Can the degree of $k$-nilpotence of a simple simply connected compact Lie group be in $(0,1)$? Let $G$ be a simple (i.e. every proper normal subgroup is discrete) simply connected compact Lie group. Define the degree of $k$-nilpotence of $G$ to be the Haar measure of the set
$$\{(x_1,\dotsc,x_{k+1}): [x_1,\dotsc,x_{k+1}]=1\}.$$
($[x,y]=x^{-1}y^{-1}xy$ and $[x_1,\dotsc,x_{k+1}]:=[[x_1,\dotsc,x_k],x_{k+1}]$.)
The following question is raised in Martino, Tointon, Valiunas, and Ventura - Probabilistic nilpotence in infinite groups:

Does a compact group with positive degree of $k$-nilpotence have an
open $k$-step nilpotent subgroup?

Is it easy to answer this question when $G$ is a simple simply connected compact Lie group?
 A: Here's a positive answer of the question for arbitrary compact Lie groups. For a group $G$, denote $W_k(G)=\{x\in G^k:[x_1,\dots,x_k]=1\}$.

Let $G$ be a compact Lie group. Then $W_k(G)$ has nonzero Haar measure for some $k\ge 1$ if and only if $G^0$ is a torus.

One direction is obvious: if $G^0$ is a torus, then $(G^0)^k\subset W_k^G$ has positive measure (for every $k\ge 2$).
Conversely, we assume $W_k$ has nonzero Haar measure. If $k=1$, it follows that $G$ is finite and we are done. Assuming otherwise, $k\ge 2$.
The subset $W_k(G)\subset G^k$ is Zariski-closed. Since it has nonzero Haar measure, it therefore contains $\prod_{i=1}^k C_i$ for cosets $C_i$ of $G^0$. [If $G$ is connected, this is $G^k$, so $G$ is nilpotent, hence is a torus and the proof is finished.]
I assume the conventions $[x,y]=xyx^{-1}y^{-1}$ and that the iterated commutator is defined by $[x]=x$ and $[x_1,\dots,x_k]=[x_1,[x_2,\dots,x_k]]$ for $k\ge 2$. Also, by $[X_1,\dots,X_k]$ I mean the set $\{[x_1,\dots,x_k]:x_i\in X_i\}$ (rather than the group it generates).
Now assume that $G^0$ is semisimple and nontrivial. I need an inductive argument.  Namely, to reach a contradiction, choose $k\ge 2$ minimal for the property that there are cosets $C_1,\dots,C_k$ of $G^0$ such that $[C_1,\dots,C_k]$ is a singleton, say $w$. Necessarily $k\ge 2$.
Define $Y=\{[x_2,\dots,x_k]:x_i\in C_i,\forall i\ge 2\}$. Then $[x,y]=w$ for all $x\in C_1$ and $y\in Y$. Then for all $s\in G^0$, $x\in C_1,y\in Y$, we have $$w=[x,y]=[xs,y]=x[s,y]x^{-1}[x,y]= x[s,y]x^{-1}w,$$
and hence $[s,y]=1$. Thus, $Y$ is contained in the centralizer of $G^0$. Since $G^0$ is semisimple, its centralizer is finite. Hence $Y=[C_2,\dots,C_k]$ is finite, hence a singleton by connectedness. This contradicts the minimality of $k$.
Now the result is proved when $G^0$ is semisimple (i.e., when $G^0$ is semisimple, if $W_k(G)$ has nonempty interior implies $G^0=1$). In general, let $Z$ be the center of $G^0$. If $W_k(G)$ has nonempty interior, since the projection $G\to G/Z$ is open and so is $G^k\to (G/Z)^k$, the subset $W_k(G/Z)$ (which contains the projection of $W_k(G)$) has nonempty interior. So, by the semisimple case $(G/Z)^0$ is trivial. That is, $Z=G^0$.
[Edit: I fixed my previous argument]
