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Pick a random rotation uniformly in SO(3), where the uniform distribution is the unique distribution that's invariant by the action of any group element.

Apply this rotation twice to the north pole of the unit sphere. What is the distribution of the resulting point?

Naturally, the distribution of the north pole rotated once is uniform over the sphere, by a symmetry argument, and numerical analysis confirms that. However, applying the rotation twice does not, a priori, yield a uniform distribution and, again, numerical analysis confirms that this distribution is not uniform over the sphere.

However, the following argument would suggest that applying the rotation twice should still yield a uniform distribution, and I can't tell what's wrong with it. It goes as follows:

A rotation can be described by a rotation axis and a rotation angle. By symmetry, the rotation axis can be picked by picking a random point uniformly on the sphere and drawing a line through the center, and the rotation angle can be picked by picking a random angle $\theta$ between $0$ and $2 \pi$ independently from the axis. Applying this random rotation twice is equivalent to rotating by $2 \theta$, but a rotation of $2 \theta$ around an axis chosen randomly and independently should have the same distribution as a random rotation with angle $\theta$ and therefore, the distribution of the image of the north pole via this rotation should be the same.

  1. Clearly something is wrong with the above explanation, but I can't find what it is.
  2. What is the actual distribution we're looking for?
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  • $\begingroup$ If applying the first random rotation yields a uniform distribution, then obviously applying a second random rotation must also yield a uniform distribution $-$ there is nothing special about the north pole. Therefore either (i) applying the first rotation doesn't yield a uniform distribution, or (ii) applying the second rotation does. My bet is on (ii). How are you generating your random rotations? $\endgroup$
    – TonyK
    Commented Nov 26, 2021 at 23:26
  • $\begingroup$ It's the same rotation, there's only one random draw. $\endgroup$
    – Arthur B.
    Commented Nov 27, 2021 at 7:54
  • $\begingroup$ Ah, now I understand. $\endgroup$
    – TonyK
    Commented Nov 27, 2021 at 12:16

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