# Probability of satisfying a word in a compact group

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 $$w = x^n$$.

• 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. Oct 13 '18 at 16:19
• 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. Oct 13 '18 at 16:27
• @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. Oct 13 '18 at 16:33
• Sean, yes. This is indeed due to connectedness. I posted an answer, for clarity. Oct 13 '18 at 17:56

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.

• 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 $[F_k,F_k]$.
– YCor
Oct 13 '18 at 21:31