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If a random vector $u \in\mathbb{R}^n$ has $\|u\|_2=1$, and for any orthogonal matrix $Q$, it has $Qu \overset{d}{=}u$, does this imply $u \sim Unif(S^{n-1})$?

Zifeng Zhang
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  • What is the definition of $\mathsf{Unif}(\mathbf{S}^{n-1})$? I ask because it seems that what you wrote is exactly that. – William M. Mar 14 '22 at 16:22
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    It is uniform distribution on the unit sphere. I actually don't know the strict definition very clearly. The probability mass is distributed uniformly on the unit sphere. It has some density but I think the corresponding measure is not Lebesgue measure. If $x$ follows normal distribution $N(0, I_n)$, then $x/||x||_2$ follows $Unif(S^{n-1})$. – Zifeng Zhang Mar 14 '22 at 16:25
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    It is just that in a sense, what one does to define a unit sphere is to take a projection $p:\mathbf{R}^n \to \mathbf{S}^{n-1}.$ We realise that $\mathbf{S}^{n-1}$ has a canonical action (is invariant under rotations, similar as how $\mathbf{R}^n$ is invarian under translations and rotations) and then we project Lebesgue measure in a way that preserves these actions. This program is actually carried to any compact group that is a projection from $\mathbf{R}^n$ and that has a nice action under some group. Unfortunately, I've never have the time to read this fully even though I have wanted to. – William M. Mar 14 '22 at 16:35
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    At any rate, uniform measure on such a manifold ought to be the projection of Lebesgue measure (since this projected measure will be invariant under the actions). – William M. Mar 14 '22 at 16:36
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    You can get proper double norm bars using \|. – joriki Mar 14 '22 at 20:26
  • Thank you! I just know this fact. – Zifeng Zhang Mar 15 '22 at 00:33
  • answer here: https://math.stackexchange.com/questions/1183734/haars-theorem-for-the-rotation-invariant-distribution-on-the-sphere – MMH May 03 '22 at 21:03

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