You can rotate a circle so that every point on it (just the perimeter, not the interior) moves "equally". That is, every point moves with the same speed and even has the same "acceleration" (first-order derivative of velocity, which is the first-order derivative of a point's movement wrt time).

However, there is no way to rotate a sphere with this same character: the farthest points from the axis of rotation (equator) move much faster and have greater acceleration than the points closest to the axis (poles).

AFAIK, there is no closed (finite, no edges) 3-dimensional surface with this property. You can do it with a cylinder, but that is not a closed shape.

You could also do it with a torus by rotating it through the center (rather than around the center) the way a smoke ring does, but that's not a true "rigid" rotation (all points maintain the same distance to all other points throughout the rotation). The points on the inner part of the ring are closer together than those on the outside.

What I've been wondering is how this plays out in higher dimensions? Can the volume-surface of a 4D sphere be rigidly rotated equally like a 2D circle? Or does it fall to the same problems as 3D spheres? What about even higher dimensional spheres?

For that matter, are there any higher dimensional closed shapes that can be so rotated?

For clarity, I am only looking for "smooth" shapes, no significant discontinuities, and not just a disconnected set of points.


One way to make sense of what you wrote is: one can rotate odd-dimensional spheres without any fixed points.

A way to do this is: if you identify $\mathbb R^{2n}$ with $\mathbb C^n$, then at time $t$ you multiply by $\exp(2\pi i t)$.

On the other hand, this is impossible for every even-dimensional sphere, due to the hairy ball theorem.

  • $\begingroup$ This even gives makes every point on the $(2n-1)$-sphere move with the same absolute velocity, which is probably how "moves equally" in the question should be interpreted. $\endgroup$ – Henning Makholm May 27 '16 at 20:14
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    $\begingroup$ Ah, I see. Because the 3D sphere is a 2-sphere. Well that's a confusing convention... $\endgroup$ – RBarryYoung May 27 '16 at 20:29
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    $\begingroup$ @RBarryYoung This is one of those conventions that's confusing at first, but eventually turns out to be right. Specifically, we naively think of the 2-sphere as three-dimensional because that's where it lives: the 2-sphere naturally embeds into $\mathbb{R}^3$, which is three-dimensional. However, if we think of the 2-sphere "on its own", it is a two-dimensional space. It turns out that this latter approach - viewing spaces on their own, as opposed to where they embed - is the one that (most of the time) gives the right intuition for things. $\endgroup$ – Noah Schweber May 27 '16 at 20:31
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    $\begingroup$ @RBarryYoung: For an example of why the above convention makes more sense than the alternative, consider the torus $T^2 = S^1\times S^1$. The most obvious way of embedding it in $\mathbb{R}^4$ is by taking the product of two embeddings $S^1 \to \mathbb{R}^2$: thus $(\theta_1, \theta_2) \to (\cos \theta_1, \sin \theta_1, \cos \theta_2, \sin \theta_2)$. Or there's the familiar embedding in $\mathbb{R}^3$. The space it happens to live in is not much of an invariant; the dimension of the neighborhood of a point on the torus is. $\endgroup$ – anomaly May 27 '16 at 22:49
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    $\begingroup$ \begin{pmatrix}\cos t&\sin t&0&0\\ -\sin t&\cos t&0&0\\ 0&0&\cos t&\sin t\\ 0&0&-\sin t&\cos t \end{pmatrix} does the trick in $\mathbb R^4$, and you can guess how it goes in higher even dimensions. $\endgroup$ – Mariano Suárez-Álvarez May 28 '16 at 15:35

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