This question is inspired by [Probability of commutation in a compact group](https://mathoverflow.net/questions/312652/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 $xyx^{-1}y^{-1}$.



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> **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 $1$?
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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 $w = x^n$.



Probability of satisfying a word in a compact group
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.




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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.


