Probability of commutation in a compact group

It is well known that if $$G$$ is a finite group, then the probability that two elements commutte is either $$1$$ (if $$G$$ is abelian) or less than or equal to $$\frac58$$.

If instead $$K$$ is a compact group, there exists a unique probability over $$K$$ that is left-invariant (Haar measure). The same problem as above still makes sense: What is the probability that two elements commutte ? In general, this probability can be non-trivial, because $$K$$ may have several connected components, being a (semi-)direct product of its neutral component $$K_0$$ with a finite group. If $$K_0$$ is abelian (a torus), then we are led back to the finite-case result.

So let me assume that $$K=K_0$$, that is $$K$$ is connected. Is it possible that the probability that two elements commutte be non-trivial, namely $$0 ?

• About the remark: when $K$ is a compact Lie group, $K_0$ is central and $K/K_0$ is abelian but $K$ is not abelian, it probably holds that the probability is $\le 5/8$, but this does not follow from the finite case (possibly it works with the same ideas).
– YCor
Oct 12 '18 at 9:11
• I checked (see math.stackexchange.com/a/953077/35400) the 1-line proof: the $5/8$ bound carries over the case of compact groups.
– YCor
Oct 12 '18 at 15:38
• @DenisSerre The following recent paper may be relevant cambridge.org/core/journals/… Jun 26 at 20:06

Edit: let me reformulate my answer since I apparently didn't answer the right question.

(a) if $$K$$ is a compact Lie group, the commuting probability is positive iff $$K_0$$ is abelian. As you noticed, $$\Leftarrow$$ is trivial.

Conversely, assume that $$K_0$$ is not abelian. We can view $$K$$ as Zariski-closed in matrix group, all its (Lie) components are also Zariski-closed, and are the irreducible components. Hence, if by contradiction the set of commuting pairs has positive measure (or equivalently has nonempty interior, or equivalently has dimension $$2\dim K$$), then it contains a product of two cosets: $$aK_0\times bK_0$$ for some $$a,b\in K$$. So $$agbh=bhag$$ for all $$g,h\in K_0$$.

Putting $$g,h=1$$, we get $$ab=ba$$. Putting $$h=1$$, we get $$agb=bag=1$$ for all $$h$$ which using $$ab=ba$$ yields $$gb=bg$$ for all $$g\in K_0$$. Putting $$g=1$$, we get $$abh=bha$$ for all $$h$$, which using $$ab=ba$$ yields $$ah=ha$$ for all $$h\in K_0$$. Thus $$a,b$$ commute and centralize $$K_0$$. So the formula simplifies as $$hg=gh$$ for all $$g,h\in K_0$$.

(b) let $$K$$ be a compact connected group with positive commuting probability. The property passes to Lie quotients of $$K$$, which are therefore abelian (the argument in the connected Lie case is contained in the above: the set of commuting pairs is a proper subset). Since $$K$$ is projective limit of its Lie quotients, it follows that $$K$$ is abelian.

If $$K/K_0$$ is finite, we reach the conclusion that $$K_0$$ is abelian.

• Note: the simple computation shows, in an arbitrary group $G$ with subgroup $H$, if $A,B$ are left-or-right cosets of $H$ and $[A,B]=1$, then $H$ is abelian and $A,B$ are both included in the centralizer of $H$.
– YCor
Oct 13 '18 at 8:18

Here is a proof that avoids Lie theory. Suppose $$G$$ is a compact connected group in which the probability that two elements commute is $$p>0$$. We claim $$G$$ is abelian.

Step 1 is to prove that the elements of $$G$$ have boundedly many conjugates. Let $$X_n$$ be the set of elements of $$G$$ with at most $$n$$ conjugates. Clearly $$X_n$$ is closed, and $$p = \int 1/|x^G| d\mu(x) \leq 1/n + \mu(X_n),$$ so $$\mu(X_n) > 0$$ provided $$n > 1/p$$. This implies that $$X_n$$ generates an open subgroup, which must be $$G$$ since $$G$$ is connected.

Step 2: $$[G,G]$$ is finite. In general a theorem of Bernhard Neumann asserts that if the elements of a group have boundedly many conjugates then $$[G,G]$$ is finite.

Step 3 is to look at the commutator map $$[,]: G\times G \to [G,G]$$. The codomain is discrete, and $$[G,1]=1$$, so there is some neighbourhood of the identity $$U$$ such that $$[G,U]=1$$. This $$U$$ generates an open central subgroup of $$G$$, which again must be $$G$$ since $$G$$ is connected.

Admittedly, not obviously simpler than YCor's proof. (All this and a little more is written up at https://randompermutations.com/2015/02/06/commuting-probability-of-compact-groups/.)

