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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 $w = x^n$.

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  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Oct 13 '18 at 16:19
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    $\begingroup$ 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 '18 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. $\endgroup$ – Sean Eberhard Oct 13 '18 at 16:33
  • $\begingroup$ Sean, yes. This is indeed due to connectedness. I posted an answer, for clarity. $\endgroup$ – Uri Bader Oct 13 '18 at 17:56
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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.

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    $\begingroup$ 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 '18 at 21:31

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