22
$\begingroup$

This question is inspired by Probability of commutation in a compact group, which asked whether $P(xyx^{-1}y^{-1} = 1)$ could take values strictly between $0$ and $1$ on a compact connected group. That question turned out to have a rather easy answer, but one that was quite particular to the word $xyx^{-1}y^{-1}$.

Question: Let $w$ be an arbitrary word in $k$ letters, and let $G$ be a compact connected group. Let $x_1, \dots, x_k$ be drawn uniformly from $G$. Can $P(w(x_1, \dots, x_k)=1)$ be strictly between $0$ and $1$?

By the Peter--Weyl theorem we may assume that $G$ is a compact connected Lie group.

There is some low-hanging fruit, like $w = [[x,y],z]$, but that's still very specialized. Admittedly I am not even clear on the words $w = x^n$.

$\endgroup$
4
  • $\begingroup$ Particularly in Lie groups, there's lots of other measure-theoretic and potential-theoretic notions of "smallness" that one could ask about. If it does have measure zero, what's its Hausdorff dimension? Is it a polar set? Et cetera. $\endgroup$ Oct 13, 2018 at 16:19
  • 3
    $\begingroup$ the solution space for a $k$ letters word will be a closed subvariety of $G^k$, and as such it will have either measure 0 or 1. Morover, if $G$ is not commutative then it contains a free group hence this subvariety will be proper, thus of 0 meaure. $\endgroup$
    – Uri Bader
    Oct 13, 2018 at 16:27
  • $\begingroup$ @UriBader Why does full dimension imply measure 1? I'm thinking of for example the solution space to $x^2=1$ in $O(2)$. Of course $O(2)$ is not connected, so it's not a counterexample. $\endgroup$ Oct 13, 2018 at 16:33
  • $\begingroup$ Sean, yes. This is indeed due to connectedness. I posted an answer, for clarity. $\endgroup$
    – Uri Bader
    Oct 13, 2018 at 17:56

1 Answer 1

20
$\begingroup$

The negative answer follows easily from a very useful fact that should be better known, so I am writing it explicitly.

Fact (Tannaka, Chevalley): Every (connected) compact Lie group is isomorphic as a topological group to the group of real points of a (connected) reductive affine real algebraic group. Moreover, the Haar measure on such a group is given by a volume form on the corrseponding variety.

We thus may view (for a given word $w$ in the rank $k$ free group $F_k$) the word map $w:G^k\to G$ as a morphism of real varieties and its solution space (the fiber over the identity element) as a the real points of a closed real subvariety. Note that $G^k$ is an irreducible variety, as $G$ is by its connectedness, thus this subvariety will be either $G^k$ itself, or a lower dimensional one. Its measure, accordingly, will be either 0 or 1.


We could be more precise:

  • If $G$ is trivial then the measure is 1.

  • If $G$ is non-trivial abelian then it is isomprphic to a torus $\text{SO}(2)^n$, and it is easy to see that the measure is 1 iff $w$ is in the commutator group of $F_k$, otherwise the measure is 0.

  • If $G$ is non-abelian then it contains a free group on $k$ generators (by Tits alternative, if you want to hit it with a hammer), thus if $w\neq 1$ the corresponding subvariety is proper and its measure is necessarily 0. If $w=1$ then the measure of 1.

$\endgroup$
1
  • 3
    $\begingroup$ This is an answer for compact connected Lie groups. For arbitrary compact connected groups $G$, you need an additional argument using Peter-Weyl. Namely, if $w$ holds with positive measure, then it holds with positive measure in every quotient, and hence it holds identically in every Lie quotient, and by Peter-Weyl $G$ is projective limit of such Lie quotients, and hence $w$ holds in $G$. The argument also shows that $G$ is either abelian or contains a free subgroup, and if it's abelian and nontrivial, it satisfies $w$ iff $w$ belongs to $[F_k,F_k]$. $\endgroup$
    – YCor
    Oct 13, 2018 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.