Actually, this is quite easy. We just need this, which is very easily proved:

Lemma: Let $$G$$ be a compact group and $$H\leq G$$ a closed subgroup. Then the following are equivalent:

1. $$\mu(H)>0$$;
2. $$H$$ has finite index;
3. $$H$$ is open.

In particular, if $$G$$ is connected then $$\mu(H) > 0$$ iff $$H=G$$.

Suppose $$G$$ is compact and connected. Let $$x \in G$$. Then the centralizer $$C_x$$ is a closed subgroup of $$G$$, so by the lemma $$\mu(C_x) > 0$$ iff $$C_x = G$$, i.e., iff $$x \in Z(G)$$. Therefore we have $$P(xyx^{-1}y^{-1} = 1) = \int \mu(C_x) \,d\mu(x) = \int 1_{x \in Z(G)} \, d\mu(x) = [G:Z(G)].$$ Now $$Z(G)$$ is also closed, so by the lemma again if the LHS is positive we must have $$Z(G)=G$$.

• I didn't use Lie theory, but indeed used both Peter-Weyl, and facts of real algebraic geometry, so this is indeed more elementary.
– YCor
Oct 12 '18 at 19:20

If the compact connected group $$K$$ is second-countable (hence metrizable, hence Polish), one can prove it with pretty much straight measure theory. (I am not sure if this argument remains valid for non-metrizable groups.)

It's based on the following elementary lemma which is due (in some form) to Steinhaus:

Lemma. Let $$G$$ be a locally compact Polish group with a left-invariant Haar measure $$\mu$$, and $$A \subset G$$ be a Borel set with $$\mu(A) > 0$$. Then the set $$A^{-1} A$$ contains an open neighborhood of the identity. In particular, if $$A$$ is a subgroup of $$G$$ then $$A$$ is open.

You can find it as Exercise 17.13 (ii) in Kechris, Classical Descriptive Set Theory. You begin by showing that the function $$x \mapsto \mu(xA \triangle A)$$ is continuous, which uses the regularity of the measure or else Lusin's theorem. (The proof I know uses dominated convergence and hence only shows that this function is sequentially continuous, which is why I am unsure about the non-metrizable case.)

So let $$E = \{(x,y) : xy = yx\} \subset K \times K$$. This set is closed, and in particular, measurable. The probability of two elements commuting is $$p = (\mu \times \mu)(E)$$ where $$\mu$$ is the left Haar probability measure on $$K$$. Let $$C_x$$ denote the centralizer of $$x$$, which is closed. We have $$(x,y) \in E$$ iff $$y \in C_x$$, and thus by Fubini's theorem $$p = \iint 1_E(x,y)\,\mu(dy)\,\mu(dx) = \iint 1_{C_x}(y)\,\mu(dy)\,\mu(dx) = \int \mu(C_x) \,\mu(dx).$$

If we suppose $$p >0$$, then the set $$B = \{x : \mu(C_x) > 0\}$$ must have positive measure. But if $$\mu(C_x) > 0$$, then by our lemma, $$C_x$$ is an open subgroup of $$K$$. It was already closed, and $$K$$ is connected, so $$C_x = K$$, i.e. $$x$$ is in the center $$Z$$ of $$K$$. Thus $$B \subset Z$$, so $$Z$$ is a closed subgroup with positive measure. Applying our lemma again, $$Z$$ is open and so $$Z=K$$, i.e. $$K$$ is abelian.

There is also be a "Baire category" analogue of this statement in the Polish case: if $$E$$ is nonmeager in $$K \times K$$, then $$K$$ is abelian. You could prove it by a nearly identical argument: use the Pettis lemma (Kechris 9.9) in place of Steinhaus above, and the Kuratowski-Ulam Theorem (Kechris 8.41) in place of Fubini. (Though maybe there's a simpler argument, since a nonmeager closed set has to have nonempty interior.)

• I think that the lemma (at least the version for the real line) is due to Steinhaus: wikipedia page. Oct 12 '18 at 21:20
• As far as I know, the analog of Fubini for Baire category is called the Kuratowski-Ulam theorem (and there's a good chance it's due to Kuratowski and Ulam). If it's not in Kechris's book,the next place I'd look for it is Oxtoby's book "Measure and Category". Oct 13 '18 at 0:02
• @AndreasBlass: That's the one, thanks. I had Lusin-Novikov stuck in my head. I'll update my answer. Oct 13 '18 at 0:10
• Probably I should have just commented on your answer, rather than edit my own. You don't need Steinhaus here, because $\mu(C_x)>0$ implies that $C_x$ has finite index, so $C_x^c$ is a finite union of cosets of $C_x$, hence closed, so $C_x$ is open. Baire category ditto I think. Oct 13 '18 at 0:54
• @SeanEberhard: Very nice, thanks. I'm upvoting your answer. I guess that all I have is a hammer, as they say... Oct 13 '18 at 4:41