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Let $\mu$ be a Borel probability measure on $\mathcal S^{n-1}$. Suppose $\mu$ is rotation-invariant, that is, $\mu(UA) = \mu(A)$ for any Borel set $A\subset\mathcal S^{n-1}$ and orthogonal matrix $U\in\mathbb R^{n\times n}$. Then $\mu$ is the uniform measure.

Could you recommend a reference (preferably a book) for this simple version of Haar's theorem?

Yoni Rozenshein
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3 Answers3

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Uniqueness of Haar measure is generally easier to prove than existence of Haar measure, particularly for compact groups. For instance in this case one can argue as follows. Let $\sigma$ be the left-invariant measure on $\text{SO}(n)$ which you know and love, and let $\mu$ be the usual rotation-invariant measure on $S^{n-1}$. Then for every $x\in S^{n-1}$ the map $g\mapsto gx$ sends the measure $\sigma$ to $\mu$, so if $\nu$ is an arbitrary normalized rotation-invariant measure on $S^{n-1}$ then for an arbitrary continuous function $f:S^{n-1}\to\mathbf{R}$ we have, by Fubini's theorem,

$$\int_{S^{n-1}} f(x) \,d\nu(x) = \int_{\text{SO}(n)} \int_{S^{n-1}} f(gx) \,d\nu(x) \,d\sigma(g)\\ = \int_{S^{n-1}} \int_{\text{SO}(n)} f(gx) \,d\sigma(g)\,d\nu(x)\\ = \int_{S^{n-1}} \int_{S^{n-1}} f(y) \,d\mu(y) \,d\nu(x)\\ = \int_{S^{n-1}} f(y) \,d\mu(y),$$ so $\nu=\mu$.

It wasn't important in the above argument to know that $\sigma$ is the unique Haar measure on $\text{SO}(n)$, only that it exists, which you can prove with Lie theory. However you could prove uniqueness of $\sigma$ with a similar argument.

Sean Eberhard
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  • Thanks for the answer, but I am looking for something more elementary, i.e. without Lie theory or the general theory of compact groups, just this case of the sphere and orthogonal-matrix-invariance. Preferably in a book that I can just quote as in "it is a well-known fact that the Lebesgue measure is the unique rotation-invariant measure on the sphere [1]" :) – Yoni Rozenshein Apr 07 '15 at 13:15
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    Surprising that such a masterful solution has so little attention. – caffeinemachine Sep 27 '17 at 17:22
  • You don't need Lie theory to show the existence of the Haar measure on $O(n)$ (or $SO(n)$), you can recover it from the usual Lebesgue measure. Consider $G$ a random matrix with i.i.d. entries, the distribution of $G$ is left and right invariant under the action of $O(n)$ and therefore the distribution of $\sqrt{G^*G}G^{-1}$ (the polar part in the polar decomposition) is the Haar measure. – Guillaume Aubrun Jan 22 '19 at 16:45
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    @GuillaumeAubrun Nice. One can also embed $SO(n)$ in $\mathbf{R}^{n^2}$ in the obvious way and define the measure of $X\subset SO(n)$ with reference to the cone connecting $0$ to $X$ in $\mathbf{R}^{n^2}$. Or one could use the isomorphism $SO(n) / SO(n-1) \cong S^{n-1}$ and induction. Many interesting approaches. – Sean Eberhard Jan 22 '19 at 20:00
  • @SeanEberhard never thought of the "cone" approach, that's even more elementary! – Guillaume Aubrun Jan 23 '19 at 08:32
  • Doesn't the third equality require also right-invariance of $\sigma$? Since we are using transitivity of the action to say $x = g_x x_0$ for some $g_x \in \DeclareMathOperator{\SO}{SO} \SO(n)$, and then would want $\int_{\SO(n)} f(gg_xx_0) ,d\sigma(g) = \int_{\SO(n)} f(gx_0) ,d\sigma(g) = \int_{S^{n-1}} f ,d\mu$. Of course since $\SO(n)$ is compact we can easily obtain a bi-invariant measure from left- and right-invariant ones. – Olius Nov 19 '23 at 12:07
  • @Olius The third equality uses the fact (stated earlier) that for every fixed $x \in S^{n-1}$ the pushforward of $\sigma$ by the map $g \mapsto gx$ is $\mu$. In any case, in fact for any compact group $G$ any left-invariant (regular) measure is also right-invariant. This follows from the uniqueness of left-invariant measure, since the pushforward of such a measure $\mu$ along a right translation is again a left-invariant measure, and the two agree on the measure of $G$. – Sean Eberhard Nov 20 '23 at 09:37
  • @SeanEberhard Yes, I meant that right invariance was necessary to ensure that the pushforward of $\sigma$ along $g\mapsto gx$ is indeed a rotation-invariant measure on $S^{n-1}$. Thank you for pointing out that in any case we already have right invariance. – Olius Nov 21 '23 at 11:00
  • I'm sorry, I meant that right invariance is necessary to assert that the pushforward is indeed the same measure no matter which $x$ we choose. – Olius Nov 22 '23 at 12:09
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I've found an answer for my question: Theorem 3.4 of Pertti Mattila's Geometry of sets and measures in Euclidean spaces: Fractals and rectifiability.

Yoni Rozenshein
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You could use the uniqueness of the Lebesgue measure on $\mathbb{R}^n$.

If $\mu$ is your rotation-invariant measure, equip $\mathbb{R}^n\setminus\{0\} = S^{n-1} \times (0,\infty) $ with the product measure $\mu \times m$ where $m$ is the Lebesgue measure. Then this measure is invariant under the group of rigid motions and is hence the Lebesgue measure on $\mathbb{R}^n$.

Now, for $A \subset S^{n-1}$, this will imply that $\mu(A) = m(A\times (0,1))$, which gives uniqueness.

Edit: as pointed out below, the correct measure should be $\mu \times m'$, where $m'((0,r)) = r^{n-1}$.

  • Oh, I like this approach, but I don't understand something. Clearly $\mu \times m$ is invariant under orthogonal transformations. Why is it invariant under translations? – Yoni Rozenshein Apr 13 '15 at 17:10
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    I don't believe that $\mu\times m$ is Lebesgue measure on $\mathbb{R}^n$. If it were, the Lebesgue measure of the unit sphere in $\mathbb{R}^n$ would be half the measure of the radius two sphere, which is clearly absurd. A measure must be invariant under translations as well as rotations for it to be Lebesgue. – user24142 Apr 13 '15 at 18:48
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    Thanks for your comment @user24142, you are right. In fact, the Lebesgue measure on $\mathbb R^n$ splits to $\mu \times$ (a measure on $\mathbb R$ with density $r^{n-2}$) or something like that. – Yoni Rozenshein Apr 13 '15 at 20:59