3
$\begingroup$

At the right column of the page 654 of the paper, R. Palais, The Visualization of Mathematics: Towards a mathematical Exploratorium, Notice AMS it is written "There can be no eversion of a circle (or any odd dimensional spheres)".

In that paper the eversion is defined as a regular homotopy between identity and the antipodal map.

I wonder what is my mistake to think that $f_t(z)=e^{it}z, \quad t\in [0, \pi]$ is a regular homotopy between the identity and the antipodal map on the circle?

$\endgroup$
13
  • 4
    $\begingroup$ I think there is a problem of terminology. In even dimension, an eversion is a regular homotopy between the identity and the antipodal map. But in odd dimension, an eversion should be a homotopy between the identity and $(x_1, \dots, x_{2n}) \mapsto (-x_1, \dots, -x_{2n-1}, x_{2n})$. Otherwise the antipodal map is just a rotation and it doesn't turn the circle "inside out". $\endgroup$ Commented Oct 4, 2018 at 12:33
  • 1
    $\begingroup$ An eversion is when you turn the sphere inside out. Try to draw pictures, a rotation of $S^1$ does not turn it inside out. $\endgroup$ Commented Oct 4, 2018 at 12:46
  • 1
    $\begingroup$ A regular homotopy doesn't need to preserve the degree, actually. Indeed, there do exist eversions of even-dimensional spheres. $\endgroup$ Commented Oct 4, 2018 at 12:58
  • 1
    $\begingroup$ It's homotopy invariant for maps $S^2 \to S^2$, but a regular homotopy is about the embedding $S^2 \to \mathbb{R}^3$. There will be self-intersections during the eversion. $\endgroup$ Commented Oct 4, 2018 at 13:01
  • 2
    $\begingroup$ It provides a different kind of invariant, called the "turning number". It is an invariant of regular homotopies. The turning number of $S^1 \to \mathbb{R}^2$ is $1$ while it is $-1$ for the (almost) antipodal, so they cannot be regularly homotopic. But for even spheres, the turning number is the same (and indeed, there exists an eversion). $\endgroup$ Commented Oct 4, 2018 at 13:04

1 Answer 1

7
$\begingroup$

Let me summarize the comments. An "eversion" of $S^n$ is a regular homotopy between, on the one part the "identity" embedding $S^n \to \mathbb{R}^{n+1}$, and on the other part an embedding $S^n \to \mathbb{R}^{n+1}$ that turns the sphere inside out.

For an even sphere, the antipodal map turns the sphere inside out (AKA reverses orientation), thus we speak of an eversion for a regular homotopy between the identity and the antipodal map. But for an odd sphere, the antipodal map preserves orientation, so we must use something else. For example, the map $S^{2n+1} \to \mathbb{R}^{2n+2}$ given by $(x_1,\dots,x_{2n+2} \mapsto (-x_1,\dots,-x_{2n+1},x_{2n+2})$. I suspect that this is essentially a typo in Palais's paper. People aren't used of talking about eversion of odd spheres, because, well, they don't exist...

Let me also point out that an eversion is not a homotopy of self-maps $S^n \to S^n$. So notions such as the degree of a map are not preserved. It is not a contradiction that there exists an eversion for even spheres, even though the degree of the identity is $1$ and of the antipodal map $-1$.

However, there exists a different invariant, called the "turning number". This invariant is basically the degree of the self-map of $S^n$ induced by the differential of an embedding $f : S^n \to \mathbb{R}^{n+1}$ (where we used the fact that a regular homotopy is an immersion at each moment, to get a self map of $S^n$: the Gauss map). The turning number of the identity is $1$. The turning number of the "not antipodal" map $S^{2n+1} \to S^{2n+1}$ is $-1$, so there can be no eversion of an odd sphere. However, the turning number of the antipodal map of $S^{2n}$ is $1$, so there is no contradiction.

The legends say – but it was before my birth, so others may confirm/infirm/clarify – that Smale's advisor's, Bott, thought that Smale's proof of the existence of an eversion of the $2$-sphere(1) was wrong; without even looking at the proof, just based on general expectations. Apparently Bott either because he believed an eversion preserved the degree, or (more likely?) because he thought the turning number of the antipodal was $-1$. As we now know, Smale's result was correct: there exists an eversion for $S^2$. It was actually a shock at the time, I gather. It was especially frustrating since the proof was not constructive, and people had to wait years for an explicit eversion.

(1) and of $S^6$, and the proof of inexistence of an eversion for other even spheres – thanks to Mike Miller for this clarification!

$\endgroup$
4
  • 6
    $\begingroup$ Smale's result is stronger: the only spheres $S^n$ that admit eversions in $\Bbb R^{n+1}$ are the tautological $n = 0$, the famous $n = 2$, and the less-known $n=6$, tied to when spheres are parallelizable. $\endgroup$
    – mme
    Commented Oct 4, 2018 at 20:56
  • 4
    $\begingroup$ @MikeMiller Thanks, I didn't know that actually! $\endgroup$ Commented Oct 4, 2018 at 20:56
  • $\begingroup$ I am sorry if my question is elementary. According to the argument in the paper, using Gauss normal map, we conclude that a regular homotopy gives us a self map homotopy. So in the case of $S^2$, existence of an eversion implies that the identity and the amtipodal maps are homotopic self map. What is my mistake? $\endgroup$ Commented Oct 6, 2018 at 12:37
  • 1
    $\begingroup$ Let $f,g : S^n \to \mathbb{R}^{n+1}$. A regular homotopy between $f$ and $g$ induces a homotopy between the two Gauss maps $G_f, G_g : S^n \to S^n$. But the Gauss map of the antipodal is not the antipodal. If $f : S^n \to \mathbb{R}^{n+1}$ is the antipodal, then the degree of the Gauss map is $(-1)^n$. So there is no contradiction... $\endgroup$ Commented Oct 6, 2018 at 13:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .