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Why does rotating through a great $n$-sphere of a $(n+1)$-sphere not produce the whole sphere $(*)$ for $n \ge 2$?

More precisely: Let $X$ be the $(n+1)$-sphere. A great sphere is the intersection of the surface of $X$ with some hyperplane through the origin, i.e. $Y: = v^{\perp} \cap \partial X$ for some $v \ne \mathbf{0}.$ "Rotating through" $Y$ means rotating $v$ through a 2D subspace. Algebraically, the rotations of $Y$ will be $\{Y_{\theta}\}_{\theta \in [0, 2\pi]}$ where $Y_{\theta} = ((\cos \theta)u + (\sin \theta) v)^{\perp} \cap \partial X$ for some fixed orthogonal $u, v.$

Note that rotating a diameter around the origin produces a circle and rotating a great circle around an axis inside it produces a 2-sphere, so $(*)$ is true for $n = 0,1.$

Motivation: The rotation trick carries over to convex bodies iff we expect it to work with simple spheres. As a result, using $(*)$ gives a positive answer to the Busemann-Petty problem for $n+2$ dimensions. Thus, we have a positive answer for $n \le 3$ and potentially a positive answer for all $n.$ In reality, the problem has a positive answer for $n \le 4$ and negative answer for $n \ge 5.$

When I gave my approach to the problem and challenged people to find the mistake, everything was scrutinized and everyone agreed with it except that one person question the trick of rotating to cover the body. He said this already fails in $4$ dimensions, specifically stating that $(*)$ is false for $n = 2$ and telling me to imagine it. I can't imagine 4 dimensions easily. Going through the algebra starting with $w^2+x^2+y^2+z^2 = 1$ is uninspiring; is there a more intuitive way to see why?

Unfortunate that the failure couldn't wait an extra dimension, as a new proof for $4$ dimensions, something which took until 1999 for Zhang to prove and required a lot of work, would be nice. At least I can walk away with a simple proof for $3$ dimensions, something which made Gardner sweat in 1994.

For reference, the Buseman-Petty problem.

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  • You may want to link the page https://en.wikipedia.org/wiki/Busemann%E2%80%93Petty_problem . Some users here may be sceptical of a claim that (*) would give an easy proof of Zhang's result. – Mikhail Katz Jan 09 '24 at 12:19
  • @MikhailKatz Added the link – Display name Jan 10 '24 at 06:15
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    After the first reading I have no idea what exactly the problem is. The only hunch I have is that may be you are confused about what rotations in higher dimensions really are? For me, a simple rotation (or a Givens rotation) is a linear transformation that rotates a 2-dimensional plane, and keeps everything in its orthogonal complement fixed. In 3D that orthogonal complement is 1-dimensional, and usually referred to as the axis of the rotation. In higher dimensions the axis = the complement is an $(n-2)$-dimensional subspace. What is a rotation in, say 4D, for you? – Jyrki Lahtonen Jan 10 '24 at 06:27
  • (cont'd) Basically, your phrase rotating $v$ around the line $\text{Span}{w}.$ is meaningless to me. In higher dimensions you cannot easily define what it means to rotate around a line. To me that is a strictly 3D thing. – Jyrki Lahtonen Jan 10 '24 at 06:31
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    Here I try and elaborate on what I think rotations in higher dimensions are. – Jyrki Lahtonen Jan 10 '24 at 06:37
  • @JyrkiLahtonen I'm treating $v$ as a point, and as the point moves the hyperplane rotates. Your link was informative. For this question, I mean the strict definition that Henning mentions, where a point just travels around in a circle. – Display name Jan 10 '24 at 11:44
  • Still needing to calibrate what the question is about. Take that first group of matrices from Henning's answer as the version of $S^1$ acting on $X={(x,y,z,w)\mid x^2+y^2+x^2+w^2=1}$. Rotation by angle $\theta$ then takes $$(x,y,z,w)\mapsto (\cos\theta x+\sin\theta y, -\sin\theta x+\cos\theta y, z,w).$$ You see that nothing at all happens to $z$ and $w$ coordinates. In other words, the plane consisting of the point $(0,0,z,w)$ is the "axis" of this group of rotations. What would $Y$, and the vectors $v$, $w$ be here? – Jyrki Lahtonen Jan 10 '24 at 16:18
  • If you simply want to claim that every orbit of the above action contains at least one point with, say, $x=0$, then that's true. Typically there are two such points (much like each latitude intersecting every great circle via the poles on the surface of the Earth at two points), but a difference is that the poles (=the end points of the axis of rotation) become circles: each point of the form $(0,0,\cos\alpha,\sin\alpha)$, $0\le\alpha<2\pi$ stays fixed under that action. – Jyrki Lahtonen Jan 10 '24 at 16:32
  • @JyrkiLahtonen You've convinced me that rotations must have a $(n-2)$-dimensional axis, so I updated my definition. In 2D that the axis is a point (typically the origin), that's where I should've stopped when originally writing a definition of rotation. The rest of my question remains the same. – Display name Jan 16 '24 at 01:08

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