is known as the "clutching map" for $TS^{n+1}$.

So the answer is, you can turn $S^n$ inside-out in $\mathbb R^{n+1}$ if and only if the tangent bundle for the next higher-dimensional sphere $S^{n+1}$ is trivial. Since $S^0$, $S^1$, $S^3$ and $S^7$ are the only spheres with trivial tangent bundles, that answers your question. i.e. you can only turn $S^0$, $S^2$, and $S^6$ inside-out.

  • $\begingroup$ Isn't it possible that the clutching map could be non-trivial but still null-homotopic? It feels like there's a step right at the end that I'm missing... $\endgroup$ – Steven Stadnicki Aug 29 '13 at 20:53
  • $\begingroup$ What do you mean by a non-trivial clutching map, that it describes a non-trivial vector bundle? If so, the answer is no. Bundles are classified by homotopy-classes of maps to Grassmannians, and this is one formulation of that equivalence. $\endgroup$ – Ryan Budney Aug 29 '13 at 20:55

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