Given a compact Hausdorff topological group $G$ and probability measures $\mu$ and $\tau$ on the Borel sets of $G$, their convolution is the probability measure


A useful interpretation of this is that $G$ is a permutation group of some deck of cards, $\mu$ describes the distribution of arrangements the deck could currently be in, $\tau$ describes a shuffling scheme i.e. the distribution of possible shuffles a dealer may apply to randomize the deck. The convolution $\tau*\mu$ is then the distribution of the arrangements of the deck after the dealer has shuffled it. For example if $\tau$ and $\mu$ are Dirac measures supported at $g_1$ and $g_2$ respectively then $\tau*\mu$ is the Dirac measure supported at $g_1g_2$.

It is helpful to note that if $\tau_0$ is the unique Haar probability measure on $G$ and $\mu$ is another probability measure, then $\tau_0*\mu=\tau_0$. The interpretation here is that $\tau_0$ is the best shuffling scheme: every arrangement of the deck is equally likely after applying it, independent of how the deck is initially arranged. Obviously it's desirable for a shuffler to use such a scheme, or at least a scheme which converges to $\tau_0$. This is my motivation.

I am interested in the following questions:

  1. For which distributions $\tau$ does the sequence of convolutions $\tau^{*n}=\tau*\tau*\ldots*\tau$ converge, let's say weakly, to the Haar distribution $\tau_0$? My first guess is these are at least the distributions whose support generates $G$. Related to this, can the speed of convergence be measured? I know of such a result for the usual central limit theorem due to Berry and Esseen.

  2. Which distributions $\mu$ on $G$ are infinitely divisible in the sense that for any $n\geq 1$ there is a distribution $\tau_n$ such that $\mu=\tau_n^{*n}$? For example, the Haar distribution is infinitely divisible, but there are more: if $g\in G$ always has an $n$-th root $g_n$, then the Dirac measure supported at $g$ is infinitely divisible: $\delta_g = \delta_{g_n}^{*n}$. What are all the infinitely divisible distributions for $G=S^1$?

If someone can point me toward a reference for any of this I would much appreciate it, I don't know much harmonic analysis.


For instance, in the recent paper by Harremoes, page 12 one finds this:

Corollary 20. Let $P$ be a probability measure on the compact group $G$ with Haar probability measure $U$. Assume that the support of $P$ is not contained in any coset of a proper subgroup of $G$. Then $P^{*n}$ converges to $U$ in the weak topology.

(See also the note in first paragraph on page 5 of that paper.)

Some results on a geometric rate of the convergence are given in Section 6 of the same paper.

It also contains a number of useful references.

  • $\begingroup$ Excellent, I believe this sufficiently answers my first question. $\endgroup$ – Jess Boling Aug 8 at 21:13
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    $\begingroup$ It is competely misleading to attribute this result to Harremoes as it was explicitly formulated and proved by Stromberg in 1960. It is amazing that Harremoes actually quotes Stromberg's paper (in a somewhat vague form), but fails to mention that his Corollary 20 is due to Stromberg. $\endgroup$ – R W Aug 8 at 21:34
  • $\begingroup$ Thank you for that comment, I was able to find the paper: Stromberg. I agree it looks like all credit belongs to Stromberg, but I wasn't able to find a comment on the speed of convergence in this paper. $\endgroup$ – Jess Boling Aug 8 at 22:06
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    $\begingroup$ As for the exponential speed of convergence in the total variation metric - it has also been known for ages and goes essentially back to Doeblin. Harremoes actually quotes (in Corollary 25) a paper of Kloss from 1959 where this was explicitly done for compact groups. In what concerns the speed of convergence in the weak topology, it can be measured by using the transportation metric, and it is known that the convergence can be arbitrarily slow, see mathscinet.ams.org/mathscinet-getitem?mr=816292 $\endgroup$ – R W Aug 9 at 0:18
  • $\begingroup$ Thank you these have been very helpful. $\endgroup$ – Jess Boling Aug 9 at 0:46

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