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The group of orthogonal transformations carries an invariant probability measure. This means that we can average a function over the group in a natural way. In particular, if $f$ is a function on the sphere and $θ$ is some point on the sphere, the average over orthogonal $U$ of the value $f(Uθ)$ is just the average of $f$ on the sphere: averaging over $U$ mimics averaging over the sphere: $$\text{avg}_U f(Uθ) = \int_{S^{n-1}}f (\phi) dσ(\phi)$$

See Pg. $22-23$ of these notes for context.

Is there an easy way to understand/prove this? I'm not able to figure out what it means!

My intuition:
Consider $\theta$ on the sphere $S^{n-1}$. Orthogonal transformations rotate this without changing the length. All orthogonal transformations send this point to someplace on the sphere, and for any two points on the sphere, we can always find an orthogonal transformation that relates them. Hence, averaging over all orthogonal transformations is the same as averaging over the sphere.

Is that correct?

stoic-santiago
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  • Not my area of expertise, but I'd bet it has something to do with the sphere being exactly the orbit of $\theta$ under $U$. – Vercassivelaunos Jan 22 '21 at 15:30
  • What does that mean? Orbit? – stoic-santiago Jan 22 '21 at 15:33
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    $U$ acts on $\theta$, and you get an element $\sigma(\theta)$ for all $\sigma\in U$. The set of all these elements is the orbit of $\theta$ under $U$, and it is exactly the sphere (rotating a vector, which is what orthogonal matrices do, allows you to reach all the points on a sphere, and no other points) – Vercassivelaunos Jan 22 '21 at 15:37
  • That's interesting! I guess it has something to do with it – stoic-santiago Jan 22 '21 at 15:43

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Let $\sigma$ be the usual rotation-invariant surface measure on the sphere $S^{n-1}$. We know that in fact $\sigma$ is the only rotation-invariant measure on the sphere. Now, if we let $m$ denote the rotation-invariant (Haar) measure on the group $SO(n)$, let us define a new measure $\nu$ on the sphere, defined by duality via $$ \int_{S^{n-1}} f \,d\nu := \int_{SO(n)} f(U \theta) \,dm(U) $$ for $f: S^{n-1} \to \mathbb{C}$ continuous, where $\theta \in S^{n-1}$ is a fixed arbitrary point. Then the rotation invariance of $m$ on $SO(n)$ implies that $\nu$ is also a rotation-invariant measure on $S^{n-1}$: $$ \int f \circ M \,d\nu = \int_{SO(n)} f(MU \theta) \,dm(U) = \int_{SO(n)} f(U\theta) \,dm(U) = \int_{S^{n-1}} f \,d\nu $$ for any $M \in SO(n)$.

Therefore we conclude that $\nu$ and $\sigma$ are the same measure, i.e. $$ \int_{S^{n-1}} f \,d\nu = \int_{SO(n)} f(U\theta) \,dm(U) = \int_{S^{n-1}} f \,d\sigma $$

Adam
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  • I think $\sigma$ is the only rotationally invariant measure on the sphere that is finite on (relatively) open sets. For instance, I would think that $\mathcal{H}^{k} \restriction_{S^{n-1}}$ (Hausdorff measure restricted to sphere) is still rotationally invariant independent of the choice of $k \in {0,1,2,n - 1}$. At any rate, the $k = 0$ case is clear (counting measure). The question is how to prove $\sigma$ is so characterized. –  Jan 22 '21 at 23:03
  • Hm, that's a good point. There's a theorem in Mattila's book that says that any two uniform measures (a measure is uniform if it gives the same measure to all balls of the same radius) are constant multiples of each other, so that probably implies that $\sigma$ is the only rotation-invariant probability measure on the sphere – Adam Jan 22 '21 at 23:56
  • Hmm, I didn't quite understand this. Is there any way you can possibly tone down this answer for someone only familiar with some real analysis and very introductory measure theory? – stoic-santiago Jan 23 '21 at 03:28
  • The rough idea is: suppose that we know that $\sigma$ is the only rotation-invariant probability measure on the sphere (this is true but maybe not so obvious, as the previous comments show). But then the "averaging over orthogonal matrices $U$" operation also defines another rotation-invariant measure on the sphere -- just let the measure of a set $E$ be the Haar measure of the set of matrices that map your base point $\theta$ into $E$. Since there's only one rotation-invariant probability measure on the sphere, this new measure has to be the same as $\sigma$ – Adam Jan 23 '21 at 07:19
  • I see, thank you! Could you see my edited post and let me know if my intuition is correct? – stoic-santiago Jan 24 '21 at 07:33
  • Yes I think it is – Adam Jan 24 '21 at 07:35